Solutions to Abbott’s Understanding Analysis

(a) Both are irrational, and in each case the proof of Theorem 1.1.1 runs unchanged except for the divisibility lemma that drives it.

For \(\sqrt{3}\) the lemma needed is that \(3 \mid p^2\) implies \(3 \mid p\): writing \(p = 3k+s\) with \(s \in \{0,1,2\}\),

\begin{equation*} p^2 = 3(3k^2 + 2ks) + s^2, \qquad s^2 \in \{0,1,4\}, \end{equation*}

so \(p^2\) leaves remainder \(0\) on division by \(3\) only when \(s = 0\).

Now assume for contradiction that \(\sqrt{3} = p/q\) with \(p,q \in \mathbf{Z}\) having no common factor. Then

\begin{equation*} \begin{aligned} p^2 &= 3q^2 &&\Rightarrow\ 3 \mid p, \text{ say } p = 3r,\\ 9r^2 &= 3q^2 &&\Rightarrow\ q^2 = 3r^2 \ \Rightarrow\ 3 \mid q, \end{aligned} \end{equation*}

so \(3\) divides both \(p\) and \(q\), contradicting that the fraction is in lowest terms.

For \(\sqrt{6}\) the same strategy works, now run on the prime \(2\). Assume \(\sqrt{6} = p/q\) in lowest terms, so \(p^2 = 6q^2\). Then \(p^2\) is even, hence \(p\) is even (the square of an odd number is odd); writing \(p = 2r\) gives \(4r^2 = 6q^2\), that is,

\begin{equation*} 2r^2 = 3q^2 . \end{equation*}

The left side is even, and \(3\) is odd, so \(q^2\) is even and therefore \(q\) is even. Again \(p\) and \(q\) share the factor \(2\), a contradiction.

(b) It breaks at the final substitution, where the contradiction is supposed to appear. Assume \(\sqrt{4} = p/q\) in lowest terms, so \(p^2 = 4q^2\). As before \(p^2\) is even, so \(p = 2r\) for some integer \(r\), and substituting yields

\begin{equation*} 4r^2 = 4q^2, \qquad \text{i.e.} \qquad r^2 = q^2 . \end{equation*}

In Theorem 1.1.1 the corresponding substitution left \(2r^2 = q^2\), forcing \(q^2\), and hence \(q\), to be even. Here the \(4\) on the right cancels the \(4\) produced by \(p = 2r\) exactly, and \(r^2 = q^2\) says nothing about the parity of \(q\), so no contradiction is reachable.

Clearing the denominator of \(r\) turns \(2^r = 3\) into an even integer equal to an odd one.

Assume for contradiction that \(r = p/q\) with \(p, q \in \mathbf{Z}\) and \(q \ge 1\). Raising \(2^{p/q} = 3\) to the \(q\)-th power gives

\begin{equation*} 2^p = \left(2^{p/q}\right)^q = 3^q . \end{equation*}

If \(p \le 0\) then \(2^p \le 1 < 3 \le 3^q\), so in fact \(p \ge 1\) and both sides are positive integers. The left side is then even; the right side is a product of \(q \ge 1\) odd numbers, hence odd. An integer cannot be both, so no such \(r\) exists.

(a) False, (b) true, (c) false, (d) true, (e) true.

(a) Take the tails of \(\mathbf{N}\):

\begin{equation*} A_n = \{n, n+1, n+2, \ldots\}. \end{equation*}

Each \(A_n\) is infinite and \(A_1 \supseteq A_2 \supseteq \cdots\), but \(\bigcap_{n=1}^{\infty} A_n = \emptyset\), since any \(m \in \mathbf{N}\) fails to lie in \(A_{m+1}\).

(b) True. Because \(A_n \subseteq A_1\) and \(A_1\) is finite, the cardinalities

\begin{equation*} |A_1| \ \ge\ |A_2| \ \ge\ |A_3| \ \ge\ \cdots \ \ge\ 1 \end{equation*}

form a nonincreasing sequence of positive integers, which must therefore be eventually constant: there is an \(N\) with \(|A_n| = |A_N|\) for all \(n \ge N\). For such \(n\) the inclusion \(A_n \subseteq A_N\) between finite sets of equal cardinality forces \(A_n = A_N\). Hence

\begin{equation*} \bigcap_{n=1}^{\infty} A_n \;=\; \bigcap_{n=1}^{N} A_n \;=\; A_N , \end{equation*}

the last equality by the nesting. So the intersection equals \(A_N\), which is finite and nonempty.

(c) False. Take \(A = B = \{1\}\) and \(C = \{2\}\):

\begin{equation*} \begin{aligned} A \cap (B \cup C) &= \{1\} \cap \{1,2\} = \{1\},\\ (A \cap B) \cup C &= \{1\} \cup \{2\} = \{1,2\}. \end{aligned} \end{equation*}

(d) True, and it is just associativity of \(\cap\): each side consists precisely of those \(x\) belonging to all three of \(A\), \(B\), \(C\), since

\begin{equation*} \begin{aligned} x \in A \cap (B \cap C) &\iff x \in A \ \wedge\ (x \in B \ \wedge\ x \in C)\\ &\iff (x \in A \ \wedge\ x \in B) \ \wedge\ x \in C\\ &\iff x \in (A \cap B) \cap C . \end{aligned} \end{equation*}

(e) True, and it is the distributive law for \(\wedge\) over \(\vee\):

\begin{equation*} \begin{aligned} x \in A \cap (B \cup C) &\iff x \in A \ \wedge\ (x \in B \ \vee\ x \in C)\\ &\iff (x \in A \wedge x \in B) \ \vee\ (x \in A \wedge x \in C)\\ &\iff x \in (A \cap B) \cup (A \cap C). \end{aligned} \end{equation*}

Sort each natural number by the power of \(2\) it contains: set

\begin{equation*} A_i \;=\; \bigl\{\, 2^{\,i-1}(2k-1) \;:\; k \in \mathbf{N} \,\bigr\}, \end{equation*}

so that \(A_1 = \{1,3,5,7,\ldots\}\), \(A_2 = \{2,6,10,14,\ldots\}\), \(A_3 = \{4,12,20,28,\ldots\}\), and in general \(A_i\) consists of the natural numbers whose largest power-of-two divisor is exactly \(2^{i-1}\).

Each \(A_i\) is infinite because \(k \mapsto 2^{\,i-1}(2k-1)\) is injective on \(\mathbf{N}\). The other two properties are both the statement that every \(n \in \mathbf{N}\) can be written as

\begin{equation*} n = 2^{a} m, \qquad a \ge 0, \ m \text{ odd}, \end{equation*}

in exactly one way. Existence: divide \(n\) by \(2\) repeatedly until an odd quotient \(m\) appears, which happens after finitely many steps since the quotients strictly decrease. Uniqueness: if \(2^a m = 2^b m_1\) with \(a < b\) and \(m, m_1\) odd, then \(m = 2^{\,b-a} m_1\) is even, a contradiction; so \(a = b\) and then \(m = m_1\).

Consequently \(n\) lies in \(A_{a+1}\) and in no other \(A_i\), which gives simultaneously \(\bigcup_{i=1}^{\infty} A_i = \mathbf{N}\) and \(A_i \cap A_j = \emptyset\) for \(i \ne j\).

Because failing a conjunction is failing one of its two clauses, and failing a disjunction is failing both.

(a) Let \(x \in (A \cap B)^c\), and suppose \(x\) lies in neither \(A^c\) nor \(B^c\). Then \(x \in A\) and \(x \in B\), so

\begin{equation*} x \in A \cap B, \end{equation*}

contradicting \(x \notin A \cap B\). Hence \(x \in A^c\) or \(x \in B^c\), that is, \(x \in A^c \cup B^c\); as \(x\) was arbitrary, \((A \cap B)^c \subseteq A^c \cup B^c\).

(b) Let \(x \in A^c \cup B^c\), so \(x \notin A\) or \(x \notin B\). Membership in \(A \cap B\) demands both, so in either case \(x \notin A \cap B\), that is, \(x \in (A \cap B)^c\). Hence \(A^c \cup B^c \subseteq (A \cap B)^c\), and combining with (a),

\begin{equation*} (A \cap B)^c = A^c \cup B^c . \end{equation*}

(c) Both inclusions.

(\(\subseteq\)) If \(x \in (A \cup B)^c\) then \(x \notin A \cup B\); since \(x \in A\) alone would already put \(x\) in \(A \cup B\), and likewise for \(B\), we get \(x \notin A\) and \(x \notin B\), i.e. \(x \in A^c \cap B^c\).

(\(\supseteq\)) If \(x \in A^c \cap B^c\) then \(x \notin A\) and \(x \notin B\). Membership in \(A \cup B\) would require \(x\) to lie in at least one of them, so \(x \notin A \cup B\), i.e. \(x \in (A \cup B)^c\).

Therefore \((A \cup B)^c = A^c \cap B^c\).

(a) When \(a\) and \(b\) share a sign the triangle inequality holds with equality.

(i) \(a \ge 0\) and \(b \ge 0\). Then \(a + b \ge 0\), so

\begin{equation*} |a+b| = a + b = |a| + |b| . \end{equation*}

(ii) \(a \le 0\) and \(b \le 0\). Then \(a + b \le 0\), so

\begin{equation*} |a+b| = -(a+b) = (-a) + (-b) = |a| + |b| . \end{equation*}

(b) Every real \(x\) satisfies \(x \le |x|\) and \(x^2 = |x|^2\), and \(|ab| = |a||b|\) by property (i) of Example 1.2.5, whose sign-by-sign verification is independent of the property (ii) being proved here. Hence \(ab \le |ab| = |a||b|\), and

\begin{equation*} \begin{aligned} (a+b)^2 &= a^2 + 2ab + b^2\\ &\le |a|^2 + 2|a||b| + |b|^2\\ &= \bigl(|a| + |b|\bigr)^2 . \end{aligned} \end{equation*}

Now put \(s = |a+b| \ge 0\) and \(t = |a| + |b| \ge 0\). The display says \(s^2 = (a+b)^2 \le t^2\), and for nonnegative numbers \(s > t\) would force \(s^2 > t^2\); therefore \(s \le t\), which is

\begin{equation*} |a + b| \le |a| + |b| . \end{equation*}

(c) Insert \(c\) and \(d\) and apply (b) twice:

\begin{equation*} \begin{aligned} |a - b| &= \bigl|(a-c) + (c-d) + (d-b)\bigr|\\ &\le |a-c| + \bigl|(c-d) + (d-b)\bigr|\\ &\le |a-c| + |c-d| + |d-b| . \end{aligned} \end{equation*}

(d) Apply (b) to the identity \(a = (a-b) + b\), and then to \(b = (b-a) + a\):

\begin{equation*} \begin{aligned} |a| \le |a-b| + |b| \ &\Rightarrow\ |a| - |b| \le |a-b|,\\ |b| \le |b-a| + |a| \ &\Rightarrow\ |b| - |a| \le |b-a| = |a-b| . \end{aligned} \end{equation*}

The number \(\bigl||a|-|b|\bigr|\) equals \(|a|-|b|\) or \(|b|-|a|\) according to the sign of \(|a|-|b|\), and both are bounded by \(|a-b|\); hence \(\bigl||a|-|b|\bigr| \le |a-b|\).

(a) \(f(A) = [0,4]\) and \(f(B) = [1,16]\); both equalities hold in this case.

Since \(x \mapsto x^2\) is increasing on \([0,\infty)\) and both intervals lie there, it carries each interval onto the interval between the squares of its endpoints. With \(A \cap B = [1,2]\) and \(A \cup B = [0,4]\),

\begin{equation*} \begin{aligned} f(A \cap B) &= [1,4] = [0,4] \cap [1,16] = f(A) \cap f(B),\\ f(A \cup B) &= [0,16] = [0,4] \cup [1,16] = f(A) \cup f(B). \end{aligned} \end{equation*}

(b) Take \(A = \{1\}\) and \(B = \{-1\}\), still with \(f(x) = x^2\). Then \(A \cap B = \emptyset\), so

\begin{equation*} f(A \cap B) = \emptyset \ne \{1\} = \{1\} \cap \{1\} = f(A) \cap f(B). \end{equation*}

(c) Let \(y \in g(A \cap B)\), say \(y = g(x)\) with \(x \in A \cap B\). Then \(x \in A\) gives \(y \in g(A)\), and \(x \in B\) gives \(y \in g(B)\); hence \(y \in g(A) \cap g(B)\).

(d) Conjecture: \(g(A \cup B) = g(A) \cup g(B)\) for every \(g\) and all \(A, B \subseteq \mathbf{R}\). Unlike in (c), a single point of \(A \cup B\) already lies in one of the two sets, so every step reverses:

\begin{equation*} \begin{aligned} y \in g(A \cup B) \ &\iff\ y = g(x) \text{ for some } x \in A \cup B\\ &\iff\ y = g(x) \text{ for some } x \in A,\\ &\qquad\ \text{ or } y = g(x) \text{ for some } x \in B\\ &\iff\ y \in g(A) \ \vee\ y \in g(B)\\ &\iff\ y \in g(A) \cup g(B). \end{aligned} \end{equation*}

All three are possible. (Throughout, \(\mathbf{N} = \{1, 2, 3, \ldots\}\).)

(a) \(f(n) = n + 1\). It is 1–1 since \(f(a_1) = f(a_2)\) forces \(a_1 = a_2\), and it is not onto because \(f(n) \geq 2\) for every \(n\), so \(1\) has no preimage.

(b)

\begin{equation*} f(n) = \begin{cases} 1 & \text{if } n = 1, \\ n - 1 & \text{if } n \geq 2. \end{cases} \end{equation*}

Onto: given \(m \in \mathbf{N}\), \(f(m+1) = m\). Not 1–1: \(f(1) = f(2) = 1\).

(c) Interleave the nonnegative and negative integers:

\begin{equation*} f(n) = \begin{cases} (n-1)/2 & \text{if } n \text{ is odd}, \\ -n/2 & \text{if } n \text{ is even}, \end{cases} \end{equation*}

so that \((f(1), f(2), f(3), f(4), \ldots) = (0, -1, 1, -2, \ldots)\). The map

\begin{equation*} g(m) = \begin{cases} 2m + 1 & \text{if } m \geq 0, \\ -2m & \text{if } m < 0 \end{cases} \end{equation*}

sends \(\mathbf{Z}\) into \(\mathbf{N}\) and satisfies \(f(g(m)) = m\) and \(g(f(n)) = n\) (Check!), the branches never crossing because odd \(n\) give \(f(n) \geq 0\) and even \(n\) give \(f(n) < 0\). A map with a two-sided inverse is 1–1 and onto.

(a) \(f^{-1}(A) = [-2, 2]\) and \(f^{-1}(B) = [-1, 1]\): since \(x^2 \geq 0\) always, \(x^2 \in [0,4]\) says exactly \(|x| \leq 2\), and \(x^2 \in [-1,1]\) says exactly \(|x| \leq 1\). Both identities hold here, with \(A \cap B = [0,1]\) and \(A \cup B = [-1,4]\):

\begin{equation*} \begin{aligned} f^{-1}(A \cap B) &= \{x : 0 \le x^2 \le 1\} = [-1,1] = f^{-1}(A) \cap f^{-1}(B), \\ f^{-1}(A \cup B) &= \{x : -1 \le x^2 \le 4\} = [-2,2] = f^{-1}(A) \cup f^{-1}(B). \end{aligned} \end{equation*}

(b) Each identity is one chain of equivalences, since \(x \in g^{-1}(S)\) just relocates the membership test to \(g(x) \in S\). For every \(x \in \mathbf{R}\),

\begin{equation*} \begin{aligned} x \in g^{-1}(A \cap B) &\iff g(x) \in A \cap B \\ &\iff g(x) \in A \ \wedge \ g(x) \in B \\ &\iff x \in g^{-1}(A) \ \wedge \ x \in g^{-1}(B) \\ &\iff x \in g^{-1}(A) \cap g^{-1}(B), \end{aligned} \end{equation*}

and verbatim with \(\cap\) replaced by \(\cup\) and \(\wedge\) by \(\vee\):

\begin{equation*} \begin{aligned} x \in g^{-1}(A \cup B) &\iff g(x) \in A \ \vee \ g(x) \in B \\ &\iff x \in g^{-1}(A) \ \vee \ x \in g^{-1}(B) \\ &\iff x \in g^{-1}(A) \cup g^{-1}(B). \end{aligned} \end{equation*}

(a) False, (b) False, (c) True.

(a) The forward implication is fine (\(a < b \le b + \epsilon\) for \(\epsilon > 0\)), but the converse fails: take \(a = b = 0\). Then \(a < b + \epsilon\) holds for every \(\epsilon > 0\), while \(a < b\) does not.

(b) This is precisely the failing converse just exhibited: \(a = b = 0\) satisfies the hypothesis and violates the conclusion.

(c) (\(\Rightarrow\)) If \(a \le b\) then \(a \le b < b + \epsilon\) for every \(\epsilon > 0\).

(\(\Leftarrow\)) Contrapositively, suppose \(a > b\) and put \(\epsilon = a - b > 0\). Then

\begin{equation*} b + \epsilon = b + (a - b) = a, \end{equation*}

so \(a < b + \epsilon\) fails for that one \(\epsilon\). Hence if the inequality holds for every \(\epsilon > 0\), we must have \(a \le b\).

(a) Negation: There exist real numbers \(a < b\) such that \(a + 1/n \geq b\) for every \(n \in \mathbf{N}\). Symbolically the claim reads

\begin{equation*} (\forall a < b)(\exists n \in \mathbf{N}) \ a + 1/n < b, \end{equation*}

so its negation is \((\exists a < b)(\forall n \in \mathbf{N}) \ a + 1/n \geq b\). The claim is the true one: it says \(1/n\) can be made smaller than the gap \(b - a > 0\), which is the Archimedean Property (Theorem 1.4.2 (ii)).

(b) Negation: For every real number \(x > 0\) there exists an \(n \in \mathbf{N}\) satisfying \(1/n \leq x\). The negation is the true one: it is Theorem 1.4.2 (ii) again, applied with \(y = x\).

(c) Negation: There exist two distinct real numbers \(a < b\) such that every rational number \(r\) satisfies \(r \leq a\) or \(r \geq b\). The claim is the true one: it is the density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3).

(a) Base case: \(y_1 = 6 > -6\). Assuming \(y_n > -6\),

\begin{equation*} y_{n+1} = \frac{2y_n - 6}{3} > \frac{2(-6) - 6}{3} = \frac{-18}{3} = -6, \end{equation*}

the strict inequality surviving because \(t \mapsto (2t-6)/3\) is increasing. By induction \(y_n > -6\) for all \(n \in \mathbf{N}\).

(b) The recursion is affine, so consecutive differences contract by the factor \(2/3\):

\begin{equation*} y_{n+2} - y_{n+1} = \frac{2y_{n+1} - 6}{3} - \frac{2y_n - 6}{3} = \frac{2}{3}\,(y_{n+1} - y_n). \end{equation*}

Base case: \(y_2 = (2 \cdot 6 - 6)/3 = 2 < 6 = y_1\). Assuming \(y_{n+1} < y_n\), the display gives \(y_{n+2} - y_{n+1} = \frac{2}{3}(y_{n+1} - y_n) < 0\), i.e. \(y_{n+2} < y_{n+1}\). By induction \(y_{n+1} < y_n\) for every \(n \in \mathbf{N}\), so \((y_n)\) is decreasing.

(a) Induct on \(n\), the two-set law \((A \cup B)^c = A^c \cap B^c\) of Exercise 1.2.5 (c) supplying both the base case (\(n = 2\); the case \(n = 1\) is the identity \(A_1^c = A_1^c\)) and the inductive step. Assume the statement for some \(n \geq 2\). Grouping the first \(n\) sets as a single set and applying Exercise 1.2.5 (c) to that set and \(A_{n+1}\),

\begin{equation*} \begin{aligned} (A_1 \cup \cdots \cup A_n \cup A_{n+1})^c &= \bigl( (A_1 \cup \cdots \cup A_n) \cup A_{n+1} \bigr)^c \\ &= (A_1 \cup \cdots \cup A_n)^c \cap A_{n+1}^c \\ &= (A_1^c \cap \cdots \cap A_n^c) \cap A_{n+1}^c , \end{aligned} \end{equation*}

the last step by the induction hypothesis. This is the statement for \(n+1\), so it holds for all \(n \in \mathbf{N}\).

(b) Take the tails of \(\mathbf{N}\): \(B_i = \{i, i+1, i+2, \ldots\}\). These are nested (\(B_1 \supseteq B_2 \supseteq \cdots\)), so for each \(n\)

\begin{equation*} \bigcap_{i=1}^{n} B_i = B_n \ni n, \qquad \text{hence } \bigcap_{i=1}^{n} B_i \neq \emptyset . \end{equation*}

But \(\bigcap_{i=1}^{\infty} B_i = \emptyset\): any candidate \(m \in \mathbf{N}\) satisfies \(m < m+1\), so \(m \notin B_{m+1}\).

(c) One chain of equivalences, driven by the quantifier duality \(\neg \exists = \forall \neg\), which is indifferent to the size of the index set. For any \(x\),

\begin{equation*} \begin{aligned} x \in \left( \bigcup_{i=1}^{\infty} A_i \right)^{c} &\iff \neg \bigl( \exists\, i \in \mathbf{N} : x \in A_i \bigr) \\ &\iff \forall\, i \in \mathbf{N}, \ x \notin A_i \\ &\iff \forall\, i \in \mathbf{N}, \ x \in A_i^c \\ &\iff x \in \bigcap_{i=1}^{\infty} A_i^c . \end{aligned} \end{equation*}

(a) A real number \(i\) is the greatest lower bound (or infimum) for a set \(A \subseteq \mathbf{R}\) if it meets the following two criteria:

(i) \(i\) is a lower bound for \(A\);

(ii) if \(l\) is any lower bound for \(A\), then \(l \le i\).

We write \(i = \inf A\).

(b) Lemma. Assume \(i \in \mathbf{R}\) is a lower bound for a set \(A \subseteq \mathbf{R}\). Then \(i = \inf A\) if and only if, for every choice of \(\epsilon > 0\), there exists an element \(a \in A\) satisfying \(a < i + \epsilon\).

(\(\Rightarrow\)) Assume \(i = \inf A\) and let \(\epsilon > 0\). Since \(i < i + \epsilon\), criterion (ii) shows \(i + \epsilon\) is not a lower bound for \(A\); that is, some \(a \in A\) satisfies \(a < i + \epsilon\).

(\(\Leftarrow\)) Let \(l\) be any lower bound for \(A\) and suppose, for contradiction, that \(i < l\). Taking \(\epsilon = l - i > 0\) produces \(a \in A\) with

\begin{equation*} a < i + \epsilon = l, \end{equation*}

contradicting the fact that \(l\) is a lower bound. Hence \(l \le i\), which is criterion (ii), and \(i = \inf A\).

(a) \(B = \{3\}\), for which \(\inf B = \sup B = 3\). Equality is the only way this can happen: if \(B\) is nonempty, picking any \(b \in B\) gives

\begin{equation*} \inf B \le b \le \sup B , \end{equation*}

so the strict inequality \(\inf B > \sup B\) is impossible.

(b) Impossible. A finite nonempty set \(F = \{x_1, \dots, x_n\}\) contains a largest element \(x_M\) (induction on \(n\): for \(n = 1\) take \(x_1\), and otherwise take the larger of \(x_n\) and the maximum of \(\{x_1,\dots,x_{n-1}\}\)). That \(x_M\) is an upper bound lying in \(F\), so \(x_M = \sup F\) by Exercise 1.3.7, and \(F\) contains its supremum. (The empty set has no infimum to contain either.)

(c) \(B = \{1/n : n \in \mathbf{N}\} \subseteq \mathbf{Q}\), the set of Example 1.3.3. It is bounded, \(1 \in B\) is an upper bound so \(\sup B = 1 \in B\) by Exercise 1.3.7, and \(\inf B = 0 \notin B\): zero is a lower bound, and no \(\epsilon > 0\) is, because the Archimedean Property (Theorem 1.4.2) supplies \(n \in \mathbf{N}\) with \(1/n < \epsilon\).

(a) The Axiom of Completeness applies to \(B\): it is nonempty because \(A\) is bounded below, and it is bounded above because every \(a \in A\) is an upper bound for \(B\) (each \(b \in B\) satisfies \(b \le a\) by definition of lower bound). So \(s = \sup B\) exists; we check the two criteria of Exercise 1.3.1(a) for \(s = \inf A\).

(i) \(s\) is a lower bound for \(A\). Fix \(a \in A\). As just noted \(a\) is an upper bound for \(B\), so Definition 1.3.2(ii) gives

\begin{equation*} s \le a \qquad \text{for every } a \in A . \end{equation*}

(ii) If \(l\) is any lower bound for \(A\), then \(l \in B\) by the definition of \(B\), and \(s\) is an upper bound for \(B\), so \(l \le s\).

Hence \(\sup B = s = \inf A\).

(b) Part (a) derives the existence of greatest lower bounds from the Axiom of Completeness: given any nonempty \(A\) bounded below, the set \(B\) of its lower bounds is nonempty and bounded above, the Axiom produces \(\sup B\), and (a) identifies that number as \(\inf A\). An assertion provable from the axiom already stated is a theorem, not an axiom, so adding it would be redundant.

(a) \(\sup(A_1 \cup A_2) = \max\{\sup A_1, \sup A_2\}\). Write \(s_k = \sup A_k\) and \(s = \max\{s_1, s_2\}\) (a maximum of two reals always exists), and verify Definition 1.3.2.

(i) Any \(x \in A_1 \cup A_2\) lies in some \(A_k\), whence \(x \le s_k \le s\).

(ii) If \(b\) is an upper bound for \(A_1 \cup A_2\), it is in particular an upper bound for each \(A_k\), so \(s_k \le b\) for \(k = 1, 2\) and therefore \(s = \max\{s_1, s_2\} \le b\).

Induction extends this to any finite union: assuming the formula for \(n - 1\) sets and applying the two-set case to \(\left(\bigcup_{k=1}^{n-1} A_k\right) \cup A_n\),

\begin{equation*} \sup\left(\bigcup_{k=1}^{n} A_k\right) = \max\{\sup A_1, \dots, \sup A_n\} . \end{equation*}

(b) No. The union of infinitely many bounded sets need not be bounded above, and even when it is, the set of suprema need not have a maximum. Both failures are visible at once:

(i) \(A_k = \{k\}\) gives \(\bigcup_{k=1}^{\infty} A_k = \mathbf{N}\), which has no supremum at all (Theorem 1.4.2).

(ii) \(A_k = \{1 - 1/k\}\) gives a bounded union whose suprema \(1 - 1/k\) have no largest element, so \(\max_k \sup A_k\) does not exist; nevertheless \(\sup\left(\bigcup_{k=1}^{\infty} A_k\right) = 1\).

The correct extension replaces the maximum by a supremum: if \(\bigcup_{k=1}^{\infty} A_k\) is bounded above, then so is \(S = \{\sup A_k : k \in \mathbf{N}\}\) (any upper bound \(b\) for the union bounds each \(A_k\), hence each \(\sup A_k\)), and

\begin{equation*} \sup\left(\bigcup_{k=1}^{\infty} A_k\right) = \sup S . \end{equation*}

Indeed \(\sup S\) is an upper bound for the union, since \(x \in A_k\) forces \(x \le \sup A_k \le \sup S\); and any upper bound \(b\) for the union satisfies \(\sup A_k \le b\) for every \(k\), so \(\sup S \le b\).

(a) Set \(s = \sup A\). If \(c = 0\) then \(cA = \{0\}\) and both sides equal \(0\), so assume \(c > 0\) and check the two parts of Definition 1.3.2.

(i) For every \(a \in A\) we have \(a \le s\), and multiplying by \(c > 0\) preserves the inequality:

\begin{equation*} ca \le cs \qquad \text{for all } a \in A , \end{equation*}

so \(cs\) is an upper bound for \(cA\).

(ii) Let \(b\) be an arbitrary upper bound for \(cA\), so \(ca \le b\) for all \(a \in A\). Dividing by \(c > 0\) gives \(a \le b/c\) for all \(a \in A\), i.e. \(b/c\) is an upper bound for \(A\). Because \(s\) is the least upper bound, \(s \le b/c\), and multiplying by \(c > 0\) yields \(cs \le b\).

Hence \(\sup(cA) = cs = c \sup A\).

(b) For \(c < 0\), \(\sup(cA) = c \inf A\) whenever \(A\) is also bounded below. Multiplication by \(c\) reverses inequalities, so it trades suprema for infima, and the identity available from the standing hypothesis (bounded above only) is the reflected one:

\begin{equation*} \inf(cA) = c \sup A . \end{equation*}

That one holds by the criteria of Exercise 1.3.1(a) with \(s = \sup A\). Every \(a \in A\) satisfies \(a \le s\), hence \(ca \ge cs\), so \(cs\) is a lower bound for \(cA\). If \(l\) is any lower bound for \(cA\), then \(ca \ge l\) for all \(a \in A\) gives \(a \le l/c\), so \(l/c\) is an upper bound for \(A\), so \(s \le l/c\) and therefore \(cs \ge l\). And \(\sup(cA) = c \inf A\) follows by applying this identity to the set \(cA\) with the scalar \(1/c < 0\), legitimate because \(A\) bounded below makes \(cA\) bounded above.

(a) A typical element of \(A + B\) is \(a + b\) with \(a \in A\) and \(b \in B\), and adding the inequalities \(a \le s\) and \(b \le t\) gives

\begin{equation*} a + b \le s + t . \end{equation*}

(b) With \(a \in A\) fixed, \(a + b \in A + B\) for every \(b \in B\), so \(a + b \le u\), i.e.

\begin{equation*} b \le u - a \qquad \text{for all } b \in B . \end{equation*}

Thus \(u - a\) is an upper bound for \(B\), and since \(t\) is the least upper bound, \(t \le u - a\).

(c) The inequality from (b) holds for every \(a \in A\) (the choice of \(a\) was arbitrary), and it rearranges to \(a \le u - t\). So \(u - t\) is an upper bound for \(A\), whence \(s \le u - t\), i.e.

\begin{equation*} s + t \le u . \end{equation*}

By (a), \(s + t\) is an upper bound for \(A + B\); by the display, it is no larger than any other upper bound \(u\). Both criteria of Definition 1.3.2 hold, so \(\sup(A + B) = s + t\).

(d) Method (2): \(s + t\) is an upper bound for \(A + B\) by (a), so Lemma 1.3.8 reduces the claim to producing, for each \(\epsilon > 0\), an element of \(A + B\) exceeding \((s + t) - \epsilon\). Applying the lemma to \(A\) and to \(B\) with \(\epsilon/2\) gives \(a \in A\) and \(b \in B\) with \(s - \epsilon/2 < a\) and \(t - \epsilon/2 < b\); adding,

\begin{equation*} (s + t) - \epsilon < a + b \in A + B . \end{equation*}

Lemma 1.3.8 now yields \(\sup(A + B) = s + t\).

Criterion (i) of Definition 1.3.2 is the hypothesis that \(a\) is an upper bound for \(A\), so only (ii) needs checking. Let \(b\) be any upper bound for \(A\). Since \(a \in A\), the defining property of an upper bound applied to the element \(a\) gives

\begin{equation*} a \le b . \end{equation*}

Thus \(a\) is an upper bound no larger than any other, and \(a = \sup A\).

(a) \(\sup = 1\) and \(\inf = 0\), neither attained: the set is exactly \(\mathbf{Q} \cap (0,1)\).

(b) \(\sup = 1\) and \(\inf = -1\), both attained at \(n = 1\) (\(m\) even, resp. \(m\) odd), hence a maximum and a minimum.

(c) \(\sup = 1/3\), not attained; \(\inf = 1/4\), attained at \(n = 1\), since \(n/(3n+1) = 1/(3 + 1/n)\) increases in \(n\).

(d) \(\sup = 1\) and \(\inf = 0\), neither attained, since \(m/(m+n) = 1/(1 + n/m)\) with \(n/m\) ranging over all positive rationals.

(a) Apply Lemma 1.3.8 to \(B\) with \(\epsilon = \sup B - \sup A > 0\) (legitimate precisely because the inequality is strict): there is \(b \in B\) with

\begin{equation*} \sup A = \sup B - \epsilon < b . \end{equation*}

Then every \(a \in A\) satisfies \(a \le \sup A < b\), so \(b\) is an upper bound for \(A\).

(b) Take \(A = \{0\}\) and \(B = \{-1/n : n \in \mathbf{N}\}\). Here \(\sup A = 0 = \sup B\), yet every \(b \in B\) is negative and hence fails to bound \(0 \in A\).

(a) Take \(c = \sup A\). It exists by the Axiom of Completeness: \(A\) is nonempty, and each of the (at least one) elements \(b \in B\) is an upper bound for \(A\) because \(a < b\) for every \(a \in A\). The two required properties are then exactly the two halves of Definition 1.3.2:

(i) \(x \le c\) for all \(x \in A\), since \(c\) is an upper bound for \(A\);

(ii) \(c \le x\) for all \(x \in B\), since each such \(x\) is an upper bound for \(A\) and \(c\) is the least one.

(b) Split \(\mathbf{R}\) at \(E\): let

\begin{equation*} \begin{aligned} B &= \{b \in \mathbf{R} : b \text{ is an upper bound for } E\}, \\ A &= \mathbf{R} \setminus B . \end{aligned} \end{equation*}

These are disjoint with union \(\mathbf{R}\) by construction; \(B \ne \emptyset\) because \(E\) is bounded above, and \(A \ne \emptyset\) because picking \(x_0 \in E\) (possible as \(E \ne \emptyset\)) makes \(x_0 - 1\) fail to be an upper bound. If \(a \in A\) and \(b \in B\), then \(a\) is not an upper bound, so some \(x \in E\) has \(x > a\), whence \(b \ge x > a\). The Cut Property supplies \(c\) with \(x \le c\) on \(A\) and \(c \le x\) on \(B\).

Then \(c = \sup E\). For (i), if \(c\) were not an upper bound for \(E\) there would be \(x \in E\) with \(x > c\); the midpoint \(t = (c+x)/2\) satisfies \(t < x\), so \(t\) is not an upper bound, i.e. \(t \in A\), forcing \(t \le c\) and contradicting \(t > c\). For (ii), any upper bound \(b\) for \(E\) lies in \(B\), so \(c \le b\).

(c) Cut \(\mathbf{Q}\) at \(\sqrt{2}\):

\begin{equation*} \begin{aligned} A &= \{x \in \mathbf{Q} : x \le 0\} \cup \{x \in \mathbf{Q} : x > 0,\ x^2 < 2\}, \\ B &= \{x \in \mathbf{Q} : x > 0,\ x^2 > 2\}. \end{aligned} \end{equation*}

They are nonempty (\(1 \in A\), \(2 \in B\)), disjoint, and \(A \cup B = \mathbf{Q}\) because no rational satisfies \(x^2 = 2\) (Theorem 1.1.1); and \(a < b\) for \(a \in A\), \(b \in B\), since \(a \le 0 < b\) in the first case and \(a^2 < 2 < b^2\) with both positive in the second. A cut point \(c\) would lie in \(A\) or in \(B\), and would be a maximum of \(A\) or a minimum of \(B\) respectively. Neither exists: for \(c \in A\) with \(c > 0\), choose \(n \in \mathbf{N}\) with \(n > (2c+1)/(2-c^2)\) (a rational bound, so such an \(n\) is found by inspection of its numerator, with no appeal to completeness), so that

\begin{equation*} \Big(c + \tfrac{1}{n}\Big)^2 \le c^2 + \frac{2c+1}{n} < 2 \end{equation*}

and \(c + 1/n \in A\) exceeds \(c\) (and if \(c \le 0\), already \(1 \in A\) exceeds \(c\)); for \(c \in B\), choose \(n > \max\{2c/(c^2-2),\, 1/c\}\), so that \(c - 1/n > 0\) and

\begin{equation*} \Big(c - \tfrac{1}{n}\Big)^2 > c^2 - \frac{2c}{n} > 2, \end{equation*}

putting \(c - 1/n \in B\) below \(c\). So the Cut Property fails over \(\mathbf{Q}\).

(a) True. \(\sup B\) is an upper bound for \(B\), hence for the subset \(A\), and \(\sup A\) is the least upper bound for \(A\) (Definition 1.3.2 (ii)), so \(\sup A \le \sup B\). Both suprema exist by the Axiom of Completeness, the sets being nonempty and bounded above.

(b) True. Take the midpoint \(c = (\sup A + \inf B)/2\), which satisfies \(\sup A < c < \inf B\). Then for all \(a \in A\) and \(b \in B\),

\begin{equation*} a \le \sup A < c < \inf B \le b . \end{equation*}

(c) False. Let \(A = (0,1)\), \(B = (1,2)\), and \(c = 1\): every \(a \in A\) and \(b \in B\) satisfies \(a < 1 < b\), yet

\begin{equation*} \sup A = 1 = \inf B , \end{equation*}

so the strict inequality fails.

(a) Write \(a = p/q\) and \(b = r/s\) with \(p, r \in \mathbf{Z}\) and \(q, s \in \mathbf{N}\). Then

\begin{equation*} a + b = \frac{ps + rq}{qs}, \qquad ab = \frac{pr}{qs}, \end{equation*}

and in each case the numerator lies in \(\mathbf{Z}\) and the denominator \(qs\) lies in \(\mathbf{N}\) (in particular \(qs \neq 0\)), because \(\mathbf{Z}\) is closed under addition and multiplication. Hence \(a + b, ab \in \mathbf{Q}\).

(b) Both are contrapositives of (a). If \(a + t = r\) were rational, then \(-a = (-1)a \in \mathbf{Q}\) and so

\begin{equation*} t = r + (-a) \in \mathbf{Q} \end{equation*}

by (a), contradicting \(t \in \mathbf{I}\). Likewise, if \(a \neq 0\) and \(at = r\) were rational, then \(1/a \in \mathbf{Q}\) (if \(a = p/q\) with \(p \neq 0\), then \(1/a = q/p\)) and

\begin{equation*} t = \frac{1}{a} \cdot r \in \mathbf{Q} \end{equation*}

by (a), again a contradiction.

(c) No, \(\mathbf{I}\) is closed under neither operation, and in general nothing can be said: \(s + t\) and \(st\) may be rational or irrational. With \(s = \sqrt{2}\) (irrational by Theorem 1.1.1, and a real number by Theorem 1.4.5):

\begin{equation*} \begin{aligned} t = -\sqrt{2}: &\quad s + t = 0 \in \mathbf{Q}, \\ t = \sqrt{2}: &\quad s + t = 2\sqrt{2} \in \mathbf{I}, \quad st = 2 \in \mathbf{Q}, \\ t = 1 + \sqrt{2}: &\quad st = 2 + \sqrt{2} \in \mathbf{I}. \end{aligned} \end{equation*}

(Part (b), applied to the rationals \(-1\), \(1\) and \(2\), puts each of \(-\sqrt{2}\), \(1 + \sqrt{2}\), \(2\sqrt{2}\) and \(2 + \sqrt{2}\) in \(\mathbf{I}\).)

\(s\) is an upper bound for \(A\): if some \(a \in A\) had \(a > s\), then \(a - s > 0\), and the Archimedean Property (Theorem 1.4.2 (ii)) supplies \(n \in \mathbf{N}\) with

\begin{equation*} \frac{1}{n} < a - s, \qquad \text{i.e.} \qquad s + \frac{1}{n} < a, \end{equation*}

contradicting the hypothesis that \(s + \frac{1}{n}\) is an upper bound for \(A\).

For leastness apply Lemma 1.3.8: let \(\varepsilon > 0\) and choose (again by Theorem 1.4.2 (ii)) an \(n \in \mathbf{N}\) with \(1/n < \varepsilon\). Since \(s - \frac{1}{n}\) is not an upper bound for \(A\), there exists \(a \in A\) with

\begin{equation*} s - \varepsilon < s - \frac{1}{n} < a . \end{equation*}

Thus \(s\) is an upper bound admitting an element of \(A\) within every \(\varepsilon\) of it, so \(s = \sup A\).

Suppose \(x \in \bigcap_{n=1}^{\infty} (0, 1/n)\). Then \(x > 0\), so the Archimedean Property (Theorem 1.4.2 (ii)) applied to \(y = x\) gives an \(n_0 \in \mathbf{N}\) with

\begin{equation*} \frac{1}{n_0} < x , \end{equation*}

whence \(x \notin (0, 1/n_0)\) and \(x\) is not in the intersection after all. No such \(x\) exists, so the intersection is empty.

The sets \((0, 1/n)\) are nonempty, bounded and nested, \((0,1) \supseteq (0,1/2) \supseteq \cdots\), so the only hypothesis of the Nested Interval Property (Theorem 1.4.1) that fails is closedness of the intervals.

\(b\) is an upper bound for \(T\) because \(T \subseteq [a, b]\); the content is leastness, which is Lemma 1.3.8 fed by the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3).

Let \(\varepsilon > 0\) and set

\begin{equation*} c = \max\{a, \, b - \varepsilon\} . \end{equation*}

Then \(c < b\) (both \(a < b\) and \(b - \varepsilon < b\) hold), so Theorem 1.4.3 applied to the pair \(c < b\) produces a rational \(r\) with \(c < r < b\). From \(c \ge a\) we get \(a < r < b\), hence \(r \in \mathbf{Q} \cap [a,b] = T\); from \(c \ge b - \varepsilon\) we get

\begin{equation*} b - \varepsilon \le c < r . \end{equation*}

So every \(\varepsilon\)-neighborhood below \(b\) catches a point of \(T\) (in particular \(T \neq \emptyset\)), and Lemma 1.3.8 gives \(\sup T = b\).

Take \(t = r + \sqrt{2}\), where \(r \in \mathbf{Q}\) is chosen by the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3) applied to the pair \(a - \sqrt{2} < b - \sqrt{2}\), so that

\begin{equation*} a - \sqrt{2} < r < b - \sqrt{2}. \end{equation*}

Adding \(\sqrt{2}\) throughout gives \(a < t < b\). And \(t \in \mathbf{I}\): \(\sqrt{2}\) is irrational by Theorem 1.1.1 (it exists as a real number by Theorem 1.4.5), so \(r + \sqrt{2} \in \mathbf{I}\) by Exercise 1.4.1 (b) with the rational \(r\).

Only (b) is dense.

(a) Not dense: the interval \((0, 1/11)\) contains no element. Indeed, if \(p/q > 0\) with \(p \in \mathbf{Z}\), \(q \in \mathbf{N}\), \(q \le 10\), then \(p \ge 1\) and

\begin{equation*} \frac{p}{q} \ge \frac{1}{q} \ge \frac{1}{10} > \frac{1}{11}. \end{equation*}

(b) Dense. Let \(a < b\). By the Archimedean Property (Theorem 1.4.2 (ii)) pick \(n \in \mathbf{N}\) with \(1/n < b - a\); since \(2^n \ge n\) for all \(n \in \mathbf{N}\) (induction: \(2^1 = 2 \ge 1\), and \(2^{n+1} = 2 \cdot 2^n \ge 2n \ge n+1\)), the denominator \(q = 2^n\) satisfies

\begin{equation*} \frac{1}{2^n} \le \frac{1}{n} < b - a . \end{equation*}

Now run the argument of Theorem 1.4.3 with this \(q\): let \(p \in \mathbf{Z}\) be the smallest integer greater than \(2^n a\), so that

\begin{equation*} p - 1 \le 2^n a < p . \end{equation*}

The right inequality gives \(a < p/2^n\), and the left one gives

\begin{equation*} \begin{aligned} p &\le 2^n a + 1 \\ &< 2^n a + 2^n (b - a) \\ &= 2^n b , \end{aligned} \end{equation*}

so \(p/2^n < b\). Hence \(a < p/2^n < b\) with \(p/2^n\) in the set.

(c) Not dense: the interval \((0, 1/10)\) contains no element, since \(10|p| \ge q\) forces

\begin{equation*} \left| \frac{p}{q} \right| = \frac{|p|}{q} \ge \frac{1}{10} . \end{equation*}

(Note \(p = 0\) is excluded outright, as \(10|p| \ge q \ge 1\) fails.)

Assume \(\alpha^2 > 2\). Note first \(\alpha \ge 1 > 0\), since \(1 \in T\) and \(\alpha\) is an upper bound for \(T\), so \((\alpha^2 - 2)/(2\alpha) > 0\) and the Archimedean Property (Theorem 1.4.2 (ii)) provides \(n_0 \in \mathbf{N}\) with

\begin{equation*} \frac{1}{n_0} < \frac{\alpha^2 - 2}{2\alpha}, \qquad \text{i.e.} \qquad \frac{2\alpha}{n_0} < \alpha^2 - 2 . \end{equation*}

Feeding this into the displayed estimate of the proof,

\begin{equation*} \begin{aligned} \left( \alpha - \frac{1}{n_0} \right)^2 &> \alpha^2 - \frac{2\alpha}{n_0} \\ &> \alpha^2 - (\alpha^2 - 2) \\ &= 2 . \end{aligned} \end{equation*}

Also \(\alpha - 1/n_0 > 0\), because \(1/n_0 < (\alpha^2 - 2)/(2\alpha) < \alpha^2/(2\alpha) = \alpha/2\).

Now \(\alpha - 1/n_0\) is an upper bound for \(T\): given \(t \in T\), either \(t \le 0 < \alpha - 1/n_0\), or \(t > 0\) and then

\begin{equation*} t^2 < 2 < \left( \alpha - \frac{1}{n_0} \right)^2 \end{equation*}

forces \(t < \alpha - 1/n_0\) (for positive reals \(x, y\), \(x \ge y\) implies \(x^2 \ge y^2\)). So \(T\) has an upper bound strictly smaller than \(\alpha\), contradicting the leastness half of \(\alpha = \sup T\). Hence \(\alpha^2 > 2\) is impossible, and with \(\alpha^2 < 2\) already excluded, \(\alpha^2 = 2\).

(a), (b), (c) are possible; (d) is impossible.

(a) Take

\begin{equation*} A = \left\{1 - \tfrac{1}{2n} : n \in \mathbf{N}\right\}, \qquad B = \left\{1 - \tfrac{1}{2n-1} : n \in \mathbf{N}\right\}. \end{equation*}

Since \(m \mapsto 1 - 1/m\) is injective and no integer is both even and odd, \(A \cap B = \emptyset\). Both sets are bounded above by \(1\) and miss it. Given \(c < 1\), the Archimedean Property (Theorem 1.4.2) supplies \(m\) with \(1/m < 1 - c\), hence also \(1/(m+1) < 1 - c\); one of \(m, m+1\) is even and the other odd, so each of \(A\), \(B\) contains a point exceeding \(c\). Thus \(\sup A = \sup B = 1 \notin A \cup B\).

(b) Take \(J_n = (-1/n, 1/n)\), which is nested since \(1/(n+1) < 1/n\). Then

\begin{equation*} \bigcap_{n=1}^{\infty} J_n = \{0\}, \end{equation*}

a single element: \(0 \in J_n\) for all \(n\), while for \(x \neq 0\) the Archimedean Property supplies \(n\) with \(1/n < |x|\), so \(x \notin J_n\).

(c) Take \(L_n = [n, \infty)\), nested because \(n < n+1\). Given \(x \in \mathbf{R}\), the Archimedean Property supplies \(n \in \mathbf{N}\) with \(n > x\), so \(x \notin L_n\); hence \(\bigcap_{n=1}^{\infty} L_n = \emptyset\).

(d) Impossible. Write \(I_n = [a_n, b_n]\) and set \(A_N = \bigcap_{n=1}^{N} I_n\), so

\begin{equation*} A_N = \Big[\max_{1 \le n \le N} a_n, \ \min_{1 \le n \le N} b_n\Big] \end{equation*}

is a closed bounded interval, nonempty by hypothesis, and \(A_{N+1} = A_N \cap I_{N+1}\) makes the \(A_N\) nested. So the Nested Interval Property (Theorem 1.4.1) applies to \((A_N)\), giving

\begin{equation*} \begin{aligned} \bigcap_{n=1}^{\infty} I_n &= \bigcap_{N=1}^{\infty} \bigcap_{n=1}^{N} I_n \\ &= \bigcap_{N=1}^{\infty} A_N \neq \emptyset. \end{aligned} \end{equation*}

Having chosen \(n_1 < n_2 < \cdots < n_k\), set

\begin{equation*} n_{k+1} \;=\; \min\{\, n \in \mathbf{N} : n > n_k \ \text{and}\ f(n) \in A \,\}, \qquad g(k+1) = f(n_{k+1}). \end{equation*}

The set being minimized is nonempty: \(f\) is 1–1 and onto \(B \supseteq A\), so \(f\) restricts to a bijection of \(S = \{n : f(n) \in A\}\) onto \(A\), whence \(S\) is infinite and meets \(\{n : n > n_k\}\). The minimum then exists by the Well-Ordering Property of \(\mathbf{N}\), so \(g : \mathbf{N} \to A\) is defined on all of \(\mathbf{N}\), and \(g(k) = f(n_k) \in A\) for every \(k\).

\(g\) is 1–1. The sequence \((n_k)\) is strictly increasing by construction, so \(j \ne k\) gives \(n_j \ne n_k\), and \(f\) being 1–1 gives \(g(j) = f(n_j) \ne f(n_k) = g(k)\).

\(g\) is onto \(A\). Fix \(a \in A\) and let \(m = f^{-1}(a)\), so \(m \in S\). Since \(n_k \ge k\) for all \(k\) (a strictly increasing sequence in \(\mathbf{N}\)), the set \(\{k : n_k \ge m\}\) is nonempty; let \(k\) be its least element. Two cases:

(i) \(k = 1\). Then \(n_1 = \min S \le m\) because \(m \in S\), and \(n_1 \ge m\), so \(n_1 = m\).

(ii) \(k > 1\). Then \(n_{k-1} < m \le n_k\) by minimality of \(k\), so \(m\) is a competitor in the minimum defining \(n_k\); hence \(n_k \le m\), and therefore \(n_k = m\).

In both cases \(g(k) = f(m) = a\). Thus \(g\) is a 1–1 correspondence between \(\mathbf{N}\) and \(A\), i.e. \(\mathbf{N} \sim A\), so \(A\) is countable by Definition 1.5.5.

The step “our construction implies \(\bigcap_{n=1}^{\infty} I_n = \emptyset\)” is false. What the construction actually gives is

\begin{equation*} \Big(\bigcap_{n=1}^{\infty} I_n\Big) \cap \mathbf{Q} = \emptyset , \end{equation*}

since the only points excluded are the \(r_n\), and those exhaust \(\mathbf{Q}\), not \(\mathbf{R}\).

In Theorem 1.5.6 (ii) the list \(\{x_1, x_2, x_3, \ldots\}\) was assumed to be all of \(\mathbf{R}\), so ruling out every listed point ruled out every point whatsoever and forced the intersection to be empty. Here the intervals \(I_n\) are still intervals of real numbers, and NIP (Theorem 1.4.1) supplies some \(x \in \bigcap_{n=1}^{\infty} I_n\). The displayed line says only that this \(x\) is irrational — a true and unsurprising statement, consistent with \(\mathbf{Q}\) being countable (Theorem 1.5.6 (i)). No contradiction arises.

(a) Interleave the two lists. Let \(B_2 = A_2 \setminus A_1\), so \(A_1 \cup A_2 = A_1 \cup B_2\) with \(A_1 \cap B_2 = \emptyset\). Since \(B_2 \subseteq A_2\) and \(A_2\) is countable, Theorem 1.5.7 says \(B_2\) is countable or finite.

(i) \(B_2\) countable. Take bijections \(f : \mathbf{N} \to A_1\) and \(g : \mathbf{N} \to B_2\) and define

\begin{equation*} h(n) = \begin{cases} f\!\left(\tfrac{n+1}{2}\right) & \text{if } n \text{ is odd},\\[2pt] g\!\left(\tfrac{n}{2}\right) & \text{if } n \text{ is even}, \end{cases} \end{equation*}

exactly the odd/even split used for \(\mathbf{N} \sim \mathbf{Z}\) in Example 1.5.3 (ii). Then \(h\) is onto because \(f\) and \(g\) are, and \(h\) is 1–1 because \(f\) and \(g\) are and their ranges \(A_1\), \(B_2\) are disjoint.

(ii) \(B_2 = \{b_1, \ldots, b_k\}\) finite (possibly empty). Put \(h(n) = b_n\) for \(1 \le n \le k\) and \(h(n) = f(n-k)\) for \(n > k\); the same two clauses certify 1–1 and onto.

Either way \(A_1 \cup A_2 \sim \mathbf{N}\). For general \(m\), induct: \(A_1 \cup \cdots \cup A_{m+1} = (A_1 \cup \cdots \cup A_m) \cup A_{m+1}\) is the union of a countable set (inductive hypothesis) with a countable set, hence countable by the two-set case.

(b) Induction proves a statement \(P(m)\) for every \(m \in \mathbf{N}\), and here \(P(m)\) reads “\(A_1 \cup \cdots \cup A_m\) is countable” — a statement about a finite union. The set \(\bigcup_{n=1}^{\infty} A_n\) is not \(A_1 \cup \cdots \cup A_m\) for any \(m\), so it is never the subject of any \(P(m)\); there is no final “\(m = \infty\)” step to take. The gap is genuine, not cosmetic: with \(A_n = \{n\}\), every \(A_1 \cup \cdots \cup A_m\) is finite, yet \(\bigcup_{n=1}^{\infty} A_n = \mathbf{N}\) is not.

(c) The array exhibits a 1–1, onto map \(\mathbf{N} \times \mathbf{N} \to \mathbf{N}\); counting along reverse diagonals, the entry in row \(n\), column \(m\) is

\begin{equation*} \sigma(n,m) \;=\; \frac{(n+m-1)(n+m)}{2} \;-\; (n-1). \end{equation*}

Each diagonal \(n + m - 1 = d\) contributes the \(d\) consecutive integers from \(\tfrac{d(d-1)}{2}+1\) to \(\tfrac{d(d+1)}{2}\), read bottom-to-top, and these blocks tile \(\mathbf{N}\) without overlap — so \(\sigma\) is a bijection. (Check: \(\sigma(1,1)=1\), \(\sigma(2,1)=2\), \(\sigma(1,2)=3\), \(\sigma(3,1)=4\), \(\sigma(2,2)=5\).)

Now let \(A_n\) be countable for each \(n \in \mathbf{N}\) and write \(A_n = \{a_{n1}, a_{n2}, a_{n3}, \ldots\}\) using a bijection \(\mathbf{N} \to A_n\). Put \(A = \bigcup_{n=1}^{\infty} A_n\) and define \(\varphi : A \to \mathbf{N}\) by

\begin{equation*} \varphi(x) = \sigma(n,m), \quad \begin{aligned} &n = \min\{k : x \in A_k\},\\ &m = \text{the index of } x \text{ in the list for } A_n . \end{aligned} \end{equation*}

Both \(n\) and \(m\) are uniquely determined by \(x\) (the first by the Well-Ordering Property, the second because the list for \(A_n\) repeats nothing), so \(\varphi\) is well defined; and it is 1–1 because \(\sigma\) is injective and the pair \((n,m)\) recovers \(x\) as \(a_{nm}\). Hence \(A \sim \varphi(A) \subseteq \mathbf{N}\), and Theorem 1.5.7 makes \(\varphi(A)\) — so also \(A\) — countable or finite. Since \(A \supseteq A_1\) is infinite, \(A\) is countable.

(a) Straighten \((a,b)\) onto \((-1,1)\) and quote Example 1.5.4. The affine map

\begin{equation*} h(x) = \frac{2x - a - b}{b - a} \end{equation*}

is strictly increasing with inverse \(h^{-1}(t) = \tfrac{1}{2}\big((b-a)t + a + b\big)\), and \(h(a) = -1\), \(h(b) = 1\), so \(h\) carries \((a,b)\) 1–1 onto \((-1,1)\). Example 1.5.4 gives \((-1,1) \sim \mathbf{R}\) via \(f(x) = x/(x^2-1)\), so \(x \mapsto f(h(x))\) is a 1–1 correspondence \((a,b) \to \mathbf{R}\) by the transitivity of \(\sim\) (Exercise 1.5.5 (c)).

(b) Take

\begin{equation*} k(x) = \frac{1}{x - a + 1}, \qquad x \in (a,\infty). \end{equation*}

For \(x > a\) we have \(x - a + 1 > 1\), so \(k(x) \in (0,1)\); and \(k\) is 1–1 onto \((0,1)\) because \(y \mapsto a - 1 + 1/y\) inverts it, mapping \((0,1)\) back into \((a,\infty)\) (indeed \(1/y > 1\) exactly when \(0 < y < 1\)). Thus \((a,\infty) \sim (0,1) \sim \mathbf{R}\) by (a) and transitivity.

(c) Shift a sequence one place along (Hilbert’s hotel). Let \(D = \{0\} \cup \{1/n : n \ge 2\}\) and define \(g : [0,1) \to (0,1)\) by

\begin{equation*} g(x) = \begin{cases} 1/2 & \text{if } x = 0,\\[2pt] 1/(n+1) & \text{if } x = 1/n,\ n \ge 2,\\[2pt] x & \text{if } x \in [0,1) \setminus D . \end{cases} \end{equation*}

Then \(g(D) = \{1/n : n \ge 2\}\) is a 1–1 image of \(D\) (the map \(0 \mapsto 1/2\), \(1/n \mapsto 1/(n+1)\) is injective), \(g\) is the identity on \([0,1) \setminus D = (0,1) \setminus \{1/n : n \ge 2\}\), and these two image sets are disjoint and have union \((0,1)\). Hence \(g\) is 1–1 and onto, and \([0,1) \sim (0,1)\).

(a) The identity \(i : A \to A\), \(i(x) = x\), is 1–1 (\(i(x) = i(y)\) says \(x = y\)) and onto (\(x = i(x)\)), so \(A \sim A\) by Definition 1.5.2.

(b) If \(f : A \to B\) is 1–1 and onto, then each \(y \in B\) has exactly one preimage — at least one by ontoness, at most one by injectivity — so \(f^{-1} : B \to A\), \(f^{-1}(y) = \) that preimage, is a function. It is onto (given \(x \in A\), \(f^{-1}(f(x)) = x\)) and 1–1 (if \(f^{-1}(y_1) = f^{-1}(y_2) = x\) then \(y_1 = f(x) = y_2\)). Hence \(A \sim B\) gives \(B \sim A\), and interchanging the roles of \(A\) and \(B\) gives the converse.

(c) Compose. Let \(f : A \to B\) and \(g : B \to C\) both be 1–1 and onto, and set \(h = g \circ f : A \to C\). Then

\begin{equation*} \begin{aligned} h(x_1) = h(x_2) \ &\Rightarrow\ f(x_1) = f(x_2) &&(g \text{ is } 1\text{–}1)\\ &\Rightarrow\ x_1 = x_2 &&(f \text{ is } 1\text{–}1), \end{aligned} \end{equation*}

so \(h\) is 1–1. Given \(z \in C\), ontoness of \(g\) supplies \(y \in B\) with \(g(y) = z\), and ontoness of \(f\) supplies \(x \in A\) with \(f(x) = y\); then \(h(x) = g(f(x)) = z\), so \(h\) is onto. Thus \(A \sim C\).

(a) \(\{\,(n, n+1) : n \in \mathbf{Z}\,\}\). Distinct integers \(m < n\) give \((m,m+1) \cap (n,n+1) = \emptyset\) since \(m + 1 \le n\), and the collection is countable because \(n \mapsto (n,n+1)\) is a 1–1 correspondence with \(\mathbf{Z}\), which is countable by Example 1.5.3 (ii) and Exercise 1.5.5 (c).

(b) No such collection exists: every collection of pairwise disjoint (nonempty) open intervals is countable or finite. Let \(\mathcal{I}\) be such a collection. By the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3), each \(I = (a,b) \in \mathcal{I}\) contains a rational; choose one and call it \(q_I\). The map

\begin{equation*} \mathcal{I} \longrightarrow \mathbf{Q}, \qquad I \longmapsto q_I , \end{equation*}

is 1–1: if \(q_I = q_J\) then that rational lies in \(I \cap J\), so \(I \cap J \ne \emptyset\) and disjointness forces \(I = J\). Hence \(\mathcal{I} \sim \{q_I : I \in \mathcal{I}\} \subseteq \mathbf{Q}\), and since \(\mathbf{Q}\) is countable (Theorem 1.5.6 (i)), Theorem 1.5.7 makes this subset — and therefore \(\mathcal{I}\) — countable or finite.

(a) \(f(x) = (x, 1/2)\). It lands in \(S\) because \(0 < x < 1\) and \(0 < 1/2 < 1\), and \(f(x_1) = f(x_2)\) forces \(x_1 = x_2\) on the first coordinate. (It is far from onto: its range is a single horizontal segment.)

(b) Interleave the digits. Fix once and for all the non-terminating decimal expansion of each point of \((0,1)\): every \(t \in (0,1)\) has exactly one expansion \(t = .t_1 t_2 t_3 \ldots\) whose digits are not eventually all \(0\) (so \(0.235\) is written \(.234999\ldots\)). With \(x = .a_1 a_2 a_3 \ldots\) and \(y = .b_1 b_2 b_3 \ldots\) in that form, define

\begin{equation*} g(x,y) \;=\; .\,a_1 b_1 a_2 b_2 a_3 b_3 \ldots \end{equation*}

Range. Since \(x > 0\), some \(a_i \ne 0\), so \(g(x,y) > 0\); since \(x < 1\) and \(x\) is not \(.999\ldots\), some \(a_i \ne 9\), so \(g(x,y) < 1\). Thus \(g : S \to (0,1)\).

1–1. The interleaved digit string never ends in all \(0\)’s, because that would force both \((a_i)\) and \((b_i)\) to end in all \(0\)’s, which the convention forbids. Two distinct decimal strings represent the same real only when one ends in all \(0\)’s and the other in all \(9\)’s; hence distinct interleavings give distinct reals. If \((x,y) \ne (x^{\prime},y^{\prime})\) then \((a_i) \ne (a_i^{\prime})\) or \((b_i) \ne (b_i^{\prime})\) — expansions being unique in our convention — so the interleavings differ as strings and therefore as numbers. Hence \(g\) is 1–1.

Onto? No. The number \(z = .1505050\ldots\) neither terminates nor ends in all \(9\)’s, so this is its only decimal string; de-interleaving it forces \(a = .1000\ldots\), a terminating string, which our convention assigns to no \(x \in (0,1)\). Hence \(z \notin g(S)\).

With \(f\) and \(g\) in hand, the Schroeder–Bernstein Theorem (Exercise 1.5.11) yields \((0,1) \sim S\).

Slice \(B\) by size: for each \(n \in \mathbf{N}\) put

\begin{equation*} B_n = \{\, b \in B : b \ge 1/n \,\}. \end{equation*}

Each \(B_n\) is finite. Indeed, if \(b_1, \dots, b_k\) are distinct elements of \(B_n\), then the hypothesis applied to this finite subset gives

\begin{equation*} 2 \;\ge\; b_1 + b_2 + \cdots + b_k \;\ge\; \frac{k}{n}, \end{equation*}

so \(k \le 2n\); thus \(B_n\) has at most \(2n\) elements.

Every \(b \in B\) satisfies \(b > 0\), so the Archimedean Property (Theorem 1.4.2) supplies an \(n\) with \(1/n < b\), i.e. \(b \in B_n\). Hence

\begin{equation*} B = \bigcup_{n=1}^{\infty} B_n . \end{equation*}

Theorem 1.5.8 (ii) is stated for countable sets, so pad: each \(B_n \cup \mathbf{N}\) is countable, since listing the finitely many points of \(B_n \setminus \mathbf{N}\) and then \(1, 2, 3, \dots\) enumerates it. Hence

\begin{equation*} \bigcup_{n=1}^{\infty} \bigl( B_n \cup \mathbf{N} \bigr) \;=\; B \cup \mathbf{N} \end{equation*}

countable. Since \(B \subseteq B \cup \mathbf{N}\), Theorem 1.5.7 gives that \(B\) is countable or finite.

(a) Exhibit the polynomials: \(\sqrt{2}\) is a root of \(x^2 - 2\), and \(\sqrt[3]{2}\) is a root of \(x^3 - 2\). For the third, square twice:

\begin{equation*} \begin{aligned} x = \sqrt{3} + \sqrt{2} \;&\Rightarrow\; x^2 = 5 + 2\sqrt{6}\\ &\Rightarrow\; (x^2 - 5)^2 = 24\\ &\Rightarrow\; x^4 - 10x^2 + 1 = 0, \end{aligned} \end{equation*}

so \(\sqrt{3}+\sqrt{2}\) is a root of \(x^4 - 10x^2 + 1\).

(b) First, \(\mathbf{Z}^{k}\) is countable for every \(k \in \mathbf{N}\), by induction: \(\mathbf{Z}\) is countable (Example 1.5.3 (ii)), and writing \(\mathbf{Z}\) as a sequence \(m_1, m_2, m_3, \dots\) gives

\begin{equation*} \mathbf{Z}^{k+1} = \bigcup_{j=1}^{\infty} \{m_j\} \times \mathbf{Z}^{k}, \end{equation*}

each slice a copy of \(\mathbf{Z}^{k}\): a countable union of countable sets, hence countable by Theorem 1.5.8 (ii).

Now let \(P_n\) be the set of degree-\(n\) polynomials with integer coefficients. Sending \(p(x) = a_nx^n + \cdots + a_0\) to \((a_0, \dots, a_n)\) is a 1–1 map of \(P_n\) into \(\mathbf{Z}^{n+1}\), so \(P_n\) is countable or finite by Theorem 1.5.7; it is infinite, hence countable. Enumerate \(P_n = \{p_1, p_2, p_3, \dots\}\) and let \(R_j \subseteq \mathbf{R}\) be the root set of \(p_j\), which is finite (at most \(n\) elements). Padding as in Exercise 1.5.8, each \(R_j \cup \mathbf{N}\) is countable, so

\begin{equation*} \bigcup_{j=1}^{\infty} \bigl( R_j \cup \mathbf{N} \bigr) \;=\; A_n \cup \mathbf{N} \end{equation*}

is countable by Theorem 1.5.8 (ii), and \(A_n\) is countable or finite by Theorem 1.5.7. It is infinite: each \(m \in \mathbf{Z}\) is a root of \(x^n - m x^{n-1} \in P_n\). So \(A_n\) is countable.

(c) The set of algebraic numbers is \(A = \bigcup_{n=1}^{\infty} A_n\), countable by Theorem 1.5.8 (ii). If the set \(T = \mathbf{R} \setminus A\) of transcendental numbers were countable or finite, then \(\mathbf{R} = A \cup T\) would be countable (Theorem 1.5.8 (i), after padding a finite \(T\) with \(\mathbf{N}\)), contradicting Theorem 1.5.6 (ii). Hence \(T\) is uncountable; in particular transcendental numbers exist.

(a) Some \(a = 1/n\) works. Suppose instead that \(C \cap [1/n, 1]\) is countable or finite for every \(n \ge 2\). Every \(x \in C\) with \(x > 0\) satisfies \(x \ge 1/n\) for some such \(n\) (Theorem 1.4.2), so

\begin{equation*} C \;\subseteq\; \{0\} \cup \bigcup_{n=2}^{\infty} \bigl( C \cap [1/n, 1] \bigr). \end{equation*}

Padding each piece with \(\mathbf{N}\) as in Exercise 1.5.8, the right-hand side is countable by Theorem 1.5.8 (ii), whence \(C\) is countable or finite by Theorem 1.5.7 — contradicting that \(C\) is uncountable.

(b) No — and not just occasionally: \(C \cap [\alpha, 1]\) is never uncountable. By (a) the set \(A\) is nonempty, and it is downward closed in \((0,1)\), since \(0 < a^{\prime} < a\) gives \(C \cap [a^{\prime},1] \supseteq C \cap [a,1]\). Thus \(\alpha = \sup A \in (0,1]\), and there are two cases.

(i) \(\alpha = 1\). Then \(C \cap [\alpha,1] = C \cap \{1\}\) has at most one element.

(ii) \(\alpha < 1\). Every \(a\) with \(\alpha < a < 1\) exceeds \(\sup A\), so \(a \notin A\) and \(C \cap [a,1]\) is countable or finite. Fix \(N\) with \(1/N < 1 - \alpha\); each \(x \in C\) with \(x > \alpha\) satisfies \(x \ge \alpha + 1/n\) for some \(n \ge N\), so

\begin{equation*} \begin{aligned} C \cap [\alpha, 1] \;\subseteq\; \{\alpha\} \cup \bigcup_{n=N}^{\infty} \bigl( C \cap [\alpha + 1/n, 1] \bigr), \end{aligned} \end{equation*}

again countable or finite by Theorem 1.5.8 (ii) and Theorem 1.5.7.

(c) No. Take \(C = \{1/n : n \in \mathbf{N}\} \subseteq [0,1]\), which is infinite. For any \(a \in (0,1)\),

\begin{equation*} C \cap [a,1] = \{\, 1/n : n \le 1/a \,\} \end{equation*}

is finite.

(a) Glue the two bijections. Since \(g\) is 1–1 and maps \(B^{\prime}\) onto \(A^{\prime}\), its restriction to \(B^{\prime}\) is a bijection \(B^{\prime} \to A^{\prime}\), so \(g^{-1}(x) \in B^{\prime}\) is unambiguous for \(x \in A^{\prime}\). Define

\begin{equation*} h(x) = \begin{cases} f(x), & x \in A,\\ g^{-1}(x), & x \in A^{\prime}. \end{cases} \end{equation*}

This is defined on all of \(X = A \cup A^{\prime}\) and unambiguous because \(A \cap A^{\prime} = \emptyset\). It is onto: \(h(A) = f(A) = B\) and \(h(A^{\prime}) = B^{\prime}\), and \(B \cup B^{\prime} = Y\). It is 1–1: \(f\) is 1–1 on \(A\), \(g^{-1}\) is 1–1 on \(A^{\prime}\), and the two images \(B\) and \(B^{\prime}\) are disjoint, so no collision can occur across the cases. Thus \(X \sim Y\) by Definition 1.5.2.

(b) If \(A_1 = \emptyset\) then \(g(Y) = X\), so \(g\) is itself 1–1 and onto and \(X \sim Y\) at once (Exercise 1.5.5 (b)); the construction below degenerates harmlessly to \(A = B = \emptyset\), \(A^{\prime} = X\), \(B^{\prime} = Y\), \(h = g^{-1}\). In general, set \(\varphi = g \circ f : X \to X\), which is 1–1 as a composition of 1–1 maps. Then \(A_{n+1} = \varphi(A_n)\), so

\begin{equation*} A_n = \varphi^{\,n-1}(A_1) \qquad (n \in \mathbf{N}). \end{equation*}

Two observations. First, a 1–1 map \(\psi\) satisfies \(\psi(S) \cap \psi(T) = \psi(S \cap T)\) for all sets \(S,T\) (if \(\psi(s) = \psi(t)\) then \(s = t\)). Second, for \(n \ge 2\) we have \(A_n = g(f(A_{n-1})) \subseteq g(Y)\), whereas \(A_1 \cap g(Y) = \emptyset\); hence

\begin{equation*} A_1 \cap A_n = \emptyset \qquad (n \ge 2). \end{equation*}

Now take \(1 \le m < n\) and put \(k = n - m + 1 \ge 2\). Applying the first observation to the 1–1 map \(\varphi^{\,m-1}\),

\begin{equation*} \begin{aligned} A_m \cap A_n &= \varphi^{\,m-1}(A_1) \cap \varphi^{\,m-1}(A_k)\\ &= \varphi^{\,m-1}(A_1 \cap A_k) = \emptyset . \end{aligned} \end{equation*}

So \(\{A_n\}\) is pairwise disjoint, and since \(f\) is 1–1,

\begin{equation*} f(A_m) \cap f(A_n) = f(A_m \cap A_n) = \emptyset \qquad (m \ne n), \end{equation*}

giving the same for \(\{f(A_n)\}\) in \(Y\).

(c) Images carry unions to unions:

\begin{equation*} f(A) = f\Bigl( \bigcup_{n=1}^{\infty} A_n \Bigr) = \bigcup_{n=1}^{\infty} f(A_n) = B, \end{equation*}

so \(f\) maps \(A\) onto \(B\); being a restriction of a 1–1 map, it does so 1–1.

(d) Both inclusions use injectivity of \(g\).

(i) \(g(B^{\prime}) \subseteq A^{\prime}\). Let \(y \in B^{\prime}\) and suppose \(g(y) \in A_n\) for some \(n\). Since \(g(y) \in g(Y)\) and \(A_1 \cap g(Y) = \emptyset\), necessarily \(n \ge 2\), so \(g(y) \in g(f(A_{n-1}))\), say \(g(y) = g(y^{\prime})\) with \(y^{\prime} \in f(A_{n-1})\). As \(g\) is 1–1, \(y = y^{\prime} \in f(A_{n-1}) \subseteq B\), contradicting \(y \in B^{\prime}\). Hence \(g(y) \notin A\), i.e. \(g(y) \in A^{\prime}\).

(ii) \(A^{\prime} \subseteq g(B^{\prime})\). Let \(x \in A^{\prime}\). Then \(x \notin A_1 = X \setminus g(Y)\), so \(x = g(y)\) for some \(y \in Y\). If \(y \in B\), then \(y \in f(A_n)\) for some \(n\) and therefore

\begin{equation*} x = g(y) \in g(f(A_n)) = A_{n+1} \subseteq A, \end{equation*}

contradicting \(x \in A^{\prime}\). So \(y \in B^{\prime}\) and \(x \in g(B^{\prime})\).

Thus \(g\) maps \(B^{\prime}\) onto \(A^{\prime}\), and with (c) the hypotheses of (a) hold, delivering \(h\).

Both sets are infinite and \((0,1) \sim \mathbf{R}\), so neither is countable without the other.

By Example 1.5.4 the map \(x \mapsto x/(x^2-1)\) carries \((-1,1)\) onto \(\mathbf{R}\) in a \(1\)–\(1\) fashion, and \(x \mapsto 2x-1\) carries \((0,1)\) onto \((-1,1)\) in a \(1\)–\(1\) fashion, so transitivity (Exercise 1.5.5(c)) gives

\begin{equation*} (0,1) \sim (-1,1) \sim \mathbf{R}. \end{equation*}

Hence \(\mathbf{N} \sim (0,1)\) implies \(\mathbf{N} \sim \mathbf{R}\), and \(\mathbf{N} \sim \mathbf{R}\) implies \(\mathbf{N} \sim (0,1)\) (using symmetry, Exercise 1.5.5(b)): one set is countable exactly when the other is. Both are infinite, each containing \(\{1/n : n \ge 2\}\), so Definition 1.5.5 makes “uncountable” mean “not countable” for each, and

\begin{equation*} (0,1) \text{ uncountable} \iff \mathbf{R} \text{ uncountable}. \qquad \blacksquare \end{equation*}

(a) Because \(b_1 \ne a_{11}\), and the expansion of \(x\) is the only one \(x\) has.

The rule gives \(b_1 = 3\) when \(a_{11} = 2\) and \(b_1 = 2\) when \(a_{11} \ne 2\); in both cases \(b_1 \ne a_{11}\). Every digit of \(x\) lies in \(\{2,3\}\), so the expansion \(.b_1b_2b_3\ldots\) neither terminates nor ends in repeating \(9\)’s, and is therefore the unique decimal representation of \(x\) (this is the point taken up in Exercise 1.6.3(b)). Hence if \(x = f(1)\), the expansion \(.a_{11}a_{12}a_{13}\ldots\) would have to be that unique representation, forcing \(a_{11} = b_1\) – false.

(b) Identical, one row down: \(b_2 \ne a_{22}\) by the rule, so \(x\) and \(f(2)\) disagree in the second decimal digit and \(x \ne f(2)\). In general, for every \(n \in \mathbf{N}\),

\begin{equation*} b_n \ne a_{nn} \quad\Longrightarrow\quad x \ne f(n), \end{equation*}

since equality would force the unique expansion of \(x\) to be \(.a_{n1}a_{n2}a_{n3}\ldots\), whose \(n\)th digit is \(a_{nn} \ne b_n\).

(c) The number \(x\) satisfies

\begin{equation*} \tfrac{2}{9} = .222\ldots \;\le\; x \;\le\; .333\ldots = \tfrac{1}{3}, \end{equation*}

so \(x \in (0,1)\). But by (b), \(x \ne f(n)\) for every \(n \in \mathbf{N}\), so \(x\) is not in the range of \(f\) – contradicting the assumption that \(f\) is onto. No \(1\)–\(1\), onto \(f : \mathbf{N} \to (0,1)\) exists, so \(\mathbf{N} \not\sim (0,1)\); since \((0,1)\) is infinite (it contains \(\{1/n : n \ge 2\}\)), Definition 1.5.5 declares \((0,1)\) uncountable. \(\blacksquare\)

(a) Run the argument on \(\mathbf{Q} \cap (0,1)\) and nothing breaks: the diagonal number \(x\) escapes the list, which says only that \(x\) is irrational.

Let \(f : \mathbf{N} \to \mathbf{Q} \cap (0,1)\) be \(1\)–\(1\) and onto; such an \(f\) exists because \(\mathbf{Q} \cap (0,1)\) is an infinite subset (it contains every \(1/n\), \(n \ge 2\)) of the countable set \(\mathbf{Q}\), so Theorem 1.5.6(i) and Theorem 1.5.7 make it countable. The construction still produces a real \(x \in (0,1)\) differing from \(f(n)\) in the \(n\)th digit for every \(n\), so \(x\) lies outside the range of \(f\). But that range is \(\mathbf{Q} \cap (0,1)\), not \((0,1)\), so no contradiction follows – only the true statement

\begin{equation*} x \in (0,1) \setminus \mathbf{Q}. \end{equation*}

Theorem 1.6.1 gets its contradiction because there \(f\) is assumed onto \((0,1)\), a set that visibly contains \(x\).

(b) It causes none, because the digits were chosen from \(\{2,3\}\) exactly to avoid it.

Two conventions settle the matter. First, for each \(m\) fix one decimal representation of \(f(m)\) (say, the one not ending in repeating \(9\)’s), so that the array entries \(a_{mn}\) are well defined. Second, observe that the constructed \(x = .b_1b_2b_3\ldots\) has every digit equal to \(2\) or \(3\): its expansion neither terminates nor ends in repeating \(9\)’s, so it is the only decimal representation \(x\) has. Hence

\begin{equation*} x = f(m) \;\Longrightarrow\; .a_{m1}a_{m2}a_{m3}\ldots = .b_1b_2b_3\ldots \;\Longrightarrow\; a_{mm} = b_m, \end{equation*}

contradicting \(b_m \ne a_{mm}\). \(\blacksquare\)

Diagonalize, flipping bits: the escaping sequence is \(s = (1 - a_{11},\, 1 - a_{22},\, 1 - a_{33},\, \ldots)\).

First, \(S\) is infinite, since \(n \mapsto e_n\), where \(e_n\) has a \(1\) in the \(n\)th slot and \(0\)’s elsewhere, is a \(1\)–\(1\) map of \(\mathbf{N}\) into \(S\).

Now assume, for contradiction, that \(S\) is countable, so that there is a \(1\)–\(1\), onto \(f : \mathbf{N} \to S\). Write the \(m\)th listed sequence as

\begin{equation*} f(m) = (a_{m1},\, a_{m2},\, a_{m3},\, \ldots), \qquad a_{mn} \in \{0,1\}, \end{equation*}

and set \(b_n = 1 - a_{nn}\), \(s = (b_1, b_2, b_3, \ldots)\). Each \(b_n\) is \(0\) or \(1\), so \(s \in S\). For every \(n \in \mathbf{N}\) the \(n\)th terms satisfy

\begin{equation*} b_n = 1 - a_{nn} \ne a_{nn}, \end{equation*}

and two sequences are equal precisely when all of their corresponding terms agree; hence \(s \ne f(n)\) for every \(n\). Thus \(s \in S\) is not in the range of \(f\), contradicting the assumption that \(f\) is onto.

Therefore \(\mathbf{N} \not\sim S\), and since \(S\) is infinite, \(S\) is uncountable by Definition 1.5.5. \(\blacksquare\)

(a) The eight subsets are

\begin{equation*} P(A) = \bigl\{\, \emptyset,\ \{a\},\ \{b\},\ \{c\},\ \{a,b\},\ \{a,c\},\ \{b,c\},\ \{a,b,c\} \,\bigr\}. \end{equation*}

(b) Induction on \(n\), splitting the subsets according to whether they contain a fixed element.

For \(n = 0\) we have \(A = \emptyset\) and \(P(\emptyset) = \{\emptyset\}\), which has \(1 = 2^0\) element. Assume every \(n\)-element set has a power set with \(2^n\) elements, let \(A\) have \(n+1\) elements, fix \(x \in A\), and put \(A^{\prime} = A \setminus \{x\}\), a set with \(n\) elements. Every subset of \(A\) either omits \(x\) or contains it, so

\begin{equation*} P(A) = P(A^{\prime}) \;\cup\; \bigl\{\, E \cup \{x\} : E \in P(A^{\prime}) \,\bigr\}, \end{equation*}

and the two pieces are disjoint. The first has \(2^n\) elements by the induction hypothesis, and \(E \mapsto E \cup \{x\}\) maps \(P(A^{\prime})\) onto the second in a \(1\)–\(1\) fashion (inverse \(F \mapsto F \setminus \{x\}\)), so the second has \(2^n\) elements too. Hence \(P(A)\) has \(2^n + 2^n = 2^{n+1}\) elements. \(\blacksquare\)

(a) The singleton map and a nested chain:

\begin{equation*} \begin{aligned} f_1 : \ & a \mapsto \{a\}, && b \mapsto \{b\}, && c \mapsto \{c\}; \\ f_2 : \ & a \mapsto \emptyset, && b \mapsto \{a\}, && c \mapsto \{a,b\}. \end{aligned} \end{equation*}

Each lists three distinct subsets of \(A\), so each is \(1\)–\(1\).

(b) \(g(k) = \{k\}\), that is

\begin{equation*} g(1) = \{1\}, \quad g(2) = \{2\}, \quad g(3) = \{3\}, \quad g(4) = \{4\}, \end{equation*}

four distinct elements of \(P( C)\), so \(g\) is \(1\)–\(1\).

(c) Counting: a map defined on \(A\) has at most three values while \(P(A)\) has eight, and one defined on \(C\) has at most four while \(P( C)\) has sixteen.

The range \(\{f(a) : a \in A\}\) acquires at most one element per element of \(A\), so it has at most three members; by Exercise 1.6.5(b), \(P(A)\) has \(2^3 = 8\) members and \(P( C)\) has \(2^4 = 16\). Since \(3 < 8\) and \(4 < 16\), the range is in each case a proper subset of the power set, so no such map is onto. \(\blacksquare\)

\(B = \emptyset\) for both singleton maps, and \(B = A\) for the nested chain.

(i) \(f_1 : A \to P(A)\), \(f_1(x) = \{x\}\), on \(A = \{a,b,c\}\). Every \(x\) satisfies \(x \in \{x\} = f_1(x)\), so no element qualifies:

\begin{equation*} B = \{x \in A : x \notin \{x\}\} = \emptyset, \end{equation*}

and the range of \(f_1\) is \(\{\{a\},\{b\},\{c\}\}\), which does not contain \(\emptyset\).

(ii) \(f_2 : a \mapsto \emptyset\), \(b \mapsto \{a\}\), \(c \mapsto \{a,b\}\). Test each element against its own image:

\begin{equation*} \begin{aligned} a &\notin \emptyset = f_2(a) && \Longrightarrow\ a \in B, \\ b &\notin \{a\} = f_2(b) && \Longrightarrow\ b \in B, \\ c &\notin \{a,b\} = f_2( c) && \Longrightarrow\ c \in B, \end{aligned} \end{equation*}

so \(B = \{a,b,c\} = A\), while the range of \(f_2\) is \(\{\emptyset, \{a\}, \{a,b\}\}\) – no \(A\) there.

(iii) \(g : C \to P( C)\), \(g(k) = \{k\}\), on \(C = \{1,2,3,4\}\). As in (i), \(k \in g(k)\) for every \(k\), so

\begin{equation*} B = \emptyset, \end{equation*}

which is not among \(\{1\}, \{2\}, \{3\}, \{4\}\). \(\blacksquare\)

Both cases fail, because substituting \(f(a^{\prime}) = B\) into the rule defining \(B\) produces the self-contradiction

\begin{equation*} a^{\prime} \in B \iff a^{\prime} \notin f(a^{\prime}) \iff a^{\prime} \notin B. \end{equation*}

(a) If \(a^{\prime} \in B\), then by the definition of \(B\) we have \(a^{\prime} \notin f(a^{\prime}) = B\), contradicting \(a^{\prime} \in B\).

(b) If \(a^{\prime} \notin B = f(a^{\prime})\), then \(a^{\prime}\) meets the defining condition \(a^{\prime} \notin f(a^{\prime})\), so \(a^{\prime} \in B\), again a contradiction.

Hence no \(a^{\prime} \in A\) satisfies \(f(a^{\prime}) = B\), and since \(B \in P(A)\) the arbitrary \(f\) is not onto.

Chain three equivalences: \(P(\mathbf{N}) \sim S \sim (0,1) \sim \mathbf{R}\), where \(S\) is the set of sequences of \(0\)’s and \(1\)’s from Exercise 1.6.4. Since \(\sim\) is transitive (Exercise 1.5.5 (c)), this suffices.

For \(P(\mathbf{N}) \sim S\), send a subset to its characteristic sequence:

\begin{equation*} \chi(A) = (a_1, a_2, a_3, \ldots), \qquad a_n = \begin{cases} 1 & \text{if } n \in A, \\ 0 & \text{if } n \notin A. \end{cases} \end{equation*}

The map \((a_n) \mapsto \{n \in \mathbf{N} : a_n = 1\}\) is a two-sided inverse, so \(\chi\) is a 1–1, onto map from \(P(\mathbf{N})\) to \(S\).

For \(S \sim (0,1)\), build a 1–1 map each way and apply the Schroder–Bernstein Theorem (Exercise 1.5.11), whose only hypothesis is that pair of injections.

Define \(g : S \to (0,1)\) by

\begin{equation*} g\big((a_n)\big) = .b_1 b_2 b_3 \ldots, \qquad b_n = \begin{cases} 1 & \text{if } a_n = 0, \\ 2 & \text{if } a_n = 1. \end{cases} \end{equation*}

Every digit lies in \(\{1,2\}\), so the value sits between \(.111\ldots = 1/9\) and \(.222\ldots = 2/9\) and hence in \((0,1)\). Decimal ambiguity is confined to terminating expansions versus repeating \(9\)’s (Exercise 1.6.3 (b)), and these digit strings are neither, so each is the only expansion of its value. Distinct sequences therefore give distinct reals, and \(g\) is 1–1.

Define \(h : (0,1) \to S\) by taking the decimal expansion \(x = .c_1 c_2 c_3 \ldots\), choosing the non-terminating representation when there is a choice, and setting \(h(x)\) to be the \(0\)–\(1\) sequence

\begin{equation*} \underbrace{1 \cdots 1}_{c_1 + 1} \, 0 \, \underbrace{1 \cdots 1}_{c_2 + 1} \, 0 \, \underbrace{1 \cdots 1}_{c_3 + 1} \, 0 \, \cdots . \end{equation*}

Each block of \(1\)’s has length \(c_n + 1 \ge 1\), so the blocks are nonempty and the zeros separating them are unambiguous: reading \(h(x)\) left to right recovers \(c_1, c_2, c_3, \ldots\) and hence \(x\). So \(h\) is 1–1. Schroder–Bernstein now yields a 1–1, onto map between \(S\) and \((0,1)\).

Finally \((0,1) \sim \mathbf{R}\) is Exercise 1.5.4 (a) with \((a,b) = (0,1)\). Composing the three, \(P(\mathbf{N}) \sim \mathbf{R}\).

(a) Countable. The map \(f \mapsto (f(0), f(1))\) is a 1–1 correspondence with \(\mathbf{N} \times \mathbf{N}\), since a function on the two-point domain \(\{0,1\}\) is exactly a choice of its two values. And

\begin{equation*} \mathbf{N} \times \mathbf{N} = \bigcup_{m=1}^{\infty} \big( \{m\} \times \mathbf{N} \big) \end{equation*}

is countable by Theorem 1.5.8 (ii): the union is indexed by \(m \in \mathbf{N}\) and each \(\{m\} \times \mathbf{N} \sim \mathbf{N}\) via \((m,n) \mapsto n\), so both hypotheses hold.

(b) Uncountable. The map \(f \mapsto \big(f(1), f(2), f(3), \ldots\big)\) is a 1–1 correspondence with the set \(S\) of sequences of \(0\)’s and \(1\)’s of Exercise 1.6.4, since a function into \(\{0,1\}\) is nothing but the list of its values; and \(S\) is uncountable by Exercise 1.6.4.

(c) Yes. Let \(F\) be the set of all nonempty finite strings of \(0\)’s and \(1\)’s, and for a sequence \(s = (s_1, s_2, s_3, \ldots) \in S\) let

\begin{equation*} A_s = \{\, s_1 s_2 \cdots s_k \in F : k \in \mathbf{N} \,\} \end{equation*}

be the set of initial segments of \(s\). Then \(\mathcal{A}_0 = \{A_s : s \in S\}\) is an uncountable antichain in \(P(F)\).

It is an antichain, and \(s \mapsto A_s\) is 1–1, for the same one-line reason. If \(s \neq t\), let \(k\) be the first index with \(s_k \neq t_k\). The string \(s_1 s_2 \cdots s_k\) lies in \(A_s\), while the only length-\(k\) string in \(A_t\) is \(t_1 t_2 \cdots t_k\), which differs from it in the \(k\)th place. Hence

\begin{equation*} A_s \not\subseteq A_t \quad \text{and, symmetrically,} \quad A_t \not\subseteq A_s . \end{equation*}

So distinct sequences give incomparable (in particular distinct) sets, \(s \mapsto A_s\) is a 1–1 correspondence of \(S\) onto \(\mathcal{A}_0\), and \(\mathcal{A}_0\) is uncountable because \(S\) is (Exercise 1.6.4).

Transfer this to \(\mathbf{N}\) along the explicit bijection \(\phi : F \to \mathbf{N}\),

\begin{equation*} \phi(w_1 w_2 \cdots w_k) = 2^k + \sum_{i=1}^{k} w_i 2^{k-i} - 1, \end{equation*}

which reads \(1 w_1 w_2 \cdots w_k\) as a binary numeral and subtracts \(1\). It carries the \(2^k\) strings of length \(k\) onto the block \(\{2^k - 1, \ldots, 2^{k+1} - 2\}\), and these blocks partition \(\mathbf{N}\), so \(\phi\) is 1–1 and onto. (Check!) Put

\begin{equation*} \mathcal{A} = \big\{ \phi(A_s) : s \in S \big\} \subseteq P(\mathbf{N}), \qquad \phi(A_s) = \{ \phi(w) : w \in A_s \}. \end{equation*}

Because \(\phi\) is a bijection, \(\phi(A_s) \subseteq \phi(A_t)\) holds precisely when \(A_s \subseteq A_t\); so \(\mathcal{A}\) is again an antichain, and \(A_s \mapsto \phi(A_s)\) carries \(\mathcal{A}_0\) onto \(\mathcal{A}\) in a 1–1 fashion. Thus \(\mathcal{A} \sim S\) is an uncountable antichain in \(P(\mathbf{N})\).

Vercongence to \(x\) is just boundedness: \((x_n)\) verconges to \(x\) if and only if \((x_n)\) is a bounded sequence, and in that case it verconges to every real number.

Taking \(N = 1\) in the definition collapses it, since the clause must hold for all \(N\):

\begin{equation*} \begin{aligned} (x_n) \text{ verconges to } x &\iff \exists\, \epsilon > 0 \text{ with } |x_n - x| < \epsilon \\ &\qquad\qquad \text{for all } n \in \mathbf{N}. \end{aligned} \end{equation*}

So the whole sequence lies in the single neighborhood \(V_\epsilon(x)\) (Definition 2.2.4); no tail behaviour is constrained at all. If such an \(\epsilon\) exists then \(|x_n| < |x| + \epsilon\) for every \(n\), so \((x_n)\) is bounded; conversely if \(|x_n| \le M\) for all \(n\), then for any \(y \in \mathbf{R}\) the choice \(\epsilon = M + |y| + 1\) gives \(|x_n - y| \le |x_n| + |y| < \epsilon\).

Examples, in order:

• \(x_n = 1\) verconges to \(1\) (any \(\epsilon > 0\) works).
• \(x_n = (-1)^n\) verconges to \(0\) with \(\epsilon = 2\), and is divergent: for any candidate limit \(a\), one of \(|1 - a|\), \(|-1 - a|\) is at least \(1\), and that value recurs past every \(N\), so \(\epsilon = 1\) has no response.
• Yes: \((-1)^n\) verconges to \(0\) and to \(17\) (take \(\epsilon = 19\)), and indeed to every real number.

Take \(N > 3/(25\epsilon)\) in (a), \(N > 2/\epsilon\) in (b), \(N > 1/\epsilon^3\) in (c); each proof then follows the template on p. 45.

(a) Let \(\epsilon > 0\) and choose \(N \in \mathbf{N}\) with \(N > 3/(25\epsilon)\). For \(n \ge N\),

\begin{equation*} \begin{aligned} \left| \frac{2n+1}{5n+4} - \frac{2}{5} \right| &= \left| \frac{5(2n+1) - 2(5n+4)}{5(5n+4)} \right| \\ &= \frac{3}{25n + 20} < \frac{3}{25n} \le \frac{3}{25N} < \epsilon . \end{aligned} \end{equation*}

(b) Let \(\epsilon > 0\) and choose \(N > 2/\epsilon\). For \(n \ge N\), using \(n^3 + 3 > n^3\),

\begin{equation*} \begin{aligned} \left| \frac{2n^2}{n^3+3} - 0 \right| &= \frac{2n^2}{n^3+3} < \frac{2n^2}{n^3} \\ &= \frac{2}{n} \le \frac{2}{N} < \epsilon . \end{aligned} \end{equation*}

(c) Let \(\epsilon > 0\) and choose \(N > 1/\epsilon^3\). For \(n \ge N\), since \(|\sin(n^2)| \le 1\),

\begin{equation*} \left| \frac{\sin(n^2)}{\sqrt[3]{n}} - 0 \right| \le \frac{1}{\sqrt[3]{n}} \le \frac{1}{\sqrt[3]{N}} < \epsilon , \end{equation*}

the last step because \(N > \epsilon^{-3}\) gives \(\sqrt[3]{N} > 1/\epsilon\).

(a) Exhibit a single college in the United States at which every student is shorter than seven feet — by the rule on p. 46, the negation of “for all \(P\), there exists \(Q\)” is “for at least one \(P\), no \(Q\) is possible.”

(b) Exhibit a single college and then, for each professor there, produce one student of that professor who received a grade other than A or B. The negation runs two levels deep:

\begin{equation*} \begin{aligned} &\neg(\forall\,\text{colleges}\ \exists\,\text{prof}\ \forall\,\text{students}) \\ &\qquad \equiv\ \exists\,\text{college}\ \forall\,\text{profs}\ \exists\,\text{student}. \end{aligned} \end{equation*}

(c) Show that at every college in the United States there is some student shorter than six feet: the original claim is an existence claim, so a single college will not do, and one short student must be supplied for each college in the country.

(a) Possible: \(a_n = 1\) for \(n\) odd, \(a_n = 0\) for \(n\) even, i.e. \((1, 0, 1, 0, 1, 0, \dots)\). There are infinitely many ones, but taking \(\epsilon = 1/2\), no \(N\) works, since every \(N\) is followed by an even index \(n \ge N\) with \(|a_n - 1| = 1 > 1/2\).

(b) Impossible. Suppose \((a_n) \to a\) with \(a \ne 1\) and \(a_n = 1\) for infinitely many \(n\). Apply Definition 2.2.3 with \(\epsilon = |a - 1| / 2 > 0\) to get \(N\) with \(|a_n - a| < |a-1|/2\) for all \(n \ge N\). Some index \(n \ge N\) has \(a_n = 1\) (infinitely many do), and for that index

\begin{equation*} |a - 1| = |a_n - a| < \frac{|a-1|}{2}, \end{equation*}

forcing \(|a - 1| < 0\), a contradiction.

(c) Possible: string together blocks of ones of increasing length, separated by single zeros,

\begin{equation*} (1,\; 0,\; 1, 1,\; 0,\; 1, 1, 1,\; 0,\; 1, 1, 1, 1,\; 0,\; \dots), \end{equation*}

whose \(k\)-th block consists of \(k\) consecutive ones. Given \(n \in \mathbf{N}\), the \(n\)-th block supplies \(n\) consecutive ones. The sequence diverges: both the value \(1\) and the value \(0\) occur at arbitrarily large indices, so for any candidate limit \(a\) the choice \(\epsilon = 1/2\) admits no \(N\) — one of \(|1 - a|\), \(|0 - a|\) is at least \(1/2\), and that value recurs past every \(N\).

(a) \(\lim a_n = 0\); (b) \(\lim a_n = 1\). Both sequences are eventually constant, so a single \(N\) answers every \(\epsilon\) at once.

(a) The first terms are \(5, 2, 1, 1, 1, 0, 0, \dots\), and for \(n \ge 6\) we have \(0 < 5/n \le 5/6 < 1\), hence \(a_n = 0\). Let \(\epsilon > 0\) and take \(N = 6\). For \(n \ge N\),

\begin{equation*} |a_n - 0| = 0 < \epsilon . \end{equation*}

(b) Write \(a_n = [\![\,4/n + 4/3\,]\!]\). The first terms are \(5, 3, 2, 2, 2, 2, 1, 1, \dots\); for \(n \ge 7\),

\begin{equation*} 1 < \frac{4}{3} \le \frac{4}{n} + \frac{4}{3} \le \frac{4}{7} + \frac{4}{3} = \frac{40}{21} < 2 , \end{equation*}

so \(a_n = 1\). Let \(\epsilon > 0\) and take \(N = 7\). For \(n \ge N\), \(|a_n - 1| = 0 < \epsilon\). (At \(n = 6\) the value is exactly \(2\), so \(N = 7\) is the smallest choice that works.)

The quoted remark says only that \(N\) may have to grow, and these sequences are the extreme case where it does not: \(N = 6\) (resp. \(N = 7\)) answers \(\epsilon = 1\) and \(\epsilon = 10^{-100}\) alike. All that is guaranteed is that an \(N\) working for \(\epsilon\) also works for every \(\epsilon^{\prime} > \epsilon\), so the least admissible \(N\) is nondecreasing in \(1/\epsilon\) — not strictly increasing, and not necessarily unbounded.

Two limits \(a\) and \(b\) satisfy \(|a - b| < \epsilon\) for every \(\epsilon > 0\), so \(a = b\) by Theorem 1.2.6.

Assume \((a_n) \to a\) and \((a_n) \to b\), and fix \(\epsilon > 0\). Definition 2.2.3 applied with \(\epsilon/2 > 0\) supplies \(N_1\) with \(|a_n - a| < \epsilon/2\) for all \(n \ge N_1\), and \(N_2\) with \(|a_n - b| < \epsilon/2\) for all \(n \ge N_2\). Taking \(n = \max\{N_1, N_2\}\), the triangle inequality (Example 1.2.5) gives

\begin{equation*} |a - b| \le |a - a_n| + |a_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, Theorem 1.2.6 yields \(a = b\).

(a) Frequently, but not eventually. Given any \(N\), the even index \(n = 2N \ge N\) has \((-1)^n = 1 \in \{1\}\); on the other hand no \(N\) works for “eventually,” since every \(N\) is followed by an odd index \(n\) with \((-1)^n = -1 \notin \{1\}\).

(b) Eventually is stronger, and eventually implies frequently. If \(a_n \in A\) for all \(n \ge N_0\), then given any \(N \in \mathbf{N}\) the index \(n = \max\{N, N_0\}\) satisfies \(n \ge N\) and \(a_n \in A\). The converse fails by part (a).

(c) Definition 2.2.3B becomes: a sequence \((a_n)\) converges to \(a\) if, for every \(\epsilon > 0\), \((a_n)\) is eventually in the \(\epsilon\)-neighborhood \(V_\epsilon(a)\).

“Eventually” is the term we want. Replacing it by “frequently” gives a strictly weaker and useless condition — by (a), \(((-1)^n)\) is frequently in \(V_\epsilon(1)\) for every \(\epsilon > 0\), yet it does not converge to \(1\).

(d) Not necessarily eventually, but always frequently. For the first, \(x_n = 2\) for \(n\) odd and \(x_n = 0\) for \(n\) even has infinitely many terms equal to \(2\), while every \(N\) is followed by an even index \(n\) with \(x_n = 0 \notin (1.9, 2.1)\). For the second, let \(N \in \mathbf{N}\); the set \(\{n : x_n = 2\}\) is infinite, hence not bounded above by \(N\), so some \(n \ge N\) has \(x_n = 2 \in (1.9, 2.1)\).

(a) Yes, with \(M = 1\). Here \(x_n = 0\) for odd \(n\) and \(x_n = 1\) for even \(n\), and every pair of consecutive indices \(\{N, N+1\}\) contains an odd one: take \(n = N\) if \(N\) is odd, \(n = N+1\) if \(N\) is even. Either way \(N \le n \le N+1\) and \(x_n = 0\).

(b) Yes. Let \(M\) witness zero-heaviness and set \(N_k = k(M+1) + 1\) for \(k = 0, 1, 2, \ldots\). Applying the definition to each \(N_k\) produces an index \(n_k\) with

\begin{equation*} k(M+1) + 1 \;\le\; n_k \;\le\; (k+1)(M+1), \qquad x_{n_k} = 0 . \end{equation*}

These windows are pairwise disjoint, so the \(n_k\) are distinct and \(\{n : x_n = 0\}\) is infinite.

(c) No: the zeros can thin out faster than any fixed window. Take

\begin{equation*} x_n = \begin{cases} 0, & n = 2^k \text{ for some } k \in \mathbf{N},\\ 1, & \text{otherwise}, \end{cases} \end{equation*}

which has infinitely many zeros. Given any \(M \in \mathbf{N}\), choose \(k\) with \(2^k > M + 1\) and put \(N = 2^k + 1\). The zeros nearest \(N\) sit at \(2^k < N\) and at \(2^{k+1}\), and

\begin{equation*} 2^{k+1} \;=\; (2^k + 1) + (2^k - 1) \;>\; N + M , \end{equation*}

so no \(n\) with \(N \le n \le N + M\) has \(x_n = 0\). Hence no \(M\) works and \((x_n)\) is not zero-heavy.

(d) Negating the quantifier string \((\exists M)(\forall N)(\exists n)\) term by term: a sequence \((x_n)\) is not zero-heavy if for all \(M \in \mathbf{N}\) there exists \(N \in \mathbf{N}\) such that for all \(n\) satisfying \(N \le n \le N + M\) we have \(x_n \ne 0\).

(a) Given \(\epsilon > 0\), apply Definition 2.2.3 with the positive number \(\epsilon^2\): there is an \(N\) such that \(n \ge N\) implies \(x_n = |x_n - 0| < \epsilon^2\), and hence

\begin{equation*} |\sqrt{x_n} - 0| = \sqrt{x_n} < \epsilon . \end{equation*}

(b) First, \(x \ge 0\) by the Order Limit Theorem 2.3.4(i) (the hypothesis \(x_n \ge 0\) holds for all \(n\)), so \(\sqrt{x}\) is defined.

(i) \(x = 0\): this is part (a).

(ii) \(x > 0\): rationalizing the numerator gives, for every \(n\),

\begin{equation*} |\sqrt{x_n} - \sqrt{x}| = \frac{|x_n - x|}{\sqrt{x_n} + \sqrt{x}} \le \frac{|x_n - x|}{\sqrt{x}}, \end{equation*}

the inequality because \(\sqrt{x_n} \ge 0\). Given \(\epsilon > 0\), choose \(N\) so that \(n \ge N\) implies \(|x_n - x| < \epsilon \sqrt{x}\); then \(|\sqrt{x_n} - \sqrt{x}| < \epsilon\) for all \(n \ge N\).

(a) Given \(\epsilon > 0\), choose \(N\) with \(|x_n - 2| < \tfrac{3}{2}\epsilon\) for \(n \ge N\). Then for such \(n\),

\begin{equation*} \left| \frac{2x_n - 1}{3} - 1 \right| = \left| \frac{2x_n - 4}{3} \right| = \frac{2}{3}\,|x_n - 2| < \epsilon . \end{equation*}

(b) The quantity to control is

\begin{equation*} \left| \frac{1}{x_n} - \frac{1}{2} \right| = \frac{|2 - x_n|}{2\,|x_n|}, \end{equation*}

so we need \(|x_n|\) bounded away from \(0\). Applying Definition 2.2.3 with \(\epsilon_0 = 1\) gives an \(N_1\) such that \(|x_n - 2| < 1\) for \(n \ge N_1\); then \(x_n > 1\), so \(|x_n| > 1\).

Now let \(\epsilon > 0\) and choose \(N_2\) with \(|x_n - 2| < 2\epsilon\) for \(n \ge N_2\). Put \(N = \max\{N_1, N_2\}\). For \(n \ge N\),

\begin{equation*} \left| \frac{1}{x_n} - \frac{1}{2} \right| = \frac{|2 - x_n|}{2\,|x_n|} < \frac{2\epsilon}{2} = \epsilon . \end{equation*}

Given \(\epsilon > 0\), Definition 2.2.3 supplies \(N_1\) with \(|x_n - l| < \epsilon\) for \(n \ge N_1\), and \(N_2\) with \(|z_n - l| < \epsilon\) for \(n \ge N_2\). Set \(N = \max\{N_1, N_2\}\); for \(n \ge N\) the hypothesis \(x_n \le y_n \le z_n\) gives

\begin{equation*} l - \epsilon < x_n \le y_n \le z_n < l + \epsilon, \end{equation*}

that is, \(|y_n - l| < \epsilon\). Hence \(\lim y_n = l\).

(a) \(1\). By Theorem 2.3.3(iii), \((a_n^2) \to 0\); then by (i) and (ii),

\begin{equation*} (1 + 2a_n) \to 1, \qquad (1 + 3a_n - 4a_n^2) \to 1 . \end{equation*}

Since the limit of the denominator is \(1 \ne 0\), Theorem 2.3.3(iv) applies and the quotient converges to \(1/1 = 1\).

(b) \(4\). The fractions are assumed defined, so \(a_n \ne 0\) and we may cancel before taking limits:

\begin{equation*} \frac{(a_n + 2)^2 - 4}{a_n} = \frac{a_n^2 + 4a_n}{a_n} = a_n + 4 . \end{equation*}

By Theorem 2.3.3(ii) this tends to \(0 + 4 = 4\).

(c) \(2\). Again \(a_n \ne 0\), so multiplying numerator and denominator by \(a_n\) is legitimate termwise:

\begin{equation*} \frac{\frac{2}{a_n} + 3}{\frac{1}{a_n} + 5} = \frac{2 + 3a_n}{1 + 5a_n} . \end{equation*}

By Theorem 2.3.3(i),(ii) the numerator tends to \(2\) and the denominator to \(1 \ne 0\), so by (iv) the limit is \(2\).

Write the shuffle explicitly: \(z_{2n-1} = x_n\) and \(z_{2n} = y_n\) for all \(n \in \mathbf{N}\).

(\(\Rightarrow\)) Suppose \((z_n) \to l\) and let \(\epsilon > 0\). Choose \(N\) so that \(|z_m - l| < \epsilon\) for all \(m \ge N\). If \(n \ge N\) then both \(2n - 1 \ge n \ge N\) and \(2n \ge n \ge N\), so

\begin{equation*} |x_n - l| = |z_{2n-1} - l| < \epsilon, \qquad |y_n - l| = |z_{2n} - l| < \epsilon . \end{equation*}

Hence \((x_n)\) and \((y_n)\) both converge, and both to the same value \(l\).

(\(\Leftarrow\)) Suppose \(\lim x_n = \lim y_n = l\), and let \(\epsilon > 0\). Choose \(N_1\) with \(|x_n - l| < \epsilon\) for \(n \ge N_1\) and \(N_2\) with \(|y_n - l| < \epsilon\) for \(n \ge N_2\), and set

\begin{equation*} N = \max\{2N_1 - 1,\ 2N_2\}. \end{equation*}

Let \(m \ge N\). If \(m = 2n - 1\) is odd, then \(2n - 1 \ge 2N_1 - 1\) gives \(n \ge N_1\), so \(|z_m - l| = |x_n - l| < \epsilon\). If \(m = 2n\) is even, then \(2n \ge 2N_2\) gives \(n \ge N_2\), so \(|z_m - l| = |y_n - l| < \epsilon\). Thus \((z_n) \to l\).

\(\lim b_n = -1\). Rationalize the numerator and then divide through by \(n\):

\begin{equation*} \begin{aligned} b_n &= \frac{n^2 - (n^2 + 2n)}{n + \sqrt{n^2 + 2n}} = \frac{-2n}{n + \sqrt{n^2 + 2n}} \\[4pt] &= \frac{-2}{1 + \sqrt{1 + 2/n}} , \end{aligned} \end{equation*}

the last step using \(\sqrt{n^2 + 2n} = n\sqrt{1 + 2/n}\) (valid since \(n > 0\)).

Now \((1 + 2/n) \to 1\) by Theorem 2.3.3(i),(ii), and \(1 + 2/n \ge 0\), so Exercise 2.3.1(b) gives \(\left(\sqrt{1 + 2/n}\right) \to 1\). Hence the denominator \(\left(1 + \sqrt{1 + 2/n}\right) \to 2 \ne 0\), and Theorem 2.3.3(iv) yields

\begin{equation*} \lim b_n = \frac{-2}{2} = -1 . \end{equation*}

(a) Possible: \(x_n = (-1)^n\) and \(y_n = (-1)^{n+1}\). Each diverges: consecutive terms differ by \(2\), so no single \(l\) can have all terms eventually within \(\epsilon = 1\) of it. Meanwhile \(x_n + y_n = 0\) for all \(n\), so the sum converges to \(0\).

(b) Impossible. If \((x_n) \to x\) and \((x_n + y_n) \to s\), then by Theorem 2.3.3(i),(ii) applied to \(y_n = (x_n + y_n) + (-1)x_n\),

\begin{equation*} \lim y_n = s - x , \end{equation*}

so \((y_n)\) converges, contradicting its divergence.

(c) Possible: \(b_n = 1/n\). Then \(b_n \ne 0\) and \((b_n) \to 0\), while \(1/b_n = n\) is unbounded, hence divergent by Theorem 2.3.2. (No contradiction with Theorem 2.3.3(iv), whose hypothesis \(b \ne 0\) fails here.)

(d) Impossible. A convergent \((b_n)\) is bounded by Theorem 2.3.2, say \(|b_n| \le M_1\); if also \(|a_n - b_n| \le M_2\) for all \(n\), then

\begin{equation*} |a_n| \le |a_n - b_n| + |b_n| \le M_2 + M_1 \end{equation*}

for all \(n\), so \((a_n)\) is bounded, contradicting the hypothesis.

(e) Possible: \(a_n = 0\) for all \(n\) and \(b_n = (-1)^n\). Then \((a_n) \to 0\) and \(a_n b_n = 0 \to 0\), while \((b_n)\) diverges as in part (a). (Theorem 2.3.3(iv) cannot be used to recover \(b_n = (a_n b_n)/a_n\), since \(\lim a_n = 0\).)

(a) Write \(p(t) = c_k t^k + c_{k-1} t^{k-1} + \cdots + c_1 t + c_0\) and run the Algebraic Limit Theorem (Theorem 2.3.3) three times.

First, \(x_n^j \to x^j\) for every \(j \ge 0\), by induction on \(j\): the case \(j = 0\) is the constant sequence \(1\), and if \(x_n^{j} \to x^{j}\) then Theorem 2.3.3 (iii) applied to the two convergent sequences \((x_n^{j})\) and \((x_n)\) gives

\begin{equation*} x_n^{j+1} = x_n^{j}\, x_n \longrightarrow x^{j} x = x^{j+1}. \end{equation*}

Next, Theorem 2.3.3 (i) gives \(c_j x_n^j \to c_j x^j\) for each \(j\), and finitely many applications of Theorem 2.3.3 (ii) add these \(k+1\) convergent sequences:

\begin{equation*} p(x_n) = \sum_{j=0}^{k} c_j x_n^{j} \longrightarrow \sum_{j=0}^{k} c_j x^{j} = p(x). \end{equation*}

(b) Take

\begin{equation*} f(t) = \begin{cases} 0 & \text{if } t \ne 0,\\ 1 & \text{if } t = 0,\end{cases} \qquad x_n = \tfrac{1}{n}. \end{equation*}

Then \((x_n) \to 0\) and \(f(x_n) = 0\) for every \(n\), so \(f(x_n) \to 0\), while \(f(0) = 1\).

(a) Pick \(M > 0\) with \(|a_n| \le M\) for all \(n\) (Definition 2.3.1). Given \(\epsilon > 0\), convergence of \((b_n)\) to \(0\) supplies \(N\) with \(|b_n| < \epsilon/M\) for all \(n \ge N\), and then for all \(n \ge N\)

\begin{equation*} |a_n b_n - 0| = |a_n|\,|b_n| \le M |b_n| < M \cdot \frac{\epsilon}{M} = \epsilon . \end{equation*}

The Algebraic Limit Theorem is unavailable because its hypothesis is that both sequences converge; here \((a_n)\) is only assumed bounded, so the symbol \(\lim a_n\) in the conclusion \(\lim(a_n b_n) = (\lim a_n)(\lim b_n)\) need not name anything.

(b) Only this: with \(b \ne 0\), the sequence \((a_n b_n)\) converges if and only if \((a_n)\) does. One direction is Theorem 2.3.3 (iii); for the other, note \(b \ne 0\) forces \(|b_n| > |b|/2 > 0\) for all \(n \ge N_0\) (take \(\epsilon = |b|/2\) in the definition of \(\lim b_n = b\)), so the quotient below is defined from \(N_0\) on, and Theorem 2.3.3 (iv) applies since \(b \ne 0\):

\begin{equation*} a_n = \frac{a_n b_n}{b_n} \longrightarrow \frac{\lim (a_n b_n)}{b}. \end{equation*}

In particular convergence of \((a_n b_n)\) is not automatic: \(a_n = (-1)^n\), \(b_n = 1\) gives \(b = 1 \ne 0\) and \((a_n b_n) = ((-1)^n)\) divergent.

(c) This is the Algebraic Limit Theorem’s product rule, \(\lim(a_n b_n) = ab\), in the case \(a = 0\). Since \((b_n)\) converges it is bounded (Theorem 2.3.2), and \((a_n) \to a = 0\), so part (a) with the roles of the two sequences interchanged gives

\begin{equation*} \lim (a_n b_n) = 0 = 0 \cdot b = ab . \end{equation*}

(a) False. Take \(a_n = b_n = n\). Then \(a_n - b_n = 0 \to 0\), but neither \(\lim a_n\) nor \(\lim b_n\) exists, so the asserted equation has no meaning.

(b) True, and no \(\epsilon\) is needed beyond the reverse triangle inequality of Exercise 1.2.6 (d):

\begin{equation*} \bigl| |b_n| - |b| \bigr| \le |b_n - b| . \end{equation*}

Given \(\epsilon > 0\), any \(N\) that works for \((b_n) \to b\) works here.

(c) True. Both \((a_n)\) and \((b_n - a_n)\) converge, so Theorem 2.3.3 (ii) applies to their sum:

\begin{equation*} b_n = a_n + (b_n - a_n) \longrightarrow a + 0 = a . \end{equation*}

(d) True. The hypothesis forces \(a_n \ge |b_n - b| \ge 0\). Given \(\epsilon > 0\), choose \(N\) with \(|a_n - 0| < \epsilon\) for all \(n \ge N\); then for those \(n\)

\begin{equation*} |b_n - b| \le a_n = |a_n| < \epsilon . \end{equation*}

(a) Say \((x_n) \to x\); split the average at the index past which the tail is already \(\epsilon/2\)-close. Since \((x_n)\) converges it is bounded (Theorem 2.3.2), say \(|x_k| \le B\) for all \(k\), so the set \(\{\,|x_k - x| : k \in \mathbf{N}\,\}\) is nonempty and bounded above by \(B + |x|\); by the Axiom of Completeness it has a supremum

\begin{equation*} M = \sup\{\, |x_k - x| : k \in \mathbf{N} \,\} . \end{equation*}

Let \(\epsilon > 0\) and choose \(N_1\) with \(|x_k - x| < \epsilon/2\) for all \(k \ge N_1\). For \(n \ge N_1\), writing \(y_n - x = \frac{1}{n}\sum_{k=1}^{n}(x_k - x)\) and applying the triangle inequality,

\begin{equation*} \begin{aligned} |y_n - x| &\le \frac{1}{n}\sum_{k=1}^{N_1 - 1} |x_k - x|

• \frac{1}{n}\sum_{k=N_1}^{n} |x_k - x|\\ &\le \frac{(N_1 - 1)M}{n} + \frac{1}{n} \cdot n \cdot \frac{\epsilon}{2}\\ &= \frac{(N_1 - 1)M}{n} + \frac{\epsilon}{2}. \end{aligned} \end{equation*}

The numerator \((N_1 - 1)M\) is a constant, so by the Archimedean Property (Theorem 1.4.2) there is \(N_2 \in \mathbf{N}\) with \((N_1 - 1)M / N_2 < \epsilon/2\). Setting \(N = \max\{N_1, N_2\}\) gives \(|y_n - x| < \epsilon\) for all \(n \ge N\).

(b) Take \(x_n = (-1)^n\), which diverges. The partial sums alternate between \(-1\) and \(0\), so

\begin{equation*} y_n = \begin{cases} 0 & \text{if } n \text{ is even},\\[2pt] -\tfrac{1}{n} & \text{if } n \text{ is odd}, \end{cases} \end{equation*}

whence \(|y_n| \le 1/n\) and \((y_n) \to 0\) by Exercise 2.3.10 (d).

(a) True. Fix \(b \in B\). Then \(b \le a_n\) for every \(n \in \mathbf{N}\), so the Order Limit Theorem (Theorem 2.3.4 (iii), with the constant \(c = b\)) gives \(b \le a\). As \(b \in B\) was arbitrary, \(a\) is an upper bound for \(B\).

(b) True. Suppose instead \(a \in (0,1)\) and set

\begin{equation*} \epsilon = \min\{a,\, 1 - a\} > 0 . \end{equation*}

Convergence supplies \(N\) with \(|a_N - a| < \epsilon\), and then

\begin{equation*} a_N > a - \epsilon \ge 0 , \qquad a_N < a + \epsilon \le 1 , \end{equation*}

so \(a_N \in (0,1)\), contradicting the hypothesis.

(c) False. By the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3) choose a rational \(a_n\) with

\begin{equation*} \sqrt{2} - \frac{1}{n} < a_n < \sqrt{2}, \end{equation*}

where \(\sqrt{2}\) denotes the real number with square \(2\) supplied by Theorem 1.4.5, and \(\sqrt{2} \notin \mathbf{Q}\) by Theorem 1.1.1. Then \(|a_n - \sqrt{2}| < 1/n\), so \((a_n) \to \sqrt{2}\) by Exercise 2.3.10 (d) while every \(a_n\) is rational.

It should represent a single number \(a\) that every entry \(a_{mn}\) approximates once both indices are large – precisely the \(\epsilon\)-\(N\) definition quoted in part (a), in which one \(N\) must serve for all pairs \((m,n)\) with \(m, n \ge N\).

(a) The two iterated limits are \(1\) and \(0\). For fixed \(n\), dividing by \(m\) gives \(a_{mn} = 1/(1 + n/m) \to 1\) as \(m \to \infty\) (Theorem 2.3.3 (iv), the denominator tending to \(1 \ne 0\)), so the first iterated limit is \(\lim_{n \to \infty} 1 = 1\). For fixed \(m\), \(a_{mn} = m/(m+n)\) satisfies \(0 < a_{mn} \le m/n \to 0\), so \(\lim_{n \to \infty} a_{mn} = 0\) and the second iterated limit is \(0\).

(b) For \(a_{mn} = 1/(m+n)\) all three limits exist and equal \(0\). Given \(\epsilon > 0\) pick \(N > 1/(2\epsilon)\) (Theorem 1.4.2); if \(m, n \ge N\) then

\begin{equation*} |a_{mn} - 0| = \frac{1}{m+n} \le \frac{1}{2N} < \epsilon , \end{equation*}

so \(\lim_{m,n \to \infty} a_{mn} = 0\); and each inner limit is \(0\) by the same bound with one index fixed, so both iterated limits are \(\lim 0 = 0\).

For \(a_{mn} = mn/(m^2+n^2)\) the two iterated limits exist and equal \(0\), but the double limit does not exist. For fixed \(m\), \(0 < a_{mn} \le mn/n^2 = m/n \to 0\), so \(\lim_{n \to \infty} a_{mn} = 0\) and hence \(\lim_{m}\lim_{n} a_{mn} = 0\); by symmetry of \(a_{mn}\) in \(m\) and \(n\) the other iterated limit is also \(0\). If the double limit were some \(a\), take \(\epsilon = 1/40\) and the corresponding \(N\); the pairs \((N,N)\) and \((2N,N)\) both qualify, and

\begin{equation*} a_{NN} = \tfrac{1}{2}, \qquad a_{2N,N} = \frac{2N^2}{5N^2} = \tfrac{2}{5}, \end{equation*}

so \(\tfrac{1}{10} = |a_{NN} - a_{2N,N}| \le |a_{NN} - a| + |a - a_{2N,N}| < \tfrac{1}{20}\), a contradiction.

(c) Take

\begin{equation*} a_{mn} = \frac{(-1)^m}{n} + \frac{(-1)^n}{m}. \end{equation*}

The double limit is \(0\): if \(m, n \ge N\) then \(|a_{mn}| \le 1/n + 1/m \le 2/N\), which is \(< \epsilon\) once \(N > 2/\epsilon\). But for each fixed \(m\) the inner limit in \(n\) fails to exist: \((-1)^m/n \to 0\) as \(n \to \infty\), so if \((a_{mn})_n\) converged, then by Theorem 2.3.3 (i) and (ii) so would

\begin{equation*} a_{mn} - \frac{(-1)^m}{n} = \frac{(-1)^n}{m}, \end{equation*}

whose terms take the two values \(\pm 1/m\) for arbitrarily large \(n\); no single \(L\) lies within \(1/m\) of both, so this sequence has no limit. Since \(a_{mn}\) is symmetric in \(m\) and \(n\), the inner limit in \(m\) fails as well, and neither iterated limit can be formed.

(d) Let \(\epsilon > 0\) and choose \(N \in \mathbf{N}\) with \(|a_{mn} - a| < \epsilon/2\) whenever \(m, n \ge N\). Fix \(m \ge N\). Since \((a_{mn})_n \to b_m\), Theorem 2.3.3 (ii) gives \(a_{mn} - a \to b_m - a\), and then Exercise 2.3.10 (b) gives

\begin{equation*} |a_{mn} - a| \longrightarrow |b_m - a| \qquad (n \to \infty). \end{equation*}

Every term of this convergent sequence with index \(n \ge N\) satisfies \(|a_{mn} - a| \le \epsilon/2\), and discarding the finitely many terms with \(n < N\) changes neither the sequence’s limit nor the hypothesis of the Order Limit Theorem (Exercise 2.2.7), so Theorem 2.3.4 (iii) yields

\begin{equation*} |b_m - a| \le \frac{\epsilon}{2} < \epsilon . \end{equation*}

As \(m \ge N\) was arbitrary, \(\lim_{m \to \infty} b_m = a\).

(e) Write \(a = \lim_{m,n \to \infty} a_{mn}\). Saying that the iterated limit \(\lim_{m}\lim_{n} a_{mn}\) exists includes the existence of \(b_m = \lim_{n \to \infty} a_{mn}\) for each fixed \(m\), so the hypotheses of (d) hold verbatim and

\begin{equation*} \lim_{m \to \infty} \Bigl( \lim_{n \to \infty} a_{mn} \Bigr) = \lim_{m \to \infty} b_m = a . \end{equation*}

For the other order, existence of the second iterated limit supplies \(c_k = \lim_{m \to \infty} a_{mk}\) for each fixed \(k\); apply (d) to the transposed array \(\tilde a_{mn} = a_{nm}\). Its double limit is again \(a\), since the defining condition \(|a_{mn} - a| < \epsilon\) for all \(m, n \ge N\) is symmetric in the two indices, and for each fixed \(m\) its inner limit is \(\lim_{n \to \infty} \tilde a_{mn} = \lim_{n \to \infty} a_{nm} = c_m\). Hence (d) gives

\begin{equation*} \lim_{n \to \infty} \Bigl( \lim_{m \to \infty} a_{mn} \Bigr) = \lim_{n \to \infty} c_n = a . \end{equation*}

\(\lim x_n = 2 - \sqrt{3}\).

(a) The interval \(I = [2-\sqrt{3},\,3]\) is invariant. Indeed \(t \mapsto 1/(4-t)\) is increasing on \(I\) (the denominator stays in \([1, 2+\sqrt{3}]\), in particular never \(0\)), so \(x_n \in I\) gives

\begin{equation*} \begin{aligned} x_{n+1} &\ge \frac{1}{4-(2-\sqrt3)} = \frac{1}{2+\sqrt3} = 2-\sqrt3,\\ x_{n+1} &\le \frac{1}{4-3} = 1 \le 3 . \end{aligned} \end{equation*}

Since \(x_1 = 3 \in I\), induction gives \(x_n \in I\) for all \(n\). Monotonicity now follows from the sign of

\begin{equation*} x_{n+1} - x_n = \frac{1 - x_n(4-x_n)}{4-x_n} = \frac{(x_n - (2-\sqrt3))(x_n-(2+\sqrt3))}{4-x_n}, \end{equation*}

whose denominator is positive and whose numerator has a nonnegative first factor and a negative second factor on \(I\) (as \(2+\sqrt3 > 3\)). So \((x_n)\) is decreasing and bounded below by \(2-\sqrt3\), and the Monotone Convergence Theorem (Theorem 2.4.2) gives convergence.

(b) Write \(L = \lim x_n\) and let \(\varepsilon > 0\). Choose \(N\) with \(|x_n - L| < \varepsilon\) for all \(n \ge N\); if \(n \ge N\) then also \(n+1 \ge N\), so \(|x_{n+1} - L| < \varepsilon\). Hence \(\lim x_{n+1} = L\). (Equivalently, \((x_{n+1})\) is a subsequence of \((x_n)\), so Theorem 2.5.2 applies.)

(c) Taking limits across \(x_{n+1}(4 - x_n) = 1\) and using the Algebraic Limit Theorem (Theorem 2.3.3) together with (b),

\begin{equation*} L(4-L) = 1, \qquad L^2 - 4L + 1 = 0, \qquad L = 2 \pm \sqrt3 . \end{equation*}

Since \(x_n \le 1\) for every \(n \ge 2\), the Order Limit Theorem (Theorem 2.3.4) forces \(L \le 1\), ruling out \(2+\sqrt3\). Thus \(\lim x_n = 2 - \sqrt{3}\).

(a) The argument assumes what it must prove: that \(\lim y_n\) exists. Here it does not. The recursion produces

\begin{equation*} 1,\ 2,\ 1,\ 2,\ 1,\ 2,\ \ldots \end{equation*}

so \(y_n = 1\) for odd \(n\) and \(y_n = 2\) for even \(n\), and no \(\varepsilon\)-neighbourhood of radius \(1/2\) about any candidate limit contains a tail. The algebra \(y = 3-y\) is a valid consequence of the recursion only once convergence is known; the number \(3/2\) it produces is merely the fixed point of \(t \mapsto 3-t\), which this sequence never approaches.

(b) Yes, and now the strategy is legitimate, because convergence can be established first. Put \(L = (3+\sqrt5)/2\), the positive root of \(t^2 - 3t + 1 = 0\), so that \(3 - 1/L = L\). The map \(t \mapsto 3 - 1/t\) is increasing on \([1, L]\), and \(y_1 = 1\), so induction gives \(y_n \in [1, L]\) for all \(n\):

\begin{equation*} 1 \le y_n \le L \implies 2 = 3 - \tfrac11 \le y_{n+1} \le 3 - \tfrac1L = L . \end{equation*}

On this range \(y_n^2 - 3y_n + 1 \le 0\), since its roots are \((3\pm\sqrt5)/2\) and \((3-\sqrt5)/2 < 1 \le y_n \le L\). Hence

\begin{equation*} y_{n+1} - y_n = \frac{3y_n - 1 - y_n^2}{y_n} = \frac{-(y_n^2 - 3y_n + 1)}{y_n} \ge 0 . \end{equation*}

Thus \((y_n)\) is increasing and bounded above, and the Monotone Convergence Theorem (Theorem 2.4.2) gives \(y = \lim y_n\). Now the strategy applies: \(y \ge 1 > 0\), so by the Algebraic Limit Theorem (Theorem 2.3.3),

\begin{equation*} y = 3 - \frac1y, \qquad y^2 - 3y + 1 = 0, \qquad y = \frac{3 \pm \sqrt5}{2}. \end{equation*}

The root \((3-\sqrt5)/2 < 1\) is excluded by \(y \ge 1\) (Theorem 2.3.4), so \(\lim y_n = \frac{3+\sqrt5}{2}\).

Both converge to \(2\).

(a) The sequence is \(x_1 = \sqrt2\), \(x_{n+1} = \sqrt{2 + x_n}\); every term is positive. It is bounded above by \(2\) by induction: \(x_1 = \sqrt2 < 2\), and \(x_n < 2\) gives \(x_{n+1} = \sqrt{2+x_n} < \sqrt4 = 2\). It is increasing because for \(0 < x_n < 2\),

\begin{equation*} x_{n+1}^2 - x_n^2 = 2 + x_n - x_n^2 = -(x_n - 2)(x_n + 1) > 0, \end{equation*}

and both terms are positive, so \(x_{n+1} > x_n\). By the Monotone Convergence Theorem (Theorem 2.4.2), \(L = \lim x_n\) exists, and \(L \ge \sqrt2 > 0\). Squaring the recursion and taking limits (Theorem 2.3.3, with \(\lim x_{n+1} = L\) as in Exercise 2.4.1(b)),

\begin{equation*} L^2 = 2 + L, \qquad (L-2)(L+1) = 0, \qquad L = 2 . \end{equation*}

(b) Here \(x_1 = \sqrt2\), \(x_{n+1} = \sqrt{2x_n}\), and the argument of (a) repeats verbatim. Induction gives \(0 < x_n < 2\) (from \(\sqrt{2\cdot 2} = 2\)), and then

\begin{equation*} x_{n+1}^2 - x_n^2 = 2x_n - x_n^2 = x_n(2 - x_n) > 0, \end{equation*}

so \(x_{n+1} > x_n\), both terms being positive. Theorem 2.4.2 applies, and \(L = \lim x_n\) satisfies \(L^2 = 2L\) with \(L \ge \sqrt2 > 0\), whence \(L = 2\).

Method (2): the recursion has the closed form

\begin{equation*} x_n = 2^{\,1 - 2^{-n}} \quad\text{since}\quad \bigl(2\cdot 2^{\,1-2^{-n}}\bigr)^{1/2} = 2^{\,1 - 2^{-(n+1)}} , \end{equation*}

so \(x_n \to 2^1 = 2\) by continuity of \(t \mapsto 2^t\) (Chapter 4) and \(2^{-n} \to 0\).

Throughout, the Monotone Convergence Theorem (Theorem 2.4.2) is taken as the axiom; the Algebraic and Order Limit Theorems (Theorems 2.3.3, 2.3.4) are proved straight from the definition of convergence and use no AoC.

(a) Suppose, for contradiction, that \(\mathbf{N}\) is bounded above in \(\mathbf{R}\). The sequence \(a_n = n\) is then increasing and bounded, so by Theorem 2.4.2 it converges, say \(a_n \to a\); and \(a_{n+1} \to a\) as well (Exercise 2.4.1(b)). By the Algebraic Limit Theorem,

\begin{equation*} 1 = \lim (a_{n+1} - a_n) = a - a = 0, \end{equation*}

a contradiction. Hence \(\mathbf{N}\) is unbounded above, which is statement (i) of Theorem 1.4.2: for any \(x \in \mathbf{R}\) there is \(n \in \mathbf{N}\) with \(n > x\). For statement (ii), given \(y > 0\) pick \(n > 1/y\); then \(1/n < y\).

(b) Let \(I_n = [a_n, b_n]\) with \(I_{n+1} \subseteq I_n\) for every \(n\). Nesting says precisely

\begin{equation*} a_1 \le a_2 \le a_3 \le \cdots \le b_3 \le b_2 \le b_1 , \end{equation*}

so \((a_n)\) is increasing and bounded above by \(b_1\). By Theorem 2.4.2 it converges; set \(a = \lim a_n\). Fix \(n\). For every \(m \ge n\) we have \(I_m \subseteq I_n\), hence

\begin{equation*} a_n \le a_m \le b_n . \end{equation*}

Applying the Order Limit Theorem to the tail \((a_m)_{m \ge n}\), which has the same limit \(a\), gives \(a_n \le a \le b_n\), i.e. \(a \in I_n\). As \(n\) was arbitrary,

\begin{equation*} a \in \bigcap_{n=1}^{\infty} I_n \ne \emptyset . \qquad \blacksquare \end{equation*}

(a) All terms are positive (\(x_1 = 2 > 0\), and \(x_n > 0\) makes \(x_{n+1}\) an average of positive numbers), so the recursion is well posed. The key identity is

\begin{equation*} x_{n+1}^2 - 2 = \frac{1}{4}\left(x_n^2 + 4 + \frac{4}{x_n^2}\right) - 2 = \frac{1}{4}\left(x_n - \frac{2}{x_n}\right)^{2} \ \ge\ 0 , \end{equation*}

so \(x_n^2 \ge 2\) for every \(n \ge 2\), and \(x_1^2 = 4 \ge 2\) as well. Consequently

\begin{equation*} x_n - x_{n+1} = x_n - \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) = \frac{x_n^2 - 2}{2x_n} \ \ge\ 0 . \end{equation*}

Thus \((x_n)\) is decreasing and bounded below by \(\sqrt2\), and Theorem 2.4.2 gives \(L = \lim x_n\) with \(L \ge \sqrt2 > 0\) (Theorem 2.3.4). Multiplying the recursion by \(2x_n\) and passing to the limit (Theorem 2.3.3, with \(\lim x_{n+1} = L\)),

\begin{equation*} 2L^2 = L^2 + 2, \qquad L^2 = 2, \qquad L = \sqrt2 . \end{equation*}

(b) For \(c > 0\) take \(x_1 = c\) and

\begin{equation*} x_{n+1} = \frac{1}{2}\left( x_n + \frac{c}{x_n} \right). \end{equation*}

The same two displays hold verbatim with \(2\) replaced by \(c\):

\begin{equation*} x_{n+1}^2 - c = \frac{1}{4}\left(x_n - \frac{c}{x_n}\right)^{2} \ge 0, \qquad x_n - x_{n+1} = \frac{x_n^2 - c}{2x_n} \ge 0 \ \ (n \ge 2), \end{equation*}

so from the second term on, \((x_n)\) decreases and is bounded below by \(\sqrt c\). Theorem 2.4.2 and \(2L^2 = L^2 + c\) give \(\lim x_n = \sqrt{c}\).

(a) Square roots are real and the square of a real number is nonnegative:

\begin{equation*} 0 \le \left(\sqrt{x} - \sqrt{y}\right)^{2} = x + y - 2\sqrt{xy} \quad\Longleftrightarrow\quad \sqrt{xy} \le \frac{x+y}{2}, \end{equation*}

valid for all \(x, y \ge 0\), with equality exactly when \(x = y\).

(b) Part (a) applied to the pair \((x_n, y_n)\) says \(x_{n+1} \le y_{n+1}\), and \(x_1 \le y_1\) is given, so

\begin{equation*} 0 \le x_n \le y_n \qquad \text{for every } n \in \mathbf{N}. \end{equation*}

(Nonnegativity propagates since \(x_{n+1}\) is a square root and \(y_{n+1}\) an average of nonnegative numbers.) These inequalities make each sequence monotone:

\begin{equation*} \begin{aligned} x_{n+1} &= \sqrt{x_n y_n} \ \ge\ \sqrt{x_n\, x_n} \ =\ x_n, \\ y_{n+1} &= \frac{x_n + y_n}{2} \ \le\ \frac{y_n + y_n}{2} \ =\ y_n . \end{aligned} \end{equation*}

So \((x_n)\) is increasing and bounded above by \(x_n \le y_n \le y_1\), while \((y_n)\) is decreasing and bounded below by \(y_n \ge x_n \ge x_1\). The Monotone Convergence Theorem (Theorem 2.4.2) gives \(x = \lim x_n\) and \(y = \lim y_n\).

They agree: \(\lim y_{n+1} = y\) (Exercise 2.4.1(b)), so taking limits in \(2y_{n+1} = x_n + y_n\) via the Algebraic Limit Theorem (Theorem 2.3.3) yields

\begin{equation*} 2y = x + y, \qquad\text{hence}\qquad x = y . \qquad \blacksquare \end{equation*}

Fix \(M > 0\) with \(|a_n| \le M\) for all \(n\), and write \(A_n = \{a_k : k \ge n\}\).

(a) Each \(A_n\) is nonempty and bounded above by \(M\), so \(y_n = \sup A_n\) exists by the Axiom of Completeness. Since \(A_{n+1} \subseteq A_n\), any upper bound for \(A_n\) is one for \(A_{n+1}\), whence

\begin{equation*} y_{n+1} \le y_n \qquad\text{for all } n, \end{equation*}

and \(y_n \ge a_n \ge -M\). So \((y_n)\) is decreasing and bounded below, and the Monotone Convergence Theorem (Theorem 2.4.2) gives convergence.

(b) Define

\begin{equation*} \liminf a_n = \lim z_n, \qquad z_n = \inf\{a_k : k \ge n\} . \end{equation*}

Each \(z_n\) exists (the set \(A_n\) is bounded below by \(-M\)), and shrinking \(A_n\) to \(A_{n+1}\) can only raise the infimum, so \((z_n)\) is increasing; it is bounded above by \(M\). Theorem 2.4.2 again gives existence.

(c) For each \(n\) the set \(A_n\) is nonempty, so \(z_n \le a_n \le y_n\); in particular \(z_n \le y_n\), and the Order Limit Theorem (Theorem 2.3.4) gives

\begin{equation*} \liminf a_n = \lim z_n \ \le\ \lim y_n = \limsup a_n . \end{equation*}

For strictness take \(a_n = (-1)^n\): here \(A_n = \{-1, 1\}\) for every \(n\), so \(z_n = -1\) and \(y_n = 1\), giving \(\liminf a_n = -1 < 1 = \limsup a_n\).

(d) (\(\Leftarrow\)) Suppose \(a_n \to a\) and let \(\varepsilon > 0\). Choose \(N\) so that \(a - \varepsilon < a_k < a + \varepsilon\) for all \(k \ge N\). For \(n \ge N\) the number \(a + \varepsilon\) is an upper bound for \(A_n\) and \(a - \varepsilon\) a lower bound, so

\begin{equation*} \begin{aligned} a - \varepsilon &< a_n \le y_n \le a + \varepsilon, \\ a + \varepsilon &> a_n \ge z_n \ge a - \varepsilon . \end{aligned} \end{equation*}

Hence \(|y_n - a| \le \varepsilon\) and \(|z_n - a| \le \varepsilon\) for all \(n \ge N\), so \(\lim y_n = \lim z_n = a\); the two agree and share the value of \(\lim a_n\).

(\(\Rightarrow\)) Suppose \(\lim z_n = \lim y_n = L\) and let \(\varepsilon > 0\). Choose \(N_1, N_2\) with \(|y_n - L| < \varepsilon\) for \(n \ge N_1\) and \(|z_n - L| < \varepsilon\) for \(n \ge N_2\). For \(n \ge \max\{N_1, N_2\}\), the sandwich \(z_n \le a_n \le y_n\) gives

\begin{equation*} L - \varepsilon < z_n \le a_n \le y_n < L + \varepsilon , \end{equation*}

so \(|a_n - L| < \varepsilon\). Thus \(\lim a_n = L\), and all three quantities equal \(L\). \(\blacksquare\)

(a) \(s_m = 1 - 1/2^m\), so the series converges to \(1\). Indeed \(2s_m - s_m\) telescopes:

\begin{equation*} \begin{aligned} 2s_m &= 1 + \tfrac12 + \cdots + \tfrac{1}{2^{m-1}},\\ s_m &= \tfrac12 + \cdots + \tfrac{1}{2^{m-1}} + \tfrac{1}{2^{m}}, \end{aligned} \end{equation*}

and subtracting gives \(s_m = 1 - 2^{-m}\). Since \(0 < 2^{-m} \le 1/m\), the Archimedean Property (Theorem 1.4.2) forces \(2^{-m} \to 0\), so \(s_m \to 1\).

(b) \(s_m = 1 - \dfrac{1}{m+1}\), so the series converges to \(1\). Partial fractions give \(\dfrac{1}{n(n+1)} = \dfrac1n - \dfrac{1}{n+1}\), and the sum telescopes:

\begin{equation*} s_m = \sum_{n=1}^{m}\left(\frac1n - \frac{1}{n+1}\right) = 1 - \frac{1}{m+1}. \end{equation*}

(c) \(s_m = \log(m+1)\), so the series diverges. Since \(\log\) carries products to sums,

\begin{equation*} s_m = \sum_{n=1}^{m}\bigl(\log(n+1)-\log n\bigr) = \log(m+1), \end{equation*}

again a telescoping sum. The sequence \((\log(m+1))\) is unbounded (given \(M>0\), take \(m+1 > e^{M}\)), and convergent sequences are bounded by Theorem 2.3.2, so \((s_m)\) diverges.

The partial sums satisfy \(s_{2^{k}} \ge \tfrac12 t_{k}\), which is the whole content. Here \((b_n)\) is decreasing with \(b_n \ge 0\) (the hypotheses of Theorem 2.4.6), and

\begin{equation*} \begin{aligned} s_m &= b_1 + b_2 + \cdots + b_m,\\ t_k &= \sum_{n=0}^{k} 2^{n}b_{2^{n}} = b_1 + 2b_2 + 4b_4 + \cdots + 2^{k}b_{2^{k}}. \end{aligned} \end{equation*}

Group the first \(2^{k}\) terms of \((s_m)\) exactly as in Example 2.4.5, and bound each block \(b_{2^{j-1}+1} + \cdots + b_{2^{j}}\) (which has \(2^{j-1}\) terms, each \(\ge b_{2^{j}}\) because \((b_n)\) is decreasing) below by \(2^{j-1}b_{2^{j}}\):

\begin{equation*} \begin{aligned} s_{2^{k}} &= b_1 + b_2 + (b_3+b_4) + \cdots + (b_{2^{k-1}+1} + \cdots + b_{2^{k}})\\ &\ge b_1 + b_2 + (b_4+b_4) + \cdots + (b_{2^{k}} + \cdots + b_{2^{k}})\\ &= b_1 + b_2 + 2b_4 + 4b_8 + \cdots + 2^{k-1}b_{2^{k}}\\ &\ge \tfrac12\bigl(b_1 + 2b_2 + 4b_4 + \cdots + 2^{k}b_{2^{k}}\bigr) = \tfrac12\,t_{k}. \end{aligned} \end{equation*}

(Only the leading term is weakened: \(b_1 \ge \tfrac12 b_1\), while every later term matches.)

Now suppose \(\sum_{n=0}^{\infty}2^{n}b_{2^{n}}\) diverges. The partial sums \((t_k)\) increase (each \(2^{n}b_{2^{n}} \ge 0\)), so the Monotone Convergence Theorem (Theorem 2.4.2) forbids them from being bounded. Hence \(s_{2^{k}} \ge \tfrac12 t_{k}\) makes the subsequence \((s_{2^{k}})\), and with it \((s_m)\), unbounded; convergent sequences are bounded (Theorem 2.3.2), so \(\sum_{n=1}^{\infty}b_n\) diverges.

(a) \(p_m = m+1\), which diverges. The product telescopes:

\begin{equation*} p_m = \prod_{n=1}^{m}\left(1 + \frac1n\right) = \prod_{n=1}^{m}\frac{n+1}{n} = \frac{2}{1}\cdot\frac{3}{2}\cdots\frac{m+1}{m} = m+1, \end{equation*}

unbounded, hence divergent by Theorem 2.3.2.

For \(a_n = 1/n^2\) the partial products \(p_m = \prod_{n=1}^{m}\frac{n^2+1}{n^2}\) are

\begin{equation*} 2,\quad \frac52,\quad \frac{25}{9},\quad \frac{425}{144},\quad \frac{221}{72},\quad \frac{8177}{2592}, \end{equation*}

that is, approximately \(2,\ 2.500,\ 2.778,\ 2.951,\ 3.069,\ 3.155\). The increments shrink rapidly; conjecture: this sequence converges.

(b) The sequence \((p_m)\) is increasing, since \(p_{m+1} = p_m(1+a_{m+1}) \ge p_m\) and \(p_m \ge 1 > 0\). So by the Monotone Convergence Theorem (Theorem 2.4.2), \((p_m)\) converges if and only if it is bounded, and likewise the increasing partial sums \(s_m = \sum_{n=1}^{m}a_n\) converge if and only if they are bounded. It therefore suffices to show \((p_m)\) is bounded if and only if \((s_m)\) is.

(\(\Leftarrow\)) Using \(1 + x \le 3^{x}\) for \(x \ge 0\) termwise,

\begin{equation*} p_m = \prod_{n=1}^{m}(1+a_n) \le \prod_{n=1}^{m} 3^{a_n} = 3^{\,s_m} \le 3^{\,s}, \end{equation*}

where \(s = \lim s_m = \sup_m s_m\) is finite when \(\sum a_n\) converges (the last step uses that \(x \mapsto 3^{x}\) is increasing). So \((p_m)\) is bounded.

(\(\Rightarrow\)) In the other direction,

\begin{equation*} p_m \ge 1 + s_m \qquad \text{for every } m, \end{equation*}

by induction: equality holds at \(m = 1\), and if \(p_m \ge 1 + s_m\) then

\begin{equation*} \begin{aligned} p_{m+1} = p_m(1+a_{m+1}) &\ge (1+s_m)(1+a_{m+1})\\ &= 1 + s_m + a_{m+1} + s_m a_{m+1}\\ &\ge 1 + s_{m+1}, \end{aligned} \end{equation*}

since \(s_m a_{m+1} \ge 0\). If \((p_m)\) converges it is bounded by some \(M\) (Theorem 2.3.2), whence \(s_m \le M - 1\) for all \(m\) and \((s_m)\) is bounded.

(a) Impossible. A bounded subsequence \((x_{n_k})\) is itself a bounded sequence, so the Bolzano–Weierstrass Theorem (Theorem 2.5.5) produces a convergent subsequence \((x_{n_{k_j}})\) of it; since \(j \mapsto n_{k_j}\) is strictly increasing, this is a subsequence of \((x_n)\) as well.

(b) Take

\begin{equation*} x_n = \begin{cases} \dfrac{1}{n+1}, & n \text{ odd},\\[4pt] \dfrac{n}{n+1}, & n \text{ even}. \end{cases} \end{equation*}

Every term lies in \((0,1)\), so no term equals \(0\) or \(1\); the odd-indexed subsequence \((1/2, 1/4, 1/6, \ldots) \to 0\) and the even-indexed subsequence \((2/3, 4/5, 6/7, \ldots) \to 1\).

(c) List the finite blocks \(1;\ 1, \tfrac12;\ 1, \tfrac12, \tfrac13;\ \ldots\) one after another:

\begin{equation*} (x_n) = \Bigl(1,\ 1, \tfrac12,\ 1, \tfrac12, \tfrac13,\ 1, \tfrac12, \tfrac13, \tfrac14,\ \ldots\Bigr). \end{equation*}

For each \(m\) the value \(1/m\) occurs in every block of length \(\ge m\), hence at infinitely many indices; the constant subsequence built from those indices converges to \(1/m\).

(d) Impossible. Suppose for each \(m \in \mathbf{N}\) some subsequence of \((x_n)\) converges to \(1/m\); then for each \(m\) the inequality \(|x_n - 1/m| < 1/m\) holds for infinitely many \(n\). Choose indices recursively: having picked \(n_{m-1}\), pick \(n_m > n_{m-1}\) with \(|x_{n_m} - 1/m| < 1/m\). Then

\begin{equation*} |x_{n_m} - 0| \le |x_{n_m} - \tfrac1m| + \tfrac1m < \tfrac{2}{m}, \end{equation*}

so \((x_{n_m}) \to 0\). This is a subsequence of \((x_n)\) converging to \(0\), which is not a point of \(\{1, 1/2, 1/3, \ldots\}\).

All four are true.

(a) The tail \((x_2, x_3, x_4, \ldots)\) is a proper subsequence, so it converges to some \(L\): given \(\epsilon > 0\) there is \(K\) with \(|x_{k+1} - L| < \epsilon\) for all \(k \ge K\). Then \(n \ge K+1\) gives \(|x_n - L| < \epsilon\), so \((x_n) \to L\) by Definition 2.2.3.

(b) Contrapositive of Theorem 2.5.2: if \((x_n)\) converged, every subsequence would converge (to the same limit).

(c) By Bolzano–Weierstrass (Theorem 2.5.5) the bounded sequence \((x_n)\) has a subsequence \((x_{n_k}) \to L\). Since \((x_n)\) does not converge to \(L\), there is an \(\epsilon_0 > 0\) with \(|x_n - L| \ge \epsilon_0\) for infinitely many \(n\); those indices give a bounded subsequence \((x_{m_j})\), and Theorem 2.5.5 again yields a further subsequence converging to some \(L^{\prime}\). Along it \(|x_{m_j} - L| \to |L^{\prime} - L|\) (Exercise 2.3.10(b) with Theorem 2.3.3), so the Order Limit Theorem (Theorem 2.3.4) gives \(|L^{\prime} - L| \ge \epsilon_0 > 0\) and \(L \ne L^{\prime}\).

(d) Say \((x_n)\) is increasing and \((x_{n_k}) \to L\). The subsequence is increasing too, so \(x_{n_k} \le L\) for every \(k\) (Theorem 2.3.4 applied to \(x_{n_k} \le x_{n_j}\) for \(j \ge k\)). Given \(n\), pick \(k\) with \(n_k \ge n\); then \(x_n \le x_{n_k} \le L\). So \((x_n)\) is increasing and bounded above, hence convergent by the Monotone Convergence Theorem (Theorem 2.4.2) — and its limit is \(L\) by Theorem 2.5.2. The decreasing case is identical with the inequalities reversed.

(a) The regrouped series has partial sums \(t_k = s_{n_k}\), a subsequence of \((s_n)\), so \((t_k) \to L\) by Theorem 2.5.2. Writing \(n_0 = 0\) and \(b_j = a_{n_{j-1}+1} + \cdots + a_{n_j}\) for the \(j\)th grouped term,

\begin{equation*} \begin{aligned} t_k \;=\; \sum_{j=1}^{k} b_j &\;=\; \sum_{j=1}^{k} \ \sum_{i=n_{j-1}+1}^{n_j} a_i \\ &\;=\; \sum_{i=1}^{n_k} a_i \;=\; s_{n_k}, \end{aligned} \end{equation*}

the middle step being a regrouping of a finite sum. The indices satisfy \(n_1 < n_2 < n_3 < \cdots\), so \((s_{n_k})\) is a subsequence of \((s_n)\) in the sense of Definition 2.5.1 and Theorem 2.5.2 applies.

(b) The hypothesis of (a) — that \((s_n)\) converges to some \(L\) — fails there. The example is \(\sum_{n=1}^{\infty} (-1)^n\), whose two groupings

\begin{equation*} \begin{aligned} (-1+1) + (-1+1) + \cdots &= 0,\\ -1 + (1-1) + (1-1) + \cdots &= -1, \end{aligned} \end{equation*}

are exactly the subsequences \((s_{2k}) = (0,0,0,\ldots)\) and \((s_{2k-1}) = (-1,-1,-1,\ldots)\) of the partial sums \((s_n) = (-1, 0, -1, 0, \ldots)\). Two subsequences with different limits force divergence by Theorem 2.5.2, so no \(L\) exists for the argument in (a) to transport.

Bisect, keeping an upper bound on the right and a non-upper-bound on the left. Let \(A \subseteq \mathbf{R}\) be nonempty and bounded above; we produce \(\sup A\).

Fix \(x_0 \in A\) and an upper bound \(b_1\) of \(A\), and set \(a_1 = x_0 - 1\), so that \(a_1\) fails to be an upper bound of \(A\) (it is less than \(x_0 \in A\)). Put \(I_1 = [a_1, b_1]\). Recursively, given \(I_n = [a_n, b_n]\) with \(b_n\) an upper bound of \(A\) and \(a_n\) not one, let \(m = (a_n + b_n)/2\) and set

\begin{equation*} I_{n+1} = \begin{cases} [a_n, m], & \text{if } m \text{ is an upper bound of } A,\\ [m, b_n], & \text{otherwise.} \end{cases} \end{equation*}

In either case \(I_{n+1} \subseteq I_n\), the right endpoint of \(I_{n+1}\) is an upper bound of \(A\), the left endpoint is not one, and the lengths satisfy

\begin{equation*} \ell_n := b_n - a_n = \frac{b_1 - a_1}{2^{n-1}} \longrightarrow 0 \end{equation*}

by the assumption \((1/2^n) \to 0\) together with the Algebraic Limit Theorem (Theorem 2.3.3). The Nested Interval Property (Theorem 1.4.1) supplies \(s \in \bigcap_{n=1}^{\infty} I_n\). We claim \(s = \sup A\).

(i) \(s\) is an upper bound. If some \(a \in A\) had \(a > s\), choose \(n\) with \(\ell_n < a - s\). Since \(s, b_n \in I_n\) and \(b_n \ge s\),

\begin{equation*} b_n \le s + \ell_n < s + (a - s) = a, \end{equation*}

contradicting that \(b_n\) is an upper bound of \(A\).

(ii) \(s\) is the least upper bound. If \(u\) were an upper bound with \(u < s\), choose \(n\) with \(\ell_n < s - u\). Since \(a_n \le s\) and \(s - a_n \le \ell_n\),

\begin{equation*} a_n \ge s - \ell_n > s - (s-u) = u. \end{equation*}

But \(a_n\) is not an upper bound of \(A\), so some \(a \in A\) satisfies \(a > a_n > u\), contradicting that \(u\) is an upper bound.

Hence \(\sup A\) exists, which is the Axiom of Completeness.

Why the assumption is needed: steps (i) and (ii) require \(\ell_n \to 0\), and \((1/2^n) \to 0\) is equivalent to the Archimedean Property (Theorem 1.4.2). Indeed \(2^n \ge n\) turns the unboundedness of \(\mathbf{N}\) into \(1/2^n \to 0\), while conversely \(1/2^n < \epsilon\) for large \(n\) exhibits the natural number \(2^n\) exceeding \(1/\epsilon\). Abbott derives Theorem 1.4.2 from AoC, so proving \((1/2^n) \to 0\) inside this argument would invoke the very axiom being established — and the hypothesis cannot be dropped, since NIP alone cannot prove \(\mathbf{N}\) is unbounded in \(\mathbf{R}\) (Section 2.6).

Suppose not. Then there is an \(\epsilon_0 > 0\) for which no \(N\) works, i.e. \(|a_n - a| \ge \epsilon_0\) for infinitely many \(n\); list those indices as \(n_1 < n_2 < n_3 < \cdots\), so

\begin{equation*} |a_{n_k} - a| \ge \epsilon_0 \qquad \text{for all } k. \end{equation*}

The subsequence \((a_{n_k})\) inherits the bound on \((a_n)\), so Bolzano–Weierstrass (Theorem 2.5.5) gives a convergent subsequence \((a_{n_{k_j}}) \to L\), and this is again a subsequence of \((a_n)\) (the indices \(n_{k_j}\) are strictly increasing in \(j\)). By hypothesis \(L = a\). But \(|a_{n_{k_j}} - a| \to |L - a|\) (Theorem 2.3.3 and Exercise 2.3.10(b)), so the Order Limit Theorem (Theorem 2.3.4) applied to \(|a_{n_{k_j}} - a| \ge \epsilon_0\) gives

\begin{equation*} 0 = |L - a| = \lim_{j \to \infty} |a_{n_{k_j}} - a| \ge \epsilon_0 > 0, \end{equation*}

a contradiction. Hence \((a_n) \to a\).

The limit is \(1\) for every \(b > 0\), and \(0\) for \(b = 0\).

For \(b = 0\) the sequence is constantly \(0\), and for \(b = 1\) it is constantly \(1\).

Case \(b > 1\). The sequence \((b^{1/n})\) is decreasing and bounded below by \(1\): raising to the power \(n(n+1)\) (strictly increasing on \([0,\infty)\)) turns \(b^{1/n} > b^{1/(n+1)}\) into \(b^{n+1} > b^{n}\), which holds since \(b > 1\), and \(b^{1/n} \ge 1\) because \(b \ge 1\). By the Monotone Convergence Theorem (Theorem 2.4.2) it converges to some \(l\), and \(l \ge 1\) by the Order Limit Theorem (Theorem 2.3.4).

Now identify \(l\) the way Example 2.5.3 identifies \(\lim b^n\). The sequence \((b^{1/(2n)})_{n \ge 1}\) is the subsequence of \((b^{1/n})\) with indices \(n_k = 2k\), so it converges to \(l\) by Theorem 2.5.2. On the other hand \(b^{1/(2n)} = \sqrt{b^{1/n}}\), so Exercise 2.3.1(b) gives \(b^{1/(2n)} \to \sqrt{l}\). Limits are unique (Theorem 2.2.7), hence

\begin{equation*} l = \sqrt{l} \quad\Longrightarrow\quad l^2 = l \quad\Longrightarrow\quad l \in \{0, 1\}, \end{equation*}

and \(l \ge 1\) forces \(l = 1\).

Case \(0 < b < 1\). Then \(1/b > 1\), so \((1/b)^{1/n} \to 1\) by the previous case, and the Algebraic Limit Theorem (Theorem 2.3.3) gives

\begin{equation*} b^{1/n} = \frac{1}{(1/b)^{1/n}} \longrightarrow \frac{1}{1} = 1 \end{equation*}

(the quotient rule applies since the limit \(1\) of the denominator is nonzero).

Both directions run off the identity \(|b^n - 0| = |b|^n\).

(\(\Leftarrow\)) Assume \(-1 < b < 1\). If \(b = 0\) the sequence is constantly \(0\). Otherwise \(0 < |b| < 1\), so Example 2.5.3 gives \(|b|^n \to 0\): for \(\epsilon > 0\) there is \(N\) with \(|b|^n < \epsilon\) whenever \(n \ge N\). Since \(|b^n - 0| = |b|^n\), the same \(N\) shows \((b^n) \to 0\) by Definition 2.2.3.

(\(\Rightarrow\)) Contrapositive: assume \(|b| \ge 1\). Then \(|b|^n \ge 1\) for all \(n\) (induction: \(|b|^1 \ge 1\), and \(|b|^{n+1} = |b|^n |b| \ge 1 \cdot 1\)), so

\begin{equation*} |b^n - 0| = |b|^n \ge 1 \qquad \text{for every } n, \end{equation*}

and no \(N\) answers the challenge \(\epsilon = 1\). Hence \((b^n)\) does not converge to \(0\), and \(\lim(b^n) = 0\) forces \(|b| < 1\), i.e. \(-1 < b < 1\).

(a) Four sequences, each listed with its peak set.

\begin{equation*} \begin{aligned} \text{zero peaks:}\quad & x_n = n, &&\text{peaks: none,}\\ \text{one peak:}\quad & 1,\ \tfrac12,\ \tfrac23,\ \tfrac34,\ \tfrac45,\ \dots, &&\text{peaks: } x_1,\\ \text{two peaks:}\quad & 2,\ 1,\ \tfrac12,\ \tfrac23,\ \tfrac34,\ \dots, &&\text{peaks: } x_1,x_2,\\ \text{infinitely many:}\quad & x_n = \frac{(-1)^n}{n}, &&\text{peaks: } x_2,x_4,x_6,\dots \end{aligned} \end{equation*}

In the first, \(x_{m+1} > x_m\) for every \(m\), so nothing is a peak. In the second, \(x_n = 1 - 1/n < 1 = x_1\) for \(n \ge 2\) and the tail is strictly increasing, so \(x_1\) is the only peak; the third prepends \(2\) to it, making \(x_1\) and \(x_2 = 1\) peaks and leaving that tail peak-free. In the fourth, \(x_{2k} = 1/(2k) \ge x_n\) for all \(n \ge 2k\) (later terms are negative or at most \(1/(2k+2)\)), so every even index is a peak, while the alternating sign rules out monotonicity.

(b) Claim: every sequence \((x_n)\) contains a monotone subsequence. Split on the number of peak terms.

(i) Infinitely many peaks. Let \(n_1 < n_2 < n_3 < \cdots\) list their indices. Since \(x_{n_k}\) is a peak and \(n_{k+1} > n_k\),

\begin{equation*} x_{n_1} \ge x_{n_2} \ge x_{n_3} \ge \cdots, \end{equation*}

a decreasing subsequence.

(ii) Finitely many peaks. Choose \(N\) larger than every peak index (take \(N = 1\) if there are none) and set \(n_1 = N\). Given \(n_k \ge N\), the term \(x_{n_k}\) is not a peak, so some \(n_{k+1} > n_k\) has \(x_{n_{k+1}} > x_{n_k}\); and \(n_{k+1} > N\), so the construction repeats. This yields

\begin{equation*} x_{n_1} < x_{n_2} < x_{n_3} < \cdots, \end{equation*}

an increasing subsequence.

Now let \((x_n)\) be bounded. By the claim it has a monotone subsequence \((x_{n_k})\), which inherits the bound, so \((x_{n_k})\) converges by the Monotone Convergence Theorem (Theorem 2.4.2). Every bounded sequence therefore has a convergent subsequence, which is the Bolzano–Weierstrass Theorem (Theorem 2.5.5).

For each \(k \in \mathbf{N}\) the index set \(\{n : s - 1/k < a_n \le s + 1/k\}\) is infinite; picking \(n_1 < n_2 < n_3 < \cdots\) out of these sets gives the subsequence.

First, \(s\) exists. Fix \(M\) with \(|a_n| \le M\) for all \(n\). Then \(-M-1 < a_n\) for every \(n\), so \(-M-1 \in S\) and \(S \neq \emptyset\); and if \(x \ge M\) then \(a_n \le M \le x\) for every \(n\), so \(x \notin S\), making \(M\) an upper bound for \(S\). The Axiom of Completeness supplies \(s = \sup S\).

Two facts about \(s\), for arbitrary \(\epsilon > 0\):

(i) \(s - \epsilon \in S\). Since \(s - \epsilon\) is not an upper bound of \(S\), there is \(x \in S\) with \(s - \epsilon < x\). That \(x\) satisfies \(x < a_n\) for infinitely many \(n\), and each such \(n\) has \(s - \epsilon < x < a_n\). So

\begin{equation*} \{n \in \mathbf{N} : a_n > s - \epsilon\} \ \text{is infinite.} \end{equation*}

(ii) \(s + \epsilon \notin S\), because \(s + \epsilon > \sup S\). By the definition of \(S\) this says

\begin{equation*} \{n \in \mathbf{N} : a_n > s + \epsilon\} \ \text{is finite.} \end{equation*}

Subtracting the finite set in (ii) from the infinite set in (i),

\begin{equation*} E_\epsilon := \{n : s - \epsilon < a_n \le s + \epsilon\} \end{equation*}

is infinite for every \(\epsilon > 0\).

Construct the indices inductively: let \(n_1\) be any element of \(E_1\), and having chosen \(n_1 < \cdots < n_{k-1}\), use the infinitude of \(E_{1/k}\) to pick \(n_k \in E_{1/k}\) with \(n_k > n_{k-1}\). Then \((a_{n_k})\) is a genuine subsequence and

\begin{equation*} |a_{n_k} - s| \le \frac{1}{k} \qquad \text{for all } k . \end{equation*}

Given \(\epsilon > 0\), the Archimedean Property (Theorem 1.4.2) gives \(K \in \mathbf{N}\) with \(1/K < \epsilon\); for \(k \ge K\) we get \(|a_{n_k} - s| \le 1/k \le 1/K < \epsilon\). Hence \((a_{n_k}) \to s\).

Split the prescribed \(\epsilon\) in half and insert the limit.

Assume \((x_n) \to x\) and let \(\epsilon > 0\). By Definition 2.2.3 applied to \(\epsilon/2\), there is \(N \in \mathbf{N}\) with \(|x_n - x| < \epsilon/2\) for all \(n \ge N\). If \(m, n \ge N\), the triangle inequality gives

\begin{equation*} \begin{aligned} |x_n - x_m| &= |(x_n - x) + (x - x_m)| \\ &\le |x_n - x| + |x_m - x| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

This is precisely Definition 2.6.1, so \((x_n)\) is Cauchy.

(a) \(a_n = (-1)^n/n\). It converges to \(0\), hence is Cauchy by Theorem 2.6.2, and \(a_1 = -1 < a_2 = 1/2 > a_3 = -1/3\), so it is neither increasing nor decreasing.

(b) Impossible. Lemma 2.6.3 says a Cauchy sequence \((x_n)\) is bounded, say \(|x_n| \le M\) for all \(n\); every subsequence \((x_{n_k})\) satisfies the same bound \(|x_{n_k}| \le M\).

(c) Impossible. Suppose \((a_n)\) is monotone — say increasing — and has a Cauchy subsequence \((a_{n_k})\). By Lemma 2.6.3 there is \(M\) with \(a_{n_k} \le M\) for all \(k\). Given any \(n\), choose \(k\) with \(n_k \ge n\); monotonicity gives

\begin{equation*} a_n \le a_{n_k} \le M , \end{equation*}

so \((a_n)\) is bounded above, and bounded below by \(a_1\); the Monotone Convergence Theorem (Theorem 2.4.2) makes it convergent, not divergent. If \((a_n)\) is decreasing, apply the same argument to \((-a_n)\).

(d) Possible:

\begin{equation*} a_n = \begin{cases} n & \text{if } n \text{ is odd},\\ 0 & \text{if } n \text{ is even}. \end{cases} \end{equation*}

The odd terms show \((a_n)\) is unbounded, while the subsequence \(a_2, a_4, a_6, \dots\) is identically \(0\) and therefore Cauchy.

(a) Let \(\epsilon > 0\). Definition 2.6.1 applied to each sequence with \(\epsilon/2\) supplies \(N_1\) and \(N_2\); put \(N = \max\{N_1, N_2\}\). For \(m, n \ge N\),

\begin{equation*} \begin{aligned} |(x_n + y_n) - (x_m + y_m)| &\le |x_n - x_m| + |y_n - y_m| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

(b) Boundedness does the work. By Lemma 2.6.3 there is \(M > 0\) with \(|x_n| \le M\) and \(|y_n| \le M\) for all \(n\). Given \(\epsilon > 0\), choose \(N\) so that both \(|x_n - x_m| < \epsilon/(2M)\) and \(|y_n - y_m| < \epsilon/(2M)\) whenever \(m, n \ge N\). Then for such \(m, n\), inserting the cross term \(x_n y_m\) gives

\begin{equation*} \begin{aligned} |x_n y_n - x_m y_m| &= |x_n(y_n - y_m) + y_m(x_n - x_m)| \\ &\le |x_n|\,|y_n - y_m| + |y_m|\,|x_n - x_m| \\ &< M \cdot \tfrac{\epsilon}{2M} + M \cdot \tfrac{\epsilon}{2M} = \epsilon . \end{aligned} \end{equation*}

(a) Cauchy. The reverse triangle inequality (Exercise 1.2.6 (d)) applied to \(u = a_n - b_n\) and \(v = a_m - b_m\) gives

\begin{equation*} \begin{aligned} \bigl| |a_n - b_n| - |a_m - b_m| \bigr| &\le |(a_n - b_n) - (a_m - b_m)| \\ &\le |a_n - a_m| + |b_n - b_m| , \end{aligned} \end{equation*}

so choosing \(N\) that makes each of the last two terms less than \(\epsilon/2\) for \(m, n \ge N\) makes \(|c_n - c_m| < \epsilon\).

(b) Not Cauchy in general. Take \(a_n = 1\) for all \(n\), which is Cauchy; then \(c_n = (-1)^n\) and \(|c_{n+1} - c_n| = 2\) for every \(n\), so no \(N\) works for \(\epsilon = 2\).

(c) Not Cauchy in general. Take \(a_n = (-1)^n/n\), which converges to \(0\) and so is Cauchy by Theorem 2.6.2. For even \(n\) we have \(0 < a_n < 1\) and \([[a_n]] = 0\); for odd \(n\) we have \(-1 \le a_n < 0\) and \([[a_n]] = -1\). Thus \(|c_{n+1} - c_n| = 1\) for all \(n\), and \(\epsilon = 1\) admits no \(N\).

(ii) is the true one.

Proof of (ii). Let \(\epsilon > 0\) and choose \(N_1\), \(N_2\) so that \(|x_{n+1} - x_n| < \epsilon/2\) for \(n \ge N_1\) and \(|y_{n+1} - y_n| < \epsilon/2\) for \(n \ge N_2\). For \(n \ge N = \max\{N_1, N_2\}\),

\begin{equation*} \begin{aligned} |(x_{n+1} + y_{n+1}) - (x_n + y_n)| &\le |x_{n+1} - x_n| + |y_{n+1} - y_n| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

Counterexample to (i). Let \(s_n = \sum_{k=1}^{n} 1/k\) be the partial sums of the harmonic series. Then

\begin{equation*} \begin{aligned} |s_{n+1} - s_n| &= \frac{1}{n+1} < \epsilon \\ &\text{whenever } n \ge N > 1/\epsilon, \end{aligned} \end{equation*}

such an \(N\) existing by the Archimedean Property (Theorem 1.4.2), so \((s_n)\) is pseudo-Cauchy. But \((s_n)\) is unbounded: Example 2.4.5 shows the harmonic series diverges, and since \((s_n)\) is increasing, a bound would force convergence by the Monotone Convergence Theorem (Theorem 2.4.2).

(a) \(a_n = (-1)^n/n\). For \(n > m \ge N\),

\begin{equation*} a_n - a_m \ge -\frac{1}{n} - \frac{1}{m} > -\frac{2}{N}, \end{equation*}

so any \(N > 2/\epsilon\) (Theorem 1.4.2) works. It is monotone on no tail because

\begin{equation*} a_{n+1} - a_n = (-1)^{n+1}\Bigl(\tfrac{1}{n+1} + \tfrac{1}{n}\Bigr) \end{equation*}

changes sign at every \(n\).

(b) \(a_n = \sqrt{n} + (-1)^n n^{-1/4}\).

Quasi-increasing: for \(n > m \ge N\) we have \(\sqrt{n} - \sqrt{m} > 0\), hence

\begin{equation*} a_n - a_m > -n^{-1/4} - m^{-1/4} \ge -2N^{-1/4}, \end{equation*}

and any \(N > (2/\epsilon)^4\) gives \(2N^{-1/4} < \epsilon\), hence \(a_n > a_m - \epsilon\).

Divergent: \(a_n \ge \sqrt{n} - 1\), which is unbounded, so \((a_n)\) cannot converge (Theorem 2.3.2).

Nowhere monotone: writing

\begin{equation*} a_{n+1} - a_n = (\sqrt{n+1} - \sqrt{n})

• (-1)^{n+1}\bigl[(n+1)^{-1/4} + n^{-1/4}\bigr], \end{equation*}

the alternating bracket dominates, since for every \(n \ge 1\)

\begin{equation*} \begin{aligned} \sqrt{n+1} - \sqrt{n} &= \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \\ &< n^{-1/4} < (n+1)^{-1/4} + n^{-1/4}, \end{aligned} \end{equation*}

the middle step because \(2n^{1/4} > 1\). Hence \(a_{n+1} - a_n\) has the sign of \((-1)^{n+1}\), which alternates.

(c) Yes — every bounded quasi-increasing sequence converges.

Let \((a_n)\) be bounded and quasi-increasing. By the Bolzano–Weierstrass Theorem (Theorem 2.5.5) there is a subsequence \((a_{n_k}) \to a\). Fix \(\epsilon > 0\). Choose \(N\) from the quasi-increasing property for \(\epsilon/3\), and then \(K\) with \(n_K \ge N\) and \(|a_{n_k} - a| < \epsilon/3\) for all \(k \ge K\). Let \(n \ge n_K\).

(i) Lower bound: if \(n > n_K\) then \(n > n_K \ge N\) gives \(a_n > a_{n_K} - \epsilon/3 > a - 2\epsilon/3\).

(ii) Upper bound: choose \(k \ge K\) with \(n_k > n\) (possible since \(n_k \to \infty\)); then \(n_k > n \ge n_K \ge N\) gives \(a_{n_k} > a_n - \epsilon/3\), so

\begin{equation*} a_n < a_{n_k} + \tfrac{\epsilon}{3} < a + \tfrac{2\epsilon}{3}. \end{equation*}

Case (ii) already covers \(n = n_K\), and there the lower bound is \(a_{n_K} > a - \epsilon/3\) directly. So \(|a_n - a| < \epsilon\) for all \(n \ge n_K\), i.e. \((a_n) \to a\).

(a) Let \((a_n)\) be increasing and bounded (the decreasing case follows by applying this one to \((-a_n)\)). By BW (Theorem 2.5.5) there is a subsequence \((a_{n_k}) \to a\); we show \((a_n) \to a\).

First, \(a_n \le a\) for every \(n\). Otherwise \(a_m > a\) for some \(m\), and since \(a_{n_k} \ge a_m\) for every \(n_k \ge m\), the Order Limit Theorem (Theorem 2.3.4 (iii), in the eventual form licensed by the remark on tails that follows it) gives \(a = \lim a_{n_k} \ge a_m > a\), a contradiction.

Now let \(\epsilon > 0\) and pick \(K\) with \(|a_{n_K} - a| < \epsilon\). For \(n \ge n_K\), monotonicity and the previous paragraph give

\begin{equation*} a - \epsilon < a_{n_K} \le a_n \le a < a + \epsilon , \end{equation*}

so \(|a_n - a| < \epsilon\). Every \(\epsilon\) is handled by an index supplied by the subsequence itself, so no statement of the form \(1/N < \epsilon\) — and hence no appeal to the Archimedean Property — occurs.

(b) Let \((a_n)\) be bounded, say \(a_n \in I_0 = [-M, M]\) for all \(n\). Bisect \(I_0\); at least one closed half contains \(a_n\) for infinitely many indices \(n\), and call it \(I_1\). Repeating, we obtain nested intervals

\begin{equation*} I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots, \qquad |I_k| = \frac{2M}{2^k}, \end{equation*}

each containing infinitely many terms of the sequence. Choose \(n_1 < n_2 < n_3 < \cdots\) with \(a_{n_k} \in I_k\), which is possible precisely because each \(I_k\) captures infinitely many indices.

This subsequence is Cauchy: if \(l \ge k \ge K\) then \(a_{n_k}, a_{n_l} \in I_K\) by nesting, so

\begin{equation*} |a_{n_k} - a_{n_l}| \le |I_K| = \frac{2M}{2^{K}} . \end{equation*}

Given \(\epsilon > 0\) we need a \(K\) with \(2M/2^{K} < \epsilon\), i.e. with \(2^{K} > 2M/\epsilon\); since \(2^K \ge K\), it suffices that \(\mathbf{N}\) contain an element exceeding \(2M/\epsilon\). This last sentence is the implicit use of the Archimedean Property (Theorem 1.4.2): without the guarantee that \(\mathbf{N}\) is unbounded in \(\mathbf{R}\), the lengths \(2M/2^k\) need not be eventually smaller than \(\epsilon\). With it, \((a_{n_k})\) is Cauchy, and the Cauchy Criterion (Theorem 2.6.4) makes it convergent.

(c) Because \(\mathbf{Q}\) is a counterexample. The rationals form an ordered field in which the Archimedean Property holds (given \(x = p/q > 0\) and \(y = r/s\), the integer \(n = |r|q + 1\) satisfies \(nx > y\)), but AoC fails there: the set \(S = \{x \in \mathbf{Q} : x^2 < 2\}\) is nonempty and bounded above yet has no least upper bound in \(\mathbf{Q}\): the trichotomy argument used to prove Theorem 1.4.5 rules out \(\alpha^2 < 2\) and \(\alpha^2 > 2\) for a least upper bound \(\alpha\) of \(S\), leaving \(\alpha^2 = 2\), which no rational satisfies (Theorem 1.1.1). A derivation of AoC from the Archimedean Property together with only the field and order axioms would be valid verbatim in \(\mathbf{Q}\), and would therefore prove a false statement.

All three proofs run off one estimate. Lemma. With \(s_n = \sum_{k=1}^n (-1)^{k+1} a_k\), where \(a_1 \ge a_2 \ge \cdots \ge 0\) and \((a_n) \to 0\), for \(n > m \ge 0\)

\begin{equation*} 0 \;\le\; a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \;\le\; a_{m+1}. \end{equation*}

Call the middle quantity \(T\). Bracketing from the left,

\begin{equation*} T = (a_{m+1}-a_{m+2}) + (a_{m+3}-a_{m+4}) + \cdots , \end{equation*}

a sum of nonnegative pairs, plus a final \(+a_n \ge 0\) when \(n-m\) is odd; hence \(T \ge 0\). Bracketing after the first term,

\begin{equation*} T = a_{m+1} - (a_{m+2}-a_{m+3}) - (a_{m+4}-a_{m+5}) - \cdots , \end{equation*}

where every bracket is nonnegative and a final \(-a_n \le 0\) appears when \(n-m\) is even; hence \(T \le a_{m+1}\).

(a) Since \(s_n - s_m = \pm\,(a_{m+1} - a_{m+2} + \cdots \pm a_n)\) for \(n > m\), the Lemma gives

\begin{equation*} |s_n - s_m| \le a_{m+1} \qquad (n > m). \end{equation*}

Let \(\epsilon > 0\). As \((a_n) \to 0\), choose \(N\) with \(a_{N+1} < \epsilon\); monotonicity gives \(a_{m+1} \le a_{N+1}\) for \(m \ge N\), so \(n > m \ge N\) forces \(|s_n - s_m| \le a_{m+1} < \epsilon\). Thus \((s_n)\) is Cauchy and converges by the Cauchy Criterion (Theorem 2.6.4).

(b) Set \(I_n = [\,s_{2n},\, s_{2n-1}\,]\). These are genuine intervals and they nest:

\begin{equation*} \begin{aligned} s_{2n-1} - s_{2n} &= a_{2n} \ge 0,\\ s_{2n+2} - s_{2n} &= a_{2n+1} - a_{2n+2} \ge 0,\\ s_{2n+1} - s_{2n-1} &= a_{2n+1} - a_{2n} \le 0, \end{aligned} \end{equation*}

so \(I_{n+1} \subseteq I_n\). By the Nested Interval Property (Theorem 1.4.1) there is some \(L \in \bigcap_{n=1}^{\infty} I_n\).

Fix \(n\). For every \(k \ge n\) the endpoints \(s_{2k}, s_{2k-1}\) of \(I_k\) lie in \(I_n\), so every \(s_m\) with \(m \ge 2n-1\) lies in \(I_n\), as does \(L\). Hence

\begin{equation*} |s_m - L| \le \operatorname{length}(I_n) = a_{2n} \qquad (m \ge 2n-1). \end{equation*}

Given \(\epsilon > 0\), pick \(n\) with \(a_{2n} < \epsilon\) and take \(N = 2n-1\); then \((s_n) \to L\).

(c) The three displayed inequalities in (b) say \((s_{2n})\) is increasing and \((s_{2n-1})\) is decreasing, while \(s_{2n} \le s_{2n-1} \le s_1 = a_1\) and \(s_{2n-1} \ge s_{2n} \ge s_2\). So \((s_{2n})\) is increasing and bounded above, \((s_{2n+1})\) is decreasing and bounded below, and the Monotone Convergence Theorem (Theorem 2.4.2) yields

\begin{equation*} s_{2n} \to L, \qquad s_{2n+1} \to L^{\prime}. \end{equation*}

Since \(s_{2n+1} - s_{2n} = a_{2n+1} \to 0\), the Algebraic Limit Theorem (Theorem 2.3.3) gives \(L^{\prime} = L\).

The two subsequences exhaust \((s_n)\), so taking \(N\) past both of their \(\epsilon\)-thresholds gives \((s_n) \to L\); hence \(\sum_{n=1}^{\infty} (-1)^{n+1} a_n = L\).

(a), (b) converge; (c), (d), (e) diverge.

(a) Converges. For all \(n \in \mathbf{N}\),

\begin{equation*} 0 \le \frac{1}{2^n + n} \le \frac{1}{2^n} = \left(\tfrac12\right)^n, \end{equation*}

and \(\sum (1/2)^n\) is geometric with \(|r| = 1/2 < 1\), hence convergent (Example 2.7.5). The Comparison Test (Theorem 2.7.4 (i)) applies.

(b) Converges. Since \(|\sin(n)| \le 1\),

\begin{equation*} 0 \le \left|\frac{\sin(n)}{n^2}\right| \le \frac{1}{n^2}, \end{equation*}

and \(\sum 1/n^2\) converges by Corollary 2.4.7 (\(p = 2 > 1\)). So \(\sum |\sin(n)/n^2|\) converges by Theorem 2.7.4 (i), and the Absolute Convergence Test (Theorem 2.7.6) finishes it.

(c) Diverges. The \(n\)th term is \((-1)^{n+1}\dfrac{n+1}{2n}\) (check \(n=1\): \(2/2 = 1\)), and

\begin{equation*} \left|\frac{n+1}{2n}\right| = \frac{1}{2} + \frac{1}{2n} \longrightarrow \frac12 \ne 0, \end{equation*}

so the terms do not tend to \(0\). By the contrapositive of Theorem 2.7.3, the series diverges.

(d) Diverges. The terms come in blocks of three with signs \(+,+,-\), so for \(m \in \mathbf{N}\)

\begin{equation*} \begin{aligned} s_{3m} &= \sum_{k=0}^{m-1}\left(\frac{1}{3k+1} + \frac{1}{3k+2} - \frac{1}{3k+3}\right)\\ &\ge \sum_{k=0}^{m-1} \frac{1}{3k+1} \;\ge\; \frac13 \sum_{k=0}^{m-1}\frac{1}{k+1}, \end{aligned} \end{equation*}

using \(1/(3k+2) \ge 1/(3k+3)\) for the first inequality and \(1/(3k+1) \ge 1/(3k+3)\) for the second. The harmonic partial sums are unbounded (Example 2.4.5), so \((s_{3m})\) is unbounded and \((s_n)\) cannot converge, since convergent sequences are bounded (Theorem 2.3.2).

(e) Diverges. Pairing an odd term with the following even term,

\begin{equation*} \begin{aligned} s_{2m} &= \sum_{k=1}^{m}\left(\frac{1}{2k-1} - \frac{1}{(2k)^2}\right)\\ &\ge \frac12\sum_{k=1}^{m}\frac{1}{k} \;-\; \sum_{k=1}^{\infty}\frac{1}{k^2}, \end{aligned} \end{equation*}

because \(1/(2k-1) \ge 1/(2k)\) and \(\sum_{k=1}^m 1/(2k)^2 \le \sum_{k=1}^{\infty} 1/k^2 =: M < \infty\) by Corollary 2.4.7. The right side tends to \(\infty\), so \((s_{2m})\) is unbounded and \((s_n)\) diverges (Theorem 2.3.2).

Part (ii) is the contrapositive of part (i), so each proof need only establish (i).

(a) Let \(\epsilon > 0\). Since \(\sum_{k=1}^{\infty} b_k\) converges, the Cauchy Criterion for Series (Theorem 2.7.2) supplies \(N \in \mathbf{N}\) with

\begin{equation*} |b_{m+1} + b_{m+2} + \cdots + b_n| < \epsilon \qquad \text{whenever } n > m \ge N . \end{equation*}

For those same \(n > m \ge N\), nonnegativity lets us drop the absolute values and the hypothesis \(a_k \le b_k\) gives

\begin{equation*} \begin{aligned} |a_{m+1} + \cdots + a_n| &= a_{m+1} + \cdots + a_n\\ &\le b_{m+1} + \cdots + b_n\\ &= |b_{m+1} + \cdots + b_n| \;<\; \epsilon . \end{aligned} \end{equation*}

So \(\sum a_k\) satisfies the Cauchy Criterion for Series, and the sufficiency direction of Theorem 2.7.2 gives convergence.

(b) Write \(s_n = a_1 + \cdots + a_n\) and \(t_n = b_1 + \cdots + b_n\). Because \(a_k \ge 0\) and \(b_k \ge 0\), both \((s_n)\) and \((t_n)\) are increasing. Assume \(\sum b_k\) converges; then \((t_n)\) converges, hence is bounded (Theorem 2.3.2), say \(t_n \le M\) for all \(n\). Term-by-term comparison gives

\begin{equation*} s_n = \sum_{k=1}^n a_k \;\le\; \sum_{k=1}^n b_k = t_n \;\le\; M . \end{equation*}

Thus \((s_n)\) is increasing and bounded above, so it converges by the Monotone Convergence Theorem (Theorem 2.4.2); that is, \(\sum a_k\) converges.

(a) Take \(x_n = y_n = n^{-2/3}\). Then \(\sum n^{-2/3}\) diverges and \(\sum x_n y_n = \sum n^{-4/3}\) converges, both by Corollary 2.4.7 (\(p = 2/3 \le 1\), \(p = 4/3 > 1\)).

(b) Take \(x_n = (-1)^n/n\) and \(y_n = (-1)^n\). Since \((1/n)\) decreases to \(0\), the Alternating Series Test (Theorem 2.7.7) applies to \(\sum (-1)^{n+1}/n\), and \(\sum x_n\) is its \((-1)\)-multiple, so it converges too (Theorem 2.7.1 (i)). The sequence \((y_n)\) is bounded by \(1\), and

\begin{equation*} \sum_{n=1}^{\infty} x_n y_n = \sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n} = \sum_{n=1}^{\infty}\frac{1}{n}, \end{equation*}

the harmonic series, which diverges (Example 2.4.5).

(c) Impossible. Writing \(y_n = (x_n + y_n) + (-1)x_n\), the Algebraic Limit Theorem for Series (Theorem 2.7.1, parts (i) and (ii)) gives

\begin{equation*} \sum_{n=1}^{\infty} y_n = \sum_{n=1}^{\infty}(x_n+y_n) - \sum_{n=1}^{\infty} x_n , \end{equation*}

so \(\sum y_n\) converges whenever the other two do.

(d) Take

\begin{equation*} x_n = \begin{cases} 1/n, & n \text{ even},\\ 0, & n \text{ odd},\end{cases} \end{equation*}

which satisfies \(0 \le x_n \le 1/n\). All odd terms of \(\sum (-1)^n x_n\) vanish and all even terms are \(+1/n\), so

\begin{equation*} s_{2m} = \sum_{k=1}^{m} \frac{1}{2k} = \frac12 \sum_{k=1}^{m}\frac1k , \end{equation*}

which is unbounded (Example 2.4.5); hence \((s_n)\) diverges by Theorem 2.3.2.

Condense, then read off a geometric series.

(i) \(p \le 0\). Then \(1/n^p = n^{-p} \ge 1\) for every \(n \in \mathbf{N}\), so the terms do not converge to \(0\) and the series diverges by the contrapositive of Theorem 2.7.3. This is consistent with the claim, since such \(p\) are not \(> 1\).

(ii) \(p > 0\). Set \(b_n = 1/n^p\). Then \(b_n > 0\) and \((b_n)\) is decreasing, so the Cauchy Condensation Test (Theorem 2.4.6) applies: \(\sum_{n=1}^{\infty} b_n\) converges if and only if \(\sum_{n=0}^{\infty} 2^n b_{2^n}\) converges. Now

\begin{equation*} \sum_{n=0}^{\infty} 2^n b_{2^n} = \sum_{n=0}^{\infty} \frac{2^n}{(2^n)^p} = \sum_{n=0}^{\infty} \left(2^{1-p}\right)^n , \end{equation*}

a geometric series with \(a = 1\) and \(r = 2^{1-p} > 0\). By Example 2.7.5 it converges if and only if \(|r| < 1\), and since \(x \mapsto 2^x\) is strictly increasing,

\begin{equation*} 2^{1-p} < 1 = 2^0 \iff 1 - p < 0 \iff p > 1 . \end{equation*}

Combining (i) and (ii), \(\sum_{n=1}^{\infty} 1/n^p\) converges precisely when \(p > 1\).

(a) false, (b) true, (c) true, (d) false.

(a) False. Take \(a_n = 1\), a bounded sequence. Then \(s_n = n\), and every subsequence \((s_{n_k}) = (n_k)\) is unbounded, hence divergent (Theorem 2.3.2). So \(\sum a_n\) does not subverge.

(b) True. If \(\sum a_n\) converges, then \((s_n)\) converges, and \((s_n)\) is a subsequence of itself (Definition 2.5.1 with \(n_k = k\)).

(c) True. Let \(t_n = |a_1| + \cdots + |a_n|\), an increasing sequence, and suppose some subsequence \((t_{n_k})\) converges; being convergent it is bounded, say \(t_{n_k} \le M\) for all \(k\). Given any \(n\), choose \(k\) with \(n_k \ge n\); then \(t_n \le t_{n_k} \le M\). So \((t_n)\) is increasing and bounded, hence convergent by the Monotone Convergence Theorem (Theorem 2.4.2). Thus \(\sum |a_n|\) converges, so \(\sum a_n\) converges by the Absolute Convergence Test (Theorem 2.7.6), and therefore subverges by (b).

(d) False. Take

\begin{equation*} (a_n) = 1,\,-1,\,2,\,-2,\,3,\,-3,\,\dots, \qquad \text{i.e. } a_{2k-1} = k,\ a_{2k} = -k . \end{equation*}

Then \(s_{2k} = 0\) for every \(k\), so the subsequence \((s_{2k})\) converges and \(\sum a_n\) subverges. But \(|a_n| \to \infty\), so every subsequence of \((a_n)\) is unbounded and hence divergent (Theorem 2.3.2).

(a) Compare with the harmonic series. Since \(n a_n > 0\) for all \(n\), the Order Limit Theorem (Theorem 2.3.4) gives \(l \ge 0\), and \(l \ne 0\) forces \(l > 0\). Apply the definition of \(\lim (na_n) = l\) with \(\epsilon = l/2\): there is \(N \in \mathbf{N}\) such that

\begin{equation*} \begin{aligned} n a_n &> l - \tfrac{l}{2} = \tfrac{l}{2},\\ \text{hence}\qquad a_n &\ge \frac{l}{2}\cdot\frac{1}{n} \ge 0, \end{aligned} \end{equation*}

for all \(n \ge N\). The series \(\sum (l/2)(1/n)\) diverges, since \(\sum 1/n\) does (Corollary 2.4.7 with \(p = 1\)) and a nonzero constant multiple of a divergent series diverges (Theorem 2.7.1 (i), applied to the factor \(2/l\)). By the Comparison Test (Theorem 2.7.4 (ii)) applied to the tails from index \(N\) on, \(\sum_{n \ge N} a_n\) diverges, and since convergence is unaffected by the finitely many terms \(a_1, \dots, a_{N-1}\), the series \(\sum a_n\) diverges.

(b) Compare with \(\sum 1/n^2\). The sequence \((n^2 a_n)\) converges by hypothesis, hence is bounded (Theorem 2.3.2): there is \(M > 0\) with \(n^2 a_n \le M\) for all \(n\). Therefore

\begin{equation*} 0 < a_n \le \frac{M}{n^2} \qquad \text{for all } n \in \mathbf{N}. \end{equation*}

Now \(\sum 1/n^2\) converges by Corollary 2.4.7 (\(p = 2 > 1\)), so \(\sum M/n^2\) converges by Theorem 2.7.1 (i), and the Comparison Test (Theorem 2.7.4 (i)) gives convergence of \(\sum a_n\).

(a) True, (b) false, (c) true.

(a) Because \(\sum |a_n|\) converges, Theorem 2.7.3 gives \((|a_n|) \to 0\), so there is an \(N\) with \(|a_n| < 1\) for all \(n \ge N\). For such \(n\),

\begin{equation*} |a_n^2| \;=\; |a_n|^2 \;\le\; |a_n|, \end{equation*}

and the Comparison Test (Theorem 2.7.4) applied to the tail \(\sum_{n \ge N} |a_n^2|\) finishes it; the finitely many terms before \(N\) do not affect convergence.

(b) Take

\begin{equation*} a_n = b_n = \frac{(-1)^n}{\sqrt{n}}. \end{equation*}

Then \(\sum a_n\) converges by the Alternating Series Test (Theorem 2.7.7), since \(1/\sqrt{n}\) decreases to \(0\), and \((b_n) \to 0\) converges. But

\begin{equation*} \sum_{n=1}^{\infty} a_n b_n \;=\; \sum_{n=1}^{\infty} \frac{1}{n} \end{equation*}

is the harmonic series, which diverges (Corollary 2.4.7).

(c) Suppose \(\sum n^2 a_n\) converged. Then \((n^2 a_n) \to 0\) by Theorem 2.7.3, so there is an \(N\) with \(|n^2 a_n| \le 1\), i.e.

\begin{equation*} |a_n| \;\le\; \frac{1}{n^2} \qquad \text{for all } n \ge N . \end{equation*}

Since \(\sum 1/n^2\) converges (Corollary 2.4.7), the Comparison Test (Theorem 2.7.4) makes \(\sum |a_n|\) converge, contradicting the hypothesis that \(\sum a_n\) converges only conditionally (Definition 2.7.8). Hence \(\sum n^2 a_n\) diverges.

(a) Apply the definition of the limit with \(\epsilon = r^{\prime} - r > 0\): there is an \(N \in \mathbf{N}\) such that

\begin{equation*} \begin{aligned} n \ge N \ &\Longrightarrow\ \left| \left| \frac{a_{n+1}}{a_n} \right| - r \right| < r^{\prime} - r \\ &\Longrightarrow\ \left| \frac{a_{n+1}}{a_n} \right| < r^{\prime} , \end{aligned} \end{equation*}

and multiplying through by \(|a_n| > 0\) gives \(|a_{n+1}| \le |a_n| r^{\prime}\) for all \(n \ge N\).

(b) Note \(0 \le r < r^{\prime} < 1\), since \(r\) is a limit of absolute values. So \(\sum (r^{\prime})^n\) is a geometric series with ratio of absolute value less than \(1\), hence convergent (Example 2.7.5), and multiplying by the constant \(|a_N|\) preserves convergence by Theorem 2.7.1(i).

(c) Iterating the inequality in (a) from \(n = N\) upward gives, by induction,

\begin{equation*} |a_{N+k}| \;\le\; |a_N| (r^{\prime})^{k} \qquad \text{for all } k \ge 0 . \end{equation*}

The series \(\sum_{k=0}^{\infty} |a_N| (r^{\prime})^k\) converges by (b), so the Comparison Test (Theorem 2.7.4), whose positivity hypothesis holds since all terms here are nonnegative, shows \(\sum_{k=0}^{\infty} |a_{N+k}|\) converges. Adding back the finitely many terms \(|a_1|, \dots, |a_{N-1}|\) changes only a partial sum, so \(\sum_{n=1}^{\infty} |a_n|\) converges. By the Absolute Convergence Test (Theorem 2.7.6), \(\sum a_n\) converges.

(a) Yes. The \(n\)-th factor is \(1 + 2^{-(n-1)}\), so the product is \(\prod_{n=1}^{\infty}(1 + a_n)\) with \(a_n = 2^{-(n-1)} \ge 0\) and

\begin{equation*} \sum_{n=1}^{\infty} 2^{-(n-1)} \;=\; 2 < \infty \end{equation*}

(a geometric series, Example 2.7.5). By Exercise 2.4.10(b) the product converges.

(b) It converges because its partial products

\begin{equation*} p_n \;=\; \prod_{k=1}^{n} \frac{2k-1}{2k} \end{equation*}

are decreasing (every factor lies in \((0,1)\)) and bounded below by \(0\), so the Monotone Convergence Theorem (Theorem 2.4.2) applies. The limit is zero. Indeed,

\begin{equation*} \frac{1}{p_n} \;=\; \prod_{k=1}^{n} \frac{2k}{2k-1} \;=\; \prod_{k=1}^{n} \left( 1 + \frac{1}{2k-1} \right), \end{equation*}

and \(\sum_{k=1}^{\infty} \frac{1}{2k-1}\) diverges, since \(\frac{1}{2k-1} \ge \frac{1}{2k}\) and the harmonic series diverges (Corollary 2.4.7). So Exercise 2.4.10(b) says the product \(\prod (1 + \frac{1}{2k-1})\) does not converge; being an increasing sequence, \((1/p_n)\) must therefore be unbounded, and hence \(p_n \to 0\).

(c) The \(n\)-th factor is

\begin{equation*} \frac{(2n)(2n)}{(2n-1)(2n+1)} \;=\; \frac{4n^2}{4n^2-1} \;=\; 1 + \frac{1}{4n^2-1}, \end{equation*}

so again the product has the form \(\prod (1 + a_n)\) with \(a_n \ge 0\). Since \(4n^2 - 1 \ge 3n^2\) for every \(n \ge 1\),

\begin{equation*} 0 \;\le\; \frac{1}{4n^2-1} \;\le\; \frac{1}{3n^2}, \end{equation*}

and \(\sum 1/n^2\) converges by Corollary 2.4.7, so \(\sum a_n\) converges by the Comparison Test (Theorem 2.7.4). Exercise 2.4.10(b) then gives convergence of the Wallis product.

Take

\begin{equation*} a_n = \begin{cases} 1 & n \text{ odd} \\[2pt] 1/n^2 & n \text{ even} \end{cases} \qquad b_n = \begin{cases} 1/n^2 & n \text{ odd} \\[2pt] 1 & n \text{ even.} \end{cases} \end{equation*}

Neither \((a_n)\) nor \((b_n)\) tends to \(0\), so both series diverge by Theorem 2.7.3, while \(\min\{a_n, b_n\} = 1/n^2\) and \(\sum 1/n^2\) converges (Corollary 2.4.7).

For the strictly positive decreasing version, interleave long blocks. Define \(n_1 = 1\) and \(n_{k+1} = 2^{\,n_k^2} n_k\), and let \(I_k = \{ n \in \mathbf{N} : n_k \le n < n_{k+1} \}\), so that \(\mathbf{N}\) is partitioned into the blocks \(I_1, I_2, I_3, \ldots\). Put

\begin{equation*} c_n = \frac{1}{n^2}, \qquad d_n = \frac{1}{n_k \, n} \ \ \text{for } n \in I_k, \end{equation*}

and set, for \(n \in I_k\),

\begin{equation*} a_n = \begin{cases} d_n & k \text{ odd} \\[2pt] c_n & k \text{ even,} \end{cases} \qquad b_n = \begin{cases} c_n & k \text{ odd} \\[2pt] d_n & k \text{ even.} \end{cases} \end{equation*}

The minimum. For \(n \in I_k\) we have \(n \ge n_k\), hence \(n_k n \le n^2\) and so \(c_n \le d_n\). Therefore \(\min\{a_n, b_n\} = c_n = 1/n^2\) for every \(n\), and \(\sum \min\{a_n,b_n\}\) converges by Corollary 2.4.7.

Strictly decreasing. Both \(c_n\) and \(d_n\) are strictly decreasing in \(n\) within a block. At the left endpoint of a block the two agree,

\begin{equation*} d_{n_{k+1}} = \frac{1}{n_{k+1} \cdot n_{k+1}} = c_{n_{k+1}}, \end{equation*}

so whichever branch is used, \(a_{n_{k+1}} = b_{n_{k+1}} = 1/n_{k+1}^2\). The preceding term is either \(c_{n_{k+1}-1} = 1/(n_{k+1}-1)^2\) or \(d_{n_{k+1}-1} = 1/\big(n_k (n_{k+1}-1)\big)\), and both exceed \(1/n_{k+1}^2\) because \((n_{k+1}-1)^2 < n_{k+1}^2\) and \(n_k(n_{k+1}-1) < n_{k+1}^2\). So \((a_n)\) and \((b_n)\) are strictly decreasing and strictly positive.

Divergence. Write \(m_k = n_k^2\), so \(n_{k+1} = 2^{m_k} n_k\). Splitting \(I_k\) into the \(m_k\) ranges \([2^j n_k, 2^{j+1} n_k)\), each of which has \(2^j n_k\) terms every one of which exceeds \(1/(2^{j+1} n_k)\), gives

\begin{equation*} \begin{aligned} \sum_{n \in I_k} \frac{1}{n} &\;>\; \frac{m_k}{2}, \\ \sum_{n \in I_k} d_n = \frac{1}{n_k} \sum_{n \in I_k} \frac{1}{n} &\;>\; \frac{n_k}{2}. \end{aligned} \end{equation*}

Since all terms are positive, the partial sums of \(\sum a_n\) exceed \(n_k/2\) for every odd \(k\) and the partial sums of \(\sum b_n\) exceed \(n_k/2\) for every even \(k\); both are unbounded, so both series diverge.

Substitute \(x_j = s_j - s_{j-1}\) and shift the index in the second sum:

\begin{equation*} \begin{aligned} \sum_{j=m}^{n} x_j y_j &= \sum_{j=m}^{n} s_j y_j - \sum_{j=m}^{n} s_{j-1} y_j \\ &= \sum_{j=m}^{n} s_j y_j - \sum_{j=m-1}^{n-1} s_j y_{j+1} . \end{aligned} \end{equation*}

The second sum runs over \(j = m-1, \ldots, n-1\); peel off its \(j = m-1\) term and add in the \(j = n\) term that is missing, which costs \(-s_{m-1}y_m + s_n y_{n+1}\):

\begin{equation*} \begin{aligned} \sum_{j=m}^{n} x_j y_j &= \sum_{j=m}^{n} s_j y_j

• \left( \sum_{j=m}^{n} s_j y_{j+1} + s_{m-1}y_m - s_n y_{n+1} \right) \\ &= s_n y_{n+1} - s_{m-1} y_m + \sum_{j=m}^{n} s_j (y_j - y_{j+1}), \end{aligned} \end{equation*}

which is the asserted identity.

(a) Put \(m = 1\) in the identity of Exercise 2.7.12. The boundary term becomes \(-s_{m-1}y_m = -s_0 y_1 = 0\), since \(s_0 = 0\) by definition, leaving

\begin{equation*} \sum_{k=1}^{n} x_k y_k \;=\; s_n y_{n+1} + \sum_{k=1}^{n} s_k (y_k - y_{k+1}). \end{equation*}

(b) Because \(\sum x_k\) converges, the sequence \((s_n)\) of partial sums converges and is therefore bounded (Theorem 2.3.2): fix \(M > 0\) with \(|s_k| \le M\) for all \(k\). Monotonicity of \((y_k)\) makes \(y_k - y_{k+1} \ge 0\), so

\begin{equation*} \big| s_k (y_k - y_{k+1}) \big| \;\le\; M (y_k - y_{k+1}) \qquad (k \in \mathbf{N}), \end{equation*}

and the majorant telescopes:

\begin{equation*} \sum_{k=1}^{n} M (y_k - y_{k+1}) \;=\; M (y_1 - y_{n+1}). \end{equation*}

The sequence \((y_k)\) is decreasing and bounded below by \(0\), so it converges to some \(y \ge 0\) by the Monotone Convergence Theorem (Theorem 2.4.2); hence these partial sums converge to \(M(y_1 - y)\), i.e. \(\sum M(y_k - y_{k+1})\) converges. Both sides being nonnegative, the Comparison Test (Theorem 2.7.4) gives convergence of \(\sum |s_k(y_k - y_{k+1})|\), and then the Absolute Convergence Test (Theorem 2.7.6) gives convergence of \(\sum s_k (y_k - y_{k+1})\) itself.

Now let \(n \to \infty\) in (a). The first term satisfies \(s_n y_{n+1} \to s y\), where \(s = \lim s_n\), by the Algebraic Limit Theorem (Theorem 2.3.3), and the second term converges by the previous paragraph. Hence the partial sums of \(\sum x_k y_k\) converge, which is Abel’s Test.

(a) The trade is one hypothesis for another: Abel demands that \((s_n)\) actually converge but lets \((y_k)\) decrease to any limit \(y \ge 0\); Dirichlet demands only that \((s_n)\) be bounded but insists that \(y_k \to 0\). Since convergence implies boundedness (Theorem 2.3.2), Dirichlet weakens the hypothesis on \((x_k)\) and strengthens the one on \((y_k)\), so neither test contains the other.

The proof runs through the same identity from Exercise 2.7.13(a),

\begin{equation*} \sum_{k=1}^{n} x_k y_k \;=\; s_n y_{n+1} + \sum_{k=1}^{n} s_k (y_k - y_{k+1}), \end{equation*}

and the estimate on the sum is unchanged, since it only ever used boundedness. Fix \(M > 0\) with \(|s_k| \le M\) for all \(k\), now by hypothesis. Since \((y_k)\) is decreasing, \(y_k - y_{k+1} \ge 0\) and

\begin{equation*} \begin{aligned} \big| s_k (y_k - y_{k+1}) \big| &\;\le\; M (y_k - y_{k+1}), \\ \sum_{k=1}^{n} M(y_k - y_{k+1}) &= M(y_1 - y_{n+1}) \to M y_1 , \end{aligned} \end{equation*}

using \(y_k \to 0\). So the majorizing series converges, the Comparison Test (Theorem 2.7.4) makes \(\sum |s_k(y_k - y_{k+1})|\) converge, and the Absolute Convergence Test (Theorem 2.7.6) makes \(\sum s_k(y_k - y_{k+1})\) converge.

The one point that changes is the boundary term. Abel’s argument used \(\lim s_n\); here \((s_n)\) need not converge, but instead

\begin{equation*} 0 \;\le\; |s_n y_{n+1}| \;\le\; M \, y_{n+1} \;\longrightarrow\; 0 , \end{equation*}

so \(s_n y_{n+1} \to 0\) by the Squeeze Theorem (Exercise 2.3.3). Letting \(n \to \infty\) in the identity, the partial sums of \(\sum x_k y_k\) converge, proving Dirichlet’s Test.

(b) Let \((a_n)\) satisfy the hypotheses of Theorem 2.7.7: \(a_1 \ge a_2 \ge a_3 \ge \cdots\) and \((a_n) \to 0\) (so \(a_n \ge 0\)). Apply Dirichlet’s Test with

\begin{equation*} x_k = (-1)^{k+1}, \qquad y_k = a_k . \end{equation*}

The partial sums of \(\sum x_k\) are

\begin{equation*} s_n = \sum_{k=1}^{n} (-1)^{k+1} = \begin{cases} 1 & n \text{ odd} \\[2pt] 0 & n \text{ even,} \end{cases} \end{equation*}

hence bounded by \(1\), and \((y_k)\) decreases to \(0\). Dirichlet’s Test therefore gives convergence of

\begin{equation*} \sum_{k=1}^{\infty} x_k y_k \;=\; \sum_{k=1}^{\infty} (-1)^{k+1} a_k , \end{equation*}

which is exactly the Alternating Series Test.

\(\lim_{n \to \infty} s_{nn} = -2\): the square limit agrees with the column-first iterated value \(-2\) and differs from the row-first value \(0\).

Fix \(n\) and sum row \(i \le n\) across the first \(n\) columns. The entries left of the diagonal vanish, the diagonal entry is \(-1\), and the remaining entries form a finite geometric sum:

\begin{equation*} \begin{aligned} \sum_{j=1}^{n} a_{ij} &= -1 + \sum_{j=i+1}^{n} \frac{1}{2^{\,j-i}} = -1 + \sum_{k=1}^{n-i} \frac{1}{2^{k}} \\ &= -1 + \left( 1 - \frac{1}{2^{\,n-i}} \right) = -\frac{1}{2^{\,n-i}} . \end{aligned} \end{equation*}

Summing these \(n\) row-sums,

\begin{equation*} \begin{aligned} s_{nn} &= -\sum_{i=1}^{n} \frac{1}{2^{\,n-i}} = -\frac{1}{2^{n}} \sum_{i=1}^{n} 2^{i} = -\frac{2^{n+1} - 2}{2^{n}} \\ &= -2 + \frac{1}{2^{\,n-1}} . \end{aligned} \end{equation*}

Hence \(\lim_{n \to \infty} s_{nn} = -2\) by the Algebraic Limit Theorem (Theorem 2.3.3), since \((1/2^{\,n-1}) \to 0\).

Each row sums absolutely, and the row sums are dominated by the \(b_i\).

Fix \(i\). Since \(\sum_{j=1}^{\infty} |a_{ij}|\) converges, the Absolute Convergence Test (Theorem 2.7.6) gives that \(\sum_{j=1}^{\infty} a_{ij}\) converges; call its value \(r_i\), so that the inner series in the second display is meaningful for every \(i\). For each \(n\) the triangle inequality gives

\begin{equation*} \left| \sum_{j=1}^{n} a_{ij} \right| \le \sum_{j=1}^{n} |a_{ij}| \le b_i , \end{equation*}

and letting \(n \to \infty\) the Order Limit Theorem (Theorem 2.3.4) yields

\begin{equation*} |r_i| \le b_i \qquad \text{for every } i \in \mathbf{N}. \end{equation*}

Now \(\sum_{i=1}^{\infty} b_i\) converges with \(0 \le |r_i| \le b_i\), so the Comparison Test (Theorem 2.7.4) shows \(\sum_{i=1}^{\infty} |r_i|\) converges, and a second application of Theorem 2.7.6 shows \(\sum_{i=1}^{\infty} r_i\) converges. Since \(r_i = \sum_{j=1}^{\infty} a_{ij}\), this is precisely the statement that the iterated series \(\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{ij}\) converges.

(a) \((t_{nn})\) is increasing and bounded above by \(\sum_{i=1}^{\infty} b_i\).

Passing from \(t_{nn}\) to \(t_{n+1,n+1}\) only adds terms \(|a_{ij}| \ge 0\), so \((t_{nn})\) is increasing. For the bound, each inner sum is a partial sum of the convergent nonnegative series defining \(b_i\), so \(\sum_{j=1}^{n} |a_{ij}| \le b_i\), whence

\begin{equation*} t_{nn} = \sum_{i=1}^{n} \sum_{j=1}^{n} |a_{ij}| \le \sum_{i=1}^{n} b_i \le \sum_{i=1}^{\infty} b_i , \end{equation*}

the last step because \((b_i)\) is nonnegative and \(\sum_{i=1}^{\infty} b_i\) converges by hypothesis. The Monotone Convergence Theorem (Theorem 2.4.2) now gives convergence of \((t_{nn})\).

(b) Write \(Q_n = \{(i,j) \in \mathbf{N} \times \mathbf{N} : 1 \le i \le n, \ 1 \le j \le n\}\). For \(n > m\) the terms of \(s_{nn}\) not appearing in \(s_{mm}\) are exactly those indexed by the L-shaped region \(Q_n \setminus Q_m\), so

\begin{equation*} \begin{aligned} |s_{nn} - s_{mm}| &= \left| \sum_{(i,j) \in Q_n \setminus Q_m} a_{ij} \right| \le \sum_{(i,j) \in Q_n \setminus Q_m} |a_{ij}| \\ &= t_{nn} - t_{mm} = |t_{nn} - t_{mm}| , \end{aligned} \end{equation*}

the last equality because \((t_{nn})\) is increasing. By (a) and Theorem 2.6.2, \((t_{nn})\) is a Cauchy sequence: given \(\epsilon > 0\) there is an \(N \in \mathbf{N}\) with \(|t_{nn} - t_{mm}| < \epsilon\) whenever \(n, m \ge N\). The displayed inequality then gives \(|s_{nn} - s_{mm}| < \epsilon\) for all \(n, m \ge N\), so \((s_{nn})\) is Cauchy and hence converges by the Cauchy Criterion (Theorem 2.6.4).

(a) Take \(N_1 = \max\{m_0, n_0\}\), where \(t_{m_0 n_0} > B - \epsilon/2\).

Such a pair exists because \(B - \epsilon/2 < B\), so \(B - \epsilon/2\) is not an upper bound for \(\{t_{mn}\}\) (Lemma 1.3.8). Since all the \(|a_{ij}|\) are nonnegative, \(t_{mn}\) is increasing in each index separately; hence for \(m, n \ge N_1\) we have \(m \ge m_0\) and \(n \ge n_0\), so

\begin{equation*} B - \frac{\epsilon}{2} < t_{m_0 n_0} \le t_{mn} \le B , \end{equation*}

the final inequality because \(B\) is an upper bound. (That \(B\) exists at all is the Axiom of Completeness applied to \(\{t_{mn}\}\), which is bounded above by \(\sum_{i=1}^{\infty} b_i\) as in Exercise 2.8.3(a).)

(b) Choose \(N \ge N_1\) with \(|s_{pp} - S| < \epsilon/2\) for all \(p \ge N\); this \(N\) works.

Let \(m, n \ge N\) and set \(p = \max\{m, n\}\). Writing \(R_{mn} = \{(i,j) : 1 \le i \le m, \ 1 \le j \le n\}\) and \(Q_p = R_{pp}\), we have the nesting

\begin{equation*} Q_{N_1} \subseteq R_{mn} \subseteq Q_p , \end{equation*}

the first inclusion because \(m, n \ge N \ge N_1\). Therefore \(Q_p \setminus R_{mn} \subseteq Q_p \setminus Q_{N_1}\), and since \(p \ge N_1\),

\begin{equation*} \begin{aligned} |s_{pp} - s_{mn}| &\le \sum_{(i,j) \in Q_p \setminus R_{mn}} |a_{ij}| \le \sum_{(i,j) \in Q_p \setminus Q_{N_1}} |a_{ij}| \\ &= t_{pp} - t_{N_1 N_1} < B - \left( B - \frac{\epsilon}{2} \right) = \frac{\epsilon}{2} , \end{aligned} \end{equation*}

using part (a) on both \(t_{pp} \le B\) and \(t_{N_1 N_1} > B - \epsilon/2\). Since \(p \ge N\), the triangle inequality finishes it:

\begin{equation*} \begin{aligned} |s_{mn} - S| &\le |s_{mn} - s_{pp}| + |s_{pp} - S| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

(a) Fix \(m \ge N\) and let \(n \to \infty\) in \(|s_{mn} - S| < \epsilon\).

Written out by rows,

\begin{equation*} \begin{aligned} s_{mn} &= \sum_{j=1}^{n} a_{1j} + \sum_{j=1}^{n} a_{2j}

• \cdots + \sum_{j=1}^{n} a_{mj} , \end{aligned} \end{equation*}

a sum of exactly \(m\) sequences in \(n\), the \(i\)th of which converges to \(r_i\). The Algebraic Limit Theorem (Theorem 2.3.3), applied \(m-1\) times, gives

\begin{equation*} \lim_{n \to \infty} s_{mn} = r_1 + r_2 + \cdots + r_m . \end{equation*}

By Exercise 2.8.4(b), \(-\epsilon < s_{mn} - S < \epsilon\) for every \(n \ge N\), so the Order Limit Theorem (Theorem 2.3.4) applied to the sequence \((s_{mn} - S)_{n=N}^{\infty}\) yields

\begin{equation*} -\epsilon \le (r_1 + \cdots + r_m) - S \le \epsilon , \end{equation*}

which is the assertion.

To conclude, let \(S_m = r_1 + \cdots + r_m\) denote the \(m\)th partial sum of \(\sum_{i=1}^{\infty} r_i\). Given an arbitrary \(\alpha > 0\), run Exercise 2.8.4 with \(\epsilon = \alpha/2\) to produce an \(N\) for which \(m \ge N\) implies \(|S_m - S| \le \alpha/2 < \alpha\). Thus \((S_m) \to S\), i.e.

\begin{equation*} \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{ij} = \sum_{i=1}^{\infty} r_i = S . \end{equation*}

(b) The columns converge absolutely, and the estimate of Exercise 2.8.4(b) is symmetric in \(m\) and \(n\).

Fix \(j\). The partial sums \(\sum_{i=1}^{m} |a_{ij}| \le t_{mj} \le B\) are increasing in \(m\) and bounded, so \(\sum_{i=1}^{\infty} |a_{ij}|\) converges by the Monotone Convergence Theorem (Theorem 2.4.2) and \(\sum_{i=1}^{\infty} a_{ij}\) converges to some \(c_j\) by Theorem 2.7.6. Now grouping the finite sum \(s_{mn}\) by columns instead of rows,

\begin{equation*} \begin{aligned} s_{mn} &= \sum_{i=1}^{m} a_{i1} + \sum_{i=1}^{m} a_{i2}

• \cdots + \sum_{i=1}^{m} a_{in} , \end{aligned} \end{equation*}

so for fixed \(n\) the Algebraic Limit Theorem gives \(\lim_{m \to \infty} s_{mn} = c_1 + \cdots + c_n\). Fixing \(n \ge N\) and letting \(m \to \infty\) in \(-\epsilon < s_{mn} - S < \epsilon\), the Order Limit Theorem gives

\begin{equation*} |(c_1 + c_2 + \cdots + c_n) - S| \le \epsilon \qquad (n \ge N), \end{equation*}

and the same halving of \(\epsilon\) as in (a) gives \(\sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_{ij} = \sum_{j=1}^{\infty} c_j = S\).

(a) The partial sums of \(\sum |d_k|\) are bounded by \(B = \sup\{t_{mn}\}\).

Write \(T_K = \{(i,j) \in \mathbf{N} \times \mathbf{N} : i + j \le K\}\), the \(K\)th triangle; the terms making up \(d_2, d_3, \ldots, d_K\) are precisely the \(a_{ij}\) with \((i,j) \in T_K\), each occurring once. Since \(i + j \le K\) with \(i, j \ge 1\) forces \(i \le K - 1\) and \(j \le K - 1\), we have \(T_K \subseteq Q_K\), and therefore

\begin{equation*} \sum_{k=2}^{K} |d_k| \le \sum_{k=2}^{K} \sum_{i=1}^{k-1} |a_{i,k-i}| = \sum_{(i,j) \in T_K} |a_{ij}| \le t_{KK} \le B . \end{equation*}

These partial sums are increasing and bounded above, so \(\sum_{k=2}^{\infty} |d_k|\) converges by the Monotone Convergence Theorem (Theorem 2.4.2); that is, \(\sum_{k=2}^{\infty} d_k\) converges absolutely, and in particular converges (Theorem 2.7.6).

(b) The triangle \(T_K\) traps the square \(Q_{\lfloor K/2 \rfloor}\), so \(D_K = \sum_{k=2}^{K} d_k\) has the same limit as \(s_{KK}\).

Let \(\epsilon > 0\) and take \(N_1\) as in Exercise 2.8.4(a), so that \(m, n \ge N_1\) implies \(B - \epsilon/2 < t_{mn} \le B\). Put \(n_K = \lfloor K/2 \rfloor\). If \((i,j) \in Q_{n_K}\) then \(i + j \le 2n_K \le K\), so

\begin{equation*} Q_{n_K} \subseteq T_K \subseteq Q_K , \end{equation*}

and hence \(Q_K \setminus T_K \subseteq Q_K \setminus Q_{n_K}\). For every \(K \ge 2N_1\) we have \(n_K \ge N_1\), and since \(D_K = \sum_{(i,j) \in T_K} a_{ij}\) while \(s_{KK} = \sum_{(i,j) \in Q_K} a_{ij}\),

\begin{equation*} \begin{aligned} |D_K - s_{KK}| &\le \sum_{(i,j) \in Q_K \setminus T_K} |a_{ij}| \le \sum_{(i,j) \in Q_K \setminus Q_{n_K}} |a_{ij}| \\ &= t_{KK} - t_{n_K n_K} < B - \left( B - \frac{\epsilon}{2} \right) = \frac{\epsilon}{2} . \end{aligned} \end{equation*}

Now choose \(N \ge 2N_1\) large enough that \(|s_{KK} - S| < \epsilon/2\) for all \(K \ge N\), which is possible since \((s_{KK}) \to S\). For \(K \ge N\),

\begin{equation*} \begin{aligned} |D_K - S| &\le |D_K - s_{KK}| + |s_{KK} - S| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon , \end{aligned} \end{equation*}

so \(\sum_{k=2}^{\infty} d_k = \lim_{K \to \infty} D_K = S\).

(a) The iterated sum of absolute values is \(\alpha \beta\), where \(\alpha = \sum_{i=1}^{\infty} |a_i|\) and \(\beta = \sum_{j=1}^{\infty} |b_j|\), both finite by hypothesis.

Apply Theorem 2.8.1 to the array \(a_{ij} = a_i b_j\). For fixed \(i\) the constant \(|a_i|\) factors out of the partial sums, so by the Algebraic Limit Theorem for Series (Theorem 2.7.1),

\begin{equation*} \sum_{j=1}^{\infty} |a_i b_j| = |a_i| \sum_{j=1}^{\infty} |b_j| = |a_i| \beta , \end{equation*}

which converges; this is the number \(b_i\) in the hypothesis of Theorem 2.8.1. Applying Theorem 2.7.1 once more,

\begin{equation*} \sum_{i=1}^{\infty} |a_i| \beta = \beta \sum_{i=1}^{\infty} |a_i| = \alpha \beta , \end{equation*}

which converges as well. So \(\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_i b_j|\) converges and Theorem 2.8.1 applies.

(b) \(s_{nn}\) factors.

Because the double sum is finite, it splits as a product of two partial sums:

\begin{equation*} s_{nn} = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i b_j = \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{j=1}^{n} b_j \right) . \end{equation*}

The two factors converge to \(A\) and \(B\) respectively, so the product rule of the Algebraic Limit Theorem (Theorem 2.3.3) gives

\begin{equation*} \lim_{n \to \infty} s_{nn} = AB . \end{equation*}

By part (a) the hypothesis of Theorem 2.8.1 is met, and its conclusion is that both iterated sums equal \(\lim_{n \to \infty} s_{nn}\):

\begin{equation*} \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_i b_j = \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_i b_j = AB . \end{equation*}

Finally, Exercise 2.8.6(b) applies under the same hypothesis and shows that the diagonal sum converges to \(\lim_{n \to \infty} s_{nn}\) too. For this array the \(k\)th diagonal is

\begin{equation*} \begin{aligned} d_k &= \sum_{i=1}^{k-1} a_i b_{k-i} = a_1 b_{k-1} + a_2 b_{k-2} + \cdots + a_{k-1} b_1 , \end{aligned} \end{equation*}

so \(\sum_{k=2}^{\infty} d_k = AB\).

(a) At the single line \(\epsilon = \min\{\epsilon_1, \epsilon_2, \ldots, \epsilon_N\}\). A finite set of positive reals has a minimum, and that minimum is again positive; an infinite set of positive reals need only have an infimum, possibly \(0\), and \(\epsilon = 0\) produces no neighborhood at all. Everything before that line (\(a \in \bigcap_{k} O_k\) yields \(V_{\epsilon_k}(a) \subseteq O_k\) for each \(k\)) is valid for an arbitrary index set.

(b) Take \(O_n = (-1/n,\, 1/n)\), each open by Example 3.2.2 (ii). Then

\begin{equation*} \bigcap_{n=1}^{\infty}\left(-\tfrac1n, \tfrac1n\right) = \{0\}, \end{equation*}

since \(|x| < 1/n\) for all \(n \in \mathbf{N}\) forces \(x = 0\) by the Archimedean Property (Theorem 1.4.2), while \(0\) lies in every \(O_n\). The singleton \(\{0\}\) has no limit points, hence is closed; it is nonempty and is not \(\mathbf{R}\).

(a) The limit points of \(A\) are \(-1\) and \(1\); the limit points of \(B\) are all of \([0,1]\).

Write \(a_n = (-1)^n + 2/n\). The odd terms \(a_{2k-1} = -1 + \frac{2}{2k-1}\) decrease strictly to \(-1\) and the even terms \(a_{2k} = 1 + \frac{1}{k}\) decrease strictly to \(1\), so \(-1\) and \(1\) are limit points by Theorem 3.2.5 (neither sequence takes its limit as a value). The odd terms lie in \([-1,1]\) with \(a_1 = 1\) and the even terms in \((1,2]\), so all the \(a_n\) are distinct and any \(x \notin \{-1,1\}\) has a neighborhood containing only finitely many \(a_n\); shrinking it further leaves at most \(x\) itself.

For \(B\): if \(y \in [0,1]\) and \(\epsilon > 0\), then \((y-\epsilon, y+\epsilon) \cap (0,1)\) is a nonempty open interval, so by Theorem 1.4.3 it contains a rational \(r \neq y\), and \(r \in B\). No \(y \notin [0,1]\) qualifies, since \(V_\epsilon(y) \cap (0,1) = \emptyset\) for small \(\epsilon\).

(b) Neither set is open: every \(\epsilon\)-neighborhood is uncountable, so it cannot sit inside the countable set \(A\); and every \(\epsilon\)-neighborhood of a point of \(B\) contains an irrational (Corollary 1.4.4), so it cannot sit inside \(B\). Neither set is closed: \(-1\) is a limit point of \(A\) with \(-1 \notin A\) (all terms satisfy \(a_n > -1\)), and \(0\) is a limit point of \(B\) with \(0 \notin B\).

(c) \(A\): every point of \(A\) except \(a_1 = 1\) is isolated. Indeed \(1 = a_1 \in A\) is a limit point by (a), hence not isolated (Definition 3.2.6); for any other \(a_n\), the distance to the nearest other term of the sequence is positive, and a neighborhood of that radius meets \(A\) only in \(a_n\). \(B\): no isolated points, since by (a) every point of \(B\) is a limit point of \(B\).

(d) By Definition 3.2.11,

\begin{equation*} \overline{A} = A \cup \{-1, 1\} = A \cup \{-1\}, \qquad \overline{B} = B \cup [0,1] = [0,1]. \end{equation*}

(a) Neither. Not open: no \(V_\epsilon(0)\) lies in \(\mathbf{Q}\), since every interval contains an irrational (Corollary 1.4.4). Not closed: \(\sqrt{2}\) is a limit point of \(\mathbf{Q}\) (Example 3.2.9 (iii): every real is) and \(\sqrt{2} \notin \mathbf{Q}\) by Theorem 1.1.1.

(b) Closed, not open. Not open: \(V_\epsilon(1)\) contains \(1 + \epsilon/2 \notin \mathbf{N}\) for \(\epsilon < 1\). Closed vacuously: \(\mathbf{N}\) has no limit points, because \(V_{1/2}(x)\) contains at most one natural number, hence meets \(\mathbf{N}\) in no point other than possibly \(x\) itself.

(c) Open, not closed. Open: given \(x \neq 0\), take \(\epsilon = |x|\); then \(V_\epsilon(x) \subseteq \mathbf{R}\setminus\{0\}\). (Equivalently, \(\{0\}\) is closed and Theorem 3.2.13 applies.) Not closed: \(0\) is a limit point, since \(V_\epsilon(0)\) contains \(\epsilon/2 \neq 0\), and \(0\) is not in the set.

(d) Neither. Let \(s_n = \sum_{k=1}^{n} 1/k^2\). These are exactly the partial sums of Example 2.4.4: \((s_n)\) is increasing and bounded above (by \(2\)), so \(s_n \to L = \sup_n s_n\) by the Monotone Convergence Theorem (Theorem 2.4.2). Since \(s_n < s_{n+1} \le L\), no term equals \(L\), so by Theorem 3.2.5 \(L\) is a limit point of the set that the set does not contain: not closed. Not open, since a neighborhood is uncountable and the set is countable.

(e) Closed, not open. Let \(h_n = \sum_{k=1}^{n} 1/k\). The harmonic series diverges (Example 2.4.5), so \(h_n \to \infty\); hence for any \(x \in \mathbf{R}\) only finitely many \(h_n\) lie in \((x-1, x+1)\), and taking \(\epsilon\) smaller than the distance from \(x\) to the nearest of those finitely many terms other than \(x\) gives \(V_\epsilon(x) \cap \{h_n\} \subseteq \{x\}\). So the set has no limit points and is closed. Not open, for the countability reason in (d) (concretely, no \(V_\epsilon(1)\) is contained in it).

(a) If \(s \in A\) we are done, since \(\overline{A} = A \cup L \supseteq A\) (Definition 3.2.11). So assume \(s \notin A\) and let \(\epsilon > 0\). Because \(s - \epsilon\) is not an upper bound of \(A\) (Lemma 1.3.8), there is \(a \in A\) with

\begin{equation*} s - \epsilon < a \le s, \end{equation*}

and \(a \neq s\) because \(s \notin A\). Thus \(V_\epsilon(s)\) meets \(A\) at a point other than \(s\) for every \(\epsilon > 0\), so \(s \in L\) by Definition 3.2.4, and again \(s \in \overline{A}\).

(b) No. Suppose \(O\) is open and \(s = \sup O\) exists with \(s \in O\). By Definition 3.2.1 there is \(\epsilon > 0\) with \(V_\epsilon(s) \subseteq O\), and then \(s + \epsilon/2 \in O\) exceeds the upper bound \(s\), a contradiction.

(\(\Rightarrow\)) Let \(F\) be closed and let \((x_n) \subseteq F\) be Cauchy. By the Cauchy Criterion (Theorem 2.6.4) \(x_n \to x\) for some \(x \in \mathbf{R}\). If \(x_n = x\) for some \(n\), then \(x \in F\) immediately. Otherwise \(x_n \neq x\) for all \(n\), so \(x\) is a limit point of \(F\) by Theorem 3.2.5, and \(x \in F\) because \(F\) contains its limit points (Definition 3.2.7).

(\(\Leftarrow\)) Assume every Cauchy sequence in \(F\) has its limit in \(F\), and let \(x\) be a limit point of \(F\). By Theorem 3.2.5 there is a sequence \((x_n) \subseteq F\) with \(x_n \neq x\) and \(x_n \to x\). Convergent sequences are Cauchy (Theorem 2.6.2), so \((x_n)\) is a Cauchy sequence contained in \(F\) and therefore \(x = \lim x_n \in F\). Thus \(F\) contains its limit points and is closed.

(a) False. \(O = \mathbf{R} \setminus \{\sqrt{2}\}\) contains \(\mathbf{Q}\) (Theorem 1.1.1), is open by the argument of Exercise 3.2.3 (c) applied at \(\sqrt2\), and is not \(\mathbf{R}\).

(b) False. Put \(F_n = [n, \infty)\). Each \(F_n\) is closed (a limit of a sequence in \(F_n\) satisfies \(x \ge n\) by the Order Limit Theorem, Theorem 2.3.4), and \(F_1 \supseteq F_2 \supseteq F_3 \supseteq \cdots\), but

\begin{equation*} \bigcap_{n=1}^{\infty} F_n = \emptyset, \end{equation*}

since any \(x \in \mathbf{R}\) fails \(x \ge n\) for \(n\) large (Archimedean Property, Theorem 1.4.2).

(c) True. Let \(O \neq \emptyset\) be open and pick \(a \in O\). By Definition 3.2.1 there is \(\epsilon > 0\) with \(V_\epsilon(a) = (a - \epsilon, a + \epsilon) \subseteq O\), and Theorem 1.4.3 supplies a rational \(r\) with \(a - \epsilon < r < a + \epsilon\). Then \(r \in O\).

(d) False. Take

\begin{equation*} A = \left\{\sqrt{2}\left(1 + \tfrac1n\right) : n \in \mathbf{N}\right\} \cup \{\sqrt{2}\}. \end{equation*}

Every element is irrational: \(\sqrt{2}\,q\) with \(q \in \mathbf{Q}\), \(q > 0\), cannot be rational, or else \(\sqrt2\) would be. \(A\) is infinite, and \(A \subseteq [\sqrt2, 2\sqrt2]\) is bounded. It is closed: the only limit point is \(\sqrt2\) (the terms decrease strictly to \(\sqrt2\), so each is isolated), and \(\sqrt2 \in A\).

(e) True. The Cantor set is \(C = \bigcap_{n=0}^{\infty} C_n\), where each \(C_n\) is a union of \(2^n\) closed intervals. A finite union of closed sets is closed (Theorem 3.2.14 (i)), so each \(C_n\) is closed; an arbitrary intersection of closed sets is closed (Theorem 3.2.14 (ii)), so \(C\) is closed.

\((\star)\) If \(y \in L\) and \(0 < |x-y| < \epsilon\), then \(V_\epsilon(x)\) contains a point of \(A\) other than \(x\). Indeed, set

\begin{equation*} \delta = \min\{\,|x-y|,\ \epsilon - |x-y|\,\} > 0. \end{equation*}

Because \(y \in L\), Definition 3.2.4 gives a point \(a \in A \cap V_\delta(y)\) with \(a \neq y\). Then

\begin{equation*} \begin{aligned} |a - x| &\le |a - y| + |y - x| < \delta + |x-y| \le \epsilon, \\ |a - y| &< \delta \le |x - y| \ \Longrightarrow\ a \neq x. \end{aligned} \end{equation*}

Hence \(a \in A \cap V_\epsilon(x)\) with \(a \neq x\), proving \((\star)\).

(a) Let \(x\) be a limit point of \(L\) and let \(\epsilon > 0\). Then \(V_\epsilon(x)\) contains some \(y \in L\) with \(y \neq x\), and \((\star)\) produces a point of \(A\) in \(V_\epsilon(x)\) different from \(x\). As \(\epsilon\) was arbitrary, \(x \in L\), so \(L\) contains its limit points and is closed (Definition 3.2.7).

(b) Let \(x\) be a limit point of \(A \cup L\) and let \(\epsilon > 0\). Choose \(z \in (A \cup L) \cap V_\epsilon(x)\) with \(z \neq x\). Two cases:

(i) \(z \in A\): then \(V_\epsilon(x)\) already meets \(A\) at a point other than \(x\).

(ii) \(z \in L\): apply \((\star)\) with \(y = z\).

Either way \(V_\epsilon(x)\) meets \(A\) away from \(x\), so \(x \in L\).

Theorem 3.2.12 now follows. Closed: by the previous paragraph every limit point of \(\overline{A} = A \cup L\) lies in \(L \subseteq \overline{A}\), so \(\overline{A}\) is closed. Smallest: let \(F\) be any closed set with \(A \subseteq F\) and let \(x \in L\). Every \(V_\epsilon(x)\) meets \(A\) at a point other than \(x\), hence meets \(F\) at a point other than \(x\), so \(x\) is a limit point of \(F\) and thus \(x \in F\). Therefore

\begin{equation*} \overline{A} = A \cup L \subseteq F, \end{equation*}

and \(\overline{A}\) is the smallest closed set containing \(A\).

(a) Definitely closed, not definitely open. Every closure is closed (Theorem 3.2.12). With \(A = (0,1)\) and \(B = \{2\}\) we get \(\overline{A \cup B} = [0,1] \cup \{2\}\), and no \(V_\epsilon(0)\) lies inside this set.

(b) Definitely open, not definitely closed. Indeed \(A \setminus B = A \cap B^c\), where \(B^c\) is open by Theorem 3.2.13, so the intersection of two open sets is open by Theorem 3.2.3 (ii). With \(A = (0,1)\) and \(B = \{1/2\}\) we get \((0,1/2) \cup (1/2,1)\), which omits its limit point \(1/2\).

(c) Definitely open, not definitely closed: this is the same set as (b), since

\begin{equation*} (A^c \cup B)^c = (A^c)^c \cap B^c = A \cap B^c = A \setminus B . \end{equation*}

(d) Definitely closed, not definitely open, because the set is exactly \(B\):

\begin{equation*} (A \cap B) \cup (A^c \cap B) = (A \cup A^c) \cap B = \mathbf{R} \cap B = B . \end{equation*}

Taking \(B = [0,1]\) shows it need not be open.

(e) Definitely open, not definitely closed, because the set is exactly \(\overline{A}^{\,c}\). From \(A \subseteq \overline{A}\) and \(A^c \subseteq \overline{A^c}\) we get

\begin{equation*} \overline{A}^{\,c} \subseteq A^c \subseteq \overline{A^c}, \qquad\text{hence}\qquad \overline{A}^{\,c} \cap \overline{A^c} = \overline{A}^{\,c}, \end{equation*}

which is open by Theorems 3.2.12 and 3.2.13. Taking \(A = (0,1)\) gives \(\overline{A}^{\,c} = (-\infty,0) \cup (1,\infty)\), which omits its limit point \(0\).

(a) Both identities are a chain of equivalences on membership. For the first,

\begin{equation*} \begin{aligned} x \in \Bigl( \bigcup_{\lambda} E_\lambda \Bigr)^{c} &\iff \neg\bigl(\exists\, \lambda \in \Lambda \text{ with } x \in E_\lambda\bigr)\\ &\iff \forall\, \lambda \in \Lambda,\ x \notin E_\lambda\\ &\iff \forall\, \lambda \in \Lambda,\ x \in E_\lambda^{c} \iff x \in \bigcap_{\lambda} E_\lambda^{c}, \end{aligned} \end{equation*}

and for the second, replacing the existential quantifier by a universal one,

\begin{equation*} \begin{aligned} x \in \Bigl( \bigcap_{\lambda} E_\lambda \Bigr)^{c} &\iff \neg\bigl(\forall\, \lambda \in \Lambda,\ x \in E_\lambda\bigr)\\ &\iff \exists\, \lambda \in \Lambda \text{ with } x \in E_\lambda^{c} \iff x \in \bigcup_{\lambda} E_\lambda^{c} . \end{aligned} \end{equation*}

(b) Theorem 3.2.14 asserts that (i) the union of a finite collection of closed sets is closed and (ii) the intersection of an arbitrary collection of closed sets is closed. Let \(\{F_\lambda : \lambda \in \Lambda\}\) be closed sets, so each \(F_\lambda^{c}\) is open by Theorem 3.2.13.

For (ii), part (a) gives

\begin{equation*} \Bigl( \bigcap_{\lambda \in \Lambda} F_\lambda \Bigr)^{c} = \bigcup_{\lambda \in \Lambda} F_\lambda^{c}, \end{equation*}

an arbitrary union of open sets, hence open by Theorem 3.2.3 (i); so \(\bigcap_\lambda F_\lambda\) is closed, again by Theorem 3.2.13.

For (i), with \(F_1, \ldots, F_N\) closed, part (a) gives

\begin{equation*} \Bigl( \bigcup_{k=1}^{N} F_k \Bigr)^{c} = \bigcap_{k=1}^{N} F_k^{c}, \end{equation*}

a finite intersection of open sets – finiteness being precisely the hypothesis Theorem 3.2.3 (ii) demands – hence open; so \(\bigcup_{k=1}^N F_k\) is closed.

Only (ii) is realizable, and \(A = \mathbf{Q} \cap [0,1]\) does it. It is an infinite subset of the countable set \(\mathbf{Q}\) (Theorem 1.5.6 (i)), hence countable by Theorem 1.5.7, and no \(q \in A\) is isolated: given \(\epsilon > 0\), the set \(V_\epsilon(q) \cap [0,1]\) contains a nondegenerate interval with endpoint \(q\) (because \(0 \le q \le 1\)), so by the density of \(\mathbf{Q}\) (Theorem 1.4.3) it contains a rational \(r \neq q\), whence \(V_\epsilon(q)\) meets \(A\) in a point other than \(q\) and \(q\) is a limit point of \(A\) (Definition 3.2.4, Definition 3.2.6).

(i) is impossible. By Definition 1.5.5 a countable set is infinite, so write \(A = \{a_1, a_2, a_3, \ldots\}\) with the \(a_n\) distinct. The sequence \((a_n)\) lies in \([0,1]\) and is therefore bounded, so the Bolzano–Weierstrass Theorem (Theorem 2.5.5) supplies a convergent subsequence \(a_{n_k} \to x\). The terms are distinct, so at most one \(a_{n_k}\) equals \(x\); deleting that one term leaves a sequence in \(A\) with \(a_{n_k} \neq x\) still converging to \(x\), and Theorem 3.2.5 makes \(x\) a limit point of \(A\).

(iii) is impossible: the isolated points of any set \(A \subseteq \mathbf{R}\) inject into \(\mathbf{Q} \times \mathbf{Q}\). If \(x\) is isolated, Definitions 3.2.4 and 3.2.6 give an \(\epsilon > 0\) with \(V_\epsilon(x) \cap A = \{x\}\), and Theorem 1.4.3 provides rationals

\begin{equation*} p_x \in (x - \epsilon, x), \qquad q_x \in (x, x + \epsilon), \qquad\text{so}\qquad (p_x, q_x) \cap A = \{x\} . \end{equation*}

The map \(x \mapsto (p_x, q_x)\) is one-to-one on the isolated points: if \(x \neq y\) were isolated with \((p_x,q_x) = (p_y,q_y)\), then \(y \in (p_y,q_y) \cap A = (p_x,q_x) \cap A = \{x\}\), forcing \(y = x\). Now \(\mathbf{Q} \times \mathbf{Q} = \bigcup_{q \in \mathbf{Q}} (\{q\} \times \mathbf{Q})\) is a countable union of countable sets (re-indexed by \(\mathbf{N}\) using Theorem 1.5.6 (i)), hence countable by Theorem 1.5.8 (ii). Its subset \(\{(p_x,q_x) : x \text{ isolated}\}\) is therefore countable or finite by Theorem 1.5.7, and the injection puts the isolated points of \(A\) in \(1\)–\(1\) correspondence with it.

(a) Both inclusions are minimality statements read off Theorem 3.2.12, which says \(\overline{E}\) is the smallest closed set containing \(E\).

\((\subseteq)\) The set \(\overline{A} \cup \overline{B}\) is a union of two closed sets, hence closed by Theorem 3.2.14 (i), and it contains \(A \cup B\). Minimality of \(\overline{A \cup B}\) gives \(\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}\).

\((\supseteq)\) The set \(\overline{A \cup B}\) is closed and contains \(A\), so minimality of \(\overline{A}\) gives \(\overline{A} \subseteq \overline{A \cup B}\); symmetrically \(\overline{B} \subseteq \overline{A \cup B}\). Hence

\begin{equation*} \overline{A} \cup \overline{B} \subseteq \overline{A \cup B} . \end{equation*}

(b) No. Only the inclusion \(\bigcup_{n} \overline{A_n} \subseteq \overline{\bigcup_n A_n}\) survives, by the \((\supseteq)\) argument above, which used nothing about the number of sets. For the failure, enumerate \(\mathbf{Q} \cap [0,1] = \{r_1, r_2, r_3, \ldots\}\) (an infinite subset of \(\mathbf{Q}\), so countable by Theorems 1.5.6 (i) and 1.5.7) and set \(A_n = \{r_n\}\). A singleton has no limit points, hence is closed by Definition 3.2.7, so \(\overline{A_n} = A_n\), while \(\bigcup_n A_n = \mathbf{Q} \cap [0,1]\) has closure \([0,1]\) by the density of \(\mathbf{Q}\) (Theorem 1.4.3):

\begin{equation*} \bigcup_{n=1}^{\infty} \overline{A_n} = \mathbf{Q} \cap [0,1] \neq [0,1] = \overline{\,\bigcup_{n=1}^{\infty} A_n\,} . \end{equation*}

Write \(B = U \cap V\) where

\begin{equation*} U = \{ s \in \mathbf{R} : A \cap (-\infty, s) \text{ is uncountable} \}, \quad V = \{ s \in \mathbf{R} : A \cap (s, \infty) \text{ is uncountable} \}, \end{equation*}

and call a set small if it is finite or countable, so that uncountable means not small (Definition 1.5.5). Two facts are used throughout. A subset of a small set is small (Theorem 1.5.7). And if \(S_1, S_2, S_3, \ldots\) are small, so is \(\bigcup_n S_n\): each \(S_n \cup \mathbf{N}\) is countable (by Theorem 1.5.8 (i) when \(S_n\) is countable; by listing the finitely many points of \(S_n \setminus \mathbf{N}\) ahead of \(\mathbf{N}\) when \(S_n\) is finite), so \(\bigcup_n (S_n \cup \mathbf{N})\) is countable by Theorem 1.5.8 (ii), and Theorem 1.5.7 passes this to the subset \(\bigcup_n S_n\).

Openness. Suppose \(s \in U\). Any \(x < s\) satisfies \(1/n < s - x\) for some \(n\) (Theorem 1.4.2 (ii)), so

\begin{equation*} A \cap (-\infty, s) = \bigcup_{n=1}^{\infty} A \cap (-\infty, s - 1/n) , \end{equation*}

and the left side is not small, so some term \(A \cap (-\infty, s - 1/n_0)\) is not small, i.e. \(s - 1/n_0 \in U\). But \(U\) is closed upward: if \(t > u \in U\) then \(A \cap (-\infty, u) \subseteq A \cap (-\infty, t)\), and a superset of an uncountable set is uncountable, so \(t \in U\). Hence

\begin{equation*} V_{1/n_0}(s) \subseteq (s - 1/n_0, \infty) \subseteq U , \end{equation*}

and \(U\) is open by Definition 3.2.1. Symmetrically \(V\) is open and closed downward. Consequently \(B = U \cap V\) is open by Theorem 3.2.3 (ii).

Nonemptiness. First, \(U \neq \emptyset\): by the Archimedean Property (Theorem 1.4.2) every real lies below some \(n \in \mathbf{N}\), so \(A = \bigcup_{n=1}^{\infty} A \cap (-\infty, n)\), and were every term small \(A\) would be small. Some \(n\) therefore has \(n \in U\), and symmetrically \(V \neq \emptyset\).

Now suppose, for contradiction, that \(U \cap V = \emptyset\). Every \(v \in V\) satisfies \(v < u\) for every \(u \in U\), since \(v = u\) or \(v > u\) would each place \(v\) in \(U\) (upward closure), hence in \(U \cap V\). Thus each \(v \in V\) is a lower bound for \(U\) and each \(u \in U\) an upper bound for \(V\); both sets being nonempty, \(c = \inf U\) and \(\sup V\) exist (Axiom of Completeness and Exercise 1.3.3), and \(\sup V \le \inf U = c\). For each \(n \in \mathbf{N}\),

\begin{equation*} \begin{aligned} c - 1/n < \inf U &\ \Longrightarrow\ c - 1/n \notin U &&\Longrightarrow\ A \cap (-\infty, c - 1/n) \text{ is small},\\ c + 1/n > \sup V &\ \Longrightarrow\ c + 1/n \notin V &&\Longrightarrow\ A \cap (c + 1/n, \infty) \text{ is small}. \end{aligned} \end{equation*}

Taking countable unions over \(n\) makes \(A \cap (-\infty, c)\) and \(A \cap (c, \infty)\) small, whence

\begin{equation*} A \subseteq (A \cap (-\infty,c)) \cup \{c\} \cup (A \cap (c,\infty)) \end{equation*}

is small – contradicting the hypothesis that \(A\) is uncountable. Hence \(B = U \cap V \neq \emptyset\).

Both \(\mathbf{R}\) and \(\emptyset\) do qualify: each is open by Example 3.2.2 (i), and each is the complement of the other, so each is closed by Theorem 3.2.13. For the converse, suppose \(E\) is both open and closed with \(E \neq \emptyset\) and \(E \neq \mathbf{R}\), and pick \(a \in E\) and \(b \in E^c\); assume \(a < b\), the case \(b < a\) being the mirror image with \(\inf\bigl(E \cap [b,a]\bigr)\) in place of the supremum below.

Let

\begin{equation*} S = E \cap [a,b], \qquad s = \sup S, \end{equation*}

which exists because \(S\) is nonempty (\(a \in S\)) and bounded above by \(b\). By Exercise 3.2.4 (a), \(s \in \overline{S}\), and \(\overline{S} \subseteq \overline{E} = E\) since \(\overline{E}\) is a closed set containing \(S\) (Theorem 3.2.12) and \(E\) is closed (Exercise 3.2.14 (a)). So \(s \in E\), while \(b \notin E\); since also \(a \in S\), this gives \(a \le s < b\).

Now use that \(E\) is open: there is \(\epsilon > 0\) with \(V_\epsilon(s) \subseteq E\), and shrinking \(\epsilon\) we may assume \(\epsilon < b - s\). Then \(a \le s < s + \epsilon/2 < s + \epsilon < b\), so

\begin{equation*} s + \epsilon/2 \in V_\epsilon(s) \cap [a,b] \subseteq E \cap [a,b] = S, \end{equation*}

while \(s + \epsilon/2 > s = \sup S\) – a contradiction.

(a) Let \(L\) be the set of limit points of \(E\), so \(\overline{E} = E \cup L\) by Definition 3.2.11. Then

\begin{equation*} \overline{E} = E \iff L \subseteq E \iff E \text{ contains its limit points} \iff E \text{ is closed}, \end{equation*}

the last step being Definition 3.2.7.

For the interior, \(E^{\circ} \subseteq E\) holds by definition, so the content is the reverse inclusion. If \(E\) is open, then every \(x \in E\) admits \(V_\epsilon(x) \subseteq E\) (Definition 3.2.1), which is exactly the requirement for \(x \in E^{\circ}\); hence \(E \subseteq E^{\circ}\) and \(E^{\circ} = E\). Conversely, if \(E^{\circ} = E\) then every \(x \in E\) has some \(V_\epsilon(x) \subseteq E\), which is Definition 3.2.1 verbatim.

(b) The first identity is a membership chain. For \(x \in \mathbf{R}\),

\begin{equation*} \begin{aligned} x \in \overline{E}^{\,c} &\iff x \notin E \ \wedge\ x \text{ is not a limit point of } E\\ &\iff x \notin E \ \wedge\ \exists\, \epsilon > 0 : V_\epsilon(x) \cap E \subseteq \{x\}\\ &\iff \exists\, \epsilon > 0 : V_\epsilon(x) \cap E = \emptyset\\ &\iff \exists\, \epsilon > 0 : V_\epsilon(x) \subseteq E^{c} \iff x \in (E^{c})^{\circ} . \end{aligned} \end{equation*}

The third equivalence is where \(x \notin E\) is spent: it upgrades \(V_\epsilon(x) \cap E \subseteq \{x\}\) to \(V_\epsilon(x) \cap E = \emptyset\), and conversely \(V_\epsilon(x) \cap E = \emptyset\) forces \(x \notin E\) since \(x \in V_\epsilon(x)\).

The second identity now follows by applying the first to the set \(E^{c}\):

\begin{equation*} \overline{E^{c}}^{\,c} = \bigl((E^{c})^{c}\bigr)^{\circ} = E^{\circ}, \qquad\text{so}\qquad \overline{E^{c}} = (E^{\circ})^{c} . \end{equation*}

(a) Fatten symmetrically:

\begin{equation*} [a,b] \;=\; \bigcap_{n=1}^{\infty}\Bigl(a-\tfrac1n,\; b+\tfrac1n\Bigr), \end{equation*}

a countable intersection of open sets. The inclusion \(\subseteq\) is immediate; conversely, if \(a-1/n < x < b+1/n\) for every \(n\), then \(x > b\) would force \(0 < x-b < 1/n\) for all \(n \in \mathbf{N}\), contradicting the Archimedean Property (Theorem 1.4.2), and symmetrically \(x < a\) is impossible, so \(x \in [a,b]\).

(b) Fatten only on the right, and exhaust from the left (here \(a < b\)):

\begin{equation*} \begin{aligned} (a,b] &= \bigcap_{n=1}^{\infty}\Bigl(a,\; b+\tfrac1n\Bigr)\\ (a,b] &= \bigcup_{n=1}^{\infty}\Bigl[a+\tfrac{b-a}{n},\; b\Bigr]. \end{aligned} \end{equation*}

In the first, each set is open and a point of the intersection has \(x > a\), while \(x > b\) is excluded exactly as in (a), so \(a < x \le b\). In the second, each \([a+(b-a)/n,\,b]\) is closed (Example 3.2.9 (ii)) and sits inside \((a,b]\); conversely \(a < x \le b\) lands in the \(n\)th set once the Archimedean Property (Theorem 1.4.2) supplies \(n\) with \((b-a)/n < x-a\).

(c) Enumerate \(\mathbf{Q} = \{q_1,q_2,q_3,\dots\}\), possible since \(\mathbf{Q}\) is countable (Theorem 1.5.6 (i)). A singleton has no limit points at all (for \(x \ne q_n\) take \(\epsilon = |x-q_n|\); for \(x = q_n\) every \(V_\epsilon(x)\) meets \(\{q_n\}\) only in \(x\)), so it is vacuously closed (Definition 3.2.7) and

\begin{equation*} \mathbf{Q} \;=\; \bigcup_{n=1}^{\infty}\{q_n\} \end{equation*}

is an \(F_\sigma\) set. Complementing and applying De Morgan’s Laws (Exercise 3.2.9),

\begin{equation*} \mathbf{I} \;=\; \mathbf{Q}^{c} \;=\; \Bigl(\bigcup_{n=1}^{\infty}\{q_n\}\Bigr)^{c} \;=\; \bigcap_{n=1}^{\infty}\{q_n\}^{c}, \end{equation*}

where each \(\{q_n\}^{c}\) is open by Theorem 3.2.13; thus \(\mathbf{I}\) is a \(G_\delta\) set.

Both are attained, because each is the limit of a sequence drawn from \(K\) and \(K\) is closed.

By Theorem 3.3.4 a compact \(K\) is closed and bounded. Since \(K\) is nonempty and bounded above, the Axiom of Completeness supplies \(s = \sup K\). For each \(n \in \mathbf{N}\) the number \(s - 1/n\) is not an upper bound for \(K\), so we may pick \(x_n \in K\) with

\begin{equation*} s - \frac{1}{n} < x_n \le s . \end{equation*}

Squeezing gives \((x_n) \to s\), so \((x_n)\) is a Cauchy sequence contained in \(K\), and Theorem 3.2.8 (the closedness criterion) forces \(s \in K\).

Symmetrically, \(K\) is bounded below, so \(i = \inf K\) exists; choosing \(y_n \in K\) with \(i \le y_n < i + 1/n\) gives \((y_n) \to i\) and hence \(i \in K\) by the same theorem.

Compact: (c) and (e). Not compact: (a), (b), (d).

(a) Take \(x_n = n\). Any subsequence satisfies \(x_{n_k} = n_k \ge k\), so it is unbounded and therefore divergent (Theorem 2.3.2: convergent sequences are bounded). Definition 3.3.1 fails at the very first step.

(b) Take rationals \(r_n \in [0,1]\) with \((r_n) \to 1/\sqrt{2}\), available by the density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 3.2.10). Every subsequence converges to the same limit \(1/\sqrt{2}\) (Theorem 2.5.2), and \(1/\sqrt{2} \notin \mathbf{Q}\). The set is bounded but not closed.

(c) Compact. With \(C = \bigcap_{n=0}^{\infty} C_n\) as in Section 3.1, each \(C_n\) is a union of \(2^n\) closed intervals, hence closed by Theorem 3.2.14 (i), and an arbitrary intersection of closed sets is closed by Theorem 3.2.14 (ii). Since \(C \subseteq [0,1]\) is bounded, Theorem 3.3.4 gives compactness.

(d) Write \(s_n = \sum_{k=1}^{n} 1/k^2\), so the set is \(S = \{s_n : n \in \mathbf{N}\}\). The sequence \((s_n)\) is strictly increasing and bounded, hence convergent by Theorem 2.4.2 (this is exactly Example 2.4.4), with

\begin{equation*} L = \lim s_n = \sup S . \end{equation*}

Strict monotonicity gives \(s_n < s_{n+1} \le L\) for every \(n\), so \(L \notin S\). By Theorem 2.5.2 every subsequence of \((s_n)\) converges to \(L\), so this sequence in \(S\) has no subsequence with limit in \(S\): \(S\) is bounded but not closed.

(e) Compact. The set is \(E = \{1\} \cup \{(n-1)/n : n \ge 2\}\), so \(E \subseteq [0,1]\) is bounded. A sequence of distinct points of \(E\) must use indices \(n \to \infty\), since only finitely many elements of \(E\) lie below any given \((N-1)/N\); such a sequence converges to \(1\). So by Theorem 3.2.5 the only limit point of \(E\) is \(1 \in E\), making \(E\) closed (Definition 3.2.7), and Theorem 3.3.4 gives compactness.

Bolzano–Weierstrass produces the subsequence and closedness catches its limit; together with the argument in the text this completes the proof of Theorem 3.3.4.

Let \((a_n)\) be an arbitrary sequence contained in \(K\). Boundedness of \(K\) (Definition 3.3.3) gives an \(M > 0\) with \(|a_n| \le M\) for all \(n\), so \((a_n)\) is a bounded sequence and Theorem 2.5.5 (Bolzano–Weierstrass) supplies a convergent subsequence

\begin{equation*} (a_{n_k}) \to a . \end{equation*}

A convergent sequence is Cauchy, and \((a_{n_k})\) is a Cauchy sequence contained in the closed set \(K\), so \(a \in K\) by Theorem 3.2.8. Thus every sequence in \(K\) has a subsequence converging to a limit in \(K\), which is Definition 3.3.1.

(a) both; (b), (c), (d) neither.

(a) \(K\) is closed and bounded by Theorem 3.3.4, so \(K \cap F\) is closed by Theorem 3.2.14 (ii) and is bounded because \(K \cap F \subseteq K\); being closed and bounded it is compact, again by Theorem 3.3.4.

(b) De Morgan gives \(F^{c} \cup K^{c} = (F \cap K)^{c}\), which is open by Theorem 3.2.13. It is never compact: with \(M > 0\) chosen so that \(K \subseteq [-M, M]\),

\begin{equation*} F^{c} \cup K^{c} \supseteq K^{c} \supseteq (M, \infty) , \end{equation*}

so the set is unbounded. Nor need it be closed: \(K = F = [0,1]\) gives \((-\infty, 0) \cup (1, \infty)\), which omits its limit point \(0\).

(c) The set \(K \backslash F \subseteq K\) is bounded, but \(K = [0,1]\) and \(F = \{0\}\) give \(K \backslash F = (0,1]\), which omits its limit point \(0\) and so is not closed (Definition 3.2.7), hence not compact (Theorem 3.3.4).

(d) \(K \cap F^{c}\) is literally the set \(K \backslash F\) of part (c), so the same example settles it: neither.

(a) True; (b), (c), (d) false.

(a) Let \(\{K_{\lambda} : \lambda \in \Lambda\}\) be a nonempty collection of compact sets, so each \(K_{\lambda}\) is closed and bounded (Theorem 3.3.4). The intersection is closed by Theorem 3.2.14 (ii), and fixing any \(\lambda_0 \in \Lambda\) gives

\begin{equation*} \bigcap_{\lambda \in \Lambda} K_{\lambda} \subseteq K_{\lambda_0} , \end{equation*}

so it is bounded. Theorem 3.3.4 returns compactness.

(b) False. Each singleton \(\{n\}\) is compact (it is closed and bounded), but

\begin{equation*} \bigcup_{n \in \mathbf{N}} \{n\} = \mathbf{N} \end{equation*}

is unbounded, hence not compact.

(c) False. Take \(A = (0,1)\) and \(K = [0,1]\). Then \(A \cap K = (0,1)\), and the sequence \((1/n)_{n \ge 2}\) lies in it while every subsequence converges to \(0 \notin (0,1)\) (Theorem 2.5.2), contradicting Definition 3.3.1.

(d) False. Put \(F_n = [n, \infty)\). Each is nonempty and closed (its complement \((-\infty, n)\) is open, Theorem 3.2.13) and the sequence is nested, yet the Archimedean Property gives

\begin{equation*} \bigcap_{n=1}^{\infty} [n, \infty) = \emptyset , \end{equation*}

since no real number exceeds every natural number.

All three hold for “finite” and all three hold for “compact”; all three fail for “closed.” (Read the sets in (a) as nonempty: the empty set is finite and compact but has no maximum.)

(a) Finite: induct on \(|A|\); a set with one element is its own maximum, and \(\max(A \cup \{x\}) = \max\{\max A, x\}\). Compact: Exercise 3.3.1 shows \(\sup K \in K\), and an element of \(K\) that is an upper bound for \(K\) is exactly \(\max K\). Closed: false — \(\mathbf{N}\) is closed (it has no limit points, so Definition 3.2.7 holds vacuously) and has no maximum.

(b) Finite: \(A + B\) is the image of \(A \times B\) under addition, so \(|A + B| \le |A| \cdot |B| < \infty\).

Compact: let \((a_n + b_n)\) be a sequence in \(A + B\). Definition 3.3.1 applied to \(A\) gives a subsequence \((a_{n_k}) \to a \in A\); applied to \((b_{n_k})\) in \(B\) it gives a further subsequence \((b_{n_{k_j}}) \to b \in B\). Since \((a_{n_{k_j}})\) still tends to \(a\) (Theorem 2.5.2), the Algebraic Limit Theorem (Theorem 2.3.3) yields

\begin{equation*} a_{n_{k_j}} + b_{n_{k_j}} \longrightarrow a + b \in A + B . \end{equation*}

Closed: false. Take

\begin{equation*} A = \mathbf{N}, \qquad B = \Big\{ -n + \tfrac{1}{n} : n \ge 2 \Big\} . \end{equation*}

Consecutive points of \(B\) satisfy \(b_n - b_{n+1} = 1 + \tfrac{1}{n} - \tfrac{1}{n+1} > 1\), so \(B\) has no limit points and is closed, as is \(\mathbf{N}\). Now \(n + (-n + 1/n) = 1/n \in A + B\) for every \(n \ge 2\), so \(0\) is a limit point of \(A + B\); but \(m - n + 1/n = 0\) would force \(1/n = n - m \in \mathbf{Z}\), impossible for \(n \ge 2\). Hence \(0 \notin A + B\) and \(A + B\) is not closed.

(c) Set \(B_N = \bigcap_{n=1}^{N} A_n\), so \((B_N)\) is nested and each \(B_N \ne \emptyset\) by hypothesis, and \(\bigcap_{n=1}^{\infty} A_n = \bigcap_{N=1}^{\infty} B_N\).

Finite: each \(B_N\) is finite, so the integers \(|B_N|\) are non-increasing and at least \(1\), hence constant for \(N \ge N_0\); with \(B_N \subseteq B_{N_0}\) this forces \(B_N = B_{N_0}\) for all \(N \ge N_0\), and the intersection equals \(B_{N_0} \ne \emptyset\).

Compact: each \(B_N\) is compact by part (a) of Exercise 3.3.5, so Theorem 3.3.5 (Nested Compact Set Property) applies to the nested nonempty compact sets \(B_1 \supseteq B_2 \supseteq \cdots\) and gives \(\bigcap_{N} B_N \ne \emptyset\).

Closed: false — with \(A_n = [n, \infty)\) every finite subcollection meets in \([\max_i n_i, \infty) \ne \emptyset\), while \(\bigcap_{n=1}^{\infty} A_n = \emptyset\).

(a) \(C_n + C_n = [0,2]\) for every \(n \ge 0\), so the required \(x_n, y_n\) always exist.

For \(n = 1\), use \([a,b] + [c,d] = [a+c, b+d]\) on each of the three pairings of the two components of \(C_1 = [0,\frac13] \cup [\frac23, 1]\):

\begin{equation*} \begin{aligned} {[0,\tfrac13]} + [0,\tfrac13] &= [0, \tfrac23], \\ {[0,\tfrac13]} + [\tfrac23, 1] &= [\tfrac23, \tfrac43], \\ {[\tfrac23, 1]} + [\tfrac23, 1] &= [\tfrac43, 2], \end{aligned} \end{equation*}

whose union is \([0,2]\). In particular \(s \in [0,2]\) lies in one of these three sets, which hands us \(x_1, y_1 \in C_1\) with \(x_1 + y_1 = s\).

For general \(n\), induct, using the self-similarity built into the construction of Section 3.1 — passing from \(C_n\) to \(C_{n+1}\) removes middle thirds inside \([0,\frac13]\) and inside \([\frac23,1]\) in exactly the scaled-by-\(\frac13\) pattern, so

\begin{equation*} C_{n+1} = \tfrac{1}{3} C_n \;\cup\; \Big( \tfrac{2}{3} + \tfrac{1}{3} C_n \Big) . \end{equation*}

The base case is \(C_0 + C_0 = [0,1] + [0,1] = [0,2]\). Assuming \(C_n + C_n = [0,2]\), the three pairings of the two pieces of \(C_{n+1}\) give

\begin{equation*} \begin{aligned} \tfrac13 C_n + \tfrac13 C_n &= \tfrac13 [0,2] = [0, \tfrac23], \\ \tfrac13 C_n + (\tfrac23 + \tfrac13 C_n) &= \tfrac23 + \tfrac13 [0,2] = [\tfrac23, \tfrac43], \\ (\tfrac23 + \tfrac13 C_n) + (\tfrac23 + \tfrac13 C_n) &= \tfrac43 + \tfrac13 [0,2] = [\tfrac43, 2], \end{aligned} \end{equation*}

and their union is \(C_{n+1} + C_{n+1} = [0,2]\).

(b) Extract a convergent subsequence from \((x_n)\) and let \((y_n)\) follow it.

Fix \(s \in [0,2]\) and choose \(x_n, y_n \in C_n\) with \(x_n + y_n = s\) as in (a). The sequence \((x_n)\) lies in \([0,1]\), so Theorem 2.5.5 (Bolzano–Weierstrass) supplies a convergent subsequence \((x_{n_k}) \to x\), and then by the Algebraic Limit Theorem (Theorem 2.3.3)

\begin{equation*} y_{n_k} = s - x_{n_k} \longrightarrow s - x =: y , \end{equation*}

so \(x + y = s\).

It remains to see that \(x, y \in C\). Fix \(m \in \mathbf{N}\). The sets \(C_n\) are nested downward, so for every \(k\) with \(n_k \ge m\) we have \(x_{n_k} \in C_{n_k} \subseteq C_m\). That tail of \((x_{n_k})\) is a convergent, hence Cauchy, sequence contained in the closed set \(C_m\) (a finite union of closed intervals, closed by Theorem 3.2.14 (i)), so \(x \in C_m\) by Theorem 3.2.8. The same argument applies verbatim to \(y_{n_k} \in C_{n_k} \subseteq C_m\), giving \(y \in C_m\). Since \(m\) was arbitrary,

\begin{equation*} x, y \in \bigcap_{m=0}^{\infty} C_m = C , \end{equation*}

and \(x + y = s\). Hence \([0,2] \subseteq C + C\), and with the reverse inclusion noted in the statement, \(C + C = [0,2]\).

(a) The infimum is attained, and disjointness then forces it to be positive.

The set \(\{|x-y| : x \in K,\ y \in L\}\) is nonempty and bounded below by \(0\), so \(d\) exists (Exercise 1.3.3). By the infimum version of Lemma 1.3.8 (Exercise 1.3.1(b)): no \(d + 1/n\) is a lower bound, so pick for each \(n \in \mathbf{N}\) points \(x_n \in K\), \(y_n \in L\) with

\begin{equation*} d \le |x_n - y_n| < d + \tfrac1n . \end{equation*}

By Definition 3.3.1 pass to a subsequence along which \(x_n \to x_0 \in K\), then to a further subsequence along which \(y_n \to y_0 \in L\); the first limit survives the second passage by Theorem 2.5.2. Relabelling, the Algebraic Limit Theorem (Theorem 2.3.3) together with \(\bigl||u|-|v|\bigr| \le |u-v|\) gives

\begin{equation*} |x_0 - y_0| = \lim_n |x_n - y_n| = d, \end{equation*}

the second equality by the squeeze above. Because \(K \cap L = \emptyset\) we have \(x_0 \ne y_0\), hence \(d = |x_0 - y_0| > 0\).

(b) Take

\begin{equation*} K = \mathbf{N}, \qquad L = \{\, n + 1/n : n \ge 2 \,\}. \end{equation*}

Consecutive points of \(K\) differ by \(1\) and consecutive points of \(L\) by \(\bigl(n+1+\tfrac{1}{n+1}\bigr) - \bigl(n+\tfrac1n\bigr) = 1 - \tfrac{1}{n(n+1)} \ge \tfrac56\), so each \(V_{1/4}(y)\) meets each set in at most one point: neither set has a limit point, so both contain their limit points vacuously and are closed (Definition 3.2.7). They are disjoint, since \(n + 1/n\) is not an integer for \(n \ge 2\). Finally

\begin{equation*} 0 \le d \le \bigl| (n + 1/n) - n \bigr| = 1/n \quad \text{for every } n \ge 2, \end{equation*}

so \(d = 0\) although \(K \cap L = \emptyset\).

(a) Bisect repeatedly. Since \(K\) is bounded, a closed interval \(I_0 \supseteq K\) exists, and \(I_0 \cap K = K\) is not finitely covered by hypothesis. Given \(I_n\) with \(I_n \cap K\) not finitely coverable, write \(I_n = A \cup B\) where \(A, B\) are the two closed halves of \(I_n\), so that

\begin{equation*} I_n \cap K = (A \cap K) \cup (B \cap K). \end{equation*}

If both \(A \cap K\) and \(B \cap K\) admitted finite subcovers, the union of those two finite collections would be a finite subcover of \(I_n \cap K\). Hence at least one half does not; call it \(I_{n+1}\). By construction \(I_0 \supseteq I_1 \supseteq \cdots\) and

\begin{equation*} |I_n| = \frac{|I_0|}{2^{\,n}} \longrightarrow 0 . \end{equation*}

(b) Each \(I_n \cap K \ne \emptyset\) (the empty set is covered by the empty finite subcollection), so pick \(x_n \in I_n \cap K\). The Nested Interval Property (Theorem 1.4.1) supplies \(x \in \bigcap_{n=0}^{\infty} I_n\), and since \(x, x_n \in I_n\),

\begin{equation*} |x_n - x| \le |I_n| \to 0, \qquad \text{so } (x_n) \to x . \end{equation*}

The sequence \((x_n) \subseteq K\) is convergent, hence Cauchy (Theorem 2.6.2), so \(x \in K\) by Theorem 3.2.8. Thus \(x \in K\) and \(x \in I_n\) for all \(n\).

(c) Take \(O_{\lambda_0}\) from the cover with \(x \in O_{\lambda_0}\). Openness gives \(\epsilon > 0\) with \(V_\epsilon(x) \subseteq O_{\lambda_0}\), and since \(|I_n| \to 0\) we may fix \(n\) with \(|I_n| < \epsilon\). Every \(y \in I_n\) satisfies \(|y - x| \le |I_n| < \epsilon\) because \(x \in I_n\), so

\begin{equation*} I_n \cap K \subseteq I_n \subseteq V_\epsilon(x) \subseteq O_{\lambda_0}. \end{equation*}

The single set \(O_{\lambda_0}\) is therefore a finite subcover of \(I_n \cap K\), contradicting (a). Hence a finite subcover of \(K\) exists after all.

(a) \(a \in S\), because \(a \in [a,b] \subseteq \bigcup_\lambda O_\lambda\) puts \(a\) in some \(O_{\lambda_1}\), and the single set \(O_{\lambda_1}\) covers \([a,a] = \{a\}\). Also \(S \subseteq [a,b]\) is bounded above by \(b\). So \(s = \sup S\) exists by the Axiom of Completeness, and \(a \le s \le b\).

(b) Since \(s \in [a,b]\), fix \(\lambda_0\) with \(s \in O_{\lambda_0}\) and then \(\epsilon > 0\) with \(V_\epsilon(s) \subseteq O_{\lambda_0}\). By Lemma 1.3.8 there is \(x \in S\) with \(x > s - \epsilon\), and by definition of \(S\) a finite subcover \(O_{\lambda_1},\dots,O_{\lambda_n}\) of \([a,x]\). For any \(t\) with \(x \le t < s + \epsilon\) and \(t \le b\),

\begin{equation*} [a,t] \subseteq [a,x] \cup (s-\epsilon,\, s+\epsilon) \subseteq O_{\lambda_1} \cup \cdots \cup O_{\lambda_n} \cup O_{\lambda_0}, \end{equation*}

since a point of \([a,t]\) exceeding \(x\) lies in \((s-\epsilon, s+\epsilon)\). Hence every such \(t\) belongs to \(S\).

Taking \(t = s\) gives \(s \in S\). If \(s < b\), then \(t = \min\{b,\, s + \epsilon/2\}\) satisfies \(x \le s < t \le b\) and \(t < s+\epsilon\), so \(t \in S\) with \(t > s = \sup S\) — impossible. Therefore \(s = b\), and \(b = s \in S\) means \([a,b]\) has a finite subcover.

(c) Let \(K\) be closed and bounded and let \(\{O_\lambda\}\) be an open cover of \(K\). Boundedness gives \(K \subseteq [a,b]\) for some \(a \le b\), and closedness makes \(K^{c}\) open (Theorem 3.2.13). Then

\begin{equation*} \{O_\lambda : \lambda \in \Lambda\} \cup \{K^{c}\} \end{equation*}

is an open cover of \([a,b]\), since every point of \([a,b]\) is either in \(K\) (hence in some \(O_\lambda\)) or in \(K^c\). By (b) it has a finite subcover; delete \(K^c\) from that finite list if it appears. The remaining \(O_{\lambda_1},\dots,O_{\lambda_n}\) still cover \(K\), because no point of \(K\) was covered by \(K^c\). That is a finite subcover of the original collection.

Only (a), (b), (d) need covers: the Cantor set in (c) is an intersection of closed sets, hence closed (Theorem 3.2.14), and lies in \([0,1]\); the set in (e) is \(\{1\} \cup \{(n-1)/n : n \ge 2\} \subseteq [0,1]\), whose only limit point is \(1\), an element of it. Both are closed and bounded, hence compact by Theorem 3.3.4.

(a) \(\mathbf{N}\). Take

\begin{equation*} O_n = \Bigl(n - \tfrac12,\ n + \tfrac12\Bigr), \qquad n \in \mathbf{N}. \end{equation*}

Then \(n \in O_n\), so \(\{O_n\}\) covers \(\mathbf{N}\), while \(O_n \cap \mathbf{N} = \{n\}\). Any finite subcollection \(O_{n_1},\dots,O_{n_k}\) therefore covers only the \(k\) points \(n_1,\dots,n_k\) and misses \(\max\{n_1,\dots,n_k\} + 1\).

(b) \(\mathbf{Q} \cap [0,1]\). Let \(\alpha = \sqrt{2}/2 \in (0,1)\), irrational because \(2\alpha = \sqrt{2}\) is (Theorem 1.1.1), and set

\begin{equation*} O_n = (-\infty,\ \alpha - 1/n) \cup (\alpha + 1/n,\ \infty) = \{\, y : |y - \alpha| > 1/n \,\}. \end{equation*}

Each \(O_n\) is open, and \(\bigcup_n O_n = \mathbf{R} \setminus \{\alpha\} \supseteq \mathbf{Q} \cap [0,1]\), since any \(y \ne \alpha\) lies in \(O_n\) once \(1/n < |y - \alpha|\). The \(O_n\) increase, so a finite subcollection has union \(O_N\) for \(N\) the largest index. By the density of \(\mathbf{Q}\) (Theorem 1.4.3) choose a rational

\begin{equation*} q \in \Bigl(\alpha,\ \min\bigl\{1,\ \alpha + \tfrac1N\bigr\}\Bigr), \end{equation*}

a nonempty interval because \(\alpha < 1\). Then \(q \in \mathbf{Q} \cap [0,1]\) but \(|q - \alpha| < 1/N\), so \(q \notin O_N\). No finite subcover.

(d) Write \(s_n = \sum_{k=1}^{n} 1/k^2\), so the set is \(A = \{s_n : n \in \mathbf{N}\}\) and \(s_1 < s_2 < s_3 < \cdots\). Put

\begin{equation*} O_n = \Bigl(-\infty,\ \frac{s_n + s_{n+1}}{2}\Bigr), \qquad n \in \mathbf{N}. \end{equation*}

Since \(s_j < \frac{s_n+s_{n+1}}{2}\) exactly when \(j \le n\), the set \(O_n\) contains \(s_1,\dots,s_n\) and excludes \(s_{n+1}, s_{n+2},\dots\); in particular \(s_n \in O_n\) for every \(n\), so \(\{O_n\}\) covers \(A\). The \(O_n\) are nested increasing, so any finite subcollection has union \(O_N\), which omits \(s_{N+1} \in A\). No finite subcover.

If a bounded infinite set \(A\) had no limit point, the neighborhoods isolating its points would form an open cover with no finite subcover.

Having no limit point, \(A\) contains its limit points vacuously and so is closed (Definition 3.2.7); being closed and bounded, it admits a finite subcover from every open cover by the implication (ii) \(\Rightarrow\) (iii) of Theorem 3.3.8, proved in Exercise 3.3.9 by bisection and the Nested Interval Property (Theorem 1.4.1) with no appeal to Bolzano–Weierstrass.

Since no \(a \in A\) is a limit point of \(A\), Definition 3.2.4 supplies \(\epsilon_a > 0\) with \(V_{\epsilon_a}(a) \cap A = \{a\}\), and \(a \in V_{\epsilon_a}(a)\) makes \(\{V_{\epsilon_a}(a) : a \in A\}\) an open cover of \(A\). Writing \(V_i = V_{\epsilon_{a_i}}(a_i)\) for the \(n\) members of a finite subcover, each \(V_i \cap A = \{a_i\}\), so

\begin{equation*} A = \bigcup_{i=1}^{n} \bigl(V_i \cap A\bigr) = \{a_1, \dots, a_n\} \end{equation*}

is finite, contradicting the hypothesis. Hence \(A\) has a limit point.

The clompact subsets of \(\mathbf{R}\) are precisely the finite sets (including \(\emptyset\)).

Finite sets are clompact: if \(A = \{a_1,\dots,a_n\}\) and \(\{C_\lambda\}\) is any cover of \(A\), pick \(\lambda_i\) with \(a_i \in C_{\lambda_i}\); then

\begin{equation*} A \subseteq C_{\lambda_1} \cup \cdots \cup C_{\lambda_n}, \end{equation*}

a finite subcover (for \(A = \emptyset\) the empty subcollection serves).

Infinite sets are not: the singletons \(\{a\}\), \(a \in A\), are closed (having no limit points, each contains them vacuously), and

\begin{equation*} A = \bigcup_{a \in A} \{a\} \end{equation*}

is a closed cover of \(A\). A finite subcollection \(\{a_1\},\dots,\{a_n\}\) has union \(\{a_1,\dots,a_n\}\), which cannot contain the infinite set \(A\). So no finite subcover exists.

Always compact; not always perfect.

\(P\) is closed (Definition 3.4.1) and \(K\) is closed and bounded (Theorem 3.3.4), so \(P \cap K\) is closed by Theorem 3.2.14 (ii) and bounded because \(P \cap K \subseteq K\); Theorem 3.3.4 again makes it compact.

Perfection fails for \(P = [0,1]\) and \(K = \{0\}\), the latter closed and bounded hence compact:

\begin{equation*} P \cap K = \{0\}, \qquad V_{1/2}(0) \cap \{0\} = \{0\}, \end{equation*}

so \(0\) is an isolated point of \(P \cap K\).

No, apart from the vacuous \(P = \emptyset\) (closed, with no isolated points).

A nonempty perfect set is uncountable by Theorem 3.4.3, while any \(P \subseteq \mathbb{Q}\) is countable or finite by Theorem 1.5.7, since \(\mathbb{Q}\) is countable by Theorem 1.5.6 (i). No set is both.

Take \(x_n\) to be an endpoint, other than \(x\) itself, of the component interval of \(C_n\) containing \(x\).

The one fact needed is the observation from Section 3.1: if \(y\) is an endpoint of one of the closed intervals making up \(C_n\), then \(y\) is again an endpoint of an interval of \(C_{n+1}\) (middle thirds removed are open, so endpoints are never deleted), whence \(y \in C_m\) for all \(m\) and so \(y \in C\).

(a) \(C_1 = [0,1/3] \cup [2/3,1]\), so \(x\) lies in one of these two closed intervals, say \(I = [a,b]\) with \(b - a = 1/3\). Both \(a\) and \(b\) lie in \(C\) by the observation above, and at most one of them equals \(x\); let \(x_1\) be one that does not. Then

\begin{equation*} x_1 \in C \cap C_1, \qquad x_1 \ne x, \qquad |x - x_1| \le b - a = \tfrac{1}{3}, \end{equation*}

the last inequality because \(x, x_1 \in I\).

(b) Identical for every \(n\). Since \(x \in C \subseteq C_n\), the point \(x\) lies in one of the \(2^n\) component intervals \(I = [a,b]\) of \(C_n\), and \(b - a = 1/3^n\). Both endpoints belong to \(C\), so choosing \(x_n \in \{a,b\}\) with \(x_n \ne x\) gives

\begin{equation*} x_n \in C \cap C_n, \qquad x_n \ne x, \qquad |x - x_n| \le \tfrac{1}{3^n}. \end{equation*}

Hence \((x_n)\) is a sequence in \(C \setminus \{x\}\) with \(|x - x_n| \le 1/3^n \to 0\), so \((x_n) \to x\) and \(x\) is a limit point of \(C\) (Theorem 3.2.5), i.e. not isolated. As \(x \in C\) was arbitrary and \(C\) is closed, \(C\) is perfect.

Compact and perfect; length \(0\); dimension \(\log 2 / \log(8/3) \approx 0.707\).

Write \(D_0 = [0,1]\) and let \(D_{n+1}\) arise from \(D_n\) by deleting from each component the open middle fourth of that component. Deleting the middle quarter of an interval of length \(L\) leaves two closed intervals of length \(\tfrac{3}{8}L\) each, so

\begin{equation*} D_1 = \left[0,\tfrac{3}{8}\right] \cup \left[\tfrac{5}{8},1\right], \end{equation*}

and inductively \(D_n\) is a union of \(2^n\) closed intervals each of length \((3/8)^n\). Set \(D = \bigcap_{n=0}^{\infty} D_n\).

(a) Each \(D_n\) is a finite union of closed intervals, hence closed, and \(D\) is closed by Theorem 3.2.14 (ii); \(D \subseteq [0,1]\) is bounded. By Theorem 3.3.4, \(D\) is compact.

\(D\) is perfect by Exercise 3.4.3 with \(1/3^n\) replaced by \((3/8)^n\): the removed middles are open, so endpoints of components of \(D_n\) survive every later stage and lie in \(D\), and taking \(x_n\) to be the endpoint other than \(x\) of the component \([a,b]\) of \(D_n\) containing \(x\) gives

\begin{equation*} x_n \in D \setminus \{x\}, \qquad |x - x_n| \le b - a = \left(\tfrac{3}{8}\right)^{n} \to 0 . \end{equation*}

So \(x\) is a limit point of \(D\) (Theorem 3.2.5); no point of \(D\) is isolated, and \(D\) is closed.

(b) At stage \(n \ge 1\) we delete one open interval of length \(\tfrac{1}{4}(3/8)^{n-1}\) from each of the \(2^{n-1}\) components of \(D_{n-1}\). The total length removed is

\begin{equation*} \begin{aligned} \sum_{n=1}^{\infty} 2^{n-1} \cdot \tfrac{1}{4}\left(\tfrac{3}{8}\right)^{n-1} &= \tfrac{1}{4}\sum_{n=1}^{\infty} \left(\tfrac{3}{4}\right)^{n-1}\\ &= \tfrac{1}{4}\cdot\frac{1}{1 - 3/4} = 1 , \end{aligned} \end{equation*}

so the length of \(D\) is \(1 - 1 = 0\).

For the dimension, magnify by \(8/3\), the reciprocal of a component’s length: \(D_0\) becomes \([0,8/3]\) and deleting its middle fourth leaves \([0,1] \cup [5/3,8/3]\), so the magnified set is \(2\) copies of \(D\). Following Figure 3.3, the dimension \(x\) satisfies

\begin{equation*} 2 = \left(\tfrac{8}{3}\right)^{x}, \qquad x = \frac{\log 2}{\log 8 - \log 3} = \frac{\log 2}{3\log 2 - \log 3} \approx 0.707 . \end{equation*}

A point of \(\overline{A} \cap B\) would have a neighborhood inside \(V\) meeting \(A \subseteq U\), contradicting \(U \cap V = \emptyset\).

In detail, suppose \(x \in \overline{A} \cap B\). Since \(x \in B \subseteq V\) and \(V\) is open, Definition 3.2.1 supplies \(\epsilon > 0\) with \(V_{\epsilon}(x) \subseteq V\). Since \(\overline{A} = A \cup L\) with \(L\) the set of limit points of \(A\) (Definition 3.2.11), either \(x \in A\) or \(x\) is a limit point of \(A\); in both cases \(V_{\epsilon}(x) \cap A \ne \emptyset\). Picking \(a\) in that intersection,

\begin{equation*} a \in A \subseteq U \quad\text{and}\quad a \in V_{\epsilon}(x) \subseteq V, \end{equation*}

so \(a \in U \cap V = \emptyset\), a contradiction. Hence \(\overline{A} \cap B = \emptyset\).

Interchanging the roles of \((A,U)\) and \((B,V)\) gives \(A \cap \overline{B} = \emptyset\). By Definition 3.4.4, \(A\) and \(B\) are separated.

Both directions run through Theorem 3.2.5, which trades the statement that \(x\) is a limit point of \(A\) for a sequence \((x_n)\) in \(A\) with \(x_n \ne x\) and \((x_n) \to x\).

(\(\Rightarrow\)) Let \(E\) be connected and let \(E = A \cup B\) with \(A, B\) nonempty and disjoint. If \(A\) and \(B\) were separated, \(E\) would be disconnected by Definition 3.4.4; so they are not, meaning

\begin{equation*} \overline{A} \cap B \ne \emptyset \quad\text{or}\quad A \cap \overline{B} \ne \emptyset . \end{equation*}

Say \(x \in \overline{A} \cap B\) (the other case is symmetric). Then \(x \in B\), and since \(A \cap B = \emptyset\) we have \(x \notin A\); as \(\overline{A} = A \cup L\) (Definition 3.2.11), \(x\) must be a limit point of \(A\). By Theorem 3.2.5 there is a sequence \((x_n)\) contained in \(A\) with \((x_n) \to x\), and \(x \in B\) is an element of the other set, as required.

(\(\Leftarrow\)) Contrapositive: assume \(E\) is disconnected, so \(E = A \cup B\) with \(A, B\) nonempty and separated. Separated sets are disjoint, since

\begin{equation*} A \cap B \subseteq \overline{A} \cap B = \emptyset , \end{equation*}

so \(A, B\) is an admissible partition. Suppose \((x_n)\) lies in \(A\) with \((x_n) \to x\) and \(x \in B\). Then \(x \notin A\), so every neighborhood \(V_{\epsilon}(x)\) contains some \(x_n \in A\) with \(x_n \ne x\), making \(x\) a limit point of \(A\); hence \(x \in \overline{A} \cap B = \emptyset\), absurd. Symmetrically no sequence in \(B\) converges to a point of \(A\), so this partition admits no such sequence.

Cut at a point of the complementary set: an irrational for \(\mathbb{Q}\), a rational for \(\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}\). Both sets are totally disconnected.

(a) Let \(x, y \in \mathbb{Q}\) be distinct, labelled so that \(x < y\). By Corollary 1.4.4 there is an irrational \(\alpha\) with \(x < \alpha < y\). Put

\begin{equation*} A = \mathbb{Q} \cap (-\infty,\alpha), \qquad B = \mathbb{Q} \cap (\alpha,\infty). \end{equation*}

Since \(\alpha \notin \mathbb{Q}\), no rational is omitted, so \(\mathbb{Q} = A \cup B\), and \(x \in A\), \(y \in B\) makes both nonempty. The sets \(U = (-\infty,\alpha)\) and \(V = (\alpha,\infty)\) are open and disjoint with \(A \subseteq U\) and \(B \subseteq V\), so \(A\) and \(B\) are separated by Exercise 3.4.5.

(b) Yes. Let \(x, y \in \mathbb{I}\) be distinct, labelled so that \(x < y\). By Theorem 1.4.3 (Density of \(\mathbb{Q}\) in \(\mathbb{R}\)) there is a rational \(r\) with \(x < r < y\). Put

\begin{equation*} A = \mathbb{I} \cap (-\infty,r), \qquad B = \mathbb{I} \cap (r,\infty). \end{equation*}

Since \(r \notin \mathbb{I}\) we have \(\mathbb{I} = A \cup B\), with \(x \in A\) and \(y \in B\); and \(A \subseteq (-\infty,r)\), \(B \subseteq (r,\infty)\), two disjoint open sets, so \(A\) and \(B\) are separated by Exercise 3.4.5.

(a) Any \(N\) with \(3^{-N} < \epsilon\) works, and one exists because \(3^{-N} \le 1/N\) while the Archimedean Property (Theorem 1.4.2) supplies \(N\) with \(1/N < \epsilon\). Indeed, by the construction in Section 3.1, \(C_N\) is a union of \(2^N\) pairwise disjoint closed intervals, each of length \(3^{-N}\); if \(x\) and \(y\) both belonged to one of them, say \([a,b]\), then

\begin{equation*} \epsilon = y - x \le b - a = 3^{-N} < \epsilon, \end{equation*}

which is absurd.

(b) Fix \(x, y \in C\) with \(x < y\) and take \(N\) as in (a). Let \([a,b]\) and \([c,d]\) be the intervals of \(C_N\) containing \(x\) and \(y\); by (a) they are distinct, hence disjoint, and \(d \ge y > x \ge a\) rules out \(d < a\), so \(c > b\). Among the finitely many intervals of \(C_N\), let \(b^{\prime}\) be the smallest left endpoint exceeding \(b\) (there is one, namely \(c\)), so \(b < b^{\prime} \le c \le y\). Then

\begin{equation*} (b, b^{\prime}) \cap C_N = \emptyset , \end{equation*}

since an interval \([p,q]\) of \(C_N\) meeting \((b,b^{\prime})\) has \(p \le b\) by minimality of \(b^{\prime}\) and \(q > b\), so \(b \in [p,q]\) and disjointness forces \([p,q] = [a,b]\), i.e. \(q = b\), a contradiction. Choose any \(z \in (b, b^{\prime})\). Then \(x \le b < z < y\) and \(z \notin C_N \supseteq C\). Now

\begin{equation*} A = C \cap (-\infty, z), \qquad B = C \cap (z, \infty) \end{equation*}

are nonempty (\(x \in A\), \(y \in B\)), satisfy \(A \cup B = C\) precisely because \(z \notin C\), and are separated by Exercise 3.4.5, being contained in the disjoint open sets \((-\infty, z)\) and \((z, \infty)\). So \(C\) is totally disconnected.

(a) \(O\) is open, being a union of the open sets \(V_{\epsilon_n}(r_n)\) (Theorem 3.2.3(i)), so \(F = O^c\) is closed by Theorem 3.2.13. It contains no rational number: every \(q \in \mathbf{Q}\) equals \(r_n\) for some \(n\), and \(r_n \in V_{\epsilon_n}(r_n) \subseteq O\).

For \(F \ne \emptyset\) it suffices to show \([0,3] \not\subseteq O\). Suppose it were. Since \([0,3]\) is compact, the Heine-Borel Theorem (Theorem 3.3.8) yields a finite set \(S \subseteq \mathbf{N}\) with \([0,3] \subseteq \bigcup_{n \in S} V_{\epsilon_n}(r_n)\), a union of finitely many open intervals of total length

\begin{equation*} \sum_{n \in S} 2 \epsilon_n \;\le\; \sum_{n=1}^{\infty} \frac{2}{2^{n}} \;=\; 2 . \end{equation*}

This contradicts the following fact, applied with \([a,b] = [0,3]\): for nonempty open intervals \((a_k,b_k)\),

\begin{equation*} [a,b] \subseteq \bigcup_{k=1}^{m}(a_k,b_k) \quad\Longrightarrow\quad b - a < \sum_{k=1}^{m}(b_k - a_k). \end{equation*}

Induct on \(m\). For \(m = 1\), \(a_1 < a \le b < b_1\) gives \(b - a < b_1 - a_1\). For \(m \ge 2\), relabel so that \(b \in (a_m, b_m)\). If \(a_m < a\) we are done as in the base case. Otherwise \(a \le a_m \le b\), and no point of \([a, a_m]\) lies in \((a_m, b_m)\), so \([a, a_m]\) is covered by the other \(m - 1\) intervals; the induction hypothesis and \(b - a_m < b_m - a_m\) give

\begin{equation*} b - a \;=\; (a_m - a) + (b - a_m) \;<\; \sum_{k=1}^{m}(b_k - a_k). \end{equation*}

(b) No, and yes. Any nonempty open interval \((c,d)\) contains a rational (Theorem 1.4.3), and that rational lies in \(O\), so \((c,d) \not\subseteq F\); thus \(F\) contains no nonempty open interval. For total disconnectedness, take \(x, y \in F\) with \(x < y\) and pick a rational \(r \in (x,y)\), again by Theorem 1.4.3. Then \(r \notin F\), so

\begin{equation*} A = F \cap (-\infty, r), \qquad B = F \cap (r, \infty) \end{equation*}

are nonempty (\(x \in A\), \(y \in B\)), satisfy \(A \cup B = F\), and are separated because they sit inside the disjoint open sets \((-\infty,r)\) and \((r,\infty)\) (Exercise 3.4.5).

(c) No: the construction does not determine the answer, because some enumerations leave \(F\) with an isolated point and hence not perfect. Since \(F\) is closed, perfection asks only that no point be isolated (Definition 3.4.1). Fix an irrational \(x\) and reserve in advance, for every \(k \ge 1\), rationals

\begin{equation*} \begin{aligned} r_{3k} &\quad\text{with}\quad 2^{-3k} < x - r_{3k} < (1.1)2^{-3k},\\ r_{3k+1} &\quad\text{with}\quad 2^{-(3k+1)} < r_{3k+1} - x < (1.1)2^{-(3k+1)}, \end{aligned} \end{equation*}

these being distinct, since the intervals they are drawn from are pairwise disjoint and lie on opposite sides of \(x\). To the remaining indices \(n\) assign rationals outside the reserved list, at each step taking the first unused such \(q\) (in some fixed master list) with \(|x - q| \ge 2^{-n}\); every rational is eventually used, since only finitely many precede a given \(q\) and \(q\) is eligible at every step with \(2^{-n} \le |x - q|\). Then \(x \notin O\), as \(|x - r_n| \ge \epsilon_n\) for every \(n\). Writing \(\rho = 2^{-3k}\), the choice above gives \(V_{\rho}(r_{3k}) \supseteq [x - 2\rho, x - \rho/10]\), and consecutive members of this chain overlap because \(2 \cdot 2^{-3(k+1)} = \rho/4 > \rho/10\); hence the chain covers \([x - 1/4, x)\). The chain of the \(r_{3k+1}\) covers \((x, x + 1/8]\) in the same way. So \(F \cap V_{1/8}(x) = \{x\}\) and \(F\) is not perfect for this enumeration.

We can, however, always build a nonempty perfect set of irrationals by a Cantor-style construction. Let \(K_0 = [0,1]\), and suppose \(K_{n-1}\) is a union of \(2^{n-1}\) pairwise disjoint nondegenerate closed intervals. Inside each such interval \(I\) choose two disjoint nondegenerate closed subintervals, each of length less than \(2^{-n}\) and neither containing \(r_n\) (possible because \(I\) minus the single point \(r_n\) always contains two such intervals), and let \(K_n\) be the union of the resulting \(2^n\) intervals. Put \(K = \bigcap_{n=0}^{\infty} K_n\). Each \(K_n\) is nonempty, compact and \(K_{n+1} \subseteq K_n\), so \(K \ne \emptyset\) by the Nested Compact Set Property (Theorem 3.3.5), and \(K\) is closed by Theorem 3.2.14(ii). No rational survives, since \(r_n \notin K_n\). Finally, let \(p \in K\) and \(\delta > 0\), and pick \(n \ge 1\) with \(2^{-n} < \delta\). The interval \(I\) of \(K_n\) containing \(p\) has length less than \(\delta\), so \(I \subseteq V_{\delta}(p)\), and one of the two stage-\((n+1)\) intervals \(J \subseteq I\) omits \(p\). Following a nested chain of intervals downward from \(J\) and applying the Nested Interval Property (Theorem 1.4.1) produces a point \(y \in K \cap J\), so \(y \in K \cap V_{\delta}(p)\) with \(y \ne p\). Hence \(K\) has no isolated points: \(K\) is a nonempty perfect set of irrational numbers.

De Morgan plus Theorem 3.2.13 (\(O\) is open if and only if \(O^c\) is closed).

(\(\Rightarrow\)) If \(A = \bigcap_{n=1}^{\infty} O_n\) with each \(O_n\) open (Definition 3.5.1), then

\begin{equation*} A^c = \Big(\bigcap_{n=1}^{\infty} O_n\Big)^c = \bigcup_{n=1}^{\infty} O_n^c , \end{equation*}

and each \(O_n^c\) is closed by Theorem 3.2.13, so \(A^c\) is a countable union of closed sets, i.e. an \(F_\sigma\) set.

(\(\Leftarrow\)) If \(A^c = \bigcup_{n=1}^{\infty} F_n\) with each \(F_n\) closed, then

\begin{equation*} A = (A^c)^c = \bigcap_{n=1}^{\infty} F_n^c , \end{equation*}

and each \(F_n^c\) is open by the second half of Theorem 3.2.13, so \(A\) is a \(G_\delta\) set.

(a) countable, (b) finite, (c) finite, (d) countable.

(a) Write \(A_k = \bigcup_{n=1}^{\infty} F_{k,n}\) with each \(F_{k,n}\) closed, \(k \in \mathbf{N}\). Then

\begin{equation*} \bigcup_{k=1}^{\infty} A_k = \bigcup_{(k,n) \in \mathbf{N} \times \mathbf{N}} F_{k,n}, \end{equation*}

a union indexed by a countable set (Theorem 1.5.8 (ii)), hence an \(F_\sigma\) set.

(b) Let \(A_k = \bigcup_{n=1}^{\infty} F_{k,n}\) for \(k = 1, \ldots, m\). Replacing \(F_{k,n}\) by \(E_{k,n} = F_{k,1} \cup \cdots \cup F_{k,n}\) (closed by Theorem 3.2.14 (i)) changes no union and makes each family increasing in \(n\). Then

\begin{equation*} \bigcap_{k=1}^{m} A_k = \bigcup_{n=1}^{\infty} \bigcap_{k=1}^{m} E_{k,n} : \end{equation*}

\(\supseteq\) is clear, and if \(x \in A_k\) for every \(k\), choose \(n_k\) with \(x \in E_{k,n_k}\) and set \(n = \max\{n_1, \ldots, n_m\}\), which exists because there are finitely many \(k\); monotonicity gives \(x \in E_{k,n}\) for all \(k\). Each \(\bigcap_{k=1}^{m} E_{k,n}\) is closed by Theorem 3.2.14 (ii), so the right-hand side is an \(F_\sigma\) set.

Countably many will not do in (b): each

\begin{equation*} \mathbf{R} \setminus \{q\} = \bigcup_{n=2}^{\infty}\Big(\big[q-n,\, q-\tfrac1n\big] \cup \big[q+\tfrac1n,\, q+n\big]\Big) \end{equation*}

is \(F_\sigma\), yet \(\mathbf{I} = \bigcap_{q \in \mathbf{Q}}(\mathbf{R} \setminus \{q\})\) is not (Exercise 3.5.6).

(c), (d) Take complements. By Exercise 3.5.1, \(B_k\) is \(G_\delta\) exactly when \(B_k^c\) is \(F_\sigma\), and

\begin{equation*} \Big(\bigcup_{k} B_k\Big)^{c} = \bigcap_{k} B_k^{c}, \qquad \Big(\bigcap_{k} B_k\Big)^{c} = \bigcup_{k} B_k^{c}. \end{equation*}

So (c) is the complementary statement of (b) (finite) and (d) is the complementary statement of (a) (countable). Countably many sets fail in (c) for the dual reason: \(\mathbf{Q} = \bigcup_{q \in \mathbf{Q}} \{q\}\) with \(\{q\} = \bigcap_{n=1}^{\infty}(q - \tfrac1n, q + \tfrac1n)\) a \(G_\delta\) set, yet \(\mathbf{Q}\) is not \(G_\delta\) (Exercise 3.5.6).

(a) \(\displaystyle [a,b] = \bigcap_{n=1}^{\infty}\Big(a - \tfrac1n,\; b + \tfrac1n\Big)\), an intersection of open sets. Indeed \([a,b]\) lies in every one of them, and if \(x\) is in every one then \(a - \tfrac1n < x < b + \tfrac1n\) for all \(n\), so \(a \le x \le b\) by the Order Limit Theorem (Theorem 2.3.4) applied to \(a - \tfrac1n \to a\) and \(b + \tfrac1n \to b\).

(b) For the \(G_\delta\) form,

\begin{equation*} (a,b] = \bigcap_{n=1}^{\infty}\Big(a,\; b + \tfrac1n\Big), \end{equation*}

by the same squeeze on the right endpoint. For the \(F_\sigma\) form,

\begin{equation*} (a,b] = \bigcup_{n=1}^{\infty}\Big[a + \tfrac{b-a}{n},\; b\Big] : \end{equation*}

each such closed interval sits inside \((a,b]\), and given \(x \in (a,b]\) we have \(x - a > 0\), so the Archimedean Property (Theorem 1.4.2) supplies \(n\) with \(\tfrac{b-a}{n} \le x - a\), putting \(x\) in the \(n\)-th interval.

(c) \(\mathbf{Q}\) is countable (Theorem 1.5.6 (i)), so \(\mathbf{Q} = \bigcup_{q \in \mathbf{Q}} \{q\}\) is a countable union of the closed sets \(\{q\}\) (a single point has no limit points, so it vacuously contains them, Definition 3.2.7), hence \(F_\sigma\). Since \(\mathbf{I} = \mathbf{Q}^c\), Exercise 3.5.1 makes \(\mathbf{I}\) a \(G_\delta\) set. Explicitly, enumerating \(\mathbf{Q} = \{q_1, q_2, q_3, \ldots\}\),

\begin{equation*} \mathbf{I} = \bigcap_{n=1}^{\infty} \big(\mathbf{R} \setminus \{q_n\}\big). \end{equation*}

Build each closed \(I_{n+1}\) strictly inside the interior of \(I_n\), giving it half the radius of the neighborhood that produced it; that halving is what pulls the endpoints inside, and it is the special attention the endpoints require.

Base case: \(G_1\) is dense, hence nonempty, so pick \(x_1 \in G_1\); openness (Definition 3.2.1) gives \(\varepsilon_1 > 0\) with \((x_1 - \varepsilon_1, x_1 + \varepsilon_1) \subseteq G_1\). Set

\begin{equation*} I_1 = \big[x_1 - \tfrac{\varepsilon_1}{2},\; x_1 + \tfrac{\varepsilon_1}{2}\big] \subseteq G_1 . \end{equation*}

Inductive step: suppose \(I_n = [a_n, b_n]\) with \(a_n < b_n\) has been built. Since \(G_{n+1}\) is dense, there is a point \(x_{n+1} \in G_{n+1}\) with \(a_n < x_{n+1} < b_n\). The set \(G_{n+1} \cap (a_n, b_n)\) is open (Theorem 3.2.3 (ii)) and contains \(x_{n+1}\), so there is \(\varepsilon_{n+1} > 0\) with

\begin{equation*} (x_{n+1} - \varepsilon_{n+1},\; x_{n+1} + \varepsilon_{n+1}) \subseteq G_{n+1} \cap (a_n, b_n). \end{equation*}

Set \(I_{n+1} = \big[x_{n+1} - \tfrac{\varepsilon_{n+1}}{2},\; x_{n+1} + \tfrac{\varepsilon_{n+1}}{2}\big]\), so

\begin{equation*} I_{n+1} \subseteq G_{n+1} \qquad\text{and}\qquad I_{n+1} \subseteq (a_n, b_n) \subseteq I_n . \end{equation*}

The \(I_n\) are nonempty closed bounded intervals with \(I_1 \supseteq I_2 \supseteq \cdots\), so the Nested Interval Property (Theorem 1.4.1) yields a point

\begin{equation*} x \in \bigcap_{n=1}^{\infty} I_n . \end{equation*}

For every \(n\) we have \(x \in I_n \subseteq G_n\), so \(x \in \bigcap_{n=1}^{\infty} G_n\) and the intersection is not empty.

Pass to complements and apply Theorem 3.5.2.

Assume such a decomposition exists and set \(G_n = F_n^{c}\), which is open by Theorem 3.2.13.

Each \(G_n\) is dense: given \(a < b\), the interval \((a,b)\) is a nonempty open interval, so \((a,b) \not\subseteq F_n\) by hypothesis; any \(x \in (a,b) \setminus F_n\) is a point of \(G_n\) with \(a < x < b\).

Theorem 3.5.2 applies to the countable collection \(\{G_1, G_2, G_3, \ldots\}\) of dense open sets and gives \(\bigcap_{n=1}^{\infty} G_n \ne \emptyset\). But De Morgan turns the assumed decomposition into

\begin{equation*} \bigcap_{n=1}^{\infty} G_n = \bigcap_{n=1}^{\infty} F_n^{c} = \Big(\bigcup_{n=1}^{\infty} F_n\Big)^{c} = \mathbf{R}^{c} = \emptyset . \end{equation*}

An \(F_\sigma\) decomposition of \(\mathbf{I}\) would combine with the one of \(\mathbf{Q}\) to split \(\mathbf{R}\) in the manner forbidden by Exercise 3.5.5.

Suppose \(\mathbf{I} = \bigcup_{n=1}^{\infty} F_n\) with each \(F_n\) closed. No \(F_n\) contains a nonempty open interval: any such interval \((a,b)\) contains a rational by Theorem 1.4.3 (Density of \(\mathbf{Q}\) in \(\mathbf{R}\)), while \(F_n \subseteq \mathbf{I}\). Enumerate \(\mathbf{Q} = \{q_1, q_2, q_3, \ldots\}\) (Theorem 1.5.6 (i)); each singleton \(\{q_k\}\) is closed and plainly contains no nonempty open interval. Then

\begin{equation*} \mathbf{R} = \mathbf{I} \cup \mathbf{Q} = \Big(\bigcup_{n=1}^{\infty} F_n\Big) \cup \Big(\bigcup_{k=1}^{\infty} \{q_k\}\Big) \end{equation*}

exhibits \(\mathbf{R}\) as a countable union (Theorem 1.5.8 (ii)) of closed sets none of which contains a nonempty open interval, contradicting Exercise 3.5.5. Hence \(\mathbf{I}\) is not \(F_\sigma\).

If \(\mathbf{Q}\) were \(G_\delta\), then \(\mathbf{Q}^{c} = \mathbf{I}\) would be \(F_\sigma\) by Exercise 3.5.1, which we have just ruled out.

Take

\begin{equation*} A = \big(\mathbf{Q} \cap (0,\infty)\big) \;\cup\; \big(\mathbf{I} \cap (-\infty,0)\big), \end{equation*}

the rationals to the right of the origin together with the irrationals to the left.

Lemma: neither \(\mathbf{I} \cap (-\infty,0)\) nor \(\mathbf{I} \cap (0,\infty)\) is an \(F_\sigma\) set.

Reflection \(x \mapsto -x\) carries \(V_\varepsilon(x)\) onto \(V_\varepsilon(-x)\), so it matches limit points with limit points and \(-F = \{-x : x \in F\}\) is closed whenever \(F\) is (Definition 3.2.7); also \(-\mathbf{I} = \mathbf{I}\). So if \(\mathbf{I} \cap (-\infty,0) = \bigcup_{n=1}^{\infty} F_n\) with each \(F_n\) closed, then \(\mathbf{I} \cap (0,\infty) = \bigcup_{n=1}^{\infty}(-F_n)\), and since \(0 \notin \mathbf{I}\),

\begin{equation*} \mathbf{I} = \bigcup_{n=1}^{\infty} F_n \;\cup\; \bigcup_{n=1}^{\infty} (-F_n) \end{equation*}

would be \(F_\sigma\), contradicting Exercise 3.5.6. Reflecting once more gives the same conclusion for \(\mathbf{I} \cap (0,\infty)\).

The set \(A\) is not \(F_\sigma\). The ray \((-\infty,0) = \bigcup_{n=1}^{\infty}\big[-n, -\tfrac1n\big]\) is an \(F_\sigma\) set. If \(A\) were \(F_\sigma\), then by the finite-intersection statement of Exercise 3.5.2 (b),

\begin{equation*} A \cap (-\infty,0) = \mathbf{I} \cap (-\infty,0) \end{equation*}

would be \(F_\sigma\), contradicting the Lemma.

The set \(A\) is not \(G_\delta\). If it were, \(A^{c}\) would be \(F_\sigma\) by Exercise 3.5.1. For \(x > 0\) we have \(x \in A\) precisely when \(x \in \mathbf{Q}\), so \(A^{c} \cap (0,\infty) = \mathbf{I} \cap (0,\infty)\). Since \((0,\infty) = \bigcup_{n=1}^{\infty}\big[\tfrac1n, n\big]\) is \(F_\sigma\), Exercise 3.5.2 (b) would make \(\mathbf{I} \cap (0,\infty)\) an \(F_\sigma\) set, again contradicting the Lemma.

Both conditions say: every interval \((a,b)\) with \(a<b\) contains a point of \(\overline{E}^{\,c}\).

Indeed, Definition 3.5.3 makes \(E\) nowhere-dense precisely when \(\overline{E}\) contains no nonempty open interval, and the definition of density (p. 107) makes a set \(G\) dense precisely when every interval \((a,b)\) with \(a<b\) meets \(G\). Hence

\begin{equation*} \begin{aligned} E \text{ is nowhere-dense} &\iff \neg\,\exists\,a<b \text{ with } (a,b)\subseteq\overline{E}\\ &\iff \forall\,a<b\ \exists\,x\in(a,b) \text{ with } x\notin\overline{E}\\ &\iff \forall\,a<b\ \exists\,x\in(a,b) \text{ with } x\in\overline{E}^{\,c}\\ &\iff \overline{E}^{\,c} \text{ is dense in } \mathbf{R}. \end{aligned} \end{equation*}

The second step is just the negation of a containment: \((a,b)\not\subseteq\overline{E}\) says exactly that some point of \((a,b)\) escapes \(\overline{E}\).

(a) in between; (b) nowhere-dense; (c) dense; (d) nowhere-dense. In each case the verdict is read off the closure, using that \(G\) is dense iff \(\overline{G}=\mathbf{R}\) and nowhere-dense iff \(\overline{G}\) swallows no nonempty open interval (Definition 3.5.3).

(a) \(\overline{A}=[0,5]\): the containment \(\subseteq\) holds because \([0,5]\) is closed (Example 3.2.9 (ii)) and \(\overline{A}\) is the smallest closed set containing \(A\) (Theorem 3.2.12), while \(\supseteq\) holds because each \(x\in[0,5]\) is a limit of rationals drawn from \([0,5]\), by the density of \(\mathbf{Q}\) (Theorem 1.4.3) with Theorem 3.2.5. Now \(\overline{A}\supseteq(0,5)\) rules out nowhere-dense and \(\overline{A}\neq\mathbf{R}\) rules out dense.

(b) \(\overline{B}=B\cup\{0\}\), since \(0\) is the only limit point of \(B\) (Example 3.2.9 (i)). Every element of \(\overline{B}\) is rational, while Corollary 1.4.4 puts an irrational inside each interval \((a,b)\) with \(a<b\); so no such interval lies in \(\overline{B}\), and \(B\) is nowhere-dense.

(c) Given \(a<b\), Corollary 1.4.4 supplies an irrational \(t\) with \(a<t<b\), which is the definition of dense verbatim; so the set \(\mathbf{I}\) of irrationals has \(\overline{\mathbf{I}}=\mathbf{R}\).

(d) \(C=\bigcap_{n\ge 0}C_n\) is an arbitrary intersection of closed sets, hence closed (Theorem 3.2.14 (ii)), so \(\overline{C}=C\). Suppose \((a,b)\subseteq C\) with \(a<b\). For every \(n\in\mathbf{N}\) we then have \((a,b)\subseteq C_n\), a union of \(2^n\) closed intervals of length \(3^{-n}\), any two of which are separated by a deleted open interval; an interval meeting two of these components would contain a point of the gap between them, so \((a,b)\) sits inside one component and \(b-a\le 3^{-n}\le 1/n\). Taking the \(n\) that Theorem 1.4.2 (ii) provides for the positive number \(b-a\) gives \(1/n<b-a\), a contradiction. Hence \(\overline{C}\) contains no nonempty open interval.

Take \(G_n=\overline{E_n}^{\,c}\) and feed the collection \(\{G_1,G_2,G_3,\dots\}\) into Theorem 3.5.2.

Its three hypotheses hold: the collection is countable, being indexed by \(\mathbf{N}\); each \(G_n\) is open, because \(\overline{E_n}\) is a closed set (Theorem 3.2.12) and the complement of a closed set is open (Theorem 3.2.13); and each \(G_n\) is dense in \(\mathbf{R}\), because \(E_n\) is nowhere-dense and Exercise 3.5.8 identifies that with density of the complement of \(\overline{E_n}\).

Theorem 3.5.2 therefore supplies a point \(x\in\bigcap_{n=1}^{\infty}G_n\). For this \(x\),

\begin{equation*} \begin{aligned} x\in\bigcap_{n=1}^{\infty}G_n &\implies x\notin\overline{E_n}\ \text{ for every } n\in\mathbf{N}\\ &\implies x\notin E_n\ \text{ for every } n\in\mathbf{N}\\ &\implies x\notin\bigcup_{n=1}^{\infty}E_n=\mathbf{R}, \end{aligned} \end{equation*}

the middle step because \(E_n\subseteq\overline{E_n}\) (Definition 3.2.11). But \(x\) is a real number, so \(x\in\mathbf{R}\). This contradiction refutes the assumption that \(\mathbf{R}\) is a countable union of nowhere-dense sets.

(a) Throughout, \(f,g : A \to \mathbf{R}\), \(c\) is a limit point of \(A\), \(\lim_{x \to c} f(x) = L\), and \(\lim_{x \to c} g(x) = M\). Let \((x_n) \subseteq A\) be arbitrary with \(x_n \ne c\) and \(x_n \to c\). Applying the forward implication (i) \(\Rightarrow\) (ii) of Theorem 4.2.3 to \(f\) and then to \(g\) gives \(f(x_n) \to L\) and \(g(x_n) \to M\), so by the Algebraic Limit Theorem for sequences (Theorem 2.3.3 (ii)),

\begin{equation*} (f+g)(x_n) = f(x_n) + g(x_n) \longrightarrow L + M . \end{equation*}

The sequence \((x_n)\) was arbitrary, so the reverse implication (ii) \(\Rightarrow\) (i) of Theorem 4.2.3, applied to the function \(f + g\) and the candidate limit \(L+M\), yields \(\lim_{x \to c} [f(x) + g(x)] = L + M\).

(b) Let \(\epsilon > 0\). Definition 4.2.1 applied to \(f\) with the challenge \(\epsilon/2\) produces \(\delta_1 > 0\) such that \(0 < |x - c| < \delta_1\) (and \(x \in A\)) implies \(|f(x) - L| < \epsilon/2\); applied to \(g\) with the same challenge it produces \(\delta_2 > 0\) with \(0 < |x-c| < \delta_2\) implying \(|g(x) - M| < \epsilon/2\). Set \(\delta = \min\{\delta_1, \delta_2\} > 0\). For \(x \in A\) with \(0 < |x - c| < \delta\) both estimates are in force, so

\begin{equation*} \begin{aligned} |(f+g)(x) - (L+M)| &\le |f(x) - L| + |g(x) - M|\\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

(c) Sequentially: with \((x_n)\) as in (a), Theorem 4.2.3 gives \(f(x_n) \to L\) and \(g(x_n) \to M\), and Theorem 2.3.3 (iii) gives \(f(x_n)g(x_n) \to LM\); since \((x_n)\) was arbitrary, Theorem 4.2.3 returns \(\lim_{x \to c} f(x)g(x) = LM\).

Directly: the algebraic identity

\begin{equation*} \begin{aligned} |f(x)g(x) - LM| &= |f(x)g(x) - f(x)M + f(x)M - LM|\\ &\le |f(x)|\,|g(x) - M| + |M|\,|f(x) - L| \end{aligned} \end{equation*}

reduces the problem to bounding \(|f(x)|\). Choose \(\delta_0 > 0\) so that \(0 < |x-c| < \delta_0\) implies \(|f(x) - L| < 1\), whence \(|f(x)| < |L| + 1\) there. Now let \(\epsilon > 0\) and choose \(\delta_1, \delta_2 > 0\) with

\begin{equation*} |g(x) - M| < \frac{\epsilon}{2(|L|+1)}, \qquad |f(x) - L| < \frac{\epsilon}{2(|M|+1)} \end{equation*}

whenever \(0 < |x - c| < \delta_1\), respectively \(0 < |x-c| < \delta_2\). With \(\delta = \min\{\delta_0, \delta_1, \delta_2\}\) and \(0 < |x - c| < \delta\),

\begin{equation*} \begin{aligned} |f(x)g(x) - LM| &< (|L|+1)\frac{\epsilon}{2(|L|+1)} + |M|\frac{\epsilon}{2(|M|+1)}\\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

(a) \(\delta = 1/5\). (b) \(\delta = 3\). (c) \(\delta = \pi - 3\). (d) \(\delta = \pi - 3\).

In each part, \(S = \{x : |f(x) - L| < \epsilon\}\) is computed exactly and \(\delta\) is the distance from \(c\) to the nearest point outside \(S\).

(a) \(|(5x-6) - 9| = 5|x - 3| < 1\) holds precisely when \(|x - 3| < 1/5\), so \(S = V_{1/5}(3)\) and the largest response is \(\delta = 1/5\).

(b) \(|\sqrt{x} - 2| < 1\) says \(1 < \sqrt{x} < 3\), i.e. \(S = (1,9)\). The largest \(\delta\)-neighborhood of \(4\) inside \((1,9)\) has radius \(\min\{4 - 1,\ 9 - 4\} = 3\), so \(\delta = 3\).

(c) Since \([[x]]\) takes only integer values, \(|[[x]] - 3| < 1\) forces \([[x]] = 3\), i.e. \(S = [3,4)\). Then

\begin{equation*} \delta = \min\{\pi - 3,\ 4 - \pi\} = \pi - 3 \approx 0.14159, \end{equation*}

since \(\pi - 3 < 1/2 < 4 - \pi\).

(d) The condition \(|[[x]] - 3| < .01\) again forces the integer \([[x]]\) to equal \(3\), so \(S = [3,4)\) and \(\delta = \pi - 3\) are unchanged: \([[x]]\) attains no value strictly between \(3\) and \(4\) for the smaller \(\epsilon\) to exclude.

(a) Take

\begin{equation*} x_n = 1 + \frac{1}{n}, \qquad y_n = 1 + \frac{\sqrt{2}}{n}, \qquad z_n = 1 - \frac{1}{n+1}, \end{equation*}

all of which converge to \(1\) and none of whose terms equals \(1\).

(b) All three limits are \(0\). Indeed \(x_n = (n+1)/n\) with \(\gcd(n+1,n) = 1\), so \(t(x_n) = 1/n \to 0\); each \(y_n\) is irrational (a rational \(y_n\) would make \(\sqrt{2} = n(y_n - 1)\) rational), so \(t(y_n) = 0\) for every \(n\); and \(z_n = n/(n+1)\) with \(\gcd(n, n+1) = 1\), so \(t(z_n) = 1/(n+1) \to 0\).

(c) Conjecture: \(\lim_{x \to 1} t(x) = 0\).

Fix \(\epsilon > 0\) and put \(S_\epsilon = \{x \in \mathbf{R} : t(x) \ge \epsilon\}\). If \(x \in S_\epsilon\) then \(t(x) > 0\), so \(x\) is rational; writing \(x = m/n\) in lowest terms with \(n > 0\) (and \(0 = 0/1\)), the value \(t(x) = 1/n\) satisfies \(1/n \ge \epsilon\), i.e.

\begin{equation*} n \le 1/\epsilon . \end{equation*}

Hence

\begin{equation*} S_\epsilon \cap (0,2) \subseteq \Big\{\tfrac{m}{n} : 1 \le n \le 1/\epsilon,\ 0 < m < 2n \Big\}, \end{equation*}

a finite set: at most \(\lfloor 1/\epsilon \rfloor\) denominators are admissible and fewer than \(2n\) numerators for each. The same count on any bounded interval shows every point of \(S_\epsilon\) is isolated.

Now let \(F = \big(S_\epsilon \cap (0,2)\big) \setminus \{1\}\), a finite set not containing \(1\), and set

\begin{equation*} \delta = \min\big(\{1\} \cup \{|x - 1| : x \in F\}\big) > 0, \end{equation*}

a minimum of finitely many strictly positive numbers. If \(x \in V_\delta(1)\) and \(x \ne 1\), then \(x \in (0,2)\) and \(x \notin F\), so \(x \notin S_\epsilon\); that is, \(t(x) < \epsilon\), and since \(t \ge 0\) always, \(t(x) \in V_\epsilon(0)\). This is exactly Definition 4.2.1B for \(\lim_{x \to 1} t(x) = 0\).

(a) \(\delta = 8\). (b) \(\delta = 1\). (c) \(\epsilon = 1/90\).

Everything follows from one table. The domain of \(f(x) = 1/[[x]]\) excludes \([0,1)\), and on the block \([n, n+1)\) with \(n \in \mathbf{Z}\setminus\{0\}\) the function is constantly \(1/n\), so the error is constantly

\begin{equation*} E(n) = \left|\frac{1}{n} - \frac{1}{10}\right| = \frac{|10 - n|}{10\,|n|} . \end{equation*}

The relevant values are

and \(E\) decreases on \(1 \le n \le 10\) and increases on \(n \ge 10\), with \(E(n) < 1/10\) for every \(n \ge 10\).

(a) With \(\epsilon = 1/2\) the blocks that fail are exactly those with \(E(n) \ge 1/2\), and by the monotonicity just noted that means \(n = 1\) alone: \(E(1) = 9/10\) while \(E(2) = 2/5 < 1/2\). So the admissible \(x\) form \([2,\infty)\), and the largest \(\delta\)-neighborhood of \(10\) inside it is \(V_8(10) = (2,18)\). Thus \(\delta = 8\); any larger \(\delta\) admits a point of \([1,2)\), where the error is \(9/10\).

(b) With \(\epsilon = 1/50 = 0.02\) the block \(n = 9\) survives (\(1/90 < 1/50\)) but \(n = 8\) fails (\(1/40 > 1/50\)), forcing \(10 - \delta \ge 9\); on the right \(n = 12\) survives (\(1/60 < 1/50\)) but \(n = 13\) fails (\(3/130 > 1/50\)), forcing \(10 + \delta \le 13\). Hence \(\delta \le \min\{1, 3\} = 1\), and \(\delta = 1\) does work since \(V_1(10) = (9,11)\) meets only the blocks \(n = 9, 10\).

(c) Every punctured neighborhood of \(10\) contains points of \([9,10)\), where the error is the fixed number \(1/90\). So a \(\delta\) response exists if and only if \(\epsilon > 1/90\), and the challenges that cannot be met are exactly \(0 < \epsilon \le 1/90\). The largest of these is

\begin{equation*} \epsilon = \frac{1}{90} . \end{equation*}

In each part let \(\epsilon > 0\) be given; the displayed \(\delta\) is the response demanded by Definition 4.2.1.

(a) Take \(\delta = \epsilon/3\). If \(0 < |x - 2| < \delta\) then

\begin{equation*} |(3x+4) - 10| = 3|x - 2| < 3\cdot\tfrac{\epsilon}{3} = \epsilon . \end{equation*}

(b) Take \(\delta = \epsilon^{1/3}\). If \(0 < |x - 0| < \delta\) then

\begin{equation*} |x^3 - 0| = |x|^3 < \delta^3 = \epsilon . \end{equation*}

(c) Take \(\delta = \min\{1,\ \epsilon/6\}\). Factoring,

\begin{equation*} |(x^2 + x - 1) - 5| = |x^2 + x - 6| = |x + 3|\,|x - 2| , \end{equation*}

and \(|x - 2| < \delta \le 1\) gives \(1 < x < 3\), hence \(|x + 3| < 6\). Therefore \(0 < |x - 2| < \delta\) implies

\begin{equation*} |x + 3|\,|x - 2| < 6\cdot\tfrac{\epsilon}{6} = \epsilon . \end{equation*}

(d) Take \(\delta = \min\{1,\ 6\epsilon\}\). For \(x \ne 0\),

\begin{equation*} \left|\frac{1}{x} - \frac{1}{3}\right| = \frac{|3 - x|}{3|x|} , \end{equation*}

and \(|x - 3| < \delta \le 1\) gives \(x > 2\), so \(3|x| > 6\) and the fraction is at most \(|x - 3|/6\). Hence \(0 < |x-3| < \delta\) implies

\begin{equation*} \left|\frac{1}{x} - \frac{1}{3}\right| < \frac{6\epsilon}{6} = \epsilon . \end{equation*}

(a) True. (b) False. (c) True. (d) False.

(a) If \(0 < \delta^{\prime} \le \delta\) and \(0 < |x - a| < \delta^{\prime}\), then \(0 < |x - a| < \delta\), so the conclusion \(|f(x) - L| < \epsilon\) already guaranteed by \(\delta\) applies verbatim.

(b) False. Definition 4.2.1 constrains \(f\) only at points with \(0 < |x - a|\), so the value \(f(a)\) is invisible to the limit. Take

\begin{equation*} f(x) = \begin{cases} 0 & x \ne 0\\ 1 & x = 0,\end{cases} \end{equation*}

for which \(\lim_{x \to 0} f(x) = 0\) (any \(\delta\) works) while \(f(0) = 1\).

(c) True, by three applications of Corollary 4.2.4. The constant function \(-2\) has \(\lim_{x \to a}(-2) = -2\) straight from Definition 4.2.1 (any \(\delta\) responds), so part (ii) gives \(f(x) - 2 \to L - 2\); part (iii) applied to this function against itself gives \([f(x)-2]^2 \to (L-2)^2\); part (i) with \(k = 3\) gives \(3[f(x)-2]^2 \to 3(L-2)^2\).

(d) False, because an unbounded \(g\) can outrun \(f\). On \(A = \mathbf{R}\setminus\{0\}\) (for which \(a = 0\) is a limit point) let \(f(x) = x\) and \(g(x) = 1/x\). Then \(\lim_{x \to 0} f(x) = 0\), yet \(f(x)g(x) = 1\) for every \(x \in A\), so \(\lim_{x \to 0} f(x)g(x) = 1 \ne 0\).

Respond to \(\epsilon\) with the \(\delta\) that \(g\) supplies for the challenge \(\epsilon/M\).

Let \(\epsilon > 0\). Since \(M > 0\), the number \(\epsilon/M\) is a legitimate challenge, so Definition 4.2.1 applied to \(\lim_{x \to c} g(x) = 0\) furnishes \(\delta > 0\) such that \(x \in A\) and \(0 < |x - c| < \delta\) imply \(|g(x) - 0| < \epsilon/M\). For such \(x\),

\begin{equation*} \begin{aligned} |g(x)f(x) - 0| &= |g(x)|\,|f(x)|\\ &\le |g(x)|\,M\\ &< \frac{\epsilon}{M}\cdot M = \epsilon , \end{aligned} \end{equation*}

the middle step using \(|f(x)| \le M\), which holds at every point of \(A\).

(a) Does not exist. Put \(x_n=2+\frac1n\) and \(y_n=2-\frac1n\); both lie in the domain, avoid \(2\), and converge to \(2\), while

\begin{equation*} \frac{|x_n-2|}{x_n-2}=\frac{1/n}{1/n}=1, \qquad \frac{|y_n-2|}{y_n-2}=\frac{1/n}{-1/n}=-1 . \end{equation*}

Since \(1\neq -1\), the Divergence Criterion (Corollary 4.2.5) applies.

(b) The limit is \(-1\). For every \(x\) with \(\frac32<x<2\) we have \(|x-2|=2-x\), hence \(\frac{|x-2|}{x-2}=-1\) identically on that interval. So given \(\epsilon>0\), take \(\delta=\frac14\): if \(0<|x-\frac74|<\frac14\) then \(\frac32<x<2\) and

\begin{equation*} \left|\frac{|x-2|}{x-2}-(-1)\right|=0<\epsilon . \end{equation*}

(c) Does not exist. Take \(x_n=\frac1{2n}\) and \(y_n=\frac1{2n+1}\), both nonzero with \(x_n\to 0\) and \(y_n\to 0\). Then \([{[}1/x_n{]}]=2n\) and \([{[}1/y_n{]}]=2n+1\), so

\begin{equation*} (-1)^{[{[}1/x_n{]}]}=1\to 1, \qquad (-1)^{[{[}1/y_n{]}]}=-1\to -1 . \end{equation*}

Corollary 4.2.5 gives divergence.

(d) The limit is \(0\). The factor \(f(x)=(-1)^{[{[}1/x{]}]}\) is bounded by \(M=1\) on \(\mathbf{R}\setminus\{0\}\), and \(g(x)=\sqrt[3]{x}\) has \(\lim_{x\to 0}g(x)=0\) — given \(\epsilon>0\), take \(\delta=\epsilon^{3}\):

\begin{equation*} 0<|x|<\delta \ \Longrightarrow\ |\sqrt[3]{x}-0|=|x|^{1/3}<\delta^{1/3}=\epsilon . \end{equation*}

Exercise 4.2.7 then gives \(\lim_{x\to 0}g(x)f(x)=0\).

(a) Take \(\delta=1/\sqrt{M}\). Given \(M>0\), if \(0<|x-0|<\delta\) then \(x\neq 0\) and \(x^{2}<\delta^{2}=1/M\), whence

\begin{equation*} \frac{1}{x^{2}}>M . \end{equation*}

(b) Definition: for \(f\) defined on a domain unbounded above, \(\lim_{x\to\infty}f(x)=L\) means that for all \(\epsilon>0\) there exists \(M>0\) such that whenever \(x>M\) (and \(x\) is in the domain of \(f\)), it follows that \(|f(x)-L|<\epsilon\).

For \(f(x)=1/x\) on \((0,\infty)\) and a given \(\epsilon>0\), take \(M=1/\epsilon\):

\begin{equation*} x>M>0 \ \Longrightarrow\ \left|\frac1x-0\right|=\frac1x<\frac1M=\epsilon . \end{equation*}

(c) Definition: \(\lim_{x\to\infty}f(x)=\infty\) means that for all \(M>0\) there exists \(N>0\) such that whenever \(x>N\), it follows that \(f(x)>M\). Example: \(\lim_{x\to\infty}x^{2}=\infty\), since \(N=\sqrt{M}\) answers the challenge \(M>0\) — if \(x>\sqrt{M}\) then \(x^{2}>M\).

(a) Let \(f:A\to\mathbf{R}\). Then \(\lim_{x\to a^{+}}f(x)=L\) means: for all \(\epsilon>0\) there exists \(\delta>0\) such that whenever \(a<x<a+\delta\) (and \(x\in A\)), it follows that \(|f(x)-L|<\epsilon\). Likewise \(\lim_{x\to a^{-}}f(x)=M\) means: for all \(\epsilon>0\) there exists \(\delta>0\) such that whenever \(a-\delta<x<a\) (and \(x\in A\)), it follows that \(|f(x)-M|<\epsilon\). (Each statement presumes \(a\) is a limit point of \(A\cap(a,\infty)\), resp. \(A\cap(-\infty,a)\), so it has content.)

(b) Everything rests on the set identity

\begin{equation*} \{x:0<|x-a|<\delta\}=(a,a+\delta)\cup(a-\delta,a). \end{equation*}

\((\Rightarrow)\) Assume \(\lim_{x\to a}f(x)=L\) and let \(\epsilon>0\). Definition 4.2.1 supplies \(\delta>0\) with \(0<|x-a|<\delta\Rightarrow|f(x)-L|<\epsilon\). The same \(\delta\) answers both one-sided challenges, since each of \(a<x<a+\delta\) and \(a-\delta<x<a\) forces \(0<|x-a|<\delta\).

\((\Leftarrow)\) Assume both one-sided limits equal \(L\) and let \(\epsilon>0\). Pick \(\delta_{1}>0\) from the right-hand definition and \(\delta_{2}>0\) from the left-hand definition, and set \(\delta=\min\{\delta_{1},\delta_{2}\}>0\). If \(0<|x-a|<\delta\), the identity above puts \(x\) in \((a,a+\delta)\subseteq(a,a+\delta_{1})\) or in \((a-\delta,a)\subseteq(a-\delta_{2},a)\), and in either case

\begin{equation*} |f(x)-L|<\epsilon . \end{equation*}

Take \(\delta=\min\{\delta_{1},\delta_{2}\}\), where \(\delta_{1},\delta_{2}\) answer the challenge \(\epsilon\) for \(f\) and for \(h\) respectively via Definition 4.2.1 (legitimate, since \(c\) is a limit point of \(A\)). For \(x\in A\) with \(0<|x-c|<\delta\) both estimates hold at once, so

\begin{equation*} L-\epsilon<f(x)\le g(x)\le h(x)<L+\epsilon , \end{equation*}

that is, \(|g(x)-L|<\epsilon\). Hence \(\lim_{x\to c}g(x)=L\).

Method (2): Let \((x_{n})\subseteq A\) with \(x_{n}\neq c\) and \(x_{n}\to c\). The Sequential Criterion (Theorem 4.2.3) gives \(f(x_{n})\to L\) and \(h(x_{n})\to L\), while \(f(x_{n})\le g(x_{n})\le h(x_{n})\) for every \(n\); the Squeeze Theorem for sequences (Exercise 2.3.3) yields \(g(x_{n})\to L\). As \((x_{n})\) was arbitrary, Theorem 4.2.3 returns \(\lim_{x\to c}g(x)=L\).

(a) Take \(\delta = \epsilon^3\). If \(|x - 0| < \epsilon^3\) then

\begin{equation*} |g(x) - g(0)| = |x|^{1/3} < (\epsilon^3)^{1/3} = \epsilon . \end{equation*}

(b) Take \(\delta = \tfrac{3}{4}|c|^{2/3}\epsilon\). Writing \(a = \sqrt[3]{x}\) and \(b = \sqrt[3]{c}\), the suggested identity gives \(x - c = (a-b)(a^2+ab+b^2)\), and the second factor is bounded below away from \(0\):

\begin{equation*} \begin{aligned} a^2 + ab + b^2 &= \Bigl(a + \tfrac{b}{2}\Bigr)^2 + \tfrac{3}{4}b^2 \\ &\ge \tfrac{3}{4}|c|^{2/3} > 0 , \end{aligned} \end{equation*}

using \(c \ne 0\). Hence for every \(x\),

\begin{equation*} \begin{aligned} |g(x) - g( c)| &= |a - b| = \frac{|x-c|}{a^2+ab+b^2} \\ &\le \frac{|x-c|}{\tfrac{3}{4}|c|^{2/3}} , \end{aligned} \end{equation*}

so \(|x - c| < \delta\) forces \(|g(x) - g( c)| < \epsilon\).

(a) Any constant function, say \(f(x) = 5\); then \(|f(x) - f( c)| = 0 < \epsilon\) for every \(x\), so \(\delta = 1\) always works. (These are the only examples: onetinuity at \(c\) says \(|f(x)-f( c)| < \epsilon\) for all \(\epsilon > 0\) whenever \(|x-c|<1\), i.e. \(f \equiv f( c)\) on \((c-1,c+1)\), and chaining such unit intervals across \(\mathbf{R}\) makes \(f\) constant.)

(b) \(f(x) = x\). Unwinding the definition, \(f\) is equaltinuous at \(c\) exactly when \(|f(x) - f( c)| \le |x - c|\) for all \(x\): letting \(\epsilon \downarrow |x-c|\) in the hypothesis gives the inequality, and the inequality clearly returns the hypothesis. The identity function satisfies it with equality, hence is equaltinuous on \(\mathbf{R}\); and it is constant on no interval, so by the parenthetical in (a) it is nowhere onetinuous.

(c) \(f(x) = 2x\). Given \(\epsilon > 0\) choose \(\delta = \epsilon/2\), which satisfies \(0 < \delta < \epsilon\), and \(|x - c| < \epsilon/2\) gives \(|f(x)-f( c)| = 2|x-c| < \epsilon\). It is nowhere equaltinuous because \(|f(x) - f( c)| = 2|x-c| > |x-c|\) for every \(x \ne c\), violating the criterion in (b).

(d) Yes to both, so the two notions coincide. A lesstinuous \(f\) produces, for each \(\epsilon\), a \(\delta > 0\) meeting the requirement of Definition 4.3.1 — the extra demand \(\delta < \epsilon\) is simply ignored. Conversely, if \(f\) is continuous at \(c\) and \(\epsilon > 0\) is given, let \(\delta\) be as in Definition 4.3.1 and replace it by

\begin{equation*} \delta^{\prime} = \min\{\delta,\ \epsilon/2\} , \end{equation*}

which satisfies \(0 < \delta^{\prime} < \epsilon\); since \(\delta^{\prime} \le \delta\), the implication \(|x-c| < \delta^{\prime} \Rightarrow |f(x)-f( c)| < \epsilon\) still holds.

(a) Let \(\epsilon > 0\). Continuity of \(g\) at \(f( c) \in B\) supplies \(\gamma > 0\) such that

\begin{equation*} y \in B,\ |y - f( c)| < \gamma \;\Longrightarrow\; |g(y) - g(f( c))| < \epsilon , \end{equation*}

and continuity of \(f\) at \(c\), applied with this \(\gamma\) in the role of the challenge, supplies \(\delta > 0\) such that

\begin{equation*} x \in A,\ |x - c| < \delta \;\Longrightarrow\; |f(x) - f( c)| < \gamma . \end{equation*}

Because \(f(A) \subseteq B\), the value \(y = f(x)\) is a legitimate input for the first implication, and the two combine: for \(x \in A\) with \(|x-c| < \delta\),

\begin{equation*} |g(f(x)) - g(f( c))| < \epsilon . \end{equation*}

(b) Let \((x_n) \to c\) with \((x_n) \subseteq A\). Theorem 4.3.2 (iii) applied to \(f\) at \(c\) gives \(f(x_n) \to f( c)\). The terms \(f(x_n)\) lie in \(B\) and \(f( c) \in B\), so Theorem 4.3.2 (iii) applied to \(g\) at the point \(f( c)\) gives

\begin{equation*} g(f(x_n)) \to g(f( c)) . \end{equation*}

Since \((x_n) \to c\) in \(A\) was arbitrary, the same theorem read in the other direction shows \(g \circ f\) is continuous at \(c\).

(a) Take \(p = q = r = 0\) with

\begin{equation*} f(x) = 0 \quad \text{for all } x, \qquad g(y) = \begin{cases} 0 & \text{if } y \ne 0 \\ 1 & \text{if } y = 0. \end{cases} \end{equation*}

Then \(\lim_{x \to 0} f(x) = 0 = q\) and \(\lim_{y \to 0} g(y) = 0 = r\) (the value \(g(0)\) is irrelevant to the limit, by Definition 4.2.1). But \(g(f(x)) = g(0) = 1\) for every \(x\), so \(\lim_{x \to 0} g(f(x)) = 1 \ne r\).

(b) Continuity of \(f\) at \(p\) and of \(g\) at \(q\) turns the two hypotheses into \(f(p) = q\) and \(g(q) = r\). Theorem 4.3.9 makes \(g \circ f\) continuous at \(p\), so

\begin{equation*} \lim_{x \to p} g(f(x)) = g(f(p)) = g(q) = r . \end{equation*}

(c) Assuming only \(f\) continuous does not suffice: the \(f\) of part (a) is constant, hence continuous, and the conclusion fails there.

Assuming only \(g\) continuous does suffice. Then \(g(q) = r\). Let \((x_n) \to p\) with \(x_n \ne p\); the sequential criterion for functional limits (Theorem 4.2.3) gives \(f(x_n) \to q\), and Theorem 4.3.2 (iii) for \(g\) at \(q\) gives

\begin{equation*} g(f(x_n)) \to g(q) = r . \end{equation*}

As this holds for every such sequence, Theorem 4.2.3 returns \(\lim_{x \to p} g(f(x)) = r\).

Take the \(\delta\) that isolates \(c\). By definition of isolated point there is a \(\delta > 0\) with

\begin{equation*} V_\delta( c) \cap A = \{c\} , \end{equation*}

and this one \(\delta\) answers every \(\epsilon > 0\): if \(x \in A\) satisfies \(|x - c| < \delta\) then \(x = c\), whence

\begin{equation*} |f(x) - f( c)| = 0 < \epsilon . \end{equation*}

(a) Let \(f\) be Dirichlet’s function and \(g = 1 - f\):

\begin{equation*} f(x) = \begin{cases} 1 & x \in \mathbf{Q} \\ 0 & x \notin \mathbf{Q} \end{cases} \qquad g(x) = \begin{cases} 0 & x \in \mathbf{Q} \\ 1 & x \notin \mathbf{Q}. \end{cases} \end{equation*}

Neither is continuous at \(0\) — \(f\) by Exercise 4.3.7 (a), and \(g\) because otherwise \(f = 1 - g\) would be by Theorem 4.3.4 (i) and (ii). But \(f + g \equiv 1\) and \(fg \equiv 0\), since one factor vanishes at each \(x\); constants are continuous.

(b) Impossible. If \(f\) and \(f + g\) were both continuous at \(0\), then by Theorem 4.3.4 (i) and (ii) so is

\begin{equation*} g = (f+g) + (-1)f , \end{equation*}

contradicting the hypothesis on \(g\).

(c) Take \(f \equiv 0\) and \(g\) Dirichlet’s function. Then \(fg \equiv 0\) is continuous at \(0\) while \(g\) is not.

(d) Take

\begin{equation*} f(x) = \begin{cases} 2 & x \in \mathbf{Q} \\ 1/2 & x \notin \mathbf{Q} , \end{cases} \end{equation*}

whose values are the reciprocal pair \(\{2, 1/2\}\), so that

\begin{equation*} f(x) + \frac{1}{f(x)} = 2 + \tfrac{1}{2} = \tfrac{5}{2} \end{equation*}

for every \(x\) — a constant, hence continuous. That \(f\) itself is discontinuous at \(0\) follows as in Exercise 4.3.7 (a): rational and irrational sequences tending to \(0\) give limits \(2\) and \(1/2\).

(e) Impossible. The cube root \(\psi(y) = \sqrt[3]{y}\) is continuous on \(\mathbf{R}\) (Exercise 4.3.1), so if \(h = f^3\) were continuous at \(0\), Theorem 4.3.9 applied to \(\psi \circ h\) would make

\begin{equation*} \psi(h(x)) = \sqrt[3]{[f(x)]^3} = f(x) \end{equation*}

continuous at \(0\).

(a) Fix \(c \in \mathbf{R}\). Because \(\mathbf{Q}\) and \(\mathbf{I}\) are both dense in \(\mathbf{R}\) (Theorem 1.4.3 and Corollary 1.4.4), we may choose \(x_n \in \mathbf{Q} \cap (c - 1/n,\, c + 1/n)\) and \(y_n \in \mathbf{I} \cap (c - 1/n,\, c + 1/n)\), so that \(|x_n - c| < 1/n\) and \(|y_n - c| < 1/n\), whence \((x_n) \to c\) and \((y_n) \to c\). Then

\begin{equation*} g(x_n) = 1 \to 1 , \qquad g(y_n) = 0 \to 0 . \end{equation*}

Since \(g( c)\) cannot equal both \(1\) and \(0\), one of these sequences has \(g\)-image failing to converge to \(g( c)\), and Corollary 4.3.3 declares \(g\) discontinuous at \(c\).

(b) Let \(c \in \mathbf{Q}\), so \(t( c) > 0\) (it is \(1\) if \(c = 0\) and \(1/n > 0\) otherwise). Picking \(y_n \in \mathbf{I}\) with \((y_n) \to c\) exactly as in (a),

\begin{equation*} t(y_n) = 0 \to 0 \ne t( c) , \end{equation*}

so Corollary 4.3.3 gives discontinuity at every rational point.

(c) Let \(c \in \mathbf{I}\), so \(t( c) = 0\), and let \((x_n) \to c\); by Theorem 4.3.2 (iii) it suffices to prove \(t(x_n) \to 0\). Fix \(\epsilon > 0\) and set

\begin{equation*} A_\epsilon = \{x \in \mathbf{R} : t(x) \ge \epsilon\} . \end{equation*}

Every \(x \in A_\epsilon\) is rational, and is either \(0\) or of the form \(m/n\) in lowest terms with \(1/n \ge \epsilon\), i.e. \(1 \le n \le 1/\epsilon\). Consequently

\begin{equation*} F = A_\epsilon \cap [c-1,\, c+1] \end{equation*}

is finite: only finitely many denominators \(n \le 1/\epsilon\) are available, and for each the interval \([c-1,c+1]\) of length \(2\) contains at most \(2n + 1\) fractions \(m/n\).

Since \(c \notin F\) (as \(t( c) = 0 < \epsilon\)), the number

\begin{equation*} \delta = \min\bigl(\{1\} \cup \{|y - c| : y \in F\}\bigr) \end{equation*}

is the minimum of a finite set of strictly positive numbers, hence \(\delta > 0\). Choose \(N\) with \(|x_n - c| < \delta\) for all \(n \ge N\). Such an \(x_n\) lies in \([c-1,c+1]\) and is not a point of \(F\), so \(x_n \notin A_\epsilon\); that is,

\begin{equation*} |t(x_n) - t( c)| = t(x_n) < \epsilon \qquad \text{for all } n \ge N . \end{equation*}

Thus \(t(x_n) \to t( c)\), and \(t\) is continuous at \(c\).

All three are true.

(a) Take \(x_n = 1 - 1/n\). Then \((x_n) \to 1\) with \(x_n < 1\), so \(g(x_n) \ge 0\) for every \(n\), and Theorem 4.3.2 (iii) together with the Order Limit Theorem 2.3.4 gives

\begin{equation*} g(1) \;=\; \lim_{n \to \infty} g(x_n) \;\ge\; 0 . \end{equation*}

(b) Fix \(x \in \mathbf{R}\). By the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3) choose \(r_n \in \mathbf{Q}\) with \(x < r_n < x + 1/n\), so that \((r_n) \to x\). Since \(g\) is continuous at \(x\), Theorem 4.3.2 (iii) yields

\begin{equation*} g(x) \;=\; \lim_{n \to \infty} g(r_n) \;=\; \lim_{n \to \infty} 0 \;=\; 0 . \end{equation*}

(c) Apply continuity at \(x_0\) with the particular choice \(\epsilon = g(x_0) > 0\): there is a \(\delta > 0\) such that

\begin{equation*} |x - x_0| < \delta \;\Longrightarrow\; |g(x) - g(x_0)| < g(x_0) \;\Longrightarrow\; g(x) > 0 . \end{equation*}

Thus \(g\) is strictly positive on the whole interval \(V_\delta(x_0) = (x_0 - \delta,\, x_0 + \delta)\), and \((x_0-\delta, x_0+\delta) \sim \mathbf{R}\) by Exercise 1.5.4 (a), which is uncountable by Theorem 1.5.6 (ii).

\(K\) contains its limit points, which is Definition 3.2.7. If \(x\) is a limit point of \(K\), Theorem 3.2.5 supplies a sequence \((x_n) \subseteq K\) with \((x_n) \to x\), and continuity of \(h\) at \(x\) gives, via Theorem 4.3.2 (iii),

\begin{equation*} h(x) \;=\; \lim_{n \to \infty} h(x_n) \;=\; \lim_{n \to \infty} 0 \;=\; 0 , \end{equation*}

so \(x \in K\).

(a) Induct on \(n\); the case \(n = 1\) is trivial. For \(n = 2\) the displayed identity gives

\begin{equation*} \max\{f_1,f_2\} = \tfrac{1}{2}(f_1+f_2) + \tfrac{1}{2}\bigl|f_1-f_2\bigr| , \end{equation*}

and every piece is continuous: the sums and scalar multiples by Theorem 4.3.4 (i), (ii), and \(|f_1 - f_2| = a \circ (f_1 - f_2)\) by Theorem 4.3.9, where \(a(t) = |t|\) is continuous on all of \(\mathbf{R}\) because \(\bigl||t| - |c|\bigr| \le |t - c|\) makes \(\delta = \epsilon\) work. The inductive step is

\begin{equation*} \max\{f_1,\ldots,f_{n+1}\} = \max\bigl\{\max\{f_1,\ldots,f_n\},\, f_{n+1}\bigr\}. \end{equation*}

(b) The supremum is

\begin{equation*} h(x) \;=\; \begin{cases} 1 & \text{if } x \ne 0, \\ 0 & \text{if } x = 0 . \end{cases} \end{equation*}

Indeed \(f_n(0) = n \cdot 0 = 0\) for every \(n\), so \(h(0) = 0\). If \(x \ne 0\), the Archimedean Property (Theorem 1.4.2) supplies \(N \in \mathbf{N}\) with \(1/N < |x|\), whence \(f_N(x) = 1\); and every \(f_n\) satisfies \(f_n(x) \le 1\) (either \(f_n(x) = 1\), or \(f_n(x) = n|x| < 1\)). Thus \(\sup_n f_n(x) = 1\).

So the extension fails: each \(f_n\) is continuous, but taking \(x_n = 1/n \to 0\) gives \(h(x_n) = 1 \not\to 0 = h(0)\), and \(h\) is discontinuous at \(0\) by Corollary 4.3.3.

(a) Given \(\epsilon > 0\) and \(p \in \mathbf{R}\), take \(\delta = \epsilon\). Then \(|x - p| < \delta\) implies

\begin{equation*} |f(x) - f(p)| \;\le\; c\,|x - p| \;<\; c\,\epsilon \;<\; \epsilon , \end{equation*}

since \(c < 1\). This is Definition 4.3.1, so \(f\) is continuous at every \(p\).

(b) Iterating the contraction bound along the sequence gives

\begin{equation*} |y_{n+1} - y_n| \;=\; |f(y_n) - f(y_{n-1})| \;\le\; c\,|y_n - y_{n-1}| \;\le\; \cdots \;\le\; c^{\,n-1}|y_2 - y_1| . \end{equation*}

Hence for \(m > n\), by the triangle inequality and the geometric series,

\begin{equation*} \begin{aligned} |y_m - y_n| &\le \sum_{k=n}^{m-1} |y_{k+1} - y_k| \;\le\; |y_2 - y_1| \sum_{k=n}^{m-1} c^{\,k-1} \\ &\le\; |y_2 - y_1|\,\frac{c^{\,n-1}}{1 - c} . \end{aligned} \end{equation*}

Since \(0 < c < 1\) we have \(c^{\,n-1} \to 0\) (Example 2.5.3), so given \(\epsilon > 0\) we may pick \(N\) with \(|y_2 - y_1| c^{\,N-1}/(1-c) < \epsilon\), and then \(|y_m - y_n| < \epsilon\) for all \(m > n \ge N\). Thus \((y_n)\) is Cauchy and converges by the Cauchy Criterion (Theorem 2.6.4); set \(y = \lim y_n\).

(c) \(f\) is continuous at \(y\) and \((y_n) \to y\), so \(f(y_n) \to f(y)\) by Theorem 4.3.2 (iii). But \(f(y_n) = y_{n+1}\), and \((y_{n+1})\) is a subsequence of \((y_n)\), hence also converges to \(y\) (Theorem 2.5.2). Uniqueness of limits (Theorem 2.2.7) forces

\begin{equation*} f(y) \;=\; y . \end{equation*}

If \(z\) also satisfies \(f(z) = z\), then

\begin{equation*} |y - z| \;=\; |f(y) - f(z)| \;\le\; c\,|y - z| \quad\Longrightarrow\quad (1 - c)\,|y - z| \le 0 , \end{equation*}

and since \(1 - c > 0\) this gives \(|y - z| = 0\), i.e. \(z = y\).

(d) Let \(x_1 = x\) and \(x_{n+1} = f(x_n)\). Using \(f(y) = y\) and iterating,

\begin{equation*} |x_n - y| \;=\; |f(x_{n-1}) - f(y)| \;\le\; c\,|x_{n-1} - y| \;\le\; \cdots \;\le\; c^{\,n-1}|x - y| . \end{equation*}

Because \(c^{\,n-1} \to 0\), the Squeeze Theorem (Exercise 2.3.3) gives \(|x_n - y| \to 0\), i.e. \((x_n) \to y\).

\(g\) is \(1\)-Lipschitz, so \(\delta = \epsilon\) proves continuity at every point.

(Each infimum exists: \(\{|x-a| : a \in F\}\) is nonempty since \(F \ne \emptyset\) and is bounded below by \(0\), so the greatest-lower-bound form of the Axiom of Completeness — Exercise 1.3.3 — applies.)

Fix \(x, y \in \mathbf{R}\). For every \(a \in F\) the triangle inequality gives

\begin{equation*} g(x) \;\le\; |x - a| \;\le\; |x - y| + |y - a| , \end{equation*}

so \(g(x) - |x-y|\) is a lower bound for \(\{|y - a| : a \in F\}\), and therefore

\begin{equation*} g(x) - |x - y| \;\le\; g(y) , \qquad\text{i.e.}\qquad g(x) - g(y) \;\le\; |x - y| . \end{equation*}

Interchanging \(x\) and \(y\) gives \(g(y) - g(x) \le |x-y|\), hence

\begin{equation*} |g(x) - g(y)| \;\le\; |x - y| . \end{equation*}

Given \(\epsilon > 0\), choose \(\delta = \epsilon\); then \(|x - y| < \delta\) implies \(|g(x) - g(y)| < \epsilon\), which is Definition 4.3.1.

Now suppose \(x \notin F\) but \(g(x) = 0\). For each \(n \in \mathbf{N}\) the number \(1/n\) is then not a lower bound for \(\{|x-a| : a \in F\}\) (Exercise 1.3.1 (b)), so there exists \(a_n \in F\) with

\begin{equation*} 0 \;\le\; |x - a_n| \;<\; 1/n . \end{equation*}

Thus \((a_n) \to x\), and \(a_n \ne x\) for every \(n\) because \(x \notin F\). By Theorem 3.2.5 this makes \(x\) a limit point of \(F\), and \(F\) is closed, so \(x \in F\) by Definition 3.2.7 — a contradiction. Hence \(g(x) > 0\), and in particular \(g(x) \ne 0\), for all \(x \notin F\).

(a) Setting \(x = y = 0\) in the additive condition gives \(f(0) = f(0) + f(0)\), so \(f(0) = 0\). Then for any \(x\),

\begin{equation*} 0 \;=\; f(0) \;=\; f\bigl(x + (-x)\bigr) \;=\; f(x) + f(-x) , \end{equation*}

hence \(f(-x) = -f(x)\).

(b) \(f(1) = k \cdot 1\), and if \(f(n) = kn\) then

\begin{equation*} f(n+1) \;=\; f(n) + f(1) \;=\; kn + k \;=\; k(n+1), \end{equation*}

so \(f(n) = kn\) for all \(n \in \mathbf{N}\) by induction. For \(z \in \mathbf{Z}\): the case \(z \in \mathbf{N}\) is done, \(f(0) = 0 = k \cdot 0\), and for \(z = -n\) with \(n \in \mathbf{N}\) part (a) gives \(f(-n) = -f(n) = -kn = k(-n)\).

For \(r = p/q\) with \(p \in \mathbf{Z}\) and \(q \in \mathbf{N}\), additivity applied to \(q\) copies of \(p/q\) gives

\begin{equation*} \begin{aligned} kp &= f(p) = f\Bigl(\tfrac{p}{q} + \cdots + \tfrac{p}{q}\Bigr) \\ &= q\,f\!\left(\tfrac{p}{q}\right), \end{aligned} \end{equation*}

so \(f( r) = f(p/q) = kp/q = kr\).

(c) Fix \(c \in \mathbf{R}\) and let \(\epsilon > 0\). Continuity at \(0\) (with \(f(0) = 0\)) supplies \(\delta > 0\) such that \(|t| < \delta\) implies \(|f(t)| < \epsilon\). Additivity gives \(f(x) - f( c) = f(x - c)\), so

\begin{equation*} |x-c| < \delta \ \Longrightarrow \ |f(x)-f( c)| = |f(x-c)| < \epsilon . \end{equation*}

The same \(\delta\) works at every \(c\), so \(f\) is continuous on \(\mathbf{R}\).

Now set \(h(x) = f(x) - kx\). It is continuous on \(\mathbf{R}\) by Theorem 4.3.4 (i), (ii) and Example 4.3.5, and \(h( r) = 0\) for every \(r \in \mathbf{Q}\) by part (b). By Exercise 4.3.8 (b), \(h \equiv 0\); that is,

\begin{equation*} f(x) \;=\; kx \qquad \text{for all } x \in \mathbf{R}. \end{equation*}

(a) Take the signed Dirichlet function supported on \(F\):

\begin{equation*} f(x) \;=\; \begin{cases} 1 & \text{if } x \in F \cap \mathbf{Q}, \\ -1 & \text{if } x \in F \setminus \mathbf{Q}, \\ 0 & \text{if } x \notin F . \end{cases} \end{equation*}

Write \(F^\circ\) for the interior (Exercise 3.2.14) and check the three cases.

(i) \(x \notin F\). Since \(F\) is closed, \(F^c\) is open (Theorem 3.2.13), so there is a \(V_\delta(x) \subseteq F^c\) on which \(f\) is identically \(0\). Then \(|f(y) - f(x)| = 0 < \epsilon\) whenever \(|y - x| < \delta\), for any \(\epsilon > 0\): \(f\) is continuous at \(x\).

(ii) \(x \in F^\circ\). Choose \(\delta > 0\) with \(V_\delta(x) \subseteq F\). For each \(n\), Theorem 1.4.3 and Corollary 1.4.4 give a rational \(r_n\) and an irrational \(s_n\) inside the interval \(\bigl(x,\, x + \min\{\delta, 1/n\}\bigr)\); both lie in \(F\), so

\begin{equation*} (r_n) \to x \ \text{ with } f(r_n) = 1, \qquad (s_n) \to x \ \text{ with } f(s_n) = -1 . \end{equation*}

Since \(f(x)\) equals \(1\) or \(-1\), one of these two sequences has \(f(\cdot\,) \not\to f(x)\), and Corollary 4.3.3 makes \(f\) discontinuous at \(x\).

(iii) \(x \in F \setminus F^\circ\). For each \(n\) the neighborhood \(V_{1/n}(x)\) is not contained in \(F\) (otherwise \(x\) would be interior), so pick \(t_n \in V_{1/n}(x)\) with \(t_n \notin F\). Then \((t_n) \to x\) and \(f(t_n) = 0\), while \(|f(x)| = 1\); Corollary 4.3.3 again gives discontinuity.

So \(f\) is discontinuous exactly on \(F^\circ \cup (F \setminus F^\circ) = F\).

(b) If \(O = \mathbf{R}\), Dirichlet’s function from Section 4.1 is discontinuous at every point and we are done; so assume \(O \ne \mathbf{R}\). Then \(F = O^c\) is a nonempty closed set (Theorem 3.2.13), and Exercise 4.3.12 makes

\begin{equation*} d(x) \;=\; \inf\{|x - a| : a \in F\} \end{equation*}

continuous on \(\mathbf{R}\), with \(d(x) > 0\) exactly when \(x \notin F\), i.e. exactly when \(x \in O\) (for \(x \in F\) the choice \(a = x\) forces \(d(x) = 0\)). Define

\begin{equation*} g(x) \;=\; \begin{cases} d(x) & \text{if } x \in \mathbf{Q}, \\ 0 & \text{if } x \notin \mathbf{Q} . \end{cases} \end{equation*}

Discontinuity on \(O\): let \(x \in O\), so \(d(x) > 0\). Choose rationals \((r_n) \to x\) and irrationals \((s_n) \to x\) (Theorem 1.4.3, Corollary 1.4.4). Then

\begin{equation*} g(r_n) = d(r_n) \to d(x) > 0, \qquad g(s_n) = 0 \to 0 , \end{equation*}

the first limit by continuity of \(d\) and Theorem 4.3.2 (iii). The two limits differ, so at least one disagrees with \(g(x)\), and \(g\) is discontinuous at \(x\) by Corollary 4.3.3.

Continuity off \(O\): let \(x \notin O\), so \(d(x) = 0\) and hence \(g(x) = 0\) in either branch. Given \(\epsilon > 0\), continuity of \(d\) at \(x\) supplies \(\delta > 0\) with \(d(y) = |d(y) - d(x)| < \epsilon\) whenever \(|y - x| < \delta\). Since \(0 \le g(y) \le d(y)\) for every \(y\),

\begin{equation*} |y-x| < \delta \ \Longrightarrow \ |g(y)-g(x)| = g(y) \le d(y) < \epsilon . \end{equation*}

Thus the set of discontinuities of \(g\) is precisely \(O\).

(a) The identity \(g(x) = x\) is continuous on \(\mathbf{R}\) (given \(\epsilon > 0\), take \(\delta = \epsilon\)), and by the Algebraic Continuity Theorem (Theorem 4.3.4 (iii)) products of continuous functions are continuous; hence \(f = g \cdot g \cdot g\) is continuous on \(\mathbf{R}\).

(b) Take \(\epsilon_0 = 3\) and the sequences

\begin{equation*} x_n = n + \frac{1}{n}, \qquad y_n = n . \end{equation*}

Then \(|x_n - y_n| = 1/n \to 0\), while

\begin{equation*} \begin{aligned} |f(x_n) - f(y_n)| &= \Big(n + \tfrac{1}{n}\Big)^{3} - n^{3} \\ &= 3n^{2}\cdot\tfrac{1}{n} + 3n\cdot\tfrac{1}{n^{2}} + \tfrac{1}{n^{3}} \\ &= 3n + \tfrac{3}{n} + \tfrac{1}{n^{3}} \;\ge\; 3 = \epsilon_0 \end{aligned} \end{equation*}

for every \(n \in \mathbf{N}\). By Theorem 4.4.5, \(f\) fails to be uniformly continuous on \(\mathbf{R}\).

(c) Let \(A \subseteq \mathbf{R}\) be bounded, say \(A \subseteq [-M, M]\) with \(M > 0\). The interval \([-M,M]\) is closed and bounded, hence compact (Theorem 3.3.4), and \(f\) is continuous on it by part (a); so \(f\) is uniformly continuous on \([-M,M]\) by Theorem 4.4.7. A \(\delta\) that works for all pairs in \([-M,M]\) works in particular for all pairs in \(A\), so \(f\) is uniformly continuous on \(A\).

Method (2): for \(x,y \in A\),

\begin{equation*} |x^{3} - y^{3}| = |x-y|\,|x^{2}+xy+y^{2}| \le 3M^{2}|x-y| , \end{equation*}

so \(\delta = \epsilon/(3M^{2})\) answers any \(\epsilon > 0\).

(a) No. Take \(\epsilon_0 = 1\) and

\begin{equation*} x_n = \frac{1}{n}, \qquad y_n = \frac{1}{2n} \qquad (n \ge 2), \end{equation*}

both in \((0,1)\). Then \(|x_n - y_n| = 1/(2n) \to 0\) while

\begin{equation*} |f(x_n) - f(y_n)| = |n - 2n| = n \ge 1 = \epsilon_0 , \end{equation*}

so Theorem 4.4.5 shows \(f\) is not uniformly continuous on \((0,1)\).

(b) Yes — \(g\) is Lipschitz with constant \(1\). For \(x,y \in (0,1)\), rationalizing gives

\begin{equation*} \begin{aligned} |g(x) - g(y)| &= \frac{|x^{2} - y^{2}|}{\sqrt{x^{2}+1} + \sqrt{y^{2}+1}} \\ &\le \frac{|x-y|\,|x+y|}{2} \;\le\; |x - y| , \end{aligned} \end{equation*}

using \(\sqrt{x^{2}+1} \ge 1\) in the denominator and \(|x+y| < 2\) in the numerator. Given \(\epsilon > 0\), take \(\delta = \epsilon\).

(c) Yes. Define \(\tilde h\) on \([0,1]\) by \(\tilde h(x) = x\sin(1/x)\) for \(x \ne 0\) and \(\tilde h(0) = 0\). On \((0,1]\) the function is continuous by the Algebraic Continuity Theorem (Theorem 4.3.4) and Theorem 4.3.9, and at \(0\) the bound

\begin{equation*} |\tilde h(x) - \tilde h(0)| = |x|\,|\sin(1/x)| \le |x| \end{equation*}

shows \(\delta = \epsilon\) works. So \(\tilde h\) is continuous on the compact set \([0,1]\) (Theorem 3.3.4), hence uniformly continuous there by Theorem 4.4.7. Restricting to \((0,1)\), where \(\tilde h = h\), preserves the \(\delta\).

On \([1,\infty)\) the function is Lipschitz with constant \(2\): for \(x,y \ge 1\),

\begin{equation*} \begin{aligned} \left|\frac{1}{x^{2}} - \frac{1}{y^{2}}\right| &= \frac{|y-x|\,(x+y)}{x^{2}y^{2}} = |x-y|\left(\frac{1}{x y^{2}} + \frac{1}{x^{2} y}\right) \\ &\le 2|x-y| , \end{aligned} \end{equation*}

since \(x,y \ge 1\) makes each of \(1/(xy^{2})\) and \(1/(x^{2}y)\) at most \(1\). Given \(\epsilon > 0\), choose \(\delta = \epsilon/2\); then \(|x-y| < \delta\) forces \(|f(x)-f(y)| < \epsilon\).

On \((0,1]\) take \(\epsilon_0 = 1\) and

\begin{equation*} x_n = \frac{1}{n}, \qquad y_n = \frac{1}{n+1} . \end{equation*}

Then \(|x_n - y_n| = \dfrac{1}{n(n+1)} \to 0\), while

\begin{equation*} |f(x_n) - f(y_n)| = (n+1)^{2} - n^{2} = 2n + 1 \ge 1 = \epsilon_0 \end{equation*}

for all \(n \in \mathbf{N}\). By Theorem 4.4.5, \(f\) is not uniformly continuous on \((0,1]\).

(a) True. The set \([a,b]\) is compact (Theorem 3.3.4), so by the Extreme Value Theorem (Theorem 4.4.2) \(f\) attains a minimum value \(m = f(x_0)\) for some \(x_0 \in [a,b]\), and \(m > 0\) by hypothesis. Hence

\begin{equation*} 0 < \frac{1}{f(x)} \le \frac{1}{m} \qquad \text{for all } x \in [a,b]. \end{equation*}

(b) True. Suppose \(f(A)\) were unbounded, and pick \(x_n \in A\) with \(|f(x_n)| \ge n\). Since \(A\) is bounded, the Bolzano–Weierstrass Theorem (Theorem 2.5.5) supplies a convergent subsequence \((x_{n_k})\), which is therefore Cauchy. Uniformly continuous functions preserve Cauchy sequences (Exercise 4.4.13 (a)), so \(\big(f(x_{n_k})\big)\) is Cauchy and hence bounded (Lemma 2.6.3). This contradicts \(|f(x_{n_k})| \ge n_k \to \infty\).

(c) False. Define

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x = 0, \\ 0 & \text{if } x \ne 0. \end{cases} \end{equation*}

For any compact \(K \subseteq \mathbf{R}\) the image \(f(K)\) is one of \(\{0\}\), \(\{1\}\), \(\{0,1\}\); a finite set is closed and bounded, hence compact (Theorem 3.3.4). But \(f\) is discontinuous at \(0\): with \(x_n = 1/n \to 0\) we get \(f(x_n) = 0 \to 0 \ne 1 = f(0)\), so continuity fails by the sequential criterion (Corollary 4.3.3).

Take \(\delta = \min\{\delta_1, \delta_2\}\), where \(\delta_1\) and \(\delta_2\) answer \(\epsilon/2\) on \((a,b]\) and on \([b,c)\) respectively; the point \(b\) lies in both pieces, and that is what lets the two estimates be chained.

Let \(\epsilon > 0\) and fix such a \(\delta\). Let \(x,y \in (a,c)\) with \(|x-y| < \delta\), and assume without loss of generality \(x \le y\).

(i) Both \(x,y \le b\). Then \(x,y \in (a,b]\) and \(|g(x)-g(y)| < \epsilon/2 < \epsilon\).

(ii) Both \(x,y \ge b\). Then \(x,y \in [b,c)\) and \(|g(x)-g(y)| < \epsilon/2 < \epsilon\).

(iii) \(x < b < y\). Then \(x,b \in (a,b]\) with \(|x-b| < |x-y| < \delta_1\), and \(b,y \in [b,c)\) with \(|b-y| < |x-y| < \delta_2\), so

\begin{equation*} \begin{aligned} |g(x) - g(y)| &\le |g(x) - g(b)| + |g(b) - g(y)| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

In every case \(|x-y| < \delta\) implies \(|g(x)-g(y)| < \epsilon\), which is Definition 4.4.4 on \((a,c)\).

(a) Take \(f(x) = 1/x\) on \((0,1)\) and \(x_n = 1/n\). The sequence \((x_n)\) converges to \(0\), hence is Cauchy (Theorem 2.6.2), and \(f\) is continuous on \((0,1)\) by the Algebraic Continuity Theorem (Theorem 4.3.4 (iv)), the denominator being nonzero there. But \(f(x_n) = n\) is unbounded, so it is not Cauchy (Lemma 2.6.3).

(b) Impossible: uniformly continuous functions preserve Cauchy sequences. Given \(\epsilon > 0\), choose \(\delta\) as in Definition 4.4.4 and then \(N\) with \(|x_n - x_m| < \delta\) for all \(m,n \ge N\); for such \(m,n\),

\begin{equation*} |f(x_n) - f(x_m)| < \epsilon . \end{equation*}

(This is Exercise 4.4.13 (a).)

(c) Impossible. A Cauchy sequence \((x_n) \subseteq [0,\infty)\) converges to some \(x \in \mathbf{R}\) by the Cauchy Criterion (Theorem 2.6.4), and \(x \ge 0\) by the Order Limit Theorem (Theorem 2.3.4), so \(x\) lies in the domain \([0,\infty)\). Continuity of \(f\) at \(x\) then gives \(f(x_n) \to f(x)\) (Theorem 4.3.2 (iii)), and convergent sequences are Cauchy (Theorem 2.6.2).

Given \(\epsilon > 0\), take \(\delta = \epsilon^{2}\); the work is the inequality

\begin{equation*} |\sqrt{x} - \sqrt{y}\,| \le \sqrt{|x-y|} \qquad (x,y \ge 0). \end{equation*}

To see it, assume without loss of generality \(x \ge y \ge 0\). Then

\begin{equation*} \begin{aligned} \big(\sqrt{y} + \sqrt{x-y}\,\big)^{2} &= y + 2\sqrt{y}\sqrt{x-y} + (x - y) \\ &\ge x = (\sqrt{x}\,)^{2}, \end{aligned} \end{equation*}

and since both bases are nonnegative, \(\sqrt{y} + \sqrt{x-y} \ge \sqrt{x}\), i.e. \(\sqrt{x} - \sqrt{y} \le \sqrt{x-y}\).

Now if \(x,y \in [0,\infty)\) satisfy \(|x-y| < \delta = \epsilon^{2}\), then

\begin{equation*} \begin{aligned} |f(x) - f(y)| &= |\sqrt{x} - \sqrt{y}\,| \\ &\le \sqrt{|x-y|} < \sqrt{\epsilon^{2}} = \epsilon , \end{aligned} \end{equation*}

the last step using that \(t \mapsto \sqrt{t}\) is increasing on \([0,\infty)\). Since \(\delta\) depends only on \(\epsilon\), Definition 4.4.4 is satisfied.

(a) Impossible. \([0,1]\) is closed and bounded, hence compact (Theorem 3.3.4), so \(f([0,1])\) is compact by Theorem 4.4.1; but \((0,1)\) is not closed and therefore not compact. (Equivalently: the Extreme Value Theorem 4.4.2 forces \(f\) to attain \(\sup f([0,1])\), which \((0,1)\) never contains.)

(b) Possible. Take

\begin{equation*} f(x) = \tfrac{1}{2}\bigl(1 + \sin(2\pi x)\bigr), \qquad x \in (0,1). \end{equation*}

Then \(f\) is continuous with \(f(x) \in [0,1]\), and \(f(1/4) = 1\), \(f(3/4) = 0\); the Intermediate Value Theorem (Theorem 4.5.1) applied on \([1/4, 3/4]\) supplies every value in between. Hence the range is exactly \([0,1]\).

(c) Possible. Take

\begin{equation*} f(x) = \tfrac{1}{2}\Bigl(1 + (1-x)\sin(1/x)\Bigr), \qquad x \in (0,1]. \end{equation*}

This is continuous on \((0,1]\) (Theorems 4.3.4 and 4.3.9), and

\begin{equation*} \begin{aligned} \Bigl| f(x) - \tfrac{1}{2} \Bigr| &= \tfrac{1}{2}(1-x)\,|\sin(1/x)| \\ &\le \tfrac{1-x}{2} \;<\; \tfrac{1}{2}, \end{aligned} \end{equation*}

so \(f(x) \in (0,1)\) for every \(x \in (0,1]\): the range is contained in \((0,1)\) and never reaches the endpoints. For surjectivity put

\begin{equation*} x_n = \frac{1}{\pi/2 + 2n\pi}, \qquad y_n = \frac{1}{3\pi/2 + 2n\pi}, \end{equation*}

so that \(y_n < x_n\) and

\begin{equation*} f(x_n) = 1 - \tfrac{x_n}{2} \to 1, \qquad f(y_n) = \tfrac{y_n}{2} \to 0 . \end{equation*}

Given \(t \in (0,1)\), choose \(n\) with \(f(y_n) < t < f(x_n)\) and apply the Intermediate Value Theorem on \([y_n, x_n] \subseteq (0,1]\) to get \(c\) with \(f( c) = t\). Hence the range is exactly \((0,1)\).

(a) Given \(\epsilon > 0\), take \(\delta = \epsilon / M\). The Lipschitz bound is exactly the statement

\begin{equation*} |f(x) - f(y)| \le M\,|x - y| \qquad \text{for all } x, y \in A \end{equation*}

(the case \(x = y\) being trivial), so \(|x - y| < \delta\) gives \(|f(x) - f(y)| \le M|x-y| < M\delta = \epsilon\). Since \(\delta\) depends only on \(\epsilon\), Definition 4.4.4 is satisfied.

(b) No. Take \(f(x) = \sqrt{x}\) on \(A = [0,1]\).

Uniformly continuous. \(f\) is continuous on \([0,1]\) (Theorem 4.3.2 (iii) together with Exercise 2.3.1), and \([0,1]\) is compact (Theorem 3.3.4), so Theorem 4.4.7 gives uniform continuity.

Not Lipschitz. Taking \(y = 0\),

\begin{equation*} \left| \frac{f(x) - f(0)}{x - 0} \right| = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} \longrightarrow \infty \end{equation*}

as \(x \to 0^{+}\), so no bound \(M\) can hold for all \(x \ne y\) in \([0,1]\).

Only the sum and the composition.

Sum: yes. Given \(\epsilon > 0\), pick \(\delta_1\) for \(f\) and \(\delta_2\) for \(g\), each answering \(\epsilon/2\), and set \(\delta = \min\{\delta_1, \delta_2\}\). For \(x,y \in A\) with \(|x-y| < \delta\),

\begin{equation*} \begin{aligned} |(f+g)(x) - (f+g)(y)| &\le |f(x)-f(y)| + |g(x)-g(y)| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

Composition: yes. Given \(\epsilon > 0\), choose \(\delta_1 > 0\) so that \(|u - v| < \delta_1\) implies \(|f(u) - f(v)| < \epsilon\), and then choose \(\delta > 0\) so that \(|x-y| < \delta\) implies \(|g(x) - g(y)| < \delta_1\). Then \(|x - y| < \delta\) gives \(|f(g(x)) - f(g(y))| < \epsilon\).

Product: no. Take \(A = \mathbf{R}\) and \(f(x) = g(x) = x\), which is uniformly continuous (\(\delta = \epsilon\)). The product is \(x^{2}\), which fails to be uniformly continuous on \(\mathbf{R}\): with \(x_n = n + 1/n\) and \(y_n = n\) we have \(|x_n - y_n| \to 0\) while

\begin{equation*} |x_n^{2} - y_n^{2}| = 2 + \tfrac{1}{n^{2}} \ge 2 , \end{equation*}

so Theorem 4.4.5 applies with \(\epsilon_0 = 2\).

Quotient: no. Take \(A = (0,1)\), \(f(x) = 1\) and \(g(x) = x\); both are uniformly continuous on \(A\) and \(g\) never vanishes there. The quotient is \(1/x\), and with \(x_n = 1/n\), \(y_n = 1/(n+1)\) we get \(|x_n - y_n| \to 0\) while

\begin{equation*} \left| \frac{1}{x_n} - \frac{1}{y_n} \right| = |n - (n+1)| = 1 , \end{equation*}

so Theorem 4.4.5 applies with \(\epsilon_0 = 1\).

Both directions are the \(\epsilon\)–\(\delta\) definition read through Theorem 4.3.2 (ii): continuity at \(c\) says every \(\epsilon\)-neighborhood \(V_\epsilon(g( c))\) pulls back to contain a \(\delta\)-neighborhood \(V_\delta( c)\).

(\(\Rightarrow\)) Assume \(g\) is continuous and let \(O\) be open. Let \(x \in g^{-1}(O)\), so \(g(x) \in O\); since \(O\) is open there is \(\epsilon > 0\) with \(V_\epsilon(g(x)) \subseteq O\) (Definition 3.2.1). Continuity at \(x\) produces \(\delta > 0\) with

\begin{equation*} |y - x| < \delta \implies |g(y) - g(x)| < \epsilon , \end{equation*}

that is, \(g(V_\delta(x)) \subseteq V_\epsilon(g(x)) \subseteq O\), and hence \(V_\delta(x) \subseteq g^{-1}(O)\). So every point of \(g^{-1}(O)\) is interior, and \(g^{-1}(O)\) is open.

(\(\Leftarrow\)) Assume preimages of open sets are open, and fix \(c \in \mathbf{R}\) and \(\epsilon > 0\). The set \(O = V_\epsilon(g( c))\) is open, so \(g^{-1}(O)\) is open; it contains \(c\) because \(g( c) \in O\). Therefore there is \(\delta > 0\) with \(V_\delta( c) \subseteq g^{-1}(O)\), which says precisely

\begin{equation*} |x - c| < \delta \implies |g(x) - g( c)| < \epsilon . \end{equation*}

Thus \(g\) is continuous at \(c\), and \(c\) was arbitrary.

Only (d) is true; the constant function refutes (a), (b) and (c) simultaneously.

(a) False. Let \(f(x) = 0\) for all \(x\). Then \(B = \{0\}\) is finite but \(f^{-1}(B) = \mathbf{R}\).

(b) False. Same \(f\): \(K = \{0\}\) is compact (closed and bounded, Theorem 3.3.4) while \(f^{-1}(K) = \mathbf{R}\) is unbounded, hence not compact.

(c) False. Same \(f\) and \(A = \{0\}\): \(f^{-1}(A) = \mathbf{R}\).

(d) True. Let \(F\) be closed. Then \(F^{c}\) is open (Theorem 3.2.13), so \(f^{-1}(F^{c})\) is open by Exercise 4.4.11. Because \(f\) is defined on all of \(\mathbf{R}\), every \(x\) satisfies exactly one of \(f(x) \in F\), \(f(x) \in F^{c}\), so

\begin{equation*} f^{-1}(F^{c}) = \bigl( f^{-1}(F) \bigr)^{c} . \end{equation*}

Thus \(\bigl( f^{-1}(F) \bigr)^{c}\) is open, and \(f^{-1}(F)\) is closed by Theorem 3.2.13 again.

(a) Let \(\epsilon > 0\) and take \(\delta > 0\) from Definition 4.4.4. Since \((x_n)\) is Cauchy there is \(N\) with \(|x_n - x_m| < \delta\) for all \(m, n \ge N\), and then

\begin{equation*} |f(x_n) - f(x_m)| < \epsilon \qquad \text{for all } m, n \ge N . \end{equation*}

So \((f(x_n))\) is Cauchy (Definition 2.6.1).

(b) (\(\Leftarrow\)) If \(g\) extends to a continuous function on \([a,b]\), then that extension is continuous on a compact set (Theorem 3.3.4), hence uniformly continuous on \([a,b]\) by Theorem 4.4.7. A \(\delta\) that answers \(\epsilon\) for all pairs in \([a,b]\) answers it in particular for all pairs in \((a,b)\), so the original \(g\) is uniformly continuous there.

(\(\Rightarrow\)) Assume \(g\) is uniformly continuous on \((a,b)\). Pick any sequence \((x_n) \subseteq (a,b)\) with \(x_n \to a\). Convergent sequences are Cauchy (Theorem 2.6.2), so \((g(x_n))\) is Cauchy by part (a) and hence converges by the Cauchy Criterion (Theorem 2.6.4). Define

\begin{equation*} g(a) = \lim_{n \to \infty} g(x_n), \qquad g(b) = \lim_{n \to \infty} g(y_n) , \end{equation*}

the latter for any \((y_n) \subseteq (a,b)\) with \(y_n \to b\).

To see the extension is continuous at \(a\), let \(\epsilon > 0\) and take \(\delta > 0\) from uniform continuity answering \(\epsilon/2\). Fix \(x \in (a,b)\) with \(|x - a| < \delta\). Since \(x_n \to a\) and \(g(x_n) \to g(a)\), choose \(n\) large enough that \(|x_n - a| < \delta\) and \(|g(x_n) - g(a)| < \epsilon/2\). Then \(|x - x_n| < \delta\) because both lie in \((a, a+\delta)\), so

\begin{equation*} \begin{aligned} |g(x) - g(a)| &\le |g(x) - g(x_n)| + |g(x_n) - g(a)| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

Since \(|g(a) - g(a)| = 0 < \epsilon\) as well, the same \(\delta\) witnesses continuity of the extended \(g\) at \(a\); the argument at \(b\) is identical. Continuity at the interior points is unaffected, so the extended \(g\) is continuous on \([a,b]\).

Fix \(\epsilon > 0\); a finite subcover by half-neighborhoods produces the uniform \(\delta\). (Theorem 4.4.7: \(f\) continuous on a compact \(K\) is uniformly continuous on \(K\).)

For each \(x \in K\), continuity of \(f\) at \(x\) supplies \(\delta(x) > 0\) with

\begin{equation*} y \in K, \; |y - x| < \delta(x) \implies |f(y) - f(x)| < \tfrac{\epsilon}{2} . \end{equation*}

The half-radius neighborhoods

\begin{equation*} \mathcal{O} = \bigl\{\, V_{\delta(x)/2}(x) \; : \; x \in K \,\bigr\} \end{equation*}

form an open cover of \(K\) (each \(x\) lies in its own set). Since \(K\) is compact, Theorem 3.3.8 (iii) yields finitely many points \(x_1, \dots, x_n \in K\) with

\begin{equation*} K \subseteq \bigcup_{i=1}^{n} V_{\delta(x_i)/2}(x_i) , \end{equation*}

and we set

\begin{equation*} \delta = \min\bigl\{ \delta(x_1)/2, \dots, \delta(x_n)/2 \bigr\} > 0 , \end{equation*}

positive precisely because the minimum is over a finite set.

Now let \(y, z \in K\) satisfy \(|y - z| < \delta\). Choose \(i\) with \(y \in V_{\delta(x_i)/2}(x_i)\). Then \(|y - x_i| < \delta(x_i)/2 < \delta(x_i)\), and

\begin{equation*} |z - x_i| \le |z - y| + |y - x_i| < \delta + \tfrac{\delta(x_i)}{2} \le \delta(x_i) . \end{equation*}

Both \(y\) and \(z\) therefore lie within \(\delta(x_i)\) of \(x_i\), so

\begin{equation*} \begin{aligned} |f(y) - f(z)| &\le |f(y) - f(x_i)| + |f(x_i) - f(z)| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

Since \(\delta\) was manufactured from \(\epsilon\) alone, \(f\) is uniformly continuous on \(K\) by Definition 4.4.4.

Apply Theorem 4.5.2 to \(E = [a,b]\), connected by Theorem 3.4.7.

Let \(f : [a,b] \to \mathbf{R}\) be continuous with \(f(a) < L < f(b)\) (the case \(f(a) > L > f(b)\) is the same with \(a\) and \(b\) exchanged). Whenever \(x < w < y\) with \(x,y \in [a,b]\) we have \(w \in [a,b]\), so Theorem 3.4.7 makes \([a,b]\) connected, and Theorem 4.5.2 makes \(f([a,b])\) connected. Since \(f(a), f(b) \in f([a,b])\) and \(f(a) < L < f(b)\), the converse half of Theorem 3.4.7 forces \(L \in f([a,b])\): say \(L = f( c)\) with \(c \in [a,b]\). Finally \(f( c) = L\) differs from both \(f(a)\) and \(f(b)\), so \(c \in (a,b)\).

(a) Possible: \(f(x) = \cos(4\pi x)\) on \((0,1)\). As \(x\) runs over \((0,1)\) the argument \(4\pi x\) runs over \((0,4\pi)\), which contains \(\pi\), \(2\pi\) and \(3\pi\); hence \(f(1/4) = -1\) and \(f(1/2) = 1\) are attained, and the range is exactly \([-1,1]\).

(b) Impossible. A closed interval \([a,b]\) is compact, so Theorem 4.4.1 makes \(f([a,b])\) compact, hence closed and bounded (Theorem 3.3.4). A nonempty open interval \((c,d)\) is neither: if \(d < \infty\) then \(d\) is a limit point of \((c,d)\) outside it, and if \(d = \infty\) the set is unbounded.

(c) Possible: \(f(x) = \dfrac{1}{x(1-x)}\) on \((0,1)\), continuous there by Example 4.3.5 since the denominator never vanishes. From \(x(1-x) \le 1/4\) we get \(f \ge 4\), with \(f(1/2) = 4\); and \(f(x) \to \infty\) as \(x \to 1^{-}\). Given \(M \ge 4\), pick \(x_0 \in (1/2,1)\) with \(f(x_0) \ge M\); Theorem 4.5.1 on \([1/2,x_0]\) attains \(M\). So the range is exactly \([4,\infty)\).

(d) Impossible. \(\mathbf{R}\) is connected, so by Theorem 4.5.2 the range \(f(\mathbf{R})\) is a connected subset of \(\mathbf{R}\). But \(\mathbf{Q}\) is disconnected: Example 3.4.5 (ii) writes

\begin{equation*} \begin{aligned} \mathbf{Q} &= A \cup B, \\ A &= \mathbf{Q} \cap (-\infty,\sqrt{2}), \\ B &= \mathbf{Q} \cap (\sqrt{2},\infty) \end{aligned} \end{equation*}

as a union of two nonempty separated sets. So \(f(\mathbf{R}) = \mathbf{Q}\) is impossible.

Monotonicity turns a single value handed over by the intermediate value property into control on a whole one-sided neighborhood.

Fix \(c \in [a,b]\) and \(\varepsilon > 0\); we verify Theorem 4.3.2 (i) by building one-sided \(\delta_{\pm}\) and taking \(\delta = \min\{\delta_{+},\delta_{-}\}\). To the right (take \(\delta_{+} = 1\) if \(c = b\)):

(i) \(f(x) < f( c) + \varepsilon\) for every \(x \in (c,b]\). Take \(\delta_{+} = 1\); monotonicity gives \(f( c) \le f(x) < f( c) + \varepsilon\) there.

(ii) Some \(y \in (c,b]\) has \(f(y) \ge f( c) + \varepsilon\). Then \(L = f( c) + \varepsilon/2\) lies strictly between \(f( c)\) and \(f(y)\), so Definition 4.5.3 applied to \(c < y\) in \([a,b]\) supplies \(d \in (c,y)\) with \(f(d) = L\). Set \(\delta_{+} = d - c > 0\); for \(c \le x < d\),

\begin{equation*} f( c) \le f(x) \le f(d) = f( c) + \tfrac{\varepsilon}{2} < f( c) + \varepsilon . \end{equation*}

The left side mirrors this: if some \(y \in [a,c)\) has \(f(y) \le f( c) - \varepsilon\), Definition 4.5.3 on \([y,c]\) gives \(d \in (y,c)\) with \(f(d) = f( c) - \varepsilon/2\), and \(\delta_{-} = c - d\) works because \(d < x \le c\) forces \(f( c) - \varepsilon < f(d) \le f(x) \le f( c)\); otherwise \(\delta_{-} = 1\). So \(x \in [a,b]\) with \(|x-c| < \delta\) gives \(|f(x) - f( c)| < \varepsilon\), for every \(c \in [a,b]\).

One coincidence \(g(x_1) = g(x_2)\) already breeds a continuum of them, so a nonempty \(F\) is uncountable. (The printed display writes \(f\) for the function the statement calls \(g\); we use \(g\) throughout.)

Assume \(F \ne \emptyset\): there are \(x_1 < x_2\) in \(A\) with \(g(x_1) = g(x_2)\), and \([x_1,x_2] \subseteq A\) because \(A\) is an interval.

(i) \(g\) constant on \([x_1,x_2]\). Every \(x\) there has a partner, so \([x_1,x_2] \subseteq F\), and a nondegenerate interval is uncountable (Theorem 1.6.1, transported by the bijection \(t \mapsto x_1 + t(x_2 - x_1)\) of \((0,1)\) onto \((x_1,x_2)\)).

(ii) \(g\) not constant on \([x_1,x_2]\). Some point of \([x_1,x_2]\) carries a value other than \(g(x_1)\), and it is neither endpoint (both have value \(g(x_1)\)), so there is \(z \in (x_1,x_2)\) with \(g(z) \ne g(x_1)\). Let \(J\) be the nondegenerate open interval with endpoints \(g(x_1)\) and \(g(z)\), and fix \(L \in J\). Theorem 4.5.1 applied to the continuous \(g\) on \([x_1,z]\) gives

\begin{equation*} u(L) \in (x_1,z) \quad\text{with}\quad g(u(L)) = L ; \end{equation*}

since \(g(x_2) = g(x_1)\), that same \(L\) lies strictly between \(g(z)\) and \(g(x_2)\), and Theorem 4.5.1 on \([z,x_2]\) gives

\begin{equation*} v(L) \in (z,x_2) \quad\text{with}\quad g(v(L)) = L . \end{equation*}

Then \(u(L) < z < v(L)\) with equal \(g\)-values, so \(u(L) \in F\). The map \(L \mapsto u(L)\) is injective, since \(L = g(u(L))\) recovers \(L\); hence \(u(J) \subseteq F\) is in bijection with the uncountable interval \(J\), and a countable \(F\) would make \(u(J)\) countable by Theorem 1.5.7.

(a) \(c = \sup K\) satisfies \(f( c) = 0\); both alternatives are killed by the least-upper-bound property.

\(K = \{x \in [a,b] : f(x) \le 0\}\) contains \(a\) and is bounded above by \(b\), so the Axiom of Completeness gives \(c = \sup K \in [a,b]\).

(i) \(f( c) > 0\). Continuity at \(c\) (Theorem 4.3.2 (i)) with \(\varepsilon = f( c)\) gives \(\delta > 0\) with \(f(x) > 0\) for every \(x \in (c-\delta,c+\delta) \cap [a,b]\). No such \(x\) lies in \(K\), and no \(x > c\) lies in \(K\) because \(c\) bounds \(K\) above; hence \(c - \delta\) also bounds \(K\) above and \(\sup K \le c - \delta < c\), a contradiction.

(ii) \(f( c) < 0\). Then \(c < b\), since \(f(b) > 0\). Continuity at \(c\) with \(\varepsilon = -f( c) > 0\) gives \(\delta > 0\) with \(f(x) < 0\) for every \(x \in (c-\delta,c+\delta) \cap [a,b]\). Any \(x_0\) with \(c < x_0 < \min\{c+\delta,\,b\}\) therefore lies in \(K\) and exceeds \(c\), contradicting that \(c\) bounds \(K\) above.

Hence \(f( c) = 0\), and \(c \in (a,b)\) because \(f(a) < 0 < f(b)\).

For the general statement of Theorem 4.5.1: if \(f(a) < L < f(b)\), apply the above to \(g = f - L\), which is continuous (Theorem 4.3.4) with \(g(a) < 0 < g(b)\); if \(f(a) > L > f(b)\), apply it to \(g = L - f\).

(b) The nested intervals close down on the required point, and the Order Limit Theorem squeezes \(f\) there.

The construction produces \(I_n = [a_n,b_n]\) with

\begin{equation*} I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots, \qquad b_n - a_n = \frac{b-a}{2^{n}}, \end{equation*}

and, for every \(n\),

\begin{equation*} f(a_n) < 0 \le f(b_n). \end{equation*}

By the Nested Interval Property (Theorem 1.4.1) there is a point \(c \in \bigcap_{n=0}^{\infty} I_n\); in particular \(c \in [a,b]\). Since \(a_n, c \in I_n\),

\begin{equation*} |a_n - c| \le b_n - a_n = \frac{b-a}{2^{n}} \longrightarrow 0, \end{equation*}

so \((a_n) \to c\), and identically \((b_n) \to c\). Continuity of \(f\) at \(c\) and the sequential characterization Theorem 4.3.2 (iii) give

\begin{equation*} f(a_n) \to f( c) \qquad\text{and}\qquad f(b_n) \to f( c). \end{equation*}

Applying the Order Limit Theorem (Theorem 2.3.4) to \(f(a_n) < 0\) and to \(f(b_n) \ge 0\) yields

\begin{equation*} f( c) \le 0 \qquad\text{and}\qquad f( c) \ge 0, \end{equation*}

so \(f( c) = 0\). As in (a), \(c \ne a\) and \(c \ne b\), so \(c \in (a,b)\), and the general case follows by replacing \(f\) with \(f - L\) or \(L - f\).

(a) Apply the Intermediate Value Theorem to \(g(x) = f(x + 1/2) - f(x)\) on \([0,1/2]\).

\(g\) is continuous there by Theorem 4.3.4 (and Theorem 4.3.9 for the composition with \(x \mapsto x + 1/2\)), and

\begin{equation*} \begin{aligned} g\bigl(\tfrac12\bigr) &= f(1) - f\bigl(\tfrac12\bigr) \\ &= f(0) - f\bigl(\tfrac12\bigr) = -g(0). \end{aligned} \end{equation*}

If \(g(0) = 0\), take \(x = 0\), \(y = 1/2\). Otherwise \(g(0)\) and \(g(1/2)\) are nonzero of opposite sign, so \(0\) lies strictly between them and Theorem 4.5.1 gives \(c \in (0,1/2)\) with \(g( c) = 0\); take \(x = c\), \(y = c + 1/2\).

(b) Same device with step \(1/n\), plus a telescoping sum.

Let \(g(x) = f(x + 1/n) - f(x)\), continuous on \([0, 1 - 1/n]\). Then

\begin{equation*} \begin{aligned} \sum_{k=0}^{n-1} g\!\left(\frac{k}{n}\right) &= \sum_{k=0}^{n-1} \left[ f\!\left(\tfrac{k+1}{n}\right)

• f\!\left(\tfrac{k}{n}\right) \right] \\ &= f(1) - f(0) = 0 . \end{aligned} \end{equation*}

If some term vanishes, say \(g(k/n) = 0\), take \(x_n = k/n\) and \(y_n = (k+1)/n\). Otherwise the \(n\) nonzero terms sum to zero, so they cannot all have the same sign: there are indices \(k, l\) with

\begin{equation*} g\!\left(\frac{k}{n}\right) > 0 > g\!\left(\frac{l}{n}\right). \end{equation*}

Both \(k/n\) and \(l/n\) lie in \([0, 1-1/n]\), so applying Theorem 4.5.1 to \(g\) on the closed interval with these two endpoints produces \(c\) strictly between them with \(g( c) = 0\). Take \(x_n = c\), \(y_n = c + 1/n\); then \(|x_n - y_n| = 1/n\) and \(f(x_n) = f(y_n)\).

(c) Take

\begin{equation*} f(x) = \sin^{2}\!\left(\frac{5\pi x}{2}\right) - x . \end{equation*}

This is continuous on \([0,1]\), and \(f(0) = 0\), \(f(1) = \sin^{2}(5\pi/2) - 1 = 1 - 1 = 0\), so \(f(0) = f(1)\). For any \(x\) with \(x, x + 2/5 \in [0,1]\),

\begin{equation*} \begin{aligned} f\!\left(x + \tfrac25\right) - f(x) &= \sin^{2}\!\left(\tfrac{5\pi x}{2} + \pi\right)

• \sin^{2}\!\left(\tfrac{5\pi x}{2}\right) - \tfrac{2}{5} \\ &= -\frac{2}{5}, \end{aligned} \end{equation*}

since \(\sin(\theta + \pi) = -\sin\theta\) makes \(\sin^{2}\) periodic with period \(\pi\). Hence \(f(x) \ne f(y)\) whenever \(|x - y| = 2/5\) with \(x,y \in [0,1]\).

Apply the Intermediate Value Theorem to \(g(x) = f(x) - x\).

\(g\) is continuous on \([0,1]\) by Theorem 4.3.4, and because \(f\) takes values in \([0,1]\),

\begin{equation*} g(0) = f(0) - 0 \ge 0, \qquad g(1) = f(1) - 1 \le 0 . \end{equation*}

(i) If \(g(0) = 0\) then \(x = 0\) is a fixed point; if \(g(1) = 0\) then \(x = 1\) is.

(ii) Otherwise \(g(1) < 0 < g(0)\), so \(0\) lies strictly between \(g(0)\) and \(g(1)\) and Theorem 4.5.1 supplies \(c \in (0,1)\) with \(g( c) = 0\), i.e. \(f( c) = c\).

Sequentially: if \(y_n \to y\), every subsequential limit of \(\left(f^{-1}(y_n)\right)\) equals \(f^{-1}(y)\), because \(f\) is continuous and one-to-one.

Write \(g := f^{-1}\). The Extreme Value Theorem (Theorem 4.4.2, applicable since \([a,b]\) is compact and \(f\) is continuous on it) gives a minimum \(c\) and a maximum \(d\) of \(f\), attained at some \(p, q \in [a,b]\); the Intermediate Value Theorem (Theorem 4.5.1) applied to \(f\) on the interval with endpoints \(p\) and \(q\) attains every value in between. Hence

\begin{equation*} g : [c,d] \longrightarrow [a,b], \qquad f([a,b]) = [c,d]. \end{equation*}

Fix \(y \in [c,d]\) and \((y_n) \subseteq [c,d]\) with \(y_n \to y\), and put

\begin{equation*} x_n = g(y_n) \in [a,b], \qquad x = g(y) \in [a,b], \end{equation*}

so \(f(x_n) = y_n\) and \(f(x) = y\). If \(x_n \not\to x\), there are \(\epsilon_0 > 0\) and a subsequence \((x_{n_k})\) with

\begin{equation*} |x_{n_k} - x| \ge \epsilon_0 \quad \text{for all } k. \end{equation*}

This subsequence is bounded (it lies in \([a,b]\)), so Bolzano-Weierstrass (Theorem 2.5.5) supplies \(x_{n_{k_l}} \to x^{\prime}\), with \(x^{\prime} \in [a,b]\) because \([a,b]\) is closed; the displayed inequality passes to the limit, giving \(|x^{\prime} - x| \ge \epsilon_0\) and hence \(x^{\prime} \ne x\). But \(f\) is continuous at \(x^{\prime}\), so by Theorem 4.3.2 (iii), together with Theorem 2.5.2 for the subsequence of \((y_n)\),

\begin{equation*} \begin{aligned} f(x^{\prime}) &= \lim_{l \to \infty} f\bigl(x_{n_{k_l}}\bigr) \\ &= \lim_{l \to \infty} y_{n_{k_l}} = y = f(x), \end{aligned} \end{equation*}

contradicting injectivity. Therefore \(x_n \to x\), i.e. \(g(y_n) \to g(y)\) for every sequence \(y_n \to y\) in \([c,d]\), and Theorem 4.3.2 (iii), now applied to \(g\), makes \(g = f^{-1}\) continuous at every \(y \in [c,d]\).

Method (2): \(f\) is strictly monotone. Say \(f(a) < f(b)\). Then \(f(a) < f(t) < f(b)\) for every \(t \in (a,b)\): if \(f(t) > f(b)\), Theorem 4.5.1 on \([a,t]\) returns \(x_1 \in (a,t)\) with \(f(x_1) = f(b)\), and if \(f(t) < f(a)\) it returns \(x_2 \in (t,b)\) with \(f(x_2) = f(a)\); either one contradicts injectivity. Since this conclusion needs only \(f(a) < f(b)\), it may be re-applied on \([t,b]\), where \(f(t) < f(b)\), giving \(f(t) < f(s)\) for \(t < s < b\); so \(f\) is strictly increasing on \([a,b]\). (If \(f(a) > f(b)\), run this on \(-f\), so \(f\) is strictly decreasing.)

Take \(f\) increasing. Then \(g\) is increasing on \([c,d]\) with \(g([c,d]) = [a,b]\), so \(g\) has the intermediate value property of Definition 4.5.3: for \(y_1 < y_2\) in \([c,d]\) and \(L\) strictly between \(g(y_1)\) and \(g(y_2)\), the point \(y_0 = f(L)\) lies in \((y_1,y_2)\) and \(g(y_0) = L\). Exercise 4.5.3 (increasing plus the intermediate value property gives continuity) finishes. If \(f\) decreases, \(-f\) increases with inverse \(y \mapsto g(-y)\), so \(g\) is continuous by Theorem 4.3.9.

The template is \(f = \varphi \cdot \mathbf{1}_{\mathbf{Q}}\) for a continuous \(\varphi\): such an \(f\) is continuous exactly where \(\varphi\) vanishes. If \(\varphi( c) = 0\) then \(|f(x) - f( c)| = |f(x)| \le |\varphi(x)| \to 0\); if \(\varphi( c) \ne 0\), rationals \(x_n \to c\) give \(f(x_n) = \varphi(x_n) \to \varphi( c)\) while irrationals \(y_n \to c\) give \(f(y_n) \to 0\), so one of these two sequences violates Corollary 4.3.3. Hence

\begin{equation*} D_{\varphi \cdot \mathbf{1}_{\mathbf{Q}}} = \{x \in \mathbf{R} : \varphi(x) \ne 0\}. \end{equation*}

(a) Take

\begin{equation*} f(x) = \begin{cases} \sin(\pi x) & \text{if } x \in \mathbf{Q} \\ 0 & \text{if } x \notin \mathbf{Q}. \end{cases} \end{equation*}

Since \(\sin(\pi x)\) is continuous and vanishes precisely on \(\mathbf{Z}\), the displayed identity gives \(D_f = \mathbf{Z}^c\).

(b) The zero set of a continuous \(\varphi\) is closed, so no such template yields the half-open \((0,1]\); break the template at the right endpoint instead. Take

\begin{equation*} f(x) = \begin{cases} x & \text{if } x \in \mathbf{Q} \cap (0,1] \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

For \(c < 0\) or \(c > 1\) the function vanishes on a whole neighborhood of \(c\), and at \(c = 0\) we have \(|f(x) - f(0)| \le |x|\), so \(\delta = \epsilon\) serves in Theorem 4.3.2 (i). For \(0 < c < 1\), rationals \(x_n \to c\) lying in \((0,1)\) give \(f(x_n) \to c \ne 0\) while irrationals give \(f \to 0\); and at \(c = 1\), irrationals \(y_n \to 1\) give \(f(y_n) = 0 \ne 1 = f(1)\). Corollary 4.3.3 applies at each such \(c\), so

\begin{equation*} D_f = \{x : 0 < x \le 1\}. \end{equation*}

\(D_f = A\).

\(A \subseteq D_f\): fix \(a_n\). Since \(A\) is countable it contains no interval, so \(A^c\) is dense and we may pick \(x_k \in A^c\) with \(x_k \to a_n\). Then \(f(x_k) = 0\) for every \(k\), while \(f(a_n) = 1/n \ne 0\), so \(f\) is discontinuous at \(a_n\) by Corollary 4.3.3.

\(A^c \subseteq D_f^c\): fix \(c \notin A\), so \(f( c) = 0\), and let \(\epsilon > 0\). Choose \(N \in \mathbf{N}\) with \(1/N < \epsilon\) (Archimedean Property, Theorem 1.4.2) and set

\begin{equation*} \delta = \min\{|c - a_1|, |c - a_2|, \ldots, |c - a_N|\} > 0, \end{equation*}

positive because it is a minimum of finitely many nonzero numbers (\(c \ne a_n\) for all \(n\)). If \(|x - c| < \delta\) then \(x \notin \{a_1, \ldots, a_N\}\), so either \(x \notin A\) and \(f(x) = 0\), or \(x = a_m\) with \(m > N\) and \(f(x) = 1/m < 1/N < \epsilon\). In both cases

\begin{equation*} |f(x) - f( c)| = |f(x)| < \epsilon, \end{equation*}

so \(f\) is continuous at \(c\) by Theorem 4.3.2 (i).

Given a limit point \(c\) of \(A\) and \(f : A \to \mathbf{R}\), write \(\lim_{x \to c^-} f(x) = L\) if for all \(\epsilon > 0\) there exists a \(\delta > 0\) such that

\begin{equation*} |f(x) - L| < \epsilon \quad \text{whenever} \quad 0 < c - x < \delta \end{equation*}

(with \(x \in A\), as always): the neighborhood \((c, c+\delta)\) of Definition 4.6.2 is replaced by \((c-\delta, c)\). Equivalently, in terms of sequences, \(\lim_{x \to c^-} f(x) = L\) if \(\lim f(x_n) = L\) for all sequences \((x_n) \subseteq A\) satisfying \(x_n < c\) and \(\lim(x_n) = c\).

Both directions are the observation that the deleted neighborhood \(\{x : 0 < |x - c| < \delta\}\) is the disjoint union of \((c-\delta, c)\) and \((c, c+\delta)\).

(\(\Rightarrow\)) Let \(\epsilon > 0\) and take \(\delta > 0\) from Definition 4.2.1 so that \(|f(x) - L| < \epsilon\) whenever \(0 < |x - c| < \delta\). Since

\begin{equation*} 0 < x - c < \delta \ \Rightarrow \ 0 < |x-c| < \delta, \qquad 0 < c - x < \delta \ \Rightarrow \ 0 < |x-c| < \delta, \end{equation*}

the same \(\delta\) witnesses both one-sided limits.

(\(\Leftarrow\)) Let \(\epsilon > 0\). Pick \(\delta_1 > 0\) with \(|f(x) - L| < \epsilon\) whenever \(0 < x - c < \delta_1\), and \(\delta_2 > 0\) with \(|f(x) - L| < \epsilon\) whenever \(0 < c - x < \delta_2\). Put \(\delta = \min\{\delta_1, \delta_2\} > 0\). If \(0 < |x - c| < \delta\) then \(x \ne c\), so exactly one of

\begin{equation*} 0 < x - c < \delta \le \delta_1 \qquad \text{or} \qquad 0 < c - x < \delta \le \delta_2 \end{equation*}

holds, and in either case \(|f(x) - L| < \epsilon\). Hence \(\lim_{x \to c} f(x) = L\).

(If \(c\) fails to be a limit point of \(A\) from one side, the corresponding one-sided condition holds vacuously for every \(L\), and the argument is unaffected.)

At every \(c\) a monotone \(f\) has both one-sided limits, and they straddle \(f( c)\); this rules out (i) and (iii) at once. Assume \(f : \mathbf{R} \to \mathbf{R}\) is increasing (Definition 4.6.1); the decreasing case follows by applying the result to \(-f\), which is increasing and has the same discontinuities.

Fix \(c \in \mathbf{R}\) and set

\begin{equation*} L^- = \sup\{f(x) : x < c\}, \qquad L^+ = \inf\{f(x) : x > c\}. \end{equation*}

Both sets are nonempty, and \(f( c)\) is an upper bound for the first and a lower bound for the second, so \(L^-\) and \(L^+\) exist in \(\mathbf{R}\) by the Axiom of Completeness and its greatest-lower-bound form (Exercise 1.3.3), and

\begin{equation*} L^- \le f( c) \le L^+ . \end{equation*}

Moreover \(\lim_{x \to c^-} f(x) = L^-\): given \(\epsilon > 0\), Lemma 1.3.8 supplies \(x_0 < c\) with \(f(x_0) > L^- - \epsilon\), and then \(\delta = c - x_0 > 0\) works, since \(0 < c - x < \delta\) forces \(x_0 < x < c\) and monotonicity gives

\begin{equation*} L^- - \epsilon < f(x_0) \le f(x) \le L^-. \end{equation*}

The symmetric argument (Lemma 1.3.8 for infima, Exercise 1.3.1 (b)) gives \(\lim_{x \to c^+} f(x) = L^+\).

Both one-sided limits therefore exist, so category (iii) is impossible. If in addition \(L^- = L^+\), then Theorem 4.6.3 says \(\lim_{x \to c} f(x)\) exists and equals that common value, which by the straddling inequality is forced to be \(f( c)\):

\begin{equation*} L^- \le f( c) \le L^+ = L^- \ \Rightarrow \ \lim_{x \to c} f(x) = f( c), \end{equation*}

so \(f\) is continuous at \(c\) and category (i) is impossible as well. Hence every discontinuity of \(f\) has \(L^- \ne L^+\), i.e. is a jump discontinuity.

Assign to each jump the rational number sitting inside the gap it opens. Take \(f : \mathbf{R} \to \mathbf{R}\) increasing (replace \(f\) by \(-f\) if decreasing), and for each \(c\) let

\begin{equation*} L^-( c) = \sup_{x < c} f(x), \qquad L^+( c) = \inf_{x > c} f(x), \end{equation*}

the one-sided limits produced in Exercise 4.6.5. By that exercise \(D_f = \{c : L^-( c) < L^+( c)\}\), so for \(c \in D_f\) the open interval \(J_c = (L^-( c), L^+( c))\) is nonempty, and by the Density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3) we may choose \(q( c) \in J_c \cap \mathbf{Q}\).

The map \(q : D_f \to \mathbf{Q}\) is injective, because the intervals \(J_c\) are pairwise disjoint. Indeed, let \(c_1 < c_2\) lie in \(D_f\) and pick any \(x\) with \(c_1 < x < c_2\). Then \(x\) belongs to \(\{t : t > c_1\}\) and to \(\{t : t < c_2\}\), so

\begin{equation*} L^+(c_1) \le f(x) \le L^-(c_2), \end{equation*}

and therefore every point of \(J_{c_1}\) is \(< L^+(c_1) \le L^-(c_2) <\) every point of \(J_{c_2}\). Hence \(q(c_1) \ne q(c_2)\).

Being injective, \(q\) is a bijection from \(D_f\) onto its range \(q(D_f) \subseteq \mathbf{Q}\), which is the required bijection with a subset of \(\mathbf{Q}\). Since \(\mathbf{Q}\) is countable (Theorem 1.5.6 (i)), Theorem 1.5.7 makes \(q(D_f)\) either finite or countable, and a set in bijective correspondence with a finite or countable set is finite or countable. Thus \(D_f\) is finite or countable, never uncountable.

(a) Exhibit the unions.

\begin{equation*} \begin{aligned} \mathbf{R} &= \bigcup_{n=1}^{\infty} [-n, n], \\ \mathbf{R}\setminus\{0\} &= \bigcup_{n=1}^{\infty} \Big( [-n, -\tfrac1n] \cup [\tfrac1n, n] \Big), \\ \mathbf{Q} &= \bigcup_{q \in \mathbf{Q}} \{q\}. \end{aligned} \end{equation*}

Closed intervals are closed (Example 3.2.9 (ii)), a finite union of closed sets is closed (Theorem 3.2.14 (i)), and singletons are closed (a one-point set has no limit points, so Definition 3.2.7 holds vacuously); the third union is indexed by the countable set \(\mathbf{Q}\) (Theorem 1.5.6 (i)). For the middle line, \(x \ne 0\) gives \(1/|x| < n\) and \(|x| < n\) for some \(n\) by the Archimedean Property (Theorem 1.4.2), so \(x\) lies in the \(n\)th set; conversely no set in the union contains \(0\).

(b) Every open interval is already a countable union of closed intervals:

\begin{equation*} (a,b) = \bigcup_{n = N}^{\infty} \Big[ a + \tfrac1n, \ b - \tfrac1n \Big], \qquad 2/N < b - a, \end{equation*}

such an \(N\) existing by Theorem 1.4.2. Each \(x \in (a,b)\) satisfies \(1/n < \min\{x - a, b - x\}\) for all large \(n\), so \(x\) lies in the union; the reverse inclusion is clear. Hence for part (a) of Exercise 4.6.1,

\begin{equation*} \mathbf{Z}^c = \bigcup_{k \in \mathbf{Z}} (k, k+1) \end{equation*}

is a countable union of countable unions of closed sets, which is again a countable union of closed sets (Theorem 1.5.8 (ii)), so \(\mathbf{Z}^c \in F_\sigma\). For part (b),

\begin{equation*} \{x : 0 < x \le 1\} = \{1\} \cup \bigcup_{n=3}^{\infty} \Big[ \tfrac1n, \ 1 - \tfrac1n \Big] \end{equation*}

is a countable union of closed sets outright, hence in \(F_\sigma\).

The complement is open, because a \(\delta\) that witnesses \(\alpha\)-continuity at \(x\) witnesses it at every point of \((x - \delta, x + \delta)\).

Let \(x \notin D_f^{\alpha}\) and pick \(\delta > 0\) as in Definition 4.6.5, so that

\begin{equation*} |f(y) - f(z)| < \alpha \quad \text{for all } y, z \in (x - \delta, x + \delta). \end{equation*}

Given \(w \in (x - \delta, x + \delta)\), set \(\delta_w = \delta - |w - x| > 0\). If \(|y - w| < \delta_w\) then

\begin{equation*} |y - x| \le |y - w| + |w - x| < \delta_w + |w - x| = \delta, \end{equation*}

so \((w - \delta_w, w + \delta_w) \subseteq (x - \delta, x + \delta)\) and the displayed estimate applies to every pair \(y, z \in (w - \delta_w, w + \delta_w)\). Hence \(f\) is \(\alpha\)-continuous at \(w\), i.e. \(w \notin D_f^{\alpha}\).

Thus \((D_f^{\alpha})^c\) contains a neighborhood of each of its points and so is open, and \(D_f^{\alpha}\) is closed by Theorem 3.2.13.

A \(\delta\) that keeps the variation below \(\alpha\) keeps it below the larger \(\alpha^{\prime}\), so \(\alpha\)-continuity implies \(\alpha^{\prime}\)-continuity; take contrapositives.

In detail, suppose \(x \notin D_f^{\alpha}\), and let \(\delta > 0\) be as in Definition 4.6.5 for \(\alpha\). For all \(y, z \in (x - \delta, x + \delta)\),

\begin{equation*} |f(y) - f(z)| < \alpha < \alpha^{\prime}, \end{equation*}

so the same \(\delta\) shows \(f\) is \(\alpha^{\prime}\)-continuous at \(x\), i.e. \(x \notin D_f^{\alpha^{\prime}}\). Therefore \(x \in D_f^{\alpha^{\prime}}\) forces \(x \in D_f^{\alpha}\), which is the asserted inclusion.

Run the definition of continuity with \(\varepsilon = \alpha/2\) and use the triangle inequality through \(f(x)\).

Since \(f\) is continuous at \(x\) (Definition 4.3.1) and \(\alpha/2 > 0\), there is a \(\delta > 0\) such that \(|y - x| < \delta\) implies \(|f(y) - f(x)| < \alpha/2\). Then for any \(y, z \in (x - \delta, x + \delta)\),

\begin{equation*} \begin{aligned} |f(y) - f(z)| &\le |f(y) - f(x)| + |f(x) - f(z)| \\ &< \tfrac{\alpha}{2} + \tfrac{\alpha}{2} = \alpha . \end{aligned} \end{equation*}

So this \(\delta\) meets the requirement of Definition 4.6.5 and \(f\) is \(\alpha\)-continuous at \(x\).

For the inclusion, take contrapositives of what was just proved: if \(x \notin D_f\), i.e. \(f\) is continuous at \(x\), then \(f\) is \(\alpha\)-continuous at \(x\), i.e. \(x \notin D_f^{\alpha}\). Hence \(D_f^{\alpha} \subseteq D_f\).

Take \(\alpha = \varepsilon_0\), the \(\varepsilon\) for which continuity fails at \(x\).

Failure of continuity at \(x\) (the negation of Definition 4.3.1) supplies an \(\varepsilon_0 > 0\) such that for every \(\delta > 0\) there is a point \(y_{\delta}\) with

\begin{equation*} |y_{\delta} - x| < \delta \quad \text{and} \quad |f(y_{\delta}) - f(x)| \ge \varepsilon_0 . \end{equation*}

Put \(\alpha = \varepsilon_0\). Given any \(\delta > 0\), the two points \(y_{\delta}, x\) both lie in \((x - \delta, x + \delta)\) yet \(|f(y_{\delta}) - f(x)| \ge \alpha\), so no \(\delta\) can satisfy Definition 4.6.5. Thus \(f\) is not \(\alpha\)-continuous at \(x\), i.e. \(x \in D_f^{\varepsilon_0}\).

Now the two inclusions. If \(x \in D_f\), produce \(\varepsilon_0 > 0\) as above and use the Archimedean Property (Theorem 1.4.2) to choose \(n \in \mathbf{N}\) with \(\alpha_n = 1/n < \varepsilon_0\). Exercise 4.6.9, applied with \(\alpha = 1/n < \alpha^{\prime} = \varepsilon_0\), gives

\begin{equation*} x \in D_f^{\varepsilon_0} \subseteq D_f^{1/n} = D_f^{\alpha_n}, \end{equation*}

so \(D_f \subseteq \bigcup_{n=1}^{\infty} D_f^{\alpha_n}\). Conversely \(D_f^{\alpha_n} \subseteq D_f\) for every \(n\) by Exercise 4.6.10, so the union is contained in \(D_f\). Hence

\begin{equation*} D_f = \bigcup_{n=1}^{\infty} D_f^{\alpha_n}. \end{equation*}

Both are the Algebraic Limit Theorem for Functional Limits (Corollary 4.2.4) applied to difference quotients, whose limits exist by hypothesis.

(i) For \(x \in A\) with \(x \neq c\),

\begin{equation*} \frac{(f+g)(x) - (f+g)( c)}{x - c} = \frac{f(x) - f( c)}{x - c} + \frac{g(x) - g( c)}{x - c}. \end{equation*}

Each summand on the right tends to \(f^{\prime}( c)\), respectively \(g^{\prime}( c)\), as \(x \to c\) (Definition 5.2.1), so both limits exist and Corollary 4.2.4 (ii) gives

\begin{equation*} (f+g)^{\prime}( c) = \lim_{x \to c} \frac{(f+g)(x) - (f+g)( c)}{x-c} = f^{\prime}( c) + g^{\prime}( c). \end{equation*}

(ii) For \(x \neq c\),

\begin{equation*} \frac{(kf)(x) - (kf)( c)}{x - c} = k \cdot \frac{f(x) - f( c)}{x - c}, \end{equation*}

and Corollary 4.2.4 (i) lets the constant \(k\) pass through the limit, so the limit exists and

\begin{equation*} (kf)^{\prime}( c) = k \lim_{x \to c} \frac{f(x) - f( c)}{x - c} = k f^{\prime}( c). \end{equation*}

(c) is the impossible one.

(a) Possible: \(f(x) = g(x) = |x|\). Neither is differentiable at zero (Example 5.2.2 (ii)), while \((fg)(x) = x^2\) has \((fg)^{\prime}(0) = 0\) by Example 5.2.2 (i).

(b) Possible: \(f(x) = |x|\) and \(g(x) = 0\). Then \(g\) is differentiable at zero and \(fg\) is identically zero, hence differentiable at zero.

(c) Impossible. If \(f+g\) and \(g\) are both differentiable at zero, then so is

\begin{equation*} f = (f+g) + (-1)g, \end{equation*}

by Theorem 5.2.4 (i) and (ii), contradicting the requirement that \(f\) not be differentiable at zero.

(d) Possible: take

\begin{equation*} f(x) = \begin{cases} x^2 & \text{if } x \in \mathbf{Q},\\ 0 & \text{if } x \notin \mathbf{Q}.\end{cases} \end{equation*}

At zero, \(|f(x) - f(0)|/|x - 0| = |f(x)|/|x| \le |x|\) for all \(x \neq 0\), so the difference quotient is squeezed to \(0\) and \(f^{\prime}(0) = 0\). At any \(c \neq 0\), \(f\) fails to be continuous: rationals \(x_n \to c\) give \(f(x_n) = x_n^2 \to c^2\) and irrationals \(y_n \to c\) give \(f(y_n) \to 0\), and since \(c^2 \neq 0\) at least one of these limits differs from \(f( c)\), so Corollary 4.3.3 applies. By Theorem 5.2.3, \(f\) is therefore not differentiable at any \(c \neq 0\).

(a) \(h^{\prime}( c) = -1/c^2\) on any interval \(A\) with \(0 \notin A\). Indeed, for \(x, c \in A\) with \(x \neq c\),

\begin{equation*} \frac{h(x) - h( c)}{x - c} = \frac{\frac{1}{x} - \frac{1}{c}}{x - c} = \frac{c - x}{xc\,(x-c)} = \frac{-1}{xc}, \end{equation*}

and \(\lim_{x \to c} (-1/(xc)) = -1/c^2\) by Corollary 4.2.4, since \(c \neq 0\).

(b) Since \(g\) is differentiable at \(c\) it is continuous there (Theorem 5.2.3), so with \(\varepsilon = |g( c)|/2\) there is \(\delta > 0\) with \(g(x) \in B := V_{|g( c)|/2}(g( c))\) for every \(x\) in the interval \(A^{\prime} := V_{\delta}( c) \cap A\). Here \(B\) is an interval each of whose points has modulus exceeding \(|g( c)|/2\), so \(0 \notin B\) and \(h(x) = 1/x\) is defined on \(B\); restricting to \(A^{\prime}\) changes no derivative at \(c\). The Chain Rule (Theorem 5.2.5) applied to \(h \circ g\) with part (a) gives

\begin{equation*} \left(\frac{1}{g}\right)^{\prime}( c) = h^{\prime}(g( c))\, g^{\prime}( c) = \frac{-g^{\prime}( c)}{[g( c)]^2}. \end{equation*}

Now \(f/g = f \cdot (1/g)\), so the Product Rule, Theorem 5.2.4 (iii), yields

\begin{equation*} \begin{aligned} \left(\frac{f}{g}\right)^{\prime}( c) &= f^{\prime}( c)\,\frac{1}{g( c)} + f( c)\,\frac{-g^{\prime}( c)}{[g( c)]^2}\\[2pt] &= \frac{g( c)f^{\prime}( c) - f( c)g^{\prime}( c)}{[g( c)]^2}. \end{aligned} \end{equation*}

(c) Keep the interval \(A^{\prime}\) from (b), on which \(g\) never vanishes. For \(x \in A^{\prime}\), \(x \neq c\), add and subtract \(f( c)g( c)\) in the numerator:

\begin{equation*} \begin{aligned} \frac{(f/g)(x) - (f/g)( c)}{x-c} &= \frac{f(x)g( c) - f( c)g(x)}{g(x)g( c)\,(x-c)}\\[2pt] &= \frac{g( c)\bigl(f(x)-f( c)\bigr) - f( c)\bigl(g(x)-g( c)\bigr)}{g(x)g( c)\,(x-c)}\\[2pt] &= \frac{1}{g(x)g( c)}\left[ g( c)\frac{f(x)-f( c)}{x-c} - f( c)\frac{g(x)-g( c)}{x-c}\right]. \end{aligned} \end{equation*}

As \(x \to c\) the two difference quotients tend to \(f^{\prime}( c)\) and \(g^{\prime}( c)\), and \(g(x) \to g( c) \neq 0\) by Theorem 5.2.3, so Corollary 4.2.4 (its quotient part (iv) is legitimate because the limit of \(g(x)g( c)\) is \([g( c)]^2 \neq 0\)) gives

\begin{equation*} \left(\frac{f}{g}\right)^{\prime}( c) = \frac{g( c)f^{\prime}( c) - f( c)g^{\prime}( c)}{[g( c)]^2}. \end{equation*}

(a) The witness is the difference quotient itself, filled in at \(a\) by the derivative:

\begin{equation*} l(x) = \begin{cases} \dfrac{h(x) - h(a)}{x-a} & \text{if } x \neq a,\\[6pt] h^{\prime}(a) & \text{if } x = a.\end{cases} \end{equation*}

(\(\Rightarrow\)) Assume \(h^{\prime}(a)\) exists and define \(l\) as above. The identity \(h(x)-h(a) = l(x)(x-a)\) holds for \(x \neq a\) by construction and at \(x = a\) because both sides are \(0\). Continuity at \(a\) is exactly Definition 5.2.1: \(\lim_{x \to a} l(x) = h^{\prime}(a) = l(a)\).

(\(\Leftarrow\)) Assume such an \(l\) exists. Dividing the identity by \(x - a\) gives

\begin{equation*} \frac{h(x)-h(a)}{x-a} = l(x) \quad \text{for all } x \in A,\ x \neq a, \end{equation*}

and continuity of \(l\) at \(a\) forces \(\lim_{x\to a} l(x) = l(a)\). Hence the limit defining \(h^{\prime}(a)\) exists and equals \(l(a)\).

(b) Apply (\(\Rightarrow\)) twice. Since \(f\) is differentiable at \(c\), there is \(l : A \to \mathbf{R}\), continuous at \(c\) with \(l( c) = f^{\prime}( c)\), such that

\begin{equation*} f(x) - f( c) = l(x)(x-c) \qquad \text{for all } x \in A. \end{equation*}

Since \(g\) is differentiable at \(f( c)\), there is \(m : B \to \mathbf{R}\), continuous at \(f( c)\) with \(m(f( c)) = g^{\prime}(f( c))\), such that

\begin{equation*} g(y) - g(f( c)) = m(y)\bigl(y - f( c)\bigr) \qquad \text{for all } y \in B. \end{equation*}

Because \(f(A) \subseteq B\), we may substitute \(y = f(x)\) for any \(x \in A\) and then use the first identity:

\begin{equation*} \begin{aligned} (g \circ f)(x) - (g \circ f)( c) &= m(f(x))\bigl(f(x)-f( c)\bigr)\\ &= \underbrace{m(f(x))\,l(x)}_{=:\,L(x)}\,(x-c), \end{aligned} \end{equation*}

valid for all \(x \in A\). It remains to see that \(L\) is continuous at \(c\). The function \(f\) is continuous at \(c\) (Theorem 5.2.3) and \(m\) is continuous at \(f( c)\), so \(m \circ f\) is continuous at \(c\) by Theorem 4.3.9; \(l\) is continuous at \(c\); and products of functions continuous at a point are continuous there (Theorem 4.3.4). By the (\(\Leftarrow\)) direction of part (a), \(g \circ f\) is differentiable at \(c\) and

\begin{equation*} (g \circ f)^{\prime}( c) = L( c) = m(f( c))\,l( c) = g^{\prime}(f( c))\,f^{\prime}( c). \end{equation*}

(a) \(a > 0\). The left-hand limit is \(0\) for every \(a\), so continuity at zero is the statement \(\lim_{x \to 0^+} x^a = 0\). For \(a > 0\) this holds, since given \(\varepsilon > 0\) the choice \(\delta = \varepsilon^{1/a}\) gives \(0 < x < \delta \Rightarrow x^a < \varepsilon\). For \(a = 0\) the right-hand limit is \(1 \neq 0\), and for \(a < 0\) we have \(x^a \to +\infty\); in both cases \(f_a\) is discontinuous at zero.

(b) \(a > 1\), and then \(f_a^{\prime}\) is continuous on all of \(\mathbf{R}\). The difference quotient at zero is

\begin{equation*} \frac{f_a(x) - f_a(0)}{x - 0} = \begin{cases} x^{a-1} & \text{if } x > 0,\\ 0 & \text{if } x < 0,\end{cases} \end{equation*}

whose left-hand limit is \(0\). So the limit exists exactly when \(\lim_{x\to 0^+} x^{a-1} = 0\), which by the computation in (a) happens exactly when \(a - 1 > 0\); then \(f_a^{\prime}(0) = 0\). (For \(a = 1\) the right-hand limit is \(1 \neq 0\), and for \(a < 1\) it is \(+\infty\).) For such \(a\),

\begin{equation*} f_a^{\prime}(x) = \begin{cases} a x^{a-1} & \text{if } x > 0,\\ 0 & \text{if } x \le 0,\end{cases} \end{equation*}

that is, \(f_a^{\prime} = a\,f_{a-1}\). Since \(a - 1 > 0\), part (a) says \(f_{a-1}\) is continuous at zero, and away from zero \(f_a^{\prime}\) is continuous by the usual rules. So yes: whenever \(f_a\) is differentiable at zero, \(f_a^{\prime}\) is continuous.

(c) \(a > 2\). By (b), \(f_a^{\prime} = a f_{a-1}\), and this is differentiable at zero precisely when \(f_{a-1}\) is, i.e. precisely when \(a - 1 > 1\). (At \(a = 2\), \(f_2^{\prime}(x) = 2x\) for \(x > 0\) and \(0\) for \(x \le 0\), whose difference quotient at zero has right-hand limit \(2\) and left-hand limit \(0\).) Away from zero \(f_a^{\prime\prime}(x) = a(a-1)x^{a-2}\) for \(x > 0\) and \(0\) for \(x < 0\) always exists, so \(f_a\) is twice-differentiable on \(\mathbf{R}\) exactly for \(a > 2\).

(a) The substitution \(h = x - c\) carries one \(\varepsilon\)-\(\delta\) statement to the other verbatim, so the two limits exist together and agree. Write \(F(x) = \bigl(g(x)-g( c)\bigr)/(x-c)\) for \(x \in A\), \(x \neq c\), and \(G(h) = F(c+h)\), defined for exactly those \(h \neq 0\) with \(c + h \in A\). Fix \(L \in \mathbf{R}\). By Definition 4.2.1, \(\lim_{x \to c} F(x) = L\) says: for every \(\varepsilon > 0\) there is \(\delta > 0\) with \(|F(x) - L| < \varepsilon\) whenever \(x \in A\) and \(0 < |x - c| < \delta\). Under \(h = x - c\) the condition \(0 < |x-c| < \delta\) reads \(0 < |h| < \delta\) and \(|F(x)-L| = |G(h)-L|\), which is word for word \(\lim_{h\to 0} G(h) = L\). Either limit may therefore serve as the definition of \(g^{\prime}( c)\).

(b) Openness of \(A\) gives \(\delta_0 > 0\) with \(V_{\delta_0}( c) \subseteq A\), so both \(c+h\) and \(c-h\) lie in \(A\) for \(0 < |h| < \delta_0\) and the symmetric quotient is defined there. Split it:

\begin{equation*} \begin{aligned} \frac{g(c+h) - g(c-h)}{2h} &= \frac{\bigl(g(c+h)-g( c)\bigr) + \bigl(g( c)-g(c-h)\bigr)}{2h}\\[2pt] &= \frac12\cdot\frac{g(c+h)-g( c)}{h} + \frac12\cdot\frac{g(c-h)-g( c)}{-h}. \end{aligned} \end{equation*}

By part (a) the first quotient tends to \(g^{\prime}( c)\) as \(h \to 0\). For the second, put \(k = -h\): as \(h \to 0\) we have \(k \to 0\), and the quotient is \(\bigl(g(c+k)-g( c)\bigr)/k\), which tends to \(g^{\prime}( c)\) as well by part (a): any \(\delta\) that works for the variable \(k\) works for \(h\), since \(|k| = |h|\). Corollary 4.2.4 now gives

\begin{equation*} \lim_{h\to 0}\frac{g(c+h)-g(c-h)}{2h} = \tfrac12 g^{\prime}( c) + \tfrac12 g^{\prime}( c) = g^{\prime}( c). \end{equation*}

(a) \(a = 3/2\); (b) \(a = 5/2\); (c) \(a = 7/2\). For noninteger \(a\) read \(x^a\) as \(|x|^a\), so that \(g_a\) is defined on all of \(\mathbf{R}\); every estimate below bounds \(|g_a^{\prime}|\) or \(|g_a^{\prime\prime}|\) by powers of \(|x|\) and so is valid for both signs, and every witnessing sequence below is positive.

Two computations govern all three parts. First, for \(x > 0\) the Chain and Product Rules (Theorems 5.2.5, 5.2.4 (iii)) give

\begin{equation*} g_a^{\prime}(x) = a x^{a-1}\sin(1/x) - x^{a-2}\cos(1/x), \end{equation*}

\begin{equation*} \begin{aligned} g_a^{\prime\prime}(x) = \;& a(a-1)x^{a-2}\sin(1/x)\\ & - (2a-2)x^{a-3}\cos(1/x) - x^{a-4}\sin(1/x). \end{aligned} \end{equation*}

Second, at the origin the difference quotient of \(g_a\) is

\begin{equation*} \frac{g_a(x) - g_a(0)}{x-0} = x^{a-1}\sin(1/x), \end{equation*}

which is bounded by \(|x|^{a-1}\) and hence tends to \(0\) when \(a > 1\); when \(a \le 1\) it fails to converge, since along \(x_n = 1/(2\pi n + \pi/2)\) it equals \(x_n^{a-1} \not\to 0\) while along \(x_n = 1/(2\pi n)\) it is \(0\). So

\begin{equation*} g_a \text{ is differentiable on } \mathbf{R} \iff a > 1, \quad\text{and then } g_a^{\prime}(0) = 0. \end{equation*}

(a) Take \(a = 3/2 > 1\), so \(g_a\) is differentiable on \(\mathbf{R}\). Along \(x_n = 1/(2\pi n) \in [0,1]\) we have \(\sin(1/x_n) = 0\) and \(\cos(1/x_n) = 1\), so

\begin{equation*} g_{3/2}^{\prime}(x_n) = -x_n^{-1/2} = -\sqrt{2\pi n} \longrightarrow -\infty . \end{equation*}

Hence \(g_{3/2}^{\prime}\) is unbounded on \([0,1]\).

(b) Take \(a = 5/2\). Since \(a > 1\), \(g_a\) is differentiable on \(\mathbf{R}\) with \(g_a^{\prime}(0) = 0\), and for \(x \neq 0\)

\begin{equation*} |g_{5/2}^{\prime}(x)| \le \tfrac52 |x|^{3/2} + |x|^{1/2} \longrightarrow 0 = g_{5/2}^{\prime}(0), \end{equation*}

so \(g_{5/2}^{\prime}\) is continuous at zero, and it is continuous elsewhere by the usual rules. But its difference quotient at zero, for \(x > 0\),

\begin{equation*} \frac{g_{5/2}^{\prime}(x) - g_{5/2}^{\prime}(0)}{x - 0} = \tfrac52 x^{1/2}\sin(1/x) - x^{-1/2}\cos(1/x), \end{equation*}

is unbounded along \(x_n = 1/(2\pi n)\) (the first term vanishes there and the second equals \(-\sqrt{2\pi n}\)), so the limit does not exist and \(g_{5/2}^{\prime}\) is not differentiable at zero.

(c) Take \(a = 7/2\). Since \(a > 1\), \(g_a\) is differentiable on \(\mathbf{R}\), and its derivative is differentiable at zero because

\begin{equation*} \left|\frac{g_{7/2}^{\prime}(x) - g_{7/2}^{\prime}(0)}{x - 0}\right| \le \tfrac72 |x|^{3/2} + |x|^{1/2} \longrightarrow 0, \end{equation*}

thus \(g_{7/2}^{\prime\prime}(0) = 0\), and \(g_{7/2}^{\prime\prime}\) exists away from zero by the formula above. However, along \(x_n = 1/(2\pi n + \pi/2)\), where \(\sin(1/x_n) = 1\) and \(\cos(1/x_n) = 0\),

\begin{equation*} g_{7/2}^{\prime\prime}(x_n) = \tfrac{35}{4}x_n^{3/2} - x_n^{-1/2} \longrightarrow -\infty, \end{equation*}

while \(g_{7/2}^{\prime\prime}(0) = 0\). So \(g_{7/2}^{\prime\prime}\) is not continuous at zero.

(a) Yes for \(f(x) = x^2\), no for \(g(x) = x^3\). For \(x \ne y\),

\begin{equation*} \frac{x^2 - y^2}{x-y} - 2y = (x+y) - 2y = x - y, \end{equation*}

so \(\delta = \epsilon\) works. For the cube,

\begin{equation*} \frac{x^3 - y^3}{x - y} - 3y^2 = x^2 + xy - 2y^2 = (x-y)(x+2y), \end{equation*}

which is unbounded once \(x - y\) is pinned down. Take \(\epsilon = 1\): given any \(\delta > 0\), set \(x - y = \delta/2\) and let \(y = 2/\delta\); then

\begin{equation*} |(x-y)(x+2y)| = \frac{\delta}{2}\left| \frac{\delta}{2} + \frac{6}{\delta} \right| > 3 > 1 , \end{equation*}

while \(0 < |x - y| < \delta\). So \(g\) fails the definition on \(\mathbf{R}\).

(b) Uniform differentiability on \(A\) makes \(f^{\prime}\) uniformly continuous on \(A\), hence continuous. Given \(\epsilon > 0\), choose \(\delta > 0\) for the value \(\epsilon/2\) in the definition. If \(x, y \in A\) satisfy \(0 < |x - y| < \delta\), then both orderings of the pair may be fed into the definition, and since the difference quotient is symmetric,

\begin{equation*} \frac{f(x) - f(y)}{x - y} = \frac{f(y) - f(x)}{y - x} , \end{equation*}

the triangle inequality gives

\begin{equation*} \begin{aligned} |f^{\prime}(x) - f^{\prime}(y)| &\le \left| f^{\prime}(x) - \frac{f(y) - f(x)}{y - x} \right| \\ &\quad + \left| \frac{f(x) - f(y)}{x - y} - f^{\prime}(y) \right| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

The case \(x = y\) is trivial, so \(|x - y| < \delta\) implies \(|f^{\prime}(x) - f^{\prime}(y)| < \epsilon\) for all \(x,y \in A\), which is Definition 4.4.4 for \(f^{\prime}\).

(c) No, differentiability on a closed interval is not enough; the correct analogue of Theorem 4.4.7 requires a continuous derivative. For the negative half, part (b) says a uniformly differentiable function has continuous derivative, and

\begin{equation*} h(x) = \begin{cases} x^2 \sin(1/x) & \text{if } x \ne 0 \\ 0 & \text{if } x = 0 \end{cases} \end{equation*}

is differentiable on \([-1,1]\) with \(h^{\prime}(x) = 2x\sin(1/x) - \cos(1/x)\) for \(x \ne 0\) and \(h^{\prime}(0) = 0\) (Exercise 5.2.7), and \(h^{\prime}\) is not continuous at \(0\) since \(h^{\prime}\bigl(1/(2\pi n)\bigr) = -1\) for all \(n\). So \(h\) is not uniformly differentiable on \([-1,1]\).

For the positive half: if \(f\) is differentiable on \([a,b]\) and \(f^{\prime}\) is continuous there, then \(f\) is uniformly differentiable on \([a,b]\). Indeed \([a,b]\) is compact, so Theorem 4.4.7 makes \(f^{\prime}\) uniformly continuous: given \(\epsilon > 0\) there is \(\delta > 0\) with \(|f^{\prime}(s) - f^{\prime}(t)| < \epsilon\) whenever \(|s - t| < \delta\). For \(x \ne y\) in \([a,b]\) the Mean Value Theorem (Theorem 5.3.2, applicable since \(f\) is continuous on the closed interval with endpoints \(x,y\) and differentiable on its interior) supplies a \(c\) strictly between \(x\) and \(y\) with

\begin{equation*} \frac{f(x) - f(y)}{x - y} = f^{\prime}( c), \qquad |c - y| < |x - y| , \end{equation*}

so \(0 < |x - y| < \delta\) forces \(|f^{\prime}( c) - f^{\prime}(y)| < \epsilon\), which is exactly the required estimate.

(a) True; (b) false; (c) true.

(a) Since \(f^{\prime}\) is not constant there are points \(a < b\) of the interval with \(f^{\prime}(a) \ne f^{\prime}(b)\), and \([a,b]\) lies inside the interval. Corollary 1.4.4 supplies an irrational \(\alpha\) strictly between \(f^{\prime}(a)\) and \(f^{\prime}(b)\); whichever of the two orderings holds, it is one of the two hypotheses \(f^{\prime}(a) < \alpha < f^{\prime}(b)\), \(f^{\prime}(a) > \alpha > f^{\prime}(b)\) of Darboux’s Theorem (Theorem 5.2.7), which applies because \(f\) is differentiable on \([a,b]\) and produces \(c \in (a,b)\) with

\begin{equation*} f^{\prime}( c) = \alpha \notin \mathbf{Q}. \end{equation*}

(b) The counterexample is the function of Exercise 5.2.10,

\begin{equation*} g(x) = \begin{cases} x/2 + x^2 \sin(1/x) & \text{if } x \ne 0 \\ 0 & \text{if } x = 0, \end{cases} \end{equation*}

which is differentiable on \(\mathbf{R}\) with \(g^{\prime}(0) = 1/2 > 0\) and, for \(x \ne 0\),

\begin{equation*} g^{\prime}(x) = \tfrac{1}{2} + 2x \sin(1/x) - \cos(1/x). \end{equation*}

Every neighborhood \(V_{\delta}(0)\) contains \(x_n = 1/(2\pi n)\) for all large \(n\), and there

\begin{equation*} g^{\prime}(x_n) = \tfrac{1}{2} + 0 - 1 = -\tfrac{1}{2} < 0 . \end{equation*}

So no \(\delta\)-neighborhood of \(c = 0\) has \(g^{\prime} > 0\) throughout.

(c) Fix \(h \ne 0\) small enough that \([0,h]\) (or \([h,0]\)) lies in the interval. The Mean Value Theorem (Theorem 5.3.2; \(f\) is differentiable, hence continuous by Theorem 5.2.3, on that closed interval) gives a point \(c_h\) strictly between \(0\) and \(h\) with

\begin{equation*} \frac{f(h) - f(0)}{h} = f^{\prime}(c_h), \qquad 0 < |c_h| < |h| . \end{equation*}

Let \((h_n)\) be any sequence tending to \(0\) with \(h_n \ne 0\). Then \(c_{h_n} \to 0\) with \(c_{h_n} \ne 0\), so by the Sequential Criterion for Functional Limits (Theorem 4.2.3) applied to \(f^{\prime}\),

\begin{equation*} \frac{f(h_n) - f(0)}{h_n} = f^{\prime}(c_{h_n}) \longrightarrow L . \end{equation*}

Theorem 4.2.3 in the other direction now gives \(f^{\prime}(0) = \lim_{h \to 0} (f(h) - f(0))/h = L\).

\(g^{\prime}(0) = 1/2\), yet \(g^{\prime} = -1/2\) at every \(x_n = 1/(2\pi n)\), and a point of negative derivative kills monotonicity nearby.

Away from the origin \(g\) is a composition and product of functions differentiable on \(\mathbf{R} \setminus \{0\}\), so by the Algebraic Differentiability Theorem (Theorem 5.2.4) and the Chain Rule (Theorem 5.2.5),

\begin{equation*} g^{\prime}(x) = \tfrac{1}{2} + 2x\sin(1/x) - \cos(1/x) \qquad (x \ne 0). \end{equation*}

At the origin, Definition 5.2.1 and \(|h\sin(1/h)| \le |h|\) give

\begin{equation*} g^{\prime}(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0}\left( \tfrac{1}{2} + h\sin(1/h) \right) = \tfrac{1}{2} > 0 . \end{equation*}

So \(g\) is differentiable on all of \(\mathbf{R}\).

Now let \(I\) be any open interval containing \(0\), and choose \(c > 0\) with \((-c,c) \subseteq I\). Fix \(n \in \mathbf{N}\) large enough that \(x_n = 1/(2\pi n) < c/2\). Since \(\sin(2\pi n) = 0\) and \(\cos(2\pi n) = 1\),

\begin{equation*} g^{\prime}(x_n) = \tfrac{1}{2} + 0 - 1 = -\tfrac{1}{2} . \end{equation*}

Apply Definition 5.2.1 at \(x_n\) with \(\epsilon = 1/4\): there is \(\delta > 0\), which we may shrink so that \(x_n + \delta < c\), such that

\begin{equation*} \left| \frac{g(t) - g(x_n)}{t - x_n} + \frac{1}{2} \right| < \frac14 \quad \text{whenever } 0 < |t - x_n| < \delta . \end{equation*}

Take \(t = x_n + \delta/2\). Then \(x_n < t < c\), so both points lie in \(I\); the quotient is \(< -1/4 < 0\) while \(t - x_n > 0\), so

\begin{equation*} g(t) - g(x_n) < 0, \qquad \text{i.e. } g(t) < g(x_n) \text{ with } x_n < t . \end{equation*}

This violates the definition of increasing on \(I\).

(a) Both points come straight out of Definition 5.2.1 with \(\epsilon\) half the relevant slope.

At \(a\): apply the definition of \(g^{\prime}(a)\) with \(\epsilon = -g^{\prime}(a)/2 > 0\) to get \(\delta > 0\) such that

\begin{equation*} \left| \frac{g(t) - g(a)}{t - a} - g^{\prime}(a) \right| < \frac{-g^{\prime}(a)}{2} \quad \text{for } t \in [a,b], \ 0 < |t-a| < \delta, \end{equation*}

so the quotient is \(< g^{\prime}(a)/2 < 0\) there. Choose any \(x\) with \(a < x < \min\{a + \delta, b\}\); since \(x - a > 0\), the displayed sign forces

\begin{equation*} g(x) - g(a) < 0, \qquad \text{i.e. } g(a) > g(x). \end{equation*}

At \(b\): apply the definition of \(g^{\prime}(b)\) with \(\epsilon = g^{\prime}(b)/2 > 0\) to get \(\delta^{\prime} > 0\) with the quotient \(> g^{\prime}(b)/2 > 0\) for \(t \in [a,b]\), \(0 < |t - b| < \delta^{\prime}\). Choose \(y\) with \(\max\{a, b - \delta^{\prime}\} < y < b\). Now \(y - b < 0\), so

\begin{equation*} g(y) - g(b) = \underbrace{\frac{g(y)-g(b)}{y-b}}_{>\,0} \cdot \underbrace{(y - b)}_{<\,0} < 0, \end{equation*}

i.e. \(g(y) < g(b)\).

(b) Continue with \(g(x) = f(x) - \alpha x\), which is differentiable on \([a,b]\) with \(g^{\prime} = f^{\prime} - \alpha\) and \(g^{\prime}(a) < 0 < g^{\prime}(b)\). Differentiability implies continuity (Theorem 5.2.3), so \(g\) is continuous on the compact set \([a,b]\) and the Extreme Value Theorem (Theorem 4.4.2) gives a point \(c \in [a,b]\) with

\begin{equation*} g( c) \le g(t) \qquad \text{for all } t \in [a,b]. \end{equation*}

Part (a) rules out both endpoints: \(g(x) < g(a)\) for some \(x \in (a,b)\) shows \(c \ne a\), and \(g(y) < g(b)\) for some \(y \in (a,b)\) shows \(c \ne b\). Hence \(c \in (a,b)\), and \(g( c)\) is in particular a minimum value of \(g\) on the open interval \((a,b)\), so the Interior Extremum Theorem (Theorem 5.2.6, whose hypotheses hold since \(g\) is differentiable on \((a,b)\)) yields

\begin{equation*} 0 = g^{\prime}( c) = f^{\prime}( c) - \alpha, \qquad f^{\prime}( c) = \alpha . \end{equation*}

For the remaining case \(f^{\prime}(a) > \alpha > f^{\prime}(b)\), apply what was just proved to \(-f\) and \(-\alpha\): \((-f)^{\prime}(a) = -f^{\prime}(a) < -\alpha < -f^{\prime}(b) = (-f)^{\prime}(b)\), so some \(c \in (a,b)\) has \(-f^{\prime}( c) = -\alpha\).

Write \(g = f^{-1}\). The whole proof is the algebraic identity

\begin{equation*} \frac{g(y) - g(y_0)}{y - y_0} = \left( \frac{f(x) - f(x_0)}{x - x_0} \right)^{-1}, \qquad x = g(y), \ x_0 = g(y_0), \end{equation*}

run through the Sequential Criterion for Functional Limits.

The domain of \(g\) is \(f([a,b]) = [c,d]\) with \(c < d\): differentiability gives continuity (Theorem 5.2.3), the Extreme Value Theorem (Theorem 4.4.2) and the Intermediate Value Theorem (Theorem 4.5.1) make the range a closed interval, and \(c \ne d\) since \(f\) is one-to-one on \([a,b]\). So every point of \([c,d]\) is a limit point of it, and \(g\) is continuous there by Exercise 4.5.8.

Fix \(y_0 \in [c,d]\) and put \(x_0 = g(y_0)\), so \(y_0 = f(x_0)\). Let \((y_n)\) be any sequence in \([c,d] \setminus \{y_0\}\) with \(y_n \to y_0\), and set \(x_n = g(y_n)\). Then:

(i) \(x_n \ne x_0\), since \(g\) is one-to-one (\(f\) is a function) and \(y_n \ne y_0\);

(ii) \(x_n \to x_0\), by continuity of \(g\) at \(y_0\) and Theorem 4.3.2 (iii).

Because \(f\) is differentiable at \(x_0\), Theorem 4.2.3 applied to the difference-quotient function \(h(t) = (f(t) - f(x_0))/(t - x_0)\) along the sequence \((x_n)\) gives

\begin{equation*} \frac{f(x_n) - f(x_0)}{x_n - x_0} \longrightarrow f^{\prime}(x_0) \ne 0 . \end{equation*}

Since \(f(x_n) = y_n\) and \(f(x_0) = y_0\), the reciprocal of this quotient is exactly \((g(y_n) - g(y_0))/(y_n - y_0)\), so the Algebraic Limit Theorem (Theorem 2.3.3 (iv), whose hypothesis \(f^{\prime}(x_0) \ne 0\) is our standing assumption) yields

\begin{equation*} \frac{g(y_n) - g(y_0)}{y_n - y_0} = \left( \frac{f(x_n) - f(x_0)}{x_n - x_0} \right)^{-1} \longrightarrow \frac{1}{f^{\prime}(x_0)} . \end{equation*}

The sequence \((y_n)\) was arbitrary, so Theorem 4.2.3 in the converse direction gives

\begin{equation*} \left( f^{-1} \right)^{\prime}(y_0) = \lim_{y \to y_0} \frac{g(y) - g(y_0)}{y - y_0} = \frac{1}{f^{\prime}(x_0)}, \end{equation*}

which is the claimed formula with \(y_0 = f(x_0)\).

(a) Take \(M = \sup_{t \in [a,b]} |f^{\prime}(t)|\), which is finite and attained because \(|f^{\prime}|\) is continuous on the compact set \([a,b]\) (Extreme Value Theorem, Theorem 4.4.2); if this supremum is \(0\) replace it by \(M = 1\) so that \(M > 0\). For \(x \ne y\) in \([a,b]\), \(f\) is continuous on the closed interval between them and differentiable on its interior, so the Mean Value Theorem (Theorem 5.3.2) supplies a \(c\) strictly between \(x\) and \(y\) with

\begin{equation*} \left| \frac{f(x) - f(y)}{x - y} \right| = |f^{\prime}( c)| \le M . \end{equation*}

(b) Yes, with (a)’s standing hypothesis that \(f^{\prime}\) is continuous. That hypothesis turns the bound into a maximum rather than a mere supremum: \(|f^{\prime}|\) is continuous on the compact set \([a,b]\), so by the Extreme Value Theorem (Theorem 4.4.2) it is attained at some \(x_0 \in [a,b]\), whence

\begin{equation*} k := \max_{t \in [a,b]} |f^{\prime}(t)| = |f^{\prime}(x_0)| < 1 . \end{equation*}

If \(k = 0\) then \(f\) is constant and any constant in \((0,1)\) serves; otherwise \(0 < k < 1\), and the display in (a) run with \(k\) in place of \(M\) gives \(|f(x) - f(y)| \le k|x - y|\) for all \(x, y \in [a,b]\), the contractive condition of Exercise 4.3.11.

Rolle’s Theorem, contrapositively. Suppose \(x < y\) in \(A\) with \(f(x) = f(y)\). Since \(A\) is an interval, \([x,y] \subseteq A\), so \(f\) is differentiable — hence continuous (Theorem 5.2.3) — on \([x,y]\) and differentiable on \((x,y)\). Rolle’s Theorem (Theorem 5.3.1) produces \(c \in (x,y) \subseteq A\) with

\begin{equation*} f^{\prime}( c) = 0, \end{equation*}

contradicting the hypothesis. Thus \(f(x) \ne f(y)\) whenever \(x \ne y\).

The converse fails: \(f(x) = x^3\) on \(A = \mathbf{R}\) is strictly increasing, hence one-to-one, but \(f^{\prime}(0) = 0\).

(a) Apply the Intermediate Value Theorem to \(g(x) = h(x) - x\), which is continuous on \([1,3]\) because \(h\) is differentiable there (Theorem 5.2.3). Its endpoint values straddle zero:

\begin{equation*} g(1) = 2 - 1 = 1 > 0, \qquad g(3) = 2 - 3 = -1 < 0 , \end{equation*}

so Theorem 4.5.1 gives \(d \in (1,3) \subseteq [0,3]\) with \(g(d) = 0\), i.e. \(h(d) = d\).

(b) The Mean Value Theorem (Theorem 5.3.2) on \([0,3]\) — where \(h\) is continuous and differentiable on \((0,3)\) — gives \(c \in (0,3)\) with

\begin{equation*} h^{\prime}( c) = \frac{h(3) - h(0)}{3 - 0} = \frac{2 - 1}{3} = \frac{1}{3}. \end{equation*}

(c) Produce a second derivative value and interpolate. The Mean Value Theorem on \([1,3]\) gives \(c_2 \in (1,3)\) with

\begin{equation*} h^{\prime}(c_2) = \frac{h(3) - h(1)}{3 - 1} = \frac{2 - 2}{2} = 0 . \end{equation*}

Together with the point \(c\) of (b) we have two values of \(h^{\prime}\) with

\begin{equation*} h^{\prime}(c_2) = 0 < \tfrac{1}{4} < \tfrac{1}{3} = h^{\prime}( c) . \end{equation*}

Since \(h\) is differentiable on the closed interval with endpoints \(c\) and \(c_2\), Darboux’s Theorem (Theorem 5.2.7) — derivatives have the intermediate value property, no continuity of \(h^{\prime}\) required — yields a point \(x\) strictly between \(c_2\) and \(c\) with \(h^{\prime}(x) = 1/4\).

(a) Feed the sequence \((x_n)\) into the Sequential Criterion for Functional Limits (Theorem 4.2.3) twice. First, \(f\) is continuous at \(0\) (Theorem 5.2.3) and \(x_n \to 0\), so

\begin{equation*} f(0) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} 0 = 0 . \end{equation*}

Second, \(f^{\prime}(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}\) exists by hypothesis, and \(0\) is a limit point of \(A \setminus \{0\}\) approached by the \(x_n\) (each \(x_n \ne 0\)), so Theorem 4.2.3 lets us evaluate that limit along \((x_n)\):

\begin{equation*} f^{\prime}(0) = \lim_{n \to \infty} \frac{f(x_n) - f(0)}{x_n} = \lim_{n \to \infty} \frac{0 - 0}{x_n} = 0 . \end{equation*}

(b) The same argument applied to \(f^{\prime}\), once we manufacture zeros of \(f^{\prime}\) accumulating at \(0\). Since \(x_n \to 0\) with every \(x_n \ne 0\), infinitely many terms share a sign; say infinitely many are positive (the negative case is identical with the inequalities reversed). Choose a subsequence strictly decreasing to \(0\),

\begin{equation*} x_{n_1} > x_{n_2} > x_{n_3} > \cdots > 0, \qquad x_{n_k} \to 0 , \end{equation*}

which is possible because for each \(k\) there are infinitely many indices \(n\) with \(0 < x_n < x_{n_k}\). On \([x_{n_{k+1}}, x_{n_k}] \subseteq A\) the function \(f\) is continuous and differentiable with

\begin{equation*} f(x_{n_{k+1}}) = 0 = f(x_{n_k}), \end{equation*}

so Rolle’s Theorem (Theorem 5.3.1) supplies \(y_k \in (x_{n_{k+1}}, x_{n_k})\) with \(f^{\prime}(y_k) = 0\). Then \(0 < y_k < x_{n_k} \to 0\) forces \(y_k \to 0\) with \(y_k \ne 0\). By (a), \(f^{\prime}(0) = 0\); and since \(f^{\prime\prime}(0) = \lim_{y \to 0} \frac{f^{\prime}(y) - f^{\prime}(0)}{y}\) exists, Theorem 4.2.3 evaluates it along \((y_k)\):

\begin{equation*} f^{\prime\prime}(0) = \lim_{k \to \infty} \frac{f^{\prime}(y_k) - f^{\prime}(0)}{y_k} = \lim_{k \to \infty} \frac{0 - 0}{y_k} = 0 . \end{equation*}

(a) Apply the Mean Value Theorem to the auxiliary function named in the book’s proof,

\begin{equation*} h(x) = [f(b) - f(a)]\,g(x) - [g(b) - g(a)]\,f(x). \end{equation*}

As a linear combination of \(f\) and \(g\), \(h\) is continuous on \([a,b]\) and differentiable on \((a,b)\) (Algebraic Differentiability Theorem, Theorem 5.2.4), and the cross terms cancel at the endpoints:

\begin{equation*} \begin{aligned} h(a) &= f(b)g(a) - f(a)g(a) - g(b)f(a) + g(a)f(a) = f(b)g(a) - g(b)f(a), \\ h(b) &= f(b)g(b) - f(a)g(b) - g(b)f(b) + g(a)f(b) = g(a)f(b) - f(a)g(b). \end{aligned} \end{equation*}

Since \(h(a) = h(b)\), the Mean Value Theorem (Theorem 5.3.2) gives \(c \in (a,b)\) with \(h^{\prime}( c) = (h(b) - h(a))/(b-a) = 0\), i.e.

\begin{equation*} [f(b) - f(a)]g^{\prime}( c) - [g(b) - g(a)]f^{\prime}( c) = 0, \end{equation*}

which is the asserted identity. If in addition \(g^{\prime} \ne 0\) on \((a,b)\), then \(g(b) \ne g(a)\) — otherwise Rolle’s Theorem (Theorem 5.3.1) applied to \(g\) alone would produce a zero of \(g^{\prime}\) — so dividing the identity by \(g^{\prime}( c)[g(b) - g(a)]\), both factors nonzero, gives

\begin{equation*} \frac{f^{\prime}( c)}{g^{\prime}( c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \end{equation*}

(b) Read \(t \mapsto (g(t), f(t))\), \(t \in [a,b]\), as a parametrized curve in the plane, with \(g\) the horizontal and \(f\) the vertical coordinate. The chord joining its endpoints \((g(a), f(a))\) and \((g(b), f(b))\) has slope

\begin{equation*} \frac{f(b) - f(a)}{g(b) - g(a)}, \end{equation*}

while the tangent vector at parameter \(c\) is \((g^{\prime}( c), f^{\prime}( c))\), of slope \(f^{\prime}( c)/g^{\prime}( c)\). The theorem says some tangent to the curve is parallel to that chord — the picture of Figure 5.4, now for a curve traced parametrically rather than a graph — and the unquotiented form \([f(b)-f(a)]g^{\prime}( c) = [g(b)-g(a)]f^{\prime}( c)\) records that parallelism as a vanishing cross product, valid even where a slope is infinite.

(a) Fix \(x \in (0,a]\) (the case \(x = 0\) is trivial) and apply the Mean Value Theorem (Theorem 5.3.2) to \(g\) on \([0,x]\), where \(g\) is continuous and differentiable on \((0,x)\): there is \(c \in (0,x)\) with

\begin{equation*} |g(x)| = |g(x) - g(0)| = |g^{\prime}( c)|\,x \le Mx . \end{equation*}

(b) Compare \(h\) against \(x^2/2\) using the Generalized Mean Value Theorem rather than the plain one, which is what buys the factor \(1/2\). Part (a) applied to \(g = h^{\prime}\) (differentiable, \(h^{\prime}(0)=0\), \(|(h^{\prime})^{\prime}| = |h^{\prime\prime}| \le M\)) gives

\begin{equation*} |h^{\prime}(t)| \le Mt \qquad \text{for all } t \in [0,a]. \end{equation*}

Fix \(x \in (0,a]\) and set \(G(t) = t^2/2\), so \(G^{\prime}(t) = t \ne 0\) on \((0,x)\). Both \(h\) and \(G\) are continuous on \([0,x]\) and differentiable on \((0,x)\), so Theorem 5.3.5 yields \(c \in (0,x)\) with

\begin{equation*} \frac{h(x)}{x^2/2} = \frac{h(x) - h(0)}{G(x) - G(0)} = \frac{h^{\prime}( c)}{G^{\prime}( c)} = \frac{h^{\prime}( c)}{c}, \end{equation*}

and the displayed bound on \(|h^{\prime}|\) makes \(|h^{\prime}( c)|/c \le M\). Hence \(|h(x)| \le Mx^2/2\).

(c) Conjecture: if \(k : [0,a] \to \mathbf{R}\) is three times differentiable with

\begin{equation*} k(0) = k^{\prime}(0) = k^{\prime\prime}(0) = 0 \qquad \text{and} \qquad |k^{\prime\prime\prime}(x)| \le M \ \text{ on } [0,a], \end{equation*}

then \(|k(x)| \le Mx^3/6\) for all \(x \in [0,a]\).

Proof: part (b) applied to \(h = k^{\prime}\) (twice differentiable, \(k^{\prime}(0) = (k^{\prime})^{\prime}(0) = k^{\prime\prime}(0) = 0\), \(|(k^{\prime})^{\prime\prime} | = |k^{\prime\prime\prime}| \le M\)) gives

\begin{equation*} |k^{\prime}(t)| \le \frac{Mt^2}{2} \qquad \text{for all } t \in [0,a]. \end{equation*}

Fix \(x \in (0,a]\) and set \(G(t) = t^3/6\), so \(G^{\prime}(t) = t^2/2 \ne 0\) on \((0,x)\). Theorem 5.3.5 on \([0,x]\) (again both functions continuous there, differentiable inside) gives \(c \in (0,x)\) with

\begin{equation*} \frac{k(x)}{x^3/6} = \frac{k(x) - k(0)}{G(x) - G(0)} = \frac{k^{\prime}( c)}{c^2/2}, \end{equation*}

whose modulus is at most \(M\) by the previous display. Hence \(|k(x)| \le Mx^3/6\).

Two fixed points would force the chord slope to be \(1\). Suppose \(x < y\) are both fixed points of \(f\) in the interval \(A\). Since \(A\) is an interval, \([x,y] \subseteq A\), so \(f\) is continuous on \([x,y]\) (Theorem 5.2.3) and differentiable on \((x,y)\), and the Mean Value Theorem (Theorem 5.3.2) gives \(c \in (x,y)\) with

\begin{equation*} f^{\prime}( c) = \frac{f(y) - f(x)}{y - x} = \frac{y - x}{y - x} = 1, \end{equation*}

contradicting \(f^{\prime}( c) \ne 1\). Hence \(f\) has at most one fixed point.

Method (2): apply Exercise 5.3.2 to \(g(x) = f(x) - x\), whose derivative \(g^{\prime}(x) = f^{\prime}(x) - 1\) is never zero; \(g\) is therefore one-to-one, so \(g(x) = 0\) for at most one \(x\).

Every difference quotient at \(0\) equals a value of \(f^{\prime}\) nearby, so it inherits the limit \(L\).

Let \(\epsilon > 0\), and choose \(\delta > 0\) with \((-\delta,\delta)\) inside the domain and

\begin{equation*} 0 < |t| < \delta \quad \Longrightarrow \quad |f^{\prime}(t) - L| < \epsilon . \end{equation*}

Fix \(x\) with \(0 < |x| < \delta\). On the closed interval with endpoints \(0\) and \(x\) the function \(f\) is continuous, and it is differentiable on the corresponding open interval, which omits \(0\); so the Mean Value Theorem (Theorem 5.3.2) supplies a point \(c_x\) strictly between \(0\) and \(x\) with

\begin{equation*} \frac{f(x) - f(0)}{x - 0} = f^{\prime}(c_x), \qquad 0 < |c_x| < |x| < \delta , \end{equation*}

and therefore

\begin{equation*} \left| \frac{f(x) - f(0)}{x - 0} - L \right| = |f^{\prime}(c_x) - L| < \epsilon . \end{equation*}

Hence \(f^{\prime}(0)\) exists and equals \(L\).

Divide the two difference quotients at \(a\) and quote the Algebraic Limit Theorem; no Mean Value Theorem is needed.

The standing hypothesis \(g^{\prime}(x) \ne 0\) for \(x \ne a\) already forces \(g(x) \ne 0\) for \(x \ne a\) (else Rolle’s Theorem, Theorem 5.3.1, applied between \(a\) and \(x\) would zero \(g^{\prime}\) somewhere strictly between), so \(f/g\) is defined near \(a\). There, using \(f(a) = g(a) = 0\),

\begin{equation*} \begin{aligned} \frac{f(x)}{g(x)} &= \frac{f(x)-f(a)}{g(x)-g(a)} \\ &= \frac{\bigl(f(x)-f(a)\bigr)/(x-a)}{\bigl(g(x)-g(a)\bigr)/(x-a)} . \end{aligned} \end{equation*}

Numerator and denominator tend to \(f^{\prime}(a)\) and \(g^{\prime}(a)\), and \(g^{\prime}(a) \ne 0\) is exactly the nonvanishing hypothesis of the quotient part of the Algebraic Limit Theorem for Functional Limits (Corollary 4.2.4 (iv)), so

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f^{\prime}(a)}{g^{\prime}(a)} . \end{equation*}

Since \(f^{\prime}\) and \(g^{\prime}\) are continuous at \(a\) with \(g^{\prime}(a) \ne 0\), the same corollary gives

\begin{equation*} L = \lim_{x \to a} \frac{f^{\prime}(x)}{g^{\prime}(x)} = \frac{f^{\prime}(a)}{g^{\prime}(a)} , \end{equation*}

and the two displays agree.

The first three limits are all \(0\); the fourth does not exist (finite or infinite).

Both functions are defined for \(x \ne 0\) and extended continuously by the value \(0\) at \(x = 0\).

(i) \(\lim_{x \to 0} g(x) = 0\), since \(1/x^2 \to \infty\) and \(e^{-t} \to 0\) as \(t \to \infty\).

(ii) \(\lim_{x \to 0} f(x) = 0\), by the squeeze \(|f(x)| \le |x| e^{-1/x^2} \le |x|\).

(iii) For \(x \ne 0\) the exponentials cancel:

\begin{equation*} \frac{f(x)}{g(x)} = x \sin(1/x^4), \qquad \left| x \sin(1/x^4) \right| \le |x| , \end{equation*}

so \(\lim_{x \to 0} f(x)/g(x) = 0\).

(iv) Differentiating,

\begin{equation*} g^{\prime}(x) = \frac{2}{x^3} e^{-1/x^2} , \end{equation*}

\begin{equation*} \begin{aligned} f^{\prime}(x) = e^{-1/x^2}\Bigl[ &\sin\frac{1}{x^4} - \frac{4}{x^4}\cos\frac{1}{x^4} \\ &+ \frac{2}{x^2}\sin\frac{1}{x^4} \Bigr] , \end{aligned} \end{equation*}

and dividing (the exponentials cancel again),

\begin{equation*} \begin{aligned} \frac{f^{\prime}(x)}{g^{\prime}(x)} &= \frac{x^3}{2}\sin\frac{1}{x^4} + x\sin\frac{1}{x^4} \\ &\quad - \frac{2}{x}\cos\frac{1}{x^4} . \end{aligned} \end{equation*}

The first two terms tend to \(0\); the third oscillates unboundedly. Take the null sequences \(x_n = (2\pi n)^{-1/4}\), where \(\sin(1/x_n^4) = 0\) and \(\cos(1/x_n^4) = 1\), and \(y_n = (\pi/2 + 2\pi n)^{-1/4}\), where the two values are reversed \((n \ge 1)\):

\begin{equation*} \frac{f^{\prime}(x_n)}{g^{\prime}(x_n)} = -\frac{2}{x_n} , \qquad 0 < \frac{f^{\prime}(y_n)}{g^{\prime}(y_n)} = \frac{y_n^3}{2} + y_n < 1 . \end{equation*}

No finite limit \(L\) is possible: by the Sequential Criterion for Functional Limits (Theorem 4.2.3) it would force \(-2/x_n \to L\), and that sequence is unbounded. Nor is the limit \(\infty\) or \(-\infty\), since every punctured neighborhood of \(0\) contains a point \(y_n\) where the quotient lies strictly between \(0\) and \(1\), defeating Definition 5.3.7 (and its \(-\infty\) analogue) at \(M = 1\).

The pair satisfies every hypothesis of Theorem 5.3.6 at \(a = 0\): both continuous on an interval containing \(0\), both differentiable off \(0\), \(f(0) = g(0) = 0\), and \(g^{\prime}(x) = (2/x^3)e^{-1/x^2} \ne 0\) for \(x \ne 0\). The surprise is that \(f/g\) converges while \(f^{\prime}/g^{\prime}\) does not, so L’Hospital’s Rule is useless here and the divergence of \(f^{\prime}/g^{\prime}\) says nothing about \(f/g\). There is no conflict: Theorem 5.3.6 is a one-way implication whose hypothesis fails in this example, and its converse is false.

The Generalized Mean Value Theorem turns \(f(x)/g(x)\) into a value of \(f^{\prime}/g^{\prime}\) nearby; in (b) the answer is yes, by the same identity with \(M\) in place of \(L \pm \epsilon\).

(a) Let \(f, g\) be as in Theorem 5.3.6 on an interval \(I\) containing \(a\), and assume \(\lim_{x \to a} f^{\prime}(x)/g^{\prime}(x) = L\).

First, \(g(x) \ne 0\) for every \(x \in I\) with \(x \ne a\). Otherwise \(g(a) = g(x) = 0\), and Rolle’s Theorem (Theorem 5.3.1) on the closed interval with endpoints \(a\) and \(x\) (there \(g\) is continuous, and differentiable on the interior, which omits \(a\)) would give \(c\) strictly between them with \(g^{\prime}( c) = 0\), contradicting the hypothesis. So \(f/g\) is defined on \(I \setminus \{a\}\).

Fix such an \(x\). Both \(f\) and \(g\) are continuous on the closed interval with endpoints \(a\) and \(x\) and differentiable on the corresponding open interval, which omits \(a\), so the Generalized Mean Value Theorem (Theorem 5.3.5) supplies \(c\) strictly between \(a\) and \(x\) with

\begin{equation*} \bigl[f(x) - f(a)\bigr]g^{\prime}( c) = \bigl[g(x) - g(a)\bigr]f^{\prime}( c) . \end{equation*}

Since \(g^{\prime}\) is never zero on that open interval and \(g(x) - g(a) = g(x) \ne 0\), this can be divided, and \(f(a) = g(a) = 0\) gives

\begin{equation*} \frac{f^{\prime}( c)}{g^{\prime}( c)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f(x)}{g(x)} . \end{equation*}

Let \(\epsilon > 0\) and choose \(\delta > 0\) so that

\begin{equation*} 0 < |t - a| < \delta \quad \Longrightarrow \quad \left| \frac{f^{\prime}(t)}{g^{\prime}(t)} - L \right| < \epsilon . \end{equation*}

If \(0 < |x - a| < \delta\), the point \(c\) above lies strictly between \(a\) and \(x\), so \(0 < |c - a| < |x - a| < \delta\) and

\begin{equation*} \left| \frac{f(x)}{g(x)} - L \right| = \left| \frac{f^{\prime}( c)}{g^{\prime}( c)} - L \right| < \epsilon . \end{equation*}

Hence \(\lim_{x \to a} f(x)/g(x) = L\).

(b) Yes. Keeping the same standing hypotheses, the identity

\begin{equation*} \frac{f(x)}{g(x)} = \frac{f^{\prime}( c)}{g^{\prime}( c)}, \qquad 0 < |c - a| < |x - a| , \end{equation*}

established in (a) makes no reference to \(L\). Given \(M > 0\), Definition 5.3.7 applied to \(f^{\prime}/g^{\prime}\) provides \(\delta > 0\) with

\begin{equation*} 0 < |t - a| < \delta \quad \Longrightarrow \quad \frac{f^{\prime}(t)}{g^{\prime}(t)} \ge M . \end{equation*}

For any \(x\) with \(0 < |x - a| < \delta\) the associated \(c\) satisfies \(0 < |c - a| < \delta\), so \(f(x)/g(x) = f^{\prime}( c)/g^{\prime}( c) \ge M\). Since \(M > 0\) was arbitrary, \(\lim_{x \to a} f(x)/g(x) = \infty\).

One application of L’Hospital’s Rule in the variable \(h\) turns the quotient into a symmetric difference quotient for \(f^{\prime}\) at \(a\).

Choose \(r > 0\) with \((a - r, a + r)\) inside the interval on which \(f\) is twice differentiable, and for \(h \in (-r, r)\) set

\begin{equation*} F(h) = f(a+h) - 2f(a) + f(a-h), \qquad G(h) = h^2 . \end{equation*}

Then \(F\) and \(G\) are continuous on \((-r,r)\) and differentiable there, with

\begin{equation*} F(0) = G(0) = 0, \qquad F^{\prime}(h) = f^{\prime}(a+h) - f^{\prime}(a-h), \qquad G^{\prime}(h) = 2h , \end{equation*}

and \(G^{\prime}(h) = 2h \ne 0\) for \(h \ne 0\): every hypothesis of Theorem 5.3.6 holds at the point \(0\), so it suffices to evaluate \(\lim_{h \to 0} F^{\prime}(h)/G^{\prime}(h)\).

Since \(f^{\prime}\) is differentiable at \(a\) with derivative \(f^{\prime\prime}(a)\), split the symmetric quotient into two ordinary ones:

\begin{equation*} \begin{aligned} \frac{F^{\prime}(h)}{G^{\prime}(h)} &= \frac{f^{\prime}(a+h) - f^{\prime}(a-h)}{2h} \\ &= \frac{1}{2}\cdot\frac{f^{\prime}(a+h) - f^{\prime}(a)}{h}

• \frac{1}{2}\cdot\frac{f^{\prime}(a-h) - f^{\prime}(a)}{-h} . \end{aligned} \end{equation*}

Both increments \(h\) and \(-h\) tend to \(0\), so each quotient converges to \(f^{\prime\prime}(a)\) and \(\lim_{h \to 0} F^{\prime}(h)/G^{\prime}(h) = f^{\prime\prime}(a)\). By Theorem 5.3.6,

\begin{equation*} \lim_{h \to 0} \frac{f(a+h) - 2f(a) + f(a-h)}{h^2} = f^{\prime\prime}(a) . \end{equation*}

Method (2), the route through the continuity of \(f^{\prime\prime}\): apply Theorem 5.3.6 a second time instead of splitting. The pair \(F^{\prime}, G^{\prime}\) again satisfies \(F^{\prime}(0) = G^{\prime}(0) = 0\), both are differentiable on \((-r,r)\), and \(G^{\prime\prime}(h) = 2 \ne 0\), while

\begin{equation*} \frac{F^{\prime\prime}(h)}{G^{\prime\prime}(h)} = \frac{f^{\prime\prime}(a+h) + f^{\prime\prime}(a-h)}{2} \longrightarrow f^{\prime\prime}(a) \end{equation*}

because \(f^{\prime\prime}\) is continuous at \(a\).

\((1/2)h(2x) = h_1(x)\) is the same sawtooth compressed by a factor of \(2\) in both directions: it is piecewise linear with slopes \(\pm 1\), vanishes at every integer, and rises to the value \(1/2\) at every half-integer. On \([-2,3]\) its corners are

with straight segments joining consecutive corners: since \(h_1\) has period \(1\), that is five complete teeth over the length-\(5\) interval \([-2,3]\).

In general \(h_n\) is the sawtooth scaled by \(2^{-n}\) in both coordinates: it is \(2^{1-n}\)-periodic, vanishes exactly at the dyadic points \(k/2^{n-1}\), attains its maximum \(2^{-n}\) exactly at the odd multiples \((2k+1)/2^n\), and is affine with slope \(\pm 1\) between consecutive points of \(2^{-n}\mathbf{Z}\). So as \(n\) grows the teeth get shorter and narrower at the same rate: the amplitude \(2^{-n}\) tends to \(0\) while the number of teeth per unit length doubles, yet \(|h_n^{\prime}| = 1\) wherever the derivative exists.

Compare with the geometric series. Since \(h\) takes values in \([0,1]\),

\begin{equation*} 0 \;\le\; \frac{1}{2^n}h(2^nx) \;\le\; \frac{1}{2^n} \qquad \text{for every } n \ge 0, \end{equation*}

and \(\sum_{n=0}^{\infty}(1/2)^n\) converges (geometric, ratio \(1/2 < 1\); Example 2.7.5). The Comparison Test (Theorem 2.7.4), whose hypothesis \(0 \le a_n \le b_n\) is exactly the display above, gives convergence of \(\sum_{n=0}^{\infty}2^{-n}h(2^nx)\), so \(g(x)\) is defined for every \(x \in \mathbf{R}\).

Composition, then the Algebraic Continuity Theorem, then induction on \(m\).

Fix \(n\) and set \(\ell_n(x) = 2^nx\). The identity function is continuous on \(\mathbf{R}\), so \(\ell_n\) is continuous by Theorem 4.3.4 (ii) (a constant multiple of a continuous function). Since \(h\) is continuous at every point of \(\mathbf{R}\) – in particular at \(\ell_n( c)\) for each \(c\) – Theorem 4.3.9 (composition of continuous functions) applies and \(h \circ \ell_n\) is continuous on \(\mathbf{R}\). Theorem 4.3.4 (ii) again gives continuity of

\begin{equation*} h_n(x) = \frac{1}{2^n}(h \circ \ell_n)(x). \end{equation*}

Finally \(g_0 = h_0\) is continuous, and if \(g_{m-1}\) is continuous then \(g_m = g_{m-1} + h_m\) is continuous by Theorem 4.3.4 (i) (sums of continuous functions). By induction \(g_m\) is continuous on \(\mathbf{R}\) for every \(m \in \mathbf{N} \cup \{0\}\).

(a) \(M = 4/3\). Existence is the Extreme Value Theorem (Theorem 4.4.2): \(g\) is continuous by assumption and \([0,2]\) is a closed bounded interval, hence compact, so \(g\) attains a maximum on it.

For the value, note first that \(h(t) = \operatorname{dist}(t, 2\mathbf{Z})\), since this distance function is \(2\)-periodic and equals \(|t|\) on \([-1,1]\). Pair consecutive terms of the series:

\begin{equation*} g(x) = \sum_{k=0}^{\infty} 4^{-k}\Big[h(y_k) + \tfrac12 h(2y_k)\Big], \qquad y_k = 4^kx . \end{equation*}

Lemma: \(h(y) + \tfrac12 h(2y) \le 1\) for all \(y\), with equality exactly when \(y\) lies in \([1/2, 3/2]\) modulo \(2\).

Both sides are \(2\)-periodic, so it suffices to check \(y \in [0,2]\):

\begin{equation*} \begin{aligned} &\text{(i) } y \in [0,\tfrac12]: && y + \tfrac12(2y) = 2y \le 1,\\ &\text{(ii) } y \in [\tfrac12,1]: && y + \tfrac12(2-2y) = 1,\\ &\text{(iii) } y \in [1,\tfrac32]: && (2-y) + \tfrac12(2y-2) = 1,\\ &\text{(iv) } y \in [\tfrac32,2]: && (2-y) + \tfrac12(4-2y) = 4-2y \le 1. \end{aligned} \end{equation*}

In (i) equality forces \(y = 1/2\) and in (iv) it forces \(y = 3/2\), which proves the equality clause.

Summing the lemma against the weights \(4^{-k}\) gives

\begin{equation*} g(x) \;\le\; \sum_{k=0}^{\infty}4^{-k} \;=\; \frac{4}{3} \qquad \text{for every } x \in \mathbf{R}, \end{equation*}

and the bound is attained at \(x = 2/3\): for every \(n \ge 0\) one has \(2^{n}\cdot\frac23 \equiv \frac23\) or \(\frac43 \pmod 2\) according as \(n\) is even or odd, and \(h(2/3) = h(4/3) = 2/3\), so

\begin{equation*} g\!\left(\tfrac23\right) = \sum_{n=0}^{\infty}\frac{1}{2^n}\cdot\frac{2}{3} = \frac{2}{3}\cdot 2 = \frac{4}{3}. \end{equation*}

(b) \(x = 2/3 \in D\), by the computation just made.

(c) \(D\) is uncountable. By the equality clause of the lemma, \(g(x) = 4/3\) precisely when

\begin{equation*} 4^{k}x \in \left[\tfrac12, \tfrac32\right] \pmod 2 \qquad \text{for every } k \ge 0 . \end{equation*}

Now build such \(x\) in base \(4\). For a sequence \((d_k)_{k \ge 1}\) with each \(d_k \in \{1,2\}\), set

\begin{equation*} x = 2z, \qquad z = \sum_{k=1}^{\infty}\frac{d_k}{4^{k}} . \end{equation*}

For each \(j \ge 0\), \(4^{j}z\) differs by an integer from \(\sum_{k \ge 1} d_{k+j}4^{-k}\), and this tail lies in \(\left[\tfrac13, \tfrac23\right]\) because \(\sum_{k\ge1}4^{-k} = 1/3\) and the digits are between \(1\) and \(2\). Hence \(4^{j}z \equiv t_j \pmod 1\) with \(t_j \in [1/3,2/3] \subset [1/4,3/4]\), so

\begin{equation*} 4^{j}x \equiv 2t_j \pmod 2, \qquad 2t_j \in \left[\tfrac23,\tfrac43\right] \subset \left[\tfrac12,\tfrac32\right], \end{equation*}

and therefore \(g(x) = 4/3\). Also \(x = 2z \in [2/3,4/3] \subset [0,2]\), so \(x \in D\).

Distinct digit sequences give distinct points: if \((d_k)\) and \((e_k)\) first differ at index \(k\), with \(d_k > e_k\) say, then

\begin{equation*} z - w \;\ge\; \frac{1}{4^{k}} - \sum_{j > k}\frac{1}{4^{j}} \;=\; \frac{1}{4^{k}} - \frac{1}{3 \cdot 4^{k}} \;>\; 0 . \end{equation*}

So \(D\) contains a one-to-one image of the set of all sequences of \(1\)s and \(2\)s, which is uncountable by Exercise 1.6.4 (relabel \(1 \mapsto 0\), \(2 \mapsto 1\)). Hence \(D\) is uncountable.

Only the first \(m+1\) terms of the series survive, and each contributes exactly \(1/2^m\). Since \(g(0) = 0\),

\begin{equation*} g(x_m) = \sum_{n=0}^{\infty}\frac{1}{2^n}h\!\left(2^{n-m}\right), \end{equation*}

and the terms split at \(n = m\): (i) \(0 \le n \le m\): here \(2^{n-m} \in (0,1]\), so \(h(2^{n-m}) = 2^{n-m}\) and the term equals \(2^{-n}2^{n-m} = 2^{-m}\).

(ii) \(n > m\): here \(2^{n-m}\) is an even integer, so \(h(2^{n-m}) = 0\) and the term vanishes. Adding the \(m+1\) surviving terms,

\begin{equation*} \frac{g(x_m) - g(0)}{x_m - 0} = \frac{(m+1)2^{-m}}{2^{-m}} = m+1 . \end{equation*}

Now \(x_m \to 0\) with \(x_m \ne 0\), so if the limit defining

\begin{equation*} g^{\prime}(0) = \lim_{x \to 0}\frac{g(x) - g(0)}{x - 0} \end{equation*}

(Definition 5.2.1) existed and equalled \(L \in \mathbf{R}\), the sequential characterization of functional limits (Theorem 4.2.3) would force \(m+1 \to L\). The sequence \((m+1)\) is unbounded, hence divergent, so no such \(L\) exists and \(g\) is not differentiable at \(0\).

(a) Approach from the right along \(x_m = 1 + 1/2^m\); the difference quotients equal \(m-1\).

First \(g(1) = h(1) = 1\), since \(h(2^n) = 0\) for \(n \ge 1\). Next, for \(n \ge 1\) the number \(2^n\) is an even integer, so \(2\)-periodicity gives \(h(2^n + 2^{n-m}) = h(2^{n-m})\), which equals \(2^{n-m}\) for \(1 \le n \le m\) and \(0\) for \(n > m\). Since \(1 + 2^{-m} \in [1,2]\) we also have \(h(1+2^{-m}) = 1 - 2^{-m}\). Hence

\begin{equation*} g\!\left(1 + \tfrac{1}{2^m}\right) = \left(1 - \tfrac{1}{2^m}\right) + \sum_{n=1}^{m}\frac{1}{2^n}\cdot\frac{2^{n}}{2^{m}} = 1 + \frac{m-1}{2^{m}}, \end{equation*}

so that

\begin{equation*} \frac{g(1 + 2^{-m}) - g(1)}{2^{-m}} = m-1 \longrightarrow \infty . \end{equation*}

As in Exercise 5.4.5, Theorem 4.2.3 rules out a finite limit for the difference quotients, so \(g^{\prime}(1)\) does not exist.

At \(x = 1/2\): \(g(1/2) = h(1/2) + \tfrac12 h(1) = \tfrac12 + \tfrac12 = 1\), the terms with \(n \ge 2\) vanishing because \(2^{n-1}\) is then an even integer. Take \(m \ge 2\) and \(x_m = 1/2 + 1/2^m\). Then \(h(1/2 + 2^{-m}) = 1/2 + 2^{-m}\) (the argument lies in \((1/2,1)\)), while \(n = 1\) contributes \(\tfrac12 h(1 + 2^{1-m}) = \tfrac12(1 - 2^{1-m}) = \tfrac12 - 2^{-m}\), and each \(2 \le n \le m\) contributes \(2^{-n}h(2^{n-m}) = 2^{-m}\) with the rest zero. Adding,

\begin{equation*} g\!\left(\tfrac12 + \tfrac{1}{2^m}\right) = 1 + \frac{m-1}{2^m}, \qquad \frac{g(x_m) - g(1/2)}{x_m - 1/2} = m - 1 \longrightarrow \infty, \end{equation*}

so \(g^{\prime}(1/2)\) does not exist.

(b) Fix \(x = p/2^{k}\) and again take \(x_m = x + 1/2^{m}\) with \(m > k\). Both series converge by Exercise 5.4.2, so Theorem 2.7.1 lets us subtract them termwise; since \(2^{-n}/2^{-m} = 1/2^{n-m}\),

\begin{equation*} \frac{g(x_m) - g(x)}{x_m - x} = \sum_{n=0}^{\infty}\frac{h\!\left(2^{n}x + 2^{n-m}\right) - h\!\left(2^{n}x\right)}{2^{n-m}} , \end{equation*}

each summand being a difference quotient of \(h\) across an interval of length \(2^{n-m}\). Sort the terms:

(i) \(n > m\): \(2^{n-m}\) is an even integer, so by \(2\)-periodicity the numerator is \(0\).

(ii) \(k < n \le m\): \(2^{n}x = 2^{n-k}p\) is an even integer, so \(h(2^nx) = 0\), while \(h(2^nx + 2^{n-m}) = h(2^{n-m}) = 2^{n-m}\) because \(2^{n-m} \in (0,1]\). Each such term equals \(+1\), and there are \(m-k\) of them.

(iii) \(0 \le n \le k\): since \(h(t) = \operatorname{dist}(t, 2\mathbf{Z})\) we have \(h(s) \le |s-t| + h(t)\) for all \(s,t\), so \(h\) is \(1\)-Lipschitz and each of these \(k+1\) terms lies in \([-1,1]\).

Consequently

\begin{equation*} \frac{g(x_m) - g(x)}{x_m - x} \;\ge\; (m-k) - (k+1) \longrightarrow \infty \qquad (m \to \infty). \end{equation*}

The difference quotients along \(x_m \to x\) with \(x_m \ne x\) are unbounded, hence divergent, so Theorem 4.2.3 rules out a finite value for \(g^{\prime}(x)\).

(a) The two-sided quotient is a convex combination of the two one-sided ones. Put

\begin{equation*} \lambda_n = \frac{b_n - a}{b_n - a_n} \in (0,1), \qquad D_n = \frac{f(b_n) - f(a)}{b_n - a}, \qquad E_n = \frac{f(a) - f(a_n)}{a - a_n}, \end{equation*}

all defined because \(a_n < a < b_n\) forces \(b_n - a > 0\) and \(a - a_n > 0\), and because \(J\) is open with \(a_n, b_n \to a \in J\), so \(a_n, b_n \in J\) for all large \(n\) (discard the earlier terms). A one-line check gives \(1 - \lambda_n = (a - a_n)/(b_n - a_n)\) and

\begin{equation*} \lambda_n D_n + (1-\lambda_n)E_n = \frac{f(b_n) - f(a)}{b_n - a_n} + \frac{f(a) - f(a_n)}{b_n - a_n} = \frac{f(b_n) - f(a_n)}{b_n - a_n}. \end{equation*}

Since \(f\) is differentiable at \(a\) and \(b_n \to a\) with \(b_n \ne a\), and \(a_n \to a\) with \(a_n \ne a\), the sequential characterization of functional limits (Theorem 4.2.3) gives \(D_n \to f^{\prime}(a)\) and \(E_n \to f^{\prime}(a)\). Using \(0 < \lambda_n < 1\),

\begin{equation*} \begin{aligned} \left|\frac{f(b_n) - f(a_n)}{b_n - a_n} - f^{\prime}(a)\right| &= \left|\lambda_n\big(D_n - f^{\prime}(a)\big) + (1-\lambda_n)\big(E_n - f^{\prime}(a)\big)\right|\\ &\le \left|D_n - f^{\prime}(a)\right| + \left|E_n - f^{\prime}(a)\right| \longrightarrow 0 . \end{aligned} \end{equation*}

(b) The quotients across \([x_m, y_m]\) are integers whose parity alternates, so they cannot converge.

Write \(Q_m = \dfrac{g(y_m) - g(x_m)}{y_m - x_m}\) with \(y_m - x_m = 2^{-m}\). Since \(h_n(t) = 2^{-n}h(2^nt)\) has corners exactly at the points of \(2^{-n}\mathbf{Z}\) and is affine with slope \(\pm 1\) between them, and since \(2^{-n}\mathbf{Z} \subseteq 2^{-m}\mathbf{Z}\) for \(n \le m\) while the open interval \((x_m, y_m)\) contains no point of \(2^{-m}\mathbf{Z}\), each \(h_n\) with \(n \le m\) is affine of slope \(\varepsilon_n \in \{-1,+1\}\) on \([x_m, y_m]\). For \(n > m\) the increment \(2^{n}(y_m - x_m) = 2^{n-m}\) is an even integer, so \(h_n(y_m) = h_n(x_m)\) and that term drops out. Therefore

\begin{equation*} Q_m = \sum_{n=0}^{\infty}\frac{h_n(y_m) - h_n(x_m)}{y_m - x_m} = \sum_{n=0}^{m}\varepsilon_n, \end{equation*}

a sum of \(m+1\) terms each equal to \(\pm 1\). (Term-by-term evaluation is legitimate: the series for \(g(y_m)\) and \(g(x_m)\) both converge by Exercise 5.4.2, so the Algebraic Limit Theorem for Series, Theorem 2.7.1, lets us subtract them termwise.) In particular \(Q_m\) is an integer with

\begin{equation*} Q_m \equiv m+1 \pmod 2 , \end{equation*}

so \(|Q_{m+1} - Q_m| \ge 1\) for every \(m\) and \((Q_m)\) is not Cauchy; by Theorem 2.6.4 it diverges.

Now suppose \(g\) were differentiable at \(x\). Applying part (a) with \(f = g\), \(J = \mathbf{R}\), \(a = x\), \(a_m = x_m\) and \(b_m = y_m\) – the hypotheses \(x_m < x < y_m\) and \(\lim x_m = \lim y_m = x\) hold because \(x\) is not dyadic and \(y_m - x_m = 2^{-m}\) – we would get \(Q_m \to g^{\prime}(x)\), contradicting divergence. Hence \(g^{\prime}(x)\) does not exist.

Yes for \(\sum_{n=0}^{\infty}(1/2^{n})h(3^{n}x)\) (bracket \(x\) by triadic rather than dyadic points); no for \(\sum_{n=0}^{\infty}(1/3^{n})h(2^{n}x)\), whose difference quotients all lie in \([-3,3]\).

Review. For \(x\) nondyadic put \(x_{m}=p_{m}/2^{m}<x<y_{m}=(p_{m}+1)/2^{m}\). Two facts drive everything. (i) For \(n>m\), \(2^{n}x_{m}=2^{n-m}p_{m}\) and \(2^{n}y_{m}=2^{n-m}(p_{m}+1)\) are even integers, where \(h\) vanishes, so those terms cancel out of the difference. (ii) For \(n\le m\) the image interval \([2^{n}x_{m},2^{n}y_{m}]\) joins consecutive multiples of \(2^{n-m}\le 1\), hence contains no integer in its interior, so \(t\mapsto h(2^{n}t)\) is affine on \([x_{m},y_{m}]\) with slope \(\epsilon_{n}2^{n}\), \(\epsilon_{n}=\pm 1\). Hence

\begin{equation*} \frac{g(y_{m})-g(x_{m})}{y_{m}-x_{m}}=\sum_{n=0}^{m}\epsilon_{n}, \end{equation*}

an integer of the same parity as \(m+1\). Were \(g^{\prime}(x)\) to exist, Exercise 5.4.7 (a) would force this sequence to converge; a convergent integer sequence is eventually constant, while the parity alternates.

Case \(f(x)=\sum_{n=0}^{\infty}(1/2^{n})h(3^{n}x)\) (here \(ab=3/2>1\)). Take \(x\) not a triadic rational and bracket it by \(x_{m}=p_{m}/3^{m}<x<y_{m}=(p_{m}+1)/3^{m}\). Fact (ii) survives verbatim: for \(n\le m\), \([3^{n}x_{m},3^{n}y_{m}]\) joins consecutive multiples of \(3^{n-m}\le 1\), so \(h(3^{n}t)\) is affine on \([x_{m},y_{m}]\) with slope \(\epsilon_{n}3^{n}\), \(\epsilon_{n}=\pm 1\); these intervals are nested, so \(\epsilon_{n}\) does not depend on \(m\). Fact (i) fails: \(3^{n-m}\) is odd, so for \(n>m\) the integers \(3^{n}x_{m}\) and \(3^{n}y_{m}\) have the parities of \(p_{m}\) and \(p_{m}+1\), giving \(h(3^{n}y_{m})-h(3^{n}x_{m})=\sigma_{m}:=(-1)^{p_{m}}\). Also \(\epsilon_{m}=\sigma_{m}\), since \(h\) increases on \([p_{m},p_{m}+1]\) exactly when \(p_{m}\) is even. So with \(A_{m}=\sum_{n=0}^{m}\epsilon_{n}(3/2)^{n}\) and \(A_{-1}=0\),

\begin{equation*} \begin{aligned} D_{m}:=\frac{f(y_{m})-f(x_{m})}{y_{m}-x_{m}} &=A_{m}+3^{m}\sigma_{m}\sum_{n=m+1}^{\infty}2^{-n}\\ &=A_{m}+\sigma_{m}(3/2)^{m}=A_{m-1}+2\sigma_{m}(3/2)^{m}, \end{aligned} \end{equation*}

the last step because \(A_{m}-A_{m-1}=\epsilon_{m}(3/2)^{m}=\sigma_{m}(3/2)^{m}\). Consecutive difference quotients therefore satisfy, for \(m\ge 1\),

\begin{equation*} D_{m}-D_{m-1}=2\sigma_{m}(3/2)^{m}-\sigma_{m-1}(3/2)^{m-1}, \end{equation*}

so \(|D_{m}-D_{m-1}|\ge 2(3/2)^{m}-(3/2)^{m-1}=2(3/2)^{m-1}\to\infty\). But if \(f^{\prime}(x)\) existed then \(D_{m}\to f^{\prime}(x)\) by Exercise 5.4.7 (a) (its hypotheses hold: \(x_{m}<x<y_{m}\) and \(x_{m},y_{m}\to x\)), forcing \(D_{m}-D_{m-1}\to 0\). So \(f^{\prime}(x)\) does not exist.

Case \(f(x)=\sum_{n=0}^{\infty}(1/3^{n})h(2^{n}x)\) (here \(ab=2/3<1\)). The argument collapses. Bracketing a nondyadic \(x\) by \(x_{m}=p_{m}/2^{m}\) and \(y_{m}=(p_{m}+1)/2^{m}\), facts (i) and (ii) hold exactly as in the review, so

\begin{equation*} \frac{f(y_{m})-f(x_{m})}{y_{m}-x_{m}}=\sum_{n=0}^{m}\epsilon_{n}(2/3)^{n}, \end{equation*}

which converges as \(m\to\infty\) by comparison with \(\sum(2/3)^{n}\): no contradiction. Nor can any variant of it succeed, since \(h\) is \(1\)-Lipschitz and therefore

\begin{equation*} |f(s)-f(t)|\le\sum_{n=0}^{\infty}\frac{|h(2^{n}s)-h(2^{n}t)|}{3^{n}} \le\sum_{n=0}^{\infty}(2/3)^{n}|s-t|=3|s-t|, \end{equation*}

so every difference quotient of \(f\) lies in \([-3,3]\).

(a) \(f(x) = 1/x\). Dividing numerator and denominator by \(n\),

\begin{equation*} f_n(x) = \frac{x}{\tfrac{1}{n} + x^2} \longrightarrow \frac{x}{x^2} = \frac{1}{x} \end{equation*}

for each fixed \(x > 0\), by the Algebraic Limit Theorem (Theorem 2.3.3).

The error, which serves (b)-(d): for \(x > 0\),

\begin{equation*} \begin{aligned} \left| f_n(x) - \frac{1}{x} \right| &= \left| \frac{nx^2 - (1+nx^2)}{x(1+nx^2)} \right| \\ &= \frac{1}{x(1+nx^2)}. \end{aligned} \end{equation*}

(b) No. Evaluating at \(x = 1/n\) gives

\begin{equation*} \left| f_n(1/n) - n \right| = \frac{n}{1 + \tfrac{1}{n}} \ge \frac{n}{2}, \end{equation*}

so \(\sup_{x \in (0,\infty)} |f_n(x) - f(x)| = \infty\) for every \(n\); the definition of uniform convergence (Definition 6.2.3) fails for, say, \(\epsilon = 1\).

(c) No. The points \(x = 1/n\) lie in \((0,1)\), so the computation in (b) applies verbatim.

(d) Yes. For \(x > 1\) we have \(x \ge 1\) and \(1 + nx^2 \ge 1 + n\), hence

\begin{equation*} \left| f_n(x) - \frac{1}{x} \right| = \frac{1}{x(1+nx^2)} \le \frac{1}{1+n} \end{equation*}

for all \(x \in (1,\infty)\). Given \(\epsilon > 0\), choose \(N > 1/\epsilon\); then \(n \ge N\) forces \(|f_n(x) - f(x)| \le 1/(1+n) < \epsilon\) for every \(x \in (1,\infty)\).

The three answers are (a) yes, no, no; (b) yes, yes, yes; (c) yes, no, yes. Write \(H = \{1, \tfrac12, \tfrac13, \ldots\}\). Each of \(f_n\), \(g_n\), \(h_n\) vanishes off \(\{1, \tfrac12, \ldots, \tfrac1n\} \subseteq [1/n, \infty)\), hence is identically zero on \((-1/n, 1/n)\) and continuous at zero – that settles the first question in all three parts.

(a) For fixed \(x\), \(f_n(x)\) is eventually constant, giving the pointwise limit

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x \in H \\ 0 & \text{otherwise.}\end{cases} \end{equation*}

The convergence is not uniform: \(f(1/(n+1)) - f_n(1/(n+1)) = 1 - 0 = 1\), so \(\sup_{x \in \mathbf{R}} |f_n(x) - f(x)| = 1\) for every \(n\). And \(f\) is not continuous at zero, since \(f(0) = 0\) while \(f(1/k) = 1\) for all \(k\), so the sequence \(1/k \to 0\) has \(f(1/k) \to 1 \ne f(0)\) (Theorem 4.3.2).

This is consistent with Theorem 6.2.6: the theorem concludes continuity of the limit only under uniform convergence, and here the convergence is merely pointwise.

(b) The pointwise limit is

\begin{equation*} g(x) = \begin{cases} x & \text{if } x \in H \\ 0 & \text{otherwise,}\end{cases} \end{equation*}

and \(g - g_n\) is supported on \(\{1/k : k > n\}\), where it equals \(1/k\). Hence

\begin{equation*} \sup_{x \in \mathbf{R}} |g_n(x) - g(x)| = \frac{1}{n+1} \longrightarrow 0, \end{equation*}

so \(g_n \to g\) uniformly. Continuity of \(g\) at zero now follows from Theorem 6.2.6 (each \(g_n\) is continuous at \(0\)), and is directly visible from \(|g(x)| \le |x|\).

(c) For fixed \(x = 1/k\) and \(n > k\) we have \(h_n(1/k) = 1/k\), so the pointwise limit is again \(h = g\). The convergence is not uniform:

\begin{equation*} \left| h_n(1/n) - h(1/n) \right| = \left| 1 - \frac{1}{n} \right| \ge \frac{1}{2} \end{equation*}

for \(n \ge 2\). Nevertheless \(h = g\) is continuous at zero.

Part (c) is consistent with Theorem 6.2.6 because that theorem is a one-way implication: uniform convergence is sufficient for the limit to inherit continuity, not necessary.

(a) The limits are

\begin{equation*} g(x) = \begin{cases} x & \text{if } 0 \le x < 1 \\ 1/2 & \text{if } x = 1 \\ 0 & \text{if } x > 1, \end{cases} \qquad h(x) = \begin{cases} 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0. \end{cases} \end{equation*}

For \(g\): if \(0 \le x < 1\) then \(x^n \to 0\) and \(g_n(x) \to x/(1+0) = x\); at \(x = 1\), \(g_n(1) = 1/2\) for every \(n\); if \(x > 1\) then \(x^n \to \infty\), and \(0 < g_n(x) \le x/x^n = x^{1-n} \to 0\). For \(h\): \(h_n(0) = 0\) for all \(n\), while for fixed \(x > 0\) every \(n > 1/x\) gives \(h_n(x) = 1\).

(b) Each \(g_n\) is continuous on \([0,\infty)\) (a quotient of continuous functions with \(1 + x^n \ge 1 > 0\), Theorem 4.3.4), and each \(h_n\) is continuous on \([0,\infty)\) (the two branches agree at the junction: \(n \cdot \tfrac1n = 1\)). But \(g\) is discontinuous at \(x = 1\) and \(h\) is discontinuous at \(x = 0\): taking \(x_k = 1 + 1/k \to 1\) gives \(g(x_k) = 0 \ne 1/2 = g(1)\), and \(x_k = 1/k \to 0\) gives \(h(x_k) = 1 \ne 0 = h(0)\). By the contrapositive of the Continuous Limit Theorem (Theorem 6.2.6), neither convergence can be uniform on \([0,\infty)\).

(c) For \((g_n)\) take \([0,a]\) with \(0 < a < 1\). For \(x \in [0,a]\),

\begin{equation*} |g_n(x) - x| = \frac{x\,|1 - (1+x^n)|}{1+x^n} = \frac{x^{n+1}}{1+x^n} \le a^{n+1}, \end{equation*}

and \(a^{n+1} \to 0\) because \(0 < a < 1\) (Example 2.5.3). Given \(\epsilon > 0\), any \(N\) with \(a^{N+1} < \epsilon\) works for all \(x \in [0,a]\) simultaneously, so \(g_n \to g\) uniformly on \([0,a]\).

For \((h_n)\) take \([a,\infty)\) with \(a > 0\). If \(n > 1/a\) then \(x \ge a > 1/n\) for every \(x\) in the set, so \(h_n(x) = 1 = h(x)\); that is,

\begin{equation*} \sup_{x \in [a,\infty)} |h_n(x) - h(x)| = 0 \quad\text{for all } n > 1/a, \end{equation*}

and the convergence is uniform (indeed eventually exact) on \([a,\infty)\).

\(f^{\prime}\) is a uniform limit of continuous difference-quotient functions, so Theorem 6.2.6 gives the conclusion. Continuity is local, so fix \(c \in A\) and produce a relative neighborhood of \(c\) on which \(f^{\prime}\) is continuous.

Since \(A\) is a nondegenerate interval there is a \(d \in A\) with \(d \ne c\); take \(s = +1\) and some \(d > c\) unless \(c\) is the right endpoint of \(A\), in which case take \(s = -1\) and some \(d < c\). Put \(\eta = |d-c|/2\) and \(J = A \cap [c-\eta,\, c+\eta]\), a neighborhood of \(c\) relative to \(A\). For \(y \in J\) and \(0 < t \le \eta\) the point \(y + st\) lies between \(y\) and \(d\), hence in \(A\):

\begin{equation*} y + st \in A \quad\text{for all } y \in J \text{ and } 0 < t \le \eta . \end{equation*}

For \(n > 1/\eta\) set

\begin{equation*} f_n(y) = \frac{f(y + \tfrac{s}{n}) - f(y)}{s/n} \qquad (y \in J), \end{equation*}

which is continuous on \(J\): \(f\) is continuous (Theorem 5.2.3, \(f\) being differentiable), and translation, subtraction and a nonzero scalar are covered by Theorem 4.3.4.

Given \(\epsilon > 0\), take \(\delta > 0\) from uniform differentiability. For \(n > \max\{1/\eta,\, 1/\delta\}\) and \(y \in J\), the pair \(x = y + \tfrac{s}{n}\), \(y\) has \(0 < |x-y| = \tfrac1n < \delta\), so \(|f_n(y) - f^{\prime}(y)| < \epsilon\), a bound independent of \(y\). Thus \(f_n \to f^{\prime}\) uniformly on \(J\) (Definition 6.2.3), and \(f^{\prime}\) is continuous on \(J\) by Theorem 6.2.6, in particular at \(c\).

(\(\Leftarrow\)) The candidate is \(f(x) = \lim_{n \to \infty} f_n(x)\): for each fixed \(x \in A\) the uniform Cauchy condition says in particular that the real sequence \((f_n(x))\) is Cauchy, hence convergent by Theorem 2.6.4. This defines \(f\) on all of \(A\).

Now let \(\epsilon > 0\) and choose \(N\) so that

\begin{equation*} |f_n(x) - f_m(x)| < \frac{\epsilon}{2} \qquad \text{for all } m,n \ge N \text{ and all } x \in A . \end{equation*}

Fix \(n \ge N\) and \(x \in A\). For every \(m \ge N\) the displayed bound reads

\begin{equation*} -\tfrac{\epsilon}{2} \;<\; f_n(x) - f_m(x) \;<\; \tfrac{\epsilon}{2}, \end{equation*}

and \(f_n(x) - f_m(x) \to f_n(x) - f(x)\) as \(m \to \infty\) by the Algebraic Limit Theorem (Theorem 2.3.3). Comparing with the constant sequences \(\pm \epsilon/2\), the Order Limit Theorem (Theorem 2.3.4) carries both bounds into the limit as weak inequalities:

\begin{equation*} |f_n(x) - f(x)| \le \frac{\epsilon}{2} < \epsilon . \end{equation*}

The single \(N\) works for every \(x \in A\), so \(f_n \to f\) uniformly on \(A\) (Definition 6.2.3).

(\(\Rightarrow\)) Assume \(f_n \to f\) uniformly on \(A\) and let \(\epsilon > 0\). Choose \(N\) with \(|f_n(x) - f(x)| < \epsilon/2\) for all \(n \ge N\) and all \(x \in A\). Then for \(m,n \ge N\) and \(x \in A\),

\begin{equation*} \begin{aligned} |f_n(x) - f_m(x)| &\le |f_n(x) - f(x)| + |f(x) - f_m(x)| \\ &< \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon . \end{aligned} \end{equation*}

Under uniform convergence (a), (b), (d) and (e) are true and (c) is false. Pointwise counterexamples first.

(a) On \(A = [0,1]\) let \(f_n(x) = x^n\). Each \(f_n\) is continuous on a compact set, hence uniformly continuous (Theorem 4.4.7). The pointwise limit is \(f(x) = 0\) for \(0 \le x < 1\) and \(f(1) = 1\), which is not even continuous at \(1\), let alone uniformly continuous.

(b) On \(A = (0,1]\) let \(f_n(x) = \min\{n,\, 1/x\}\). Each \(f_n\) is bounded by \(n\), and for fixed \(x\) every \(n > 1/x\) gives \(f_n(x) = 1/x\), so the pointwise limit \(f(x) = 1/x\) is unbounded on \((0,1]\).

(c) Covered by the counterexample in the uniform case below, which converges uniformly and so a fortiori pointwise.

(d) On \(A = \mathbf{R}\) let \(f_n(x) = x^{2n}/(1+x^{2n})\) and \(M = 2\). Each \(f_n\) is continuous, so has \(0 < 2\) discontinuities, but

\begin{equation*} f(x) = \begin{cases} 0 & |x| < 1 \\ 1/2 & |x| = 1 \\ 1 & |x| > 1 \end{cases} \end{equation*}

has exactly \(2\) discontinuities, at \(x = \pm 1\).

(e) On \(A = \mathbf{R}\) fix an enumeration \(\mathbf{Q} = \{q_1, q_2, q_3, \ldots\}\) (possible by Theorem 1.5.6(i)) and let \(f_n = \mathbf{1}_{\{q_1,\ldots,q_n\}}\). Each \(f_n\) has \(n\) discontinuities. The pointwise limit is Dirichlet’s function \(\mathbf{1}_{\mathbf{Q}}\), which is discontinuous at every real number – an uncountable set of discontinuities.

Now the uniform case. Parts (d) and (e) run on a sharpening of Theorem 6.2.6: if \(f_n \to f\) uniformly on \(A\) and \(f_n\) is continuous at \(c\) for infinitely many \(n\), then \(f\) is continuous at \(c\) – apply Theorem 6.2.6 to the subsequence of those \(f_n\), which still converges uniformly to \(f\). Contrapositively, writing \(D(g)\) for the discontinuity set of \(g\): if \(c \in D(f)\) then \(c \in D(f_n)\) for all but finitely many \(n\), so in particular \(D(f) \subseteq \bigcup_n D(f_n)\).

(a) True. Given \(\epsilon > 0\), choose \(N\) with \(|f_N(x) - f(x)| < \epsilon/3\) for all \(x \in A\), then choose \(\delta > 0\) from the uniform continuity of \(f_N\) so that \(|x-y| < \delta\) implies \(|f_N(x) - f_N(y)| < \epsilon/3\). For \(x,y \in A\) with \(|x-y| < \delta\),

\begin{equation*} \begin{aligned} |f(x)-f(y)| \le\;& |f(x)-f_N(x)| + |f_N(x)-f_N(y)| \\ &+ |f_N(y)-f(y)| \\ <\;& \tfrac{\epsilon}{3}+\tfrac{\epsilon}{3}+\tfrac{\epsilon}{3} = \epsilon . \end{aligned} \end{equation*}

(b) True. Choose \(N\) with \(|f_N(x) - f(x)| < 1\) for all \(x \in A\), and let \(M_0\) bound \(|f_N|\). Then \(|f(x)| \le |f(x)-f_N(x)| + |f_N(x)| < 1 + M_0\) for all \(x \in A\).

(c) False even for uniform convergence. Take \(g_n\) from Exercise 6.2.2(b): \(g_n(x) = x\) on \(\{1,\tfrac12,\ldots,\tfrac1n\}\) and \(0\) elsewhere. Each \(g_n\) has exactly \(n\) discontinuities, and \(\sup_x |g_n(x) - g(x)| = 1/(n+1) \to 0\), so the convergence is uniform; but the limit \(g\) is discontinuous at each of the infinitely many points \(1/k\).

(d) True. Suppose \(f\) had at least \(M\) discontinuities \(c_1, \ldots, c_M\). By the sharpening above, for each \(i\) there is an \(N_i\) with \(n \ge N_i\) implying \(c_i \in D(f_n)\). Any \(n \ge \max\{N_1,\ldots,N_M\}\) then gives an \(f_n\) with at least \(M\) discontinuities, contradicting the hypothesis. Hence \(f\) has fewer than \(M\).

(e) True. By the containment above, \(D(f) \subseteq \bigcup_{n=1}^{\infty} D(f_n)\). Each \(D(f_n)\) is countable or finite, so listing the elements of the nonempty ones row by row indexes the union by a subset of \(\mathbf{N} \times \mathbf{N}\); the union is thus countable or finite by the diagonal argument of Theorem 1.5.8(ii), and so is its subset \(D(f)\) (Theorem 1.5.7).

The displacement \(|(x + \tfrac1n) - x| = \tfrac1n\) is the same at every \(x\), which is exactly what uniform continuity is built to exploit.

Let \(\epsilon > 0\). By uniform continuity (Definition 4.4.4) there is a \(\delta > 0\) such that \(|u - v| < \delta\) implies \(|f(u) - f(v)| < \epsilon\) for all \(u,v \in \mathbf{R}\). Choose \(N > 1/\delta\) by the Archimedean Property (Theorem 1.4.2). Then for every \(n \ge N\) and every \(x \in \mathbf{R}\), the points \(u = x + \tfrac1n\) and \(v = x\) satisfy \(|u - v| = \tfrac1n \le \tfrac1N < \delta\), so

\begin{equation*} |f_n(x) - f(x)| = \left| f\!\left(x + \tfrac1n\right) - f(x) \right| < \epsilon . \end{equation*}

Since \(N\) depends on \(\epsilon\) alone, \(f_n \to f\) uniformly on \(\mathbf{R}\) (Definition 6.2.3).

For the counterexample take \(f(x) = x^2\), which is continuous on \(\mathbf{R}\) but not uniformly continuous there (Theorem 4.4.5 with \(x_n = n\), \(y_n = n + \tfrac1n\), as noted following Example 4.4.6). Here

\begin{equation*} f_n(x) - f(x) = \left(x + \tfrac1n\right)^2 - x^2 = \frac{2x}{n} + \frac{1}{n^2}, \end{equation*}

which is unbounded in \(x\) for each fixed \(n\). Concretely \(f_n(n) - f(n) = 2 + 1/n^2 > 2\), so no \(N\) can force \(|f_n(x) - f(x)| < 1\) for all \(x\).

The whole exercise is the constant \(m = \inf_{x \in K}|g(x)|\), and the point is that \(m > 0\): the Continuous Limit Theorem (Theorem 6.2.6) makes \(g\), hence \(|g|\), continuous on \(K\), and \(K\) is compact, so the Extreme Value Theorem (Theorem 4.4.2) supplies \(x_0 \in K\) with \(|g(x_0)| = m\); since \(g(x_0) \neq 0\), \(m > 0\).

Choose \(N_1\) so that \(|g_n(x) - g(x)| < m/2\) for all \(x \in K\) and \(n \ge N_1\). Then for such \(n\) and \(x\),

\begin{equation*} |g_n(x)| \ge |g(x)| - |g(x) - g_n(x)| > m - \frac{m}{2} = \frac{m}{2} > 0 , \end{equation*}

so \(1/g_n\) is defined on all of \(K\) once \(n \ge N_1\) (uniform convergence is a statement about a tail, so this suffices), and

\begin{equation*} \begin{aligned} \left| \frac{1}{g_n(x)} - \frac{1}{g(x)} \right| &= \frac{|g(x) - g_n(x)|}{|g_n(x)|\,|g(x)|} \\ &\le \frac{2}{m^2}\, |g_n(x) - g(x)| . \end{aligned} \end{equation*}

Given \(\epsilon > 0\), pick \(N \ge N_1\) with \(|g_n(x) - g(x)| < m^2\epsilon/2\) for all \(x \in K\) and \(n \ge N\). Then \(|1/g_n(x) - 1/g(x)| < \epsilon\) for all \(x \in K\) and \(n \ge N\).

(a) Say \(f_n \to f\) and \(g_n \to g\) uniformly on the common domain \(A\). Given \(\epsilon > 0\), choose \(N\) beyond which both \(|f_n(x) - f(x)| < \epsilon/2\) and \(|g_n(x) - g(x)| < \epsilon/2\) hold for every \(x \in A\). Then for \(n \ge N\) and every \(x \in A\),

\begin{equation*} |(f_n + g_n)(x) - (f + g)(x)| \le |f_n(x) - f(x)| + |g_n(x) - g(x)| < \epsilon , \end{equation*}

so \(f_n + g_n \to f + g\) uniformly.

(b) Take \(A = \mathbf{R}\), \(f_n(x) = x\) and \(g_n(x) = 1/n\). The sequence \((f_n)\) is constant, hence uniformly convergent to \(f(x) = x\), and \(g_n \to 0\) uniformly since \(\sup_{\mathbf{R}}|g_n - 0| = 1/n\). But \(f_n(x)g_n(x) = x/n\) converges pointwise to \(0\) while

\begin{equation*} \sup_{x \in \mathbf{R}} \left| \frac{x}{n} - 0 \right| = \infty \quad \text{for every } n , \end{equation*}

so the convergence of \((f_n g_n)\) is not uniform.

(c) The bound \(|f_n| \le M\) passes to the pointwise limit by the Order Limit Theorem (Theorem 2.3.4), so \(|f| \le M\) and likewise \(|g| \le M\). For every \(x \in A\),

\begin{equation*} \begin{aligned} |f_n(x)g_n(x) - f(x)g(x)| &\le |f_n(x)|\,|g_n(x) - g(x)| \\ &\quad + |g(x)|\,|f_n(x) - f(x)| \\ &\le M\bigl(|g_n(x) - g(x)| + |f_n(x) - f(x)|\bigr). \end{aligned} \end{equation*}

Given \(\epsilon > 0\), choose \(N\) beyond which \(|f_n - f| < \epsilon/(2M)\) and \(|g_n - g| < \epsilon/(2M)\) uniformly on \(A\); then \(|f_n g_n - fg| < \epsilon\) on \(A\) for all \(n \ge N\).

Monotonicity lets the two endpoints of a short subinterval control \(f_n\) on all of it. First, \(f\) is itself increasing: if \(x \le y\) then \(f_n(x) \le f_n(y)\) for every \(n\), and the inequality survives the pointwise limit (Order Limit Theorem, Theorem 2.3.4).

Since \(f\) is continuous on the compact set \([a,b]\), it is uniformly continuous there (Theorem 4.4.7). Let \(\epsilon > 0\) and choose \(\delta > 0\) so that \(|f(x) - f(y)| < \epsilon/2\) whenever \(|x - y| < \delta\) in \([a,b]\). Fix a partition

\begin{equation*} a = x_0 < x_1 < \cdots < x_m = b, \qquad x_j - x_{j-1} < \delta , \end{equation*}

so that \(f(x_j) - f(x_{j-1}) < \epsilon/2\) for each \(j\). Because the partition is finite, pointwise convergence gives an \(N\) with

\begin{equation*} |f_n(x_j) - f(x_j)| < \frac{\epsilon}{2} \quad \text{for all } n \ge N,\ 0 \le j \le m . \end{equation*}

Now take any \(x \in [a,b]\) and \(n \ge N\); say \(x \in [x_{j-1}, x_j]\). Since \(f\) is increasing, \(f(x_{j-1}) \le f(x) \le f(x_j)\), so both \(f(x_j) \le f(x) + \epsilon/2\) and \(f(x_{j-1}) \ge f(x) - \epsilon/2\). Monotonicity of \(f_n\) then gives

\begin{equation*} \begin{aligned} f_n(x) &\le f_n(x_j) < f(x_j) + \frac{\epsilon}{2} \le f(x) + \epsilon , \\ f_n(x) &\ge f_n(x_{j-1}) > f(x_{j-1}) - \frac{\epsilon}{2} \ge f(x) - \epsilon . \end{aligned} \end{equation*}

Hence \(|f_n(x) - f(x)| < \epsilon\) for all \(x \in [a,b]\) and all \(n \ge N\).

(a) Each \(g_n = f - f_n\) is continuous on \(K\) (Algebraic Continuity Theorem, Theorem 4.3.4), and for each fixed \(x \in K\):

\begin{equation*} g_n(x) \ge 0, \qquad g_{n+1}(x) \le g_n(x), \qquad g_n(x) \to 0 . \end{equation*}

Nonnegativity and monotone decrease both come from \(f_n(x) \uparrow f(x)\); the limit is \(0\) by definition of pointwise convergence. The goal is now: \(g_n \to 0\) uniformly on \(K\), i.e. for each \(\epsilon > 0\) there is an \(N\) with \(g_n(x) < \epsilon\) for all \(x \in K\) and \(n \ge N\).

(b) Fix \(\epsilon > 0\). Each \(K_n\) is compact: it is bounded, being a subset of the bounded set \(K\); and it is closed, because if \(x_k \in K_n\) with \(x_k \to x\), then \(x \in K\) (\(K\) is closed, by Theorem 3.3.4) and \(g_n(x) = \lim_k g_n(x_k) \ge \epsilon\) by continuity (Theorem 4.3.2) and the Order Limit Theorem (Theorem 2.3.4), so \(x \in K_n\). Theorem 3.3.4 then gives compactness. Nesting is immediate from \(g_{n+1} \le g_n\):

\begin{equation*} x \in K_{n+1} \implies \epsilon \le g_{n+1}(x) \le g_n(x) \implies x \in K_n . \end{equation*}

Suppose, for contradiction, that every \(K_n\) is nonempty. The Nested Compact Set Property (Theorem 3.3.5) then produces

\begin{equation*} x_0 \in \bigcap_{n=1}^{\infty} K_n , \qquad \text{so } g_n(x_0) \ge \epsilon \text{ for all } n , \end{equation*}

contradicting \(g_n(x_0) \to 0\). Therefore \(K_N = \emptyset\) for some \(N\), and by nesting \(K_n = \emptyset\) for all \(n \ge N\). That is exactly

\begin{equation*} 0 \le g_n(x) < \epsilon \quad \text{for all } x \in K,\ n \ge N , \end{equation*}

so \(g_n \to 0\) uniformly and \(f_n \to f\) uniformly on \(K\).

(a) \(f_0\) is the diagonal of the unit square; \(f_1\) is the piecewise linear graph rising with slope \(3/2\) from \((0,0)\) to \((1/3, 1/2)\), running flat at height \(1/2\) across \((1/3,2/3)\), then rising with slope \(3/2\) from \((2/3,1/2)\) to \((1,1)\). The three formulas agree at the joints (\((3/2)(1/3) = 1/2\) and \((3/2)(2/3) - 1/2 = 1/2\)), so \(f_1\) is continuous; it is increasing because each piece is increasing or constant and the pieces match; and it equals \(1/2\) throughout \((1/3,2/3) = [0,1] \setminus C_1\).

(b) The process is the recursion

\begin{equation*} f_{n+1}(x) = \begin{cases} \tfrac{1}{2} f_n(3x) & 0 \le x \le 1/3, \\ 1/2 & 1/3 < x < 2/3, \\ \tfrac{1}{2} f_n(3x - 2) + \tfrac{1}{2} & 2/3 \le x \le 1, \end{cases} \end{equation*}

which reproduces the printed \(f_1\) from \(f_0\) and the printed \(f_2\) from \(f_1\) (for \(n \ge 1\) the middle clause \(f_n(x)\) equals \(1/2\) on \((1/3,2/3)\), so the two forms coincide).

An induction using the recursion gives \(f_n(0) = 0\), \(f_n(1) = 1\), each \(f_n\) continuous and increasing on \([0,1]\): the left piece maps \([0,1/3]\) increasingly onto \([0,1/2]\), the right piece maps \([2/3,1]\) increasingly onto \([1/2,1]\), and the endpoint values \(\tfrac12 f_n(1) = \tfrac12\) and \(\tfrac12 f_n(0) + \tfrac12 = \tfrac12\) match the middle constant.

For the convergence, subtract two consecutive instances of the recursion. On \((1/3,2/3)\) both \(f_{n+1}\) and \(f_n\) equal \(1/2\) (for \(n \ge 1\)), and on the two outer thirds the difference is \(\tfrac12\) times a difference of \(f_n\) and \(f_{n-1}\) evaluated at \(3x\) or \(3x-2\). Hence, writing \(\|h\| = \sup_{[0,1]}|h|\),

\begin{equation*} \|f_{n+1} - f_n\| \le \tfrac{1}{2}\,\|f_n - f_{n-1}\| \quad (n \ge 1). \end{equation*}

Since \(\|f_1 - f_0\| = 1/6\) (Check! the three pieces give \(x/2\), \(1/2 - x\), \(x/2 - 1/2\)), induction yields

\begin{equation*} \|f_{n+1} - f_n\| \le \frac{1}{6}\cdot \frac{1}{2^{n}} , \end{equation*}

and therefore, for \(m > n\),

\begin{equation*} \|f_m - f_n\| \le \sum_{k=n}^{m-1} \frac{1}{6 \cdot 2^{k}} < \frac{1}{3 \cdot 2^{n}} . \end{equation*}

Given \(\epsilon > 0\), choose \(N\) with \(3^{-1}2^{-N} < \epsilon\); then \(|f_m(x) - f_n(x)| < \epsilon\) for all \(x \in [0,1]\) and all \(m,n \ge N\). By the Cauchy Criterion for Uniform Convergence (Theorem 6.2.5), \((f_n)\) converges uniformly on \([0,1]\).

(c) Each \(f_n\) is continuous and the convergence is uniform, so \(f\) is continuous by the Continuous Limit Theorem (Theorem 6.2.6). Each \(f_n\) is increasing, so \(x \le y\) gives \(f_n(x) \le f_n(y)\), and the Order Limit Theorem (Theorem 2.3.4) passes this to \(f(x) \le f(y)\). Since \(f_n(0) = 0\) and \(f_n(1) = 1\) for every \(n\), also \(f(0) = 0\) and \(f(1) = 1\).

For the derivative, the claim is:

(i) for every \(n \ge 1\) and every \(m \ge n\), \(f_m\) is constant on each connected component of \([0,1] \setminus C_n\).

Induct on \(n\). For \(n = 1\) the only component is \((1/3,2/3)\), where \(f_m = 1/2\) for all \(m \ge 1\). Assume (i) for \(n\). The components of \([0,1]\setminus C_{n+1}\) are \((1/3,2/3)\) together with the images of the components of \([0,1]\setminus C_n\) under \(x \mapsto x/3\) and \(x \mapsto (x+2)/3\). Let \(m \ge n+1\). On \((1/3,2/3)\), \(f_m = 1/2\). If \(I\) is a component of \([0,1]\setminus C_n\) and \(x \in I/3\), then \(f_m(x) = \tfrac12 f_{m-1}(3x)\) with \(3x \in I\) and \(m - 1 \ge n\), so \(f_{m-1}\) is constant on \(I\) by the inductive hypothesis and \(f_m\) is constant on \(I/3\); the right-hand copy is identical with \(f_m(x) = \tfrac12 f_{m-1}(3x-2) + \tfrac12\). This proves (i).

Now let \(x \in [0,1] \setminus C\). Then \(x \notin C_n\) for some \(n\), so \(x\) lies in an open component \(I\) of \([0,1] \setminus C_n\). By (i), \(f_m\) is constant on \(I\) for all \(m \ge n\), hence \(f = \lim_m f_m\) is constant on \(I\). Since \(I\) is an open interval containing \(x\), the difference quotients \((f(y) - f(x))/(y-x)\) vanish for all \(y \in I\) with \(y \neq x\), so \(f^{\prime}(x) = 0\).

(a) The real sequence \(\bigl(f_n(x_1)\bigr)_{n}\) satisfies \(|f_n(x_1)| \le M\), so it is bounded, and the Bolzano-Weierstrass Theorem (Theorem 2.5.5) supplies a convergent subsequence \(\bigl(f_{n_k}(x_1)\bigr)_k\). Relabel the underlying functions \(f_{1,k} = f_{n_k}\).

(b) The same bound applies at \(x_2\): \(|f_{1,k}(x_2)| \le M\) for all \(k\), so \(\bigl(f_{1,k}(x_2)\bigr)_k\) is a bounded real sequence and Theorem 2.5.5 applies again.

(c) Construct rows inductively. Row \(1\) is \((f_{1,k})_k\) from (a). Given row \(m\), apply Theorem 2.5.5 to the bounded real sequence \(\bigl(f_{m,k}(x_{m+1})\bigr)_k\) and let row \(m+1\), \((f_{m+1,k})_k\), be the corresponding subsequence of \((f_{m,k})_k\). Two properties hold by construction:

\begin{equation*} \begin{aligned} &(\text{i}) \ (f_{m+1,k})_k \text{ is a subsequence of } (f_{m,k})_k , \\ &(\text{ii}) \ \bigl(f_{m,k}(x_j)\bigr)_k \text{ converges for every } j \le m . \end{aligned} \end{equation*}

Property (ii) follows from (i) together with Theorem 2.5.2: row \(m\) was chosen to converge at \(x_m\), and every later row is a subsequence of it, so convergence at \(x_1,\ldots,x_{m-1}\) inherited from earlier rows is not lost.

Now take the diagonal: \(g_k = f_{k,k}\). It is a genuine subsequence of \((f_n)\): row \(k+1\) is a subsequence of row \(k\), so its \((k+1)\)-st term sits at position \(\ge k+1\) in row \(k\), whence \(f_{k+1,k+1}\) carries a strictly larger original index than \(f_{k,k}\).

Fix \(m \in \mathbf{N}\). For \(k \ge m\), the function \(g_k = f_{k,k}\) is a term of row \(k\), which is a subsequence of row \(m\); moreover the terms \(g_m, g_{m+1}, g_{m+2}, \ldots\) occur in row \(m\) in strictly increasing order of position. Hence \((g_k)_{k \ge m}\) is a subsequence of \((f_{m,k})_k\), and by (ii) and Theorem 2.5.2 the sequence \(\bigl(g_k(x_m)\bigr)_k\) converges. Since \(m\) was arbitrary, \((g_k)\) converges at every point of \(A\).

(a) The position of the quantifier on \(n\). Uniform continuity of each \(f_n\) separately says

\begin{equation*} \begin{aligned} &\forall \epsilon > 0 \ \ \forall n \ \ \exists \delta_n > 0 : \\ &\qquad |x-y| < \delta_n \Rightarrow |f_n(x) - f_n(y)| < \epsilon , \end{aligned} \end{equation*}

so \(\delta\) is allowed to depend on \(n\). Equicontinuity says

\begin{equation*} \begin{aligned} &\forall \epsilon > 0 \ \ \exists \delta > 0 \ \ \forall n : \\ &\qquad |x-y| < \delta \Rightarrow |f_n(x) - f_n(y)| < \epsilon , \end{aligned} \end{equation*}

one \(\delta\) serving the whole family at once. Equicontinuity therefore implies each \(f_n\) is uniformly continuous, and the converse fails as soon as the admissible \(\delta_n\) can only be chosen with \(\inf_n \delta_n = 0\).

(b) Yes, each \(g_n\) is uniformly continuous, being continuous on the compact set \([0,1]\) (Theorem 4.4.7). But the family is not equicontinuous: near \(x = 1\) the graphs steepen without bound (\(g_n^{\prime}(1) = n\)), so the \(\delta\) that works for \(g_n\) shrinks to \(0\) as \(n\) grows. Explicitly, with \(\epsilon_0 = 1/4\),

\begin{equation*} y_n = 2^{-1/n} , \qquad g_n(y_n) = \tfrac{1}{2}, \qquad g_n(1) = 1 , \end{equation*}

and \(y_n \to 1\), so every proposed \(\delta > 0\) admits an \(n\) with \(|1 - y_n| < \delta\) while \(|g_n(1) - g_n(y_n)| = 1/2 > \epsilon_0\).

(a) Apply Exercise 6.2.13 with \(A = \mathbf{Q} \cap [0,1]\), enumerated as \(A = \{q_1, q_2, q_3, \ldots\}\): this \(A\) is countable, being an infinite subset of the countable set \(\mathbf{Q}\) (Theorem 1.5.6(i) and Theorem 1.5.7), and the hypothesis \(|f_n(x)| \le M\) for all \(n \in \mathbf{N}\) and \(x \in [0,1]\) restricts to the uniform bound on \(A\) that 6.2.13 requires. It therefore supplies a subsequence \((f_{n_k})\) converging at every point of \(A\). Set \(g_k = f_{n_k}\); equicontinuity passes to the subsequence unchanged, since each \(g_k\) is one of the \(f_n\) and the \(\delta\) from the definition is common to the whole family.

(b) Fix \(\epsilon > 0\) and let \(\delta > 0\) be the equicontinuity constant for \(\epsilon/3\), so that

\begin{equation*} |x-y| < \delta \ \Longrightarrow \ |g_k(x) - g_k(y)| < \frac{\epsilon}{3} \quad \text{for all } k \in \mathbf{N}. \end{equation*}

The collection \(\{V_\delta( r) : r \in \mathbf{Q} \cap [0,1]\}\) is an open cover of \([0,1]\): for \(x \in [0,1]\) the interval \((x - \delta, x + \delta) \cap (0,1)\) is nonempty and open, so density of \(\mathbf{Q}\) in \(\mathbf{R}\) (Theorem 1.4.3) puts a rational \(r\) inside it, and \(|x - r| < \delta\) gives \(x \in V_\delta( r)\). As \([0,1]\) is closed and bounded, Heine-Borel (Theorem 3.3.8, (ii) \(\Rightarrow\) (iii)) extracts a finite subcover

\begin{equation*} [0,1] \subseteq V_\delta(r_1) \cup V_\delta(r_2) \cup \cdots \cup V_\delta(r_m), \qquad r_i \in \mathbf{Q} \cap [0,1]. \end{equation*}

Each \(r_i\) is rational, so by part (a) the real sequence \((g_k(r_i))_{k=1}^\infty\) converges, hence is Cauchy (Theorem 2.6.2): for each \(i \in \{1,\ldots,m\}\) there is an \(N_i \in \mathbf{N}\) with \(|g_s(r_i) - g_t(r_i)| < \epsilon/3\) for all \(s,t \ge N_i\). Put

\begin{equation*} N = \max\{N_1, N_2, \ldots, N_m\}, \end{equation*}

which then works simultaneously for all \(m\) points.

Finiteness is what makes this last step legal: a maximum of finitely many natural numbers is again a natural number, whereas the infinite list \(N_1, N_2, N_3, \ldots\) attached to all of \(\mathbf{Q} \cap [0,1]\) need not be bounded above, and then no single \(N\) would serve every rational at once.

(c) Let \(x \in [0,1]\) be arbitrary and choose \(i\) with \(x \in V_\delta(r_i)\), i.e. \(|x - r_i| < \delta\). For \(s, t \ge N\),

\begin{equation*} \begin{aligned} |g_s(x) - g_t(x)| &\le |g_s(x) - g_s(r_i)| \\ &\quad + |g_s(r_i) - g_t(r_i)| \\ &\quad + |g_t(r_i) - g_t(x)| \\ &< \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} = \epsilon, \end{aligned} \end{equation*}

the outer two terms by the choice of \(\delta\) in (b) (applied to the functions \(g_s\) and \(g_t\)) and the middle term by the choice of \(N\). The index \(N\) depends only on \(\epsilon\), not on \(x\), so the estimate holds for every \(x \in [0,1]\) at once. By the Cauchy Criterion for Uniform Convergence (Theorem 6.2.5), \((g_k) = (f_{n_k})\) converges uniformly on \([0,1]\).

(a) \(g \equiv 0\) and \(g^{\prime} \equiv 0\); the convergence is uniform because

\begin{equation*} \sup_{x \in [0,1]} \left| \frac{x^n}{n} - 0 \right| = \frac{1}{n} \longrightarrow 0 . \end{equation*}

The zero function is differentiable with \(g^{\prime}(x) = 0\) for every \(x \in [0,1]\) (one-sided at the endpoints).

(b) \(g_n^{\prime}(x) = x^{n-1}\), which converges pointwise on \([0,1]\) to

\begin{equation*} h(x) = \begin{cases} 0 & 0 \le x < 1, \\ 1 & x = 1. \end{cases} \end{equation*}

The convergence is not uniform: each \(g_n^{\prime}\) is continuous on \([0,1]\) while \(h\) is not, so the Continuous Limit Theorem (Theorem 6.2.6) forbids uniform convergence.

So \(h\) and \(g^{\prime}\) agree on \([0,1)\) but not at the endpoint: \(h(1) = 1 \ne 0 = g^{\prime}(1)\). They are not the same.

(a) \(h_n \to h\) uniformly on \(\mathbf{R}\), where \(h(x) = |x|\). Rationalizing the difference,

\begin{equation*} \begin{aligned} \left| \sqrt{x^2 + \tfrac1n} - |x| \right| &= \frac{1/n}{\sqrt{x^2 + \tfrac1n} + |x|} \\ &\le \frac{1/n}{\sqrt{1/n}} = \frac{1}{\sqrt n}, \end{aligned} \end{equation*}

uniformly in \(x\), and \(1/\sqrt n \to 0\).

(b) Since \(x^2 + 1/n > 0\) for all \(x\), each \(h_n\) is differentiable with

\begin{equation*} h_n^{\prime}(x) = \frac{x}{\sqrt{x^2 + \tfrac1n}} . \end{equation*}

For \(x \ne 0\) the denominator tends to \(|x|\), so \(h_n^{\prime}(x) \to x/|x|\); and \(h_n^{\prime}(0) = 0\) for every \(n\). Hence

\begin{equation*} g(x) = \lim h_n^{\prime}(x) = \begin{cases} 1 & x > 0, \\ 0 & x = 0, \\ -1 & x < 0. \end{cases} \end{equation*}

Each \(h_n^{\prime}\) is continuous, but \(g\) is discontinuous at \(0\). By the Continuous Limit Theorem (Theorem 6.2.6), uniform convergence on a set forces the limit to be continuous there; so the convergence cannot be uniform on any neighborhood of \(0\).

(a) The extrema sit at \(x = \pm 1/\sqrt{n}\), and \(f_n \to 0\) uniformly. Indeed

\begin{equation*} f_n^{\prime}(x) = \frac{(1 + nx^2) - x(2nx)}{(1 + nx^2)^2} = \frac{1 - nx^2}{(1 + nx^2)^2}, \end{equation*}

which vanishes exactly at \(x = \pm 1/\sqrt n\), is positive on \((-1/\sqrt n, 1/\sqrt n)\) and negative outside. So \(f_n\) has its minimum at \(x = -1/\sqrt n\) and its maximum at \(x = 1/\sqrt n\), with

\begin{equation*} f_n\!\left(\tfrac{1}{\sqrt n}\right) = \frac{1/\sqrt n}{2} = \frac{1}{2\sqrt n}, \qquad f_n\!\left(\tfrac{-1}{\sqrt n}\right) = \frac{-1}{2\sqrt n}. \end{equation*}

(Also \(f_n(x) \to 0\) as \(x \to \pm\infty\), so these are global extrema.) Hence

\begin{equation*} \sup_{x \in \mathbf{R}} |f_n(x) - 0| = \frac{1}{2\sqrt n} \longrightarrow 0, \end{equation*}

so \(f_n \to f \equiv 0\) uniformly on \(\mathbf{R}\).

(b) With \(f_n^{\prime}\) as computed above, fix \(x \ne 0\): the numerator grows like \(-nx^2\) while the denominator grows like \(n^2 x^4\), so

\begin{equation*} \lim_{n \to \infty} \frac{1 - nx^2}{(1 + nx^2)^2} = 0 = f^{\prime}(x). \end{equation*}

At \(x = 0\), however, \(f_n^{\prime}(0) = 1\) for every \(n\), so \(\lim f_n^{\prime}(0) = 1 \ne 0 = f^{\prime}(0)\).

Thus \(f^{\prime}(x) = \lim f_n^{\prime}(x)\) for all \(x \ne 0\), and only there.

Uniform convergence is the bound \(|\sin(nx)| \le 1\):

\begin{equation*} \sup_{x \in \mathbf{R}} \left| \frac{\sin(nx)}{\sqrt n} - 0 \right| \le \frac{1}{\sqrt n} \longrightarrow 0 . \end{equation*}

For the derivatives, \(h_n^{\prime}(x) = \sqrt n \cos(nx)\). Fix \(x \in \mathbf{R}\) and suppose, for contradiction, that \((h_n^{\prime}(x))\) converges. A convergent sequence is bounded (Theorem 2.3.2), so there is \(M > 0\) with \(\sqrt n\,|\cos(nx)| \le M\) for all \(n\), whence

\begin{equation*} |\cos(nx)| \le \frac{M}{\sqrt n} \longrightarrow 0, \qquad\text{so}\qquad \cos(nx) \to 0 . \end{equation*}

Now read off the same limit two ways along the even indices. On the one hand \((\cos(2nx))\) is a subsequence of \((\cos(nx))\), so \(\cos(2nx) \to 0\) (Theorem 2.5.2). On the other hand the double-angle identity gives

\begin{equation*} \cos(2nx) = 2\cos^2(nx) - 1 \longrightarrow 2 \cdot 0 - 1 = -1 . \end{equation*}

Uniqueness of limits (Theorem 2.2.7) makes \(0 = -1\), a contradiction. Hence \((h_n^{\prime}(x))\) diverges, and \(x\) was arbitrary.

(a) \(g(x) = x/2\), so \(g^{\prime}(x) = 1/2\). Indeed \(g_n(x) = \dfrac{x}{2} + \dfrac{x^2}{2n} \to \dfrac{x}{2}\) for each fixed \(x\).

(b) \(g_n^{\prime}(x) = \dfrac12 + \dfrac{x}{n}\), and on \([-M,M]\)

\begin{equation*} \sup_{|x| \le M} \left| g_n^{\prime}(x) - \tfrac12 \right| = \frac{M}{n} \longrightarrow 0, \end{equation*}

so \(g_n^{\prime} \to 1/2\) uniformly there. Each \(g_n\) is a polynomial, hence differentiable on \([-M,M]\), and \(g_n(0) = 0\) converges, so Theorem 6.3.3 applies with \(x_0 = 0\): \(g = \lim g_n\) is differentiable on \([-M,M]\) with \(g^{\prime}(x) = \lim g_n^{\prime}(x) = 1/2\). Since \(M\) was arbitrary and differentiability is local, this holds on all of \(\mathbf{R}\).

(c) \(f(x) = x^2/2\) and \(f^{\prime}(x) = x\). For (a), fix \(x\) and divide through by \(n\):

\begin{equation*} f_n(x) = \frac{x^2 + 1/n}{2 + x/n} \longrightarrow \frac{x^2}{2} . \end{equation*}

For (b), fix \(M > 0\) and restrict to \([-M,M]\) with \(n > M\), so that \(2n + x \ge 2n - M > n > 0\) and \(f_n\) is a well-defined differentiable rational function there. The quotient rule gives

\begin{equation*} \begin{aligned} f_n^{\prime}(x) &= \frac{2nx(2n+x) - (nx^2+1)}{(2n+x)^2} \\ &= \frac{4n^2x + nx^2 - 1}{(2n+x)^2}, \end{aligned} \end{equation*}

and subtracting \(x\) (using \(x(2n+x)^2 = 4n^2x + 4nx^2 + x^3\)),

\begin{equation*} f_n^{\prime}(x) - x = \frac{-\left(3nx^2 + x^3 + 1\right)}{(2n+x)^2}. \end{equation*}

Hence for \(|x| \le M\) and \(n > M\),

\begin{equation*} \left| f_n^{\prime}(x) - x \right| \le \frac{3nM^2 + M^3 + 1}{n^2} \longrightarrow 0, \end{equation*}

so \((f_n^{\prime})\) converges uniformly on \([-M,M]\) to the identity function. With \(f_n(0) = 1/(2n) \to 0\) convergent, Theorem 6.3.3 gives that \(f = \lim f_n\) is differentiable on \([-M,M]\) with \(f^{\prime}(x) = \lim f_n^{\prime}(x) = x\), matching (a).

(a) Possible: take \(f_n = \dfrac{1}{n} g\), where \(g\) is the continuous nowhere-differentiable function of Section 5.4,

\begin{equation*} g(x) = \sum_{k=0}^{\infty} \frac{1}{2^k}\, h(2^k x), \end{equation*}

with \(h\) the \(2\)-periodic extension of \(|x|\) on \([-1,1]\). Each \(f_n\) is nowhere differentiable (if \(f_n\) were differentiable at \(c\), so would be \(n f_n = g\)). Since \(0 \le h \le 1\) we have \(|g| \le \sum_k 2^{-k} = 2\), so

\begin{equation*} \sup_{x \in \mathbf{R}} |f_n(x) - 0| \le \frac{2}{n} \longrightarrow 0, \end{equation*}

giving \(f_n \to f \equiv 0\) uniformly, and \(f\) is everywhere differentiable.

(b) Possible: take the constant functions \(f_n(x) = n\). Then \(f_n^{\prime} \equiv 0\), so \((f_n^{\prime})\) converges uniformly to \(0\), while \((f_n(x)) = (n)\) diverges for every \(x \in \mathbf{R}\).

(c) Impossible. If \((f_n)\) converges uniformly then in particular \(f_n \to f\) pointwise, and \((f_n^{\prime})\) converges uniformly to some \(g\). Fix any \(c \in \mathbf{R}\) and apply the Differentiable Limit Theorem (Theorem 6.3.1) on the closed interval \([c-1, c+1]\): its hypotheses hold there (pointwise convergence of \((f_n)\), differentiability of each \(f_n\), uniform convergence of \((f_n^{\prime})\) to \(g\)), so \(f\) is differentiable on \([c-1,c+1]\) with \(f^{\prime} = g\). In particular \(f\) is differentiable at \(c\), and \(c\) was arbitrary.

We show \((f_n)\) is uniformly Cauchy on \([a,b]\); the Cauchy Criterion for Uniform Convergence (Theorem 6.2.5) then gives the conclusion. Assume \(a < b\), the case \(a = b\) being trivial. Let \(\epsilon > 0\).

(i) \((f_n(x_0))\) converges, hence is Cauchy, so there is \(N_1\) with

\begin{equation*} |f_n(x_0) - f_m(x_0)| < \frac{\epsilon}{2} \qquad \text{for all } m, n \ge N_1 . \end{equation*}

(ii) \((f_n^{\prime})\) converges uniformly on \([a,b]\), hence is uniformly Cauchy (Theorem 6.2.5 again), so there is \(N_2\) with

\begin{equation*} |f_n^{\prime}(t) - f_m^{\prime}(t)| < \frac{\epsilon}{2(b-a)} \qquad \text{for all } t \in [a,b],\ m,n \ge N_2 . \end{equation*}

Set \(N = \max\{N_1, N_2\}\) and fix \(m, n \ge N\) and \(x \in [a,b]\). Put \(d = f_n - f_m\), differentiable and hence continuous on \([a,b]\). If \(x \ne x_0\), the Mean Value Theorem (Theorem 5.3.2), applied to \(d\) on the closed interval with endpoints \(x_0\) and \(x\), supplies \(\alpha\) strictly between them with

\begin{equation*} d(x) - d(x_0) = d^{\prime}(\alpha)(x - x_0) = \bigl( f_n^{\prime}(\alpha) - f_m^{\prime}(\alpha) \bigr)(x - x_0). \end{equation*}

Since \(\alpha \in [a,b]\) and \(|x - x_0| \le b - a\), step (ii) gives

\begin{equation*} \bigl| (f_n(x) - f_m(x)) - (f_n(x_0) - f_m(x_0)) \bigr| < \frac{\epsilon}{2(b-a)} (b-a) = \frac{\epsilon}{2}, \end{equation*}

and this bound holds trivially when \(x = x_0\), where the left side is \(0\).

Feeding this and (i) into the displayed triangle inequality,

\begin{equation*} |f_n(x) - f_m(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{equation*}

for all \(x \in [a,b]\) whenever \(m, n \ge N\).

Verify the Cauchy Criterion for Uniform Convergence of Series (Theorem 6.4.4) directly from the Cauchy Criterion for the convergent numerical series \(\sum M_n\).

Let \(\epsilon > 0\). Since \(\sum_{n=1}^{\infty} M_n\) converges, its sequence of partial sums is Cauchy, so by the Cauchy Criterion for Series (Theorem 2.7.2) there is \(N \in \mathbf{N}\) with

\begin{equation*} M_{m+1} + M_{m+2} + \cdots + M_n < \epsilon \qquad \text{whenever } n > m \ge N . \end{equation*}

(The terms \(M_k\) are positive, so the absolute value bars in 2.7.2 may be dropped.) For this same \(N\), any \(n > m \ge N\), and any \(x \in A\), the triangle inequality gives

\begin{equation*} \begin{aligned} |f_{m+1}(x) + \cdots + f_n(x)| &\le |f_{m+1}(x)| + \cdots + |f_n(x)| \\ &\le M_{m+1} + \cdots + M_n \\ &< \epsilon . \end{aligned} \end{equation*}

The bound is independent of \(x\), so the hypothesis of Theorem 6.4.4 holds, and \(\sum_{n=1}^{\infty} f_n\) converges uniformly on \(A\). \(\square\)

(a) True, (b) true, (c) false.

(a) Run the Cauchy Criterion (Theorem 6.4.4) at \(n = m+1\), where the sum collapses to a single term: given \(\epsilon > 0\), the \(N\) supplied by 6.4.4 satisfies

\begin{equation*} |g_{m+1}(x)| < \epsilon \qquad \text{for all } m \ge N \text{ and all } x \in A, \end{equation*}

which is uniform convergence of \((g_n)\) to \(0\).

(b) Again Theorem 6.4.4, with the bars free of charge since \(0 \le f_k \le g_k\). Given \(\epsilon > 0\), pick \(N\) so that \(|g_{m+1}(x) + \cdots + g_n(x)| < \epsilon\) for all \(n > m \ge N\) and all \(x\). Then for those same \(m, n, x\),

\begin{equation*} \begin{aligned} \left| \sum_{k=m+1}^{n} f_k(x) \right| &= \sum_{k=m+1}^{n} f_k(x) \le \sum_{k=m+1}^{n} g_k(x) \\ &= \left| \sum_{k=m+1}^{n} g_k(x) \right| < \epsilon . \end{aligned} \end{equation*}

So \(\sum f_n\) meets the Cauchy Criterion and converges uniformly.

(c) False. On \(A = [0,1]\) take the constant functions \(f_n(x) = (-1)^n/n\). The numerical series \(\sum (-1)^n/n\) converges by the Alternating Series Test (Theorem 2.7.7: the terms \(1/n\) decrease to \(0\)), and since every partial sum \(s_k\) is constant, \(\sup_{x \in A} |s_k(x) - f(x)|\) is just the numerical error \(|s_k - f| \to 0\), so the convergence is uniform. But any admissible bound obeys

\begin{equation*} M_n \ge \sup_{x \in A} |f_n(x)| = \frac{1}{n} , \end{equation*}

and \(\sum 1/n\) diverges (Example 2.4.5), so \(\sum M_n\) diverges by comparison (Theorem 2.7.4).

(a) The M-Test with \(M_n = 1/2^n\). For every \(x \in \mathbf{R}\),

\begin{equation*} \left| \frac{\cos(2^n x)}{2^n} \right| \le \frac{1}{2^n} = M_n , \qquad \sum_{n=0}^{\infty} \frac{1}{2^n} = 2 < \infty \end{equation*}

(geometric series, Example 2.7.5), so by Corollary 6.4.5 the series converges uniformly on \(\mathbf{R}\). Each term \(x \mapsto \cos(2^n x)/2^n\) is continuous, so the Term-by-term Continuity Theorem (Theorem 6.4.2) makes \(g\) continuous on all of \(\mathbf{R}\).

(b) The attempt fails at the one hypothesis that matters: the differentiated series does not converge uniformly on any nondegenerate interval. Writing \(f_n(x) = \cos(2^n x)/2^n\), the Chain Rule (Theorem 5.2.5) gives

\begin{equation*} f_n^{\prime}(x) = \frac{-2^n \sin(2^n x)}{2^n} = -\sin(2^n x), \end{equation*}

so Theorem 6.4.3 asks for uniform convergence of \(\sum_{n=0}^{\infty} -\sin(2^n x)\). Fix any interval \([a,b]\) with \(a < b\) and choose \(N\) with \(2^N (b-a) \ge 2\pi\). For every \(n \ge N\) the image of \([a,b]\) under \(x \mapsto 2^n x\) is an interval of length at least \(2\pi\), hence contains a full period of the sine, and therefore

\begin{equation*} \sup_{x \in [a,b]} |f_n^{\prime}(x)| = 1 \quad \text{for all } n \ge N . \end{equation*}

Thus \((f_n^{\prime})\) does not converge uniformly to \(0\) on \([a,b]\), so by Exercise 6.4.2(a) the series \(\sum f_n^{\prime}\) cannot converge uniformly on \([a,b]\). Its sole hypothesis unmet, Theorem 6.4.3 yields no information about the differentiability of \(g\).

The series converges exactly on \((-1,1)\), and \(g\) is continuous there.

Write \(t = x^2 \ge 0\), so the \(n\)th term is \(a_n = t^n/(1+t^n)\).

(i) \(|x| \ge 1\). Then \(t \ge 1\), so \(t^n \ge 1\) and

\begin{equation*} a_n = \frac{t^n}{1+t^n} = \frac{1}{1 + t^{-n}} \ge \frac{1}{2} , \end{equation*}

so \(a_n \not\to 0\) and the series diverges by Theorem 2.7.3.

(ii) \(|x| < 1\). Then \(1 + t^n \ge 1\) gives \(0 \le a_n \le t^n = x^{2n}\), and \(\sum_{n=0}^{\infty} x^{2n}\) is a convergent geometric series (\(0 \le x^2 < 1\)), so the series converges by the Comparison Test (Theorem 2.7.4).

For continuity, fix any \(b\) with \(0 < b < 1\). For all \(x \in [-b,b]\) and all \(n \ge 0\),

\begin{equation*} \left| \frac{x^{2n}}{1+x^{2n}} \right| \le x^{2n} \le b^{2n} =: M_n , \qquad \sum_{n=0}^{\infty} b^{2n} = \frac{1}{1-b^2} < \infty , \end{equation*}

so by the Weierstrass M-Test (Corollary 6.4.5) the series converges uniformly on \([-b,b]\). Each term is continuous (the denominator \(1 + x^{2n} \ge 1\) never vanishes), so Theorem 6.4.2 makes \(g\) continuous on \([-b,b]\). Given \(c \in (-1,1)\), choosing \(b\) with \(|c| < b < 1\) places \(c\) in the interior of \([-b,b]\), so continuity there is continuity of \(g\) at \(c\); hence \(g\) is continuous on \((-1,1)\).

(a) The M-Test with \(M_n = 1/n^2\). For \(|x| \le 1\),

\begin{equation*} \left| \frac{x^n}{n^2} \right| = \frac{|x|^n}{n^2} \le \frac{1}{n^2} = M_n , \end{equation*}

and \(\sum_{n=1}^{\infty} 1/n^2\) converges (p-series with \(p = 2 > 1\), Corollary 2.4.7). By Corollary 6.4.5 the series converges uniformly on \([-1,1]\); each term \(x^n/n^2\) is a polynomial, hence continuous; so \(h\) is continuous on \([-1,1]\) by Theorem 6.4.2.

(b) Localize: run the M-Test not on \([-1,1)\) but on \([-b,b]\) with \(|x_0| < b < 1\), say \(b = (1 + |x_0|)/2\). (On all of \([-1,1)\) it must fail, since \(\sup_{|x|<1} |x^n/n| = 1/n\) and \(\sum 1/n\) diverges.) For every \(x \in [-b,b]\) and every \(n \in \mathbf{N}\),

\begin{equation*} \left| \frac{x^n}{n} \right| \le \frac{b^n}{n} \le b^n =: M_n , \qquad \sum_{n=1}^{\infty} b^n = \frac{b}{1-b} < \infty \end{equation*}

since \(0 < b < 1\). By Corollary 6.4.5 the series \(\sum x^n/n\) converges uniformly on \([-b,b]\), and by Theorem 6.4.2 its sum \(f\) is continuous on \([-b,b]\).

Since \(|x_0| < b\), some neighborhood \(V_{\delta}(x_0)\) lies inside \([-b,b]\), so the \(\delta^{\prime} \le \delta\) furnished by continuity on \([-b,b]\) serves verbatim for \(f\) on \([-1,1)\): \(f\) is continuous at \(x_0\).

\(f\) is defined on all of \((0,\infty)\), and it is continuous and differentiable there, with

\begin{equation*} f^{\prime}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(x+n)^2} . \end{equation*}

The M-Test is unavailable (\(\sum 1/(x+n)\) diverges); the alternating-series tail estimate supplies the uniform control instead.

Put \(f_n(x) = (-1)^n/(x+n)\) and \(s_k = f_0 + \cdots + f_k\). For each fixed \(x > 0\) the numbers \(b_n = 1/(x+n)\) are positive, decrease in \(n\), and tend to \(0\), so \(\sum_{n=0}^{\infty} (-1)^n b_n\) converges by the Alternating Series Test (Theorem 2.7.7): \(f(x)\) exists for every \(x > 0\). The nested-interval proof of 2.7.7 (Exercise 2.7.1(b)) traps the limit between consecutive partial sums, whence

\begin{equation*} |f(x) - s_k(x)| \le |s_{k+1}(x) - s_k(x)| = \frac{1}{x+k+1} \le \frac{1}{k+1} \end{equation*}

for all \(x > 0\). The bound is independent of \(x\) and tends to \(0\), so \(s_k \to f\) uniformly on \((0,\infty)\), and \(f\) is continuous there by Theorem 6.4.2 (each \(s_k\) is a finite sum of continuous functions).

Each \(f_n\) is differentiable on \((0,\infty)\) with

\begin{equation*} f_n^{\prime}(x) = \frac{(-1)^{n+1}}{(x+n)^2} , \end{equation*}

and \(\sum_{n=0}^{\infty} f_n^{\prime}\) is again alternating with terms \(1/(x+n)^2\) positive, decreasing, and tending to \(0\); the same estimate bounds its tails by \(1/(k+1)^2\) uniformly in \(x > 0\), so it converges uniformly on \((0,\infty)\) to some \(g\).

Fix any \([a,b] \subset (0,\infty)\). There the \(f_n\) are differentiable, \(\sum f_n^{\prime}\) converges uniformly, and \(\sum f_n(a)\) converges – every hypothesis of the Term-by-term Differentiability Theorem (Theorem 6.4.3). So \(f\) is differentiable on \([a,b]\) with \(f^{\prime} = g\), and since each point of \((0,\infty)\) is interior to such an interval, \(f\) is differentiable on all of \((0,\infty)\). \(\square\)

(a) \(f^{\prime}(x) = \sum_{k=1}^{\infty} \cos(kx)/k^2\), continuous on \(\mathbf{R}\).

Put \(f_k(x) = \sin(kx)/k^3\), so \(f_k^{\prime}(x) = k\cos(kx)/k^3 = \cos(kx)/k^2\). Two applications of the M-Test (Corollary 6.4.5), both with convergent \(p\)-series (Corollary 2.4.7):

\begin{equation*} \left| \frac{\sin(kx)}{k^3} \right| \le \frac{1}{k^3} , \qquad \left| \frac{\cos(kx)}{k^2} \right| \le \frac{1}{k^2} \qquad (x \in \mathbf{R}), \end{equation*}

so both \(\sum f_k\) and \(\sum f_k^{\prime}\) converge uniformly on \(\mathbf{R}\).

On any \([a,b]\) the \(f_k\) are differentiable, \(\sum f_k^{\prime}\) converges uniformly, and \(\sum f_k(a)\) converges – all hypotheses of Theorem 6.4.3 – so \(f\) is differentiable there with \(f^{\prime} = \sum f_k^{\prime}\); as \([a,b]\) was arbitrary this holds on \(\mathbf{R}\). Finally \(f^{\prime}\) is continuous by Theorem 6.4.2, being a uniformly convergent sum of the continuous functions \(\cos(kx)/k^2\).

(b) No – not with Theorem 6.4.3, because the next differentiated series genuinely fails to converge uniformly near \(x = 0\).

Differentiating once more gives \(f_k^{\prime\prime}(x) = -\sin(kx)/k\), and the M-Test is useless: \(\sup_x |\sin(kx)/k| = 1/k\) and \(\sum 1/k\) diverges. Uniform convergence genuinely fails on any set \(A \supseteq (0,\delta)\), hence on every interval with \(0\) in its interior – precisely what 6.4.3 would need to reach \(x = 0\). Suppose \(\sum_{k=1}^{\infty} \sin(kx)/k\) did converge uniformly on such an \(A\). By the Cauchy Criterion (Theorem 6.4.4) with \(\epsilon = \sqrt{2}/4\) there would be \(N\) with

\begin{equation*} \left| \sum_{k=m+1}^{n} \frac{\sin(kx)}{k} \right| < \frac{\sqrt{2}}{4} \qquad \text{for all } n > m \ge N,\ x \in A . \end{equation*}

Take \(m \ge N\) large enough that \(x_m = \pi/(4m) < \delta\) lies in \(A\), and set \(n = 2m\). For \(m+1 \le k \le 2m\) we have \(k x_m \in [\pi/4, \pi/2]\), where \(\sin \ge \sqrt{2}/2\), so

\begin{equation*} \begin{aligned} \sum_{k=m+1}^{2m} \frac{\sin(k x_m)}{k} &\ge \frac{\sqrt{2}}{2} \sum_{k=m+1}^{2m} \frac{1}{k} \\ &\ge \frac{\sqrt{2}}{2} \cdot m \cdot \frac{1}{2m} = \frac{\sqrt{2}}{4} , \end{aligned} \end{equation*}

contradicting the displayed bound. So the hypothesis of Theorem 6.4.3 fails, and it tells us nothing about whether \(f^{\prime}\) is differentiable at \(0\). \(\square\)

\(f\) is defined on all of \(\mathbf{R}\), where it is continuous, differentiable, and twice-differentiable.

Write \(g_k(x) = \sin(x/k)/k\). Since \(|\sin t| \le |t|\), for \(|x| \le M\)

\begin{equation*} |g_k(x)| = \left| \frac{\sin(x/k)}{k} \right| \le \frac{|x|}{k^2} \le \frac{M}{k^2}, \end{equation*}

and \(\sum M/k^2\) converges, so the Weierstrass M-Test (Corollary 6.4.5) gives uniform convergence of \(\sum g_k\) on \([-M,M]\). Each \(g_k\) is continuous, so the Term-by-term Continuity Theorem (Theorem 6.4.2) makes \(f\) continuous on \([-M,M]\); since \(M > 0\) is arbitrary and continuity is local, \(f\) is defined and continuous on all of \(\mathbf{R}\).

For the derivative, \(g_k^{\prime}(x) = \cos(x/k)/k^2\) satisfies

\begin{equation*} |g_k^{\prime}(x)| \le \frac{1}{k^2} \quad \text{for all } x \in \mathbf{R}, \end{equation*}

so \(\sum g_k^{\prime}\) converges uniformly on \(\mathbf{R}\) by the M-Test. Each \(g_k\) is differentiable and \(\sum g_k(0) = 0\) converges, so the Term-by-term Differentiability Theorem (Theorem 6.4.3), applied on \(A = \mathbf{R}\), gives that \(f\) is differentiable with

\begin{equation*} f^{\prime}(x) = \sum_{k=1}^{\infty} \frac{\cos(x/k)}{k^2}. \end{equation*}

Run the same argument on this series. Here \(g_k^{\prime\prime}(x) = -\sin(x/k)/k^3\) with

\begin{equation*} |g_k^{\prime\prime}(x)| \le \frac{1}{k^3} \quad \text{for all } x \in \mathbf{R}, \end{equation*}

so \(\sum g_k^{\prime\prime}\) converges uniformly on \(\mathbf{R}\), while \(\sum g_k^{\prime}(0) = \sum 1/k^2\) converges. Theorem 6.4.3 applies again: \(f^{\prime}\) is differentiable on \(\mathbf{R}\), so \(f\) is twice-differentiable there, with

\begin{equation*} f^{\prime\prime}(x) = -\sum_{k=1}^{\infty} \frac{\sin(x/k)}{k^3}. \end{equation*}

(a) Each term \(h_n(x) = 1/(x^2+n^2)\) is defined and continuous on all of \(\mathbf{R}\) (the denominator is never zero), and

\begin{equation*} 0 < \frac{1}{x^2+n^2} \le \frac{1}{n^2} \quad \text{for all } x \in \mathbf{R}. \end{equation*}

Since \(\sum 1/n^2\) converges, the Weierstrass M-Test (Corollary 6.4.5) gives that \(\sum h_n\) converges uniformly on \(\mathbf{R}\); in particular \(h\) is defined on all of \(\mathbf{R}\), and the Term-by-term Continuity Theorem (Theorem 6.4.2) makes \(h\) continuous there.

(b) Yes, and \(h^{\prime}\) is continuous. We have

\begin{equation*} h_n^{\prime}(x) = \frac{-2x}{(x^2+n^2)^2}, \end{equation*}

and this is bounded uniformly in \(x\) by a summable constant: for \(x \ne 0\), the inequality \(x^2 + n^2 \ge 2|x|n\) gives

\begin{equation*} \begin{aligned} |h_n^{\prime}(x)| &= \frac{2|x|}{(x^2+n^2)^2} \\ &\le \frac{2|x|}{\bigl(2|x|n\bigr)(x^2+n^2)} \\ &= \frac{1}{n(x^2+n^2)} \le \frac{1}{n^3}, \end{aligned} \end{equation*}

and \(|h_n^{\prime}(0)| = 0 \le 1/n^3\) as well. Since \(\sum 1/n^3\) converges, the M-Test makes \(\sum h_n^{\prime}\) converge uniformly on \(\mathbf{R}\). Each \(h_n\) is differentiable and \(\sum h_n(0) = \sum 1/n^2\) converges, so Theorem 6.4.3 gives that \(h\) is differentiable on \(\mathbf{R}\) with

\begin{equation*} h^{\prime}(x) = -\sum_{n=1}^{\infty} \frac{2x}{(x^2+n^2)^2}. \end{equation*}

This last series converges uniformly on \(\mathbf{R}\) and its terms are continuous, so \(h^{\prime}\) is continuous by Theorem 6.4.2.

Everything follows from the single bound \(0 \le u_n(x) \le 1/2^n\), valid for every \(x \in \mathbf{R}\).

Defined on all of \(\mathbf{R}\): since \(\sum_{n=1}^{\infty} 1/2^n = 1\) converges, the Weierstrass M-Test (Corollary 6.4.5) gives that \(\sum u_n\) converges uniformly on \(\mathbf{R}\). Thus \(h\) is defined on all of \(\mathbf{R}\), with \(0 \le h(x) \le 1\).

Monotone: each \(u_n\) is nondecreasing. Indeed, let \(x < y\). If \(u_n(x) = 1/2^n\) then \(x > r_n\), hence \(y > x > r_n\) and \(u_n(y) = 1/2^n\); otherwise \(u_n(x) = 0 \le u_n(y)\). Either way \(u_n(x) \le u_n(y)\), so for every \(N\)

\begin{equation*} \sum_{n=1}^{N} u_n(x) \le \sum_{n=1}^{N} u_n(y), \end{equation*}

and both sides converge, so the Order Limit Theorem (Theorem 2.3.4) gives \(h(x) \le h(y)\). Hence \(h\) is monotone increasing on \(\mathbf{R}\).

Continuity at each irrational: fix \(c \notin \mathbf{Q}\). Each \(u_n\) is continuous at every point of \(\mathbf{R} \setminus \{r_n\}\), because \(u_n\) is constant on the open set \((-\infty, r_n)\) and constant on the open set \((r_n, \infty)\). Since \(c\) is irrational, \(c \ne r_n\) for every \(n\), so every \(u_n\) is continuous at \(c\), and therefore so is each partial sum

\begin{equation*} s_N(x) = \sum_{n=1}^{N} u_n(x) \end{equation*}

(Theorem 4.3.4, algebraic continuity). Since \(s_N \to h\) uniformly on \(\mathbf{R}\) and each \(s_N\) is continuous at \(c\), the Continuous Limit Theorem (Theorem 6.2.6) gives that \(h\) is continuous at \(c\).

\(g\) is defined and continuous exactly on \((-1,1]\), and \(g^{\prime}(x) = 1/(1+x)\) on \((-1,1)\).

Write \(g(x) = \sum_{n=1}^{\infty} (-1)^{n+1} x^n/n\).

(a) On \((-1,1)\): for \(|x| < 1\) the Ratio Test (Exercise 2.7.9) gives

\begin{equation*} \left| \frac{(-1)^{n+2}x^{n+1}/(n+1)}{(-1)^{n+1}x^{n}/n} \right| = |x|\,\frac{n}{n+1} \longrightarrow |x| < 1, \end{equation*}

so the series converges absolutely and \(g\) is defined on \((-1,1)\). Given \(x \in (-1,1)\), choose \(c\) with \(|x| < c < 1\); the series converges absolutely at \(c\), so by Theorem 6.5.2 it converges uniformly on \([-c,c]\), and a uniform limit of continuous functions is continuous (Theorem 6.4.2). Hence \(g\) is continuous at \(x\), and so on all of \((-1,1)\).

At \(x = 1\) the series becomes the alternating harmonic series \(\sum (-1)^{n+1}/n\), which converges by the Alternating Series Test (Theorem 2.7.7). So \(g\) is defined on \((-1,1]\). Continuity on \((-1,1]\): the series converges pointwise on \(A = (-1,1]\), so by Theorem 6.5.5 it converges uniformly on every compact \(K \subseteq A\). Any point of \(A\) lies inside such a compact subinterval (for \(x_0 \in (-1,1]\) take \(K = [-c,1]\) with \(|x_0| \le c < 1\), or \(K = [0,1]\) when \(x_0 = 1\)), so \(g\) is continuous on \((-1,1]\).

At \(x = -1\) the series becomes

\begin{equation*} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^n}{n} = -\sum_{n=1}^{\infty}\frac{1}{n}, \end{equation*}

the harmonic series, which diverges (Example 2.4.5). So \(g\) is not defined on \([-1,1]\).

For points \(|x| > 1\): no. If the series converged at some \(x_0\) with \(|x_0| > 1\), then Theorem 6.5.1 would force absolute convergence at every \(x\) with \(|x| < |x_0|\), in particular at \(x = -1\) — contradicting the divergence just observed.

(b) \(g\) is a power series with radius of convergence \(R = 1\), so by Theorem 6.5.7 it is differentiable on \((-1,1)\) with

\begin{equation*} g^{\prime}(x) = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = \sum_{k=0}^{\infty}(-x)^{k} = \frac{1}{1+x}, \end{equation*}

the last equality being the geometric series (Example 2.7.5), valid for \(|x| < 1\). At \(x = 1\) the differentiated series is \(1 - 1 + 1 - \cdots\), whose terms do not tend to \(0\), so it diverges and Theorem 6.5.7 supplies nothing beyond \((-1,1)\).

(a) \(a_n = 1/n!\). For any \(x\), the Ratio Test (Exercise 2.7.9) gives \(|x^{n+1}/(n+1)!| \cdot |n!/x^n| = |x|/(n+1) \to 0 < 1\).

(b) Impossible. Every power series converges at \(x = 0\), where the sum is \(a_0\).

(c) \(a_n = 1/n^2\) (with \(a_0 = 0\)), i.e. \(\sum_{n=1}^{\infty} x^n/n^2\). For \(|x| \le 1\),

\begin{equation*} \sum_{n=1}^{\infty}\left|\frac{x^n}{n^2}\right| \le \sum_{n=1}^{\infty}\frac{1}{n^2} < \infty \end{equation*}

by the Comparison Test (Theorem 2.7.4) and the \(p\)-series with \(p = 2\) (Corollary 2.4.7). For \(|x| > 1\) we have \(|x|^n/n^2 \to \infty\), so the terms do not tend to \(0\) and the series diverges (Theorem 2.7.3).

(d) Impossible. Absolute convergence at \(x = 1\) means \(\sum |a_n| < \infty\). But then

\begin{equation*} \sum_{n=0}^{\infty} |a_n(-1)^n| = \sum_{n=0}^{\infty}|a_n| < \infty, \end{equation*}

so the series converges absolutely at \(x = -1\) as well, not conditionally.

(e) Take the even-index series

\begin{equation*} \sum_{m=1}^{\infty} \frac{(-1)^m}{m}\,x^{2m}, \end{equation*}

that is, \(a_{2m} = (-1)^m/m\) for \(m \ge 1\) and \(a_n = 0\) for all odd \(n\) and for \(n = 0\). Since \(x^{2m}\) takes the value \(1\) at both \(x = 1\) and \(x = -1\), the series at either endpoint is the alternating harmonic series \(\sum_{m}(-1)^m/m\), which converges (Alternating Series Test, Theorem 2.7.7), while

\begin{equation*} \sum_{m=1}^{\infty}\left|\frac{(-1)^m}{m}\right| = \sum_{m=1}^{\infty}\frac{1}{m} = \infty . \end{equation*}

So the convergence is conditional at both endpoints.

Take \(M_n = |a_n|c^n\).

For every \(x \in [-c,c]\) and every \(n \ge 0\),

\begin{equation*} |a_n x^n| = |a_n|\,|x|^n \le |a_n| c^n = M_n , \end{equation*}

since \(|x| \le c\). The hypothesis is precisely that

\begin{equation*} \sum_{n=0}^{\infty} M_n = \sum_{n=0}^{\infty} |a_n|\,|x_0|^n = \sum_{n=0}^{\infty} |a_n x_0^n| \end{equation*}

converges. Both hypotheses of the Weierstrass M-Test (Corollary 6.4.5) are therefore met by the functions \(f_n(x) = a_n x^n\) on the set \([-c,c]\), and the M-Test yields that \(\sum_{n=0}^{\infty} a_n x^n\) converges uniformly on \([-c,c]\). (Corollary 6.4.5 asks for \(M_n > 0\); if some \(a_n\) or \(c\) itself vanishes, run the test with \(M_n + 2^{-n}\) instead, which is still summable.)

(a) Fix \(x \in (-R,R)\) and choose \(t\) with \(|x| < t < R\). The original series converges at \(t\), so by Theorem 6.5.1 it converges absolutely at \(x\): \(\sum_{n=0}^{\infty}|a_n x^n| < \infty\). Then for every \(n\),

\begin{equation*} \left| \frac{a_n}{n+1}x^{n+1} \right| = \frac{|x|}{n+1}\,|a_n x^n| \le |x| \cdot |a_n x^n| , \end{equation*}

so the series defining \(F\) converges absolutely by the Comparison Test (Theorem 2.7.4). Hence \(F\) is defined on all of \((-R,R)\).

\(F\) is itself a power series, \(F(x) = \sum_{m=0}^{\infty} b_m x^m\) with \(b_0 = 0\) and \(b_m = a_{m-1}/m\) for \(m \ge 1\), and we have just shown it converges on \((-R,R)\). Theorem 6.5.7 therefore makes \(F\) differentiable on \((-R,R)\) with derivative obtained term by term:

\begin{equation*} \begin{aligned} F^{\prime}(x) &= \sum_{m=1}^{\infty} m\,b_m x^{m-1} = \sum_{m=1}^{\infty} m \cdot \frac{a_{m-1}}{m}\,x^{m-1} \\ &= \sum_{m=1}^{\infty} a_{m-1}x^{m-1} = \sum_{n=0}^{\infty} a_n x^n = f(x). \end{aligned} \end{equation*}

(b) \(g(x) = g(0) + \sum_{n=0}^{\infty} \dfrac{a_n}{n+1}x^{n+1}\).

Indeed \((g - F)^{\prime}(x) = f(x) - f(x) = 0\) for every \(x \in (-R,R)\), so \(g - F\) is constant on that interval by Corollary 5.3.4 (a consequence of the Mean Value Theorem). Evaluating at \(x = 0\), where \(F(0) = 0\), identifies the constant as \(g(0)\).

(a) Put \(x_n = n s^{n-1}\) and fix \(r\) with \(s < r < 1\). Since

\begin{equation*} \frac{x_{n+1}}{x_n} = \frac{(n+1)s^{n}}{n s^{n-1}} = s\left(1 + \frac{1}{n}\right) \longrightarrow s < r, \end{equation*}

there is an \(N \in \mathbf{N}\) with \(x_{n+1} \le r\,x_n\) for all \(n \ge N\). Induction gives \(x_n \le x_N r^{\,n-N} \le x_N\) for all \(n \ge N\), so

\begin{equation*} n s^{n-1} \le \max\{x_1, x_2, \ldots, x_{N}\} =: M \qquad \text{for all } n \ge 1 . \end{equation*}

(b) Fix \(x \in (-R,R)\). If \(x = 0\) the differentiated series reduces to the single term \(a_1\) and there is nothing to prove, so assume \(x \ne 0\) and pick \(t\) with \(|x| < t < R\). Set \(s = |x|/t\), so \(0 < s < 1\), and let \(M\) be the bound from part (a).

First, \(\sum_{n=0}^{\infty}|a_n t^n| < \infty\): choose \(t^{\prime}\) with \(t < t^{\prime} < R\); the series converges at \(t^{\prime}\) by hypothesis, and \(|t| < |t^{\prime}|\), so Theorem 6.5.1 gives absolute convergence at \(t\).

Now estimate the terms of the differentiated series, using \(|x|^{n-1} = s^{n-1}t^{n-1}\):

\begin{equation*} \begin{aligned} \left| n a_n x^{n-1} \right| &= \left(n s^{n-1}\right) |a_n|\,t^{n-1} \\ &\le M\,|a_n|\,t^{n-1} = \frac{M}{t}\,\left| a_n t^{n} \right| . \end{aligned} \end{equation*}

Since \(\sum (M/t)|a_n t^n|\) converges, the Comparison Test (Theorem 2.7.4) shows \(\sum_{n=1}^{\infty} n a_n x^{n-1}\) converges absolutely. As \(x \in (-R,R)\) was arbitrary, the differentiated series converges at every point of \((-R,R)\).

For the last clause: \(\sum_{n=1}^{\infty} n a_n x^{n-1}\) is itself a power series which we have just shown converges pointwise on the set \(A = (-R,R)\). Theorem 6.5.5 then gives uniform convergence on every compact \(K \subseteq (-R,R)\).

\begin{equation*} \sum_{n=1}^{\infty}\frac{n}{2^n} = 2, \qquad \sum_{n=1}^{\infty}\frac{n^2}{2^n} = 6 . \end{equation*}

The geometric series has radius of convergence \(R = 1\), so Theorem 6.5.7 permits term-by-term differentiation on \((-1,1)\):

\begin{equation*} \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}, \qquad \text{hence} \qquad \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n} . \end{equation*}

Multiplying a power series by \(x\) shifts the exponents and changes nothing about convergence, so the right-hand series is again a power series valid on \((-1,1)\). Setting \(x = 1/2\):

\begin{equation*} \sum_{n=1}^{\infty}\frac{n}{2^n} = \frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = 2 . \end{equation*}

Apply Theorem 6.5.7 once more to \(\sum_{n=1}^{\infty} n x^n = x/(1-x)^2\) on \((-1,1)\):

\begin{equation*} \begin{aligned} \sum_{n=1}^{\infty} n^2 x^{n-1} &= \frac{d}{dx}\left[\frac{x}{(1-x)^2}\right] \\ &= \frac{1}{(1-x)^2} + \frac{2x}{(1-x)^3} = \frac{1+x}{(1-x)^3}, \end{aligned} \end{equation*}

so \(\sum_{n=1}^{\infty} n^2 x^n = x(1+x)/(1-x)^3\). At \(x = 1/2\),

\begin{equation*} \sum_{n=1}^{\infty}\frac{n^2}{2^n} = \frac{(1/2)(3/2)}{(1/2)^3} = \frac{3/4}{1/8} = 6 . \end{equation*}

(a) Fix \(x\) with \(|x| < 1/L\); for \(x = 0\) the series trivially converges to \(a_0\), so take \(x \ne 0\) and set \(c_n = a_n x^n\), which is nonzero. Then

\begin{equation*} \left|\frac{c_{n+1}}{c_n}\right| = \left|\frac{a_{n+1}}{a_n}\right|\,|x| \longrightarrow L|x| < 1 \end{equation*}

by the Algebraic Limit Theorem (Theorem 2.3.3). The limiting ratio \(L|x|\) is strictly less than \(1\), which is exactly the hypothesis of the Ratio Test (Exercise 2.7.9), so \(\sum c_n = \sum a_n x^n\) converges absolutely.

(b) Same computation. For any \(x \ne 0\),

\begin{equation*} \left|\frac{a_{n+1}x^{n+1}}{a_n x^n}\right| = |x|\left|\frac{a_{n+1}}{a_n}\right| \longrightarrow |x|\cdot 0 = 0 < 1, \end{equation*}

so the Ratio Test gives absolute convergence at every \(x \in \mathbf{R}\) (and \(x = 0\) is trivial). Equivalently, \(L = 0\) means the radius of convergence is \(R = \infty\).

(c) One argument covers both, with the convention \(1/L^{\prime} = \infty\) when \(L^{\prime} = 0\). The hypothesis that \(L^{\prime}\) exists forces the ratios \(|a_{k+1}/a_k|\) to be bounded, so Exercise 2.4.7(a) applies: \(L^{\prime} = \limsup |a_{n+1}/a_n| \ge 0\), and \(L^{\prime} = L\) whenever the plain limit \(L\) exists (Exercise 2.4.7(d)).

Fix \(x \ne 0\) with \(|x| < 1/L^{\prime}\), i.e. \(L^{\prime}|x| < 1\), and pick \(r\) with \(L^{\prime}|x| < r < 1\). Since \(s_n \to L^{\prime}\) and \(r/|x| > L^{\prime}\), there exists \(N \in \mathbf{N}\) with

\begin{equation*} s_N = \sup\left\{\left|\frac{a_{k+1}}{a_k}\right| : k \ge N\right\} < \frac{r}{|x|} . \end{equation*}

The supremum dominates each of its members, so for every \(k \ge N\),

\begin{equation*} \left|\frac{a_{k+1}}{a_k}\right| < \frac{r}{|x|}, \qquad \text{i.e.} \qquad \left| a_{k+1}x^{k+1} \right| < r \left| a_k x^{k} \right| . \end{equation*}

Iterating this inequality from \(k = N\) gives

\begin{equation*} \left| a_n x^n \right| \le \left| a_N x^N \right| r^{\,n-N} \qquad \text{for all } n \ge N, \end{equation*}

and \(\sum_{n \ge N} |a_N x^N| r^{n-N}\) is a convergent geometric series (\(0 < r < 1\)). By the Comparison Test (Theorem 2.7.4), \(\sum a_n x^n\) converges absolutely.

Thus the series converges on \((-1/L^{\prime}, 1/L^{\prime})\) when \(L^{\prime} \ne 0\), which is statement (a); and when \(L^{\prime} = 0\) the constraint \(L^{\prime}|x| < 1\) holds for every \(x \in \mathbf{R}\), so the series converges on all of \(\mathbf{R}\), which is statement (b).

(a) The coefficients are read off the derivatives at the origin: \(a_k = f^{(k)}(0)/k!\).

Let \(f\) denote the common value of the two series on \((-R,R)\). By Theorem 6.5.7 the function \(f\) is infinitely differentiable on \((-R,R)\) and every derivative is obtained by term-by-term differentiation of the first series, so for each \(k \ge 0\),

\begin{equation*} f^{(k)}(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)\, a_n x^{n-k}, \qquad x \in (-R,R). \end{equation*}

Setting \(x = 0\) kills every term except \(n = k\) (whose factor \(x^{n-k} = x^0 = 1\)), leaving \(f^{(k)}(0) = k!\,a_k\). Running the identical computation on \(\sum b_n x^n\), which represents the same function \(f\), gives \(f^{(k)}(0) = k!\,b_k\). Hence \(k!\,a_k = k!\,b_k\), and \(a_k = b_k\) for all \(k = 0,1,2,\ldots\).

(b) \(a_n = 1/n!\) for every \(n \ge 0\).

From \(f(0) = 1\) we get \(a_0 = 1\). By Theorem 6.5.7 again,

\begin{equation*} f^{\prime}(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n , \end{equation*}

a power series converging on all of \((-R,R)\). The hypothesis \(f^{\prime} = f\) makes this the same function as \(\sum_{n=0}^{\infty} a_n x^n\) on \((-R,R)\), so part (a) applies and matches coefficients:

\begin{equation*} (n+1)a_{n+1} = a_n \quad \text{for all } n \ge 0, \qquad\text{i.e.}\qquad a_{n+1} = \frac{a_n}{n+1}. \end{equation*}

With \(a_0 = 1\) this recursion gives \(a_1 = 1\), \(a_2 = 1/2\), \(a_3 = 1/6\), and by induction

\begin{equation*} a_n = \frac{1}{n!}, \qquad n = 0, 1, 2, \ldots \end{equation*}

The whole proof is the identity \(h(x) = f(x)g(x)\) on \((-1,1)\) followed by \(x \to 1^-\).

Assume \(\sum a_n = A\), \(\sum b_n = B\), and \(\sum d_n = D\) all converge. Each of the three power series therefore converges at \(x = 1\), so Abel’s Theorem (Theorem 6.5.4, with \(R = 1\)) gives uniform convergence on \([0,1]\); the partial sums are polynomials, hence by the Term-by-term Continuity Theorem (Theorem 6.4.2) \(f\), \(g\), \(h\) are continuous on \([0,1]\). In particular

\begin{equation*} \lim_{x \to 1^-} f(x) = A, \qquad \lim_{x \to 1^-} g(x) = B, \qquad \lim_{x \to 1^-} h(x) = D . \end{equation*}

Now fix \(x\) with \(|x| < 1\). Since \(\sum a_n \cdot 1^n\) converges, Theorem 6.5.1 says \(\sum a_n x^n\) converges absolutely; likewise \(\sum b_n x^n\). So Exercise 2.8.7 applies to the two absolutely convergent series \(\sum (a_n x^n)\) and \(\sum (b_n x^n)\): their Cauchy product converges to the product of the sums. The \(n\)-th Cauchy term of these two series is

\begin{equation*} \sum_{k=0}^{n} (a_k x^k)(b_{n-k} x^{n-k}) = \Big( \sum_{k=0}^{n} a_k b_{n-k} \Big) x^n = d_n x^n , \end{equation*}

so Exercise 2.8.7 reads

\begin{equation*} h(x) = \sum_{n=0}^{\infty} d_n x^n = f(x) g(x) \qquad \text{for all } |x| < 1 . \end{equation*}

(The conclusion of 2.8.7 includes the convergence of the Cauchy product, so \(h\) needs no separate convergence check on \((-1,1)\).)

Let \(x \to 1^-\) in this identity: the left side tends to \(D\) and the right side to \(AB\), by the three limits above and Corollary 4.2.4. Hence

\begin{equation*} \sum_{n=0}^{\infty} d_n = D = AB . \end{equation*}

Factor out the lowest surviving power: \(g(x) = x^m h(x)\) with \(h\) continuous and \(h(0) = b_m\), and the vanishing along \((x_n)\) forces \(h(0) = 0\).

If every \(b_n = 0\) there is nothing to prove, so assume some coefficient is nonzero and let

\begin{equation*} m = \min\{\, n \ge 0 : b_n \neq 0 \,\}. \end{equation*}

Define

\begin{equation*} h(x) = \sum_{k=0}^{\infty} b_{m+k}\, x^k . \end{equation*}

This power series converges on \((-R,R)\): for \(0 < |x| < R\) the series \(\sum_{k} b_{m+k} x^{m+k}\) is a tail of the convergent series \(\sum_n b_n x^n\), and dividing its partial sums by the nonzero constant \(x^m\) shows \(\sum_k b_{m+k} x^k\) converges; at \(x = 0\) convergence is trivial. Moreover, since \(b_0 = \cdots = b_{m-1} = 0\),

\begin{equation*} g(x) = \sum_{n=m}^{\infty} b_n x^n = x^m h(x) \qquad \text{for all } x \in (-R,R). \end{equation*}

By Theorem 6.5.7, \(h\) is continuous on \((-R,R)\). For each \(n\) we have \(x_n^m\, h(x_n) = g(x_n) = 0\) with \(x_n \neq 0\), so \(h(x_n) = 0\). Continuity of \(h\) at \(0\) together with \(x_n \to 0\) (Theorem 4.3.2, the sequential characterization of continuity) gives

\begin{equation*} b_m = h(0) = \lim_{n \to \infty} h(x_n) = 0, \end{equation*}

contradicting \(b_m \neq 0\). Hence every \(b_n = 0\) and \(g \equiv 0\) on \((-R,R)\).

(a) Abel’s Theorem makes \(f\) continuous at the endpoint, so \(\lim_{x\to 1^-} f(x) = f(1) = L\).

Assume \(\sum_{n=0}^{\infty} a_n = L\); that is, \(f(x) = \sum a_n x^n\) converges at \(x = 1\) with value \(L\). Theorem 6.5.1 then gives absolute convergence for every \(|x| < 1\), so \(f\) is defined on all of \([0,1]\). Abel’s Theorem (Theorem 6.5.4 with \(R = 1\), whose hypothesis “converges at \(x = R\)” is exactly our assumption) makes the convergence uniform on \([0,1]\), and the partial sums are polynomials, so the Term-by-term Continuity Theorem (Theorem 6.4.2) makes \(f\) continuous on \([0,1]\). Continuity at \(x = 1\) gives

\begin{equation*} \lim_{x \to 1^-} f(x) = f(1) = \sum_{n=0}^{\infty} a_n = L . \end{equation*}

Thus \(\sum a_n\) is Abel-summable to \(L\).

(b) The sum is \(1/2\).

Here \(a_n = (-1)^n\), and for \(x \in [0,1)\) the series \(\sum_{n=0}^{\infty} (-1)^n x^n = \sum_{n=0}^{\infty} (-x)^n\) is geometric with ratio \(-x\) satisfying \(|-x| = x < 1\), so by Example 2.7.5 it converges with

\begin{equation*} f(x) = \sum_{n=0}^{\infty} (-x)^n = \frac{1}{1-(-x)} = \frac{1}{1+x} . \end{equation*}

Since \(x \mapsto 1/(1+x)\) is continuous at \(x = 1\),

\begin{equation*} \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{1}{1+x} = \frac{1}{2}, \end{equation*}

so \(\sum_{n=0}^{\infty} (-1)^n\) is Abel-summable to \(1/2\).

The value is \(\arctan(1) = \pi/4\), and the identity is Leibniz’s series

\begin{equation*} 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}. \end{equation*}

Let \(S(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n+1}/(2n+1)\). At \(x = 1\) it reads \(\sum (-1)^n/(2n+1)\), whose terms decrease monotonically to zero, so the Alternating Series Test (Theorem 2.7.7) applies and \(S\) converges on a set \(A \supseteq (-1,1]\). Theorem 6.5.7 makes \(S\) continuous on all of \(A\), the endpoint \(x = 1\) included.

Since \(S(x) = \arctan(x)\) on \((-1,1)\) by equation (2) of the section, and \(\arctan\) is continuous at \(1\) by hypothesis, letting \(x \to 1^{-}\) gives

\begin{equation*} \begin{aligned} S(1) &= \lim_{x \to 1^{-}} S(x) = \lim_{x \to 1^{-}} \arctan(x)\\ &= \arctan(1) = \frac{\pi}{4}. \end{aligned} \end{equation*}

Writing out \(S(1)\) gives the stated identity.

(a) \(x\cos(x^2) = \sum_{n=0}^{\infty} (-1)^n x^{4n+1}/(2n)!\), valid for all \(x \in \mathbf{R}\).

The series \(x - x^3/3! + x^5/5! - \cdots\) converges uniformly to \(\sin(x)\) on every \([-R,R]\) (Example 6.6.4), so Theorem 6.5.7 permits term-by-term differentiation on all of \(\mathbf{R}\):

\begin{equation*} \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}, \qquad x \in \mathbf{R}. \end{equation*}

Substituting \(x^2\) for \(x\) is legitimate at every real \(x\) since \(x^2 \in \mathbf{R}\), and pulling the constant \(x\) through the sum is Theorem 2.7.1:

\begin{equation*} \begin{aligned} x\cos(x^2) &= x\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{4n}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{4n+1}. \end{aligned} \end{equation*}

(b) \(\dfrac{x}{(1+4x^2)^2} = \sum_{n=0}^{\infty} (-1)^n (n+1)4^n x^{2n+1}\), valid precisely for \(|x| < 1/2\).

Example 6.6.1 gives \(1/(1-x)^2 = \sum_{n=0}^{\infty}(n+1)x^n\) for \(|x| < 1\). Substituting \(-4x^2\) for \(x\), which requires \(|-4x^2| < 1\), i.e. \(|x| < 1/2\),

\begin{equation*} \frac{1}{(1+4x^2)^2} = \sum_{n=0}^{\infty}(n+1)(-4)^n x^{2n}, \end{equation*}

and multiplying by \(x\) gives the displayed series. At \(|x| = 1/2\) the \(n\)th term has absolute value \((n+1)4^n 2^{-(2n+1)} = (n+1)/2\), which does not tend to zero, so the series diverges there and hence for all \(|x| \ge 1/2\).

(c) \(\log(1+x^2) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}x^{2n}\), valid precisely for \(|x| \le 1\).

Replacing \(x\) by \(-x\) in equation (1) gives \(1/(1+x) = \sum_{n=0}^{\infty}(-1)^n x^n\) on \((-1,1)\); term-by-term antidifferentiation (Exercise 6.5.4) produces \(F\) on \((-1,1)\) with \(F^{\prime} = 1/(1+x)\) and \(F(0) = 0\), and \(\log(1+x)\) has the same derivative and the same value at \(0\), so the two agree by Corollary 5.3.4:

\begin{equation*} \log(1+u) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}u^{n}, \qquad u \in (-1,1), \end{equation*}

and Exercise 6.6.4 extends this to \(u = 1\). Since \(u = x^2\) ranges over \([0,1]\) exactly when \(|x| \le 1\), the substitution is legitimate there and gives the stated series. For \(|x| > 1\) the terms \(x^{2n}/n\) are unbounded, so the series diverges (Theorem 2.7.3).

Differentiate the series \(k\) times and set \(x = 0\): the terms below \(x^k\) have been annihilated and those above still carry a factor of \(x\), so only \(k!\,a_k\) survives.

The interval of convergence contains some \((-R,R)\) with \(R > 0\), and on it Theorem 6.5.7 makes \(f\) infinitely differentiable with every derivative computed term by term. Applying that \(k\) times gives, for \(x \in (-R,R)\),

\begin{equation*} f^{(k)}(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)\,a_n x^{n-k}, \end{equation*}

the terms with \(n < k\) having been annihilated one at a time. (Formally: the case \(k = 0\) is the hypothesis, and differentiating the displayed formula for \(k\) term by term multiplies \(a_n x^{n-k}\) by \((n-k)\) and lowers the exponent, which is exactly the formula for \(k+1\).)

Evaluate at \(x = 0\). Every summand with \(n > k\) carries a factor \(x^{n-k}\) with \(n - k \ge 1\) and so vanishes, while the \(n = k\) summand is the constant \(k(k-1)\cdots 1 \cdot a_k = k!\,a_k\). Hence

\begin{equation*} f^{(k)}(0) = k!\,a_k, \qquad\text{i.e.}\qquad a_k = \frac{f^{(k)}(0)}{k!}, \end{equation*}

for every \(k = 0,1,2,\ldots\) (with \(0! = 1\) and \(f^{(0)} = f\), which gives \(a_0 = f(0)\)).

The modification is to allow \(x\) to be an endpoint. Theorem 6.6.3 is stated for \(x\) in the open interval \((-R,R)\), but its proof only ever applies the Generalized Mean Value Theorem (Theorem 5.3.5) to \(E_N\) and \(x^{N+1}\) on the closed interval \([0,x]\) and its successive subintervals, so it holds verbatim in the form: if \(f\) is \(N+1\) times differentiable on an open interval containing \([0,b]\), \(a_n = f^{(n)}(0)/n!\) for \(0 \le n \le N\), and \(S_N(x) = a_0 + a_1x + \cdots + a_Nx^N\), then each \(x \in (0,b]\) admits \(c \in (0,x)\) with

\begin{equation*} E_N(x) = f(x) - S_N(x) = \frac{f^{(N+1)}( c)}{(N+1)!}x^{N+1}. \end{equation*}

Apply this to \(f(x) = \log(1+x)\) with \(b = 1\) and \(x = 1\); the hypothesis holds because \(f\) is infinitely differentiable on \((-1,\infty) \supseteq [0,1]\), with

\begin{equation*} f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} \qquad (n \ge 1), \end{equation*}

so \(a_0 = 0\) and \(a_n = f^{(n)}(0)/n! = (-1)^{n-1}/n\), making

\begin{equation*} S_N(1) = 1 - \frac{1}{2} + \frac{1}{3} - \cdots + \frac{(-1)^{N-1}}{N} \end{equation*}

the \(N\)th partial sum of the series in question. For the error there is \(c \in (0,1)\) with

\begin{equation*} \begin{aligned} E_N(1) &= \frac{f^{(N+1)}( c)}{(N+1)!}\cdot 1^{N+1}\\ &= \frac{(-1)^{N}N!}{(1+c)^{N+1}(N+1)!} = \frac{(-1)^{N}}{(N+1)(1+c)^{N+1}}. \end{aligned} \end{equation*}

Since \(c > 0\) we have \((1+c)^{N+1} > 1\), whence

\begin{equation*} |E_N(1)| \le \frac{1}{N+1} \longrightarrow 0. \end{equation*}

Therefore \(S_N(1) \to f(1) = \log 2\), which is the asserted identity.

(a) \(a_n = 1/n!\), so the Taylor series is \(\sum_{n=0}^{\infty} x^n/n!\).

Indeed \(f^{(n)}(x) = e^x\) for every \(n\), so \(a_n = f^{(n)}(0)/n! = e^0/n! = 1/n!\) by Theorem 6.6.2.

Fix \(R > 0\). Since \(f\) is infinitely differentiable on all of \(\mathbf{R}\), Lagrange’s Remainder Theorem (Theorem 6.6.3) applies on the interval \((-(R+1), R+1)\), which contains \([-R,R]\): for each \(x \in [-R,R]\) with \(x \ne 0\) there is \(c\) with \(|c| < |x| \le R\) and

\begin{equation*} E_N(x) = e^x - S_N(x) = \frac{e^{c}}{(N+1)!}x^{N+1}. \end{equation*}

Because \(e^c \le e^R\) and \(|x| \le R\), and because \(E_N(0) = 0\),

\begin{equation*} \sup_{x \in [-R,R]}|E_N(x)| \le \frac{e^{R}R^{N+1}}{(N+1)!}. \end{equation*}

The series \(\sum R^n/n!\) converges by the Ratio Test (Exercise 2.7.9), since the ratio of consecutive terms is \(R/(n+1) \to 0\); hence its terms satisfy \(R^{N+1}/(N+1)! \to 0\) (Theorem 2.7.3). The right-hand bound is a constant multiple of that, so it tends to \(0\), and \(S_N \to e^x\) uniformly on \([-R,R]\).

(b) Every \(x \in \mathbf{R}\) lies in some \([-R,R]\), so by (a) the series \(\sum x^n/n!\) converges to \(e^x\) on all of \(\mathbf{R}\). Theorem 6.5.7 then licenses term-by-term differentiation:

\begin{equation*} \begin{aligned} f^{\prime}(x) &= \sum_{n=1}^{\infty} n\,\frac{x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}\\ &= \sum_{m=0}^{\infty} \frac{x^{m}}{m!} = e^x. \end{aligned} \end{equation*}

(c) Substituting \(-x\) for \(x\) (valid for all \(x \in \mathbf{R}\) since \(-x \in \mathbf{R}\)),

\begin{equation*} e^{-x} = \sum_{n=0}^{\infty}\frac{(-x)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n. \end{equation*}

Collecting the coefficient of \(x^n\) in the product of the two series means summing \(\bigl(1/k!\bigr)\bigl((-1)^{n-k}/(n-k)!\bigr)\) over \(0 \le k \le n\):

\begin{equation*} \begin{aligned} \sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!\,(n-k)!} &= \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}1^{k}(-1)^{n-k}\\ &= \frac{(1-1)^n}{n!} = 0 \qquad (n \ge 1), \end{aligned} \end{equation*}

by the binomial theorem, while for \(n = 0\) the coefficient is \(1\). So the product series is the constant \(1\), in agreement with \(e^x \cdot e^{-x} = 1\).

One limit settles all three parts:

\begin{equation*} (\ast)\qquad \lim_{x \to 0} q(1/x)\,e^{-1/x^2} = 0 \quad\text{for every polynomial } q. \end{equation*}

Put \(t = 1/x\), so \(|t| \to \infty\); by the Algebraic Limit Theorem it suffices to show \(|t|^m e^{-t^2} \to 0\) for each fixed \(m \in \mathbf{N}\). For \(|t| \ge 1\) we have \(|t|^m \le t^{2m} = s^m\) with \(s = t^2 \to \infty\), and the series of Exercise 6.6.5(a) has all positive terms for \(s > 0\), so \(e^s > s^{m+1}/(m+1)!\) and therefore

\begin{equation*} 0 \le \frac{|t|^m}{e^{t^2}} \le \frac{s^m}{e^s} < \frac{(m+1)!}{s} \longrightarrow 0 . \end{equation*}

(a) For \(x \ne 0\) the chain rule gives

\begin{equation*} g^{\prime}(x) = e^{-1/x^2}\cdot\frac{d}{dx}\Bigl(-\frac{1}{x^2}\Bigr) = \frac{2}{x^3}e^{-1/x^2}. \end{equation*}

Since \(g^{\prime}(0) = 0\) was computed in the text, the definition of the derivative gives

\begin{equation*} \begin{aligned} g^{\prime\prime}(0) &= \lim_{x \to 0}\frac{g^{\prime}(x) - g^{\prime}(0)}{x - 0}\\ &= \lim_{x \to 0}\frac{2}{x^4}e^{-1/x^2} = 0 \end{aligned} \end{equation*}

by \((\ast)\) with \(q(t) = 2t^4\).

(b) Differentiating again for \(x \ne 0\),

\begin{equation*} \begin{aligned} g^{\prime\prime}(x) &= \Bigl(\frac{4}{x^6} - \frac{6}{x^4}\Bigr)e^{-1/x^2},\\ g^{\prime\prime\prime}(x) &= \Bigl(\frac{8}{x^9} - \frac{36}{x^7} + \frac{24}{x^5}\Bigr)e^{-1/x^2}. \end{aligned} \end{equation*}

The pattern is that each derivative is a polynomial in \(1/x\) times \(e^{-1/x^2}\). Precisely: there are polynomials \(p_n\) with

\begin{equation*} g^{(n)}(x) = p_n(1/x)\,e^{-1/x^2} \qquad (x \ne 0), \end{equation*}

defined recursively by \(p_0(t) = 1\) and

\begin{equation*} p_{n+1}(t) = -t^2 p_n^{\prime}(t) + 2t^3 p_n(t). \end{equation*}

Indeed, differentiating \(p_n(1/x)e^{-1/x^2}\) by the product and chain rules gives

\begin{equation*} \Bigl[p_n^{\prime}(1/x)\Bigl(-\frac{1}{x^2}\Bigr) + p_n(1/x)\frac{2}{x^3}\Bigr]e^{-1/x^2}, \end{equation*}

which is \(p_{n+1}(1/x)e^{-1/x^2}\); and \(p_{n+1}\) is again a polynomial. (The recursion returns \(p_1(t) = 2t^3\), \(p_2(t) = 4t^6 - 6t^4\), \(p_3(t) = 8t^9 - 36t^7 + 24t^5\), matching the computations above.)

(c) Induct on \(n\), the induction hypothesis being that \(g^{(n)}\) exists on all of \(\mathbf{R}\), that \(g^{(n)}(0) = 0\), and that \(g^{(n)}(x) = p_n(1/x)e^{-1/x^2}\) for \(x \ne 0\). The case \(n = 0\) is the definition of \(g\) (with \(p_0 = 1\)). Assuming it for \(n\), part (b) supplies the formula for \(g^{(n+1)}\) away from zero, and at zero the definition of the derivative gives

\begin{equation*} \begin{aligned} g^{(n+1)}(0) &= \lim_{x \to 0}\frac{g^{(n)}(x) - g^{(n)}(0)}{x - 0}\\ &= \lim_{x \to 0}\frac{1}{x}\,p_n(1/x)\,e^{-1/x^2} = 0 \end{aligned} \end{equation*}

by \((\ast)\) applied to the polynomial \(q(t) = t\,p_n(t)\). Thus \(g^{(n+1)}\) exists everywhere and vanishes at the origin, completing the induction: \(g\) is infinitely differentiable with \(g^{(n)}(0) = 0\) for all \(n\).

All three exist.

(a) \(g(x) = \dfrac{1}{1+x^2}\).

It is infinitely differentiable on \(\mathbf{R}\) (the denominator never vanishes), and Example 6.6.1 substitutes \(-x^2\) into equation (1) to give

\begin{equation*} g(x) = \sum_{n=0}^{\infty}(-1)^n x^{2n} \qquad \text{for } |x| < 1 . \end{equation*}

By Theorem 6.6.2 this power series is precisely the Taylor series of \(g\). For \(|x| \ge 1\) the terms satisfy \(|(-1)^nx^{2n}| \ge 1\), so they do not tend to zero and the series diverges (Theorem 2.7.3). Hence the Taylor series converges to \(g\) exactly on \((-1,1)\).

(b) \(h(x) = \sin(x) + g(x)\), where \(g\) is the flat function of Exercise 6.6.6.

By Exercise 6.6.6(c), \(g^{(n)}(0) = 0\) for every \(n\), so

\begin{equation*} h^{(n)}(0) = \sin^{(n)}(0) + g^{(n)}(0) = \sin^{(n)}(0), \end{equation*}

and \(h\) has exactly the Taylor series \(x - x^3/3! + x^5/5! - \cdots\) of \(\sin(x)\). Yet for \(x \ne 0\),

\begin{equation*} h(x) - \sin(x) = e^{-1/x^2} > 0, \end{equation*}

so \(h(x) \ne \sin(x)\) at every nonzero point. (\(h\) is infinitely differentiable as a sum of two such functions.)

(c) Take

\begin{equation*} f(x) = \begin{cases} e^{-1/x^2} & \text{for } x > 0,\\ 0 & \text{for } x \le 0. \end{cases} \end{equation*}

On \((0,\infty)\) and on \((-\infty,0)\) the function is visibly infinitely differentiable, with

\begin{equation*} f^{(n)}(x) = p_n(1/x)e^{-1/x^2} \ (x>0), \qquad f^{(n)}(x) = 0 \ (x<0), \end{equation*}

for the polynomials \(p_n\) of Exercise 6.6.6(b). At the origin, induction as in Exercise 6.6.6(c) gives \(f^{(n)}(0) = 0\) for all \(n\): the case \(n = 0\) is \(f(0) = 0\), and assuming it for \(n\), the left-hand difference quotient of \(f^{(n)}\) at \(0\) is identically \(0\), while the right-hand quotient is \((1/x)p_n(1/x)e^{-1/x^2} \to 0\) by limit \((\ast)\) of Exercise 6.6.6; the two one-sided limits agree, so \(f^{(n+1)}(0) = 0\).

Consequently every Taylor coefficient of \(f\) is \(0\), so the Taylor series is the zero series, which converges for all \(x \in \mathbf{R}\) with sum \(0\). Its sum equals \(f(x)\) precisely when \(f(x) = 0\), that is, precisely when \(x \le 0\).

(a) Apply the Mean Value Theorem (Theorem 5.3.2) to \(\varphi = h - g\). For \(t \in (0,x]\) there is \(d \in (0,t)\) with

\begin{equation*} \varphi(t) = \varphi(t) - \varphi(0) = \varphi^{\prime}(d)\,t = \bigl(h^{\prime}(d) - g^{\prime}(d)\bigr)t \ge 0, \end{equation*}

since \(g^{\prime}(d) \le h^{\prime}(d)\) and \(t > 0\); and \(\varphi(0) = 0\). Hence \(g(t) \le h(t)\) on \([0,x]\).

(b) Because \(S_N\) is a polynomial of degree at most \(N\) matching the first \(N\) derivatives of \(f\) at \(0\) (as in the proof of Theorem 6.6.3),

\begin{equation*} E_N^{(n)}(0) = 0 \quad (0 \le n \le N), \qquad E_N^{(N+1)}(t) = f^{(N+1)}(t). \end{equation*}

Claim: for each \(k = 0,1,\dots,N+1\),

\begin{equation*} \bigl|E_N^{(N+1-k)}(t)\bigr| \le \frac{M t^{k}}{k!} \quad \text{for all } t \in [0,x]. \end{equation*}

The case \(k = 0\) is the hypothesis \(|f^{(N+1)}(t)| \le M\). Assume the claim for some \(k \le N\) and write \(m = N - k\), so \(0 \le m \le N\) and \(E_N^{(m)}(0) = 0\). Two applications of the lemma on \([0,x]\), with the pairs

\begin{equation*} \begin{aligned} g(t) &= E_N^{(m)}(t), & h(t) &= \frac{M t^{k+1}}{(k+1)!},\\ g(t) &= \frac{-M t^{k+1}}{(k+1)!}, & h(t) &= E_N^{(m)}(t), \end{aligned} \end{equation*}

are legitimate: both pairs agree at \(t = 0\) (all four values are \(0\)), and in each case the derivative inequality is exactly the inductive hypothesis

\begin{equation*} -\frac{M t^{k}}{k!} \le E_N^{(m+1)}(t) \le \frac{M t^{k}}{k!}. \end{equation*}

The two conclusions combine to give \(|E_N^{(m)}(t)| \le M t^{k+1}/(k+1)!\) on \([0,x]\), which is the claim for \(k+1\).

Taking \(k = N+1\) (so \(N+1-k = 0\)) and then \(t = x\):

\begin{equation*} |E_N(x)| \le \frac{M x^{N+1}}{(N+1)!}. \end{equation*}

(a) \(E_N(x,x) = 0\): every term of \(S_N(x,x)\) with \(n \ge 1\) carries the factor \((x-x)^n = 0\), so \(S_N(x,x) = c_0 = f(x)\).

(b) Differentiability: \(f\) is \(N+1\) times differentiable on \((-R,R)\), so each \(f^{(n)}\) with \(n \le N\) is differentiable there, and \(a \mapsto (x-a)^n\) is a polynomial; \(S_N(x,\cdot)\) is a finite sum of products of differentiable functions, and \(f(x)\) is a constant in \(a\). Differentiating term by term with the product rule,

\begin{equation*} \frac{d}{da}\left[\frac{f^{(n)}(a)}{n!}(x-a)^n\right] = \frac{f^{(n+1)}(a)}{n!}(x-a)^n - \frac{f^{(n)}(a)}{(n-1)!}(x-a)^{n-1}, \end{equation*}

the second term being absent when \(n = 0\). Writing \(A_n = f^{(n+1)}(a)(x-a)^n/n!\), the subtracted term for index \(n\) is exactly \(A_{n-1}\), so the sum telescopes:

\begin{equation*} \begin{aligned} \frac{\partial}{\partial a} S_N(x,a) &= \sum_{n=0}^{N}\bigl(A_n - A_{n-1}\bigr)\\ &= A_N = \frac{f^{(N+1)}(a)}{N!}(x-a)^N. \end{aligned} \end{equation*}

Hence

\begin{equation*} E_N^{\prime}(x,a) = -\frac{\partial}{\partial a}S_N(x,a) = \frac{-f^{(N+1)}(a)}{N!}(x-a)^N. \end{equation*}

(c) Apply the Mean Value Theorem (Theorem 5.3.2) to \(a \mapsto E_N(x,a)\) on the closed interval with endpoints \(0\) and \(x\); by (b) this function is differentiable on all of \((-R,R)\), hence continuous on that closed interval and differentiable on its interior. So there is \(c\) between \(0\) and \(x\) with

\begin{equation*} \begin{aligned} E_N(x,x) - E_N(x,0) &= E_N^{\prime}(x,c)\,(x-0)\\ &= \frac{-f^{(N+1)}( c)}{N!}(x-c)^N x. \end{aligned} \end{equation*}

Since \(E_N(x,x) = 0\) by (a),

\begin{equation*} E_N(x) = E_N(x,0) = \frac{f^{(N+1)}( c)}{N!}(x-c)^N x . \end{equation*}

The series is

\begin{equation*} \sum_{n=0}^{\infty} c_n x^n, \qquad c_n = \frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)} = \frac{1}{4^n}\binom{2n}{n}, \end{equation*}

with \(c_0 = 1\); an induction on \(n\) gives

\begin{equation*} f^{(n)}(t) = \frac{1\cdot 3 \cdots (2n-1)}{2^n}\,(1-t)^{-(2n+1)/2} \end{equation*}

(differentiate once; each step multiplies by \((2n+1)/2\) and lowers the exponent by \(1\)), whence \(c_n = f^{(n)}(0)/n! = 1\cdot 3\cdots (2n-1)/(2^n n!)\) and \(2^n n! = 2\cdot 4 \cdots (2n)\).

Two facts used throughout. First, \(0 < c_n \le 1\), and in fact

\begin{equation*} c_n^2 \le \prod_{k=1}^{n}\frac{2k-1}{2k}\cdot\frac{2k}{2k+1} = \frac{1}{2n+1}, \qquad\text{so}\qquad c_n \le \frac{1}{\sqrt{2n+1}}, \end{equation*}

because \(\frac{2k-1}{2k} < \frac{2k}{2k+1}\) for every \(k\). Second,

\begin{equation*} \frac{f^{(N+1)}(t)}{(N+1)!} = c_{N+1}\,(1-t)^{-(2N+3)/2}. \end{equation*}

(a) Fix \(0 < x \le 1/2\) (the case \(x=0\) being trivial). Lagrange’s Remainder Theorem (Theorem 6.6.3) applies, since \(f\) is infinitely differentiable on \((-1,1)\) and \(x\) lies in \((-R,R)\) for \(R = 3/4\); it gives \(c\) with \(|c| < x\), so

\begin{equation*} |E_N(x)| = c_{N+1}\,\frac{x^{N+1}}{(1-c)^{N+1}\,(1-c)^{1/2}} \le c_{N+1}\,\frac{1}{\sqrt{1-x}}\left(\frac{x}{1-x}\right)^{N+1}, \end{equation*}

using only \(0 < 1-x < 1-c\), which follows from \(c < x\).

(i) \(0 < x < 1/2\). Then \(r = x/(1-x) < 1\), so \(|E_N(x)| \le (1-x)^{-1/2} r^{N+1} \to 0\) (Example 2.5.3), using \(c_{N+1} \le 1\).

(ii) \(x = 1/2\). Now \(r = 1\) and the bound degenerates to the constant \(\sqrt{2}\,c_{N+1}\); the extra care is supplied by \(c_{N+1} \le 1/\sqrt{2N+3}\):

\begin{equation*} |E_N(1/2)| \le \sqrt{2}\,c_{N+1} \le \frac{\sqrt{2}}{\sqrt{2N+3}} \longrightarrow 0 . \end{equation*}

In both cases \(E_N(x) \to 0\), so \(S_N(x) \to f(x)\) on \([0,1/2]\).

For \(1/2 < x < 1\) the attempt fails. Theorem 6.6.3 pins \(c\) down no further than \(|c| < x\), and the resulting uniform estimate is the one displayed above, whose factor \((x/(1-x))^{N+1}\) grows geometrically since \(x/(1-x) > 1\), while \(c_{N+1}\) decays only like \(N^{-1/2}\). The bound therefore tends to \(\infty\) rather than \(0\) and yields no information.

(b) Fix \(0 < x < 1\) (\(x = 0\) again trivial). Cauchy’s Remainder Theorem (Exercise 6.6.9, applied on \((-R,R)\) with \(x < R < 1\), where \(f\) is \(N+1\) times differentiable) gives \(c\) between \(0\) and \(x\) with

\begin{equation*} E_N(x) = \frac{f^{(N+1)}( c)}{N!}(x-c)^N x = \frac{2N+1}{2}\,c_N\,\frac{(x-c)^N x}{(1-c)^{N+3/2}}, \end{equation*}

since \(f^{(N+1)}( c)/N! = (N+1)c_{N+1}(1-c)^{-(2N+3)/2}\) and \((N+1)c_{N+1} = \tfrac{2N+1}{2}c_N\).

The decisive estimate is that for \(0 < c < x < 1\),

\begin{equation*} \frac{x-c}{1-c} \le x \iff x - c \le x - xc \iff xc \le c, \end{equation*}

which holds because \(x < 1\) and \(c > 0\). Combining this with \((1-c)^{-3/2} \le (1-x)^{-3/2}\) and \(c_N \le 1/\sqrt{2N+1}\):

\begin{equation*} |E_N(x)| \le \frac{2N+1}{2}c_N\,\frac{x^{N+1}}{(1-x)^{3/2}} \le \frac{\sqrt{2N+1}}{2}\cdot\frac{x^{N+1}}{(1-x)^{3/2}} . \end{equation*}

The right side tends to \(0\), since \(x^{N+1}\) decays geometrically while \(\sqrt{2N+1}\) grows only polynomially: writing \(t_N = \sqrt{2N+1}\,x^{N+1}\) we have \(t_{N+1}/t_N \to x < 1\), so \(\sum t_N\) converges by the Ratio Test (Exercise 2.7.9) and in particular \(t_N \to 0\) (Theorem 2.7.3). Hence \(E_N(x) \to 0\) and the series represents \(f\) on all of \([0,1)\).

Run WAT once for each tolerance \(\epsilon = 1/n\).

For each \(n \in \mathbf{N}\), apply Theorem 6.7.1 to the continuous function \(f\) with \(\epsilon = 1/n\); this produces a polynomial \(p_n\) with

\begin{equation*} |f(x) - p_n(x)| < \frac{1}{n} \quad \text{for all } x \in [a,b]. \end{equation*}

Given \(\epsilon > 0\), choose \(N \in \mathbf{N}\) with \(N > 1/\epsilon\) (Archimedean Property, Theorem 1.4.2). Then for every \(n \ge N\) and every \(x \in [a,b]\),

\begin{equation*} |f(x) - p_n(x)| < \frac{1}{n} \le \frac{1}{N} < \epsilon . \end{equation*}

Since \(N\) depends on \(\epsilon\) alone and not on \(x\), this is precisely Definition 6.2.3: \(p_n \to f\) uniformly on \([a,b]\).

Take \(\phi\) to be the polygonal interpolation of \(f\) at the nodes of any partition of mesh smaller than a uniform-continuity \(\delta\) for \(\epsilon/2\).

Since \([a,b]\) is compact (Theorem 3.3.4) and \(f\) is continuous, \(f\) is uniformly continuous on \([a,b]\) by Theorem 4.4.7: there is \(\delta > 0\) such that

\begin{equation*} |x - y| < \delta \ \Longrightarrow \ |f(x) - f(y)| < \frac{\epsilon}{2} \qquad (x, y \in [a,b]). \end{equation*}

Fix \(n \in \mathbf{N}\) with \((b-a)/n < \delta\) and let \(x_i = a + i(b-a)/n\) for \(i = 0, 1, \ldots, n\), so that \(a = x_0 < x_1 < \cdots < x_n = b\) is a partition of mesh \((b-a)/n < \delta\). Define \(\phi\) on each \([x_{i-1}, x_i]\) by

\begin{equation*} \phi(x) = f(x_{i-1}) + \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}\,(x - x_{i-1}). \end{equation*}

On each subinterval this is a linear function, and the two formulas attached to adjacent subintervals agree at the shared endpoint (both give \(f(x_i)\)), so \(\phi\) is well defined and continuous on \([a,b]\); it is polygonal in the sense of Definition 6.7.2.

Now fix \(x \in [a,b]\) and choose \(i\) with \(x \in [x_{i-1}, x_i]\). Writing \(t = (x - x_{i-1})/(x_i - x_{i-1}) \in [0,1]\), the displayed formula says exactly that \(\phi(x)\) is the convex combination

\begin{equation*} \phi(x) = (1-t)f(x_{i-1}) + t\,f(x_i). \end{equation*}

Both \(|x - x_{i-1}|\) and \(|x - x_i|\) are at most \(x_i - x_{i-1} < \delta\), so

\begin{equation*} \begin{aligned} |f(x) - \phi(x)| &= \bigl| (1-t)\bigl(f(x) - f(x_{i-1})\bigr) + t\bigl(f(x) - f(x_i)\bigr) \bigr| \\ &\le (1-t)\,|f(x) - f(x_{i-1})| + t\,|f(x) - f(x_i)| \\ &\le (1-t)\frac{\epsilon}{2} + t\,\frac{\epsilon}{2} \ = \ \frac{\epsilon}{2} \ < \ \epsilon . \end{aligned} \end{equation*}

Since \(x \in [a,b]\) was arbitrary, \(\phi\) is the required polygonal function.

(a) \(p(x) = x^2\).

The three interpolation conditions read

\begin{equation*} \begin{aligned} p(0) &= q_0 = 0, \\ p(1) &= q_0 + q_1 + q_2 = 1, \\ p(-1) &= q_0 - q_1 + q_2 = 1 . \end{aligned} \end{equation*}

Subtracting the third equation from the second gives \(2q_1 = 0\), so \(q_1 = 0\) and then \(q_2 = 1\).

Sketch: \(g(x) = |x|\) is the V with vertex at the origin and endpoints \((\pm 1, 1)\); \(p(x) = x^2\) is the parabola meeting it exactly at \(x = -1, 0, 1\) and lying strictly below it in between (since \(x^2 < |x|\) for \(0 < |x| < 1\)), sagging to a maximum gap of \(|x| - x^2 = 1/4\) at \(x = \pm 1/2\).

(b) \(p(x) = \dfrac{7}{3}x^2 - \dfrac{4}{3}x^4\).

The five nodes are symmetric about \(0\) and \(g\) is even, so the interpolating polynomial of degree at most four is even as well: if \(p\) interpolates the data then so does \(x \mapsto p(-x)\), and the interpolant through five points is unique among polynomials of degree \(\le 4\), forcing \(p(-x) = p(x)\). Write \(p(x) = b_0 + b_2 x^2 + b_4 x^4\). Then \(p(0) = 0\) gives \(b_0 = 0\), and the remaining two conditions are

\begin{equation*} \begin{aligned} p(1/2) &= \tfrac14 b_2 + \tfrac{1}{16} b_4 = \tfrac12, \\ p(1) &= b_2 + b_4 = 1 . \end{aligned} \end{equation*}

Multiplying the first by \(16\) gives \(4b_2 + b_4 = 8\); subtracting the second yields \(3b_2 = 7\), so \(b_2 = 7/3\) and \(b_4 = -4/3\). (Check: \(p(\pm 1/2) = 7/12 - 1/12 = 1/2\).)

Sketch: this quartic meets \(|x|\) at all five nodes and hugs the V much more closely than \(x^2\) did. It still sags below \(|x|\) on \((0,1/2)\) (for instance \(p(1/4) = 9/64\) against \(|1/4| = 1/4\)), but since \(p^{\prime}(x) = \tfrac{2x}{3}(7 - 8x^2)\) it has interior maxima at \(x = \pm\sqrt{7/8} \approx \pm 0.935\), where

\begin{equation*} p\bigl(\pm\sqrt{7/8}\bigr) = \frac{49}{48} \approx 1.021 > 0.935 = |x|, \end{equation*}

so it bulges above the V near the endpoints.

Differentiate \((1-x)^{1/2}\) \(n\) times and feed the result into the defining formula \(a_n = f^{(n)}(0)/n!\) for Taylor coefficients (p. 200; Theorem 6.6.2 is unavailable here, since a power series representation for \(f\) is exactly what is not yet known).

Since \(\frac{d}{dx}(1-x)^{\alpha} = -\alpha(1-x)^{\alpha - 1}\), an induction gives, for all \(n \ge 0\) and \(x < 1\),

\begin{equation*} f^{(n)}(x) = (-1)^n \Bigl(\prod_{k=0}^{n-1}(\tfrac12-k)\Bigr)(1-x)^{1/2-n}, \end{equation*}

the empty product for \(n = 0\) being \(1\). Setting \(x = 0\) leaves

\begin{equation*} f^{(n)}(0) = (-1)^n \prod_{k=0}^{n-1}\bigl(\tfrac12 - k\bigr). \end{equation*}

For \(n \ge 1\) pull a factor \(1/2\) out of each of the \(n\) terms:

\begin{equation*} \prod_{k=0}^{n-1}\bigl(\tfrac12 - k\bigr) = \frac{1 \cdot (-1) \cdot (-3) \cdots \bigl(-(2n-3)\bigr)}{2^n} = \frac{(-1)^{n-1}\, 1 \cdot 3 \cdots (2n-3)}{2^n}, \end{equation*}

since exactly \(n-1\) of the \(n\) numerators carry a minus sign. Hence

\begin{equation*} f^{(n)}(0) = (-1)^n \cdot \frac{(-1)^{n-1}\, 1\cdot 3 \cdots (2n-3)}{2^n} = -\,\frac{1 \cdot 3 \cdots (2n-3)}{2^n}. \end{equation*}

Dividing by \(n!\) and absorbing \(2^n n! = 2 \cdot 4 \cdot 6 \cdots 2n\) into the denominator,

\begin{equation*} \begin{aligned} a_n &= \frac{f^{(n)}(0)}{n!} = \frac{-1 \cdot 3 \cdot 5 \cdots (2n-3)}{2 \cdot 4 \cdot 6 \cdots 2n} \qquad (n \ge 1), \end{aligned} \end{equation*}

while \(a_0 = f(0) = 1\). (For \(n = 1\) the product \(1 \cdot 3 \cdots (2n-3)\) is empty, hence \(1\), and \(a_1 = -1/2\); for \(n = 2\), \(a_2 = -1/8\).)

(a) Apply the Mean Value Theorem to \(a \mapsto E_N(x,a)\) on the interval between \(0\) and \(x\).

Fix \(x \ne 0\) in \((-R,R)\). First, every \(n \ge 1\) term of \(S_N(x,a)\) carries a factor \((x-a)^n\), so

\begin{equation*} E_N(x,x) = f(x) - S_N(x,x) = f(x) - f(x) = 0 . \end{equation*}

Second, \(a \mapsto E_N(x,a)\) is differentiable on \((-R,R)\) because each \(f^{(n)}\), \(n \le N\), is differentiable there (\(f\) is \(N+1\) times differentiable). Differentiating termwise in \(a\), the \(n = 0\) term contributes \(f^{\prime}(a)\) and each \(n \ge 1\) term contributes

\begin{equation*} \frac{f^{(n+1)}(a)}{n!}(x-a)^n - \frac{f^{(n)}(a)}{(n-1)!}(x-a)^{n-1}, \end{equation*}

so the sum telescopes and only the last positive piece survives:

\begin{equation*} \frac{\partial}{\partial a} S_N(x,a) = \frac{f^{(N+1)}(a)}{N!}(x-a)^N, \qquad E_N^{\prime}(x,a) = -\,\frac{f^{(N+1)}(a)}{N!}(x-a)^N . \end{equation*}

Now \(a \mapsto E_N(x,a)\) is continuous on the closed interval with endpoints \(0\) and \(x\) and differentiable on its interior, so the Mean Value Theorem (Theorem 5.3.2) supplies a \(c\) strictly between \(0\) and \(x\) with

\begin{equation*} E_N(x,x) - E_N(x,0) = E_N^{\prime}(x,c)\,(x - 0). \end{equation*}

Since \(E_N(x,x) = 0\) and \(E_N(x,0) = E_N(x)\), this reads

\begin{equation*} -E_N(x) = -\,\frac{f^{(N+1)}( c)}{N!}(x-c)^N x, \qquad\text{i.e.}\qquad E_N(x) = \frac{f^{(N+1)}( c)}{N!}(x-c)^N x . \end{equation*}

(For \(x = 0\) the formula is trivially true, both sides being \(0\).)

(b) Part (a) applies to \(f(x) = \sqrt{1-x}\) with \(R = 1\), since \(f\) is infinitely differentiable on \((-1,1)\) (every \((1-x)^{1/2-n}\) is defined there). Its derivative formula from Exercise 6.7.4 gives, for \(c < 1\),

\begin{equation*} \bigl| f^{(N+1)}( c) \bigr| = \frac{1 \cdot 3 \cdots (2N-1)}{2^{N+1}} \, (1-c)^{-N-1/2}, \end{equation*}

because \(\prod_{k=0}^{N} \bigl|\tfrac12 - k\bigr| = 2^{-(N+1)}\, 1 \cdot 3 \cdots (2N-1)\). Substituting into (a) and writing \(2^{N+1} N! = 2\,(2 \cdot 4 \cdots 2N)\),

\begin{equation*} \bigl| E_N(x) \bigr| = \frac{1 \cdot 3 \cdots (2N-1)}{2 \cdot 4 \cdots 2N} \cdot \frac{|x|}{2} \cdot \frac{1}{\sqrt{1-c}} \cdot \left| \frac{x-c}{1-c} \right|^{N}, \end{equation*}

in which the first factor is a ratio of \(N\) matched terms and so is at most \(1\). The whole point is that the last factor is controlled: for \(c\) between \(0\) and \(x\) with \(|x| < 1\),

\begin{equation*} \left| \frac{x-c}{1-c} \right| \le |x| . \end{equation*}

Indeed, (i) if \(0 < c < x < 1\) then \(x - c \ge 0\) and \(x - c \le x - xc = x(1-c)\) since \(xc \le c\); (ii) if \(-1 < x < c < 0\) then \(|x-c| = c-x\) and \(c - x \le -x + xc = -x(1-c) = |x|(1-c)\) since multiplying \(x \le 1\) by the negative number \(c\) reverses it to \(xc \ge c\). Finally \(1 - c\) lies between \(\min\{1, 1-x\}\) and \(\max\{1,1-x\}\), so \((1-c)^{-1/2} \le M_x := \max\{1, (1-x)^{-1/2}\}\), a constant depending only on \(x\). Therefore

\begin{equation*} \bigl| E_N(x) \bigr| \ \le \ \frac{M_x}{2}\, |x|^{N+1} \ \longrightarrow \ 0 \end{equation*}

as \(N \to \infty\), since \(|x| < 1\) (Example 2.5.3). Hence the partial sums \(\sum_{n=0}^{N} a_n x^n\) converge to \(\sqrt{1-x}\), which is equation (1), for every \(x \in (-1,1)\).

(a) In fact \(c_n < 1/\sqrt{2n+1}\), which is half the stated bound. Square and telescope:

\begin{equation*} c_n^2 = \prod_{k=1}^{n} \frac{(2k-1)^2}{(2k)^2} < \prod_{k=1}^{n} \frac{2k-1}{2k+1} = \frac{1}{2n+1}, \end{equation*}

where the middle inequality is termwise, being equivalent to \((2k-1)(2k+1) < (2k)^2\), i.e. \(4k^2 - 1 < 4k^2\). Taking square roots,

\begin{equation*} c_n < \frac{1}{\sqrt{2n+1}} < \frac{2}{\sqrt{2n+1}} . \end{equation*}

(b) The two products differ by a single factor: for \(n \ge 1\),

\begin{equation*} |a_n| = \frac{1 \cdot 3 \cdots (2n-3)}{2 \cdot 4 \cdots 2n} = \frac{c_n}{2n-1}, \end{equation*}

(the \(n = 1\) case reading \(|a_1| = c_1/1 = 1/2\)). Hence, using (a) together with \(2n-1 \ge n\) and \(\sqrt{2n+1} > \sqrt{2n}\),

\begin{equation*} |a_n| < \frac{2}{(2n-1)\sqrt{2n+1}} \le \frac{2}{n \cdot \sqrt{2n}} = \frac{\sqrt{2}}{n^{3/2}} . \end{equation*}

Since \(\sum 1/n^{3/2}\) converges (\(p\)-series with \(p = 3/2 > 1\), Corollary 2.4.7), the Comparison Test (Theorem 2.7.4) gives \(\sum_{n=0}^{\infty} |a_n| < \infty\); absolute convergence implies convergence of \(\sum a_n\) (Theorem 2.7.6).

(c) Set \(M_n = |a_n|\). For every \(x \in [-1,1]\) we have \(|a_n x^n| \le |a_n| = M_n\), and \(\sum M_n\) converges by (b), so the Weierstrass M-Test (Corollary 6.4.5) says

\begin{equation*} g(x) = \sum_{n=0}^{\infty} a_n x^n \end{equation*}

converges uniformly on \([-1,1]\). Each partial sum is a polynomial, hence continuous, so \(g\) is continuous on all of \([-1,1]\) by the Term-by-term Continuity Theorem (Theorem 6.4.2). The function \(\sqrt{1-x}\) is also continuous on \([-1,1]\), and by Exercise 6.7.5(b) the two agree on \((-1,1)\).

Two continuous functions agreeing on \((-1,1)\) agree on the closure: taking \(x_k \to 1^{-}\) with \(x_k \in (-1,1)\), sequential continuity (Theorem 4.3.2) gives

\begin{equation*} g(1) = \lim_{k \to \infty} g(x_k) = \lim_{k \to \infty} \sqrt{1 - x_k} = 0 = \sqrt{1-1}, \end{equation*}

and the same argument at \(x = -1\) with \(x_k \to -1^{+}\) gives \(g(-1) = \sqrt{2}\). So equation (1) holds at both endpoints as well, hence for all \(x \in [-1,1]\).

(a) Substitute \(1-x^2\) for \(x\) in equation (1) and truncate:

\begin{equation*} q(x) = \sum_{n=0}^{N} a_n \,(1-x^2)^n \end{equation*}

for \(N\) large enough, with \(a_n\) the coefficients of Exercise 6.7.4.

This is legitimate because \(x \in [-1,1]\) forces \(x^2 \in [0,1]\) and therefore \(1 - x^2 \in [0,1] \subseteq [-1,1]\), exactly the range on which Exercise 6.7.6(c) established equation (1). Evaluating (1) at the point \(1-x^2\),

\begin{equation*} |x| = \sqrt{x^2} = \sqrt{1 - (1-x^2)} = \sum_{n=0}^{\infty} a_n (1-x^2)^n . \end{equation*}

Each \(q(x)\) above is a polynomial in \(x\) (of degree \(2N\)). Since \(0 \le 1-x^2 \le 1\), the tail estimate is uniform in \(x\):

\begin{equation*} \bigl| \,|x| - q(x) \,\bigr| = \left| \sum_{n=N+1}^{\infty} a_n (1-x^2)^n \right| \le \sum_{n=N+1}^{\infty} |a_n| . \end{equation*}

By Exercise 6.7.6(b) the series \(\sum |a_n|\) converges, so its tails tend to \(0\); choose \(N\) with \(\sum_{n > N} |a_n| < \epsilon\) and the resulting \(q\) works for every \(x \in [-1,1]\) simultaneously.

(b) Rescale: let \(M = \max\{|a|, |b|\}\), which is positive since \(a < b\), and put

\begin{equation*} p(x) = M \, q(x/M), \end{equation*}

where \(q\) is chosen by part (a) for the tolerance \(\epsilon/M\) on \([-1,1]\).

Then \(p\) is a polynomial, and \(x \in [a,b]\) implies \(|x| \le M\), i.e. \(x/M \in [-1,1]\), so

\begin{equation*} \bigl| \,|x| - p(x) \,\bigr| = M \left| \, \left| \frac{x}{M} \right| - q\!\left( \frac{x}{M} \right) \right| < M \cdot \frac{\epsilon}{M} = \epsilon \end{equation*}

for all \(x \in [a,b]\).

(a) \(h_a\) is the ramp that switches on at \(a\):

\begin{equation*} h_a(x) = \begin{cases} \tfrac{1}{2}\bigl((a-x)+(x-a)\bigr) = 0, & -1 \le x \le a,\\[2pt] \tfrac{1}{2}\bigl((x-a)+(x-a)\bigr) = x-a, & a \le x \le 1 . \end{cases} \end{equation*}

Its graph is the segment along the axis from \((-1,0)\) to \((a,0)\) followed by the segment of slope \(1\) from \((a,0)\) to \((1,1-a)\), so \(h_a\) is polygonal for the partition \(-1 < a < 1\) (and is a single line segment when \(a = \pm 1\)).

(b) Because \(h_a\) differs from a multiple of an absolute value by a linear function. Let \(\epsilon > 0\). For \(x \in [-1,1]\) and \(a \in [-1,1]\) we have \(x - a \in [-2,2]\), so by Exercise 6.7.7(b) applied to the interval \([-2,2]\) there is a polynomial \(q\) with

\begin{equation*} \bigl|\,|t| - q(t)\,\bigr| < 2\epsilon \quad \text{for all } t \in [-2,2]. \end{equation*}

Put \(p(x) = \tfrac{1}{2}\bigl(q(x-a) + (x-a)\bigr)\), a polynomial in \(x\). Then for every \(x \in [-1,1]\),

\begin{equation*} |h_a(x) - p(x)| = \tfrac{1}{2}\bigl|\,|x-a| - q(x-a)\,\bigr| < \epsilon . \end{equation*}

(c) Take \(b_0 = m_1\) and \(b_i = m_{i+1} - m_i\) for \(1 \le i \le n-1\), where \(m_i\) is the slope of \(\phi\) on \([a_{i-1}, a_i]\). Set

\begin{equation*} \psi(x) = \phi(-1) + \sum_{i=0}^{n-1} b_i h_{a_i}(x). \end{equation*}

Fix \(k\) with \(1 \le k \le n\) and let \(x \in [a_{k-1}, a_k]\). By (a), \(h_{a_i}(x) = x - a_i\) for \(i \le k-1\) and \(h_{a_i}(x) = 0\) for \(i \ge k\), so \(\psi\) is linear on \([a_{k-1}, a_k]\) with slope

\begin{equation*} \sum_{i=0}^{k-1} b_i = m_1 + \sum_{i=1}^{k-1}(m_{i+1} - m_i) = m_k , \end{equation*}

the sum telescoping. Thus \(\psi\) is polygonal for the same partition and has the same slope as \(\phi\) on each subinterval. Also \(h_{a_i}(-1) = 0\) for every \(i\) (since \(a_i \ge -1\)), so \(\psi(-1) = \phi(-1)\). Now induct on \(k\): if \(\psi(a_{k-1}) = \phi(a_{k-1})\) then on \([a_{k-1},a_k]\) both functions equal \(\phi(a_{k-1}) + m_k(x - a_{k-1})\), which gives \(\psi = \phi\) there and in particular \(\psi(a_k) = \phi(a_k)\). Hence \(\psi = \phi\) on all of \([-1,1]\).

(d) Let \(f : [-1,1] \to \mathbf{R}\) be continuous and let \(\epsilon > 0\). By Theorem 6.7.3 there is a polygonal \(\phi\) with \(|f(x) - \phi(x)| < \epsilon/2\) on \([-1,1]\); its partition runs from \(-1\) to \(1\) (Definition 6.7.2), so (c) writes it with constants \(b_0,\ldots,b_{n-1}\). Set \(M = 1 + \sum_{i=0}^{n-1}|b_i|\). By (b) choose polynomials \(p_i\) with \(|h_{a_i}(x) - p_i(x)| < \epsilon/(2M)\) on \([-1,1]\), and put \(p(x) = \phi(-1) + \sum_{i=0}^{n-1} b_i p_i(x)\), a polynomial. Then

\begin{equation*} \begin{aligned} |\phi(x) - p(x)| &\le \sum_{i=0}^{n-1} |b_i|\,\bigl|h_{a_i}(x) - p_i(x)\bigr|\\ &< \frac{\epsilon}{2M} \sum_{i=0}^{n-1} |b_i| < \frac{\epsilon}{2}, \end{aligned} \end{equation*}

and the triangle inequality gives \(|f(x) - p(x)| < \epsilon\) for all \(x \in [-1,1]\).

For a general interval \([a,b]\), let \(L(t) = a + \tfrac{b-a}{2}(t+1)\), the increasing affine bijection of \([-1,1]\) onto \([a,b]\), with affine inverse \(L^{-1}(x) = \tfrac{2(x-a)}{b-a} - 1\). Given \(f\) continuous on \([a,b]\), the composition \(f \circ L\) is continuous on \([-1,1]\), so the case just proved supplies a polynomial \(p\) with \(|f(L(t)) - p(t)| < \epsilon\) for all \(t \in [-1,1]\). Then \(P = p \circ L^{-1}\) is again a polynomial, and substituting \(t = L^{-1}(x)\) gives \(|f(x) - P(x)| < \epsilon\) for all \(x \in [a,b]\).