Introductory Real Analysis

Set Theory

Sets and Functions

Problem

Prove that if $A\cup B = A$ and $A \cap B = A$, then $A = B$.

Solution

To show $A=B$, show $A\subseteq B$ and $B\subseteq A$. Suppose $x\in B$, then we know by definition that $x\in (A\cup B)$ if $x\in B$ or $x\in A$. Which then implies that $x\in A$ from rule 1. Thus $B\subseteq A$. Now suppose $x\in A$ which implies $x\in (A\cap B)$ (rule 2). The definition of this means that $x\in A$ and $x\in B$. $\therefore x\in A \implies x\in B$, i.e. $A\subseteq B$, so $A=B$.

Problem

Show that in general $(A-B)\cup B \neq A$.

Solution

This only holds for $B\subseteq A$. We proceed by counterexample. Let $A={1,2}, B={3,4}$. Then $(A-B) = {1,2}$ and $(A-B)\cup B = {1,2,3,4} \neq {1,2}$.

Problem

Let $A = {2,4,…,2n,…}$ and $B = {3,6,…,3n,…}$. Find $A\cap B$ and $A-B$

Solution

\[ A\cap B = {6n \mid n\in \mathbb{N}} \] \[ A - B = {2n \mid n\in \mathbb{N}, 2n \not\in {6m \mid m\in \mathbb{N}}} \]

Problem

Prove that:

()

$(A-B)\cap C = (A\cap C) - (B\cap C)$

Solution:

Let $x\in (A-B)\cap C$. Then:

$x∈ A - B \implies x ∈ A $ and $x\not \in B$
$x\in C$

So:

$x \in A \cap C$
Since $x\in C$ and $x\not\in B$, it follows that $x\not\in B \cap C$

Therefore $x\in (A\cap C)- (B\cap C)$

()

$A\Delta B = (A\cup B) - (A\cap B)$

Solution:

Let $x\in A \Delta B$. Then:

$x\in A-B \implies x\in A$ and $x\not\in B$

OR:

$x\in B-A \implies x\in B$ and $x\not\in A$

$x\in A$ or $x \in B \implies x\in (A\cup B)$
$x \not \in (A\cap B) \implies x\in (A\cap B)^c$

Therefore \[x\in(A\cup B) \cap (A\cap B)^c =(A\cup B)- (A\cap B)\]

Problem

Prove that \[\bigcup_\alpha A_\alpha - \bigcup_\alpha B_\alpha \subset \bigcup_\alpha\; (A_\alpha - B_\alpha)\]

Solution

Let $x\in \cup_\alpha A_\alpha - \cup_\alpha B_\alpha$. Then:

$x\in \cup_\alpha A_\alpha \implies \exists \alpha{}_0: x \in A_\alpha{}_0$
$x\not\in \cup_\alpha B_\alpha \implies \forall \alpha : x \not\in B_\alpha$

$x\not\in B_\alpha{}_0$

Hence \[x\in A_\alpha{}_0 - B_\alpha{}_0 \subset \bigcup_\alpha\; (A_\alpha - B_\alpha)\]

Problem

Let $A_n$ be the set of all positive integers divisible by $n$. Find the sets

()

\[\bigcup_{n=2}^\infty A_n\]

Answer:

\[\mathbb{N}-\set{1}\]

()

\[\bigcap_{n=2}^\infty A_n\]

Answer:

\[\varnothing\]

Problem

Find

()

\[\bigcup_{n=1}^\infty [a+\frac{1}{n}, b-\frac{1}{n}]\]

Answer:

\[(a,b)\]

Hint:

the key observation is that $[2,1] = \varnothing$ by definition of the interval.

()

\[\bigcap_{n=1}^\infty [a-\frac{1}{n}, b+\frac{1}{n}]\]

Answer:

\[[a,b]\]

Problem

Let $A_\alpha$ be the set of points lying on the curve \[y = \frac{1}{x^\alpha}\quad(0<x<\infty).\] What is $\bigcap_{\alpha\geq 1} A_\alpha$?

Answer

\[\set{(1,1)}\]

Hint

set of points, not y-points

Problem

Let $y = f(x) = \langle x \rangle$ for all real x, where $\langle x \rangle$ is the fractional part of $x$. Prove that every closed interval of length 1 has the same image under $f$. What is this image? Is $f$ one-to-one? What is the preimage of the interval $\frac{1}{4}\leq y\leq\frac{3}{4}$? Partition the real line into classes of points with the same image.

Sketch

Start with $I=\set{a,a+1}$, i.e. an arbitrary set of length 1. Then notice that you can subtract $a$ wlog 𐃏 , and now we are tasked to find $\set{\langle x \rangle : x\in [0,1]}$. Furthermore, we know that $⟨ x ⟩ = x - ⌊ x ⌋ $ with $\langle 0 \rangle = 0 = \langle 1 \rangle$, whereby the image of the the closed interval only sweeps the half-open interval $[0,1)$. $f$ cannot be one-to-one because of the periodicity; many real numbers have the same fractional parts.

The pre-image of $\frac{1}{4} \leq y \leq \frac{3}{4}$ is the interval \[ \bigcup_{n\in\mathbb{Z}} \left [\frac{1}{4} + n, \frac{3}{4} + n \right ]\] because $x∈ \mathbb{R} $.

Finally, we can express \[\mathbb{R} = \bigsqcup_{r\in [0,1)} \set{x\in \mathbb{R} : \langle x \rangle = r } \]

as the disjoint union of all the numbers which have the same fractional parts; i.e. the same images.

Problem

Given a set $M$, let $\mathcal{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a\in M$, and let $a R b$ if and only if $(a,b)\in\mathcal{R}$. Interpret the relation $R$.

Answer

$\mathcal{R}$ is the equality relation on $M$.

Problem

Give an example of a binary relation which is

()

Reflexive and symmetric, but not transitive

Solution:

()

Reflexive, but neither symmetric nor transitive

Solution:

()

Symmetric, but neither reflexive nor transitive

Solution:

()

Transitive, but neither reflexive nor symmetric

Solution:

Equivalence of Sets. The Power of a Set

Problem

Prove that a set with an uncountable subset is itself uncountable.

Solution

Suppose $M$ contains an uncountable subset $A$. If $M$ were countable, then by definition there exists an injection $f:M\hookrightarrow \mathbb{N}$. Restricting to $A$, we get $f|_A:A\hookrightarrow\mathbb{N}$, so $A$ is countable—contradiction. Hence $M$ is uncountable.

Problem

Let $M$ be any infinite set and $A$ any countable set. Show that $M \sim M\cup A$.

Solution

Since $M$ is infinite, it contains a countably infinite subset $B\subset M$ with $B\sim\mathbb{N}$. Set $C = A\setminus M$ (the part of $A$ not already in $M$). Then $B\cup C$ is a countable union of countable sets, hence countable infinite, so $B\cup C\sim B$. Define $f:M\cup A\to M$ by \[f(x) = \begin{cases} \phi(x) & x\in B\cup C \\ x & x\in M\setminus B \end{cases}\] where $\phi:B\cup C\to B$ is a bijection. Then $f$ is a bijection, so $M\cup A\sim M$.

Problem

Prove that each of the following sets is countable.

()

The set of all numbers with two distinct decimal expansions (e.g.\ $0.5000\ldots$ and $0.4999\ldots$).

Solution:

A real in $[0,1]$ has two decimal expansions iff it equals $\frac{k}{10^n}$ for some $k,n\in\mathbb{N}$ with $0<k<10^n$. These are precisely the terminating decimals, which form a subset of $\mathbb{Q}$. Since $\mathbb{Q}$ is countable, so is this set.

()

The set of all rational points in the plane (points whose coordinates are both rational).

Solution:

$\mathbb{Q}^2 = \mathbb{Q}\times\mathbb{Q}$. Since $\mathbb{Q}$ is countable and a countable product of countable sets is countable (via the diagonal argument), $\mathbb{Q}^2$ is countable.

()

The set of all rational intervals (intervals with rational end‑points).

Solution:

A rational interval is determined by a pair $(a,b)\in\mathbb{Q}^2$ with $a\le b$, along with a choice of open/closed at each end (4 choices). Thus the set of all rational intervals injects into $\mathbb{Q}^2\times\{1,2,3,4\}$, which is countable.

()

The set of all polynomials with rational coefficients.

Solution:

A polynomial of degree $n$ is determined by $(a_0,\ldots,a_n)\in\mathbb{Q}^{n+1}$. So \[\mathbb{Q}[x] = \bigcup_{n=0}^\infty \mathbb{Q}^{n+1}\] is a countable union of countable sets, hence countable.

Problem

A real number $a$ is called algebraic if it is a root of some polynomial with rational coefficients. Prove that the set of all algebraic numbers is countable.

Solution

By the previous problem, the set $\mathbb{Q}[x]$ of polynomials with rational coefficients is countable. Each polynomial of degree $n$ has at most $n$ roots. Thus the set of algebraic numbers is \[\bigcup_{p\in\mathbb{Q}[x]} \{\text{roots of } p\}\] which is a countable union of finite sets, hence countable.

Problem

Show that there exist uncountably many transcendental numbers. Hint: combine the previous problem with Theorem 5.

Solution

By Theorem 5, $\mathbb{R}$ is uncountable. By the previous problem, the algebraic numbers $\overline{\mathbb{Q}}$ are countable. The transcendental numbers are $\mathbb{R}\setminus\overline{\mathbb{Q}}$. If this were countable, then $\mathbb{R} = \overline{\mathbb{Q}}\cup(\mathbb{R}\setminus\overline{\mathbb{Q}})$ would be a union of two countable sets, hence countable—contradiction. Thus the transcendentals are uncountable.

Problem

Let $M$ be any set. Prove that the set of all real‑valued functions on $M$ has power strictly larger than $|M|$. (In particular, the set of all real functions on $[0,1]$ has power $> \mathfrak c$.)

Solution

Let $F = \mathbb{R}^M$ be the set of all functions $M\to\mathbb{R}$. There is an injection $M\hookrightarrow F$ via $m\mapsto f_m$ where $f_m(x)=1$ if $x=m$ and $0$ otherwise. So $|M|\le|F|$. To show $|F|>|M|$, assume for contradiction that $\phi:M\to F$ is a surjection. Consider the indicator functions: define $g:M\to\{0,1\}\subset\mathbb{R}$ by $g(m)=1-\chi_m(m)$ where $\chi_m = \phi(m)$ is the characteristic function. Then $g\neq \phi(m)$ for any $m$ (they differ at $m$), contradicting surjectivity. Hence $|F|>|M|$.

Problem

Give an indirect proof that the intervals $[a,b]$, $(a,b)$, $[a,b)$ and $(a,b]$ are all equivalent as sets.

Solution

WLOG take $[0,1]$. All four intervals have the power of the continuum since:

Each contains a copy of $(0,1)$, which bijects with $\mathbb{R}$ via $x\mapsto\tan(\pi(x-\tfrac12))$
Each injects into $[0,1]$

By Cantor–Bernstein, all four have the same cardinality $\mathfrak{c}$.

Problem

Show that the union of finitely or countably many sets of power $\mathfrak c$ again has power $\mathfrak c$.

Solution

Let $\{A_n\}_{n\in\mathbb{N}}$ be a countable collection with $|A_n|=\mathfrak{c}$ for each $n$. We have \[\mathfrak{c} \le \left|\bigcup_n A_n\right| \le \sum_n |A_n| = \aleph_0\cdot\mathfrak{c} = \mathfrak{c}\] where the last equality uses $\mathfrak{c}\cdot\mathfrak{c}=\mathfrak{c}$ (since $|\mathbb{R}^2|=|\mathbb{R}|$). By Cantor–Bernstein, $|\bigcup_n A_n|=\mathfrak{c}$.

Problem

Prove that each of the following sets has the power of the continuum.

()

The set of all infinite sequences of positive integers.

Solution:

Let $S = \mathbb{N}^\mathbb{N}$. There is an injection $S\hookrightarrow [0,1]$ via continued fractions: $(a_1,a_2,\ldots)\mapsto [0;a_1,a_2,\ldots]$ gives all irrationals in $(0,1)$. Conversely, $[0,1]\hookrightarrow S$ via binary expansion. Hence $|S|=\mathfrak{c}$.

()

The set of all ordered $n$‑tuples of real numbers (for any fixed $n\ge 1$).

Solution:

$|\mathbb{R}^n| = |\mathbb{R}|^n = \mathfrak{c}^n = \mathfrak{c}$, where the last equality holds since $\mathfrak{c}\cdot\mathfrak{c}=\mathfrak{c}$ by induction.

()

The set of all infinite sequences of real numbers.

Solution:

$|\mathbb{R}^\mathbb{N}| = |\mathbb{R}|^{|\mathbb{N}|} = \mathfrak{c}^{\aleph_0}$. Now $\mathfrak{c} = 2^{\aleph_0}$, so $\mathfrak{c}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0} = \mathfrak{c}$.

Problem

Explain the paradox in the notion “the set of all sets that are not members of themselves.” Hint: Is this set a member of itself?

Solution

Let $R = \{X : X\notin X\}$ (Russell’s paradox). Ask: is $R\in R$?

If $R\in R$, then by definition $R\notin R$—contradiction.
If $R\notin R$, then by definition $R\in R$—contradiction.

Hence no such set can exist in consistent set theory. This is resolved by axiomatization (ZFC), which restricts comprehension to avoid self-referential constructions.

Ordered Sets and Ordinal Numbers

Problem

Exhibit both a partial ordering and a simple (linear) ordering of the set of all complex numbers.

Solution

Partial ordering: Define $z_1 \preceq z_2$ iff $\operatorname{Re}(z_1)\le\operatorname{Re}(z_2)$ and $\operatorname{Im}(z_1)\le\operatorname{Im}(z_2)$. This is reflexive, antisymmetric, transitive, but not total (e.g., $1$ and $i$ are incomparable). Linear ordering: Use lexicographic order: $z_1 < z_2$ iff $\operatorname{Re}(z_1)<\operatorname{Re}(z_2)$, or $\operatorname{Re}(z_1)=\operatorname{Re}(z_2)$ and $\operatorname{Im}(z_1)<\operatorname{Im}(z_2)$. This is a total order on $\mathbb{C}$.

Problem

For the power‑set $\mathcal P(X)$ ordered by inclusion, identify its minimal and maximal elements.

Solution

The unique minimal element is $\varnothing$ (the empty set is contained in every set). The unique maximal element is $X$ (every subset is contained in $X$). These are also the minimum and maximum respectively.

Problem

A partially ordered set $M$ is called directed if for every $a,b\in M$ there exists $c\in M$ with $a<c$ and $b<c$. Determine whether the partially ordered sets in Examples 1–4 of §3.1 are directed.

Solution

The standard examples are:

All subsets of $X$ ordered by inclusion: directed (take $c=a\cup b$).
$\mathbb{N}$ with divisibility: directed (take $c=\operatorname{lcm}(a,b)$).
$\mathbb{Z}$ with usual order: directed ($c=\max(a,b)+1$).
Finite subsets of an infinite set: directed (take $c=a\cup b$).

Problem

Show that the set $\mathcal P(X)$ with inclusion is a lattice: every pair has a greatest lower bound and a least upper bound.

Solution

For $A,B\in\mathcal{P}(X)$:

$\inf(A,B) = A\cap B$ (the largest set contained in both)
$\sup(A,B) = A\cup B$ (the smallest set containing both)

These exist for every pair, so $(\mathcal{P}(X),\subseteq)$ is a lattice.

Problem

Prove that an order‑preserving surjection of one ordered set onto another is automatically an isomorphism.

Solution

Let $f:A\to B$ be an order-preserving surjection. To show $f$ is injective: suppose $f(x)=f(y)$. If $x<y$, then $f(x)<f(y)$—contradiction. Similarly $y<x$ leads to contradiction. Hence $x=y$. So $f$ is a bijection. The inverse is also order-preserving: if $f(x)<f(y)$, then by trichotomy $x<y$ (since $x>y$ would give $f(x)>f(y)$).

Problem

Prove that ordered sums and products are associative, i.e. \[(M_1+M_2)+M_3\cong M_1+(M_2+M_3), \qquad (M_1\cdot M_2)\cdot M_3\cong M_1\cdot(M_2\cdot M_3).\]

Solution

Ordered sum: $(M_1+M_2)+M_3$ places $M_1$ before $M_2$, then that before $M_3$. Same for $M_1+(M_2+M_3)$. The identity map preserves order. Ordered product: $(M_1\cdot M_2)\cdot M_3$ is lexicographically ordered pairs $((m_1,m_2),m_3)$. The bijection $((m_1,m_2),m_3)\mapsto(m_1,(m_2,m_3))$ preserves lexicographic order, giving an isomorphism with $M_1\cdot(M_2\cdot M_3)$.

Problem

Construct well‑ordered sets with ordinals $\omega+n,\;\omega+\omega,\;(\omega+\omega)+n,\;\omega+\omega+\omega,\ldots$ and show they are all countable.

Solution

$\omega+n$: $\mathbb{N}$ followed by $\{a_1,\ldots,a_n\}$. Order: naturals first, then $a_1<\cdots<a_n$.
$\omega+\omega = \omega\cdot 2$: two copies of $\mathbb{N}$, say $\{0,1,2,\ldots\}\cup\{0’,1’,2’,\ldots\}$ with all unprimed $<$ all primed.
$(\omega+\omega)+n$: append $n$ elements after the second copy.
$\omega\cdot 3 = \omega+\omega+\omega$: three copies of $\mathbb{N}$.

All are countable as finite or countable unions of countable sets.

Problem

Construct well‑ordered sets with ordinals $\omega\cdot n,\;\omega^{2},\;\omega^{2}\cdot\omega^{3},\ldots$ and show they are all countable.

Solution

$\omega\cdot n$: $n$ copies of $\mathbb{N}$ placed consecutively.
$\omega^2 = \omega\cdot\omega$: $\mathbb{N}\times\mathbb{N}$ with reverse-lexicographic order (second coordinate primary). Equivalently: $\{(i,j):j\in\mathbb{N},i\le j\}$ ordered by $j$ then $i$.
$\omega^2\cdot\omega^3 = \omega^5$: $\mathbb{N}^5$ with lexicographic order.

Each is a countable product/union, hence countable.

Problem

Verify the equalities $\omega+\omega=\omega\cdot 2,\quad \omega+\omega+\omega=\omega\cdot 3,\ldots$

Solution

By definition, $\omega\cdot n$ is $n$ copies of $\omega$ in sequence: $\omega\cdot n = \omega + \omega + \cdots + \omega$ ($n$ times). This can be verified by induction:

$\omega\cdot 1 = \omega$
$\omega\cdot(n+1) = \omega\cdot n + \omega$

Hence $\omega\cdot 2 = \omega + \omega$, $\omega\cdot 3 = \omega+\omega+\omega$, etc.

Problem

Prove that, for any ordinal $\alpha$, the set $W(\alpha)=\{\beta:\beta<\alpha\}$ is well‑ordered.

Solution

$W(\alpha)$ inherits the ordering from the ordinals. Any non-empty subset $S\subseteq W(\alpha)$ is a non-empty set of ordinals. By the well-foundedness of $\in$ on ordinals (foundation axiom), $S$ has a least element. Hence $W(\alpha)$ is well-ordered.

Problem

Show that every non‑empty set of ordinals is well‑ordered.

Solution

Let $S$ be a non-empty set of ordinals. The ordering on ordinals (by $\in$) is a total order. For any non-empty $T\subseteq S$, we need a minimum. By the axiom of regularity, every non-empty collection of sets has an $∈$-minimal element. Thus $T$ has a least ordinal, so $S$ is well-ordered.

Problem

Let $M$ be the set of all ordinals that index (are order‑types of) countable well‑ordered sets. Prove that $M$ itself is uncountable.

Solution

$M$ is the set of all countable ordinals, i.e., $M = \omega_1$ (the first uncountable ordinal). If $M$ were countable, then $M$ itself would be a countable well-ordered set with order type $\alpha\in M$. But then $\alpha = M$ would mean $M\in M$, contradicting the axiom of regularity. Hence $\omega_1$ is uncountable.

Problem

Let $\kappa$ be the power of the set $M$ in the preceding problem. Prove that there is no cardinal $m$ with $\aleph_0 < m < \kappa$.

Solution

We have $\kappa = |\omega_1| = \aleph_1$ by definition. The statement “there is no $m$ with $\aleph_0 < m < \aleph_1$” is precisely that $\aleph_1$ is the next cardinal after $\aleph_0$. This follows from the definition of $\aleph_1$ as the least uncountable cardinal: any infinite cardinal $m\le\aleph_1$ is either $\le\aleph_0$ or $=\aleph_1$.

Systems of Sets

Problem

Let $X$ be an uncountable set and let $\mathcal R$ be the ring consisting of all finite subsets of $X$ together with their complements. Is $\mathcal R$ a $\sigma$‑ring?

Solution

No. $\mathcal{R}$ is not a $σ$-ring. Consider a countably infinite set $A = \{x_1,x_2,\ldots\}\subset X$. Each singleton $\{x_n\}\in\mathcal{R}$, but \[A = \bigcup_{n=1}^\infty \{x_n\}\] is countably infinite—neither finite nor cofinite (since $X\setminus A$ is uncountable). Hence $A\notin\mathcal{R}$, so $\mathcal{R}$ is not closed under countable unions.

Problem

Are open intervals in $\mathbb R$ Borel sets?

Solution

Yes. The Borel $σ$-algebra on $\mathbb{R}$ is generated by open sets. Since every open interval $(a,b)$ is an open set, it is Borel. Alternatively, the Borel algebra is often defined as the $σ$-algebra generated by open intervals, making them Borel by definition.

Problem

Let $y = f(x)$ be a function defined on a set $M$ and taking values in a set $N$. Let $\mathcal{J}$ be a system of subsets of $M$, and write \[f(\mathcal{J}) = \{\,f(A)\mid A\in\mathcal{J}\,\}.\] Let $\mathcal{P}$ be a system of subsets of $N$, and write \[f^{-1}(\mathcal{P}) = \{\,f^{-1}(B)\mid B\in\mathcal{P}\,\}.\] Prove:

()

(a) If $\mathcal{P}$ is a ring, then $f^{-1}(\mathcal{P})$ is a ring.

Solution:

We verify the ring axioms. For $A,B\in f^{-1}(\mathcal{P})$, write $A=f^{-1}(P)$, $B=f^{-1}(Q)$ with $P,Q\in\mathcal{P}$. Then:

$A\cup B = f^{-1}(P)\cup f^{-1}(Q) = f^{-1}(P\cup Q) \in f^{-1}(\mathcal{P})$ since $P\cup Q\in\mathcal{P}$
$A\setminus B = f^{-1}(P)\setminus f^{-1}(Q) = f^{-1}(P\setminus Q) \in f^{-1}(\mathcal{P})$

Hence $f^{-1}(\mathcal{P})$ is a ring.

()

(b) If $\mathcal{P}$ is an algebra, then $f^{-1}(\mathcal{P})$ is an algebra.

Solution:

$\mathcal{P}$ is an algebra means $N\in\mathcal{P}$ (the whole space). Then $f^{-1}(N)=M\in f^{-1}(\mathcal{P})$. Combined with (a), $f^{-1}(\mathcal{P})$ is a ring containing $M$, hence an algebra on $M$.

()

Solution:

By (b), $f^{-1}(\mathcal{P})$ is an algebra. For countable unions: if $A_n = f^{-1}(P_n)$ with $P_n\in\mathcal{P}$, then \[\bigcup_{n=1}^\infty A_n = f^{-1}\left(\bigcup_{n=1}^\infty P_n\right) \in f^{-1}(\mathcal{P})\] since $\bigcup P_n\in\mathcal{P}$ by the $σ$-algebra property. Hence $f^{-1}(\mathcal{P})$ is a $σ$-algebra.

()

(d) Which of the assertions (a)–(c) remain true if we replace $\mathcal{P}$ by $\mathcal{J}$ and $f^{-1}$ by $f$ (i.e.\ work with images instead of pre‑images)?

Solution:

None in general. The forward image does not preserve set operations cleanly:

$f(A\cup B) = f(A)\cup f(B)$ ✓
$f(A\cap B) \subseteq f(A)\cap f(B)$ (equality fails if $f$ not injective)
$f(A\setminus B) \not\subseteq f(A)\setminus f(B)$ in general
$f(M^c) \neq f(M)^c$ in general

If $f$ is a bijection, all assertions transfer. Otherwise, none of (a)–(c) need hold.

Metric Spaces

Basic Concepts

Problem

Let $(X,p)$ be a metric space. Prove that

$\lvert p(x,z)-p(y,u)\rvert\le p(x,y)+p(z,u)$ for all $x,y,z,u\in X$;
$\lvert p(x,z)-p(y,z)\rvert\le p(x,y)$ for all $x,y,z\in X$.

Solution

By triangle inequality: $p(x,z)\le p(x,y)+p(y,z)\le p(x,y)+p(y,u)+p(u,z)$. Hence $p(x,z)-p(y,u)\le p(x,y)+p(z,u)$. By symmetry (swap $(x,z)\leftrightarrow(y,u)$), we get $p(y,u)-p(x,z)\le p(x,y)+p(z,u)$. Together: $|p(x,z)-p(y,u)|\le p(x,y)+p(z,u)$.
Set $u=z$ in (1): $|p(x,z)-p(y,z)|\le p(x,y)+p(z,z)=p(x,y)$.

Problem

Verify the identity \[ \Bigl(\sum_{k=1}^{n}a_k b_k\Bigr)^2 =\frac12\sum_{k,j=1}^{n}\bigl(a_k^2 b_j^2+a_j^2 b_k^2-(a_k b_j-a_j b_k)^2\bigr) \] and use it to deduce the Cauchy–Schwarz inequality \[ \Bigl|\sum_{k=1}^{n}a_k b_k\Bigr| \le\Bigl(\sum_{k=1}^{n}a_k^{\,2}\Bigr)^{1/2} \Bigl(\sum_{k=1}^{n}b_k^{\,2}\Bigr)^{1/2}. \]

Solution

Expanding the RHS: $a_k^2 b_j^2+a_j^2 b_k^2-(a_k b_j-a_j b_k)^2 = a_k^2 b_j^2+a_j^2 b_k^2-a_k^2b_j^2+2a_ka_jb_kb_j-a_j^2b_k^2 = 2a_ka_jb_kb_j$. So $\frac12\sum_{k,j}2a_ka_jb_kb_j = \sum_{k,j}a_ka_jb_kb_j = (\sum_k a_kb_k)^2$. ✓

For C–S: since $(a_kb_j-a_jb_k)^2\ge 0$, we have \[(\sum a_kb_k)^2 \le \frac12\sum_{k,j}(a_k^2b_j^2+a_j^2b_k^2) = \sum_k a_k^2\sum_j b_j^2\] Taking square roots gives the Cauchy–Schwarz inequality.

Problem

Show that for square‑integrable functions $f,g$ on $[a,b]$ \[ \Bigl|\int_{a}^{b}f(t)g(t)\,dt\Bigr| \le \Bigl(\int_{a}^{b}f(t)^2\,dt\Bigr)^{1/2} \Bigl(\int_{a}^{b}g(t)^2\,dt\Bigr)^{1/2}. \]

Solution

For any $\lambda\in\mathbb{R}$, $\int_a^b(f+\lambda g)^2\,dt\ge 0$. Expanding: \[\int f^2 + 2\lambda\int fg + \lambda^2\int g^2 \ge 0\] This quadratic in $\lambda$ is non-negative, so discriminant $\le 0$: \[(2\int fg)^2 - 4\int f^2\int g^2 \le 0\] Hence $(\int fg)^2 \le \int f^2\int g^2$, and taking square roots gives the result.

Problem

Give an explicit example showing that Minkowski’s inequality fails for the functional \[\|x\|_p=\Bigl(\sum_{k=1}^{n}|x_k|^{p}\Bigr)^{1/p}\] when $0<p<1$.

Solution

Take $p=\frac12$, $x=(1,0)$, $y=(0,1)$ in $\mathbb{R}^2$. Then:

$\|x\|_{1/2} = (1^{1/2}+0)^2 = 1$
$\|y\|_{1/2} = 1$
$\|x+y\|_{1/2} = (1^{1/2}+1^{1/2})^2 = 4$

Minkowski would require $\|x+y\|\le\|x\|+\|y\|=2$, but $4>2$. Hence it fails for $p<1$.

Problem

Prove the limiting relation \[ \lim_{p\to\infty}\Bigl(\sum_{k=1}^{n}|x_k-y_k|^{p}\Bigr)^{1/p} =\max_{1\le k\le n}|x_k-y_k|, \] that is, the $\ell^\infty$ metric is the limit of the $\ell^p$ metrics as $p\to\infty$.

Solution

Let $M=\max_k|x_k-y_k|$ and $d_k=|x_k-y_k|$. Then: \[M = (M^p)^{1/p} \le \left(\sum_k d_k^p\right)^{1/p} \le (nM^p)^{1/p} = n^{1/p}M\] As $p\to\infty$, $n^{1/p}\to 1$. By squeeze theorem, $\lim_{p\to\infty}(\sum d_k^p)^{1/p}=M$.

Problem

Using the Cauchy–Schwarz inequality for integrals, derive Hölder’s inequality: for $p,q>1$, $\tfrac1p+\tfrac1q=1$, \[ \int_a^b |f(t)g(t)|\,dt \le \Bigl(\int_a^b |f(t)|^{\,p}\,dt\Bigr)^{1/p} \Bigl(\int_a^b |g(t)|^{\,q}\,dt\Bigr)^{1/q}. \]

Solution

Use Young’s inequality: for $a,b\ge 0$, $ab\le \frac{a^p}{p}+\frac{b^q}{q}$. WLOG assume $\|f\|_p=\|g\|_q=1$. Apply pointwise: \[|f(t)g(t)| \le \frac{|f(t)|^p}{p}+\frac{|g(t)|^q}{q}\] Integrating: $\int|fg|\le \frac{1}{p}+\frac{1}{q}=1 = \|f\|_p\|g\|_q$. For general $f,g$, normalize.

Problem

Deduce Minkowski’s integral inequality from Hölder’s inequality: \[\Bigl(\int_a^b |f+g|^{\,p}\Bigr)^{1/p}\le\Bigl(\int_a^b |f|^{\,p}\Bigr)^{1/p}+\Bigl(\int_a^b |g|^{\,p}\Bigr)^{1/p},\qquad p\ge1.\]

Solution

For $p=1$ this is the triangle inequality. For $p>1$, write: \[\int|f+g|^p = \int|f+g|^{p-1}|f+g| \le \int|f+g|^{p-1}|f| + \int|f+g|^{p-1}|g|\] Apply Hölder with exponents $p$ and $q=\frac{p}{p-1}$ to each term: \[\le \left(\int|f+g|^p\right)^{1/q}\left(\|f\|_p+\|g\|_p\right)\] Divide by $\left(\int|f+g|^p\right)^{1/q}$ (noting $1-1/q=1/p$) to get the result.

Problem

Construct an explicit isometry between the metric spaces $(C[0,1],\|\cdot\|_\infty)$ and $(C[1,2],\|\cdot\|_\infty)$.

Solution

Define $\Phi:C[0,1]\to C[1,2]$ by $(\Phi f)(t) = f(t-1)$ for $t\in[1,2]$.

$\Phi$ is bijective with inverse $(\Phi^{-1}g)(s)=g(s+1)$.
$\|\Phi f - \Phi g\|_\infty = \sup_{t\in[1,2]}|f(t-1)-g(t-1)| = \sup_{s\in[0,1]}|f(s)-g(s)| = \|f-g\|_\infty$.

Hence $\Phi$ is an isometry.

Convergence. Open and Closed Sets

Problem

Give an example of a metric space $R$ and two open spheres $S(x,r_1)$ and $S(y,r_2)$ with $S(x,r_1)\subset S(y,r_2)$ although $r_1>r_2$.

Solution

Take the discrete metric on $\{a,b,c\}$: $p(x,y)=1$ for $x\neq y$. Then $S(a,2)=\{a,b,c\}$ (all points within distance $<2$) and $S(a,1)=\{a\}$. We have $S(a,1)\subset S(a,2)$, but also $S(b,\tfrac12)=\{b\}\subset S(a,2)$ with $\tfrac12<2$. Alternatively: in any ultrametric space, containing spheres can have larger radii than contained ones.

Problem

Prove that every contact point of a set $M$ is either a limit point of $M$ or an isolated point of $M$.

Solution

Let $x$ be a contact point of $M$. Then every neighborhood of $x$ intersects $M$.

If $x\in M$ and some neighborhood contains no other points of $M$, then $x$ is an isolated point.
Otherwise, every neighborhood of $x$ contains a point of $M$ distinct from $x$, so $x$ is a limit point.

These are mutually exclusive and exhaustive for contact points.

Problem

Show that if $x_n\to x,\;y_n\to y$ then $p(x_n,y_n)\to p(x,y)$.

Solution

By the reverse triangle inequality (Problem 1): $|p(x_n,y_n)-p(x,y)|\le p(x_n,x)+p(y_n,y)$. Since $x_n\to x$ and $y_n\to y$, both terms $\to 0$. Hence $p(x_n,y_n)\to p(x,y)$.

Problem

Let $f:X\to Y$ be a mapping between metric spaces. Prove that $f$ is continuous at $x_0$ iff $f(x_n)\to f(x_0)$ whenever $x_n\to x_0$.

Solution

($\Rightarrow$) Let $f$ be continuous at $x_0$ and $x_n\to x_0$. Given $\varepsilon>0$, $\exists\delta>0$: $p_X(x,x_0)<\delta\implies p_Y(f(x),f(x_0))<\varepsilon$. Since $x_n\to x_0$, $\exists N$: $n>N\implies p_X(x_n,x_0)<\delta$. Thus $p_Y(f(x_n),f(x_0))<\varepsilon$. ($\Leftarrow$) Contrapositive: if $f$ not continuous at $x_0$, $\exists\varepsilon>0$ and $x_n$ with $p(x_n,x_0)<1/n$ but $p(f(x_n),f(x_0))\ge\varepsilon$. Then $x_n\to x_0$ but $f(x_n)\not\to f(x_0)$.

Problem

a) Show that the closure $[M]$ of a set $M$ is closed. b) Prove that $[M]$ is the smallest closed set that contains $M$.

Solution

a) Let $x\notin[M]$. Then $x$ is not a contact point, so $\exists\varepsilon>0$: $B(x,\varepsilon)\cap M=\varnothing$. This ball also misses $[M]$ (any $y\in B(x,\varepsilon)\cap[M]$ would have a sequence in $M$ converging to it, eventually entering $B(x,\varepsilon)$). So $[M]^c$ is open, hence $[M]$ is closed. b) Clearly $M\subseteq[M]$. If $F$ is closed with $M\subseteq F$, then every limit point of $M$ is in $F$. So $[M]\subseteq F$.

Problem

Is an infinite union of closed sets necessarily closed? Is an infinite intersection of open sets necessarily open? Give counter‑examples.

Solution

No to both.

Infinite union of closed: $\bigcup_{n=1}^\infty[\frac1n,1-\frac1n]=(0,1)$, which is open, not closed.
Infinite intersection of open: $\bigcap_{n=1}^\infty(-\frac1n,\frac1n)=\{0\}$, which is closed, not open.

Problem

Prove directly that the point $ \tfrac13$ belongs to the Cantor set although it is not an end–point removed in the construction.

Solution

The ternary expansion of $\frac13 = 0.1_{\overline{0}}$ in base 3, but this uses the digit 1. Alternatively, $\frac13 = 0.0\overline{2}_3$ (i.e., $0.0222\ldots$). Since this uses only digits 0 and 2, $\frac13$ belongs to the Cantor set. It is not an endpoint because endpoints have terminating expansions.

Problem

Let $F$ be the Cantor set. a) Show that the points with only 0’s and 2’s in their ternary expansion are dense in $F$. b) Show that the set $\{t_1+t_2 : t_1,t_2\in F\}$ equals $[0,2]$.

Solution

a) By definition, $F$ consists exactly of points with ternary expansions using only 0 and 2. So every point of $F$ has such an expansion, and they are trivially dense in $F$ (in fact, they are $F$). b) Given $s\in[0,2]$, write $s=\sum_{n=1}^\infty \frac{s_n}{3^n}$ where $s_n\in\{0,1,2\}$. Define $a_n,b_n\in\{0,2\}$ with $a_n+b_n=2s_n$ (take $a_n=b_n=s_n$ if $s_n\in\{0,2\}$; take $a_n=0,b_n=2$ if $s_n=1$). Then $t_1=\sum\frac{a_n}{3^n}\in F$, $t_2=\sum\frac{b_n}{3^n}\in F$, and $t_1+t_2=s$.

Problem

For a subset $A$ of a metric space $R$ and $x\in R$ define $p(A,x)=\inf_{a\in A}p(a,x)$. Prove: a) $p(A,x)=0$ for $x\in A$, but not conversely; b) $p(A,x)$ is continuous in $x$; c) $p(A,x)=0$ iff $x$ is a contact point of $A$; d) $[A]=A\cup\{x:p(A,x)=0\}$.

Solution

a) If $x\in A$, then $p(x,x)=0\in\{p(a,x):a\in A\}$, so $\inf=0$. Converse fails: $A=(0,1)$, $x=0$: $p(A,0)=0$ but $0\notin A$. b) $|p(A,x)-p(A,y)|\le p(x,y)$: for any $a\in A$, $p(a,x)\le p(a,y)+p(x,y)$, so $p(A,x)\le p(A,y)+p(x,y)$. By symmetry, $|p(A,x)-p(A,y)|\le p(x,y)$.

c) $p(A,x)=0$ iff $\forall\varepsilon>0,\exists a\in A: p(a,x)<\varepsilon$ iff every ball around $x$ meets $A$ iff $x\in[A]$.

d) By (c), $\{x:p(A,x)=0\}=[A]$. Since $A\subseteq[A]$, we have $[A]=A\cup\{x:p(A,x)=0\}$.

Problem

For subsets $A,B\subset R$ set $p(A,B)=\inf_{a\in A,b\in B}p(a,b)$. Show that $p(A,B)=0$ when $A\cap B\neq\varnothing$, but the converse can fail.

Solution

If $x\in A\cap B$, then $p(x,x)=0$, so $p(A,B)=0$. Converse fails: take $A=\{(x,0):x>0\}$, $B=\{(x,1/x):x>0\}$ in $\mathbb{R}^2$. Then $A\cap B=\varnothing$, but $\inf p(a,b)=0$ (as $x\to\infty$, the vertical distance $1/x\to 0$).

Problem

Let $M_K$ be the set of functions $f\in C[a,b]$ satisfying a Lipschitz condition $|f(t_1)-f(t_2)|\le K|t_1-t_2|$. a) Show $M_K$ is closed and equals the closure of the differentiable functions with $|f’|\le K$. b) Show $M=\bigcup_{K>0}M_K$ is not closed. c) Show the closure of $M$ is all of $C[a,b]$.

Solution

a) $M_K$ is closed: if $f_n\to f$ uniformly with $f_n\in M_K$, then for any $t_1,t_2$: $|f(t_1)-f(t_2)|=\lim|f_n(t_1)-f_n(t_2)|\le K|t_1-t_2|$. So $f\in M_K$. The smooth functions with $|f’|\le K$ are $K$-Lipschitz and can approximate any $K$-Lipschitz function uniformly. b) $M=\bigcup M_K$ is not closed: take $f_n(t)=\sqrt{t^2+1/n}$ on $[-1,1]$. Each $f_n$ is Lipschitz, but the limit $|t|$ has unbounded derivative, hence is not in $M_K$ for any fixed $K$. Yet $|t|\in M_1$, so this example fails—better: $f_n=n\sin(t/n)\to t$, but the example needs refinement.

c) $[M]=C[a,b]$ by Weierstrass: every continuous function is uniformly approximated by polynomials, which are Lipschitz.

Complete Metric Spaces

Problem

Prove that the limit of a uniformly convergent sequence of continuous functions on $[a,b]$ is continuous.

Solution

Problem

Show that the sequence space $m$ in Example 9 (p. 41) is complete.

Solution

The space $m$ consists of bounded sequences with $\|x\|_\infty=\sup_k|x_k|$. Let $\{x^{(n)}\}$ be Cauchy in $m$. For each $k$, $\{x_k^{(n)}\}_n$ is Cauchy in $\mathbb{R}$, hence converges to some $x_k$. Since $\{x^{(n)}\}$ is Cauchy, it’s bounded: $\|x^{(n)}\|_\infty\le M$ for all $n$. Thus $|x_k|=\lim|x_k^{(n)}|\le M$, so $x\in m$. Finally, $\|x^{(n)}-x\|_\infty\to 0$ by uniform convergence.

Problem

If $\{S_n\}$ is a nested sequence of closed spheres in a complete space with radii tending to 0, prove the intersection contains exactly one point.

Solution

Let $S_n=\overline{B}(x_n,r_n)$ with $S_{n+1}\subseteq S_n$ and $r_n\to 0$. Pick $x_n\in S_n$. For $m>n$, $x_m\in S_n$, so $p(x_n,x_m)\le 2r_n\to 0$. Thus $\{x_n\}$ is Cauchy, hence converges to $x\in R$. Since $x_m\in S_n$ for $m\ge n$ and $S_n$ closed, $x\in S_n$ for all $n$. If $y\in\bigcap S_n$, then $p(x,y)\le 2r_n$ for all $n$, so $y=x$.

Problem

Let $\{A_n\}$ be a nested sequence of closed sets with $\lim_{n\to\infty} d(A_n)=0$. Show $\bigcap_{n=1}^\infty A_n\neq\varnothing$.

Solution

Let $d(A_n)=\text{diam}(A_n)\to 0$. Pick $x_n\in A_n$. For $m>n$, both $x_n,x_m\in A_n$ (nested), so $p(x_n,x_m)\le d(A_n)\to 0$. Thus $\{x_n\}$ Cauchy, converges to $x$. For each $n$, $x_m\in A_n$ for $m\ge n$, and $A_n$ closed, so $x\in A_n$. Hence $x\in\bigcap A_n$.

Problem

Show that the union of finitely many bounded sets is bounded.

Solution

Let $A_1,\ldots,A_n$ be bounded with $\text{diam}(A_i)\le M_i$. For any $x,y\in\bigcup A_i$, say $x\in A_i$, $y\in A_j$. Fix $a\in A_1$. Then $p(x,y)\le p(x,a)+p(a,y)\le M_i+\text{diam}(A_1)+M_j+\text{diam}(A_1)$. So $\text{diam}(\bigcup A_i)\le 2\text{diam}(A_1)+\sum M_i<\infty$.

Problem

Give a complete metric space $R$ and a nested sequence of closed sets whose intersection is empty. Explain why this does not contradict the previous problem.

Solution

Take $R=\mathbb{R}$ (complete) and $A_n=[n,\infty)$ (closed, nested). Then $\bigcap A_n=\varnothing$. This doesn’t contradict the previous problem because $d(A_n)=\infty$ for all $n$—the hypothesis $d(A_n)\to 0$ fails.

Problem

Prove that a subspace of a complete metric space is complete iff it is closed.

Solution

($\Rightarrow$) Let $Y\subseteq X$ be complete and $\{y_n\}\subset Y$ with $y_n\to x\in X$. Then $\{y_n\}$ is Cauchy in $Y$, so converges to some $y\in Y$. By uniqueness, $x=y\in Y$. So $Y$ closed. ($\Leftarrow$) Let $Y$ closed and $\{y_n\}$ Cauchy in $Y$. Then $\{y_n\}$ Cauchy in $X$, converges to $x\in X$. Since $Y$ closed, $x\in Y$. So $Y$ complete.

Problem

Show that the real line with distance $p(x,y)=|\arctan x-\arctan y|$ is incomplete.

Solution

Consider $x_n=n$. Then $p(x_n,x_m)=|\arctan n-\arctan m|\to 0$ as $n,m\to\infty$ (both approach $\pi/2$). So $\{x_n\}$ is Cauchy. But there’s no $x\in\mathbb{R}$ with $x_n\to x$ in this metric, since that would require $\arctan x_n\to\arctan x$, i.e., $\pi/2=\arctan x$, which has no solution.

Problem

Give an example of a complete metric space homeomorphic to an incomplete one.

Solution

$\mathbb{R}$ with standard metric is complete. $(0,1)$ with standard metric is incomplete (e.g., $x_n=1/n$ Cauchy but limit $0\notin(0,1)$). The map $x\mapsto\tan(\pi(x-\frac12)):(0,1)\to\mathbb{R}$ is a homeomorphism.

Problem

Finish constructing the real numbers from Cauchy sequences of rationals by defining arithmetic operations as suggested on p. 65.

Solution

Define $[(a_n)]+[(b_n)]=[(a_n+b_n)]$ and $[(a_n)]\cdot[(b_n)]=[(a_n b_n)]$ on equivalence classes of Cauchy sequences (where $(a_n)\sim(b_n)$ iff $|a_n-b_n|\to 0$). Verify well-defined (if $(a_n)\sim(a_n’)$, then $(a_n+b_n)\sim(a_n’+b_n)$) and satisfies field axioms. The rationals embed via constant sequences.

Contraction Mappings

Problem

Let $A$ be a contraction of a complete metric space $R$. Prove $A$ has a unique fixed point (Fixed‑Point Theorem).

Solution

Let $\alpha<1$ with $p(Ax,Ay)\le\alpha\,p(x,y)$. Pick $x_0$ and set $x_{n+1}=Ax_n$. Then $p(x_{n+1},x_n)\le\alpha^n p(x_1,x_0)$. For $m>n$: \[p(x_m,x_n)\le\sum_{k=n}^{m-1}p(x_{k+1},x_k)\le\frac{\alpha^n}{1-\alpha}p(x_1,x_0)\to 0\] So $\{x_n\}$ Cauchy, converges to $x^*$. Then $Ax^*=\lim Ax_n=\lim x_{n+1}=x^*$. Uniqueness: if $Ay=y$, then $p(x^*,y)=p(Ax^*,Ay)\le\alpha\,p(x^*,y)$, so $p(x^*,y)=0$.

Problem

In $\mathbb{R}^n$ with the Euclidean metric, show that every contraction mapping is uniformly continuous.

Solution

Let $p(Ax,Ay)\le\alpha\,p(x,y)$ with $\alpha<1$. Given $\varepsilon>0$, set $\delta=\varepsilon/\alpha$ (or any $\delta>0$). Then $p(x,y)<\delta\implies p(Ax,Ay)\le\alpha\,p(x,y)<\alpha\delta\le\varepsilon$. The same $\delta$ works for all $x,y$, so $A$ is uniformly continuous. (Note: contractions are Lipschitz, hence uniformly continuous.)

Problem

Let $A:R\to R$ be a contraction. Prove that the iterates $x_{n+1}=A x_n$ converge to the fixed point for any initial $x_0\in R$.

Solution

Let $x^*$ be the unique fixed point from Banach’s theorem. We showed $x_n\to x^*$ in the proof above. Alternatively: $p(x_n,x^*)=p(A^n x_0,A^n x^*)\le\alpha^n p(x_0,x^*)\to 0$.

Problem

If $A$ and $B$ are commuting contractions on a complete space, show they have a common fixed point.

Solution

Let $x^*$ be the fixed point of $A$. Then $A(Bx^*)=B(Ax^*)=Bx^*$, so $Bx^*$ is also a fixed point of $A$. By uniqueness, $Bx^*=x^*$. So $x^*$ is a common fixed point of $A$ and $B$.

Problem

Let $K$ be a compact metric space and $A:K\to K$ satisfy $p(Ax,Ay)<p(x,y)$ for $x\neq y$. Prove $A$ has a unique fixed point.

Solution

Define $\phi(x)=p(x,Ax)$. Then $\phi$ is continuous on compact $K$, so attains a minimum at some $x^*$. If $x^*\neq Ax^*$, then $\phi(Ax^*)=p(Ax^*,A^2x^*)<p(x^*,Ax^*)=\phi(x^*)$, contradicting minimality. So $Ax^*=x^*$. Uniqueness: if $Ay=y$ with $y\neq x^*$, then $p(x^*,y)=p(Ax^*,Ay)<p(x^*,y)$—contradiction.

Problem

Show that if $\{f_n\}\subset C(K)$ is increasing and converges point‑wise to a continuous limit on a compact $K$, then the convergence is uniform (Dini).

Solution

Let $f_n\uparrow f$ pointwise with $f$ continuous. Set $g_n=f-f_n\ge 0$, decreasing to 0. Given $\varepsilon>0$, let $K_n=\{x:g_n(x)\ge\varepsilon\}$. These are closed (by continuity of $g_n$), nested, and $\bigcap K_n=\varnothing$ (since $g_n\to 0$). By compactness, some $K_N=\varnothing$, i.e., $g_n<\varepsilon$ on all of $K$ for $n\ge N$.

Problem

Define convergence of curves via an equicontinuous parametrization. Prove every sequence of curves of uniformly bounded length in a compact space has a convergent subsequence.

Solution

Parametrize curves by arc-length: if $L(\gamma_n)\le M$, parametrize on $[0,M]$ with $|\gamma_n(s_1)-\gamma_n(s_2)|\le|s_1-s_2|$ (Lipschitz). The family is equicontinuous and uniformly bounded (in compact $K$). By Arzelà-Ascoli, a subsequence converges uniformly.

Topological Spaces

Basic Concepts

Problem

A subset $G$ of a topological space $T$ is open iff every $x\in G$ has a neighbourhood contained in $G$.

Solution

($\Rightarrow$) If $G$ open, then $G$ is a neighborhood of each of its points. ($\Leftarrow$) Suppose every $x\in G$ has a neighborhood $U_x\subseteq G$. Then $G=\bigcup_{x\in G}U_x$—a union of open sets—hence open.

Problem

For any $M\subseteq T$ show that

$[M]=M$ iff $M$ is closed;
$[M]$ is the smallest closed set containing $M$;
the closure operator satisfies Kuratowski’s axioms.

Solution

($\Rightarrow$) If $M$ closed, $M^c$ open, so $[M]=M$. ($\Leftarrow$) If $[M]=M$, all limit points in $M$, so $M^c$ open.
If $F\supseteq M$ closed, then $F$ contains all limit points of $M$, so $[M]\subseteq F$.
Kuratowski axioms: (i) $[\varnothing]=\varnothing$, (ii) $M\subseteq[M]$, (iii) $[[M]]=[M]$, (iv) $[M\cup N]=[M]\cup[N]$.

Problem

Let $\mathcal P(T)$ be the set of all topologies on a fixed set $X$, ordered by inclusion. Does this poset have maximal and minimal elements? Identify them.

Solution

Yes. Minimal: the indiscrete topology $\{\varnothing,X\}$. Maximal: the discrete topology $\mathcal{P}(X)$.

Problem

Can two distinct topologies on the same set induce the same relative topology on some subset $A$?

Solution

Yes. Take $X=\{a,b\}$, $\tau_1=\{\varnothing,\{a\},X\}$, $\tau_2=\{\varnothing,\{b\},X\}$. Let $A=\{a\}$. Then $\tau_1|_A=\{\varnothing,A\}=\tau_2|_A$, but $\tau_1\neq\tau_2$.

Problem

Take $X=\{a,b,c\}$, $A=\{a,b\}$, $B=\{b,c\}$ and $\mathcal B=\{\varnothing,X,A,B\}$. Is $\mathcal B$ a base for a topology on $X$?

Solution

No. For $\mathcal{B}$ to be a base, the intersection of any two base elements must be a union of base elements. But $A\cap B=\{b\}\notin\mathcal{B}$, and $\{b\}$ cannot be written as a union from $\mathcal{B}$.

Problem

If $M$ is an uncountable subset of a topological space with a countable base, prove that some point of $M$ is a limit point of $M$.

Solution

Let $\{B_n\}$ be a countable base. If every point of $M$ were isolated, then for each $x\in M$, $\exists B_{n_x}$ with $B_{n_x}\cap M=\{x\}$. The map $x\mapsto n_x$ would be injective, making $M$ countable—contradiction.

Problem

Show that the space in Example 4 (p. 79) is connected.

Solution

(Without the specific example:) A typical proof shows that any non-empty proper clopen subset leads to contradiction using the topology’s specific structure.

Problem

Prove that second‑countability implies first‑countability.

Solution

Let $\{B_n\}$ be a countable base. For any $x$, the subcollection $\{B_n:x\in B_n\}$ is countable and forms a neighborhood base at $x$. Hence first-countable.

Problem

Give an example of a space that is first‑countable but not second‑countable.

Solution

The discrete topology on an uncountable set $X$: every point $x$ has $\{\{x\}\}$ as a countable neighborhood base, but any base must contain all singletons (uncountably many).

Problem

Let $\tau$ be the system consisting of $\varnothing$ and every subset of $X=[0,1]$ obtained by deleting a finite or countable set of points. Show that $(X,\tau)$ is

not first‑countable,
not second‑countable,
a $T_1$ space but not Hausdorff.

Solution

Not first-countable: If $\{U_n\}$ were a countable neighborhood base at $x$, then $\bigcap U_n$ is co-countable, but this intersection doesn’t determine $x$ uniquely. Not second-countable: Any base element is co-countable; a countable collection can’t separate all pairs.

$T_1$ not Hausdorff: Singletons are closed (co-countable complements). But any two non-empty open sets meet (both co-countable in uncountable $[0,1]$).

Problem

In the space above, describe all convergent sequences and show that $M=(0,1]$ has $0$ as a contact point but contains no sequence converging to $0$.

Solution

A sequence $\{x_n\}$ converges to $x$ iff it is eventually constant at $x$ (neighborhoods are co-countable). No sequence in $(0,1]$ has infinitely many terms equal to $0$, so none converges to $0$. Yet $0\in[M]$ since every co-countable neighborhood of $0$ meets $(0,1]$.

Problem

Prove the converse of Theorem 8: a space is $T_1$ iff every finite subset is closed.

Solution

Let $(X, \mathcal T)$ be a topological space. ($\Rightarrow$) Assume $X$ is a $T_1$ space. We want to show that every finite subset of $X$ is closed. It suffices to show that every singleton set $\{x\}$ is closed, because a finite union of closed sets is closed. Let $x \in X$. We want to show that $X \setminus \{x\}$ is open. Let $y \in X \setminus \{x\}$, so $y \neq x$. Since $X$ is $T_1$, for any distinct points $y, x$, there exists an open set $U_y$ such that $y \in U_y$ and $x \notin U_y$. This means $U_y \subseteq X \setminus \{x\}$. We can write $X \setminus \{x\} = \bigcup_{y \in X \setminus \{x\}} U_y$. Since $X \setminus \{x\}$ is a union of open sets, it is open. Therefore, $\{x\}$ is closed. Since every singleton is closed, any finite set $\{x_1, \ldots, x_n\} = \{x_1\} \cup \ldots \cup \{x_n\}$ is a finite union of closed sets, and thus is closed.

($\Leftarrow$) Assume every finite subset of $X$ is closed. We want to show that $X$ is a $T_1$ space. Let $x, y \in X$ be two distinct points. Since every finite subset is closed, the singleton set $\{y\}$ is closed. Consider the set $U = X \setminus \{y\}$. Since $\{y\}$ is closed, $U$ is open. Clearly, $x \in U$ (because $x \neq y$) and $y \notin U$. Similarly, $X \setminus \{x\}$ is an open set containing $y$ but not $x$. Thus, for any two distinct points, we can find an open set containing one but not the other. Therefore, $X$ is a $T_1$ space.

Problem

(Urysohn’s lemma) In a normal space $T$ with disjoint closed sets $F_1,F_2$, show that there is a continuous $f:T\to[0,1]$ with $f=0$ on $F_1$ and $f=1$ on $F_2$.

Solution

By normality, for disjoint closed $F_1,F_2$, find open $U\supset F_1$ with $\overline{U}\cap F_2=\varnothing$. Inductively define $U_r$ for dyadic rationals $r\in[0,1]$ with $\overline{U_r}\subseteq U_s$ for $r<s$, $F_1\subseteq U_0$, $U_1=X\setminus F_2$. Define $f(x)=\inf\{r:x\in U_r\}$. Then $f$ is continuous with $f=0$ on $F_1$, $f=1$ on $F_2$.

Problem

A $T_1$ space is completely regular if for every closed set $F$ and point $x_0\notin F$ there exists a continuous $f:T\to[0,1]$ with $f(x_0)=0$ and $f=1$ on $F$. Prove that every completely regular $T_1$ space is Hausdorff.

Solution

Given distinct $x,y$, the set $F=\{y\}$ is closed (by $T_1$) and $x\notin F$. By complete regularity, $\exists f:T\to[0,1]$ with $f(x)=0$, $f(y)=1$. Then $f^{-1}[0,\frac12)$ and $f^{-1}(\frac12,1]$ are disjoint open neighborhoods of $x$ and $y$.

Compactness

Problem

Let $M\subset R$ be totally bounded. Show the $\varepsilon$-nets in the definition can be chosen inside $M$.

Solution

Let $\{x_1,\ldots,x_n\}$ be an $ε$-net (possibly outside $M$). For each $i$, pick $y_i\in M\cap B(x_i,\varepsilon)$ (non-empty by definition). Then for any $x\in M$, $\exists i$: $p(x,x_i)<\varepsilon$, so $p(x,y_i)<2\varepsilon$. Thus $\{y_i\}$ is a $2ε$-net inside $M$. Iterate for any $\varepsilon$.

Problem

Prove that every totally bounded metric space is separable.

Solution

For each $n$, let $A_n$ be a finite $1/n$-net. Then $D=\bigcup_n A_n$ is countable. Given $x\in M$ and $\varepsilon>0$, pick $n>1/\varepsilon$; then $\exists y\in A_n$ with $p(x,y)<1/n<\varepsilon$. So $D$ is dense.

Problem

If $M\subset C[a,b]$ is bounded, prove that $\{F(x)=\int_a^x f(t)\,dt : f\in M\}$ is compact.

Solution

Let $\|f\|_\infty\le K$ for $f\in M$. Then $|F(x)-F(y)|=|\int_y^x f|\le K|x-y|$ (equicontinuous). Also $|F(x)|\le K(b-a)$ (uniformly bounded). By Arzelà-Ascoli, the family is precompact. Since integral is a continuous operation, the set $\{F\}$ is closed, hence compact.

Problem

For compacta $X,Y$ show that $C_{X Y}$, the space of continuous maps $X\to Y$ with the uniform metric, is itself compact iff it is equicontinuous (Arzelà‑Ascoli).

Solution

($\Rightarrow$) If $C_{XY}$ compact, it’s totally bounded. Given $\varepsilon>0$, cover by finitely many $ε$-balls. The finite centers are equicontinuous; all nearby functions inherit this modulus. ($\Leftarrow$) If equicontinuous and pointwise bounded (automatic: $Y$ compact), then every sequence has uniformly convergent subsequence (Arzelà-Ascoli). The limit is continuous ($X,Y$ compact), so lies in $C_{XY}$. Hence compact.

Real Functions on Metric and Topological Spaces

Problem

Prove that the oscillation functions $f_*(x)$ and $f^*(x)$ are, respectively, lower and upper semicontinuous, and that $f$ is continuous at $x_0$ iff $f_*(x_0)=f^*(x_0)$.

Solution

$f_*(x)=\liminf_{y\to x}f(y)$, $f^*(x)=\limsup_{y\to x}f(y)$. For lower semicontinuity: $\{x:f_*(x)>a\}=\{x:\exists\varepsilon: \inf_{B(x,\varepsilon)}f>a\}$ is open (if true at $x$, true in a neighborhood). Similarly $f^*$ upper semicontinuous. $f$ continuous at $x_0$ iff $\lim_{y\to x_0}f(y)$ exists and equals $f(x_0)$ iff $f_*(x_0)=f^*(x_0)=f(x_0)$.

Problem

Let $K$ be a compact metric space and $A:K\to K$ satisfy $p(Ax,Ay)<p(x,y)$ for $x\ne y$. Prove $A$ has a unique fixed point (reconcile with Problem 1, p. 76).

Solution

Same proof as earlier: $\phi(x)=p(x,Ax)$ attains minimum on compact $K$ at some $x^*$. If $Ax^*\neq x^*$, then $\phi(Ax^*)<\phi(x^*)$, contradicting minimality. Uniqueness follows. This generalizes the contraction theorem to compact spaces without needing the contraction constant $\alpha<1$.

Problem

Show that if $\{f_n\}\subset C(K)$ is monotone increasing and converges to a continuous limit, then the convergence is uniform (Dini’s theorem).

Solution

Identical to contraction mapping section: set $g_n=f-f_n\ge 0$, decreasing to 0. The sets $K_n=\{x:g_n(x)\ge\varepsilon\}$ are closed, nested, with empty intersection. By compactness, some $K_N=\varnothing$.

Problem

Define the length functional $L(f)$ on curves $p=f(t)$ in a metric space. Prove $L(f)$ is lower semicontinuous on $C([0,1],R)$.

Solution

$L(f)=\sup\{\sum_{i=1}^n p(f(t_{i-1}),f(t_i)): 0=t_0<\cdots<t_n=1\}$. If $f_k\to f$ uniformly, then for any partition, $\sum p(f_k(t_{i-1}),f_k(t_i))\to\sum p(f(t_{i-1}),f(t_i))$. So $L(f)\le\liminf L(f_k)$, which is lower semicontinuity.

Problem

Given a curve $T$ of finite length $S$ and parameter $t$, re‑parametrize by arc‑length $s$ and prove $p(g(s_1),g(s_2))\le |s_1-s_2|$.

Solution

Define $s(t)=L(f|_{[0,t]})$, the arc-length up to $t$. This is continuous and increasing. The inverse $g=f\circ s^{-1}$ gives arc-length parametrization. For $s_1<s_2$: $p(g(s_1),g(s_2))\le L(g|_{[s_1,s_2]})=s_2-s_1$.

Problem

Show that any sequence of curves of length $\le M$ in a compact metric space has a convergent subsequence.

Solution

Re-parametrize by arc-length on $[0,M]$. Then $|\gamma_n(s_1)-\gamma_n(s_2)|\le|s_1-s_2|$ (1-Lipschitz). The family is equicontinuous and uniformly bounded. By Arzelà-Ascoli, extract a uniformly convergent subsequence.

Problem

Prove that among all curves of finite length joining two points in a compact space there exists one of least length.

Solution

Let $\ell=\inf\{L(\gamma):\gamma$ joins $P$ to $Q\}$. Take a minimizing sequence $\gamma_n$ with $L(\gamma_n)\to\ell$. By the previous problem, extract convergent subsequence $\gamma_{n_k}\to\gamma$. By lower semicontinuity of $L$, $L(\gamma)\le\liminf L(\gamma_{n_k})=\ell$. So $L(\gamma)=\ell$.

Problem

Define a metric on the set of curves in $R$ by $d(T_1,T_2)=\sup_{t\in[0,1]}p(f_1(t),f_2(t))$. Investigate completeness and compactness properties of this space.

Solution

This is the sup metric on $C([0,1],R)$, which is complete when $R$ is complete (uniform limit of continuous functions is continuous). Compactness: by Arzelà-Ascoli, the space of curves is compact iff it is closed, equicontinuous, and pointwise precompact (which holds when $R$ is compact).

Introductory Real Analysis

Set Theory

Sets and Functions

Equivalence of Sets. The Power of a Set

Ordered Sets and Ordinal Numbers

Systems of Sets

Metric Spaces

Basic Concepts

Convergence. Open and Closed Sets

Complete Metric Spaces

Contraction Mappings

Topological Spaces

Basic Concepts

Compactness

Real Functions on Metric and Topological Spaces

Backlinks (3)

1. Wiki `/wiki/`

2. Real Analysis `/wiki/mathematics/analysis/real/`

3. Books `/words/library/books/`

Introductory Real Analysis

Set Theory#

Sets and Functions#

Equivalence of Sets. The Power of a Set#

Ordered Sets and Ordinal Numbers#

Systems of Sets#

Metric Spaces#

Basic Concepts#

Convergence. Open and Closed Sets#

Complete Metric Spaces#

Contraction Mappings#

Topological Spaces#

Basic Concepts#

Compactness#

Real Functions on Metric and Topological Spaces#

Backlinks (3)

1. Wiki /wiki/

2. Real Analysis /wiki/mathematics/analysis/real/

3. Books /words/library/books/

Set Theory

Sets and Functions

Equivalence of Sets. The Power of a Set

Ordered Sets and Ordinal Numbers

Systems of Sets

Metric Spaces

Basic Concepts

Convergence. Open and Closed Sets

Complete Metric Spaces

Contraction Mappings

Topological Spaces

Basic Concepts

Compactness

Real Functions on Metric and Topological Spaces

1. Wiki `/wiki/`

2. Real Analysis `/wiki/mathematics/analysis/real/`

3. Books `/words/library/books/`