Introductory Real Analysis

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 $A=B$.

This only holds for $B\subseteq A$. We proceed by counterexample.

Let $A={1,2}, B={3,4}$. Then $(A-B) = {1,2}$ and $(A-B)\cup B = {1,2,3,4} \neq {1,2}$.

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

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

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

So

• $x\not\in B_\alpha{}_0$

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

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.

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 $M\cup A\sim M$.

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.

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.

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 $|F|>|M|$.

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 $\mathfrak{c}$.

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, $|\bigcup_n A_n|=\mathfrak{c}$.

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.

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 $\mathbb{C}$.

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.

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 $c=a\cup b$).

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.

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 $f(x)>f(y)$).

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 $M_1\cdot(M_2\cdot M_3)$.

• $\omega+n$: $\mathbb{N}$ 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.

• $\omega\cdot n$: $n$ 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.

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.

$W(\alpha)$ 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.

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 $\in$-minimal element. Thus $T$ has a least ordinal, so $S$ is well-ordered.

$M$ 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.

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 $=\aleph_1$.

No. $\mathcal{R}$ is not a $\sigma$-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.

Yes. The Borel $\sigma$-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 $\sigma$-algebra generated by open intervals, making them Borel by definition.

1. 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: $|p(x,z)-p(y,u)|\le p(x,y)+p(z,u)$.
2. Set $u=z$ in (1): $|p(x,z)-p(y,z)|\le p(x,y)+p(z,z)=p(x,y)$.

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.

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.

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 $p<1$.

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, $\lim_{p\to\infty}(\sum d_k^p)^{1/p}=M$.

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.

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.

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.

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.

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.

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 $p(x_n,y_n)\to p(x,y)$.

($\Rightarrow$) 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 $f(x_n)\not\to f(x_0)$.

1. 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.
2. Clearly $M\subseteq[M]$. If $F$ is closed with $M\subseteq F$, then every limit point of $M$ is in $F$. So $[M]\subseteq F$.

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.

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.

1. 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 $F$).
2. 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 $t_1+t_2=s$.

1. 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 $0\notin A$.
2. $|p(A,x)-p(A,y)|\le p(x,y)$: 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, $|p(A,x)-p(A,y)|\le p(x,y)$.
3. $p(A,x)=0$ iff $\forall\varepsilon>0,\exists a\in A: p(a,x)<\varepsilon$ iff every ball around $x$ meets $A$ iff $x\in[A]$.
4. By (c), $\{x:p(A,x)=0\}=[A]$. Since $A\subseteq[A]$, we have $[A]=A\cup\{x:p(A,x)=0\}$.

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 $1/x\to 0$).

1. $M_K$ 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.
2. $M=\bigcup M_K$ 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.
3. $[M]=C[a,b]$ by Weierstrass: every continuous function is uniformly approximated by polynomials, which are Lipschitz.

Let $f_n\to f$ uniformly and each $f_n$ continuous. Fix $x_0$ and $\varepsilon>0$. Choose $N$: $\|f_N-f\|_\infty<\varepsilon/3$. By continuity of $f_N$, $\exists\delta$: $|x-x_0|<\delta\implies|f_N(x)-f_N(x_0)|<\varepsilon/3$. Then: $$|f(x)-f(x_0)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\varepsilon$$

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.

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 $y=x$.

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 $x\in\bigcap A_n$.

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 $\text{diam}(\bigcup A_i)\le 2\text{diam}(A_1)+\sum M_i<\infty$.

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.

($\Rightarrow$) 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.

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.

$\mathbb{R}$ 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.

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.

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 $p(x^*,y)=0$.

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.)

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

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 $B$.

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.

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 $n\ge N$.

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.

($\Rightarrow$) 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.

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

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

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 $\tau_1\neq\tau_2$.

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 $\mathcal{B}$.

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.

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

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.

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).

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 $[0,1]$).

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 $(0,1]$.

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.

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 $F_2$.

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 $y$.

Let $\{x_1,\ldots,x_n\}$ be an $\varepsilon$-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 $2\varepsilon$-net inside $M$. Iterate for any $\varepsilon$.

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.

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.

($\Rightarrow$) If $C_{XY}$ compact, it's totally bounded. Given $\varepsilon>0$, cover by finitely many $\varepsilon$-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.

$f_*(x)=\liminf_{y\to x}f(y)$, $f^*(x)=\limsup_{y\to x}f(y)$. 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 $f_*(x_0)=f^*(x_0)=f(x_0)$.

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 $\alpha<1$.

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 $K_N=\varnothing$.

$L(f)=\sup\{\sum_{i=1}^n p(f(t_{i-1}),f(t_i)): 0=t_0<\cdots<t_n=1\}$. 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.

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 $s_1<s_2$: $p(g(s_1),g(s_2))\le L(g|_{[s_1,s_2]})=s_2-s_1$.

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.

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 $L(\gamma)=\ell$.

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).