\subsection{Sheet 1} \tutorial{02}{2023-10-24}{} % Points: 15 / 16 \nr 1 Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$. Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$. \begin{itemize} \item $d(-,A)$ is uniformly continuous: Clearly $|d(x,A) - d(y,A)| \le d(x,y)$. \todo{Add details} \item $d(x,A) = 0 \iff x \in \overline{A}$. $d(x,A) = 0$ iff there is a sequence in $A$ converging towards $x$ iff $x \in \overline{A}$. \end{itemize} \nr 2 Let $X$ be a discrete space. For $f,g \in X^{\N}$ define \[ d(f,g) \coloneqq \begin{cases} (1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\ 0 &: f= g. \end{cases} \] \begin{enumerate}[(a)] \item $d$ is an \vocab{ultrametric}, i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ : Let $f,g,h \in X^{\N}$. We need to show that $d(f,g) \le \max(d(f,h), d(g,h))$. If $f = g$ this is trivial. Otherwise let $n$ be minimal such that $f(n) \neq g(n)$. Then $h(n) \neq f(n)$ or $ h(n) \neq g(n)$ must be the case. W.l.o.g.~$h(n) \neq f(n)$. Then $d(f,g) = \frac{1}{1+n} \le d(f,h)$. \item $d$ induces the product topology on $X^{\N}$: It suffices to show that the $\epsilon$-balls with respect to $d$ are exactly the basic open set of the product topology, i.e.~the sets of the form \[ \{x_1\} \times \ldots \times \{x_n\} \times X^{\N} \] for some $n \in \N$, $x_1,\ldots,x _n \in X$. Let $\epsilon > 0$. Let $n$ be minimal such that $\frac{1}{1+n} \ge \epsilon$. Then $B_{\epsilon}((x_i)_{i \in \N}) = \{x_1\} \times \{x_n\} \times X^{\N}$. Since $\N \ni n \mapsto \frac{1}{1+n}$ is injective, every basic open set of the product topology can be written in this way. \item $d$ is complete: Let $(f_n)_{n \in \N}$ be a Cauchy sequence with respect to $d$. For $n \in \N$ take $N_n \in \N$ such that $d(f_i, f_j) < \frac{1}{1 + n}$. Clearly $f_i(n) = f_j(n)$ for all $n > N_n$. Define $f \in X^\N$ by $f(n) \coloneqq f_{N_n}(n)$. Then $ (f_n)_{n \in \N}$ converges to $f$, since for all $n > N_n$ $f_n$ \item If $X$ is countable, then $X^{\N}$ with the product topology is a Polish space: (We assume that $X$ is non-empty, as otherwise the claim is wrong) We need to show that there exists a countable dense subset. To this end, pick some $x_0 \in X$ and consider the set $D \coloneqq \bigcup_{n\in \N} (X^n \times \{x_{0}\}^{\N})$. Since $X$ is countable, so is $D$. Take some $(a_n)_{n \in \N} \in X^{\N}$ and consider $B \coloneqq B_{\epsilon}((a_n)_{n \in \N})$. Let $m$ be such that $\frac{1}{1+m} < \epsilon$. Then $(b_{n})_{n \in \N} \in B \cap D$, where $b_n \coloneqq a_n$ for $n \le m$ and $b_n \coloneqq x_0$ otherwise. Hence $D$ is dense. \end{enumerate} \nr 3 Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology. Let \[ S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}. \] \begin{enumerate}[(a)] \item $S_{\infty}$ is a Polish space: From \yaref{s1e2} we know that $\N^{\N}$ is Polish. Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$ with respect to $\N^\N$. Consider the sets $I \coloneqq \bigcap_{(i,j) \in \N^2, i < j} \{f \in \N^{\N} | f(i) \neq f(j)\}$ and $S \coloneqq \bigcap_{n \in \N} \{f \in \N^\N | n \in \im f\}$. We have that $\{f \in \N^\N | f(i) \neq f(j)\} = \bigcup_{n \in \N} \N^{i-1} \times \{n\} \times \N^{i - j -1 } \times (\N \setminus \{n\} ) \times \N^\N$ is open. Hence $I$ is $G_{\delta}$. Furthermore $\{f \in \N^{\N} | n \in \im f\} = \bigcup_{k \in \N} \N^k \times \{n\} \times \N^\N$j is open, thus $S$ is $G_\delta$ as well. In particular $S \cap G$ is $G_\delta$. Since $I$ is the subset of injective functions and $S$ is the subset of surjective functions, we have that $S_{\infty} = I \cap S$. \item $S_{\infty}$ is not locally compact: Consider the point $x = (i)_{i \in \N} \in S_{\infty}$. Let $x \in B$ be open. We need to show that there is no closed compact set $C \supseteq B$ W.l.o.g.~let $B = (\{0\} \times \ldots \times \{n\} \times \N^\N) \cap S_\infty$ for some $n \in \N$. Let $C \supseteq B$ be some closed set. Consider the open covering \[ \{S_{\infty} \setminus B\} \cup \{ B_j | j > n\}. \] where \[ B_j \coloneqq (\{0\} \times \ldots \times \{n\} \times \{j\} \times \N^{\N}) \cap S_\infty. \] Clearly there cannot exist a finite subcover as $B$ is the disjoint union of the $B_j$. \end{enumerate} \nr 4 \begin{fact} Let $X $ be a compact Hausdorff space. Then the following are equivalent: \begin{enumerate}[(i)] \item $X$ is Polish, \item $X$ is metrisable, \item $X$ is second countable. \end{enumerate} \end{fact} \begin{proof} (i) $\implies$ (ii) clear (i) $\implies$ (iii) clear (ii) $\implies$ (i) Consider the cover $\{B_{\epsilon}(x) | x \in X\}$ for every $\epsilon \in \Q$ and chose a finite subcover. Then the midpoints of the balls from the cover form a countable dense subset. The metric is complete as $X$ is compact. (For metric spaces: compact $\iff$ seq.~compact $\iff$ complete and totally bounded) (iii) $\implies$ (ii) Use Urysohn's metrisation theorem and the fact that compact Hausdorff spaces are normal \end{proof} Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish} and $Y $ Polish. Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$. Consider the \vocab{uniform metric} $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$. Clearly $d_u$ is a metric. \begin{claim} $d_u$ is complete. \end{claim} \begin{subproof} Let $(f_n)$ be a Cauchy sequence in $\cC(X,Y)$. A $Y$ is complete, there exists a pointwise limit $f$. $f_n$ converges uniformly to $f$: \[ d(f_n(x), f(x)) \le \overbrace{d(f_n(x), f_m(x))}^{\mathclap{\text{$(f_n)$ is Cauchy}}} + \underbrace{d(f_m(x), f(x))}_{\mathclap{\text{small for appropriate $m$}}}. \] $f$ is continuous by the uniform convergence theorem. \end{subproof} \begin{claim} There exists a countable dense subset. \end{claim} \begin{subproof} Fix a metric $d_X$ on $X$ defining its topology. Let \[ C_{m,n} \coloneqq \{f \in \cC(X,Y) : \forall x,y \in X.~\left( d_X(x,y) < \frac{1}{m+1} \implies d(f(x), f(y)) <\frac{1}{n+1}\right) \}. \] Choose $X_m \subseteq X$ finite with $X \subseteq \bigcup_{x \in X_m} B_{\frac{1}{m+1}}(x)$. Let $D_{m,n} \subseteq C_{m,n}$ be countable, such that for every $f \in C_{m,n}$ and every $\eta > 0$, there is $g \in D_{m,n}$ with $d(f(y), g(y)) < \frac{\eta}{3}$ for each $y \in X_m$. Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$: Indeed if $f \in \cC(X,Y)$ and $\eta > 0$, we find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$, since $f$ is uniformly continuous. Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$. We have $d_u(f,g) \le \eta$, since for every $x \in X$, we find $y \in X_m$ with $d_X(x,y) < \frac{1}{m+1}$, hence \begin{IEEEeqnarray*}{rCl} d_Y(f(x), g(x)) &\le& d_Y(f(x), f(y)) + d_Y(f(y), g(y)) + d_Y(g(y), g(x))\\ &\le& \frac{1}{n+1} + \frac{1}{n+1} + \frac{1}{n+1} \le \eta. \end{IEEEeqnarray*} \end{subproof}