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