\subsection{Sheet 1}
\tutorial{02}{2023-10-24}{}
% Points: 15 / 16

\nr 1
\todo{handwritten solution}

\nr 2

\begin{enumerate}[(a)]
    \item $d$ is an ultrametric:

        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

\begin{enumerate}[(a)]
    \item $S_{\infty}$ is a Polish space:

        From (2) 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 $B_j$.

         % TODO Think about this
\end{enumerate}

\nr 4

% (uniform metric)
% 
% \begin{enumerate}[(a)]
%     \item $d_u$ is a metric on $\cC(X,Y)$:
% 
%         It is clear that $d_u(f,f) = 0$.
% 
%         Let  $f \neq g$. Then there exists $x \in X$ with
%         $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
% 
%         Since $d$ is symmetric, so is $d_u$.
% 
%         Let $f,g,h \in \cC(X,Y)$.
%         Take some $\epsilon > 0$
%         choose $x_1, x_2 \in X$
%         with $d_u(f,g) \le   d(f(x_1), g(x_1)) + \epsilon$,
%         $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
% 
%         Then for all $x \in X$
%         \begin{IEEEeqnarray*}{rCl}
%             d(f(x), h(x)) &\le &
%             d(f(x), g(x)) + d(g(x), h(x))\\
%                           &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
%                           &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
%         \end{IEEEeqnarray*}
%         Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
%         Taking $\epsilon \to 0$ yields the triangle inequality.
% 
%     \item $\cC(X,Y)$ is a Polish space:
%         \todo{handwritten solution}
% 
%         \begin{itemize}
%             \item $d_u$ is a complete metric:
% 
%                 Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
% 
%                 Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
%                 to  $d$ for every $x$.
%                 Hence there exists a pointwise limit $f$ of the $f_n$.
%                 We need to show that $f$ is continuous.
% 
%                 %\todo{something something uniform convergence theorem}
% 
%             \item $(\cC(X,Y), d_u)$ is separable:
% 
%                 Since $Y$ is separable, there exists a countable
%                 dense subset $S \subseteq Y$.
% 
%                 Consider $\cC(X,S) \subseteq  \cC(X,Y)$.
%                 Take some $f \in \cC(X,Y)$.
%                 Since $X$ is compact,
% 
% 
%                 % TODO
% 
%         \end{itemize}
% \end{enumerate}

\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 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}
\todo{handwritten solution}