\tutorial{02}{2023-10-24}{}

% Points: 15 / 16

\subsubsection{Exercise 4}

\begin{fact}
    Let $X $ be a compact Hausdorffspace.
    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 (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}
