tutorial 02

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\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}