From 7478dfd30d052449caf05c174dbaf8d1afec4008 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <git@jrpie.de>
Date: Tue, 31 Oct 2023 10:12:58 +0100
Subject: [PATCH] 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}