tutorial 03

inputs/tutorial_03.tex

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+\ctutorial{03}{2023-10-31}{}
+
+% 15 / 16
+
+\begin{remark}
+    $F_\sigma$
+    stands for ferm\'e sum denumerable.
+\end{remark}
+
+
+\subsection{Exercise 2}
+
+(b)
+Let $f(x^{(i)})$ be a sequence in $f(X)$.
+Suppose that $f(x^{(i)}) \to y$.
+We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
+Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
+Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
+
+
+\subsection{Exercise 3}
+
+\begin{example}
+    Consider
+    \begin{IEEEeqnarray*}{rCl}
+        f\colon \R &\longrightarrow & [0,1] \\
+        \frac{p}{q} &\longmapsto & \frac{1}{q}\\
+        \R \setminus \Q \ni x &\longmapsto & 0
+    \end{IEEEeqnarray*}
+    Then $\osc_f(\frac{p}{q}) = \frac{1}{q}$
+    and $\osc_f(x) = 0$ for $x \not\in \Q$.
+\end{example}
+
+
+\begin{definition}
+    We say that $f\colon X \to Y$ is continuous
+    at $a \in X$,
+    if for $N$ a neighbourhood of $f(a)$ (i.e.~there exists
+    $f(a) \in U \overset{\text{open}}{\subseteq}  N$,
+    then $f^{-1}(N)$ is a neighbourhood of $a$.
+\end{definition}
+
+
+
+\begin{theorem}[Kuratowski]
+    Let $X$ be metrizable, $Y$ completely metrizable,
+    $f\colon S \to Y$ continuous.
+    Then $f$ can be extended to a continuous fnuction $f_n$
+    on a $G_\delta$ set  $G$ with $S \subseteq  G \subseteq \overline{G}$.
+\end{theorem}
+\begin{proof}
+    Let $G \coloneqq  \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
+    Clearly $S \subseteq G$ as $f$ is continouos on $f$.
+    \begin{claim}
+        $G$ is $G_\delta$
+    \end{claim}
+    \begin{subproof}
+        $\overline{S}$ is closed
+        and
+        \[
+        \bigcap_{n \ge 1} \{x : \osc_f(x) <\frac{1}{n}\}
+        \]
+        is an intersection of open sets.
+    \end{subproof}
+    is an intersection of open sets.
+
+    For $x \in G$, as $x \in  \overline{S}$,
+    there exists $(x_n)_{x_n < \omega}$, $x_n \in S_$
+    such that $x_n \to x$.
+    We have that $(f(x_n))_n$ is Cauchy,
+    as $\osc_f(x) = 0$.
+    % TODO
+
+    \todo{Something is missing here}
+
+
+\end{proof}
+
+\begin{corollary}
+    Let $X$ be Polish and $Y \subseteq X$ Polish.
+    Then $Y$ is $G_{\delta}$.
+\end{corollary}
+\begin{proof}
+    Consider the identity $f\colon Y \hookrightarrow X$.
+    Then $f$ can be extended to a $G_{\delta}$ set.
+    $f$ and $\id_G$ agree on $Y$.
+    Hence $Y \subseteq G \subseteq  \overline{Y}$.
+    $Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????}
+    So $f = \id_G$, i.e.~$G = Y$.
+\end{proof}
+
+\subsection{Exercise 4}
+
+Let $C$ be the subspace of $2^{\omega}$ consisting
+of sequences with finitely many $1$s.
+We want to show that $C \cong \Q$.
+
+Go to the right in the even digits, go to the left for the odd digits,
+i.e.~let $C = (1,-1,1,-1, \ldots)$
+and set $x < y$ iff $C \cdot  x <_{\text{lex}} C \cdot y$.