\ctutorial{03}{2023-10-31}{}

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\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 $C \cdot  x <_{\text{lex}} C \cdot y$.