tutorial 03
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\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 $C \cdot x <_{\text{lex}} C \cdot y$.