\lecture{02}{2023-10-13}{Subsets of Polish spaces}

\begin{theorem}
    \label{subspacegdelta}
    A subspace of a Polish space is Polish
    iff it is $G_{\delta}$.
\end{theorem}

\begin{remark}
    Closed subsets of a metric space $(X, d )$
    are $G_{\delta}$.
\end{remark}
\begin{proof}
\gist{
    Let $C \subseteq X$ be closed.
    Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
    Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
    Let $x \in \bigcap U_{\frac{1}{n}}$.
    Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
    The $x_n$ converge to $x$ and since $C$ is closed,
    we get $x \in C$.
    Hence $C = \bigcap U_{\frac{1}{n}}$ 
    is $G_{\delta}$.
}{%
    For $C \overset{\text{closed}}{\subseteq} X$,
    we have $C = \bigcap_{n \in \N} \{x | d(x,C) < \frac{1}{n}\}$.
}
\end{proof}

\gist{
\begin{example}
    Let $ X$ be Polish.
    Let $d$ be a complete metric on $X$.
    \begin{enumerate}[a)]
        \item If $Y \subseteq  X$ is closed,
            then $(Y,d\defon{Y})$ is complete.
        \item $Y = (0,1) \subseteq \R$
           with the usual metric $d(x,y) = |x-y|$.
           Then  $x_n \to 0$ is Cauchy in $((0,1), d)$.

           But
           \[
               d_1(x,y) \coloneqq  | x -y|
                   + \left|\frac{1}{\min(x, 1- x)}
                   - \frac{1}{\min(y, 1-y)}
                   \right|
           \]
           also is a complete metric on $(0,1)$
           which is compatible with $d$.

           We want to generalize this idea.
    \end{enumerate}
\end{example}
}{}

\begin{refproof}{subspacegdelta}
    \begin{claim}
        \label{psubspacegdelta:c1}
        If $Y \subseteq (X,d)$ is $G_{\delta}$,
        then there exists a complete metric on $Y$.
    \end{claim}
    \begin{refproof}{psubspacegdelta:c1}
        Let $Y = U$ be open in $X$.
        Consider the map
        \begin{IEEEeqnarray*}{rCl}
            f_U\colon U &\longrightarrow &
                \underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
            x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
        \end{IEEEeqnarray*}

        Note that $X \times  \R$ with the
        \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
        metric is complete.

        $f_U$ is an embedding of $U$ into $X \times \R$%
        \gist{:
            \begin{itemize}
                \item It is injective because of the first coordinate.
                \item It is continuous since $d(x, U^c)$ is continuous
                    and only takes strictly positive values. % TODO
                \item The inverse is continuous because projections
                    are continuous.
            \end{itemize}
        }{.}

        \gist{%
            So we have shown that $U$ and
            the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
            are homeomorphic.
            The graph is closed \gist{in $U \times  \R$,
            because $\tilde{f_U}$ is continuous.
            It is closed}{} in $X \times \R$ \gist{because
            $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.

            Therefore we identified $U$ with a closed subspace of
            the Polish space $(X \times  \R, d_1)$.
        }{%
            So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
            and the RHS is a close subspace of the Polish space
            $(X \times \R, d_1)$.
        }

    \end{refproof}

    Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
    Consider
    \begin{IEEEeqnarray*}{rCl}
        f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
             x &\longmapsto &
                 \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
    \end{IEEEeqnarray*}

    \gist{
        As for an open $U$, $f_Y$ is an embedding.
        Since $X \times \R^{\N}$
        is completely metrizable,
        so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
    }{}

    \begin{claim}
        \label{psubspacegdelta:c2}
        If $Y \subseteq  (X,d)$ is completely metrizable,
        then $Y$ is a $G_{\delta}$ subspace.
    \end{claim}
    \begin{refproof}{psubspacegdelta:c2}
        There exists a complete metric  $d_Y$ on $Y$.
        For every $n$,
        let $V_n \subseteq  X$ be the union
        of all open sets $U \subseteq  X$ such that
        \begin{enumerate}[(i)]
            \item $U \cap Y \neq  \emptyset$,
            \item \gist{$\diam_d(U) \le \frac{1}{n}$,%
                \footnote{The proof gets a little easier if we bound by $\frac{1}{2^n}$ instead of $\frac{1}{n}$,
                    as that allows to simply take $U'_n \coloneqq \bigcup_{m > n} U_m$,
                    but both bounds work.}
                }{$\diam_d(U) \le \frac{1}{2^n}$.}
            \item \gist{%
                $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.}{%
                $\diam_{d_Y}(U \cap Y) \le \frac{1}{2^n}$.}
        \end{enumerate}
        \gist{%
            We want to show that $Y = \bigcap_{n \in \N} V_n$.
            For $x \in Y$, $n \in \N$ we have $x \in V_n$,
            as we can choose two neighbourhoods
            $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
            such that $\diam_{d_Y}(U) < \frac{1}{n}$
            and $U_2 \cap Y = U_1$.
            Additionally choose $x \in U_3$ open in $X$
            with $\diam_{d}(U_3) < \frac{1}{n}$.
            Then consider $U_2 \cap U_3 \subseteq V_n$.
            Hence $Y \subseteq  \bigcap_{n \in \N}  V_n$.

            Now let $x \in \bigcap_{n \in \N} V_n$.
            For each $n$ pick $x \in U_n \subseteq X$ open
            satisfying (i), (ii), (iii).
            From (i) and (ii) it follows that $x \in \overline{Y}$,
            since we can consider a sequence of points $y_n \in U_n \cap Y$
            and get $y_n \xrightarrow{d} x$.
            For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
            is an open set containing $x$,
            hence $U_n' \cap Y \neq \emptyset$.
            Thus we may assume that the $U_i$ form a decreasing sequence.
            We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
            If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
            since $\diam(U_n \cap Y) \xrightarrow{d_Y}  0$
            and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y}  0$.
            The sequence $y_n$ converges to the unique point in
            $\bigcap_{n} \overline{U_n \cap Y}$.
            Since the topologies agree, this point is $x$.
        }{Then $Y = \bigcap_n U_n$.}
    \end{refproof}
\end{refproof}