\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}
     Let $C \subseteq X$ be closed.
     Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
     \todo{Exercise}
     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 $G_{\delta}$.
 \end{proof}

 \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 $X \times \R$:
         \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}

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

         Therefore we identified $U$ with a closed subspace of
         the Polish space $(X \times  \R, d_1)$.
     \end{refproof}

     Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
     Take
     \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*}

     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 $\diam_d(U) \le \frac{1}{n}$,
             \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
         \end{enumerate}

         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 $x$.
     \end{refproof}
 \end{refproof}

