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\lecture{02}{20231013}{Subsets of Polish spaces}





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\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 $d(x,y) = xy$.




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, 1y)}




\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.





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$f_U$ 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}




}{.}

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\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




$(X \times \R, d_1)$.




}

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\end{refproof}








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

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Consider

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\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*}





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\gist{




As for an open $U$, $f_Y$ is an embedding.




Since $X \times \R^{\N}$




is completely metrizable,




so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.




}{}

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\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 $U \cap Y \neq \emptyset$,

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\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{%




$\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.}{%




$\diam_{d_Y}(U \cap Y) \le \frac{1}{2^n}$.}

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\end{enumerate}

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\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 $Y \subseteq \bigcap_{n \in \N} V_n$.

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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 $Y = \bigcap_n U_n$.}

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\end{refproof}




\end{refproof}
