Josia Pietsch 3df55b6516
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\lecture{02}{2023-10-13}{Subsets of Polish spaces}
A subspace of a Polish space is Polish
iff it is $G_{\delta}$.
Closed subsets of a metric space $(X, d )$
are $G_{\delta}$.
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 $C = \bigcap_{n \in \N} \{x | d(x,C) < \frac{1}{n}\}$.
Let $ X$ be Polish.
Let $d$ be a complete metric on $X$.
\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 $((0,1), d)$.
d_1(x,y) \coloneqq | x -y|
+ \left|\frac{1}{\min(x, 1- x)}
- \frac{1}{\min(y, 1-y)}
also is a complete metric on $(0,1)$
which is compatible with $d$.
We want to generalize this idea.
If $Y \subseteq (X,d)$ is $G_{\delta}$,
then there exists a complete metric on $Y$.
Let $Y = U$ be open in $X$.
Consider the map
f_U\colon U &\longrightarrow &
\underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
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 $X \times \R$%
\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.
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
$(X \times \R, d_1)$.
Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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)
As for an open $U$, $f_Y$ is an embedding.
Since $X \times \R^{\N}$
is completely metrizable,
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
If $Y \subseteq (X,d)$ is completely metrizable,
then $Y$ is a $G_{\delta}$ subspace.
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
\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{%
$\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.}{%
$\diam_{d_Y}(U \cap Y) \le \frac{1}{2^n}$.}
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 $Y = \bigcap_n U_n$.}