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4 changed files with 158 additions and 145 deletions

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@ -12,7 +12,7 @@ jobs:
- name: Prepare pages
run: |
mkdir public
mv build/logic3.pdf build/logic3.log README.md public
mv build/*.pdf build/*.log README.md public
- name: Deploy to pages
uses: actions/pages@v1
with:

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@ -198,18 +198,19 @@ suffices to show that open balls in one metric are unions of open balls in the o
such that $f: X \to f(X)$ is a homeomorphism.
\end{proposition}
\begin{proof}
$X$ is separable, so it has some countable dense subset,
which we order as a sequence $(x_n)_{n \in \omega}$.
\gist{$X$ is separable, so it has some countable dense subset,
which we order as a sequence $(x_n)_{n \in \omega}$.}%
{Let $(x_n)_{n \in \omega}$ be a countable dense subset.}
Let $d$ be a metric on $X$ which is compatible with the topology.
W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).
Let $d$ be the metric of $X$ and define
\gist{Let $d$ be a metric on $X$ which is compatible with the topology.
W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}%
{Let $d \le 1$ be a metric of $X$.}
Define
\begin{IEEEeqnarray*}{rCl}
f\colon X &\longrightarrow & [0,1]^{\omega} \\
x&\longmapsto & (d(x,x_n))_{n < \omega}
\end{IEEEeqnarray*}
\gist{
\begin{claim}
$f$ is injective.
\end{claim}
@ -237,17 +238,21 @@ suffices to show that open balls in one metric are unions of open balls in the o
Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
is open\footnote{as a subset of $f(X)$!}.
\end{subproof}
}{}
\end{proof}
\begin{proposition}
Countable disjoint unions of Polish spaces are Polish.
\end{proposition}
\gist{
\begin{proof}
Define a metric in the obvious way.
\end{proof}
}{}
\begin{proposition}
Closed subspaces of Polish spaces are Polish.
\end{proposition}
\gist{}{
\begin{proof}
Let $X$ be Polish and $V \subseteq X$ closed.
Let $d$ be a complete metric on $X$.
@ -255,15 +260,14 @@ suffices to show that open balls in one metric are unions of open balls in the o
Subspaces of second countable spaces
are second countable.
\end{proof}
}
\begin{definition}
Let $X$ be a topological space.
A subspace $A \subseteq X$ is called
$G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets.
\end{definition}
\gist{
Next time: Closed sets are $G_\delta$.
A subspace of a Polish space is Polish iff it is $G_{\delta}$
}{}

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@ -1,148 +1,158 @@
\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{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 $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
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{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}
\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 $((0,1), d)$.
\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 $d$.
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}
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*}
\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.
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}
$f_U$ 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}
}{.}
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}
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 $d(x, U^c) \to 0$}{}.
\todo{Make this precise}
Therefore we identified $U$ with a closed subspace of
the Polish space $(X \times \R, d_1)$.
\end{refproof}
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*}
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 $f_Y(Y) \subseteq X \times \R^\N$.
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}
\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}
\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 $Y \subseteq \bigcap_{n \in \N} V_n$.
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}
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}

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@ -163,7 +163,6 @@ For this we define
is dense in $\overline{x} \mapsto f(\overline{x})$.
Since the flow is distal, it suffices to show
that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
% TODO REF Distal flow can be decomposed into disjoint minimal flows
\item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
Consider the flows we get from $(f_i)_{i < j}$