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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:



@ 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 $G_{\delta}$








}{}



@ 1,148 +1,158 @@


\lecture{02}{20231013}{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 $d(x,y) = xy$.


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 $d(x,y) = xy$.


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 $d$.


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}


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}





@ 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 $(f_i)_{i < j}$



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