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E70B571D66986A2D
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@ -12,7 +12,7 @@ jobs:
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- name: Prepare pages
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run: |
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mkdir public
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mv build/logic3.pdf build/logic3.log README.md public
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mv build/*.pdf build/*.log README.md public
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- name: Deploy to pages
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uses: actions/pages@v1
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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
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such that $f: X \to f(X)$ is a homeomorphism.
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\end{proposition}
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\begin{proof}
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$X$ is separable, so it has some countable dense subset,
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which we order as a sequence $(x_n)_{n \in \omega}$.
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\gist{$X$ is separable, so it has some countable dense subset,
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which we order as a sequence $(x_n)_{n \in \omega}$.}%
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{Let $(x_n)_{n \in \omega}$ be a countable dense subset.}
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Let $d$ be a metric on $X$ which is compatible with the topology.
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W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).
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Let $d$ be the metric of $X$ and define
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\gist{Let $d$ be a metric on $X$ which is compatible with the topology.
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W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}%
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{Let $d \le 1$ be a metric of $X$.}
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Define
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\begin{IEEEeqnarray*}{rCl}
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f\colon X &\longrightarrow & [0,1]^{\omega} \\
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x&\longmapsto & (d(x,x_n))_{n < \omega}
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\end{IEEEeqnarray*}
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\gist{
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\begin{claim}
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$f$ is injective.
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\end{claim}
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@ -237,17 +238,21 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
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is open\footnote{as a subset of $f(X)$!}.
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\end{subproof}
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}{}
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\end{proof}
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\begin{proposition}
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Countable disjoint unions of Polish spaces are Polish.
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\end{proposition}
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\gist{
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\begin{proof}
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Define a metric in the obvious way.
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\end{proof}
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}{}
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\begin{proposition}
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Closed subspaces of Polish spaces are Polish.
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\end{proposition}
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\gist{}{
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\begin{proof}
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Let $X$ be Polish and $V \subseteq X$ closed.
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Let $d$ be a complete metric on $X$.
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@ -255,15 +260,14 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Subspaces of second countable spaces
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are second countable.
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\end{proof}
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}
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\begin{definition}
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Let $X$ be a topological space.
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A subspace $A \subseteq X$ is called
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$G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets.
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\end{definition}
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\gist{
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Next time: Closed sets are $G_\delta$.
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A subspace of a Polish space is Polish iff it is $G_{\delta}$
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}{}
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@ -1,148 +1,158 @@
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\lecture{02}{2023-10-13}{Subsets of Polish spaces}
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\begin{theorem}
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\label{subspacegdelta}
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A subspace of a Polish space is Polish
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iff it is $G_{\delta}$.
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\end{theorem}
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\begin{theorem}
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\label{subspacegdelta}
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A subspace of a Polish space is Polish
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iff it is $G_{\delta}$.
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\end{theorem}
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\begin{remark}
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Closed subsets of a metric space $(X, d )$
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are $G_{\delta}$.
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\end{remark}
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\begin{proof}
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Let $C \subseteq X$ be closed.
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Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
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Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
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Let $x \in \bigcap U_{\frac{1}{n}}$.
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Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
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The $x_n$ converge to $x$ and since $C$ is closed,
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we get $x \in C$.
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Hence $C = \bigcap U_{\frac{1}{n}}$
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is $G_{\delta}$.
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\end{proof}
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\begin{remark}
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Closed subsets of a metric space $(X, d )$
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are $G_{\delta}$.
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\end{remark}
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\begin{proof}
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\gist{
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Let $C \subseteq X$ be closed.
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Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
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Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
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Let $x \in \bigcap U_{\frac{1}{n}}$.
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Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
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The $x_n$ converge to $x$ and since $C$ is closed,
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we get $x \in C$.
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Hence $C = \bigcap U_{\frac{1}{n}}$
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is $G_{\delta}$.
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}{%
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For $C \overset{\text{closed}}{\subseteq} X$,
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we have $C = \bigcap_{n \in \N} \{x | d(x,C) < \frac{1}{n}\}$.
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}
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\end{proof}
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\begin{example}
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Let $ X$ be Polish.
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Let $d$ be a complete metric on $X$.
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\begin{enumerate}[a)]
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\item If $Y \subseteq X$ is closed,
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then $(Y,d\defon{Y})$ is complete.
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\item $Y = (0,1) \subseteq \R$
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with the usual metric $d(x,y) = |x-y|$.
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Then $x_n \to 0$ is Cauchy in $((0,1), d)$.
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\gist{
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\begin{example}
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Let $ X$ be Polish.
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Let $d$ be a complete metric on $X$.
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\begin{enumerate}[a)]
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\item If $Y \subseteq X$ is closed,
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then $(Y,d\defon{Y})$ is complete.
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\item $Y = (0,1) \subseteq \R$
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with the usual metric $d(x,y) = |x-y|$.
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Then $x_n \to 0$ is Cauchy in $((0,1), d)$.
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But
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\[
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d_1(x,y) \coloneqq | x -y|
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+ \left|\frac{1}{\min(x, 1- x)}
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- \frac{1}{\min(y, 1-y)}
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\right|
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\]
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also is a complete metric on $(0,1)$
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which is compatible with $d$.
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But
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\[
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d_1(x,y) \coloneqq | x -y|
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+ \left|\frac{1}{\min(x, 1- x)}
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- \frac{1}{\min(y, 1-y)}
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\right|
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\]
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also is a complete metric on $(0,1)$
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which is compatible with $d$.
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We want to generalize this idea.
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\end{enumerate}
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\end{example}
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We want to generalize this idea.
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\end{enumerate}
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\end{example}
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}{}
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\begin{refproof}{subspacegdelta}
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\begin{claim}
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\label{psubspacegdelta:c1}
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If $Y \subseteq (X,d)$ is $G_{\delta}$,
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then there exists a complete metric on $Y$.
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\end{claim}
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\begin{refproof}{psubspacegdelta:c1}
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Let $Y = U$ be open in $X$.
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Consider the map
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\begin{IEEEeqnarray*}{rCl}
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f_U\colon U &\longrightarrow &
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\underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
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x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
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\end{IEEEeqnarray*}
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\begin{refproof}{subspacegdelta}
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\begin{claim}
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\label{psubspacegdelta:c1}
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If $Y \subseteq (X,d)$ is $G_{\delta}$,
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then there exists a complete metric on $Y$.
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\end{claim}
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\begin{refproof}{psubspacegdelta:c1}
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Let $Y = U$ be open in $X$.
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Consider the map
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\begin{IEEEeqnarray*}{rCl}
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f_U\colon U &\longrightarrow &
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\underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
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x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
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\end{IEEEeqnarray*}
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Note that $X \times \R$ with the
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\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
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metric is complete.
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Note that $X \times \R$ with the
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\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
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metric is complete.
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$f_U$ is an embedding of $U$ into $X \times \R$:
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\begin{itemize}
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\item It is injective because of the first coordinate.
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\item It is continuous since $d(x, U^c)$ is continuous
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and only takes strictly positive values. % TODO
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\item The inverse is continuous because projections
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are continuous.
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\end{itemize}
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$f_U$ is an embedding of $U$ into $X \times \R$\gist{:
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\begin{itemize}
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\item It is injective because of the first coordinate.
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\item It is continuous since $d(x, U^c)$ is continuous
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and only takes strictly positive values. % TODO
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\item The inverse is continuous because projections
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are continuous.
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\end{itemize}
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}{.}
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So we have shown that $U$ is homeomorphic to % TODO with ?
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the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$.
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The graph is closed in $U \times \R$,
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because $\tilde{f_U}$ is continuous.
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It is closed in $X \times \R$ because
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$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$.
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\todo{Make this precise}
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So we have shown that $U$ and
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the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
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are homeomorphic.
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The graph is closed \gist{in $U \times \R$,
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because $\tilde{f_U}$ is continuous.
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It is closed}{} in $X \times \R$ \gist{because
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$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
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\todo{Make this precise}
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Therefore we identified $U$ with a closed subspace of
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the Polish space $(X \times \R, d_1)$.
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\end{refproof}
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Therefore we identified $U$ with a closed subspace of
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the Polish space $(X \times \R, d_1)$.
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\end{refproof}
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Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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Take
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\begin{IEEEeqnarray*}{rCl}
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f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
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x &\longmapsto &
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\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
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\end{IEEEeqnarray*}
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Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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Take
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\begin{IEEEeqnarray*}{rCl}
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f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
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x &\longmapsto &
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\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
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\end{IEEEeqnarray*}
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As for an open $U$, $f_Y$ is an embedding.
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Since $X \times \R^{\N}$
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is completely metrizable,
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
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As for an open $U$, $f_Y$ is an embedding.
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Since $X \times \R^{\N}$
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is completely metrizable,
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
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\begin{claim}
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\label{psubspacegdelta:c2}
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If $Y \subseteq (X,d)$ is completely metrizable,
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then $Y$ is a $G_{\delta}$ subspace.
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\end{claim}
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\begin{refproof}{psubspacegdelta:c2}
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There exists a complete metric $d_Y$ on $Y$.
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For every $n$,
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let $V_n \subseteq X$ be the union
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of all open sets $U \subseteq X$ such that
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\begin{enumerate}[(i)]
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\item $U \cap Y \neq \emptyset$,
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\item $\diam_d(U) \le \frac{1}{n}$,
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\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
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\end{enumerate}
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\begin{claim}
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\label{psubspacegdelta:c2}
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If $Y \subseteq (X,d)$ is completely metrizable,
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then $Y$ is a $G_{\delta}$ subspace.
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\end{claim}
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\begin{refproof}{psubspacegdelta:c2}
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There exists a complete metric $d_Y$ on $Y$.
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For every $n$,
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let $V_n \subseteq X$ be the union
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of all open sets $U \subseteq X$ such that
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\begin{enumerate}[(i)]
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\item $U \cap Y \neq \emptyset$,
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\item $\diam_d(U) \le \frac{1}{n}$,
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\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
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\end{enumerate}
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\gist{
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We want to show that $Y = \bigcap_{n \in \N} V_n$.
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For $x \in Y$, $n \in \N$ we have $x \in V_n$,
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as we can choose two neighbourhoods
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$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
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such that $\diam_{d_Y}(U) < \frac{1}{n}$
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and $U_2 \cap Y = U_1$.
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Additionally choose $x \in U_3$ open in $X$
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with $\diam_{d}(U_3) < \frac{1}{n}$.
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Then consider $U_2 \cap U_3 \subseteq V_n$.
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Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
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We want to show that $Y = \bigcap_{n \in \N} V_n$.
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For $x \in Y$, $n \in \N$ we have $x \in V_n$,
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as we can choose two neighbourhoods
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$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
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such that $\diam_{d_Y}(U) < \frac{1}{n}$
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and $U_2 \cap Y = U_1$.
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Additionally choose $x \in U_3$ open in $X$
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with $\diam_{d}(U_3) < \frac{1}{n}$.
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Then consider $U_2 \cap U_3 \subseteq V_n$.
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Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
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Now let $x \in \bigcap_{n \in \N} V_n$.
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For each $n$ pick $x \in U_n \subseteq X$ open
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satisfying (i), (ii), (iii).
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From (i) and (ii) it follows that $x \in \overline{Y}$,
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since we can consider a sequence of points $y_n \in U_n \cap Y$
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and get $y_n \xrightarrow{d} x$.
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For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
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is an open set containing $x$,
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hence $U_n' \cap Y \neq \emptyset$.
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Thus we may assume that the $U_i$ form a decreasing sequence.
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We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
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If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
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and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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The sequence $y_n$ converges to the unique point in
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$\bigcap_{n} \overline{U_n \cap Y}$.
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Since the topologies agree, this point is $x$.
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\end{refproof}
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\end{refproof}
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Now let $x \in \bigcap_{n \in \N} V_n$.
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For each $n$ pick $x \in U_n \subseteq X$ open
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satisfying (i), (ii), (iii).
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From (i) and (ii) it follows that $x \in \overline{Y}$,
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since we can consider a sequence of points $y_n \in U_n \cap Y$
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and get $y_n \xrightarrow{d} x$.
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For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
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is an open set containing $x$,
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hence $U_n' \cap Y \neq \emptyset$.
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Thus we may assume that the $U_i$ form a decreasing sequence.
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We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
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If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
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and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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The sequence $y_n$ converges to the unique point in
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$\bigcap_{n} \overline{U_n \cap Y}$.
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Since the topologies agree, this point is $x$.
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}{Then $Y = \bigcap_n U_n$.}
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\end{refproof}
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\end{refproof}
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@ -163,7 +163,6 @@ For this we define
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is dense in $\overline{x} \mapsto f(\overline{x})$.
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Since the flow is distal, it suffices to show
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that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
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% TODO REF Distal flow can be decomposed into disjoint minimal flows
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\item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
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Consider the flows we get from $(f_i)_{i < j}$
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