**Josia Pietsch**82f0dfd4de

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271 lines
8.4 KiB
TeX

271 lines

8.4 KiB

TeX

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\subsection{The Ellis semigroup}
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\lecture{17}{2023-12-12}{The Ellis semigroup}
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Let $(X, d)$ be a compact metric space
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and $(X, T)$ a flow.
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Let $X^{X} \coloneqq \{f\colon X \to X\}$
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be the set of all functions.\footnote{We take all the functions,
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they need not be continuous.}
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We equip this with the product topology,
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i.e.~a subbasis
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is given by sets
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\[
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U_{\epsilon}(x,y) \coloneqq \{f \in X^X : d(x,f(y)) < \epsilon\}.
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\]
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for all $x,y \in X$, $\epsilon > 0$.
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$X^{X}$ is a compact Hausdorff space.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}}
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\begin{remark}%
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\footnote{cf.~\yaref{s11e1}}
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Let $f_0 \in X^X$ be fixed.
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\begin{itemize}
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\item $X^X \ni f \mapsto f \circ f_0$
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is continuous:
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Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
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We have $f \circ f_0 \in U_{\epsilon}(x,y)$
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iff $f \in U_\epsilon(x,f_0(y))$.
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\item Fix $x_0 \in X$.
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Then $f \mapsto f(x_0)$ is continuous.
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\item In general $f \mapsto f_0 \circ f$ is not continuous,
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but if $f_0$ is continuous, then the map is continuous.
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\end{itemize}
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\end{remark}
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\begin{definition}
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\gist{%
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Let $(X,T)$ be a flow.
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Then the \vocab{Ellis semigroup}
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is defined by
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$E(X,T) \coloneqq \overline{T} \subseteq X^X$,
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i.e.~identify $t \in T$ with $x \mapsto tx$
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and take the closure in $X^X$.
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}{%
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The \vocab{Ellis semigroup} of a flow $(X,T)$
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is $E(X,T) \coloneqq \overline{T} \subseteq X^X$.
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}
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\end{definition}
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$E(X,T)$ is compact and Hausdorff,
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since $X^X$ has these properties.
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\gist{
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Properties of $(X,T)$ translate to properties of $E(X,T)$:
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\begin{goal}
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We want to show that if $(X,T)$ is distal,
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then $E(X,T)$ is a group.
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\end{goal}
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}{}
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\begin{proposition}
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$E(X,T)$ is a semigroup,
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i.e.~closed under composition.
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\end{proposition}
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\begin{proof}
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\gist{
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Let $G \coloneqq E(X,T)$.
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Take $t \in T$. We want to show that $tG \subseteq G$,
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i.e.~for all $h \in G$ we have $th \in G$.
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We have that $t^{-1}G$ is compact,
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since $t^{-1}$ is continuous
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and $G$ is compact.
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It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
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So $G = \overline{T} \subseteq t^{-1}G$.
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Hence $tG \subseteq G$.
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\begin{claim}
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If $g \in G$, then
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\[
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\overline{T} g = \overline{Tg}.
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\]
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\end{claim}
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\begin{subproof}
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Cf.~\yaref{s11e1}
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\end{subproof}
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Let $g \in G$.
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We need to show that $Gg \subseteq G$.
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It is
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\[
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Gg = \overline{T}g = \overline{Tg}.
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\]
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Since $G$ compact,
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and $Tg \subseteq G$,
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we have $ \overline{Tg} \subseteq G$.
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}{
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$G \coloneqq E(X,T)$.
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\begin{itemize}
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\item $\forall t \in T. ~ tG \subseteq G$:
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\begin{itemize}
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\item $t^{-1}G$ is compact.
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\item $T \subseteq t^{-1}G$,
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\item $\leadsto G = \overline{T} \subseteq t^{-1}G$,
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i.e.~$tG \subseteq G$.
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\end{itemize}
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\item $\forall g \in G.~\overline{T}g = \overline{Tg}$ (cts.~map from compact to Hausdorff)
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\item $\forall g \in G.~Gg \subseteq G$ :
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$Gg = \overline{T}g = \overline{Tg} \overset{G \text{ compact}, Tg \subseteq G}{\subseteq} G$.
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\end{itemize}
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}
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\end{proof}
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\begin{definition}
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A \vocab{compact semigroup} $S$
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is a nonempty semigroup\footnote{may not contain inverses or the identity}
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with a compact Hausdorff topology,
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such that $S \ni x \mapsto xs$ is continuous for all $s$.
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\end{definition}
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\gist{
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\begin{example}
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The Ellis semigroup is a compact semigroup.
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\end{example}
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}{}
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\begin{lemma}[Ellis–Numakura]
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\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
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Every non-empty compact semigroup
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contains an \vocab{idempotent} element,
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i.e.~$f$ such that $f^2 = f$.
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\end{lemma}
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\begin{proof}
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\gist{
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Using Zorn's lemma, take a $\subseteq$-minimal
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compact subsemigroup $R$ of $S$
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and let $s \in R$.
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Then $Rs$ is also a compact subsemigroup
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and $Rs \subseteq R$.
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By minimality of $R$, $R = Rs$.
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Let $P \coloneqq \{ x \in R : xs = s\}$.
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Then $P \neq \emptyset$,
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since $s \in Rs$
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and $P$ is a compact semigroup,
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since $x \overset{\alpha}\mapsto xs$
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is continuous and $P = \alpha^{-1}(s) \cap R$.
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Thus $P = R$ by minimality,
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so $s \in P$,
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i.e.~$s^2 = s$.
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}{
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\begin{itemize}
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\item Take $R \subseteq S$ minimal compact subsemigroup (Zorn),
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$s \in R$.
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\item $Rs \subseteq R \implies Rs = R$.
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\item $P \coloneqq \{x \in R : xs = s\}$:
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\begin{itemize}
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\item $P \neq \emptyset$, since $s \in Rs$
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\item $P$ compact, since $P = \alpha^{-1}(s) \cap R$,
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$\alpha: x \mapsto xs$ cts.
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\item $P = R \implies s^2 = s$.
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\end{itemize}
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\end{itemize}%
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}
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\end{proof}
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\gist{
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The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
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since we already know that it has an identity.
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%in fact we might have chosen $R = \{1\}$ in the proof.
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But it is interesting for other semigroups.
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}{}
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\begin{theorem}[Ellis]
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$(X,T)$ is distal iff $E(X,T)$ is a group.
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\end{theorem}
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\begin{proof}
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Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.
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\gist{
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For all $g \in G$ we need to show that $x \mapsto gx$ is injective.
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If we had $gx = gy$, then $d(gx,gy) = 0$.
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Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal,
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hence $x = y$.
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Let $g \in G$. Consider the compact semigroup $\Gamma \coloneqq Gg$.
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By the \yaref{lem:ellisnumakura},
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there is $f \in \Gamma$ such that $f^2 = f$,
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i.e.~for all $x \in X$ we have $f^2(x) = f(x)$.
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Since $f$ is injective, we get that $x = f(x)$,
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i.e.~$f = \id$.
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Since $f \in Gg$, there exists $g' \in G$ such that $f = g' \circ g$.
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It is $g' = g'gg'$,
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so $\forall x .~g'(x) = g'(g g'(x))$.
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Hence $g'$ is injective
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and $x = gg'(x)$,
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i.e.~$g g' = \id$.
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}{
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\begin{itemize}
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\item $x \mapsto gx$ injective for all $g \in G$:
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\[gx = gy
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\implies d(gx,gy) = 0
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\implies \inf_{t \in T} d(tx, ty) = 0
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\overset{\text{distal}}{\implies} x = y.
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\]
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\item Fix $g \in G$.
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\begin{itemize}
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\item $\Gamma \coloneqq Gg$ is a compact semigroup.
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\item $\exists f\in \Gamma.~f^2 = f$ (\yaref{lem:ellisnumakura})
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\item $f$ is injective, hence $f = \id$.
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\item Take $g'$ such that $g'g = f$. $g'gg' = g' \implies gg' = \id$.
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\end{itemize}
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\end{itemize}
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}
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\gist{
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On the other hand if $(x_0,x_1)$ is proximal,
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then there exists $g \in G$ such that $gx_0 = gx_1$.%
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\footnote{cf.~\yaref{s11e1} (e)}
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It follows that an inverse to $g$ can not exist.
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}{If $(x_0,x_1)$ is proximal, there is $g \in G$ with $gx_0 = gx_1$, i.e.~no inverse to $g$.}
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\end{proof}
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% Let $(X,T)$ be a flow.
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% Then by Zorn's lemma, there exists $X_0 \subseteq X$
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% such that $(X_0, T)$ is minimal.
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% In particular,
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% for $x \in X$ and $\overline{Tx} = Y$
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% we have that $(Y,T)$ is a flow.
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% However if we pick $y \in Y$, $Ty$ might not be dense.
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% TODO: question!
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% TODO: think about this!
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% We want to a minimal subflow in a nice way:
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\begin{theorem}
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\label{thm:distalflowpartition}
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If $(X,T)$ is distal, then $X$ is the disjoint union of minimal subflows.
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In fact those disjoint sets
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will be orbits of $E(X,T)$.
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\end{theorem}
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\begin{proof}
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Let $G = E(X,T)$.
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\gist{
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Note that for all $x \in X$,
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we have that $Gx \subseteq X$ is compact
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and invariant under the action of $G$.
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Since $G$ is a group, the orbits partition $X$.%
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\footnote{Note that in general this does not hold for semigroups.}
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We need to show that $(Gx, T)$ is minimal.
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Suppose that $y \in Gx$, i.e.~$Gx = Gy$.
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Since $g \mapsto gy$ is continuous,
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we have $Gx = Gy = \overline{T}y = \overline{Ty}$,
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so $Ty$ is dense in $Gx$.
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}{
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\begin{itemize}
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\item $G$ is a group, so the $G$-orbits partition $X$.
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\item Let $y \in Gx$, then $Gx \overset{y \in Gx}{=} Gy \overset{\text{def}}{=} \overline{T}y \overset{g \mapsto gy \text{ cts.}}{=} \overline{Ty}$,
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i.e.~$(Gx,T)$ is minimal.
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\end{itemize}
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}
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\end{proof}
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\begin{corollary}
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If $(X,T)$ is distal and minimal,
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then $E(X,T) \acts X$ is transitive.
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\end{corollary}
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