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@ -233,6 +233,7 @@ Recall:
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correspond to metrics witnessing that the flow is isometric.
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\end{remark}
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\begin{proposition}
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\label{prop:isomextdistal}
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An isometric extension of a distal flow is distal.
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\end{proposition}
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\begin{proof}
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@ -263,11 +264,12 @@ Recall:
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% TODO THE inverse limit is A limit
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of $\Sigma$ iff
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\[
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\forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
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\forall x_1 \neq x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
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\]
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\end{definition}
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\begin{proposition}
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\label{prop:limitdistal}
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A limit of distal flows is distal.
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\end{proposition}
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\begin{proof}
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@ -1,5 +1,4 @@
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\lecture{16}{2023-12-08}{}
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% TODO ANKI-MARKER
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$X$ is always compact metrizable.
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@ -18,16 +17,19 @@ $X$ is always compact metrizable.
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% and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
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% \end{example}
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\begin{proof}
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% TODO TODO TODO Think!
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The action of $1$ determines $h$.
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Consider
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\[
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\{h^n : n \in \Z\} \subseteq \cC(X,X) = \{f\colon X \to X : f \text{ continuous}\},
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\{h^n : n \in \Z\} \subseteq \cC(X,X)\gist{ = \{f\colon X \to X : f \text{ continuous}\}}{},
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\]
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where the topology is the uniform convergence topology. % TODO REF EXERCISE
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Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
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Since
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Since the family $\{h^n : n \in \Z\}$ is uniformly equicontinuous,
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i.e.~
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\[
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\forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon
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\forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon,
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% Here we use isometric
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\]
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we have by the Arzel\`a-Ascoli-Theorem % TODO REF
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that $G$ is compact.
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@ -126,8 +128,8 @@ $X$ is always compact metrizable.
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Every quasi-isometric flow is distal.
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\end{corollary}
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\begin{proof}
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\todo{TODO}
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% The trivial flow is distal.
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The trivial flow is distal.
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Apply \yaref{prop:isomextdistal} and \yaref{prop:limitdistal}.
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\end{proof}
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\begin{theorem}[Furstenberg]
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@ -199,7 +201,8 @@ The Hilbert cube $\bH = [0,1]^{\N}$
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embeds all compact metric spaces.
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Thus we can consider $K(\bH)$,
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the space of compact subsets of $\bH$.
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$K(\bH)$ is a Polish space.\todo{Exercise}
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$K(\bH)$ is a Polish space.\footnote{cf.~\yaref{s9e2}, \yaref{s12e4}}
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% TODO LEARN EXERCISES
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Consider $K(\bH^2)$.
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A flow $\Z \acts X$ corresponds to the graph of
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\begin{IEEEeqnarray*}{rCl}
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@ -1,4 +1,5 @@
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\subsection{The Ellis semigroup}
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% TODO ANKI-MARKER
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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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@ -68,61 +68,62 @@ This will follow from the following lemma:
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we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
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\end{refproof}
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\begin{refproof}{lem:ftophelper}%
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\footnote{This was not covered in class.}
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\notexaminable{\footnote{This was not covered in class.}
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Let $T = \bigcup_n T_n$,% TODO Why does this exist?
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$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
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let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
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Take $b$ such that $F(x,x') < b < a$.
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Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
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is open in $G(x,x')$
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and since $F(x,x') < b$ we have $U \neq \emptyset$.
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\begin{claim}
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There exists $n$ such that
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\[
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\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
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\]
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\end{claim}
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\begin{subproof}
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Suppose not.
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Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
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with
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\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
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Note that the RHS is closed.
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For $m > n$ we have
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$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
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since $T_n \subseteq T_m$.
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By compactness of $X$,
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there exists $v,v'$ and some subsequence
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such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
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Let $T = \bigcup_n T_n$,% TODO Why does this exist?
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$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
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let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
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Take $b$ such that $F(x,x') < b < a$.
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Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
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is open in $G(x,x')$
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and since $F(x,x') < b$ we have $U \neq \emptyset$.
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\begin{claim}
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There exists $n$ such that
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\[
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\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
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\]
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\end{claim}
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\begin{subproof}
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Suppose not.
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Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
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with
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\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
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Note that the RHS is closed.
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For $m > n$ we have
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$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
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since $T_n \subseteq T_m$.
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By compactness of $X$,
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there exists $v,v'$ and some subsequence
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such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
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So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
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hence $T(v,v') \cap U = \emptyset$,
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so $G(v,v') \cap U = \emptyset$.
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But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
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\end{subproof}
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The map
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\begin{IEEEeqnarray*}{rCl}
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T\times X&\longrightarrow & X \\
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(t,x) &\longmapsto & tx
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\end{IEEEeqnarray*}
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is continuous.
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Since $T_n$ is compact,
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we have that $\{(x,t) \mapsto tx : t \in T_n\}$
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is equicontinuous.\todo{Sheet 11}
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So there is $\epsilon > 0$ such that
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$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
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for all $t \in T_n$.
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So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
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hence $T(v,v') \cap U = \emptyset$,
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so $G(v,v') \cap U = \emptyset$.
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But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
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\end{subproof}
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The map
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\begin{IEEEeqnarray*}{rCl}
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T\times X&\longrightarrow & X \\
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(t,x) &\longmapsto & tx
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\end{IEEEeqnarray*}
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is continuous.
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Since $T_n$ is compact,
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we have that $\{(x,t) \mapsto tx : t \in T_n\}$
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is equicontinuous.\todo{Sheet 11}
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So there is $\epsilon > 0$ such that
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$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
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for all $t \in T_n$.
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Suppose now that $F(x', x'') < \epsilon$.
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Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
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hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
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Since $(t_0x, t_0x') \in G(x,x')$,
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there is $t_1 \in T_n$
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with $(t_1t_0x, t_1t_0x') \in U$,
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i.e.~$d(t_1t_0x, t_1t_0x') < b$
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and therefore
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$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
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Suppose now that $F(x', x'') < \epsilon$.
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Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
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hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
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Since $(t_0x, t_0x') \in G(x,x')$,
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there is $t_1 \in T_n$
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with $(t_1t_0x, t_1t_0x') \in U$,
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i.e.~$d(t_1t_0x, t_1t_0x') < b$
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and therefore
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$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
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}
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\end{refproof}
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Now assume $Z = \{\star\}$.
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@ -145,48 +145,54 @@ For this we define
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This is the same as for iterated skew shifts.
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% TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
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% $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
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\item Minimality:
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\item Minimality:%
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\gist{%
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\footnote{This is not relevant for the exam.}
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Let $\langle E_n : n < \omega \rangle$
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be an enumeration of a countable basis for $\mathbb{K}^I$.
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Let $\langle E_n : n < \omega \rangle$
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be an enumeration of a countable basis for $\mathbb{K}^I$.
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For all $n$ let
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\[
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U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
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\]
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where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
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For all $n$ let
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\[
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U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
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\]
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where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
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Beleznay and Foreman showed that $U_n$ is open
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and dense for all $n$.
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Beleznay and Foreman showed that $U_n$ is open
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and dense for all $n$.
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So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$
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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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So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$
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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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}{ Not relevant for the exam.}
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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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resp.~$(f_i)_{i \le j}$
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denoted by $X_{<j}$ resp.~$X_{\le j}$.
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We aim to show that $X_{\le j} \to X_{<j}$
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is a maximal isometric extension for comeagerly many $\overline{f}$.
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\item The order of the flow is $\eta$:%
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\gist{%
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\footnote{This is not relevant for the exam.}
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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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resp.~$(f_i)_{i \le j}$
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denoted by $X_{<j}$ resp.~$X_{\le j}$.
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We aim to show that $X_{\le j} \to X_{<j}$
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is a maximal isometric extension for comeagerly many $\overline{f}$.
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The following open dense sets are used to make sure that all isometric extensions
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are maximal and hence the order of the flow is $\eta$:
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The following open dense sets are used to make sure that all isometric extensions
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are maximal and hence the order of the flow is $\eta$:
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Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
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For $\epsilon \in \Q$ let
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\begin{IEEEeqnarray*}{rCl}
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V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\
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&&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
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&&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
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&&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
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&&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
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&&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
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&&\}
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\end{IEEEeqnarray*}
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Beleznay and Foreman show that this is open and dense.%
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\footnote{This is not relevant for the exam.}
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% TODO similarities to the lemma used today
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Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
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For $\epsilon \in \Q$ let
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\begin{IEEEeqnarray*}{rCl}
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V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\
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&&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
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&&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
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&&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
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&&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
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&&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
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&&\}
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\end{IEEEeqnarray*}
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Beleznay and Foreman show that this is open and dense.%
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% TODO similarities to the lemma used today
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}{ Not relevant for the exam.}
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\end{itemize}
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\end{proof}
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@ -24,47 +24,49 @@ Let $I$ be a linear order
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\end{theorem}
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\begin{proof}[sketch]
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Consider $\WO(\N) \subset \LO(\N)$.
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We know that this is $\Pi_1^1$-complete. % TODO ref
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\notexaminable{%
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Consider $\WO(\N) \subset \LO(\N)$.
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We know that this is $\Pi_1^1$-complete. % TODO ref
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Let
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\begin{IEEEeqnarray*}{rCll}
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S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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\end{IEEEeqnarray*}
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\todo{Exercise sheet 12}
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$S$ is Borel.
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Let
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\begin{IEEEeqnarray*}{rCll}
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S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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\end{IEEEeqnarray*}
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\todo{Exercise sheet 12}
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$S$ is Borel.
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We will % TODO ?
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construct a reduction
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\begin{IEEEeqnarray*}{rCl}
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M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
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% \alpha &\longmapsto & M(\alpha)
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\end{IEEEeqnarray*}
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We want that $\alpha \in \WO(\N) \iff M(\alpha)$
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codes a distal minimal flow of rank $\alpha$.
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We will % TODO ?
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construct a reduction
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\begin{IEEEeqnarray*}{rCl}
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M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
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% \alpha &\longmapsto & M(\alpha)
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\end{IEEEeqnarray*}
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We want that $\alpha \in \WO(\N) \iff M(\alpha)$
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codes a distal minimal flow of rank $\alpha$.
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\begin{enumerate}[1.]
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\item For any $\alpha \in S$, $M(\alpha)$ is a code for
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a flow which is coded by a generic $(f_i)_{i \in I}$.
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Specifically we will take a flow
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corresponding to some $(f_i)_{i \in I}$
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which is in the intersection of all
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$U_n$, $V_{j,m,n,\frac{p}{q}}$
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(cf.~proof of \yaref{thm:distalminimalofallranks}).
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\begin{enumerate}[1.]
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\item For any $\alpha \in S$, $M(\alpha)$ is a code for
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a flow which is coded by a generic $(f_i)_{i \in I}$.
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Specifically we will take a flow
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corresponding to some $(f_i)_{i \in I}$
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which is in the intersection of all
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$U_n$, $V_{j,m,n,\frac{p}{q}}$
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(cf.~proof of \yaref{thm:distalminimalofallranks}).
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\item If $\alpha \in \WO(\N)$,
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then additionally $(f_i)_{i \in I}$ will code
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a distal minimal flow of ordertype $\alpha$.
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\end{enumerate}
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One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$,
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such that $T^{\alpha}_n$ is closed,
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$T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$,
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$T^\alpha_{n+1} \subseteq T^\alpha_n$,
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$T^{\alpha}_n \subseteq W^{\alpha}_n$,
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where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$.
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Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
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\item If $\alpha \in \WO(\N)$,
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then additionally $(f_i)_{i \in I}$ will code
|
||||
a distal minimal flow of ordertype $\alpha$.
|
||||
\end{enumerate}
|
||||
|
||||
One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$,
|
||||
such that $T^{\alpha}_n$ is closed,
|
||||
$T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$,
|
||||
$T^\alpha_{n+1} \subseteq T^\alpha_n$,
|
||||
$T^{\alpha}_n \subseteq W^{\alpha}_n$,
|
||||
where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$.
|
||||
Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
|
||||
}
|
||||
\end{proof}
|
||||
\begin{lemma}
|
||||
Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal
|
||||
|
|
|
|||
|
|
@ -156,5 +156,6 @@
|
|||
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
|
||||
\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
|
||||
\newcommand\nr[1]{\subsubsection{Exercise #1}\yalabel{Sheet \arabic{subsection}, Exercise #1}{E \arabic{subsection}.#1}{s\arabic{subsection}e#1}}
|
||||
\newcommand\notexaminable[1]{\gist{\footnote{Not relevant for the exam.}#1}{Not relevant for the exam.}}
|
||||
|
||||
\usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}
|
||||
|
|
|
|||
Loading…
Reference in New Issue
Block a user