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7 changed files with 149 additions and 133 deletions

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@ -233,6 +233,7 @@ Recall:
correspond to metrics witnessing that the flow is isometric.
\end{remark}
\begin{proposition}
\label{prop:isomextdistal}
An isometric extension of a distal flow is distal.
\end{proposition}
\begin{proof}
@ -263,11 +264,12 @@ Recall:
% TODO THE inverse limit is A limit
of $\Sigma$ iff
\[
\forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
\forall x_1 \neq x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
\]
\end{definition}
\begin{proposition}
\label{prop:limitdistal}
A limit of distal flows is distal.
\end{proposition}
\begin{proof}

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@ -1,5 +1,4 @@
\lecture{16}{2023-12-08}{}
% TODO ANKI-MARKER
$X$ is always compact metrizable.
@ -18,16 +17,19 @@ $X$ is always compact metrizable.
% and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
% \end{example}
\begin{proof}
% TODO TODO TODO Think!
The action of $1$ determines $h$.
Consider
\[
\{h^n : n \in \Z\} \subseteq \cC(X,X) = \{f\colon X \to X : f \text{ continuous}\},
\{h^n : n \in \Z\} \subseteq \cC(X,X)\gist{ = \{f\colon X \to X : f \text{ continuous}\}}{},
\]
where the topology is the uniform convergence topology. % TODO REF EXERCISE
Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
Since
Since the family $\{h^n : n \in \Z\}$ is uniformly equicontinuous,
i.e.~
\[
\forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon
\forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon,
% Here we use isometric
\]
we have by the Arzel\`a-Ascoli-Theorem % TODO REF
that $G$ is compact.
@ -126,8 +128,8 @@ $X$ is always compact metrizable.
Every quasi-isometric flow is distal.
\end{corollary}
\begin{proof}
\todo{TODO}
% The trivial flow is distal.
The trivial flow is distal.
Apply \yaref{prop:isomextdistal} and \yaref{prop:limitdistal}.
\end{proof}
\begin{theorem}[Furstenberg]
@ -199,7 +201,8 @@ The Hilbert cube $\bH = [0,1]^{\N}$
embeds all compact metric spaces.
Thus we can consider $K(\bH)$,
the space of compact subsets of $\bH$.
$K(\bH)$ is a Polish space.\todo{Exercise}
$K(\bH)$ is a Polish space.\footnote{cf.~\yaref{s9e2}, \yaref{s12e4}}
% TODO LEARN EXERCISES
Consider $K(\bH^2)$.
A flow $\Z \acts X$ corresponds to the graph of
\begin{IEEEeqnarray*}{rCl}

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@ -1,4 +1,5 @@
\subsection{The Ellis semigroup}
% TODO ANKI-MARKER
\lecture{17}{2023-12-12}{The Ellis semigroup}
Let $(X, d)$ be a compact metric space

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@ -68,61 +68,62 @@ This will follow from the following lemma:
we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
\end{refproof}
\begin{refproof}{lem:ftophelper}%
\footnote{This was not covered in class.}
\notexaminable{\footnote{This was not covered in class.}
Let $T = \bigcup_n T_n$,% TODO Why does this exist?
$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
Take $b$ such that $F(x,x') < b < a$.
Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
is open in $G(x,x')$
and since $F(x,x') < b$ we have $U \neq \emptyset$.
\begin{claim}
There exists $n$ such that
\[
\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
\]
\end{claim}
\begin{subproof}
Suppose not.
Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
with
\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
Note that the RHS is closed.
For $m > n$ we have
$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
since $T_n \subseteq T_m$.
By compactness of $X$,
there exists $v,v'$ and some subsequence
such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
Let $T = \bigcup_n T_n$,% TODO Why does this exist?
$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
Take $b$ such that $F(x,x') < b < a$.
Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
is open in $G(x,x')$
and since $F(x,x') < b$ we have $U \neq \emptyset$.
\begin{claim}
There exists $n$ such that
\[
\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
\]
\end{claim}
\begin{subproof}
Suppose not.
Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
with
\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
Note that the RHS is closed.
For $m > n$ we have
$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
since $T_n \subseteq T_m$.
By compactness of $X$,
there exists $v,v'$ and some subsequence
such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
hence $T(v,v') \cap U = \emptyset$,
so $G(v,v') \cap U = \emptyset$.
But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
\end{subproof}
The map
\begin{IEEEeqnarray*}{rCl}
T\times X&\longrightarrow & X \\
(t,x) &\longmapsto & tx
\end{IEEEeqnarray*}
is continuous.
Since $T_n$ is compact,
we have that $\{(x,t) \mapsto tx : t \in T_n\}$
is equicontinuous.\todo{Sheet 11}
So there is $\epsilon > 0$ such that
$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
for all $t \in T_n$.
So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
hence $T(v,v') \cap U = \emptyset$,
so $G(v,v') \cap U = \emptyset$.
But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
\end{subproof}
The map
\begin{IEEEeqnarray*}{rCl}
T\times X&\longrightarrow & X \\
(t,x) &\longmapsto & tx
\end{IEEEeqnarray*}
is continuous.
Since $T_n$ is compact,
we have that $\{(x,t) \mapsto tx : t \in T_n\}$
is equicontinuous.\todo{Sheet 11}
So there is $\epsilon > 0$ such that
$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
for all $t \in T_n$.
Suppose now that $F(x', x'') < \epsilon$.
Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
Since $(t_0x, t_0x') \in G(x,x')$,
there is $t_1 \in T_n$
with $(t_1t_0x, t_1t_0x') \in U$,
i.e.~$d(t_1t_0x, t_1t_0x') < b$
and therefore
$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
Suppose now that $F(x', x'') < \epsilon$.
Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
Since $(t_0x, t_0x') \in G(x,x')$,
there is $t_1 \in T_n$
with $(t_1t_0x, t_1t_0x') \in U$,
i.e.~$d(t_1t_0x, t_1t_0x') < b$
and therefore
$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
}
\end{refproof}
Now assume $Z = \{\star\}$.

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@ -145,48 +145,54 @@ For this we define
This is the same as for iterated skew shifts.
% TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
% $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
\item Minimality:
\item Minimality:%
\gist{%
\footnote{This is not relevant for the exam.}
Let $\langle E_n : n < \omega \rangle$
be an enumeration of a countable basis for $\mathbb{K}^I$.
Let $\langle E_n : n < \omega \rangle$
be an enumeration of a countable basis for $\mathbb{K}^I$.
For all $n$ let
\[
U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
\]
where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
For all $n$ let
\[
U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
\]
where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
Beleznay and Foreman showed that $U_n$ is open
and dense for all $n$.
Beleznay and Foreman showed that $U_n$ is open
and dense for all $n$.
So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$
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}).
So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$
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}).
}{ Not relevant for the exam.}
\item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
Consider the flows we get from $(f_i)_{i < j}$
resp.~$(f_i)_{i \le j}$
denoted by $X_{<j}$ resp.~$X_{\le j}$.
We aim to show that $X_{\le j} \to X_{<j}$
is a maximal isometric extension for comeagerly many $\overline{f}$.
\item The order of the flow is $\eta$:%
\gist{%
\footnote{This is not relevant for the exam.}
Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
Consider the flows we get from $(f_i)_{i < j}$
resp.~$(f_i)_{i \le j}$
denoted by $X_{<j}$ resp.~$X_{\le j}$.
We aim to show that $X_{\le j} \to X_{<j}$
is a maximal isometric extension for comeagerly many $\overline{f}$.
The following open dense sets are used to make sure that all isometric extensions
are maximal and hence the order of the flow is $\eta$:
The following open dense sets are used to make sure that all isometric extensions
are maximal and hence the order of the flow is $\eta$:
Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
For $\epsilon \in \Q$ let
\begin{IEEEeqnarray*}{rCl}
V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\
&&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
&&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
&&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
&&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
&&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
&&\}
\end{IEEEeqnarray*}
Beleznay and Foreman show that this is open and dense.%
\footnote{This is not relevant for the exam.}
% TODO similarities to the lemma used today
Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
For $\epsilon \in \Q$ let
\begin{IEEEeqnarray*}{rCl}
V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\
&&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
&&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
&&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
&&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
&&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
&&\}
\end{IEEEeqnarray*}
Beleznay and Foreman show that this is open and dense.%
% TODO similarities to the lemma used today
}{ Not relevant for the exam.}
\end{itemize}
\end{proof}

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@ -24,47 +24,49 @@ Let $I$ be a linear order
\end{theorem}
\begin{proof}[sketch]
Consider $\WO(\N) \subset \LO(\N)$.
We know that this is $\Pi_1^1$-complete. % TODO ref
\notexaminable{%
Consider $\WO(\N) \subset \LO(\N)$.
We know that this is $\Pi_1^1$-complete. % TODO ref
Let
\begin{IEEEeqnarray*}{rCll}
S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
\end{IEEEeqnarray*}
\todo{Exercise sheet 12}
$S$ is Borel.
Let
\begin{IEEEeqnarray*}{rCll}
S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
\end{IEEEeqnarray*}
\todo{Exercise sheet 12}
$S$ is Borel.
We will % TODO ?
construct a reduction
\begin{IEEEeqnarray*}{rCl}
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
% \alpha &\longmapsto & M(\alpha)
\end{IEEEeqnarray*}
We want that $\alpha \in \WO(\N) \iff M(\alpha)$
codes a distal minimal flow of rank $\alpha$.
We will % TODO ?
construct a reduction
\begin{IEEEeqnarray*}{rCl}
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
% \alpha &\longmapsto & M(\alpha)
\end{IEEEeqnarray*}
We want that $\alpha \in \WO(\N) \iff M(\alpha)$
codes a distal minimal flow of rank $\alpha$.
\begin{enumerate}[1.]
\item For any $\alpha \in S$, $M(\alpha)$ is a code for
a flow which is coded by a generic $(f_i)_{i \in I}$.
Specifically we will take a flow
corresponding to some $(f_i)_{i \in I}$
which is in the intersection of all
$U_n$, $V_{j,m,n,\frac{p}{q}}$
(cf.~proof of \yaref{thm:distalminimalofallranks}).
\begin{enumerate}[1.]
\item For any $\alpha \in S$, $M(\alpha)$ is a code for
a flow which is coded by a generic $(f_i)_{i \in I}$.
Specifically we will take a flow
corresponding to some $(f_i)_{i \in I}$
which is in the intersection of all
$U_n$, $V_{j,m,n,\frac{p}{q}}$
(cf.~proof of \yaref{thm:distalminimalofallranks}).
\item If $\alpha \in \WO(\N)$,
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 $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
\item If $\alpha \in \WO(\N)$,
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

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@ -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}