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4
inputs/lecture_15.tex
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@ -233,6 +233,7 @@ Recall:
correspond to metrics witnessing that the flow is isometric. correspond to metrics witnessing that the flow is isometric.
\end{remark} \end{remark}
\begin{proposition} \begin{proposition}
\label{prop:isomextdistal}
An isometric extension of a distal flow is distal. An isometric extension of a distal flow is distal.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
@ -263,11 +264,12 @@ Recall:
% TODO THE inverse limit is A limit % TODO THE inverse limit is A limit
of $\Sigma$ iff 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} \end{definition}
\begin{proposition} \begin{proposition}
\label{prop:limitdistal}
A limit of distal flows is distal. A limit of distal flows is distal.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}

17
inputs/lecture_16.tex
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@ -1,5 +1,4 @@
\lecture{16}{2023-12-08}{} \lecture{16}{2023-12-08}{}
% TODO ANKI-MARKER
$X$ is always compact metrizable. $X$ 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 $\C$.} % and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
% \end{example} % \end{example}
\begin{proof} \begin{proof}
% TODO TODO TODO Think!
The action of $1$ determines $h$. The action of $1$ determines $h$.
Consider 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 where the topology is the uniform convergence topology. % TODO REF EXERCISE
Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$. 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 we have by the Arzel\`a-Ascoli-Theorem % TODO REF
that $G$ is compact. that $G$ is compact.
@ -126,8 +128,8 @@ $X$ is always compact metrizable.
Every quasi-isometric flow is distal. Every quasi-isometric flow is distal.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
\todo{TODO} The trivial flow is distal.
% The trivial flow is distal. Apply \yaref{prop:isomextdistal} and \yaref{prop:limitdistal}.
\end{proof} \end{proof}
\begin{theorem}[Furstenberg] \begin{theorem}[Furstenberg]
@ -199,7 +201,8 @@ The Hilbert cube $\bH = [0,1]^{\N}$
embeds all compact metric spaces. embeds all compact metric spaces.
Thus we can consider $K(\bH)$, Thus we can consider $K(\bH)$,
the space of compact subsets of $\bH$. 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 $K(\bH^2)$. Consider $K(\bH^2)$.
A flow $\Z \acts X$ corresponds to the graph of A flow $\Z \acts X$ corresponds to the graph of
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}

1
inputs/lecture_17.tex
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@ -1,4 +1,5 @@
\subsection{The Ellis semigroup} \subsection{The Ellis semigroup}
% TODO ANKI-MARKER
\lecture{17}{2023-12-12}{The Ellis semigroup} \lecture{17}{2023-12-12}{The Ellis semigroup}
Let $(X, d)$ be a compact metric space Let $(X, d)$ be a compact metric space

105
inputs/lecture_18.tex
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@ -68,61 +68,62 @@ This will follow from the following lemma:
we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$. we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
\end{refproof} \end{refproof}
\begin{refproof}{lem:ftophelper}% \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? Let $T = \bigcup_n T_n$,% TODO Why does this exist?
$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and $T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$. let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
Take $b$ such that $F(x,x') < b < a$. Take $b$ such that $F(x,x') < b < a$.
Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$ Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
is open in $G(x,x')$ is open in $G(x,x')$
and since $F(x,x') < b$ we have $U \neq \emptyset$. and since $F(x,x') < b$ we have $U \neq \emptyset$.
\begin{claim} \begin{claim}
There exists $n$ such that There exists $n$ such that
\[ \[
\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset. \forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
\] \]
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Suppose not. Suppose not.
Then for all $n$, there is $(u_n, u_n') \in G(x,x')$ Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
with with
\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\] \[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
Note that the RHS is closed. Note that the RHS is closed.
For $m > n$ we have For $m > n$ we have
$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$ $T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
since $T_n \subseteq T_m$. since $T_n \subseteq T_m$.
By compactness of $X$, By compactness of $X$,
there exists $v,v'$ and some subsequence there exists $v,v'$ and some subsequence
such that $(u_{n_k}, u'_{n_k}) \to (v,v')$. such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$, So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
hence $T(v,v') \cap U = \emptyset$, hence $T(v,v') \cap U = \emptyset$,
so $G(v,v') \cap U = \emptyset$. so $G(v,v') \cap U = \emptyset$.
But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$. But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
\end{subproof} \end{subproof}
The map The map
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
T\times X&\longrightarrow & X \\ T\times X&\longrightarrow & X \\
(t,x) &\longmapsto & tx (t,x) &\longmapsto & tx
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
is continuous. is continuous.
Since $T_n$ is compact, Since $T_n$ is compact,
we have that $\{(x,t) \mapsto tx : t \in T_n\}$ we have that $\{(x,t) \mapsto tx : t \in T_n\}$
is equicontinuous.\todo{Sheet 11} is equicontinuous.\todo{Sheet 11}
So there is $\epsilon > 0$ such that So there is $\epsilon > 0$ such that
$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$ $d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
for all $t \in T_n$. for all $t \in T_n$.
Suppose now that $F(x', x'') < \epsilon$. Suppose now that $F(x', x'') < \epsilon$.
Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$, 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 $t \in T_n$. hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
Since $(t_0x, t_0x') \in G(x,x')$, Since $(t_0x, t_0x') \in G(x,x')$,
there is $t_1 \in T_n$ there is $t_1 \in T_n$
with $(t_1t_0x, t_1t_0x') \in U$, with $(t_1t_0x, t_1t_0x') \in U$,
i.e.~$d(t_1t_0x, t_1t_0x') < b$ i.e.~$d(t_1t_0x, t_1t_0x') < b$
and therefore and therefore
$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$. $F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
}
\end{refproof} \end{refproof}
Now assume $Z = \{\star\}$. Now assume $Z = \{\star\}$.

78
inputs/lecture_22.tex
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@ -145,48 +145,54 @@ For this we define
This is the same as for iterated skew shifts. This is the same as for iterated skew shifts.
% TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$, % TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
% $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$. % $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
\item Minimality: \item Minimality:%
\gist{%
\footnote{This is not relevant for the exam.}
Let $\langle E_n : n < \omega \rangle$ Let $\langle E_n : n < \omega \rangle$
be an enumeration of a countable basis for $\mathbb{K}^I$. be an enumeration of a countable basis for $\mathbb{K}^I$.
For all $n$ let For all $n$ let
\[ \[
U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\} U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
\] \]
where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$. where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
Beleznay and Foreman showed that $U_n$ is open Beleznay and Foreman showed that $U_n$ is open
and dense for all $n$. and dense for all $n$.
So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$ So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$
is dense in $\overline{x} \mapsto f(\overline{x})$. is dense in $\overline{x} \mapsto f(\overline{x})$.
Since the flow is distal, it suffices to show Since the flow is distal, it suffices to show
that one orbit is dense (cf.~\yaref{thm:distalflowpartition}). that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
}{ Not relevant for the exam.}
\item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$. \item The order of the flow is $\eta$:%
Consider the flows we get from $(f_i)_{i < j}$ \gist{%
resp.~$(f_i)_{i \le j}$ \footnote{This is not relevant for the exam.}
denoted by $X_{<j}$ resp.~$X_{\le j}$. Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
We aim to show that $X_{\le j} \to X_{<j}$ Consider the flows we get from $(f_i)_{i < j}$
is a maximal isometric extension for comeagerly many $\overline{f}$. 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 The following open dense sets are used to make sure that all isometric extensions
are maximal and hence the order of the flow is $\eta$: are maximal and hence the order of the flow is $\eta$:
Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$. Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
For $\epsilon \in \Q$ let For $\epsilon \in \Q$ let
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\ 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{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}\\ &&\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,\\ &&\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_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\\ &&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
&&\} &&\}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Beleznay and Foreman show that this is open and dense.% 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
% TODO similarities to the lemma used today }{ Not relevant for the exam.}
\end{itemize} \end{itemize}
\end{proof} \end{proof}

74
inputs/lecture_23.tex
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@ -24,47 +24,49 @@ Let $I$ be a linear order
\end{theorem} \end{theorem}
\begin{proof}[sketch] \begin{proof}[sketch]
Consider $\WO(\N) \subset \LO(\N)$. \notexaminable{%
We know that this is $\Pi_1^1$-complete. % TODO ref Consider $\WO(\N) \subset \LO(\N)$.
We know that this is $\Pi_1^1$-complete. % TODO ref
Let Let
\begin{IEEEeqnarray*}{rCll} \begin{IEEEeqnarray*}{rCll}
S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\ S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\} &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\todo{Exercise sheet 12} \todo{Exercise sheet 12}
$S$ is Borel. $S$ is Borel.
We will % TODO ? We will % TODO ?
construct a reduction construct a reduction
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\ M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
% \alpha &\longmapsto & M(\alpha) % \alpha &\longmapsto & M(\alpha)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
We want that $\alpha \in \WO(\N) \iff M(\alpha)$ We want that $\alpha \in \WO(\N) \iff M(\alpha)$
codes a distal minimal flow of rank $\alpha$. codes a distal minimal flow of rank $\alpha$.
\begin{enumerate}[1.] \begin{enumerate}[1.]
\item For any $\alpha \in S$, $M(\alpha)$ is a code for \item For any $\alpha \in S$, $M(\alpha)$ is a code for
a flow which is coded by a generic $(f_i)_{i \in I}$. a flow which is coded by a generic $(f_i)_{i \in I}$.
Specifically we will take a flow Specifically we will take a flow
corresponding to some $(f_i)_{i \in I}$ corresponding to some $(f_i)_{i \in I}$
which is in the intersection of all which is in the intersection of all
$U_n$, $V_{j,m,n,\frac{p}{q}}$ $U_n$, $V_{j,m,n,\frac{p}{q}}$
(cf.~proof of \yaref{thm:distalminimalofallranks}). (cf.~proof of \yaref{thm:distalminimalofallranks}).
\item If $\alpha \in \WO(\N)$, \item If $\alpha \in \WO(\N)$,
then additionally $(f_i)_{i \in I}$ will code then additionally $(f_i)_{i \in I}$ will code
a distal minimal flow of ordertype $\alpha$. a distal minimal flow of ordertype $\alpha$.
\end{enumerate} \end{enumerate}
One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$, One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$,
such that $T^{\alpha}_n$ is closed, such that $T^{\alpha}_n$ is closed,
$T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$, $T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$,
$T^\alpha_{n+1} \subseteq T^\alpha_n$, $T^\alpha_{n+1} \subseteq T^\alpha_n$,
$T^{\alpha}_n \subseteq W^{\alpha}_n$, $T^{\alpha}_n \subseteq W^{\alpha}_n$,
where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$. where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$.
Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$. Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
}
\end{proof} \end{proof}
\begin{lemma} \begin{lemma}
Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal

1
logic.sty
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@ -156,5 +156,6 @@
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}} \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #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\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} \usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}