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8
inputs/lecture_01.tex
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@ -185,11 +185,11 @@ suffices to show that open balls in one metric are unions of open balls in the o
\begin{definition}[Our favourite Polish spaces]
\leavevmode
\begin{itemize}
\item $2^{\omega}$ is called the \vocab{Cantor set}.
\item $2^{\N}$ is called the \vocab{Cantor set}.
(Consider $2$ with the discrete topology)
\item $\omega^{\omega}$ is called the \vocab{Baire space}.
($\omega$ with descrete topology)
\item $[0,1]^{\omega}$ is called the \vocab{Hilbert cube}.
\item $\cN \coloneqq \N^{\N}$ is called the \vocab{Baire space}.
($\N$ with descrete topology)
\item $\mathbb{H} \coloneqq [0,1]^{\N}$ is called the \vocab{Hilbert cube}.
($[0,1] \subseteq \R$ with the usual topology)
\end{itemize}
\end{definition}

4
inputs/lecture_03.tex
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@ -205,8 +205,8 @@
\[
D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN
\]
and a continuous bijection from
$D$ onto $X$ (the inverse does not need to be continuous).
and a continuous bijection $f\colon D \to X$
(the inverse does not need to be continuous).
Moreover there is a continuous surjection $g: \cN \to X$
extending $f$.

2
inputs/lecture_04.tex
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@ -64,7 +64,7 @@
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x\defon{n} = s\}$.
Clearly $S$ is a pruned tree.
Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
Moreover, since $D$ is closed, we have that\footnote{cf.~\yaref{s3e1}}
\[
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
\]

13
inputs/lecture_08.tex
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@ -78,18 +78,17 @@ where $X$ is a metrizable, usually second countable space.
we also have
$A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$.
We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set
We construct a $(2^{\omega \times \omega}) \cong 2^\omega$-universal set
for $\Sigma^0_\xi(X)$.
For $(y_n) \in (2^\omega)^\omega$
For $(y_{m,n}) \in (2^{\omega \times \omega})$
and $x \in X$
we set $((y_n), x) \in \cU$
iff $\exists n.~(y_n, x) \in U_{\xi_n}$,
i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$.
we set $((y_{m,n}), x) \in \cU$
iff $\exists n.~((y_{m,n})_{m < \omega}, x) \in U_{\xi_n}$,
i.e.~iff $\exists n.~x \in (U_{\xi_n})_{(y_{m,n})_{m < \omega}}$.
Let $A \in \Sigma^0_\xi(X)$.
Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
% TODO
Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
\end{proof}
\begin{remark}
Since $2^{\omega}$ embeds

7
inputs/lecture_09.tex
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@ -45,9 +45,8 @@ We will see that not every analytic set is Borel.
\begin{remark}
In the definition we can replace the assertion that
$f$ is continuous
by the weaker assertion of $f$ being Borel.
\todo{Copy exercise from sheet 5}
% TODO WHY?
by the weaker assertion of $f$ being Borel.%
\footnote{use \yaref{thm:clopenize}, cf.~\yaref{s6e2}}
\end{remark}
\begin{theorem}
@ -155,7 +154,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
\begin{proof}
Take $\cU \subseteq Y \times X \times \cN$
which is $Y$-universal for $\Pi^0_1(X \times \cN)$.
Let $\cV \coloneqq \proj_{Y \times Y}(\cU)$.
Let $\cV \coloneqq \proj_{Y \times X}(\cU)$.
Then $\cV$ is $Y$-universal for $\Sigma^1_1(X)$:
\begin{itemize}
\item $\cV \in \Sigma^1_1(Y \times X)$

7
inputs/lecture_10.tex
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@ -34,7 +34,8 @@ we need the following definition:
\begin{lemma}
\label{lem:lusinsephelp}
If $P = \bigcup_{m < \omega} P_m$, $Q = \bigcup_{n < \omega} Q_n$ are such that
for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable.
for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable,
then $P$ and $Q$ are Borel separable.
\end{lemma}
\begin{proof}
For all $m, n$ pick $R_{m,n}$ Borel,
@ -42,7 +43,7 @@ we need the following definition:
and $Q_n \cap R_{m,n} = \emptyset$.
Then $R = \bigcup_m \bigcap_n R_{m,n}$
has the desired property
that $R \subseteq R$ and $R \cap Q = \emptyset$.
that $P \subseteq R$ and $R \cap Q = \emptyset$.
\end{proof}
\begin{notation}
@ -60,7 +61,7 @@ we need the following definition:
Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
Note that $A_s = \bigcup_m A_{s\concat m}$
and $B_ns = \bigcup_{n < \omega} B_{s\concat n}$.
and $B_s = \bigcup_{n < \omega} B_{s\concat n}$.
In particular
$A = \bigcup_{m < \omega} A_{\underbrace{\langle m \rangle}_{\in \omega^1}}$

2
inputs/lecture_13.tex
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@ -183,7 +183,7 @@ i.e.}{}
Let $X$ be Polish and $C \subseteq X$ coanalytic.
Then $\phi\colon C \to \Ord$
is a \vocab[Rank!$\Pi^1_1$-rank]{$\Pi^1_1$-rank}
provided that $\le^\ast$ and $<^\ast$ are coanalytic subsets of $X \times X$,
provided that $\le^\ast_\phi$ and $<^\ast_\phi$ are coanalytic subsets of $X \times X$,
where
$x \le^\ast_{\phi} y$
iff

2
inputs/lecture_14.tex
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@ -104,7 +104,7 @@
\forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast).
\]
We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
\footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
\footnote{Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{picture}.}
\end{theorem}
\begin{proof}
Let $\phi\colon R \to \Ord$

6
inputs/lecture_15.tex
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@ -55,7 +55,7 @@
then $\xi = \alpha$
and
\[
E_\xi = E_\alpha = \bigcup_{\eta < \alpha}
E_\xi = E_\alpha = \bigcup_{\eta < \alpha} E_\eta
\]
is a countable union of Borel sets by the previous case.
\end{itemize}
@ -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}

19
inputs/lecture_16.tex
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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]
@ -138,7 +140,7 @@ By Zorn's lemma, this will follow from
\begin{theorem}[Furstenberg]
\label{thm:l16:3}
Let $(X, T)$ be a minimal distal flow
and let $(Y, T)$ be a proper factor.
and let $(Y, T)$ be a proper factor.%
\footnote{i.e.~$(X,T)$ and $(Y,T)$ are not isomorphic}
Then there is another factor $(Z,T)$ of $(X,T)$
which is a proper isometric extension of $Y$.
@ -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}

136
inputs/lecture_17.tex
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@ -15,16 +15,16 @@ U_{\epsilon}(x,y) \coloneqq \{f \in X^X : d(x,f(y)) < \epsilon\}.
\]
for all $x,y \in X$, $\epsilon > 0$.
$X^{X}$ is a compact Hausdorff space.
\begin{remark}
\todo{Copy from exercise sheet 10}
$X^{X}$ is a compact Hausdorff space.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}}
\begin{remark}%
\footnote{cf.~\yaref{s11e1}}
Let $f_0 \in X^X$ be fixed.
\begin{itemize}
\item $X^X \ni f \mapsto f \circ f_0$
is continuous:
Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
We have $ff_0 \in U_{\epsilon}(x,y)$
We have $f \circ f_0 \in U_{\epsilon}(x,y)$
iff $f \in U_\epsilon(x,f_0(y))$.
\item Fix $x_0 \in X$.
Then $f \mapsto f(x_0)$ is continuous.
@ -34,27 +34,35 @@ $X^{X}$ is a compact Hausdorff space.
\end{remark}
\begin{definition}
\gist{%
Let $(X,T)$ be a flow.
Then the \vocab{Ellis semigroup}
is defined by
$E(X,T) \coloneqq \overline{T} \subseteq X^X$,
i.e.~identify $t \in T$ with $x \mapsto tx$
and take the closure in $X^X$.
}{%
The \vocab{Ellis semigroup} of a flow $(X,T)$
is $E(X,T) \coloneqq \overline{T} \subseteq X^X$.
}
\end{definition}
$E(X,T)$ is compact and Hausdorff,
since $X^X$ has these properties.
\gist{
Properties of $(X,T)$ translate to properties of $E(X,T)$:
\begin{goal}
We want to show that if $(X,T)$ is distal,
then $E(X,T)$ is a group.
\end{goal}
}{}
\begin{proposition}
$E(X,T)$ is a semigroup,
i.e.~closed under composition.
\end{proposition}
\begin{proof}
\gist{
Let $G \coloneqq E(X,T)$.
Take $t \in T$. We want to show that $tG \subseteq G$,
i.e.~for all $h \in G$ we have $th \in G$.
@ -63,7 +71,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
since $t^{-1}$ is continuous
and $G$ is compact.
Then $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
So $G = \overline{T} \subseteq t^{-1}G$.
Hence $tG \subseteq G$.
@ -74,7 +83,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
\]
\end{claim}
\begin{subproof}
\todo{Homework}
Cf.~\yaref{s11e1}
\end{subproof}
Let $g \in G$.
@ -87,17 +96,34 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
Since $G$ compact,
and $Tg \subseteq G$,
we have $ \overline{Tg} \subseteq G$.
}{
$G \coloneqq E(X,T)$.
\begin{itemize}
\item $\forall t \in T. ~ tG \subseteq G$:
\begin{itemize}
\item $t^{-1}G$ is compact.
\item $T \subseteq t^{-1}G$,
\item $\leadsto G = \overline{T} \subseteq t^{-1}G$,
i.e.~$tG \subseteq G$.
\end{itemize}
\item $\forall g \in G.~\overline{T}g = \overline{Tg}$ (cts.~map from compact to Hausdorff)
\item $\forall g \in G.~Gg \subseteq G$ :
$Gg = \overline{T}g = \overline{Tg} \overset{G \text{ compact}, Tg \subseteq G}{\subseteq} G$.
\end{itemize}
}
\end{proof}
\begin{definition}
A \vocab{compact semigroup} $S$
is a nonempty semigroup with a compact
Hausdorff topology,
is a nonempty semigroup\footnote{may not contain inverses or the identity}
with a compact Hausdorff topology,
such that $S \ni x \mapsto xs$ is continuous for all $s$.
\end{definition}
\gist{
\begin{example}
The Ellis semigroup is a compact semigroup.
\end{example}
}{}
\begin{lemma}[Ellis–Numakura]
\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
@ -106,6 +132,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
i.e.~$f$ such that $f^2 = f$.
\end{lemma}
\begin{proof}
\gist{
Using Zorn's lemma, take a $\subseteq$-minimal
compact subsemigroup $R$ of $S$
and let $s \in R$.
@ -122,25 +149,39 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
Thus $P = R$ by minimality,
so $s \in P$,
i.e.~$s^2 = s$.
}{
\begin{itemize}
\item Take $R \subseteq S$ minimal compact subsemigroup (Zorn),
$s \in R$.
\item $Rs \subseteq R \implies Rs = R$.
\item $P \coloneqq \{x \in R : xs = s\}$:
\begin{itemize}
\item $P \neq \emptyset$, since $s \in Rs$
\item $P$ compact, since $P = \alpha^{-1}(s) \cap R$,
$\alpha: x \mapsto xs$ cts.
\item $P = R \implies s^2 = s$.
\end{itemize}
\end{itemize}%
}
\end{proof}
\gist{
The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
since we already know that it has an identity,
in fact we have chosen $R = \{1\}$ in the proof.
since we already know that it has an identity.
%in fact we might have chosen $R = \{1\}$ in the proof.
But it is interesting for other semigroups.
}{}
\begin{theorem}[Ellis]
$(X,T)$ is distal iff $E(X,T)$ is a group.
\end{theorem}
\begin{proof}
Let $G \coloneqq E(X,T)$.
Let $d$ be a metric on $X$.
Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.
\gist{
For all $g \in G$ we need to show that $x \mapsto gx$ is bijective.
If we had $gx = gy$, then $d(gx,gy) = 0$.
Then $\inf d(tx,ty) = 0$, but the flow is distal,
Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal,
hence $x = y$.
Let $g \in G$. Consider the compact semigroup $\Gamma \coloneqq Gg$.
@ -150,25 +191,46 @@ But it is interesting for other semigroups.
Since $f$ is injective, we get that $x = f(x)$,
i.e.~$f = \id$.
Take $g' \in G$ such that $f = g' \circ g$.%
%\footnote{This exists since $f \in Gg$.}
Since $f \in Gg$, there exists $g' \in G$ such that $f = g' \circ g$.
It is $g' = g'gg'$,
so $\forall x .~g'(x) = g'(g g'(x))$.
Hence $g'$ is bijective
and $x = gg'(x)$,
i.e.~$g g' = \id$.
}{
\begin{itemize}
\item $x \mapsto gx$ injective for all $g \in G$:
\[gx = gy
\implies d(gx,gy) = 0
\implies \inf_{t \in T} d(tx, ty) = 0
\overset{\text{distal}}{\implies} x = y.
\]
\item Fix $g \in G$.
\begin{itemize}
\item $\Gamma \coloneqq Gg$ is a compact semigroup.
\item $\exists f\in \Gamma.~f^2 = f$ (\yaref{lem:ellisnumakura})
\item $f$ is injective, hence $f = \id$.
\item Take $g'$ such that $g'g = f$. $g'gg' = g' \implies gg' = \id$.
\end{itemize}
\end{itemize}
}
\todo{The other direction is left as an easy exercise.}
\gist{
On the other hand if $(x_0,x_1)$ is proximal,
then there exists $g \in G$ such that $gx_0 = gx_1$.%
\footnote{cf.~\yaref{s11e1} (e)}
It follows that an inverse to $g$ can not exist.
}{If $(x_0,x_1)$ is proximal, there is $g \in G$ with $gx_0 = gx_1$, i.e.~no inverse to $g$.}
\end{proof}
Let $(X,T)$ be a flow.
Then by Zorn's lemma, there exists $X_0 \subseteq X$
such that $(X_0, T)$ is minimal.
In particular,
for $x \in X$ and $\overline{Tx} = Y$
we have that $(Y,T)$ is a flow.
However if we pick $y \in Y$, $Ty$ might not be dense.
% Let $(X,T)$ be a flow.
% Then by Zorn's lemma, there exists $X_0 \subseteq X$
% such that $(X_0, T)$ is minimal.
% In particular,
% for $x \in X$ and $\overline{Tx} = Y$
% we have that $(Y,T)$ is a flow.
% However if we pick $y \in Y$, $Ty$ might not be dense.
% TODO: question!
% TODO: think about this!
% We want to a minimal subflow in a nice way:
@ -181,6 +243,7 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
\end{theorem}
\begin{proof}
Let $G = E(X,T)$.
\gist{
Note that for all $x \in X$,
we have that $Gx \subseteq X$ is compact
and invariant under the action of $G$.
@ -188,17 +251,18 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
Since $G$ is a group, the orbits partition $X$.%
\footnote{Note that in general this does not hold for semigroups.}
% Clearly the sets $Gx$ cover $X$. We want to show that they
% partition $X$.
% It suffices to show that $y \in Gx \implies Gy = Gx$.
% Take some $y \in Gx$.
% Recall that $\overline{Ty} = \overline{T} y = Gy$.
% We have $\overline{Ty} \subseteq Gx$,
% so $Gy \subseteq Gx$.
% Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$,
% hence $Gx \subseteq Gy$
% TODO: WHY?
We need to show that $(Gx, T)$ is minimal.
Suppose that $y \in Gx$, i.e.~$Gx = Gy$.
Since $g \mapsto gy$ is continuous,
we have $Gx = Gy = \overline{T}y = \overline{Ty}$,
so $Ty$ is dense in $Gx$.
}{
\begin{itemize}
\item $G$ is a group, so the $G$-orbits partition $X$.
\item Let $y \in Gx$, then $Gx \overset{y \in Gx}{=} Gy \overset{\text{def}}{=} \overline{T}y \overset{g \mapsto gy \text{ cts.}}{=} \overline{Ty}$,
i.e.~$(Gx,T)$ is minimal.
\end{itemize}
}
\end{proof}
\begin{corollary}
If $(X,T)$ is distal and minimal,

36
inputs/lecture_18.tex
View file

@ -3,14 +3,6 @@
The goal for this lecture is to give a very rough
sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
% \begin{theorem}[Furstenberg]
% Let $(X, T)$ be a minimal distal flow
% and let $(Z,T)$ be a proper factor of $X$%
% \footnote{i.e.~$(X,T)$ and $(Z,T)$ are not isomorphic.}
% Then three is another factor $(Y,T)$ of $(X,T)$
% which is a proper isometric extension of $Z$.
% \end{theorem}
Let $(X,T)$ be a distal flow.
Then $G \coloneqq E(X,T)$ is a group.
@ -26,7 +18,9 @@ F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
\item $F(x,x') = F(x', x)$,
\item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$.
\item $F(gx, gx') = F(x,x')$ since $G$ is a group.
\item $F$ is an upper semi-continuous function on $X^2$,
\item $F$ is an \vocab{upper semi-continuous}\footnote{%
Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/c/c0/Upper_semi.svg}{picture}.}
function on $X^2$,
i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$.
This holds because $F$ is the infimum of continuous functions
@ -52,23 +46,28 @@ This will follow from the following lemma:
\begin{lemma}
\label{lem:ftophelper}
Let $F(x,x') < a$.
\gist{%
Then there exists $\epsilon > 0$ such that
whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.
}{Then $\exists \epsilon > 0.~\forall x''.~F(x',x'') < \epsilon \implies F(x,x'') < a$.}
\end{lemma}
\begin{refproof}{def:ftop}
\gist{%
We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$,
then this intersection is the union
of sets of this kind.
Let $x' \in U_a(x_1)$.
}{}
Let $x' \in U_a(x_1) \cap U_b(x_2)$.
Then by \yaref{lem:ftophelper},
there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$.
Similarly there exists $\epsilon_2 > 0$
such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.
Similarly there exists $\epsilon_2 > 0$\gist{
such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.}{.}
So for $\epsilon \le \epsilon_1, \epsilon_2$,
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.}
% TODO: maybe learn?
Let $T = \bigcup_n T_n$,% TODO Why does this exist?
$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
@ -123,6 +122,7 @@ This will follow from the following lemma:
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 $Z = \{\star\}$.
@ -168,8 +168,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
\item One can show that $H$ is a topological group and $(M,H)$
is a flow.\footnote{This is non-trivial.}
\item Since $H$ is compact,
$(M,H)$ is equicontinuous, %\todo{We didn't define this}
i.e.~it is isometric.
$(M,H)$ is equicontinuous, i.e.~it is isometric.
In particular, $(M,T)$ is isometric.
\end{enumerate}
\item $M \neq \{\star\}$, i.e.~$(M,T)$ is non-trivial:
@ -213,9 +212,9 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
Let $X$ be a metric space
and $\Gamma\colon X \to \R$ be upper semicontinuous.
Then the set of continuity points of $\Gamma$ is comeager.
\todo{Missing figure: upper semicontinuous function}
\end{theorem}
\begin{proof}
\notexaminable{
Take $x$ such that $\Gamma$ is not continuous at $x$.
Then there is an $\epsilon > 0$
and $x_n \to x$ such that
@ -227,10 +226,11 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
\]
$X \setminus B_q = \{a \in X : \Gamma(a) < q\}$
is open, i.e.~$B_q$ is closed.
Note that $x \in F_q \coloneqq B_q \setminus B_q^\circ$
and $B_q \setminus B_q^\circ$ is nwd
Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$
and $B_q \setminus \inter(B_q)$ is nwd
as it is closed and has empty interior,
so $\bigcup_{q \in \Q} F_q$ is meager.
}
\end{proof}

29
inputs/lecture_19.tex
View file

@ -1,5 +1,5 @@
\subsection{The order of a flow}
\lecture{19}{2023-12-19}{Orders of flows}
\subsection{The Order of a Flow}
\lecture{19}{2023-12-19}{Orders of Flows}
See also \cite[\href{https://terrytao.wordpress.com/2008/01/24/254a-lecture-6-isometric-systems-and-isometric-extensions/}{Lecture 6}]{tao}.
@ -71,7 +71,6 @@ equicontinuity coincide.
i.e.~if $(Y_\alpha, f_{\alpha,\beta})$,
$\beta < \alpha \le \Theta$
are isometric, then the inverse limit $Y$ is isometric.%
\todo{Why does an inverse limit exist?}
% https://q.uiver.app/#q=WzAsNCxbMSwwLCJZX1xcYWxwaGEiXSxbMSwxLCJZX1xcYmV0YSJdLFswLDAsIlkiXSxbMiwwLCJYIl0sWzAsMSwiZl97XFxhbHBoYSwgXFxiZXRhfSJdLFsyLDAsImZfXFxhbHBoYSJdLFsyLDEsImZfXFxiZXRhIiwyXSxbMywwLCJcXHBpX1xcYWxwaGEiLDJdLFszLDEsIlxccGlfXFxiZXRhIl1d
\[\begin{tikzcd}
Y & {Y_\alpha} & X \\
@ -193,8 +192,8 @@ More generally we can show:
on the fibers of $Y$ over $Z_2$
and invariant under $T$.
$\sigma$ is a metric, since if
if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
$\sigma$ is a metric,
since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
\end{proof}
@ -221,6 +220,7 @@ More generally we can show:
For this, we show that for all $\xi < \eta$,
$(X_\xi', T)$ is a factor of $(X_\xi ,T)$
using transfinite induction.
% https://q.uiver.app/#q=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
\[\begin{tikzcd}
{X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\
@ -242,7 +242,8 @@ More generally we can show:
\arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4]
\arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}\]
% TODO: induction start?
We'll only show the successor step:
Suppose we have
$(X'_\xi, T) = \theta((X_\xi, T)$.
@ -251,18 +252,21 @@ More generally we can show:
\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}.\]
Then
% https://q.uiver.app/#q=WzAsNSxbMCwwLCIoWF97XFx4aSsxfSxUKSJdLFsyLDAsIihZLFQpIl0sWzMsMSwiKFgnX3tcXHhpKzF9LFQpIl0sWzIsMiwiKFgnX1xceGksVCkiXSxbMSwxLCIoWF9cXHhpLFQpIl0sWzAsNCwiXFx0ZXh0e21heC5+aXNvfSIsMV0sWzQsMywiXFx0aGV0YSIsMV0sWzIsMywie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsNCwie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsMl0sWzAsMSwiIiwwLHsiY3VydmUiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
\[\begin{tikzcd}
% https://q.uiver.app/#q=WzAsNSxbMCwwLCIoWF97XFx4aSsxfSxUKSJdLFsyLDAsIihZLFQpIl0sWzMsMSwiKFgnX3tcXHhpKzF9LFQpIl0sWzIsMiwiKFgnX1xceGksVCkiXSxbMSwxLCIoWF9cXHhpLFQpIl0sWzAsNCwiXFx0ZXh0e21heC5+aXNvfSIsMV0sWzQsMywiXFx0aGV0YSIsMV0sWzIsMywie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsNCwie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsMl0sWzAsMSwiIiwwLHsiY3VydmUiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSwyLCJcXHBpJyIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6Im5vbmUifSwiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs3LDMsIlxcdGhldGEnIiwwLHsibGV2ZWwiOjEsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6Im5vbmUifSwiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsxLDgsIlxccGkiLDIseyJsZXZlbCI6MSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoibm9uZSJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d
\begin{tikzcd}
{(X_{\xi+1},T)} && {(Y,T)} \\
& {(X_\xi,T)} && {(X'_{\xi+1},T)} \\
&& {(X'_\xi,T)}
\arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2]
\arrow["\theta"{description}, from=2-2, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
\arrow[""{name=0, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
\arrow[from=1-3, to=2-4]
\arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3]
\end{tikzcd}\]
\arrow["{\pi'}", draw=none, from=1-3, to=2-4]
\arrow["{\theta'}", draw=none, from=0, to=3-3]
\arrow["\pi"', draw=none, from=1-3, to=1]
\end{tikzcd}
The diagram commutes, since all maps are the induced maps.
By definition of $Y$ is clear that $\pi$ and $\pi'$ separate points in $Y$.
@ -275,6 +279,9 @@ More generally we can show:
In particular,
$(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
\end{proof}
% TODO ANKI-MARKER
\begin{example}[{\cite[p. 513]{Furstenberg}}]
\label{ex:19:inftorus}
Let $X$ be the infinite torus

2
inputs/lecture_20.tex
View file

@ -1,5 +1,6 @@
\lecture{20}{2024-01-09}{The Infinite Torus}
\gist{
\begin{example}
\footnote{This is the same as \yaref{ex:19:inftorus},
but with new notation.}
@ -17,6 +18,7 @@
In the lecture both notations were used. % to make things extra confusing.
Here I'll try to only use multiplicative notation.
\end{remark}
}{}
We will be studying projections to the first $d$ coordinates,
i.e.
\[

16
inputs/lecture_22.tex
View file

@ -78,7 +78,7 @@
is Borel for all $\alpha < \omega_1$.
\end{enumerate}
\end{theorem}
\todo{This was already stated as \yaref{thm:beleznayforeman} in lecture 16
\todo{This was already stated as \yaref{thm:beleznay-foreman} in lecture 16
and should not have two numbers.}
A few words on the proof:
@ -86,7 +86,7 @@ Let $\mathbb{K} = S^1$
and $I$ a countable linear order.
Let $\mathbb{K}^I$ be the product of $|I|$ many $\mathbb{K}$,
$\mathbb{K}^{<i} \coloneqq \mathbb{K}^{\{j : j < i\}}$
and $\pi_{i}\colon \mathbb{K}^{I} \to \mathbb{K}^{<i}$\todo{maybe call it $\pi_{<i}$?}
and $\pi_{i}\colon \mathbb{K}^{I} \to \mathbb{K}^{<i}$% \todo{maybe call it $\pi_{<i}$?}
the projection.
Let $\mathbb{K}_I \coloneqq \prod_{i \in I} C(\mathbb{K}^{<i}, \mathbb{K})$.
@ -145,7 +145,9 @@ 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 $\mathbb{K}^I$.
@ -163,8 +165,12 @@ For this we define
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 $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
\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.~$X_{\le j}$.
@ -186,7 +192,7 @@ For this we define
&&\}
\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
}{ Not relevant for the exam.}
\end{itemize}
\end{proof}

12
inputs/lecture_23.tex
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@ -1,5 +1,9 @@
\lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman}
% TODO read notes
% TODO def. almost distal
% From Lecture 23, you need to know the proposition on page 7 (with the proof), but I won't ask you for other proofs from that lecture
\begin{notation}
Let $X$ be a Polish space and $\cP$ a property of elements of $X$,
then we say that $x_0 \in X$ is \vocab{generic}
@ -24,6 +28,7 @@ Let $I$ be a linear order
\end{theorem}
\begin{proof}[sketch]
\notexaminable{%
Consider $\WO(\N) \subset \LO(\N)$.
We know that this is $\Pi_1^1$-complete. % TODO ref
@ -65,6 +70,7 @@ Let $I$ be a linear order
$T^{\alpha}_n \subseteq W^{\alpha}_n$,
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
@ -102,6 +108,8 @@ Let $I$ be a linear order
The order %TODO (Furstenberg rank)
is a $\Pi^1_1$-rank.
\end{theorem}
\begin{proof}[sketch]
\notexaminable{
For the proof one shows that $\le^\ast$ and $<^\ast$
are $\Pi^1_1$, where
\begin{enumerate}[(1)]
@ -123,7 +131,8 @@ in $X$,
for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$,
there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$,
$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
}
\end{proof}
\begin{proposition}
The order of a minimal distal flow on a separable,
metric space is countable.
@ -169,4 +178,3 @@ $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
Then $\alpha \mapsto U_\alpha$ is an injection.
\end{proof}

14
inputs/lecture_24.tex
View file

@ -17,17 +17,19 @@
we have $X \in \cU \lor \N \setminus X \in \cU$.
\end{enumerate}
\end{definition}
\gist{
\begin{remark}
\begin{itemize}
\item If $X \cup Y \in \cU$ then $X \in \cU \lor Y$ or $Y \in \cU$:
\item If $X \cup Y \in \cU$ then $X \in \cU$ or $Y \in \cU$:
Consider $((\N \setminus X) \cap (\N \setminus Y) = \N \setminus (X \cup Y)$.
\item Every filter can be extended to an ultrafilter.
(Zorn's lemma)
\end{itemize}
\end{remark}
}{}
\begin{definition}
An ultrafilter is called \vocab[Ultrafilter!principal]{principal} or \vocab[Ultrafilter!trivial]{trivial}
if it is of the form
iff it is of the form
\[
\hat{n} = \{X \subseteq \N : n \in X\}.
\]
@ -46,9 +48,9 @@
Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
\begin{enumerate}[(1)]
\item $(\cU n) (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$.
\item $(\cU n) (\phi(n) \lor \psi(m)) \iff (\cU n) \phi(n) \lor (\cU n) \psi(n)$.
\item $(\cU n) \lnot \phi(n) \iff \lnot (\cU n) \phi(n)$.
\item $(\cU n) ~ (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$.
\item $(\cU n) ~ (\phi(n) \lor \psi(m)) \iff (\cU n) ~ \phi(n) \lor (\cU n) ~ \psi(n)$.
\item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
\end{enumerate}
\end{observe}
\begin{lemma}
@ -59,7 +61,7 @@
there is a unique $x \in X$,
such that
\[
(\cU n) (x_n \in G)
(\cU n)~(x_n \in G)
\]
for every neighbourhood%
\footnote{$G \subseteq X$ is a neighbourhood iff $x \in \inter G$.}

6
inputs/tutorial_06.tex
View file

@ -108,13 +108,13 @@ This requires the use of the axiom of choice.
\item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$.
Then $B$ is comeager in $\R$ and $|B| = \fc$.
We have $|B| = \fc$, since $B$ contains a comeager
$G_\delta$ set, $B'$:
We have $|B| = \fc$:
$B$ contains a comeager $G_\delta$ set, say $B'$.
$B'$ is Polish,
hence $B' = P \cup C$
for $P$ perfect and $C$ countable,
and $|P| \in \{\fc, 0\}$.
But $B'$ can't contain isolated point.
But $B'$ can't contain an isolated point.
\item We use $B$ to find a suitable point $a_i$:
To ensure that (i) holds, it suffices to chose

2
inputs/tutorial_07.tex
View file

@ -57,7 +57,7 @@ form a $\sigma$-algebra).
Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$.
Each $U_n$ is open, hence Borel,
so by a theorem from the lecture$^{\text{tm}}$
so by \hyperref[thm:clopenize]{a theorem from the lecture™}
there exists a Polish topology $\tau_n$
such that $U_n$ is clopen, preserving Borel sets.

2
inputs/tutorial_09.tex
View file

@ -116,7 +116,7 @@ amounts to a finite number of conditions on the preimage.
\]
is closed as an intersection of clopen sets.
Clearly $\pr_{LO(\N)}(\cF)$ is the complement
Clearly $\proj_{LO(\N)}(\cF)$ is the complement
of $WO(\N)$, hence $WO(\N)$ is coanalytic.
\end{itemize}
\nr 4

4
inputs/tutorial_12.tex
View file

@ -59,6 +59,8 @@
and since $B$ is Hausdorff, compact subsets of $B$ are closed.
\end{subproof}
\nr 1
Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup.
Let $d$ be a compatible metric on $X$.
\begin{enumerate}[(a)]
@ -100,7 +102,7 @@ Let $d$ be a compatible metric on $X$.
Consider $\ev_x \colon X^X \to X$.
$X^X$ is compact and $X$ is Hausdorff.
Hence we can apply
\label{fact:t12:2}.
\yaref{fact:t12:2}.
\item Let $x_0 \neq x_1 \in X$.
Then $(x_0,x_1)$ is a proximal pair iff $d(gx_0,gx_1) = 0$

8
inputs/tutorial_12b.tex
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@ -17,18 +17,12 @@ Then $t_n y \to (0) = t_n x$.
% TODO this is redundant
\begin{refproof}{fact:isometriciffequicontinuous}.
$d$ and $d'(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$
induce the same topology.
Let $\tau, \tau'$ be the corresponding topologies.
$\tau \subseteq \tau'$ easy,
$\tau' \subseteq \tau'$ : use equicontinuity.
\end{refproof}

1
logic.sty
View file

@ -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}