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@misc{tao,
ISSN = {0003486X},
URL = {https://terrytao.wordpress.com/category/teaching/254a-ergodic-theory/},
author = {Terence Tao},
title = {254A Ergodic Theory},
urldate = {2024-01-01},
year = {2008},
}
@misc{coanalyticranks,
AUTHOR = {Christian Rosendal},
URL = {https://homepages.math.uic.edu/~rosendal/WebpagesMathCourses/MATH511-notes/DST%20notes%20-%20CoanalyticRanks05.pdf},
TITLE = {MATH511 - 5. Coanalytic Ranks and the Reflection Theorems},
YEAR = {2012},
}
@MISC{801106,
TITLE = {Closure of a topological group},
AUTHOR = {Mariano Suarez-Alvarez},
HOWPUBLISHED = {Mathematics Stack Exchange},
URL = {https://math.stackexchange.com/q/801106},
}
@article{Furstenberg,
author = {Furstenberg, H.},
journal = {American journal of mathematics},
issn = {0002-9327},
number = {3},
keywords = {Continuous functions ; Eigenfunctions ; Equivalence relation ; Geometry ; Integers ; Mathematical functions ; Mathematics ; Myelinated nerve fibers ; Structure (category theory) ; Topological compactness ; Topological spaces ; Topological theorems ; Topology},
language = {eng},
pages = {477-515},
publisher = {Johns Hopkins Press},
volume = {85},
year = {1963},
title = {The Structure of Distal Flows},
}

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\subsection{Topological Dynamics}
\begin{fact}[\url{https://math.stackexchange.com/a/801106}]
\begin{fact}[\cite{801106}]
\label{fact:topsubgroupclosure}
Let $H$ be a topological group
and $G \subseteq H$ a subgroup.

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@ -161,6 +161,7 @@ Recall:
A flow is \vocab{distal} iff
it has no proximal pair.
\end{definition}
\begin{definition}+
Let $(T,X)$ and $(T,Y)$ be flows.
A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$

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@ -135,9 +135,10 @@ $X$ is always compact metrizable.
\end{theorem}
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,
i.e.~$(X,T)$ and $(Y,T)$ are note isomorphic.
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 $Y$.

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\lecture{17}{2023-12-12}{The Ellis semigroup}
\subsection{The Ellis semigroup}
\lecture{17}{2023-12-12}{The Ellis semigroup}
Let $(X, d)$ be a compact metric space
and $(X, T)$ a flow.
@ -25,11 +23,11 @@ $X^{X}$ is a compact Hausdorff space.
\item $X^X \ni f \mapsto f \circ f_0$
is continuous:
Consider $\{f : f f_0 \in U_{\epsilon}(x,y)\}$.
Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
We have $ff_0 \in U_{\epsilon}(x,y)$
iff $f \in U_\epsilon(x,f_0(y))$.
\item Fix $x_0 \in X$.
Then $f \mapsto f(x)$ is continuous.
Then $f \mapsto f(x_0)$ is continuous.
\item In general $f \mapsto f_0 \circ f$ is not continuous,
but if $f_0$ is continuous, then the map is continuous.
\end{itemize}
@ -40,23 +38,24 @@ 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 $T$ with $x \mapsto tx$
i.e.~identify $t \in T$ with $x \mapsto tx$
and take the closure in $X^X$.
\end{definition}
$E(X,T)$ is compact and Hausdorff,
since $X^X$ has these properties.
Properties of $(X,T)$ translate to
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}
$G$ is a semigroup,
$E(X,T)$ is a semigroup,
i.e.~closed under composition.
\end{proposition}
\begin{proof}
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 $th \in G$.
@ -97,7 +96,7 @@ Properties of $(X,T)$ translate to
such that $S \ni x \mapsto xs$ is continuous for all $s$.
\end{definition}
\begin{example}
Ellis semigroup is a compact semigroup.
The Ellis semigroup is a compact semigroup.
\end{example}
\begin{lemma}[Ellis–Numakura]
@ -181,10 +180,10 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
\begin{proof}
Let $G = E(X,T)$.
Note that for all $x \in X$,
we have $Gx \subseteq X$ is compact
we have that $Gx \subseteq X$ is compact
and invariant under the action of $G$.
Since $G$ is a group, we have that the orbits partition $X$.%
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

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\subsection{Sketch of proof of \yaref{thm:l16:3}}
\lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}
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.
\begin{definition}
For $x, x' \in X$ define
\[
F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
\]
\end{definition}
\begin{fact}
\begin{enumerate}[(a)]
\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 $X^2$,
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
\begin{IEEEeqnarray*}{rCl}
f_g\colon X^2 &\longrightarrow & \R \\
(x,x') &\longmapsto & d(gx, gx')
\end{IEEEeqnarray*}
for $g \in G$.
\end{enumerate}
\end{fact}
\begin{theoremdef}
\label{def:ftop}
The sets
\[
U_a(x) \coloneqq \{x' : F(x,x') < a\}
\]
form the basis of a topology in $X$.
This topology is called the \vocab{F-topology} on $X$.
In this setting, the original topology
is also called the \vocab{E-topology}.
\end{theoremdef}
This will follow from the following lemma:
\begin{lemma}
\label{lem:ftophelper}
Let $F(x,x') < a$.
Then there exists $\epsilon > 0$ such that
whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.
\end{lemma}
\begin{refproof}{def:ftop}
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 $x' \in U_a(x_1)$.
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 $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.
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.}
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 $t \in T_n$.
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 $Z = \{\star\}$.
We want to sketch a proof of \yaref{thm:l16:3} in this case,
i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
$(X,T)$ then there is another factor $(Y,T)$ of $(X,T)$
which is a proper isometric extension of $Z$.
\begin{proof}[sketch] % TODO: Think about this
\leavevmode
\begin{enumerate}[1.]
\item For $x \in X$ define
\begin{IEEEeqnarray*}{rCl}
F_x\colon X &\longrightarrow & \R \\
x' &\longmapsto & F(x,x').
\end{IEEEeqnarray*}
\item Define an equivalence relation on $X$,
by $x_1 \sim x_2 :\iff \{x \in X : F_{x_1}(x) = F_{x_2}(x)\}$
is comeager in $X$\footnote{with respect to the E-topology}.
Then for all $g \in G$ we have
$x_1 \sim x_2 \implies gx_1 \sim ~ gx_2$.
Let $M \coloneqq \{[x]_{\sim } : x \in X\} = \faktor{X}{\sim}$
bet the quotient space.
It is compact, second countable and Hausdorff.
Let $\pi\colon X\to M$ denote the quotient map.
\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
is an isometric flow:
\begin{enumerate}
\item For $a > 0$, $x,x' \in X$ let
\[
W(x,x') \coloneqq \{g \in G : F(x, gx') < a\}.
\]
This turns out to be a subbasis of a topology
which is coarser than the original topology on $G$.
The new topology makes $G$ compact.
\item Let $\theta(g)$ be the transformation of $M$
defined by $\theta(g) \pi(x) = \pi(gx)$.
This is well defined.
Let $H = \theta(G)$.
This is just a quotient of $G$, $g \mapsto \theta(g)$
may not be injective.
\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.
In particular, $(M,T)$ is isometric.
\end{enumerate}
\item $M \neq \{\star\}$, i.e.~$(M,T)$ is non-trivial:
Suppose towards a contradiction that $M = \{\star\}$,
i.e.~$x_1 \sim x_2$ for all $x_1,x_2 \in X$.
Fix $x_2$. For every $x_1 \in X$
we have that
\[
\{x : F(x_1,x) = F(x_2,x)\}
\]
is comeager.
Let $x_1$ be a point of continuity of $F_{x_2}$.
Let $\langle a_n : n < \omega \rangle$ be a sequence
of elements that set, i.e.~$F(x_1, a_n) = F(x_2, a_n)$,
such that $a_n \to x_1$.
So by the continuity of $F_{x_2}$ at $x_1$
\begin{IEEEeqnarray*}{rCl}
\lim_{n \to \infty} F(x_2, a_n) &=& F(x_2, x_1)
\end{IEEEeqnarray*}
and by the definition of $F$
\begin{IEEEeqnarray*}{rCl}
\lim_{n \to \infty} F(x_1,a_n) &=& F(x_1,x_1) = 0.
\end{IEEEeqnarray*}
So
\[
F(x_2,x_1) = \lim_{n \to \infty} F(x_2, a_n) = \lim_{n \to \infty}
F(x_1,a_n) = 0
\]
and by distality we get $x_1 = x_2$.
Since almost all points of $X$
are points of continuity of $F_{x_2}$
(\yaref{thm:usccomeagercont})
this implies that $X \setminus \{x_2\}$ is meager.
But then $X = \{\star\} \lightning$.
\end{enumerate}
\end{proof}
\begin{theorem}\footnote{Not covered in class}
\label{thm:usccomeagercont}
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}
Take $x$ such that $\Gamma$ is not continuous at $x$.
Then there is an $\epsilon > 0$
and $x_n \to x$ such that
$\Gamma(x_n) + \epsilon \le \Gamma(x)$.
Take $q \in \Q$ such that $\Gamma(x) - \epsilon < q < \Gamma(x)$.
Then let
\[
B_q \coloneqq \{a \in X : \Gamma(a) \ge q\}.
\]
$X \setminus B_q = \{a \in X : \Gamma(a) < q\}$
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
as it is closed and has empty interior,
so $\bigcup_{q \in \Q} F_q$ is meager.
\end{proof}

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@ -0,0 +1,292 @@
\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}.
\begin{definition}+
Let $X,Y$ be metric spaces. A family $F$ of functions $X \to Y$
is called \vocab{equicontinuous} at $x_0 \in X$
iff
\[
\forall \epsilon > 0.~\exists \delta > 0.~ \forall f \in F.~d_X(x_0, x) < \delta \implies d_Y(f(x_0),f(x)) < \epsilon.
\]
It is called equicontinuous iff it is equicontinuous at every point.
It is called \vocab{uniformly equicontinuous}
iff
\[
\forall \epsilon > 0.~\exists \delta > 0.~ \forall x_0 \in X.~\forall f \in F.~d_X(x_0, x) < \delta \implies d_Y(f(x_0),f(x)) < \epsilon.
\]
A flow $(X,T)$ is called equicontinuous iff $T$ is equicontinuous.
\end{definition}
Note that since $X$ compact the notions of equicontinuity and uniform
equicontinuity coincide.
\begin{fact}+[{\cite[Lecture 6, Exercise 1]{tao}}]
A flow $(X,T)$ is isometric iff it is equicontinuous.
\end{fact}
\begin{proof}
Clearly an isometric flow is equicontinuous.
On the other hand suppose that $T$ is uniformly equicontinuous.
Define a metric $\tilde{d}$ on $X$ by setting
$\tilde{d}(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$.
By equicontinuity of $T$ we get that $\tilde{d}$ and $d$
induce the same topology on $X$.
\end{proof}
\begin{question}
What is the minimal number of steps required
when building the tower to reach the flow
as in \yaref{thm:l16:3}?
\end{question}
\begin{definition}[{\cite[{}13.1]{Furstenberg}}]
Let $(X,T)$ be a quasi isometric flow,
and let $\eta$ be the smallest ordinal
such that there exists a quasi-isometric system $\{(X_\xi, T), \xi \le \eta\}$
with $(X,T) = (X_\xi, T)$.
Then $\eta$ is called the \vocab{order} of the flow.
\end{definition}
\begin{theorem}[Maximal isometric factor]
\label{thm:maxisomfactor}
For every flow $(X,T)$ there is a maximal factor $(Y,T)$, $\pi\colon X\to Y$,
i.e.~if $(Y',T), \pi'\colon X \to Y'$ is any isometric factor of $(X,T)$,
then $(Y',T)$ is a factor of $(Y,T)$.
% https://q.uiver.app/#q=WzAsMyxbMCwwLCIoWCxUKSJdLFsxLDEsIihZLFQpXFxcXFxcdGV4dHttYXhpbWFsIGlzb21ldHJpY30iXSxbMCwyLCIoWScsVCkiXSxbMCwyLCJcXHRleHR7aXNvbWV0cmljfSIsMl0sWzAsMV0sWzEsMiwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
\[\begin{tikzcd}
{(X,T)} \\
& {\substack{(Y,T)\\\text{maximal isometric}}} \\
{(Y',T)}
\arrow["{\text{isometric}}"', from=1-1, to=3-1]
\arrow[from=1-1, to=2-2]
\arrow["\exists", dashed, from=2-2, to=3-1]
\end{tikzcd}\]
\end{theorem}
\begin{proof}
% TODO Think about this
We want to apply Zorn's lemma.
If suffices to show that isometric flows are closed under inverse limits,
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
\[\begin{tikzcd}
Y & {Y_\alpha} & X \\
& {Y_\beta}
\arrow["{f_{\alpha, \beta}}", from=1-2, to=2-2]
\arrow["{f_\alpha}", from=1-1, to=1-2]
\arrow["{f_\beta}"', from=1-1, to=2-2]
\arrow["{\pi_\alpha}"', from=1-3, to=1-2]
\arrow["{\pi_\beta}", from=1-3, to=2-2]
\end{tikzcd}\]
Consider
\[
\Delta_\alpha \coloneqq \{(y,y') \in Y^2 : f_{\alpha}(y) = f_\alpha(y')\}.
\]
Let $d$ be a metric on $Y$ and $d_{\alpha}$ a metric on $Y_{\alpha}$,
wlog.~$d, d_\alpha \le 1$.
Note that $\beta < \alpha \implies \Delta_\beta \supseteq \Delta_\alpha$
and
\[
\bigcap_{\alpha \le \theta}\Delta_\alpha = \{(y,y) : y \in Y\}.
\]
Consider
\[\{(y,y') \in \Delta_\alpha : d(y,y') \ge \epsilon\}\]
for any $\epsilon > 0$.
By the finite intersection property % TODO WHY? TODO what is this TODO for compact?
we get
\[
\exists \alpha.~f_\alpha(y) = f_\alpha(y') \implies d(y,y') < \epsilon,
\]
i.e.~$\forall z \in Y_\alpha.~\diam(f^{-1}_\alpha(z)) \le \epsilon$.
Towards a contradiction assume that $Y$ is not isometric,
i.e.~not equicontinuous.
Then there are $(y_j), (y'_j) \in Y$
such that $d(y_j,y'_j) \to 0$
and $\epsilon > 0, t_j \in T$
such that $d(t_jy_j, t_jy'_j) > \epsilon$.
By compactness wlog.~$(y_j)$ and $(y'_j)$
converge (to the same point).
Find $\alpha$ such that $f_\alpha(y) = f_{\alpha}(y') \implies d(y,y') < \frac{\epsilon}{4}$.
Let $z_j \coloneqq f_{\alpha}(y_j)$ and $z'_j \coloneqq f_\alpha(y'_j)$.
Then $(z_j)$ and $(z'_j)$ converge to the same point $z \in Y_\alpha$.
By equicontinuity of $(Y_\alpha, T)$,
$d_{Y_{\alpha}}(t_jz_j, t_jz'_j) \to 0$.
Wlog.~$(t_jz_j)$ and $(t_jz'_j)$ converge.
Let $z^\ast$ be their limit.
On the one hand, by the triangle inequality we get
\[
d(f^{-1}_\alpha(t_jz_j), f^{-1}_\alpha(t_jz_j')) > \underbrace{\epsilon}_{\mathclap{< d(t_jy_j, t_jy_j')}} - \overbrace{\frac{\epsilon}{4}}^{\mathclap{\text{Diameter of fiber}}}- \frac{\epsilon}{4} = \frac{\epsilon}{2}.
\]
On the other hand, from
\begin{IEEEeqnarray*}{rCl}
d(f^{-1}_\alpha(t_jz_j), f^{-1}_{\alpha}(z^\ast)) &\to & 0,\\
d(f^{-1}_\alpha(t_jz'_j), f^{-1}_{\alpha}(z^{\ast})) &\to & 0,\\
\diam f^{-1}_\alpha(\{z^\ast\}) & <& \frac{\epsilon}{4}
\end{IEEEeqnarray*}
we obtain
\[
d(f^{-1}_\alpha(t_jz_j), f^{-1}_\alpha(t_jz'_j)) < \frac{\epsilon}{2} \lightning.
\]
\end{proof}
More generally we can show:
\begin{theorem}[{\cite[13.1]{Furstenberg}}]
Let $(X,T)$ be a distal flow
and $(Y,T) = \pi(X,T)$ a factor.
Then there exists an isometric extension $(Y,T)$ of $(Z,T)$
which is a factor of $(X,T)$,
such that $(Y,T)$ is maximal among such extensions,
i.e.~if $(Y',T)$ is any flow with these two properties,
then $(Y',T)$ is a factor of $(Y,T)$.
% https://q.uiver.app/#q=WzAsMyxbMCwwLCIoWCxUKSJdLFswLDIsIihaLFQpIl0sWzEsMSwiKFksVCkiXSxbMCwyXSxbMCwxLCJcXHBpIl0sWzIsMV1d
\[\begin{tikzcd}
{(X,T)} \\
& {(Y,T)} \\
{(Z,T)}
\arrow[from=1-1, to=2-2]
\arrow["\pi", from=1-1, to=3-1]
\arrow[from=2-2, to=3-1]
\end{tikzcd}\]
\end{theorem}
\begin{lemma}
\label{lec19:lem1}
Let four flows be given as in
% https://q.uiver.app/#q=WzAsNCxbMSwwLCIoWSxUKSJdLFsyLDEsIihaXzIsIFQpIl0sWzAsMSwiKFpfMSwgVCkiXSxbMSwyLCIoVyxUKSJdLFsyLDMsIndfMSJdLFsxLDMsIndfMiIsMl0sWzAsMiwiXFxwaV8xIl0sWzAsMSwiXFxwaV8yIiwyXV0=
\[\begin{tikzcd}
& {(Y,T)} \\
{(Z_1, T)} && {(Z_2, T)} \\
& {(W,T)}
\arrow["{w_1}", from=2-1, to=3-2]
\arrow["{w_2}"', from=2-3, to=3-2]
\arrow["{\pi_1}", from=1-2, to=2-1]
\arrow["{\pi_2}"', from=1-2, to=2-3]
\end{tikzcd}\]
Suppose that whenever $y \neq y' \in Y$,
then either % TODO REALLY?
$\pi_1(y) \neq \pi(y')$
or $\pi_2(y) \neq \pi_2(y')$.
If $(Z_1,T)$ is an isometric extension of $(W,T)$,
then $(Y,T)$ is an isometric extension of $(Z_2, T)$.
\end{lemma}
\begin{proof}
% TODO TODO TODO Think about this
For $z_1,z_1' \in Z_1$ with
$w_1(z_1) = w_1(z_1')$ let
$\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$.
Set $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$ whenever $\pi_2(y) = \pi_2(y')$.
In this case $w_2 \circ \pi_2(y) = w_2 \circ \pi_2(y')$
and $w_1 \circ \pi_1(y) = w_1 \circ \pi_1(y')$,
so $\sigma$ is well defined.
$\sigma$ is a semi-metric\footnote{Like a metric, but the distinct points can have distance $0$.}
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 $\sigma(y,y') = 0$,
then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
\end{proof}
\begin{definition}
A quasi-isometric system
$\{(X_\xi, T) : \xi \le \eta\}$
is called \vocab{normal} if $(X_{\xi+1}, T)$ is the maximal
isometric extension of $(X_\xi,T)$ in $(X_\eta, T)$
for all $\xi < \eta$.
\end{definition}
\begin{theorem}[{\cite[{}13.2]{Furstenberg}}]
If $\{(X_\xi, T), \xi \le \eta\}$
is a normal quasi-isometric
system, then $(X_\eta, T)$ has order $\eta$.
\end{theorem}
\begin{proof}
We only sketch the proof here.
Details can be found in \cite{Furstenberg}, section 13.
Let $\{(X_\xi', T), \xi \le \eta'\} $ be
another quasi-isometric system
terminating with $(X_\eta, T) = (X'_{\eta'}, T)$.
We want to show that $\eta' \ge \eta$.
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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
\[\begin{tikzcd}
{X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\
X \\
{X_\eta} & \dots & {X_3} & {X_2} & {X_1}
\arrow[Rightarrow, no head, from=1-1, to=2-1]
\arrow[Rightarrow, no head, from=2-1, to=3-1]
\arrow[from=3-3, to=3-4]
\arrow[from=3-4, to=3-5]
\arrow[from=1-4, to=1-5]
\arrow[from=1-3, to=1-4]
\arrow[dotted, from=3-3, to=1-3]
\arrow[dotted, from=3-4, to=1-4]
\arrow[dotted, from=3-5, to=1-5]
\arrow["{\pi_3}"', curve={height=6pt}, from=3-1, to=3-3]
\arrow["{\pi_2}"', curve={height=18pt}, from=3-1, to=3-4]
\arrow["{\pi_1}"', curve={height=30pt}, from=3-1, to=3-5]
\arrow["{\pi'_3}", curve={height=-6pt}, from=1-1, to=1-3]
\arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4]
\arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}\]
% TODO: induction start?
Suppose we have
$(X'_\xi, T) = \theta((X_\xi, T)$.
Let $\pi_\xi$ and $\pi'_\xi$ denote the maps from $X$ to $X_\xi$ resp.~$X'_\xi$.
Set
\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}.\]
Then
% https://q.uiver.app/#q=WzAsNSxbMCwwLCIoWF97XFx4aSsxfSxUKSJdLFsyLDAsIihZLFQpIl0sWzMsMSwiKFgnX3tcXHhpKzF9LFQpIl0sWzIsMiwiKFgnX1xceGksVCkiXSxbMSwxLCIoWF9cXHhpLFQpIl0sWzAsNCwiXFx0ZXh0e21heC5+aXNvfSIsMV0sWzQsMywiXFx0aGV0YSIsMV0sWzIsMywie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsNCwie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsMl0sWzAsMSwiIiwwLHsiY3VydmUiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
\[\begin{tikzcd}
{(X_{\xi+1},T)} && {(Y,T)} \\
& {(X_\xi,T)} && {(X'_{\xi+1},T)} \\
&& {(X'_\xi,T)}
\arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2]
\arrow["\theta"{description}, from=2-2, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
\arrow[from=1-3, to=2-4]
\arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3]
\end{tikzcd}\]
The diagram commutes, since all maps are the induced maps.
By definition of $Y$ is clear that $\pi$ and $\pi'$ separate points in $Y$.
Thus \yaref{lec19:lem1} can be applied.
Since $\theta'$ is an isometric extension, so is $\pi$.
Then $(Y,T)$ is a factor of $(X_{\xi+1}, T)$ by
the maximality of the isometric extension
$(X_{\xi+1 }, T) \to (X_\xi, T)$.
In particular,
$(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
\end{proof}
\begin{example}[{\cite[p. 513]{Furstenberg}}]
Let $X$ be the infinite torus
\[
X \coloneqq \{(\xi_1, \xi_2, \ldots) : \xi_i \in \C, |\xi_i| = 1\}.
\]
Let $\pi_n$ be the projection to the first $n$ coordinates
and $X_n \coloneqq \pi_n(X)$.
Let $\tau_1(\xi_1,\xi_2, \ldots, \xi_n, \ldots) = (e^{\i \alpha} \xi_1, \xi_1\xi_2, \ldots, \xi_{n-1}\xi_n, \ldots)$
where $\frac{\alpha}{\pi}$ is irrational.
Let $T = \langle \tau_1 \rangle \cong \Z$.
We will show that $(X_n,T)$ is minimal for all $n$,
and so $(X,T)$ is minimal.
Furthermore $(X_{n+1},T)$ is the maximal isometric extension of $(X_n,T)$
so $(X,T)$ has order $\omega$.
\end{example}

26
inputs/tutorial_02.tex
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@ -229,4 +229,28 @@ Clearly $d_u$ is a metric.
\begin{claim}
There exists a countable dense subset.
\end{claim}
\todo{handwritten solution}
\begin{subproof}
Fix a metric $d_X$ on $X$ defining its topology.
Let
\[
C_{m,n} \coloneqq \{f \in \cC(X,Y) : \forall x,y \in X.~\left( d_X(x,y) < \frac{1}{m+1} \implies d(f(x), f(y)) <\frac{1}{n+1}\right) \}.
\]
Choose $X_m \subseteq X$ finite with $X \subseteq \bigcup_{x \in X_m} B_{\frac{1}{m+1}}(x)$.
Let $D_{m,n} \subseteq C_{m,n}$ be countable,
such that for every $f \in C_{m,n}$ and every $\eta > 0$,
there is $g \in D_{m,n}$ with $d(f(y), g(y)) < \frac{\eta}{3}$
for each $y \in X_m$.
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
we finde $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
since $f$ is uniformly continuous.
Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
We have $d_u(f,g) \le \eta$,
since for every $x \in X$, we find $y \in X_m$ with $d_X(x,y) < \frac{1}{m+1}$,
hence
\begin{IEEEeqnarray*}{rCl}
d_Y(f(x), g(x)) &\le& d_Y(f(x), f(y)) + d_Y(f(y), g(y)) + d_Y(g(y), g(x))\\
&\le& \frac{1}{n+1} + \frac{1}{n+1} + \frac{1}{n+1} \le \eta.
\end{IEEEeqnarray*}
\end{subproof}

5
inputs/tutorial_09.tex
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@ -16,9 +16,8 @@
Material on topological dynamics:
\begin{itemize}
\item Terence Tao's notes on ergodic theory 254A:
\url{https://terrytao.wordpress.com/category/teaching/254a-ergodic-theory/}
\item Furstenberg (uses very different notation!).
\item Terence Tao's notes on ergodic theory 254A: \cite{tao}
\item \cite{Furstenberg} (uses very different notation!).
\end{itemize}

202
inputs/tutorial_10.tex Normal file
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@ -0,0 +1,202 @@
\tutorial{10}{2023-12-19}{Sheet 9}
\subsection{Sheet 9}
\nr 1
$(X, \tau') \xrightarrow{x \mapsto x} (X, \tau)$ is Borel
(by one of the equivalent definitions of being Borel).
Thus $\cB(X, \tau) \subseteq \cB(X, \tau')$
(by the other equivalent definition of being Borel).
Let $U \subseteq (X, \tau')$ be Borel.
$\id\defon{U}$ is injective,
hence $U$ is Borel in $(X, \tau)$ by Lusin-Suslin.
\paragraph{Related stuff}
\begin{fact}
Let $X,Y$ be Polish.
$f\colon X \to Y$
is Borel iff its graph $\Gamma_f$ is Borel.
\end{fact}
\begin{proof}
Take a countable open base
$V_0, V_1, \ldots$ of $Y$.
Then $\Gamma_f = \{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}$
(because the space is Hausdorff).
If $f$ is Borel,
then clearly the RHS is Borel
since
\begin{IEEEeqnarray*}{rCl}
&&\{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}\\
&=& \bigcap_{n < \omega} (f^{-1}(V_n)^{c}Y \cup f^{-1}(V_n) \times V_n\}\\
\end{IEEEeqnarray*}
On the other hand suppose that $\Gamma_f$ is Borel.
Then
\[
f^{-1}(B) = \pi_X(X \times B \cap \Gamma_f)
\]
is analytic.\footnote{Note that the projection of a Borel set is not necessarily Borel.
Moreover note that we only used that $\Gamma_f$ is analytic.}
On the other hand
\[
f^{-1}(B)^c = f^{-1}(B^c)
\]
is analytic
and we know that $\Sigma_1^1 \cap \Pi_1^1 = \cB$
by the \yaref{cor:lusinseparation}.
\end{proof}
In fact we have shown
\begin{fact}
The following are equivalent
\begin{itemize}
\item $f$ is Borel,
\item $\Gamma_f$ is Borel,
\item $\Gamma_f$ is analytic.
\end{itemize}
\end{fact}
\nr 2
\begin{definition}
Let $X$ be a topological space.
Let $K(X)$ be the set of all compact subspaces of $X$.
The \vocab{Vietoris Topology}, $\tau_V$, on $K(X)$
is the topology with basic open sets
\[
[U_0; U_1, \ldots, U_n] = \{K \in K(X) : K \subseteq U_0 \land \forall 1 \le i \le n .~K \cap U_i \neq \emptyset\}
\]
for $U_i \overset{\text{open}}{\subseteq} X$.
\end{definition}
\begin{definition}
Let $(X,d)$ be a matric space with $d \le 1$.
We define a metric $d_H$ on $K(X)$ as follows:
$d_H(\emptyset, \emptyset) \coloneqq 0$, $d_H(K, \emptyset) \coloneqq 1$ for $K \neq \emptyset$
and
\[
d_H(K_0, K_1) \coloneqq \max \{\max_{x \in K_0} d(x,K_1), \max_{x \in K_1} d(x,K_0)\}
\]
for $K_0, K_1 \neq \emptyset$.
\end{definition}
\begin{fact}
$d_H$ is indeed a metric.
\end{fact}
\begin{proof}
Let $\delta(K, L) \coloneqq \max_{x \in K} d(x,L)$.
It suffices to show
$\delta(X,Z) \le \delta(X,Y) + \delta(Y,Z)$,
since then
\begin{IEEEeqnarray*}{rCl}
d_H(X,Z) &\le & \max \{\delta(X,Y) + \delta(Y,Z), \delta(Z,Y) + \delta(Y,X)\}\\
&\le & d_H(X,Y) + d_H(Y,Z).
\end{IEEEeqnarray*}
Using the fact that $d(\cdot , Z)$ is uniformly continuous,
specifically
\[
|d(x,Z) - d(y,Z)| \le d(x,y), % TODO REF SHEET 1
\]
we get
\begin{IEEEeqnarray*}{lrCl}
&d(x,Z) &\le & d(x,y) + d(y, Z)\\
& &\le & d(x,y) + \delta(Y,Z)\\
\implies& d(x,Z) - \delta(Y,Z) &\le & d(x,Y)\\
\implies& d(x,Z) &\le & \delta(X,Y) + \delta(Y,Z)\\
\implies & \delta(X,Z) &\le & \delta(X,Y) + \delta(Y,Z).
\end{IEEEeqnarray*}
\end{proof}
\begin{itemize}
\item We have
\begin{IEEEeqnarray*}{rCl}
d_H(K_0, K_1) < \epsilon & \iff & \max \{\max_{x \in K_0}d(x, K_1), \max_{x \in K_1} d(x,K_0)\} < \epsilon\\
&\iff& \max_{x \in K_0} d(x, K_1) < \epsilon \land \max_{x \in K_1} d(x, K_0) < \epsilon\\
&\iff& K_0 \subseteq B_{\epsilon}(K_1) \land K_1 \subseteq B_\epsilon(K_0).
\end{IEEEeqnarray*}
\item Note that a subbase of $\tau_V$
is given by $[U]$ and $\langle U \rangle \coloneqq [X;U]$ for $U \overset{\text{open}}{\subseteq} X$.
Let $K \in [U]$.
Then $d(\cdot , U^c)\colon U \to \R_{\ge 0}$
is always non-zero and continuous.
So $d(K,U^c)$ attains a minimum $\epsilon > 0$.
Then $B_{\epsilon}^H(K) \subseteq U$,
so $[U]$ is open in $\tau_V$.
Let $K \in \langle U \rangle$.
Take some $k \in K \cap U$.
Then there is some $\epsilon > 0$
such that $B_\epsilon(k) \subseteq U$.
Then $K \in B_{\epsilon}^H(K) \subseteq \langle U \rangle$.
\todo{Other direction}
% $\tau_H \subseteq \tau_V$
\item Consider a countable dense subset of $X$.
Let $\cK$ be the set of finite subsets of that countable dense subset.
Then $\cK \subseteq K(X)$ is dense:
Take $K \in K(X)$ an let $\epsilon > 0$.
$K$ can be covered with finitely many $\epsilon$-balls
with centers from the countable dense subsets.
Let $K' \in \cK$ be the set of the centers.
Then $d_H(K, K') \le \epsilon$.
\end{itemize}
\nr 3
\begin{itemize}
\item By transfinite induction we get that $\alpha$ is an ordinal,
since $\prec$ is well-founded and the supremum of a sets
of ordinals is an ordinal.
Since $\rho_{\prec}\colon X \to \alpha$
is a surjection, it follows that $\alpha \le |X|$,
i.e.~$\alpha < |X|^+$.
\item By induction on $\rho_{\prec_X}(x)$ we show that
$\rho_{\prec_{X}}(x) \le \rho_{\prec_Y}(f(x))$.
For $0$ this is trivial.
Suppose that $\rho_{\prec_{X}}(x) = \alpha$
and the statement was shown for all $\beta < \alpha$.
Then
\begin{IEEEeqnarray*}{rCl}
\rho_{\prec_Y}(f(x)) &=& \sup \{\rho_{\prec_Y}(y') + 1 | y' \prec f(x)\}\\
&\ge& \sup \{\rho_{\prec_Y}(f(x')) + 1 | f(x') \prec f(x)\}\\
&\ge & \sup \{\rho_{\prec_Y}(f(x')) + 1 | x' \prec x\}\\
&\ge & \sup \{\rho_{\prec_X}(x') + 1 | x' \prec x\}\\
&=& \rho_{\prec_X}(x).
\end{IEEEeqnarray*}
\item Infinite branches of $T_\prec$ correspond
to infinite descending chain of $\prec$,
hence $T_{\prec}$ is well-founded iff $\prec$ is well-founded.
% Unwarp$^{\text{tm}}$ definitions.
Suppose that $\prec$ is well-founded.
Note that $\rho_T(s)$ depends only on the last element of $s$,
as for $s, s' \in T$ with the same last element,
we have $s \concat x \in T \iff s' \concat x \in T$.
Let $s = (s_0, \ldots, s_n)$.
Let us show that $\rho_T(s) = \rho_{\prec}(s_n)$.
We use induction on $\rho_T(s)$.
For leaves this is immediate.
From the last exercise sheet we know that
\[
\rho_T(s) = \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}.
\]
Hence
\begin{IEEEeqnarray*}{rCl}
\rho_T(s) &=& \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}\\
&=& \sup \{\rho_{\prec}(a) + 1 | s \concat a \in T\}\\
&=& \sup \{\rho_{\prec}(a) + 1 | a \prec s_n\}\\
&=& \rho_\prec(s_n).
\end{IEEEeqnarray*}
\end{itemize}
\nr 4
A solution can be found in \cite{coanalyticranks}.
% TODO Copy relevant points

3
logic.sty
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@ -56,6 +56,7 @@
\usepackage{imakeidx}
\makeindex[name = ccode, title = \texttt{C} functions and macros]
\PassOptionsToPackage{hyphens}{url}%
\usepackage{hyperref}
\usepackage[quotation]{knowledge}[22/02/12]
@ -153,3 +154,5 @@
\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}}
\usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}

4
logic3.tex
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@ -42,6 +42,8 @@
\input{inputs/lecture_15}
\input{inputs/lecture_16}
\input{inputs/lecture_17}
\input{inputs/lecture_18}
\input{inputs/lecture_19}
@ -70,5 +72,7 @@
\PrintVocabIndex
\printbibliography
\end{document}