bibliography/references.bib

@@ -1,27 +0,0 @@
-@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{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},
-}

inputs/facts.tex

@@ -1,6 +1,6 @@
 \subsection{Topological Dynamics}
 
-\begin{fact}[\cite{801106}]
+\begin{fact}[\url{https://math.stackexchange.com/a/801106}]
     \label{fact:topsubgroupclosure}
     Let $H$ be a topological group
     and $G \subseteq H$ a subgroup.

inputs/lecture_15.tex

@@ -161,7 +161,6 @@ 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} $\pi\colon (T,X) \to (T,Y)$

inputs/lecture_16.tex

@@ -135,10 +135,9 @@ $X$ 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.
+    and let $(Y, T)$ be a proper factor,
-    \footnote{i.e.~$(X,T)$ and $(Y,T)$ are not isomorphic}
+    i.e.~$(X,T)$ and $(Y,T)$ are note isomorphic.
     Then there is another factor $(Z,T)$ of $(X,T)$
     which is a proper isometric extension of  $Y$.
 

inputs/lecture_17.tex

@@ -1,6 +1,8 @@
-\subsection{The Ellis semigroup}
 \lecture{17}{2023-12-12}{The Ellis semigroup}
 
+
+\subsection{The Ellis semigroup}
+
 Let $(X, d)$ be a compact metric space
 and $(X, T)$ a flow.
 
@@ -27,7 +29,7 @@ $X^{X}$ is a compact Hausdorff space.
             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  $f \mapsto f(x_0)$ is continuous.
+            Then  $f \mapsto f(x)$ 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}
@@ -38,24 +40,23 @@ 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 $x \mapsto tx$
+i.e.~identify $T$ 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 $E(X,T)$:
+Properties of $(X,T)$ translate to 
 \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,
+    $G$ 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 $th \in G$.
 
@@ -96,7 +97,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
     such that $S \ni x \mapsto xs$ is continuous for all  $s$.
 \end{definition}
 \begin{example}
-    The Ellis semigroup is a compact semigroup.
+    Ellis semigroup is a compact semigroup.
 \end{example}
 
 \begin{lemma}[Ellis–Numakura]
@@ -180,10 +181,10 @@ 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 that $Gx \subseteq X$ is compact
+    we have $Gx \subseteq X$ is compact
     and invariant under the action of $G$.
 
-    Since $G$ is a group, the orbits partition $X$.%
+    Since $G$ is a group, we have that 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

inputs/lecture_18.tex

@@ -1,236 +0,0 @@
-\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 for all $n$.
-    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}
-
-
-

inputs/lecture_19.tex

@@ -1,292 +0,0 @@
-\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}

inputs/tutorial_02.tex

@@ -229,28 +229,4 @@ Clearly $d_u$ is a metric.
 \begin{claim}
     There exists a countable dense subset.
 \end{claim}
-\begin{subproof}
+\todo{handwritten solution}
-    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}

inputs/tutorial_09.tex

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

logic.sty

@@ -56,7 +56,6 @@
 \usepackage{imakeidx}
 \makeindex[name = ccode, title = \texttt{C} functions and macros]
 
-\PassOptionsToPackage{hyphens}{url}%
 \usepackage{hyperref}
 
 \usepackage[quotation]{knowledge}[22/02/12]
@@ -154,5 +153,3 @@
 \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}

logic3.tex

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