Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3

inputs/lecture_01.tex

@@ -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}.
+        \item $\cN \coloneqq \N^{\N}$ is called the \vocab{Baire space}.
-            ($\omega$ with descrete topology)
+            ($\N$ with descrete topology)
-        \item $[0,1]^{\omega}$ is called the \vocab{Hilbert cube}.
+        \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}

inputs/lecture_03.tex

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

inputs/lecture_04.tex

@@ -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}}
     \[
     D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
     \]

inputs/lecture_08.tex

@@ -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 $((y_n), x) \in \cU$
+    we set $((y_{m,n}), x) \in \cU$
-    iff $\exists n.~(y_n, x) \in U_{\xi_n}$,
+    iff $\exists n.~((y_{m,n})_{m < \omega}, x) \in U_{\xi_n}$,
-    i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$.
+    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 $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
-    Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
 \end{proof}
 \begin{remark}
     Since $2^{\omega}$ embeds

inputs/lecture_09.tex

@@ -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.
+    by the weaker assertion of $f$ being Borel.%
-    \todo{Copy exercise from sheet 5}
+    \footnote{use \yaref{thm:clopenize}, cf.~\yaref{s6e2}}
-    % TODO WHY?
 \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 $\cV \in \Sigma^1_1(Y \times X)$

inputs/lecture_10.tex

@@ -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
     $A = \bigcup_{m < \omega} A_{\underbrace{\langle m \rangle}_{\in \omega^1}}$

inputs/lecture_13.tex

@@ -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

inputs/lecture_14.tex

@@ -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 $\phi\colon R \to  \Ord$

inputs/lecture_15.tex

@@ -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}

inputs/lecture_16.tex

@@ -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}

inputs/lecture_17.tex

@@ -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.
+$X^{X}$ is a compact Hausdorff space.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}}
-\begin{remark}
+\begin{remark}%
-    \todo{Copy from exercise sheet 10}
+    \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}
-Let $(X,T)$ be a flow.
+\gist{%
-Then the \vocab{Ellis semigroup}
+    Let $(X,T)$ be a flow.
-is defined by
+    Then the \vocab{Ellis semigroup}
-$E(X,T) \coloneqq \overline{T} \subseteq X^X$,
+    is defined by
-i.e.~identify $t \in T$ with $x \mapsto tx$
+    $E(X,T) \coloneqq \overline{T} \subseteq X^X$,
-and take the closure in  $X^X$.
+    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.
 
-Properties of $(X,T)$ translate to properties of $E(X,T)$:
+\gist{
-\begin{goal}
+    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}
+    \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
+    is a nonempty semigroup\footnote{may not contain inverses or the identity}
-    Hausdorff topology,
+    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}
 
-The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
+\gist{
-since we already know that it has an identity,
+    The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
-in fact we have chosen $R = \{1\}$ in the proof.
+    since we already know that it has an identity.
-But it is interesting for other semigroups.
+    %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 $G \coloneqq  E(X,T)$ and let $d$ be a metric on $X$.
-    Let $G \coloneqq  E(X,T)$.
+\gist{
-    Let $d$ be a metric on $X$.
-
     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 $f = g' \circ g$.%
+    Since $f \in Gg$, there exists $g' \in G$ such that $f = g' \circ g$.
-    %\footnote{This exists since $f \in Gg$.}
 
     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.
+% Let $(X,T)$ be a flow.
-Then by Zorn's lemma, there exists $X_0 \subseteq X$
+% Then by Zorn's lemma, there exists $X_0 \subseteq X$
-such that $(X_0, T)$ is minimal.
+% such that $(X_0, T)$ is minimal.
-In particular,
+% In particular,
-for $x \in X$ and $\overline{Tx} = Y$
+% for $x \in X$ and $\overline{Tx} = Y$
-we have that $(Y,T)$ is a flow.
+% we have that $(Y,T)$ is a flow.
-However if we pick $y \in Y$, $Ty$ might not be dense.
+% 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
+    We need to show that $(Gx, T)$ is minimal.
-    % partition $X$.
+    Suppose that $y \in Gx$, i.e.~$Gx = Gy$.
-    % It suffices to show that $y \in Gx \implies Gy = Gx$.
+    Since $g \mapsto gy$ is continuous,
-
+    we have $Gx = Gy = \overline{T}y = \overline{Ty}$,
-%     Take some $y \in Gx$.
+    so $Ty$ is dense in $Gx$.
-%     Recall that $\overline{Ty} = \overline{T} y = Gy$.
+}{
-%     We have $\overline{Ty} \subseteq  Gx$,
+    \begin{itemize}
-%     so $Gy \subseteq Gx$.
+        \item $G$ is a group, so the $G$-orbits partition $X$.
-%     Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$,
+        \item Let $y \in Gx$, then $Gx \overset{y \in Gx}{=} Gy \overset{\text{def}}{=} \overline{T}y \overset{g \mapsto gy \text{ cts.}}{=} \overline{Ty}$,
-%     hence $Gx \subseteq  Gy$
+            i.e.~$(Gx,T)$ is minimal.
-%     TODO: WHY?
+    \end{itemize}
+}
 \end{proof}
 \begin{corollary}
     If $(X,T)$ is distal and minimal,

inputs/lecture_18.tex

@@ -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  $\epsilon_2 > 0$
+    Similarly there exists  $\epsilon_2 > 0$\gist{
-    such that $U_{\epsilon_2}(x') \subseteq  U_b(x_2)$.
+    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}
+                $(M,H)$ is equicontinuous, i.e.~it is isometric.
-                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 $x \in F_q \coloneqq B_q \setminus B_q^\circ$
+    Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$
-    and $B_q \setminus B_q^\circ$ is nwd
+    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}
 
 

inputs/lecture_19.tex

@@ -1,5 +1,5 @@
-\subsection{The order of a flow}
+\subsection{The Order of a Flow}
-\lecture{19}{2023-12-19}{Orders of flows}
+\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
+    $\sigma$ is a metric,
-    if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
+    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=
+    % https://q.uiver.app/#q=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
-\[\begin{tikzcd}
+\begin{tikzcd}
 	{(X_{\xi+1},T)} && {(Y,T)} \\
 	& {(X_\xi,T)} && {(X'_{\xi+1},T)} \\
 	&& {(X'_\xi,T)}
 	\arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2]
 	\arrow["\theta"{description}, from=2-2, to=3-3]
-	\arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
+	\arrow[""{name=0, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
-	\arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
+	\arrow[""{name=1, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
 	\arrow[from=1-3, to=2-4]
 	\arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3]
-\end{tikzcd}\]
+	\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

inputs/lecture_20.tex

@@ -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.}
@@ -14,9 +15,10 @@
 \begin{remark}+
     Note that we can identify $S^1$ with a subset of  $\C$ (and use multiplication)
     or with  $\faktor{\R}{\Z}$ (and use addition).
-    In the lecture both notations were used.% to make things extra confusing.
+    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.
 \[

inputs/lecture_22.tex

@@ -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}

inputs/lecture_23.tex

@@ -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
@@ -102,9 +108,11 @@ Let $I$ be a linear order
     The order %TODO (Furstenberg rank)
     is a $\Pi^1_1$-rank.
 \end{theorem}
-For the proof one shows that $\le^\ast$ and $<^\ast$
+\begin{proof}[sketch]
-are  $\Pi^1_1$, where
+    \notexaminable{
-\begin{enumerate}[(1)]
+        For the proof one shows that $\le^\ast$ and $<^\ast$
+        are  $\Pi^1_1$, where
+        \begin{enumerate}[(1)]
             \item $p_1 \le^\ast p_2$ iff $p_1$ codes
                 a distal minimal flow and if
                 $p_2$ also codes a distal minimal flow,
@@ -113,17 +121,18 @@ are  $\Pi^1_1$, where
                 a distal minimal flow and if
                 $p_2$ also codes a distal minimal flow,
                 then $\mathop{order}(p_1) < \mathop{order}(p_2)$.
-\end{enumerate}
+        \end{enumerate}
-
-One uses that $(Y_{i+1}, T)$ is a maximal
-isometric extension of $(Y_i,T)$
-ind $(X,T)$
-iff for all $x_1,x_2$ from a fixed countable dense set
-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 $\pi_i(z_k) = \pi_i(x_1)$,
-$F(z_k, x_1) \to  0$, $F(z_k, x_2) \to 0$.
 
+        One uses that $(Y_{i+1}, T)$ is a maximal
+        isometric extension of $(Y_i,T)$
+        ind $(X,T)$
+        iff for all $x_1,x_2$ from a fixed countable dense set
+        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}
 
-

inputs/lecture_24.tex

@@ -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 $(\cU n) ~ (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$.
-        \item $(\cU n) (\phi(n) \lor \psi(m)) \iff (\cU n) \phi(n) \lor (\cU n) \psi(n)$.
+        \item $(\cU n) ~ (\phi(n) \lor \psi(m)) \iff (\cU n) ~ \phi(n) \lor (\cU n) ~ \psi(n)$.
-        \item $(\cU n) \lnot \phi(n) \iff \lnot (\cU n) \phi(n)$.
+        \item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
     \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 $x \in \inter G$.}

inputs/tutorial_06.tex

@@ -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
+        We have $|B| = \fc$:
-        $G_\delta$ set, $B'$:
+        $B$ 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

inputs/tutorial_07.tex

@@ -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.
 

inputs/tutorial_09.tex

@@ -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

inputs/tutorial_12.tex

@@ -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 $d(gx_0,gx_1) = 0$

inputs/tutorial_12b.tex

@@ -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}

logic.sty

@@ -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}