improved lecture 17

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_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

@@ -154,7 +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_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}

inputs/lecture_16.tex

@@ -140,7 +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  $Y$.

inputs/lecture_17.tex

@@ -15,7 +15,7 @@ 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}%
     \footnote{cf.~\yaref{s11e1}}
     Let $f_0 \in X^X$ be fixed.
@@ -34,16 +34,20 @@ $X^{X}$ is a compact Hausdorff space.
 \end{remark}
 
 \begin{definition}
+\gist{%
     Let $(X,T)$ be a flow.
     Then the \vocab{Ellis semigroup}
     is defined by
     $E(X,T) \coloneqq \overline{T} \subseteq X^X$,
     i.e.~identify $t \in T$ with $x \mapsto tx$
     and take the closure in  $X^X$.
+}{%
+    The \vocab{Ellis semigroup} of a flow $(X,T)$
+    is  $E(X,T) \coloneqq \overline{T} \subseteq X^X$.
+}
 \end{definition}
 $E(X,T)$ is compact and Hausdorff,
 since $X^X$ has these properties.
-% TODO THINK ABOUT THIS
 
 \gist{
     Properties of $(X,T)$ translate to properties of $E(X,T)$:
@@ -58,15 +62,14 @@ since $X^X$ has these properties.
     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 $th \in G$.
 
-    \gist{
     We have that $t^{-1}G$ is compact,
     since $t^{-1}$ is continuous
     and $G$ is compact.
-    }{$t^{-1}G$ is compact.}
 
     It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
     
@@ -93,6 +96,21 @@ since $X^X$ has these properties.
     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}
@@ -101,9 +119,11 @@ since $X^X$ has these properties.
     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}
@@ -112,6 +132,7 @@ since $X^X$ has these properties.
     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 $s \in R$.
@@ -128,25 +149,39 @@ since $X^X$ has these properties.
     Thus $P = R$ by minimality,
     so $s \in P$,
     i.e.~$s^2 = s$.
+}{
+    \begin{itemize}
+        \item Take $R \subseteq S$ minimal compact subsemigroup (Zorn),
+            $s \in R$.
+        \item $Rs \subseteq R \implies Rs = R$.
+        \item $P \coloneqq \{x \in R : xs = s\}$:
+            \begin{itemize}
+                \item $P \neq \emptyset$, since $s \in Rs$
+                \item $P$ compact, since $P = \alpha^{-1}(s) \cap R$,
+                    $\alpha: x \mapsto xs$ cts.
+                \item $P = R \implies s^2 = s$.
+            \end{itemize}
+    \end{itemize}%
+}
 \end{proof}
 
+\gist{
     The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
     since we already know that it has an identity.
     %in fact we 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 $\Gamma \coloneqq  Gg$.
@@ -164,19 +199,39 @@ But it is interesting for other semigroups.
     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}
 
-% TODO ANKI-MARKER
+% Let $(X,T)$ be a flow.
-
+% Then by Zorn's lemma, there exists $X_0 \subseteq X$
-Let $(X,T)$ be a flow.
+% such that $(X_0, T)$ is minimal.
-Then by Zorn's lemma, there exists $X_0 \subseteq X$
+% In particular,
-such that $(X_0, T)$ is minimal.
+% for $x \in X$ and $\overline{Tx} = Y$
-In particular,
+% we have that $(Y,T)$ is a flow.
-for $x \in X$ and $\overline{Tx} = Y$
+% However if we pick $y \in Y$, $Ty$ might not be dense.
-we have that $(Y,T)$ is a flow.
-However if we pick $y \in Y$, $Ty$ might not be dense.
 % TODO: question!
 % TODO: think about this!
 % We want to a minimal subflow in a nice way:
@@ -189,6 +244,7 @@ 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 $G$.
@@ -196,17 +252,18 @@ 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

@@ -1,16 +1,11 @@
 \subsection{Sketch of proof of \yaref{thm:l16:3}}
 \lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}
 
+% TODO ANKI-MARKER
+
+
 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.

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:

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_12.tex

@@ -102,7 +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 $d(gx_0,gx_1) = 0$