more gist

inputs/lecture_01.tex

@@ -15,7 +15,7 @@
 \end{definition}
 Note that Polishness is preserved under homeomorphisms,
 i.e.~it is really a topological property.
-
+\gist{%
 \begin{example}
     \begin{itemize}
         \item $\R$ is a Polish space,
@@ -26,6 +26,7 @@ i.e.~it is really a topological property.
             considered as a topological space.
     \end{itemize}
 \end{example}
+}{}
 \gist{%
 Polish spaces behave very nicely.
 We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum
@@ -75,11 +76,13 @@ However the converse of this does not hold.
         \item \vocab{Lindelöf} (every open cover has a countable subcover).
     \end{itemize}
 \end{fact}
+\gist{%
 \begin{fact}
     Compact Hausdorff spaces are \vocab{normal} (T4)
     i.e.~two disjoint closed subsets can be separated
     by open sets.
 \end{fact}
+}{}
 \begin{fact}
     For a metric space, the following are equivalent:
     \begin{itemize}
@@ -172,7 +175,8 @@ suffices to show that open balls in one metric are unions of open balls in the o
         D\left( (x_n), (y_n) \right) \coloneqq
         \sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
     \]
-    Clearly $D \le 1$.
+\gist{%
+        Clearly $D \le 1$.
     It is also clear, that $D$ is a metric.
 
     We need to check that $D$ is complete:
@@ -180,6 +184,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
     Consider the pointwise limit $(a_n)$.
     This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
     Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
+}{Clearly $D$ is a complete metric.}
 \end{proof}
 
 \begin{definition}[Our favourite Polish spaces]

inputs/lecture_03.tex

@@ -59,7 +59,6 @@
     define extension, initial segments
     and concatenation of a finite sequence with an infinite one.
 \end{notation}
-}{}
 
 \begin{definition}
     A \vocab{tree}
@@ -87,6 +86,7 @@
     \forall  t\in T.\exists  s \supsetneq t.~s \in T.
     \]
 \end{definition}
+}{}
 
 \begin{definition}
     A \vocab{Cantor scheme}

inputs/lecture_06.tex

@@ -139,7 +139,7 @@
 
 \subsection{The hierarchy of Borel sets}
 
-Let $\omega_1$ be the first uncountable ordinal.
+\gist{Let $\omega_1$ be the first uncountable ordinal.}{}
 For every $d < \omega_1$,
 we define by transfinite recursion
 classes $\Sigma^0_\alpha$
@@ -201,6 +201,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
+\gist{%
     \begin{enumerate}[(a)]
         \item \begin{observe}
                 \label{ob:sigmasuffices}
@@ -215,7 +216,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
             since $\Delta^0_\eta(X)$ is closed under complements.
 
             Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$,
-            by \yaref{ob:sigmasuffices}
+            by the observation
             (since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$
             and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$).
 
@@ -246,6 +247,17 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
             Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$.
             It is clear that $\cB_0$ is closed under complements.
     \end{enumerate}
+}{
+    \begin{enumerate}[(a)]
+        \item It suffices to show that $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$
+            for all $1 \le  \eta < \xi < \omega_1$.
+            For $\eta = 1, \xi = 2$ this holds,
+            since open sets of a metrizable space are  $F_\sigma$.
+            Induction.
+        \item Let  $\cB_0 \coloneqq  \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
+            This is a $\sigma$-algebra containing all open sets.
+    \end{enumerate}
+}
 
 \end{proof}
 

inputs/lecture_07.tex

@@ -6,6 +6,7 @@
     % $\fc := 2^{\aleph_0}$
 \end{proposition}
 \begin{proof}
+\gist{%
     We use strong induction on  $\xi < \omega_1$.
     We have $\Sigma^0_1(X) \le \fc$
     (for every element of the basis, we can decide
@@ -27,6 +28,9 @@
     \[
     |\cB(X)| \le  \omega_1 \cdot  \fc = \fc.
     \]
+}{Use strong induction. $|\Sigma^0_1(X)| \le \fc$, since  $X$ is second countable.
+    \[|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\]
+}
 \end{proof}
 
 \begin{proposition}[Closure properties]

inputs/lecture_08.tex

@@ -41,6 +41,7 @@ where $X$ is a metrizable, usually second countable space.
     similarly for $\Pi^0_\xi(X)$.
 \end{theorem}
 \begin{proof}
+\gist{%
     Note that if $\cU$ is $2^{\omega}$ universal for
     $\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
     is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
@@ -89,6 +90,18 @@ where $X$ is a metrizable, usually second countable space.
     Let $A \in \Sigma^0_\xi(X)$.
     Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
     Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
+}{
+    \begin{itemize}
+        \item Suffices for $\Sigma^0_\xi$ (complement is $\Pi^0_\xi$-universal).
+        \item $2^\omega$-universal set for $\Sigma^0_1(X)$, since
+            $X$ is second countable ($(y,x) \in \cU \iff x \in \bigcup_n \{V_n : y_n = 1\}$).
+        \item Induction: Take $\xi_k \to \xi$,
+             and $\cU_{\xi_k}$ $\Sigma^0_{\xi_k}$-universal.
+             Construct $2^{ \omega \times  \omega}$-universal:
+
+             $(y_{m,n}, x) \in \cU :\iff \exists n.~((y_{m,n}), x) \in \cU_{\xi_n}$.
+    \end{itemize}
+}
 \end{proof}
 \begin{remark}
     Since $2^{\omega}$ embeds
@@ -96,6 +109,6 @@ where $X$ is a metrizable, usually second countable space.
     % such that the image is closed,
     we can replace $2^{\omega}$ by $Y$
     in the statement of the theorem.%
-    \footnote{By definition of the subspace topology
-    and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
+    \gist{\footnote{By definition of the subspace topology
+    and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}}{}
 \end{remark}

inputs/lecture_14.tex

@@ -108,14 +108,14 @@
 \end{theorem}
 \begin{proof}
     Let $\phi\colon R \to  \Ord$
-    by a $\Pi^1_1$-rank.
-    Set
+    be a $\Pi^1_1$-rank.
+    \gist{Set
     \begin{IEEEeqnarray*}{rCl}
         (x,n) \in R^\ast &:\iff& (x,n) \in R\\
                          &&\land \forall m.~(x,n) \le^\ast_\phi (x,m)\\
                          &&\land \forall m.~\left( (x,n) <^\ast_\phi (x,m) \lor n \le m \right),
     \end{IEEEeqnarray*}
-    i.e.~take the element with minimal rank
+    i.e.~take}{Take} the element with minimal rank
     that has the minimal second coordinate among those elements.
 \end{proof}
 \gist{
@@ -196,8 +196,8 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
     \begin{itemize}
         \item $n < m$ and
-        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.%
-            \footnote{sic! (there is a typo in the official notes)}
+        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
+            % \footnote{sic! (there was a typo in the official notes)}
     \end{itemize}
     \begin{claim}
         $\prec^\ast$ is well-founded.

inputs/lecture_15.tex

@@ -66,7 +66,7 @@
 
 % \subsection*{Basic Definitions}
 % TODO: move to appendix?
-
+\gist{%
 Recall:
 \begin{definition}+
     Let $X$ be a set.
@@ -112,6 +112,7 @@ Recall:
         g&\longmapsto & (x \mapsto g \cdot x).
     \end{IEEEeqnarray*}
 \end{remark}
+}{}
 
 \begin{definition}+
     A group $G$ with a topology
@@ -185,13 +186,13 @@ Recall:
     a homeomorphism $X \leftrightarrow Y$
     commuting with the group action.
 \end{definition}
+
+\gist{%
 \begin{warning}+
     What is called ``factor'' here is called ``subflow''
     by Furstenberg.
 \end{warning}
 
-
-
 \begin{example}
     Recall that $S_1 = \{z \in \C : |z| = 1\}$.
     Let $X = S_1$, $T = S_1$
@@ -200,6 +201,7 @@ Recall:
         and $\alpha + \beta$ denotes the addition of \emph{angles},
         i.e.~$\alpha \cdot \beta$ in complex numbers.}
 \end{example}
+}{}
 
 \begin{definition}
     \label{def:isometricextension}
@@ -237,6 +239,7 @@ Recall:
     An isometric extension of a distal flow is distal.
 \end{proposition}
 \begin{proof}
+    \gist{%
     Let $\pi\colon  X\to Y$ be an isometric extension.
     Towards a contradiction,
     suppose that $x_1,x_2 \in X$ are proximal.
@@ -253,6 +256,11 @@ Recall:
     we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
     Therefore $\rho(x_1,x_2) = 0$.
     Hence $x_1 = x_2$ $\lightning$.
+    }{Let $\pi\colon X \to Y$ isometric extension.
+        Suppose $x_1,x_2 \in X$ is proximal.
+        Then $\pi(x_1) = \pi(x_2)$.
+        But there exists a $T$-equivariant metric on the fibers.
+    }
 \end{proof}
 
 \begin{definition}

inputs/lecture_18.tex

@@ -149,6 +149,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of  a minimal distal flow
         bet the quotient space.
         It is compact, second countable and Hausdorff.
         Let $\pi\colon X\to M$ denote the quotient map.
+\gist{%
     \item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
         is an isometric flow:
         \begin{enumerate}
@@ -204,6 +205,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of  a minimal distal flow
         (\yaref{thm:usccomeagercont})
         this implies that $X \setminus \{x_2\}$ is meager.
         But then $X = \{\star\} \lightning$.
+}{}
 \end{enumerate}
 \end{proof}
 

inputs/lecture_19.tex

@@ -173,6 +173,7 @@ More generally we can show:
     then $(Y,T)$ is an isometric extension of $(Z_2, T)$.
 \end{lemma}
 \begin{proof}
+\gist{%
     % TODO TODO TODO Think about this
     For $z_1,z_1' \in Z_1$ with
     $w_1(z_1) = w_1(z_1')$ let
@@ -185,9 +186,16 @@ More generally we can show:
     on the fibers of $Y$ over $Z_2$
     and invariant under $T$.
 
-    $\sigma$ is a metric,
+    $\sigma$ is a metric on fibers,
     since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
     then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
+}{%
+    \begin{itemize}
+        \item Let $\rho\colon Z_1 \times_W Z_1 \to \R$.
+        \item Consider $\sigma\colon Y \times_{Z_2} Y \to \R$
+            given by $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$.
+    \end{itemize}
+}
 \end{proof}
 
 
@@ -242,7 +250,7 @@ More generally we can show:
     $(X'_\xi, T) = \theta((X_\xi, T)$.
     Let $\pi_\xi$ and $\pi'_\xi$ denote the maps from $X$ to $X_\xi$ resp.~$X'_\xi$.
     Set
-    \[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times  X'_{\xi+ 1}: x \in X\}.\]
+    \[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times  X'_{\xi+ 1}: x \in X\} \subseteq X \times X'_{\xi+1}.\]
     Then
 
     % https://q.uiver.app/#q=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

inputs/lecture_20.tex

@@ -46,9 +46,6 @@ Let $\pi_n\colon X \to  (S^1)^n$ be the projection to the first $n$
 coordinates.
 
 
-% TODO ANKI-MARKER
-
-
 \begin{lemma}
     \label{lem:lec20:1}
     Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
@@ -66,7 +63,7 @@ coordinates.
     where $d$ is the metric on $X$,
     $d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
 \end{lemma}
-\begin{proof}
+\begin{refproof}{lem:lec20:1}
     Let
     \begin{IEEEeqnarray*}{rCl}
         x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
@@ -74,7 +71,7 @@ coordinates.
     \end{IEEEeqnarray*}
     We will choose $x_k$ of the form
     \[
-        (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1} \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
+        (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
     \]
     where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
     and $|\beta_k| < 2^{-k}$.
@@ -89,7 +86,6 @@ coordinates.
     $u' = (\xi'_n)_{n \in \N}$,
     let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
     ($X$ is a group).
-    
     We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
     but it is easier to consider the distance between
     their quotient and $1$.
@@ -97,6 +93,7 @@ coordinates.
     \[
         w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
     \]
+\gist{%
     \begin{claim}
         $F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
         where  $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
@@ -121,7 +118,8 @@ coordinates.
                                   &\le & 2^{-n} | e^{\i \beta_k}- 1|\\
                                   &<& 2^{-n-k}.
     \end{IEEEeqnarray*}
-\end{proof}
+}{[some technical details omitted]}
+\end{refproof}
 
 
 \begin{definition}

inputs/lecture_22.tex

@@ -1,6 +1,7 @@
 \lecture{22}{2024-01-16}{}
 
 \begin{refproof}{thm:21:xnmaxiso}
+    % TODO TODO TODO
     We have the following situation:
     % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
     \[\begin{tikzcd}
@@ -17,7 +18,7 @@
     
     We want to show that this tower is normal,
     i.e.~the isometric extensions are maximal isometric extension.
-    
+\gist{%
     Let $Y$ be a maximal isometric extension of $X_{n-1}$ in $X$
     and let $\overline{g} = \pi^n_{n-1} \circ h$. % factor map?
     We need to show that $h$ is an isomorphism.
@@ -62,6 +63,25 @@
     But $x$ and $x'$ don't depend on $k$,
     hence $R(x,x') = 0$.
     It follows that $\pi'(x) = \pi'(x')$ $\lightning$.
+}{
+    \begin{itemize}
+        \item $Y$ max.~isometric extension of $X_{n-1}$ in $X$
+            and $\overline{g} = \pi^n_{n-1} \circ h$.
+        \item $h$ isomorphism.
+            Suppose not, then $\exists  y_0,y_1 \in X.~\pi'(y_0) \neq  \pi'(y_1),
+            \pi_n(y_0) = \pi_n(y_1) = t$.
+
+        \item Apply \yaref{lem:lec20:1} $\leadsto$ sequence $(x_k)$ in  $X$,
+            such that $\pi_{n-1}(x_k) = \pi_{n-1}(y_i)$,
+            $F(x_k,y_i) \to 0$.
+        \item $\rho\colon  Y \times_{X_{n-1}} Y \to \R$ witnessing isometric.
+        \item $R(a,b) \coloneqq  \rho(\pi'(a), \pi'(b))$ for $a,b \in X$ with
+            $\overline{g}(\pi'(a)) = \overline{g}(\pi'(b))$.
+            (defined for any two of $x_k$, $y_0$, $y_1$, $\tau$-equivariant)
+        \item $F(y_0,x_{k}) \to 0$, so $d(\tau^{m_k} y_0, \tau^{m_k} x_k) \to 0$.
+        \item $R(y_0,x_k) \to 0$, hence $\underbrace{R(y_0,y_1)}_{\text{no } k} \to 0$ $\lightning$.
+    \end{itemize}
+}
 \end{refproof}