some changes

anki

@@ -0,0 +1,74 @@
+Hindman (Furstenberg)
+
+[latex]
+\begin{theorem}[Hindman]
+    \label{thm:hindman}
+    \label{thm:hindmanfurstenberg}
+    If $\N$ is partitioned into finitely many
+    sets,
+    then there is is an infinite subset $H \subseteq \N$
+    such that all finite sums of distinct
+    elements of $H$
+    belong to the same set of the partition.
+\end{theorem}
+[/latex]
+Use:
+[latex]
+\begin{theorem}
+    \label{thm:unifrprox}
+    Let $X$ be a compact Hausdorff space and $T\colon  X \to X$
+    continuous.
+    Consider $(X,T)$.%TODO different notations
+    Then for every $x \in X$
+    there is a uniformly recurrent $y \in X$
+    such that $y $ is proximal to $x$.
+\end{theorem}
+[/latex]
+[latex]
+\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
+    \begin{itemize}
+        \item View partition as $f\colon  \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
+            Let $x \in X$ be the given partition.
+        \item $T\colon  X \to X$ shift: $T(y)(n) \coloneqq  y(n+1)$.
+        \item Let $y$ proximal to $x$, uniformly recurrent.
+        \begin{itemize}
+            \item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon_N = T^n(y)\defon_N$
+                for infinitely many $n$.
+            \item uniform recurrence $\leadsto$
+                \[
+                    \forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
+                    \text{ contains } $y\defon{\{0,\ldots,n\}}$ \text{ as a subsequence.}
+                \]
+                (consider neighbourhood $G_n = \{z \in X : z\defon{n}  = y\defon{n} \}$).
+        \end{itemize}
+        \item Consider $c \coloneqq y(0)$. This color works:
+            \begin{itemize}
+                \item $G_0 \coloneqq  y\defon{\{0\}}$,
+                take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
+                contains $y(0)$ for all $r$ (unif.~recurrence).
+                $y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
+                for infinitely many $r$ (proximality).
+                Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
+                \item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
+                take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
+                contains $y\defon{\{0,\ldots,h_0\} }$
+                for all $r$ (unif.~recurrence).
+                So among ever $N_1$ terms, there are two of distance $h_0$
+                where $y$ has value  $c$.
+                So $\exists  h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
+                (proximality).
+
+            \item Repeat:
+                Choose $h_i$ such that
+                for all sums $s$ of subsets of  $\{h_0,\ldots, h_{i-1}\}$,
+                $x(s+h_i) = y(s+h_i) = c$:
+                Find $N_i$ such that every $N_i$ consecutive
+                terms of $y$ contain a segment that coincides
+                with the initial segment of $y$
+                up  to the largest $s$,
+                then find a segment of length $N_i$ beyond $h_{i-1}$
+                where $x$ and $y$ coincide.
+            \end{itemize}
+    \end{itemize}
+\end{refproof}
+[/latex]

inputs/lecture_19.tex

@@ -204,7 +204,6 @@ Such a factor $(Y,T)$ is called a \vocab{maximal isometric extension} of $(Z,T)$
 \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
     $\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$.

inputs/lecture_22.tex

@@ -1,7 +1,6 @@
 \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}

inputs/lecture_24.tex

@@ -1,7 +1,5 @@
 \lecture{24}{2024-01-23}{Combinatorics!}
 
-% ANKI 2
-
 
 \subsection{Applications to Combinatorics} % Ramsey Theory}
 

inputs/lecture_25.tex

@@ -191,6 +191,7 @@ to obtain
 
 \begin{theorem}[Hindman]
     \label{thm:hindman}
+    \label{thm:hindmanfurstenberg}
     If $\N$ is partitioned into finitely many
     sets,
     then there is is an infinite subset $H \subseteq \N$
@@ -198,7 +199,7 @@ to obtain
     elements of $H$
     belong to the same set of the partition.
 \end{theorem}
-\begin{proof}[Galvin,Glazer]
+\begin{refproof}{thm:hindman}[Galvin,Glazer]
     Let $\cU \in \beta\N \setminus \N$
     be such that $\cU + \cU = \cU$.
     Let $P$ be the piece of the partition
@@ -234,7 +235,7 @@ to obtain
             Chose $x_n > x_{n-1}$ that satisfies this.
     \end{itemize}
     Set $H \coloneqq \{x_i : i < \omega\}$.
-\end{proof}
+\end{refproof}
 
 
 Next time we'll see another proof of this theorem.

inputs/lecture_26.tex

@@ -36,7 +36,8 @@ This gives $\N \acts X$.
 
 
 We do a second proof of \yaref{thm:hindman}:
-\begin{proof}[Furstenberg]
+\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
+\gist{%
     A partition of $\N$ into $k$-many pieces can be viewed
     as a function $f\colon \N \to k$.
 
@@ -99,10 +100,10 @@ We do a second proof of \yaref{thm:hindman}:
             Let $G_0 \coloneqq  [y(0)]$
             and let $N_0$ be such that
             \[
-            \forall  r.~(y(r), \ldots, y(r + N-0 - 1)) \text{ contains $y(0)$.}
+            \forall  r.~(y(r), \ldots, y(r + N_0 - 1)) \text{ contains $y(0)$.}
             \]
             By proximality, there exist infinitely many $r$
-            such that  $(y(r), \ldots, (y(r+N-1)) = (x(r), \ldots, x(r+N-1))$.
+            such that  $(y(r), \ldots, (y(r+N_0-1)) = (x(r), \ldots, x(r+N_0-1))$.
             Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
 
         \item%
@@ -133,19 +134,68 @@ We do a second proof of \yaref{thm:hindman}:
             $(y(0), \ldots, y(h_0+h_1))$
             is contained in $(x(r), \ldots, x(r+N-1))$.
             Let $(y(0), \ldots, y(h_0+h_1)) = (x(r+s), \ldots, x(r+s+N-1))$.
-        \item $\ldots$
+        \item Repeat this:
+            Inductively choose $h_i$ such that $x(s+h_i) = y(s+h_i) = c$
+            for all sums  $s$ of subsets of $\{h_0,\ldots,h_{i-1}\}$.
+            To do this, find $N_i$ such that every $N_i$ consecutive
+            terms of $y$ contain $(y(0), \ldots, y(\sum_{j < i} h_j))$.
+            Then find $h_i > h_{i-1}$ such that
+            $(x(h_i), \ldots, x(\sum_{j < i} h_j)) = (y(0), \ldots, y(\sum_{j < i} h_j))$.
     \end{itemize}
-\end{proof}
-% TODO ultrafilter extension continuous
+}{
+    \begin{itemize}
+        \item View partition as $f\colon  \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
+            Let $x \in X$ be the given partition.
+        \item $T\colon  X \to X$ shift: $T(y)(n) \coloneqq  y(n+1)$.
+        \item Let $y$ proximal to $x$, uniformly recurrent.
+        \begin{itemize}
+            \item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon{N} = T^n(y)\defon{N}$
+                for infinitely many $n$.
+            \item uniform recurrence $\leadsto$
+                \[
+                    \forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
+                    \text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
+                \]
+                (consider neighbourhood $G_n = \{z \in X : z\defon{n}  = y\defon{n} \}$).
+        \end{itemize}
+        \item Consider $c \coloneqq y(0)$. This color works:
+            \begin{itemize}
+                \item $G_0 \coloneqq  y\defon{\{0\}}$,
+                take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
+                contains $y(0)$ for all $r$ (unif.~recurrence).
+                $y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
+                for infinitely many $r$ (proximality).
+                Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
+                \item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
+                take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
+                contains $y\defon{\{0,\ldots,h_0\} }$
+                for all $r$ (unif.~recurrence).
+                So among ever $N_1$ terms, there are two of distance $h_0$
+                where $y$ has value  $c$.
+                So $\exists  h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
+                (proximality).
+
+            \item Repeat:
+                Choose $h_i$ such that
+                for all sums $s$ of subsets of  $\{h_0,\ldots, h_{i-1}\}$,
+                $x(s+h_i) = y(s+h_i) = c$:
+                Find $N_i$ such that every $N_i$ consecutive
+                terms of $y$ contain a segment that coincides
+                with the initial segment of $y$
+                up  to the largest $s$,
+                then find a segment of length $N_i$ beyond $h_{i-1}$
+                where $x$ and $y$ coincide.
+            \end{itemize}
+    \end{itemize}
+}
+\end{refproof}
 
 In order to prove \yaref{thm:unifrprox},
-we need the following:
+we need to rephrase the problem in terms of $\beta\N$:
 \begin{theorem}
     \label{thm:unifrprox:helper}
     Let $X$ be a compact Hausdorff space. % ?
     Let $T\colon  X \to X$ be continuous.
-    Let us rephrase the problem in terms of $\beta\N$:
-    \todo{remove duplicate}
     \begin{enumerate}[(1)]
         \item $x \in X$ is recurrent
             iff $T^\cU(x) = x$ for some  $\cU \in  \beta\N \setminus \N$.
@@ -161,7 +211,7 @@ we need the following:
 \end{theorem}
 
 
-
+\gist{%
 \begin{refproof}{thm:unifrprox:helper}[sketch]
     We only prove (2) here, as it is the most interesting point.%
     \todo{other parts will be in the official notes}
@@ -176,7 +226,7 @@ we need the following:
 
         Let $M$ be such that
         \[
-        \forall  n .¨ \exists  k < M.~T^{n+k}(x) \in G.
+        \forall  n .~ \exists  k < M.~T^{n+k}(x) \in G.
         \]
         So there is a $k < M$ such that
         \[
@@ -212,3 +262,4 @@ we need the following:
 
     \phantom\qedhere
 \end{refproof}
+}{}

inputs/lecture_27.tex

@@ -2,6 +2,7 @@
 
 
 \begin{refproof}{thm:unifrprox:helper}
+\gist{%
     \begin{subproof}[(2), $\impliedby$, sketch]
         Assume that $x $ is not uniformly recurrent.
         Then there is a neighbourhood $G \ni x$
@@ -41,7 +42,18 @@
         % TODO Why? Think about this.
 
     \end{subproof}
-   
+}{%
+    \begin{itemize}
+        \item 2, $\implies$:
+            \begin{itemize}
+                \item TODO
+            \end{itemize}
+        \item 2, $\impliedby$:
+            \begin{itemize}
+                \item TODO
+            \end{itemize}
+    \end{itemize}
+}
 \end{refproof}
 Take $X = \beta\N$,
 $S \colon \beta\N \to \beta\N$,
@@ -51,7 +63,6 @@ Then
 S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) =
 \ulim{\cV}_n \hat{n} + \cU = \cV + \cU.
 \]
-% TODO check
 
 \begin{corollary}
     $\cU$ is recurrent
@@ -175,12 +186,10 @@ S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) =
     Then $\cU + \cU = \cW + \cV + \cU = \cU$.
 
     (2) $\implies$ (3):
-    \todo{TODO}
-    % TODO
-    %  Let $\cU$ be an idempotent.
-    %  We want to find $\cV$ such that $\cV + \cU = \cU$.
-    %   $\cV'$ such that $\cV' + \cU = \cV' + 0$ proximal to  $0$?
-    % TAKE $\cV = \cV' = \cU$.
+    Let $\cU$ be an idempotent.
+    Then $\cV + \cU = \cV$ (proximal to  $0$)
+    and  $\cV + \cU = \cU$ (recurrent)
+    are satisfied for  $\cV \coloneqq  \cU$.
 \end{proof}
 
 \begin{corollary}