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

@@ -195,7 +195,7 @@ Such a factor $(Y,T)$ is called a \vocab{maximal isometric extension} of $(Z,T)$
 \end{tikzcd}\]
 
     Suppose that whenever $y \neq y' \in Y$,
-    then % either % TODO REALLY?
+    then either % TODO REALLY?
     $\pi_1(y) \neq \pi(y')$
     or $\pi_2(y) \neq  \pi_2(y')$.