From dcf4851177d2c86fc9164cec546afa290f7640a3 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <git@jrpie.de>
Date: Thu, 8 Feb 2024 18:15:11 +0100
Subject: [PATCH] removed junk

---
 anki | 74 ------------------------------------------------------------
 1 file changed, 74 deletions(-)
 delete mode 100644 anki

diff --git a/anki b/anki
deleted file mode 100644
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-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]