From de89e2dc1d40810f23cb7d9b7b7115d7d2b4fec4 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 8 Feb 2024 17:31:10 +0100 Subject: [PATCH] some changes --- anki | 74 +++++++++++++++++++++++++++++++++++++++++++ inputs/lecture_19.tex | 1 - inputs/lecture_22.tex | 1 - inputs/lecture_24.tex | 2 -- inputs/lecture_25.tex | 5 +-- inputs/lecture_26.tex | 73 +++++++++++++++++++++++++++++++++++------- inputs/lecture_27.tex | 25 ++++++++++----- 7 files changed, 156 insertions(+), 25 deletions(-) create mode 100644 anki diff --git a/anki b/anki new file mode 100644 index 0000000..be64ca4 --- /dev/null +++ b/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] diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex index 54f11a1..569a5f8 100644 --- a/inputs/lecture_19.tex +++ b/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 $W$. diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index ae8fdba..2c3c431 100644 --- a/inputs/lecture_22.tex +++ b/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} diff --git a/inputs/lecture_24.tex b/inputs/lecture_24.tex index 958fa5f..5e410b3 100644 --- a/inputs/lecture_24.tex +++ b/inputs/lecture_24.tex @@ -1,7 +1,5 @@ \lecture{24}{2024-01-23}{Combinatorics!} -% ANKI 2 - \subsection{Applications to Combinatorics} % Ramsey Theory} diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex index 27acec3..5b1e269 100644 --- a/inputs/lecture_25.tex +++ b/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. diff --git a/inputs/lecture_26.tex b/inputs/lecture_26.tex index d075850..25e1a9a 100644 --- a/inputs/lecture_26.tex +++ b/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 \[ - \forall n .ยจ \exists k < M.~T^{n+k}(x) \in G. + \forall n .~ \exists k < M.~T^{n+k}(x) \in G. \] So there is a $k < M$ such that \[ @@ -212,3 +262,4 @@ we need the following: \phantom\qedhere \end{refproof} +}{} diff --git a/inputs/lecture_27.tex b/inputs/lecture_27.tex index e825c2a..0fd0756 100644 --- a/inputs/lecture_27.tex +++ b/inputs/lecture_27.tex @@ -2,6 +2,7 @@ \begin{refproof}{thm:unifrprox:helper} +\gist{% \begin{subproof}[(2), $\impliedby$, sketch] Assume that $x $ is not uniformly recurrent. Then there is a neighbourhood $G \ni x$ @@ -41,7 +42,18 @@ % TODO Why? Think about this. \end{subproof} - +}{% + \begin{itemize} + \item 2, $\implies$: + \begin{itemize} + \item TODO + \end{itemize} + \item 2, $\impliedby$: + \begin{itemize} + \item TODO + \end{itemize} + \end{itemize} +} \end{refproof} Take $X = \beta\N$, $S \colon \beta\N \to \beta\N$, @@ -51,7 +63,6 @@ Then S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) = \ulim{\cV}_n \hat{n} + \cU = \cV + \cU. \] -% 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}