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Hindman (Furstenberg)
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[latex]
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\begin{theorem}[Hindman]
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\label{thm:hindman}
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\label{thm:hindmanfurstenberg}
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If $\N$ is partitioned into finitely many
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sets,
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then there is is an infinite subset $H \subseteq \N$
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such that all finite sums of distinct
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elements of $H$
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belong to the same set of the partition.
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\end{theorem}
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[/latex]
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Use:
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[latex]
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\begin{theorem}
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\label{thm:unifrprox}
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Let $X$ be a compact Hausdorff space and $T\colon X \to X$
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continuous.
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Consider $(X,T)$.%TODO different notations
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Then for every $x \in X$
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there is a uniformly recurrent $y \in X$
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such that $y $ is proximal to $x$.
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\end{theorem}
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[/latex]
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[latex]
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\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
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\begin{itemize}
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\item View partition as $f\colon \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
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Let $x \in X$ be the given partition.
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\item $T\colon X \to X$ shift: $T(y)(n) \coloneqq y(n+1)$.
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\item Let $y$ proximal to $x$, uniformly recurrent.
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\begin{itemize}
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\item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon_N = T^n(y)\defon_N$
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for infinitely many $n$.
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\item uniform recurrence $\leadsto$
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\[
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\forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
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\text{ contains } $y\defon{\{0,\ldots,n\}}$ \text{ as a subsequence.}
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\]
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(consider neighbourhood $G_n = \{z \in X : z\defon{n} = y\defon{n} \}$).
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\end{itemize}
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\item Consider $c \coloneqq y(0)$. This color works:
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\begin{itemize}
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\item $G_0 \coloneqq y\defon{\{0\}}$,
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take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
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contains $y(0)$ for all $r$ (unif.~recurrence).
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$y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
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for infinitely many $r$ (proximality).
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Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
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\item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
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take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
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contains $y\defon{\{0,\ldots,h_0\} }$
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for all $r$ (unif.~recurrence).
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So among ever $N_1$ terms, there are two of distance $h_0$
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where $y$ has value $c$.
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So $\exists h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
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(proximality).
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\item Repeat:
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Choose $h_i$ such that
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for all sums $s$ of subsets of $\{h_0,\ldots, h_{i-1}\}$,
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$x(s+h_i) = y(s+h_i) = c$:
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Find $N_i$ such that every $N_i$ consecutive
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terms of $y$ contain a segment that coincides
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with the initial segment of $y$
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up to the largest $s$,
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then find a segment of length $N_i$ beyond $h_{i-1}$
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where $x$ and $y$ coincide.
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\end{itemize}
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\end{itemize}
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\end{refproof}
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[/latex]
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@ -204,7 +204,6 @@ Such a factor $(Y,T)$ is called a \vocab{maximal isometric extension} of $(Z,T)$
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\end{lemma}
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\begin{proof}
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\gist{%
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% TODO TODO TODO Think about this
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For $z_1,z_1' \in Z_1$ with
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$w_1(z_1) = w_1(z_1')$ let
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$\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$.
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@ -1,7 +1,6 @@
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\lecture{22}{2024-01-16}{}
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\begin{refproof}{thm:21:xnmaxiso}
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% TODO TODO TODO
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We have the following situation:
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% https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
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\[\begin{tikzcd}
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@ -1,7 +1,5 @@
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\lecture{24}{2024-01-23}{Combinatorics!}
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% ANKI 2
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\subsection{Applications to Combinatorics} % Ramsey Theory}
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@ -191,6 +191,7 @@ to obtain
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\begin{theorem}[Hindman]
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\label{thm:hindman}
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\label{thm:hindmanfurstenberg}
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If $\N$ is partitioned into finitely many
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sets,
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then there is is an infinite subset $H \subseteq \N$
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@ -198,7 +199,7 @@ to obtain
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elements of $H$
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belong to the same set of the partition.
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\end{theorem}
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\begin{proof}[Galvin,Glazer]
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\begin{refproof}{thm:hindman}[Galvin,Glazer]
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Let $\cU \in \beta\N \setminus \N$
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be such that $\cU + \cU = \cU$.
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Let $P$ be the piece of the partition
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@ -234,7 +235,7 @@ to obtain
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Chose $x_n > x_{n-1}$ that satisfies this.
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\end{itemize}
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Set $H \coloneqq \{x_i : i < \omega\}$.
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\end{proof}
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\end{refproof}
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Next time we'll see another proof of this theorem.
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@ -36,7 +36,8 @@ This gives $\N \acts X$.
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We do a second proof of \yaref{thm:hindman}:
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\begin{proof}[Furstenberg]
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\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
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\gist{%
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A partition of $\N$ into $k$-many pieces can be viewed
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as a function $f\colon \N \to k$.
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@ -99,10 +100,10 @@ We do a second proof of \yaref{thm:hindman}:
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Let $G_0 \coloneqq [y(0)]$
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and let $N_0$ be such that
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\[
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\forall r.~(y(r), \ldots, y(r + N-0 - 1)) \text{ contains $y(0)$.}
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\forall r.~(y(r), \ldots, y(r + N_0 - 1)) \text{ contains $y(0)$.}
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\]
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By proximality, there exist infinitely many $r$
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such that $(y(r), \ldots, (y(r+N-1)) = (x(r), \ldots, x(r+N-1))$.
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such that $(y(r), \ldots, (y(r+N_0-1)) = (x(r), \ldots, x(r+N_0-1))$.
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Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
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\item%
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@ -133,19 +134,68 @@ We do a second proof of \yaref{thm:hindman}:
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$(y(0), \ldots, y(h_0+h_1))$
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is contained in $(x(r), \ldots, x(r+N-1))$.
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Let $(y(0), \ldots, y(h_0+h_1)) = (x(r+s), \ldots, x(r+s+N-1))$.
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\item $\ldots$
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\item Repeat this:
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Inductively choose $h_i$ such that $x(s+h_i) = y(s+h_i) = c$
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for all sums $s$ of subsets of $\{h_0,\ldots,h_{i-1}\}$.
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To do this, find $N_i$ such that every $N_i$ consecutive
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terms of $y$ contain $(y(0), \ldots, y(\sum_{j < i} h_j))$.
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Then find $h_i > h_{i-1}$ such that
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$(x(h_i), \ldots, x(\sum_{j < i} h_j)) = (y(0), \ldots, y(\sum_{j < i} h_j))$.
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\end{itemize}
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\end{proof}
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% TODO ultrafilter extension continuous
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}{
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\begin{itemize}
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\item View partition as $f\colon \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
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Let $x \in X$ be the given partition.
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\item $T\colon X \to X$ shift: $T(y)(n) \coloneqq y(n+1)$.
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\item Let $y$ proximal to $x$, uniformly recurrent.
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\begin{itemize}
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\item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon{N} = T^n(y)\defon{N}$
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for infinitely many $n$.
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\item uniform recurrence $\leadsto$
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\[
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\forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
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\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
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\]
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(consider neighbourhood $G_n = \{z \in X : z\defon{n} = y\defon{n} \}$).
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\end{itemize}
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\item Consider $c \coloneqq y(0)$. This color works:
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\begin{itemize}
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\item $G_0 \coloneqq y\defon{\{0\}}$,
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take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
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contains $y(0)$ for all $r$ (unif.~recurrence).
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$y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
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for infinitely many $r$ (proximality).
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Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
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\item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
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take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
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contains $y\defon{\{0,\ldots,h_0\} }$
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for all $r$ (unif.~recurrence).
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So among ever $N_1$ terms, there are two of distance $h_0$
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where $y$ has value $c$.
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So $\exists h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
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(proximality).
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\item Repeat:
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Choose $h_i$ such that
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for all sums $s$ of subsets of $\{h_0,\ldots, h_{i-1}\}$,
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$x(s+h_i) = y(s+h_i) = c$:
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Find $N_i$ such that every $N_i$ consecutive
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terms of $y$ contain a segment that coincides
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with the initial segment of $y$
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up to the largest $s$,
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then find a segment of length $N_i$ beyond $h_{i-1}$
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where $x$ and $y$ coincide.
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\end{itemize}
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\end{itemize}
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}
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\end{refproof}
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In order to prove \yaref{thm:unifrprox},
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we need the following:
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we need to rephrase the problem in terms of $\beta\N$:
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\begin{theorem}
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\label{thm:unifrprox:helper}
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Let $X$ be a compact Hausdorff space. % ?
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Let $T\colon X \to X$ be continuous.
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Let us rephrase the problem in terms of $\beta\N$:
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\todo{remove duplicate}
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\begin{enumerate}[(1)]
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\item $x \in X$ is recurrent
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iff $T^\cU(x) = x$ for some $\cU \in \beta\N \setminus \N$.
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@ -161,7 +211,7 @@ we need the following:
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\end{theorem}
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\gist{%
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\begin{refproof}{thm:unifrprox:helper}[sketch]
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We only prove (2) here, as it is the most interesting point.%
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\todo{other parts will be in the official notes}
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@ -176,7 +226,7 @@ we need the following:
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Let $M$ be such that
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\[
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\forall n .¨ \exists k < M.~T^{n+k}(x) \in G.
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\forall n .~ \exists k < M.~T^{n+k}(x) \in G.
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\]
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So there is a $k < M$ such that
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\[
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@ -212,3 +262,4 @@ we need the following:
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\phantom\qedhere
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\end{refproof}
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}{}
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\begin{refproof}{thm:unifrprox:helper}
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\gist{%
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\begin{subproof}[(2), $\impliedby$, sketch]
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Assume that $x $ is not uniformly recurrent.
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Then there is a neighbourhood $G \ni x$
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@ -41,7 +42,18 @@
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% TODO Why? Think about this.
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\end{subproof}
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}{%
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\begin{itemize}
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\item 2, $\implies$:
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\begin{itemize}
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\item TODO
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\end{itemize}
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\item 2, $\impliedby$:
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\begin{itemize}
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\item TODO
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\end{itemize}
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\end{itemize}
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}
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\end{refproof}
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Take $X = \beta\N$,
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$S \colon \beta\N \to \beta\N$,
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@ -51,7 +63,6 @@ Then
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S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) =
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\ulim{\cV}_n \hat{n} + \cU = \cV + \cU.
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\]
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% TODO check
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\begin{corollary}
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$\cU$ is recurrent
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@ -175,12 +186,10 @@ S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) =
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Then $\cU + \cU = \cW + \cV + \cU = \cU$.
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(2) $\implies$ (3):
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\todo{TODO}
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% TODO
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% Let $\cU$ be an idempotent.
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% We want to find $\cV$ such that $\cV + \cU = \cU$.
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% $\cV'$ such that $\cV' + \cU = \cV' + 0$ proximal to $0$?
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% TAKE $\cV = \cV' = \cU$.
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Let $\cU$ be an idempotent.
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Then $\cV + \cU = \cV$ (proximal to $0$)
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and $\cV + \cU = \cU$ (recurrent)
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are satisfied for $\cV \coloneqq \cU$.
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\end{proof}
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\begin{corollary}
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