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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]

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@ -204,7 +204,6 @@ Such a factor $(Y,T)$ is called a \vocab{maximal isometric extension} of $(Z,T)$
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
\gist{% \gist{%
% TODO TODO TODO Think about this
For $z_1,z_1' \in Z_1$ with For $z_1,z_1' \in Z_1$ with
$w_1(z_1) = w_1(z_1')$ let $w_1(z_1) = w_1(z_1')$ let
$\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$. $\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$.

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@ -1,7 +1,6 @@
\lecture{22}{2024-01-16}{} \lecture{22}{2024-01-16}{}
\begin{refproof}{thm:21:xnmaxiso} \begin{refproof}{thm:21:xnmaxiso}
% TODO TODO TODO
We have the following situation: We have the following situation:
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0= % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
\[\begin{tikzcd} \[\begin{tikzcd}

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

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

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@ -36,7 +36,8 @@ This gives $\N \acts X$.
We do a second proof of \yaref{thm:hindman}: 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 A partition of $\N$ into $k$-many pieces can be viewed
as a function $f\colon \N \to k$. as a function $f\colon \N \to k$.
@ -99,10 +100,10 @@ We do a second proof of \yaref{thm:hindman}:
Let $G_0 \coloneqq [y(0)]$ Let $G_0 \coloneqq [y(0)]$
and let $N_0$ be such that 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 $r$ 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 $x(h_0) = y(0)$. Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
\item% \item%
@ -133,19 +134,68 @@ We do a second proof of \yaref{thm:hindman}:
$(y(0), \ldots, y(h_0+h_1))$ $(y(0), \ldots, y(h_0+h_1))$
is contained in $(x(r), \ldots, x(r+N-1))$. is contained in $(x(r), \ldots, x(r+N-1))$.
Let $(y(0), \ldots, y(h_0+h_1)) = (x(r+s), \ldots, x(r+s+N-1))$. 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{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}, In order to prove \yaref{thm:unifrprox},
we need the following: we need to rephrase the problem in terms of $\beta\N$:
\begin{theorem} \begin{theorem}
\label{thm:unifrprox:helper} \label{thm:unifrprox:helper}
Let $X$ be a compact Hausdorff space. % ? Let $X$ be a compact Hausdorff space. % ?
Let $T\colon X \to X$ be continuous. Let $T\colon X \to X$ be continuous.
Let us rephrase the problem in terms of $\beta\N$:
\todo{remove duplicate}
\begin{enumerate}[(1)] \begin{enumerate}[(1)]
\item $x \in X$ is recurrent \item $x \in X$ is recurrent
iff $T^\cU(x) = x$ for some $\cU \in \beta\N \setminus \N$. iff $T^\cU(x) = x$ for some $\cU \in \beta\N \setminus \N$.
@ -161,7 +211,7 @@ we need the following:
\end{theorem} \end{theorem}
\gist{%
\begin{refproof}{thm:unifrprox:helper}[sketch] \begin{refproof}{thm:unifrprox:helper}[sketch]
We only prove (2) here, as it is the most interesting point.% We only prove (2) here, as it is the most interesting point.%
\todo{other parts will be in the official notes} \todo{other parts will be in the official notes}
@ -176,7 +226,7 @@ we need the following:
Let $M$ be such that 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 So there is a $k < M$ such that
\[ \[
@ -212,3 +262,4 @@ we need the following:
\phantom\qedhere \phantom\qedhere
\end{refproof} \end{refproof}
}{}

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