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@ 89,7 +89,6 @@ We do a second proof of \yaref{thm:hindman}:


(y(0), y(1), \ldots, y(n))


\text{ as a subsequence.}


\]




\end{itemize}




Consider $y(0)$.



@ 139,10 +138,14 @@ We do a second proof of \yaref{thm:hindman}:


\end{proof}


% TODO ultrafilter extension continuous






\begin{refproof}{thm:unifrprox}[sketch]


In order to prove \yaref{thm:unifrprox},


we need the following:


\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 $\cU \in \beta\N \setminus \N$.



@ 155,10 +158,15 @@ We do a second proof of \yaref{thm:hindman}:


such that $T^\cU(x) = T^\cU(y)$.


% TODO compare with the statement for the ellis semigroup.


\end{enumerate}


We only to (2) here, as it is the most interesting point.%


\end{theorem}








\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}




\begin{subproof}[(2)]


\begin{subproof}[(2), $\implies$]


Suppose that $ x$ is uniformly recurrent.


Take some $\cV \in \beta\N$.


Let $G_0$ be a neighbourhood of $x$.



@ 200,18 +208,7 @@ We do a second proof of \yaref{thm:hindman}:




Since we get this for every neighbourhood,


it follows that $T^\cU ( T^\cV(x)) = x$.


















\end{subproof}












\phantom\qedhere


\end{refproof}




248
inputs/lecture_27.tex
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248
inputs/lecture_27.tex
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@ 0,0 +1,248 @@


\lecture{27}{20240202}{}






\begin{refproof}{thm:unifrprox:helper}


\begin{subproof}[(2), $\impliedby$, sketch]


Assume that $x $ is not uniformly recurrent.


Then there is a neighbourhood $G \ni x$


such that for all $M \in \N$


\[


Y_M = \{ n \in \N : \forall k < M.~T^{n+k}(x) \not\in G\} \neq \emptyset.


\]


Note that $Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \ldots$


Take $\cV \in \beta\N$ containing all $Y_n$.




We aim to show that there is no $\cU\in \beta\N$ such that $T_\cU(T_\cV(x)) = x$.


Towards a contradiction suppose that such $\cU$ exists.




For every $k + 1$ we have $Y_{k+1} \in \cV$.


In particular


\[


\{n \in \N : T^{n+k}(x) \not\in G\} \supseteq Y_{k+1},


\]


so


\[


(\cV n) T^{n+k}(x) \not\in G,


\]


i.e.~


\[


(\cV n) T^n(x) \not\in \underbrace{T^{k}(G)}_{\text{open}}.


\]


Thus


\[


\underbrace{\cV\lim_n T^n(x)}_{T^\cV(x)} \not\in T^{k}(G).


\]


We get that


\[


\forall k.~T^k(T^\cV(x)) \not\in G.


\]


It follows that


$\forall \cU \in \beta\N.~T^{\cU}(T^\cV(x)) \not\in G$.


% TODO Why? Think about this.




\end{subproof}




\end{refproof}


Take $X = \beta\N$,


$S \colon \beta\N \to \beta\N$,


$S(\cU ) = \hat{1}+ \cU$.


Then


\[


S^\cV(\cU) = \cV\lim_n S^n(\cU) = \cV\lim_n(\hat{n} + \cU) =


\cV\lim_n \hat{n} + \cU = \cV + \cU.


\]


% TODO check




\begin{corollary}


$\cU$ is recurrent


iff


\[


\exists \cV \in \beta\N \setminus \N .~S^\cV(\cU) = \cU.


\]




$\cU$ is uniformly recurrent iff


\[


\forall \cV.~\exists \cW.~\cW + \cV + \cU = \cU.


\]


$\cU_1$ and $\cU_2$ are proximal


iff $\exists \cV.~\cV + \cU_1 = \cV + \cU_2$.


\end{corollary}




\begin{definition}


We say that $I \subseteq \beta\N$


is a \vocab{left ideal} ,


if


\[


\forall \cU \in I.~\forall \cV \in \beta\N.~\cV + \cU \in I.


\]






\end{definition}




\begin{theorem}


\yalabel{thm:unifrprox:helper2}


\begin{enumerate}[(1)]


\item $\cU$ is uniformly recurrent in $\beta\N$


iff $\cU$ belongs to a minimal\footnote{wrt.~$\subseteq $} (closed)


left ideal in $\beta\N$.


\item $\cU$ is an idempotent in $\beta\N$


iff $\cU$ belong to a minimal closed


subsemigroup of $\beta\N$.


\end{enumerate}


\end{theorem}


\begin{proof}


\begin{enumerate}[(1)]


\item


\gist{


Note that any $\cU \in \beta\N$ yields


%gives rise to


a left ideal $\beta\N + \cU$.


It is closed, since it is the image


of $\beta\N$ under the continuous maps


$\cV \mapsto \cV + \cU$


and $\beta\N$ is compact.


}{%


Note that $\beta\N + \cU$ is closed,


since $\beta\N$ is compact and $\cdot + \cU$ continuous.


}


$\cU$ belongs to a minimal left ideal


iff $\beta\N + \cU$ is minimal%


\gist{,


since every ideal containing $\cU$


contains $\beta\N + \cU$.


}{.}


\gist{%


Note that $\beta\N + \cV + \cU \subseteq \beta\N + \cU$


and if $I \subsetneq \beta\N + \cU$,


we have $\cV_0 = \cV + \cU \in I$


and $\beta\N + \cV + \cU \subseteq \beta\N + \cU$.


So $\cU$ belongs to a minimal left ideal iff


}{Equivalently}


\[


\forall \cV \in \beta\N .~\beta\N + \cV + \cU = \beta\N + \cU.


\]




This is the case iff


\[


\underbrace{\forall \cV .~\exists \cW.~ \cW + \cV + \cU = \cU.}_%


{\cV \text{ uniformly recurrent}}


\]


\gist{(For one direction take $\cW$ such that $\cW + \cV + \cU= \hat{0} +\cU$.


For the other direction note that


for every $\cV_0 $, $\cV_0 + \cU$


can be written as $\cV_0 + \cW + (\cV + \cU)$.


Where we take $\cW$ such that $\cW + \cV + \cU = \cU$.


}{}




\item This is very similar to the proof of the \yaref{lem:ellisnumakura}.




If $\cU$ is idempotent, then $\{\cU\}$


is a semigroup.


Let $C$ be a minimal closed subsemigroup of $\beta\N$.


Then $C + \cU$ is a closed subsemigroup.


By minimality, we get $C = C + \cU$.




Let $D = \{ \cV \in C .~ \cV + \cU = \cU\}$.


We have $D \neq \emptyset$.


$D$ is a closed semigroup,


so $D = C$ be minimality.


Hence $\cU + \cU = \cU$.


\end{enumerate}


\end{proof}


\begin{corollary}


Idempotent and uniformly recurrent elements exist.


\end{corollary}


\begin{proof}


Use \yaref{thm:unifrprox:helper2}


and Zorn's lemma.


\end{proof}


\begin{theorem}


(1) $\implies$ (2) $\implies$ (3)


where


\begin{enumerate}[(1)]


\item $\cU$ is uniformly recurrent and proximal to $\hat{0}$.


\item $\cU$ is an idempotent.


\item $\cU$ is recurrent and proximal to $\hat{0}$.


\end{enumerate}


\end{theorem}


\begin{proof}


(1) $\implies$ (2):


Let $\cU$ be uniformly recurrent and proximal to $ \hat{0}$.


Take $\cV$ such that $\cV + \cU = \cV + \hat{0} = \cV$.


% TODO REF beginning of lecture




Since $\cU$ is uniformly recurrent,


there exists $\cW$ such that $\cW + \cV + \cU = \cU$,


i.e.~$\cW + \cV = \cU$.


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 $\cV = \cV' = \cU$.


\end{proof}




\begin{corollary}


$\cU$ is uniformly recurrent and proximal to $0$


iff $\cU$ is an idempotent


and belongs to some minimal left ideal of $\beta\N$.


\end{corollary}




Finally:


\begin{refproof}{thm:unifrprox}


Let $T\colon X \to X$ and $x \in X$.


We want to find $y \in X$


such that $y$ is uniformly recurrent


and proximal to $x$.




We first prove a version for ultrafilters and then


transfer it to $X$.




There exists a uniformly recurrent $\cV \in \beta\N$.


So for any $\cW$,


$\cW + \cV$ is also uniformly recurrent\gist{:


Take $\cV_0$.


We need to find $\cX$ such that $\cX + \cV_0 + \cW +\cV = \cW + \cV$.


By uniform recurrence of $\cV$ we find $\cX'$


such that $\cX' + (\cV_0 + \cW) + \cV = \cV$.


Then $\cX = \cW + \cX'$ works.


}{.}


So all elements of $\beta\N + \cV$


are uniformly recurrent.


It is a closed ideal and hence a closed semigroup.


So $\beta\N + \cV$ contains a minimal closed


semigroup.


In particular, there exists an idempotent $\cU \in \beta\N + \cV$.




$\cU$ is idempotent and uniformly recurrent


hence it is proximal to $0$.






Now let us consider $X$.


Take $y = T^\cU(x)$.




\begin{claim}


$y$ uniformly recurrent.


\end{claim}


\begin{subproof}


Recall that $T^{\cV_1 + \cV_2} = T^{\cV_1} \circ T^{\cV_2}$.




Since $\cU$ is uniformly recurrent,


$\forall \cV .~\exists \cW.~\cW+ \cV+\cU= \cU$,


i.e.~$T^{\cW + \cV + \cU} (x) = T^\cW(T^\cV(y)) = T^\cU(x) = y$.


\end{subproof}


\begin{claim}


$y$ is proximal to $x$.


\end{claim}


\begin{subproof}


$\cU$ is proximal to $0$.


So $\exists \cV.~\cV + \cU = \cV + \hat{0} = \cV$,


i.e.~$T^{\cV}(y) = T^{\cV + \cU}(x) = T^\cV(x)$.


Thus $x$ and $y$ are proximal.%TODO REF


\end{subproof}


\end{refproof}


% Office hours wednesday 15:30  18:30 office 805


% Exam: First question: present favorite theorem (78 minutes, moderate length proof)


@ 53,6 +53,7 @@


\input{inputs/lecture_24}


\input{inputs/lecture_25}


\input{inputs/lecture_26}


\input{inputs/lecture_27}




\cleardoublepage





@ 80,10 +81,8 @@


\input{inputs/facts}








\PrintVocabIndex




\printbibliography






\end{document}




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