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@ -89,7 +89,6 @@ We do a second proof of \yaref{thm:hindman}:
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(y(0), y(1), \ldots, y(n))
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\text{ as a subsequence.}
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\]
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\end{itemize}
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Consider $y(0)$.
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@ -139,10 +138,14 @@ We do a second proof of \yaref{thm:hindman}:
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\end{proof}
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% TODO ultrafilter extension continuous
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\begin{refproof}{thm:unifrprox}[sketch]
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In order to prove \yaref{thm:unifrprox},
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we need the following:
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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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@ -155,10 +158,15 @@ We do a second proof of \yaref{thm:hindman}:
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such that $T^\cU(x) = T^\cU(y)$.
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% TODO compare with the statement for the ellis semigroup.
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\end{enumerate}
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We only to (2) here, as it is the most interesting point.%
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\end{theorem}
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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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\begin{subproof}[(2)]
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\begin{subproof}[(2), $\implies$]
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Suppose that $ x$ is uniformly recurrent.
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Take some $\cV \in \beta\N$.
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Let $G_0$ be a neighbourhood of $x$.
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@ -200,18 +208,7 @@ We do a second proof of \yaref{thm:hindman}:
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Since we get this for every neighbourhood,
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it follows that $T^\cU ( T^\cV(x)) = x$.
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\end{subproof}
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\phantom\qedhere
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\end{refproof}
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248
inputs/lecture_27.tex
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248
inputs/lecture_27.tex
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@ -0,0 +1,248 @@
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\lecture{27}{2024-02-02}{}
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\begin{refproof}{thm:unifrprox:helper}
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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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such that for all $M \in \N$
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\[
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Y_M = \{ n \in \N : \forall k < M.~T^{n+k}(x) \not\in G\} \neq \emptyset.
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\]
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Note that $Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \ldots$
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Take $\cV \in \beta\N$ containing all $Y_n$.
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We aim to show that there is no $\cU\in \beta\N$ such that $T_\cU(T_\cV(x)) = x$.
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Towards a contradiction suppose that such $\cU$ exists.
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For every $k + 1$ we have $Y_{k+1} \in \cV$.
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In particular
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\[
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\{n \in \N : T^{n+k}(x) \not\in G\} \supseteq Y_{k+1},
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\]
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so
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\[
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(\cV n) T^{n+k}(x) \not\in G,
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\]
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i.e.~
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\[
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(\cV n) T^n(x) \not\in \underbrace{T^{-k}(G)}_{\text{open}}.
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\]
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Thus
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\[
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\underbrace{\cV-\lim_n T^n(x)}_{T^\cV(x)} \not\in T^{-k}(G).
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\]
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We get that
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\[
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\forall k.~T^k(T^\cV(x)) \not\in G.
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\]
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It follows that
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$\forall \cU \in \beta\N.~T^{\cU}(T^\cV(x)) \not\in G$.
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% TODO Why? Think about this.
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\end{subproof}
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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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$S(\cU ) = \hat{1}+ \cU$.
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Then
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\[
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S^\cV(\cU) = \cV-\lim_n S^n(\cU) = \cV-\lim_n(\hat{n} + \cU) =
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\cV-\lim_n \hat{n} + \cU = \cV + \cU.
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\]
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% TODO check
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\begin{corollary}
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$\cU$ is recurrent
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iff
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\[
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\exists \cV \in \beta\N \setminus \N .~S^\cV(\cU) = \cU.
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\]
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$\cU$ is uniformly recurrent iff
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\[
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\forall \cV.~\exists \cW.~\cW + \cV + \cU = \cU.
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\]
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$\cU_1$ and $\cU_2$ are proximal
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iff $\exists \cV.~\cV + \cU_1 = \cV + \cU_2$.
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\end{corollary}
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\begin{definition}
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We say that $I \subseteq \beta\N$
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is a \vocab{left ideal} ,
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if
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\[
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\forall \cU \in I.~\forall \cV \in \beta\N.~\cV + \cU \in I.
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\]
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\end{definition}
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\begin{theorem}
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\yalabel{thm:unifrprox:helper2}
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\begin{enumerate}[(1)]
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\item $\cU$ is uniformly recurrent in $\beta\N$
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iff $\cU$ belongs to a minimal\footnote{wrt.~$\subseteq $} (closed)
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left ideal in $\beta\N$.
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\item $\cU$ is an idempotent in $\beta\N$
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iff $\cU$ belong to a minimal closed
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subsemigroup of $\beta\N$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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\begin{enumerate}[(1)]
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\item
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\gist{
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Note that any $\cU \in \beta\N$ yields
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%gives rise to
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a left ideal $\beta\N + \cU$.
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It is closed, since it is the image
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of $\beta\N$ under the continuous maps
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$\cV \mapsto \cV + \cU$
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and $\beta\N$ is compact.
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}{%
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Note that $\beta\N + \cU$ is closed,
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since $\beta\N$ is compact and $\cdot + \cU$ continuous.
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}
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$\cU$ belongs to a minimal left ideal
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iff $\beta\N + \cU$ is minimal%
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\gist{,
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since every ideal containing $\cU$
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contains $\beta\N + \cU$.
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}{.}
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\gist{%
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Note that $\beta\N + \cV + \cU \subseteq \beta\N + \cU$
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and if $I \subsetneq \beta\N + \cU$,
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we have $\cV_0 = \cV + \cU \in I$
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and $\beta\N + \cV + \cU \subseteq \beta\N + \cU$.
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So $\cU$ belongs to a minimal left ideal iff
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}{Equivalently}
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\[
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\forall \cV \in \beta\N .~\beta\N + \cV + \cU = \beta\N + \cU.
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\]
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This is the case iff
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\[
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\underbrace{\forall \cV .~\exists \cW.~ \cW + \cV + \cU = \cU.}_%
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{\cV \text{ uniformly recurrent}}
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\]
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\gist{(For one direction take $\cW$ such that $\cW + \cV + \cU= \hat{0} +\cU$.
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For the other direction note that
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for every $\cV_0 $, $\cV_0 + \cU$
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can be written as $\cV_0 + \cW + (\cV + \cU)$.
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Where we take $\cW$ such that $\cW + \cV + \cU = \cU$.
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}{}
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\item This is very similar to the proof of the \yaref{lem:ellisnumakura}.
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If $\cU$ is idempotent, then $\{\cU\}$
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is a semigroup.
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Let $C$ be a minimal closed subsemigroup of $\beta\N$.
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Then $C + \cU$ is a closed subsemigroup.
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By minimality, we get $C = C + \cU$.
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Let $D = \{ \cV \in C .~ \cV + \cU = \cU\}$.
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We have $D \neq \emptyset$.
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$D$ is a closed semigroup,
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so $D = C$ be minimality.
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Hence $\cU + \cU = \cU$.
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\end{enumerate}
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\end{proof}
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\begin{corollary}
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Idempotent and uniformly recurrent elements exist.
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\end{corollary}
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\begin{proof}
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Use \yaref{thm:unifrprox:helper2}
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and Zorn's lemma.
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\end{proof}
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\begin{theorem}
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(1) $\implies$ (2) $\implies$ (3)
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where
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\begin{enumerate}[(1)]
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\item $\cU$ is uniformly recurrent and proximal to $\hat{0}$.
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\item $\cU$ is an idempotent.
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\item $\cU$ is recurrent and proximal to $\hat{0}$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(1) $\implies$ (2):
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Let $\cU$ be uniformly recurrent and proximal to $ \hat{0}$.
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Take $\cV$ such that $\cV + \cU = \cV + \hat{0} = \cV$.
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% TODO REF beginning of lecture
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Since $\cU$ is uniformly recurrent,
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there exists $\cW$ such that $\cW + \cV + \cU = \cU$,
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i.e.~$\cW + \cV = \cU$.
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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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\end{proof}
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\begin{corollary}
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$\cU$ is uniformly recurrent and proximal to $0$
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iff $\cU$ is an idempotent
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and belongs to some minimal left ideal of $\beta\N$.
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\end{corollary}
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Finally:
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\begin{refproof}{thm:unifrprox}
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Let $T\colon X \to X$ and $x \in X$.
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We want to find $y \in X$
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such that $y$ is uniformly recurrent
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and proximal to $x$.
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We first prove a version for ultrafilters and then
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transfer it to $X$.
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There exists a uniformly recurrent $\cV \in \beta\N$.
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So for any $\cW$,
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$\cW + \cV$ is also uniformly recurrent\gist{:
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Take $\cV_0$.
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We need to find $\cX$ such that $\cX + \cV_0 + \cW +\cV = \cW + \cV$.
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By uniform recurrence of $\cV$ we find $\cX'$
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such that $\cX' + (\cV_0 + \cW) + \cV = \cV$.
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Then $\cX = \cW + \cX'$ works.
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}{.}
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So all elements of $\beta\N + \cV$
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are uniformly recurrent.
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It is a closed ideal and hence a closed semigroup.
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So $\beta\N + \cV$ contains a minimal closed
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semigroup.
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In particular, there exists an idempotent $\cU \in \beta\N + \cV$.
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$\cU$ is idempotent and uniformly recurrent
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hence it is proximal to $0$.
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Now let us consider $X$.
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Take $y = T^\cU(x)$.
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\begin{claim}
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$y$ uniformly recurrent.
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\end{claim}
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\begin{subproof}
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Recall that $T^{\cV_1 + \cV_2} = T^{\cV_1} \circ T^{\cV_2}$.
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Since $\cU$ is uniformly recurrent,
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$\forall \cV .~\exists \cW.~\cW+ \cV+\cU= \cU$,
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i.e.~$T^{\cW + \cV + \cU} (x) = T^\cW(T^\cV(y)) = T^\cU(x) = y$.
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\end{subproof}
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\begin{claim}
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$y$ is proximal to $x$.
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\end{claim}
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\begin{subproof}
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$\cU$ is proximal to $0$.
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So $\exists \cV.~\cV + \cU = \cV + \hat{0} = \cV$,
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i.e.~$T^{\cV}(y) = T^{\cV + \cU}(x) = T^\cV(x)$.
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Thus $x$ and $y$ are proximal.%TODO REF
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\end{subproof}
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\end{refproof}
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% Office hours wednesday 15:30 - 18:30 office 805
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% Exam: First question: present favorite theorem (7-8 minutes, moderate length proof)
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@ -53,6 +53,7 @@
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\input{inputs/lecture_24}
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\input{inputs/lecture_25}
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\input{inputs/lecture_26}
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\input{inputs/lecture_27}
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\cleardoublepage
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\input{inputs/facts}
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\PrintVocabIndex
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\printbibliography
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\end{document}
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