\lecture{26}{2024-01-30}{} Let $T\colon X \to X$ be a continuous map. This gives $\N \acts X$. \begin{definition} \label{def:unifrec} A point $x \in X$ is called \vocab{uniformly recurrent} iff for each neighbourhood $G$ of $x$, there is $M \in \N_+$, such that \[ \forall n \in \N.~\exists k < m.~T^{n+k}(x) \in G. \] \end{definition} \begin{definition} A pair $x,y \in X$ is \vocab{proximal}% \footnote{see also \yaref{def:flow}, where we defined proximal for metric spaces} iff for all neighbourhoods $G$ of the diagonal% \gist{\footnote{recall that the diagonal is defined to be $\Delta \coloneqq \{(x,x) : x \in X\}$}}{} infinitely many $n$ satisfy $(T^n(x), T^n(y)) \in G$. \end{definition} \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} We do a second proof of \yaref{thm:hindman}: \begin{proof}[Furstenberg] A partition of $\N$ into $k$-many pieces can be viewed as a function $f\colon \N \to k$. Let $X = k^\N$ be the set of all such functions. Equip $X$ with the product topology. Then $X$ is compact and Hausdorff. Let $T\colon X \to X$ be the shift given by \gist{% \begin{IEEEeqnarray*}{rCl} T\colon k^{\N} &\longrightarrow & k^{\N} \\ (y\colon \N \to k)&\longmapsto & \begin{pmatrix*}[l] \N &\longrightarrow & k \\ n &\longmapsto & y(n+1), \end{pmatrix*} \end{IEEEeqnarray*} i.e.~}{}% $T(y)(n) = y(n+1)$. Let $x $ be the given partition. We want to find an infinite set $H$ for $x$ as in the theorem. Let $y$ be uniformly recurrent and proximal to $x$. \begin{itemize} \item % Gist: proximal Since $x$ and $y$ are proximal, we get that for every $N \in \N$, there are infinitely many $n$ such that $T^n(x)\defon{N} = T^n(y)\defon{N}$.% \footnote{% Consider $G_N = \{(a,b) \in X^2 : a\defon N = b\defon N\}$ This is a neighbourhood of the diagonal.% } \item % Gist: unif. recurrent Consider the neighbourhood \[ G_n \coloneqq \{z \in X: z\defon{n} = y\defon{n}\} \] of $y$. By the uniform recurrence of $y$, we get that% \footnote{Note that here we might need to choose a bigger $N$ than the $M$ in \yaref{def:unifrec}, but $2M$ suffices.}% \[ \forall n.~\exists N% \gg n .~\forall r.~(y(r), y(r+1), \ldots, y(r+N - 1) \text{ contains } (y(0), y(1), \ldots, y(n)) \text{ as a subsequence.} \] \end{itemize} Consider $y(0)$. We will prove that this color works and construct a corresponding $H$. \begin{itemize} \item % Step 1 Let $G_0 \coloneqq [y(0)]$ and let $N_0$ be such that \[ \forall r.~(y(r), \ldots, y(r + N-0 - 1)) \text{ contains $y(0)$.} \] By proximality, there exist infinitely many $r$ such that $(y(r), \ldots, (y(r+N-1)) = (x(r), \ldots, x(r+N-1))$. Fix $h_0 \in \N$ such that $x(h_0) = y(0)$. \item% % Step 2 Let $G_{n_0} = [(y(0), \ldots, y(h_0)]$. Choose $N_1$. For all $r$, $(y(r), \ldots, y(r+N-1))$ contains $(\underbrace{y(0)}_{= C}, \ldots, \underbrace{y(h_0)}_{= C})$. Pick $r > h_0$ such that $(x(r), \ldots, x(r+N-1))$ contains $(y(0), \ldots y(h_0))$. Let $(x(r+s), \ldots, x(r+s+h_0)) = (y(0), \ldots, y(h_0))$. Then set $h_1 = r + s$. Then $x(h_0) = c$, $x(h_1) = y(0) = c$ and $x(h_0+h_1) = y(h_0) = c$. \item% % Step 3 Let $G_{h_0 + h_1} = [y(0), \ldots, y(h_0+h_1)]$. Let $r > h_0 + h_1$. Choose $N_2$ large enough such that $(y(0), \ldots, y(h_0+h_1))$ 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 $\ldots$ \end{itemize} \end{proof} % TODO ultrafilter extension continuous 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 $\cU \in \beta\N \setminus \N$. \item $x \in X$ is uniformly recurrent iff for every $\cV \in \beta\N$, there is $\cU \in \beta\N$ with $T^{\cU}(T^{\cV}(x)) = x$. \item $x, y \in X$ are proximal iff there is $\cU \in \beta\N$ such that $T^\cU(x) = T^\cU(y)$. % TODO compare with the statement for the ellis semigroup. \end{enumerate} \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), $\implies$] Suppose that $ x$ is uniformly recurrent. Take some $\cV \in \beta\N$. Let $G_0$ be a neighbourhood of $x$. Then $x \in G \subseteq G_0$, where $G$ is a closed neighbourhood, i.e.~$X \in \inter G$. Let $M$ be such that \[ \forall n .ยจ \exists k < M.~T^{n+k}(x) \in G. \] So there is a $k < M$ such that \[ (\cV n) T^{n +k}(x) \in G. \] Hence \[ (\cV n) T^n(x) \in \underbrace{T^{-k}(\overbrace{G}^{\text{closed}})}_{\text{closed}}. \] Therefore $\cV-\lim_n T^n(x) \in T^{-k}(G)$. So $T^k(T^\cV(x)) \in G \subseteq G_0$. We have shown that for every open neighbourhood $G$ of $x$, the set $Y_G = \{k \in \N : T^k(T^\cV(x)) \in G\} \neq \emptyset$. The sets $\{Y_G : G \text{ open neighbourhood of } G\}$ form a filter basis,\footnote{The sets and their supersets form a filter.} since $Y_{G_1} \cap Y_{G_2} = Y_{G_1 \cap G_2}$. Let $\cU$ be an ultrafilter containing all the $Y_G$. Then \[ (\cU k) R^k(T^\cV)(x)) \in G \] i.e.~$T^\cU(T^\cV(x)) \in \overline{G}$. Since we get this for every neighbourhood, it follows that $T^\cU ( T^\cV(x)) = x$. \end{subproof} \phantom\qedhere \end{refproof}