\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{refproof}{thm:hindmanfurstenberg}[Furstenberg] \gist{% 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 $H$. % def power(s): % if len(s) == 0: % return [[]] % else: % p = power(s[0:-1]) % return [q + [s[-1]] for q in p] + [q for q in p] % % % def draw(hs): % s = "\\begin{tikzpicture}" % s += "\n\t\\node at (-0.5,0.5) {$x$};"; % for (i,h) in enumerate(hs): % s += "\n\t\\node at (" + str(hs[i]) + ",0.5) {$h_"+str(i)+"$};"; % % for subset in power(range(0,len(hs))): % if len(subset) <= 1: % continue % c = sum([hs[i] for i in subset]) % c2 = 0.9 if 0 in subset else 0.7 % s += "\n\t\\node[black!40!white] at (" + str(c) + ", " + str(c2) + ") {\\tiny{$" + " + ".join(map(lambda x : "h_{" + str(x) + "}", subset)) + "$}};" % s += "\n\t\\draw[black!40!white, very thin] (" + str(c) + ", " + str(c2 - 0.1) + ") -- (" + str(c) + ",-0.1);" % if subset != list(range(0,len(subset))): % s += "\n\t\\node[blue!40!white] at (" + str(c) + ", " + str(c2-2) + "){\\tiny{$y(" + " + ".join(map(lambda x: "h_{" + str(x) + "}", subset[0:-1])) + ")$}};" % s += "\n\t\\draw[blue!40!white, very thin] (" + str(c) + ", -0.1) -- (" + str(c) + ", " + str(c2-1.8) + ");" % for (i,h) in enumerate(hs): % hsum = sum(hs[0:i+1]) % s += '\n\t\\draw[blue] ('\ % + str(h) + ', 0) -- ('\ % + str(h) + ', -0.2) -- ('\ % + str(hsum) + ', -0.2) -- ('\ % + str(hsum) + ', 0);' % s += "\n\t\\node[blue] at (" + str(h) + ", -0.5) {$y(0)$};" % if i > 0: % s += "\n\t\\node[blue] at (" + str(hsum)\ % + ", -0.5) {$y(" + ("\ % + ".join(list(map(lambda x : "h_{" + str(x) + "}", range(0,i)))))\ % + ")$};"; % s += "\n\t\\draw[thick] (0,0) -- ("+ str(sum(hs) + 2) + ",0);" % s += "\n\\end{tikzpicture}" % return s % print(draw(np.cumsum([0.8,1,2.4,4.5]))) \adjustbox{scale=0.7,center}{% \begin{tikzpicture} \node at (-0.5,0.5) {$x$}; \node at (0.8,0.5) {$h_0$}; \node at (1.8,0.5) {$h_1$}; \node at (4.2,0.5) {$h_2$}; \node at (8.7,0.5) {$h_3$}; \node[black!40!white] at (15.5, 0.9) {\tiny{$h_{0} + h_{1} + h_{2} + h_{3}$}}; \draw[black!40!white, very thin] (15.5, 0.8) -- (15.5,-0.1); \node[black!40!white] at (14.7, 0.7) {\tiny{$h_{1} + h_{2} + h_{3}$}}; \draw[black!40!white, very thin] (14.7, 0.6) -- (14.7,-0.1); \node[blue!40!white] at (14.7, -1.3){\tiny{$y(h_{1} + h_{2})$}}; \draw[blue!40!white, very thin] (14.7, -0.1) -- (14.7, -1.1); \node[black!40!white] at (13.7, 0.9) {\tiny{$h_{0} + h_{2} + h_{3}$}}; \draw[black!40!white, very thin] (13.7, 0.8) -- (13.7,-0.1); \node[blue!40!white] at (13.7, -1.1){\tiny{$y(h_{0} + h_{2})$}}; \draw[blue!40!white, very thin] (13.7, -0.1) -- (13.7, -0.9); \node[black!40!white] at (12.899999999999999, 0.7) {\tiny{$h_{2} + h_{3}$}}; \draw[black!40!white, very thin] (12.899999999999999, 0.6) -- (12.899999999999999,-0.1); \node[blue!40!white] at (12.899999999999999, -1.3){\tiny{$y(h_{2})$}}; \draw[blue!40!white, very thin] (12.899999999999999, -0.1) -- (12.899999999999999, -1.1); \node[black!40!white] at (11.299999999999999, 0.9) {\tiny{$h_{0} + h_{1} + h_{3}$}}; \draw[black!40!white, very thin] (11.299999999999999, 0.8) -- (11.299999999999999,-0.1); \node[blue!40!white] at (11.299999999999999, -1.1){\tiny{$y(h_{0} + h_{1})$}}; \draw[blue!40!white, very thin] (11.299999999999999, -0.1) -- (11.299999999999999, -0.9); \node[black!40!white] at (10.5, 0.7) {\tiny{$h_{1} + h_{3}$}}; \draw[black!40!white, very thin] (10.5, 0.6) -- (10.5,-0.1); \node[blue!40!white] at (10.5, -1.3){\tiny{$y(h_{1})$}}; \draw[blue!40!white, very thin] (10.5, -0.1) -- (10.5, -1.1); \node[black!40!white] at (9.5, 0.9) {\tiny{$h_{0} + h_{3}$}}; \draw[black!40!white, very thin] (9.5, 0.8) -- (9.5,-0.1); \node[blue!40!white] at (9.5, -1.1){\tiny{$y(h_{0})$}}; \draw[blue!40!white, very thin] (9.5, -0.1) -- (9.5, -0.9); \node[black!40!white] at (6.800000000000001, 0.9) {\tiny{$h_{0} + h_{1} + h_{2}$}}; \draw[black!40!white, very thin] (6.800000000000001, 0.8) -- (6.800000000000001,-0.1); \node[black!40!white] at (6.0, 0.7) {\tiny{$h_{1} + h_{2}$}}; \draw[black!40!white, very thin] (6.0, 0.6) -- (6.0,-0.1); \node[blue!40!white] at (6.0, -1.3){\tiny{$y(h_{1})$}}; \draw[blue!40!white, very thin] (6.0, -0.1) -- (6.0, -1.1); \node[black!40!white] at (5.0, 0.9) {\tiny{$h_{0} + h_{2}$}}; \draw[black!40!white, very thin] (5.0, 0.8) -- (5.0,-0.1); \node[blue!40!white] at (5.0, -1.1){\tiny{$y(h_{0})$}}; \draw[blue!40!white, very thin] (5.0, -0.1) -- (5.0, -0.9); \node[black!40!white] at (2.6, 0.9) {\tiny{$h_{0} + h_{1}$}}; \draw[black!40!white, very thin] (2.6, 0.8) -- (2.6,-0.1); \draw[blue] (0.8, 0) -- (0.8, -0.2) -- (0.8, -0.2) -- (0.8, 0); \node[blue] at (0.8, -0.5) {$y(0)$}; \draw[blue] (1.8, 0) -- (1.8, -0.2) -- (2.6, -0.2) -- (2.6, 0); \node[blue] at (1.8, -0.5) {$y(0)$}; \node[blue] at (2.6, -0.5) {$y(h_{0})$}; \draw[blue] (4.2, 0) -- (4.2, -0.2) -- (6.800000000000001, -0.2) -- (6.800000000000001, 0); \node[blue] at (4.2, -0.5) {$y(0)$}; \node[blue] at (6.800000000000001, -0.5) {$y(h_{0} + h_{1})$}; \draw[blue] (8.7, 0) -- (8.7, -0.2) -- (15.5, -0.2) -- (15.5, 0); \node[blue] at (8.7, -0.5) {$y(0)$}; \node[blue] at (15.5, -0.5) {$y(h_{0} + h_{1} + h_{2})$}; \draw[thick] (0,0) -- (17.5,0); \end{tikzpicture} } \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 $(y(r), \ldots, (y(r+N_0-1)) = (x(r), \ldots, x(r+N_0-1))$. 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 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} }{ \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 $\leadsto$ \begin{IEEEeqnarray*}{rl} \forall n .~\exists N.~\forall r.~&y\defon{\{r,\ldots,r+N-1\}}\\ &\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.} \end{IEEEeqnarray*} (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}, we need to rephrase the problem in terms of $\beta\N$: \begin{theorem} \label{thm:unifrprox:helper} Let $X$ be a compact Hausdorff space. % ? Let $T\colon X \to X$ be continuous. \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} \gist{% \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 \[ \forall n .~ \exists k < M.~T^{n+k}(x) \in G. \] 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 $\ulim{\cV}_n T^n(x) \in T^{-k}(G)$. 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} }{}