From d17e25f4d0cc41c603f654d14272ff93b5ba68df Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 30 Jan 2024 11:55:30 +0100 Subject: [PATCH] lecture 26 --- inputs/lecture_09.tex | 2 +- inputs/lecture_15.tex | 3 +- inputs/lecture_17.tex | 2 +- inputs/lecture_25.tex | 13 ++- inputs/lecture_26.tex | 217 ++++++++++++++++++++++++++++++++++++++++++ 5 files changed, 227 insertions(+), 10 deletions(-) create mode 100644 inputs/lecture_26.tex diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index f91347a..d3b2af2 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -13,7 +13,7 @@ there is a $X$-universal set $\cU$ for $\Sigma^0_\xi(X)$. Take $A \coloneqq \{y \in X : (y,y) \not\in \cU\}$. - Then $A \in \Pi^0_\xi(X)$.\todo{Needs a small proof} + Then $A \in \Pi^0_\xi(X)$.\footnote{cf.~\yaref{s7e1} and use that $\{(x,x) \in X^2\} \cong X$.} By assumption $A \in \Sigma^0_\xi(X)$, i.e.~there exists some $z \in X$ such that $A = \cU_z$. We have diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 6f3f804..8b97525 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -129,6 +129,7 @@ Recall: \end{definition} \begin{definition} + \label{def:flow} Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology} and let $X$ be a compact metrizable space. @@ -181,7 +182,7 @@ Recall: \end{definition} \begin{warning}+ What is called ``factor'' here is called ``subflow'' - by Fürstenberg. + by Furstenberg. \end{warning} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index 8f6f36e..f4caad5 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -118,7 +118,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: since $s \in Rs$ and $P$ is a compact semigroup, since $x \overset{\alpha}\mapsto xs$ - is continuous and $P = \alpha^-1(s) \cap R$. + is continuous and $P = \alpha^{-1}(s) \cap R$. Thus $P = R$ by minimality, so $s \in P$, i.e.~$s^2 = s$. diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex index 2b8ea9a..0703780 100644 --- a/inputs/lecture_25.tex +++ b/inputs/lecture_25.tex @@ -6,10 +6,10 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. \item This is a topological space, where a basis consist of sets $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$. - + (For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$ and $\beta\N = V_\N$.) - + \item Note also that for $A, B \subseteq \N$, $V_{A \cup B} = V_A \cup V_B$, $V_{A^c} = \beta\N \setminus V_A$. @@ -102,7 +102,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. Since $X$ is compact, there exists a finite subcover $G_{x_1}, \ldots, G_{x_m}$. - + But then \begin{IEEEeqnarray*}{rCl} \N &=& \{ n \in \N : x_n \in \bigcup_{i=1}^m G_{x_i}\}\\ @@ -161,10 +161,10 @@ This is associative, but not commutative. \end{IEEEeqnarray*} \end{proof} \begin{corollary} - $(\beta\N,+)$ + $(\beta\N,+)$ is a %(left) \vocab{compact semigroup}, - i.e.~it is comapct, Hausdorff, associative and left-continuous.% + i.e.~it is compact, Hausdorff, associative and left-continuous.% %\footnote{There is no convention on left and right.} \end{corollary} So we can apply the \yaref{lem:ellisnumakura} @@ -181,6 +181,7 @@ to obtain \end{observe} \begin{theorem}[Hindman] + \label{thm:hindman} If $\N$ is partitioned into finitely many sets, then there is is an infinite subset $H \subseteq \N$ @@ -225,8 +226,6 @@ to obtain Chose $x_n > x_{n-1}$ that satisfies this. \end{itemize} Set $H \coloneqq \{x_i : i < \omega\}$. - - \end{proof} diff --git a/inputs/lecture_26.tex b/inputs/lecture_26.tex new file mode 100644 index 0000000..0b1b5a8 --- /dev/null +++ b/inputs/lecture_26.tex @@ -0,0 +1,217 @@ +\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 + + +\begin{refproof}{thm:unifrprox}[sketch] + Let $T\colon X \to X$ be continuous. + Let us rephrase the problem in terms of $\beta\N$: + \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} + We only to (2) here, as it is the most interesting point.% + \todo{other parts will be in the official notes} + + \begin{subproof}[(2)] + 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 $\cV-\lim_n T^n(x) \in T^{-k}(G)$. + 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} + + + + + +\end{refproof}