lecture 26

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

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}
 
 

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.~$s^2 = s$.

inputs/lecture_25.tex

@@ -164,7 +164,7 @@ This is associative, but not commutative.
     $(\beta\N,+)$
     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}
 
 

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}