w23-logic-3/inputs/lecture_26.tex

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\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
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In order to prove \yaref{thm:unifrprox},
we need the following:
\begin{theorem}
\label{thm:unifrprox:helper}
Let $X$ be a compact Hausdorff space. % ?
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Let $T\colon X \to X$ be continuous.
Let us rephrase the problem in terms of $\beta\N$:
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\todo{remove duplicate}
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\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}
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\end{theorem}
\begin{refproof}{thm:unifrprox:helper}[sketch]
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), $\implies$]
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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}
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\phantom\qedhere
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\end{refproof}