w23-logic-3/inputs/lecture_26.tex
Josia Pietsch 6c5eda59e4
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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{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.}%
\begin{IEEEeqnarray*}{rl}
\forall n.~\exists 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{IEEEeqnarray*}
\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}
}{}