This commit is contained in:
parent
8f86b71d92
commit
d17e25f4d0
5 changed files with 227 additions and 10 deletions
|
@ -13,7 +13,7 @@
|
||||||
there is a $X$-universal set $\cU$ for $\Sigma^0_\xi(X)$.
|
there is a $X$-universal set $\cU$ for $\Sigma^0_\xi(X)$.
|
||||||
|
|
||||||
Take $A \coloneqq \{y \in X : (y,y) \not\in \cU\}$.
|
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 $A \in \Sigma^0_\xi(X)$,
|
By assumption $A \in \Sigma^0_\xi(X)$,
|
||||||
i.e.~there exists some $z \in X$ such that $A = \cU_z$.
|
i.e.~there exists some $z \in X$ such that $A = \cU_z$.
|
||||||
We have
|
We have
|
||||||
|
|
|
@ -129,6 +129,7 @@ Recall:
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
|
\label{def:flow}
|
||||||
Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
|
Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
|
||||||
and let $X$ be a compact metrizable space.
|
and let $X$ be a compact metrizable space.
|
||||||
|
|
||||||
|
@ -181,7 +182,7 @@ Recall:
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{warning}+
|
\begin{warning}+
|
||||||
What is called ``factor'' here is called ``subflow''
|
What is called ``factor'' here is called ``subflow''
|
||||||
by Fürstenberg.
|
by Furstenberg.
|
||||||
\end{warning}
|
\end{warning}
|
||||||
|
|
||||||
|
|
||||||
|
|
|
@ -118,7 +118,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
|
||||||
since $s \in Rs$
|
since $s \in Rs$
|
||||||
and $P$ is a compact semigroup,
|
and $P$ is a compact semigroup,
|
||||||
since $x \overset{\alpha}\mapsto xs$
|
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,
|
Thus $P = R$ by minimality,
|
||||||
so $s \in P$,
|
so $s \in P$,
|
||||||
i.e.~$s^2 = s$.
|
i.e.~$s^2 = s$.
|
||||||
|
|
|
@ -164,7 +164,7 @@ This is associative, but not commutative.
|
||||||
$(\beta\N,+)$
|
$(\beta\N,+)$
|
||||||
is a %(left)
|
is a %(left)
|
||||||
\vocab{compact semigroup},
|
\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.}
|
%\footnote{There is no convention on left and right.}
|
||||||
\end{corollary}
|
\end{corollary}
|
||||||
So we can apply the \yaref{lem:ellisnumakura}
|
So we can apply the \yaref{lem:ellisnumakura}
|
||||||
|
@ -181,6 +181,7 @@ to obtain
|
||||||
\end{observe}
|
\end{observe}
|
||||||
|
|
||||||
\begin{theorem}[Hindman]
|
\begin{theorem}[Hindman]
|
||||||
|
\label{thm:hindman}
|
||||||
If $\N$ is partitioned into finitely many
|
If $\N$ is partitioned into finitely many
|
||||||
sets,
|
sets,
|
||||||
then there is is an infinite subset $H \subseteq \N$
|
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.
|
Chose $x_n > x_{n-1}$ that satisfies this.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
Set $H \coloneqq \{x_i : i < \omega\}$.
|
Set $H \coloneqq \{x_i : i < \omega\}$.
|
||||||
|
|
||||||
|
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
||||||
|
|
||||||
|
|
217
inputs/lecture_26.tex
Normal file
217
inputs/lecture_26.tex
Normal file
|
@ -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}
|
Loading…
Reference in a new issue