lecture 27
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@ -89,7 +89,6 @@ We do a second proof of \yaref{thm:hindman}:
(y(0), y(1), \ldots, y(n))
\text{ as a subsequence.}
Consider $y(0)$.
@ -139,10 +138,14 @@ We do a second proof of \yaref{thm:hindman}:
% TODO ultrafilter extension continuous
In order to prove \yaref{thm:unifrprox},
we need the following:
Let $X$ be a compact Hausdorff space. % ?
Let $T\colon X \to X$ be continuous.
Let us rephrase the problem in terms of $\beta\N$:
\todo{remove duplicate}
\item $x \in X$ is recurrent
iff $T^\cU(x) = x$ for some $\cU \in \beta\N \setminus \N$.
@ -155,10 +158,15 @@ We do a second proof of \yaref{thm:hindman}:
such that $T^\cU(x) = T^\cU(y)$.
% TODO compare with the statement for the ellis semigroup.
We only to (2) here, as it is the most interesting point.%
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 $x$.
@ -200,18 +208,7 @@ We do a second proof of \yaref{thm:hindman}:
Since we get this for every neighbourhood,
it follows that $T^\cU ( T^\cV(x)) = x$.

inputs/lecture_27.tex Normal file
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@ -0,0 +1,248 @@
\begin{subproof}[(2), $\impliedby$, sketch]
Assume that $x $ is not uniformly recurrent.
Then there is a neighbourhood $G \ni x$
such that for all $M \in \N$
Y_M = \{ n \in \N : \forall k < M.~T^{n+k}(x) \not\in G\} \neq \emptyset.
Note that $Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \ldots$
Take $\cV \in \beta\N$ containing all $Y_n$.
We aim to show that there is no $\cU\in \beta\N$ such that $T_\cU(T_\cV(x)) = x$.
Towards a contradiction suppose that such $\cU$ exists.
For every $k + 1$ we have $Y_{k+1} \in \cV$.
In particular
\{n \in \N : T^{n+k}(x) \not\in G\} \supseteq Y_{k+1},
(\cV n) T^{n+k}(x) \not\in G,
(\cV n) T^n(x) \not\in \underbrace{T^{-k}(G)}_{\text{open}}.
\underbrace{\cV-\lim_n T^n(x)}_{T^\cV(x)} \not\in T^{-k}(G).
We get that
\forall k.~T^k(T^\cV(x)) \not\in G.
It follows that
$\forall \cU \in \beta\N.~T^{\cU}(T^\cV(x)) \not\in G$.
% TODO Why? Think about this.
Take $X = \beta\N$,
$S \colon \beta\N \to \beta\N$,
$S(\cU ) = \hat{1}+ \cU$.
S^\cV(\cU) = \cV-\lim_n S^n(\cU) = \cV-\lim_n(\hat{n} + \cU) =
\cV-\lim_n \hat{n} + \cU = \cV + \cU.
% TODO check
$\cU$ is recurrent
\exists \cV \in \beta\N \setminus \N .~S^\cV(\cU) = \cU.
$\cU$ is uniformly recurrent iff
\forall \cV.~\exists \cW.~\cW + \cV + \cU = \cU.
$\cU_1$ and $\cU_2$ are proximal
iff $\exists \cV.~\cV + \cU_1 = \cV + \cU_2$.
We say that $I \subseteq \beta\N$
is a \vocab{left ideal} ,
\forall \cU \in I.~\forall \cV \in \beta\N.~\cV + \cU \in I.
\item $\cU$ is uniformly recurrent in $\beta\N$
iff $\cU$ belongs to a minimal\footnote{wrt.~$\subseteq $} (closed)
left ideal in $\beta\N$.
\item $\cU$ is an idempotent in $\beta\N$
iff $\cU$ belong to a minimal closed
subsemigroup of $\beta\N$.
Note that any $\cU \in \beta\N$ yields
%gives rise to
a left ideal $\beta\N + \cU$.
It is closed, since it is the image
of $\beta\N$ under the continuous maps
$\cV \mapsto \cV + \cU$
and $\beta\N$ is compact.
Note that $\beta\N + \cU$ is closed,
since $\beta\N$ is compact and $\cdot + \cU$ continuous.
$\cU$ belongs to a minimal left ideal
iff $\beta\N + \cU$ is minimal%
since every ideal containing $\cU$
contains $\beta\N + \cU$.
Note that $\beta\N + \cV + \cU \subseteq \beta\N + \cU$
and if $I \subsetneq \beta\N + \cU$,
we have $\cV_0 = \cV + \cU \in I$
and $\beta\N + \cV + \cU \subseteq \beta\N + \cU$.
So $\cU$ belongs to a minimal left ideal iff
\forall \cV \in \beta\N .~\beta\N + \cV + \cU = \beta\N + \cU.
This is the case iff
\underbrace{\forall \cV .~\exists \cW.~ \cW + \cV + \cU = \cU.}_%
{\cV \text{ uniformly recurrent}}
\gist{(For one direction take $\cW$ such that $\cW + \cV + \cU= \hat{0} +\cU$.
For the other direction note that
for every $\cV_0 $, $\cV_0 + \cU$
can be written as $\cV_0 + \cW + (\cV + \cU)$.
Where we take $\cW$ such that $\cW + \cV + \cU = \cU$.
\item This is very similar to the proof of the \yaref{lem:ellisnumakura}.
If $\cU$ is idempotent, then $\{\cU\}$
is a semigroup.
Let $C$ be a minimal closed subsemigroup of $\beta\N$.
Then $C + \cU$ is a closed subsemigroup.
By minimality, we get $C = C + \cU$.
Let $D = \{ \cV \in C .~ \cV + \cU = \cU\}$.
We have $D \neq \emptyset$.
$D$ is a closed semigroup,
so $D = C$ be minimality.
Hence $\cU + \cU = \cU$.
Idempotent and uniformly recurrent elements exist.
Use \yaref{thm:unifrprox:helper2}
and Zorn's lemma.
(1) $\implies$ (2) $\implies$ (3)
\item $\cU$ is uniformly recurrent and proximal to $\hat{0}$.
\item $\cU$ is an idempotent.
\item $\cU$ is recurrent and proximal to $\hat{0}$.
(1) $\implies$ (2):
Let $\cU$ be uniformly recurrent and proximal to $ \hat{0}$.
Take $\cV$ such that $\cV + \cU = \cV + \hat{0} = \cV$.
% TODO REF beginning of lecture
Since $\cU$ is uniformly recurrent,
there exists $\cW$ such that $\cW + \cV + \cU = \cU$,
i.e.~$\cW + \cV = \cU$.
Then $\cU + \cU = \cW + \cV + \cU = \cU$.
(2) $\implies$ (3):
% Let $\cU$ be an idempotent.
% We want to find $\cV$ such that $\cV + \cU = \cU$.
% $\cV'$ such that $\cV' + \cU = \cV' + 0$ proximal to $0$?
% TAKE $\cV = \cV' = \cU$.
$\cU$ is uniformly recurrent and proximal to $0$
iff $\cU$ is an idempotent
and belongs to some minimal left ideal of $\beta\N$.
Let $T\colon X \to X$ and $x \in X$.
We want to find $y \in X$
such that $y$ is uniformly recurrent
and proximal to $x$.
We first prove a version for ultrafilters and then
transfer it to $X$.
There exists a uniformly recurrent $\cV \in \beta\N$.
So for any $\cW$,
$\cW + \cV$ is also uniformly recurrent\gist{:
Take $\cV_0$.
We need to find $\cX$ such that $\cX + \cV_0 + \cW +\cV = \cW + \cV$.
By uniform recurrence of $\cV$ we find $\cX'$
such that $\cX' + (\cV_0 + \cW) + \cV = \cV$.
Then $\cX = \cW + \cX'$ works.
So all elements of $\beta\N + \cV$
are uniformly recurrent.
It is a closed ideal and hence a closed semigroup.
So $\beta\N + \cV$ contains a minimal closed
In particular, there exists an idempotent $\cU \in \beta\N + \cV$.
$\cU$ is idempotent and uniformly recurrent
hence it is proximal to $0$.
Now let us consider $X$.
Take $y = T^\cU(x)$.
$y$ uniformly recurrent.
Recall that $T^{\cV_1 + \cV_2} = T^{\cV_1} \circ T^{\cV_2}$.
Since $\cU$ is uniformly recurrent,
$\forall \cV .~\exists \cW.~\cW+ \cV+\cU= \cU$,
i.e.~$T^{\cW + \cV + \cU} (x) = T^\cW(T^\cV(y)) = T^\cU(x) = y$.
$y$ is proximal to $x$.
$\cU$ is proximal to $0$.
So $\exists \cV.~\cV + \cU = \cV + \hat{0} = \cV$,
i.e.~$T^{\cV}(y) = T^{\cV + \cU}(x) = T^\cV(x)$.
Thus $x$ and $y$ are proximal.%TODO REF
% Office hours wednesday 15:30 - 18:30 office 805
% Exam: First question: present favorite theorem (7-8 minutes, moderate length proof)

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@ -80,10 +81,8 @@