some small changes
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Josia Pietsch 2024-02-05 02:14:32 +01:00
parent dfd9be7925
commit c9212aefdd
Signed by: josia
GPG Key ID: E70B571D66986A2D
2 changed files with 22 additions and 14 deletions

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@ -24,7 +24,7 @@ $X^{X}$ is a compact Hausdorff space.
is continuous:
Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
We have $ff_0 \in U_{\epsilon}(x,y)$
We have $f \circ f_0 \in U_{\epsilon}(x,y)$
iff $f \in U_\epsilon(x,f_0(y))$.
\item Fix $x_0 \in X$.
Then $f \mapsto f(x_0)$ is continuous.
@ -43,12 +43,15 @@ and take the closure in $X^X$.
\end{definition}
$E(X,T)$ is compact and Hausdorff,
since $X^X$ has these properties.
% TODO THINK ABOUT THIS
Properties of $(X,T)$ translate to properties of $E(X,T)$:
\begin{goal}
We want to show that if $(X,T)$ is distal,
then $E(X,T)$ is a group.
\end{goal}
\gist{
Properties of $(X,T)$ translate to properties of $E(X,T)$:
\begin{goal}
We want to show that if $(X,T)$ is distal,
then $E(X,T)$ is a group.
\end{goal}
}{}
\begin{proposition}
$E(X,T)$ is a semigroup,
@ -59,11 +62,14 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
Take $t \in T$. We want to show that $tG \subseteq G$,
i.e.~for all $h \in G$ we have $th \in G$.
We have that $t^{-1}G$ is compact,
since $t^{-1}$ is continuous
and $G$ is compact.
\gist{
We have that $t^{-1}G$ is compact,
since $t^{-1}$ is continuous
and $G$ is compact.
}{$t^{-1}G$ is compact.}
Then $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
So $G = \overline{T} \subseteq t^{-1}G$.
Hence $tG \subseteq G$.
@ -91,8 +97,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
\begin{definition}
A \vocab{compact semigroup} $S$
is a nonempty semigroup with a compact
Hausdorff topology,
is a nonempty semigroup\footnote{may not contain inverses or the identity}
with a compact Hausdorff topology,
such that $S \ni x \mapsto xs$ is continuous for all $s$.
\end{definition}
\begin{example}
@ -125,8 +131,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$:
\end{proof}
The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
since we already know that it has an identity,
in fact we have chosen $R = \{1\}$ in the proof.
since we already know that it has an identity.
%in fact we might have chosen $R = \{1\}$ in the proof.
But it is interesting for other semigroups.

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@ -59,6 +59,8 @@
and since $B$ is Hausdorff, compact subsets of $B$ are closed.
\end{subproof}
\nr 1
Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup.
Let $d$ be a compatible metric on $X$.
\begin{enumerate}[(a)]