Josia Pietsch 99ca39e52c
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lecture 18, 19
2024-01-04 19:26:34 +01:00

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\subsection{Topological Dynamics}
Let $H$ be a topological group
and $G \subseteq H$ a subgroup.
Then $\overline{G}$ is a topological
Moreover if $H$ is Hausdorff and $G$ is abelian,
then is $\overline{G}$ is abelian.
Let $g,h \in \overline{G}$. We need to show that $g\cdot h \in \overline{G}$.
Take some open neighbourhood $g \cdot h \in U \overset{\text{open}}{\subseteq} H$.
Let $V \overset{\text{open}}{\subseteq} H \times H$
be the preimage of $U$ under $(a,b) \mapsto a \cdot b$.
Let $A \times B \subseteq V$ for some open neighbourhoods of $g$ resp.~$h$.
Take $g' \in A \cap G$ and $h' \in B \cap G$.
Then $g'h' \in U \cap G$,
hence $U \cap G \neq \emptyset$.
Similarly one shows that $\overline{G}$ is closed under inverse images.
Now suppose that $H$ is Hausdorff and $G$ is abelian.
Consider $f\colon (g,h) \mapsto [g,h]$\footnote{Recall that the \vocab{commutator} is $[g,h] \coloneqq g^{-1}h gh^{-1}$.}.
Clearly this is continuous.
Since $G$ is abelian, we have $f(G\times G) = \{1\}$.
Since $H$ is Hausdorff, $\{1\}$ is closed, so
\{1\} = \overline{f(G \times G)} \supseteq f(\overline{G \times G}) = f(\overline{G} \times \overline{G}).