2023-12-09 18:23:59 +01:00
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\subsection{Topological Dynamics}
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2024-01-04 19:26:34 +01:00
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\begin{fact}[\cite{801106}]
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\label{fact:topsubgroupclosure}
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Let $H$ be a topological group
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and $G \subseteq H$ a subgroup.
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Then $\overline{G}$ is a topological
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group.
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Moreover if $H$ is Hausdorff and $G$ is abelian,
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then is $\overline{G}$ is abelian.
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\end{fact}
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\begin{proof}
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Let $g,h \in \overline{G}$. We need to show that $g\cdot h \in \overline{G}$.
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Take some open neighbourhood $g \cdot h \in U \overset{\text{open}}{\subseteq} H$.
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Let $V \overset{\text{open}}{\subseteq} H \times H$
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be the preimage of $U$ under $(a,b) \mapsto a \cdot b$.
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Let $A \times B \subseteq V$ for some open neighbourhoods of $g$ resp.~$h$.
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Take $g' \in A \cap G$ and $h' \in B \cap G$.
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Then $g'h' \in U \cap G$,
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hence $U \cap G \neq \emptyset$.
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Similarly one shows that $\overline{G}$ is closed under inverse images.
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Now suppose that $H$ is Hausdorff and $G$ is abelian.
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Consider $f\colon (g,h) \mapsto [g,h]$\footnote{Recall that the \vocab{commutator} is $[g,h] \coloneqq g^{-1}h gh^{-1}$.}.
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Clearly this is continuous.
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Since $G$ is abelian, we have $f(G\times G) = \{1\}$.
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Since $H$ is Hausdorff, $\{1\}$ is closed, so
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\[
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\{1\} = \overline{f(G \times G)} \supseteq f(\overline{G \times G}) = f(\overline{G} \times \overline{G}).
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\]
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\end{proof}
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