subgroup of topological group is either clopen or has empty interior

Theorem - Every subgroup of a topological group is either clopen or has empty interior.

$\\,$

Proof: Let $G$ be a topological group and $H\\subseteq G$ a subgroup. Suppose the interior of $H$ is nonempty, i.e. there is a non-empty open set $U$ of $G$ such that $U\\subseteq H$ . Translating $U$ around $H$ we can see that $H$ is open: if $u\\in U$ then for every $h\\in H$ the set $hu^{-1}U$ is open in $G$ , is contained in $H$ and contains $h$ , which implies that $H$ is open in $G$ .

Let us now see that $H$ is closed. Let $\\overline{H}$ denote the closure of $H$ and let $H^{2}$ be the set of elements of the form $h_{1}h_{2}$ where $h_{1},h_{2}\\in H$ . Of course, since $H$ is a subgroup of $G$ , we have that $H^{2}=H$ . Also, since $H$ is open we know that $H\\subseteq\\overline{H}\\subseteq H^{2}$ (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 5). Hence $\\overline{H}=H$ , i.e. $H$ is closed.

We have proven that a subgroup of a topological group must be clopen or it must have empty interior. Since this two topological properties can never be satisfied simultaneously, we have that every subgroup of a topological group is either clopen or it has empty interior. $\\square$

单词	SubgroupOfTopologicalGroupIsEitherClopenOrHasEmptyInterior
释义	subgroup of topological group is either clopen or has empty interior Theorem - Every subgroup of a topological group is either clopen or has empty interior. $\\,$ Proof: Let $G$ be a topological group and $H\\subseteq G$ a subgroup. Suppose the interior of $H$ is nonempty, i.e. there is a non-empty open set $U$ of $G$ such that $U\\subseteq H$ . Translating $U$ around $H$ we can see that $H$ is open: if $u\\in U$ then for every $h\\in H$ the set $hu^{-1}U$ is open in $G$ , is contained in $H$ and contains $h$ , which implies that $H$ is open in $G$ . Let us now see that $H$ is closed. Let $\\overline{H}$ denote the closure of $H$ and let $H^{2}$ be the set of elements of the form $h_{1}h_{2}$ where $h_{1},h_{2}\\in H$ . Of course, since $H$ is a subgroup of $G$ , we have that $H^{2}=H$ . Also, since $H$ is open we know that $H\\subseteq\\overline{H}\\subseteq H^{2}$ (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 5). Hence $\\overline{H}=H$ , i.e. $H$ is closed. We have proven that a subgroup of a topological group must be clopen or it must have empty interior. Since this two topological properties can never be satisfied simultaneously, we have that every subgroup of a topological group is either clopen or it has empty interior. $\\square$
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