connected locally compact topological groups are $\\sigma$ -compact

The main result of this entry is the following theorem (whose proof is given below). The result expressed in the title then follows as a corollary.

Theorem - Every locally compact topological group $G$ has an open $\\sigma$ -compact (http://planetmath.org/SigmaCompact) subgroup $H$ .

Corollary 1 - Every locally compact topological group is the topological disjoint union of $\\sigma$ -compact spaces.

Corollary 2 - Every connected locally compact topological group is $\\sigma$ -compact.

We first outline the proofs of the above corollaries:

Proof (Corollaries 1 and 2) : Let $G$ be a locally compact topological group. The main theorem implies that there is an open $\\sigma$ -compact subgroup $H$ .

It is known that every open subgroup of $G$ is also closed (see this entry (http://planetmath.org/ClosednessOfSubgroupsOfTopologicalGroups)). Therefore, each $g H$ is a clopen $\\sigma$ -compact subset of $G$ , and $G$ is the topological disjoint union $\\displaystyle\\bigcup_{g\\in G}\\;gH$ .

Of course, if $G$ is connected then $H$ must be all of $G$ . Hence, $G$ is $\\sigma$ -compact. $\\square$

$\\quad$

Proof (Theorem) : Let us fix some notation first. If $A$ is a subset of $G$ we use the notation $A^{-1}:=\\{a^{-1}:a\\in A\\}$ , $A^{n}:=\\{a_{1}\\dots a_{n}:a_{1},\\dots,a_{n}\\in A\\}$ and $\\overline{A}$ denotes the closure of $A$ .

Pick a neighborhood $W$ of $e$ (the identity element of $G$ ) with compact closure. Then $V:=W\\cap W^{-1}$ is a neighborhood of $e$ with compact closure such that $V=V^{-1}$ .

Let $H:=\\bigcup_{n=1}^{\\infty}V^{n}$ . $H$ is clearly a subgroup of $G$ . We now only have to prove that $H$ is open and $\\sigma$ -compact.

We have that (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 3, 4 and 5)

•
$V^{n}$ is open
•
$\\overline{V}^{n}$ is compact
•
$\\overline{V}^{n}\\subset V^{2n}$

So $H$ is open and also $H=\\bigcup_{n=1}^{\\infty}\\overline{V}^{n}$ , which implies that $H$ is $\\sigma$ -compact. $\\square$

单词	ConnectedLocallyCompactTopologicalGroupsAresigmacompact
释义	connected locally compact topological groups are $\\sigma$ -compact The main result of this entry is the following theorem (whose proof is given below). The result expressed in the title then follows as a corollary. Theorem - Every locally compact topological group $G$ has an open $\\sigma$ -compact (http://planetmath.org/SigmaCompact) subgroup $H$ . Corollary 1 - Every locally compact topological group is the topological disjoint union of $\\sigma$ -compact spaces. Corollary 2 - Every connected locally compact topological group is $\\sigma$ -compact. We first outline the proofs of the above corollaries: Proof (Corollaries 1 and 2) : Let $G$ be a locally compact topological group. The main theorem implies that there is an open $\\sigma$ -compact subgroup $H$ . It is known that every open subgroup of $G$ is also closed (see this entry (http://planetmath.org/ClosednessOfSubgroupsOfTopologicalGroups)). Therefore, each $g H$ is a clopen $\\sigma$ -compact subset of $G$ , and $G$ is the topological disjoint union $\\displaystyle\\bigcup_{g\\in G}\\;gH$ . Of course, if $G$ is connected then $H$ must be all of $G$ . Hence, $G$ is $\\sigma$ -compact. $\\square$ $\\quad$ Proof (Theorem) : Let us fix some notation first. If $A$ is a subset of $G$ we use the notation $A^{-1}:=\\{a^{-1}:a\\in A\\}$ , $A^{n}:=\\{a_{1}\\dots a_{n}:a_{1},\\dots,a_{n}\\in A\\}$ and $\\overline{A}$ denotes the closure of $A$ . Pick a neighborhood $W$ of $e$ (the identity element of $G$ ) with compact closure. Then $V:=W\\cap W^{-1}$ is a neighborhood of $e$ with compact closure such that $V=V^{-1}$ . Let $H:=\\bigcup_{n=1}^{\\infty}V^{n}$ . $H$ is clearly a subgroup of $G$ . We now only have to prove that $H$ is open and $\\sigma$ -compact. We have that (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 3, 4 and 5) • $V^{n}$ is open • $\\overline{V}^{n}$ is compact • $\\overline{V}^{n}\\subset V^{2n}$ So $H$ is open and also $H=\\bigcup_{n=1}^{\\infty}\\overline{V}^{n}$ , which implies that $H$ is $\\sigma$ -compact. $\\square$
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connected locally compact topological groups are σ-compact

connected locally compact topological groups are $\\sigma$ -compact