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 $gH$ 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. $\mathrm{\square }$

$$

Proof (Theorem) : Let us fix some notation first. If $A$ is a subset of $G$ we use the notation $A^{-1}:=\{a^{-1}:a\in A\}$, $A^n:=\{a_1\mathrm{\dots }{a}_n:a_1,\mathrm{\dots },a_n\in A\}$ 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 $H:={\bigcup }_{n=1}^{\mathrm{\infty }}{V}^n$. $H$ 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 $H={\bigcup }_{n=1}^{\mathrm{\infty }}{\overline{V}}^n$, which implies that $H$ is $\sigma$-compact. $\mathrm{\square }$