Locally compact topological group is paracompact

Question

Let $G$ be a locally compact, connected topological group.Show that $G$ is paracompact.

Answer 1

Let $U$ be an open (symmetric) neighbourhood of $e$ with compact closure and note that the subgroup $U$ generated by $U$, can be written as $\bigcup_n U^n$, and is an open subgroup of $G$ (as it has non-empty interior) and hence also a closed subgroup (standard fact), and so it equals $G$ by connectedness.

As $G =\bigcup_n U^n = \bigcup_n \overline{U}^n$ we see that $G$ is thus $\sigma$-compact and hence Lindelöf and a Lindelöf regular space is (strongly) paracompact.

We can even dispense with connectedness altogether: by a non-trivial structure theorem for locally compact topological groups, all such Hausdorff groups are homeomorphic to $\mathbb{R}^n \times K \times D$, where $n$ is a finite integer, $D$ a discrete space and $K$ a compact Hausdorff topological group.

As $\mathbb{R}^n \times D$ is metrisable (hence paracompact) and the product of a paracompact and a compact space is paracompact, we are done. But this theorem might be overkill, the connectedness idea is more elementary.