LCG has an open $\sigma$-compact subgroup in form of $\cup_{n=1}^{\infty}V^n$

Question

I met a lemma in the page 6 of the book Principles of Harmonic Analysis (2e) by Deitmar which states

A locally compact group has an open $\sigma$-compact subgroup in form of $\cup_{n=1}^{\infty}V^n$ where $V$ is a symmetric, relatively compact neighborhood around unit element $e$.

For information, a neighborhood $U$ around $e$ is said to be symmetric if $U=U^{-1}$. A set $U$ is said to be relatively compact if $\overline{U}$ is compact.

In the proof, it says

For each $n\in\mathbb{N}$, one has $\overline{V}^n\subset V\cdot V^n = V^{n+1}$.

I totally lost idea here. How could we get this subset inclusion so easily ? Any tips ? Thanks.

Answer 1

If $x\in\overline V$, then $xV\cap V\ne\emptyset$ as $xV$ is a neighbourhood of $x$. So $x\in VV^{-1}=V^2$ as $V$ is symmetric, so $\overline V\subseteq V^2$.

In general, $\overline V^n$ is the closure of $V^n$, so the same trick gives $\overline V^n\subseteq V^nV=V^{n+1}$.