Non-discrete locally compact group

Question

Why every non-discrete locally compact group contains a nontrivial convergent sequence?

Answer 1

This is true. The difficulty is the case in which the group is not metrizable: Pavel M (and Henno Brandsma) noted that if $G$ is a metrizable locally compact group without isolated points then every neighborhood $U$ of the neutral element is infinite, so it suffices to choose an infinite sequence in a compact $U$ and extract a convergent subsequence.

In the general case, I doubt that there is a simple proof, although I think the following argument is far from optimal.

1. Let $G$ be a locally compact group and let $U$ be a compact symmetric neighborhood of the identity. Then $H = \bigcup_{n=1}^\infty U^n$ is a $\sigma$-compact clopen subgroup of $G$.

   If $H$ is metrizable then $U$ must be metrizable Pavel M's observation does the job.

   If $H$ is not metrizable, then it contains a compact normal subgroup $K$ such that $H/K$ is metrizable. See Hewitt-Ross, Abstract Harmonic analysis I, Theorem 8.7. Note that $K$ can't be finite (or even metrizable) because otherwise $H$ would have to be metrizable. Therefore $K$ has no isolated points.

2. Kuz'minov proved the remarkable fact that a compact group $K$ is a dyadic space, i.e., a continuous image of the Cantor cube $\{0,1\}^\kappa$ for some $\kappa$. See Chapter III of Stevo Todorcevic, Topics in topology for a proof.

3. Katetov and Efimov independently proved that in a dyadic space every non-isolated point is the limit of a non-trivial sequence. See Cor. 2 on page 301 of Engelking, Cartesian products and dyadic spaces, Fund. Math. 57 (1964), 287-304.