Corollary

Suppose that K is locally compact. Then $G_f$ is the union of the compact subgroups of G.

Let $x \in G$. If x belongs to a compact subgroup of G, then $x \in G_f$ (Proposition 13). Suppose that $x \in G_f$. As K is locally compact, there exists an open subgroup $G_1$ of G which is compact. Then there exists an integer $m > 0$ such that $x^m \in G_1$. The closed subgroup $G_2$ generated by $x^m$ is contained in $G_1$ and is therefore compact. Then x commutes with the elements of $G_2$ and hence $G_2 \cup xG_2 \cup \cdots \cup x^{m-1}G_2$ is a compact subgroup of G which contains x.