Lemma 27.2.2.label Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact.
Proof. Let $K \in \cn_{G}(1)$ and $K^{(1)}= K$. For each $n \in \natp$, let $K^{(n+1)}= KK^{(n)}$, then $K^{(n+1)}$ is compact by Proposition 5.16.2 with $K^{(n+1)}\in \cn_{G}(K^{(n)})$. Let $H = \bigcup_{n \in \natp}K^{(n)}$, then $H$ is a subgroup of $G$, which is open by Lemma 5.4.3. Since $H$ admits an exhaustion by compact sets, it is $\sigma$-compact.
Finally, since
\[G \setminus H = G \setminus \bigcup_{n \in \natp}K^{(n)}= \bigcap_{n \in \natp}G \setminus K^{(n)}\]
and $K^{(n)}$ is closed for each $n \in \natp$ by Proposition 5.16.3, $G \setminus H$ is closed, and hence $H$ is open.$\square$
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