4

Theorem: If $ H $ is a locally compact subgroup of a topological group $ G $, then $ H $ is closed in $ G $.

proof: Let $ K $ be the closure of $ H $ in $G $. Then $ K $ is a subgroup of $ G $. Since $ H$ is a dense locally compact subset of $ K $, also $ H $ is open in $ K $. However, an open subgroup of a topological group is closed, so $ H = K $.

1: Why is $H $ open in $ K $?

2: Where do we use locally compactness ?

The next question is not related to above text, please help me.

We know that if $ H $ is a discrete subgroup of a topological group $ G $, then $ H $ is closed in $ G $.

So, is it right to say:

3: Every discrete subgroup $ H $ of a countably compact quasi topological group $ G $ is finite.

rae306
  • 9,850
Jak
  • 497

1 Answers1

6

Any locally compact dense subspace of a Hausdorff space is open. Indeed, suppose $H$ is locally compact and dense in a Hausdorff space $K$. Let $x\in H$, and choose an open neighborhood $U$ of $x$ in $H$ whose closure $C$ is compact. We can write $U=V\cap H$ for some $V$ that is open in $K$. Since $C$ is compact and $K$ is Hausdorff, $C$ is also closed in $K$. Thus $V\setminus C$ is open in $K$. But $V\setminus C$ does not intersect $H$ since $V\cap H=U\subseteq C$, so since $H$ is dense $V\setminus C$ must be empty. That is, $V\subseteq C$ and in particular $V\subseteq H$. Thus $H$ contains an open neighborhood (in $K$) of $x$, and since $x\in H$ was arbitrary, $H$ is open in $K$.

As for your last question, that's correct. The argument works just as well for quasitopological groups, since it's still true that an open subgroup is closed.

Eric Wofsey
  • 342,377