# Thread: Topological group/locally compact subgroup

1. ## Topological group/locally compact subgroup

Proposition: Let $G$ be a Hausdorff topological group. Then a subgroup $H$ of $G$ that is locally compact (in the subspace topology) is moreover closed. In particular, every discrete subgroup of $G$ is closed.

I've been struggling over the proof of this all day. I have been reading a proof in Fourier analysis by Ramakrishnan & Valenza and I understand it all except for this one point. It starts off by using local compactness of $H$ and letting $K\subset H$ be a compact neighbourhood of the identity element $e$ in the subspace topology of $H$ which is fine. Then it concludes therefore there is a closed neighbourhood $U$ of $e$ in $G$ such that $K = H \cap U$. How on earth does one achieve this? I mean, its easy to find a closed set $U$ with that, but how do I also make sure $U$ is a neighbourhood of $e$?

2. I've been able to provide a fix for the proof given in the book. I'm fairly sure the author made a thinko. The proof can be fixed since it is enough to prove the theorem for $H$ dense in $G$. For instance if we can prove that $H$ is closed in $\overline{H}$, then its also closed in $G$.

So assuming that $\overline{H} = G$ its easy to prove the statement I was having trouble with:

Let $K\subset H$ be a compact neighbourhood of $e\in H$ (in the subspace topology $H$).

That $K$ is a neighbourhood of $e$ in $H$ means that there exists an open neighbourhood $U \subset G$ of $e$ such that $U \cap H \subset K$.

Because $U$ is open it is easy to show that for any $X \subset G$ we have that $U \cap \overline X \subset \overline{U \cap X}$. In particular we have $U \cap \overline H \subset \overline{U \cap H} \subset \overline K = K$.

Since $H$ is dense in $G$ this means that $U \cap \overline{H} = U \cap G = U \subset K$ and so $K$ is a closed neighbourhood of $e$ in $G$ such that $K\cap H = K$.

3. I've worked out that the book was right after all - though the reasoning is not obvious (to me). The claim is: Let $G$ be a Hausdorff space and $H \subset G$ is a locally compact subspace. If $x \in H$ then there is a closed neighbourhood $C \subset G$ of $x$ such that $H\cap C$ is a compact neighbourhood of $x$ in the subspace topology of $H$.

Proof of claim: As $H$ is locally compact we have a compact n'hood $K\subset H$ of $x$. This means that $K$ is compact in both $H$ and $G$ and there is an open $U \subset G$ containing $x$ such that $H \cap U \subset K$. Now $H$ is locally compact and Hausdorff and this means that it is a regular space. So since $H \cap U$ is an open n'hood of $x$ in the regular space $H$, there is an open n'hood $V \subset G$ of $x$ such that $H \cap \overline V \subset H \cap U$. It is easy to check that $C = K \cup \overline V \subset G$ is such a closed n'hood of $x$ that we desired.

Phew.

I've worked out that the book was right after all - though the reasoning is not obvious (to me). The claim is: Let $G$ be a Hausdorff space and $H \subset G$ is a locally compact subspace. If $x \in H$ then there is a closed neighbourhood $C \subset G$ of $x$ such that $H\cap C$ is a compact neighbourhood of $x$ in the subspace topology of $H$.

Proof of claim: As $H$ is locally compact we have a compact n'hood $K\subset H$ of $x$. This means that $K$ is compact in both $H$ and $G$ and there is an open $U \subset G$ containing $x$ such that $H \cap U \subset K$. Now $H$ is locally compact and Hausdorff and this means that it is a regular space. So since $H \cap U$ is an open n'hood of $x$ in the regular space $H$, there is an open n'hood $V \subset G$ of $x$ such that $H \cap \overline V \subset H \cap U$. It is easy to check that $C = K \cup \overline V \subset G$ is our desired closed n'hood of $x$.

Phew.
This is good, but don't you meant to say that $x$ has a precompact neighborhood?

5. I'm not quite sure what you mean - I'm not asking that $C$ be compact for instance. To see where I got stuck and why I wanted to work this out see on page 8 proposition 1.6 in google books link.

I'm not quite sure what you mean - I'm not asking that $C$ be compact for instance. To see where I got stuck and why I wanted to work this out see on page 8 proposition 1.6 in google books link.