# Thread: Topological group/locally compact subgroup

1. ## Topological group/locally compact subgroup

Proposition: Let $\displaystyle G$ be a Hausdorff topological group. Then a subgroup $\displaystyle H$ of $\displaystyle G$ that is locally compact (in the subspace topology) is moreover closed. In particular, every discrete subgroup of $\displaystyle G$ 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 $\displaystyle H$ and letting $\displaystyle K\subset H$ be a compact neighbourhood of the identity element $\displaystyle e$ in the subspace topology of $\displaystyle H$ which is fine. Then it concludes therefore there is a closed neighbourhood $\displaystyle U$ of $\displaystyle e$ in $\displaystyle G$ such that $\displaystyle K = H \cap U$. How on earth does one achieve this? I mean, its easy to find a closed set $\displaystyle U$ with that, but how do I also make sure $\displaystyle U$ is a neighbourhood of $\displaystyle e$?

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 $\displaystyle H$ dense in $\displaystyle G$. For instance if we can prove that $\displaystyle H$ is closed in $\displaystyle \overline{H}$, then its also closed in $\displaystyle G$.

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

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

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

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

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

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

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

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

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

5. I'm not quite sure what you mean - I'm not asking that $\displaystyle C$ 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 $\displaystyle C$ 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.