The question is :
Let $G$ be a locally compact topological subgroup and $H$ is a discrete subgroup of $G$ (i.e. the hereditary topology of $H$ is discrete). Is $H$ always closed?
I think the answer is “yes”. I want to prove that if $G$ is a locally compact and $H$ is discrete subgroup then $H\cap K$ is finite for every compact set $K\subseteq G$. It is clear that this will implies $H$ is closed. But I still don’t know how to prove it. If you could prove this claim, it also will be a nice answer for me. Thanks.
Best Answer
This is true if $G$ is $T_1$ (which implies it is Hausdorff, see How to show that topological groups are automatically Hausdorff?)
Since $H$ is discrete, there exists an open neighborhood $U$ of $1$ such that $U\cap H=\left\{1\right\}$. Take $V$ another neighborhood of $1$ s.t. $V^{-1}V\subseteq U$.
Let $g\in G\setminus H$. We need to prove that there is a neighborhood of $g$ which does not intersect $H$. If $gV\cap H=\varnothing$ then we are done.
Suppose then $gV\cap H\neq\varnothing$. Take $h\in gV\cap H$. Let us prove that $gV\cap H=\left\{h\right\}$. Given $k\in gV\cap H$, we have \begin{align*} k^{-1}h &\in (gV)^{-1}(gV)\\ &=V^{-1}g^{-1}gV\\ &=V^{-1}V\\ &\subseteq U\end{align*} and also $k^{-1}h\in H$, so $k^{-1}h=1$, which means that $k=h$.
So $gV\cap H=\left\{h\right\}$. But $g\not\in H$ by definition, so $g\neq h$. Since $G$ is $T_1$, there exists another neighborhood $W$ of $1$ s.t. $h\not\in gW$. Therefore $gW\cap H=\varnothing$.
If $G$ is not $T_1$ then this is not true, for in this case $H=\left\{1\right\}$ is not closed but it is discrete.