Let $G$ be a topological $T_1$ group. If $H$ is a discrete subgroup of $G$, then there exists $U\in V(1)$ (here $V(1)$ is a filter of all neighborhoods of $1$) such that $U\cap H=\{1\}$.
Let $V\in V(1)$ such that $V^{-1}V \subset U$. Note that:
- $|xV\cap H|\leq1$ because $$ h = xv,h'= xv'\in xV\cap H \implies h^{-1}h'\in V^{-1}V\cap H = \{1\} \implies h = h'. $$
So, if $xV\cap H=\emptyset\implies x\notin cl(H)$. Therefore, $xV$ is a neighborhood of $x$ contained in $G\setminus H$.
How I do proof the case on what $xV\cap H = \{h\}$. I need to find a neighborhood $W$ of $x$ such that $W\cap H=\emptyset$ and concluded that $x\notin H$.
