I have just begun to learn about topological group recently and is still not familiar with combining topology and group theory together.
I have read a useful property of discrete group on the wikipedia:
every discrete subgroup of a Hausdorff group is closed
But I have no idea how to prove it. I find that it cannot be proved only considering the topological structure, since $\left\{\frac{1}{n}: n=1,2,3,...\right\}$ is a discrete subspace of $\Bbb R$, which is not closed.
I don't know how to use the group structure here. Can you please help? Thanks.
Let $H$ be a discrete subgroup of a Hausdorff group $G$. Then, of course, $H=\bigcup_{h\in H}\{h\}$. For each $h\in G$, the sets $\{h\}$ is closed, and the family of closed sets $\mathcal F=\{\{h\}:h\in H\}$ is locally finite: it follows that the union is also closed.
Locally finite means: each point of $G$ has a neighborhood $U$ such that $U$ intersects non-trivially finitely many elements of $\mathcal F$.
Later. I guess a hint was not all that was asked for... Of course, we now have to show that $\mathcal F$ is locally finite... Suppose, to reach a contradiction, that there is a $p\in G$ such that every neighborhood of $p$ contains infinitely many elements of $H$. As $H$ is discrete, there is an open subset $U\subseteq G$ such that $U\cap H=\{e\}$; since $G$ is a topological group, there exists an open set $V$ such that $e\in V$ and $V^{-1}V\subseteq U.$ Now $pV$ is an open neighborhood of $p$ so the hypothesis implies that there exist distinct $h_1$, $h_2\in H$ such that $h_1$, $h_2\in pV$. It follows that $h_1^{-1}h_2\in V^{-1}V\subseteq U$ so that $e\neq h_1^{-1}h_2\in U\cap H$. This is absurd.