Discrete Subgroup of Hausdorff Group is Closed

Theorem

Let $H$ be a discrete subgroup of $G$.

Then $H$ is closed in $G$.

Let $g \in \overline H$ be in the closure of $H$.

We will show that $g \in H$.

Aiming for a contradiction, suppose $g \notin H$.

Let $e$ be the identity of $G$.

Because $H$ is discrete, there exists an open set $U \subset G$ such that $U \cap H = \set e$.

Then $V = U \cap U^{-1}$ is an open neighborhood of $e$ in $G$.

By definition of closure, $g V \cap H \ne \O$.

Let $h \in g V \cap H$.

Because $g \ne h$ and $G$ is Hausdorff, there exists a neighborhood $W$ of $g$ in $G$ with $h \notin W$.

Then $g V \cap W$ is a neighborhood of $g$.

Let $k \in g V \cap W \cap H$.

Then $k \ne h$.

Then

$\ds k^{-1} h$

$\in$

$\ds \paren {g V \cap H}^{-1} \cdot \paren {g V \cap H}$

$\ds $

$=$

$\ds \paren {g V}^{-1}g V \cap H$

$\ds $

$=$

$\ds V^{-1} V \cap H$

$\ds $

$=$

$\ds V \cap H$

$\ds $

$\subset$

$\ds U \cap H$

Bu assumption, $k^{-1} h = e$, so $k = h$, which is a contradiction.

Thus $\overline H = H$.

Thus $H$ is closed.

$\blacksquare$