Discrete Subgroup of Hausdorff Group is Closed

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Theorem

Let $G$ be a Hausdorff topological group.

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


Then $H$ is closed in $G$.


Proof

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 $U \cap H = \left\{{e}\right\}$.

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

By Right and Left Regular Representations in Topological Group are Homeomorphisms, $g V$ is a neighborhood of $g$ in $G$.

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

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

\(\displaystyle k^{-1} h\) \(\in\) \(\displaystyle \left({g V \cap H}\right)^{-1} \cdot \left({g V \cap H}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({g V}\right)^{-1}g V \cap H\)
\(\displaystyle \) \(=\) \(\displaystyle V^{-1} V \cap H\)
\(\displaystyle \) \(=\) \(\displaystyle V \cap H\)
\(\displaystyle \) \(\subset\) \(\displaystyle U \cap H\)

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

Thus, by Proof by Contradiction, $g \in H$.

Thus $\overline H = H$.

Thus $H$ is closed.

$\blacksquare$


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