Closure of Subgroup is Group

Theorem
Let $G$ be a topological group.

Let $H\leq G$ be a subgroup.

Let $\overline H$ denote its closure.

Then $\overline H$ is a subgroup of $G$.

Proof
We use the One-Step Subgroup Test.

Because $H\subset \overline H$, $\overline H$ is non-empty.

Let $a,b \in \overline H$.

Let $U$ be a neighborhood of $ab^{-1}$.

Let the mapping $f: G\times G \to G$ be defined as:
 * $f\left({x, y}\right) = xy^{-1}$

By definition of topological group, $f$ is continuous.

By definition of product space, there exist neighborhoods $A,B$ of $a,b$ respectively such that $A\times B\subset f^{-1}(U)$.

By assumption, there exist $x \in A \cap H$ and $y \in B \cap H$.

Then $xy^{-1} \subset U \cap H$.

Because $U$ was arbitrary, $ab^{-1} \in \overline H$.

By One-Step Subgroup Test, $\overline H$ is a subgroup of $G$.