Proof of Closure of Subgroup is Group

https://proofwiki.org/wiki/Closure_of_Subgroup_is_Group

My question is that is it possible for if $a,b \in \overline{H}$, but $\overline{H}$ is not a subset of $G$. If that is the case, how can we find an open neighbourhood $U$ of $ab^{-1}$ that is in $G$? Thus, we cannot proceed by the definition of $f$ being continuous on $G$, since $a,b$ may not be in $G$ at all?

Solution 1:

$H \subseteq G$ of course and the closure $\overline{H}$ is by definition the smallest closed subset of $G$ that contains $H$, so in particular $\overline{H} \subseteq G$. The "universe" is just $G$, as it were. We're not working in a subspace or something like that.

My favourite proof: If $a,b \in \overline{H}$ we can find nets (with common domain $I$) from $H$ with $a_i \to a, b_i \to b$ and $H$ being a subgroup means $a_i \cdot b_i^{-1} \in H$ for all $i$ so, as $a\cdot b^{-1} = \lim_i a_i \cdot b_i^{-1}$ by continuity of the group operations, $a\cdot b^{-1} \in \overline{H}$, as required.