How do we know that if $H$ is discrete then $H$ is open subgroup of $\overline{H}$?

Question

I have a problem with one corollary from topological groups.

Corollary: If $H$ is discrete subgroup of quasitopological group $G$, then $H$ is closed in $G$.

Proof: There exists Proposition which say that $\overline{H}$ is subgroup of $G$. With topology inherited from $G$, $\overline{H}$ is a quasitopological group. Since $H$ is discrete in itself, it is an open subgroup of $\overline{H}$. From onother theorem it follows that $H$ is closed in $\overline{H}$. Hence, $H$ is closed in $G$. $\square$

This corollary is from Topological groups and Related Structures (Arhangel'ski, Tkachenko). You can find it on page 30, corollary 1.4.18.

Question: How do we know that if $H$ is discrete then $H$ is open subgroup of $\overline{H}$?

Thanks :D

Answer 1

I remember being confused by the same sentence when I read that part of the book, if I remember correctly this is how I sorted it out.

Since $H$ is discrete in $G$, for any $h\in H$ we can find $U_h\subseteq G$ open with $U_h\cap H=\{h\}$. But then we also have $U_h\cap\overline H=\{h\}$, since by definition any nbhd of any point in $\overline H\setminus H$ meets $H$ in infinitely many points. Now $H$ is the union of its singletons, which are all open in $\overline H$, so that $H$ itself is open in $\overline H$.