Regular open sets in topological groups Does every topological group $G$ have a local base of regular open sets at $e$?
If $V$ is an open subset of a space $X$, then is it true that $\overline{(\overline{V})^\circ}=\overline{V}$ ?
 A: A set $U$ is called regular open if it is the interior of its own closure. Dually, a closed set is called regular closed if it is the closure of its own interior.
The statement $V\text{ open}\implies\overline{\text{int}\overline V}=\overline V$, then reads as The closure of an open set is regular closed. This can be proven as follows:
Each closed set contains the closure of its interior. Applying this to the closed set $\overline V$ gives one inclusion. On the other hand, each open set is a subset of the interior of its closure (this applies to $V$). Then taking the closure on both sides gives the other inclusion.
Dually, we have that $\text{int}\left(\overline{\text{int }C}\right)=\text{int }C$ for any closed set $C$. So the interior of a closed set is regular open.
Each topological group is completely regular, thus regular. Therefore, each point has local base of closed sets. Now you can just take the interiors to get a base of regular open set.
A: If the topological group is $T_0$ then it is Tychonoff and so regular. So there is a local base of regular open sets. Not all basic elements will be regular open, but there will be a local base of them.
A: 
Does every topological group $G$ have a local base of regular open sets at $e$?

Yes. This can be easlily seen as follows. Let $G$ be a topological group (I do not assume any of separation axioms for $G$) and $U$ be an arbitrary neighborhood of the unit of the group $G$. There exists a neighborhood $V$ of the unit of the group $G$ such that $VV^{-1}\subset U$. Then $e\in V\subset \operatorname{int}\overline V\subset \overline V\subset VV^{-1}\subset U$. Indeed, if $x\in\overline V$ then $xV\cap V\not=\varnothing$. Therefore $x\in VV^{-1}$.  
