Show that every topological group is $T_3$

Question

I know that it is sufficient to show that for one point ($e$) and any neighbourhood $U$ of $e$, we have a neighbouhood $V$ with $\bar{V} \subseteq U$. Since $x \to x^{-1}$ is continuous, it follows from continuity of multiplication that we have neighbourhoods $V_e$ and $V_e^{-1}$ of $e$ such that $V_e \circ V_e^{-1} \subseteq U$. However, I don't know how to show that this implies that $\bar{V_e} \subseteq U$. Every source I have found just states that the last inclusion is true, but I don't understand why?

Answer 1

(I’m assuming that $e$ is the identity element.)

Let $x\in\operatorname{cl}V_e$; $xV_e$ is an open nbhd of $x$, so $xV_e\cap V_e\ne\varnothing$. Thus, there are $v_0,v_1\in V_e$ such that $v_1=xv_0$, and hence $x=v_1v_0^{-1}\in V_eV_e^{-1}\subseteq U$.

Answer 2

Generalizing this a bit. In a topological group $G$ we have for all sets $A\subseteq G$ the recipe $$ \overline{A}=\bigcap_{U\in D} AU, $$ where $D$ is the family of symmetric open neighborhoods of $e$, and $AU=\{au\mid u\in U,a\in A\}$. This is because all the neighborhoods of a point $x\in G$ contain a neighborhood of the form $Ux$ for some $U\in D$. Symmetricity of $U$ means that $Ux$ has a non-empty intersection with $A$ iff $x\in AU$.

In particular $\overline{A}\subseteq AU$ for all such $U$.

Apply this to $A=V_e$ and $U=V_e^{-1}$.