A more Intuitive proof of regularity of topological group

Question

The proof given in Munkres's Topology is quite unnatural at least to me. He uses the notion of symmetric neighboorhood and I'm pretty sure it's quite ad hoc. After all, is it really useful? Can anybody provide a more intuitive alternative proof? Thanks in advance.

Answer 1

Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1\in U\subseteq \bar U\subseteq V.$$ Consider the map $\mu:G\times G\to G$ defined by $$\mu(g,h)= g\cdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $\mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $U\times U\subseteq \mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2\in U$ we have that $u_1\cdot u_2^{-1}\in V$ (and in particular $U\subseteq V$ since $1\in U$). I claim that this $U$ is the one we are after; since we already have $1\in U\subseteq V$, we just need to prove $\bar U\subseteq V$.

Let $x\in \bar U$. The set $$xU=\{x\cdot u|u\in U\}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2\in U$ such that $x\cdot u_1 =u_2$. Therefore $x=u_1\cdot u_2^{-1}\in \mu(U\times U)\subseteq V$, as needed.

This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.

The main ideas are that $\mu$ is continuous and that if $W$ is an open set and $S$ is any set then $S\cdot W$ contains in the closure of $S$.