Let $G$ be a topological group, $U$ an open neighbourhood of $x$. Prove there are open neighbourhoods $V$ of $x$ and $W$ of $1$, such that $VW \subseteq U$. Neutral element is denoted as $1$.
I found this statement in a textbook without further elaboration, and I don't know how to prove it myself. This is a small lemma that one might find useful when proving that $AH$ and $HA$, for closed $A$ and compact $H$, are closed subsets of a topological group $G$. I don't really know where to start, so could you help me please?
Use that the multiplication map $m:G \times G \to G, m(g_1,g_2)=g_1 \cdot g_2$ is continuous: $(x,1) \in m^{-1}[U]$ (as $m(x,1)=x\cdot 1 =x \in U$) which is open so for some basic open $V \times W$ containing $(x,1)$ we have $V \times W \subseteq m^{-1}[U]$. The last fact says exactly $VW \subseteq U$ if you think about it.