Let $(G,\cdot)$ be a topological group and $U$ be a neighborhood of $1$. Then there exists a symmetric neighborhood of $1$, $V^{-1} = V$, such that $V\cdot V \subset U$. Having a hard time proving this. $V^{-1} = \{v^{-1}: v \in V\}$. I know that an open set times any set is also open. And a hint is that $VV^{-1}$ is symmetric and an open neighborhood of $1$ when $V$ is an open neighborhood of $1$ contained in $U$, but showing $VV^{-1}$ or an expression involving it is a subset of $U$ or $V$ requires something else.
[Math] Every neighborhood of identity in a topological group contains the product of a symmetric neighborhood of identity.
general-topologytopological-groups
Best Answer
Use that the multiplication is continuous and $1\cdot1=1$. Let $U$ be a neighborhood of $1$. By continuity there are neighborhoods $A,B\ni1$ such that $AB\subseteq U$. Let $W=A\cap B$ and then $V:=W\cap W^{-1}$ is a symmetric neighborhood of $1$ such that $VV\subseteq U$.