Topological groups question from Munkres

Question

This if from the 'Supplementary Exercises' at the end of Chapter 2 in Munkres' Topology.

If $A$ and $B$ are subsets of (a topological group) $G$, let $A \cdot B$ denote the set of all points $a \cdot b$ for $a \in A$ and $b \in B$. Let $A^{-1}$ denote the set of all points $a^{-1}$, for $a \in A$.

A neighbourhood $V$ of the identity element $e$ is said to be symmetric if $V = V^{-1}$. If $U$ is a neighbourhood of $e$, show there is a symmetric neighbourhood $V$ of $e$ such that $V \cdot V \subseteq U$. [Hint: If $W$ is a neighbourhood of $e$, then $W \cdot W^{-1}$ is symmetric.]

I've attempted this without success. The hint seems to be a large one showing us how to achieve the symmetric property, but then we require of $W$ to be such that $W \cdot W^{-1}$ remains within $U$ which I can achieve by looking at preimages of $U$ under the composition and inversion maps, but what I cannot achieve with that approach is for $W \cdot W^{-1}$ itself to be a neighbourhood. Help would be much appreciated.

Answer 1

By continuity of multiplication, there are neighbourhoods $V_1$ and $V_2$ of $e$ such that $V_1 \cdot V_2 \subseteq U$. Try $V = V_1 \cap V_1^{-1} \cap V_2 \cap V_2^{-1}$.

Answer 2

If $U$ is a neighborhood of $e$ then $U^{-1}$ is also a neighborhood of $e$. $U \cap U^{-1}$ is also a neighborhood of $e$. Choose a neighborhood $W$ of $e$ contained in $U \cap U^{-1}$. Then $W^{-1}$ is also contained in $U \cap U^{-1}$. Now let $V = W \cdot W^{-1}$. $V$ is a symmetric neighborhood of $e$. Now show that $V\cdot V$ is contained in $U$.

$W\cdot W^{-1}$ is a neighborhood because multiplication and inversion are continuous.