Originally Posted by

**huram2215** This is from Munkres, 2nd ed., Ch2, supp. exercise 7(a).

__Definitions__:

__Topological Group__: $\displaystyle ((G , \cdot), \tau_G)$ is a topological group with identity $\displaystyle e$. This means that the product function $\displaystyle \mu (g_1,g_2) = g_1 \cdot g_2$ is continuous and $\displaystyle I(g_1) = g_1^{-1}$ is a continuous map.

For $\displaystyle A, B$ subsets of $\displaystyle G, A \cdot B = \{ a \cdot b : a \in A, b \in B\}$. $\displaystyle A^{-1} = \{ a^{-1} : a \in A \}$.

__Symmetric neighborhood__: A neighborhood $\displaystyle V$ of $\displaystyle e$ such that $\displaystyle V = V^{-1}$. Note that, for U open in G and $\displaystyle e \in U$, $\displaystyle U \cap U^{-1}$ and $\displaystyle U \cdot U^{-1}$ are both symmetric.

__Claim__: Any neighborhood U of e contains a symmetric neighborhood $\displaystyle V \subseteq U$ such that $\displaystyle V \cdot V \subseteq U$.

I can show that if V is symmetric, that $\displaystyle V \cdot V$ is an open set in $\displaystyle G$ and $\displaystyle e \in V \cdot V$ trivially. But I'm not seeing how to prove that a construction of a symmetric V from U has $\displaystyle V \cdot V \subseteq U$?