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I've found some literature which would be helpful if I understood the following,

"we can choose neighborhood $U,V$ of the identity such that $VV^{-1} \subset U$ and $U \cap \Gamma = \{e\}$. "

Question: I can figure out the last, but how do you choose set such that $VV^{-1} \subset U$?

Fundamental
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  • Wow, that was too easy. And $U^{-1}$ is open since the inversion map is continuous. – Fundamental Feb 17 '16 at 03:55
  • It's deceptively easy. This is a standard trick in the theory of topological vector spaces. – Cameron Williams Feb 17 '16 at 03:58
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    I edited my answer as it was mistaken. Something had been bothering me. I neglected the fact that we could multiply elements of $U$ with each other and leave the set in my suggestion. It was nagging at me, but I couldn't quite see what the issue was until now. – Cameron Williams Feb 17 '16 at 05:42

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So that this question does not go unanswered: picking $V = U\cap U^{-1}$ does the trick. This is a standard trick used in the theory of topological vector spaces as well as in harmonic analysis when proving very technical lemmas.

Edit: The above is incorrect as I had jumped the gun a bit. Taking the intersection of a set with its inverse is necessary to the proof, but there is more to it since we have to whittle down the set a bit to keep from leaving it when we multiply by the inverse set.

Since multiplication is continuous and $U$ contains $1$, consider the inverse image $\mu^{-1}(U)$, where $\mu$ is the multiplication map. Since $U$ is open, $\mu^{-1}(U)$ is open. Moreover, $(e,e)\in \mu^{-1}(U)$. Let $V_1,V_2$ be open neighborhoods of $e$ such that $V_1V_2\subseteq U$. Taking $V_3 = V_1\cap V_2$ and $V = V_3\cap V_3^{-1}$, we have the result.

Cameron Williams
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