I was doing the following exercise:
Let $U$ be a neighbourhood of $1$ in a topological group $G$. And, let $<U>$ be the subgroup generated by $U.$ Then show that $<U>$ is open and closed in $G$.
I know already that open subgroups of topological groups are closed. So it suffices to show that $<U>$ is open. To this end, I first replace $U$ by $V=U\cup U^{-1}.$ Now observe that $<V>=<U>.$ So suffices to show that $<V>$ is open, but $<V>=\cup_{n=1}^{\infty} V^m$ where $V^m=\{ g_1...g_m: g_k \in V\}.$
Since $U.V$ is open when $U$ and $V$ are open. We can argue inductively that $V^m$ is open for each $n$. And, hence we are done.
My questions are the following: 1. Is this proof (outline) correct? 2. I understand that in the first step I am replacing $U$ by a symmetric neighborhood. Usually, we use the neighbourhood $U\cap U^{-1}$ as the symmetric neighborhood. But this is smaller than $U$ and hence might not generate the same group as $U$ does. Am I correct in my understanding or I am missing something? 3. My source of doubt is the following: I do not see any use of the fact that $U$ is a neighbourhood of identity. The whole proof still goes through. And, I do not know if it is true in general that the subgroup generated by open sets are open (and hence closed).