The subgroup $<S>$ generated by an open subset $S$ of a Lie group $G$ is open in $G$.
I was told to prove this using the fact that $<S>= \cup_{g\in<S>}gS$. But this does not feel so right, since $\cup_{g\in<S>}gS$ does not look like a group. Suppose that I have $g_1s_1$, its inverse is $s_1^{-1}g_1^{-1}$, this is not necessarily in $\cup_{g\in<S>}gS$, right?
Assume $S \ne \emptyset$, and let $s_{0} \in S$
Clearly $\bigcup \{ g S : g \in \langle S \rangle \} \subseteq \langle S \rangle$, as $ \langle S \rangle$ is a subgroup containing $S$. (Actually the smallest such subgroup.)
Conversely, if $g \in \langle S \rangle$, then $g = g s_{0}^{-1} s_{0} \in (g s_{0}^{-1}) S$, with $g s_{0}^{-1} \in \langle S \rangle$.