About connected Lie Groups How can I prove that a connected Lie Group is generated by any  neighborhood of the identity?
The result is almost trivial for $R^n$ but I tried using the open subgroup generated by this neighborhood.

An open subgroup $H$ of a topological group $G$ is closed because
$$
G \smallsetminus H = \bigcup_{g \notin H} gH
$$
is open as union of the open sets $gH$.
Now take your neighborhood $U$ of the identity, let $H = \bigcup_{n \in \mathbb{Z}} U^{n}$ and check that $H$ is an open (hence closed) subgroup of $G$. By connectedness $G = H$.
