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 $G = H$.

-
The desired result holds for every connected topological group. No need to assume a smooth structure. –  t.b. Apr 26 '12 at 14:32
Take any $g\in G$ define $l_g:G\rightarrow G, l_g(k)=gk$, Let $H$ be any open subgroup of $G$,clearly as $l_g$ is a homeomorphism, and $H$ is open so $l_g(H)=gH$ is also open, but $G$ is connected so $H=G$
Yes, this is the same question. How do you get from $H$ open and $gH$ open and connectedness of $G$ to $H = G$? You don't know that $G = H \cup gH$. –  t.b. Aug 19 '12 at 6:35