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.

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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 $G = H$.

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The desired result holds for every connected topological group. No need to assume a smooth structure. –  t.b. Apr 26 '12 at 14:32
This is one of my favourite arguments of all time. So simple, yet powerful. I first learnt this one from Sagle & Walde section 3.3 where they scale back their assumptions from analytic Lie groups and deal with general topological groups and was in awe that so much could be gleaned from these more general assumptions. Their chapter 3 is an excellent exposition of top groups - your answer brings back fond memories indeed. –  WetSavannaAnimal aka Rod Vance Jul 12 at 1:26

Is the question is similar to: Prove that any open subgroup of a connected lie group is the whole group then here is my answer, if not then please ignore it.

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 $H=G$

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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