Intuition for why an open neighborhood of the identity in a Lie group generates the whole Lie group Intuition for why an open neighborhood of the identity in a connected Lie group generates the whole Lie group.
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I think the standard proof for this is showing that the subgroup generated by any open neighborhood is both an open and closed subgroup of $G$ and thus is all of $G$ since $G$ is connected. Can somebody explain give me a more conceptual explanation of why this result must be true?
 A: Let me present an alternative proof, which sounds more intuitive to me - hopefully it'll help you. The proof should be clear on its own, but I'll add a detailed intuition explanation at then end.
A connected Lie group is path connected.
Let $U$ be your neighbourhood. Up to taking $U\cap U^{-1}$, we may assume that $U$ is symmetric.
Let $\gamma : [0,1]\to G$ be a path from $e$ to any element $x$; and for every $t\in[0,1]$, let $U_t$ be a small enough open interval of $[0,1]$ containing $t$ such that $\gamma(t)^{-1}\gamma(U_t)\subset U$. This  is of course possible, as $\gamma(t)U$ is a neighbourhood of $\gamma(t)$.
Then $\bigcup_t U_t = [0,1]$ so by compactness, there are $0<t_1<...<t_n<1$ such that $U_0\cup U_{t_1}\cup ... \cup U_{t_n} \cup U_1 = [0,1]$.
But then (with $t_0=0,t_{n+1}=1$), for each $i$, $U_{t_i}\cap U_{t_{i+1}}$ must contain some element $s_i$ (this is because $[0,1]$ is connected, and I chose intervals).
Then $x=\gamma(1)= \gamma(1)\gamma(s_n)^{-1}\gamma(s_n)= \gamma(1)\gamma(s_n)^{-1}\gamma(s_n)\gamma(t_n)^{-1}\gamma(t_n)$.
$\gamma(1)\gamma(s_n)^{-1}\in (\gamma(1)\gamma(U_1))^{-1}\subset U^{-1} = U$, and similarly, $\gamma(s_n)\gamma(t_n)^{-1}\in U$.
So $x\in \langle U\rangle \iff \gamma(t_n)\in \langle U\rangle$. Of course we may then induct on $n$ and obtain that $x\in \langle U\rangle \iff e\in \langle U\rangle$, but that's obvious: $x\in \langle U\rangle$.
Now the intuition behind this proof is that if you draw a path from $e$ to $x$, for every small enough value of $\epsilon$, $\gamma(t)$ and $\gamma(t+\epsilon)$ will only differ by something in $U$ (or $U^{-1}$).
But by compactness of $[0,1]$, the necessary value of $\epsilon$ is somehow bounded below (so we get our partition $t_1<...<t_n$), and this allows us to make big enough jumps while staying in $U$, and so, ultimately, staying in the subgroup generated by  $U$ if we just record the jumps.
This is related to how $G$ is a "uniform" space : the gaps between two elements can be seen as gaps between $e$ and some other element; so this allows one to reduce a lot of questions to local questions around $e$
