Connected manifold

Question

I need to prove a rather simple fact: let $X$ be a connected smooth (or just topological) manifold. Now we need to prove that $X$ is linearly connected. So, it seems to be obvious, because $X$ is locally like a unit open ball in $\mathbb{R}^n$, but i cannot get clear proof.

Yes, it is, i am sorry for terminology, i just translated it from russian word-to-word. — , Apr 26 '13 at 18:49
A good general strategy is to use the fact that you find it obvious to figure out how you know it's true. Here, I'd ask how being locally like an open unit ball means it should be linearly connected — Loki Clock, Apr 26 '13 at 19:09

score 0 · Answer 1 · answered Apr 26 '13 at 18:32

You have the right idea. Just show that that being linearly connected is an open condition (i.e. every point has a neighborhood that is linearly connected). Then the linearly connected components' of the space form a decomposition into disjoint open sets. Here thelinearly connected components' are the maximal subsets that are linearly connected.

Connected manifold

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