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Prove that $[0,\infty)$ is not a manifold.

Using diffeomorphisms and the implicit function theorem perhaps.

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To me manifolds seem a part of algebraic topology and differential geometry. But I'm wondering whether or not to leave the general-topology tag. – Rudy the Reindeer Nov 13 '12 at 20:47
Of course $[0,\infty)$ is a manifold with boundary. It's just one of the strange twists of mathematical terminology that a manifold with boundary may not be a manifold … – Harald Hanche-Olsen Nov 13 '12 at 21:12

A topological manifold is a space that looks locally like $\mathbb R^n$. Does $0$ in $[0, \infty)$ look like a point in $\mathbb R$?

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But I thought you could take a subset of set [0,∞) so (0,∞) to be locally diffeomorphic to R^p ? – Rebekah Nov 13 '12 at 20:41
@Rebekah: look at the definition more carefully. – Chris Eagle Nov 13 '12 at 20:41
I think you need to prove that [$0, a)$ cannot be homeomorphic to $\mathbb{R}$ for every $a > 0$. It seems obvious, but the proof does not seem to be so obvious. – Makoto Kato Nov 13 '12 at 20:58
Think connectedness. What happens when you remove a point from $\mathbb{R}$? What happens when you remove $0$ from $[0,\infty)$? – Harald Hanche-Olsen Nov 13 '12 at 21:10
@RickyDemer I know you know the proof. But I would like to prove it for other readers who don't know it. It suffices to prove that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^n$ for $n>1$. This is because $\mathbb{R}$ minus one point is not connected, while $\mathbb{R}^n$ minus one point is connected. – Makoto Kato Nov 15 '12 at 0:28

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