Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Using diffeomorphisms and the implicit function theorem perhaps.

share|improve this question
3  
What have you tried? –  Jason DeVito Nov 13 '12 at 20:34
    
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. –  Matt N. 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

1 Answer 1

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

share|improve this answer
    
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
1  
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
2  
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.