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I want to prove that if $n\neq m$ then $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$.

This deceptively simple topology question came up on an algebraic topology worksheet on which the rest of the questions centre around the Mayer-Vietoris sequence and degrees of maps. I have to admit I have little idea about how to proceed, except maybe to use homotopy types (and I'm not even sure how to start there). A hint or two would be very handy...

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@user8268: Yes I believe I can. $\mathbb{R}^n\setminus\{0\} \simeq S^{n-1}$, $\mathbb{R}^m\setminus\{0\} \simeq S^{m-1}$. But if $S^{n-1}$ and $S^{m-1}$ were homotopy equivalent, then $H_*(S^{n-1}) = H_*(S^{m-1})$, which is false. So this gives that $\mathbb{R}^n \setminus\{0\}$ and $\mathbb{R}^m\setminus \{0\}$ are not even homotopy equivalent. Ah, I think I see how to proceed with the last bit now, thanks to Aaron's answer below. Cheers! :) –  Sputnik Jun 5 '11 at 20:33
    
mathoverflow.net/questions/34232/… –  user9413 Jun 6 '11 at 3:13

3 Answers 3

up vote 12 down vote accepted

Hint: If $\varphi: \mathbb{R}^m\to \mathbb{R}^n$ is a homeomorphism, then $\mathbb{R}^m\setminus x$ is homeomorphic to $\mathbb{R}^n\setminus \varphi(x)$. Compute the homology.

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haha, my kind of thinking ;) –  gfes Jun 5 '11 at 20:29
    
@gfes Indeed! Actually, when I hit submit, I saw your answer beat mine by 36 seconds. But mine is longer, so I'm going to call it a tie. :P –  Aaron Jun 5 '11 at 20:36
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I think I see it now! A little question because I haven't had enough sleep: is $\phi(\mathbb{R}^n\setminus x) = \phi(\mathbb{R}^n) \setminus \phi(x)$ a set theoretic result or is that specifically because $\phi$ is a homeomorphism? –  Sputnik Jun 5 '11 at 20:37
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@Fahad Sperinck That is completely set theoretic, using just that $\varphi$ is a bijection. If $\varphi^{-1}(\varphi(x))$ had more than one point, the map wouldn't work. However, the fact that restricted map is still a homeomorphism comes from the definition of the subspace topology. More generally, if $U\subset X$ and $\varphi:X\to Y$ is a homeomorphism, it restricts to a homeomorphism $U\to \varphi(U)$. –  Aaron Jun 5 '11 at 20:45

Hint: Try removing a point and computing the homology.

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Thanks! This was there first and would have done the trick, but I read both at the same time. –  Sputnik Jun 5 '11 at 20:40

Another possibility is to compare the one-point compactifications of $\mathbb{R}^n$ and $\mathbb{R}^m$, ie. $\mathbb{S}^n$ and $\mathbb{S}^m$ respectively, for example by computing the homology.

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