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As part of a problem sheet I have been asked to show that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ whenever $n \neq m$.

When I first proved that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$ I did so using cut points - removing a point from $\mathbb{R}$ leaves two disconnected components, whereas $\mathbb{R}^2$ minus a point is connected, this led me to wonder about the following:

  • Could you argue that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$ since if you remove a copy of $S^1$ from $\mathbb{R}^2$ you get two disconnected components (the insde and the outside) whereas removing $S^1 \subset \mathbb{R}^3$ leaves a path-connected (hence connected) space?

  • Could you then extend this to say that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ for $m>n$ since $\mathbb{R}^n \setminus S^n$ is disconnected and $\mathbb{R}^m \setminus S^n$ is connected?

It seems that it might be difficult to show that you do get two disconnected components, this would presumably need a generalised form of the Jordan Curve Theorem, but is there any other reason that this would not work?

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  • $\begingroup$ Your method seems to be sound. $\endgroup$ – Daniel Goldman Oct 30 '14 at 17:49
  • $\begingroup$ Would be possible to use something simpler, e.g. a $n$-dimensional hyperplane (instead of $S^n$)? $\endgroup$ – Milly Oct 30 '14 at 17:51
  • $\begingroup$ "Is there any other reason this would not work": well, isn't this enough reason already? It's already insanely difficult to prove the Jordan curve theorem, so a generalized version is certainly not a cakewalk. Usually this result is proved using algebraic topology methods such as homology. $\endgroup$ – Najib Idrissi Oct 30 '14 at 17:57
  • $\begingroup$ @Milly that sounds far easier! I still have concerns over the method as I can't give justification for extrapolating from removing finitely many cut points to removing an uncountable infinite of cut points, it seems intuitive, but that is not a proof! $\endgroup$ – sjt Oct 30 '14 at 18:01
  • $\begingroup$ It isn't at all easy to scale this sort of argument up, because you have to worry about every possible embedded $S^n$ or, in Milly's version, every possible embedded $\mathbb{R}^{n-1}$-it's not enough to restrict to hyperplanes. Do you know Brouwer's fixed point theorem, or even invariance of domain? $\endgroup$ – Kevin Carlson Oct 30 '14 at 18:03

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