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So for them to not be homeomorphic the function or inverse of the function must not be continuous. Correct? Should I assume homeomorphism first, and create open balls?

Marc van Leeuwen
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user105225
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    One should look for a topological property that $\mathbb{R}^n$ ($n\gt 1$) has and $\mathbb{R}$ doesn't have. What happens when you remove a point from $\mathbb{R}$? – André Nicolas Nov 04 '13 at 07:18
  • I suppose this question must have been asked in MSE. –  Nov 04 '13 at 07:19
  • Do you know any homology theory? – user99680 Nov 04 '13 at 07:20
  • You can also look up "invariance of the domain", as this is a special case of a more general theorem. – zibadawa timmy Nov 04 '13 at 07:21
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    Connectedness is the key property to look at, specifically, the notion of a cut point. Homology theory and invariance of domain are sledgehammers that you really don’t need to know anything about in order to swat this particular fly. – Brian M. Scott Nov 04 '13 at 07:21
  • Well, if you want to compare $\mathbb R^n$ to $\mathbb R^m$ , cut points become too complicated. Because then they become cut lines, cut planes, etc. – user99680 Nov 04 '13 at 07:31
  • Sorry, I misread from $\mathbb R^n$ to $\mathbb R^m$ in general. Connectedness and cut points don't work here no more. – user99680 Nov 04 '13 at 08:03
  • Brian's idea answers the original question properly, but homology/invariance arguments do answer a more general question which future users may find interesting. Granted, most texts that could possibly lead to this question also provide the proof. – zibadawa timmy Nov 04 '13 at 08:05
  • You're right, zibadawa timmy; invariance of domain would also give a general answer by considering $\mathbb R^n $ as embedded in $\mathbb R^m$ , for $n<m$ , and then using that $\mathbb r^n$ cannot be open in $\mathbb R^m $ – user99680 Nov 04 '13 at 08:24
  • There are other arguments that allow one to show $\mathbb R^n $ is not homeo to $\mathbb R^n$ using degree thehttp://blog.plover.com/math/R3-root.html, as in – user99680 Nov 04 '13 at 09:00

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HINT: If you remove a point from $\Bbb R$, what’s left is not connected. If $n\ge 2$, does removing a point from $\Bbb R^n$ leave you with a connected or a disconnected set? (I think that we’ve had this before, but if so, I can’t immediately find it.)

Brian M. Scott
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This is a formal proof, with heavy tools if you know homology. If $\mathbb R^n $ and $\mathbb R^m$ are homeomorphic thru, say,$ h$ , then the restriction of $h$ to $\mathbb R^n-{(0^n)}$ to $\mathbb R^m- {(0^m)} $ (compose with a new homeomorphism so that $0^n$ is mapped to $0^n $ if necessary) is also a homeomorphism. But $\mathbb R^n-{0^n}$ is homotopic to $S^{n-1}$ and $\mathbb R^m -{0^m} $ is homotopically-equivalent to $S^{m-1}$ ; and any two homeomorphic spaces are homotopically-equivalent.

For a partial argument for why $\mathbb R^m $ is not homeomorphic to $\mathbb R^n$, here is an argument for why $\mathbb R^{2n+1}$ cannot be homeomorphic to $\mathbb R^{2m}$,where maybe the "guts"of the proof can be seen, we can generalize the argument here: http://blog.plover.com/math/R3-root.html , that uses degree theory (degree of a map, which must be +/- 1 for a homeomorphism, and degree has "nice"properties under composition.) which shows that $\mathbb R^3 $ is not ä square root", meaning $\mathbb R^3$ is not homeomorphic to the product of a space with itself. This proof can be generalized to any $\mathbb R^{2n+1}$ not being a product of a space with itself, while we have the trivial result that $\mathbb R^{2m}$ is homeomorphic to the product $\mathbb R^m \times \mathbb R^m $ . This shows $\mathbb R^{2n+1}$ is not homeomorphic to $\mathbb R^{2m}$

user99680
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The classical argument to prove that $\mathbb{R}$ and $\mathbb{R}^n$ ($n \geq 2$) are not homeomorphic (given by Brian M. Scott) is really simple, elementary, and it uses a key topological property of the real line: every point is a strong cut point. The discussion Topological Characterisation of the real line on MO can be mentionned as a justification.

By the way, another argument may be:

  • Any connected subset of $\mathbb{R}$ is arc-connected (since any such subset is an interval).
  • For $n \geq 2$, $\mathbb{R}^n$ contains non-arc-connected connected subsets. (See for example Topologist's sine curve.)
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Seirios
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Here's a quick and elementary proof:

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a continuous function. Take two distinct points $a,\,b \in \mathbb{R}^n$, and take two disjoint (other than at $a$ and $b$) continuous curves connecting them. The intermediate value theorem says that on each of these curves $f$ takes on every value in $[f(a),f(b)]$. But the curves are disjoint, so $f$ is not injective. So there is no such homeomorphism.

JLA
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