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It is straightforward to prove using a connectedness argument that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^n$, for $n>1$.

How do you prove that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$?

Note: I'm looking for a proof which does not use any algebraic topology,not even Brouwer's fixed point theorem. [I know that there are proofs of Brouwer's theorem using purely analytic methods, but I still do not want to include it - I'm looking for something even simpler]

Me and a few friends have been at this for a couple of weeks, but kept running in circles. Unfortunately, this was a long time ago, and I have completely forgotten what we tried. Any ideas?

Edit: Thanks for all the answers, but I still didn't get what I was looking for. I should have been more explicit : I don't want to use the Jordan curve theorem either. (The simplest proof of that which I've seen involves Brouwer's fixed point theorem). I'm looking for a proof which does not use anything apart from ideas of connectedness and compactness.

Looking at the answers, a second question came to my mind: Given that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$, can you deduce the Jordan curve theorem?

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A circle doesn't disconnect $\mathbb{R}^3$, while it disconnects $\mathbb{R}^2$. – egreg May 27 '13 at 14:22
I think one can also take the $x$-axis, which disconnects $\Bbb R^2$, but its image does not disconnect $\Bbb R^3$. But how elementary is it to prove that such a line keeps the space connected @egreg? – Stefan Hamcke May 27 '13 at 14:24
Here is a proof using the Jordan Curve Theorem. – Alex Becker May 27 '13 at 14:24
@StefanH. For a closed curve it's not elementary, because Jordan's theorem is quite difficult to prove. Unfortunately no simple method seems to apply; for the line and the plane two points suffice. – egreg May 27 '13 at 14:36
@egreg: Thanks. Do you know how easy a prove using an axis in $R^2$ and its image in $R^3$ would be? – Stefan Hamcke May 27 '13 at 14:39
up vote 3 down vote accepted

A elementary proof can be found in the following article:

An interesting proof on the nonexistence of a continuous function between $\mathbb{R}^2$ and $\mathbb{R}^n$ for $n \neq 2$, by F. Malek, H. Daneshpajouh, H.R. Daneshpajouh and J. Hahn.

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In $\mathbb{R}^2$ any closed curve has an outside and an inside (Jordan curve theorem), but in $\mathbb{R}^3$, what happens then?

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This is the same idea as the method using the Jordan curve theorem as in the link given by Alex Becker. – Dedalus May 27 '13 at 14:25

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