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.

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?

share|improve this question
2  
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
2  
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

2 Answers 2

Hint:

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

share|improve this answer
    
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

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.

share|improve this answer

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.