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?
Best Answer
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.