TITLE: Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$
QUESTION [137 upvotes]: It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.
Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.
However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.
Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?

REPLY [5 votes]: Consider the one point compactifications, $S^n$ and $S^m$, respectively. If $\mathbb R^n$ is homeomorphic to $R^m$, their one-point compactifications would be, as well.  But $H_n(S^n)=\mathbb Z$, whereas $H_n(S^m)=0$, for $n\ne m,0$.