Proof of there's no homeomorphism between Euclidean spaces I read a wonderful proof about there's no homeomorphism between Euclidean spaces of 1 and 2 dimension. If there is, then Euclidean space of 1 and 2 dimension are homeomorphic and hence have the same topological properties. But when one remove one point from 1 dimensional Euclidean space, it becomes disconnected while it is not when remove from 2 dimensional space. So I'm wondering if I can generalize to any dimension $m,n$, assuming $m<n$. Then I remove an $m-1$ dimensional subspace from both spaces. And one is connected, the other is not. So the homeomorphism doesn't exist. Am I right?

Edit: Thanks for all replies. I'm current unknown for algebraic topology. Could you please explain me for why the method I use is invalid?
 A: This only really works in the specific case of distinguishing $1$ dimensional and $n$ dimensional Eulcidean space. For higher dimensions you will need a more sophisticated approach. Normally this includes the use of techniques from algebraic topology.
See this page for one approach to solving this problem.

To comment specifically on why your approach won't immediately work, consider the proof in the case for $1$ dimension compared to $n$. You suppose there is a homeomorphism $h\colon\mathbb{R}^1\to\mathbb{R}^n$ and infer that be removing a point $p$ from $\mathbb{R}^1$, $h$ restricted to this subspace is still a homeomorphism onto the image which must be $\mathbb{R}^n\setminus \{h(p)\}$ because $h$ is a bijection. This quickly leads to a contraidction.
Now, replace $\mathbb{R}^1$ with $\mathbb{R}^m$ and suppose we remove a subspace $S\cong\mathbb{R}^{m-1}$ from $\mathbb{R}^m$. It is still true that $h$ restricted to this subspace is a homeomorphism onto its image, but now you have a much harder time of finding properties of the image, that copy of $\mathbb{R}^{m-1}$ could be wrapped up in very strange ways inside $\mathbb{R}^n$ (for instance one might imagine something like a space filling curve with minor pertubations on the usual points which are not $1-1$ so that the line is embedded in $\mathbb{R}^3$ in such a way that an entire $2$ dimensional subspace is covered by the line - this can't be the case but it's not obvious how to show that this can't happen).
The point is, it's not entirely clear that removing it from the codomain will not disconnect $\mathbb{R}^n$ as is the case with a single point.
