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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?

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For higher dimensions you need homotopy or homology groups. One needs a "higher" notion of connectivity. – Martin Brandenburg Mar 3 '14 at 12:58
Unfortunately, it is not that easy. – Jochen Mar 3 '14 at 12:59
The same proof almost works for 2 versus 3 dimensions. But replace the word connected, with simply-connected. – Tim Seguine Mar 3 '14 at 13:04
I hope my edit helps show why this approach isn't as easy as you might first think. – Dan Rust Mar 3 '14 at 13:23
up vote 6 down vote accepted

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.

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Thanks. But if I remove subspace $X=(x_1,\ldots,x_{m-1},0)$, then I can easily prove that there's no continuous path from up-half space to down-half and hence disconnected. Also, there's continuous path in higher dimensional space and hence connected. What's the problem of this statement? – zhangwfjh Mar 3 '14 at 13:32
"Also, there's continuous path in higher dimensional space and hence connected." this part. How do you know that the image of the embedded $m-1$ dimensional subspace does not disconnect the higher dimensional space after mapping it via a homeomorphism? – Dan Rust Mar 3 '14 at 13:35
Because in higher dimension, when I get stuck by removed space, I have extra dimensions which allow me to jump over the pit. – zhangwfjh Mar 3 '14 at 13:38
But what happens if the embedded subspace ends up being something like a space filling curve? It's conceivable (or at least not obviously out of the question) that an embedded line in $\mathbb{R}^3$ could fill up a planar subspace in this way and so disconnect $\mathbb{R}^3$. Proving that this is not the case is far from trivial. – Dan Rust Mar 3 '14 at 13:47
I'm confused again since I have explicitly indicated that embedded subspace is a hyperplane, which I think can avoid the case you mentioned. – zhangwfjh Mar 4 '14 at 23:36

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