Open subsets of euclidean space are not homeomorphic unless they belong to the same space

Asked Jan 18 '21 at 03:49

Active Jan 18 '21 at 03:49

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Suppose that $U \subset \mathbb{R}^m$ and $V \subset \mathbb{R}^n$ be homeomorphic open subsets, then $m=n$.

How can I show this? I think the subsets $U$ and $V$ are supposed to be non-empty. If I assume that $U$ and $V$ are non-epmty, I started by assuming that $m \neq n$. Without loss of generality, I assumed $m <n$ and now I am supposed to obtain a contradiction from here.

Any hint?

asked Jan 18 '21 at 03:49

zero2infinity

Are you familiar with the Homotopy groups? – N. S. Jan 18 '21 at 03:51
@N.S I am, but I would prefer a simpler explanation – zero2infinity Jan 18 '21 at 03:53
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If I remember right this is why the homotopy groups were invented. The problem is very easy to solve when $m=1$ or $m=2$ but, as far as I know, very hard to solve without Algebraic topology otherwise – N. S. Jan 18 '21 at 03:55
Here is a similar question which has a general topology solution: https://math.stackexchange.com/questions/24873/elementary-proof-that-mathbbrn-is-not-homeomorphic-to-mathbbrm – Pedro Amaral Jan 18 '21 at 03:58
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This is not a trivial theorem but there are lots of different proofs for this theorem. – Matías Ures Jan 18 '21 at 05:15
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modern treatment https://arxiv.org/pdf/1310.8090.pdf – Will Jagy Jan 18 '21 at 05:34
@N.S. Brouwer (who first showed this) did it without algebraic topology. My general topology course also treated this without. It’s hard but not insurmountable. – Henno Brandsma Jan 18 '21 at 05:39
This is known as the invariance of domain and it is a major theorem, far from trivial. And AFAIK we can only solve it using algebraic topology. – freakish Jan 18 '21 at 10:05

Open subsets of euclidean space are not homeomorphic unless they belong to the same space

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