A fundamental question on homeomorphism


by krete
Tags: fundamental, homeomorphism
krete
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#1
Jan15-13, 06:35 AM
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It is well known that there does NOT exist a homeomorphism between R^m and R^n if m>n. My question is whether it is possible to construct a homeomorphism between R^m (as a whole) and a subset of R^n (note that we also suppose that m>n)?

Intuitively, it is impossible. Is my intuition right? Thank you for your replying in advance!
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HallsofIvy
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Jan15-13, 07:14 AM
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Any subspace of Rn is Rk for k< n< m. And you have already said "there does NOT exist a homeomorphism between R^m and R^k if m>k" (where I have replaced your "n" with "k").
krete
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#3
Jan15-13, 07:26 AM
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Hi, HallsofIvy,

How about if the subset of R^n is not the whole R^k (k<n) but some ill-behaved set (e.g., a space filling line)?

quasar987
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#4
Jan15-13, 08:04 AM
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A fundamental question on homeomorphism


The usual tool for proving the "no homeo thm" is Brouwer's Invariance of Domain theorem:

http://en.wikipedia.org/wiki/Invariance_of_domain

It can in the same way be used to answer your question: Assume a homeo btw S (subset of R^n) and R^m exists. Consider R^n as a subset of R^m (say as R^n x {0,...,0}). Then we have a map

R^m --> S --> R^m

which is the homeomorphism of R^m with S composed with the inclusion of R^n in R^m. This map is not open since the inclusion of R^n in R^m maps any subset of R^n to a non open subset of R^m. This contradicts Brouwer's invariance of domain theorem.
krete
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#5
Jan15-13, 12:46 PM
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Quote Quote by quasar987 View Post
The usual tool for proving the "no homeo thm" is Brouwer's Invariance of Domain theorem:

http://en.wikipedia.org/wiki/Invariance_of_domain

It can in the same way be used to answer your question: Assume a homeo btw S (subset of R^n) and R^m exists. Consider R^n as a subset of R^m (say as R^n x {0,...,0}). Then we have a map

R^m --> S --> R^m

which is the homeomorphism of R^m with S composed with the inclusion of R^n in R^m. This map is not open since the inclusion of R^n in R^m maps any subset of R^n to a non open subset of R^m. This contradicts Brouwer's invariance of domain theorem.
Dear quasar987,

Thank you very much for your helpful answer. It is really a nice proof.
Bacle2
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#6
Jan19-13, 03:11 PM
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You may want to check this thread too:

http://www.physicsforums.com/showthr...60#post3650960


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