## A fundamental question on homeomorphism

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)?

 Recognitions: Gold Member Science Advisor Staff Emeritus 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").
 Hi, HallsofIvy, How about if the subset of R^n is not the whole R^k (k

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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.

 Quote by quasar987 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,