Is it true that every contractible open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$?
Are contractible open sets in $\mathbb{R}^n$ homeomorphic to $\mathbb R^n$?
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0Related MathOverflow questions: http://mathoverflow.net/questions/64192/is-there-a-classification-of-open-subsets-of-euclidean-space-up-to-homeomorphism, http://mathoverflow.net/questions/4468/what-are-the-open-subsets-of-mathbbrn-that-are-diffeomorphic-to-mathbb – 2011-08-02
1 Answers
the answer to your general question is "no"
A contractible open subset of $R^n$ need not be "simply connected at infinity". ( "X is simply connected at infinity" means that for each compact K there is a larger compact L such that the induced map on $\pi_1$ from X - L to X - K is trivial.)
A contractible open subset of $R^n$ which is simply connected at infinity is homeomorphic to $R^n$
a) if n > 4: by J. Stallings, The piecewise linear structure of Euclidean space, Proc Camb Phil Soc 58(1962) (481-88)
b) n = 4: by M. Freedman - see Topology of 4-Manifolds by Freedman and Quinn.
c) For n = 3 this is a standard exercise - I don't know who gets the credit, but you oould refer to AMS memoir 411 by Brin and Thickstun.
The ingredients are
- the Loop theorem and
Alexander's theorem
that a PL sphere in R^3 bounds a 3-ball - you could even get around that by using the generalized Schoenfliess theorem of Morton Brown.
Hope this helps
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0Great answer! ${}$ – 2011-08-02