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Is it true that every contractible open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$?

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The title is somewhat opaque... – Mark Aug 2 '11 at 13:04
Related MathOverflow questions:…,… – Jonas Meyer Aug 2 '11 at 16:14

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

  1. the Loop theorem and
  2. 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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Great answer! ${}$ – Mariano Suárez-Alvarez Aug 2 '11 at 16:36

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