Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
The title is somewhat opaque... –  Mark Aug 2 '11 at 13:04
    
Related MathOverflow questions: mathoverflow.net/questions/64192/…, mathoverflow.net/questions/4468/… –  Jonas Meyer Aug 2 '11 at 16:14
add comment

1 Answer 1

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

share|improve this answer
    
Great answer! ${}$ –  Mariano Suárez-Alvarez Aug 2 '11 at 16:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.