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 02 '11 at 13:04
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Related 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 – Jonas Meyer Aug 02 '11 at 16:14
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The whitehead manifold is a counterexample in dimension 3. https://en.wikipedia.org/wiki/Whitehead_manifold – coudy Jun 06 '23 at 14:46
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The answer to your general question is "no".
A contractible open subset of $\Bbb R^n$ 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 $\Bbb R^n$ which is simply connected at infinity is homeomorphic to $\Bbb R^n$
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 $\Bbb R^3$ bounds a 3-ball - you could even get around that by using the generalized Schoenfliess theorem of Morton Brown.
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@m-winter for n=1, 2 and n>4 they are in fact diffeomorphic. Do you know something about n=3? And clearly it is not the case for n=4. – melomm Mar 15 '22 at 12:22