TITLE: Are contractible open sets in $\mathbb{R}^n$ homeomorphic to $\mathbb R^n$?
QUESTION [38 upvotes]: Is it true that every contractible open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$?

REPLY [45 votes]: 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.