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