Are open sets in R^n always homeomorphic to R^n?

[iLoveTopology]

I know that open intervals in R are homeomorphic to R. But does this extend to any dimension of Euclidean space? (Like an open 4-ball is it homeomorphic to R^4?)

My book doesn't talk about anything general like that and only gives examples from R^2.

 

[Bacle2]

No for general open sets; look at, e.g., an open annulus, or any disconnected open set. But an open n-disk D:={x in R^n : ||x||<1 } (or any translation of it) is homeomorphic to R^n.

 

[disregardthat]

Any connected open set in R^n is homeomorphic to R^n, for any n. An open set in R^n is homeomorphic to the disjoint union of equally many R^n's as connected parts of your open set.

 

[Bacle2]

Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.

 

[disregardthat]

Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.

thanks for the correction, i meant simply connected :)

 

[micromass]

thanks for the correction, i meant simply connected :)

Also not true, then open set $B(0,1)\setminus\{0\}$ of $\mathbb{R}^3$ is simply connected but not homeomorphic to $\mathbb{R}^3$.

 

[mathwonk]

notice R^n is contractible. But even that is not enough. Look at this:

http://math.stackexchange.com/quest...en-sets-in-mathbbrn-homeomorphic-to-mathbb-rn