In three dimensions there is an embedding of S^2 (which is the surface of a sphere[0]) into R^3 such that the exterior is not homeomorphic to R^3 with B^3 missing.
Hmm, so because part of the surface is a fractal, it can't be simply connected?
I'd be interested to learn whether a linear version (straight pipes instead of curved/broken torus ones) also has the same topological properties. It is clearly fractal, and clearly also homotopy identical to S^2, but obviously the two ends are no longer interlocking, and I wonder if this is as crucial to the result as both 'ends' being fractal clearly is.
The surface is simply connected, it's the outside that ends up not being simply connected. This works equally well with piece-wise linear embeddings. And the current version of the Alexander Horned Sphere is not actually interlocking - the embedding is contractable back to S^2.
It takes a while to get your head around what this really is.
http://en.wikipedia.org/wiki/Alexander_horned_sphere (search for "jordan")
In essence, the "outside" gets very "tangled" and can't be smoothly converted into a "proper" exterior.
[0] S^2 = { (x,y,z) : , x, y, z, in R with x^2+y^2+z^2 = 1 }