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by ColinWright 4741 days ago
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.

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 }

1 comments

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.

In any case, cool! thanks :)

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.

OK, "proximate" rather than interlocking. I did understand the geometry of it, I just chose the wrong word. Thanks for your explanations! :)