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Ooh, OK, you've tickled my interest. Can you provide an explanation (or a link to something similar) with the disproof of the 3D version?
by tehwalrus 13y ago
Ooh, OK, you've tickled my interest. Can you provide an explanation (or a link to something similar) with the disproof of the 3D version?
- ColinWright 13y agoIn 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 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 }
- tehwalrus 13y agoHmm, 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 :)
- ColinWright 13y agoThe 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.
- tehwalrus 13y agoOK, "proximate" rather than interlocking. I did understand the geometry of it, I just chose the wrong word. Thanks for your explanations! :)
- deleted 13y ago[deleted]