4 ms·
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
by tehwalrus 13y ago
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 :)
- 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]