4 ms·
What often happens is the first version of a proof is long and arduous, but the new ideas that it spawns slowly circle back so that someone else can come up wit
by dbaupp 13y ago
What often happens is the first version of a proof is long and arduous, but the new ideas that it spawns slowly circle back so that someone else can come up with a much neater proof, it just takes a while.
(Proving theorems is hard work!)
- assafs 13y agoIn other cases, a very complex problem can be stated very simply, especially to a lay person. A good example is the Jordan curve theorem[1], which at first sight seems like a trivial problem but in effect requires a good deal of topology and analysis to even understand why it is so complicated in the first place. [1] https://en.wikipedia.org/wiki/Jordan%27s_theorem https://en.wikipedia.org/wiki/Jordan%27s_theorem
- ColinWright 13y agoThere is an obvious enhancement of the Jordan Curve Theorem, which is to say that the interior is effectively a disk, and the exterior is effectively a plane with a hole. There is then an obvious 3D version, with a (topological) sphere separating space into a topological ball[0] and a topological R^3 with a hole. This enhanced version, although natural and obvious, is false. I use this as an example of why "obvious" things aren't always true, need proving, and understanding their proofs can lead to useful insights. [0] Edit: corrected "sphere" to "ball"
- tehwalrus 13y agoOoh, 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]
- gohrt 13y ago(I think) this is a case were lay intuition assumes objects are piece-wise differentiable, but the formal math definitions don't. That is, most people think of physical objects, where there is a minimum "planck length", and no fractal-ish structures with infinitely fine-scale structure. It is still abstractly interesting that even 2-D fractals satisfy the Jordan Curve therorem, whie 3-D faractals do not satisfy the 3-D version of the conjecture.
- lmm 13y agoI think that's largely because mathematics abstracts more than is obvious. My favourite example is Hilbert's basis theorem, which looks like it's not even a question until you realise how general the notion of "space" that it applies to.
- gohrt 13y agoRight, same reason that the Banach-Tarski paradox is intuitively baffling. It relies on fractal-structures that require uncountably-infinite fine detail.