6 ms·
They left off some qualifiers. According to wikipedia, it's applicable to a continuous function mapping a nonempty compact convex set to itself. Which ends up m
by kadoban 3y ago
They left off some qualifiers. According to wikipedia, it's applicable to a continuous function mapping a nonempty compact convex set to itself. Which ends up making a lot more sense to me.
- mjfisher 3y agoSo is it just a fancy way of saying "every transformation has an axis or origin"?
- Elyra 3y agoThis is also how I interpreted it, which I take it is probably incorrect and would like to know why.
- nerdponx 3y agoOne of the pre-conditions of the theorem is that the domain and codomain are the same. So that's one way to satisfy the theorem. But it's not really intuitive. If you gently swirl a bottle of water, somewhere in the bottle, some water molecule did not move. It's not necessarily in the center of anything. The fixed point could be anywhere. The theorem is about existence only, not construction. (I am not a professional mathematician, I might also be wrong.)
- tomrod 3y agoNo, you are on point.
- l__l 3y agoThat's not the best way to view it. You can prove it by showing that for an n-dimensional disc, there is no contraction to its boundary; which I think is a bit more illustrative of what this FPT is doing
- tsimionescu 3y agoWell, some continuous functions are simple and it's highly intuitive why this applies to them. Others are very weird and it's not intuitive at all that they should have such a point. And of course, some functions/transformations are not continuous, and those may not have such an "axis of origin" at all.
- r0uv3n 3y agoI don't quite know what you mean by that, but consider that this is not true for e.g. a torus or a sphere (take the function mapping every point to its antipodal point). The fact that the underlying space is contractible is very important here.
- nerdponx 3y agoIsn't a container of water actually a good example? Maybe swirling gently (laminar flow) is a better mental image than shaking vigorously (possibly discontinuous).
- l33t233372 3y agoShaking a container is a good example I think, assuming the glass is convex. Shaking has to be continuous, the particles move quickly and erratically, but they trace continuous paths.
- almostnormal 3y agoErratic movement only occurs if the container isn't filled completely. If it is filled completely and the shaking doesn't include any turning/twisting motion, not much happens.
- mbeex 3y agoEven then. The required mapping is 'onto' (otherwise, a discontinuous counterexample would be trivial). The air particles are equivalent to the water.
- plonk 3y agoThe paths are continuous, but if they move two neighbouring molecules away from each other, the final transformation won't be continuous, will it?
- l33t233372 3y agoIt’s difficult to discuss this physical example because particles are discrete. In an ideal system with points instead of particles, shaking would be continuous.
- Someone 3y agoAnd then, we would not call it shaking but bending and twisting, would we? Think of the 1D variant. If you shuffle a deck of cards, but require that to be ‘continuous’, few shuffles remain (I think only the identity mapping and ‘flipping the deck upside down’). I doubt anybody would restricting the possible permutations that much stil call shuffling.