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> Brouwer’s fixed point theorem. The theorem says that for any continuous function, there is guaranteed to be one point that the function leaves unchanged — a f
by sockaddr 3y ago
> Brouwer’s fixed point theorem. The theorem says that for any continuous function, there is guaranteed to be one point that the function leaves unchanged — a fixed point, as it’s known. This is true in daily life. If you stir a glass of water, the theorem guarantees that there absolutely must be one particle of water that will end up in the same place it started from.
Wait. What if by stirring I shook the glass and at the end set it back down a foot away from its initial location. Is this author claiming that Brouwer’s theorem guarantees a particle floating somewhere where the glass used to be? Unchanged?
I can imagine an unlimited number of scenarios where I can guarantee the water particles are displaced.
This seems like a bad application of the theorem, right?
- stormfather 3y ago> If you stir a glass of water, the theorem guarantees that there absolutely must be one particle of water that will end up in the same place it started from. Wait what? If you stir a glass long enough, any configuration of particles should be possible. For instance you can imagine "cutting" the water like a deck of cards.
- kadoban 3y agoThey 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.
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- GeneralMayhem 3y agoIt only applies when you're mapping a space onto itself (plus a couple other qualifications), so you have to put the glass back in the same place. I think of it as a generalization of the intermediate value theorem - some things are going left, some things are going right, one thing must be sitting still in between.
- plank 3y agoAnd: it has to be ‘continuous’. E.g. in ‘real’ water, the molecule of H2O ‘colloquially’ known as ‘number 12345678998776165441’ which sat next to molecule known as ‘number 1’, still has to sit next to it. All be it in a different location. I think something like mayonaise would have been a better example, as water molecules that are ‘next’ to each other can in reality split up quite easily by itself. While something like mayonaise would not. *off course, quantum mechanics and all that suggesting that it would be impossible to label the individual molecules.
- dekhn 3y agoQM doesn't say that it's impossible to label the individual moleccules. However, hydrogens are labile, it makes more sense to identify the unique oxygens.
- Enginerrrd 3y agoAre you trying to say that for every point P there exists a neighborhood U around it for which the each transformed point T(P) is contained in the neighborhood T(U)?
- dataflow 3y agoWhat's the conceptual difference between this and the hairy ball theorem?
- hgsgm 3y agoLoosely, Hairy ball theorem is about the derivative of a mapping at a single point in time. (Imagine slowly deforming/mixing the input configuration t to obtain the output configuration. Brouwer's theorem can be thought of as being about the integral of a continuous family of hairy balls over a time interval. It's not exactly the same because the assumptions about differentiability are different.
- deleted 3y ago[deleted]
- T-A 3y agoAs James Bond would say, https://en.wikipedia.org/wiki/Shaken,_not_stirred https://en.wikipedia.org/wiki/Shaken,_not_stirred
- seydor 3y agoYou are not just stirring, then
- BorisTheBrave 3y agoA better example I've seen for the theorem is that if you take a paper map of a country, messily schrunch it up into a ball and drop it somewhere in the country, there will be at least one point on the map that is exactly above the corresponding real point in the country. As others have said, it's meant for mappings of the space to itself. So stirring the water, but not moving the glass. But anyway, the theorem works only for continiuous mappings. The moment they started mentioning "water particles", instead of some hypothetical fluid that is a continuous block, the theorem no longer applied. You could break it by mirroring the position of every particle. There's still a fixed point (the line of mirroring), but there's no obligation that there's a particle on that line.
- deleted 3y ago[deleted]
- chacham15 3y agoThis still doesnt make sense to me. Imagine a continuous line whose positions map to the real numbers between 0 and 1. If I "move" the line over 0.1 wrapping the end back to the beginning (i.e. x2 = (x1 + 0.1) % 1), there will be no points that are in the same position as they were in before. EDIT: If you need a continuous function, wouldnt expanding the space to a line from -Inf to +Inf and then using x2 = x1 + 0.1 do the trick?
- frutiger 3y agoThe theorem only applies to continuous functions.
- contravariant 3y agoBrouwer's fixed point theorem only applies to compact convex sets. Infinite lines don't work, as they are not compact. Similarly a circle would not work as it is not convex (you're close with your example, you just need to glue together the endpoints to turn it into a circle and make the map continuous).
- 3y ago
- sva_ 3y agoI wonder how this really applies, since IEEE 754 floating points are not continuous.
- rowanG077 3y agoI'm misunderstanding the theorem. Take f(x) = x + 1. That is a continuous function. But there does not exist an x where f(x) ~ x holds. What am I missing?
- arbitrarybits 3y agoThat's because there is also a precondition for the domain to be bounded, and since the real numbers are not bounded, the theorem does not apply here (or for any function R -> R).
- amalcon 3y agoYou can just do the same thing in the integers modulo any value greater than 1, creating a bounded domain without creating a fixed point. I imagine this runs afoul of a different precondition which I haven't identified, though -- it's obviously way more likely than that the entire field has managed to miss something this obvious. Edit: Actually it's obviously just that this set is not convex, or even continuous.
- vecter 3y agoIt must be a compact convex set. The entire reals is not compact.
- ninepoints 3y agoI do not understand this comment. In what world does "stirring" fluid in a cup imply any sort of dispacement of the cup itself?
- sorokod 3y agoThat function is from a compact convex set to itself.
- jjtheblunt 3y agoi remember the example of "you can't comb the hair on a coconut without a whorl".
- hgsgm 3y agoThat's a different fixed point theorem. https://en.m.wikipedia.org/wiki/Brouwer_fixed-point_theorem https://en.m.wikipedia.org/wiki/Brouwer_fixed-point_theorem
- jjtheblunt 3y agoThanks
- taeric 3y agoThe full theorem is a continuous mapping back on itself. So, only the water in the glass mapped back to the glass. Pour it out to another glass, and you no longer mapped back to the glass. Pour it back into original glass, and you are back to having a fixed point possible. And, obviously, only with respect to the mapping in the glass.
- KolenCh 3y agoThe theorem assumes the function has identical domain and codomain. Eg f(x) = x^3 maps [0, 1] to itself. So that's why they gave the glass stirring example, the domain here is the whole volume of water. So as long as it ends up in the same volume of water, the process of stirring is assumed to be continuous and therefore this theorem applies. Your example changes where it ends up (ie not back to the same domain) and so it cannot be applied.