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Comment by miketery | Hacker News Reader

> 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.

This seems counter intuitive. Why would this be true? I would have thought that it’s almost impossible for a particle to end up in the same place as it started.

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The paragraph leaves out an important constraint: it’s only for functions that map some compact convex set to itself. So, the theory doesn’t apply to the sibling commenter’s example of f(x) = x + 1 because it doesn’t map a compact set to itself.

One example of a compact convex set is a circle. Imagine some continuous function that maps a circle to a circle. An easy example is rotation. For rotation, the center stays in the same place. Now imagine some more complicated function that might do rotation, stretching, spiral distortions, whatever you can imagine. But it has to be continuous, meaning you can’t go off one edge and come back on the other side (among other things). Try to draw, or even just imagine a transformation like this and you’ll find that there’s always some point that stays in the same place.

For a 3D object like a glass it’s hard to see the same thing intuitively, but Brouwer’s theorem says that it holds (and it also holds for higher-dimensional sets).

Minor wording nit: rather than the circle, I would say “consider the disc”. Usually in math “circle” means just the boundary whereas “disc” means the circle together with its interior.

thank you, that did clear it up for me.

> For a 3D object like a glass

A glass is not convex so this theorem does not apply. You'd need other fixed-point theorems for that. If the glass has a handle the result is trivially false (consider rotating a doughnut, this does not have any fixed point).

I think GP meant the water in the glass (as the article).

What about a circle with 60cm diameter that I move one meter to the right? Something is missing in my understanding of the definition.

It has to map the same set onto itself.

The devil lies in the details here. Brouwer's fixed point theorem says that for a continuous function f that maps a _compact, convex_ set to itself, there must exist a point x₀ such that f(x₀) = x₀. This theorem is only valid for sets that are compact (i.e., closed and bounded) and homeomorphic to convex.

I think the 3-dimensional analogy (similar to the stirring of coffee and sugar Brouwer observed) is rather unintuitive. I think some of the proofs are quite beautiful, but I like this analogy from Wikipedia:

> Take an ordinary map of a country, and suppose that that map is laid out on a table inside that country. There will always be a "You are Here" point on the map which represents that same point in the country.

As others have mentioned, the article is missing a crucial constraint, which is essentially that it holds on the closed unit ball and spaces homeomorphic to it.

As you mentioned, it's also not intuitive. It's by now well-digested, but it's a fairly surprising theorem, even to Brouwer.

Finally, it's not constructive. Generally you can't figure out which point is fixed. If I recall correctly, frustration with this fact fueled Brouwer's interest in intuitionism, which gives us a type of logic where some propositions are neither true nor false but essentially undecidable.

I'm not sure what you mean by a "particle" in your sentence. If you mean to go down to the quantum scale, then I'm sure all sorts of things go wrong in trying to apply the theorem. For one, even talking about the position of a particle at all is very different from talking about a point in Euclidean space.

If you don't mean to go down to the quantum scale, then lots of maps have fixed points. Rotations an reflections are two large classes of examples.

> Generally you can't figure out which point is fixed.

So just thinking out loud here but like in the 1d case... The set will be an interval [a,b] Let g(x) = f(x) - x

g(a) >= 0 and g(b) <= 0. Now we can use the intermediate value value theorem and binary search to find (get arbitrary close) a point with g(x) = 0 (i.e. fix point of f).

Not possible to do such searches in higher dimensions?

You can generally approximate the fixed point. You can possibly do this to arbitrary precision, but I'm not sure.

This is useful for practical applications, but it's not the same as knowing the fixed point for mathematical applications.

E.g. you may want to prove that some function preserves the irrational number pi. But you may not be able to prove it, only show that computations make your conjecture more and more likely.

I'm not sure that even works in 1 dimension...?

I don't think you necessarily have g(a) >= 0 and g(b) <= 0. Finding the starting points for binary search might be essentially as hard as finding the fixed point.

a <= f(x) <= b

Hence a <= f(a), so 0 <= f(a) - a = g(a)

Does intuitionism go beyond Gödel’s incompleteness theorem? Or is it an unrelated thing?

They are different things, although "unrelated" might be a bit strong. I'm sure there are interesting things to say about the incompleteness theorem and intuitionistic logic.

Intuitionistic logic is a logic that is weaker than classical logic in the sense that there are some proofs in classical logic that don't work in intuitinistic logic. Specifically, the law of the excluded middle (either A is true or not A is true") does not hold.

You can convince yourself that this is a reasonable type of logic by interpreting "A is true" to mean "I have a proof of A", and likewise "not A is true" means "I have a proof of not A". Then it's pretty reasonable that you neither have a proof of A or of not A.

I haven't read Godel's proof, but my understanding is that he uses intuitionistic logic in some of the key steps. Since facts in intuitionistic logic are also facts in classical logic, then I believe the incompleteness theorem holds in both logics. But I'm not at all an expert here, so someone please correct me if I'm wrong.

"Intuitionism" also refers to a specific philosophy of mathematics. But a lot of people just use the word to refer to the logic that came out of that philosophical school.

Intuitionistic logic is also very common in type theory, so a lot of proof assistants use it.

>>> intuitionism, which gives us a type of logic where some propositions are neither true nor false but essentially undecidable.

Between math and software development, I feel I've encountered this idea at least three times (+ rediscovered it independently), under different names - "3-value logic", "3-state booleans" - without ever hearing about "intuitionism". Thanks for mentioning it, and now I wonder how it all relates (i.e. I'm about to get stuck on Wikipedia).

> interpreting "A is true" to mean "I have a proof of A", and likewise "not A is true" means "I have a proof of not A".

Makes me think of an interesting progression I've also seen when studying maths and then (classical) AI algorithms. You start with boolean logic, then figure out you need to represent lack of knowledge/certainty and get 3-state logic. Then someone notices that there are degrees to being sure, and you get fuzzy logic[0], representing logical values as real numbers. And then someone notices that this smells awful lot like probability theory, and you get Bayesian probability as applied to reasoning, with all the extra concepts like using whole probability distributions as first-class logic values.

[0] - I'm not sure how much mathematical rigour there is to it; it's something I've learned when studying AI for videogames.

This is an interesting point. I haven't really thought about the connection between intuitionistic logic and three-valued logic before.

They are different logics in general though. 3-value logic is more like the A is True, False or None/Unknown.

Another interpretation one can put on intuitionistic logic is game semantics. Here a statement is true if the verifier has a winning strategy to prove the proposition. It's false if the falsifier has a winning strategy. Then intuitionistic logic corresponds roughly to the idea that some games don't have a winning strategy for either player.

So here (and in the proofs example I gave above) is that we get an idea of "neither true nor false" that is represented by the lack of an explicit rule for concluding truth or falsity in some cases.

In contrast, 3-valued logic gets at the idea of "neither true or false" by introducing another symbol (e.g. "None" or "-1") together with new operators that operate on sentences with that symbol.

So we get one from classical logic by deleting something (namely the law of the excluded middle) and the other by adding something.

This philosophy.se post answers that question: https://philosophy.stackexchange.com/questions/7194/how-is-g...

https://www.youtube.com/watch?v=oX9aPNF6_h8

This is a great video explaining Brouwer's theorem and its (theoretical) application to stirring a cup of water.

It only works for some idealized continous fluid. Not for actual water, which is made up of a large but finite amount of molecules.

Even then it doesn’t feel like a great example since it’s not obvious to me that stirring should be a continuous function. E.g. if I put half the glass into a container and then pour the container back into the glass, that is definitely a discontinuous operation!

I wonder if there is some idealized definition of "stirring" being used here. It seems that actual stirring could conceivably result in any possible rearrangement of particles. As a simple example, which is illustrative even though it seems unlikely as a result of stirring: leave the (x, y) coordinates of each particle the same, while changing the z coordinate as z -> z+k mod h. The result is effectively a discontinuous map (violating the theorem), even though the path of each individual particle must be continuous for physical reasons. Two different notions of continuity here.

But stirring doesn’t break the connnectedness of the water like that?

I guess wherever you dip the spoon in will break it in a kind of discontinuous way?

If the spoon was one of those spoons with holes in the, non-handle-part, then I think it would have to produce a discontinuity.

But for a normal spoon shape, I don’t think dipping it in makes the map between points in the water fail to be continuous. I’m somewhat confident that there is a smooth path from an embedding of a cylinder to R^3 to an embedding of a cylinder which is a little taller and has a spoon shape removed from it, into R^3 .

I stumbled upon this theorem after trying to come up with a mapping of color space, RGB to RGB, where each color maps to a new one. (It was for a mouse cursor effect in a painting app.) I couldn't come up with one, and indeed it is impossible!

Are there any constraints on the mapping? If no, then why not map each intensity value to the immediately next value, and then circle it around? For instance, 0 -> 1, 1 -> 2, 2 -> 3, ...., 255 -> 0.

The function you are describing is not continuous, so the theorem does not apply. But yeah, it has no fixed point.

It's counter intuitive because it's obviously false as stated: f(x) = 1 + x for x in R. They left out the main assumption of the theorem, namely that the domain of the function is compact, and gets mapped (in)to itself.

>I would have thought that it’s almost impossible for a particle to end up in the same place as it started.

That's true on a short time scale. But on a longer time scale, you will eventually visit all possible states of the system.

I think this is similar to the concept of ergodicity[0].

[0]: https://en.wikipedia.org/wiki/Ergodicity

I'm with this guy. Obviously I'm missing something or not understanding what is being said here, because to my head f(x) = x+1 changes every single point and is continuous.

Or in the glass of water point, what if I pour the water out on the table. It's certainly all been moved, yes?

(To be clear I'm obviously wrong / not understanding the premise, but I would be curious to know what I'm missing)

It only applies to functions on bounded sets. The reals are not bounded. There are other restrictions, but that is one of them.

Your example function doesn't map any compact set back to itself.

Yep. I notice mention of this additional constraint from another comment. As far as I can tell the article is just incorrect since it misses this constraint up to that point, unless I'm just a clown and somehow not seeing it.

The upside here is that it seems perhaps I wasn't missing the premise, and it was just mis-stated :)

No you are right, quanta sucks. Im sure there are thousands of other people thinking of x+1 and scratching their heads

So add "mod p" to it.

Transformation is not continuous at the boundary p

I think it's because we think of the drink points as molecules which can indeed move and rearrange in ways that continuous points cannot.

Also, an example of Brouwer's fixed point theorem is the Y Combinator in the lambda calculus. See https://en.wikipedia.org/wiki/Fixed-point_combinator#Y_combi...

What is the compact space in question there? I don’t see one. Or, I don’t even see an appropriate topology to apply?

Are you sure?

watch this, it explains it well: https://www.youtube.com/watch?v=csInNn6pfT4

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