What's your opinions on the Axiom of Choice?

[George Jones]

If there were some physics device/experiment that could measure an infinite number,

How could this be done? How could one get an infinite reading on a dial or a counter?

then simply use that device/experiment to see if for infinite numbers a and b,

a + b = max{a,b}, ab = max{a,b}

is actually observed in nature. If it turns out that it doesn't, then the axiom of choice has finally failed once and for all.

 

[quadraphonics]

The stuff about infinities is a red herring: NO statement in (pure) mathematics can be experimentally falsified. It is not a natural science like physics or chemistry. You can falsify physical theories that employ mathematics, but that doesn't falsify the mathematics itself; it simply means that they aren't applicable in the way you tried to use them.

As an example, consider a simple mathematical model of balls of clay. To start with, we have 2 balls of clay. Then, we mush them together, resulting in a single ball of clay. So, we start with two balls of clay, add them together, and get one ball of clay. This does not imply that the statement 1+1=2 is false. It simply means that it does not describe the behavior of the number of balls of clay under the mushing-together operation (in contrast, it would be exactly correct if we were interested in the mass of the clay, rather than the number of balls). It's always the *application* of math to a particular scenario that is falsifiable, not the underlying mathematics itself.

 

[mathman]

One problem with AC is the Banach-Tarski theorem. Physically it is impossible to divide a ball into a finite number of pieces and reassemble them into a sphere of a different size, which is what Banach-Tarski says you can do.

 

[gel]

One problem with AC is the Banach-Tarski theorem.

Personally, I don't find that a problem at all. Just because you have a decomposition of a set into certain subsets, doesn't mean that you can physically split an object up like that. You couldn't split a ball up into its two subsets that consist of points which are algebraic and transcendental distances from its center, although you can do this mathematically.
To me, the Banach-Tarski "paradox" is just a striking demonstration of the existence of non-measurable sets.

You may prefer to throw out AC and, in its place, use an alternative axiom which implies that all subsets of Rⁿ are measurable. This might make measure theory a little easier, but it seems like you are throwing out a lot and not gaining much by doing so.

Keeping countable dependent choice, throwing out the uncountable version of AC, but adding an axiom which implies that all subsets of Rⁿ are measurable sounds like a better compromise, but I'd still prefer to just keep AC.

 

[eastside00_99]

One problem with AC is the Banach-Tarski theorem. Physically it is impossible to divide a ball into a finite number of pieces and reassemble them into a sphere of a different size, which is what Banach-Tarski says you can do.

This is why accepting the Correspondence Theory of Truth will naturally lead to thinking the axiom of choice is false. But, for by same token, accepting the correspondence theory of truth will could also lead one to think that the axiom about the empty set is false. Its best, if you do math, to adopt the Consistency Theory of Truth or at least pretend to when you do maths.

 

[Hurkyl]

One problem with AC is the Banach-Tarski theorem. Physically it is impossible to divide a ball into a finite number of pieces and reassemble them into a sphere of a different size, which is what Banach-Tarski says you can do.

No, it does not say that: the Banach-Tarski pseudoparadox talks about subsets of topological spaces, and makes no assertions about physical reality.

 

[mathman]

No, it does not say that: the Banach-Tarski pseudoparadox talks about subsets of topological spaces, and makes no assertions about physical reality.

I was trying to address the question raised by arildno, i.e. could a physics experiment be devised to test the axiom of choice? Based on Banach-Tarski, the answer would be no.

 

[Crosson]

I find myself in perfect agreement with the opinion that "the axiom of choice obviously true, the well-ordering principle is obviously false, and with Zorn's lemma who can tell?"

Does anyone know if ZF + "there is an infinite dimensional vector space without a basis" has been proved to be consistent? If so, then either the axiom of choice is true or the standard formulation of quantum mechanics is somewhat incorrect. Personally I think it is naive to think that the axiom of choice or its negation will ever be a fact (objectively confirmed in reality) just as it is naive to consider the standard formulation of QM to be such.

 

[morphism]

Does anyone know if ZF + "there is an infinite dimensional vector space without a basis" has been proved to be consistent?

That every vector space admits a basis was proven to be equivalent to the axiom of choice [Blass, 1984].

There's a two-volume book by Rubin & Rubin, called Equivalents of the Axiom of Choice, that has a large list of some pretty interesting statements that are equivalent to choice -- some of them are bizarre. One of my favorites is the following: Every nonempty set admits a binary operation that turns it into an abelian group. Another interesting one (maybe this one is weaker than, but certainly implied by, the AOC -- I don't really remember): R^3 is the disjoint union of discs.

 

[eastside00_99]

I was trying to address the question raised by arildno, i.e. could a physics experiment be devised to test the axiom of choice? Based on Banach-Tarski, the answer would be no.

I don't know about that. Some computer scientist think that is AC is false (or at least problematic) but this is not based so much on The Banach-Tarski paradox.

c.f.,

On the Computational Content of the Axiom of Choice
Stefano Berardi, Marc Bezem, Thierry Coquand
The Journal of Symbolic Logic, Vol. 63, No. 2 (Jun., 1998), pp. 600-622

That's one thing that separates mathematicians from other scientists: they don't insist on there being any relationship between nominal reality and their theories.

 

[arildno]

Hmm..all I contributed with, was that NO physical experiment can ever test the VALIDITY of any mathematical statement, only its suitability, for the purposes of the physicist in his modelling activities.

I did not specifically address the issue of the particular mathematical statement called the axiom of choice.

 

[mathboy]

I'm not sure. I don't see how, but maybe I'm too narrow-minded.



I don't see how it's possible to test infinities.



Maybe some physicists never see the axiom of choice, but many do see it, so it's not like the physics community is unaware of this piece of mathematics. I saw it, or something equivalent to it, in a number of my math courses. Because (I think that) it's not physically testable, (I think) it's not of experimental interest to physicists.

How about this: The axiom of choice implies that every infinite-dimensional vector space has a basis (and vice versa). So just test out the physical results of some infinite-dimenional vector spaces used in physics (e.g. quantum mechanics?), and observe if the mathematical predictions by those infinite-dimensional vector spaces are actually observed in the physical universe. Isn't that one way to test the validity of the axiom of choice?

 

[LukeD]

How about this: The axiom of choice implies that every infinite-dimensional vector space has a basis (and vice versa). So just test out the physical results of some infinite-dimenional vector spaces used in physics (e.g. quantum mechanics?), and observe if the mathematical predictions by those infinite-dimensional vector spaces are actually observed in the physical universe. Isn't that one way to test the validity of the axiom of choice?

But what predictions would you get for an infinite-dimensional vector space without a basis? Also, unless I am mistaken, for most spaces you use in quantum mechanics, you can actually write down a basis. The axiom of choice need not be invoked if you can actually write the basis. It's only needed when no explicit basis has been (or, in many cases, can be) found (or to prove abstract results about all of the bases, etc). Therefore, most, if not all, of the results of QM are independent of Choice, and for any that do in fact depend on it, we have no way of knowing (as far as I am aware) of how the results would differ were the negation of the Axiom of Choice assumed. (Not to mention that the negation implies that there EXISTS a vector space without a basis, so finding that all of the vector spaces used in QM do have a basis would tell us nothing about every other vector space. There could still be one without a basis.)

 

[mathboy]

I'm pretty sure there are some infinite-dimensional vector spaces used in quantum mechanics whose basis is assumed to exist (thanks to the axiom of choice). And even if there is an explicit basis used, that it actually spans the vector space probably relies on the axiom of choice--(there must be examples of this). So, use these specific infinite-dimensional vector spaces, and test them out physically. This is just one idea to test out the Axiom of Choice. I'm pretty sure there are other testable ideas too, if the physicists will simply use their physics knowledge to think of them.

 

[quadraphonics]

Isn't that one way to test the validity of the axiom of choice?

No, there is no way to experimentally determine the validity of any statement in (pure) mathematics. You can test whether it is suitable, or applicable, to the physical phenomenon in question, but this wouldn't tell you anything about it's underlying "validity." Mathematical axioms cannot be "valid" or "invalid" (at least, beyond the level of self-consistency); rather, they can be either "useful" or "useless." Unlike the natural sciences, mathematical conjectures are differentiated by aesthetic values, rather than agreement with nature.

To put it another way: if you can test it experimentally, it's a natural science, not (pure) math.

 

[Crosson]

I did not mean to imply that there was a simple empirical connection between quantum mechanics and the axiom of choice. Suppose AOC is false, then all we can conclude is that there is at least one (infinite-dimensional) vector space V without a basis (good luck constructing it). We do not know, however, that V represents the configuration space of any quantum system. Even though we often assume a quantum system has a basis without displaying it explicitly, we have not contradicted anything unless we know that the "defective" space V in fact represents a real quantum system (which would force the dichotomy I mentioned above).

To put it another way: if you can test it experimentally, it's a natural science, not (pure) math.

That all depends on what you mean by "test it." You alluded to validity above, and I agree that the validity of mathematical theorems are independent of any empirical fact. The theorems are all tautologies after all; although we often leave out the antecedent of theorems like "the square root of 2 is irrational" we could choose to make the hypothesis explicit if so desired.

But as you mentioned, another "test" of mathematical results is their aesthetic value. Why do we place so much emphasis on the ring of integers? Of course it is because the integers are suitable and applicable, as you say. But there is not only one model of arithmetic, and so there remains the question of whether we can rule out various models based on empirical testing. Similarly, there is not only one model of set theory, so this leaves open the possibility that some of these models will be shown to be incompatible with empirical facts.

 

[Ka Yan]

Why is that the Banach-Tarski Paradox (Theorem, say) ture for R^n of n$$\geq$$3, but nor for R and R^2, please?

 

[Hurkyl]

I just wanted to point out a factual error -- the notion of a vector space basis is slightly different than the notion of a Hilbert space basis. In particular, for the latter, every vector can be written uniquely as a possibly infinite sum of multiples of basis vectors.

 

[Crosson]

I just wanted to point out a factual error -- the notion of a vector space basis is slightly different than the notion of a Hilbert space basis. In particular, for the latter, every vector can be written uniquely as a possibly infinite sum of multiples of basis vectors.

Thanks for clarifying that in this context Hurkyl! I had not realized what an unintuitive theorem "every vector space has a basis" is, since I had thought that in an arbitrary Hilbert space H there is not necessarily a set of vectors B, every subset of which satisfies linear independence, such that each element of H can be written as a linear combination of the vectors in some finite subset of B. But this is what it means to be a basis in a purely algebraic sense, and since an arbitrary infinite-dimensional vector space does not necessarily have the structure to define infinite-sums this must be what is meant. Very surprising, makes AOC seem even less aesthetically satisfying for me!

 

[Hurkyl]

Well, I would expect (but do not know) that the AoC is also equivalent to the claim that every Hilbert space has a Schauder basis. AoC is clearly equivalent to the claim that every topological vector space has a Schauder basis. (Consider the discrete topology)

(the term Schauder basis denotes one where every element is an infinite linear combination. The finitary version is called a Hamel basis)


Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space). I feel like I should already know the answer, but I'm not recalling it.

 

[mathboy]

So how much does quantum mechanics rely on Axiom of Choice? And are there any observable phenomenon in QM that depend on the accept of AoC?

 

[Crosson]

Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space).

The position and momentum representations do not yield bases, but the position and momentum operators do (with respect to a particular basis). The necessary conditions for the operators $\overhat{x},\overhat{p}$ are that they be self-adjoint and obey the commutator:

$$[\hat{x},\hat{p}] = i$$

And so in the position representation we have:

$$\hat{x} = x$$

$$\hat{p} = i \frac{\partial}{\partial x}$$

And in the momentum representation we have:

$$\hat{x} = i \frac{\partial}{\partial k}$$

$$\hat{p} = k$$

Now, to try an answer to your question I would say that the eigenvectors of the position operator in the position representation are:

$$\hat{x} = x \rightarrow eigenVectors(\hat{x}) = \{\phi_{x'}:\mathbb{R} \rightarrow \mathbb{C} | \phi_{x'}(x) \mapsto \delta(x - x') , x'\in \mathbb{R}\}$$

$$\hat{p} = i \frac{\partial}{\partial x} \rightarrow eigenVectors(\hat{p}) = \{\phi_{k}:\mathbb{C} \rightarrow \mathbb{R} | \phi_{k}(x) \mapsto e^{i k x} , k\in \mathbb{R}\}$$

In physics we say that either of these sets of eigenvectors are an uncountable basis! The linear combinations become definite integrals covering the range of the eigenvalues:

[tex] f(x) = \int_{-\infty} ^{\infty} c_{x'} \phi_{x'}(x) dx' [/itex]

where the coefficients are given by Fourier's trick:

[tex] c_{x'} = \int_{-\infty} ^{\infty} f(x) \phi_{x'}(x) dx [/itex]

I'm interested to learn more about what's really going on here.

So how much does quantum mechanics rely on Axiom of Choice? And are there any observable phenomenon in QM that depend on the accept of AoC?

The standard formulation of quantum mechanics is built on the fact that the set of eigenvectors of a self-adjoint operator in a Hilbert space will be an orthonormal basis. We are trying to determine whether this theorem depends on the axiom of choice. Even then I don't know any real experiments that verify the quantum state of a system to be an infinite-dimensional vector as opposed to high-finite dimensional one, but there is hopefully there is some major qualitative difference between these cases that could one day be observed.

There was a time when it was suitable to treat matter as an uncountable continuum, an actual infinity. Even in the 1920s we knew that was far from the case so there was no fear (or joy) that the Banach-Tarski construction could be realized. But we still treat time and space as uncountable and complete in quantum field theory, and that is likely to be with us for sometime. Also as mentioned the quantum mechanical state of a single particle alone in the universe is an infinite-dimensional vector. It is only through these few remaining vestiges of believable actual infinities in physical reality that we ever have any hope of "testing" the axiom of choice.

 

[morphism]

Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space). I feel like I should already know the answer, but I'm not recalling it.

There's the (normalized) Hermite functions (Gram-Schmidt {x^n exp(-x²/2)}).

 

[mathboy]

The standard formulation of quantum mechanics is built on the fact that the set of eigenvectors of a self-adjoint operator in a Hilbert space will be an orthonormal basis. We are trying to determine whether this theorem depends on the axiom of choice. Even then I don't know any real experiments that verify the quantum state of a system to be an infinite-dimensional vector as opposed to high-finite dimensional one, but there is hopefully there is some major qualitative difference between these cases that could one day be observed.

There was a time when it was suitable to treat matter as an uncountable continuum, an actual infinity. Even in the 1920s we knew that was far from the case so there was no fear (or joy) that the Banach-Tarski construction could be realized. But we still treat time and space as uncountable and complete in quantum field theory, and that is likely to be with us for sometime. Also as mentioned the quantum mechanical state of a single particle alone in the universe is an infinite-dimensional vector. It is only through these few remaining vestiges of believable actual infinities in physical reality that we ever have any hope of "testing" the axiom of choice.

Aha! George Jones, are you reading this? There is some hope after all. Could some physicist please take the plunge !?

 

[ice109]

now i haven't read the whole thread but why again are we discussing the truth value of an axiom?

 

[Hurkyl]

Now, to try an answer to your question I would say that the eigenvectors of the position operator in the position representation are:

$$\hat{x} = x \rightarrow eigenVectors(\hat{x}) = \{\phi_{x'}:\mathbb{R} \rightarrow \mathbb{C} | \phi_{x'}(x) \mapsto \delta(x - x') , x'\in \mathbb{R}\}$$

$$\hat{p} = i \frac{\partial}{\partial x} \rightarrow eigenVectors(\hat{p}) = \{\phi_{k}:\mathbb{C} \rightarrow \mathbb{R} | \phi_{k}(x) \mapsto e^{i k x} , k\in \mathbb{R}\}$$

In physics we say that either of these sets of eigenvectors are an uncountable basis!

The linear combinations become definite integrals covering the range of the eigenvalues:

[tex] f(x) = \int_{-\infty} ^{\infty} c_{x'} \phi_{x'}(x) dx' [/itex]

where the coefficients are given by Fourier's trick:

[tex] c_{x'} = \int_{-\infty} ^{\infty} f(x) \phi_{x'}(x) dx [/itex]

I'm interested to learn more about what's really going on here.

Well, the thing here is that neither of these "bases" contain a single element of your Hilbert space! The position and momentum operators, acting on our Hilbert space, actually do not have any nonzero eigenvectors.

There are at least two ways to make sense of this. One is via a "rigged Hilbert space", where in addition to our Hilbert space, we consider a smaller subspace of "test functions", its dual space of "generalized functions". The position and momentum operators have eigenvectors in this larger space of generalized functions. (distribution is another buzzword to look for)

Another is through a "direct integral" of Hilbert spaces, which generalizes the (finite) direct sum of Hilbert spaces in a way similar to how the ordinary integral can be viewed as generalizing finite sums of numbers. We can write our Hilbert space as a direct integral of a copy of C at each point in the real line. Now, your "basis vectors" are not elements of our Hilbert space of interest, but instead a choice of basis vectors for these individual Hilbert spaces.

 

[Hurkyl]

now i haven't read the whole thread but why again are we discussing the truth value of an axiom?

Because people aren't used to separating the notion of a mathematical theory and their favorite interpretation of that theory.