What's your opinions on the Axiom of Choice?

[Ka Yan]

I, by the first time, came across with the Axiom of Choice today, found it beautiful, of course.
And I'm interested in seeing, ladies and gentlemen here, as mathematicians, what are your attitude towards that axiom, I mean, the stronger one (i.e. the infinite axiom of choice).
For example, do you think the axiom itself perfect in logical, but will sometimes make the proof loggically more satisfactory but more complicate? Or how do you compare it with the Principle of Induction?

I hope my statement to the discussion was clear enough.
And I'd love to hear your sounds.

 

[ice109]

i don't have any problem with the countable axiom of choice. the uncountable variant is a little bit harder to accept though

 

[mathboy]

Without AC, many, many useful theorems go out the window, even very innocent-looking theorems are gone. For example, every surjective map between infinite sets has a right-inverse requires AC, just to give you an idea of how even the simplest theorems depend on AC. When you study other math subjects, you will find it very useful as well. However, there are many strange results that arise from AC. Wikipedia gives some good examples. As a result some mathematicians refuse to accept AC, but I believe it is less than 1% of all mathematicians that do. It would be interesting to see a poll for that.

 

[nicksauce]

I am pro-choice.

 

[mathboy]

I'm for AC myself. However, I believe Bertrand Russell adamently stated that "there is no reason to believe in the Axiom of Choice."

 

[Dragonfall]

There's no reason to believe any of the axioms anywhere. Accepting AC or not depends on how much "power" you want in your theory over infinities.

 

[Hurkyl]

ZF+C is generally a simpler theory than ZF-C, letting you say more interesting things about ZF+C.

 

[mathman]

Is it possible to set up a system where the Banach-Tarksi theorem (paradox) is not true? In other words a system where a sphere cannot be divided into a finite number of sets, have them go through translations and rotations, and be put together into a sphere of a different size.

 

[mathboy]

Yes, get rid of the axiom of choice.

 

[Dragonfall]

What theorems of measure theory depend on AC?

 

[CRGreathouse]

Frankly, I'm suspicious of AC. Although I can't entirely avoid using it, I prefer to know when I'm using it if possible, and would prefer to see theorems proved without it if it's not needed.

 

[Hurkyl]

What theorems of measure theory depend on AC?

One notorious example is:

Thereom: There exists a nonmeasurable subset of the reals.

 

[John Creighto]

Frankly, I'm suspicious of AC. Although I can't entirely avoid using it, I prefer to know when I'm using it if possible, and would prefer to see theorems proved without it if it's not needed.

It's difficult though isn't it as it is equivalent to mathematical induction.

 

[andytoh]

It's difficult though isn't it as it is equivalent to mathematical induction.

I'm trying to decipher your statement. If you just said that mathematical induction depends on the axiom of choice, then you are mistaken. Even transfinite induction is independent of the axiom of choice. Transfinite induction only needs a well-ordered set. You only need AC when using transfinite induction if you need cannot construct a well-ordering of your set explicitly, and thus need to invoke the well-ordering theorem to well-order your set.

 

[John Creighto]

I'm trying to decipher your statement. If you just said that mathematical induction depends on the axiom of choice, then you are mistaken. Even transfinite induction is independent of the axiom of choice. Transfinite induction only needs a well-ordered set. You only need AC when using transfinite induction if you need cannot construct a well-ordering of your set explicitly, and thus need to invoke the well-ordering theorem to well-order your set.

I remember something about an equivalence but maybe there is some limits to the equivalence . Here is what I can find at the moment:

"It is to be noted that AC1 and CAC for finite collections of sets are both provable (by induction) in the usual set theories. But in the case of an infinite collection, even when each of its members is finite, the question of the existence of a choice function or a transversal is problematic[4]. For example, as already mentioned, it is easy to come up with a choice function for the collection of pairs of real numbers (simply choose the smaller element of each pair). But it is by no means obvious how to produce a choice function for the collection of pairs of arbitrary sets of real numbers."
http://plato.stanford.edu/entries/axiom-choice/
http://mathforum.org/library/drmath/view/55696.html

 

[andytoh]

No one can produce a choice function for the collection of all nonempty subsets of the reals. As Hurkyl mentioned, even taking the measure of a subset of R fails because some subsets of R are nonmeasurable. For this collection of sets, the axiom of choice is needed.

 

[Ka Yan]

I overheard there were some paradox which brought troubles of AC and even Set Theory to mathematicians, I wonder what was them, what troubles did they lead to, have mathematicians solved them already, how, or why not?

 

[Hurkyl]

I overheard there were some paradox which brought troubles of AC and even Set Theory to mathematicians, I wonder what was them, what troubles did they lead to, have mathematicians solved them already, how, or why not?

When Cantor made the first¹ attempt to formalize set theory, it led to Russell's paradox, and an assortment of other paradoxes. But Zermelo postulated his set theory a hundred years ago, so those things haven't been a foundational problem for a long time.

1: At least, I think it was the first

 

[Diffy]

From wiki

"The difficulty appears when there is no natural choice of elements from each set. If we cannot make explicit choices, how do we know that our set exists? For example, suppose that X is the set of all non-empty subsets of the real numbers. First we might try to proceed as if X were finite. If we try to choose an element from each set, then, because X is infinite, our choice procedure will never come to an end, and consequently, we will never be able to produce a choice function for all of X. So that won't work. Next we might try the trick of specifying the least element from each set. But some subsets of the real numbers don't have least elements. For example, the open interval (0,1) does not have a least element: If x is in (0,1), then so is x/2, and x/2 is always strictly smaller than x. So taking least elements doesn't work, either."


I was thinking about the example, and thought, well for each set, you could take the least upper bounds and the greatest lower bounds, average them and that could you your choice for each set within X... Wouldn't that work? I understand the the least value wouldn't work as a choice function but one can still be defined on an infinite set of real intervals... right?


Can anyone provide a better example of an infinite set that a choice function can not be defined.

 

[andytoh]

You choice function does not work for so many subsets of R. e.g. R itself, Q, Z, etc... Even if every set had a lub and glb, how do you know their average would be in the set.

 

[eastside00_99]

When Cantor made the first¹ attempt to formalize set theory, it led to Russell's paradox, and an assortment of other paradoxes. But Zermelo postulated his set theory a hundred years ago, so those things haven't been a foundational problem for a long time.

1: At least, I think it was the first

I guess you could see Russell's paradox as responding to the formalization of set theory by Cantor but the history goes like this:

Frege sent a copy of Foundation of Arithmetic (yet to be published) to Russell and Russel saw the paradox and wrote back to Frege. Frege acknowledged the paradox and published his book with an amendment stating the paradox and that as of yet there was no way to resolve it.

To me AC is only troublesome if you believe in the correspondence theory of truth in the since that you think that mathematical objects exist in other world than the physical universe. I believe in the coherence theory of truth--i.e., whether something is true or not depends on what system you are working in. By this criterion, the question is (and I believe it is still an open question) whether or not there is a formal system (as "simple" as ZF) which gives you AC. If you believe in the correspondence theory, then I think I have heard that computer scientist believe the the AC false because of some of the work they are doing. I, personally, am not that hung up on the concept of reality.

 

[mathman]

Yes, get rid of the axiom of choice.

If you do that the Banach-Tarski paradox is unprovable. I would like to see what axiom would be needed to prove it to be false. Obviously the axiom could be simply stated that the Banach-Tarski paradox is false, but there should be a more fundamental statement.

 

[Hurkyl]

If you do that the Banach-Tarski paradox is unprovable. I would like to see what axiom would be needed to prove it to be false. Obviously the axiom could be simply stated that the Banach-Tarski paradox is false, but there should be a more fundamental statement.

Why? "Fundamentality" isn't an inherent quality of any logical statement. If your sole intent of adopting a particular negation of the axiom of choice is specifically so that all subsets of R will be measurable, then the most straightforward approach is to adopt that as an axiom.

 

[Diffy]

You choice function does not work for so many subsets of R. e.g. R itself, Q, Z, etc... Even if every set had a lub and glb, how do you know their average would be in the set.

Ah I see, I was assuming that it was the set of all real intervals. But rather it was all subsets of the reals which don't have to be intervals..

 

[mathman]

Why? "Fundamentality" isn't an inherent quality of any logical statement. If your sole intent of adopting a particular negation of the axiom of choice is specifically so that all subsets of R will be measurable, then the most straightforward approach is to adopt that as an axiom.

Can this "axiom" (all subsets of R are measurable) be shown to be consistent with ZF? Otherwise its no good.

 

[Hurkyl]

Can this "axiom" (all subsets of R are measurable) be shown to be consistent with ZF? Otherwise its no good.

Yes. (At least, I'm pretty sure that I've read that it's possible)

 

[mathman]

Yes. (At least, I'm pretty sure that I've read that it's possible)

Any reference?

 

[morphism]

Any reference?

R. Solovay, A model of set theory in which every set of reals is Lebesgue measurable, Ann. of. Math. 92 (1970), 1–56.

Solovay's construction also proves that {every subset of R is Lebesgue measurable} is consistent with ZFCC (countable choice).

[Note: He assumes the existence of an inaccessible cardinal; more info:
S. Shelah, Can you take Solovay's inaccessible away?, Israel J. Math. 48 (1984), 1-47.]

 

[CRGreathouse]

[Note: He assumes the existence of an inaccessible cardinal; more info:
S. Shelah, Can you take Solovay's inaccessible away?, Israel J. Math. 48 (1984), 1-47.]

So you're saying the result is Cons(ZF + "there exists an inaccessible cardinal") = Cons(ZF + "all subsets of \mathbb{R} are measurable")?

 

[George Jones]

Platonist Roger Penrose writes

However, there are other kinds of mathematical assertion whose truth could plausibly be regarded as a 'matter of opinion'. perhaps the best known of such assertions is the axiom of choice. ... Most mathematicians would probably regard the axiom of choice as 'obviously true', while others might regard it as a somewhat questionable assertion which might even be false (and I am myself inclined, to some extent, towards this second viewpoint). Still others would take it as an assertion whose 'truth' is a mere matter of opinion or, rather, as something which can be taken one way or the other, depending on which systems of axioms and rules of procedure (a 'formal system') one choose to adhere to. Mathematicians who support this final viewpoint (but who accept the objectivity of the truth of relatively clear-cut mathematical statements, like the Fermat assertion discussed above) would be relatively weak Platonists. Those who adhere to the truth of the axiom of choice would be stronger Platonists. ...

If the axiom of choice can be settled one way or the other by some appropriate of unassailable of mathematical reasoning,⁷ then its truth is indeed an entirely objective matter, and either it belongs to the Platonic world or its negation does, in the sense that I am interpreting this term 'Platonic world'. If the axiom of choice is, on the other hand, a mere matter of opinion or of arbitrary decision, then the Platonic world of absolute mathematical forms contains neither the axiom of choice nor its negation (although it could contain assertions of the form 'such-and-such follows from the axiom of choice' or 'the axiom of choice is a theorem according to the rule of such-and-such mathematical system').

The mathematical assertion that can belong to Plato's world are precisely those that are objectively true. Indeed, I would regard mathematical objectively as really what mathematical Platonism is all about. To say that some mathematical assertion has a Platonic existence is merely to say that it is true in an objective sense. A similar comment applies to mathematical notions - such as the concept of the number 7, for example, or the rule of multiplication of integers, or the idea that some set contains infinitely many elements - all of which have a Platonic existence because they are objective notions. To my way of thinking, Platonic existence is simply a matter of objectivity and, accordingly, should certainly not be viewed as something 'mystical' or 'unscientific', despite the fact that some people regard it that way.

⁷ ... It should be made clear that the Godel-Cohen argument does not in itself establish that the axiom of choice will never be settled in one way or the other. This kind of point is stressed in the final section of Paul Cohen's book ...

 

[mathboy]

Geoge Jones, seeing that you are a physicist. Do you think it is possible that a physics experiment can be conjured up to test the validity of the axiom of choice? It's not impossible for a physics experiment to test out the infinities, is it? It's possible that no physicist have done it simply they were not interested (or don't know the axiom of choice), right?

 

[George Jones]

Geoge Jones, seeing that you are a physicist. Do you think it is possible that a physics experiment can be conjured up to test the validity of the axiom of choice?

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

It's not impossible for a physics experiment to test out the infinities, is it?

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

It's possible that no physicist have done it simply they were not interested (or don't know the axiom of choice), right?

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.

 

[gel]

To me AC is only troublesome if you believe in the correspondence theory of truth in the since that you think that mathematical objects exist in other world than the physical universe. I believe in the coherence theory of truth--i.e., whether something is true or not depends on what system you are working in. By this criterion, the question is (and I believe it is still an open question) whether or not there is a formal system (as "simple" as ZF) which gives you AC. If you believe in the correspondence theory, then I think I have heard that computer scientist believe the the AC false because of some of the work they are doing. I, personally, am not that hung up on the concept of reality.

I would agree with that. I don't think AC is either right or wrong. If you accept it then you are just working with a slightly different concept of a set (or functions between them) than if you don't.
I certainly use AC simply because it is so useful, but if a result can be proven without it then that is preferable as it makes it a stronger result.

If you want to throw out AC because it asserts the existence of things that can't be explicitly constructed -- such as a choice function for nonempty subsets of R -- then I think to be consistent with that view you should also throw out the law of the excluded middle (i.e. use constructive/intuitionistic logic).

 

[arildno]

Geoge Jones, seeing that you are a physicist. Do you think it is possible that a physics experiment can be conjured up to test the validity of the axiom of choice? It's not impossible for a physics experiment to test out the infinities, is it? It's possible that no physicist have done it simply they were not interested (or don't know the axiom of choice), right?

Hmm..a physicist might perhaps be interested in the SUITABILITY of some mathematical axiom to his "optimal" mathematical model of the universe, meaning a model whose axioms best mimick the observed properties of the universe.

But to test the "validity" of such an axiom?

That doesn't make sense at all...

 

[mathboy]

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

 

[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.