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 ...