What's your opinions on the Axiom of Choice?

1. Mar 16, 2008

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.

Last edited: Mar 16, 2008
2. Mar 16, 2008

ice109

i dont have any problem with the countable axiom of choice. the uncountable variant is a little bit harder to accept though

3. Mar 16, 2008

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.

4. Mar 16, 2008

nicksauce

I am pro-choice.

5. Mar 16, 2008

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."

Last edited: Mar 16, 2008
6. Mar 16, 2008

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.

7. Mar 16, 2008

Hurkyl

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

8. Mar 16, 2008

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.

9. Mar 16, 2008

mathboy

Yes, get rid of the axiom of choice.

10. Mar 16, 2008

Dragonfall

What theorems of measure theory depend on AC?

11. Mar 16, 2008

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.

12. Mar 16, 2008

Hurkyl

Staff Emeritus
One notorious example is:

Thereom: There exists a nonmeasurable subset of the reals.

13. Mar 16, 2008

John Creighto

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

14. Mar 16, 2008

andytoh

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.

15. Mar 16, 2008

John Creighto

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

Last edited: Mar 16, 2008
16. Mar 16, 2008

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.

17. Mar 17, 2008

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?

18. Mar 17, 2008

Hurkyl

Staff Emeritus
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

19. Mar 17, 2008

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.

20. Mar 17, 2008

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.

Last edited: Mar 17, 2008