Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it possible to prove the infamous Banach-Tarski theorem without using the Axiom of Choice?

I have never seen a proof which refutes this claim.

share|improve this question
add comment

3 Answers

up vote 14 down vote accepted

The Banach-Tarski theorem heavily uses non-measurable sets. It is consistent that without the axiom of choice all sets are measurable and therefore the theorem fails in such universe. The paradox, therefore, relies on this axiom.

It is worth noting, though, that the Hahn-Banach theorem is enough to prove it, and there is no need for the full power of the axiom of choice.

More information can be found through here:

  1. Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.

  2. Schechter, E. Handbook of Analysis and Its Foundations. Academic Press, 1997.

share|improve this answer
1  
+1 for mentioning Herrlich--lovely little book! –  trb456 Jul 28 '12 at 21:48
    
@trb456: I agree very much! –  Asaf Karagila Jul 28 '12 at 21:59
add comment

Directly from Wikipedia's page on the Paradox/Theorem

Unlike most theorems in geometry, this result depends in a critical way on the axiom of choice in set theory. This axiom allows for the construction of nonmeasurable sets, collections of points that do not have a volume in the ordinary sense and for their construction would require performing an uncountably infinite number of choices.

share|improve this answer
1  
More precisely, it's consistent with ZF that there do not exist nonmeasurable subsets of $\mathbb{R}$ (see en.wikipedia.org/wiki/Solovay_model). –  Qiaochu Yuan Jun 9 '12 at 20:10
    
@QiaochuYuan Feel free to make my copy-paste answer into a good one! –  Pedro Tamaroff Jun 9 '12 at 20:11
    
From that same page: "So Pawlikowski proved that the set theory needed to prove the Banach–Tarski paradox, while stronger than ZF, is weaker than full ZFC." –  WimC Jun 9 '12 at 20:16
    
@QiaochuYuan: It is consistent if ZF plus the existence of a inaccessible cardinal is consistent. –  Michael Greinecker Jun 9 '12 at 21:20
    
@MichaelGreinecker: You can always consider the Feferman-Levy model in which all sets are Borel, and analysis fails so bad that even finitists may cry havoc... –  Asaf Karagila Jun 9 '12 at 21:38
add comment

While I'm aware you did not ask for this I cannot resist to suggest that you have a look at Stan Wagon's book 'The Banach-Tarski Paradox', Cambridge University Press 1985.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.