Friday, June 14, 2013

The Banach-Tarski Paradox and the Axiom of Choice

The Banach-Tarski theorem (BTT) says that, given the Axiom of Choice, a continuous ball can be decomposed into a finite number of pieces that can be rearranged to form two balls of equal size. That's weird, and is taken by some to be an argument against the Axiom of Choice.

I don't think we should take it as such an argument. Sure, BTT is paradoxical. But when one looks at the proof, one notes that the proof makes use of paradoxical results that do not depend on the Axiom of Choice. For instance, a lemma in standard proofs of BTT is the surprising fact that you can take any circle that's missing a countable number of points, decompose that circle into two disjoint (messy!) pieces, and reassemble the pieces, without overlap, to get a complete circle. That lemma is about as weird as BTT, but it doesn't use the Axiom of Choice at all.[note 1] Moreover, the proof uses paradoxical decompositions of various countable sets, and these are, well, paradoxical, but do not involve the Axiom of Choice. Using the Axiom of Choice lets you put all the paradoxicality together into a neat package, but when I think about the proof of the result, I just don't see Choice as the source of paradoxicality. In fact, once one sees all the other ingredients of the proof of BTT, the Axiom of Choice step seems quite intuitive.

Another way to put the point is this: Once one reflects enough on all the pieces of the proof of BTT that do not use Choice and accepts them, BTT no longer seems very surprising. When you cut things up into strangely scattered pieces, it's not that surprising that you can put them back together in various ways.


grodrigues said...

You probably know all this, but just in case, by relatively recent results of Pawlikowski (1990's), Banach-Tarski needs considerably less than full AC, the Hahn-Banach theorem is enough. HB is weaker than the Boolean Prime Ideal theorem (BPI = existence of free ultrafilters, equivalent to the axiom of choice for families of *finite sets*), itself weaker than AC, and if memory serves me right, it is independent of (countable) dependent choice.

The Rubin & Rubin monograph (or their website) is the go-to place to settle such matters.

Alexander R Pruss said...

I think I came across that. But a lot of these AC-ish results have no intuitive plausibility over and beyond AC. The only reason I have to think BPI or HB are true is that I have reason to think AC is true--because it just seems true--and AC entails them. If I thought AC were false, I would have no reason to think BPI and HB are true.

Dependent Choice (DC), and Choice for families of two-element sets (AC2), may be different--they may have intuitive plausibility over and beyond that of full AC.

Paul Rimmer said...

I liked Feynman's commentary on this paradox. A topologist showed him the proof, and he agreed with it, saying something along the lines of "It's fine you can do it with 'continuous spheres', since there's no such thing. The important thing is you can't do it with oranges, because oranges are made of a finite number of indivisible parts."