Asaf Karagila
I don't have much choice...

Many people, more often than not these are people from analysis or worse (read: physicists, which in general are not bad, but I am bothered when they think they have a say in how theoretical mathematics should be done), pseudo-mathematical, non-mathematical, philosophical communities, and from time to time actual mathematicians, would say ridiculous things like "We need to omit the axiom of choice, and keep only Dependent Choice, since the axiom of choice is a source for constant bookkeeping in the form of non-measurable sets".

People often like to cite the paradoxical decomposition of the unit sphere given by Banach-Tarski. "Yes, it doesn't make any sense, therefore the axiom of choice needs to be omitted".

To those people I say that they know too little. The axiom of choice is not at fault here. The axiom of infinity is. Infinite objects are weird. Period. End of discussion.

Don't believe me? Here's my favorite rebuttal:

Theorem (ZF+DC). Suppose that all sets of reals are Lebesgue measurable, then there is a partition of the real line into strictly more parts than elements.

Proof. If $\aleph_1\leq2^{\aleph_0}$ then there is a non-measurable set. Therefore $\aleph_1\nleq2^{\aleph_0}$. However there is a definable surjection from $\Bbb R$ onto $\omega_1$:

Fix a bijection between $e\colon(0,1)\to \Bbb (0,1)^\omega$, if $r$ is a real number such that $e(r)$ is a well-ordered set (under the natural order of the real numbers) then map $r$ to the order type of $e(r)$. Otherwise map it to $0$. Easily we can see that this is a surjection onto $\omega_1$.

Consider the partition induced by considering the singletons in $\Bbb R\setminus(0,1)$ and the preimages of each ordinal from the surjection above. This has $2^{\aleph_0}+\aleph_1$ equivalence classes. But since $2^{\aleph_0}$ and $\aleph_1$ are incomparable as cardinals, this is a strictly larger partition. $\square$

We can do other crazy partitions too. It all depends on how much you are willing to work, and how much more you are willing to assume.

How is this not a paradoxical result? More parts than elements, all of which are non-empty? Is this not worse than the Banach-Tarski paradox, or at least comparably horrible? In fact, just the fact we can partition $\Bbb R$ into $\aleph_1$ parts, which is a number of parts incomparable with the number of elements should be alarming.

Many people will disregard that, but this act of disregarding this sort of paradox is exactly what we do when we restrict ourselves to Borel sets, or Lebesgue measurable sets. We disregard the part that bothers us. And the axiom of choice has been so good to us in so many ways, that discarding it only for the sake of not having to cope with the Banach-Tarski paradox is plain stupid.

### There are 11 comments on this post.

By
(Sep 22 2014, 16:06)

Hah! I didn't know that one. Great example. I agree with the sentiment that infinity is a (possibly more) important culprit that people (TM) tend to gloss over.

Btw I'm really enjoying these op ed pieces of yours. Keep'em coming!

By
(Sep 22 2014, 20:47 In reply to Peter Krautzberger)

Yeah, it's not the axiom of choice. Since rejecting it causes all sort of crazy things to happen too. The only way to ensure nothing crazy happen is to reject infinity altogether, or reject the power set axiom, or reject the law of excluded middle. In all cases, however, you find yourself wishing that you haven't... :-)

By the way, I thought about you when I wrote those last two posts. I figured you'd like them. :-)

By
(Sep 22 2014, 23:01)

Warning: another Banach-Tarski joke.

By
(Sep 23 2014, 05:19 In reply to Joao Marcos)

I love SMBC, but I don't see how this is a BT joke.

By Yemon Choi
(Sep 23 2014, 05:23)

I don't know which analysts you've encountered with this attitude, but I for one want the Hahn-Banach theorem (separation form)...

By
(Sep 23 2014, 06:32 In reply to Yemon Choi)

I certainly agree that Hahn-Banach might be an excellent rebuttal. But someone willing to reject choice might often be less interested in hearing about its benefits like Hahn-Banach. If you don't expect the axiom of choice to be true, then it's possible that you don't expect to be able and separate things as easily, or at all.

Yes. I agree very much that a utilitarian approach is the right approach here. But there are people objecting on philosophical merits that the Banach-Tarski paradox is "proof" that the axiom of choice is "wrong". And this here is my favorite rebuttal about paradoxical decompositions.

By A
(Aug 30 2015, 21:48)

Could you be more verbose on the first sentence in your proof, or provide a reference? Probably it is sort of obvious to somebody in this area but unfortunately not to me.

By Asaf Karagila
(Aug 30 2015, 21:59 In reply to A)

This is a very nontrivial theorem by Shelah. You can find the proof in Section 5 in Saharon Shelah, "Can you take Solovay’s inaccessible away?", Israel J. Math. 48 (1984), no. 1, 1--47. More specifically this is spelled out as Theorem 5.1B at the bottom of the first page of the section.

By
(Oct 21 2015, 01:10)

[…] had already written about anti-anti-Banach-Tarski arguments. But now the Mathematical T-Rex has something to say […]

By B
(Nov 21 2015, 18:15)

This is very nice! I have a question: does the existence of a non-measurable set automatically imply Banach-Tarski? Could still have ℵ_1 ≤ 2^{ℵ_0} (or some other condition that rules out the existence of this kind of partitions) and no paradoxical decompositions?

By Asaf Karagila
(Nov 21 2015, 22:24 In reply to B)

Well, I can't recall any specific result along these lines. But I'd be very surprised if that was the case (that BT is equivalent to having non-measurable sets).

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