Anti-anti Banach-Tarski arguments

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.