>> ... the "two copies of the natural numbers" is sorta fine, except that they're more "spread out" ...

> What do you mean "spread out"? Aren't there the same amount of even numbers as natural numbers?

Yes, there are the same number, but when you look at just the even numbers, they are each distance 2 from their neighbours, whereas the natural numbers are all distance 1 from their neighbours. So people are less surprised, because the even numbers are "spread out", they are less dense in any given area. To map the even numbers back onto the natural numbers you have to "compress" them.

But this is not the case with the Banach-Tarski Theorem. There is a set, A, and another set B, which is just A rotated around, and they are disjoint. So they have a union, C=AuB. But when you rotate C, you can get an exact copy of A. There's no squashing or spreading needed.

So we have A and B, with B=r(A), and A intersect B is empty. Then we have C=AuB. No problem here.

The challenge comes that there is a rotation, s, such that s(C)=A.

So even though C is made up of two copies of A, it's actually identical to A. So start with C, divide it into A and B, then rotate B back to become a copy of A, and then rotate each of those to become copies of C. So you start with C, do some "cutting" and rotations, and you get two copies of C.

Finally, when you take a few of these and put them together, you get a full sphere, so you can't say they have zero volume.

Does that make sense? Does that answer your question?

Does that help?