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The drawing on the page is confusing. It depicts the "pieces" as contiguous solid chunks--but in order to reassemble the sphere into two, don't you have to pick sets which are more like general predicates on the points of R^3? Those don't look like puzzle pieces.

It's like depicting the irrational numbers by coloring everything on a number line between pi and e blue.

Is it possible to do the cut-up-and-reassembly trick with a finite number of sets where the membership is computable? If so the "Banach-Tarski Algorithm" for the special case of duplicating the sphere might be helpful for understanding.

No, you can't do it with computable sets, because if you could then there'd be a proof that doesn't need the Axiom of Choice, but Banach-Tarski can fail without the Axiom of Choice. (For instance, without AC it can happen that all subsets of R^n are Lebesgue measurable, and then Banach-Tarski is impossible.)

[EDITED to add: er, actually in order to prove that "without AC it can happen that all subsets of R^n are Lebesgue measurable" you need there to be an inaccessible cardinal. If you don't know what that means, ignore it. I'm mentioning it just to avoid leaving a false statement in the record.]

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