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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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