2 ms·
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
by bcoates 14y ago
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.
- gjm11 14y agoNo, 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.]