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Question: While the standard construction of the set for Banach-Tarski requires AOC, is it proven that without AOC, Banach-Tarksi isn't true?
by davrosthedalek 4y ago
Question: While the standard construction of the set for Banach-Tarski requires AOC, is it proven that without AOC, Banach-Tarksi isn't true?
- tomp 4y agoNot sure, but if you replace AC with a weaker axiom, Axiom of Dependent Choice, you can construct a model in which all subsets of reals are measurable (called the Solovay model), which precludes the Banach-Tarski paradox (which requires non-measurable sets). Could BT be constructed without AC but with other way of ensuring there exist non-measurable sets? No idea.
- throwaway81523 4y agoBy "without AC" I think you mean "with the negation of AC". It turns out that BT+ZF is equivalent to ZF + a weaker version of AC called the Ultrafilter Lemma, iirc. There is more info in the Wikipedia article about BT. Also, the Hahn-Banach theorem (plus ZF) proves BT.
- Kranar 4y agoDepends what you mean by without AOC. If you just mean whether Banach-Tarski can be proven in ZF the answer is no. But you also don't need the full power of AOC to prove Banach-Tarski.