Is Banach–Tarski paradox false without axiom of choice?

Question

I know that you need axiom of choice to prove Banach–Tarski paradox. But what happens with paradox when we remove axiom of choice? Does theorem become false? Or is there just no proof of it without axiom of choice?

Answer 1

As explained in the last paragraph here, if you combine ZF set theory with the assumption that the axiom of choice is false, the Banach-Tarski paradox becomes undecidable rather than refutable. Indeed, ZF plus something weaker than AC called the ultrafilter lemma renders the BT a theorem; it doesn't need full ZFC. For more details, see the 1991 paper that proved this.

As a result, the models of ZF in which AC is false, i.e. the models of ZF$\neg$C, include some in which BT follows because the ultrafilter lemma is true, but also some (such as the Solovay model) in which the BT is false. This is why the BT is undecidable in ZF$\neg$C. (These specific examples are owed to @Reveillark.)

Answer 2

Removing an axiom from a consistent system of axioms can't make a theorem false: if statement $\phi$ follows from a consistent system of axioms $\Gamma$ and for some statement $\gamma \in \Gamma$, $\lnot\phi$ follows from $\Gamma \setminus \{\gamma\}$, then $\lnot\phi$ follows from $\Gamma$, so $\Gamma$ is inconsistent.

[I'm not saying anything against J.G.'s very helpful answer. I am just trying to point out a general logical principle (which is actually confounded in some bizare logical systems, but not ZF and ZFC, so let's not go into that just now).]