I am working on the Banach–Tarski paradox and the fact that the Hahn–Banach theorem implies that paradox. The proof involves the equivalence of the Hahn–Banach theorem and the fact that for every Boolean algebra $$\mathcal A$$ there is a finitely additive "measure" $$\mathrm m:\mathcal A \to [0,+\infty]$$ such that some element of $$\mathcal A$$ has finite measure. $$\mathsf{ZF}$$ is not enough to derive the Banach-Tarski paradox, nor is $$\mathsf{ZF}+\mathsf{DC}$$.

However, I am interested in reversing the roles, in order to characterize the Banach–Tarski paradox in terms of choice-like axioms. Hahn–Banach is strictly weaker than the axiom of choice or the ultrafilter lemma/boolean prime ideal theorem.

Can we find a weaker axiom to derive the Banach–Tarski paradox? Do you know any reference or have any ideas on this? Are there weaker axioms that should suffice in order to get the Banach–Tarski paradox? Is it possible to reformulate the Banach–Tarski paradox in a way that this assumption allows us in $$\mathsf{ZF}$$ to recover the Hahn–Banach extension theorem?

These questions do not seem clear to me, so I would be very pleased if they are somehow absurd. Thanks in advance.

• I believe the Banach-Tarski paradox can be also proved from well ordering of the reals, hence it does not imply the Hahn-Banach theorem. Mar 23 '16 at 22:27
• @EmilJeřábek: Indeed, like the construction of a Vitali set, the key step is to find an action of an appropriate countable group on $\mathbb{R}^3$, and then choose one element from each orbit. A well-ordering on $\mathbb{R}$ would certainly let you do that. Mar 24 '16 at 1:40
• Have you looked in Howard and Rubin, Consequences of the Axiom of Choice? That's where people always send me whenever I ask a question like this. Mar 24 '16 at 1:42
• No, well ordering of the reals does not imply the axiom of choice, at all. In general, a statement (such as well orderability of the reals) that only constrains a part of the set theoretic universe of bounded rank cannot imply a statement (such as AC or the H-B theorem) that can be violated arbitrarily high in the cumulative hierarchy. Meanwhile, did you check the source mentioned by Nate Eldredge? Mar 25 '16 at 14:44
• consequences.emich.edu/conseq.htm The Banach-Tarski paradox is form 309, the Hahn-Banach theorem is form 52, and plugging these numbers in tells me that BT holds and HB fails in model N51. Now, go back to the book if you want details about the model. Mar 25 '16 at 19:11

The Banach-Tarski paradox is about real numbers, so it is limited to special kind of sets. The Hahn-Banach theorem is much more general. Therefore, trying to recover the Hahn-Banach theorem, we have to reformulate the Banach-Tarski paradox in a reasonable way. Recall that the core of the paradox is the existence of a free group of rotations acting on the sphere $$\mathbb{S}^2$$ in the way that the set of fixed points can be eliminated (we can absorb the points and recover two spheres from the one). Clearly, we can generalize the paradoxical construction to arbitrary group actions on some abstract spaces. The point is that we must care about fixed points. So consider a group $$G$$ acting on a space $$X$$, we say that the set $$F$$ of fixed points of the $$G$$-action is absorbable if $$X$$ is $$G$$-equidecomposable to $$X \setminus F$$, that is $$X$$ can be partitioned into finitely many sets $$A_1,...A_k$$ and $$X \setminus F$$ can be partitioned into finitely many sets $$B_1,...,B_k$$ such that $$g_i(A_i) = B_i$$ for $$i=1,...,k$$ and $$g_i \in G$$. Thus we have the following generalization of the Banach-Tarski paradox:

If a free non-abelian group of rank two $$F_2$$ acts on a set $$X$$ in the way that the set of fixed points is absorbable, then $$X$$ is $$F_2$$-paradoxical (there is a Banach-Tarski decomposition of X via elements of the group $$F_2$$).

I think that this general setting should correspond with the existence/non-existence of measures that is used in the implication Hahn-Banach theorem $$\Rightarrow$$ Banach-Tarski paradox.