The Banach-Tarski paradox states that for a solid ball in 3‑dimensional space, there exists a decomposition into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original one.

Obviously it is based on AC. I was wondering if anyone here knew if analysis under the axioms of ZF has been developed to invent a version of Banach-Tarski which is independent of AC. What does the Banach-Tarski paradox look like without AC? Are there any versions of it? (For an example, one of the theorems that has been proven without AC is the Heine-Borel theorem.)

  • 5
    $\begingroup$ Depends on what you mean exactly. There are models of ZF in which every set of real numbers is measurable (en.wikipedia.org/wiki/Solovay_model), which would prevent a lot of the paradoxes you might have in mind. This question is probably too basic for MO, though. $\endgroup$ Aug 20, 2021 at 14:18
  • 1
    $\begingroup$ To add on to what @Sam wrote; many theorems in analysis either don't use choice, or their specific uses in classical analysis don't use choice (e.g. Baire Category Theorem is equivalent to Dependent Choice; but for separable spaces it is provable in ZF). Asking about development of the whole of analysis in ZF or ZF+DC is tantamount to going through Rudin (or some other book on analysis) and checking each theorem to see what you can or cannot do wit or without choice. Some places do that, to some extent (e.g. Schechter's book), but generally it's too broad of a question for a Q&A website. $\endgroup$
    – Asaf Karagila
    Aug 20, 2021 at 15:23
  • $\begingroup$ "which is independent of AC": I guess you mean "which is a theorem of ZF". The formulation that P is independent of AC might (?) mean that both ZFC+P and ZFC+(not P) are consistent. $\endgroup$
    – YCor
    Aug 20, 2021 at 17:45
  • 3
    $\begingroup$ In light of the very nice answer of Paul, I retract my insinuation that this question might be too basic. But I will leave the link to the Solovay model, which is still certainly relevant for some considerations... $\endgroup$ Aug 20, 2021 at 22:03
  • 1
    $\begingroup$ The fact that there is no finitely-additive SO(3)-invariant measure on S2 with the discrete σ-algebra has been proven in ZF + Hahn-Banach by Foreman and Wehrung: hal.archives-ouvertes.fr/hal-00004713 Their argument transports the lack of invariant measure from F2 to S2. It can then be used to transport F2's explicit paradoxical decomposition across: matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13813.pdf $\endgroup$ Aug 21, 2021 at 2:11

1 Answer 1


According to Dougherty and Foreman's 1992 PNAS paper, Banach-Tarski paradox using pieces with the property of Baire (doi:10.1073/pnas.89.22.10726), the following result can be shown without AC: the unit ball has a finite collection of disjoint open subsets that transforms by rigid motions to another collection of disjoint open sets such that the closure of the union is two unit balls.

  • 3
    $\begingroup$ The "constructive" adjective most often includes not using the law of excluded middle. So "classically constructive" would be a more precise adjective in this case. (Uses of "constructive" are generally relative to the author and authors with more a restrictive notion of "constructive" tend to use it more often, so this is a touchy subject. On MO, and also in more general settings,a search often leads to "most often used in this context", so you can see this is sometimes problematic regardless of what you or anyone else thinks is most appropriate.) [FWIW: I am not a constructivist.] $\endgroup$ Aug 20, 2021 at 19:50
  • 8
    $\begingroup$ The given citation is to the PNAS paper, and a fuller description is given (by the same authors) in: ams.org/journals/jams/1994-07-01/S0894-0347-1994-1227475-8/… one cannot explain away their paradoxical nature by blaming it on [AC]. Instead, one can note that the open sets resulting from our construction have boundaries of positive measure. When the open sets are packed into a small space, these boundaries overlap greatly; when the open sets are rearranged within a larger set, the boundary overlap is less extreme, so the total measure of the closures is greater. $\endgroup$
    – user334725
    Aug 20, 2021 at 20:38
  • 5
    $\begingroup$ Reading again, I think Feferman was right and it really is constructive, but a more careful eye is needed than mine. Some versions of constructivism have multiple not-provably-isomorphic versions of $\mathbb{R}$, so the question may be subtle. $\endgroup$ Aug 20, 2021 at 21:36
  • 7
    $\begingroup$ @EsaPulkkinen What is wrong with the elementary definition of the closure of a set as the intersection of all closed sets containing it? $\endgroup$ Aug 21, 2021 at 13:06
  • 4
    $\begingroup$ @Esa yes, they exist by the definition of a topology (the open sets are the data of a topology, the closed sets are their complements). No AC is needed. The new points in the closure are already elements of the given space. It seems you may be thinking of a completion, which is not what is happening here. $\endgroup$ Aug 22, 2021 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.