Let the base field be the real numbers or the complex numbers (I don't think it will matter).

Let $(\ell^{\infty})'$ be the continuous dual of the Banach space $\ell^{\infty}$.
Let $\: f : \ell^1 \to (\ell^{\infty})' \:$ be the obvious embedding.

Does ZF+AD prove that $f$ is surjective?
Does ZF+DC+AD prove that $f$ is surjective?

  • $\begingroup$ A related question on MO (but it doesn't answer this question as far as I know): mathoverflow.net/questions/5351/… $\endgroup$ Jan 29, 2012 at 3:32
  • $\begingroup$ @Asaf: That, on the other hand, is trivial, since AC is consistent with DC and contradicts AD. $\;$ $\endgroup$
    – user57159
    Jan 29, 2012 at 7:51
  • $\begingroup$ @Ricky: That is not trivial either, DC is much weaker than AC and I always thought that AD implied DC. The proof is not even presented in Kanamori's The Higher Infinite, he only quotes the theorems "If $V=L(\mathbb R)$ then AD implies DC and there is a forcing extension in which there is an inner model in which AD holds but Axiom of countable choice doesn't." and since DC implies ACC... $\endgroup$
    – Asaf Karagila
    Jan 29, 2012 at 7:58
  • $\begingroup$ @Asaf: For every model M of ZF+DC, M's version of L is a model of ZFC, and so in particular is a model of DC but not AD. Therefore AD is independent from DC. (Although thanks for giving that reference.) $\;$ $\endgroup$
    – user57159
    Jan 29, 2012 at 8:04
  • 1
    $\begingroup$ @Ricky: I was in Jerusalem today attending a course by Menachem Magidor about descriptive set theory. I asked him this question during the break, he couldn't give me an answer, and said he'll think about it. As luck would have it, I'm seeing him on Friday as well. If he won't have an answer by then, I'm going to bet that there isn't a written one (I did try and search for it in several places). If you are not in a rush, by the end of the academic year I should be capable to attempt this proof on my own. $\endgroup$
    – Asaf Karagila
    Jan 30, 2012 at 20:35

1 Answer 1


By a random chance, in this answer t.b. posted an answer which deals with models of $\mathrm{ZF+DC+PM_\omega}$. The latter in fact states that $\ell^1$ is reflexive.

Martin Väth, The dual space of $L^\infty$ is $L^1$, Indag. Mathem., N.S., 9 (4), 1998, 619–625.

It is mentioned that the axiom $\mathrm{PM_\omega}$ holds in Solovay's model. The paper itself cites both Solovay's original paper as well a paper by David Pincus which I was not able to find online (MR link).

Assuming $\mathrm{AD}$ holds in $L(\mathbb R)$ implies it is indeed a Solovay model, so the above should be applicable (since $\mathrm{AD}+V=L(\mathbb R)$ implies $\mathrm{DC}$).

Lastly, one of the remarks was that it is the fact that every set of real numbers has the Baire property which implies $\mathrm{PM}_\omega$, this was later shown consistent without an inaccessible cardinal, by Shelah.

Saharon Shelah, Can you take Solovay's inaccessible away?, Israel Journal of Mathematics 48 (1): 1–47.

While the above answers perfectly the second question, it can be done in a clearer way via Fremlin's Measure Theory, and in particular Vol. 5, Ch. 6 whose last section deals with $\mathrm{ZF+AD}$.

It is not clear to me whether or not the arguments brought in that chapter are sufficient to answer the first question positively, though.

(the results themselves are available in a .ps file which you can convert to .pdf here)

  • $\begingroup$ Fremlin Measure theory, 567K proves that ZF+AD+AC$(\omega)$ implies that all $L$-spaces with a weak order unit are reflexive (567K, Volume 5II, page 245). This fully answers the second question. Moreover, there's the result (567J) that AD alone implies that a finitely additive functional on a $\sigma$-complete Boolean algebra is countably additive. If it is possible to identify the dual of $\ell^\infty$ with the finitely additive functionals on $\mathcal{P}(\mathbb{N})$ without any choice, then a positive answer to the first question should be reachable by the results in that chapter. $\endgroup$
    – t.b.
    Feb 21, 2012 at 12:58
  • $\begingroup$ (Sorry it took so long for me to notice this answer. $\:$ I was busy.) $\:$ In the absence of Hahn-Banach, how is reflexivity defined? $\:$ Is it that there is an isomorphism or that the canonical map is a isomorphism? $\hspace{.5 in}$ Are isomorphisms specified to be homeomorphic or isometric? $\;\;$ $\endgroup$
    – user57159
    Feb 23, 2012 at 6:45
  • $\begingroup$ @Ricky: He $V$ to be reflexive if every element of $V^{\ast\ast}$ is an evaluation functional (that is the canonical map is surjective). Hahn-Banach simply implies it is an isometry. $\endgroup$
    – Asaf Karagila
    Feb 23, 2012 at 18:24
  • $\begingroup$ this should be a comment to Asaf's answer but I don't have the privilege to comment. @Ricky: It is always the definition of reflexivity that the canonical map $X \to X^{\ast\ast}$ given by evaluation is an isomorphism. Assuming Hahn-Banach and in many explicit examples, this reduces to checking surjectivity. But even with full choice having just some isomorphism is not good enogh. Look up the James space, which is isometrically isomorphic to its bidual but not reflexive. $\endgroup$
    – user25669
    Feb 24, 2012 at 18:40

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