Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, according to Woodin in $ZF+DC+AD_{\mathbb{R}}$ low uncountable cardinalities form a nice lattice, and according to Caicedo and Ketchersid in $ZF+AD^++V=L\big(\mathcal{P}(\mathbb{R})\big)$ if $S$ is strictly larger than a well-ordered cardinal $\kappa$, then either $\kappa^+$ or $\kappa\cup\mathbb{R}$ embeds into $S$, and "$\omega_1$ and $\mathbb{R}$ form a basis for the uncountable cardinals" (not sure what basis means in this context).

What other axioms/ZF models are known to have some taming effect on the ordering of cardinalities? Do weaker forms of choice like "every set is linearly orderable" have such effect? References appreciated.

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    $\begingroup$ I wouldn't dare calling $\sf AD$ anything close to "taming" the cardinals. Even the abstract of Woodin's paper points out that the cardinals become very complicated in that model. $\endgroup$
    – Asaf Karagila
    Jun 20, 2014 at 21:01
  • $\begingroup$ Somewhat related: is it known whether statements like "the cardinalities form a complete lattice" are consistent with ZF + $\neg$AC, or even better, ZF + AD? $\endgroup$ Jun 20, 2014 at 21:51
  • $\begingroup$ @Noah: I don't think there's too much known about the structure of the cardinals in $\sf ZF+\lnot AC$. Mostly we know what's not known. Controlling the structure of the cardinals is a lot more difficult than it is in $\sf ZFC$ using current technologies. I believe that the answer is that if there are $\omega$ many sets which are pairwise incomparable, then it is not a cardinal. But you need sets, not just cardinals. This looks like a good theme for a blog post. I'll try to write something tonight. Seems like a much better use of my time than loafing around. $\endgroup$
    – Asaf Karagila
    Jun 20, 2014 at 22:02

1 Answer 1


I don't have anything in mind which "tames" the cardinals structure. I also think that you don't fully understand the consequences of $\sf AD$ on the cardinal structure if you say that it tames the structure down. (Note that even Woodin, in the abstract you reference, points out that the cardinal structure is very complicated.)

But let me just point out the following. The axiom "Every set is linearly orderable" follows from the Boolean Prime Ideal theorem, which is shown to hold in Cohen's first model. In that model there exists a Dedekind-finite set. Therefore the example I gave you in the math.SE question about the failure of infimum of two sets holds in that model.

Now, without sitting to verify the details, here's a suggestion as to how to destroy the supremum property while still having "Every set is linearly orderable". By adding $\aleph_0$ Cohen reals, and $\aleph_1$ Cohen subsets of $\omega_1$, and taking automorphisms of the forcing which look like the ones used in Cohen's first model (namely, an automorphism of a disjoint union of $\omega$ and $\omega_1$ which doesn't "mix" the two parts), and similar ideal and so on, one should be able to get a model in which there are two incomparable Dedekind-finite sets, but every set can be linearly ordered. In that case, the generalization of the example in which $\sup$ fails can be executed.

(I should say that I don't remember at the moment whether or not the old question "If there is an infinite Dedekind-finite cardinal, then there are two incomparable Dedekind-finite cardinals" is answered. In case that the answer is positive, then Cohen's first model is a model for the failure of $\sup$ for cardinals as well.)

  • $\begingroup$ Well, at least real determinacy tames it below what Woodin calls $|A_1|$, perhaps stronger forms can do more. What does "basis" mean in Caicedo and Ketchersid, does it mean that uncountable cardinalities are generated by $\omega_1$ and $\mathbb{R}$ somehow? $\endgroup$
    – Conifold
    Jun 20, 2014 at 21:22
  • $\begingroup$ Didn't Sageev prove that it's consistent that the Dedekind-finite cardinals are a nonstandard model of arithmetic? $\endgroup$
    – bof
    Jun 20, 2014 at 21:25
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    $\begingroup$ Yes (or more accurately, equiconsistent with an inaccessible): "A model of ZF + there exists an inaccessible, in which the Dedekind cardinals constitute a natural non-standard model of arithmetic" (sciencedirect.com/science/article/pii/0003484381900176). Actually, Sageev proved "the Dedekind cardinals are linearly ordered" is equiconsistent with ZF + an inaccessible; Ellentuck (logic-library.berkeley.edu/catalog/detail/739) had proved in ZF that the Dedekind cardinals are linearly ordered iff they form an elementary extension of $\mathbb{N}$, and so satisfy true arithmetic. $\endgroup$ Jun 20, 2014 at 21:37
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    $\begingroup$ Here's the MO question I just asked: mathoverflow.net/questions/172329/…. (Frankly, I find it pretty strange that this linearity can happen at all!) $\endgroup$ Jun 20, 2014 at 21:49
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    $\begingroup$ Just to answer Conifold's easy question: To say that $\omega_1$ and the continuum are a basis for the uncountable cardinals means that every uncountable cardinal is $\geq$ one of these. In other words, every uncountable set has either a well-ordered uncountable subset or a subset in one-to-one correspondence with the reals (or both). $\endgroup$ Jun 20, 2014 at 23:22

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