Recall the definition of cardinal definable, where every set being cardinal definable is proved consistent relative to ZF + V=HOD. To re-iterate it:

$Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda_1,.., \lambda_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V_{\rho(\kappa)} \mid \phi^{V_{\rho (\kappa)}} (y,\lambda_1,..,\lambda_n)\}$

Where: $\lambda_i <^\rho \kappa \iff \rho(\lambda_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "cardinal" is defined after Scott's as an equivalence class under bijection of sets of the lowest possible rank.

Now, is the principle stating that every set is cardinal definable consistent with $\sf ZF + \neg AC$?

A related question replacing cardinal by ordinal in the above question would lead to $\sf V=HOD$, which is known to prove $\sf AC$ and so would be inconsistent with $\sf ZF + \neg AC$.

  • $\begingroup$ Do you know the answer if “cardinal” is replaced by “ordinal” in your question? If so, it would be seemly to recall what it is as part of the context for your question. If not, you might also say so. $\endgroup$
    – Gro-Tsen
    Dec 26, 2021 at 16:30
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    $\begingroup$ @Gro-Tsen The answer is no in that case. The class of ordinal-definable sets has a definable well-ordering over ZF, so if all sets are ordinal-definable, AC holds. $\endgroup$
    – Wojowu
    Dec 26, 2021 at 17:54
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    $\begingroup$ @Wojowu Actually I knew that, but my point was mostly to remind the poster that when asking a question it is in good taste to give motivation and context, including the answer to very obviously related questions when possible. $\endgroup$
    – Gro-Tsen
    Dec 26, 2021 at 18:51
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    $\begingroup$ This is a good question. I expect the answer to be yes, but one essentially has to cook up a model of ZF with cardinal structure as rich as its set structure (at least where choice fails). $\endgroup$ Dec 26, 2021 at 19:15
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    $\begingroup$ @Gro-Tsen, I see your point. I actually tried to provide some context by the linked sites, but I'll add that point to the question. Thanks. $\endgroup$ Dec 26, 2021 at 19:59

1 Answer 1


This is consistent. Kanovei constructed a model $M$ with an infinite Dedekind finite set of reals which is lightface projectively definable. By descending to $L(R),$ we can further assume it satisfies $V=L(R).$

Clearly choice fails in this model. Since it satisfies $V=L(R),$ every set is definable from an ordinal and a real. Of course, any ordinal is definable from an $\aleph,$ so we just need to check that every real is cardinal definable.

Let $A$ be the definable Dedekind finite set of reals, and fix a canonical surjection $f: A \rightarrow \omega$ (as exists for any infinite set of reals). Let $\langle q_n \rangle$ be an enumeration of the rationals. Define $A_r=f^{-1}(\{n: q_n<r\}).$ For $r<s,$ $|A_r|<|A_s|$ since these are Dedekind finite sets. Clearly $r$ is definable from the cardinality of $A_r,$ so we're done.


Kanovej, V. G., On the nonemptiness of classes in axiomatic set theory, Math. USSR, Izv. 12, 507-535 (1978). ZBL0427.03044.


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