Is every set being cardinal definable consistent with ZF + negation of Choice?

Question

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 $\sf ZF + \neg AC$.

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.

Reference:

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