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Zuhair Al-Johar
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Is every set being cardinal definable consistent with ZF + negation of Choice?

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 $\sf ZF + \neg AC$.

Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47