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Is it consistent to add the following axiom to $\sf ZF$?

Definability: $$\forall X \exists \alpha \exists \phi: X=\{y \in V_\alpha \mid V_\alpha \models \phi(y)\}$$

Where $\phi$ is a formula in one free variable only.

This is intended to capture the informal notion of every set being parameter free definable.

Zuhair
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    Given that the language of ZF doesn't even quantify over formulas, this axiom is quite a stretch – Hagen von Eitzen May 07 '21 at 13:10
  • @HagenvonEitzen, why then $\sf V=HOD$ is using that style of writing? This sentence is expressable by a first order formula I believe, much like the expression in Hamkins comment at: https://mathoverflow.net/questions/391651/can-we-have-sf-v-htd-how-it-relates-to-sf-v-hod#comment998892_391651 – Zuhair May 07 '21 at 13:24
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    @Hagen: You can use internal formulas (so you're really just quantifying over the internal object that is "formulas in the language of set theory") and then use the Reflection theorem to argue that this is the same. But we are using some very strong properties of ZF here, so it's not immediately obvious that one can do that. – Asaf Karagila May 07 '21 at 14:04

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Yes. In pointwise definable models every $X$ is definable by a formula $\phi$ with one free variable, and by applying reflection we get this schema. You can read more about pointwise definable models in

Hamkins, Joel David; Linetsky, David; Reitz, Jonas, Pointwise definable models of set theory, J. Symb. Log. 78, No. 1, 139-156 (2013). ZBL1270.03101.

On the other hand, your statement is an internal statement, so any model which is elementary equivalent to a pointwise definable model will satisfy it. But that's easy to arrange with an uncountable model which cannot be pointwise definable (since pointwise definable models are necessarily countable).

You can also read What does it really mean for a model to be pointwise definable? for more information.

Asaf Karagila
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