Is parameter free definability consistent with ZF?

99 Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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.