This posting is generally related to a prior posting titled "Are all constructible from below sets parameter free definable?"

If we work in infinitary language $\mathcal L_{\omega_1, \omega}$, then we can define parameter free definable, denoted "$D$", as:

$Dx \iff \bigvee x= \{ y \mid \Phi \}$

where $\Phi$ range over all formulas in $\mathcal L_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.

Axiom of definability: $\forall x Dx$

Let $\sf ZF + Def$ be the theory that extends $\mathcal L_{\omega_1, \omega}$ with axioms of $\sf ZF$ (written in $\mathcal L_{\omega, \omega}$) and the axiom of definability.

Is $\sf ZF +Def$ consistent ?

If so then does $\sf ZF + Def$ have all of its models being pointwise definable models?

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    $\begingroup$ @Michael Hardy, Thank you for correcting the title. $\endgroup$ Jun 8, 2023 at 16:53

1 Answer 1


Yes, the theory is consistent, if ZF is consistent, because there are pointwise definable models of ZF. Any such model is a model of your theory, which is therefore satisfiable and hence consistent.

And yes, clearly every model of your theory is pointwise definable (in the first-order language), because that is precisely what your axiom of definability asserts. The only way to make this axiom true in a model of ZF is for every set to be definable.

The models of ZF+Def are exactly the pointwise definable models of ZF.

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    $\begingroup$ So the usual upward Löwenheim–Skolem theorem fails drastically here. This is similar to the situation of characterizing the standard model of arithmetic in $\mathcal L_{\omega_1, \omega}$, where all models are countable! Thanks a lot. $\endgroup$ Jun 8, 2023 at 16:50

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