# Compactness for definable models?

In this question, I use "model" to mean a model in the language $\{\in\}$ of set theory. Call a model $M$ "definable" iff for every $x \in M$, there is a formula $\phi(\vec{y},z)$, where $\vec{y} \in M$ is a sequence of parameters, such that $M \models \forall z (z \in x \leftrightarrow \phi(\vec{y},z))$.

I want the following variation on the compactness theorem: given a theory $T$ (in the language $\{\in\}$), if every finite subset of $T$ has a definable model, then $T$ has a definable model. Is this true?

[Edit: the question above is not interesting, as explained in an answer below. A potentially interesting variant is as follows. Say that a model $M$ is "definable" iff for every $x \in M$ there is a formula $\phi(z)$, without parameters, such that $M \models \forall z (z \in x \leftrightarrow \phi(z))$. Ask the same question. Or, ask the same question using pointwise definability, as described by Andres Caicedo.]

Thank you!

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Nick, there are pointwise definable models of set theory. You may want to look at this link for some context, and a paper: jdh.hamkins.org/math-tea-argument-vienna-2011 Anyway, here is a sketch: If there is a model of ZFC, then there is one, call it M , satisfying V = L. This model admits deﬁnable Skolem functions, so we can form the Skolem hull H of the empty set. This is an elementary substructure of M , thus a model of ZFC, and it is pointwise deﬁnable. Note that it suffices that M satisfies V = HOD and that, if M is transitive, then the transitive collapse of H exists. –  Andres Caicedo Jan 28 '13 at 4:12
@AndresCaicedo: Thank you for informing me of that! I shall retract my claim that my definition is odd. –  Nick Thomas Jan 28 '13 at 4:15
This appears to be a very weak form of definability. In particular, it seems that every $\{ \in \}$-model is definable. Note that for $x \in M$ we have that $M \models ( \forall z ) ( z \in x \leftrightarrow z \in x )$, and I cannot see any reason $x$ cannot be used as a parameter for itself. In the talk Andres Caicedo linked to, Joel Hamkins uses a much stronger form of definability: $a \in M$ is definable iff there is a formula $\phi (x)$ such that $M \models \phi[b]$ iff $b = a$ for all $b \in M$. –  Arthur Fischer Jan 28 '13 at 4:31
Arthur: That's a very good point. I read Hamkins' paper and was in the process of writing that his definition was different, but now you've said it for me. Meanwhile my definition is clearly useless as it stands, and I must go back and figure out what I'm really trying to say. Thank you for identifying that error. –  Nick Thomas Jan 28 '13 at 4:36
Arthur: I've hopefully fixed the question, by adding another condition. I apologize; the problem arose because I am trying to abstract this question from a specific problem context. In that context the added condition holds, but I failed to think about it in formulating my question. –  Nick Thomas Jan 28 '13 at 4:51

This appears to be a very weak form of definability. In particular, it seems that every $\{∈\}$-model is definable. Note that for $x∈M$ we have that $M⊨(∀z)(z∈x↔z∈x)$, and I cannot see any reason $x$ cannot be used as a parameter for itself. In the talk Andres Caicedo linked to, Joel Hamkins uses a much stronger form of definability: $a∈M$ is definable iff there is a formula $ϕ(x)$ such that $M⊨ϕ[b]$ iff $b=a$ for all $b∈M$. – Arthur Fischer