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A formula $\varphi[x]$ with one free variable $x$ in the language of sets is a definition in ZF if $ZF \vdash \exists y \forall z(\varphi[z] \longleftrightarrow z = y)$.

Is there a definition in ZF of the (meta-)set of Gödel numbers of definitions in ZF?

I suspect it isn't but I can't find a contradiction in assuming this existence. One idea would be to find a trick to define the truth of sentences in the language of set theory using the definition of the fact of being a definition, but I don't see how.

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  • $\begingroup$ Your question asks, in short, we can define a class of definable sets? $\endgroup$
    – Hanul Jeon
    Dec 28, 2016 at 18:24
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    $\begingroup$ A theorem by Hamkins might be relevant. $\endgroup$
    – Hanul Jeon
    Dec 28, 2016 at 18:25
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    $\begingroup$ I doubt it. Tarski's theorem on the non-definability of truth comes to mind. Let $[S]$ be the Godel number of the sentence $ S.$ (A sentence is a formula with no free variables). There cannot exist a formula $\pi$ in one free variable such that $\pi [(S)]\iff S$ holds for every $S$.... We can define the set $T$ of all Godel numbers of the form $[\;\exists!y\;(\psi (y)].$ What you are asking for is a formula $\pi$ in 1 free variable such that $\pi (U) \iff U=\{n\in T: ZF\vdash n^*\}$ where $n^*$ is a sentence of the form $\exists! y\;(\theta (y)$ whose Godel number is $n.$ $\endgroup$
    – DanielWainfleet
    Dec 29, 2016 at 9:27
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    $\begingroup$ Continued from previous comment: We would then have a truth-formula in ZF, for all sentences of the form $\exists!y\;\phi(y),$ namely $S\iff (\;[S]\in T\land \pi ([S])\;).$ If we included sentences of every form we would be contradicted by Tarski's theorem. But (it just occurred to me) any sentence $W$ is equivalent to $\exists! y\;(\forall z \; (z\not \in y\land W)\;).$ So we would have a truth-formula for all sentences, which is precisely contrary to Tarski's theorem, $\endgroup$
    – DanielWainfleet
    Dec 29, 2016 at 9:41
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    $\begingroup$ @user254665: That trick of $\exists ! y (\forall z(z \notin y \wedge W))$ is exactly what I was looking for but couldn't find, thanks! You can post your comment as an answer I you want, and I'll accept it. $\endgroup$
    – nombre
    Dec 29, 2016 at 10:51

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Um just let $D = \{ φ : \text{$φ$ is a $1$-parameter formula over ZF} \land \text{ZF} \vdash \exists! x ( φ[x] )\}$. Then $D$ is the set you are looking for. You would not be able to get a contradiction from this unless your meta-system is inconsistent.

If ZF is inconsistent, then ZF proves every sentence, and any reasonable meta-system can see that, and so $D$ would be the set of all $1$-parameter formulae. However, if ZF is consistent, it may still be $Σ_1$-unsound and think that itself is inconsistent, in which case $D$ is again the set of all $1$-parameter formulae if you are using ZF as your meta-system.

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  • $\begingroup$ Thank you, I had misunderstood the théorème of the undefinability. $\endgroup$
    – nombre
    Jan 2, 2017 at 13:51
  • $\begingroup$ @nombre: You're welcome! $\endgroup$
    – user21820
    Jan 2, 2017 at 16:43

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