# Definition of the set of definitions.

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.

• Your question asks, in short, we can define a class of definable sets? Dec 28, 2016 at 18:24
• A theorem by Hamkins might be relevant. Dec 28, 2016 at 18:25
• 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 $n.$ Dec 29, 2016 at 9:27
• 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, Dec 29, 2016 at 9:41
• @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. Dec 29, 2016 at 10:51

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.