TITLE: A question about definable elements in a model of ZFC
QUESTION [6 upvotes]: Let $\langle M,E\rangle$ be a model of $\mathsf{ZFC}$.
Does there exist a $d\in M$ such that, for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$ without parameters?
A result of J.D. Hamkins, etc. (cf. Pointwise definable models of set theory) shows that, for some models of $\mathsf{ZFC}$, such a $d$ cannot exist. Is there a model in which such a $d$ exists?

REPLY [2 votes]: For definiteness we say that $a$ is definable in a model $\langle M,E\rangle$ of ZF if there is a formula $\varphi$ with one free variable in the language with one non-logical symbol $\in$ with the property that $\langle M,E\rangle\models\phi(x)$ iff $x=a$. (The Gödel number of such a formula must be standard.)
The condition given by Joel David Hamkins in the last sentence of (v)
of his answer is necessary and sufficient for there to be a set in $M$ consisting of the definable elements of $M$:

Theorem. For any model $\langle M,E\rangle\models\mathrm{ZF}$, the following are equivalent:

There exists a $d\in M$ such that for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$.

$M$ is an $\omega$-model, and there is $N\in M$ such that $N\prec M$.



Proof:
1 → 2: Suppose $d\in M$, and for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$.

Claim. Let $M\models\text{“$\alpha$ is the least ordinal not in $d$”}$. Then $V_\alpha^M\prec M$.

Proof: Write $V_\alpha$ for $V_\alpha^M$. We prove
$$\forall p_1,\dots,p_n\in V_\alpha\:\bigl(V_\alpha\models\phi(p_1,\dots,p_n)\iff M\models\phi(p_1,\dots,p_n)\bigr)\tag1$$
by induction on the complexity of $\phi$. For atomic formulas $\phi$, (1) holds.
Suppose that (1) holds for $\phi$ and $\psi$. Then
$$\begin{align*}
V_\alpha\models\phi\land\psi&
\iff(V_\alpha\models\phi\text{ and }V_\alpha\models\psi)\\
&\iff(M\models\phi\text{ and }M\models\psi)
\iff M\models\phi\land\psi.
\end{align*}$$
Also,
$$V_\alpha\models\neg\phi\iff ¬V_\alpha\models\phi\iff ¬M\models\phi\iff M\models\neg\phi.$$
Suppose (1) holds for $\theta(x,\dots)$. If $V_\alpha\models\exists x\,\theta(x,p_1,\dots,p_n)$, then $V_\alpha\models\theta(p,p_1,\dots,p_n)$ for some $p\mathrel{E}V_\alpha$. Therefore $M\models\theta(p,p_1,\dots,p_n)$ and so $M\models\exists x\,\theta(x,p_1,\dots,p_n)$.
Finally, suppose $M\models\exists x\,\theta(x,p_1,\dots,p_n)$ for $p_i$ in $V_\alpha$. Then there is an ordinal $\delta$ in $d$ such that $\{p_1,\dots,p_n\}$ is contained in $V_\delta$. $\beta\mathrel{E}d$ must hold for the least ordinal $\beta$ such that $\delta\mathrel{E}\beta$, $V_\beta$ reflects $\exists x\,\theta$, and $V_\beta$ reflects $\theta$. Therefore $M\models\theta(p,p_1,\dots,p_n)$ for some $p\mathrel{E}V_\beta$, and consequently $V_\alpha\models\theta(p,p_1,\dots,p_n)$, and thus $V_\alpha\models\exists x\,\theta(x,p_1,\dots,p_n)$.
This completes the proof of the Claim. It remains to prove that $M$ is an $\omega$-model.
For each $a\mathrel{E}d$, let $g_a$ be the least Gödel number of a formula which defines $a$ in $V_\alpha$. There is a set $S\in M$ consisting of the $g_a$ for which $a\mathrel{E}d$. Since $S$ is an infinite set of standard natural numbers, its union must be $\omega$. Thus, $\omega\in M$, and $M$ is an $\omega$-model.
2 → 1: Suppose $M$ is an $\omega$-model, $N\in M$ and $N\prec M$. There is a $d\in M$ such that $b\mathrel{E}d$ iff $b\mathrel{E}N$ and
$M\models\text{“$b$ is definable in $N$”}$. Then $b\mathrel{E}d$ if and only if $b$ is definable in $\langle M,E\rangle$.