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](https://doi.org/10.2178/jsl.7801090)) shows that, for some models of $\mathsf{ZFC}$, such a $d$ cannot exist. Is there a model in which such a $d$ exists?



A question about definable elements in a model of ZFC
In the paper you mention in the original post, we mention several of the possibilities as follows. Item (v) includes the particular situation you asked about.


*Hamkins, Joel David; Linetsky, David; Reitz, Jonas*, [**Pointwise definable models of set theory**](http://dx.doi.org/10.2178/jsl.7801090), J. Symb. Log. 78, No. 1, 139-156 (2013). [ZBL1270.03101](https://zbmath.org/?q=an:1270.03101). [blog post](http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/)



> 
> Let us now turn to the question of the extent to which
>  definability is first-order expressible, by presenting a
>  number of examples that illustrate the range of
>  possibility. We have already observed that the property of
>  a model being pointwise definable is not first order
>  expressible, since it is not preserved by nontrivial
>  elementary extensions. Since pointwise definability is a
>  strong generalization of the axiom $V=\newcommand\HOD{\text{HOD}}\HOD$, it is tempting
>  to introduce such notation as $V=D$ or $V=HD$ to express
>  that a model is pointwise definable, thereby maintaining a
>  parallel to the classical $V=\HOD$ notation while
>  emphasizing that the definitions need no parameters. We
>  hesitate to adopt this notation, however, because we fear
>  it would incorrectly suggest that the concept is
>  first-order expressible, which isn't the case.
> 
> 
> (i) *There is no uniform definition of the class of
>  definable elements.* Specifically, there is no formula
>  $\mathop{\rm df}(x)$ in the language of set theory that is
>  satisfied in any model $M\newcommand\satisfies{\models}\satisfies\newcommand\ZFC{\text{ZFC}}\ZFC$ exactly by the
>  definable elements. The reason is that if $M\_0$ is
>  pointwise definable and $M\_0\prec M$ is a nontrivial
>  elementary extension, then the definable elements of $M\_0$
>  and $M$ are precisely the elements of $M\_0$, and so $M\_0$
>  should satisfy $\forall x\,\mathop{\rm df}(x)$ but $M$
>  would satisfy $\exists x\,\neg \mathop{\rm df}(x)$,
>  contrary to $M\_0\prec M$.
> 
> 
> (ii) *The class of definable elements can form a
>  definable class.* Although there is no uniform definition
>  of the class of definable elements, it can sometimes happen
>  that a model enjoys a certain structure that allows it to
>  see its collection of definable elements as a definable
>  class. For example, in a pointwise definable model, the
>  class of definable elements includes every object and is
>  therefore defined by the formula $x=x$. See also (iv) and
>  (v) below.
> 
> 
> (iii) *The collection of definable elements might
>  not form a class.* Consider any model
>  $M\satisfies\ZFC$, and let $N$ be an ultrapower of
>  $M$ by an ultrafilter on the natural numbers. The
>  parameter-free definable elements of $N$ are
>  necessarily contained in the range of the
>  ultrapower map, and in particular, do not include
>  any of the newly added nonstandard natural
>  numbers. Thus, the class of definable elements of
>  $N$ is not amenable to $N$, for it would reveal
>  that its natural number are not well-founded.
> 
> 
> (iv) *The definable elements can form a definable
>  class in a model having no class function $r\mapsto\psi\_r$
>  mapping definable elements to definitions.* Suppose
>  that $M$ is a pointwise definable model of $\ZFC$.
>  The definable elements of $M$ are all of $M$,
>  which is certainly a definable class in $M$. But
>  $M$ cannot have a function $r\mapsto\psi\_r$
>  associating to each element $r$ of $M$, or even to
>  each real of $M$, a defining formula $\psi\_r$,
>  since such a map would reveal to $M$ that it has
>  only countably many reals.
> 
> 
> (v) *The definable elements can be a set in a model
>  that does have a definability map $r\mapsto\psi\_r$.* 
>  Suppose that $\kappa$ is an inaccessible cardinal (this
>  hypothesis can be reduced), and observe by a
>  Lowenheim-Skolem argument that there are numerous
>  $\gamma<\kappa$ with $V\_\gamma\prec
> V\_\kappa\satisfies\ZFC$. It follows that the definable
>  elements of $V\_\kappa$ are all in $V\_\gamma$ and satisfy
>  the same definitions there as in $V\_\kappa$. Since
>  $V\_\gamma$ is a set in $V\_\kappa$, we may construct in
>  $V\_\kappa$ the function $r\mapsto \psi\_r$ that maps
>  every definable element $r$ of $V\_\gamma$ to the
>  smallest definition $\psi\_r$ of it, and because
>  $V\_\gamma\prec V\_\kappa$, this function has the same
>  property with respect to $V\_\kappa$, as desired. The large
>  cardinal hypothesis can be reduced; it is sufficient to
>  have an $\omega$-model $M$ with some $M\_0\in M$ having
>  $M\_0\prec M$.
> 
> 
> (vi) *No model can have a **definable** definability
>  map $r\mapsto\psi\_r$.* If such a map were definable, then
>  since there are only countably many definitions $\psi\_r$,
>  we could easily diagonalize against it to produce a
>  definable real not in the domain of the map. In (v), the
>  map is definable from parameter $\gamma$.
> 
> 
> The surviving content of the math-tea argument seems to be
>  the observation that in any model with access to a
>  definability map $r\mapsto\psi\_r$, the definable reals do
>  not exhaust all the reals.
> 
> 
> 


