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\begin{multline} \mathrm{Def}(X) := \Bigl\{ \{y \mid y\in X \text{ and } \Phi(y,z_1,\ldots,z_n) \text{ is true in }(X,\in) \} \mid \\ \Phi \text{ is a first order formula and } z_1,\ldots,z_n\in X\Bigr\}. \end{multline}

(Constructible Universe, Wikipedia)

I saw somewhere at stackexchange that $\Phi$ can be defined, but was not able to find it. So, is there anyway to define what such $\Phi$ would be?

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    Have you tried to open a *book*? Wikipedia is a good place to get an overview about things, but not to study. This site is not a supplement for trying to figure things on your own after working through the first part of Jech's *Set Theory*, for example.2012-09-18

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Note that given a (first-order) formula $\Phi$ of set theory (with free variables among $u, v_1 , \ldots , v_n$) and $a_1 , \ldots , a_n \in X$ we can define the set $\{ y \in X : ( X , \in ) \models \Phi [ y, a_1 , \ldots , a_n ] \}$. The family $\mathrm{Def}(X)$ of all sets "definable" over $X$ is then the collection of all such sets as $\Phi$ varies over all formulas and $a_1 , \ldots , a_n$ vary over all elements of $X$.

There is a formula $\psi (x)$ such that a set $a$ is constructible iff $\psi[a]$ holds (and so $\mathbf{L} = \{ x : \psi[x] \}$). A common abbreviation for this formula is $( \exists \alpha \in \mathbf{Ord}) ( x \in L_\alpha )$, and I understand if this is not too helpful. For more detail please consult some of the following references:

  1. Jech, Set Theory: Third Millennium Edition, pp.175-188.
  2. Drake, Set Theory: An Introduction to Large Cardinals, pp.127-134.
  3. Kunen, Set Theory: An Introduction to Independence Proofs, pp.165-173.

For some (minimal) details.

You can fix to a Gödel numbering of the formulas of set theory, and say that a set $Y$ is "definable" over $X$ iff for some $n \in \omega$ and parameters $a_1 , \ldots , a_n \in X$ (where $n$ is determined by $n$) $Y$ is the set of all $y \in X$ such that the structure $( X , \in )$ "thinks" that the $n$th formula with the parameters $y, a_1 , \ldots , a_n$ is true. This step is definable, though perhaps tedious to set up.

Another option is to go via so-called Gödel operations, and show that for a transitive $X$, $\operatorname{Def}(X)$ is just the closure of $X \cup \{ X \}$ under these operations (restricted to the power set of $X$). Each of these operations is definable (recursive, even), and from there it is pretty straightforward to define the closure (by recusion up to $\omega$).

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    @MinghuiOuyang The set Def($X$) is not defined within $X$, but from the outside. $X$ is a *set* in the universe of set-theory $V$. In $V$ you can definitely quantify over the formulas - they are simple objects defined recursively. The problem is to have a *satisfaction relation* for all formulas, which you can't have from within the model $(X,\in)$, but $V$ does have a satisfaction relation for $X$, i.e in $V$ one can define the notion $(X,\in) \vDash \phi$ for every formula, and since these are all sets, you can also (in $V$!) quantify over the formulas.2018-03-03