 Solution

For a set $A$, its definable power set isOnly finitely many parameters occur in each definition. Satisfaction for the set structure is definable, so this specifies a uniformly definable class function assigning a set to every set $A$, with $\mathrm{Def}(A)\subseteq \mathcal{P}(A)$. Every subset definable over the structure with its parameters occurs. The full set $A$ itself occurs by the formula $x=x$. If $A$ is transitive, then $A\subseteq \mathrm{Def}(A)$, since any $p\in A$ is defined as a subset of $A$ by $x\in p$. In that case $\mathrm{Def}(A)$ is transitive as well: a member of one of its elements lies in $A$, hence in $\mathrm{Def}(A)$.

The constructible hierarchy and universe areThe displayed properties prove by induction that levels are transitive and increasing, so $L$ is transitive. It contains every ordinal: inductively $L_{\alpha }\cap \mathrm{Ord}=\alpha$, and the subset of $L_{\alpha }$ consisting of ordinals is definable there, giving $\alpha \in L_{\alpha +1}$. Ordinalhood is absolute here because the levels are transitive. Both the level function and membership in $L$ are definable in $ZF$. This is Gödel's constructible universe, which can contain fewer subsets than the full universe.