Relational Closure Exists for Set-Like Relation

Theorem
Let $A$ be a class.

Let $\prec$ be a foundational relation.

Furthermore, let every $\prec$-initial segment of $A$ be a small class.

Let $S$ be a set that is a subset of class $A$.

Let $G$ be a mapping such that $G \left({ x }\right) = A \cap \mathop \prec^{-1} \left({ x }\right)$.

Let $F$ be defined using Finite Recursion:


 * $F\left({0}\right) = S$


 * $F\left({n^+}\right) = F\left({n}\right) \cup G\left({F\left({n}\right)}\right)$

Let $\displaystyle T = \bigcup_{n \mathop \in \omega} F\left({n}\right)$.

Then: }
 * 1) $T$ is a set and satisfies:


 * $\forall x \in A \forall y \in T: \left({ x \prec y \implies x \in T }\right)$


 * In other words, $T$ is $\prec$-transitive.


 * 1) $S \subseteq T$


 * 1) If $R$ is $\prec$-transitive and $S \subseteq R$, then $T \subseteq R$.

That is, given any set $S$, there is an explicit construction for its relational closure.

Proof of First Part
Take any $x \in A$ and $y \in T$.

If $x \prec y$, then $x$ is in the initial segment of $y$.

Moreover, since $y \in T$, it is in $F\left({n}\right)$ for some $n$.

Therefore $x \in F\left({n+1}\right)$ and $x \in T$.

Proof of Second Part

 * $F\left({ 0 }\right) = S$

Therefore, if $x \in S$, then $x \in F\left({n}\right)$ for some $n$.

It follows that $x \in T$.

By the definition of subset, it follows that:


 * $S \subseteq T$