Relational Closure Exists for Set-Like Relation

Theorem
Let $A$ be a class.

Let $\prec$ be a relation on $A$.

Furthermore, let ${\prec^{-1}}\left({a}\right)$ be a small class for each $a \in A$.

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

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

Let $F$ be defined using the Principle of Recursive Definition:


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


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

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

Then:
 * $(1): \quad T$ 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.
 * $(2): \quad S \subseteq T$
 * $(3): \quad$ 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 $(1)$
Let $x \in A$ and $y \in T$ such that $x \prec y$.

Then by definition:
 * $x \in {\prec^{-1}} \left({y}\right)$

Moreover, since $y \in T$:
 * $\exists n \in \omega: y \in F \left({n}\right)$

Therefore:
 * $x \in F \left({n+1}\right)$

and so:
 * $x \in T$

Proof of $(2)$
Let $x \in S$.

We have that:
 * $F \left({0}\right) = S$

Therefore:
 * $\exists n \in \omega: x \in F \left({n}\right)$

It follows that $x \in T$.

By the definition of subset, it follows that:


 * $S \subseteq T$

Proof of $(3)$
Let $R$ be $\prec$-transitive.

Let $S \subseteq R$.

Then:


 * $F \left({n}\right) \subseteq R$ shall be proved by finite induction.

For all $n \in \omega$, let $P \left({n}\right)$ be the proposition:
 * $F \left({n}\right) \subseteq R$

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $F \left({0}\right) \subseteq R$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $F \left({k}\right) \subseteq R$

Then we need to show:
 * $F \left({k+1}\right) \subseteq R$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $F \left({n}\right) \subseteq R$ for all $n \in \omega$

Then by Indexed Union Subset:
 * $T \subseteq R$