Relational Closure Exists for Set-Like Relation

Theorem
Let $A$ be a class.

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

Furthermore, let $\map {\prec^{-1} } a$ 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:
 * $\map G x = A \cap \paren {\map {\prec^{-1} } x}$

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


 * $\map F 0 = S$


 * $\map F {n^+} = \map F n \cup \map G {\map F n}$

Let $\ds T = \bigcup_{n \mathop \in \omega} \map F n$.

Then:
 * $(1): \quad T$ is a set and satisfies:
 * $\forall x \in A: \forall y \in T: \paren {x \prec y \implies x \in T}$
 * 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:
 * $x \in \map {\prec^{-1} } y$

Moreover, since $y \in T$:
 * $\exists n \in \omega: y \in \map F n$

Therefore:
 * $x \in \map F {n + 1}$

and so:
 * $x \in T$

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

We have that:
 * $\map F 0 = S$

Therefore:
 * $\exists n \in \omega: x \in \map F n$

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:


 * $\map F n \subseteq R$ shall be proved by finite induction.

For all $n \in \omega$, let $\map P n$ be the proposition:
 * $\map F n \subseteq R$

Basis for the Induction
$\map P 0$ is the case:
 * $\map F 0 \subseteq R$

which has been proved above.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\map F k \subseteq R$

Then we need to show:
 * $\map F {k + 1} \subseteq R$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall n \in \omega: \map F n \subseteq R$

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