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 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) $S \subseteq T$
 * 2) 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 $x \in {\prec^{-1}} \left({y}\right)$.

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:


 * $S \subseteq T$

Proof of Third Part
Suppose that $S \subseteq R$ and that $R$ is $\prec$-transitive.

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:
 * $F\left({n}\right) \subseteq R$

$P \left({0}\right)$ is true, as this just says $S \subseteq R$.

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 $n \in \omega$


 * $T \subseteq R$ by Indexed Union Subset