Transitive Closure of Set-Like Relation is Set-Like

Theorem
Let $A$ be a class.

Let $\RR$ be a set-like endorelation on $A$.

Let $\RR^+$ be the transitive closure of $\RR$.

Then $\RR^+$ is also a set-like relation.

Proof
Let $x \in A$.

Let $A'$ be the class of all subsets of $A$.

For each $s \in A'$, $\RR^{-1}$ is a subset of $A$.

Hence by Inverse Image of Set under Set-Like Relation is Set and the definition of endorelation:
 * $\RR^{-1} \in A'$

Define a mapping $G: A' \to A'$ as:
 * $\forall s \in A': G \left({s}\right) = \RR^{-1} \left({s}\right)$

Recursively define a mapping $f: \N \to A'$ as follows:


 * $f \left({0}\right) = \left\{ {x}\right\}$
 * $f \left({n + 1}\right) = G \left({f \left({n}\right)}\right)$

By the Axiom of Infinity and the Axiom of Replacement:
 * $f \left({\N}\right)$ is a set.

Thus by the Axiom of Union:
 * $\bigcup f \left({\N}\right)$ is a set.

Let $y \in \left({\RR^+}\right)^{-1} \left({x}\right)$.

By the definition of transitive closure:
 * for some $n \in \N_{>0}$ there are $a_0, a_1, \dots, a_n$ such that $y = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = x$.

Then by induction (working from $n$ to $0$), $a_n, a_{n-1}, \dots, a_0 \in \bigcup f \left({\N}\right)$.

As this holds for all such $y$:
 * $\left({\RR^+}\right)^{-1} \left({x}\right) \subseteq \bigcup f \left({\N}\right)$.

By the Axiom of Specification:
 * $\left({\RR^+}\right)^{-1} \left({x}\right)$ is a set.

As this holds for all $x \in A$:
 * $\RR^+$ is a set-like relation.