Transitive Closure of Set-Like Relation is Set-Like

Theorem
Let $A$ be a class.

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

Let $\mathcal R^+$ be the transitive closure of $\mathcal R$.

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

Proof
Let $x \in A$.

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

For each $s \in A'$, $\mathcal R^{-1}$ is a subset of $A$, and hence an element of $A'$, by Inverse Image of Set under Set-Like Relation is Set and the definition of endorelation.

Define a mapping $G: A' \to A'$ by letting $G(s) = \mathcal R^{-1}(s)$.

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


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

By the Axiom of Infinity and the Axiom of Replacement, $f(\N)$ is a set.

Thus by the Axiom of Union, $\bigcup f(\N)$ is a set.

Let $y \in (\mathcal R^+)^{-1}(x)$.

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

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

As this holds for all such $y$, $(\mathcal R^+)^{-1}(x) \subseteq \bigcup f(\N)$.

By the Axiom of Subsets, $(\mathcal R^+)^{-1}(x)$ is a set.

As this holds for all $x \in A$, $\mathcal R^+$ is a set-like relation.