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:
- $\forall s \in A': \map G s = \map {\RR^{-1} } s$
Recursively define a mapping $f: \N \to A'$ as follows:
- $\map f n = \begin {cases} \set x & : n = 0 \\ \map G {\map f {n - 1} } & : n > 0 \end {cases}$
By the Axiom of Infinity and the Axiom of Replacement:
- $\map f \N$ is a set.
Thus by the Axiom of Unions:
- $\ds \bigcup \map f \N$ is a set.
Let $y \in \map {\paren {\RR^+}^{-1} } x$.
By the definition of transitive closure:
- for some $n \in \N_{>0}$ there are $a_0, a_1, \dots, a_n$ such that $y = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = x$.
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Then by induction (working from $n$ to $0$), $\ds a_n, a_{n - 1}, \dots, a_0 \in \bigcup \map f \N$.
As this holds for all such $y$:
- $\ds \map {\paren {\RR^+}^{-1} } x \subseteq \bigcup \map f \N$
By the Axiom of Specification:
- $\map {\paren {\RR^+}^{-1} } x$ is a set.
As this holds for all $x \in A$:
- $\RR^+$ is a set-like relation.
$\blacksquare$