Relational Closure from Transitive Closure

Theorem
Let $A$ be a set or class.

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

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

Let $B \subseteq A$.

Let $B' = B \cup \map {\paren {\RR^+}^{-1} } B$.

Let $C$ be an $\RR$-transitive subset or subclass of $A$ such that $B \subseteq C$.

Then:


 * $B'$ is $\RR$-transitive
 * $B' \subseteq C$
 * If $B$ is a set and $\RR$ is set-like then $B'$ is a set. That is, $B'$ is the relational closure of $B$ under $\RR$.

$B'$ is $\RR$-transitive
Let $x \in B'$ and $y \in A$, and let $y \mathrel \RR x$.

If $x \in B$, then by the definition of transitive closure:
 * $y \mathrel {\RR^+} x$

so:
 * $y \in B'$

Let $x \in \map {\paren {\RR^+}^{-1} } B$.

Then:
 * $x \mathrel {\RR^+} b$

for some $b \in B$.

Since $\RR \subseteq \RR^+$, it follows that:
 * $y \mathrel {\RR^+} x$

Since $\RR^+$ is transitive:
 * $y \mathrel {\RR^+} b$

That is:
 * $y \in \map {\paren {\RR^+}^{-1} } B$

so $y \in B'$.

As this holds for all such $x$ and $y$, $B'$ is $\RR$-transitive.

$B' \subseteq C$
Let $x \in B'$.

Then $x \in B$ or $x \in \map {\paren {\RR^+}^{-1} } B$.

Let $x \in B$.

Then because $B \subseteq C$:
 * $x \in C$

Suppose that $x \in \map {\paren {\RR^+}^{-1} } B$.

Then for some $b \in B$:
 * $x \mathrel \RR b$

By the definition of transitive closure:


 * for some $n \in \N_{>0}$ there exist $a_0, a_1, \dots, a_n$ such that:
 * $x = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = b$

Thus by the Principle of Mathematical Induction:
 * $x \in C$

Set-like implies set
Let $B$ be a set.

Let $\RR$ be set-like.

Then by Inverse Image of Set under Set-Like Relation is Set:
 * $\paren {\RR^+}^{-1}$ is a set.

Thus $B'$ is a set by the Axiom of Unions.