Transitive Closure Always Exists (Relation Theory)/Outline

Theorem
Let $\mathcal R$ be a relation on a set $S$.

Then the transitive closure $\mathcal R^+$ of $\mathcal R$ always exists.

Proof
First, note that there exists at least one transitive relation containing $\mathcal R$.

That is, the trivial relation $S \times S$, which is an equivalence and therefore transitive by definition.

Next, note that the Intersection of Transitive Relations is Transitive.

Hence the transitive closure of $\mathcal R$ is the intersection of all transitive relations containing $\mathcal R$.

Also see

 * Recursive Construction of Transitive Closure