Transitive Closure Always Exists (Relation Theory)/Outline
Jump to navigation
Jump to search
Theorem
Let $\RR$ be a relation on a set $S$.
Then the transitive closure $\RR^+$ of $\RR$ always exists.
Outline of Proof
First, note that there exists at least one transitive relation containing $\RR$.
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 $\RR$ is the intersection of all transitive relations containing $\RR$.