Transitive Closure Always Exists (Relation Theory)/Proof

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

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

Proof
Note that the trivial relation $\mathcal T = S \times S$ on $S$ contains $\mathcal R$, by definition.

Further, $\mathcal T$ is transitive by Trivial Relation is Equivalence.

Thus there is at least one transitive relation on $S$ that contains $\mathcal R$.

Now define $\mathcal R^\cap$ as the intersection of all transitive relations on $S$ that contain $\mathcal R$:


 * $\displaystyle \mathcal R^\cap := \bigcap \left\{{\mathcal R': \text{$\mathcal R'$ is transitive and $\mathcal R \subseteq \mathcal R'$}}\right\}$

By Intersection of Transitive Relations is Transitive, $\mathcal R^\cap$ is also a transitive relation on $S$.

By Set Intersection Preserves Subsets, it also holds that $\mathcal R \subseteq \mathcal R^\cap$.

Lastly, by Intersection is Subset, for any transitive relation $\mathcal R'$ containing $\mathcal R$, it must be that $\mathcal R^\cap \subseteq \mathcal R'$.

Thus $\mathcal R^\cap$ is indeed the minimal transitive relation on $S$ containing $\mathcal R$.

That is, $\mathcal R^+ = \mathcal R^\cap$, and thence the transitive closure of $\mathcal R$ exists.