Equivalence of Definitions of Transitive Closure (Relation Theory)/Intersection is Smallest/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 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 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.