Equivalence of Definitions of Reflexive Closure

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

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

Union with Diagonal is Smallest Reflexive Superset
Let $\Delta_S$ be the diagonal relation on $S$.

Let $\mathcal R^= = \mathcal R \cup \Delta_S$

By Smallest Element is Unique, at most one relation on $S$ can be the smallest reflexive superset of $\mathcal R$.

From Subset of Union:
 * $\mathcal R \subseteq \mathcal R^=$
 * $\Delta_S \subseteq \mathcal R^=$

By Relation Contains Diagonal Relation iff Reflexive, $\mathcal R^=$ is reflexive.

Thus $\mathcal R^=$ is a reflexive relation containing $\mathcal R$.

Again by Relation Contains Diagonal Relation iff Reflexive, every reflexive relation containing $\mathcal R$ must also contain $\Delta_S$.

From Union is Smallest Superset, it follows that $\mathcal R^=$ is the smallest reflexive relation on $S$ which contains $\mathcal R$.

Intersection of Reflexive Supersets is Union with Diagonal
Let $\mathcal Q$ be the set of all reflexive relations containing $\mathcal R$ as a subset.

Let $\mathcal R^= = \bigcap \mathcal Q$.

By the above proof that $\mathcal R \cup \Delta_S$ is a reflexive relation containing $\mathcal R$:
 * $\mathcal R \cup \Delta_S \in \mathcal Q$

By Intersection is Subset:
 * $\mathcal R^= \subseteq \mathcal R \cup \Delta_S$

By the above proof that $\mathcal R \cup \Delta_S$ is the smallest reflexive relation containing $\mathcal R$:
 * $\forall \mathcal P \in \mathcal Q: \mathcal R \cup \Delta_S \subseteq \mathcal P$

By Intersection is Largest Subset:
 * $\mathcal R \cup \Delta_S \subseteq \mathcal R^=$

Thus by definition of set equality:
 * $\mathcal R^= = \mathcal R \cup \Delta_S$