Equivalence of Definitions of Reflexive Closure

Theorem
Let $\RR$ be a relation on a set $S$.

Proof
Let $\RR$ be a relation on a set $S$.

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

Let $\RR^= = \RR \cup \Delta_S$

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

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

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

Thus $\RR^=$ is a reflexive relation containing $\RR$.

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

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

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

Let $\RR^= = \bigcap \QQ$.

By the above proof that $\RR \cup \Delta_S$ is a reflexive relation containing $\RR$:
 * $\RR \cup \Delta_S \in \QQ$

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

By the above proof that $\RR \cup \Delta_S$ is the smallest reflexive relation containing $\RR$:
 * $\forall \PP \in \QQ: \RR \cup \Delta_S \subseteq \PP$

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

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