Reflexive Closure is Closure Operator

Theorem

Let $S$ be a set.

Let $R$ be the set of all endorelations on $S$.

Then the reflexive closure operator on $R$ is a closure operator.

Proof 1

Let $\QQ$ be the set of reflexive relations on $S$.

By Intersection of Reflexive Relations is Reflexive, the intersection of any subset of $\QQ$ is in $Q$.

By the definition of reflexive closure as the intersection of reflexive supersets:

The reflexive closure of a relation $\RR$ on $S$ is the intersection of elements of $\QQ$ that contain $S$.

From Closure Operator from Closed Sets we conclude that reflexive closure is a closure operator.

$\blacksquare$

Proof 2

Reflexive Closure is Inflationary

Let $\RR \in R$.

The reflexive closure $\RR^=$ of $\RR$ is defined as:

$\RR^= := \RR \cup \Delta_S$

From Set is Subset of Union:

$\RR \subseteq \RR^=$

Hence the reflexive closure operator is an inflationary mapping.

$\Box$

Reflexive Closure is Order Preserving

Let $\RR, \SS \in R$.

Suppose:

$\RR \subseteq \SS$

Their respective reflexive closures $\RR^=$ and $\SS^=$ are defined as:

$\RR^= := \RR \cup \Delta_S$

$\SS^= := \SS \cup \Delta_S$

Hence by Corollary to Set Union Preserves Subsets:

$\RR^= \subseteq \SS^=$

$\Box$

Reflexive Closure is Idempotent

Let $\RR \in R$.

By the definition of reflexive closure:

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

$\paren {\RR^=}^= = \paren {\RR \cup \Delta_S} \cup \Delta_S$

By Union is Associative:

$\paren {\RR^=}^= = \RR \cup \paren {\Delta_S \cup \Delta_S}$

By Set Union is Idempotent:

$\paren {\RR^=}^= = \RR \cup \Delta_S$

Hence:

$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$

$\Box$

Thus by the definition of closure operator, reflexive closure is a closure operator.

$\blacksquare$