Reflexive Closure is Order Preserving

Theorem

Let $S$ be a set.

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

Then the reflexive closure operator is an order preserving mapping on $R$.

That is:

$\forall \RR, \SS \in R: \RR \subseteq \SS \implies \RR^= \subseteq \SS^=$

where $\RR^=$ and $\SS^=$ denote the reflexive closure of $\RR$ and $\SS$ respectively.

Proof

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:

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

$\blacksquare$