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:


 * $\forall \mathcal R, \mathcal S \in R: \mathcal R \subseteq \mathcal S \implies \mathcal R^= \subseteq S^=$

where $\mathcal R^=$ and $\mathcal S^=$ denote the reflexive closure of $\mathcal R$ and $\mathcal S$ respectively.

Proof
Let $\mathcal R, \mathcal S \in R$.

Suppose:


 * $\mathcal R \subseteq \mathcal S$

Their respective reflexive closures $\mathcal R^=$ and $\mathcal S^=$ are defined as:


 * $\mathcal R^= := \mathcal R \cup \Delta_S$


 * $\mathcal S^= := \mathcal S \cup \Delta_S$

Hence by Corollary to Set Union Preserves Subsets:


 * $\mathcal R^= \subseteq S^=$

$\blacksquare$