Reflexive Closure is Order Preserving

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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:

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

$\blacksquare$