Reflexive Closure is Closure Operator

From ProofWiki
Jump to navigation Jump to search

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 $\mathcal Q$ be the set of reflexive relations on $S$.

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

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

The reflexive closure of a relation $\mathcal R$ on $S$ is the intersection of elements of $\mathcal Q$ 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 $\mathcal R \in R$.

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

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

From Set is Subset of Union:

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

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 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.

$\blacksquare$