Equivalence of Definitions of Reflexive Closure

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Theorem

The following definitions of the concept of Reflexive Closure are equivalent:

Let $\RR$ be a relation on a set $S$.

Definition 1

Let $\RR$ be a relation on a set $S$.


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

$\RR^= := \RR \cup \set {\tuple {x, x}: x \in S}$

That is:

$\RR^= := \RR \cup \Delta_S$

where $\Delta_S$ is the diagonal relation on $S$.


Definition 2

Let $\RR$ be a relation on a set $S$.


The reflexive closure of $\RR$ is defined as the smallest reflexive relation on $S$ that contains $\RR$ as a subset.


The reflexive closure of $\RR$ is denoted $\RR^=$.


Definition 3

Let $\RR$ be a relation on a set $S$.

Let $\QQ$ be the set of all reflexive relations on $S$ that contain $\RR$.

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

$\RR^= := \bigcap \QQ$

That is:

$\RR^=$ is the intersection of all reflexive relations on $S$ containing $\RR$.


Proof

Let $\RR$ be a relation on a set $S$.

Union with Diagonal is Smallest Reflexive Superset

Let $\Delta_S$ be the diagonal relation on $S$.

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

By Smallest Element is Unique, at most one relation on $S$ can be the smallest reflexive superset of $\RR$.


From Subset of Union:

$\RR \subseteq \RR^=$
$\Delta_S \subseteq \RR^=$

By Relation Contains Diagonal Relation iff Reflexive, $\RR^=$ is reflexive.

Thus $\RR^=$ is a reflexive relation containing $\RR$.


Again by Relation Contains Diagonal Relation iff Reflexive, every reflexive relation containing $\RR$ must also contain $\Delta_S$.

From Union is Smallest Superset, it follows that $\RR^=$ is the smallest reflexive relation on $S$ which contains $\RR$.

$\Box$


Intersection of Reflexive Supersets is Union with Diagonal

Let $\QQ$ be the set of all reflexive relations containing $\RR$ as a subset.

Let $\RR^= = \bigcap \QQ$.

By the above proof that $\RR \cup \Delta_S$ is a reflexive relation containing $\RR$:

$\RR \cup \Delta_S \in \QQ$

By Intersection is Subset:

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

By the above proof that $\RR \cup \Delta_S$ is the smallest reflexive relation containing $\RR$:

$\forall \PP \in \QQ: \RR \cup \Delta_S \subseteq \PP$

By Intersection is Largest Subset:

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

Thus by definition of set equality:

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

$\blacksquare$

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