# Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation

## Theorem

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

Then $\RR$ is both antisymmetric and reflexive if and only if:

$\RR \cap \RR^{-1} = \Delta_S$

where $\Delta_S$ denotes the diagonal relation.

## Proof

### Necessary Condition

Let $\RR$ be both antisymmetric and reflexive.

Then:

 $\ds \RR \cap \RR^{-1}$ $\subseteq$ $\ds \Delta_S$ Relation is Antisymmetric iff Intersection with Inverse is Coreflexive $\ds \RR$ $\supseteq$ $\ds \Delta_S$ Definition 2 of Reflexive Relation $\ds \leadsto \ \$ $\ds \RR^{-1}$ $\supseteq$ $\ds \Delta_S$ Inverse of Reflexive Relation is Reflexive $\ds \leadsto \ \$ $\ds \RR \cap \RR^{-1}$ $\supseteq$ $\ds \Delta_S$ Intersection is Largest Subset $\ds \leadsto \ \$ $\ds \RR \cap \RR^{-1}$ $=$ $\ds \Delta_S$ Definition 2 of Set Equality

$\Box$

### Sufficient Condition

Let $\RR$ be such that:

$\RR \cap \RR^{-1} = \Delta_S$

Then by definition of set equality:

$\RR \cap \RR^{-1} \subseteq \Delta_S$

Also by definition of set equality:

$\Delta_S \subseteq \RR \cap \RR^{-1}$
$\RR \cap \RR^{-1} \subseteq \RR$
$\Delta_S \subseteq \RR$

So, by definition, $\RR$ is reflexive.

$\blacksquare$}