Reflexive Reduction is Antireflexive

Theorem

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

Let $\RR^\ne$ denote the reflexive reduction of $\RR$.


Then $\RR^\ne$ is antireflexive.


Proof

By the definition of reflexive reduction:

$\RR^\ne = \RR \setminus \Delta_S$

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

By Set Difference Intersection with Second Set is Empty Set:

$\paren {\RR \setminus \Delta_S} \cap \Delta_S = \O$

Hence by Relation is Antireflexive iff Disjoint from Diagonal Relation, $\RR^\ne$ is antireflexive.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7: 1^\circ$
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