Reflexive Reduction is Antireflexive
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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 $14.7: 1^\circ$