Null Relation is Antireflexive, Symmetric and Transitive
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Theorem
Let $S$ be a set which is non-empty.
Let $\RR \subseteq S \times S$ be the null relation.
Then $\RR$ is antireflexive, symmetric and transitive.
If $S = \O$ then Relation on Empty Set is Equivalence applies.
Proof
From the definition of null relation:
- $\RR = \O$
Antireflexivity
This follows directly from the definition:
- $\RR = \O \implies \forall x \in S: \tuple {x, x} \notin \RR$
and so $\RR$ is antireflexive.
$\Box$
Symmetry
It follows vacuously that:
- $\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$
and so $\RR$ is symmetric.
$\Box$
Transitivity
It follows vacuously that:
- $\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
and so $\RR$ is transitive.
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $24$
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $11$