Null Relation is Antireflexive, Symmetric and Transitive

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 $\text{A}.2$: $11$
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