Trivial Relation is Equivalence

Theorem

The trivial relation on $S$:

$\RR = S \times S$

is always an equivalence in $S$.

Let us verify the conditions for an equivalence in turn.

For $\RR$ to be reflexive means:

$\forall x \in S: \tuple {x, x} \in S \times S$

which is trivial by definition of the Cartesian product $S \times S$.

$\Box$

For $\RR$ to be symmetric means:

$\forall x, y \in S: \tuple {x, y} \in S \times S \land \tuple {y, x} \in S \times S$

Since we have by definition of Cartesian product that:

$\forall x, y \in S: \tuple {y, x} \in S \times S$

$\Box$

For $\RR$ to be transitive means:

$\tuple {x, y} \in S \times S \land \tuple {y, z} \in S \times S \implies \tuple {x, z} \in S \times S$

By definition of Cartesian product, we have that:

$\forall x, z \in S: \tuple {x, z} \in S \times S$

hence by True Statement is implied by Every Statement, it follows that $\RR$ is transitive.

$\Box$

Having verified all three conditions, we conclude $\RR$ is an equivalence.

$\blacksquare$

1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 7$: Relations
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Relations
1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $15$