Equivalence Relation is Circular

Theorem

Let $\RR \subseteq S \times S$ be an equivalence relation.

Then $\RR$ is also a circular relation.

Proof

Let $x, y, z \in S$ be arbitrary such that:

$\tuple {x, y} \in \RR$ and $\tuple {y, z} \in \RR$

We have a fortiori that $\RR$ is transitive.

Hence:

$\tuple {x, z} \in \RR$

We also have a fortiori that $\RR$ is symmetric.

Hence:

$\tuple {z, x} \in \RR$

We have demonstrated that:

$\tuple {x, y} \in \RR$ and $\tuple {y, z} \in \RR \implies \tuple {z, x} \in \RR$

and $\RR$ is circular by definition.

$\blacksquare$