# Reflexive Circular Relation is Equivalence

## Theorem

Let $\mathcal R \subseteq S \times S$ be a reflexive and circular relation in $S$.

Then $\mathcal R$ is an equivalence relation.

## Proof

To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.

So, checking in turn each of the criteria for equivalence:

### Reflexive

By hypothesis $\mathcal R$ is reflexive.

$\Box$

### Symmetric

By reflexivity:

- $\tuple {x, x} \in \mathcal R$

If $\tuple {x, y} \in \mathcal R$ then by the definition of circular relation $\tuple {y, x} \in \mathcal R$.

Hence $\mathcal R$ is symmetric.

$\Box$

### Transitive

Let $\tuple {x, y}, \tuple {y, z} \in \mathcal R$.

By definition of circular relation:

- $\tuple {z, x} \in \mathcal R$

By $\mathcal R$ being symmetric:

- $\tuple {x, z} \in \mathcal R$

Hence $\mathcal R$ is transitive.

$\Box$

Thus is $\mathcal R$ is reflexive, symmetric and transitive, and therefore by definition an equivalence.

$\blacksquare$

## Sources

- 1967: Saunders Mac Lane and Garrett Birkhoff:
*Algebra*: $\S 1.3$ Exercise $2$