Equivalence of Definitions of Equivalence Relation

Theorem
Let $\mathcal R \subseteq S \times S$ be a relation on the set $S$.

Then $\mathcal R \subseteq S \times S$ is an equivalence relation iff:


 * $\Delta_S \cup \mathcal R^{-1} \cup \mathcal R \circ \mathcal R \subseteq \mathcal R$

Necessary Condition
Let $\mathcal R$ be an equivalence relation.

By definition, $\mathcal R$ is reflexive, symmetric and transitive.

From Reflexive contains Diagonal Relation:
 * $\Delta_S \subseteq \mathcal R$

From Relation equals Inverse iff Symmetric:
 * $\mathcal R^{-1} = \mathcal R$

and so from Equality of Sets:
 * $\mathcal R^{-1} \subseteq \mathcal R$

From Relation contains Composite with Self iff Transitive:
 * $\mathcal R \circ \mathcal R \subseteq \mathcal R$

From Union Smallest:


 * $\Delta_S \cup \mathcal R^{-1} \cup \mathcal R \circ \mathcal R \subseteq \mathcal R$

Sufficient Condition
Suppose that $\Delta_S \cup \mathcal R^{-1} \cup \mathcal R \circ \mathcal R \subseteq \mathcal R$.

Then, by Union Smallest, we have:


 * $(1): \quad \Delta_S \subseteq \mathcal R$
 * $(2): \quad \mathcal R^{-1} \subseteq \mathcal R$
 * $(3): \quad \mathcal R \circ \mathcal R \subseteq \mathcal R$

From Reflexive contains Diagonal Relation, $(1)$ gives directly that $\mathcal R$ is reflexive.

From Inverse Relation Equal iff Subset, $(2)$ gives that $\mathcal R^{-1} = \mathcal R$.

So from Relation equals Inverse iff Symmetric $\mathcal R$ is symmetric.

From Relation contains Composite with Self iff Transitive, $(3)$ gives that $\mathcal R$ is transitive.

So $\mathcal R$ has been shown to be reflexive, symmetric and transitive, and so by definition an equivalence relation.