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:


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

Proof

 * Let $\mathcal R$ be an equivalence. Then:


 * $(1): \quad \Delta_S \subseteq \mathcal R$ from Reflexive contains Diagonal Relation
 * $(2): \quad \mathcal R^{-1} = \mathcal R \implies \mathcal R^{-1} \subseteq \mathcal R$ and Relation equals Inverse iff Symmetric
 * $(3): \quad \mathcal R \circ \mathcal R \subseteq \mathcal R$ from Transitive Relation contains Composite with Self.

From Union Smallest, this gives us that:


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


 * Now 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 $\Delta_S \subseteq \mathcal R$, $\mathcal R$ is reflexive
 * From $\mathcal R^{-1} \subseteq \mathcal R \implies \mathcal R^{-1} = \mathcal R$ from Inverse Relation Equals iff Subset, $\mathcal R$ is symmetric
 * From $\mathcal R \circ \mathcal R \subseteq \mathcal R$, $\mathcal R$ is transitive.

So $\mathcal R$ is reflexive, symmetric and transitive, and thus an equivalence relation.