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) $$\Delta_S \subseteq \mathcal R$$ from Reflexive contains Diagonal Relation;
 * 2) $$\mathcal R^{-1} = \mathcal R \implies \mathcal R^{-1} \subseteq \mathcal R$$ and Relation equals Inverse iff Symmetric;
 * 3) $$\mathcal R \circ \mathcal R \subseteq \mathcal R$$ 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:


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


 * 1) From $$\Delta_S \subseteq \mathcal R$$, $$\mathcal R$$ is reflexive;
 * 2) From $$\mathcal R^{-1} \subseteq \mathcal R \implies \mathcal R^{-1} = \mathcal R$$ from Inverse Relation Equals iff Subset, $$\mathcal R$$ is symmetric;
 * 3) 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.