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) $$\mathcal{R}^{-1} = \mathcal{R} \Longrightarrow \mathcal{R}^{-1} \subseteq \mathcal{R}$$ 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 $$\mathcal{R}^{-1} \subseteq \mathcal{R} \Longrightarrow \mathcal{R}^{-1} = \mathcal{R}$$ 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.