Union of Equivalence Classes is Whole Set

Theorem
Let $\mathcal R \subseteq S \times S$ be an equivalence on a set $S$.

Then the set of $\mathcal R$-classes constitutes the whole of $S$.

Proof
Also:

From Equality of Sets, $\displaystyle \bigcup {S / \mathcal R} = S$, and so the set of $\mathcal R$-classes constitutes the whole of $S$.

Also see

 * Fundamental Theorem on Equivalence Relations


 * Equivalence Classes are Disjoint
 * Equivalence Class is Not Empty