Union of Equivalence Classes is Whole Set

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

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

Proof
Also:

By definition of set equality:
 * $\displaystyle \bigcup S / \RR = S$

and so the set of $\RR$-classes constitutes the whole of $S$.

Also see

 * Fundamental Theorem on Equivalence Relations


 * Equivalence Classes are Disjoint
 * Equivalence Class is not Empty