> [!NOTE] Theorem (Quotient partitions the set) > Suppose that $X$ is a set. Suppose that $E$ is an [[Equivalence relations|equivalence relation]] on $X.$ Then the [[Equivalence relations|quotient]] $X/E$ is a [[Partition of a Set|partition]] of $X.$ ^277258 **Proof**: Clearly $\bigcup_{x\in X} [x]_{E} \subset X.$ Let $x\in X.$ Since $E$ is reflexive we have that $x\in [x]_{E}$ so $[x]_{E}$ is non-empty. Moreover we have that $X= \bigcup_{x\in X} [x]_{E}.$ Take $x,y\in X.$ If $[x]_{E}$ and $[y]_{E}$ are disjoint, we are done. Otherwise suppose that $z\in [x]_{E} \cap [y]_{E}.$ Thus $xEz$ and $yEz.$ So by symmetry and transitivity we have $zEy$ and $xEy.$ Suppose now that $w\in [y]_{E}$ then $yEw.$ Since $xEy,$ by transitivity we have $xEw.$ Thus $[y]_{E}\subset[x]_{E}.$ Similarly, $[x]_{E}\subset[y]_{E}.$ So $[x]_{E}=[y]_{E}.$ # Applications **Consequences**: Note, [[For every partition there is a unique equivalence relation|every partition corresponds to a unique equivalence relation and vice versa]].