Equivalence Class holds Equivalent Elements

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

Then:


 * $\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$

Necessary Condition
First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$.

Suppose:
 * $\tuple {x, y} \in \RR: x, y \in S$

Then:

So:
 * $\eqclass x \RR \subseteq \eqclass y \RR$

Now:

... so we have shown that:
 * $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$

Sufficient Condition
Next we prove that $\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$.

By definition of set equality:
 * $\eqclass x \RR = \eqclass y \RR$

means:
 * $\paren {x \in \eqclass x \RR \iff x \in \eqclass y \RR}$

So by definition of equivalence class:
 * $\tuple {y, x} \in \RR$

Hence by definition of equivalence relation: $\RR$ is symmetric
 * $\tuple {x, y} \in \RR$

So we have shown that
 * $\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$

Thus, we have:

So by equivalence:
 * $\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$