Equivalence Class holds Equivalent Elements

Theorem

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

Then:

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

Proof

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:

 $\displaystyle z$ $\in$ $\displaystyle \eqclass x \RR$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {x, z}$ $\in$ $\displaystyle \RR$ Definition of Equivalence Class $\displaystyle \leadsto \ \$ $\displaystyle \tuple {z, x}$ $\in$ $\displaystyle \RR$ Definition of Equivalence Relation: $\RR$ is symmetric $\displaystyle \leadsto \ \$ $\displaystyle \tuple {z, y}$ $\in$ $\displaystyle \RR$ Definition of Equivalence Relation: $\RR$ is transitive $\displaystyle \leadsto \ \$ $\displaystyle \tuple {y, z}$ $\in$ $\displaystyle \RR$ Definition of Equivalence Relation: $\RR$ is symmetric $\displaystyle \leadsto \ \$ $\displaystyle z$ $\in$ $\displaystyle \eqclass y \RR$ Definition of Equivalence Class

So:

$\eqclass x \RR \subseteq \eqclass y \RR$

Now:

 $\displaystyle \tuple {x, y} \in \RR$ $\implies$ $\displaystyle \eqclass x \RR \subseteq \eqclass y \RR$ (see above) $\displaystyle \tuple {x, y} \in \RR$ $\implies$ $\displaystyle \tuple {y, x} \in \RR$ Definition of Equivalence Relation: $\RR$ is symmetric $\displaystyle$ $\leadsto$ $\displaystyle \eqclass y \RR \subseteq \eqclass x \RR$ from above $\displaystyle$ $\leadsto$ $\displaystyle \eqclass y \RR = \eqclass x \RR$ Definition of Set Equality

... so we have shown that:

$\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$.

$\Box$

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:

 $\displaystyle \tuple {x, y} \in \RR$ $\implies$ $\displaystyle \eqclass x \RR = \eqclass y \RR$ $\displaystyle \eqclass x \RR = \eqclass y \RR$ $\implies$ $\displaystyle \tuple {x, y} \in \RR$

$\Box$

So by equivalence:

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

$\blacksquare$