Equivalence Classes are Disjoint/Proof 2

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Theorem

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


Then all $\mathcal R$-classes are pairwise disjoint:

$\tuple {x, y} \notin \mathcal R \iff \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O$


Proof

Suppose that for $x, y \in S$:

$\eqclass x {\RR} \cap \eqclass y \RR \ne \O$

Let:

$z \in \eqclass x \RR$
$z \in \eqclass y \RR$

Then by definition of equivalence class:

$\tuple {x, z} \in \RR$
$\tuple {y, z} \in \RR$


Let $c \in \eqclass x \RR$.

That is:

$\tuple {x, c} \in \RR$

By definition of equivalence relation, $\RR$ is symmetric so:

$\tuple {z, x} \in \RR$

By definition of equivalence relation, $\RR$ is transitive so:

$\tuple {z, x} \in \RR \land \tuple {x, c} \in \RR \implies \tuple {z, c} \in \RR$

and

$\tuple {y, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {y, c} \in \RR$

So we have $c \in \eqclass y \RR$.

By definition of subset:

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


Similarly, let $c \in \eqclass y \RR$.

That is:

$\tuple {y, c} \in \RR$

By definition of equivalence relation, $\RR$ is symmetric so:

$\tuple {z, y} \in \RR$

By definition of equivalence relation, $\RR$ is transitive so:

$\tuple {z, y} \in \RR \land \tuple {y, c} \in \RR \implies \tuple {z, c} \in \RR$

and

$\tuple {x, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {x, c} \in \RR$

So we have $c \in \eqclass x \RR$.

By definition of subset:

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


That is:

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

and

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

By definition of set equality:

$\eqclass x \RR = \eqclass y \RR$


Thus:

$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \eqclass x \RR = \eqclass y \RR$

and the result follows.

$\blacksquare$


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