Equivalence Classes are Disjoint/Proof 1

Theorem

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

Then all $\RR$-classes are pairwise disjoint:

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

Proof

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First we show that:

$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$

Suppose two $\RR$-classes are not disjoint:

$\ds \eqclass x \RR \cap \eqclass y \RR \ne \O$

$\leadsto$

$\ds \exists z: z \in \eqclass x \RR \cap \eqclass y \RR$

Definition of Empty Set

$\ds $

$\leadsto$

$\ds \exists z: z \in \eqclass x \RR \land z \in \eqclass y \RR$

Definition of Set Intersection

$\ds $

$\leadsto$

$\ds \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {y, z} \in \RR}$

Definition of Equivalence Class

$\ds $

$\leadsto$

$\ds \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {z, y} \in \RR}$

Definition of Symmetric Relation: $\RR$ is symmetric

$\ds $

$\leadsto$

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

Definition of Transitive Relation: $\RR$ is transitive

So:

$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$

By the Rule of Transposition:

$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$

$\Box$

Now we show that:

$\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$

Suppose:

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

Then:

$\ds $

$\ds y \in \eqclass y \RR$

Definition of Equivalence Class

$\ds $

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

by hypothesis

$\ds $

$\leadsto$

$\ds y \in \eqclass x \RR$

Definition of Equivalence Class

$\ds $

$\leadsto$

$\ds y \in \eqclass x \RR \land y \in \eqclass y \RR$

Rule of Conjunction

$\ds $

$\leadsto$

$\ds y \in \eqclass x \RR \cap \eqclass y \RR$

Definition of Set Intersection

$\ds $

$\leadsto$

$\ds \eqclass x \RR \cap \eqclass y \RR \ne \O$

Definition of Empty Set

So:

$\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$

By the Rule of Transposition:

$\eqclass x \RR \cap \eqclass y \RR = \O \implies \paren {x, y} \notin \RR$

Using the rule of Biconditional Introduction on these results:

$\eqclass x \RR \cap \eqclass y \RR = \O \iff \paren {x, y} \notin \RR$

and the proof is complete.

$\blacksquare$

Sources

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 2.4$. Equivalence classes: Lemma $\text{(ii)}$
1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 17.4, \ \S 17.5 \ \text{(ii)}$: Equivalence classes