Equivalence Classes are Disjoint/Proof 1
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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
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 |
Thus we have shown that $\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$.
Therefore, by the Rule of Transposition:
- $\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$
Now we show that:
- $\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$
Suppose $\tuple {x, y} \in \RR$.
\(\ds \) | \(\) | \(\ds y \in \eqclass y \RR\) | 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 |
Thus we have shown that:
- $\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$
Therefore, 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