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:

$\ds z$

$\in$

$\ds \eqclass x \RR$

$\ds \leadsto \ \ $

$\ds \tuple {x, z}$

$\in$

$\ds \RR$

Definition of Equivalence Class

$\ds \leadsto \ \ $

$\ds \tuple {z, x}$

$\in$

$\ds \RR$

Definition of Equivalence Relation: $\RR$ is symmetric

$\ds \leadsto \ \ $

$\ds \tuple {z, y}$

$\in$

$\ds \RR$

Definition of Equivalence Relation: $\RR$ is transitive

$\ds \leadsto \ \ $

$\ds \tuple {y, z}$

$\in$

$\ds \RR$

Definition of Equivalence Relation: $\RR$ is symmetric

$\ds \leadsto \ \ $

$\ds z$

$\in$

$\ds \eqclass y \RR$

Definition of Equivalence Class

So:

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

Now:

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

$\implies$

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

(see above)

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

$\implies$

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

Definition of Equivalence Relation: $\RR$ is symmetric

$\ds $

$\leadsto$

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

from above

$\ds $

$\leadsto$

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

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:

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

$\implies$

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

$\implies$

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

$\Box$

So by equivalence:

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

$\blacksquare$

Sources

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 2.4$. Equivalence classes: Lemma $\text{(i)}$
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Relations: Theorem $4$
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Equivalence Relations: $\S 17$
1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.3 \ (2)$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 17.2, \ \S 17.3$: Equivalence classes
2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text{A}$: Set Theory: Equivalence Relations