# Equivalence Class Equivalent Statements

## Theorem

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

Let $x, y \in S$.

The following statements are equivalent:

$(1): \quad x$ and $y$ are in the same $\RR$-class
$(2): \quad \eqclass x \RR = \eqclass y \RR$
$(3): \quad x \mathrel \RR y$
$(4): \quad x \in \eqclass y \RR$
$(5): \quad y \in \eqclass x \RR$
$(6): \quad \eqclass x \RR \cap \eqclass y \RR \ne \O$

## Proof

### $(1)$ Equivalent to $(2)$

$\eqclass x \RR$ is the unique $\RR$-class to which $x$ belongs

and:

$\eqclass y \RR$ is the unique $\RR$-class to which $y$ belongs.

As these are unique for each, they must be the same set.

$\blacksquare$

### 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:

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

$\Box$

So by equivalence:

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

$\blacksquare$

### $(3)$ Equivalent to $(4)$

This follows directly by the definition of equivalence class.

$\blacksquare$

### $(3)$ Equivalent to $(5)$

This follows through dint of the symmetry of $\RR$ and the definition of Equivalence Class.

$\blacksquare$

### $(3)$ Equivalent to $(6)$

Follows directly from Equivalence Classes are Disjoint.

$\blacksquare$