## Theorem 1.3.3: Equivalence Classes

*r*be an equivalence relation on a set

**A**. Then

**A**can be written as a union of disjoint sets with the following properties:

- If
*a,b*are in**A**then*a ~ b*if and only if*a*and*b*are in the same set - The subsets are non-empty and pairwise disjoint.

**equivalence classes**.

### Proof:

We have to decide what the equivalence classes should be: Since by property (1) two elements*a*and

*b*are supposed to be related if and only if they are in the same class, it seems natural to define:

- =
{
*b***A**:*b ~*}

**A**() . Now we need to check whether this is a good definition, and whether it satisfies the above properties.

Because of reflexivity of the equivalence relation, the class * A(a)*
contains the element

*a*and is therefore not empty.

Next, we will show that two classes that are different can not have any
elements in common, i.e. they are disjoint. Recall that disjoint means that
two sets have either an empty intersection or are the same sets. Take two
elements *a* and *a’* and suppose that * A(a)* and

*have a non-empty intersection; say both classes contain the element*

**A**(a’)*c*. Then

*c ~ a*, because*c*is in**A**(a)*a ~ c*, by symmetry and the above line*c ~ a’*because*c*is in**A**(a’)- Therefore, by transitivity,
*a ~ a’*

*b*in

*. Then*

**A**(a’)*b ~ a’*and

*a’ ~ a*, so that

*b ~ a*by transitivity, and hence

*b*is contained in

*. Therefore*

**A**(a)*is a subset of*

**A**(a’)*. The same argument works the other way around as well (by symmetry), showing that*

**A**(a)*is a subset of*

**A**(a)*. Hence,*

**A**(a’)*. This shows that two different classes must be disjoint.*

**A**(a) =**A**(a’)
Finally, we need to show that all classes * A(a)* make up the
original set

**A**. But that is clear, because every element

*a*in

**A**belongs to the class

*, hence the union of all classes*

**A**(a)*must be contained in*

**A**(a)**A**, and therefore must be equal to

**A**.