Recall the  relation $\equiv \text{ mod} (4)$ on the set $\mathbb{ N}.$

One of the equivalence classes is $[0] = \{ ..., -8, -4, 0, 4, 8, ...\}$ which is equivalent to writing $[0] = [4] = [-4] = [8] = [-8] ...$

We could do this because the equivalence class collects all the natural numbers that are related to zero under the relation $\equiv \text{ mod} (4)$

The following theorem generalises this idea for any relation $\equiv \text{ mod} (n)$ on the set $\mathbb{ N}:$ for the integer $n.$

Let $R$ be an equivalence relation on set $A.$ If $a, b \in A,$ then $[a] = [b] \iff aRb.$

Essentially, equivalence classes  $[a] = [b]$ are equal if the elements  $a, b \in A,$ are related under the relation $R.$ And simultaneously, knowing that elements $a, b \in A,$ are related under $R$ means their equivalence classes  $[a] = [b]$ are equal.

An equivalence class  $\equiv \text{ mod} (n)$ divides set a $A$ into $n$ equivalence classes. We call this situation a partition of set $A.$

A partition of a set $A$ is defined as a set of non-empty subsets of $A,$ such that both these conditions are simultaneously satisfied:

(i) the union of all these subsets equals $A.$

(ii) the intersection of any two different subsets is

Let’s return to our example: $\equiv \text{ mod} (4)$ on the set $\mathbb{ N}.$ We could represent this set as:

• NOTE: Each equivalence class above represents an infinite set and despite the drawing suggesting $[0]$ is larger than $[3]$ for instance, this is not true.
• The union of these equivalence classes gives the whole set:

$[0] \cup [1] \cup [2] \cup [3] = \mathbb{ Z}$

In general, $[0] \cup [1] \cup [2] \cup ... \cup [n-1] \cup [n] = A$ for the relation $\equiv \text{mod}(n)$ on a set $A.$

• The intersection of any of the equivalence classes results in an empty set:

$[0] \cap [1]$ for example is an empty set $\emptyset$ since there are no elements in $[0]$ that are also in $[1]$

In general, we write $[a] \cap [b] = \emptyset \text{ } \forall a, b \in \mathbb{N}$

• The partition formed for this set $\{ [0], [1], [2], [3] \}$

Let’s consider a new example.

Consider the partition $P = \big \{ \{ ..., -4, -2, 0, 2, 4, ... \} \text{, } \{ ..., -5, -3, -1, 1, 3, 5, ... \} \big \} \text{ of } \mathbb{ Z}.$

Let $R$ be the equivalence relation on the set of integers. The equivalence classes are the two elements of $P.$ We want to identify the equivalence relation $R.$

Solution:

• Clearly, the set $\mathbb{Z}$ has been split into odd and even integers. So we write

$R : \equiv \text{(mod)} 2$ on the set of integers.

Note the following:

• We have exactly two equivalence sets, each are infinite. Since this is a partition, the sets do not intersect.
• Since $P$ is a partition, we also know the union of the two sets gives us $\mathbb{ Z}.$
• So we can conclude that these are the equivalence classes of $\mathbb{ Z }$ if the relation $R = \equiv \text{( mod)} 2.$

Hence:

• So $[0] \cup [1] = \mathbb{ Z}$
• The equivalence classes do not intersect: $[0] \cap [1] = \emptyset$

Consider the next example:

List all the partitions of the set $X = \{ a, b\}$

Solution:

• We have the partition $\{ \{a\}, \{b\} \}$ which corresponds to the equivalence relation We have $R = \{ (a,a), (b,b)\}$
• We also have the partition $\{ \{a, b \} \}$ corresponding to the equivalence relation $R =\{ (a,a), (a,b), (b,a), (b,b)\}$
• Note in both partitions, the union of the elements give the set $X$ and the intersection of the elements is empty.

 How clear is this post?