Partitions and Equivalence Relations
I want to show how these two concepts are a single mathematical idea.
Definition 1: A
partition of a nonempty set,
X, is a disjoint class
{Xi}i∈I of non-empty subsets of
X whose union is
X. Each of the
Xi is called a
partition set.
For example, if X=R, then X can be partitioned as
X=n∈Z⋃[n,n+1). Definition 2: A
binary relation, or simply
relation,
R, in the set
X is a subset of
X×X. For x,y∈X, we denote the fact (x,y)∈R by writing xRy. A function may be defined as a special kind of binary relation.
Let's assume that a partition of our non-empty set X is given, and we associate with this partition a relation, ∼, in X defined as follows: x∼y if x and y belong to the same partition set. It can easily checked that the relation ∼ satisfies:
x∼y for every x∈X (reflexivity);
x∼y⟹y∼x (symmetry);
x∼y and y∼z ⟹ x∼z (transitivity).
Definition 3: Any relation in
X which possesses these three properties is called an
equivalence relation in
X. Let X=Z and let x∼y if 2∣x−y for x,y∈X. Then clearly ∼ is an equivalence relation in X.
Let X, Y be any non-empty sets and f be a mapping from X onto Y. Let x∼y if f(x)=f(y) for x,y∈X. This defines an equivalence relation in X. Indeed, f(x)=f(x), and so x∼x. If f(x)=f(y) then f(y)=f(x), and so x∼y⟹y∼x. Finally, if f(x)=f(y) and f(y)=f(z) then f(x)=f(z), and so x∼y and y∼z implies x∼z. The first example is a special case of this one if we take X=Z, Y={0,1} and f(x)=xmod2.
We have just seen that each partition of X has associated with it a natural equivalence relation in X. Let us now reverse the situation and show that a given equivalence relation in X determines a natural partition of X.
Let ∼ be an equivalence relation in X. For every x∈X define the set
[x]:={y∈X:y∼x} called the
equivalence set of
x. We show that the class of all distinct equivalence sets forms a partition of
X. By reflexivity, x∈[x] for every x∈X, and thus each equivalence set is non-empty and their union is X. We now need to show that any two equivalence sets [x1] and [x2] are either disjoint or identical. We prove this by showing that if [x1] and [x2] are not disjoint then they are identical. To this end, let z be a common element of [x1] and [x2]. Let y be any element of [x1]. Using transitivity,
y∼x1∼z∼x2. Therefore, y∈[x2]. Since y was an arbitrary element of [x1], we get [x1]⊆[x2]. We can similarly show that [x2]⊆[x1]. In short, [x1]=[x2].
We have shown that there is no real distinction between partitions of a set and equivalence relations in the set. They are two equivalent approaches for the same mathematical idea. The approach we choose in an application depends entirely on our own convenience.