Equivalence relations and partitions: some core basic theorems

Suppose R is an equivalence relation on a set S. For each a in S, let [a] denote the set of elements of S to which a is related under R, that is, [a] = \{ x: (a,x) \in R\}

We call [a] the equivalence class of a in S under R. The collection of all such equivalence classes is denoted by S/R, that is, S/R = \{ [a]: a \in S\}. It is called the quotient set of S by R.

The fundamental property of an equivalence relation and its quotient set is contained in the following theorem:

Theorem I:

Let R be an equivalence relation on a set S. Then, the quotient set S/R is a partition of S. Specifically,

(i) For each a \in S, we have a \in [a].

(ii) [a]=[b] if and only if (a,b) \in R.

(iii) If [a] \neq [b], then [a] and [b] are disjoint.

Proof of (i):

Since R is reflexive, (a,a) \in R for every a \in S and therefore a \in [a].

Proof of (ii):

Assume: (a,b) \in R.

we want to show that [a] = [b]. That is, we got to prove, (i) [b] \subseteq [a] and (ii) [a] \subseteq [b].

Let x \in [b]; then, (b,x) \in R. But, by hypothesis (a,b) \in R and so, by transitivity, (a,x) \in R. Accordingly, x \in [a]. Thus, [b] \subseteq [a].

To prove that [a] \subseteq [b], we observe that (a,b) \in R implies, by symmetry, that (b,a) \in R. Then, by a similar argument, we obtain [a] \subseteq [b]. Consequently, [a]=[b].

Now, assume [a] = [b].

Then by part (i) of this proof that for each a \in S, we have a \in [a]. So, also, here b \in [b]=[a]; hence, (a,b) \in R.

Proof of (iii):

Here, we prove the equivalent contrapositive of the statement (iii), that is:

If [a] \bigcap [b] \neq \emptyset then [a] = [b].

if [a] \bigcap [b] \neq \emptyset then there exists an element x \in A with x \in [a] \bigcap [b]. Hence, (a,x) \in R and (b,x) \in R. By symmetry, (x,b) \in R, and, by transitivity, (a,b) \in R. Consequently, by proof (ii), [a] = [b].

The converse of the above theorem is also true. That is,

Theorem II:

Suppose P = \{ A_{i}\} is a partition of set S. Then, there is an equivalence relation \sim on S such that the set S/\sim of equivalence classes is the same as the partition P = \{ A_{i}\}.

Specifically, for a, b \in S, the equivalence \sim in Theorem I is defined by a \sim b if a and b belong to the same cell in P.

Thus, we see that there is a one-one correspondence between the equivalence relations on a set S and the partitions of S.

Proof of Theorem II:

Let a, b \in S, define a \sim b if a and b belong to the same cell A_{k} in P. We need to show that \sim is reflexive, symmetric and transitive.

(i) Let a \in S. Since P is a partition there exists some A_{k} in P such that a \in A_{k}. Hence, a \sim a. Thus, \sim is reflexive.

(ii) Symmetry follows from the fact that if a, b \in A_{k}, then b, a \in A_{k}.

(iii) Suppose a \sim b and b \sim c. Then, a, b \in A_{i} and b, c \in A_{j}. Therefore, b \in A_{i} \bigcap A_{j}. Since P is a partition, A_{i} = A_{j}. Thus, a, c \in A_{i} and so a \sim c. Thus, \sim is transitive.

Accordingly, \sim is an equivalence relation on S.

Furthermore,

[a] = \{ x: a \sim x\}.

Thus, the equivalence classes under \sim are the same as the cells in the partition P.

More later,

Nalin Pithwa.