Composition of Equivalence Relations: A Critical Analysis

Understanding the properties and composition of equivalence relations is crucial in the field of relation theory. In particular, it is essential to explore whether the composition of two equivalence relations, denoted as R circ S, necessarily retains the properties of an equivalence relation. This essay will delve into the properties of equivalence relations, the definition of relation composition, and provide a counterexample to illustrate the failure of certain properties under composition.

Properties of Equivalence Relations

Equivalence relations on a set must satisfy three key properties:

Reflexivity

For all a, a R a and a S a must hold. This ensures that every element is related to itself.

symmetry

If a R b, then b R a and if a S b, then b S a. This ensures that the relation is symmetric.

Transitivity

If a R b and b R c, then a R c and if a S b and b S c, then a S c. This ensures that the relation is transitive.

Composition of Relations

The composition of two relations R and S, denoted as R circ S, is defined as: a R circ S c if there exists b such that a R b and b S c.

Checking if R circ S is an Equivalence Relation

Reflexivity

To check if R circ S is reflexive, we need a R circ S a for all a.

- Since R and S are reflexive, for any a, we have a R a and a S a. By taking b a, we satisfy a R b and b S a. Therefore, a R circ S a holds.

Symmetry

To check if R circ S is symmetric, we need that if a R circ S b, then b R circ S a.

- Suppose a R circ S b. Then there exists c such that a R c and c S b.

- Since R is symmetric, we have c R a. Since S is symmetric, b S c implies c S b.

- However, this does not guarantee b R circ S a, as we need to find a d such that b R d and d S a, which might not be possible.

Transitivity

To check if R circ S is transitive, if a R circ S b and b R circ S c, we need to show a R circ S c.

- Suppose a R circ S b gives us a R d and d S b and b R circ S c gives us b R e and e S c.

- There is no guarantee that we can connect d and e through R or S.

Counterexample

Consider the following counterexample:

- Let R be an equivalence relation on the set A {1, 2} defined by R {(1, 1), (2, 2), (1, 2), (2, 1)}.

- Let S be an equivalence relation on the set B {3, 4} defined by S {(3, 3), (4, 4), (3, 4), (4, 3)}.

Now consider the relations R and S on the union A cup B {1, 2, 3, 4}. The composition R circ S does not apply here as they are on different sets.

- If we attempt to form R circ S on the union, it would involve elements from A and B, which are not related in either R or S. Therefore, the composition fails to satisfy the properties of an equivalence relation.

Conclusion

The composition R circ S of two equivalence relations is not necessarily an equivalence relation. The failure of symmetry and transitivity demonstrates this. This analysis underscores the importance of understanding the properties of relations and their compositions in relation theory.