B.2 Relations

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## B.2 Relations

A binary relation R on two sets A and B is a subset of the Cartesian product A × B. If (a, b) ∈ R, we sometimes write a R b. When we say that R is a binary relation on a set A, we mean that R is a subset of A × A. For example, the "less than" relation on the natural numbers is the set {(a, b) : a, b ∈ N and a < b}. An n-ary relation on sets A1, A2,..., An is a subset of A1 × A2 × ··· × An.

A binary relation R ⊆ A × A is reflexive if

a R a

for all a ∈ A. For example, "=" and "≤" are reflexive relations on N, but "<" is not. The relation R is symmetric if

a R b implies b R a

for all a, b ∈ A. For example, "=" is symmetric, but "<" and "≤" are not. The relation R is transitive if

a R b and b R c imply a R c

for all a, b, c ∈ A. For example, the relations "<," "≤," and "=" are transitive, but the relation R = {(a, b) : a, b ∈ N and a = b - 1} is not, since 3 R 4 and 4 R 5 do not imply 3 R 5.

A relation that is reflexive, symmetric, and transitive is an equivalence relation. For example, "=" is an equivalence relation on the natural numbers, but "<" is not. If R is an equivalence relation on a set A, then for a ∈ A, the equivalence class of a is the set [a] = {b ∈ A : a R b}, that is, the set of all elements equivalent to a. For example, if we define R = {(a, b) : a, b ∈ N and a + b is an even number}, then R is an equivalence relation, since a + a is even (reflexive), a + b is even implies b + a is even (symmetric), and a + b is even and b + c is even imply a + c is even (transitive). The equivalence class of 4 is [4] = {0, 2, 4, 6,...}, and the equivalence class of 3 is [3] = {1, 3, 5, 7,...}. A basic theorem of equivalence classes is the following.

Theorem B.1: (An equivalence relation is the same as a partition)

The equivalence classes of any equivalence relation R on a set A form a partition of A, and any partition of A determines an equivalence relation on A for which the sets in the partition are the equivalence classes.

Proof For the first part of the proof, we must show that the equivalence classes of R are nonempty, pairwise-disjoint sets whose union is A. Because R is reflexive, a ∈ [a], and so the equivalence classes are nonempty; moreover, since every element a ∈ A belongs to the equivalence class [a], the union of the equivalence classes is A. It remains to show that the equivalence classes are pairwise disjoint, that is, if two equivalence classes [a] and [b] have an element c in common, then they are in fact the same set. Now a R c and b R c, which by symmetry and transitivity imply a R b. Thus, for any arbitrary element x ∈ [a], we have x R a implies x R b, and thus [a] ⊆ [b]. Similarly, [b] ⊆ [a], and thus [a] = [b].

For the second part of the proof, let be a partition of A, and define R = {(a, b) : there exists i such that a ∈ Ai and b ∈ Ai }. We claim that R is an equivalence relation on A. Reflexivity holds, since a ∈ Ai implies a R a. Symmetry holds, because if a R b, then a and b are in the same set Ai, and hence b R a. If a R b and b R c, then all three elements are in the same set, and thus a R c and transitivity holds. To see that the sets in the partition are the equivalence classes of R, observe that if a ∈ Ai, then x ∈ [a] implies x ∈ Ai, and x ∈ Ai implies x ∈ [a].

A binary relation R on a set A is antisymmetric if

a R b and b R a imply a = b.

For example, the "≤" relation on the natural numbers is antisymmetric, since a ≤ b and b ≤ a imply a = b. A relation that is reflexive, antisymmetric, and transitive is a partial order, and we call a set on which a partial order is defined a partially ordered set. For example, the relation "is a descendant of" is a partial order on these of all people (if we view individuals as being their own descendants).

In a partially ordered set A, there may be no single "maximum" element a such that b R a for all b ∈ A. Instead, there may several maximal elements a such that for no b ∈ A, where b ≠ a, is it the case that a R b. For example, in a collection of different-sized boxes there may be several maximal boxes that don't fit inside any other box, yet no single "maximum" box into which any other box will fit.[3]

A partial order R on a set A is a total or linear order if for all a, b ∈ A, we have a R b or b R a, that is, if every pairing of elements of A can be related by R. For example, the relation "≤" is a total order on the natural numbers, but the "is a descendant of" relation is not a total order on the set of all people, since there are individuals neither of whom is descended from the other.

Exercises B.2-1

Prove that the subset relation "⊆" on all subsets of Z is a partial order but not a total order.

Exercises B.2-2

Show that for any positive integer n, the relation "equivalent modulo n" is an equivalence relation on the integers. (We say that a ≡ b (mod n) if there exists an integer q such that a - b = qn.) Into what equivalence classes does this relation partition the integers?

Exercises B.2-3

Give examples of relations that are

1. reflexive and symmetric but not transitive,

2. reflexive and transitive but not symmetric,

3. symmetric and transitive but not reflexive.

Exercises B.2-4

Let S be a finite set, and let R be an equivalence relation on S × S. Show that if in addition R is antisymmetric, then the equivalence classes of S with respect to R are singletons.

Exercises B.2-5

Professor Narcissus claims that if a relation R is symmetric and transitive, then it is also reflexive. He offers the following proof. By symmetry, a R b implies b R a. Transitivity, therefore, implies a R a. Is the professor correct?

[3]To be precise, in order for the "fit inside" relation to be a partial order, we need to view a box as fitting inside itself.

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