2.5 Relations and their properties

Let A and B be sets. A binary relation from A to B is a subset of . In other words, a binary relation from A to B is a set R of ordered pairs where the first element of each ordered pair comes from A and the second element comes from B. We use the notation aRb to denote that (a, b)  R and to denote that (a, b)  R. Moreover, when (a, b) belongs to R, a is said to be related to b by R. Binary relations represent relationships between the elements of two sets. We will omit the word “binary” when there is no danger of confusion.

Example. Let A be the set of all cities, and let B be the set of the 50 states in the USA. Define the relation R by specifying that (a, b) belongs to R if city a is in state b. For instance, (Boulder, Colorado), (Chicago, Illinois), (Cupertino, California) are in R, but (Chicago, Colorado), (Cupertino, Illinois) are not in R.

Functions as relations. Recall that a function f from a set A to a set B assigns a unique element of B to each element of A. The graph of f is the set of ordered pairs (a, b) such that b = f(a). Since the graph of f is a subset of , it is a relation from A to B. Moreover, the graph of a function has the property that every element of A is the first element of exactly one ordered pair of the graph. Conversely, if R is a relation from A to B such that every element in A is the first element of exactly one ordered pair of R, then a function can be defined with R as its graph. This can be done by assigning to an element a of A the unique element such that (a, b)  R.

2.5.1 Combining relations

Since relations from A to B are subsets of , two relations from A to B can be combined in any way two sets can be combined.

Example. Let A = {1, 2, 3} and B = {1, 2, 3, 4}. The relations R₁ = {(1, 1), (2, 2), (3, 3)} and R₂ = {(1, 1), (1, 2), (1, 3), (1, 4)} can be combined to obtain , , .

There is another way that relations are combined which is analogous to the composition of functions.

Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of ordered pairs (a, c), where , and for which there exists an element such that and . We denote the composite of R and S by .

Theorem 1. The relation R on a set A is transitive iff for n = 1, 2, 3, …

n-ary relations

Let A₁, A₂, …, A_n be sets. An n-ary relation on these sets is a subset of . The sets A₁, A₂, …, A_n are called the domains of the relation, and n is called its degree.

Example. Let R be the relation consisting of triples (a, b, c), where a, b and c are integers with a < b < c. Then , but . The degree of this relation is 3. Its domains are all equal to the set of integers.

2.5.2 Representing relations using matrices

A relation between finite sets can be represented using a zero-one matrix. Suppose that R is a relation from to . The relation R can be represented by the matrix , where .

In other words, the zero-one matrix representing R has a 1 as its (i, j) entry when a_i is related to b_j, and a 0 in this position if a_i is not related to b_j.

Example. Suppose that A = {1, 2, 3} and B = {1, 2}. Let R be the relation from A to B containing (a, b) if and a > b. What is the matrix representing R if and ?

Solution: Since , the matrix for R is .

2.5.3 Representing relations using digraphs

A directed graph, or digraph, consists of a set V of vertices (or nodes) together with a set E of ordered pairs of elements of V called edges (or arcs). The vertex a is called the initial vertex of the edge (a, b), and the vertex b is called the terminal vertex of this edge.

An edge of the form (a, a) is represented using an arc from the vertex a back to itself. Such an edge is called a loop.

Example. The directed graph with vertices a, b, c and d, and edges (a, b), (a, d), (b, b), (b, d), (c, a), (c, b) and (d, b) is displayed in Figure:

2.5.4 Equivalence relations

A relation on a set A is called an equivalence relation if it is reflexive, symmetric and transitive. Two elements that are related by an equivalence relation are called equivalent.

Example. Suppose that R is the relation on the set of strings of English letters such that aRb iff l(a) = l(b), where l(x) is the length of the string x. Is R an equivalence relation?

Solution: Since l(a) = l(a), it follows that aRa whenever a is a string, so that R is reflexive. Next, suppose that aRb, so that l(a) = l(b). Then bRa, since l(b) = l(a). Hence, R is symmetric. Finally, suppose that aRb and bRc. Then l(a) = l(b) and l(b) = l(c). Hence, l(a) = l(c), so that aRc. Consequently, R is transitive. Since R is reflexive, symmetric and transitive, it is an equivalence relation.

.

The following theorem shows that the equivalence classes of two elements of A are either identical or disjoint.

Theorem 1. Let R be an equivalence relation on a set A. The following statements are equivalent:

(i) aRb; (ii) ; (iii) .

We are now in a position to show how an equivalence relation partitions a set. Let R be an equivalence relation on a set A. The union of the equivalence classes of R is all of A, since an element a of A is in its own equivalence class, namely, . In other words,

In addition, from Theorem 1, it follows that these equivalence classes are either equal or disjoint, so when . These two observations show that the equivalence classes form a partition of A, since they split A into disjoint subsets. More precisely, a partition of a set S is a collection of disjoint nonempty subsets of S that have S as their union. In other words, the collection of subsets (where I is an index set) forms a partition of S iff for , when and

Example. Suppose that S = {1, 2, 3, 4, 5, 6}. The collection of sets A₁ = {1, 2, 3}, A₂ = {4, 5} and A₃ = {6} forms a partition of S, since these sets are disjoint and their union is S.

Theorem 2. Let R be an equivalence relation on a set S. Then the equivalence classes of R form a partition of S. Conversely, given a partition of the set S, there is an equivalence relation R that has the sets , as its equivalence classes.

Example. What are the sets in the partition of the integers arising from congruence modulo 4?

Solution: There are four congruence classes, corresponding to and They are the sets , , ,

. These congruence classes are disjoint, and every integer is in exactly one of them. In other words, as Theorem 2 says, these congruence classes form a partition.

2.5.5 Lexicographic order

The words in a dictionary are listed in alphabetic, or lexicographic, order, which is based on the ordering of the letters in the alphabet. This is a special case of an ordering of strings on a set constructed from a partial ordering on the set.

The lexicographic ordering on is defined by specifying that one pair is less than a second pair if the first entry of the first pair isles than (in A₁) the first entry of the second pair, or if the first entries are equal, but the second entry of this pair is less than (in A₂) the second entry of the second pair. In other words, , either if or if both and . We obtain a partial ordering by adding equality to the ordering on .

Example. Determine whether , whether , whether in the poset , where is the lexicographic ordering constructed from the usual ≤ relation on Z.

A lexicographic ordering can be defined on the Cartesian product of n posets , , …, . Define the partial ordering on by if , or if there is an integer such that and .

We can now define lexicographic ordering of strings. Consider the strings and on a partially ordered set S. Suppose these strings are not equal. Let t be the minimum of m and n. The definition of lexicographic ordering is that the string is less than iff or and m < n where in this inequality represents the lexicographic ordering of S ^t.

2.5.6 Hasse diagrams

Many edges in the directed graph for a finite poset do not have to be shown since they must be present. For instance, consider the directed graph for the partial ordering on the set {1, 2, 3, 4}, shown in Figure 1(a). Since this relation is a partial ordering, it is reflexive, and its directed graph has loops at all vertices. Consequently, we do not have to show these loops since they must be present; in Figure 1(b) loops are not shown. Because a partial ordering is transitive, we do not have to show those edges that must be present because of transitivity. For example, in Figure 1(c) the edges (1, 3), (1, 4) and (2, 4) are not shown since they must be present. If we assume that all edges are pointed “upward” (as they are drawn in the figure), we do not have to show the directions of the edges; Figure 1(c) does not show directions. It is called the Hasse diagram for ({1, 2, 3, 4}, ).

Example. Draw the Hasse diagram for the partial ordering {(A, B) | A  B} on the power set P(S) where S = {a, b, c}.

Solution: The Hasse diagram for this partial ordering is obtained from the associated digraph by deleting all the loops and all the edges that occur from transitivity, namely, (, {a, b}), (, {a, c}), (, {b, c}), (, {a, b, c}), ({a}, {a, b, c}), ({b}, {a, b, c}), and ({c}, {a, b, c}). Finally all edges point upward and arrows are deleted. The resulting Hasse diagram is illustrated in Figure below.

2.5.7 Maximal and minimal elements

Elements of a poset that have certain extremal properties are important for many applications. An element a of a poset is called maximal if there is no such that . An element a of a poset is called minimal if there is no such that .

Example. Which elements of the poset are maximal, and which are minimal?

Solution: The Hasse diagram for this poset shows that the maximal elements are 12, 20 and 25, and the minimal elements are 2 and 5. As this example shows, a poset can have more than one maximal element and more than one minimal element.

Sometimes there is an element in a poset that is greater (or less) than every other element. An element a is called the greatest element of the poset if for all . The greatest element is unique when it exists. Likewise, an element a is called the least element of if for all . The least element is unique when it exists.

Example. Determine whether the posets represented by each of the Hasse diagrams in Figure below have a greatest element and a least element.

Solution: The least element of the poset with Hasse diagram (a) is a. This poset has no greatest element. The poset with Hasse diagram (b) has neither a least nor a greatest element. The poset with Hasse diagram (c) has no least element. Its greatest element is d. The poset with Hasse diagram (d) has least element a and greatest element d.

Sometimes it is possible to find an element that is greater than all the elements in a subset A of a poset (S, ). If u is an element of S such that a  u for all elements a  A, then u is called an upper bound of A. Likewise, there may be an element less than all the elements in A. If l is an element of S such that l  a for all elements a  A, then l is called a lower bound of A.

The element x is called the least upper bound of a subset A if x is an upper bound that is less than every other upper bound of A. That is, x is the least upper bound of A if a  x whenever a  A, and x  z whenever z is an upper bound of A. The least upper bound of A (lub(A)) is unique if it exists. Similarly, the element y is called the greatest lower bound of A if y is a lower bound of A and z  y whenever z is a lower bound of A. The greatest lower bound of A (glb(A)) is unique if it exists.

PROJECT PART