Binary Relations and Equivalence Relations

Transcription

1 Binary Relations and Equivalence Relations Intuitively, a binary relation R on a set A is a proposition such that, for every ordered pair (a, b) A A, one can decide if a is related to b or not. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. Formally, Definition 1.1 A binary relation in a set A is a subset R A A. The statement (a, b) R is written as a R b. Example 1.2 (a) For any set A the diagonal = {(a, a) a A} is the relation of equality. The relation (A A) \ is the relation of inequality. (b) The relation between two real numbers is the set {(x, y) x y} R R. (c) In P(A) the relation of inclusion is given by {(A, B) P(A) P(A) B A}. If A is a set with a binary relation R and B A then the relation R (B B) is a binary relation on the set B. This is the relation on B induced by R. Of all the relations, one of the most important is the equivalence relation. Definition 1.3 An equivalence relation on a set X is a binary relation on X which is reflexive, symmetric and transitive, i.e. (a) a A : a R a (reflexive). (b) a R b b R a (symmetric). (c) a Rb and b R c a R c (transitive). If a R b we say that a is equivalent to b. Example 1.4 (a) The relation is an equivalence relation. 1

2 (b) In Z the relation {(x, y) Z Z x y is divisible by 2} is an equivalence relation. (c) Let f : X Y be a function. Then {(x 1, x 2 ) f(x 1 ) = f(x 2 )} is an equivalence relation in X. Let us check the assertion (b). First, reflexibity. x we have x x = 0 and 0 is divisible by 2. Hence the relation is reflexive. Moreover, since y x = ( 1) (x y) it is clear that if 2 (x y then 2 (y x). Hence the relation is symmetric. Finally, if 2 (x y) and 2 (y z), then since x z = (x y) + (y z), it is clear that 2 (x z). So the relation is also transitive and hence is an equivalence relation. Suppose that R is an equivalence relation on the set X. If x X let E(x, R) denote the set of all elements y X such that xry. The set E(x, R) is called the equivalence class of x for the equivalence relation R. Since R is an equivalence relation, the equivalence classes have the following properties: 1. Each E(x, R) is non-empty for, since xrx, x E(x, R). 2. Let x and y be elements of X. Since R is symmetric, y E(x; R) if and only if x E(y; R). 3. If x, y X the equivalence classes E(x; R) and E(y; R) are either identical or they have no members in common. Indeed, suppose, first, that xry. Let z E(x; R). Then, by symmetry, since zrx we have also xrz. Hence, by transitivity, zry and so, by symmetry, yrz. This shows that E(x; R) E(y; R). By the symmetry of R we see that E(y; R) E(x; R). Hence E(x; R) = E(y; R). Finally, notice that if the points x, y X are not related then E(x; R) E(y; R) =. Indeed, if z E(x; R) E(y; R) then xrz and yrz and so xrz and zry. Therefore xry which is a contradiction. These facts lead to the following assertions concerning the family, F, of equivalence classes for the equivalence relation R: 1. Every element of the family F is non-empty. 2. Each element x X belongs to one and only one of the sets in the family F. 3. xry if and only if x and y belong to the same set in the family F. 2

3 Otherwise said, and equivalence relation subdivides a set (or partitions the set) into the union of a family of non-overlapping, non-smpty subsets. Here is an example which is perhaps the first most students see when they discuss number systems. Example 1.5 :In the construction of the rational numbers, which we will denote by Q, we first introduce ratios of integers p/q where p N and q Z. If p/q represents a point on the number line, then the ratios kp/kq must represent the same point and hence the same rational number. Thus, two ratios p/q and r/s represent the same rational number and can be treated as equal and can be substituted for one another in proofs involving rational numbers whenever the equality is true. ps = rq Now, let us define a relation on N Zby (p, q) R (r, s) if and only if ps = rq. We check that this is an equivalence relation as follows: (a) (Reflexivity): pq = pq hence (p, q) R (p, q). (b) (Symmetry): If ps = rq then rq = ps and so (pq) R (r, s) (r, s) R (p, q). (c) (Transitivity): If ps = rq and rt = vs, then (pt) s = (ps) t = (rq) t = (rt) q = (vs) q = (vq) s and thur pt = vq since s 0. (p, q) R (v, t). Hence (p, q) R (r, s) and (r, s) R (v, t) implies From this we see that the rational numbers can be viewed as equivalence classes of ratios of integers modulo the relation R given in the example. As a final example consider the following: 3

4 Example 1.6 : Consider the set Z and let n be a fixed positive integer. Define a relation R n by xr n y provided (x y) is divisible by n. This relation is called the relation of congruence modulo n. It is easy to check that this is an equivalence relation on Z. (See the special case for n = 2 treated above.) Moreover, there are n equivalence classes. Each integer x is uniquely expressible in the form x = q n + r, where q and r are integers and 0 R n 1. (The integers q and r are called the quotient and the remainder respectively. Hence each x is congruent modulo n to one of the n integers 0, 1,..., n 1. The equivalence classes are E 0 = {..., 2n, n, 0, n, 2n,...} E 1 = {..., 1 2n, 1 n, 1 + n, 1 + 2n,...}.. E n 1 = {..., n 1 2n, n 1 n, n 1, n 1 + n, n 1 + 2n,...} Formallly, the domain of a relation, R, is the set of all first coordinates of the members of R while, in this context, the range is the set of all second coordinates. Formally while dom (R) = {x X for some y Y, (x, y) R}, rng (R) = {y Y for some x X, (x, y) R}. The inverse of a relation R, denoted R 1, is obtained by reversing each of the pairs belonging to R. Thus R 1 = {(y, x) Y X (x, y) R}. Hence the domain of the inverse is the range of R and the range of R 1 is always the domain of R. If R and S are relations, then the composition R S is defined as {(x, z) X Z for some y, (x, y) S and (y, z) R}. 4

5 Example 1.7 If R = {(1, 2)} and S = {(0, 1)} then R S = {(0, 1)} while S R =. Concerning compositions and inverses we have the following result Proposition 1.8 Let R, S, and T be relations. Then (a) (R 1 ) 1 = R. (b) (R S) 1 = S 1 R 1. (c) R (S T ) = (R S) T Proof: (of (b)) We have (x, a) (R S) 1 (x, z) R S for some y, (x, y) S and (y, z) R. Consequently, (z, x) (R S) 1 if and only if (y, z) R 1 and (y, a) S 1 for some y. But this is the condition that (z, x) S 1 R 1. 5