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Binary Relations and Equivalence Relations 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) 2 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) 2 R" is written as a R b. Example 1.2 (a) For any set A the diagonal ∆ = f(a; a) j a 2 Ag is the relation of equality. The relation (A × A) n ∆ is the relation of inequality. (b) The relation ≤ between two real numbers is the set f(x; y) j x ≤ yg ⊂ R × R : (c) In P(A) the relation of inclusion is given by f(A; B) 2 P(A) × P(A) jB ⊂ Ag : 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) 8a 2 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 (b) In Z the relation f(x; y) 2 Z × Z j x − y is divisible by 2g is an equivalence relation. (c) Let f : X ! Y be a function. Then f(x1; x2) j f(x1) = f(x2)g is an equivalence relation in X. Let us check the assertion (b). First, reflexibity. 8x 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 2j(x − y then 2j(y − x). Hence the relation is symmetric. Finally, if 2j(x − y) and 2j(y − z), then since x − z = (x − y) + (y − z), it is clear that 2j(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 2 X let E(x; R) denote the set of all elements y 2 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 2 E(x; R). 2. Let x and y be elements of X. Since R is symmetric, y 2 E(x; R) if and only if x 2 E(y; R). 3. If x; y 2 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 2 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 2 X are not related then E(x; R)\E(y; R) = ;. Indeed, if z 2 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 2 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 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 2 N and q 2 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 ps = rq is true. 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 6= 0. Hence (p; q) R (r; s) and (r; s) R (v; t) implies (p; q) R (v; t). 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 Example 1.6 : Consider the set Z and let n be a fixed positive integer. Define a relation Rn by xRny 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 E0 = f:::; −2n; −n; 0; n; 2n; : : :g E1 = f:::; 1 − 2n; 1 − n; 1 + n; 1 + 2n; : : :g . En−1 = f: : : ; n − 1 − 2n; n − 1 − n; n − 1; n − 1 + n; n − 1 + 2n; : : :g 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 dom (R) = fx 2 X j for some y 2 Y; (x; y) 2 Rg ; while rng (R) = fy 2 Y j for some x 2 X; (x; y) 2 Rg : The inverse of a relation R, denoted R−1, is obtained by reversing each of the pairs belonging to R. Thus R−1 = f(y; x) 2 Y × X j (x; y) 2 Rg : 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 f(x; z) 2 X × Z j for some y; (x; y) 2 S and (y; z) 2 Rg : 4 Example 1.7 If R = f(1; 2)g and S = f(0; 1)g then R ◦ S = f(0; 1)g 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) 2 (R ◦ S)−1 , (x; z) 2 R ◦ S , for some y ; (x; y) 2 S and (y; z) 2 R: Consequently, (z; x) 2 (R ◦ S)−1 if and only if (y; z) 2 R−1 and (y; a) 2 S−1 for some y. But this is the condition that (z; x) 2 S−1 ◦ R−1. 2 5.
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