9 RELATIONS. 9.1 Reflexive, symmetric and transitive relations. MATH Foundations of Pure Mathematics

Transcription

1 MATH Foundations of Pure Mathematics 9 RELATIONS 9.1 Reflexive, symmetric and transitive relations Let A be a set with A. A relation R on A is a subset of A A. For convenience, for x, y A, write xry if (x, y) R and x Ry if (x, y) / R. Examples: (i) A = Q, R = {(x, y) : x, y Q, x < y}. Hence xry x < y, i.e., R is the relation <. 1R2 but 3 R1. (ii) A = Z. Fix n N. Let R = {(x, y) : x, y Z, n (x y)}. So xry x y mod n. (iii) A = N, R = {(x, y) : x, y N, y x is prime}. E.g., 7 R11 (since 11 7 = 4, not prime), 12R7 (since 7 12 = 5, prime). (iv) A = R, x, y A, xry x y. (v) A any set, R = A A. So x, y A, xry. (vi) A = {1, 2, 3}, R = {(1, 2)}. (vii) Let A = N 10. Let A 1 = {3, 4, 6}, A 2 = {1, 5} and A 3 = {2, 7, 8, 9, 10}. Define R by: xry iff {x, y} A i for some i. We are interested in three properties that a relation might have. Let A be a non-empty set and R a relation on A. R is reflexive if x A, xrx. R is symmetric if x, y A, xry yrx. R is transitive if x, y, z A, xry and yrz xrz. We revisit the previous examples. Examples revisited: (i) A = Q, xry x < y. 2 2, so 2 R2, so not reflexive. 1R2 but 2 R1, so not symmetric. Let x, y, z Q. Suppose xry and yrz, i.e., x < y and y < z. Then x < y < z, so x < z, i.e., (x, z) R. So transitive. (ii) A = Z, n N, xry n (x y). 1

2 x N, n (x x) = 0, so x N, xrx. Hence reflexive. Let x, y N. Then n (x y) n (y x), so symmetric. Let x, y, z N. If n (x y) and n (y z), then n (x y + y z) = x z, so transitive. (iii) A = N, R = {(x, y) : x, y N, y x is prime}. x x = 0, not prime, so x Rx. So not reflexive. If xry, then y x is prime, so x y is prime, so yrx. Hence symmetric. 8R10 and 10R17, but 8 R17, so not transitive. (iv) A = R, x, y A, xry x y. x R, x x, so xrx. So reflexive. 1 2 but 2 1, so not symmetric. Let x, y, z R. If x y and y z, then x y z, so x z. So transitive. (v) A any set, R = A A. x A, (x, x) R, so xrx. So reflexive. x, y A, xry and yrx, so symmetric. x, y, z A, xry, yrz and xrz, so transitive. (vi) A = {1, 2, 3}, R = {(1, 2)}. 1 R1, so not reflexive. 1R2 but 2 R1, so not symmetric. Conditions for transitivity trivially satisfied, so transitive. (vii) A = N 10. Let A 1 = {3, 4, 6}, A 2 = {1, 5} and A 3 = {2, 7, 8, 9, 10}. R defined by: xry iff {x, y} A i for some i. R is clearly reflexive. R is symmetric since {x, y} = {y, x}. R is transitive: Suppose {x, y} A i and {y, z} A j. Then y A i A j. But A 1, A 2 and A 3 are pairwise disjoint, so i = j. So {x, y}, {y, z} A i. So {x, y} {y, z} A i. So {x, z} {x, y, z} A i. 9.2 Equivalence relations An equivalence relation on a non-empty set A is a relation which is reflexive, symmetric and transitive. The model example of an equivalence relation is = (i.e., x, y A, xry x = y). Examples (ii), (v) and (vii) above are equivalence relations, but (i), (iii), (iv) and (vi) are not. Examples: (i) A = R, xry x = y is an equivalence relation. (ii) Let A = N, xry x = 2 k y for some k Z. Reflexive: x N, x = 2 0 x. Symmetric: Let x, y N. Suppose xry. Then x = 2 k y for some k Z. So 2 k x = y, so yrx. 2

3 Transitive: Let x, y, z N. Suppose xry and yrz. So k 1, k 2 Z such that x = 2 k 1 y and y = 2 k 2 z. Then x = 2 k 1 2 k 2 z = 2 k 1+k 2 z, so xrz. So R is an equivalence relation. 9.3 Equivalence classes Let R be an equivalence relation on a (non-empty) set A. Let a A The equivalence class of a is R a = {x A : arx}. Remarks: (i) Note that a R a since R is reflexive. (ii) R a A. (iii) Since R is symmetric, R a = {x A : arx} = {x A : xra}. Examples: (i) A = Z, xry x = y. Let a Z. R a = {x Z : a = x } = {a, a}. So R 0 = {0}, R 1 = {1, 1} = R 1, R 2 = R 2 = {2, 2}, etc. Note that Z = R 0 R 1 R 2. (ii) A = Z, xry 3 (x y). Let a Z. R a = {x Z : a x mod 3}. So R 0 = {..., 3, 0, 3, 6, 9,...} = {3k : k Z}, R 1 = {..., 2, 1, 4, 7, 10,...} = {3k + 1 : k Z}, R 2 = {..., 1, 2, 5, 8, 11,...} = {3k + 2 : k Z}. Note Z = R 0 R 1 R 2. We will show that the equivalence classes form what is called a partition of A. Theorem 9.1 Let R be an equivalence relation on a set non-empty A. Let a, b A. (i) If arb, then R a = R b. (ii) If a Rb, then R a R b =. Proof. (i) Suppose that arb. Let x R a. Then arx by definition. Since R is symmetric, bra. Hence we have bra and arx, so by transitivity brx. So x R b. We have shown R a R b. Let x R b. Then brx. So arb and brx. So arx by transitivity. Hence x R a. So R b R a. Hence R a = R b, since R a R b and R b R a. (ii) Suppose a Rb. Suppose that R a R b. Then x R a R b. So arx and brx. Since R is symmetric, xrb. So arx andxrb. Since R is transitive, this means arb, a contradiction. So R a R b =. 3

4 Definition 9.2 Let X be a non-empty set. Let {X i : i I} be a collection of non-empty subsets of X (I is called an index set) such that (i) i I X i = X and (ii) i, j I, X i = X j or X i X j =. Then call {X i : i I} a partition of X. Theorem 9.3 Let R be an equivalence relation on a non-empty set A. Then {R a : a A} is a partition of A. Proof. Since R is reflexive, a A, a R a. Hence A a A R a and each R a is non-empty. Also, a A we have R a A, so a A R a A. Hence a A R a = A. By Theorem 9.1, for a, b A, either R a = R b or R a R b =. So {R a : a A} is a partition of A. Conversely, we have: Theorem 9.4 Let A be a non-empty set and let {A i : i I} be a partition of A. Define a relation R on A by arb {a, b} A i for some i I. Then R is an equivalence relation, with equivalence classes A i for i I. Proof. Let x A. Then x A i for some i I since {A i : i I} is a partition. Hence {x, x} = {x} A i. So R is reflexive. Let x, y A. Suppose that xry. Then {y, x} = {x, y} A i for some i I. Hence yrx. So R is symmetric. Let x, y, z A. Suppose xry and yrz. Then {x, y} A i and {y, z} A j for some i, j I. Then y A i A j, and so A i A j. Since {A i : i I} is a partition, we have i = j. So {x, y}, {y, z} A i. So {x, y} {y, z} A i. So {x, z} {x, y, z} A i. Hence xrz. So R is transitive. Let a A. Then a A i for some i I. For all x A, we have x R a if and only if x A i. Hence the A j for j I form the equivalence classes of R. Examples (i) Let A = Z, xry x = y. Z = {0} {1, 1} {2, 2} {3, 3} R 0 R 1 R 2 R 3. {R i : i N {0}} is a partition of Z. We say that N {0} is a set of equivalence class representatives for R. (ii) A = N, xry x = 2 k y for some k Z. R 1 = {y N : 1Ry} = {y N : 1 = 2 k y for some k Z} = {1, 2, 4, 8, 16,...}, R 3 = {y N : 3Ry} = {y N : 3 = 2 k y for some k Z} = {3, 6, 12, 24, 48,...}, 4

5 R 5 = {y N : 5Ry} = {y N : 5 = 2 k y for some k Z} = {5, 10, 20, 40, 80,...}, etc. N = R 1 R 3 R 5 R 7 R 9. {R 2k 1 : k N} is a partition of N. The set of odd natural numbers {2k 1 : k N} is a set of equivalence class representatives for R. (iii) Let A = C and define an equivalence relation R on A by z 1 Rz 2 z 1 = z 2. For x C, we have R x = {z C : x = z }. C = x R 0 {z C : z = x}, so R 0 is a set of equivalence class representatives for R. The equivalence classes are circles in the Argand diagram with centre the origin. (iv) Definition of Q Informally, for a, b, c, d Z with b, d 0, we have a = c if and only if ad = bc. b d This motivates the definition of the rationals. Let A = Z (Z\{0}). Define a relation R on A by (a, b)r(c, d) if and only if ad = bc. By Exercise 9.2(i) R is an equivalence relation. Define Q = {R (a,b) : (a, b) A}, the set of equivalence classes. We write a for R b (a,b). Then a = c if and only if R b d (a,b) = R (c,d). 9.4 Integers modulo n Fix n N. Define the relation R on Z by arb a b mod n. The equivalence classes are R 0 = {y Z : 0 y mod n} = {..., 2n, n, 0, n, 2n,...} R 1 = {y Z : 1 y mod n} = {..., 2n + 1, n + 1, 1, n + 1, 2n + 1,...} R 2 = {y Z : 2 y mod n} = {..., 2n + 2, n + 2, 2, n + 2, 2n + 2,...}. R n 1 = {y Z : n 1 y mod n} = {..., 2n 1, n 1, 1, n 1, 2n 1,...}. We may identify the equivalence classes with the elements of the set {0, 1,..., n 1}. Define Z n = {0, 1,..., n 1}. Define addition and multiplication on Z n as follows: Let a, b Z n. By the division theorem, there are q, r Z with 0 r < n such that a +b = qn+r. Define a b = r. So a b is the unique r Z n with a + b r mod n. Similarly, a b is the unique t Z n with ab t mod n. 5

6 Examples: In Z 6 : mod 6, so 2 5 = mod 6, so 5 4 = mod 6, so 2 5 = mod 6, so 4 3 = Additional examples (i) Let A = R and R be the relation on A given by xry z Z, x = y + z. Every equivalence class of R contains a unique element in [0, 1), and every element of [0, 1) is contained in an equivalence class. Hence [0, 1) is a set of representatives of the equivalence classes. We may think of this as wrapping R around a circle of unit circumference, with elements occurring in the same equivalence class if and only if they coincide on the circle. (ii) Now let A = R R = R 2, and let R be the relation on A given by (x 1, x 2 )R(y 1, y 2 ) (z 1, z 2 ) Z Z, (x 1, x 2 ) = (y 1 + z 1, y 2 + z 2 ). In this case the equivalence classes of R are represented by elements of [0, 1) [0, 1). We may think of this as wrapping R 2 around a torus. 6