Let R be a relation on a set A, i.e., a subset of AxA.

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1 6 Relations Let R be a relation on a set A, i.e., a subset of AxA. Notation: xry iff (x, y) R AxA. Recall: A relation need not be a function. Example: The relation R 1 = {(x, y) RxR x 2 + y 2 =1} is not a function. Some definitions 1. R is reflexive if xrx x A. 2. R is symmetric if xry yrx x, y A. 3. R is antisymmetric if xry yrx x = y x, y A. 4. R is transitive if xry yrz xrz x, y, z A. We now give a number of examples. 1. R = < onz. R, S, A,T. 2. R = onz. R, S, A,T. 3. R = = onz. R,S,A,T. 75

2 4. R, S,A, T. 5. R, S, A,T. 6. R, S,A,T. 7. R, S, A, T. 8. R, S, A, T. 9. R, S, A, T. 10. R, S,A, T. 76

3 Note: 1. In a reflexive relation there is a loop at each vertex. 2. In a symmetric relation, there are either 2 arcs or no arcs between any two distinct nodes. 3. In an antisymmetric relation there is either 1 arc or no arcs between any two distinct nodes. Definition: IfR is a relation from X to Y, then the inverse of R is R 1 = {(y, x) (x, y) R}. Examples: R: R 1 : R: R 1 : R: R 1 : Theorem 6.1 R is symmetric iff R = R 1. Theorem 6.2 The reflexive, symmetric, antisymmetric and transitive properties of relations are preserved by the inverse, i.e., if R has such a property then so does R 1. 77

4 Equivalence Relations Definition: A relation R AxA is an equivalence relation if it is reflexive, symmetric and transitive. Examples: 1. = on Z. 2. The universal relation U A = AxA, i.e., the relation consisting of all elements of AxA. 3. Let A be the set of all triangles in the plane. Then T 1 RT 2 iff T 1 and T 2 are similar triangles. 4. Let A be the set of all points in the plane. Then p 1 Rp 2 iff the distance from p 1 to the origin equals the distance from p 2 to the origin. 5. A = Z, m > 0. ar m b iff m a b, i.e., c Z such that m c = a b. (a) R m is reflexive since m a a. (b) R m is symmetric since m a b m b a. (c) R m is transitive since if m a b and m b c, then m (a b)+(b c) orm a c. Notation: Ifm a b we say a is congruent to b mod m, or a b (mod m). 78

5 6. Let f : A B. Then R f given by a 1 R f a 2 iff f(a 1 )= f(a 2 ) is an equivalence relation. 7. Definition: Let R be an equivalence relation on A and b A. Then [b] ={x A xrb} is the equivalence class generated by b. Example: LetA = Z and consider ar 3 b. Thus a and b are related iff 3 a b. Then [0] = {... 6, 3, 0, 3, 6,...} [1] = {... 5, 2, 1, 4, 7,...} [2] = {... 4, 1, 2, 5, 8,...} Example: In (7) above, [1] = [2] = {1, 2} and [3] = [4] = {3, 4}. Definition: Let A = α Λ A α,whereeacha α φ and the A α s are pairwise disjoint. Then {A α α Λ} is a partition of A. Example: A 1,A 2,...,A 7 is a partition of A. 79

6 Note: A partition of a set defines an equivalence relation in a very natural way. Definition: LetP be a partition of a set A. Then the equivalence relation R(P) associated with P is given by: ar(p )b iff a and b are in the same set in P. Note: R(P ) is clearly an equivalence relation. Example: ThesetsA 1,A 2,A 3 partition Z. A 1 = {... 6, 3, 0, 3, 6,...} A 2 = {... 5, 2, 1, 4, 7,...} A 3 = {... 4, 1, 2, 5, 8,...} Thus 1R(P)7 and -4R(P)8. We now wish to show that each equivalence relation on a set A defines a partition in a natural way. 80

7 Theorem 6.3 Let R be an equivalence relation on a set A. Then 1. b [b] b A. 2. a, b A, [a] =[b] arb. 3. a, b A, either [a] =[b] or [a] [b] =φ. Proof: First recall that [b] ={x A xrb}. 1. Since R is reflexive, brb. Hence b [b]. 2. ( ) Suppose [a] =[b]. Since a [a], a [b]. Hence arb. ( ) Suppose x [b]. Then xrb. Also, since arb and R is symmetric, bra. Since R is also transitive, xra, i.e., x [a]. Hence [b] [a]. Similarly [a] [b]. 3. Suppose x [a] [b]. Then xra and xrb. Since R is symmetric, arx. Now since R is transitive, arb. Hence by (2), [a] =[b]. 2 Let P (R) ={[a] a A}. IfR is an equivalence relation on aseta, the distinct sets P (R ) partition A. Thuswehavea 1-1 correspondence between the partitions of a set A and the equivalence relations on A. Note: P (R(P )) = P and R(P (R )) = R. 81

8 Posets Let R be a relation on a set A. ThenR is a partial ordering on A if R is 1. reflexive 2. antisymmetric 3. transitive Examples: 1. R : onz. 2. R : on P(A), the power set of A. 3. R : divides on Z +. (a) a a. (b) a b and b a a = b. (c) a b and b c a c. Definition: If R is a partial ordering on A we call (A, R) a partially ordered set or poset. Note: A subset of a partially ordered set is a partially ordered set (with the same ordering). Notation: When the relation is a partial ordering, we often use a b instead of arb. 82

9 Definition: Suppose (A, R) is a poset. Elements a and b of A are said to be comparable if, and only if, either arb or bra. Otherwise they are noncomparable. Definition: Let R be a partial order relation on a set A. If any two elements a and b in A are comparable, then R is a total order relation on A. Examples: 1. R : onz is a total order. 2. R : onp (A) is not a total order if A has more than 1 element. 3. R : divides is not a total order on Z +, e.g., 3 does not divide 5 and 5 does not divide 3. Definition: Let(A, R) be a poset. A subset B of A is called a chain if, and only if, each pair of elements in B is comparable. The length of a chain is the number of elements in the chain. Note: The book has a different definition of length. Example: ThesetP ({a, b, c}) is partially ordered with respect to subset inclusion. The set S = {φ, {a}, {a, b}, {a, b, c}} is a chain of length 4 in P ({a, b, c}). 83

10 Hasse Diagrams Let A = {0, 1} and consider the poset (P (A), ). Hasse diagram Properties of Hasse Diagrams arrows are omitted - edges are directed upward self loops are omitted edges implied by transitivity are omitted More examples: 84

11 Definition: A subset of a poset (A, R) isanantichain if no two distinct elements of the subset are related. Example: {c, f, e} Definition: Let (A, ) be a poset. An element a A is a maximal element if there does not exist b A such that b a and a b. Note: minimal element is defined similarly. In the examples above a is a maximal element g is a minimal element 1, 2, 7 are maximal elements 8, 9, 10 are minimal elements Note: Any finite, nonempty partially ordered set has a minimal (and maximal) element. Theorem 6.4 Let (A, ) be a poset. If n is the length of a longest chain in (A, ), then A can be partitioned into n disjoint antichains. Proof: Later, by induction. 85

12 Well Orderings Let (A, ) beaposetandb A. An element b B is a greatest element of B if b b b B. It is a least element of B if b b b B. Example: Consider(N, ) andb = {3, 7, 12, 15}. Then3is a least element of B and 15 is a greatest element of B. Theorem 6.5 Let (A, R) be a poset and B A. Ifx and y are greatest elements of B, then x = y. Proof: If x and y are greatest elements of B, thenx y and y x. SinceR is antisymmetric, x = y. 2 Note: Least elements are also unique. Definition: A relation R on A is a well ordering if R is a total ordering and every nonempty subset of A has a least element. Examples: Any finite total ordering is a well ordering. (R + {0}) is NOT a well ordering since R + R + {0} does not have a least element. (Z, ) is NOT a well ordering since Z has no least element. (N, ) is a well ordering - this is actually taken as an axiom. 86

13 Theorem 6.6 Every set can be well ordered. Note: It is not always easy to find the ordering. Example: The integers can be well ordered as follows: Let f : Z N be defined by f(k) = 2k if k 0-2k - 1 if k<0. Note: f(0) = 0,f( 1) = 1,f(1) = 2,f( 2) = 3, etc. Note: Nowa b iff f(a) f(b) is a well ordering. 87

14 Closure Operations on Relations Example: Suppose we define a relation R on a set A of cities as follows: arb iff there is a direct communication link from city a to city b for transmission of messages. Problem: Find a relation that describes how messages can be transmitted from one city to another, either through a direct communication link, or through any number of intermediate cities. Definition: Let R be a relation on a set A. The transitive closure of R is a relation R t such that 1. R t is transitive 2. R R t 3. If R 1 is transitive and R R 1,thenR t R 1. Note: The transitive closure is unique. Note: The reflexive and symmetric closures are defined in an analogous way. Example: R: R t : 88

15 Theorem 6.7 Let {S α α Λ} be the set of all transitive relations containing a relation R. Then R t = α Λ S α. Thus the transitive closure of a relation R is the smallest transitive relation containing R. It is obtained by adding the least number of ordered pairs to ensure transitivity. Note: A similar theorem holds for reflexive and symmetric closures. 89

16 Composition of Relations Definition: Let R 1 be a relation from A to B and R 2 be a relation from B to C. Thecomposition of R 1 and R 2 is a relation from A to C given by R 1 R 2 = {(a, c) a A, c C b B such that [(a, b) R 1 (b, c) R 2 ]}. Example: Note: In general, R 1 R 2 R 2 R 1. In fact, if R 1 is a relation from A to B and R 2 is a relation from B to C, thenr 2 R 1 is not defined. Example: LetA = {0, 1, 2, 3} and consider R 1 and R 2 on A. 90

17 Theorem 6.8 Let R 1 AxB, R 2 BxC and R 3 CxD. Then (R 1 R 2 )R 3 = R 1 (R 2 R 3 ), i.e., the composition of relations is associative. Proof: ( ) Let(a, d) (R 1 R 2 )R 3.Then c C such that (a, c) R 1 R 2 and (c, d) R 3. Since (a, c) R 1 R 2 b B such that (a, b) R 1 and (b, c) R 2. Now (b, c) R 2 and (c, d) R 3 (b, d) R 2 R 3. But now (a, b) R 1 (a, d) R 1 (R 2 R 3 ). ( ) Similar. 2 Definition: LetR be a binary relation on a set A. ThenR n is defined as follows: 1. R 0 = {(x, x) x A}. 2. R n+1 = R n R. 91

18 Example: R 0 : R 1 = R: R 2 = R 1 R: R 3 = R 2 R: R 4 = R 3 R: Note: In this example, R 4 = R 2. 92

19 Theorem 6.9 Let A = n and R AxA. Then s, t, 0 s<t 2 n2, such that R s = R t. Proof: First note that AxA has n 2 elements. Hence there are 2 n2 distinct relations on A. By the pigeonhole principle, at least two of them are equal. 2 93