Discussion #15 Week of December 10, 2001 I. Equivalence Relations Example 1 Let R 0,0, 0,4, 1,1, 1,3, 2,2, 3,1, 3,3, 4,0, 4,4 be an equivalence relation on A 0,1,2,3,4. Find the equivalence classes of R. Solution We group together the elements 0 and 4 in one class since they are all related to each other. We group 1 and 3 in another class since they are all related. Finally 2 is in its own class since it is only related to itself. Thus the classes are: 0 4 0,4 1 3 1,3 2 2 Example 2 Let A R (the real numbers) and let S be the “absolute value” relation on A be defined as follows: x, y R x S y x y Show that S is an equivalence relation on A and find the equivalence classes of S. Solution Claim #1: S is reflexive. Proof: Given x A, x x , hence x, x S , x A and so S is reflexive. Claim #2: S is symmetric. Proof: Suppose x, y S for some real numbers x and y. Then x y and so y x which implies that y, x S , making S symmetric. Claim #3: S is transitive. Proof: Suppose x, y , y, z S for some real numbers x, y and z. Then x y and y z . But this implies that x z hence x, z S and S is transitive. This completes the proof that S is an equivalence relation on A. What do the equivalence classes of S look like? Part of the answer is that there are an uncountable number of them. One way to characterize them is to let there be one class for each nonnegative real number. Then all classes, save one, contain two numbers. The special equivalence class is 0 0. For any other positive real number, t, we have t t, t. Of course the characterization above is not unique since any element of an equivalence may be used to represent the class and thus t t . Example 3: Prove/disprove that S and R are equivalence relations on a set A only if S R is an equivalence relation on A. Solution: It is not hard to show that S R is reflexive and symmetric whenever S and R are equivalence relations. However S R is not necessarily transitive. Here is a counterexample. Let A 1,2,3. Let R 1,1, 2,2, 3,3, 1,2, 2,1 and let S 1,1, 2,2, 3,3, 2,3, 3,2. Then S R 1,1, 2,2, 3,3, 1,2, 2,1, 2,3, 3,2. Now note that 1,3 should be in S R for transitivity. Since it is not we have that S R is not transitive and hence not an equivalence relation on A. II. Partial Orderings Example 1: Let A a, b. Write down all partial orderings of A and show the corresponding Hasse diagram for each ordering. Solution: There are exactly three partial orderings of A, namely: R1 a, a, b, b, a, b, R2 a, a , b, b, b, a and R3 a, a , b, b . The Hasse diagrams of R1 , R2 and R3 are, respectively, R1 R2 R3 b a b a b a Example 2: Let R be the set of real numbers and let S be a relation on R R defined by: a, b , c, d R R a, b S c, d either 1) a c or 2) a c and b d Prove/disprove that R R, S is a poset. Claim #1: S is reflexive Proof: Let a, b R R for some arbitrary real numbers a and b. Then a, b, a, b S , since condition 2 is met, namely: a a and b b . Thus S is reflexive. Claim #2: S is antisymmetric. Proof: Suppose a, b, c, d S and c, d , a, b S for some real numbers a, b, c and d. We must show that a, b c, d . It suffices to show that a c and b d . Now a, b, c, d S implies that either 1) a c or else 2) a c and b d . Also, c, d , a, b S implies that either 3) c a or else 4) c a and d b . If condition 1) is true then neither of 3) or 4) can be true, since if a c then neither c a nor c a can hold. Since one of 3) and 4) must hold, it follows that 1) does not hold, i.e. 2) holds so we have a c and b d . Now if 2) holds then 3) cannot hold, thus 4) must be true. That is, c a and d b . But b d and d b implies b d . But this is what we needed to show: a c and b d . So we conclude that S is antisymmetric. Claim #3: S is transitive. Proof: Suppose that a, b, c, d S and c, d , e, f S for some real numbers a, b, c, d, e and f. We must show that a, b, e, f S . That is, we need to show that either a e or else a e and b f . Now a, b, c, d S implies that 1) a c or else 2) a c and b d . Also, c, d , e, f S implies that 3) c e or else 4) c e and d f . Case 1: 1) and 3) hold. Then a c e , i.e. a e which means a, b, e, f S . Case2: 1) and 4) hold. Then a c e and so a e and then a, b, e, f S . Case 3: 2) and 3) hold. Then a c e implying that a e so that a, b, e, f S . Case 4: 2) and 4) hold. Then a c e , b d and d f . That is, a e and b f , giving a, b, e, f S . We have now shown that S is transitive. Therefore R R, S is a poset and the proof is complete. Example #3: Prove/disprove that if R is a partial order on a set A then R 1 is also a partial order on A. We show this is true. Assume R is a partial order on A. We must show that R 1 is reflexive, antisymmetric and transitive. Claim #1: R 1 is reflexive. Proof: R a partial order on A a, a R a A (since R is reflexive) a, a R 1 a A (def. of inverse relation) Hence R 1 is reflexive. Claim #2: R 1 is antisymmetric. Proof: Suppose that a, b , b, a R 1 . We show a b . b, a , a, b R (def. of inverse relation) (since R is antisymmtric) a b is antisymmetric. a, b, b, a R 1 Thus R 1 Claim #3: R 1 is transitive. Proof: Suppose a, b, b, c R 1 for some a, b, c A . We show a, c R 1 . b, a , c, b R (def. of inverse relation) (since R is transitive) c, a R 1 (def. of inverse relation) a, c R 1 Thus R is transitive. This shows that R 1 is a partial order on A whenever R is and the proof is complete. a, b, b, c R 1 Example #4: Let A 2,3,5,6,7,11,12,35,385 and let R be the “divides” relation on A. That is, given a, b A, a, b R a b . Draw a Hasse diagram for R, find all maximal and all minimal elements. Find lub 2,3, lub 2,12 lub 7,11 glb 2,12 glb 2,3, glb 6,385. Is there a largest element? A smallest? Is A, a lattice? Solution: The Hasse diagram is 12 385 6 2 35 3 5 7 11 Maximal elements: 12 and 385 Minimal elements: 2, 3, 5, 7, and 11 lub 2,3= 6 lub 2,12 =12 lub 7,11=385 glb 2,12 =2 glb 2,3 does not exist lub 6,385 does not exist There is no largest element and no smallest element. Since not every pair of elements of A has a glb and lub, A, is not a lattice.