1.4 Equivalence Relations and Partitions

Transcription

1 24 CHAPTER 1. REVIEW 1.4 Equivalence Relations and Partitions Equivalence Relations Definition (Relation) A binary relation or a relation on a set S is a set R of ordered pairs. This is a very general definition. The ordered pairs simply list the elements which are related. If (a, b) R, we also write arb and it simply means that a is in relation with b, whatever relation R is. Here are a few examples. Example If S = N, we define a relation R by R = { (a, b) N 2 : a and b are either both even or both odd } Then, (2, 6) R but (2, 3) / R. Example If S = R, we define a relation R by R = { (a, b) R 2 : a = b }. Then, every real number a is in relation with itself that is (a, a) R since a = a. Also, ( a, a) R. Example If S = Z, we define a relation R by R = { (a, b) Z 2 : a b }. Here, our relation has its own special notation,. Example Is a function a relation? Is a relation a function? A function f is a special relation in which the following can t hap[pen: (a, b) f and (a, c) f. Not every relation is a function since to be a function, every input can only produce one output in other words we cannot have (a, b) f and (a, c) f. In its most general form, relations are not very interesting. However, relations with certain properties are more interesting. In this section, we will focus on equivalent relations. First, we list some of the properties a relation may have which are of interest. Definition Let R be a relation defined on a set S and let a, b, c be elements of S. 1. R is said to be reflexive if a S, (a, a) R. In other words, every element is in relation with itself. 2. R is said to be symmetric if a, b S, (a, b) R = (b, a) R. 3. R is said to be antisymmetric if a, b S, (a, b) R and (b, a) R = a = b. 4. R is said to be transitive if a, b, c S, (a, b) R and (b, c) R = (a, c) R.

2 1.4. EQUIVALENCE RELATIONS AND PARTITIONS 25 We have the following definitions (though both are important, only the first one will be of interest to us here): Definition (Equivalence Relation) An equivalence relation on a set S is a relation, that is a set R of ordered pairs which has the following three properties: 1. reflexive. 2. symmetric. 3. transitive. Definition (Order Relation) An order relation on a set S is a relation, that is a set R of ordered pairs which has the following three properties: 1. reflexive. 2. antisymmetric. 3. transitive. Remark If R is an equivalence relation, instead of writing (a, b) R or arb, we often write a b. Using this notation, the reflexive property becomes a a, the symmetric property becomes a b = b a, and the transitive property becomes a b and b c = a c. Definition (Equivalence Class) If is an equivalent relation on S and a S, then the equivalence class of containing a, denoted [a], is defined to be: [a] = {x S : a x} Example Consider the relation R = { (a, b) R 2 : a = b }. Show it is an equivalence relation. For an element a R, what is [a]? Reflexive: We need to prove ara for any a R. Let a R. Clearly, a = a so that ara. Symmetric: We need to prove that arb = bra. Now, arb = a = b = b = a = bra. Transitive: We need to prove that arb and brc = arc. Suppose arb and brc then, a = b and b = c = a = c = arc. [a] = {a, a} since a = b a = ±b. Example Consider the relation R = { (a, b) Z 2 : a b } Is it an equivalence relation? Is it an order relation? Reflexive: We need to prove ara for any a Z. Let a Z, then a a so that ara.

3 26 CHAPTER 1. REVIEW Symmetric: We need to prove that arb = bra. This is obviously not true. If a = 2 and b = 3, we have a b but not b a. Antisymmetric: We need to show that if arb and bra then a = b that is a b and b a = a = b. We know this to be true. Transitive: We need to prove that arb and brc = arc. Suppose arb and brc then, a b and b c = a c that is arc. Hence, it is an order relation. Example The relation we are about to define is an important one. It has a special notation:. If n is a positive integer and a and b are integers, we say that a b if a mod n = b mod n that is if n (a b) or a b is divisible by n. Show it is an equivalence relation and find its equivalence classes. Reflexive: Let a Z, we need to show that a a that is n (a a) which is true since any integer divides 0. Symmetric: Let a and b be integers. We need to show that if a b then b a. a b = n (a b) = a b = kn for some integer k = b a = ( k) n = n b a = b a Transitive: Let a, b and c be integers. We need to show that if a b and b c then a c. If a b then a b = k 1 n. Similarly, b c = k 2 n. Therefore, and therefore n (a c) or a c. a c = a b + b c = (k 2 + k 2 ) n We can also find the equivalent classes of an element a. By definition, we know that b a if n b a that is b a = kn for some integer k or b = a + kn. Hence [a] = {a + kn : k Z}. Next, we give an important property about equivalence classes. The proof of this property is left as an exercise. Lemma Let S be a set and an equivalence relation on S. If a, b S then a b = [a] = [b]. Proof. See homework.

4 1.4. EQUIVALENCE RELATIONS AND PARTITIONS Partitions Equivalent classes play an important role as we will see in the next theorem. Definition (Partition) A partition of a set S is a collection of nonempty, disjoint subsets of S whose union is S. Example The sets {2, 4, 6, 8,...} and {1, 3, 5, 7,...} form a partition of N. They are disjoint and their union is N. Example The sets {0}, {1, 2, 3,...}, {..., 3, 2, 1} form a partition of Z. They are disjoint and their union is Z. Theorem Let S be a set and be an equivalence relation on S. The equivalent classes of form a partition of S. Conversely, for any partition P of S, there is an equivalence relation on S whose equivalent classes are the elements of P. Proof. We need to prove both directions. Let be an equivalent relation on S. We need to show that the equivalent classes of form a partition of S that is they are not empty, disjoint and their union is S. Let a S. Clearly a a from the reflexive property of. Thus, [a]. Also, the union of all these classes is S. Next, let [a] and [b] be distinct equivalence classes, we must show [a] [b] =. Suppose not, that is c [a] [b].then, c a and c b. By transitivity, it implies that a b and by the lemma it implies [a] = [b] which is a contradiction since they are supposed to be distinct. Hence [a] [b] =. Let P be a partition of S that is P is a collection of non-empty disjoint subsets of S whose union is S. For a and b in S, we define a b if a and b belong to the same element of P. It is easy to check that this is an equivalence relation. Obviously, a and a belong to the same subset of P so that a a. Also, if a b then a and b belong to the same subset of P. Thus b and a belong to the same subset of P hence b a. Finally, if a b and b c then a and b belong to the same subset of P. So do b and c.it follows that a and c belong to the same subset of P hence a c. So, is reflexive, symmetric and transitive. It is an equivalence relation Exercises 1. Prove that if a [b] then [a] = [b]. Use this to prove lemma in the notes. 2. What are the equivalent classes of the relation defined by a b if a mod n = b mod n in the cases n = 1, n = 4, n = 5? 3. Let S = R. For a, b S, define the relation by a b a b Z.

5 28 CHAPTER 1. REVIEW (a) Show that is an equivalence relation on S. (b) Describe the equivalence classes of S. 4. Let S = Z. For a, b S, define the relation R by arb ab 0. Is R an equivalence relation on S? 5. Let S = Z. For a, b S, define the relation R by arb a + b is even. (a) Show that R is an equivalence relation on S. (b) Describe the equivalence classes of S. 6. Do the following problems at the end of chapter 12 of your book: B3, B4, B5, C5, D1, D3.

6 Bibliography [G] Gallian, Joseph A., Contemporary Abstract Algebra, Seventh Edition, Brooks/Cole, [P] Pinter, Charles C., A Book of Abstract Algebra, Second Edition, Dover Publications,