# Section 7.5: Equivalence Relations Def: A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive. Ex: Let.

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Section 7.5: Equivalence Relations Def: A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive. Ex: Let A be any set and define R = {(a, b) | a = b}. That is, every element of A is related to itself only. R is an equivalence relation. R is clearly reflexive, and vacuously symmetric and transitive. R is called the trivial equivalence on A. Ex: Let A be the set of real numbers and define R = {(a, b) | a – b  Z}. R is reflexive since a – a = 0 for all a  A and 0 is an integer. R is symmetric since if a – b is an integer, then –(a – b) = b – a is also an integer. R is transitive since if a – b is an integer and b – c is an integer, then a – c = (a – b) + (b – c) is also an integer as the sum of two integers.

Ex: Let A = Z and R = {(a, b) | a divides b}. R is not an equivalence relation because it is not symmetric. For example, 1 divides 2 but 2 does not divide 1. Ex: Let m  Z +. Then R = {(a, b) | a  b (mod m)} is an equivalence relation over the set of integers. Proof: Let m  Z+. Now let a, b  Z such that (a, b)  R. That is a  b (mod m) which means that m | (a – b). So a – b = km for some integer k. Then b – a = -(a – b) = -km = (-k)m. So m | (b – a). So b  a (mod m). So (b, a)  R. Hence R is symmetric. Now let a, b, c  Z such that (a, b), (b, c)  R. That is a  b (mod m) and b  c (mod m) which means that m | (a – b) and m | (b – c). Then m | ((a – b) + (b – c)). That is, m | (a – c). So a  c (mod m). So (a, c)  R. Hence R is transitive. So R is an equivalence relation.  Let a  Z. Then (a, a)  R since m | (a – a). So R is reflexive.

Def: Let R be an equivalence relation over a set A. The set of all elements that are related to an element a of A is called the equivalence class of a. The equivalence class of a with respect to R is denoted as [a] R. When only one relation is under consideration, we will use just [a] to denote the equivalence class of a with respect to R. Remark: [a] R = {s  A | (s, a)  R}. Ex: We have seen that R = {(a, b) | a  b (mod 3)} is an equivalence relation over the set of integers. What is [1], [2], [3], [4]? [1] = {s  Z | s  1 (mod 3)} = {…, -5, -2, 1, 4, 7, …} = {3k + 1 | k  Z} [2] = {s  Z | s  2 (mod 3)} = {…, -4, -1, 2, 5, 8, …} = {3k + 2 | k  Z} [3] = {s  Z | s  3 (mod 3)} = {…, -3, 0, 3, 6, 9, …} = {3k + 3 | k  Z} [4] = {s  Z | s  4 (mod 3)} = {…, -2, 1, 4, 7, 10, …} = {3k + 4 | k  Z} = {3k + 1 | k  Z} = [1]. We call 1 and 4 representatives for the equivalence class [1] or [4].

Theorem: Let R be an equivalence relation on a set A. Then TFAE: (1) (a, b)  R(2) [a] = [b](3) [a]  [b]   Proof: Let A be a set and R an equivalence relation on A. (1  2) Let a, b  A such that (a, b)  R. Now let c  [a]. Then (c, a)  R. We wish to show that c  [b] or (c, b)  R. Since R is an equivalence relation and (c, a), (a, b)  R then (c, b)  R. Now let c  [b] and we will show that c  [a]. So (c, b)  R. Then since (a, b)  R then (b, a)  R. Since (c, b), (b, a)  R then (c, a)  R. (2  3) Let a, b  A such that [a] = [b]. Since R is reflexive, then (a, a)  R. So a  [a]. Now since [a] = [b] then a  [b] also. So a  [a]  [b]. Hence [a]  [b]  . (3  1) Let a, b  A such that [a]  [b]  . Then there is some element x  [a]  [b]. So (x, a), (x, b)  R. Since R is symmetric then (a, x)  R. Now since R is transitive then (a, b)  R. Since 1  2  3  1 then 1, 2, and 3 are logically equivalent. 

Def: Let S be a set. A partition of S is a collection of non-empty, pairwise disjoint subsets of S where the union of all sets in the collection is S. In other words, if we use I as an indexing set for our collection, then a partition of S is a collection of sets {A i } satisfying the following: (1) Each A i is non-empty. That is  i  I, A i  . (2) The collection is pairwise disjoint. That is  j  k  I, A j  A k = . (3) The union of all sets in the collection is S. That is  i  I A i = S. Ex: The collection {E, O} (where E is the set of even integers and O is the set of odd integers) is another partition of the set of integers. Ex: The collection {Z -, {0}, Z + } is a partition of the set of integers. Ex: The collection {P, C} (where P is the set of primes and C is the set of composites) is a partition of the set of integers greater than 1. Ex: The collection {Q, I} is a partition of the set of real numbers.

Ex: The collection {…, {-2}, {-1}, {0}, {1}, {2}, …} is another partition of the set of integers. This is a trivial partition. Ex: The collection {Z}is another partition of the set of integers. This is another trivial partition. Theorem: Let R be an equivalence relation on a set S. Then the collection of equivalence classes of R, {[a] | a  S}, is a partition of S. Conversely, given a partition {A i } (with indexing set I), there is an equivalence relation R that has the sets in the partition as its equivalence classes. Ex: Recall the equivalence relation R = {(a, b) | a  b (mod 3)} over the set of integers. The theorem says that if we take the collection {…, [-2], [-1], [0], [1], [2], …} it is a partition of the integers. Realize first that this collection is {[0], [1], [2]} after removing duplicates. We can see that each set in the collection is non-empty and they are pairwise disjoint. Also their union is Z.

Let’s prove the forward direction of the previous theorem. Proof: Let R be an equivalence relation on a set S. We must show that the set of equivalence classes of R, {[a] | a  S}, is a partition of S. From a previous theorem we know that [a]  [b]    [a] = [b]. That is, if two equivalence classes aren’t disjoint then they are the same. This gets us that the collection of equivalence classes is pairwise disjoint. Further, every equivalence class [x] is non-empty since an equivalence relation is reflexive, so (x, x)  R, hence x  [x]. Lastly, the union of all of the equivalence classes of R gives the set S. For any element x  S, x  [x]. So we get all of S. We can get no more than S because each equivalence class of R is a subset of S.  In order to prove the converse direction, we would have to show that given an arbitrary partition of S, we can find an equivalence relation on S the has as its equivalence classes the sets of the partition. [think on it]

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