# Equivalence relations

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1 Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence specified by the pair (a, b) Z (Z {0}). But different ordered pairs (a, b) can define the same rational number a/b. In fact, a/b and c/d define the same rational number if and only if ad = bc. One way to solve this problem is to agree that we shall only look at those pairs (a, b) in lowest terms, in other words such that b > 0 is as small as possible, which happens exactly when a and b have no common factor. But this leads into complicated questions about factoring, and it is more convenient to let (a, b) be any element of Z (Z {0}) and then describe a very general mechanism for treating certain such pairs as equal. Let X be a set. A relation R is just a subset of X X. Choose some symbol such as and denote by x y the statement that (x, y) R. There are three important types of relations in mathematics: functions f : X X (we denote by y = f(x) the condition that (x, y) R), order (we use x y or x < y for (x, y) R), and equivalence relations for relations that are like equality. These are usually denoted by some special symbol such as, =, or. Here is the formal definition: Definition 0.1. An equivalence relation on a set X is a subset R X X with the following properties: denoting (x, y) R by x y, we have 1. For all x X, x x. (We say is reflexive.) 2. For all x, y X, if x y then y x. (We say is symmetric.) 3. For all x, y, z X, if x y and y z then x z. (We say is transitive.) Here (1) says that the diagonal X = {(x, x) : x X} is a subset of R. (2) says that the set R is symmetric about the diagonal, i.e. that t R = R, where t R is the set t R = {(y, x) : (x, y) R}, 1

2 and (3) is not so easy to describe geometrically. Examples. (1) The graph of a function f : X X is an equivalence relation only if it is the identity, i.e. the graph is the diagonal. (This follows since we must have (x, x) in the graph for every x X.) (2) Order is not an equivalence relation on, say, X = R: is not symmetric and < is neither reflexive nor symmetric. (3) For X = {humans}, the relation x loves y is neither reflexive, symmetric nor transitive. On the other hand, here are some equivalence relations: 1. Equality. In other words, x y x = y. Here R = X. 2. The relation where x y for all x, y X. Here R = X X. 3. The relation of congruence on the set of all plane triangles (or all plane figures); likewise similarity of triangles. 4. Let l 1 = p 1 q 1 and l 2 = p 2 q 2 be two directed line segments in the plane (i.e. l i is the line segment starting at p i and ending at q i ). Then we can define l 1 and l 2 to be equivalent if they have the same magnitude and direction, or equivalently if they define the same vectors; as we have seen, this is the same as requiring that q 1 p 1 = q 2 p 2. It is straightforward to check that this defines an equivalence relation on the set of directed line segments. 5. For X = R n {0}, where 0 = (0,..., 0) is the origin in R n, define v w there exists a t R, t 0, such that tv = w. Here v v (take t = 1). Thus is reflexive. If v w, then there exists a t R, t 0, such that tv = w. But then v = t 1 w, so that w v and is symmetric. Finally, we leave the argument that is transitive as an exercise. Clearly, v w v and w lie on the same line through the origin in R n. 6. We can define when two sets A and B have the same number of elements by saying that there is a bijection from A to B. This is an equivalence relation, provided we restrict to a set of sets (we cannot just define this as an equivalence relation on the set of all sets, since this is too big to be a set). For example, we could define this relation on a set such as P(R), the set of all subsets of the real numbers. The content of this statement is as follows: (1) given a set A, A A, i.e. there is a bijection from A to itself (the identity function Id A ); (2) If 2

3 A B, i.e. there is a bijection from A to B, say f : A B, then there is a bijection from B to A, in fact f 1 exists since f is a bijection, and f 1 : B A is a bijection from B to A since f is its inverse; (3) If A B and B C, then A C. In fact, given a bijection from A to B, say f, and a bijection from B to C, say g, then we have seen that the composition g f is a bijection from A to C. 7. Consider the following equivalence relation on integers: n and m are equivalent (write this as n m (mod 2)) if they are both even or both odd. Another way to say this is to say that n m (mod 2) if and only if n m is even, if and only if 2 divides n m. More generally, if n is a fixed positive integer, we define a b (mod n) n (b a), where the notation d k, for integers d, k, means d divides k. Then an argument shows that this is an equivalence relation: clearly a a (mod n), since n always divides a a = 0. Thus (mod n) is reflexive. Next, if a b (mod n), then n (b a), and hence n (a b) = (b a). Hence (mod n) is symmetric. Finally, to see that (mod n) is transitive, suppose that a b (mod n) and that b c (mod n). Then by definition n (b a) and n (c b), so that n ((b a) + (c b)) = c a. Then a c (mod n), so that equivalence relation. (mod n) is transitive and hence an 8. For a related example, define the following relation (mod 2π) on R: given two real numbers, which we suggestively write as θ 1 and θ 2, θ 1 θ 2 (mod 2π) θ 2 θ 1 = 2kπ for some integer k. An argument similar to that above shows that (mod 2π) is an equivalence relation. Here intuitively θ 1 θ 2 (mod 2π) if θ 1 and θ 2 define the same angle. 9. Suppose that X is a set and that f : X Y is a fixed function from X to some set Y. Define a b if f(a) = f(b). The fact that is an equivalence relation follows from the basic properties of equality on Y. For example, a a just says that f(a) = f(a). Warning: A relation R is a subset of X X, but equivalence relations say something about elements of X, not ordered pairs of elements of X. The 3

4 ordered pair part comes in because the relation R is the set of all (x, y) such that x y. It is accidental (but confusing) that our original example of an equivalence relation involved a set X (namely Z (Z {0})) which itself happened to be a set of ordered pairs. Equivalence relations are a way to break up a set X into a union of disjoint subsets. Given an equivalence relation and a X, define [a], the equivalence class of a, as follows: [a] = {x X : x a}. Thus we have a [a]. Given an equivalence class [a], a representative for [a] is an element of [a], in other words it is a b X such that b a. For example: 1. If is equality =, then [a] = {a}. 2. If corresponds to R = X X, in other words x y for all x, y X, then [a] = X for every a X. 3. If is congruence = of triangles, then the equivalence class of a triangle T is the set of all triangles which are congruent to T. This is sometimes called a congruence class. 4. An equivalence class of directed line segments is called (in physics) a vector. 5. For X = R n {0}, where 0 = (0,..., 0) is the origin in R n, and v w there exists a t R, the equivalence class [v] is L v {0}, where L v is the line through the origin containing v. 6. Here, an equivalence class is called a cardinal number. 7. For the equivalence relation on Z, (mod 2), there are two equivalence classes, [0], which is the set of even integers, and [1], which is the set of odd integers. More generally, given a positive integer n, the equivalence classes for (mod n) correspond to the possible remainders when we divide by n, in other words there are n equivalence classes which we can write as [0], [1],..., [n 1], and, given a Z with 0 a n 1, an integer k [a] k has remainder a when divided by n, i.e. there exists an integer q such that k = nq + a. We shall describe this process in more detail later. 8. As noted before, we think of equivalence classes (mod 2π) as angles. 4

5 9. Given f : X Y a function and the equivalence relation x y if f(x) = f(y), the equivalence classes are the sets of preimages f 1 (z) for z in the image of f. (Why?) Note that, in this case, [x] = f 1 (f(x)). In the above examples, two equivalence classes which are not equal are disjoint. In fact, this is a general property: Proposition 0.2. Let be an equivalence relation on a set X, and let [a] be the equivalence class of a. If [a] [b], then [a] = [b]. Thus, for every a X, a is contained in exactly one equivalence class. Proof. Suppose that there is some c [a] [b]. We will show that [a] [b]. By symmetry [b] [a] (there is nothing special about either one) so that [a] = [b]. By definition c a and c b. Using symmetry of, a c as well. Given x [a], by definition x a also. Now x a and a c, so by transitivity x c. Since c b, again by transitivity x b. Thus by definition x [b]. For an equivalence relation, we have the set of all equivalence classes {[a] : a X}. This set is sometimes written X/. It is a subset of the power set P(X) with the following property: two subsets of X which lie in X/ are either equal or disjoint (this is the statement of the proposition above) and every element of X lies in some (hence exactly one) set in X/. Put another way: X is a disjoint union of the equivalence classes. One can also define an equivalence relation by reversing this procedure: suppose that X is the union of of disjoint subsets, and define x y if x and y are in the same subset. Then one can check that is an equivalence relation whose equivalence classes are exactly the subsets we started with. Finally, we note that there is a natural surjective function from X to X/, often called projection, defined by: for all x X, π(x) = [x]. Notation: For X = Z and equal to (mod n), we write Z/nZ for Z/. For X = R and equal to (mod 2π), we write R/2πZ for R/. Sometimes equivalence classes can have a best representative. For example, for the rational number example below, a good choice of representative is to take (a, b) with b > 0 and as small as possible. For the relation on Z, (mod 2), there are two equivalence classes, the even and the odd integers, and an obvious choice is to take [0] for the equivalence class of even 5

8 (ii) To see that the function f is well-defined, we have to show that, if we chose another representative y [x], then F (y) = F (x), or equivalently that x y = F (x) = F (y). This is called showing that the function f is well-defined. For example, for X = R and the equivalence relation (mod 2π), we can define the cosine of an angle [θ], i.e. we take the function F : R R defined by F (x) = cos x. Then we define f([θ]) = F (θ). If θ [θ], i.e. if θ and θ are two different representatives for [θ], then θ = θ + 2kπ, and hence cos(θ ) = cos θ. Thus f is well-defined. Note that, if we had instead chosen the function F (x) = x, or F (x) = x 2, or F (x) = e x, then the corresponding function would not be well-defined. For example, for F (x) = x, the function F has different values on two different representatives θ and θ + 2kπ of [θ], as long as k 0, since F (θ) = θ F (θ + 2kπ) = θ + 2kπ. Often, Y = X / is also a space of equivalence classes, i.e. X is a set and is an equivalence relation on X. In many of our examples, in fact, X = X and is equal to. In this case, there is a slight variant of (i) and (ii) above that we use: (i) Start with a function F : X X and define: f([x]) = [F (x)]. In other words, we choose a representative x [x] and define f([x]) to be the equivalence class (for ) containing F (x). (ii) To see that the function f is well-defined, we have to show that, if we chose another representative y [x], then [F (y)] = [F (x)], or equivalently that x y = F (x) F (y). There are similar variations for functions f : (X/ ) (X/ ) Y or f : (X/ ) (X/ ) X / ; we have already seen examples of this. For example, we have seen that addition and multiplication are well defined functions from Q Q to Q. Here are some more examples: (1) For Z/nZ, in other words for X = Z and the equivalence relation (mod n), we shall see that addition and multiplication are well-defined functions from (Z/nZ) (Z/nZ) to Z/nZ. (2) For R/2πZ, in other words for X = R and the equivalence relation (mod 2π), it is not hard to show that addition is a well-defined function from (R/2πZ) (R/2πZ) to R/2πZ. The argument is very similar to that for addition on Z/nZ. However, multiplication is not well-defined for R/2πZ, in other words one can add angles but it is not in general possible to multiply 8

9 angles. Concretely, if t R but t / Z, then multiplication by t is not well-defined. For example, if θ, θ R are two representatives of [θ], then θ θ = 2kπ with k Z. Then tθ tθ = 2tkπ, and this is of the form 2lπ for some integer l tk = l. For multiplication to be well-defined, i.e. independent of the choice of representative, we must have tk Z for all k Z. In particular, taking k = 1, we see that t Z. 9

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