# Sets, Relations and Functions

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1 Sets, Relations and Functions Eric Pacuit Department of Philosophy University of Maryland, College Park pacuit.org ugust 26, 2014 These notes provide a very brief background in discrete mathematics. 1 asic Set Theory We often groups things together. Everyone in this class, your group of friends, your family. These are all collections of people. Set theory is a mathematical language to talk about collections. The easiest way to visualize a set is to use a Venn diagram. Venn Diagram is a geometrical visualization of a set, or collection of sets. bstractly, a set is just a collection of objects that have something in common. We will denote sets with upper-case letters, and elements of sets will be denoted with lower-case letters. There are two ways to write down the contents of a set: 1. List all the elements of the set. Each element should be separated by a comma and there entire list of elements is written between curly brackets: { and }. For example suppose is the set of the first 5 letters of the alphabet. Then = {a, b, c, d, e}. 2. Write down a property that all elements of the set have in common. For example, if is the set of all positive integers, then we can describe as follows = {x x 0 and x is an integer}. This is read is the set of all x such that x is an integer that is greater than or equal to zero. 1

2 Definition 1.1 (Set) ny collection of objects (formally, a set is a collection of elements of a universal set 1 ). Definition 1.2 (Element) member of a set. We write x to mean x is an element of the set. Definition 1.3 (Subset) set is a subset of a set, denoted, provided every element of is also an element of. Notice that every set is a subset of the universal set. The notion of subset can be pictured as follows: Definition 1.4 (Union) The union of two (or more) sets is a set that contains all the elements of each set. For two sets and, the union of and is the set = {x x or x }. The union of two sets can be pictured as follows (the gray shaded region is the union of and ): Definition 1.5 (Intersection) The intersection of two (or more) sets is the set of all items in common to each set. If and are two sets, then the intersection of and is the set = {x x and x }. 1 Typically, we start by fixing a universal set that defines the domain of discourse. 2

3 The intersection can be pictured as follows (the gray shaded region is the intersection of and ): Definition 1.6 (Set Difference) The difference between two sets and ( minus ) is all elements in but not in. The difference between and is the set = {x x and x }. The differences between and can be pictured as follows: Definition 1.7 (Complement) The Complement of a set is the set of all elements not contained in that set. Formally, the complement of the set is C = {x x is in the universal set and x } Why is the notion of a universal set necessary for this definition? Exercise 1.8 Using Venn diagrams, convince yourself that for any sets and, = C Definition 1.9 (Symmetric Difference) The symmetric difference of two sets is all the elements in either set but not in both. The symmetric difference is the set ( ) ( ). The symmetric difference can be pictured as follows: 3

4 ( ) ( ) Definition 1.10 (Product) The Cartesian product of two or more sets is all possible pairings of each elements of the sets. If and are two sets, then the cartesian product of and is the set = {(x, y) x and y } Definition 1.11 (Null Set) The null or empty set is a set that contains no elements. We write to denote the set containing no elements. Definition 1.12 (Power Set) The power set is the set of all subsets. If is a set, the power set of is the set () = { }. Notice that the empty set is a subset of every set. Definition 1.13 (Cardinality of a Set) The cardinality of a finite 2 set is the total number of elements in, and is denoted. Definition 1.14 (Partition) partition of a set S is a collection of sets S = {S 1, S 2,...} (possibly infinite) such that the sets are pairwise disjoint: if S i, S j S with i j, then S i S j = their union is S, that is, S = Si SS i. 2 Relations Definition 2.1 (inary Relation) binary relation R on two sets and is a subset of the cross product of and, i.e., R 2 The notion of cardinality can be applied to infinite sets as well. However, a discussion of this is beyond the scope of these introductory notes. 4

5 You have already studied binary relations during your mathematical eduction: =,,, <, and > are all relations on numbers (eg., the natural numbers N, real numbers R, rational numbers Q, etc.) and is a relation on the power set of a set S. For example, the binary relation N N is the set {(a, b) a, b N and a is less than or equal to b} Suppose that R is a relation. We often write a R b to mean (a, b) R. The following are properties of relations that may be of interest: Reflexivity for all a, a R a Transitivity for all a, b, c, if a R b and b R c then a R c Completeness for all a, b, either a R b or b R a (or a = b) Symmetry for all a, b, if a R b then b R a ntisymmetric for all a, b, if a R b and b R a then a = b Definition 2.2 (Equivalence Relation) relation R that is reflexive, symmetric and transitive is said to be an equivalence relation Definition 2.3 (Equivalence Class) If R is an equivalence relation on, then for each a, the equivalence class of a, denoted by [a], is the following set [a] = {b arb}. Definition 2.4 (Partial Order) relation that is reflexive, antisymmetric and transitive is said to be a partial order. The standard example of a partial order is the relation. The following is our first theorem. It is somewhat technical, but illustrates a fundamental idea about equivalence classes and partitions. Namely, that every partition has an equivalence relation associated with it, and every equivalence class has a partition associated with it. Theorem 2.5 The equivalence classes of any equivalence relation R on a set forms a partition of, and any partition of determines an equivalence relation on for which the sets in the partition are the equivalence classes. Proof. Suppose R is an equivalence relation on. We must show that the equivalence classes of R forms a partition of. 5

6 1. Each equivalence class is non-empty, since a R a for all a. 2. Clearly is the union of all the equivalence classes (since each element of belongs to at least one equivalence class) 3. We must show any two equivalence classes are disjoint. Let [a], [b] be two distinct equivalence classes. Suppose c [a] [b]. Then a R c and b R c. Hence by symmetry, c R b. nd so by transitivity, a R b. Let x [a], then x R c and by the above argument x R b (Why?), and so x [b]. Thus [a] [b]. Using a similar argument, we can show [b] [a]. Therefore [a] = [b], which contradicts the fact that [a] and [b] are distinct equivalence classes. For the second part of the theorem, suppose = { 1,..., n } is any partition of. Define R = {(a, b) a i and b i }. We must show that R is reflexive. Let a be any element. Then a i for some i, and hence by definition of R, a R a. Next we will show that R is symmetric. Suppose a R b. Then a i and b i for some i. Then clearly, b i and a i and hence b R a. We must show R is transitive. Suppose, a R b and b R c. Then a i and b i, and b j and c j for some i, j. Since b i j, i = j (since the elements of are pairwise disjoint). Therefore, a i and c i and hence a R c. qed Functions are a special type of relation: Definition 2.6 (Function) function f is a binary relation on and (i.e., f ) such that for all a, there exists a unique b such that (a, b) f. We write f : when f is a function, and if (a, b) f, then write f(a) = b. Suppose f : is a function. codomain. is said to be the domain and the Definition 2.7 (Image) The image of a set is the set: f( ) = {b b = f(a) for some a } Definition 2.8 (Range) The range of a function is the image of its domain. Suppose that f : is a function. 6

7 Definition 2.9 (Surjection) f is a surjection (or onto) if its range is equal to its codomain. I.e., f is surjective iff for each b, there exists an a such that f(a) = b Definition 2.10 (Injection) f is an injection (or 1-1) if distinct elements of the domain produce distinct elements of the codomain. I.e., f is 1-1 iff a a implies f(a) f(a ), or equivalently f(a) = f(a ) implies a = a. Definition 2.11 (ijection) f is a bijection if it is injective and surjective. In this case, f is often called a one-to-one correspondence. Definition 2.12 (Inverse Image) Suppose that f : and that Y. The inverse image of Y is the set f 1 (Y ) = {x x and f(x) Y } 3 Proofs 3.1 Introduction Learning how to write mathematical proofs takes time and lots of practice. proof of a mathematical statement is simply an explanation of that statement written in the language of mathematics. It is very important that you become comfortable with the definitions. If you don t know or understand the formal definitions, then you will not be able to write down your explanations. It would be like trying to explain something to someone in Italian without actually knowing Italian. 3.2 Proving Equality and Subset How do you prove that two sets are equal? The answer to this question depends on who you are trying to convince. In this class, we will always err on the side of caution and given fairly detailed formal proofs. In turns out that proving two sets are equal reduces to proving the sets are subsets of each other. Fact 3.1 = if and only if and Why is this true? Well, if and are equal, then they both name the same collection of objects. I.e., is another name for the collection of objects that names and vice versa. So, if and are equal then of course since is always a subset of itself and is simply another name for. Similarly, we can show. Conversely, suppose and. We want to know that 7

8 and name the same collection of objects. Suppose they didn t, then there should be some object x that is not in or some object y that is not in. Well, we know such an object x cannot exist since, and so, every element of is an element of. Similarly, the element y cannot exist. Hence, and must name the same collection of objects. What about trying to prove that two sets are not equal? This turns out to be easier. In order to show that does not equal, you need only find an element in that is not in OR an element of that is not in. Showing two sets are equal reduces to proving that the sets are subsets of each other. ut, how to show that a set is a subset of another set? The general procedure to show is to show that each element of is also and element of. This is straightforward if and are both finite sets. For example, suppose = {2, 3, 4} and = {1, 2, 3, 4, 5, 6}. How do we show that. Since is finite, we simply notice that 2, 3 and 4. What if is the set of even numbers and is the set of all integers? We would get awfully tired (and bored) if we waited around to show that each and every element of is also an element of. Imagine and are two boxes, and you would like to know whether all the elements in s box are also in s box. Suppose you reach in box and select an element, say the number 10. fter inspecting 10, you notice that 10 is in fact an integer and so must also be an element of box. ut you are not satisfied, since you cannot be sure that the next element you choose from will also be an element of. In fact, even if you have shown that the first million even integers are all members of box, you cannot be sure that the next element you select from box will in fact be an integer. Instead, you should consider the property that all elements of have in common and show that any object satisfying that property must be an element of. What property does x satisfy if it is contained in s box? The answer is x = 2 n, where n is some integer. Then, you simply notice that if n is an integer, then 2 n is also an integer; and hence, any element of must also be an element of. 3.3 Examples Theorem 3.2 = Proof. We must show and. 8

9 We will show. Suppose x. Then x or x. Suppose x then x. Then x (if x is not in then x is certainly not in both and ). Hence x. Suppose x. For similar reason, x. Hence in either case, x. Therefore,. We must show. Suppose x. Then x, and so x or 3 x. Hence either x or x. In either case, x. qed Theorem 3.3 iff =. Proof. We must show implies = and = implies. ssume that. We must show =. I.e. we must show (1) and (2). The first statement is trivial, it is always the case that. For the second statement, assume x. We must show x. Since, x. Hence x. ssume =. We must show. Let x. Then x since =. Hence x. qed ttempt to answer these questions before looking at the answers. Exercise 3.4 Suppose that f :. Prove or disprove the following: 1. If X and Y, then f(x Y ) = f(x) f(y ). 2. If X, Y and f is 1-1,then f(x Y ) = f(x) f(y ). 3. If X and Y, then f 1 (X Y ) = f 1 (X) f 1 (Y ). Claim 3.5 It not the case that if X and Y, then f(x Y ) f(x) f(y ). Proof of Claim 3.5. To prove this, we must find counterexample. Let = {1, 2, 3} and = {a, b, c}. nd f : be defined as follows: f(1) = c, f(2) = b and f(3) = c. Let X = {1, 2} and Y = {2, 3}. Then X Y = {2} and f(x Y ) = f({2}) = {f(2)} = {b}. ut, f(x) f(y ) = {f(1), f(2)} {f(2), f(3)} = {c, b} {b, c} = {b, c}. Hence, f(x Y ) f(x) f(y ). qed (of Claim) 3 Notice that x does not imply that x and x. The and in italics is should be an or. Make sure you clearly understand the logic here, since this is often misunderstood by students. 9

10 It is true that for any function f : and all subsets X, Y, f(x Y ) f(x) f(y ) (for a proof see below). Claim 3.6 If X, Y and f is 1-1,then f(x Y ) = f(x) f(y ). Proof of Claim 3.6. Suppose that f : is a 1-1 function. Let X and Y. We must show (1) f(x Y ) f(x) f(y ) and (2) f(x) f(y ) f(x Y ). Notice that (1) is true even if f is not 1-1. Let y f(x Y ). Then there is an element x X Y such that f(x) = y. Since x X Y, x X and x Y. Therefore, y = f(x) f(x) and y = f(x) f(y ). Hence, y f(x) f(y ). We now prove (2). Let y f(x) f(y ). Then y f(x) and y f(y ). Since y f(x) there is x 1 X such that f(x 1 ) = y. Since y f(y), there is x 2 Y such that f(x 2 ) = y. Since f is 1-1, x 1 = x 2. Therefore x 1 = x 2 X Y ; and so, y = f(x 1 ) = f(x 2 ) f(x Y ). qed (of Claim) Claim 3.7 If X and Y, then f 1 (X Y ) = f 1 (X) f 1 (Y ). Proof of Claim 3.7. Let f : be any function and suppose X and Y. We must show f 1 (X Y ) f 1 (X) f 1 (Y ) and f 1 (X) f 1 (Y ) f 1 (X Y ). Suppose x f 1 (X Y ). Then f(x) X Y. Hence f(x) X and f(x) Y. Since f(x) X, x f 1 (X). Since f(x) Y, x f 1 (Y ). Therefore x f 1 (X) f 1 (Y ). Suppose x f 1 (X) f 1 (Y ). Then x f 1 (X) and x f 1 (Y ). Therefore, f(x) X and f(x) Y. Hence, f(x) X Y ; and so, x f 1 (X Y ). asdfasdfasdfasdfa qed (of Claim) 10

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