Section 2.2

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## Section 2.2

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**Math in Our World**Section 2.2 Subsets and Set Operations**Learning Objectives**• Define the complement of a set. • Find all subsets of a set. • Use subset notation. • Find the number of subsets for a set. • Find intersections, unions, and differences of sets. • Find the Cartesian product of two sets.**Universal Set**A universal set, symbolized by U, is the set of all potential elements under consideration for a specific situation. Once we define a universal set in a given setting, we are restricted to considering only elements from that set. If U = {1, 2, 3, 4, 5, 6, 7, 8}, then the only elements we can use to define other sets in this setting are the integers from 1 to 8.**Complement**The complement of a set A, symbolized A, is the set of elements contained in the universal set that are not in A. Using set-builder notation, the complement of A is A = {x | xU and xA}. This Venn Diagram shows the visual representation of the sets U and A. The complement of a set A is all the things inside the rectangle, U, that are not inside the circle representing set A. U A A**U**A EXAMPLE 1 Finding the Complement of a Set Let U = {v, w, x, y, z} and A = {w, y, z}.Find Aand draw a Venn diagram that illustrates these sets. SOLUTION Using the list of elements in U, we just have to cross out the ones that are also in A. The elements left over are in A. U ={v, w, x, y, z} A={v, x} w z y v x**Subsets**If every element of a set A is also an element of a set B, then A is called a subset of B. The symbol is used to designate a subset; in this case, we write A B. • Every set is a subset of itself. Every element of a set A is of course an element of set A, so A A. • The empty set is a subset of every set. The empty set has no elements, so for any set A, you can’t find an element of that is not also in A.**EXAMPLE 2 Finding All Subsets of a Set**Find all subsets of A = {American Idol, Survivor}. SOLUTION Number of elements in Subset Subsets with that Number of Elements 2 1 0**EXAMPLE 2 Finding All Subsets of a Set**Find all subsets of A = {American Idol, Survivor}. SOLUTION Number of elements in Subset Subsets with that Number of Elements 2 {American Idol, Survivor} 1 {American Idol},{Survivor} 0 So the subsets are: {American Idol, Survivor}, {American Idol},{Survivor}, **U**Proper Subsets If a set A is a subset of a set B and is not equal to B, then we call A a proper subset of B, and write A B. The Venn diagram for a proper subset is shown below. In this case, U = {1, 2, 3, 4, 5}, A = {1, 3, 5}, and B = {1, 3}. A B 2 1 3 5 4**EXAMPLE 3 Finding Proper Subsets of a Set**Find all proper subsets of {x, y, z}. SOLUTION Number of elements in Subset Subsets with that Number of Elements 3 {x, y, z} 2 {x, y}, {x, z}, {y, z} 1 {x}, {y}, {z} 0 So the proper subsets are: {x, y}, {x, z}, {y, z}, {x}, {y}, {z}, We’ll eliminate this one since it’s equal to the original.**EXAMPLE 4 Understanding Subset Notation**State whether each statement is true or false. • {1, 3, 5} {1, 3, 5, 7} (b) {a, b} {a, b} (c) {x | x N and x > 10} N (d) {2, 10} {2, 4, 6, 8, 10} (e) {r, s, t} {t, s, r} (f ) {Lake Erie, Lake Huron} The set of Great Lakes . - “not a subset of” - “not a subset of”**EXAMPLE 4 Understanding Subset Notation**SOLUTION (a) All of 1, 3, and 5 are in the second set, so {1, 3, 5} is a subset of {1, 3, 5, 7}. The statement is true. (b) Even though {a, b} is a subset of {a, b}, it is not a proper subset, so the statement is false. (c) Every element in the first set is a natural number, but not all natural numbers are in the set, so that set is a proper subset of the natural numbers. The statement is true. (d) Both 2 and 10 are elements of the second set, so {2, 10} is a subset, and the statement is false. (e) The two sets are identical, so {r, s, t} is not a proper subset of {t, s, r}. The statement is true. (f ) Lake Erie and Lake Huron are both Great Lakes, so the statement is true.**EXAMPLE 5 Understanding Subset Notation**State whether each statement is true or false. • {5, 10, 15} (b) {u, v, w, x} {x, w, u} (c) {0} (d) **EXAMPLE 5 Understanding Subset Notation**SOLUTION (a) True: the empty set is a proper subset of every set. (b) False: v is an element of {u, v, w, x} but not {x, w, u}. (c) The set on the left has one element, 0. The empty set has no elements, so the statement is false. (d) The empty set is a subset of itself (as well as every other set), but not a proper subset of itself since it is equal to itself. The statement is false.**Number of Subsets for a Finite Set**If a finite set has n elements, then the set has 2n subsets and 2n – 1 proper subsets.**EXAMPLE 6 Finding the Number of Subsets of a Set**Find the number of subsets and proper subsets of the set {1, 3, 5, 7, 9, 11}. SOLUTION The set has n = 6 elements, so there are 2n, or 26 = 64, subsets. Of these, 2n – 1 , or 64 – 1 = 63, are proper.**Intersection of Sets**The intersection of two sets A and B, symbolized by A B, is the set of all elements that are in both sets. In set-builder notation, A B = {x | xA and xB}. Note that the word “and” means intersection. U B A The shaded area represents the intersection of sets A and B.**EXAMPLE 7 Finding Intersections**If A = {5, 10, 15, 20, 25}, B = {0, 10, 20, 30, 40}, and C = {30, 50, 70, 90}, find (a) A B (b) B C (c) AC SOLUTION (a) The elements 10 and 20 are in both sets A and B, so A B = {10, 20}. (b) The only member of both sets B and C is 30, so B C = {30}. (c) There are no elements common to sets A and C, so AC = .**U**A B Disjoint Sets When the intersection of two sets is the empty set, the sets are said to be disjoint. For example, the set of students who stop attending class midway through a term and the set of students earning A’s are disjoint, because you can’t be a member of both sets.**Union of Sets**The union of two sets A and B, symbolized by AB, is the set of all elements that are in either set A or set B (or both). In set-builder notation, A B = {x | xA or xB}. U Note that the word “or” means union. B A The shaded area represents the union of sets A and B.**EXAMPLE 8 Finding Unions**If A = {0, 1, 2, 3, 4, 5}, B = {2, 4, 6, 8, 10}, and C = {1, 3, 5, 7}, find each. (a) A B (b) A C (c) BC SOLUTION To find a union, just make a list of all the elements in either set without writing repeats. (a) AB = {0, 1, 2, 3, 4, 5, 6, 8, 10} (b) A C = {0, 1, 2, 3, 4, 5, 7} (c) B C = {1, 2, 3, 4, 5, 6, 7, 8, 10}**EXAMPLE 9 Performing Set Operations**Let A = {l, m, n, o, p}, B = {o, p, q, r}, and C = {r, s, t, u}. Find each. (a) (A B) C (b) A (BC) (c) (A B) C**EXAMPLE 9 Performing Set Operations**SOLUTION A = {l, m, n, o, p}, B = {o, p, q, r}, and C = {r, s, t, u} • (A B) C First find A B : A B = {l, m, n, o, p, q, r}. Then intersect this set with set C. The only common element is r, so (A B) C = {r}.**EXAMPLE 9 Performing Set Operations**SOLUTION A = {l, m, n, o, p}, B = {o, p, q, r}, and C = {r, s, t, u} (b) A (BC) First find BC : BC= {o, p, q, r, s, t, u}. Then intersect this set with set A. So A (BC) = {o, p}.**EXAMPLE 9 Performing Set Operations**SOLUTION A = {l, m, n, o, p}, B = {o, p, q, r}, and C = {r, s, t, u} (c) (A B) C First find A B : A B = {o, p}. Then find the union of this set with set C. So (A B) C = {o, p, r, s, t, u}.**EXAMPLE 10 Performing Set Operations**If U = {10, 20, 30, 40, 50, 60, 70, 80}, A = {10, 30, 50, 70}, B = {40, 50, 60, 70}, and C = {20, 40, 60}, find each. (a) A C • (A B) C (c) B(A C)**EXAMPLE 10 Performing Set Operations**SOLUTION U = {10, 20, 30, 40, 50, 60, 70, 80}, A = {10, 30, 50, 70} B = {40, 50, 60, 70} C = {20, 40, 60} (a) A C First find A by eliminating the elements in set A from U. So A = {20, 40, 60, 80}. Then find C by eliminating the elements in set C from U. So C = {10, 30, 50, 70, 80}. Now note that 80 is the only element common to both, so A C = {80}.**EXAMPLE 10 Performing Set Operations**SOLUTION U = {10, 20, 30, 40, 50, 60, 70, 80}, A = {10, 30, 50, 70} B = {40, 50, 60, 70} C = {20, 40, 60} (b) (A B) C First find A B since it’s in parentheses. So A B = {50, 70}. Then find (A B) by eliminating the elements in set A B from U. So (A B) = {10, 20, 30, 40, 60, 80}. Finally, we find the intersection of this set and C, so (A B) C = {20, 40, 60}.**EXAMPLE 10 Performing Set Operations**SOLUTION U = {10, 20, 30, 40, 50, 60, 70, 80}, A = {10, 30, 50, 70} B = {40, 50, 60, 70} C = {20, 40, 60} • B(A C) First find C by eliminating the elements in set C from U. C = {10, 30, 50, 70, 80} Then find A C. A C = {10, 30, 50, 70} Then find B by eliminating the elements in set B from U. B = {10, 20, 30, 80} So B(A C) = {10, 20, 30, 50, 70, 80}**Set Subtraction**The difference of set A and set B is the set of elements in set A that are not in set B. In set-builder notation, A – B = {x | xA and xB}.**EXAMPLE 11 Finding the Difference of Two Sets**Let U = {2, 4, 6, 8, 10, 12}, A = {4, 6, 8, 10}, B = {2, 6, 12}, C = {8, 10} Find each. (a) A – B (b) A – C (c) B – C**EXAMPLE 11 Finding the Difference of Two Sets**SOLUTION U = {2, 4, 6, 8, 10, 12}, A = {4, 6, 8, 10}, B = {2, 6, 12}, C = {8, 10} (a) Start with the elements in set A and take out the elements in set B that are also in set A. In this case, only 6 is removed, and A – B= {4, 8, 10}. (b) Start with the elements in set A and remove the elements in set C that are also in set A. In this case, 8 and 10 are removed, and A – C = {4, 6}. (c) Start with the elements in set B and take out the elements in set C that are also in set B. In this case, none of the elements in B are also in C. So B – C = {2, 6, 12}.**Cartesian Product**The Cartesian product (denoted A B) of two sets A and B is formed by writing all possible ordered pairs in which the first component is an element of A and the second component is an element of B. Using set-builder notation, A B = {(x, y) | xA and yB}.**Ordered Pair**An ordered pair is a pair of numbers or objects that are associated by writing them together in a set of parentheses, like (3, 5). In this ordered pair, 3 is called the first component and 5 is called the second component.**EXAMPLE 12 Finding Cartesian Products**If A = {1, 3, 5} and B = {2, 4}, find A B and B A. SOLUTION To form A B, first form ordered pairs with first component 1: (1, 2) and (1, 4). Then form pairs with first component 3: (3, 2) and (3, 4). Finally, use 5 as the first component: (5, 2) and (5, 4). A B = {(1, 2), (1, 4), (3, 2), (3, 4), (5, 2), (5, 4)}. For B A, form all possible ordered pairs with first components from B and second components from A: B A = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5)}.