64 CHAPTER 2 Sets Note, from Example 8 that parts a) and b) showed that ′ ′ = ′ F N F N ( ) < > and that parts c) and d) showed that ′ ′ = ′ F N F N ( ) . > < Such relationships between union, intersection, and complement will be discussed further in Section 2.4. The Meaning of and and or The words and and or are very important in many areas of mathematics. We use these words in several chapters in this book, including Chapter 3, Logic and Chapter 11, Probability. The word and is generally interpreted to mean intersection, whereas or is generally interpreted to mean union. Suppose = A {1, 2, 3, 5, 6, 8} and = B {1, 3, 4, 7, 9, 10}. The elements that belong to set A and set B are 1 and 3. These are the elements in the intersection of the sets. The elements that belong to set A or set B are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. These are the elements in the union of the sets. The Relationship Between n A B ( ), < n A( ), n B( ), and n A B ( ) > Having looked at unions and intersections, we can now determine a relationship between n A B n A n B ( ), ( ), (), < and n A B ( ). > Suppose set A has eight elements, set B has five elements, and A B > has two elements. How many elements are in A B? < Let’s make up some arbitrary sets that meet the criteria specified and draw a Venn diagram. If we let = A a b c d e f g h { , , , , , , , },thenset B must contain five elements, two of which are also in set A. Let = B g h i j k { , , , , }. We construct a Venn diagram by filling in the intersection first, as shown in Fig. 2.10. The number of elements in A B < is 11. The elements g and h are in both sets, and if we add + n A n B ( ) ( ), we are counting these elements twice. To determine the number of elements in the union of sets A and B, we can add the number of elements in sets A and B and then subtract the number of elements common to both sets. Solution a) We first determine F’ and N F. {MD, St, Su, Du, Do} ′ ′ = and ′ = N {CF, TB, We, Do, Ch}. Then, < < F N {MD, St, Su, Du, Do} {CF, TB, We, Do, Ch} {MD, St, Su, Du, Do, CF, TB, We, Ch} ′ ′ = = b) We first determine F N {CF, We, BK, TB, Ch} {Du, Su, MD, St, BK} > > = = {BK}. Then, ′ = F N ( ) {MD, St, CF, TB, Su, We, Du, Do, Ch}. > Although, the order of the elements is different, this is the same set we obtained in part a). c) Recall from part a) that ′ = F {MD, St, Su, Du, Do} and ′ = N {CF, TB, We, Do, Ch}. Then, ′ ′ = F N {Do} > d) We first determine = F N {CF, We, BK, TB, Ch} {Du, Su, MD, St, BK} < < = {CF, We, BK, TB, Ch, Du, Su, MD, St, BK}. Then, ′ = F N ( ) {Do} < Note, this is the same set we obtained in part c). 7 Now try Exercise 87 U A a b c d e f g h i j k B Figure 2.10
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