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THE NATURE OF SETS Copyright © Cengage Learning. All rights reserved. 2.

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Presentation on theme: "THE NATURE OF SETS Copyright © Cengage Learning. All rights reserved. 2."— Presentation transcript:

1 THE NATURE OF SETS Copyright © Cengage Learning. All rights reserved. 2

2 2.2 Operations with Sets

3 3 Fundamental Operations

4 4 Suppose we consider two general sets, B and Y, as shown in Figure 2.7. If we show the set Y using a yellow highlighter and the set B using a blue highlighter, it is easy to visualize two operations. Figure 2.7 Venn diagram showing intersection and union

5 5 Fundamental Operations The intersection of the sets is the region shown in green (the parts that are both yellow and blue). We see this is region III, and we describe this using the word “and.” The union of the sets is the part shown in any color (the parts that are yellow or blue or green). We see this is regions II, III, and IV, and we describe this using the word “or.”

6 6 Example 2 – Venn diagram with three sets Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, B = {1, 3, 5, 7}, and C = {5, 7}. A Venn diagram showing these sets is shown in Figure 2.10. a. A  C b. B  C c. B  C d. A  C Figure 2.10 Venn diagram showing three sets

7 7 Example 2 – Solution a. A  C = {2, 4, 6, 8}  {5, 7} = {2, 4, 5, 6, 7, 8} Notice that the union consists of all elements in A or in C or in both. b. B  C = {1, 3, 5, 7}  {5, 7} = {1, 3, 5, 7} Notice that, even though the elements 5 and 7 appear in both sets, they are listed only once. That is, the sets {1, 3, 5, 7} and {1, 3, 5, 5, 7, 7} are equal (exactly the same).

8 8 Example 2 – Solution Notice that the resulting set has a name (it is called B), and we write B  C = B c. B  C = {1, 3, 5, 7}  {5, 7} = {5, 7} = C d. A  C = {2, 4, 6, 8}  {5, 7} = { } These sets have no elements in common, so we write { } or. The intersection contains the elements common to both sets. cont’d

9 9 Fundamental Operations Suppose we consider the cardinality of the various sets in Example 2: | U | = 9, | A | = 4, | B | = 4, and | C | = 2 |A  C| = 6 (part a) |B  C| = 4 (part b) |B  C| = 2 (part c)

10 10 Fundamental Operations |A  C| = 0 (part d) | | = 0 (part d) Consider the set S formed from the sets in Example 2: S = {U, A, B, C, }

11 11 Fundamental Operations This is a set of sets; there are five sets in S, so | S | = 5. Furthermore, if we remove sets from S, one-by-one, we find T = {A, B, C, }, so | T | = 4 U = {B, C, }, so | U | = 3 V = {C, }, so | V | = 2 Finally, W = { }, so | W | = 1 Thus |{ }| = 1, but | | = 0, so { } ≠.

12 12 Cardinality of Intersections and Unions

13 13 Cardinality of Intersections and Unions The cardinality of an intersection is easy; it is found by looking at the number of elements in the intersection. The cardinality of a union is a bit more difficult. For sets with small cardinalities, we can find the cardinality of the unions by direct counting, but if the sets have large cardinalities, it might not be easy to find the union and then the cardinality by direct counting.

14 14 Cardinality of Intersections and Unions Some students might want to find |B  C| by adding | B | and |C|, but you can see from Example 2 that |B  C| ≠ | B | + | C |. However, if you look at the Venn diagram for the number of elements in the union of two sets, the situation becomes quite clear, as shown in Figure 2.11. Figure 2.11 Venn diagram for the number of elements in the union of two sets

15 15 Cardinality of Intersections and Unions

16 16 Example 3 – Cardinality of a union Suppose a survey indicates that 45 students are taking mathematics and 41 are taking English. How many students are taking math or English?

17 17 Example 3 – Solution At first, it might seem that all you do is add 41 and 45, but such is not the case. Let M = {persons taking math} and E = {persons taking English}. Suppose 12 students are taking both math and English.

18 18 Example 3 – Solution In this case we see that By diagram: First, fill in 12 in M  E. Then, | M | = 45, so fill in 33 since 45 – 12 = 33 | E | = 41, so fill in 29 since 41 – 12 = 29 The total number is 33 + 12 + 29 = 74. cont’d

19 19 Example 3 – Solution By formula: |M  E| = | M | + | E | – |M  E| = 45 + 41 – 12 = 74 cont’d


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