Schaum's Outline of Discrete Mathematics, Fourth Edition

CHAPTER 1

Set Theory

1.1 INTRODUCTION

The concept of a set appears in all mathematics. This chapter introduces the notation and terminology of set theory which is basic and used throughout the text. The chapter closes with the formal definition of mathematical induction, with examples.

1.2 SETS AND ELEMENTS, SUBSETS

A set may be viewed as any well-defined collection of objects, called the elements or members of the set. One usually uses capital letters, A, B, X, Y, … ⁡, to denote sets, and lowercase letters, a, b, x, y, … ⁡, to denote elements of sets. Synonyms for set are class, collection, and family.

Membership in a set is denoted as follows:

a ∈ S denotes that a belongs to a set S

a, b ∈ S denotes that a and b belong to a set S

Here ∈ is the symbol meaning is an element of. We use ∉ to mean is not an element of.

Specifying Sets

There are essentially two ways to specify a particular set. One way, if possible, is to list its members separated by commas and contained in braces {}. A second way is to state those properties which characterized the elements in the set. Examples illustrating these two ways are:

That is, A consists of the numbers 1, 3, 5, 7, 9. The second set, which reads:

B is the set of x such that x is an even integer and x is greater than 0,

denotes the set B whose elements are the positive even integers. Note that a letter, usually x, is used to denote a typical member of the set; and the vertical line | is read as such that and the comma as and.

EXAMPLE 1.1

(a)   The set A above can also be written as is an odd positive integer, x < 10}.

(b)   We cannot list all the elements of the above set B although frequently we specify the set by

where we assume that everyone knows what we mean. Observe that 8 ∈ B, but 3∉B.

(c)   Let

We emphasize that a set does not depend on the way in which its elements are displayed. A set remains the same if its elements are repeated or rearranged.

Even if we can list the elements of a set, it may not be practical to do so. That is, we describe a set by listing its elements only if the set contains a few elements; otherwise we describe a set by the property which characterizes its elements.

Subsets

Suppose every element in a set A is also an element of a set B, that is, suppose a ∈ A implies a ∈ B. Then A is called a subset of B. We also say that A is contained in B or that B contains A. This relationship is written

Two sets are equal if they both have the same elements or, equivalently, if each is contained in the other. That is:

If A is not a subset of B, that is, if at least one element of A does not belong to B, we write A B.

EXAMPLE 1.2 Consider the sets:

Then C ⊆ A and C ⊆ B since 1 and 3, the elements of C, are also members of A and B. But B A since some of the elements of B, e.g., 2 and 5, do not belong to A. Similarly, A B.

Property 1: It is common practice in mathematics to put a vertical line | or slanted line / through a symbol to indicate the opposite or negative meaning of a symbol.

Property 2: The statement A ⊆ B does not exclude the possibility that A = B. In fact, for every set A we have A ⊆ A since, trivially, every element in A belongs to A. However, if A ⊆ B and A ≠ B, then we say A is a proper subset of B (sometimes written A ⊂ B).

Property 3: Suppose every element of a set A belongs to a set B and every element of B belongs to a set C. Then clearly every element of A also belongs to C. In other words, if A ⊆ B and B ⊆ C, then A ⊆ C.

The above remarks yield the following theorem.

Theorem 1.1: Let A, B, C be any sets. Then:

(i) A ⊆ A

(ii) If A ⊆ B and B ⊆ A, then A = B

(iii) If A ⊆ B and B ⊆ C, then A ⊆ C

Special symbols

Some sets will occur very often in the text, and so we use special symbols for them. Some such symbols are:

N = the set of natural numbers or positive integers: 1, 2, 3,… ⁡

Z = the set of all integers: … ⁡,−2,−1, 0, 1, 2,… ⁡

Q = the set of rational numbers

R = the set of real numbers

C = the set of complex numbers

Observe that N ⊆ Z ⊆ Q ⊆ R ⊆ C.

Universal Set, Empty Set

All sets under investigation in any application of set theory are assumed to belong to some fixed large set called the universal set which we denote by

unless otherwise stated or implied.

Given a universal set U and a property P, there may not be any elements of U which have property P. For example, the following set has no elements:

Such a set with no elements is called the empty set or null set and is denoted by

There is only one empty set. That is, if S and T are both empty, then S = T, since they have exactly the same elements, namely, none.

The empty set ∅ is also regarded as a subset of every other set. Thus we have the following simple result which we state formally.

Theorem 1.2: For any set A, we have ∅⊆ A ⊆U.

Disjoint Sets

Two sets A and B are said to be disjoint if they have no elements in common. For example, suppose

Then A and B are disjoint, and A and C are disjoint. But B and C are not disjoint since B and C have elements in common, e.g., 5 and 6. We note that if A and B are disjoint, then neither is a subset of the other (unless one is the empty set).

1.3 VENN DIAGRAMS

A Venn diagram is a pictorial representation of sets in which sets are represented by enclosed areas in the plane. The universal set U is represented by the interior of a rectangle, and the other sets are represented by disks lying within the rectangle. If A ⊆ B, then the disk representing A will be entirely within the disk representing B as in Fig. 1-1(a). If A and B are disjoint, then the disk representing A will be separated from the disk representing B as in Fig. 1-1(b).

Fig. 1-1

However, if A and B are two arbitrary sets, it is possible that some objects are in A but not in B, some are in B but not in A, some are in both A and B, and some are in neither A nor B; hence in general we represent A and B as in Fig. 1-1(c).

Arguments and Venn Diagrams

Many verbal statements are essentially statements about sets and can therefore be described by Venn diagrams. Hence Venn diagrams can sometimes be used to determine whether or not an argument is valid.

EXAMPLE 1.3 Show that the following argument (adapted from a book on logic by Lewis Carroll, the author of Alice in Wonderland) is valid:

The statements S1, S2, and S3 above the horizontal line denote the assumptions, and the statement S below the line denotes the conclusion. The argument is valid if the conclusion S follows logically from the assumptions S1, S2, and S3.

By S1 the tin objects are contained in the set of saucepans, and by S3 the set of saucepans and the set of useful things are disjoint. Furthermore, by S2 the set of your presents is a subset of the set of useful things. Accordingly, we can draw the Venn diagram in Fig. 1-2.

The conclusion is clearly valid by the Venn diagram because the set of your presents is disjoint from the set of tin objects.

Fig. 1-2

1.4 SET OPERATIONS

This section introduces a number of set operations, including the basic operations of union, intersection, and complement.

Union and Intersection

The union of two sets A and B, denoted by A ∪ B, is the set of all elements which belong to A or to B; that is,

Here or is used in the sense of and/or. Figure 1-3(a) is a Venn diagram in which A ∪ B is shaded.

Fig. 1-3

The intersection of two sets A and B, denoted by A ∩ B, is the set of elements which belong to both A and B; that is,

Figure 1-3(b) is a Venn diagram in which A ∩ B is shaded.

Recall that sets A and B are said to be disjoint or nonintersecting if they have no elements in common or, using the definition of intersection, if A ∩ B = ∅, the empty set. Suppose

Then S is called the disjoint union of A and B.

EXAMPLE 1.4

(a)   Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6, 7}, C = {2, 3, 8, 9}. Then

(b)   Let U be the set of students at a university, and let M denote the set of male students and let F denote the set of female students. The U is the disjoint union of M of F; that is,

This comes from the fact that every student in U is either in M or in F, and clearly no student belongs to both M and F, that is, M and F are disjoint.

The following properties of union and intersection should be noted.

Property 1: Every element x in A ∩ B belongs to both A and B; hence x belongs to A and x belongs to B. Thus A ∩ B is a subset of A and of B; namely

Property 2: An element x belongs to the union A ∪ B if x belongs to A or x belongs to B; hence every element in A belongs to A ∪ B, and every element in B belongs to A ∪ B. That is,

We state the above results formally:

Theorem 1.3: For any sets A and B, we have:

The operation of set inclusion is closely related to the operations of union and intersection, as shown by the following theorem.

Theorem 1.4: The following are equivalent: A ⊆ B,  A ∩ B = A,  A ∪ B = B.

This theorem is proved in Problem 1.8. Other equivalent conditions to A ⊆ B are given in Problem 1.31.

Complements, Differences, Symmetric Differences

Recall that all sets under consideration at a particular time are subsets of a fixed universal set U. The absolute complement or, simply, complement of a set A, denoted by AC, is the set of elements which belong to U but which do not belong to A. That is,

Some texts denote the complement of A by A′ or Ā. Fig. 1-4(a) is a Venn diagram in which AC is shaded.

Fig. 1-4

The relative complement of a set B with respect to a set A or, simply, the difference of A and B, denoted by A\B, is the set of elements which belong to A but which do not belong to B; that is

The set A\B is read "A minus B." Many texts denote A\B by A − B or A ∼ B. Fig. 1-4(b) is a Venn diagram in which A\B is shaded.

The symmetric difference of sets A and B, denoted by A ⊕ B, consists of those elements which belong to A or B but not to both. That is,

Figure 1-4(c) is a Venn diagram in which A ⊕ B is shaded.

EXAMPLE 1.5 Suppose U = N = {1, 2, 3,… ⁡} is the universal set. Let

(Here E is the set of positive even integers.) Then:

That is, EC is the set of odd positive integers. Also:

Furthermore:

Fundamental Products

Consider n distinct sets A1, A2, …, An. A fundamental product of the sets is a set of the form

We note that:

(i) There are m = 2n such fundamental products.

(ii) Any two such fundamental products are disjoint.

(iii) The universal set U is the union of all fundamental products.

Thus U is the disjoint union of the fundamental products (Problem 1.60). There is a geometrical description of these sets which is illustrated below.

EXAMPLE 1.6 Figure 1-5(a) is the Venn diagram of three sets A, B, C. The following lists the m = 2³ = 8 fundamental products of the sets A, B, C:

The eight products correspond precisely to the eight disjoint regions in the Venn diagram of sets A, B, C as indicated by the labeling of the regions in Fig. 1-5(b).

Fig. 1-5

1.5 ALGEBRA OF SETS, DUALITY

Sets under the operations of union, intersection, and complement satisfy various laws (identities) which are listed in Table 1-1. In fact, we formally state this as:

Table 1-1 Laws of the algebra of sets

Theorem 1.5: Sets satisfy the laws in Table 1-1.

Remark: Each law in Table 1-1 follows from an equivalent logical law. Consider, for example, the proof of DeMorgan’s Law 10(a):

Here we use the equivalent (DeMorgan’s) logical law:

where ¬ ⁡ means not, ∨ means or, and ∧ means and. (Sometimes Venn diagrams are used to illustrate the laws in Table 1-1 as in Problem 1.17.)

Duality

The identities in Table 1-1 are arranged in pairs, as, for example, (2a) and (2b). We now consider the principle behind this arrangement. Suppose E is an equation of set algebra. The dual E∗ of E is the equation obtained by replacing each occurrence of ∪, ∩, U and ∅ in E by ∩, ∪, ∅, and U, respectively. For example, the dual of

Observe that the pairs of laws in Table 1-1 are duals of each other. It is a fact of set algebra, called the principle of duality, that if any equation E is an identity then its dual E∗ is also an identity.

1.6 FINITE SETS, COUNTING PRINCIPLE

Sets can be finite or infinite. A set S is said to be finite if S is empty or if S contains exactly m elements where m is a positive integer; otherwise S is infinite.

EXAMPLE 1.7

(a)   The set A of the letters of the English alphabet and the set D of the days of the week are finite sets. Specifically, A has 26 elements and D has 7 elements.

(b)   Let E be the set of even positive integers, and let I be the unit interval, that is,

Then both E and I are infinite.

A set S is countable if S is finite or if the elements of S can be arranged as a sequence, in which case S is said to be countably infinite; otherwise S is said to be uncountable. The above set E of even integers is countably infinite, whereas one can prove that the unit interval I = [0, 1] is uncountable.

Counting Elements in Finite Sets

The notation n(S) or |S| will denote the number of elements in a set S. (Some texts use #(S) or card(S) instead of n(S).) Thus n(A) = 26, where A is the letters in the English alphabet, and n(D) = 7, where D is the days of the week. Also n(∅) = 0 since the empty set has no elements.

The following lemma applies.

Lemma 1.6: Suppose A and B are finite disjoint sets. Then A ∪ B is finite and

This lemma may be restated as follows:

Lemma 1.6: Suppose S is the disjoint union of finite sets A and B. Then S is finite and

Proof. In counting the elements of A ∪ B, first count those that are in A. There are n(A) of these. The only other elements of A ∪ B are those that are in B but not in A. But since A and B are disjoint, no element of B is in A, so there are n(B) elements that are in B but not in A. Therefore, n(A ∪ B) = n(A) + n(B).

For any sets A and B, the set A is the disjoint union of A\B and A ∩ B. Thus Lemma 1.6 gives us the following useful result.

Corollary 1.7: Let A and B be finite sets. Then

For example, suppose an art class A has 25 students and 10 of them are taking a biology class B. Then the number of students in class A which are not in class B is:

Given any set A, recall that the universal set U is the disjoint union of A and AC. Accordingly, Lemma 1.6 also gives the following result.

Corollary 1.8: Let A be a subset of a finite universal set U. Then

For example, suppose a class U with 30 students has 18 full-time students. Then there are 30 − 18 = 12 part-time students in the class U.

Inclusion–Exclusion Principle

There is a formula for n(A ∪ B) even when they are not disjoint, called the Inclusion–Exclusion Principle. Namely:

Theorem (Inclusion–Exclusion Principle) 1.9: Suppose A and B are finite sets. Then A ∪ B and A ∩ B are finite and

That is, we find the number of elements in A or B (or both) by first adding n(A) and n(B) (inclusion) and then subtracting n(A ∩ B) (exclusion) since its elements were counted twice.

We can apply this result to obtain a similar formula for three sets:

Corollary 1.10: Suppose A, B, C are finite sets. Then A ∪ B ∪ C is finite and

Mathematical induction (Section 1.8) may be used to further generalize this result to any number of finite sets.

EXAMPLE 1.8 Suppose a list A contains the 30 students in a mathematics class, and a list B contains the 35 students in an English class, and suppose there are 20 names on both lists. Find the number of students: (a) only on list A, (b) only on list B, (c) on list A or B (or both), (d) on exactly one list.

(a)   List A has 30 names and 20 are on list B; hence 30 − 20 = 10 names are only on list A.

(b)   Similarly, 35 − 20 = 15 are only on list B.

(c)   We seek n(A ∪ B). By inclusion–exclusion,

In other words, we combine the two lists and then cross out the 20 names which appear twice.

(d)   By (a) and (b), 10 + 15 = 25 names are only on one list; that is, n(A ⊕ B) = 25.

1.7 CLASSES OF SETS, POWER SETS, PARTITIONS

Given a set S, we might wish to talk about some of its subsets. Thus we would be considering a set of sets. Whenever such a situation occurs, to avoid confusion, we will speak of a class of sets or collection of sets rather than a set of sets. If we wish to consider some of the sets in a given class of sets, then we speak of subclass or subcollection.

EXAMPLE 1.9 Suppose S = {1, 2, 3, 4}.

(a)   Let A be the class of subsets of S which contain exactly three elements of S. Then

That is, the elements of A are the sets {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}.

(b)   Let B be the class of subsets of S, each which contains 2 and two other elements of S. Then

The elements of B are the sets {1, 2, 3}, {1, 2, 4}, and {2, 3, 4}. Thus B is a subclass of A, since every element of B is also an element of A. (To avoid confusion, we will sometimes enclose the sets of a class in brackets instead of braces.)

Power Sets

For a given set S, we may speak of the class of all subsets of S. This class is called the power set of S, and will be denoted by P(S). If S is finite, then so is P(S). In fact, the number of elements in P(S) is 2 raised to the power n(S). That is,

(For this reason, the power set of S is sometimes denoted by 2S.)

EXAMPLE 1.10 Suppose S = {1, 2, 3}. Then

Note that the empty set ∅ belongs to P(S) since ∅ is a subset of S. Similarly, S belongs to P(S). As expected from the above remark, P(S) has 2³ = 8 elements.

Partitions

Let S be a nonempty set. A partition of S is a subdivision of S into nonoverlapping, nonempty subsets. Precisely, a partition of S is a collection {Ai} of nonempty subsets of S such that:

(i) Each a in S belongs to one of the Ai.

(ii) The sets of {Ai} are mutually disjoint; that is, if

The subsets in a partition are called cells. Figure 1-6 is a Venn diagram of a partition of the rectangular set S of points into five cells, A1, A2, A3, A4, A5.

Fig. 1-6

EXAMPLE 1.11 Consider the following collections of subsets of S = {1, 2,… ⁡,8, 9}:

(i)   [{1, 3, 5}, {2, 6}, {4, 8, 9}]

(ii)   [{1, 3, 5}, {2, 4, 6, 8}, {5, 7, 9}]

(iii)   [{1, 3, 5}, {2, 4, 6, 8}, {7, 9}]

Then (i) is not a partition of S since 7 in S does not belong to any of the subsets. Furthermore, (ii) is not a partition of S since {1, 3, 5} and {5, 7, 9} are not disjoint. On the other hand, (iii) is a partition of S.

Generalized Set Operations

The set operations of union and intersection were defined above for two sets. These operations can be extended to any number of sets, finite or infinite, as follows.

Consider first a finite number of sets, say, A1, A2, … ⁡,Am. The union and intersection of these sets are denoted and defined, respectively, by

That is, the union consists of those elements which belong to at least one of the sets, and the intersection consists of those elements which belong to all the sets.

Now let 𝒜 be any collection of sets. The union and the intersection of the sets in the collection 𝒜 is denoted and defined, respectively, by

That is, the union consists of those elements which belong to at least one of the sets in the collection 𝒜 and the intersection consists of those elements which belong to every set in the collection A.

EXAMPLE 1.12 Consider the sets

Then the union and intersection of the sets are as follows:

DeMorgan’s laws also hold for the above generalized operations. That is:

Theorem 1.11: Let 𝒜 be a collection of sets. Then:

1.8 MATHEMATICAL INDUCTION

An essential property of the set N = {1, 2, 3, …} of positive integers follows:

Principle of Mathematical Induction I: Let P be a proposition defined on the positive integers N; that is, P(n) is either true or false for each n ∈ N. Suppose P has the following two properties:

(i)   P(1) is true.

(ii)   P(k + 1) is true whenever P(k) is true.

Then P is true for every positive integer n ∈ N.

We shall not prove this principle. In fact, this principle is usually given as one of the axioms when N is developed axiomatically.

EXAMPLE 1.13 Let P be the proposition that the sum of the first n odd numbers is n²; that is,

(The kth odd number is 2k − 1, and the next odd number is 2k + 1.) Observe that P(n) is true for n = 1; namely,

Assuming P(k) is true, we add 2k + 1 to both sides of P(k), obtaining

which is P(k + 1). In other words, P(k + 1) is true whenever P(k) is true. By the principle of mathematical induction, P is true for all n.

There is a form of the principle of mathematical induction which is sometimes more convenient to use. Although it appears different, it is really equivalent to the above principle of induction.

Principle of Mathematical Induction II: Let P be a proposition defined on the positive integers N such that:

(i)   P(1) is true.

(ii)   P(k) is true whenever P(j) is true for all 1 ≤ j < k.

Then P is true for every positive integer n ∈ N.

Remark: Sometimes one wants to prove that a proposition P is true for the set of integers

where a is any integer, possibly zero. This can be done by simply replacing 1 by a in either of the above Principles of Mathematical Induction.

Solved Problems

SETS AND SUBSETS

1.1   Which of these sets are equal: {x, y, z}, {z, y, z, x}, {y, x, y, z}, {y, z, x, y}?

They are all equal. Order and repetition do not change a set.

1.2   List the elements of each set where N = {1, 2, 3, …}.

(a)   A consists of the positive integers between 3 and 9; hence A = {4, 5, 6, 7, 8}.

(b)   B consists of the even positive integers less than 11; hence B = {2, 4, 6, 8, 10}.

(c)   No positive integer satisfies 4 + x = 3; hence C = ∅, the empty set.

1.3   Let A = {2, 3, 4, 5}.

(a)   Show that A is not a subset of

(b)   Show that A is a proper subset of C = {1, 2, 3,… ⁡,8, 9}.

(c)   It is necessary to show that at least one element in A does not belong to B. Now 3 ∈ A and, since B consists of even numbers, 3∉B; hence A is not a subset of B.

(d)   Each element of A belongs to C so A ⊆ C. On the other hand, 1 ∈ C but 1∉A. Hence A ≠ C. Therefore A is a proper subset of C.

SET OPERATIONS

1.4   Let U = {1, 2, …, 9} be the universal set, and let

Find: (a) A ∪ B and A ∩ B; (b) A ∪ C and A ∩ C; (c) D ∪ F and D ∩ F.

Recall that the union X ∪ Y consists of those elements in either X or Y (or both), and that the intersection X ∩ Y consists of those elements in both X and Y.

(a)   A ∪ B = {1, 2, 3, 4, 5, 6, 7}and A ∩ B = {4, 5}

(b)   A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9} = U and A ∩ C = {5}

(c)   D ∪ F = {1, 3, 5, 7, 9} = D and D ∩ F = (1, 5, 9) = F

Observe that F ⊆ D, so by Theorem 1.4 we must have D ∪ F = D and D ∩ F = F.

1.5   Consider the sets in the preceding Problem 1.4. Find:

Recall that:

(1)   The complements XC consists of those elements in U which do not belong to X.

(2)   The difference X\Y consists of the elements in X which do not belong to Y.

(3)   The symmetric difference X ⊕ Y consists of the elements in X or in Y but not in both.

Therefore:

1.6   Show that we can have: (a) A ∩ B = A ∩ C without B = C; (b) A ∪ B = A ∪ C without B = C.

1.7   Prove: B\A = B ∩ AC. Thus, the set operation of difference can be written in terms of the operations of intersection and complement.

1.8   Prove Theorem 1.4. The following are equivalent: A ⊆ B, A ∩ B = A, A ∪ B = B.

Suppose A ⊆ B and let x ∈ A. Then x ∈ B, hence x ∈ A ∩ B and A ⊆ A ∩ B. By Theorem 1.3, (A ∩ B) ⊆ A. Therefore A ∩ B = A. On the other hand, suppose A ∩ B = A and let x ∈ A. Then x ∈ (A ∩ B); hence x ∈ A and x ∈ B. Therefore, A ⊆ B. Both results show that A ⊆ B is equivalent to A ∩ B = A.

Suppose again that A ⊆ B. Let x ∈ (A ∪ B). Then x ∈ A or x ∈ B. If x ∈ A, then x ∈ B because A ⊆ B. In either case, x ∈ B. Therefore A ∪ B ⊆ B. By Theorem 1.3, B ⊆ A ∪ B. Therefore A ∪ B = B. Now suppose A ∪ B = B and let x ∈ A. Then x ∈ A ∪ B by definition of the union of sets. Hence x ∈ B = A ∪ B. Therefore A ⊆ B. Both results show that A ⊆ B is equivalent to A ∪ B = B.

Thus A ⊆ B, A ∪ B = A and A ∪ B = B are equivalent.

VENN DIAGRAMS, ALGEBRA OF SETS, DUALITY

1.9   Illustrate DeMorgan’s Law (A ∪ B)C = AC ∩ BC using Venn diagrams.

Shade the area outside A ∪ B in a Venn diagram of sets A and B. This is shown in Fig. 1-7(a); hence the shaded area represents (A ∪ B)C. Now shade the area outside A in a Venn diagram of A and B with strokes in one direction (////), and then shade the area outside B with strokes in another direction (\\\\). This is shown in Fig. 1-7(b); hence the cross-hatched area (area where both lines are present) represents AC ∩ BC. Both (A ∪ B)C and AC ∩ BC are represented by the same area; thus the Venn diagram indicates (A ∪ B)C = AC ∩ BC. (We emphasize that a Venn diagram is not a formal proof, but it can indicate relationships between sets.)

Fig. 1-7

1.10   Prove the Distributive Law: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Here we use the analogous logical law p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) where ∧ denotes and and ∨ denotes or.

1.11   Write the dual of: (a) (U ∩ A) ∪ (B ∩ A) = A; (b) (A ∩U) ∩ (∅ ∪ AC) = ∅.

Interchange ∪ and ∩ and also U and ∅ in each set equation:

(a) (∅ ∪ A) ∩ (B ∪ A) = A; (b) (A ∪∅) ∪ (U ∩ AC) = U.

1.12   Prove: (A ∪ B)\(A ∩ B) = (A\B) ∪ (B\A). (Thus either one may be used to define A ⊕ B.)

Using X\Y = X ∩ Y C and the laws in Table 1.1, including DeMorgan’s Law, we obtain:

1.13   Determine the validity of the following argument:

The premises S1 and S3 lead to the Venn diagram in Fig. 1-8(a). By S2, John belongs to the set of friends which is disjoint from the set of neighbors. Thus S is a valid conclusion and so the argument is valid.

Fig. 1-8

FINITE SETS AND THE COUNTING PRINCIPLE

1.14   Each student in Liberal Arts at some college has a mathematics requirement A and a science requirement B. A poll of 140 sophomore students shows that:

60 completed A,  45 completed B,  20 completed both A and B.

Use a Venn diagram to find the number of students who have completed:

(a)   At least one of A and B; (b) exactly one of A or B; (c) neither A nor B.

Translating the above data into set notation yields:

n(A) = 60, n(B) = 45, n(A ∩ B) = 20, n(U) = 140

Draw a Venn diagram of sets A and B as in Fig. 1-1(c). Then, as in Fig. 1-8(b), assign numbers to the four regions as follows:

20 completed both A and B, so n(A ∩ B) = 20.

60 − 20 = 40 completed A but not B, so n(A\B) = 40.

45 − 20 = 25 completed B but not A, so n(B\A) = 25.

140 − 20 − 40 − 25 = 55 completed neither A nor B.

By the Venn diagram:

1.15   In a survey of 120 people, it was found that:

(a)   Find the number of people who read at least one of the three magazines.

(b)   Fill in the correct number of people in each of the eight regions of the Venn diagram in Fig. 1-9(a) where N, T, and F denote the set of people who read Newsweek, Time, and Fortune, respectively.

(c)   Find the number of people who read exactly one magazine.

Fig. 1-9

(a)   We want to find n(N ∪ T ∪ F). By Corollary 1.10 (Inclusion–Exclusion Principle),

(b)   The required Venn diagram in Fig. 1-9(b) is obtained as follows:

8 read all three magazines,

20 − 8 = 12 read Newsweek and Time but not all three magazines,

25 − 8 = 17 read Newsweek and Fortune but not all three magazines,

15 − 8 = 7 read Time and Fortune but not all three magazines,

65 − 12 − 8 − 17 = 28 read only Newsweek,

45 − 12 − 8 − 7 = 18 read only Time,

42 − 17 − 8 − 7 = 10 read only Fortune,

120 − 100 = 20 read no magazine at all.

(b)   28 + 18 + 10 = 56 read exactly one of the magazines.

1.16   Prove Theorem 1.9. Suppose A and B are finite sets. Then A ∪ B and A ∩ B are finite and

If A and B are finite then, clearly, A ∪ B and A ∩ B are finite.

Suppose we count the elements in A and then count the elements in B.

Then every element in A ∩ B would be counted twice, once in A and once in B. Thus

CLASSES OF SETS

1.17   Let A = [{1, 2, 3}, {4, 5}, {6, 7, 8}]. (a) List the elements of A; (b) Find n(A).

(a)   A has three elements, the sets {1, 2, 3}, {4, 5}, and {6, 7, 8}.

(b)   n(A) = 3.

1.18   Determine the power set P(A) of A = {a, b, c, d}.

The elements of P(A) are the subsets of A. Hence

As expected, P(A) has 2⁴ = 16 elements.

1.19   Let S = {a, b, c, d, e, f, g}. Determine which of the following are partitions of S:

(a)   P1 is not a partition of S since f ∈ S does not belong to any of the cells.

(b)   P2 is not a partition of S since e ∈ S belongs to two of the cells.

(c)   P3 is a partition of S since each element in S belongs to exactly one cell.

(d)   P4 is a partition of S into one cell, S itself.

1.20   Find all partitions of S = {a, b, c, d}.

Note first that each partition of S contains either 1, 2, 3, or 4 distinct cells. The partitions are as follows:

There are 15 different partitions of S.

1.21   Let N = {1, 2, 3,…} and, for each n ∈ N, Let An = {n, 2n, 3n,…}. Find:

(a)   Those numbers which are multiples of both 3 and 5 are the multiples of 15; hence A3 ∩ A5 = A15.

(b)   The multiples of 12 and no other numbers belong to both A4 and A6, hence A4 ∩ A6 = A12.

(c)   Every positive integer except 1 is a multiple of at least one prime number; hence

1.22   Let be an indexed class of sets and let i0 ∈ I. Prove

1.23   Prove (De Morgan’s law): For any indexed class

.

Using the definitions of union and intersection of indexed classes of sets:

MATHEMATICAL INDUCTION

1.24   Prove the proposition P(n) that the sum of the first n positive integers is that is,

The proposition holds for n = 1 since:

Assuming P(k) is true, we add k + 1 to both sides of P(k), obtaining

which is P(k + 1). That is, P(k + 1) is true whenever P(k) is true. By the Principle of Induction, P is true for all n.

1.25   Prove the following proposition (for n  ≥  0):

P(0) is true since 1 = 2¹ − 1. Assuming P(k) is true, we add 2k+1 to both sides of P(k), obtaining

which is P(k + 1). That is, P(k + 1) is true whenever P(k) is true. By the principle of induction, P(n) is true for all n.

Supplementary Problems

SETS AND SUBSETS

1.26   Which of the following sets are equal?

1.27   List the elements of the following sets if the universal set is U = {a, b, c, …, y, z}.

Furthermore, identify which of the sets, if any, are equal.

1.28   Let A = {1, 2, …, 8, 9}, B = {2, 4, 6, 8}, C = {1, 3, 5, 7, 9}, D = {3, 4, 5}, E = {3, 5}.

Which of the these sets can equal a set X under each of the following conditions?

SET OPERATIONS

1.29   Consider the universal set U = {1, 2, 3, …, 8, 9} and sets A = {1, 2, 5, 6}, B = {2, 5, 7}, C = {1, 3, 5, 7, 9}. Find:

1.30   Let A and B be any sets. Prove:

1.31   Prove the following:

(Compare the results with Theorem 1.4.)

1.32   Prove the Absorption Laws: (a) A ∪ (A ∩ B) = A; (b) A ∩ (A ∪ B) = A.

1.33   The formula A\B = A ∩ BC defines the difference operation in terms of the operations of intersection and complement. Find a formula that defines the union A ∪ B in terms of the operations of intersection and complement.

VENN DIAGRAMS

1.34   The Venn diagram in Fig. 1-5(a) shows sets A, B, C. Shade the following sets:

1.35   Use the Venn diagram in Fig. 1-5(b) to write each set as the (disjoint) union of fundamental products:

1.36   Consider the following assumptions:

S1: All dictionaries are useful.

S2: Mary owns only romance novels.

S3: No romance novel is useful.

Use a Venn diagram to determine the validity of each of the following conclusions:

(a)   Romance novels are not dictionaries.

(b)   Mary does not own a dictionary.

(c)   All useful books are dictionaries.

ALGEBRA OF SETS AND DUALITY

1.37   Write the dual of each equation:

1.38   Use the laws in Table 1-1 to prove each set identity:

FINITE SETS AND THE COUNTING PRINCIPLE

1.39   Determine which of the following sets are finite:

1.40   Use Theorem 1.9 to prove Corollary 1.10: Suppose A, B, C are finite sets. Then A ∪ B ∪ C is finite and

1.41   A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of three popular options, air-conditioning (A), radio (R), and power windows (W), were already installed. The survey found:

Find the number of cars that had: (a) only W; (b) only A; (c) only R; (d) R and W but not A; (e) A and R but not W; (f) only one of the options; (g) at least one option; (h) none of the options.

CLASSES OF SETS

1.42   Find the power set P(A) of A = {1, 2, 3, 4, 5}.

1.43   Given A = [{a, b}, {c}, {d, e, f}].

1.44   Suppose A is finite and n(A) = m. Prove the power set P(A) has 2m elements.

PARTITIONS

1.45   Let S = {1, 2, …, 8, 9}. Determine whether or not each of the following is a partition of S :

1.46   Let S = {1, 2, 3, 4, 5, 6}. Determine whether or not each of the following is a partition of S :

1.47   Determine whether or not each of the following is a partition of the set N of positive integers:

1.48   Let [A1, A2, …, Am] and [B1, B2, …, Bn] be partitions of a set S.

Show that the following collection of sets is also a partition (called the cross partition) of S :

Observe that we deleted the empty set ∅.

1.49   Let S = {1, 2, 3, …, 8, 9}. Find the cross partition P of the following partitions of S :

INDUCTION

1.50   

1.51   

1.52   

1.53   

1.54   

1.55   Prove 7n − 2n is divisible by 5 for all n ∈ N

1.56   Prove n³ − 4n + 6 is divisible by 3 for all n ∈ N

1.57   Use the identity 1 + 2 + 3 + ⋯ + n = n(n + 1)∕2 to prove that

MISCELLANEOUS PROBLEMS

1.58   Suppose N = {1, 2, 3, …} is the universal set, and

1.59   Prove the following properties of the symmetric difference:

1.60   Consider m nonempty distinct sets A1, A2, …, Am in a universal set U. Prove:

(a)   There are 2m fundamental products of the m sets.

(b)   Any two fundamental products are disjoint.

(c)   U is the union of all the fundamental products.

Answers to Supplementary Problems

1.26    B = C = E = F, A = D = G = H.

1.27    A = {a, e, i, o, u}, B = D = {l, i, t, e}, C = {a, b, c, d,  e}.

1.28   

1.29   

1.33   

1.34   See Fig.1.10

1.35   

1.36   The three premises yield the Venn diagram in Fig. 1-1 1(a). (a) and (b) are valid, but (c) is not valid.

1.37   

1.39   (a) Infinite; (b) finite; (c) infinite; (d) finite.

Fig. 1-10

Fig. 1-11

1.41   Use the data to fill in the Venn diagram in Fig. 1-11(b). Then:

(a)   5; (b) 4; (c) 2; (d) 1; (e) 6; (f) 11; (g) 23; (h) 2.

1.42   

1.43   

1.44   

1.45   (a) No, (b) no, (c) yes, (d) yes.

1.46   (a) No, (b) no, (c) yes, (d) no.

1.47   (a) No, (b) no, (c) yes.

1.49   [{1, 31, {2, 41, {5, 71, {91, {6, 81]

1.55   Hint: 7k+1 - 2k+1 = 7k+1 = 7(2k) C7(2k) ^- 2k+1 = 7(7k - 2k) C (7 - 2)2k

1.58   (a) {1, 2, 3, 7, 8, 91; (b) {1, 3, 4, 6, 81; (c) and (d) {2, 3, 4, 61.

CHAPTER 2

Relations

2.1 INTRODUCTION

The reader is familiar with many relations such as less than, is parallel to, is a subset of, and so on. In a certain sense, these relations consider the existence or nonexistence of a certain connection between pairs of objects taken in a definite order. Formally, we define a relation in terms of these ordered pairs. Functions are a subset of the (more general) relations we will discuss in Chapter 2.

An ordered pair of elements a and b, where a is designated as the first element and b as the second element, is denoted by (a, b). In particular,

if and only if a = c and b = d. Thus (a, b) ≠ (b, a) unless a = b. This contrasts with sets where the order of elements is irrelevant; for example, {3, 5} = {5, 3}.

2.2 PRODUCT SETS

Consider two arbitrary sets A and B. The set of all ordered pairs (a, b) where a ∈ A and b ∈ B is called the product, or Cartesian product, of A and B. A short designation of this product is A × B, which is read "A cross B." By definition,

One frequently writes A² instead of A × A.

EXAMPLE 2.1 R denotes the set of real numbers and so R² = R×R is the set of ordered pairs of real numbers. The reader is familiar with the geometrical representation of R² as points in the plane as in Fig. 2-1. Here each point P represents an ordered pair (a, b) of real numbers and vice versa; the vertical line through P meets the x-axis at a, and the horizontal line through P meets the y-axis at b. R² is frequently called the Cartesian plane.

EXAMPLE 2.2 Let A = {1, 2} and B = {a, b, c}. Then

Also, A × A = {(1, 1),(1, 2),(2, 1),(2, 2)}

There are two things worth noting in the above examples. First of all A × B ≠ B × A. The Cartesian product deals with ordered pairs, so naturally the order in which the sets are considered is important. Secondly, using n(S) for the number of elements in a set S, we have:

Fig. 2-1

In fact, n(A × B) = n(A)n(B) for any finite sets A and B. This follows from the observation that, for an ordered pair (a, b) in A × B, there are n(A) possibilities for a, and for each of these there are n(B) possibilities for b.

The idea of a product of sets can be extended to any finite number of sets. For any sets A1, A2, … ⁡,An, the set of all ordered n-tuples (a1, a2, … ⁡,an) where a1 ∈ A1, a2 ∈ A2, … ⁡,an ∈ An is called the product of the sets A1, … ⁡,An and is denoted by

Just as we write A² instead of A × A, so we write An instead of A × A ×⋯ × A, where there are n factors all equal to A. For example, R³ = R × R × R denotes the usual three-dimensional space.

2.3 RELATIONS

We begin with a definition.

Definition 2.1: Let A and B be sets. A binary relation or, simply, relation from A to B is a subset of A × B.

Suppose R is a relation from A to B. Then R is a set of ordered pairs where each first element comes from A and each second element comes from B. That is, for each pair a ∈ A and b ∈ B, exactly one of the following is true:

(i)   (a, b) ∈ R; we then say "a is R-related to b", written aRb.

(ii)   (a, b)∉R; we then say "a is not R-related to b", written a b.

If R is a relation from a set A to itself, that is, if R is a subset of A² = A × A, then we say that R is a relation on A.

The domain of a relation R is the set of all first elements of the ordered pairs which belong to R, and the range is the set of second elements.

Although n-ary relations, which involve ordered n-tuples, are introduced in Section 2.10, the term relation shall then mean binary relation unless otherwise stated or implied.

EXAMPLE 2.3

(a)   

The domain of R is {1, 3} and the range is {y, z}.

(b)   Set inclusion ⊆ is a relation on any collection of sets. For, given any pair of set A and B, either

(c)   A familiar relation on the set Z of integers is "m divides n." A common notation for this relation is to write m|n when m divides n. Thus 6 | 30 but 7 | 25.

(d)   Consider the set L of lines in the plane. Perpendicularity, written ⊥, is a relation on L. That is, given any pair of lines a and b, either

(e)   Let A be any set. An important relation on A is that of equality,

which is usually denoted by =. This relation is also called the identity or diagonal relation on A and it will also be denoted by ΔA or simply Δ.

(f)   Let A be any set. Then A × A and ∅ are subsets of A × A and hence are relations on A called the universal relation and empty relation, respectively.

Inverse Relation

Let R be any relation from a set A to a set B. The inverse of R, denoted by R−1, is the relation from B to A which consists of those ordered pairs which, when reversed, belong to R; that is,

For example, let A = {1, 2, 3} and B = {x, y, z}. Then the inverse of

Clearly, if R is any relation, then (R−1)−1 = R. Also, the domain and range of R−1 are equal, respectively, to the range and domain of R. Moreover, if R is a relation on A, then R−1 is also a relation on A.

2.4 PICTORIAL REPRESENTATIVES OF RELATIONS

There are various ways of picturing relations.

Relations on R

Let S be a relation on the set R of real numbers; that is, S is a subset of R² = R × R. Frequently, S consists of all ordered pairs of real numbers which satisfy some given equation E(x, y) = 0 (such as x² + y² = 25).

Since R² can be represented by the set of points in the plane, we can picture S by emphasizing those points in the plane which belong to S. The pictorial representation of the relation is sometimes called the graph of the relation. For example, the graph of the relation x² + y² = 25 is a circle having its center at the origin and radius 5. See Fig. 2-2(a).

Fig. 2-2

Directed Graphs of Relations on Sets

There is an important way of picturing a relation R