Set Operations

Set Operations

We have defined replacement set and solution set and have even represented sets symbolically, but we have not discussed the concept of “set” formally. Let us do so now.

The idea of set is too fundamental to attempt to define. Let us merely say that any collection of objects shall be called a set. We have been calling those things which belong to a set the members of the set. We may also use the word element as a synonym for member. We shall continue to use braces, { }, to set off the members of a set and commas to separate one member from another. We shall use the symbol  to stand for the phrase “is a member of.” For example, if we consider the set {1, 3, 5, 7, 9}, we observe that 1 is a member of this set, 57 is not one of its members, the set contains five elements, and 7 {is-in} {1,3,5,7,9}.

Often we shall use a single (upper-case) letter to represent a set. If we write A={2, 4, 6, 8}, then we mean that A is the set whose members are 2, 4, 6 and 8. When two sets have exactly the same members, we shall call them equal and represent this by means of the symbol =. Then we write {1, 2, 3, 4, 5} = {1, 5, 3, 2, 4} even though the members are not listed in the same order. The sets are equal because they have exactly the same members. Also {1, 2, 3, 4, 5} = {1, 1, 2, 3, 3, 3, 4, 5, 5, 5} even though the set on the right contains several (unnecessary) repetitions of some of its members. The sets are equal because each contains the numbers 1, 2, 3, 4 and 5 and nothing else.

We have already encountered a set with no members, which we represented by the symbol { }. We shall call this the empty set and shall often represent it by the symbol {empty}. Note that {0}!={empty}, for the sets do not have the same members. The set on the left has a single member, while the set on the right has no members. Sometimes we shall encounter sets which have so many members that it would be quite frustrating to have to list all of them. Consider the following set.
        {1, 3, 5, 7, 9, 11, 18, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39}

Fortunately there are several Ways to avoid having to write all the members of this set. We might use a verbal description of the set members: “The set of all odd natural numbers less than 40.” Another way to describe this set is to use a partial listing of its members: {1, 3, 5, 7, ..., 39}. In using this incomplete listing method we merely omit from our list of members those whose identity can clearly be inferred and replace them with three dots. In fact the only time we can use a partial listing is when there is no doubt of the set’s membership, and this can occur only when the pattern of membership is clear from a few example members. Thus {10, 20, 30, ..., 1000} is obviously the set of the first one hundred multiples of 10, but it is not at all clear what set is meant by {1, 50, 23, ..., 57}. Then {1, 2, 3, ..., 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, but {1, 2, 3, ..., 10} !={1, 2, 3, 10}.

Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5}. Note that every member of set A is also a member of set B. We shall describe this relationship between sets A and B by saying that A is a subset of B. To illustrate further, if C = {4, 5, 6} and D = {1, 2, 3, ..., 20}, we see that every member of C is also a member of D, and we say that C is a subset of D. On the other hand, since E = {1, 2, 3, 4} contains one member, 4, which is not a member of A = {1, 2, 3}, we cannot say that E is a subset of A. For P  to be a subset of Q  two conditions must be met: P must be, first of all, a set, and, at the same time, each member of P  must also be a member of Q.

Example 1

Let A = {1, 2, 3, 4, 5}. Decide whether each of the following is or is not a subset of A.
(a) {1, 2, 3}  (b) {4, 5, 6}  (c) {1, 2, 3, 4, 5} (d) 1  (e) {1, 2,3, ..., 20}


(a) Is (b) Is not (6 is not a member of A) (c) Is (d) Is not (1 is not a set) (e) Is not

Let R = {1, 2, 3} and S = {1, 3, 5, 7}. We can form a new set, {1, 2, 3, 5, 7}, which contains each member of R as well as each member of S, but no other members. We shall call this set the union of sets R and S. Similarly, if P = {2, 4, 6} and Q = {4, 5, 6, 7}, then the union of P and Q is the set {2, 4, 5, 6, 7}. It will be helpful to introduce a symbol to represent the union of two sets. The symbol we shall use is . Then, if M = {1, 5, 10} and N = {5, 10, 15}, we shall represent the union of M and N  by M {union} N={1,5,10,15}.

Example 2  Find:
                 (a) {1,2,3} {union} {4,5,6}    (b) {4,5,10} {union} {2,4,6}

                 (c) {5,7,9,11} {union} {7,9}


                 (a) {1,2,3,4,5,6}   (b) {2,4,5,6,10}   (c) {5,7,9,11}

Let R = {1, 2, 3} and T = {1, 3, 5, 7}. This time let us form a new set, {1, 3}, which contains all those members found in set R and at the same time, in set T, but no other members. We shall call this set the intersection of sets R and T. If P = {2, 4, 6} and Q = {3, 4, 5}, then the intersection of P and Q is the set {4}. Again, a symbol, , will be used to represent the intersection of two sets. Then if M = {1, 2, 3, 4} and N = {2, 4, 6, 8}, we shall represent the intersection of M and N by M {intersect} N={2,4}.

Example 3  Find:
                 (a) {10,20,30,40} {intersect} {5,10,15}    (b) {1,2,3,4,5} {intersect} {1,5}

                 (c) {1,2,3,...,20} {intersect} {10,20,30}    (d) {2,4,6,8} {intersect} {1,3,5,7}  


                 (a) {10}    (b) {1,5}    (c) {10, 20}    (d) {empty}

In some discussions we may wish to designate some set as a universe and consider subsets of this universe. For example, we might call set U = {1, 2, 3, 4, 5} our universe and consider one of its subsets A = {2, 3}.

We could then define the complement of set A to be the set {1, 4, 5}. Similarly, if B = {1, 3, 5}, the complement of B is {2, 4}. Note that the complement of A contains those members found in U which are not found in A, and the complement of B contains members of U which are not found in B. We shall use the symbol C  to represent the complement of C. Hence, if C = {1, 3}, then C = {2, 4, 5}. One word of caution. Unless we know what universe a given set is a subset of, we can not determine its complement. That is, the question “What is the complement of D={10, 20, 30}?” cannot be answered unless we know what universe D is a subset of.

Example 4

Let U = {0, 5, 10, 15, 20} be the universe.
(a) If A = {8, 10}, find A     (b) If B = {0, 10, 15, 20}, find B    (c) What set is U?

Solution (a) {0, 15, 20}     (b) {5}     (c) {empty}

In Example 4(c) we observe that U{empty} since the set containing all members of {0, 5, 10, 15, 20} not in U must be empty. In Example 3(d) the answer was also the empty set. These examples illustrate one reason for even considering a set with no members—it allows us to answer some questions (with a set) where we might otherwise have no available answer. Thus, any two sets have some set as their intersection and any subset of a universe has a complement. Of course, the empty set also creates some interesting results. If A = {0, 1, 2, 3}, we must conclude that {empty} is a subset of A, for it is true that {empty} is a set, and it is true that every member of {empty} is also a member of A. (For which member of {empty} is not a member of A?) Indeed, by this reasoning we are forced to conclude that the empty set is a subset of every set!

Example 5

Let V = {3, 6, 9, 12, 15} be the universe and let A = {3, 6} and B = {3, 12, 15}. Find each of the following:

(a) A ∪ B    (b) A ∩ B    (c) A ∪ B     (d) ∩ B     (e)  


(a) {3,6,12,15}    (b) {3}

(c) A ∪ B  = {9,12,15} {6,9} = {6,9,12,15}

(d) ∩ B = {3} = {6,9,12,15}

(e)   = {  } = {3,6,9,12,15} = V

Infinite Sets

Each of the sets we have considered until now has had an obvious, unstated property. It has always been possible to list every one of its members. True, when there were many members we sometimes elected to make only a partial list or to use a verbal description of the members. But it has always been possible to list them all even when we chose not to do so. Sets which have this property we shall call finite sets. Thus, each of the following is a finite set: {1, 2, 3}, {4, 8, 12,...,40}, {empty}, “the set of natural numbers less than 1000.”

There are sets which are not finite. Such sets have so many members that no matter how long we might try, we could not list them all. We shall call such nonfinite sets infinite sets. It is important to realize that the reason a set will be called infinite is not our lack of time to list its members, or our lack of space to display the list, or our unwillingness to do so. Rather, it is because no listing can possibly be complete. Let us consider some examples of infinite sets we shall need as we continue our study of algebra.

We have referred to natural numbers several times. Without trying to define carefully what a natural number is, we shall simply say that the natural numbers are the numbers used to count. Indeed, “counting number” is a synonym for “natural number.” We have agreed to use the symbols 1, 2, 3, etc. as the simplest forms of the natural numbers. When we consider the set of all natural numbers, we encounter an infinite set, which we represent by {1, 2, 3, ...}. Note that we do not end the list with some last natural number for there is no last natural number. Other examples of infinite sets and their representations are the set of odd natural numbers {1, 3, 5,7, ...}, and the set of all multiples of 3 {3, 6, 9, 12, ...}. Frequently we shall use letters to name certain infinite sets. Thus, hereafter we shall always use the letter N as a symbol for the set of all natural numbers: N = {1, 2, 3, 4, ...}.

We shall now introduce some numbers not previously considered. Let us define a number called the additive inverse of 1, represent that number by the symbol -1, and define its sum with 1 to be 0. Then 1 + -1 = 0. Similarly, we define the additive inverse of 2 to be -2 with 2 +-2 = 0. In fact we can continue in this way, defining an additive inverse for each of the natural numbers. The additive inverse of 3 will be represented by -3, and 3 + -3 = 0, and so on. Since the set of natural numbers is infinite, there will be infinitely many additive inverses of the sort we have been defining. We could express the set of all such additive inverses by writing {-1, -2, -3,-4, ...} or {...,-4, -8, -2,-1}, since these are equal sets.

Let us form a new set whose members are precisely the natural numbers, their additive inverses, and zero. One way to represent this set is to form the union of two sets, {0,1,2,3,4, ...} {union} {-1,-2,-3, ...}, but we shall often use the representation {0, 1, -1, 2,-2, 3, -3, ...} or better still {..., -3,-2,-1, 0, 1, 2, 3, ...}. We shall call this set the integers and represent it always with the letter I. Note that since every member of N = {1, 2,3,4,...} is also a member of I = {...,-3, -2,-1, 0, 1, 2, 3, ...}, then N is a subset of I. Another subset of I is the set of the additive inverses of the natural numbers; that is, {-1, -2, -3,-4, ...}. These infinite subsets of I will sometimes be referred to by other names. We will call N = {1, 2,3, ...} the set of positive integers; the set {-1, -2,-3, ...} we will call the set of negative integers. We observe that 0 is an integer, but it is not a member of either of these subsets. Hence 0 is neither a positive nor a negative integer.

Let us consider fractions such as 3/4, -5/16, 12/-7 and -5/-10, whose numerators and denominators are integer symbols. Any number which can be represented by such a fraction will be called a rational number, since it is the quotient or ratio of two integers. The letter F (for fraction) will be used to represent the set of all rational numbers. We shall see in other tutorial that a fraction whose denominator represents zero is not a number at all. But any fraction whose numerator and denominator are integer symbols, with the denominator not zero, represents a rational number. The numbers represented by the decimal expressions 0.75 and 2.5 can also be represented by the fractions 3/4 and 5/2, and hence they are also rational numbers. Decimal expressions with only a finite number of digits such as 0.75 and 2.5 shall be called terminating decimals or, more simply, decimals. It can be shown that every (terminating) decimal, positive or negative, can be represented by a fraction whose numerator and denominator are integer symbols and hence represents a rational number. We shall not at this time consider decimals with infinitely many digits. In another tutorial we shall see that such (nonterminating) decimals sometimes do represent rational numbers, but sometimes do not.

Example 1  Show that each of the following is a rational number by expressing each as a fraction whose numerator and denominator are (positive) integers.

(a) 0.125     (b) 32.2     (c) 3

(a) 0.125 = 125/1000= 1/8     (b) 32.2 = 322/10 = 161/5    (c) 3=3/1

  The set F has some interesting subsets. Since 1/1=1, 2/1=2, 3/1 =3 etc., we see that every natural number can be represented as a ratio of integers, and so every natural number is also a rational number. Then N is a subset of F. Similarly, we will show in another tutorial  that 0/1 = 0, -1/1 = -1, -2/1 = -2, -3/1 =-3 etc., and so every integer can be represented as a ratio of integers. Then every integer is also a rational number and I is a subset of F. To summarize, we can say that N is a subset of I and both N and I are subsets of F.

Example 2  Each of these is a member of one or more of the special sets, N, I, and F. Indicate which.
(a) -2     (b) -3/5     (c) 1/2+1/2

Solution  (a) I,F     (b) F     (c) N, I, F (since 1/2+1/2 = 1)

As we see, the set of rational numbers contains both positive and negative numbers. We shall call the set of all positive rational numbers the set of quotient numbers and represent it by the letter Q. Included in this set are all positive integers or natural numbers, and all numbers represented by fractions such as 2/3 and 7/4, whose numerators and denominators are natural number symbols, or by positive decimals. Since every natural number is also a quotient number, we see that N is a subset of Q. But since the negative integers are not quotient numbers, we see that I is not a subset of Q. Of course Q, the set of positive rational numbers, is a subset of F, the set of all rational numbers.

We may now summarize. There are four rather special infinite sets of numbers.

1. N = {1, 2, 3, . . .}, the set of natural numbers.

2. I={..., -2, -1, 0, 1, 2, ...}, the set of integers.

3. F = the set of rational numbers, any number which is the ratio of two integers. All fractions with numerator and denominator integer symbols (and denominator not zero), all integers, and all (terminating) decimals represent rational numbers.

4. Q = the set of all positive rational numbers. All fractions with numerator and denominator natural number symbols, all natural numbers, and all (terminating) positive decimals represent quotient numbers.

Example 3  Each of these is a member of one or more of the special sets N, I, Q, and F. Indicate which.
(a) 0     (b) 3/2     (c) -2/5    (d) 4.3     (e) -0.3     (f) 14/7

(a) I,F     (b) Q, F     (c) F     (d) Q,F     (e) F     (f) N, I, Q, F (since 14/7 = 2)

Example 4 Which of the special sets have the given set as a subset?
(a) A = {-3, 1/2}     (b) B = {1, 2}

(a) Both -3 and 1/2 are rational numbers, so A is a subset of F. Since -3 is not a quotient number A is not a subset of Q. Since 1/2 is not an integer, A is not a subset of I. Since -3 is not a natural number, A is not a subset of N.
(b) Both 1 and 2 are natural numbers, so B is a subset of N. Since 1 and 2 are also integers, also quotient numbers, and also rational numbers, B  is also a subset of I, Q, and F.

Sets of Points

All of the sets we have so far considered have been sets of numbers. Sometimes these sets were finite, sometimes infinite; one set was empty. We may also consider sets whose members are not numbers. For example, we could form the set whose members are people (such as the set of all current members of the United States Senate), nations (the set of members of the United Nations), or automobiles (the set of all automobiles licensed by the state of New York this year). We shall not want to investigate these sets in our study of algebra, but we will want to consider certain sets of points.

  The concept of a plane is difficult to define. Let us agree that the flat page of a book, the top surface of a table, and the writing surface of a classroom blackboard are all examples of a plane surface. By a point of a plane we shall mean a particular position or location in the plane. A set of points in a plane is any collection of some points of the plane. We shall often describe a set of points by means of a picture.

Set of points on a plane.

Thus in Figure 1.1, A is the set of all points on and within the drawn circle and B includes all points within, but not on, the drawn rectangle.

  Sometimes the context of a problem will make it clear what the set of points is without any shading. In Figure 1.2, A is the set of points inside and on the circle, while B is the set of points inside and on the rectangle. Since every point of A is also a point of B, we see that A is a subset of B.

Example 1  Decide whether A is or is not a subset of B.

Determining whether A is a subset of B or not.

(a) A is a subset of B     (b) A is not a subset of B    (c) A is not a subset of B (but B is a subset of A)

  Figure 1.3 shows how to represent the union and intersection of sets of points. A is the set of points on and inside the left-hand circle, and B is the set of points on and inside the right-hand circle. Notice that A {union} B, represented by shading, contains all the points which are either in A or in B, while A {intersect} B contains those points which are common to both A and B.

Union and Intersection of sets of points.

Example 2  Shade sets A {union} B and A {intersect} B.

Finding union and intersection of set A and set B in two different cases.


Shading the union and intersection of set A and set B.

We may also represent the complement of a set of points with a picture as shown in Figure 1.4.

 Remember that it is necessary to have some universe whenever the complement of a set is discussed. The universe here is the set of points inside and on the rectangle U. Note that A consists of all points which are outside of the circle but in the universe. Since the circle itself is a part of set A, then it is not a part of set A, and this is indicated by making a boundary of A a dotted circle.

Complement of a set of points.

Example 3  In each case the universe is the rectangular set, and A is the set indicated by shading. Represent A by shading.

Indicating the sets by shading.


Indicating the complementary sets by shading.

At times we shall want to consider sets of points each of which lies on some straight line. Figure 1.5 illustrates some sets of this sort. Set A represents all the points in a certain straight line. The arrowheads show that set A continues forever in both directions. We shall call set B a ray, a half-line which includes one end point and all points on a straight line which lie on only one side of that point. Set C will be called a line segment; note that it has two end points and does not continue forever in either direction. Set D is that subset of a line segment whose end points are not members of the set. We shall indicate this by using small empty circles at the ends.

Sets of points which lies on some straight line.

Example 4  Make Venn diagrams to show the subset relation of these sets of numbers.

(a)  N, Q and F    (b)  I, Q and F


Venn diagrams to show the subset relation of the sets.

Example 5  There are 100 boys in Delta Tau Delta fraternity. Of these, 75 take English, 60 take math, and 10 take neither English nor math. How many take:

(a) English, but not Math?     (b) Math, but not English?    (c) Both Math and English?

Solution We make a Venn diagram showing the sets E (those taking English) and M (those taking Math) in the universe (the set of all fraternity men). Since 10 take neither English nor Math, the size of E∪M is 10 and this we show by writing 10 outside of both E and M. But then we must account for 90 students in the set E {union} M and do so in such a way that 60 students are in M and 75 are in E. Since 90-60 = 30 we must place 30 students in E {union} M but not in M. And since 90-75 = 15 we must place 15 students in E {union} M but not in E. As shown in Figure 1.6 this requires that we place 45 students in E {intersect} M.

(a) 30     (b) 15     (c) 45

Solving word problem by Venn diagram.

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