CHAPTER 6
SET THEORY
Copyright © Cengage Learning. All rights reserved.
SECTION 6.1
Set Theory: Definitions and the
Element Method of Proof
Copyright © Cengage Learning. All rights reserved.
Set Theory: Definitions and the Element Method of Proof
The words set and element are undefined terms of set
theory just as sentence, true, and false are undefined terms
of logic.
The founder of set theory, Georg Cantor, suggested
imagining a set as a “collection into a whole M of definite
and separate objects of our intuition or our thought. These
objects are called the elements of M.”
Cantor used the letter M because it is the first letter of the
German word for set: Menge.
3
Subsets: Proof and Disproof
4
Subsets: Proof and Disproof
We begin by rewriting what it means for a set A to be a
subset of a set B as a formal universal conditional
statement:
The negation is, therefore, existential:
5
Subsets: Proof and Disproof
A proper subset of a set is a subset that is not equal to its
containing set. Thus
6
Example 1 – Testing Whether One Set Is a Subset of Another
Let A = {1} and B = {1, {1}}.
a. Is A  B?
b. If so, is A a proper subset of B?
Solution:
a. Because A = {1}, A has only one element, namely the
symbol 1.
This element is also one of the elements in set B. Hence
every element in A is in B, and so A  B.
7
Example 1 – Solution
cont’d
b. B has two distinct elements, the symbol 1 and the set {1}
whose only element is 1.
Since 1  {1}, the set {1} is not an element of A, and so
there is an element of B that is not an element of A.
Hence A is a proper subset of B.
8
Determining Subsets
Example:
Determine whether set A is a subset of set B.
A = { 3, 5, 6, 8 }
B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
All of the elements of set A are contained in set B,
so A⊆ B.
2.2-9
9
Proper Subset
Set A is a proper subset of set B, symbolized A
⊂ B, if and only if all of the elements of set A
are elements of set B and set A ≠ B (that is, set
B must contain at least one element not in set
A).
10
Determining Proper Subsets
Example:
Determine whether set A is a proper subset of set
B.
A = { dog, cat }
B = { dog, cat, bird, fish }
Solution:
All the elements of set A are contained in set B,
and sets A and B are not equal, therefore A ⊂ B.
2.2-11
11
Determining Proper Subsets
Example:
Determine whether set A is a proper subset of set
B.
A = { dog, bird, fish, cat }
B = { dog, cat, bird, fish }
Solution:
All the elements of set A are contained in set B,
but sets A and B are equal, therefore A ⊄ B.
2.2-12
12
Number of Distinct Subsets
The number of distinct subsets of a finite set
A is 2n, where n is the number of elements in
set A.
2.2-13
13
Number of Distinct Subsets
Example:
Determine the number of distinct subsets for the
given set {t, a, p, e}.
List all the distinct subsets for the given set {t, a,
p, e}.
2.2-14
14
Number of Distinct Subsets
Solution:
Since there are 4 elements in the
given set, the number of distinct
subsets is
24 = 2 • 2 • 2 • 2 = 16.
{t,a,p,e}, {t,a,p}, {t,a,e}, {t,p,e},
{a,p,e},{t,a}, {t,p}, {t,e}, {a,p},
{a,e}, {p,e}, {t}, {a}, {p}, {e}, {
}
2.2-15
15
Number of Distinct Proper Subsets
The number of distinct proper subsets of a
finite set A is 2n – 1, where n is the number of
elements in set A.
2.2-16
16
Number of Distinct Proper Subsets
Example:
Determine the number of distinct proper
subsets for the given set
{t, a, p, e}.
2.2-17
17
Number of Distinct Subsets
Solution:
The number of distinct proper subsets is
24 – 1= 2 • 2 • 2 • 2 – 1 = 15.
They are {t,a,p}, {t,a,e}, {t,p,e}, {a,p,e},{t,a}, {t,p},
{t,e}, {a,p}, {a,e}, {p,e}, {t}, {a}, {p}, {e}, { }.
Only {t,a,p,e}, is not a proper subset.
2.2-18
18
Practice Problems: 2.2
Let A = {1, 3, 5, 10}
Let B= {5,6,7,8}
Let C= {5,8}
Let U= {1,3,5,6,7,8,10}
Determine whether statement is true or false:
1) { } is a subset of C
a) false, the empty set is not a subset of any set, b) True,
C) False the empty set is Not a subset of U
2.2-19
19
Homework
Section 6.1 #’s11a 1c, 3 ,and try 6a,
20
Practice Problems: 2.2
Use subset, proper subset or both to make a true statement out of the
following:
{8,9,10}_______ {7,8,9,10}
a) subset, b) Proper subset, c) Not a proper subset d) both subset and
proper subset
If you had the following set of pizza toppings (olives, peppers, cheese,
pineapple, meatballs}
How many subsets of pizzas could you make?
How many distinct proper subsets?
2.2-21
21
Subsets: Proof and Disproof
Because we define what it means for one set to be a
subset of another by means of a universal conditional
statement, we can use the method of direct proof to
establish a subset relationship.
Such a proof is called an element argument and is the
fundamental proof technique of set theory.
22
Example 2 – Proving and Disproving Subset Relations
Define sets A and B as follows:
a. Outline a proof that A  B.
b. Prove that A  B.
c. Disprove that B  A.
23
Example 2 – Solution
a. Proof Outline:
Suppose x is a particular but arbitrarily chosen element
of A.
Therefore, x is an element of B.
b. Proof:
Suppose x is a particular but arbitrarily chosen element
of A.
24
Example 2 – Solution
cont’d
By definition of A, there is an integer r such that
x = 6r + 12.
Let s = 2r + 4.
Then s is an integer because products and sums of
integers are integers.
25
Example 2 – Solution
cont’d
Also
Thus, by definition of B, x is an element of B,
c. To disprove a statement means to show that it is false,
and to show it is false that B  A, you must find an
element of B that is not an element of A.
26
Example 2 – Solution
cont’d
By the definitions of A and B, this means that you must
find an integer x of the form 3  (some integer) that
cannot be written in the form 6  (some integer) + 12.
A little experimentation reveals that various numbers do
the job. For instance, you could let x = 3.
Then x  B because 3 = 3  1, but x  A because there is
no integer r such that 3 = 6r + 12. For if there were such
an integer, then
27
Example 2 – Solution
cont’d
but 3/2 is not an integer. Thus 3  B but 3  A, and
so B A.
28
Set Equality
29
Set Equality
We have known that by the axiom of extension, sets A and
B are equal if, and only if, they have exactly the same
elements.
We restate this as a definition that uses the language of
subsets.
30
Set Equality
This version of the definition of equality implies the
following:
To know that a set A equals a set B, you must know
that A  B and you must also know that B  A.
31
Example 3 – Set Equality
Define sets A and B as follows:
Is A = B?
Solution:
Yes. To prove this, both subset relations A  B and B  A
must be proved.
32
Example 3 – Solution
cont’d
Part 1, Proof That A  B:
Suppose x is a particular but arbitrarily chosen element of A.
[We must show that x  B. By definition of B, this means we
must show that x = 2  (some integer) – 2.]
By definition of A, there is an integer a such that x = 2a.
[Given that x = 2a, can x also be expressed as 2  (some
integer) – 2? i.e., is there an integer, say b, such that
2a = 2b – 2? Solve for b to obtain b = (2a + 2)/2 = a + 1.
Check to see if this works.]
33
Example 3 – Solution
cont’d
Let b = a + 1.
[First check that b is an integer.]
Then b is an integer because it is a sum of integers.
[Then check that x = 2b – 2.]
Also 2b – 2 = 2(a + 1) – 2 = 2a + 2 – 2 = 2a = x,
Thus, by definition of B, x is an element of B
[which is what was to be shown].
Part 2, Proof That B ⊆ A:
Similarly we can prove that B ⊆ A. Hence A = B.
34
Venn Diagrams
35
Venn Diagrams
If sets A and B are represented as regions in the plane,
relationships between A and B can be represented by
pictures, called Venn diagrams, that were introduced by
the British mathematician John Venn in 1881.
For instance, the relationship A  B can be pictured in one
of two ways, as shown in Figure 6.1.1.
A⊆B
Figure 6.1.1
36
Venn Diagrams
The relationship A B can be represented in three different
ways with Venn diagrams, as shown in Figure 6.1.2.
A
B
Figure 6.1.2
37
Example 4 – Relations among Sets of Numbers
Since Z, Q, and R denote the sets of integers, rational
numbers, and real numbers, respectively, Z is a subset of
Q because every integer is rational (any integer n can be
written in the form ).
Q is a subset of R because every rational number is real
(any rational number can be represented as a length on the
number line).
Z is a proper subset of Q because there are rational
numbers that are not integers (for example, ).
38
Example 4 – Relations among Sets of Numbers
cont’d
Q is a proper subset of R because there are real numbers
that are not rational (for example, ).
This is shown diagrammatically in Figure 6.1.3.
Figure 6.1.3
39
Operations on Sets
40
Operations on Sets
Most mathematical discussions are carried on within some
context. For example, in a certain situation all sets being
considered might be sets of real numbers.
In such a situation, the set of real numbers would be called
a universal set or a universe of discourse for the
discussion.
41
Operations on Sets
42
Operations on Sets
Venn diagram representations for union, intersection,
difference, and complement are shown in Figure 6.1.4.
Shaded region
represents A  B.
Shaded region
represents A  B.
Shaded region
represents B – A.
Shaded region
represents Ac.
Figure 6.1.4
43
Example 5 – Unions, Intersections, Differences, and Complements
Let the universal set be the set U = {a, b, c, d, e, f, g} and
let A = {a, c, e, g} and B = {d, e, f, g}. Find A  B, A  B,
B – A, and Ac.
Solution:
44
Operations on Sets
There is a convenient notation for subsets of real numbers
that are intervals.
Observe that the notation for the interval (a, b) is identical
to the notation for the ordered pair (a, b). However, context
makes it unlikely that the two will be confused.
45
Example 6 – An Example with Intervals
Let the universal set be the set R of all real numbers and
let
These sets are shown on the number lines below.
Find A  B, A  B, B – A, and Ac.
46
Example 6 – Solution
47
Example 6 – Solution
cont’d
48
Operations on Sets
The definitions of unions and intersections for more than
two sets are very similar to the definitions for two sets.
49
Operations on Sets
An alternative notation for
and an
alternative notation for
50
Example 7 – Finding Unions and Intersections of More than Two Sets
For each positive integer i, let
a.
b.
Solution:
a.
51
Example 7 – Solution
cont’d
52
Example 7 – Solution
cont’d
b.
53
The Empty Set
54
The Empty Set
We have stated that a set is defined by the elements that
compose it. This being so, can there be a set that has no
elements? It turns out that it is convenient to allow such a
set.
Because it is unique, we can give it a special name. We
call it the empty set (or null set) and denote it by the
symbol Ø.
Thus {1, 3}  {2, 4} = Ø and {x  R| x2 = –1} = Ø.
55
Example 8 – A Set with No Elements
Describe the set
Solution:
We have known that a < x < b means that a < x and x < b.
So D consists of all real numbers that are both greater than
3 and less than 2.
Since there are no such numbers, D has no elements and
so D = Ø.
56
Partitions of Sets
57
Partitions of Sets
In many applications of set theory, sets are divided up into
nonoverlapping (or disjoint) pieces. Such a division is
called a partition.
58
Example 9 – Disjoint Sets
Let A = {1, 3, 5} and B = {2, 4, 6}. Are A and B disjoint?
Solution:
Yes. By inspection A and B have no elements in common,
or, in other words, {1, 3, 5}  {2, 4, 6} = Ø.
59
Partitions of Sets
60
Example 10 – Mutually Disjoint Sets
a. Let A1 = {3, 5}, A2 = {1, 4, 6}, and A3 = {2}. Are A1, A2,
and A3 mutually disjoint?
b. Let B1 = {2, 4, 6}, B2 = {3, 7}, and B3 = {4, 5}. Are B1, B2,
and B3 mutually disjoint?
Solution:
a. Yes. A1 and A2 have no elements in common, A1 and A3
have no elements in common, and A2 and A3 have no
elements in common.
b. No. B1 and B3 both contain 4.
61
Partitions of Sets
Suppose A, A1, A2, A3, and A4 are the sets of points
represented by the regions shown in Figure 6.1.5.
A Partition of a Set
Figure 6.1.5
Then A1, A2, A3, and A4 are subsets of A, and
A = A1 U A2 U A3 U A4.
62
Partitions of Sets
Suppose further that boundaries are assigned to the
regions representing A2, A3, and A4 in such a way that
these sets are mutually disjoint.
Then A is called a union of mutually disjoint subsets, and
the collection of sets {A1, A2, A3, A4} is said to be a
partition of A.
63
Example 11 – Partitions of Sets
a. Let A = {1, 2, 3, 4, 5, 6}, A1 = {1, 2}, A2 = {3, 4}, and
A3 = {5, 6}. Is {A1, A2, A3} a partition of A?
b. Let Z be the set of all integers and let
Is {T0, T1, T2} a partition of Z?
64
Example 11 – Solution
a. Yes. By inspection, A = A1  A2  A3 and the sets A1, A2,
and A3 are mutually disjoint.
b. Yes. By the quotient-remainder theorem, every integer n
can be represented in exactly one of the three forms
for some integer k.
This implies that no integer can be in any two of the sets
T0, T1, or T2. So T0, T1, and T2 are mutually disjoint.
It also implies that every integer is in one of the sets T0,
T1, or T2. So Z = T0  T1  T2.
65
Example : Yogurt Taste Test
A yogurt company wishes to introduce a new
yogurt flavor. The company is considering
two flavors: raspberry cheesecake (R) and
orange creme (O). In a survey of 250 people
it was found that
180 people liked raspberry cheesecake.
139 people liked orange creme.
82 people liked both flavors.
2.5-66
66
Example : Yogurt Taste Test
Of those surveyed, how many people
a) did not like either raspberry cheesecake or
orange crème?
b) liked raspberry cheesecake, but not
orange creme?
c) liked orange creme, but not raspberry
cheesecake?
d) liked either raspberry cheesecake or
orange creme?
2.5-67
67
Example : Yogurt Taste Test
Solution
People surveyed is 250: n(U) = 250.
People surveyed who liked raspberry
cheesecake is 180: n(R) = 180.
People surveyed who liked orange creme is
139: n(O) = 139.
People surveyed who liked both raspberry
cheesecake and orange creme is 82: n(R ⋂
O) = 82.
2.5-68
68
Example: Yogurt Taste Test
Solution
Region II is n(R ⋂ O) = 82
I is n(R) – n(R ⋂ O) = 180 – 82 = 98
III is n(O) – n(R ⋂ O) = 139 – 82 = 57
Add I, II, & III
98 + 82 + 57 = 237
IV is
n(U) – n(R ⋃ O) =
250 – 237 = 13
2.5-69
69
Three Sets: Eight Regions
When three sets overlap, it creates
eight regions.
2.4-70
70
General Procedure for Constructing Venn Diagrams with Three
Sets, A, B, and C
Determine the
elements to be placed
in region V by finding
the elements that are
common to all three
sets,
A ∩ B ∩ C.
2.4-71
71
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the elements
to be placed in region II.
Find the elements in A ∩
B and place the
elements that are not
listed in region V in
region II.
2.4-72
72
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the elements
to be placed in region
IV. Find the elements in
A ∩ C and place the
elements that are not
listed in region V in
region IV.
2.4-73
73
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the elements
to be placed in region
VI. Find the elements in
B ∩ C and place the
elements that are not
listed in region V in
region VI.
2.4-74
74
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the elements
to be placed in region I
by determining the
elements in set A that
are not in regions II, IV,
and V.
2.4-75
75
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the
elements to be placed
in region III by
determining the
elements in set B that
are not in regions II,
V, and VI.
2.4-76
76
General Procedure for Constructing Venn
Diagrams with Three Sets, A, B, and C
Determine the
elements to be placed
in region VII by
determining the
elements in set C that
are not in regions IV,
V, and VI.
2.4-77
77
General Procedure for Constructing Venn Diagrams
with Three Sets, A, B, and C
Determine the elements
to be placed in region
VIII by finding the
elements in the
universal set that are
not in regions I through
VII.
2.4-78
78
Example: Blood Types
Human blood is classified (typed)
according to the presence or absence of
the specific antigens A, B, and Rh in the
red blood cells. Antigens are highly
specified proteins and carbohydrates
that will trigger the production of
antibodies in the blood to fight
infection. Blood containing the Rh
antigen is labeled positive, +, while
blood lacking the Rh antigen is labeled
negative, –.
2.4-79
79
Example: Blood Types
Blood lacking both A and B antigens is called type O.
Sketch a Venn diagram with three sets A, B, and Rh
and place each type of blood listed in the proper
region. A person has only one type of blood.
2.4-80
80
Example: Blood Types
2.4-81
81
Example: Blood Types
Solution
Blood containing Rh is is +
Blood not containing Rh is –
All blood in the Rh
circle is +
All blood outside the Rh
circle is –
Intersection of all 3
sets, V, is AB+
II contains only A and
B, AB–
2.4-82
82
Example: Blood Types
Solution
I contains A only, A–
III contains B only, B–
IV is A+
VI is B+
VII contains only Rh
antigen, O+
VIII lacks all three
antigens, O–
2.4-83
83
Power Sets
84
Power Sets
There are various situations in which it is useful to consider
the set of all subsets of a particular set.
The power set axiom guarantees that this is a set.
85
Example 12 – Power Set of a Set
Find the power set of the set {x, y}. That is, find
({x, y}).
Solution:
({x, y}) is the set of all subsets of {x, y}. We know that Ø
is a subset of every set, and so Ø  ({x, y}).
Also any set is a subset of itself, so {x, y}  ({x, y}). The
only other subsets of {x, y} are {x} and {y}, so
86
Cartesian Products
87
Cartesian Products
88
Example 13 – Ordered n-tuples
a.
b.
Solution:
a. No. By definition of equality of ordered 4-tuples,
But 3  4, and so the ordered 4-tuples are not equal.
89
Example 13 – Solution
cont’d
b. Yes. By definition of equality of ordered triples,
Because these equations are all true, the two ordered
triples are equal.
90
Cartesian Products
91
Example 14 – Cartesian Products
Let A1 = {x, y}, A2 = {1, 2, 3}, and A3 = {a, b}.
a.
b.
c.
Solution:
a. A1  A2 = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}
b. The Cartesian product of A1 and A2 is a set, so it may be
used as one of the sets making up another Cartesian
product. This is the case for (A1  A2)  A3.
92
Example 14 – Solution
cont’d
c. The Cartesian product A1  A2  A3 is superficially similar
to, but is not quite the same mathematical object as,
(A1  A2) × A3. (A1  A2)  A3 is a set of ordered pairs of
which one element is itself an ordered pair, whereas
A1  A2  A3 is a set of ordered triples.
93
Example 14 – Solution
cont’d
By definition of Cartesian product,
94
An Algorithm to Check Whether One
Set Is a Subset of Another (Optional)
95
An Algorithm to Check Whether One Set Is a Subset of Another (Optional)
Order the elements of both sets and successively compare
each element of the first set with each element of the
second set.
If some element of the first set is not found to equal any
element of the second, then the first set is not a subset of
the second.
But if each element of the first set is found to equal an
element of the second set, then the first set is a subset of
the second. The following algorithm formalizes this
reasoning.
96
An Algorithm to Check Whether One Set Is a Subset of Another (Optional)
Algorithm 6.1.1 Testing Whether A ⊆ B:
[Input sets A and B are represented as one-dimensional
arrays a[1], a[2], . . . , a[m] and b[1], b[2], . . . , b[n],
respectively. Starting with a[1] and for each successive a[i]
in A, a check is made to see whether a[i] is in B. To do this,
a[i] is compared to successive elements of B. If a[i] is not
equal to any element of B, then answer is given the value
“A B.”
If a[i] equals some element of B, the next successive
element in A is checked to see whether it is in B. If every
successive element of A is found to be in B, then answer
never changes from its initial value “A ⊆ B.”]
97
An Algorithm to Check Whether One Set Is a Subset of Another (Optional)
Input:
m [a positive integer], a[1], a[2], . . . , a[m]
[a one-dimensional array representing the set A], n [a
positive integer], b[1], b[2], . . . , b[n] [a one-dimensional
array representing the set B]
Algorithm Body:
i := 1, answer := “A ⊆ B”
while (i ≤ m and answer = “A ⊆ B”)
j := 1, found := “no”
while ( j ≤ n and found = “no”)
if a[i] = b[j] then found := “yes”
98
An Algorithm to Check Whether One Set Is a Subset of Another (Optional)
j := j + 1
end while
[If found has not been given the value “yes” when
execution reaches this point, then a[i]  B.]
if found = “no” then answer := “A B”
i := i + 1
end while
Output: answer [a string]
99
Example 15 – Tracing Algorithm 6.1.1
Trace the action of Algorithm 6.1.1 on the variables i, j ,
found, and answer for m = 3, n = 4, and sets A and B
represented as the arrays a[1] = u, a[2] = v, a[3] = w,
b[1] = w, b[2] = x, b[3] = y, and b[4] = u.
Solution:
100