Chapter Two
Sets and subsets.
2.1. Sets.
A fundamental concept in all branches of mathematics is that of a
set.
Definition 2.1. A set is a collection of objects. The objects in sets can be
anything: numbers, people, letters, cars, students, rivers, etc. The objects
of a set are called the elements, points , or members of the set. Sets are
usually denoted by capital letters as A , B ,C , D , X ,Y ... , and the elements
of the sets denoted by small letters as a,b ,c ,d , x , y ... . Braces as
  often used to name a set and the elements separated by commas as
A  x , y  .
Example 2.1.
1. The set of integers 1,2,7,15 and 20 is the set 1,2,7,15,20 .
2. The capital of Egypt is Cairo represented by the set Cario .
3. The countries Jordan, Saudi Arabia, Marco is the set
Jordan, Saudi Arabia, Marco .
Let P (x ) be a well-defined property of objects. Then we write
x P (x ).
Example 2.2.
1. P (x ) may be the property that the element x is a positive real
number x P (x )  x is a positive real number .
2. P (x ) is the property that x is an even integer and we write it as
x P (x )  x is a positive real number .
Example 2.3. Consider the following sets.
1. A  x x is a non-negative real numbers .
2. B  x x is a positive integer numbers .
3. C  x 1  x  4: x is a real number .
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4. D  x 1  x  3, x is an integer  1,0,1,2,3 .
Now, if A is a set, and x is an element in A , we write x  A , and we
mean that x is in A . If x is not an element in A , then we write x  A .
Example 2.4.
1. 14,1,2 .
2. 3 1,0,1,2 .
3. 20 x x is a multiple of 2 .
4. 20 x x is an odd number .
Definition 2.2. Let X be any set.
1. The empty (null) set is the set having no elements. We
denote it by   x  X x  x  .
2. The set with only one element is called the singleton set, we
denote it by x  X  .
Definition 2.3. A set is called finite if it is empty or its elements can be
matched up precisely with the elements of the set 1, 2,..., n , where n is
some positive integer. A set that is not finite is called infinite set.
Example 2.5.
1. The sets 4,1,2 and 1,0,1,2 are finite sets.
2. The sets x x is a multiple of 2 and x x is an odd number are
infinite sets.
Definition 2.4. A set A is called a subset of a set B if every element of
A is also in B , that is if (A  B )  (x  A  x  B ) , where, A  B means
that A is a subset of B , A is contained in B , or B contains A . If
A  B , then A is called a proper subset of B , that is A  B and A  B . If
A is not a subset of B , we write A  B .
Definition 2.5. Two sets A and B are called equal, written A  B , when
they have the same elements. We write A  B if A and B are not equal.
Hence , we have
1. (A  B )  (A  B and B  A ) .
2. (A  B )  (A  B or B  A ) .
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Example 2.6.
1. If A  1, 2 and B 2,1 , then A  B .
2. If A  1, 2,3 and B  1, 2,3,7 , then A  B and A  B .
3. x x is a student in Math 151  x x a student in Faculty of Science .
4. The set E x x 2  5x  6  0  2,3 .
5. E x x 2  5x  6  0  2,3 .
Theorem 1.1. Let A , B and C be sets.Then
1.   A .
2. A  A .
3. If and A  B and B  C , then A  C .
Proof .
1. For each x if x  , then x  A . This means that x   x  A and
has a false antecedent, and hence is true statement. Thus   A .
2. For each if x  A , then x  A which is a conditional statement and
it is a tautology. Thus A  A .
3. Let A  B and x  A , then x  B . If B  C , then x C . Therefore,
A C .
2.2. Operations on sets.
In this section, define the operations of union, intersection, and
difference of sets.
1. Union. The union of the sets A and B is the set of all elements
in A or in B , that is
A
B  x x  A or x  B 
read it as A union B . Thus
x A
B  x  A or x  B
x A
x A
x A
B  x A  x B
B  x  A and x  B
B  x A  x B
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Example 2.7.
 Let A  1, 2,3, 4 , and B  2,5,6 . Then
A
B  1, 2,3, 4,5,6
 a,b  c , d   a,b ,c , d  .
Theorem 2.2. Let A , B and C be sets. Then
(a) A B  B A .
(b) A  A B and B  A B .
(c) A   A .
(d) (A B ) C  A (B C ) .
(e) A B    A   and B   .
Proof. (a) Let x  A B  x  A  x  B  x  B  x  A  x  B A . Thus
A B B A .
(b) Let x  A  x  A B  A  A B . Similarly, B  A B .
(c) Let x  A   x  A  x . But x  (empty set). Thus x  A and
A   A . From (b), we have A  A  . Therefore, A   A .
(d) Let
x  (A B ) C
 x  (A B )  x C
 (x  A  x  B )  x C
 x  A  (x  B  x C )
Therefore, A (B
 x  A  (x  B C )
 x  A (B C )
C )  (A B ) C .
(e) We have from
A A
B  , and B  A
B    A   and B   .
But   A and   B . Thus A   and B   .
2. Intersection. The intersection of the sets A and B is the set of
all elements in A and in B , that is
A
B  x x  A and x  B 
read it as A intersection B . Thus
x A
B  x  A and x  B
x A
B  x A  x B
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Example 2.8.
 Let A  1, 2,3, 4 , and B  2,5,6 . Then A B  2
 a,b  c , d    .
Theorem 2.3. Let A , B and C be sets. Then
(a) A B  B A .
(b) A B  A and A B  B .
(c) A    .
(d) (A B ) C  A (B C ) .
Proof. (a) Let x  A B  x  A  x  B  x  B  x  A  x  B A . Thus
A B B A .
(b) Let
x A
B  x  A and x  B  A
B  A and A
B B .
(c) Let A    . Then there exist an element x  A  such that
But x  A  implies that x  A and x  which ia a
x  .
contradiction for x  (empty set). Thus we have a contradiction.
Therefore, A    . But   A  . Hence A    .
(d) Let
x  (A B ) C
 x  ( A B )  x C
 (x  A  x  B )  x C
 x  A  (x  B  x C )
Therefore, (A
 x  A  (x  B C )
 x  A (B C )
B ) C  A (B C ) .
Theorem 2.4. Let A , B and C be sets. Then
(a) A (B C )  (A B ) (A C ) .
(b) A (B C )  (A B ) (A C ) .
Proof . (a) Let
x A
(B
C )  x  A  x  (B
C)
 x  A  (x  B  x C )
 (x  A  x  B )  (x  A  x C )
 (x  A
B )  (x  A C )
 x  (A
B ) (A C )
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Thus A (B C )  (A B ) (A C ) .
Similarly, we can prove A (B C )  (A B ) (A C ) .
Definition 2.6. Two sets A and B are called disjoint if A B   .
3. Difference. Let A and B be two sets. The set difference of
A and B is the set of all elements in A which are not in B , that
is
A  B  x x  A and x  B  .
read it as A difference B . Some times, we use A \ B in place A  B .
x  A  B  x  A and x  B
x A  B  x A  x B
Example 2.8.
 Let A  1, 2,3, 4 , and B  2,5,6 . Then
A  B  1,3, 4
B  A  5,6
A
B  2
A A 
A   A
A  
In general, A  B  B  A and (A  B ) (B  A ) (A B )   , that is
disjoint.
 a,b  c , d    .
Theorem 2.5. Let A , B and C be sets. Then
(a) A  B  A .
(b) (A  B ) B   .
(c) A  (B C )  (A  B ) (A  C ) .
(d) A (B  C )  (A B )  (A C ) .
Proof . (a) Let
x A  B  x A  x B  x A  A  B  A .
(b) Let
x  (A  B ) B
 x  (A  B )  x  B
 (x  A  x  B )  x  B
But there is no element such that x  B  x  B . Thus (A  B ) B   .
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(c) Let
x  A  (B C )
 x  A  x  (B
C)
 x  A  (x  B  x C )
 (x  A  x  B )  (x  A  x C )
Thus A  (B
(d) Let
 (x  A  B )  (x  A  C )
 x  (A  B ) (A  C )
C )  (A  B ) (A  C ) .
x  A (B  C )
 x  A  x  (B  C )
 x  A  (x  B  x C )
 (x  A  x  B )  (x  A  x C )
 (x  A
 x  (A
B )  (x  A C )
B )  (A C )
Thus, A (B  C )  (A B )  (A C ) .
4. Complement. In any application of the theory of sets under
investigation will likely be subsets of a fixed set, this fixed set is
called the universal set. We denote it by U . Let A be a subset
of any set U . The complement of A in U is the set
A   x x U  x  A   U  A
x A  x A
x A  x A
Thus A  the set of all elements in the universal set U which do not
belong to A . Also, some times we denote the complement as
A  Ac .
Example 2.9.
 Let U  1, 2,3, 4,5,6,7 , and A  2,3,7 . Then
A   1, 4,5, 6
(A )  2,3, 7  A
 U   .
   1, 2,3, 4,5,6,7  U .
Theorem 2.6. Let A and B be subsets of a universal set U . Then
(a) (A )  A .
(b) A  B  A B  .
(c) A A    .
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(d) A A   U .
(e) U    .
(h)   U .
(g) A  B  B   A  .
Proof . (a) Let
x  (A )  x  A   x  A
Thus (A )  A .
(b) Let
x  (A  B )  x  A  x  B  x  A  x  B   x  A
B ( A )  A
(c) Let x  A A   x  A  x  A   x  A  x  A . Thus A A    .
(d) Let x  A A   x  A  x  A   x U . Thus A A   U .
(e) x U   x U  x U . Thus U    .
(h) We have   U . Then x   x U  x   x U . Hence,   U .
(g) Suppose that A  B . We want to prove that B   A . Let x  B  . Then
x  B . Since A  B and x  B we have x  A . Thus x  A  . Therefore
x  B  implies that x  A  , that is B   A .
Conversely, let B   A . We want to prove that A  B . Let x  A . Then
x  A  and x  B  for B   A  . Thus x  B . Therefore, x  A implies that
x  B , that is A  B .
Theorem 2.7. ( De Morgan's Laws) Let A and B be subsets of a
universal set U . Then
(a) (A B )  A  B  .
(b) (A B )  A  B  .
Proof . (a) Let
x  (A
B )
 x A
B
 x A  x B
 x A x B 
 x A B 
Hence, (A B )  A  B  .
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(b) Let
x  (A
B )
 x A
B
 x A  x B
 x A x B 
 x A B 
Thus, (A B )  A  B  .
Example 2.10. Let U  1, 2,3, 4,5,6,7,8,9,10 , A  1, 2,3, 4 , B  2, 4,6 ,
and C  3, 4,5,6 . Then
(A C )  (3, 4)  1, 2,5,6,7,8,9,10
A   5,6,7,8,9,10 ,C   1, 2,7,8,9,10
A  C   1, 2,5,6,7,8,9,10
Thus (A C )  A  C  .
B   1,3,5,7,8,9,10
A
B   1,3
Theorem 2.8. Let A and B be subsets of a universal set U . Then
(A
B )  C  (A  C ) (B  C )
Proof . By Theorem 2.6 (b), Theorem 2.3 (a), and Theorem 2.4 (b) , we
have
(A
B )  C  (A B ) C 
( Theorem 2.6(b) A  B  A B )
 C  (A B )
( Theorem 2.3 (a) A B  B A )
 (C  A ) (C  B ) ( Theorem 2.4 (b) A (B C )  (A
 (A C ) (B C )
B ) (A
C ))
 (A  C ) (B  C )
2.3. The power set of a set.
In order to represent intersection and union as operations, we had
to consider the set of all subsets of a given set.
Definition 2.7. Let A be any set. Then the power set, written (A ) , is
the set of all sub sets of A . In other words, (A )  B B  A  .
It is clear that the members or the elements of (A ) are sets. Thus
  (A ), A  (A ) and a (A ) for each a  A . If A is a finite set with n
elements, then (A ) is a set with 2 n elements. Thus we write (A )  2n .
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Example 2.11. (i) ()   , and ()  20  1.
ii) (1)  , 1 , and (1)  21  2 .
iii) (1, 2)  ,1 , 2 , 1, 2 , and (1, 2)  22  4 .
iv) (1, 2,3)  ,1 ,2 , 3 , 1, 21,32,3 , 1, 2,3 , and (1, 2,3)  23  8 .
2.4. Sets of numbers.
Important sets which we meet in elementary mathematics
are sets of numbers.
1. R the set of all real numbers.
2. Z  ..., 3, 2, 1,0,1, 2,3,... the set of integers.

p
q

3. Q  x x  , 0  q , p  Z  the set of rational numbers.


4. Q  x  R x  Q the set of irrational numbers.
5. Z  1, 2,3,... the set of positive integers (natural numbers).
6. Z  1, 2, 3,... the set of negative integers.
7. Z00  0,1, 2,3,... the set of non-negative integers.
8. 2Z  2, 4,6,... the set of even positive integers.
9. n Z  kn k  Z , where n  Z .
Note that
1) Z   Z 00  Z  Q  R. and Q  R .
2
 Q and 4Q .
3
3
3) 2 Q ,  Q and 3  Q .
5
3
4)   Q .
5
5) R  Q Q .
2)
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2.5. Ordered pairs and Cartesian products.
The Cartesian product is a familiar to students of calculus in a
special case, the Cartesian plane , R 2  (x , y ) x , y  R .
Definition 2.8. Let A and B be any sets and let x  A , y  B . The
ordered pair x and y , written (x , y ) , is defined to be the set
(x , y )  x  , x , y  . For the pair (x , y ) , x is called the first coordinate,
and y is called the second coordinate.
This definition allows us to distinguish between (x , y ) and ( y , x ) .
Thus (x , y )  x  , x , y  and
( y , x )   y  ,  y , x  . Note that
(x , y )  ( y , x ) , for x   (x , y ) , but x   ( y , x ) . In general,
(x , y )  (a, b )  x  a and y  b
Example 2.12.
(i) (4,3)  (3, 4) .
(ii) (3x  1, y  1)  (4,3)  3x  1  4 and y  1  3  x  1 and y  2 .
Definition 2.9. Let A and B be any sets. The Cartesian product of A
and B is the set
A  B  (a, b ) a  A and b  B  ,
that is A  B is the set of all ordered pairs (a , b ) , where, a  A and b  B .
If A  B , then we write A 2 for A  A .
Example 2.13. Let A  1, 2 and B  3, 4,5 . Then
A  B  (1,3),(1, 4),(1,5),(2,3),(2, 4),(2,5)
B  A  (3,1),(3, 2),(4,1),(4, 2),(5,1),(5, 2)
A 2  (1,1),(1, 2),(2,1),(2, 2)
B 2  (3,3),(3, 4),(3,5),(4,3),(4, 4),(4,5),(5,3),(5, 4),(5,5)
Thus, in general A  B  B  A .
The concept of the Cartesian product can be extended to more than
two sets. Thus
A1  A 2  A n  (a1 , a2 ,..., an ) a1  A1 , a2  A 2 ,..., an  A n 
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Theorem 2.9. Let A , B , C and D be sets. Then
1. A  (B C )  (A  B ) (A C ) .
2. A  (B C )  (A  B ) (A C ) .
3. (A  B ) (C  D )  (A C )  (B D ) .
4. (A  B ) (C  D )  (A C )  (B D ) .
Proof .
1. Let
(x , y )  A  (B C )
 (x  A )  ( y  B
C)
 (x  A )  ( y  B  y C )
 (x  A  y  B )  (x  A  y C )
 [(x , y )  A  B ]  [( x , y )  ( A C )]
 (x , y )  (A  B ) (A C )
C )  ( A  B ) (A C ) .
Therefore, A  (B
2. Let
(x , y )  A  (B  C )
 (x  A )  ( y  B
C)
 (x  A )  ( y  B  y C )
 (x  A  y  B )  (x  A  y C )
 [(x , y )  A  B ]  [( x , y )  ( A  C )]
 (x , y )  (A  B ) (A C )
3. Let
x  (A  B ) (C  D )
 [x  (A  B )]  [x  (C  D )]
 [x  (a, b )  A  B , wher,a  A , b  B ]  [x  (c , d ) C  D , wher,c C , d  D ]
Thus,
b  d C
a  c A C
x  (a, b )  (c , d )  (A C ) and (B  D )
x  (a, b )  (c , d )  a  c and b  d .
D.
Therefore,
x  (A C ) (B  D ) . Hence,
(A  B ) (C  D )  (A
Now, let
Hence
C )  (B
and
and
D)
x  (u ,v )  (A C )  (B D )
 (u  A C )  (v  B D )
 (u  A  u C )  (v  B  v  D
 (u  A  v  B )  (u C  v  D )
 [x  (u ,v )  A  B ]  [ x  (u ,v ) C  D ]
 x  (u ,v )  (A  B ) (C  D )
Thus,
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(A
C )  (B
D )  (A
C )  (B
D )(A  B ) (C  D )
Therefore,
(A  B ) (C  D )  (A
C )  (B
D).
(A  B ) (C  D )  (A
C )  (B
D) ,
4. To show that
let x  (A  B ) (C  D ) , then either x  A  B or x C  D . If
x A B ,
then x  (a, b )  A  B , where a  A and b  B . Since
a  A  A C and b  B  B D , we have
we,
x  (a, b )  (A
C )  (B
D)
On the other hand, if x C  D , then x  (c , d ) C  D , where,
c C , and d  D . Since c C  A C and d  D  B D , we have
x  (c , d )  (A
C )  (B
D)
Therefore, if x  (A  B ) (C  D ) , then x  (A C )  (B D ) , and hence,
(A  B ) (C  D )  (A
C )  (B
D)
Example 2.14. Let A  1, 2 , B  3 , C  4,5 and D  6 . Then
A  B  (1,3), (2,3)
C  D  (4, 6), (5, 6)
A C  1, 2, 4,5
B
D  3, 6
(A  B ) (C  D )  (1,3), (2,3), (4, 6), (5, 6)
(A C )  (B
Hence,
D )  (1,3), (1, 6), (2,3), (2, 6), (4,3), (4, 6), (5,3), (5, 6)
(A  B ) (C  D )  (A
C )  (B
D)
and
(A C )  (B
D )  (A  B ) (C  D ) .
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