A set is a well-defined collection of distinct objects.

Well-defined collection means that there

exists a rule with the help of which it is possible to tell whether a given object belongs or does
not belong to given collection. Generally sets are denoted by capital letters A, B, C, X, Y, Z etc.

Representation Of A Set

Roster method or Listing method 

In this method a set is described by listing elements, separated by commas, within braces {}.
The set of vowels of English alphabet may be described as {a, e, i, o, u}.

Example: Let B be the set of natural numbers between 3 and 9. Then,  B = {4, 5, 6, 7, 8}

Set-builder method or Rule method 

In this method, a set is described by a characterizing property P(x) of its elements x. In such a

case the set is described by {x : P(x) holds} or {x | P(x) holds}, which is read as ‘the set of
all x such that P(x) holds’. The symbol ‘|’ or ‘:’ is read as ‘such that’.

The set   can be written as  . 

Example: Let A be a set of even numbers less than 10. It can be represented by set builder
form as:

A = {x : x is an even number and x < 10}

Illustration -1

Write the solution set of the equation x2 + x – 2 = 0 in roster form.

Solution

The given equation can be written as (x – 1)  (x + 2) = 0, i. e.,   x = 1, – 2

Therefore, the solution set of the given equation can be written in roster form as {1, – 2}.

Illustration -2

Write the set A =Types Of Sets

1. Empty set/null set/void set

The set which contains no element at all is called the null set. This set is sometimes also called

the ‘empty set’ or the ‘void set’. It is denoted by the symbol   or {}.

Example: Let A be a set of VI class students in class VIII. 

A={}

2. Singleton set

A set consisting of a single element is called a singleton set. The set {5} is a singleton set.

Example: A = {1, 2, 3, 4, 5, 6}, {x/x is a factor of 24}

3. Finite set

A set is called a finite set if it is either void set or its elements can be listed (counted, labelled)
by natural number 1, 2, 3, … and the process of listing terminates at a certain natural
number n (say).

Example

i) Set of peoples in a village.

ii) A set containing students of your class.

iii) A = {x/x ∈ N, x < 500000}

Cardinal number of a finite set

The number n in the above definition is called the cardinal number or order of a finite set A and
is denoted by n(A) or O(A).

Example:

Let B = {x / x is a letter in the word ‘Mathematics’}, then B = {M, A, T, H, E, M, A, T, I, C, S}.

Hence, n(B) = 8.

4. Infinite sets

A set whose elements cannot be listed by the natural numbers 1, 2, 3, …., n, for any natural
number n is called an infinite set.

Example:

i) The multiples of 3 are 3, 6, 9, 12……… i.e., A = {x/x is a multiple of 3} is an infinite set.

ii) A set of Natural numbers.

iii) {x∈z : x < 0}

5. Equivalent sets
Two finite sets A and B are equivalent if their cardinal numbers are same i.e. n(A) = n(B).

Example :  ;   are equivalent sets,  .

6. Equal sets

Two sets A and B are said to be equal iff every element of A is an element of B and also every
element of Bis an element of A. Symbolically, A = B if x ∈ A ⇔ x ∈ B. 

Example : If and  . Then   because each element of A is an element

of B and vice-versa.

Illustration -3

Which of the following sets are empty, singleton, pair and which of them are equal.

A = {x : x2 = 9 and 2x = 3}

B = {x : x2 – 5x + 6 = 0, 2x = 6}

C = {x : x2 – 4x + 3 = 0}

D = {x : x2 = 25}

E = {x : 2x = 6 or x = 1}

Solution

{1, 4, 9, 16, 25, . . .}in set-builder form.

Solution

We may write the set A as A = {x : x is the square of a natural number}

Alternatively, we can write A = {x : x = n2, where n ∈ N}

Solution

We may write the set A as A = {x : x is the square of a natural number}

Alternatively, we can write A = {x : x = n2, where n ∈ N}

The multiples of 7 are 7, 14, 21, 28, 35… Hence, the number of elements in set A = {7, 14, 21,
28, 35…} is not definite. Hence, it is an infinite set.
The two given points are (1, 1) and (0,0) and we know that there is one and only one line
passing through two fixed points. Hence, there will be only one line that passes through the
given points.
Thus, the set contains only one element. Hence, it is a finite set.

Illustration -4

State which of the following sets are finite or infinite:

(i) {x : x ∈ N and (x – 1) (x – 2) = 0}      (ii) {x : x ∈ N and x2 = 4}

(iii) {x : x ∈ N and 2x – 1 = 0}                (iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

Solution

The multiples of 7 are 7, 14, 21, 28, 35… Hence, the number of elements in set A = {7, 14, 21,
28, 35…} is not definite. Hence, it is an infinite set.

The two given points are (1, 1) and (0,0) and we know that there is one and only one line
passing through two fixed points. Hence, there will be only one line that passes through the
given points.
Thus, the set contains only one element. Hence, it is a finite set.

Illustration -5

State whether each of the following sets is finite or infinite:

Set of multiples of 7

Set of lines passing through the point (1, 1) as well as the origin

Solution

The multiples of 7 are 7, 14, 21, 28, 35… Hence, the number of elements in set A = {7, 14, 21,
28, 35…} is not definite. Hence, it is an infinite set.

The two given points are (1, 1) and (0,0) and we know that there is one and only one line
passing through two fixed points. Hence, there will be only one line that passes through the
given points.
Thus, the set contains only one element. Hence, it is a finite set.

Illustration -6

State which of the following sets are finite or infinite:

(i) {x : x ∈ N and (x – 1) (x – 2) = 0}       (ii) {x : x ∈ N and x2 = 4}
(iii) {x : x ∈ N and 2x – 1 = 0}                 (iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

Solution

The multiples of 7 are 7, 14, 21, 28, 35… Hence, the number of elements in set A = {7, 14, 21,
28, 35…} is not definite. Hence, it is an infinite set.

The two given points are (1, 1) and (0,0) and we know that there is one and only one line
passing through two fixed points. Hence, there will be only one line that passes through the
given points.
Thus, the set contains only one element. Hence, it is a finite set.

Subset And Super Set

subset

Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B.

If A is subset of B, we write A ⊆ B, which is read as “A is a subset of B” or “A is contained in B”.

Thus, A ⊆ B ⇒ a ∈ A ⇒ a ∈ B.

superset

If  , then ,B is called superset of A and we write  .

Proper subset

If A is a subset of B and   then A is a proper subset of B. We write this as  .

The null set   is subset of every set and every set is subset of itself, i.e.,   

and   for every set A. They are called improper subsets of A. Thus every non-empty set

has two improper subsets. It should be noted that   has only one subset   which
is improper.

All other subsets of A are called its proper subsets. Thus, if    ,  ,
then A is said to be proper subset of B.
Example : Let  . Then A has   as its subsets out of which   and {1, 2}
are improper and {1} and {2} are proper subsets.

Comparable Set

In two sets one is a subset of the other ,then the sets are called comparable sets.

Properties of subsets

i) Every set is its own subset.

ii) Empty set is a subset of each set.

Let A = {    } = ϕ [i.e., empty set] and B = {1, 2, 3, 4}

It is clear that set A has no element and we can say that there is no element in ϕ which is not in
B. Hence, each element of ϕ is an element of B.

⇒ 

Therefore, empty set is a subset of each set.

iii) For any two sets A and B, A = B   A   B and B A.

Let A and B be two unequal sets, i.e., do not contain exactly same elements, like an element
a that belongs to one set but not to the other.

But if a∈A, then a ∈ B since A B and if a ∈ B, then a  ∈ A since B  A.

Hence, ‘a’ belongs to both A and B.

That is, there cannot be any element in one set that does not belong to the other set. So, and
B contain exactly the same elements i.e., A = B.

Therefore, for any two sets, A = B ⇔ A  B and B   A.

iv) If A  B and B   C, then A   C.

Let’s observe the figure  

From the figure, it is clear that x is the part of 

A (x ∈ A) and x is the part of B (x∈ B)

∵ A ⊂ B]

x is the part of C (x ∈ c) [∵ B ⊂ C]

∴ x∈ A  ⇒x∈ c

i.e., A⊂ C.

Illustration -7

List all the subsets and all the proper subsets of the set {–1, 0, 1}.

Solution

Let A = {–1, 0, 1}.

Subset of A having no element is: ϕ

Subsets of A having one elements are: {–1}, {0}, {1}.

Subsets of A having two elements are: {–1, 0}, {0, 1}, {–1, 1}.

Subsets of A having three elements: {–1, 0, 1}.

Thus, all the subsets of A are:

ϕ, {–1}, {0}, {1}, {–1, 0}, {0, 1}, {–1, 1}, {–1, 0, 1}

Proper subsets of A are:

ϕ, {–1}, {0}, {1}, {–1, 0}, {0, 1}, {–1, 1}.

Illustration -8

Let A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}. Find all sets X such that:

(i) X ⊆ B and X ⊆ C(ii) X ⊆ A and X ⊄ B

Solution

Given, A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}.

(i) Subsets of B are: ϕ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

Subsets of C are: ϕ, {2}, {4}, {2, 4}

Since X is a subset of both B and C, therefore, X = ϕ, {2}.

(ii) A = {1, 2, 3, 4} B = {1, 2, 3}

Here 1, 2, 3 are elements of both A and B and 4 is the only element of A which is not in B.

Since, X ⊆ A and X ⊄ B, therefore, 4 must belong to X.

Hence X = {4}, {1, 4}, {2, 4}, {3, 4}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}.

Illustration -9

Consider the sets ϕ, A = {1, 3 },   B = {1, 5, 9},   C = {1, 3, 5, 7, 9}. Insert the
symbol ⊂ or ⊄between each of the following pair of sets:

(i)  ϕ . . . B (ii) A . . . B (iii) A . . . C (iv) B . . . C

Solution

(i) ϕ ⊂ B as ϕ is a subset of every set.

(ii) A ⊄ B as 3 ∈ A and 3 ∉ B

No, as some of the element of set A are not present in set B, A is not a subset of B.

No, similarly, few element of set B are not present in set A, B is not a subset of A.

Illustration -11

Let A, B and C be three sets. If A∈ B and B ⊂ C, is it true that A ⊂ C? If not, give an example.

Solution

No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1} and B ⊂ C. But A ⊄ C as
1 ∈ A and 1 ∉ C.

Note that an element of a set can never be a subset of itself.

Illustration -12
Let B be a subset of A and let P(A : B) = (X ∈ P(A) : X ⊇ B}. Show that P(A : ϕ) = P(A).

Solution

Given, P(A : B) = set of all those subsets of A which are supersets of B.

∴ P(A = ϕ) = set of all those subsets of A which are supersets of ϕ.

= set of all subsets of A[∵ every subset of A is a superset of ϕ] = P(A)

Intervals as Subsets of R

A subset of the real line is called an interval. Intervals are important in solving inequalities or
in finding domains etc. If there are two numbers a, b ∈R such that a < b, following type of
intervals can be defined

Finite Intervals

Open Interval: (a, b) = {x|a < x <

b}
Close Interval:

Open-close
Interval:
Close-open
Interval:

Infinite Intervals

          

Power Set

If S is any set, then the family of all the subsets of S is called the power set of S.

The power set of S is denoted by P(S). Symbolically, P(S) = {T : T ⊆ S}. Obviously   

and S are both elements of P(S).
Example :  Let S = {a, b, c}, then P(S) = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}.

Power set of a given set is always non-empty

Universal Set

A set that contains all sets in a given context is called the universal set.

It should be noted that universal set is not unique. It may differ in problem to problem.

Example: Let A = {1, 2, 3}; B = {3, 4, 6, 9} and C = {0, 1}.

We can take μ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} as the universal set of sets A,B and C.

Venn Diagrams

The diagrams drawn to represent sets are called Venn diagrams or Euler -Venn diagrams. Here
we represent the universal set U by points within rectangle and the subset A of the set U
represented by the interior of a circle. If a set A is a subset of a set B then the circle
representing A is drawn inside the circle representing B. If A and B are no equal but they have
some common elements, then to represent A and B by.

Illustration -13

A class has 175 students. The following table shows the number of students studying one or
more of the following subjects in this case

Subjects   No. of students

Mathematics 100

Physics  70

Chemistry 46

Mathematics and Physics  30

Mathematics and Chemistry  28

Physics and Chemistry 23

Mathematics, Physics and Chemistry 18

How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone?
Are there students who have not offered any one of these subjects?
(iii) A ⊂ C as 1, 3 ∈ A also belongs to C

(iv) B  ⊂ C as each element of B is also an element of C.

Illustration -10

Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B?  Is B a subset of A? Sol.i)

A = {x/x is a square of a natural number, x ≤ 5}

ii) B = {x/x is a month whose name begins with A}

iii) C = {x/x is a multiple of 5, x ≤ 30}

iv) D = {x/x = 1/n, n ≤ 6 and n∈N}

v) E= {x/x is a multiple of 10, x ≤ 90}

4. Describe the following sets in Roster form.

i) A = {x/x is a natural number lying between 5 and 12}

ii) B = {x/x is a factor of 13}

iii) C = {x/x is a root of x + 3 = 0}

iv) D = {x/x is a prime less than 9}

v) E = {x/x is a letter in the word “Accumulator”}

Sol.i) A = {6, 7, 8, 9, 10, 11}

ii) B = {1, 13}

iii) C = {– 3}

iv) D = {2, 3, 5, 7}

v) E = {A, C, U, M, L, T, O, R}

5.Select the empty sets among the following sets.

i) A = {x/x is a root of x + 3 = 3}ii) C = {x/x is an even as well as an odd number}

iii) D = {x/x is a natural number less than 1}iv) The set all triangles with two obtuse angles.

v) The set of points in a plane.

Sol.ii), iii),  iv) are the empty sets.

6.State which of the following sets are finite and which are infinite.

i) A2 = {x / x ∈ N and x ≥ 5} ii) A3 = {12, 22, 32, ……..}

iii)B = {5, 7, 9, ……}iv)  C = {1, 2, 3, 4}

Sol.i) A2 = {x / x ∈ N; and x≥ 5}

A2 = {5, 6, 7, 8, 9, …..}

It is not possible to find out the last number.

Hence, A2 is an infinite set.

ii) A3 = {12, 22, 32, ……}

A3 is an infinite set.

iii) B = {5, 7, 9, ……}

B is an infinite set. iv) C = {1, 2, 3, 4}C set contains 4 elements.

7.A = {x / x is a prime number between 24 and 28}. Find A.

Sol.The numbers between 24 and 28 are 25, 26, 27.

Here all the numbers are composite.

⇒ There are no prime numbers.

∴ A = {x / x is a prime between 24 and 28 = ?}

∴ A is an empty set

8.Which of the following sets are equal.

i) {x / x is a letter in the word FOLLOW}

ii) The letters which appears in the word WOLF

iii) {x / x is a letter in the word FLOW}

iv) The letters F, W, O, L.

Sol.i) A = {f, o, l, w}

ii) B = {w, o, l, f}
iii) C = {f, l, o, w}

iv) D = {f, w, o, l}

The above for sets contain the same elements. Hence, all sets are equal.

9.i) If A  = {x/x is a natural number such that x < 40 and also divisible by atleast one of 5 or 7
or 11. Find n(A)

ii) Given A = {x/x is a natural number less than 9 and more than 4.}

B = {x/x is a prime number less than 14}

C = {x/x is a multiple of 5 and x is less than 15}. Find n(A), n(B) and n(C).

Sol.i) The natural numbers less than 40 divisible by 5 are: {5, 10, 15, 20, 25, 30 35}

The natural numbers less than 40 divisible by 7 are: {7, 14, 21, 28, 35}

The natural numbers less than 40 divisible by 11 are: {11, 22, 33}

∴ n(A) = 14

ii) The natural numbers less than 9 and more than 4 are:

A = {5, 6, 7, 8}

∴ n (A) = 4

The prime numbers less than 14 are:

B = {2, 3, 5, 7, 11, 13}

∴ n(B) = 6

The multiples of 5 and less than 15 are:

C = {5, 10}

∴ n(C) = 2

10. Let A = {x/x is a square}, B = {x/x is a rectangle}

C = {x/x is a quadrilateral},D = {x/x is a Rhombus}

State which sets are proper subsets of the others?

Sol.Set A is called proper subset of B, if the number of elements in super set B is greater than
the number of elements in A.

A ⊂ C, B ⊂ C, D ⊂ C, A ⊂  B, A ⊂ D

11.Is A = {2, 4, 5, 6} a subset of B = {x/x is an even number}?

Sol.No

Since, the element 5 of set A is not present in set B.

12.Write the power set p(x) of the set X = {3, 4}

Sol.The set of all possible subsets of a set A is called the power set of A, denoted by P(A).

The possible subsets of set X = {3, 4} are: ϕ, {3}, {4}, {3, 4}

∴  Power set of set X = P(X) = {ϕ, {3}, {4}, {3, 4}}

13. If A = {1, 2, 3, 4}, State which of the following statement are true;

i) 2 ∈ Aii) 2 ∈ {1, 2, 3, 4}       

iii) 2 ⊂ {1, 2, 3, 4}iv) {2, 3, 4} ⊂ {1, 2, 3, 4}

Sol.Given A = {1, 2, 3, 4}

i) 2 ∈ A (True) (∵ 2 is there in set A)

ii) 2 ∈ {1, 2, 3, 4}(True) (∵ 2 is there in set A)

iii) 2 ⊂ {1, 2, 3, 4}(False) (∵  2 is not a set)

iv) {2, 3, 4} ⊂ {1, 2, 3, 4}(True) (∵ The set {2, 3, 4} is subset of {1, 2, 3, 4}

14.Suggest a universal set for the following:

i) A = {2, 4, 6, 8}, B = {1, 3, 5, 7}, C = {4, 6, 8}

ii) A = {Parallelogram, Rhombus}, B = {Rectangles, Squares}, C = {Trapezium}

iii) A = Set of prime numbers less than 50.

B = Set of odd numbers less than 40.

C = Set of even numbers less than 60.

Sol.i) μ = Set of all natural numbers less than 9 ⇒ 

 = {1, 2, 3, 4, 5, 6, 7, 8}

ii) μ = Set of all quadrilaterals.

iii) μ = Set of all natural numbers less than 60.

15.Name the sets shaded:

Sol.(i)  A ∩ B ∩  C

(ii) ( A ∩ B ) ∪ ( A ∩ C) ∪ ( B ∩ C)

16.65% of students in a class like cartoon movies, 70% like horror movies and 75% like war
movies. What is the smallest percent of student liking all the three type of movies?

Sol.Let d, e, f be the students liking only cartoon, war, horror movies respectively.

Let ‘a’ be the students liking both cartoon, horror movies but not war movies.

Let ‘b’ be the students liking both cartoon, war movies but not horror movies.

Let ‘c’ be the students liking both horror, war movies but not cartoon movies.

Let ‘x’ be the students liking cartoon, horror and war movies.

The smallest percent of students liking all the three type of movies indicates the minimum value
of x.

x will be minimum if d = e = f = 0

∴ a + b + c + x = 100

Now, a + b + x = 65, a + c + x = 70, b + c + x = 75

x will be minimum if d = e = f = 0

∴ a + b + c + x = 100 __________ (i)

Given, a + b + x = 65 ___________ (ii)

a + c + x = 70 ___________ (iii)

b + c + x = 75 ___________ (iv)

Adding the above equations (ii),(iii) and (iv), we get

a + b + x + a + c + x + b + c + x = 65 + 70 + 75

2a + 2b + 2c + 3x = 210

2(a + b + c + x) + x = 210

2(100) + x = 210  (∵  From (i), a + b + c + x = 100)

200 + x = 210

∴ x = 10

Hence, atleast 10% of students like all the three types of movies.

17.If n (A) = n(B), find n (A ∩ B) and n (A∪B), see the following figure.

ol.Given, n(A) = n(B)

i.e., (4x – 8) + 2x = (x + 7) + 2x

6x – 8 = 3x + 7

⇒ 6x – 3x = 7 + 8 = 3x = 15

∴ x = 5

 n (A ∪ B)  = (4x – 8) + (2x) + (x + 7)

= (4×5 – 8) + (2×5) + (5 + 7) = 34

n (A ∩ B) = 2x  = 2 × 5 = 10

18.Draw a Venn diagram for three non-empty sets A, B and C so that A, B and C will have the
following property:

i) A ⊂ B, C ⊂ B, A ∩ C = ϕ       ii) If A ⊂ C, A ≠ C, B ∩ C = ϕ

iii) A ⊂ B, C ⊄ B, A ∩ C = ϕ      iv) A ⊂ (B, C), B ⊂ C, C ≠ B, A ≠ C

Operations On Sets And Their Algebraic Properties

1. Union Of Sets

Let A and B be two sets. The union of A and B is the set of all elements which are in set A or in B. We

denote the union of A and B by   , which is usually read as “A union B”.

Symbolically, 

Illustration -14

Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team.
Let 
Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football
team. Find X ∪ Y and interpret the set.

Solution

We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of students from Class XI
who are in the hockey team or the football team or both.

2. Intersection Of Sets

Let A and B be two sets. The intersection of A and B is the set of all those elements that belong to both 
A and B.

The intersection of A and B is denoted by A ∩ B (read as “A intersection B”).

Thus, A ∩ B = {x : x ∈ A and x ∈ B}.

Illustration -15

Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find A ∩ B and hence show that A ∩ B
= B.
 Solution

We have A ∩ B = {2, 3, 5, 7} = B. We note that B ⊂ A and that A ∩ B = B. ′

3. Disjoint Sets

Two sets A and B are said to be disjoint, if A ∩ B = ϕ. If A ∩ B ≠ ϕ, then A and B are said to be non-

intersecting or non-overlapping sets.

Example : Sets {1, 2}; {3, 4} are disjoint sets.

4. Difference Of Sets

Let A and B be two sets. The difference of A and B written as A – B, is the set of all those  elements
of Awhich do not belong to B.

Thus, A – B = {x : x ∈ A and x ∉ B}

Similarly, the difference  is the set of all those elements of B that do not belong

to A i.e.,  .

Example : Consider the sets   and  , then  .

Symmetric difference of two sets

Let A and B be two sets. The symmetric difference of sets A and B is the set   and is denoted by

. Thus,   =  .

Properties on symmetric difference

A,B,C  are any three sets

i)          

ii)
iii) 

iv)     

v)  

vi) 

Illustration -16

If A = {1, 2, 3, 4}, B = {2, 4, 6}

Let’s find

i) A – Bii) B – A iii) (A – B) ∪ (B – A)

Solution

i) A – B = {1, 2, 3, 4} – {2, 4, 6} = {1, 3}

ii) B – A = {2, 4, 6} – {1, 2, 3, 4} = {6}

iii) (A – B) ∪ (B – A) = {1, 3} ∪ {6} = {1, 3, 6}

Thus, set (A – B)   (B – A) contains the elements that belong to one of the sets A and B,
but not to both. This set is denoted by AΔB.

∴  AΔB = {1, 3, 6}′

5. Compliment Of A Set

Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U
which are not the elements of A. Symbolically, we write A′ to denote the complement of A with respect
to U. Thus, A′ = {x : x  ∈ U and x  ∉ A }. Obviously A′ = U – A

Example

If μ = {1, 2, 3, 4, 5, 6} and A = {1, 2, 3}, then A’ consists of all the elements that are in μ and not in A.

A’ = {4, 5, 6}

In the set-builder form:

If A = {x/x ∈ μ and x ∉ A’}; then

A’= {x/x ∈ μ and x∉  A}

Venn Diagram Representation Of Complement Set

The elements of all that are not in A constitute a new set called the complement of A and denoted by A°
or Ac and read as “A complement”

Properties Of Complement Sets

If A be a set, A’ be its complement and S be the universal set, then

(i) A ∩ A’ = ϕ            (ii) A ∪ A’ = S           (iii) ϕ’ = S           (iv) S’ = ϕ

These laws are called complement laws.

Proof

Let x ∈ A ∩ A’          ⇒ x ∈ A and x ∈ A’

                              ⇒ x ∈ A and x ∈ A’[No such x exists]

Hence A ∩ A’ contains no element ∴ A ∩ A’ = ϕ

(ii) Let x ∈ A ∪ A’ ⇔ x ∈ A or x ∈ A’

⇔ x ∈ A or x ∉ A’           [This is true for all x ∈S]

⇔ x ∈ S

Hence A ∪ A’ = S          [This is true for all x ∈ S]

(iii) Let x = ϕ’ ⇔ x ∉ ϕ

⇔ x ∈ S

Hence ϕ = S

(iv) Let x ∈S’ ⇒ x ∉ S         [No such x is possible]

Hence S’ contains no element of S

∴ S’ = ϕ
Illustration -17

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

Solution

We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Hence 
A′ = {2, 4, 6, 8,10}.

Illustration -18

Let U be universal set of all the students of Class XI of a coeducational school and A be the set
of all girls in Class XI. Find A′.

Solution

Since A is the set of all girls, A′ is clearly the set of all boys in the class.

Illustration -19

Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.

Find A′, B′ , A′ ∩  B′, A ∩ B and hence show that (A ∪ B)′ = A′ ∩ B′.

Solution

Clearly A′ = {1, 4, 5, 6}, B′ = {1, 2, 6}. Hence A′ ∩ B′ = {1, 6}

Also A ∪ B = {2, 3, 4, 5}, so that (A ∪ B)′ = {1, 6}

(A ∪ B)′ = {1, 6} = A′ ∩ B′

Solved Problems Set-2

1.If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Sol.If A and B are two sets such that A ⊂ B, then A ∪ B = B.

2.If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find

(i) A ∪ B   (ii) A ∪ C      (iii) B ∪ C    (iv) B ∪ D   (v) A ∪ B ∪ C  

(vi) A ∪ B ∪ D   (vii) B ∪ C ∪ D

Sol.A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i)  A ∪ B = {1, 2, 3, 4, 5, 6}

(ii)  A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

3.Find the intersection of each pair of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = {a, e, i, o, u} B = {a, b, c}

(iii) A = {x: x is a natural number and multiple of 3}

B = {x: x is a natural number less than 6}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = ϕ

Sol.(i) X = {1, 3, 5}, Y = {1, 2, 3}

X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A ∩ B = ϕ
ol.Given, n(A) = n(B)

i.e., (4x – 8) + 2x = (x + 7) + 2x

6x – 8 = 3x + 7

⇒ 6x – 3x = 7 + 8 = 3x = 15

∴ x = 5

 n (A ∪ B)  = (4x – 8) + (2x) + (x + 7)

= (4×5 – 8) + (2×5) + (5 + 7) = 34

n (A ∩ B) = 2x  = 2 × 5 = 10

18.Draw a Venn diagram for three non-empty sets A, B and C so that A, B and C will have the
following property:

i) A ⊂ B, C ⊂ B, A ∩ C = ϕ       ii) If A ⊂ C, A ≠ C, B ∩ C = ϕ

iii) A ⊂ B, C ⊄ B, A ∩ C = ϕ      iv) A ⊂ (B, C), B ⊂ C, C ≠ B, A ≠ C

Operations On Sets And Their Algebraic Properties

1. Union Of Sets

Let A and B be two sets. The union of A and B is the set of all elements which are in set A or in B. We

denote the union of A and B by   , which is usually read as “A union B”.

Symbolically, 

Illustration -14

Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team.
Let 
Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football
team. Find X ∪ Y and interpret the set.

Solution
We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of students from Class XI
who are in the hockey team or the football team or both.

2. Intersection Of Sets

Let A and B be two sets. The intersection of A and B is the set of all those elements that belong to both 
A and B.

The intersection of A and B is denoted by A ∩ B (read as “A intersection B”).

Thus, A ∩ B = {x : x ∈ A and x ∈ B}.

Illustration -15

Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find A ∩ B and hence show that A ∩ B
= B.

 Solution

We have A ∩ B = {2, 3, 5, 7} = B. We note that B ⊂ A and that A ∩ B = B. ′

3. Disjoint Sets

Two sets A and B are said to be disjoint, if A ∩ B = ϕ. If A ∩ B ≠ ϕ, then A and B are said to be non-

intersecting or non-overlapping sets.

Example : Sets {1, 2}; {3, 4} are disjoint sets.

4. Difference Of Sets

Let A and B be two sets. The difference of A and B written as A – B, is the set of all those  elements
of Awhich do not belong to B.

Thus, A – B = {x : x ∈ A and x ∉ B}

Similarly, the difference  is the set of all those elements of B that do not belong

to A i.e.,  .

Example : Consider the sets   and  , then  .

Symmetric difference of two sets

Let A and B be two sets. The symmetric difference of sets A and B is the set   and is denoted by

. Thus,   =  .

Properties on symmetric difference

A,B,C  are any three sets

i)          

ii)

iii) 

iv)     

v)  

vi) 

Illustration -16

If A = {1, 2, 3, 4}, B = {2, 4, 6}

Let’s find

i) A – Bii) B – A iii) (A – B) ∪ (B – A)

Solution

i) A – B = {1, 2, 3, 4} – {2, 4, 6} = {1, 3}

ii) B – A = {2, 4, 6} – {1, 2, 3, 4} = {6}

iii) (A – B) ∪ (B – A) = {1, 3} ∪ {6} = {1, 3, 6}

Thus, set (A – B)   (B – A) contains the elements that belong to one of the sets A and B,
but not to both. This set is denoted by AΔB.

∴  AΔB = {1, 3, 6}′

5. Compliment Of A Set

Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U
which are not the elements of A. Symbolically, we write A′ to denote the complement of A with respect
to U. Thus, A′ = {x : x  ∈ U and x  ∉ A }. Obviously A′ = U – A

Example

If μ = {1, 2, 3, 4, 5, 6} and A = {1, 2, 3}, then A’ consists of all the elements that are in μ and not in A.

A’ = {4, 5, 6}

In the set-builder form:

If A = {x/x ∈ μ and x ∉ A’}; then

A’= {x/x ∈ μ and x∉  A}

Venn Diagram Representation Of Complement Set

The elements of all that are not in A constitute a new set called the complement of A and denoted by A°
or Ac and read as “A complement”

Properties Of Complement Sets

If A be a set, A’ be its complement and S be the universal set, then

(i) A ∩ A’ = ϕ            (ii) A ∪ A’ = S           (iii) ϕ’ = S           (iv) S’ = ϕ

These laws are called complement laws.

Proof

Let x ∈ A ∩ A’          ⇒ x ∈ A and x ∈ A’

                              ⇒ x ∈ A and x ∈ A’[No such x exists]

Hence A ∩ A’ contains no element ∴ A ∩ A’ = ϕ

(ii) Let x ∈ A ∪ A’ ⇔ x ∈ A or x ∈ A’

⇔ x ∈ A or x ∉ A’           [This is true for all x ∈S]

⇔ x ∈ S

Hence A ∪ A’ = S          [This is true for all x ∈ S]

(iii) Let x = ϕ’ ⇔ x ∉ ϕ

⇔ x ∈ S

Hence ϕ = S

(iv) Let x ∈S’ ⇒ x ∉ S         [No such x is possible]

Hence S’ contains no element of S

∴ S’ = ϕ

Illustration -17

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

Solution

We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Hence 
A′ = {2, 4, 6, 8,10}.

Illustration -18

Let U be universal set of all the students of Class XI of a coeducational school and A be the set
of all girls in Class XI. Find A′.

Solution

Since A is the set of all girls, A′ is clearly the set of all boys in the class.

Illustration -19

Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.

Find A′, B′ , A′ ∩  B′, A ∩ B and hence show that (A ∪ B)′ = A′ ∩ B′.

Solution

Clearly A′ = {1, 4, 5, 6}, B′ = {1, 2, 6}. Hence A′ ∩ B′ = {1, 6}
Also A ∪ B = {2, 3, 4, 5}, so that (A ∪ B)′ = {1, 6}

(A ∪ B)′ = {1, 6} = A′ ∩ B′

Solved Problems Set-2

1.If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Sol.If A and B are two sets such that A ⊂ B, then A ∪ B = B.

2.If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find

(i) A ∪ B   (ii) A ∪ C      (iii) B ∪ C    (iv) B ∪ D   (v) A ∪ B ∪ C  

(vi) A ∪ B ∪ D   (vii) B ∪ C ∪ D

Sol.A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i)  A ∪ B = {1, 2, 3, 4, 5, 6}

(ii)  A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

3.Find the intersection of each pair of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = {a, e, i, o, u} B = {a, b, c}

(iii) A = {x: x is a natural number and multiple of 3}

B = {x: x is a natural number less than 6}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = ϕ

Sol.(i) X = {1, 3, 5}, Y = {1, 2, 3}

X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A ∩ B = ϕ

Solution