1
Set Theory and
Real Number System
Set
A set is a collection of well defined objects. The objects of a set are called the members
or elements of the set and their membership is defined by the certain conditions. The
elements of the set can be anything pencil, apple, rubber, Sun etc.
Suppose, S be the set i.e., a collection of objects and x is an object which belongs to S
i.e., member of S, then we will write x ∈ S. We can also write S = { x : Q( x )}. It means S is
the set of objects for which the statement Q( x ) involving x is true.
Syllabus
Elementary Set Theory
Finite Countable and
Uncountable Sets
Real Number System as a
Complete Ordered Field
Note (a) Sets are generally denoted by capital letters A, B, C, P, Q, X, Y etc.
Archimedean Property
(b) Elements of the set are denoted by the small letters a, b, c, p, q, x, y etc.
(c) If x is not the member of the set S, then it is written as x ∉S and read as x
does not belong to S.
Supremum, Infimum
Subsets
If every element of a set A is also an element of a set B, then A is called a subset of B and
it is denoted by A ⊆ B.
Superset
If A is a subset of B means A is contained in B, we can also say that B contains A or B is
superset of A, it can be written as B ⊃ A.
Equality of Sets
If two sets A and B are equal, then symbollically it is written as A = B. If every element of
A belongs to B and every element of B belongs to A. i.e., A = B if and only if
x ∈ A ⇔ x ∈ B or A ⊂ B, B ⊂ A ⇔ A = B
Proper Set
If every element of the set A is an element of the set B and B contains atleast one element
which does not belong to A, i.e., if A ⊂ B and A ≠ B, then we can say that A is proper
subset of B and it is denoted by A ⊂ B.
e.g., {2, 3, 6, 4} is proper subset of {4, 3, 2, 6, 5, 8}
2
UGC-CSIR NET Tutor
Mathematical Sciences
Universal Set
We consider all the sets to be the subsets of a given fixed set
known as universal set or universe of discourse. It is denoted
by U or X.
A
A
B
Finite Set
If a set consist of finite number of elements, it is called a finite
set.
e.g., {7, 9, 11} is a finite set.
Infinite Set
If a set consist of an infinite number of elements, it is called,
infinite set.
e.g., Set of Natural numbers N, Set of Rational numbers R, etc.
In the diagram shaded portion represents A ∪ B.
∴ A ∪ B = { x : x ∈ A or x ∈ B or x ∈ both A and B}
Intersection
The intersection of two sets A and B written as A ∩ B. It means
the set of points which belongs to both A and B.
Power Set
Power set of the set is defined as the family consisting of all
subsets of a set. It is denoted by P.
e.g., { A : A ⊂ B} is power set of B. It can be written as
P = { A : A ⊂ B}
Null Set
A set consisting of no points is called the empty set or null set.
It is denoted by φ or {}.
Indexed Set and Index Set
Let St be a non-empty set for each t in a set ∆. The sets
A1, A2 , A3 , ....., An are called indexed sets and the set
∆ = {1, 2, 3, ..., n} is called index set. The suffix t ∈ ∆ of At is
called an index and such a family of sets is denoted by
{ A t : t ∈ ∆} or { A t }, t ∈ ∆.
Singleton Set
A set consisting of a single element is called a singleton set.
e.g., {1}, {2}, { a } etc., are singleton sets.
Pairwise Disjoint Sets
A family { An } of sets is said to be pairwise disjoint, if
A t ∩ AS = φ, ∀ t, s ∈ ∆ s.t. t ≠ s.
Hereditary Property
A non-empty family A = { A t } of sets is said to have hereditary
property, if
A t ⊂ As, A s ∈ A ⇒ A t ∈ A
A
B
In the diagram shaded portion represents A ∩ B.
∴
A ∩ B = { x : x ∈ A and x ∈ B}
If A ∩ B = φ, it means there is no common element in A and
B. In this case, the sets A and B are said to be disjoint.
Note (a) Sometimes A + B can be written in place of
A ∪ B.
(b) Sometimes A ⋅ B can be written in place of A ∩ B .
(c) If A = {1, 2 }, B = { 3, 4,5 }, then A ∪ B = {1, 2, 3, 4,5 }.
(d) If A = {1, 2 },B = {2, 3, 4 }, then A ∩ B = {2 }.
(e) If x ∈ A ⇒ x ∈ A ∪ B and any x ∈ B ⇒ x ∈ A ∪ B
∴ A ⊂ A ∪ B, B ⊂ A ∪ B
(f) If x ∈ A ∩ B ⇒ x ∈ A and any x ∈ A ∩ B ⇒ x ∈ B
∴ A ∩ B ⊂ A, A ∩ B ⊂ B
Complement of a Set
The complement of a set A is denoted by A C or A′ i.e., the set
of all points in the universal set U which do not belong to A.
Symbollically, it can be written as
A C = U − A = { x : x ∈ U and x ∉ A}
i.e., for any x ∈ A ′ ⇔ x ∉ A
Difference of Sets
Set Operations
The difference A − B between two sets A and B is the set of
points in A which do not belong to B, i.e.,
A − B = A ∩ BC
Union
Symmetric Difference of Sets
Union of two sets A and B written as A ∪ B. It means the set of
points which belongs to one of the sets A and B i.e., which
belongs to A or to B or to both.
The symmetric difference of two sets A and B is denoted by
A ∆ B. It is defined as
A ∆ B = ( A − B) ∪ ( B − A )
Set Theory and Real Number System
Note (a) φ′ = U for φ′ = U − φ = U
(b) U ′ = φ for U ′ = U − U = φ
(c) A − B = A ∩ B C , B − A = B ∩ AC
For A − B = {x : x ∈ A, x ∉B }
= {x : x ∈ A, x ∈ B ′ }
= A ∩B′
(d) A ∪ A′ = U , A ∩ A′ = φ
(e) x ∈ A ∪ B ⇒ x ∈ A or x ∈ B
x ∉ A ∪ B ⇒ x ∉ A and x ∉B
e.g., If A = {2, 3, 4, 5} and B = {4, 5, 6, 7}, then
A − B = {2, 3}, B − A = {6, 7}
A ∆ B = ( A − B) ∪ ( B − A ) = {2, 3} ∪ {6, 7}
= {2, 3, 6, 7}
Important Laws
(vii)
(viii)
(ix)
(x)
3
( A ∩ B) × ( C ∩ D ) = ( A × C ) ∩ ( B × D )
A⊆ A× A⇒A=φ
A= A×B ⇒ A=φ
If A contains n elements and B contains m elements,
then A × B contains n ⋅ m elements.
Example 1. If | A ∩ B| = r. Then, find|( A × B) ∩ (B × A)|.
Solution. Let
and
⇒
A ∩ B = D. Then,
D ×D ⊆ A ×B
D ×D ⊆ B× A
|D × D| = r × r = r 2
⇒ |( A × B) ∩ (B × A)| = r 2
Relation
If A, B, C be any sets, then
1. Commutative Laws
A∪ B= B∪ A
A∩ B= B∩ A
2. Associative Laws
( A ∪ B) ∪ C = A ∪ ( B ∪ C )
( A ∩ B) ∩ C = A ∩ ( B ∩ C )
3. Distributive Laws
A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
A relation from set A to set B is a subset of A × B. Similarly A
relation from set B to set A is a subset of B × A.
e.g., Let A = { a, b} and B = {1, 2}. Then,
A × B = {( a, 1), ( a, 2), ( b, 1), ( b, 2)} is a relation from A to B.
Thus, every subset of A × B is a relation from A to B.
Similarly,
B × A = {(1, a), (1, b), (2, a), (2, b)} is a relation from B to A.
Thus, every subset of B × A is a relation from B to A.
Since, φ is a subset of every set. Hence, φ is a relation from A
to B (and B to A) known as empty relation.
A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )
Theorem If a set A has n element and set B has m
4. De-Morgan’s Laws
( A ∪ B)′ = A ′ ∩ B ′
element, then total number of relations from set A to set
B is 2n⋅m = 2| A| × |B| .
( A ∩ B)′ = A ′∪ B′
5.
A − ( B ∪ C ) = ( A − B) ∩ ( A − C )
Relation on a Set
A − ( B ∩ C ) = ( A − B) ∪ ( A − C )
A relation on a set A is a subset of A × A.
e.g., φ and A × A are relation on A.
e.g., ∆ = {( a, a) : a ∈ A} is a relation on A known as the
diagonal relation on A.
e.g., Let A = { a, b, c}. Then,
R = {( a, b), ( b, a), ( a, c )} is a relation on A.
e.g., Let A be a set. Let R = {( X , Y ) : X , Y ∈ P( A ) and X ⊆ Y } is a
relation on P( A ).
Now, let R and S be relations on A. Then, R ∪ S, R ∩ S and
R − S are all subsets of A × A and hence they are also relations
on A. The relation
RoS = {( a1, a2 ) ∈ A × A : ∃ a3 ∈ X such that ( a1, a3 ) ∈ S and
( a3 , a2 ) ∈ R} is called the composition of R and S.
Cartesian Product
Let A and B be two sets. Then,
A × B = {( x, y ) : x ∈ A, y ∈ B}
is known as cartesian product of A and B.
In general
A× B≠ B× A
where
B × A = {( y, x ) : y ∈ B, x ∈ A}
Properties of Cartesian Product
Let A, B, C and D are sets. Then,
(i) ( A ∪ B) × C = ( A × C ) ∪ ( B × C )
(ii) ( A ∩ B) × C = ( A × C ) ∩ ( B × C )
(iii) ( A − B) × C = ( A × C ) − ( B × C )
(iv) A × B = φ ⇔ A = φ or B = φ
(v) A × C ⊆ B × C, C ≠ φ ⇒ A ⊆ B
(vi) ( A × B = B × A ) ⇔ ( A = φ or B = φ or A = B)
Properties of Relation
Let R, S and T be relations on X. Then,
(i) ( RoS )oT = Ro( SoT )
(ii) Ro∆ = R = ∆oR
(iii) Ro( S ∪ T ) = ( RoS ) ∪ ( RoT )
4
UGC-CSIR NET Tutor
Mathematical Sciences
(iv) Ro( S ∩ T ) ⊆ ( RoS ) ∩ ( RoT )
(v) ( R ∪ S )oT = ( RoT ) ∪ ( SoT )
(vi) ( R ∩ S )oT ⊆ ( RoT ) ∩ ( SoT )
i.e.,
U= A
X ∈P
(ii) X , Y ∈ P, X ≠ Y ⇒ X ∩ Y = φ
Inverse Relation
Function
Let R be a relation on A. Then, the relation
R −1 = { ( a1, a2 ) ∈ A × A |( a2 , a1) ∈ R }
is called the inverse relation of R.
Let A and B be two sets. A subset f of A × B is called a function
or a mapping (or a map) from A to B, if
(i) ∀ a ∈ A, ∃ b ∈ B such that ( a, b) ∈ f
(ii) ( a, b1) ∈ f and ( a, b2 ) ∈ f
⇒ b1 = b2
A is called the domain and B is called the codomain of f. If
( a, b) ∈ f , we write b = f ( a) and call it the image of the element
a under the map f. Thus,
f = {( a, f ( a)) : a ∈ A}
we also adopt the notations f : A → B to say that f is a map
from A to B.
Let f : A → B and g : B → C be two maps.
Then, gof :{( a, c ) :( a, b) ∈ f and ( b, c ) ∈ g}
is a map from A to C called the composition of f and g, given
by
( gof )( a) = g( f ( a)), ∀ a ∈ A.
Definition The subset ∆ of A × A is also a map from A to A
called the identity map on A and is denoted by I A. Thus,
I A( a) = a, ∀ a ∈ A
and
foI A = f = IBof
where f : A → B is a map.
Theorem Let R and S be relations on A. Then,
(i) ( R −1) −1 = R
(ii) ( RoS) −1 = S−1oR −1
Types of Relations
Let R be a relation on A. Then,
(i) R is called reflexive, if ( a, a) ∈ R, ∀ a ∈ A or
equivalently ∆ ⊆ R.
(ii) R is called symmetric, if ( a1, a2 ) ∈ R ⇒ ( a2 , a1) ∈ R or
equivalently R −1 = R.
(iii) R is called anti-symmetric, if ( a1, a2 ) ∈ R and ( a2 , a1) ∈ R
⇒ a1 = a2 or equivalently R ∩ R −1 ⊆ ∆.
(iv) R is called transitive, if ( a1, a2 ) ∈ R and ( a2 , a3 ) ∈ R
⇒ ( a1, a3 ) ∈ R or equivalently RoR ⊆ R.
e.g., Let A = { a, b, c} and
R = {( a, a), ( b, b), ( c, c ), ( a, b), ( b, c ), ( c, b)}
Then, R is reflexive but none of the rest of three.
e.g., Let A = { a, b, c} and
R = {( c, b), ( a, c )}
Then, R is a anti-symmetric but none of the rest three.
e.g., Let A = ( a, b, c ) and
R = {( a, a), ( b, b), ( c, c ), ( a, b), ( b, a), ( b, c ), ( c, b)}
Then, R is reflexive , symmetric but neither anti-symmetric nor
transitive.
Equivalence Relation
A relation R on a set A, which is reflexive, symmetric and
transitive is called equivalence relation on A.
Let R be an equivalence relation on A and a ∈ A. The subset
Ra = { b ∈ A :( a, b) ∈ R}
is called the equivalence class of A modulo R determined by a.
Theorem Let R be an equivalence relation on A. Then,
(i) Ra = Rb ⇔ ( a, b) ∈ R
(ii) Ra ≠ Rb ⇔ Ra ∩ Rb = φ
Partition of Set
Let A be a set. A subset P of the power set of A is called
partition of A, if
(i) union of member of P is A
Definition Let X ⊆ A. Then, IX = {( x, x ) : x ∈ X } is a map from X
to A called the inclusion map from X to A.
Definition Let A and B be two sets and b ∈ B. Then, A × { b} is
a map f from A to B such that
f ( a) = y, ∀ a ∈ A
called constant map.
Definition Let A and B be two sets.
The map
p1 : A × B → A
defined by
p1( a, b) = a
is called the first projection and
p2 : A × B → B
defined by
p2 ( a, b) = b
is called the second projection map.
Theorem Let f : A → B, g : B → C and h : C → D be maps.
Then,
(hog)of = ho(gof )
i.e., The set of maps follows the associative law.
Definition Let f : X → Y be a map. Then,
f −1 = { ( b, a) :( a, b) ∈ f }
⊆ B× A
Set Theory and Real Number System
need not be a map from B to A. But f −1 will be a map iff
(i) ∀ b ∈ B, ∃ a ∈ A such that ( a, b) ∈ f
(ii) ( a1, b) ∈ f and ( a2 , b) ∈ f
⇒ a1 = a2
called inverse map of f.
Definition A map f : A → B is called injective or one-one, if
f ( a1) = f ( a2 )
⇒
a1 = a2
or equivalently
a1 ≠ a2
⇒
f ( a1) ≠ f ( a2 )
or equivalently
( a1, b) ∈ f , ( a2 , b) ∈ f
⇒
a1 = a2
Definition A map f : A → B is called surjective map or onto
map, if ∀ b ∈ B, ∃ a ∈ A such that ( a, b) ∈ f . Thus, f is surjective,
if ∀ b ∈ B, ∃ a ∈ A such that f ( a) = b.
Definition A map f which is injective as well as surjective is
called a bijective or one-one-onto map.
Note f −1 is a map iff f is bijective.
Example 2. Show that a surjective map need be injective.
Solution. Let A = {a, b, c} and B = {x, y}
Define the map f : A → B by f ( a) = x = f ( b), f ( c) = y.
Then, f is surjective but not injective for f ( a) = f ( b) but a ≠ b.
Example 3. Show that an injective map need not be
surjective.
Solution. Let A = {a, b} and B = {x, y , z}
Define the map f : A → B by f ( a) = x, f ( b) = y. Then, f is
injective but not surjective for there is no element in A
whose image is z.
Properties of Map
Inverse of a bijective map is a bijective map.
Composition of any two injective maps is injective.
Composite of two surjective maps is surjective.
Composite of two bijective maps is bijective.
Let f : A → B and g : B → C be maps. Then,
(a) If gof is surjective, then g is surjective.
(b) If gof is injective, then f is injective.
(c) If gof is bijective, then g is surjective and f is
injective.
(vi) A map f : A → B is injective iff f can be left cancelled
in the sense that
fog = foh ⇒ g = h
A map f : A → B is surjective iff f can be right
cancelled in the sense that
(i)
(ii)
(iii)
(iv)
(v)
5
gof = hof
⇒
g=h
(vii) A map f : A → B is bijective iff it can be left as well as
right cancelled.
(viii) Let f : A → B be a map. Then, f is bijective iff ∃ a map
g : B → A such that
gof = I A and
fog = IB and then g = f −1
(ix) Let f : A → B and g : B → C be bijective maps. Then
( gof )−1 = f −1og −1
Example 4. Show that there is no surjective map from any
set A to its power set P( A).
Solution. Let f : A → P ( A) be a map. Consider the set
X = {a ∈ A : a ∉ f ( a)}
Then, X ∈ P( A). Suppose that f ( b) = X for some b ∈ A, if
b ∉ X = f ( b), then b ∈ X . If b ∈ X = f ( b), then b ∉ f ( b) = X .
Hence, supposition that f ( b) = X for some b ∈ A is false. This
shows that f cannot be surjective.
Theorem Let f be a map from A to B and X ⊆ A. Then,
X ⊆ f −1( f ( X)). Also,
X = f −1( f ( X)),∀ X ⊆ A iff f is injective.
Theorem Let f : A → B be a map and Y ⊆ B. Then,
f ( f −1(Y )) ⊆ Y . Also,
Y = f ( f −1(Y )),∀ Y ∈ B iff f is surjective.
Theorem Let f : A → B be a map. Let X1 and X2 be
subsets of A. Then,
(i) f ( X1 ∪ X2) = f ( X1) ∪ f ( X2)
(ii) f ( X1 ∩ X2) ⊆ f ( X1) ∩ f ( X2)
Theorem Let f : A → B be a map. Let Y1 and Y2 be
subsets of Y. Then,
(i) f −1( B1 ∩ B2) = f −1( B1) ∩ f −1( B 2)
(ii) f −1( B1 ∪ B2) = f −1( B1) ∪ f −1( B 2)
(iii)f −1( B1 − B2) = f −1( B1) − f −1( B 2)
Ordered Pair
Order pair is an element of the form ( a, b). The element a is
called first element and the element b is called the second
element of the ordered pair.
Equality of Ordered Pairs
Let ( a, b) and ( c, d ) be any two ordered pairs, then
( a, b) = ( c, d ) ⇔ a = c, b = d
6
UGC-CSIR NET Tutor
Mathematical Sciences
Cardinally Equivalent or Equivalent
Sets
A set A is said to be equivalent or cardinally equivalent to the
set B if there exist one-one map from A to B. This relation is
denoted by the symbol ~. Therefore, A ~ B ⇔ A and B are
equivalent.
e. g., The set of natural numbers N = {1, 2, ...} and set of all
odd natural numbers O = {1, 3, 5, ...} are equivalent because f
is a map f : N → O by the formula f ( n) = (2n − 1), ∀ n ∈ N
which is one-one from N onto O.
Note (a) The relation ~ is an equivalent relation.
(b) The relation A ~ B in the family of sets is an
equivalent relation.
Cardinal Numbers
(or Power or Potency)
The cardinal number of a equivalence sets is any
representative of the class or we can say that every
equivalence class defines a unique cardinal number.
If S is any set consisting of s elements, then the cardinal
number is s.
The cardinal number of φ is defined as zero.
If A ~ {1, 2, 3, ..., n}, then n is called the cardinal number of A.
The cardinal number of the any set A is denoted by card A or
| A|. Therefore,| A| or card A = n.
Note (a) If A ~ B , then by the definition of cardinal
number | A | = | B |.
(b) Cardinal number corresponding to a finite set is
called a finite cardinal number.
(c) Cardinal number corresponding to an infinite set
is called transfinite cardinal number.
(d) All transfinite cardinal numbers are greater than
any finite cardinal numbers.
Sum of Cardinal Numbers
Let A and B be any two sets with cardinality n and m s.t. their
intersection is empty. Let
| A| = n, | B| = m and A ∩ B = φ
Then, the sum of n and m is defined as n + m =| A ∪ B|.
e.g.,
If
A = { a, b, c, d}, B = { a , p, q, r, s, t}
Then, A ∪ B = { a, b, c, d} ∪ { a, p, q, r, s, t}
= { a, b, c, d, p, q, r, s, t}
∴
| A ∪ B| = 9
and
| A| = 4,| B| = 6, A ∩ B ≠ φ
∴ By the definition
| A ∪ B| ≠| A | + | B |
Note This example shows that A ∩ B = φ is a
necessary condition for the rule | A ∪ B | = | A | + | B |.
Product of Cardinal Numbers
Suppose A and B be any two sets s.t.| A| = n and| B| = m, then
the product of cardinal numbers n and m is defined as
n × m =| A × B|
e.g., Let
A = { a, b, c}, B = { a, 2, 3, 4}
Then,
| A| = 3 and | B| = 4
∴
| A × B| = 3 × 4 = 12
Now,| A × B| =|{( a, a), ( a, 2), ( a, 3), ( a, 4 ), ( b, a), ( b, 2), ( b, 3),
( b, 4 ), ( c, a), ( c, 2), ( c, 3), ( c, 4 )}|
= 12
∴
| A × B| =| A| ×| B| = n × m
Denumerable (or Enumerable or
Countably Infinite) Countable
and Uncountable Sets
A set A is called a denumerable set if ∃ a one-one mapping
from the set N of all natural numbers onto the set A i.e., if
A ~ N.
A set A is said to be countable set if either A is denumerable
or A is finite i.e., if either A ~ N or A is finite.
If the set A is not countable, then it is said to be uncountable.
i.e., A set is called uncountable set, if either A is an infinite or
A is not cardinally equivalent to N.
Uncountable
sets
are
also
called
non-countable, non-denumerable, non-enumerable.
Note
Decimal Representation
The digits 0, 1, 2, ..., 9 are called the decimal digits. When the
series
d1
d
d
d
d
+ 2 + 3 + 4 + ... + nn + ...
10 102 103 104
10
converges to x with each dn as a decimal digit. Therefore, the
decimal representation of x is given by
x = 0 ⋅ d1d2 d3 d4 ...
3
6
8
e.g.,
+ 2 + 3 + ... represents 0.368…
10 10
10
Ternary Representation
The digits 0, 1, 2 are said to be ternary digits when the series
t1 t2
t
t
t
+
+ 3 + 4 + ... + nn + ...
3 32 33 34
3
converges to x with each tn as a ternary digit, then a ternary
representation of x is given by
x = 0 ⋅ t1t2t3t4 ...
7
Set Theory and Real Number System
Important Theorems on Countable and Uncountable Sets
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
Every subset of a finite set is finite.
Every superset of an infinite set is infinite.
If A and B are finite sets, then A ∩ B is also a finite set.
If A and B are finite sets, then A ∪ B is also a finite set.
Every subset of a countable set is countable.
Every infinite subset of a denumerable set is denumerable.
If A and B are countable sets, then A ∩ B is also a countable set.
Every superset of an uncountable set is uncountable.
Every infinite set has a denumerable subset.
Countably infinite sets are the smallest infinite sets.
The countable union of countable sets is countable.
Every infinite set is equivalent to one of its proper subsets.
The set of all rational numbers is countable.
The set of all rational numbers in [0 , 1] is countable.
The unit interval [0 , 1] is uncountable i.e., the set of all real numbers in the closed interval [0 ,1] is not enumerable
i.e., the set of real numbers x s.t. 0 ≤ x ≤ 1 is not countable.
The set of real numbers is uncountable.
The set of irrational numbers is not countable.
Any open interval ] a, b [ is equivalent to any other open interval ] c, d [.
The set N × N is countable, where N is the set of natural numbers.
The cartesian product of two countable sets is countable. i.e., If A and B are countable sets, then A × B is
countable.
(i) The intervals ] 0 , 1[ and [0 , 1] are equivalent.
(ii) The intervals [0 , 1] and [ 0 , 1[ are equivalent.
(iii) The intervals [0 , 1] and ] 0 , 1[ are equivalent.
(iv) Any open interval is equivalent to [0 , 1.]
Any interval is equivalent to the set R of real numbers. In particular, R is equivalent to [0 ,1.]
Binary Representation
The digits 0 and 1 are said to be the binary digits when the
b1 b2 b3 b4
b
series
+
+
+
+ .... + nn + ...
2 22 23 24
2
converges to x with each bn as a binary digit, then the binary
expression of x is given by
x = 0 ⋅ b1b2 b3 b4 ...
1
0
1
0
1
1
e.g.,
+ 2 + 3 + 4 + ... = 4 =
1
2 2
3
2
2
1−
4
1
Therefore,
= 0 ⋅ 0101010101...
3
1
and similarly,
= 0 ⋅ 010000 ...
4
Note Suppose A be a set of all sequences whose
elements are the digits 0 and 1, then the set A is
countable.
Cantor Set
0
1
1
32
2
32
First step
Second step Removed
1
3
2
3
7
32
8
32
Removed
Removed
Consider the closed interval [0, 1.] Divide this closed interval
into three equal parts and remove the middle one i.e. remove
1 2
the open interval  ,  . This is our first step of the
 3 3
construction of cantor set.
Now, again we divide the each of the remaining two intervals
into three equal parts and remove the middle part means
1 2
7 8
remove the open intervals  2 , 2  and  ,  . This is the
3 3 
9 9
second step of construction. On proceeding in this way
infinitely many steps, at the rth step we have 2r − 1 open
1
intervals are removed each of the length r . The remaining
3
set constitutes cantor set or cantor-ternary set.
8
UGC-CSIR NET Tutor
Mathematical Sciences
Thus, the cantor set contains the points 0, 1,
1 2 1 2 7
, , , , ,
3 3 9 9 9
etc. Therefore, cantor set is non-empty.
Definition of Cantor Set
The cantor set is the set of all numbers in the interval [0, 1]
which have a ternary expansion without the digit 1 i.e. ternary
expansion involves only two digits 0 or 2.
Cantor set is the complement of an open set, therefore, cantor
set is closed.
Important Properties of Cantor Set
1. Cantor ternary set is measurable and its measure is zero.
2. Cantor set is equivalent to [0, 1.]
3. Cantor set is uncountable.
Real Number System as a
Complete Ordered Field
The set R of real numbers is an ordered field.
Order Completeness of R
The set of upper bounds of a non-empty set of real numbers
which is bounded above has a smallest number or we can also
say that every non-empty set of real numbers which is
bounded above admits of a least upper bound i.e.,
supremum.
This property of the set R of real numbers is referred to as its
order-completeness. This property states that if A be the set of
real numbers which is bounded above there exists the smallest
of the upper bounds of A. The fact of a number s being the
smallest of the upper bounds of A can be described by the
following properties:
1. The number s is an upper bound of A i.e., no member of
A is greater than s.
x ∈ A ⇒ x ≤ s ⇔ x ≤ s, ∀ x ∈ A
2. No number less than s is an upper bound of A i.e., if s′ be
the number less than s so that s′ is not an upper bound of
A, ∃ atleast one number x ∈ A s.t. x > s′. Therefore, if s′ < s,
then there exist x ∈ A s.t. x > s′.
Thus, we have completed the description of the set of real
numbers as a complete ordered field.
Note The ordered field of rational numbers is not
order-complete.
Note If we have a positive real number m and a
natural number n, then there exist one and only one
positive real number b s.t. bn = m.
Important Theorems on Real Numbers
Theorem 1 A non-empty subset of real numbers which
is bounded below has the greatest lower bound (infimum)
in R, the set of all real numbers.
Theorem 2 (Dedekind Property) Suppose L and U be
two non-empty subsets of the ordered field R s.t.
(i) L ∪ U = R
(ii) each member of L is less than each member of U i.e.
x ∈ L ∧ y ∈U ⇒ x < y
Then, either the subset L has a greatest member or the
subset U has a smallest member.
Subtraction and Division in R
1. a − b = a + ( − b)
a
2. a ÷ b = or a( b−1), b ≠ 0
b
Note The multiplicative inverse of b ≠ 0 may be
denoted as b −1 or
1
.
b
Archimedean Property of Real
Numbers
Theorem 1 If x and y are two given real numbers w ith
x > 0, then ∃ a natural number n s.t. nx > y .
Theorem 2 For any real number b ∃ a positive integer n
s.t. n > b.
Theorem 3 For any ε > 0 ∃ a positive integer n s.t.
1
<ε
n
Theorem 4 For any real number y, ∃ two integers p and
n s.t. p < x < n.
Theorem 5 For any real number x, ∃ a unique integer n
s.t. n ≤ x < n + 1
Theorem 6 For any real number x, ∃ a unique integer n
s.t. x − 1 < n < x.
Irrational Number
Archimedean Ordered Field
A real number which is not rational is called irrational. e.g.,
3, 10 etc.
Archimedean ordered field is an ordered field which has
Archimedean property. For example, the field R of the set of
real numbers is an Archimedean ordered field.
Set Theory and Real Number System
Important Theorems
Denseness Property of Real Number System
Theorem 1 (The Density Theorem)
Between any two distinct numbers, there lies atleast one
rational number and hence there lie infinitely many
rational numbers.
Theorem 2 Between any two distinct real numbers,
there lies atleast one irrational number also, and hence
there lies infinitely many irrational numbers also.
Theorem 3 Between any two distinct rational numbers,
there lies atleast one and hence infinite number of
rational numbers.
Supremum or least upper bound (lub)
If the set of all upper bounds of a set S of real numbers has a
smallest member k, then k is said to be a least upper bound or
supremum of S and it is denoted by lub or Sup S.
9
Greatest Member of a Set Bounded
above
A number η is called the greatest member of the set S, if
(i) η ∈ S ⇒ η is itself a member of S.
(ii) No member of S is greater than η
⇒ η is an upper bound of S, therefore
x ∈S ⇒x ≤ η
Hence, the greatest member of a set is a member of the set
bounded above and as well as an upper bound of the set.
Smallest Number of a Set Bounded
below
A number ξ is called the smallest number of a set S, if
(i) ξ ∈ S ⇒ ξ is itself a member of S.
(ii) No member of S is smaller than ξ i.e., ξ is a lower
bound of S
∴
x ∈S ⇒ x ≥ ξ
Hence, the smallest member of a set is a member of the set
bounded below and as well as lower bound of the set.
Infimum or greatest lower bound (glb)
If the set of all lower bounds of a set S of real numbers has a
greatest member K, then K is said to be a infimum or greatest
lower bound of S and it is denoted by Inf S or glb.
Important Theorems on Supremum
and Infimum
Theorem 1 lub or supremum of a non-empty set S of
real numbers, whenever it exists is unique.
Theorem 2 glb or infimum of a non-empty set S of real
numbers whenever it exists, is unique.
Theorem 3 The necessary and sufficient condition for a
real number ‘ t ’ to be the glb or infimum of a bounded
below set S is that ‘ t ’ must satisfy the follow ing
conditions:
(i) x ≥ t , ∀ x ∈ S
(ii) For each positive real number ε, ∃ a real number
x ∈ S s.t. x < 1 + ε.
Theorem 4 The necessary and sufficient condition for a
real number ‘ s’ to be the lub or supremum of a bounded
above set S is that ‘ s’ must satisfy the follow ing
conditions:
(i) x ≤ s, ∀ x ∈ S
(ii) For each positive real number ε, ∃ a real number x ∈ S
s.t. x > s − ε.
Note The supremum or infimum if exist they are
unique.
Note (a) The smallest member of a set, if it exists, is
the infimum or glb of the set.
(b) The greatest member of a set, if it exists, is the
supremum or lub of the set.
Bounded Set
A set S of real numbers is said to be bounded if it is bounded
above as well as below. When the set S is bounded, ∃ two real
numbers l and m s.t.
l ≤ x ≤ m, ∀ x ∈ S
Note By saying that the set is bounded it means that
∃ an interval I s.t. S ⊂ I .
e.g. (i) The sets I, Q are neither bounded above nor bounded
below.
(ii) The set N of natural numbers is bounded below but not
bounded above. The smallest number is 1.
Therefore, Inf N = 1.
Limit Point of a Set
Adherent Point
A point p ∈ R is said to be an adherent point of a set A ⊂ R if
every neighbourhood of p contains a point of A. The set of all
adherent points of A is called the adherence or closure of A
and it is denoted by Adh( A ). Adherent point is also called
closure point.
10
UGC-CSIR NET Tutor
Mathematical Sciences
Limit Point
(i) A point p ∈ R is said to be a limit point of the set
A ⊂ R if every neighbourhood of p contains atleast
one point of A other than p. Or we can say that a
point p ∈ R is a limit point of A ⊂ R if and only if for
each neighbourhood N of p, we have
( N ∩ A ) ~{ p} ≠ φ
(ii) A point p ∈ R is called a limit point of a set A ⊂ R if for
each ε > 0, the open interval ] p − ε, p + ε [ contains a
point of A, other than p.
Important Theorems
Theorem 1 Interior of a set is an open set.
Theorem 2 A point p ∈ R is a limit point of a set A ⊂ R iff
every neighbourhood of p contains infinitely many points of
A.
Theorem 3 If a non-empty subset A of R which is
bounded above and has no maximum member, then its
supremum is a limit point of the set A.
Theorem 4 The finite set has no limit point.
Derived Set
The set of all limit points of a set A ⊂ R is said to be derived
set and it is denoted by D ( A ) or A′. Therefore,
A ′ = D( A ) = [ p : p is a limit point of A ]
The derived set of A′ i.e., ( A′ )′ or A ′′.
Isolated Point
A point p ∈ R is called an isolated point of A, if it is not the
limit point of A i.e., if there exist a neighbourhood of p which
contains no point of A other than p.
Note (a) A set A is called the discrete set if all its
points are isolated points.
(b) Each point of a set A is either an isolated point of
A or a limit point of A.
Closed Set
1. A set A is said to be closed set if it contains all its limit
point.
2. A set A is said to be closed if its complement is open.
e.g.,
(i) The null set φ is closed.
(ii) The set Q is not a closed set.
(iii) Finite set is always closed.
Theorem 5 If a non-empty subset A of R, which is
bounded below has no maximum member, then its
infimum is a limit point of the set A.
Theorem 6
(Bolzano-Weierstrass Theorem) Every
infinite bounded set of real numbers has a limit point.
Theorem 7 Every bounded infinite set has the smallest
and the greatest limit point.
Important Definitions
Condensation Point
If every neighbourhood of a point contains an infinite number
of points of A, then that point is called the condensation point
of A. Thus, a finite set has no condensation point.
Interior Point
The point x ∈ A is called an interior point of A if there exist a
neighbourhood containing x and contained in A. Evidently, x
must belong to A.
Closed Set
If every limit point of a set is contained in the set itself, then
such a set is called closed set. Thus, a closed set has the
derived set as its subset, D( A ) ⊆ A, A is a closed set.
Closure of a Set
Open Set
The smallest closed set containing A, is called the closure of A
and is denoted by A.
If every point of set A is an interior point of A, then A is always
called an open set. Open interval is always an open set.
Dense Set
1. A set A is said to be dense or everywhere dense in R, if
A = R.
2. A set A is said to be nowhere dense in R if interior of the
closure of A is empty i.e., ( A )° = φ .
Note (a) The cover G is called an open cover, if
every member of G is an open set.
(b) If ∃ G1 ⊂ G s.t. G1 is a cover of A, then G1 is called
a subcover of G. If every member of G1 is open s.t. G1
is finite set, then G1 is called a finite open subcover of
G.
Set Theory and Real Number System
11
Important Theorems
1. The finite intersection of open subsets of R is an open subset of R.
2. Suppose A ⊂ R , then
(i) A° is an open subset of R.
(ii) A° is the largest open set contained in A.
(iii) A° is open if and only if A° = A
where A° represents the set of all interior point of A.
3. A subset A of R is closed if and only if its complement Ac is open.
4. The arbitrary intersection of closed sets is a closed set.
5. The finite union of closed sets is a closed set.
6. (Heine-Borel Theorem) Suppose, A ⊂ R and G be the collection of subsets of R. Then, G is called a cover of A if
A ⊂ ∪ G.
G ∈G
SOLVED EXAMPLES
Type I (Only one correct option)
Example 1. If A = {a, b, c} and
R = {( a, a), ( a, b), ( b, c ), ( b, b), (c , c ), (c , a)} is a binary relation
on A, then which one of the following is correct?
(a) R is reflexive and symmetric, but not transitive
(b) R is reflexive and transitive, but not symmetric
(c) R is reflexive, but neither symmetric not transitive
(d) R is reflexive, symmetric and transitive
Solution. (c) Q ( a, a), ( b, b), ( c, c) ∈ R
∴ R is a reflexive relation
But ( a, b) ∈ R and ( b, a) ∉ R
∴R is not a symmetric relation
Also, ( a, b), ( b, c) ∈ R
⇒
( a, c) ∉ R
∴ R is not a transitive relation.
Example 2. If the cardinality of a set A is 4 and that of a set
B is 3, then what is the cardinality of the set A ∆ B?
(a) 1
(b) 5
(c) 7
(d) Cannot be determined as the sets A and B are not given.
Solution. (d) Since, the sets A and B are not known, thus
cardinality of the set A ∆ B cannot be determined.
Example 3. Assertion (A) If events, A, B, C and D are
(b) A and R are both correct, but R is not the correct
explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Solution. (a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
Q
( A ∪ B ∪ C) c = Ac ∩ Bc ∩ C c = D
Example 4. Elements of a population are classified
according to the presence or absence of each of 3 attributes
A, B and C. What is the number of smallest ultimate classes
into which the population is divided?
(a) 5
(c) 8
Solution. (d) Elements of a population are classified according
to the presence or absence of each of 3 attributes A, B and C.
Then, the smallest number of smallest ultimate classes into
which the population is divided, is 9.
Example 5. If A and B are subsets of a set X, then what is
{A ∩ ( X − B)} ∪ B equal to?
(a) A ∪ B
(c) A
excluded from the sets A, B and C, then it is included in D.
(a) A and R are both correct, also R is the correct
explanation of A.
(b) A ∩ B
(d) B
Solution. (a) Q A ⊆ X and B ⊆ X
∴
mutually exhaustive, then ( A ∪ B ∪ C ) c = D
Reason (R) ( A ∪ B ∪ C ) c = D implies if any element is
(b) 6
(d) 9
{( A ∩ ( X − B))} ∪ B
(Q X − B = B′ )
= ( A ∩ B′ ) ∪ B = ( A ∪ B) ∩ (B′∪B) = ( A ∪ B) ∩ X
= A∪B
Example 6. The total number of subsets of a finite set A has
56 more elements, then the total number of subsets of another
finite set B. What is the number of elements in the set A?
(a) 5
(b) 6
(c) 7
(d) 8
UGC-CSIR NET Tutor
12
Mathematical Sciences
Solution. (b) Let sets A and B have m and n elements,
Example 10. If A and B are disjoint sets, then A ∩ ( A ′∪B) is
respectively.
(According to question)
∴
2 m − 2 n = 56
n m− n
3
⇒
2 (2
− 1) = 8 × 7 = (2) × 7 = 23 (23 − 1)
equal to which one of the following?
On comparing
⇒
n = 3 and m − n = 3
⇒
m = 6 and n = 3
Number of subsets of A = 23 + 56 = 64 = 2 6
⇒ Number of elements in A = 6
Solution. (a) Q A ∩ B = φ
1, if x is a rational number
0 , if x is an irrational number
Example 7. If f ( x) = 
what is the value of ( fof )( 3)?
(a) 0
(c) Both 0 and 1
Solution.
(b) 1
(d) None of these
1, if x is a rational number
(b) Q f ( x) = 
0 , if x is an irrational number
( fof ) 3 = f {f ( 3)}
= f(0)
(Q 3 is an irrational number)
=1
(Q 0 is a rational number)
Example 8. Consider the following statement.
I. Parallelism of lines is an equivalence relation.
II. xRy, if x is a father of y, is an equivalence relation.
Which of the statements given above is/are correct?
(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II
Solution. (a) I. Let l , m, n are parallel lines and R is a relation.
∴ l || l, then R is reflexive.
and l || m and m|| l, the R is symmetric.
Also, l || m, m|| n ⇒ l || n,
then R is transitive.
Hence, R is an equivalence relation.
II. If x is father of y and y is not father of x, then relation is
not symmetric, thus relation is not equivalence.
Example 9. The function f : R → R defined by f ( x) = ( x2 + 1)35
for all x ∈ R is
(a) φ
(c) A ∪ B
(b) A
(d) A − B′
(given)
A ∩ ( A′∪B) = ( A ∩ A′ ) ∪ ( A ∩ B) = φ ∪ φ = φ
Example 11. Let U = {1, 2, 3, ..., 20}. Let A, B, C be the
subsets of U. Let A be the set of all numbers, which are
perfect squares, B be the set of all numbers which are
multiples of 5 and C be the set of all numbers, which are
divisible by 2 and 3.
Consider the following statements.
I. A, B, C are mutually exclusive.
II. A, B, C are mutually exhaustive.
III. The number of elements in the complement set of
A ∪ B is 12.
Which of the statements given above the correct?
(a) I and II
(c) II and III
(b) I and III
(d) I, II and III
Solution. (b) U = {1, 2, 3,... , 20}
A = Set of all natural numbers which are perfect square
= {1, 4, 9, 16}
B = Set of all natural numbers which are multiples of 5
= {5, 10 , 15, 20}
C = Set of all natural numbers which are divisible by 2 and
3 = {6, 12, 18}
Q A∩B∩C = φ
and
A ∪ B = {1, 4, 9, 16, 5, 10 , 15, 20}
⇒
n( A ∪ B) = 8
⇒
n( A ∪ B)′ = 20 − 8 = 12
∴A, B, C are mutually exclusive and the number of elements
in the complement set of A ∪ B is 12.
Example 12. The function f ( x) = e x, x ∈ R is
(a) onto but not one-one
(b) one-one onto
(c) one-one but not onto
(d) neither one-one nor onto
Solution. (c) It is clear from the graph that f ( x) = ex , ∀ x ∈ R is
(a) one-one but not onto
(b) onto but not one-one
(c) neither one-one nor onto
(d) both one-one and onto
one-one but not onto. Since, range ≠ codomain, so f ( x) is
into.
y
y = ex
Solution. (c) Since, f ( −1) = f (1) = 2
35
i.e., Two real number 1 and −1 have the same image. So,
the function is not one-one and let
y = ( x2 + 1)35
⇒
x = (y)1 35 − 1
Thus, every real number has no pre-image. So, the function
is not onto.
Hence, the function is neither one-one nor onto.
x
x′
y′
Set Theory and Real Number System
Example 13. If A = P {1, 2} where P denotes the power set,
then which one of the following is correct?
(a) {1, 2} ⊂ A
(b) 1 ∈ A
(c) φ ∉ A
(d) {1, 2} ∈ A
Solution. (d) A = P{1, 2} = {φ , {1}, {2}, {1, 2}}
From above, it is clear that
{1, 2 } ∈ A
Example 14. Let R and S be two equivalence relations on a
set A. Then,
(a) R ∪ S is an equivalence relation on A
(b) R ∩ S is an equivalence relation on A
(c) R − S is an equivalence relation on A
(d) None of the above
Solution. (b) Given, R and S are relations on set A.
∴ R ⊆ A × A and S ⊆ A × A ⇒ R ∩ C ⊆ A × A
⇒ R ∩ S is also a relation on A.
Reflexivity Let a be an arbitrary element of A. Then,
a ∈ A ⇒ ( a, a) ∈ R and ( a, a) ∈ S ,
[Q R and S are reflexive]
⇒ ( a, a) ∈ R ∩ S
Thus, ( a, a) ∈ R ∩ S for all a ∈ A.
So, R ∩ S is a reflexive relation on A.
Symmetry Let a, b ∈ A such that ( a, b) ∈ R ∩ S.
Then, ( a, b) ∈ R ∩ S ⇒ ( a, b) ∈ R and ( a, b) ∈ S
⇒
( b, a) ∈ R and ( b, a) ∈ S
[Q R and S are symmetric]
⇒
( b, a) ∈ R ∩ S
Thus, ( a, b) ∈ R ∩ S
⇒
( b, a) ∈ R ∩ S for all ( a, b) ∈ R ∩ S.
So, R ∩ S is symmetric on A.
Transitivity Let a, b, c ∈ A such that ( a, b) ∈ R ∩ S and
( b, c) ∈ R ∩ S. Then, ( a, b) ∈ R ∩ S and ( b, c) ∈ R ∩ S
⇒ {(( a, b) ∈ R and ( a, b) ∈ S)}
and {(( b, c) ∈ R and ( b, c) ∈ S)}
⇒ {( a, b) ∈ R , ( b, c) ∈ R} and {( a, b) ∈ S , ( b, c) ∈ S}
⇒ ( a, c) ∈ R and ( a, c) ∈ S
 Q R and S are transitive. So
( a, b) ∈ R and ( b, c) ∈ R ⇒ ( a, c) ∈ R

( a, b) ∈ S and ( b, c) ∈ S ⇒ ( a, c) ∈ S
⇒ ( a, c) ∈ R ∩ S
Thus, ( a, b) ∈ R ∩ S and ( b, c) ∈ R ∩ S ⇒ ( a, c) ∈ R ∩ S.
So, R ∩ S is transitive on A.
Hence, R is an equivalence relation on A.
Example 15. If (1 + 3 + 5 + ... + p) + (1 + 3 + 5 + ... + q)
= (1 + 3 + 5 + ... + r) where each set of parentheses contains
the sum of consecutive odd integers as shown, what is the
smallest possible value of ( p + q + r) where p > 6?
(a) 12
(b) 21
(c) 45
Solution. (b) Since, Tn = a + (n −1)d
∴
p +1
,
p = 1 + (n − 12
) ⇒n =
2
q +1
q = 1 + (n − 12
) ⇒n =
2
and r = 1 + (n − 12
) ⇒n =
r +1
2
n
× [ 2a + (n − 1)d ]
2
 q + 1
p +1


 2  
p
+
1




q +1  
− 1 2 +
∴ 2 2 × 1 + 
2 × 1 +  2 − 1 2
 2
 
2 
2


r +1
r +1  
=
− 1 2
2 ×1+ 
 
 2
4 
p +1
q +1
r +1
[ 2 + ( p − 1)] +
[ 2 + (q − 1)] =
[ 2 + r − 1]
⇒
4
4
4
⇒ ( p + 1) 2 + (q + 1) 2 = (r + 1) 2
Q Sum of n terms of an AP =
This the possible only when p = 7, q = 5, r = 9
[Q p > 6 (given)]
∴
p+q +r =7+5+9
= 21
Example 16. Let A = {x| x ≤ 9, x ∈ N}. Let B = {a, b, c} be the
subset of A where ( a + b + c ) is a multiple of 3. What is the
largest possible number of subsets like B?
(a) 12
(b) 21
(c) 27
(d) 30
Solution. (d) A = {x : x ≤ 9, x ∈ N}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
Total possible multiple of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27
But 3 and 27 are not possible.
6 → 1+ 2 + 3
9 → 2 + 3 + 4, 5 + 3 + 1, 6 + 2 + 1
12 → 9 + 2 + 1, 8 + 3 + 1, 7 + 1 + 4, 7 + 2 + 3, 6 + 4 + 2,
6 + 5 + 1, 5 + 4 + 3
15 → 9 + 4 + 2, 9 + 5 + 1, 8 + 6 + 1, 8 + 5 + 2, 8 + 4 + 3,
7 + 6 + 2, 7 + 5 + 3, 6 + 5 + 4
18 → 9 + 8 + 1, 9 + 7 + 2, 9 + 6 + 3, 9 + 5 + 4, 8 + 7 + 3,
8 + 6 + 4, 7 + 6 + 5
21 → 9 + 8 + 4, 9 + 7 + 5, 8 + 7 + 6
24 → 9 + 8 + 7
Hence, total number of largest possible subsets are 30.
Example 17. A mapping f : R → R which is defined as
f ( x) = cos x; x ∈ R is
(a) one-one only
(c) one-one onto
(b) onto only
(d) neither one-one nor onto
Solution. (d) Given, f ( x) = cos x
y
(d) 54
(Q nth term of an AP)
13
f (x) = cos x
x
x′
y′
UGC-CSIR NET Tutor
14
Mathematical Sciences
It is clear from the figure that f ( x) is neither one-one nor a
onto function.
Since, whenever we drawn a line parallel to x-axis, then it
intersects at infinite points to the curve. So, f ( x) is not
one-one.
And, range of f ( x) = [ −1, 1]
Codomain of f ( x) = R
Range of f ( x) ≠ codomain of f ( x)
∴f ( x) is not onto.
Example 18. If n( A) = 115, n(B) = 326, n( A − B) = 47, then
what is n( A ∪ B) equal to?
(a) 373
(c) 370
(b) 165
(d) 394
Solution. (a) Now, n( A − B) = n( A) − n( A ∩ B)
⇒
⇒
47 = 115 − n( A ∩ B)
n( A ∩ B) = 68
n( A ∪ B) = n( A) + n(B) − n( A ∩ B)
= 115 + 326 − 68 = 373
Example 19. In a town of 10000 families it was found that
40% family buy newspaper A, 20% buy newspaper B and 10%
families buy newspaper C, 5% families buy A and B, 3% buy B
and C and 4% buy A and C. If 2% families buy all the three
newspaper, then number of families which buy A only is
(a) 3100
(c) 2900
(b) 3300
(d) 1400
Solution. (b) n( A) = 40% of 10000 = 4000
n(B) = 20% of 10000 = 2000
n(C) = 10% of 10000 = 1000
n( A ∩ B) = 5% of 10000 = 500
n(B ∩ C) = 3% of 10000 = 300
n(C ∩ A) = 4% of 10000 = 400
n( A ∩ B ∩ C) = 2% of 10000 = 200
We want to find n( A ∩ Bc ∩ C c ) = n[ A ∩ (B ∪ C) c ]
= n( A) − n[ A ∩ (B ∪ C)] = n( A) − n[( A ∩ B) ∪ ( A ∩ C)]
= n( A) − [n( A ∩ B) + n( A ∩ C) − n( A ∩ B ∩ C)]
= 4000 − [500 + 400 − 200 ] = 4000 − 700 = 3300
Example 20. A survey shows that 63% of the Americans like
cheese whereas 76% like apples. If x% of the Americans like
both cheese and apples, then
(a) x = 39
(c) 39 ≤ x ≤ 63
(b) x = 63
(d) None of these
Solution. (c) Let A denote the set of Americans who like
cheese and let B denote the set of Americans who like
apples.
Let population of American be 100.
Then,
n( A) = 63, n(B) = 76
Now,
n( A ∪ B) = n( A) + n(B) − n( A ∩ B)
= 63 + 76 − n( A ∩ B)
∴ n( A ∪ B) + n( A ∩ B) = 139
⇒
n( A ∩ B) = 139 − n( A ∪ B)
But
n( A ∪ B) ≤ 100
∴
−n( A ∪ B) ≥ −100
∴
139 − n( A ∪ B) ≥ 139 − 100 = 39
...(i)
∴
n( A ∩ B) ≥ 39 i.e., 39 ≤ n( A ∩ B)
Again,
A ∩ B ⊆ A, A ∩ B ⊆ B
∴
n( A ∩ B) ≤ n( A) = 63 and n( A ∩ B) ≤ n(B) = 76
...(ii)
∴
n( A ∩ B) ≤ 63
Then,
39 ≤ n( A ∩ B) ≤ 63 ⇒ 39 ≤ x ≤ 63
Example 21. The set of all limit point of the set
1 1

S =  + : m ∈ù , n ∈ ù  is
m n

(a) φ
1

(c)  : m ∈ ù 
m


(b) {0}
(d) None of these
Solution. (c) Let us fixed m ∈ ù. Then, for every δ > 0.
1
1

 − δ , + δ contains infinitely many elements of ù.
m
m 

1
Hence, the set of all limit point of S is  : m ∈ ù .

m
Example 22. The set of all limit points of the set

1
S =  : n ∈ ù is

n
(a) φ
(c) N
(b) {0}
(d) None of these
Solution. (b) For every δ > 0 , ( − δ , δ) contains infinite number of
elements of set S. Hence, {0} is the set of all limit point of
the set S.
Type II (One or more than one correct option)
Example 23. Which of the following(s) is/are correct?
(a)
(b)
(c)
(d)
A non-empty finite set is not a nbd of any point
Every point of a non-empty finite set is an interior point
No point of a non-empty finite set is an interior point
A non-empty finite set is a nbd of each of its points
Solution. (a), (c)
A set can be a nbd of a point if it contains an open interval
containing the point. Since, an interval necessarily contains
an infinite number of points. Therefore in order that a set be
a nbd of a point it necessarily contain an infinity of points.
Thus, a finite set cannot be a nbd of any points. Hence, no
points of finite non-empty set is an interior point.
Example 24. Which of the following(s) is/are correct?
(a) The set of interior points of N is empty
(b) The set of interior points of I is empty
(c) The set of interior points of Z is empty
(d) The set of interior points of R is empty
Set Theory and Real Number System
Solution. (a), (b), (c)
Since, for the set ù, I and Z there is no open interval which is
contained in ù or I or Z as every open interval contains
rational and irrational number but the set of interior points of
ú is ú.
Example 25. Which of the following(s) set is/are open?
(a) ú
(b) Q
(d) φ
(c) Z
Solution. (a), (d)
The set of real number ú is an open set as every real number
is an interior point of ú. Similarly the set φ is open as the set
φ is a nbd of each of its points in the sense that there is no
point in φ of which it is not nbd. But the set Q and Z are not
open as no points of Q or Z is an interior points.
Example 26. Which of the following(s) is/are correct?
(a) Every open interval is an open set
(b) Every open interval is a nbd of each of its points
(c) Every point of an open interval is an interior point
1

(d) The set  : n ∈ ù  is not open
n

Solution. (a), (b), (c), (d)
Let ( a, b) be an open interval and x ∈( a, b) which implies that
a < x < b.
c
a
d
x
b
Let c, d be two numbers such that a < c < x and n < d < b
⇒
a < c < x < d < b ⇒ x ∈ ( c, d) ⊂ ( a, b)
Thus, the given interval ( a, b) contains an open interval
containing the point x and is therefore a nbd of x.
Hence, the open interval is a nbd of each of its points and is
therefore an open set.
Hence, every point of an open interval is an interior point.
1

Now ‘ o ’ is an interior point of the set  : n ∈ ù  which does
n


not lie in this set.
1

Hence, the set  : n ∈ ù  is not open.
n

Example 27. Which of the following(s) is/are correct?
(a) The intersection of any finite number of open sets is
open
(b) The intersection of an arbitrary number of open sets is
open
(c) The union of an arbitrary family of open sets is open
(d) Exactly one of the above is true
Solution. (a), (c)
Let G1 and G 2 be two open sets. Then,
if G1 ∩ G 2 = φ, it is open
if G1 ∩ G 2 ≠ φ, let x ∈ G1 ∩ G 2
⇒ x ∈ G1 and x ∈ G 2
⇒ G1 , G 2 are nbd of x.
⇒ G1 ∩ G 2 is a nbd of x.
but since x is any point of G1 ∩ G 2, therefore G1 ∩ G 2 is a nbd
of each of its points. Hence, G1 ∩ G 2 is open.
15
Now, consider the open sets
 1 1
Gn =  − ,  , n ∈ ù
 n n
⇒
I G n = {0}, which is not an open set.
n∈ ù
⇒ The intersection of arbitrary number of open set need
not be open.
Next, let G be the union of an arbitrary family
F = {G λ : λ ∈ Λ} of open sets, Λ being an index set. To prove
that G is an open set, we shall show that for any point x ∈ G,
it contains an open interval containing x.
Let x ∈ G
⇒ ∃ atleast one member, say
G λ 1 of F such that
x ∈G λ1
Since, G λ 1 is an open set, ∃ an open interval In such that
x ∈ In ⊆ G λ 1 ⊆ G
Thus, the set G contains an open interval containing any
point x of G. Hence, G is an open set.
Example 28. Which of the following(s) is/are correct?
(a) The set I has no limit point
(b) The set ù has no limit point
(c) Every point of the set Q is a limit point
(d) Every point of the set ú is a limit point
Solution. (a), (b), (c), (d)
1
1

The set I has no limit point, for a nbd m − , m +  of m ∈ I,

2
2
contains no point of I other than m. Thus the derived set of I
is the null set φ.
1
1

The set ù has no limit point, for a nbd m − , m +  of

2
2
m ∈ù, contains no points of ù other than m. Thus, the
derived set of ù is null set φ. Every point of the set Q of
rationals is a limit point, for between any two rationals there
exist an infinity of rational.
Further every point of ú is a limit point, for every nbd of any
of its points contains an infinite member of ú.
Example 29. If S denotes the closure of the set S, then
which of the following(s) is/are correct?
(a) I = I
(b) Q = ú
(c) R = ú
1

1

(d) If S =  : n ∈ ù , then S =  : n ∈ ù  ∪ {0}
n
n




Solution. (a), (b), (c), (d)
Since, I′ = φ, Q′ = ú, ú′ = ú and S′ = {0}
Hence,
I = I ∪ I′ = I ∪ φ = I
Q = Q ∪ Q′ = Q ∪ ú = ú
ú = ú ∪ ú′ = ú ∪ ú = ú
1

S = S ∪ S′ =  : n ∈ ù  ∪ {0}
n


UGC-CSIR NET Tutor
16
Mathematical Sciences
Example 30. Which of the following(s) is/are countable?
(a) The set ú of real number is uncountable
(b) The set of rational number in [0 , 1] is countable
(c) The set of all rational numbers is countable
(d) None of the above
Solution. (a), (b), (c)
Let us suppose that the set of real number ú is countable,
then ú = {x1 , x2 ,... , xn ,....}.
Enclose each member xn of ú in an open interval
1
1 

In =  xn − n + 1 , xn + n + 1 

2
2 
1
of length n , n = 1, 2, 3,...
2
∞
∞
n =1
n =1
xn ∈ ú and R = U {xn} ⊆ U In
⇒ The whole real line is contained in the union of intervals
whose lengths add up to 1. Which is a contradiction. Hence,
ú is uncountable.
Arrange the set of rationals according to increasing
1 1 2 1 3 1 2 3 4 1 5
denominators as 0 ,1, , , , , , , , , , , ,....
2 3 3 4 4 5 5 5 5 6 6
Then the one-one correspondence can be indicated as
2
2
5↔
9↔
1↔ 0
3
5
1
3
6↔
10 ↔
2 ↔1
4
5
1
3
4
3↔
7↔
11 ↔
2
4
5
1
1
..........
4↔
8↔
3
5
..........
which shows that the set of rational numbers in [0 ,1] is
countable.
∞
Now, the set of all rational numbers is the union U Ai , where
i=1
Ai is the set of rationals which can be written with
denominator i. That is
0 −1 1 −2 2

Ai =  , , ,
, ,...
i i i i i

Each Ai is equivalent to the set of all positive integers and
hence countable.
Example 31. Let A = [1, 2, 3, 4] and R be a relation in A given
by R = {(1, 1), (2, 2), (3, 3), ( 4, 4), (1, 2), (2, 1), (3, 1), (1, 3)}.
Then, R is
(a) reflexive
(c) transitive
(1, 1), (2, 2), (3, 3), ( 4, 4) ∈R; ∴R is reflexive.
Q (1, 2), (3, 1) ∈R and also (2, 1), (1, 3) ∈R.
Hence, R is symmetric. But clearly R is not transitive.
Example 32. Let X = {1, 2, 3, 4, 5} and Y = {1, 3, 5, 7, 9}.
Which of the following(s) is/are relations from X to Y?
(a) R1 = {( x, y)| y = 2 + x, x ∈ X , y ∈ Y}
(b) R2 = {(1, 1), (2, 1), (3, 3), ( 4, 3), (5, 5)}
(c) R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}
(d) R4 = {(1, 3), (2, 5), (2, 4), (7, 9)}
1 1
1
where the sum of lengths of In′ s is + 2 + 3 + ...
2 2
2
But
Solution. (a), (b)
(b) symmetric
(d) an equivalence relation
Solution. (a), (b), (c)
R4 is not a relation from X to Y, because (7, 9) ∈R4 but
(7, 9) ∉ X × Y.
Example 33. In a class of 55 students, the number of
students studying different subjects are 23 in Mathematics, 24
in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9
in Mathematics and Chemistry, 7 in Physics and Chemistry
and 4 in all the three subjects. The number of students who
have taken exactly one subject is
(a) 6
(b) 9
(c) 7
(d) None of these
Solution. (a), (b), (c)
n(M) = 23, n(P) = 24, n(C) = 19
n(M ∩ P) = 12, n(M ∩ C) = 9, n(P ∩ C) = 7
n(M ∩ P ∩ C) = 4
We have, to find
n(M ∩ P′∩C′ ), n(P ∩ M′∩C′ ), n(C ∩ M′∩P′ )
Now, n(M ∩ P′∩C′ ) = n[M ∩ (P ∪ C)′ ]
= n(M) − n(M ∩ (P ∪ C))
= n(M) − n[(M ∩ P) ∪ (M ∩ C)]
= n(M) − n(M ∩ P) − n(M ∩ C) + n(M ∩ P ∩ C)
= 23 − 12 − 9 + 4 = 27 − 21 = 6
n(P ∩ M′∩C′ ) = n[P ∩ (M ∪ C)′ ]
= n(P) − n[P ∩ (M ∪ C)] = n(P) − n[(P ∩ M) ∪ (P ∩ C)]
= n(P) − n(P ∩ M) − n(P ∩ C) + n(P ∩ M ∩ C)
= 24 − 12 − 7 + 4 = 9
n(C ∩ M′∩P′ ) = n(C) − n(C ∩ P) − n(C ∩ M) + n (C ∩ P ∩ M)
= 19 − 7 − 9 + 4 = 23 − 16 = 7