set theory and real numbers

8/20/2019 Set Theory and Real Numbers

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Set A set is a collection of well defined objects. The objects of a set are called the membersor elements of the set and their membership is defined by the certain conditions. Theelements 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 Si.e., member of S, then we will write x S∈ . We can also write S x Q x= { : ( )}. It means S isthe set of objects for which the statement Q x( ) involving x is true.

Note (a) Sets are generally denoted by capital letters A, B, C, P, Q, X, Y etc.

(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 .

Subsets

If every element of a set A is also an element of a set B, then A is called a subset of B andit 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 propersubset 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

Set Theory

Real Number System

and1

Syllabus

Elementary Set Theory

Finite Countable andUncountable Sets

Real Number System as a

Archimedean Property

Supremum, Infimum

Complete Ordered Field


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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 denotedby U or X .

Finite Set

If a set consist of finite number of elements, it is called a finiteset.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.

Power Set

Power set of the set is defined as the family consisting of allsubsets of a set. It is denoted by P.

e.g., { : } A A B⊂ is power set of B. It can be written asP 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 A A A An1 2 3, , , ....., are called indexed sets and the set ∆ = { , , , . .. , }1 2 3 n is called index set. The suffix t ∈ ∆ of At iscalled an index and such a family of sets is denoted by

{ : } A tt ∈ ∆ or { }, A tt ∈ ∆.

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 A t st S∩ = ∀ ∈φ, , ∆ s.t. t s≠ .

Hereditary Property

A non-empty family A At= { } of sets is said to have hereditaryproperty, if

A A A A A At s s t⊂ ∈ ⇒ ∈,

Set Operations

Union

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., whichbelongs to A or to B or to both.

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 meansthe set of points which belongs to both A and 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 andB. In this case, the sets A and B are said to be disjoint.

Note (a) Sometimes A B + can be written in place ofA B ∪ .(b) Sometimes A B ⋅ can be written in place of A B ∩ .(c) If A B = ={ , }, { , , }1 2 3 4 5 , then A B ∪ = { , , , , }12 3 4 5 .(d) If A B = ={ , }, { , , }1 2 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 AC 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 U A x x UC = − = ∈{ : and x A∉ }i.e., for any x A x A∈ ′ ⇔ ∉

Difference of Sets

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− = ∩

Symmetric Difference of Sets

The symmetric difference of two sets A and B is denoted by A B∆ . It is defined as

A B A B B A∆ = − ∪ −( ) ( )

2 UGC-CSIR NET Tutor Mathematical Sciences

B A

A A AB


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Note (a) φ′= U for φ φ′= − =U U

(b) U

′= φ for

′ = − =U U U

φ(c) A B A B B A B AC C − = ∩ − = ∩,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 B A− = − ={ , }, { , }2 3 6 7 A B A B B A∆ = − ∪ − = ∪( ) ( ) { , } { , }2 3 6 7

= { , , , }2 3 6 7

Important Laws

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 B C A B A C∪ ∩ = ∪ ∩ ∪( ) ( ) ( )4. De-Morgan’s Laws

( ) A B A B∪ ′ = ′ ∩ ′( ) A B A B∩ ′ = ′∪ ′

5. A B C A B A C− ∪ = − ∩ −( ) ( ) ( ) A B C A B A C− ∩ = − ∪ −( ) ( ) ( )

Cartesian ProductLet 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 A× = ⇔ =φ φ or B = φ

(v) A C B C C A B× ⊆ × ≠ ⇒ ⊆, φ(vi) ( ) ( A B B A A× = × ⇔ = φ or B = φ or A B= )

(vii) ( ) ( ) ( ) ( ) A B C D A C B D∩ × ∩ = × ∩ ×(viii) A A A A⊆ × ⇒ = φ

(ix) A A B A= × ⇒ = φ(x) 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 A B D∩ = . Then,D D A B× ⊆ ×

and D D B A× ⊆ ×

⇒ | |D D r r × = × = r 2

⇒ |( ) ( )| A B B A r × ∩ × = 2

Relation A relation from set A to set B is a subset of A B× . Similarly Arelation 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 a b b× = {( , ), ( , ), ( , ), ( , )}1 2 1 2 is a relation from A to B.Thus, every subset of A B× is a relation from A to B.Similarly,

B A a b a b× = {( , ), ( , ), ( , ), ( , )}1 1 2 2 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 Ato B (and B to A) known as empty relation.

Theorem If a set A has n element and set B has melement, then total number of relations from set A to set B is 2 2n m A B⋅ ×= | | | |.

Relation on a Set

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 thediagonal 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 arelation on P A( ).

Now, let R and S be relations on A. Then, R S∪ , R S∩ andR S− are all subsets of A A× and hence they are also relationson A. The relation

RoS a a A A a X = ∈ × ∃ ∈{( , ) :1 2 3 such that ( , )a a S1 3 ∈ and( , ) }a a R3 2 ∈ is called the composition of R and S.

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 ( ) ( ) ( )∪ = ∪

Set Theory and Real Number System 3


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(iv) Ro S T RoS RoT ( ) ( ) ( )∩ ⊆ ∩(v) ( ) ( ) ( )R S oT RoT SoT ∪ = ∪

(vi) ( ) ( ) ( )R S oT RoT SoT ∩ ⊆ ∩

Inverse Relation

Let R be a relation on A. Then, the relation

R a a A A a a R− = ∈ × ∈1 1 2 2 1{ ( , ) |( , ) }is called the inverse relation of R.

Theorem Let R and S be relations on A. Then,

(i) ( ) R R− − =1 1

(ii) ( ) RoS S oR− − −=1 1 1

Types of RelationsLet R be a relation on A. Then,

(i) R is called reflexive, if ( , ) ,a a R a A∈ ∀ ∈ orequivalently ∆ ⊆ R.

(ii) R is called symmetric, if ( , ) ( , )a a R a a R1 2 2 1∈ ⇒ ∈ orequivalently R R− =1 .

(iii) R is called anti-symmetric, if ( , )a a R1 2 ∈ and ( , )a a R2 1 ∈⇒ a a1 2= or equivalently R R∩ ⊆

−1 ∆.

(iv) R is called transitive, if ( , )a a R1 2 ∈ and ( , )a a R2 3 ∈⇒ ( , )a a R1 3 ∈ 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= ( , , ) andR a a b b c c a b b a b c c b= {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}

Then, R is reflexive , symmetric but neither anti-symmetric nortransitive.

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

R b A a b Ra = ∈ ∈{ :( , ) }is called the equivalence class of A modulo R determined by a.

Theorem Let R be an equivalence relation on A. Then,

(i) R R a b Ra b= ⇔ ∈( , )(ii) R R R Ra b a b≠ ⇔ ∩ = φ

Partition of Set

Let A be a set. A subset P of the power set of A is calledpartition of A, if

(i) union of member of P is A

i.e., =∈

A X PU

(ii) X Y P X Y , ,∈ ≠ ⇒ ∩ = X Y φ

FunctionLet A and B be two sets. A subset f of A B× is called a functionor a mapping (or a map) from A to B, if

(i) ∀ ∈ ∃ ∈a A b B, such that ( , )a b f ∈(ii) ( , )a b f 1 ∈ and ( , )a b f 2 ∈

⇒ b b1 2= 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 mapfrom 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, givenby

( )( ) ( ( )), gof a g f a a A= ∀ ∈ .Definition The subset ∆ of A A× is also a map from A to Acalled the identity map on A and is denoted by I A. Thus,

I a a a A A( ) ,= ∀ ∈and foI f I of A B= =where f A B: → is a map.

Definition Let X A⊆ . Then, I x 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× { } isa 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

p A B A1 : × →defined by

p a b a1( , ) =

is called the first projection and p A B B2 : × →

defined by

p a b b2 ( , ) =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 b a a b f − = ∈1 { ( , ) :( , ) }

⊆ ×B A



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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) ( , )a b f 1 ∈ a nd ( , )a b f 2 ∈⇒ a a1 2=

called inverse map of f .

Definition A map f A B: → is called injective or one-one, if f a f a( ) ( )1 2=

⇒ a a1 2=or equivalently

a a1 2≠⇒ f a f a( ) ( )1 2≠or equivalently

( , ) , ( , )a b f a b f 1 2∈ ∈

⇒ a a1 2=Definition A map f A B: → is called surjective map or ontomap, 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 iscalled 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 besurjective.

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 isinjective but not surjective for there is no element in Awhose image is z.

Properties of Map

(i) Inverse of a bijective map is a bijective map.

(ii) Composition of any two injective maps is injective.(iii) Composite of two surjective maps is surjective.

(iv) Composite of two bijective maps is bijective.

(v) 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 isinjective.

(vi) A map f A B: → is injective iff f can be left cancelledin the sense that

fog foh g h= ⇒ = A map f A B: → is surjective iff f can be right

cancelled in the sense that

gof hof =⇒ g h=

(vii) A map f A B: → is bijective iff it can be left as well asright cancelled.

(viii) Let f A B: → be a map. Then, f is bijective iff ∃ a map g B A: → such that

gof I A= andfog IB= and then g f =

−1

(ix) Let f A B: → and g B C: → be bijective maps. Then( ) gof f og− − −=1 1 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. Thisshows that f cannot be surjective.

Theorem Let f be a map from A to B and X A⊆ . Then,

X f f X ⊆ −1( ( )). Also,

X f f X X A= ∀ ⊆−1( ( )), iff f is injective.

Theorem Let f A B: → be a map and Y B⊆ . Then,

f f Y Y ( ( ))− ⊆1 . Also,

Y f f Y Y B= ∀ ∈−( ( )),1 iff f is surjective.

Theorem Let f A B: → be a map. Let X 1 and X 2 besubsets of A. Then,

(i) f X X f X f X ( ) ( ) ( )1 2 1 2∪ = ∪

(ii) f X X f X f X ( ) ( ) ( )1 2 1 2∩ ⊆ ∩

Theorem Let f A B: → be a map. Let Y 1 and Y 2 besubsets of Y . Then,

(i) f B B f B f B− − −∩ = ∩1 1 21

11

2( ) ( ) ( )

(ii) f B B f B f B− − −∪ = ∪1 1 21

11

2( ) ( ) ( )

(iii) f B B f B f B− − −− = −1 1 21

11

2( ) ( ) ( )

Ordered PairOrder pair is an element of the form ( , )a b . The element a iscalled first element and the element b is called the secondelement 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= ⇔ = =



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Cardinally Equivalent or Equivalent

Sets A set A is said to be equivalent or cardinally equivalent to theset B if there exist one-one map from A to B. This relation isdenoted by the symbol ~. Therefore, A B A~ ⇔ and B areequivalent.

e g. ., The set of natural numbers N = { , , . . .}1 2 and set of allodd natural numbers O = { , , , . ..}1 3 5 are equivalent because f is a map f N O: → by the formula f n n n N( ) ( ),= − ∀ ∈2 1which 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 anequivalent relation.

Cardinal Numbers(or Power or Potency)The cardinal number of a equivalence sets is anyrepresentative of the class or we can say that everyequivalence class defines a unique cardinal number.If S is any set consisting of s elements, then the cardinalnumber is s.

The cardinal number of φ is defined as zero.If A n~ { , , , . .., }1 2 3 , 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 cardinalnumber | | | |A B = .(b) Cardinal number corresponding to a finite set iscalled a finite cardinal number.

(c) Cardinal number corresponding to an infinite setis called transfinite cardinal number.

(d) All transfinite cardinal numbers are greater thanany finite cardinal numbers.

Sum of Cardinal Numbers

Let A and B be any two sets with cardinality n and m s.t. theirintersection is empty. Let

| | , | | A n B m A B= = ∩ =and φThen, 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∪ = 9and | | ,| | , A B A B= = ∩ ≠4 6 φ∴ By the definition

| | | | | | A B A B∪ ≠ +

Note This example shows that A B ∩ = φ is anecessary 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= , thenthe 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 4Then, | | A = 3 and | |B = 4∴ | | A B× = × =3 4 12Now,| | |{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ) A B a a a a a b a b b× = 2 3 4 2 3 ,

( , ), ( , ), ( , ), ( , ), ( , )}|b c a c c c4 2 3 4

= 12∴ | | | | | | A B A B n m× = × = ×

Denumerable (or Enumerable orCountably Infinite) Countableand 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 denumerableor 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.

Note Uncountable sets are also callednon-countable, non-denumerable, non-enumerable.

Decimal Representation

The digits 0 12 9, , , ..., are called the decimal digits. When theseries

d d d d dnn

1 22

33

4410 10 10 10 10

+ + + + + +... ...

converges to x with each dn as a decimal digit. Therefore, the

decimal representation of x is given by x d d d d= ⋅0 1 2 3 4 .. .

e.g., 3

10

6

10

8

102 3+ + + ... represents 0.368…

Ternary Representation

The digits 0, 1, 2 are said to be ternary digits when the seriest t t t tn

n1 2

233

443 3 3 3 3

+ + + + + +... ...

converges to x with each tn as a ternary digit, then a ternaryrepresentation of x is given by

x t t t t= ⋅0 12 3 4 .. .



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Binary Representation

The digits 0 and 1 are said to be the binary digits when the

series b b b b bn

n1 2

233

442 2 2 2 2

+ + + + + +.... ...

converges to x with each bn as a binary digit, then the binary

expression of x is given by

x b b b b= ⋅0 1 2 3 4.. .

e.g., 0

2

1

2

0

2

1

2

1

4

1 1

4

1

32 3 4+ + + + =

−=...

Therefore, 1

30 0101010101= ⋅ ...

and similarly, 1

40 010000= ⋅ ...

Note Suppose A be a set of all sequences whoseelements are the digits 0 and 1, then the set A iscountable.

Cantor Set

Consider the closed interval [ , ]0 1. Divide this closed intervalinto three equal parts and remove the middle one i.e. remove

the open interval 1

3

2

3, .

This is our first step of the

construction of cantor set.

Now, again we divide the each of the remaining two intervalsinto three equal parts and remove the middle part means

remove the open intervals 1

3

2

32 2,

and

7

9

8

9, .

This is the

second step of construction. On proceeding in this wayinfinitely many steps, at the rth step we have 2 1r − open

intervals are removed each of the length 1

3r . The remaining

set constitutes cantor set or cantor-ternary set.


0 1

1

322

321

3

2

3

7

328

32

Removed

RemovedRemoved

First step

Second step

Important Theorems on Countable and Uncountable Sets

1. Every subset of a finite set is finite.

2. Every superset of an infinite set is infinite.

3. If A and B are finite sets, then A B∩ is also a finite set.4. If A and B are finite sets, then A B∪ is also a finite set.5. Every subset of a countable set is countable.

6. Every infinite subset of a denumerable set is denumerable.

7. If A and B are countable sets, then A B∩ is also a countable set.8. Every superset of an uncountable set is uncountable.

9. Every infinite set has a denumerable subset.

10. Countably infinite sets are the smallest infinite sets.

11. The countable union of countable sets is countable.

12. Every infinite set is equivalent to one of its proper subsets.13. The set of all rational numbers is countable.

14. The set of all rational numbers in [ , ]0 1 is countable.

15. The unit interval [ , ]0 1 is uncountable i.e., the set of all real numbers in the closed interval [ , ]0 1 is not enumerablei.e., the set of real numbers x s.t. 0 1≤ ≤ x is not countable.

16. The set of real numbers is uncountable.

17. The set of irrational numbers is not countable.

18. Any open interval ] , [a b is equivalent to any other open interval ] , [c d .

19. The set N N× is countable, where N is the set of natural numbers.20. The cartesian product of two countable sets is countable. i.e., If A and B are countable sets, then A B× is

countable.

21. (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.

22. Any interval is equivalent to the set R of real numbers. In particular, R is equivalent to [ , ]0 1.


8/16

Thus, the cantor set contains the points 0 1 1

3

2

3

1

9

2

9

7

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 1which have a ternary expansion without the digit 1 i.e. ternaryexpansion involves only two digits 0 or 2.

Cantor set is the complement of an open set, therefore, cantorset 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 FieldThe set R of real numbers is an ordered field.

Order Completeness of RThe set of upper bounds of a non-empty set of real numberswhich is bounded above has a smallest number or we can also

say that every non-empty set of real numbers which isbounded above admits of a least upper bound i.e.,supremum.

This property of the set R of real numbers is referred to as itsorder-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 thesmallest of the upper bounds of A can be described by thefollowing 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 realnumbers as a complete ordered field.

Note The ordered field of rational numbers is notorder-complete.

Irrational Number

A real number which is not rational is called irrational. e.g.,

3 10, etc.

Note If we have a positive real number m and anatural number n , then there exist one and only one

positive real number b s.t. b m n

= .

Important Theorems on Real Numbers

Theorem 1 A non-empty subset of real numbers whichis bounded below has the greatest lower bound (infimum)in R, the set of all real numbers.

Theorem 2 (Dedekind Property) Suppose L and U betwo 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 thesubset U has a smallest member.

Subtraction and Division in R

1. a b a b− = + −( )

2. a b a

b÷ = or a b b( ),− ≠1 0

Note The multiplicative inverse of b ≠ 0 may be

denoted as b −1 or 1

b .

Archimedean Property of RealNumbers

Theorem 1 If x and y are two given real numbers withx > 0, then ∃ a natural number n s.t. nx y > .

Theorem 2 For any real number b ∃ a positive integer ns.t. n b> .

Theorem 3 For any ε > 0 ∃ a positive integer n s.t. 1

n< ε

Theorem 4 For any real number y , ∃ two integers p andn s.t. p x n< < .

Theorem 5 For any real number x, ∃ a unique integer ns.t. n x n≤ < + 1

Theorem 6 For any real number x, ∃ a unique integer ns.t. x n x− <


9/16

Important Theorems

Denseness Property of Real Number System

Theorem 1 (The Density Theorem)Between any two distinct numbers, there lies atleast onerational number and hence there lie infinitely manyrational numbers.

Theorem 2 Between any two distinct real numbers,there lies atleast one irrational number also, and hencethere 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 asmallest member k, then k is said to be a least upper bound orsupremum of S and it is denoted by lub or Sup S.

Infimum or greatest lower bound (glb)If the set of all lower bounds of a set S of real numbers has agreatest 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 Supremumand 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 realnumbers whenever it exists, is unique.

Theorem 3 The necessary and sufficient condition for areal number ‘ ’t to be the glb or infimum of a boundedbelow set S is that ‘ ’t must satisfy the following

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 areal number ‘ ’s to be the lub or supremum of a boundedabove set S is that ‘ ’s must satisfy the followingconditions:

(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 areunique.

Greatest Member of a Set Bounded

above A number η is called the greatest member of the set S, if

(i) η η∈ ⇒S 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 Boundedbelow

A number ξ is called the smallest number of a set S, if (i) ξ ξ∈ ⇒S is itself a member of S.

(ii) No member of S is smaller than ξ i.e., ξ is a lowerbound of S

∴ x S x∈ ⇒ ≥ ξHence, the smallest member of a set is a member of the set bounded below and as well as lower bound of the set.

Note (a) The smallest member of a set, if it exists, isthe infimum or glb of the set.

(b) The greatest member of a set, if it exists, is thesupremum or lub of the set.

Bounded Set A set S of real numbers is said to be bounded if it is boundedabove as well as below. When the set S is bounded, ∃ two realnumbers 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 boundedbelow.

(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 alladherent points of A is called the adherence or closure of Aand it is denoted by Adh( ). A Adherent point is also called

closure point.



10/16

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 apoint p R∈ is a limit point of A R⊂ if and only if foreach 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 p− +ε ε contains apoint of A, other than p.

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( ) o r 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 thelimit point of A i.e., if there exist a neighbourhood of p whichcontains no point of A other than p.

Note (a) A set A is called the discrete set if all itspoints 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.

Closure of a Set

The smallest closed set containing A, is called the closure of Aand is denoted by A.

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 theclosure of A is empty i.e., ( ) A ° = φ .

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 isbounded above and has no maximum member, then itssupremum is a limit point of the set A.

Theorem 4 The finite set has no limit point.

Theorem 5 If a non-empty subset A of R, which isbounded below has no maximum member, then its

infimum is a limit point of the set A.

Theorem 6 (Bolzano-Weierstrass Theorem) Everyinfinite bounded set of real numbers has a limit point.

Theorem 7 Every bounded infinite set has the smallestand 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 aneighbourhood containing x and contained in A. Evidently, xmust belong to A.

Closed Set

If every limit point of a set is contained in the set itself, thensuch a set is called closed set. Thus, a closed set has thederived set as its subset, D A A( ) ⊆ , A is a closed set.

Open SetIf every point of set A is an interior point of A, then A is alwayscalled an open set. Open interval is always an open set.

Note (a) The cover G is called an open cover, ifevery member of G is an open set.

(b) If ∃ ⊂G G 1 s.t. G 1 is a cover of A, then G 1 is calleda subcover of G . If every member of G 1 is open s.t. G 1is finite set, then G 1 is called a finite open subcover ofG .



11/16

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 E XAMPLES

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 relationon 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 relationBut ( , )a b R∈ and ( , )b a R∉∴R is not a symmetric relationAlso, ( , ), ( , )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, thuscardinality of the set A B∆ cannot be determined.

Example 3. Assertion (A) If events, A B C , , and D aremutually exhaustive, then ( ) A B C Dc∪ ∪ =

Reason (R) ( ) A B C Dc∪ ∪ = implies if any element isexcluded from the sets A B, and C , then it is included in D.

(a) A and R are both correct, also R is the correctexplanation of A.

(b) A and R are both correct, but R is not the correctexplanation 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 correctexplanation of (A).

Q ( ) A B C A B C Dc c c c∪ ∪ = ∩ ∩ =

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 classesinto which the population is divided?

(a) 5 (b) 6

(c) 8 (d) 9

Solution. (d) Elements of a population are classified accordingto 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∪ (b) A B∩(c) A (d) B

Solution. (a)Q A X ⊆ and B X ⊆

∴ {( ( ))} 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 has56 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



12/16

Solution. (b) Let sets A and B have m and n elements,respectively.

∴ 2 2 56m n− = (According to question)⇒ 2 2 1 8 7 2 7 2 2 13 3 3n m n( ) ( ) ( )− − = × = × = −

On comparing

⇒ n = 3 and m n− = 3⇒ m = 6 and n = 3Number of subsets of A = + = =2 56 64 23 6

⇒ Number of elements in A = 6

Example 7. If f x , if x is a rational number

, if x is an irr ( ) =

1

0 ational number

what is the value of ( )( )fof 3 ?

(a) 0 (b) 1

(c) Both 0 and 1 (d) None of these

Solution. (b)Q f x , x

, x( ) =

1

0

if is a rational number

if is an irrational number

( ) { ( )}fof f f 3 3=

= f ( )0 (Q 3 is an irrational number)

=1 (Q0 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 isnot symmetric, thus relation is not equivalence.

Example 9. The function f R R: → defined by f x x( ) ( )= +2 351for all 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

Solution. (c) Since, f f ( ) ( )− = =1 1 235

i.e., Two real number 1 and −1have the same image. So,the function is not one-one and let

y x= +( )2 351

⇒ x y = −( )135 1

Thus, every real number has no pre-image. So, the functionis not onto.

Hence, the function is neither one-one nor onto.

Example 10. If A and B are disjoint sets, then A A B∩ ′∪( ) isequal to which one of the following?

(a) φ (b) A(c) A B∪ (d) A B− ′

Solution. (a)Q A B∩ = φ (given)

A A B A A A B∩ ′∪ = ∩ ′ ∪ ∩ = ∪ =( ) ( ) ( ) φ φ φ

Example 11. Let U = { , , , ..., }1 2 3 20 . Let A B C , , be thesubsets of U. Let A be the set of all numbers, which are

perfect squares, B be the set of all numbers which aremultiples of 5 and C be the set of all numbers, which aredivisible 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 (b) I and III

(c) II 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 18Q A B C ∩ ∩ = φand 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 elementsin the complement set of A B∪ is12.

Example 12. The function f x e x R x( ) ,= ∈ 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 e x R x( ) ,= ∀ ∈ isone-one but not onto. Since, range ≠ codomain, so f x( ) isinto.


x

y ′

x ′

y y = e x


13/16

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 ∈ ASolution. (d) A P = ={ , } { , { }, { }, { , }}1 2 1 2 1 2φ

From above, it is clear that

{ , }1 2 ∈ A

Example 14. Let R and S be two equivalence relations on aset 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∈ [QR 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∈ [QR 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∈

QR S

a b R b c R a c R

and

( , ) and ( , ) ( , )

(

are transitive. So

∈ ∈ ⇒ ∈a b S b c S a c S, ) and ( , ) ( , )∈ ∈ ⇒ ∈

⇒ ( , )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 1 3 5+ + + + + + + + + p q= + + + +( ... )1 3 5 r where each set of parentheses containsthe sum of consecutive odd integers as shown, what is thesmallest possible value of ( ) p q r + + where p > 6?

(a) 12 (b) 21 (c) 45 (d) 54

Solution. (b) Since, T a n d n = + −( )1 (Qnth term of an AP)

∴ p n= + −1 12( ) ⇒ n p=

+ 12

,

q n= + −1 12( ) ⇒ n q= + 12

and r n= + −1 12( ) ⇒ n r =

+12

Q Sum of n terms of an AP = × + −n a n d 2

2 1[ ( ) ]

∴

p p

q

q+

× + +

−

+

+

× + +

−

1

22

2 1 1

21 2

1

2

22 1

1

21

2

= +

× + +

−

r r 1

42 1

1

21 2

⇒ p

p q

q r

r +

+ − + +

+ − = +

+ −1

42 1

1

42 1

1

42 1[ ( )] [ ( )] [ ]

⇒ ( ) ( ) ( ) p q r + + + = +1 1 12 2 2

This the possible only when p q r = = =7 5 9, ,[Q p > 6 (given)]

∴ p q r + + = + +7 5 9= 21

Example 16. Let A x x x N= ≤ ∈{ | , }9 . Let B a b c= { , , } be thesubset of A where ( )a b c+ + is a multiple of 3. What is thelargest possible number of subsets like B?

(a) 12 (b) 21 (c) 27 (d) 30

Solution. (d) A x x x N= ≤ ∈{ : , }9

= { , , , , , , , , }1 2 3 4 5 6 7 8 9Total 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 asf x x x R( ) cos ;= ∈ is

(a) one-one only (b) onto only(c) one-one onto (d) neither one-one nor onto

Solution. (d) Given, f x x( ) cos=


f x x ( ) = cos

y

x

y ′

x ′


14/16

It is clear from the figure that f x( ) is neither one-one nor aonto function.

Since, whenever we drawn a line parallel to x-axis, then itintersects at infinite points to the curve. So, f x( ) is notone-one.

And, range of f x( ) [ , ]= −1 1Codomain of f x R( ) =Range of f x( ) ≠ codomain of f x( )∴f x( ) is not onto.

Example 18. If n A n B n A B( ) , ( ) , ( )= = − =115 326 47 , thenwhat is n A B( )∪ equal to?

(a) 373 (b) 165

(c) 370 (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 that40% family buy newspaper A, %20 buy newspaper B and 10%families buy newspaper C , %5 families buy A and B , 3% buy Band C and 4% buy A and C. If 2% families buy all the threenewspaper, then number of families which buy A only is

(a) 3100 (b) 3300

(c) 2900 (d) 1400

Solution. (b) n A( ) %= 40 of 10000 4000=n B( ) %= 20 of 10000 2000=n C ( ) %=10 of10000 1000=

n A B( ) %∩ = 5 of 1 0000 500=n B C ( ) %∩ = 3 of 1 0000 300=

n C A( ) %∩ = 4 of 1 0000 400=n A B C ( ) %∩ ∩ = 2 of 1 0000 200=

We want to find n A B C n A B C c 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 likecheese whereas 76% like apples. If x% of the Americans likeboth cheese and apples, then

(a) x = 39 (b) x = 63(c) 39 63≤ ≤ x (d) None of these

Solution. (c) Let A denote the set of Americans who likecheese and let B denote the set of Americans who likeapples.

Let population of American be 100.

Then, n A n B( ) , ( )= =63 76Now, 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 n A B( ) ( )∩ = − ∪139But n A B( )∪ ≤100∴ − ∪ ≥ −n A B( ) 100∴ 139 139 100 39− ∪ ≥ − =n A B( )

∴ n A B( )∩ ≥ 39 i.e., 39 ≤ ∩n A B( ) ...(i)Again, A B A A B B∩ ⊆ ∩ ⊆,∴ n A B n A( ) ( )∩ ≤ = 63 and n A B n B( ) ( )∩ ≤ = 76∴ n A B( )∩ ≤ 63 ...(ii)Then, 39 63≤ ∩ ≤n A B( ) ⇒ 39 63≤ ≤ x

Example 21. The set of all limit point of the set

Sm n

m n= + ∈ ∈

1 1: , is

(a) φ (b) { }0

(c) 1

mm: ∈

(d) None of these

Solution. (c) Let us fixed m ∈ . Then, for every δ > 0.1 1

m m− +

δ δ, contains infinitely many elements of .

Hence, the set of all limit point of S is 1

mm: ∈

.

Example 22. The set of all limit points of the set

Sn

n= ∈

1: is

(a) φ (b) { }0(c) N (d) None of these

Solution. (b) For every δ δ δ> −0 , ( , ) 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) A non-empty finite set is not a nbd of any point(b) Every point of a non-empty finite set is an interior point(c) No point of a non-empty finite set is an interior point(d) 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 intervalcontaining the point. Since, an interval necessarily containsan 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, nopoints 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



15/16

Solution. (a), (b), (c)Since, for the set , I and Z there is no open interval which is

contained in or I or Z as every open interval containsrational and irrational number but the set of interior points of is .

Example 25. Which of the following(s) set is/are open? (a) (b) Q (c) Z (d) φ

Solution. (a), (d)The set of real number is an open set as every real number is an interior point of . Similarly the set φ is open as the setφ is a nbd of each of its points in the sense that there is nopoint in φ of which it is not nbd . But the set and Z are notopen as no points of 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

(d) The set 1

nn: ∈

is not open

Solution. (a), (b), (c), (d)Let ( , )a b be an open interval and x a b∈( , ) which implies thata 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 intervalcontaining the point x and is therefore a nbd of x.Hence, the open interval is a nbd of each of its points and istherefore an open set.Hence, every point of an open interval is an interior point.

Now ‘ ’o is an interior point of the set 1

nn: ∈

which does

not lie in this set.

Hence, the set 1

nn: ∈

is not open.

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 isopen

(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 G2 be two open sets. Then,

if G G1 2∩ = φ, it is openif G G1 2∩ ≠ φ, let x G G∈ ∩1 2⇒ x G∈ 1 and x G∈ 2⇒ G G1 2, are nbd of x.⇒ G G1 2∩ is a nbd of x.but since x is any point of G G

1 2

∩ , therefore G G1 2

∩ is a nbd of each of its points. Hence, G G1 2∩ is open.

Now, consider the open sets

Gn n

nn = −

∈

1 1, ,

⇒ Gnn∈

=

I { }0 , which is not an open set.

⇒ The intersection of arbitrary number of open set neednot be open.

Next, let G be the union of an arbitrary family = ∈{ : }Gλ λ Λ of open sets, Λ being an index set. To provethat 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

Gλ1 of such that

x G∈ λ1Since, Gλ1 is an open set, ∃ an open interval In such that

x I G Gn∈ ⊆ ⊆λ1Thus, the set G contains an open interval containing anypoint 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 has no limit point

(c) Every point of the set Q is a limit point

(d) Every point of the set is a limit point

Solution. (a), (b), (c), (d)

The set I has no limit point, for a nbd m m− +

1

2

1

2, of m I∈ ,

contains no point of I other than m. Thus the derived set of Iis the null set φ.

The set has no limit point, for a nbd m m− +

1

2

1

2, of

m ∈, contains no points of other than m. Thus, thederived set of is null set φ. Every point of the set Q of rationals is a limit point, for between any two rationals thereexist an infinity of rational.

Further every point of is a limit point, for every nbd of anyof its points contains an infinite member of .

Example 29. If S denotes the closure of the set S , thenwhich of the following(s) is/are correct?

(a) I I=(b) Q =

(c) R =

(d) If Sn

n= ∈

1: , then S

nn=

1∈

∪: { } 0

Solution. (a), (b), (c), (d)Since, I Q′ = ′ = ′ =φ, , and S′ = { }0Hence,

I I I I I= ∪ ′ = ∪ φ =Q Q Q Q= ∪ ′ = ∪ = = ∪ ′ = ∪ =

S S Sn

n= ∪ ′ = ∈

∪

1

0: { }


a

c

x

d

b


16/16

Example 30. Which of the following(s) is/are countable? (a) The set 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 is countable,then = { , ,. .. , ,. .. .} x x xn1 2 .Enclose each member xn of in an open interval

I x xn n n n n= − +

+ +

1

2

1

21 1,

of length 1

21 2 3

n n, , , ,...=

where the sum of lengths of I sn′ is

1

2

1

2

1

2

2 3+ + + .. .

But xn ∈ and R xnn

==

∞{ }

1U ⊆

=

∞

Inn 1U

⇒ The whole real line is contained in the union of intervalswhose lengths add up to 1. Which is a contradiction. Hence, is uncountable.

Arrange the set of rationals according to increasing

denominators as 0 1 1

2

1

3

2

3

1

4

3

4

1

5

2

5

3

5

4

5

1

6

5

6, , , , , , , , , , , , ,....

Then the one-one correspondence can be indicated as

1 0↔ 5 2

3↔ 9

2

5↔

2 1↔ 6 1

4

↔ 10 3

5

↔

3 1

2↔ 7

3

4↔ 11

4

5↔

4 1

3↔ 8

1

5↔ ..........

..........which shows that the set of rational numbers in [ , ]0 1 iscountable.

Now, the set of all rational numbers is the union Aii=

∞

1U , where

Ai is the set of rationals which can be written withdenominator i. That is

Ai i i i i

i = − −

0 1 1 2 2, , , , ,...

Each Ai is equivalent to the set of all positive integers andhence countable.

Example 31. Let A = [ , , , ]1 2 3 4 and R be a relation in A givenby R = {( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )1 1 2 2 3 3 4 4 1 2 2 1 3 1 1 3 }.Then, R is

(a) reflexive (b) symmetric

(c) transitive (d) an equivalence relation

Solution. (a), (b)

( , ), ( , ), ( , ), ( , ) ;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) R x y y x x X y Y 1 2= = + ∈ ∈{( , )| , , }

(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= {( , ), ( , ), ( , ), ( , )}

Solution. (a), (b), (c)

R4

is not a relation from X to Y , because ( , )7 9 4

∈R but( , )7 9 ∉ × X Y .

Example 33. In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students whohave taken exactly one subject is

(a) 6 (b) 9

(c) 7 (d) None of these

Solution. (a), (b), (c)

n M n P n C ( ) , ( ) , ( )= = =23 24 19n M P n M C n P C ( ) , ( ) , ( )∩ = ∩ = ∩ =12 9 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


set theory and real numbers

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