Iit Jam - Real Analysis

INDEX
S. No. Contents Page No.
1. SET THEORY & COUNTABILITY 1

1.1 Sets………………………………………………………………………………………………………………….... 1
1.2 Relation……………………………………………………………………………………………………………… 2
1.3 Function…………………………………………………………………………………………………………….. 3
1.4 Classification of Functions………………………………………………………………………………….. 4
1.5 Similar/Equivalent Sets………………………………………………………………………………………. 7
1.6 Countable & Uncountable Sets…………………………………………………………………………… 8
1.7 Inequality……………………………………………………………………………………………………………. 12
2. POINT SET TOPOLOGY 16

2.1 Archimedean Property…………………………………………………………………………………………16
2.2 Bounds of A Set…………………………………………………………………………………………………… 16
2.3 Intervals……………………………………………………………………………………………………………… 18
2.4 Connectedness…………………..………………………………………………………………………………. 18
2.5 Neighbourhood of A Point……………………………………………………………………………………18
2.6 Some Important Type of Points & Its Properties…………………………………….. …………..19
3. SEQUENCES 29
3.1 Definitions…………………………………………………………………………………………………………… 29
3.2 Bounded & Monotonic Sequences……………………………………………………………………… 29
3.3 Subsequence’s of A Sequence…………………………………………………………………………….. 31
3.4 Limit of A Sequence……………………………………………………………………………………………..33
3.5 Non Convergent Sequence…………………………………………………………………………………..39
3.6 Some Important Limits……………………………………………………………………………………….. 42
3.7 Limit Superior & Limit Inferior…………………………………………………………………………….. 49
3.8 Sequence of Natural Number’s…………………………………………………………………………… 54
3.9 Sequence of Rational Numbers…………………………………………………………………………… 55
3.10 Cauchy Sequences………………………………………………………………………………………………. 55
4. SERIES OF REAL NUMBERS 64
4.1 Sequence of Partial Sum…………………………………………………………………………………….. 64
4.2 Positive Term Series…………………………………………………………………………………………… 65
4.3 Some Important Series & Their Convergence…………………………………………………….. 67
4.4 Important Test for Convergence………………………………………………………………………… 70
4.5 Alternating Term Series……………………………………………………………………………………… 77
4.6 Series of Arbitrary Terms……………………………………………………………………………………. 82
4.7 Sum of Series……………………………………………………………………………………………………… 84
5. ELEMENTARY FUNCTIONS 97
5.1 Function……………………………………………………………………………………………………………… 97
5.2 Wavy Curve Method…………………………………………………………………………………………… 98
5.3 Some Well Known Functions………………………………………………………………………………. 99
5.4 Trigonometric Functions…………………………………………………………………………………….. 100
5.5 Inverse Trignometric Function.…………………………………………………………………………… 101
5.6 Graphical Transformations ………………………………………………………………………………… 102
5.7 Algebraic & Transcendental Function…………………………………………………………………. 105
5.8 Bounded Function………………………………………………………………………………………………. 106
6. LIMIT & CONTINUITY 109

6.1 Limit of A Function……………………………………………………………………………………………… 109
6.2 Left Hand Limit & Right Hand Limit……………………………………………………………………… 111
6.3 Continuity…………………………………………………………………………………………………………… 114
6.4 Discontinuity………………………………………………………………………………………………………. 115
6.5 Cauchy Functional Equation………………………………………………………………………………… 117
6.6 Properties of Continuous Functions……………………………………………………………………. 118
6.7 Roots of A Function…………………………………………………………………………………………….. 121
6.8 Lipschitz Function……………………………………………………………………………………………….. 124
6.9 Uniform Continuity……………………………………………………………………………………………… 124

7. DIFFERENTIABILITY 138
7.1 Differentiability…………………………………………………………………………………………………… 138
7.2 Monotonic Functions & Differentiability……………………………………………………………… 143
7.3 Fixed Point………………………………………………………………………………………………………….. 144
7.4 Local Extrema of a Function…..……………………………………………………………………………. 146
7.5 Application of Differentiability……………………………………………………………………………. 147
8. RIEMANN INTEGRATION & BOUNDED VARIATION 159

8.1 Riemann Integration…………………………………………………………………………………………… 159
8.2 Definite Integral as Limit of Sum…………………………………………………………………………. 161
9. IMPROPER INTEGRAL 166
10. FUNCTION OF SEVERAL VARIABLE 171

10.1 Some Important Definitions……………………………………………………………………………….. 171
10.2 Limit of A Function……………………………………………………………………………………………… 172
10.3 Continuity…………………………………………………………………………………………………………… 176
10.4 Directional Derivatives………………………………………………………………………………………… 178
10.5 Differentiability…………………………………………………………………………………………………… 180
10.6 Partial Derivatives of Higher Order …………………………………………………………………….. 183
10.7 Extreme Values…………………………………………………………………………………………………… 184
10.8 Function of 𝑛 Variables……………………………………………………………………………………….. 186
10.9 Vector Valued Function………………………………………………………………………………………. 186

1
CHAPTER
SET THEORY & COUNTABILITY
SET THEORY Then 𝐴 is called subset of 𝐵.

1.1 SETS : OR,
1.1.1 DEFINITION: ∄ 𝑥 ∈ 𝐴 such that 𝑥 ∉ 𝐵.
A Set is a well-defined collection of distinct objects. Notation:
The individual objects of the set are called member or (1) If 𝐴 is subset of 𝐵, then it is denoted by 𝐴 ⊆ 𝐵(𝐵 is
elements of the set. superset of 𝐴)
Set is denoted by Capital letters 𝐴, 𝐵, 𝐶, 𝑋, 𝑌, 𝑍 … (2) If 𝐴 is proper subset of 𝐵, then it is denoted by 𝐴 ⊊ 𝐵 (i.e.
Examples :- ∃ 𝑥 ∈ 𝐵 such that 𝑥 ∉ 𝐴).
(1) Set of all vowels in English Alphabet X = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢}.
(2) Some Important sets : ℕ = set of all natural numbers Example :
ℤ = set of all integers 1) If 𝑥 ∈ 𝐴 ⇒ 𝑥 ∈ 𝐴
ℝ = set of all real numbers ⇒ 𝐴 is subset of 𝐴 i.e 𝐴 ⊆ 𝐴
ℂ = set of all complex numbers ∴ Every set is a subset of itself.
ℚ = set of all rational numbers 2) ϕ (empty set) is a subset of 𝑋 i.e. ϕ ⊆ 𝑋.
ℚ𝑐 = set of all irrational numbers
(3) The set of all natural numbers which are strictly less than 4. 1.1.4 Power Set :
𝑋 = {𝑥 ∈ ℕ | 𝑥 < 4} = {1,2,3} The collection of all subsets of a set 𝐴 is called Power set of 𝐴.
It is denoted by 𝑃(𝐴).
Note:
(1) Set guarantees that the Collection will contains distinct Example :
elements but collection doesn't guarantee this. (1) 𝐴 = {1,2,3}
(2) The word 'Well-defined' means there is no ambiguity or 𝑃(𝐴) = {𝜙, A, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}}.
confusion as to whether an object belongs to collection or Note:- for any set 𝐴
not. (1) 𝜙 ∈ P(A)
1.1.2 Equality of Two sets : (2) 𝜙 is element as well as subset of P(A).
Two sets are said to be equal when they consist of exactly the
same elements. 1.1.5 Cartesian Product :
(i) {𝑎, 𝑏, 𝑐} = {𝑏, 𝑐, 𝑎} = {𝑐, 𝑏, 𝑎} Let 𝐴 & 𝐵 are two sets. Then Cartesian product of 𝐴 & 𝐵 is
(ii) {𝑥 ∈ ℕ | 𝑥 < 4} = {𝑥 ∈ ℕ|𝑥 ≤ 3} = {1,2,3} denoted by 𝐴 × 𝐵 and it is defined as
Cardinality of the set : 𝐴 × 𝐵 = {(𝑎, 𝑏)| 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵} , where (𝑎, 𝑏) is an order pair.
The number of elements in the set 𝐴 is called cardinality of the
set 𝐴 Example:-
Cardinality of the set 𝐴 is denoted by |𝐴|. (1) 𝐴 = {1,2}, 𝐵 = {3,4}
Russel’s Paradox : 𝐴 × 𝐵 = {(1,3), (1,4), (2,3), (2,4)}
Let 𝑅 be the collection of all sets that are not members of (2) 𝐴 = {1,2}, 𝐵 = 𝜙, then
themselves. 𝐴 ×𝐵=𝜙
𝑅 = {𝐴| 𝐴 ∉ 𝐴}
1.1.6 Union & Intersection of two sets :
Exercise: Show that 𝑅 ∉ 𝑅 ⇔ 𝑅 ∈ 𝑅. (1) Union :
If 𝐴 & 𝐵 are two sets, then the set consisting of all elements
1.1.3 Subset : which belongs to either 𝐴 or 𝐵 is defined as union of 𝐴 & 𝐵.
If 𝐴 & 𝐵 are two sets such that each member of 𝐴 is also a It is denoted by 𝐴 ∪ 𝐵
member of 𝐵 𝐴 ∪ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵}
i.e. 𝑥 ∈ 𝐴 ⇒ 𝑥 ∈ 𝐵
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(2) Intersection : Example :
If 𝐴 & 𝐵 are two sets, then the set consisting of all elements 𝐴 = {1, 2, 3} & 𝑅 = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1)} is
which belongs to both 𝐴 & 𝐵 is defined as intersection of symmetric relation.
𝐴&𝐵
It is denoted by 𝐴 ∩ 𝐵 3) Transitive Relation :
𝐴 ∩ 𝐵 = {𝑥| 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵} Let 𝑅 be a binary relation on a set 𝐴 i.e. 𝑅 ⊆ 𝐴 × 𝐴.
Note: Then R is called a transitive relation if
(1) Two Set 𝐴 & 𝐵 are said to be disjoint if they have no (𝑎, 𝑏) & (𝑏, 𝑐) ∈ 𝑅 then (𝑎, 𝑐) ∈ 𝑅
common elements. Example :
i.e. 𝐴 ∩ 𝐵 = 𝜙 𝐴1 = {1, 2, 3} , 𝑅1 = {(1, 1), (1, 2), (2, 3), (1, 3)} is
(2) 𝐴 ∪ 𝜙 = 𝐴 , 𝐴 ∪ 𝐴 = 𝐴 , 𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴 transitive relation.
𝐴 ∩𝜙 = 𝜙, 𝐴 ∩𝐴 =𝐴, 𝐴 ∩𝐵 =𝐵∩𝐴.
Equivalence Relation :
1.1.7 Union & Intersection of an arbitrary family of sets : A binary relation 𝑅 ⊆ 𝐴 × 𝐴 is called equivalence relation if it
Index Set : is reflexive, symmetric & transitive.
An index set is a set whose members labels (or index) members
of another set. It is denoted by ⋀ 1.2.2 Equivalence class :
If 𝑅 ⊆ 𝐴 × 𝐴 is an equivalence relation. Then equivalence class
Arbitrary Union & Intersection of 𝐴 with respect to 𝑅 is denoted by [𝑎] & is defined as
Let ⋀ be an index set, then [𝑎] = {𝑥 ∈ 𝐴| (𝑎, 𝑥) ∈ 𝑅}
⋃𝑖 ∈⋀ 𝐴𝑛 = {𝑥| 𝑥 ∈ 𝐴𝑖 , for some 𝑖 ∈ ⋀}
Result :-
and
(1) Two equivalence class are either equal or disjoint.
⋂𝑖 ∈⋀ 𝐴𝑛 = {𝑥| 𝑥 ∈ 𝐴𝑖 , for each 𝑖 ∈ ⋀}
(2) Fundamental Theorems of equivalence relation.
An equivalence relation on a set 𝐴 form a partition of 𝐴
1.2 Relations :
conversely, Corresponding to any partition of 𝐴, there
1.2.1 Definition:
exist an equivalence relation on 𝐴.
If 𝐴 & 𝐵 be two non-empty sets, then a relation from 𝐴 to 𝐵 is
(3) The number of equivalence relation on a set 𝐴 is equals to
a subset of 𝐴 × 𝐵 & every subset of 𝐴 × 𝐵 is a relation from 𝐴
the number of partition of a set 𝐴.
to 𝐵
A relation from 𝐴 to 𝐴 is called binary relation.
1.2.3 Number of Equivalence relations :
Let 𝐴 be any set with cardinality |𝐴|, then
Examples :
|𝐴| Number of equivalence relation
1) Null relation 𝜙 ⊆ 𝐴 × 𝐵
0 1
2) 𝐴 = {1, 2}, & 𝐵 = {0, 1, 2}, then 1 1
𝑅1 = {(1, 1), (2, 2)} is a relation. 2 2
𝑅2 = {(1, 0), (1, 2), (2, 1), (2, 2)} is a relation 3 5
𝑅3 = {(1, 0), (0, 1)} is not a relation. 4 15
5 52
1) Reflexive Relation :
Let 𝑅 be a binary relation on a set 𝐴 i.e. 𝑅 ⊆ 𝐴 × 𝐴. Definition :
Then R is called a reflexive relation if Let 𝑆 = 𝜙, & 𝑅 ⊆ 𝑆 × 𝑆 be a relation on 𝑆, then 𝑅 is null
(𝑎, 𝑎) ∈ 𝑅, ∀ 𝑎 ∈ 𝐴 relation & it is an equivalence relation.
Example :
Anti-symmetric Relation:
𝐴1 = {1, 2, 3} & 𝑅1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
𝐴2 = {𝑎, 𝑏} & 𝑅2 = {(𝑎, 𝑎), (𝑏, 𝑏), (𝑎, 𝑏), (𝑏, 𝑎)} A binary relation 𝑅 is anti-symmetric relation if
Both are reflexive relation. (𝑎, 𝑏) ∈ 𝑅 & (𝑏, 𝑎) ∈ 𝑅 ⇒ 𝑎 = 𝑏
1.2.4 Partial order Relation :

2) Symmetric Relation:
A relation 𝑅 on a set 𝐴 is said to be partial order relation if it is
Let 𝑅 be a binary relation on a set 𝐴 i.e. 𝑅 ⊆ 𝐴 × 𝐴.
(a) Reflexive
Then R is called a symmetric relation if
(b) Anti symmetric
(𝑎, 𝑏) ∈ 𝑅 ⇒ (𝑏, 𝑎) ∈ 𝑅
(c) Transitive
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Chapter – 1 Set Theory & Countability 3
Example: (f) Partial order relation
Let 𝑥, 𝑦 ∈ ℝ, define 𝑥𝑅𝑦 iff 𝑥 − 𝑦 + √2 ∈ ℚ . Then 𝑅 is 𝑐 (If 𝑅 is an equivalence relation, find the equivalence classes)
(a) Reflexive relation
(b) Symmetric relation 1.3 Function:
(c) Transitive relation 1.3.1 Definition:
(d) Equivalence relation Let 𝐴 & 𝐵 be any two non-empty sets. Then a rule which assign
(e) Anti-symmetric relation each member 𝑥 of 𝐴 to unique member 𝑦 of 𝐵 is called a
(f) Partial order relation function and 𝑦 is the image of 𝑥 under 𝑓.
Solution:
(a) If 𝑦 = 𝑥 then 𝑥 − 𝑦 + √2 = 𝑥 − 𝑥 + √2 = √2 ∈ ℚ𝑐 Notation:
⇒ 𝑥 − 𝑦 + √2 ∈ ℚ𝑐 𝑓 ∶ 𝐴 → 𝐵 i.e. 𝑓 is a mapping or a function from 𝐴 to 𝐵.
⇒𝑥𝑅𝑥
⇒ 𝑅 is a reflexive relation. Note :
(i) A function is a special type of relation in which each element
(b) 𝑥 = 1 + √2, 𝑦 = 1 of 𝐴 is pair with exactly one element of 𝐵.
𝑥 − 𝑦 + √2 = 1 + √2 − 1 + √2 = 2√2 ∈ ℚ𝑐 i.e. if (𝑎, 𝑏) ∈ 𝑓 ⇒ 𝑓(𝑎) = 𝑏 ,
⇒ 𝑥𝑅𝑦
Consider 𝑦 − 𝑥 + √2 = 1 − (1 + √2) + √2 = 1 − 1 − 1.3.2 Definition:
Domain, Co-domain & Range of :
√2 + √2 = 0 ∉ ℚ𝑐
Let 𝑓 ∶ 𝐴 → 𝐵 be any function from 𝐴 to 𝐵 then
⇒ 𝑦 is not related to 𝑥
i) The set 𝐴 is called the domain of the function 𝑓.
⇒ 𝑅 is not a symmetric relation.
ii) The set 𝐵 is called the co-domain of the function 𝑓.
−√2 √2 ii) The set of all elements in 𝐵 which are the image of the
(c) 𝑥 = , 𝑦 = 1, 𝑧 =
2 2 elements of the domain A is called the range set of the
−√2 −√2−2+2√2 −2+√2
𝑥 − 𝑦 + √2 = − 1 + √2 = = ∈ ℚ𝑐 function f range of 𝑓 = 𝑅(𝑓) = {𝑦 ∈ 𝐵 | 𝑦 =
2 2 2
√2 2−√2+2√2 2+√2 𝑓(𝑥) for some 𝑥 ∈ 𝐴}
𝑦 − 𝑧 + √2 = 1 − 2
+ √2 = 2
= 2
∈ ℚ𝑐
⇒ 𝑥𝑅𝑦 & 𝑦𝑅𝑧
−√2 √2 2√2
But 𝑥 − 𝑧 + √2 = − + √2 = − + √2 =
2 2 2
𝑐
−√2 + √2 = 0 ∉ ℚ
⇒ 𝑅 is not a transitive relation
(d) Clearly 𝑅 is not an equivalence relation
(e) 𝑥 = √2, 𝑦 = √3
𝑥 − 𝑦 + √2 = √2 − √3 + √2 = 2√2 − √3 ∈ ℚ𝑐
⇒ 𝑥𝑅𝑦
𝑦 − 𝑥 + √2 = √3 − √2 + √2 = √3 ∈ ℚ𝑐
⇒ 𝑦𝑅𝑥 1.3.3 Counting Number of functions from 𝑨 to 𝑩 :
But 𝑥 ≠ 𝑦 Let |𝐴| = 𝑚 & |𝐵 | = 𝑛, then the number of functions from 𝐴
⇒ 𝑅 is not an anti-symmetric relation to 𝐵 are |𝐵 ||𝐴| = 𝑛𝑚
(f) Clearly, 𝑅 is not a partial order relation. Example :

Q. If |𝑨| = 𝟑 & |𝑩| = 𝟓, then find number of functions from
EXERCISE 𝑨 to 𝑩.
Let 𝑥, 𝑦 ∈ ℤ. Define 𝑥𝑅𝑦 iff 𝑥 = 𝑦 (mod 3). Then 𝑅 is Solution :
(a) Reflexive relation If |𝐴| = 𝑚, |𝐵 | = 𝑛 &
(b) Symmetric relation 𝐹 = {𝑓|𝑓 ∶ 𝐴 → 𝐵} then
(c) Transitive relation |𝐹 | = 𝑛𝑚
(d) Equivalence relation Here 𝑚 = 3 and 𝑛 = 5
(e) Anti-symmetric relation ∴ |𝐹 | = 𝑛𝑚 = 53 = 125
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Q. Let |𝑨| = 𝒎 & |𝑩| = 𝒏 defline Example :-
𝑭𝟏 = {𝒇|𝒇 ∶ 𝑨 → 𝑩 𝐢𝐬 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧} & Q. Let 𝑨 = {𝟏, 𝟐, 𝟑} & 𝑩 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓}
𝑭𝟐 = {𝒇|𝒇 ∶ 𝑩 → 𝑨 𝐢𝐬 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧}. Then total number of suppose 𝑭𝟏 = { 𝒇|𝒇 ∶ 𝑨 →
functions form 𝑭𝟐 to 𝑭𝟏 are 𝑩 𝐛𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝒇(𝟏) = 𝟏 & 𝒇(𝟐) ≠ 𝟐 } & 𝑭𝟐 =
𝑚 𝑛+1
(a) 𝑚𝑛 (b) 𝑛𝑚 {𝒇|𝒇 ∶ 𝑨 → 𝑩 𝐛𝐞 𝐚 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝒇(𝟏) =
(c) 𝑚𝑛
𝑚+1
(d) 𝑛𝑚
𝑛 𝟏 & 𝒇(𝟐) ≠ 𝟐 } then the cardinality of sets 𝑭𝟏 & 𝑭𝟐
Solution: Here |𝐴| = 𝑚, |𝐵 | = 𝑛 respectively are ?
(1) 9, 9 (2) 20, 10
∴ |𝐹1 | = |𝐵 ||𝐴| = 𝑛𝑚
|𝐹2 | = |𝐴||𝐵| = 𝑚𝑛 (3) 20, 9 (4) 18, 9
𝐹 = {𝑓|𝑓 ∶ 𝐹2 → 𝐹1 𝑏𝑒 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛} Solution:- Given |𝐴| = 3 & |𝐵 | = 5
Then (i) Given 𝑓(1) = 1 & 𝑓(2) ≠ 2
𝑛
|𝐹 | = |𝐹1 ||𝐹2| = (𝑛𝑚 )(𝑚 ) i.e. image of element 1 is fixed
𝑛 𝑛+1 𝑓(2) have 4 choices (i.e. 1, 3, 4, 5)
= 𝑛𝑚×𝑚 = 𝑛𝑚
𝑛+1 𝑓(3) have 5 choices (i.e. 1, 2, 3, 4, 5)
|𝐹 | = 𝑛𝑚
𝑛+1 Hence, total number of functions in 𝐹1 are 1 × 4 × 5 = 20
∴ Number of functions from 𝐹2 to 𝐹1 are 𝑛𝑚
(ii) Here 𝑛 = |𝐵 | = 5 > 3 = 𝑚 = |𝐴|
1.4 Classification of function:- 𝑓(1) = 1 & 𝑓(2) ≠ 2
On the basis of mapping of elements, there are five types of ∴ 𝑓(2) has three choices (i.e. 3,4,5)
functions & 𝑓(3) has three choices of other then 1 and 𝑓 (2).
(1) one to one function (Injective function) So,
(2) Many to one function Total number of functions in set 𝐹2 is 1 × 3 × 3 = 9
(3) Onto function (Surjective function) Hence, Option (3) is correct option
(4) Into function
(5) One-one & onto function (Bijective function or Invertible Q. The number of one-one function from set 𝑨 to set 𝑩 where
function) 𝑨 = {𝟎, 𝟏, 𝟐, 𝟑} & 𝑩 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓} are
(a) 60 (b) 40
1.4.1 One-one function (Injective function): (c) 100 (d) 120
A function 𝑓 ∶ 𝐴 → 𝐵 is said to be one-one function, if two Solution:- Here 𝐴 = 𝑚 = 4 and |𝐵 | = 𝑛 = 5
| |
distinct elements of the domain have distinct images in the co- ⇒𝑚≤𝑛
domain. ∴ Total number of one-one functions from 𝐴 to 𝐵 is 𝑛𝑝𝑚
i.e. f is one-one iff for any 𝑥1 , 𝑥2 ∈ 𝐴 if 𝑥1 ≠ 𝑥2 ⇒ 𝑓(𝑥1 ) ≠ 5! 5!
= 5𝑝4 = (5−4)! = 1! = 5! = 120
𝑓(𝑥2 )
OR
If 𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⇒ 𝑥1 = 𝑥2 1.4.2 Many-one function:
A function 𝑓: 𝐴 → 𝐵 is said to be many to one function, if
there is at least two elements in the domain 𝐴 has same
image in 𝐵.
Note: A function that is not one-one is called to be many-one

function.
Counting number of many to one function:

If |𝐴| = 𝑚 & |𝐵 | = 𝑛 then
Total number of many-one function =
Counting number of one-one function from 𝑨 to 𝑩: Total number of functions – Total number of one-one
If |𝐴| = 𝑚 & |𝐵 | = 𝑛. then total number of one-one functions function
If 𝑚 > 𝑛, then
from 𝐴 to 𝐵 are
𝑛
Number of many to one function = 𝑛𝑚 − 𝑜 = 𝑛𝑚
𝑝 𝑚≤𝑛 𝑛!
={ 𝑚 where 𝑛𝑝𝑚 = (𝑛−𝑚)! If 𝑚 ≤ 𝑛, then
0 𝑚>𝑛 Number of many to one function = 𝑛𝑚 − 𝑛𝑃𝑚
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1.4.3 Onto (surjective) function: Note:
A function 𝑓: 𝐴 → 𝐵 is said to be onto if each element of 𝐵 (1) If 𝑓: 𝐴 → 𝐵 be a bijective function
has atleast one preimage in 𝐴. ⇒ (i) 𝑓 is one-one ⇒ |𝐴| ≤ |𝐵 |
& (ii) 𝑓 is onto ⇒ |𝐴| ≥ |𝐵|
Preimage: If 𝑓 is function such that 𝑦 = 𝑓(𝑥). Then we say ∴ If 𝑓: 𝐴 → 𝐵 be a bijective function then |𝐴| = |𝐵|
(i) 𝑦 is image of 𝑥 under 𝑓
(ii) 𝑥 is preimage of 𝑦 under 𝑓 Converse need not be true.
i.e. if |𝐴| = |𝐵 | ⇏ 𝑓: 𝐴 → 𝐵 is a bijective function
Note: 𝑓: 𝐴 → 𝐵 is said to be onto if ∀ 𝑦 ∈ 𝐵, ∃ 𝑥 ∈ 𝐴 such
that Example: If 𝑓: [0, 1] → [0, 1] such that (𝑥) = 1 ∀𝑥 ∈ 𝐴.
𝑓(𝑥) = 𝑦 ∄ any 𝑥 ∈ [0, 1] such that 𝑓(𝑥) = 0
⇒ 𝑓 is not onto function.
Counting number of onto functions form 𝑨 to 𝑩: ⇒ 𝑓 is not a bijective function.
1. If |𝐴| = 𝑚 & |𝐵 | = 𝑛, then the number of onto functions
from 𝐴 to 𝐵 is 2. If 𝑓: 𝐴 → 𝐵 be a function where |𝐴| = |𝐵 | = 𝑛 then 𝑓 is
𝑛𝑚 − [∑𝑛𝑟=1(−1)𝑟+1 𝑛𝐶𝑟 (𝑛 − 𝑟)𝑚 ], 𝑚 ≥ 𝑛 one-one iff 𝑓 is onto iff 𝑓 is bijective
={
0 𝑚<𝑛 The above statement is not necessarily true when the
Note: cardinality of 𝐴 & 𝐵 are infinite.
(i) If range set is proper subset of 𝐵 then 𝑓 cannot be an
onto function. Ex. 𝑓: ℕ → ℕ such that 𝑓(𝑛) = 𝑛 + 1
(ii) If 𝑅(𝑓) = 𝐵, then 𝑓 is onto Here |ℕ| = |ℕ| but f is not onto
∵ ∄ any 𝑛 ∈ ℕ such that 𝑓(𝑛) = 1
Examples:
The number of onto functions (surjective functions) from set Counting Number of bijective functions:
𝑿 = {𝟏, 𝟐, 𝟑, 𝟒} (i) If |𝐴| = |𝐵 | = 𝑛, then number of bijective functions from
to set 𝒀 = {𝒂, 𝒃, 𝒄} is A to B is equals to 𝑛!
(a) 36 (b) 64 (ii) If |𝐴| = |𝐵 | = 𝑛, then number of functions from 𝐴 to 𝐵
(c) 81 (d) 72 which is neither one-one nor onto is equals to 𝑛𝑛 − 𝑛!
Solution: |𝑋| = 𝑚 = 4, |𝑌| = 𝑛 = 3
∴ total number of onto functions from 𝑋 to 𝑌 Example:
= 𝑛𝑚 − 𝑛𝐶1 (𝑛 − 1)𝑚 + 𝑛𝐶2 (𝑛 − 2)𝑚 − 𝑛𝐶3 (𝑛 − 3)𝑚 + ⋯ If 𝐴 = {1,2,3.4,5} & 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}. Then
= 34 − 3𝐶1 (3 − 1)4 + 3𝐶2 (3 − 2)4 − 3𝐶3(3 − 3)4 + ⋯ (a) The number of bijective functions
= 81 − 3(16) + 3(1)4 = 81 − 48 + 3 = 36 Answer: 5!
Total number of onto functions are 36 (b) The number of function which is one-on e but not onto
Hence, option (a) is correct. Answer: 0
(c) The number of function which is onto but not noe-one
1.4.4 Into function: Answer: 0
A function 𝑓: 𝐴 → 𝐵 is said to be into, if there exist an (d) The number of function which is neither one-one nor
element in 𝐵 which does not have preimage in 𝐴. onto
i.e. 𝑓(𝐴) ⊊ 𝐵 55 − 5!
Note:
NOTE: If 𝑓: ℝ → ℝ is a function then
(i) A function which is not onto is an into function.
(1) 𝑓 is not one-one if graph of 𝑓 cuts some line parallel to
(ii) Number of into functions + number of onto functions =
Total number of functions 𝑥 −axis at least two points.
(2) 𝑓 is one-one if graph of 𝑓 cuts every lines parallel to
Important Note: 𝑥 −axis at most one point.
(i) If |𝐴| = 𝑚 & |𝐵 | = 𝑛, then total number of constant (3) 𝑓 is not onto if graph of 𝑓 cuts some lines parallel to
functions are 𝑛. 𝑥 −axis at no point.
(ii) If |𝐴| = 𝑚 & |𝐵 | = 2, then any function from 𝐴 to 𝐵 is (4) 𝑓 is onto if graph of 𝑓 cuts every lines parallel to 𝑥 −axis
into iff it is constant function.
at atleast one point.
⇒ Total number of into function from 𝐴 to 𝐵 is 2
Total number of functions from 𝐴 to 𝐵 is 2𝑚
∴ Total number of onto functions from 𝐴 to 𝐵 is 2𝑚 − 2. Examples:
Q. If the functions 𝒇, 𝒈: ℤ → ℤ defined as (𝒏) = 𝟑𝒏 +
1.4.5 One to one and onto OR Bijective OR invertible 𝟐 & 𝒈(𝒏) = 𝒏𝟐 − 𝟓 , then
function: (1) Neither 𝑓 nor 𝑔 is one-one function
A function 𝑓: 𝐴 → 𝐵 is bijective if it is one-one function and
(2) The function 𝑓 is one-one but not 𝑔
onto function.
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(3) The function 𝑔 is one-one but not 𝑓
(4) Both 𝑓 & 𝑔 are one-one functions
Solution:-
(i) 𝑓(𝑛) = 3𝑛 + 2
Let 𝑛1 , 𝑛2 𝜖 𝕫 such that 𝑓( 𝑛1 ) = 𝑓(𝑛2 )
⇒ 3𝑛1 + 2 = 3𝑛2 + 2
⇒ 3𝑛1 = 3𝑛2
⇒ 𝑛1 = 𝑛2
⇒ 𝑓 is one-one function
(ii) 𝑔(𝑛) = 𝑛2 − 5 For any 𝑦 ∈ (−∞, 1), ∄ any 𝑥 ∈ ℝ such that 𝑓(𝑥) = 𝑦
Here for any 𝑛 ∈ ℤ Thus, 𝑓 is not onto function.
𝑔(𝑛) = 𝑛2 − 5 & 𝑔(−𝑛) = (−𝑛)2 − 5 = 𝑛2 − 5
But 𝑛 ≠ −𝑛 Q. Let ℤ denote the set of integers and ℤ≥𝟎 denotes the set
⇒ 𝑔 is not one-one function {𝟎, 𝟏, 𝟐, 𝟑, … . }. Consider the map 𝒇: ℤ≥𝟎 × ℤ → ℤ Given
Hence, option (2) is correct by 𝒇(𝒎, 𝒏) = 𝟐𝒎 . (𝟐𝒏 + 𝟏). Then the map 𝒇 is
(a) Onto (surjective) but not one-one (injective)
Q. Let 𝒇: 𝑿 → 𝑿 such that 𝒇(𝒇(𝒙)) = 𝒙 for all 𝑿 ∈ 𝑿. Then (b) One-one (injective) but not onto (surjective)
(1) 𝑓 is one-one and onto (c) Both one- one and onto
(2) 𝑓 is one-one but not onto (d) Neither one-one nor onto
(3) 𝑓 is onto but not one-one Solution :
(4) 𝑓 need not be either one-one or onto Given, 𝑓: ℤ≥0 × ℤ ⟶ ℤ defined by 𝑓((𝑚, 𝑛)) = 2𝑚 (2𝑛 +
Solution:- for any 𝑥 ∈ 𝑋, 𝑓(𝑓(𝑥)) = 𝑥 1)
Let 𝑥1 , 𝑥2 ∈ 𝑋 such that 𝑓(𝑥1 ) = 𝑓(𝑥2 ) Let (𝑚1 , 𝑛1 ), (𝑚2 , 𝑛2 ) ∈ ℤ≥0 × ℤ
Apply 𝑓 on above equation Then, 𝑓((𝑚1 , 𝑛1 )) = 𝑓((𝑚2 , 𝑛2 ))
⇒ 𝑓(𝑓(𝑥1 )) = 𝑓(𝑓(𝑥2 )) (∵ 𝑓 is a function) ⇒ 2𝑚1 (2𝑛1 + 1) = 2𝑚2 (2𝑛2 + 1)
⇒ If 𝑚1 = 𝑚2 ⇒ 2𝑛1 + 1 = 2𝑛2 + 1 ⇒ 𝑛1 = 𝑛2
⇒ 𝑥1 = 𝑥2
⇒ 𝑓 is one-one function If 𝑚1 ≠ 𝑚2 , Then, without loss of generality we can
Onto : consider that 𝑚1 > 𝑚2
(i) for any 𝑥 ∈ 𝑋 (codomain), there is an element ⇒ 2𝑚1−𝑚2 (2𝑛1 + 1) = (2𝑛2 + 1),
Which is a contradiction because term on left hand side is
𝑓(𝑥) ∈ 𝑋 such that 𝑓(𝑓(𝑥)) = 𝑥
a multiple of 2 i.e even number and term on right hand
i.e. for any 𝑖 ∈ 𝑋 (Codomain), 𝑓(𝑖) is a Preimage of 𝑖
side is an odd number.
Hence, 𝑓 is onto function
So, 𝑚1 = 𝑚2 and 𝑛1 = 𝑛2
OR,
∴ 𝑓 is an one-one function.
If 𝑓(𝑔(𝑥)) = 𝑥 ∀ 𝑥 ∈ 𝑋 then 𝑓 is onto & g is one-one
As, 0 ∈ ℤ and 𝑓(𝑚, 𝑛) = 2𝑚 (2𝑛 + 1) ≠ 0 ∀ (𝑚, 𝑛) ∈
function.
ℤ≥0 × ℤ
Consider 𝑓(𝑓(𝑥)) = 𝑥 1
(∵ 2𝑚 (2𝑛 + 1) = 0 ⇒ 𝑛 = − ∉ ℤ)
⇒ 𝑓 is onto & 𝑓 is one-one function 2
Hence 𝑓 is bijective map. ⇒ 𝑓 is not an onto function.

Thus, option (1) is correct. Hence option (b) is correct
Q. The function 𝒇: ℝ → ℝ defined by 𝒇(𝒙) = (𝒙𝟐 + 𝟏)𝟐𝟎𝟐𝟏 is Q. Let 𝑨 = {𝒙𝟐 : 𝟎 < 𝒙 < 𝟏} and 𝑩 = {𝒙𝟑 : 𝟏 < 𝒙 < 𝟐} Which
(1) Onto but not one-one of the following statements is true?
(2) One-one but not onto (a) There is a one to one, onto function from 𝐴 to 𝐵
(3) Both one-one & onto (b) There is no one to one, onto function from 𝐴 to 𝐵
(4) Neither one-one nor onto taking rational to rational.
Solution:- Here 𝑓(𝑥) = (𝑥 2 + 1)2021 (c) there is no one to one function from 𝐴 to 𝐵 which is
𝑓(−𝑥) = ((−𝑥)2 + 1)2021 but 𝑥 ≠ −𝑥 onto
So, 𝑓 is not one-one function (d) There is no onto function from 𝐴 to 𝐵 Which is one to
one
Consider ∀𝑥 ∈ ℝ, 𝑥 2 + 1 ≥ 1
⇒ (𝑥 2 + 1)2021 ≥ (1)2021 = 1
Hence
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Solution: 𝐴 = {𝑥 2 : 0 < 𝑥 < 1} and 𝐵 = {𝑥 3 : 1 < 𝑥 < 2}
𝐴 and 𝐵 are open intervals and in metric space (𝑅, 𝑑) open
intervals are homeomorphic
So, there exists a homeomorphism of 𝐴 onto 𝐵.
∴ there is one-one and onto function from 𝐴 to 𝐵.
∴ option (a) is correct.
(b)
Q. How many onto function from set A = { Let 𝑋 = { 𝑎, 𝑏, 𝑐, 𝑑 } , 𝑌 = { 1,4 } and let 𝑓: 𝑋 → 𝑌 such that
1,2,3,4,5,6,7,8,9,10} to the set B = { a, b } 𝑓( 𝑎) = 1 , 𝑓(𝑏) = 1 , 𝑓( 𝑐) = 4 , 𝑓( 𝑑) = 4
(a) 1024 (b) 1022 𝐴 = {𝑎, 𝑐} ⇒ 𝑓(𝐴) = {1,4}
(c) 100 (d) 98 𝐵 = {𝑏, 𝑑} ⇒ 𝑓(𝐵) = {1, 4}
Answer : (2) ⇒ 𝑓(𝐴) ∩ 𝑓(𝐵) = {1,4}
But 𝐴 ∩ 𝐵 = 𝜙
⇒ 𝑓(𝐴 ∩ 𝐵) = 𝜙
Q. Let 𝒇(𝒏) = 𝒏𝟑 − 𝟑𝒏 defined on set of integers is
⇒ 𝑓(𝐴 ∩ 𝐵) ≠ 𝑓(𝐴) ∩ 𝑓(𝐵)
injective function. [ T/F]
So, Option (b) is incorrect.
Answer : False
If 𝑥 ∈ 𝐴 ∩ 𝐵 ⇒ 𝑥 ∈ 𝐴 & 𝑥 ∈ 𝐵 ⇒ 𝑓(𝑥) ∈ 𝑓(𝐴) &𝑓(𝑥) ∈ 𝑓(𝐵)
𝑓(𝑥) ∈ 𝑓(𝐴 ∩ 𝐵)
Note:
⇒ 𝑓(𝐴 ∩ 𝐵) ⊆ 𝑓(𝐴) ∩ 𝑓(𝐵)
(1) If 𝑓: 𝐴 → 𝐵 is an invertible function then 𝑓 −1 : 𝐵 → 𝐴 is
Hence, option (c) is correct
also invertible function where 𝑓 −1 (𝑦) = 𝑥, whenever
𝑓(𝑥) = 𝑦
Let 𝑋 = { 𝑎, 𝑏, 𝑐, 𝑑 } , 𝑌 = { 1,4 }
(2) Let 𝑓: 𝐴 → 𝐵 is bijective function then
𝐴 = {𝑎, 𝑏} and 𝐵 = {𝑐, 𝑑} and let 𝑓: 𝑋 → 𝑌 such that 𝑓( 𝑥) =
(i) 𝑓 ∘ 𝑓 −1 = 𝐼 where 𝑓 ∘ 𝑓 −1 : 𝐵 → 𝐵
1 ∀𝑥 ∈ {𝑎, 𝑏, 𝑐, 𝑑}
(ii) 𝑓 −1 ∘ 𝑓 = 𝐼 where 𝑓 −1 ∘ 𝑓: 𝐴 → 𝐴
Then
𝑓(𝐴) = {1} and 𝑓(𝐵) = {1}
Result:
Therefore, 𝑓(𝐴\𝐵) = 𝒇({𝑎, 𝑏}{𝑐, 𝑑}) = 𝒇({𝑎, 𝑏}) = {𝟏}
1. If 𝑓: 𝑋 → 𝑌 is one-one function & 𝐴, 𝐵 ⊆ 𝑋, then
And 𝑓(𝐴)\(𝐵) = {1}\{1} = 𝜙
(i) 𝑓(𝐴 ∪ 𝐵) = 𝑓(𝐴) ∪ 𝑓(𝐵)
Hence, option (d) is incorrect
(ii) 𝑓(𝐴 ∩ 𝐵) = 𝑓(𝐴) ∩ 𝑓(𝐵)
1.4.6 Cantor’s Theorem:

2. If 𝑓: 𝑋 → 𝑌 be a function such that 𝑓 −1 exist & 𝐴, 𝐵 ⊆ 𝑌
There is no onto map from any non-empty set to its power set.
then
Note:- There exist a one-one map from set 𝐴 to 𝑃(𝐴)
(i) 𝑓 −1 (𝐴 ∪ 𝐵) = 𝑓 −1 (𝐴) ∪ 𝑓 −1 (𝐵)
Since 𝑓: 𝐴 → 𝑃(𝐴) is not an onto map.
(ii) 𝑓 −1 (𝐴 ∩ 𝐵) = 𝑓 −1 (𝐴) ∩ 𝑓 −1 (𝐵)
𝑐 i.e. potential of power set of any set is more than set itself.
(iii) 𝑓 −1 (𝐴𝑐 ) = (𝑓 −1 (𝐴))
i.e. (potential of 𝑃(ℕ) is more than ℕ)
Remark:-
Example:- For any infinite set A, |𝐴| ≪ |𝑃(𝐴)| (𝑖. 𝑒 |ℤ| < |𝑃(ℤ)|, |ℚ| <
1. If 𝒇: 𝑿 → 𝒀 is a function & 𝑨, 𝑩, 𝑪 ⊆ 𝑿. If 𝒇(𝜙) = 𝜙 then |𝑃(ℚ)|, 𝑒𝑡𝑐)
which of the following is/are true? ∴ We need to define symbols to denote cardinality of infinite
(1) 𝑓(𝐴 ∪ 𝐵) = 𝑓(𝐴) ∪ 𝑓(𝐵) sets of different potential.
(2) 𝑓(𝐴 ∩ 𝐵) = 𝑓(𝐴) ∩ 𝑓(𝐵) These symbols are called transfinite cardinals.
(3) 𝑓(𝐴 ∩ 𝐵) ⊆ 𝑓(𝐴) ∩ 𝑓(𝐵) |ℕ| = ℵ0 (aleph null, aleph nought)
(4) 𝑓(𝐴\𝐵) = 𝑓(𝐴)\(𝐵) |𝑃(ℕ)| = 𝑐 (continum)
Solution:
(a) 𝐴, 𝐵 ⊆ 𝐴 ∪ 𝐵 ⇒ 𝑓(𝐴), 𝑓(𝐵) ⊆ 𝑓(𝐴 ∪ 𝐵) ⇒ 𝑓(𝐴) ∪ 1.5 Similar /Equivalent Sets:
𝑓(𝐵) ⊆ 𝑓(𝐴 ∪ 𝐵) 1.5.1 Definition:
𝑥 ∈ 𝐴 ∪ 𝐵 ⇒ 𝑓(𝑥) ∈ 𝑓(𝐴 ∪ 𝐵) Two sets 𝐴 & 𝐵 are said to be equivalent or similar if there is
Hence 𝑓(𝐴 ∪ 𝐵) ⊆ 𝑓(𝐴) ∪ 𝑓(𝐵) one to one correspondence between their elements.
⇒ 𝑓(𝐴 ∪ 𝐵) = 𝑓(𝐴) ∪ 𝑓(𝐵) i.e. there exist a bijective map from 𝐴 to 𝐵 and it is denoted by
𝐴~𝐵.
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When two infinite sets are similar, the same transfinite (2) Define 𝑓: ℕ → ℤ such that
cardinals are used to denote their cardinality & conversely. 𝑛
𝑓(𝑛) = (−1)𝑛 [ ]
2
𝑛−1
The Initial segment: −( ) if 𝑛 is odd
2
Or, 𝑓(𝑛) = { 𝑛
Let 𝑛 ∈ ℕ if 𝑛 is odd
2
The initial segment of the natural numbers determined by 𝑛 𝑓(1) = 0, 𝑓(2) = 1, 𝑓(3) = −1, 𝑓(4) = 2, 𝑓(5) = −2
{1, 2, 3, … , 𝑛} & is denoted by 𝑆𝑛 ∴ 𝑓 is a bijective map
(We will use 𝑆𝑛 ) Hence, ℕ ~ℤ
Note: Result :- ′ ∼ ′ is an equivalence relation

1. Two sets 𝐴 & 𝐵 are similar iff |𝐴| = |𝐵| Proof:
2. Two finite set 𝐴 & 𝐵 are similar iff both 𝐴 & 𝐵 are similar i) Let 𝐴 be any set
to 𝑆𝑛 for some 𝑛 ∈ ℕ Define 𝑓: 𝐴 → 𝐴 such that 𝑓(𝑥) = 𝑥
⇒ 𝑓 is bijective
Example: ⇒ 𝐴~𝐴 ⇒ ~ is reflexive
Which of the following sets equivalent to 𝑺𝟒 = {𝟏, 𝟐, 𝟑, 𝟒} ii) Let 𝐴 ~ 𝐵 ⇒ ∃ a bijective map 𝑓: 𝐴 → 𝐵
(a) 𝐴1 = {𝑎, 𝑏, 𝑐, 𝑑} Take, 𝑓 −1 : 𝐵 → 𝐴
(b) 𝐴2 = {2,3,4,1} As 𝑓 is bijective map ⇒ 𝑓 −1 is also bijective map
(c) 𝐴3 = {{1}, {2}, {3}, {4}} ⇒ 𝐵~𝐴
(d) 𝐴4 = {{1,2}, 𝜙, {2,3}, 𝑆4 } ⇒ ′~′ is a symmetric.
(e) 𝐴5 = {ℕ, 𝑃(ℕ), ℝ, ℝ × ℝ} iii) Let 𝐴 ~ 𝐵 & 𝐵 ~ 𝐶
Solution : All sets are similar to 𝑆4 then ∃ two bijective maps 𝑓: 𝐴 → 𝐵 & 𝑔: 𝐵 → 𝐶
∵ |𝐴1 | = |𝐴2 | = |𝐴3 | = |𝐴4 | = |𝐴5 | = 4 As 𝑓 & 𝑔 are bijective maps
⇒ 𝑔𝑜𝑓 is also bijective map, where 𝑔𝑜𝑓: 𝐴 → 𝐶
Note : ⇒𝐴~𝐶
i) There does not exist any transfinite Cardinal number ⇒ ′~ ′ is transitive
which is strictly less than ℵ0 ⇒′ ~ ′ is an equivalence relation.
ii) There does not exist any transfinite Cardinal number
1. Q. Is ℕ~ [0,1] ?
between ℵ0 and 𝑐.
Solution:-
iii) There is infinitely many transfinite Cardinal number
Suppose ℕ~ [0,1] ⇒ ∃ one-one onto function ′𝑓′ from ℕ to
which is strictly greater than 𝑐.
[0,1]
Let the mapping is as follows
Examples :-
𝑓(1) = 0. 𝑎11 𝑎12 𝑎13 … … . . 𝑎1𝑛 … … .
Which of the following is/are correct ?
𝑓(2) = 0. 𝑎21 𝑎22 𝑎23 … … . . 𝑎2𝑛 … … .
(1) ℕ~𝑘ℕ, whose 𝑘 ∈ ℝ − {0} 𝑓(3) = 0. 𝑎31 𝑎32 𝑎33 … … . . 𝑎3𝑛 … … .
(2) ℕ ~ℤ .
(3) ℕ~ ℕ × ℕ .
(4) None of the above .
{𝑓(𝑛) = 0. 𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 … … . . 𝑎𝑛𝑛 … … . .
Solution:-
Take a number 𝑥 in [0,1] such that
(1) Define 𝑓: ℕ → 𝑘. ℕ such that
𝑥 = 0. 𝑏1 𝑏2 𝑏3 … . . 𝑏𝑛 … ..such that 𝑏𝑖 ≠ 𝑎𝑖𝑖
𝑓(𝑛) = 𝑘𝑛
⇒ 𝑥 is different from 𝑓(𝑖) ∀ 𝑖 ∈ ℕ
Clearly, 𝑓 is well defined
⇒ ∄ any 𝑖 ∈ ℕ such that 𝑓(𝑖) = 𝑥
i) Let 𝑓(𝑛1 ) = 𝑓(𝑛2 )
⇒ 𝑓 is not an onto map.
⇒ 𝑘𝑛1 = 𝑘𝑛2
Hence, our assumption is wrong
⇒ 𝑛1 = 𝑛2
⇒ ℕ ≁ [0,1].
⇒ 𝑓 is one one map
ii) for any 𝑘𝑛 ∈ 𝑘ℕ, take 𝑛 ∈ ℕ such that 𝑓(𝑛) = 𝑘𝑛 1.6 Countable & Uncountable Sets
⇒ 𝑓 is onto map 1.6.1 Countably Infinite set:
⇒ 𝑓 is bijective map A set 𝑆 is said to be countably infinite set or denumerable set
Hence, ℕ~𝑘. ℕ ∀ 𝑘 ∈ ℝ − {0} or enumerable set if it is similar to ℕ.
i.e. 𝑆 is countably infinite iff 𝑆 ~ℕ.
IFAS Publications
Examples: Q: Let 𝑨 be any subset of real number. Then which of the
i) ℕ, ℤ , 𝑘. ℕ, 𝑘. ℤ (where 𝑘 ∈ ℝ − {0} ) following is possible?
ii) ℝ is not countable infinite set. a) 𝑃(𝐴) is an infinite set
Reason: [0,1] ⊆ ℝ⇒ |[0,1]| ≤ |ℝ| b) 𝑃(𝐴) is empty set
But , we know that ℕ ≁ [0,1] implies |[0,1]| > ℵ0 c) 𝑃(𝐴) is countably infinite set
⇒ |ℝ| > ℵ0 d) 𝑃(𝐴) is countable set
⇒ ℝ is not countably infinite set. e) 𝑃(𝐴) is uncountable set
Observation :- A set 𝑆 is countably infinite ⇔ |𝑆| = ℵ0 Solution:
a) If 𝐴 =ℕ ⇒ |𝑃(𝐴)| = 2ℵ0 = 𝐶
1.6.2 Countable set: 𝑃(𝐴) is an infinite set.
A set 𝑆 is said to be countable set if it is either finite or b) for any non-empty set 𝐴 such that |𝐴| ≥ 1
countably infinite. ⇒ |𝑃(𝐴)| ≥ 21 = 2
Ex: If 𝐴 = ∅ ⇒ 𝑃(𝐴) = {∅} ⇒ |𝑃(𝐴)| = 1
i) Any finite set is countable set. ∴ for any set 𝐴, |𝑃(𝐴)| ≥ 1
ii) Any set similar to set of natural number is countable set. c)
iii) Empty set is countable set (|∅| = 0 ⟹ ∅ is finite set.) i) If 𝐴 is finite set ⟹ |𝐴| = 𝑛 for some 𝑛 ∈ ℕ
⇒ |𝑃(𝐴)| = 2𝑛 (finite)
1.6.3 Uncountable set: i.e If set ‘𝐴’ is finite then 𝑃(𝐴) is also finite set.
A set 𝑆 is said to be uncountable set if it is not countable. ii) If 𝐴 is infinite set ⇒ |𝐴| ≥ ℵ°
⟹ |𝑃(𝐴)| ≥ 2ℵ0 = 𝐶
Observation: ⇒ If 𝐴 is an infinite set, then cardinality of power set of 𝐴
If the set ‘𝑆’ has cardinality 𝐶 (continum) then 𝑆 is is at least continum.
uncountable. Converse need not be true. Thus, For any set 𝐴, |𝑃(𝐴)| ≠ ℵ0
i.e for any set 𝐴, 𝑃(𝐴) cannot be countably infinite.
Example: 𝑃(𝑃(𝑃(ℕ))) is an uncountable set, but | P(P(P(ℕ)))| d) As 𝑃(𝐴) is finite for any finite set A
> 2𝑐 > 𝑐 ⇒ 𝑃 is countable set.
Ex. e) 𝐴 =ℕ ⇒ |𝑃(𝐴)| = 2ℵ0 = 𝐶
1. [𝑎, 𝑏], [𝑎, 𝑏), (𝑎, 𝑏], (𝑎, 𝑏) are uncountable sets (a< 𝑏) ⇒ 𝑃(𝐴) is uncountable
2. Set of real number is uncountable set.
3. 𝑃(ℕ) is an uncountable set 1.6.4 Result:
1) Every infinite set has a countably infinite Subset.
Algebra of Transfinite cardinal numbers: 2) Every infinite set is similar to at least one of it's proper
subset.
Let ℵ0 denotes the cardinality of ℕ & 𝐶 denotes the
3) If 𝑆 is infinite subset of ℝ & 𝑋 is any finite subset of 𝑆, then
cordinality of 𝑃(ℕ), then
|𝑆| = |𝑆 − 𝑋|
1) ℵ0 + ℵ0 + ℵ0 + ⋯ + ℵ0 = ℵ0
Example : Show that [𝑎, 𝑏]~ [𝑐, 𝑑], (𝑎, 𝑏) ~ (𝑐, 𝑑) &
2) 𝐶 + 𝐶 + ⋯ + 𝐶 = 𝐶
[𝑎, 𝑏]~(𝑎, 𝑏)
3) 𝑛 + ℵ0 = ℵ0 ∀ 𝑛 ∈ ℕ Solution: Consider, two sets [𝑎, 𝑏] & [𝑐, 𝑑] , 𝑎 < 𝑏 , 𝑐 < 𝑑
There is a line which map [𝑎, 𝑏] to [𝑐, 𝑑]
4) ℵ0 + 𝐶 = 𝐶 , ℵ0 ∙ 𝐶 = 𝐶
𝑑−𝑐
𝑦=( ) (𝑥 − 𝑎) + 𝑐
5) ℵ0 ∙ ℵ0 ∙∙∙ ℵ0 = ℵ𝑛0 = ℵ0 ∀ 𝑛 ∈ ℕ 𝑏−𝑎
Which is one-one & onto map ⇒ [𝑎, 𝑏] ~ [𝑐, 𝑑]
6) 𝐶 ∙ 𝐶 ∙∙∙ 𝐶 = 𝐶 𝑛 = 𝐶 ∀ 𝑛 ∈ ℕ By using the same map, we get (𝑎, 𝑏) ~ (𝑐, 𝑑)
7) 2ℵ0 = 𝐶 > ℵ0 We know that If 𝑆 is infinite subset of ℝ & 𝑋 is any finite
subset of 𝑆, then |𝑆| = |𝑆 − 𝑋|
ℵ
8) ℵ00 = 2ℵ0 = 𝐶 If 𝑆 = [𝑎, 𝑏] and 𝑋 = {𝑎, 𝑏} then |𝑆| = |𝑆 − 𝑋|
i.e. |[𝑎, 𝑏]| = |(𝑎, 𝑏)|
9) ℵ𝐶0 = 2𝐶
10) 𝐶 ℵ0 = 𝐶 −𝜋 𝜋
Example: Show that 𝐴 = ( 2 , 2 )& 𝐵 =ℝ are similar
viii) 𝑛 < ℵ0 < 𝐶 ∀ 𝑛 ∈ ℕ Define 𝑓: 𝐴 → 𝐵
IFAS Publications
Such that 𝑓(𝑥) = 𝑡𝑎𝑛𝑥 Q. Which of the following set is countable?
Clearly, 𝑓 is bijective map. 1] 𝑺𝟏 = {𝒇|𝒇: {𝟏, 𝟐} → ℕ is a function}
⇒A~B Solution: Set 𝑆1 contains all functions 𝑓 from {1,2} to ℕ
−𝜋 𝜋 −𝜋 𝜋
⇒ ℝ~ ( , ) ~[ , ] Thus, |𝑆1 | = |ℕ||{1,2}| = ℵ20 = ℵ∘ × ℵ∘ = ℵ∘
2 2 2 2
⇒ 𝑆1 is countable set.
2) 𝑺𝟐 = {𝒇 |𝒇: ℕ → {𝟏, 𝟐} is a function}

Solution: Set 𝑆2 contains all functions 𝑓 from ℕ to {1,2}
Thus, |𝑆2 | = |{1,2}||ℕ| = 2ℵ∘ = 𝐶
⇒ 𝑆2 is an uncountable set.
3] 𝑺𝟑 = {𝒇|𝒇: {𝟏, 𝟐} → ℕ 𝐢𝐬 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 & 𝒇(𝟏) ≤ 𝒇(𝟐)}

Solution: We know that 𝑆1 is countable set.
Here, 𝑆3 ⊆ 𝑆1
A subset of a countable set is countable set.
⇒ S3 is countable set.
Some important results on Countability:
1. Countable union of countable set is countable. 4] 𝑺𝟒 = {𝒇|𝒇: ℕ → {𝟎, 𝟏} is a function & 𝒇(𝟏) ≤ 𝒇(𝟐)}
2. Family of all finite subsets of countable set is countable. Solution: Here the possible choices for 𝑛 = 1 &𝑛 = 2
3. Family of all infinite subset of countably infinite set is are
uncountable. i) 𝑓(1) = 1 , 𝑓(2) = 1
4. Finite Cartesian product of countable set is countable. ii) 𝑓(1) = 0, 𝑓(2) = 1
5. Infinite Cartesian product of countable set 𝐴1 , 𝐴2 , … iii) 𝑓(1) = 0, 𝑓(2) = 0
(where |𝐴𝑖 | ≥ 2 for infinitely many 𝑖) is an uncountable The choices for all other elements 3,4,5,6,……… are
set {0,1}
Total number of functions 𝑓: ℕ →
Example: Show that Set of rational number is countable set. {0,1} such that 𝑓(1) ≤ 𝑓(2) are
Solution: The given set ℚ can be written as 3 × 2 × 2 × 2 × 2 … … … … … = 3 × 2ℵ∘ = 𝐶
ℚ = ℚ+ ∪ ℚ− ∪ {0} ⇒ S4 is an uncountable set.
Consider the following set
1
𝐸1 = {𝑛 | 𝑛 ∈ ℕ} 5] 𝑺𝟓 = {𝒇|𝒇: ℕ → {𝟏, 𝟐, 𝟑 … . . 𝒏} be a function}
2 Solution: Set 𝑆5 contains all functions 𝑓 from
𝐸2 = {𝑛 | 𝑛 ∈ ℕ }
ℕ to {1,2,3 … . . 𝑛}
3
𝐸3 = {𝑛 |𝑛 ∈ ℕ } … Thus, |𝑆5 | = |{1,2,3 … . . 𝑛}||ℕ| = 𝑛ℵ∘ = 𝐶
𝑘
𝐸𝑘 = {𝑛 | 𝑛 ∈ ℕ } ⇒ 𝑆5 is an uncountable set.
Here each 𝐸𝑘 is countable set ∀ 𝑘 ∈ ℕ 6] 𝑺𝟔 = {𝒇|𝒇: {𝟏, 𝟐, 𝟑, … . 𝒏} → ℕ be a function}

𝑘
(Define 𝑓: ℕ → 𝐸2 such that 𝑓(𝑛) = ) Solution: Set 𝑆6 contains all functions 𝑓 from
𝑛
We know that, countable union of countable set is countable. {1,2,3, … . 𝑛} to ℕ
⇒ ⋃∞ ∞
𝑘=1 𝐸𝑘 is countable where ⋃𝑛=1 𝐸𝑘 = ℚ
+
Thus, |𝑆6 | = |ℕ||{1,2,3,….𝑛}| = ℵ𝑛0 = ℵ∘
⇒ ℚ+ is countable set ⇒ 𝑆6 is countable set.
Define 𝑓: ℚ+ → ℚ− such that 𝑓(𝑥) = −𝑥
Clearly, 𝑓 is bijective function. 7] 𝑺𝟕 = {𝒂𝒏 |𝒂𝒏 𝐢𝐬 𝐚 𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 & 𝒂𝒏 ∈ {𝟎, 𝟏} ∀𝒏 ∈ ℕ}
⇒ ℚ− is also countable set Solution: We know that,
Hence as ℚ+ , ℚ− {0} all are countable set. Sequence is a special type of function having domain
⇒ ℚ = ℚ+ ∪ ℚ− ∪ {0} is countable set set of natural numbers.
Now, ℝ = ℚ ∪ ℚ𝒄 Set 𝑆7 contains all functions 𝑓 from ℕ to {0,1}
If ℚ𝑐 is countable set |𝑆7 | = |{0,1}||ℕ| = 2ℵ∘ = 𝐶
⇒ ℚ ∪ ℚ𝑐 is also Countable (∵ ℚ is Countable)
⇒ ℝ is Countable set
Which is a contradiction.
Hence, ℚ𝑐 is an Uncountable Set.
IFAS Publications
Solved Examples : 𝑆 ′ = ⋃𝑛∈ℕ 𝑆𝑛 is countable
Now for any 𝑦 ∈ ℚ,
1. Which of the following sets is uncountable?
𝑝 We have at most two solutions for 𝑥, So for 𝑛 = 𝑘 then
(a) {𝑥 ∈ ℝ| log(𝑥) = 𝑞 for some 𝑝, 𝑞 ∈ ℕ}
the set
(b) { 𝑥 ∈ ℝ|(cos(𝑥))𝑛 + (sin(𝑥))𝑛 = 1 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑛 ∈ 1
𝑆𝑘 = {(𝑥, 𝑦): 𝑥 2 + 𝑦 2 = , 𝑦 ∈ ℚ} is countable
ℕ} 𝑘2
′′
𝑝 So, 𝑆 = ⋃𝑛∈ℕ 𝑆𝑛 is countable
(c) { 𝑥 ∈ ℝ|𝑥 = log ( ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 = 𝑝, 𝑞 ∈ ℕ}
𝑞 𝑆 = 𝑆′ ∪ 𝑆 ′′ is countable.
𝑝
(d) { 𝑥 ∈ ℝ| cos(𝑥) = 𝑞
𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑝, 𝑞 ∈ ℕ}
Solution: 3. Which of the following is necessarily true for a function
(a) we have log 𝒙 = 𝒑/𝒒 𝒇: 𝑿 → 𝒀 ?
⇒ 𝑥 = 𝑒 𝑝/𝑞 (a) If 𝑓 is injective, then there exists 𝑔: 𝑌 → 𝑋 such that
= 𝑒𝑟 𝑓(𝑔(𝑦)) = 𝑦 for all 𝑦 ∈ 𝑌
𝑝, 𝑞 ∈ ℕ (b) If 𝑓 is surjective, then there exists 𝑔: 𝑌 → 𝑋 such that
𝑟∈ℚ 𝑓(𝑔(𝑦)) = 𝑦 for all 𝑦 ∈ 𝑌
∵ |ℚ| = ℵ0 i.e. ℚ is countable set. (c) If 𝑓 is injective and 𝑌 is countable then 𝑋 is finite
∴ choices of 𝑒 𝑟 is also countable. (d) If 𝑓 is surjective and 𝑋 is uncountable then 𝑌 is
Therefore (a) is incorrect countably infinite
(b) (𝐜𝐨𝐬 𝒙)𝒏 + (𝐬𝐢𝐧 𝒙)𝒏 = 𝟏; 𝒏 ∈ ℕ Solution:
Take 𝑛 = 2, 𝑐𝑜𝑠 2 𝑥 + 𝑠𝑖𝑛2 𝑥 = 1 ∀ 𝑥 ∈ ℝ For option (a),
⇒ 𝑆 = {𝑥 ∈ ℝ|𝑐𝑜𝑠 2 𝑥 + 𝑠𝑖𝑛2 𝑥 = 1} = ℝ Let 𝑋 = {0,1}, 𝑌 = {0,1,2}
∴ |𝑆| = |ℝ| = 𝐶 Take, 𝑓: 𝑋 ⟶ 𝑌 be defined as 𝑓(𝑥) = 𝑥
⇒ option (b) is correct Clearly, 𝑓 is injective but for function 𝑔: 𝑌 ⟶ 𝑋 such that
𝒑
(c) 𝒙 = 𝐥𝐨𝐠 (𝒒) 𝑓(𝑔(𝑦)) = 𝑦 for all 𝑦 ∈ 𝑌
⇒ 𝑒 𝑥 = 𝑝/𝑞, for some 𝑝, 𝑞 ∈ ℕ There does not exist pre-image of 2 under 𝑓 ∘ 𝑔.
= 𝑝⁄𝑞 ∈ ℚ ∴ Option (a) is incorrect.
∵ |ℚ| = ℵ0 For option (b),
Option (c) is incorrect. As 𝑓 is surjective
(d) 𝐜𝐨𝐬 𝒙 =
𝑷
= 𝒓 ∈ ℚ, 𝐟𝐨𝐫 𝐬𝐨𝐦𝐞 𝒑, 𝒒 ∈ ℕ ∴ 𝑓: 𝑋 ⟶ 𝑌 for any 𝑦 ∈ 𝑌 there exists at least one 𝑥 ∈ 𝑋
𝒒
for which 𝑓(𝑥) = 𝑦
|ℚ| = ℵ0
Define 𝑔: 𝑌 ⟶ 𝑋 in such a way that 𝑔(𝑦) = 𝑥
Choices for 𝑥 such that cos 𝑥 = 𝑟 also countable
𝑓 ∘ 𝑔(𝑦) = 𝑓(𝑥) = 𝑦
Option (d) is incorrect.
So, option (b) is correct.
For option (c),
𝟏
2. Let 𝑺 = {(𝒙, 𝒚)| 𝒙𝟐 + 𝒚𝟐 = 𝒏𝟐 , where 𝒏 ∈ ℕ and either Take, 𝑋 = ℕ, 𝑌 = ℕ
𝒙 ∈ ℚ 𝒐𝒓 𝒚 ∈ ℚ}. Here ℚ is the set of rational numbers Let 𝑓: ℕ ⟶ ℕ defined by 𝑓(𝑥) = 𝑥, clearly 𝑓 is injective
and ℕ is the set of positive integers. Which of the and 𝑌 is countable but 𝑋 is also countable
following is true? Thus, option (c) is incorrect.
(a) S is a finite non-empty set For option (d), Take, 𝑋 = ℝ, 𝑌 = ℝ
(b) S is countable Let 𝑓: 𝑋 ⟶ 𝑌 defined by 𝑓(𝑥) = 𝑥
(c) S is uncountable Clearly 𝑓 is surjective and 𝑋 is uncountable but Y is also
(d) S is empty uncountable
Solution: ∴ option (d) is incorrect.
1
𝑆 = {(𝑥, 𝑦)|𝑥 2 + 𝑦 2 = where 𝑛 ∈ ℕ and either 𝑥 ∈
𝑛2
ℚ 𝑜𝑟 𝑦 ∈ ℚ}. 1.6.5 Algebraic & Transcendental Number:
1 If 𝑃 = {𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ 𝑎𝑛 𝑥 𝑛 /𝑎𝑖 ∈ ℚ, 𝑛 ∈ ℕ U {0}} =
Solving equation 𝑥 2 + 𝑦 2 = 𝑛2 for 𝑦 with 𝑛 = 1, we get
the set of all polynomials with rational coefficients
𝑦 = ±√1 − 𝑥 2 Define 𝑓 ∶ 𝑃 → 𝐸 such that
For any 𝑥 ∈ ℚ, we have at most two solution for 𝑦
Where 𝑓: (𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) = (𝑎0 , 𝑎1 , … . , 𝑎𝑛 )
In general, for 𝑛 = 𝑘
Clearly, 𝑓 is one-one & onto
1
𝑆𝑘 = {(𝑥, 𝑦): 𝑥 2 + 𝑦 2 = 𝑘 2 , 𝑥 ∈ ℚ} is countable ⇒ 𝑓 is bijective
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⇒ 𝑃 ~𝐸 𝐴1 = {{1}, {2}, {3}, … … , {𝑛}, … … }
Where 𝐸 = ⏟
ℚ × ℚ × ℚ × …× ℚ ⇒ 𝐴1 ~ℕ ⇒ 𝐴1 is countable
𝑛+1 times
𝐴2 = {𝑋 ⊆ ℕ| |𝑋| = 2}
As ℚ is Countable and finite Cartesian product of countable set
𝐴2 = {{𝑎, 𝑏}| 𝑎, 𝑏 ∈ ℕ}
is countable.
Define,
⇒ 𝐸 is Countable
𝑓2 : 𝐴2 → ℕ × ℕ such that 𝑓2 (𝑥) =
⇒ 𝑃 is Countable
(𝑎, 𝑏), 𝑎 < 𝑏
The set 𝑃 can be written as { 𝑤ℎ𝑒𝑟𝑒 𝑋 = {𝑎, 𝑏}
(𝑏, 𝑎), 𝑏 < 𝑎
𝑃 = {𝛼1 (𝑥), 𝛼2 (𝑥), 𝛼3 (𝑥), … , 𝛼𝑛 (𝑥), … } U ℚ. ⇒ 𝑓 is one-one & onto
Where each 𝛼𝑖 (𝑥) is a polynomial over ℚ. ⇒ 𝐴2 ~ℕ × ℕ~ℕ ⇒ 𝐴2 is Countable
𝑆1 = {𝑎 ∈ ℂ| 𝛼(𝑎) = 0}
𝐴𝑛 = {𝑋 ⊆ ℕ||𝑋| = 𝑛}
𝑆1 contains all roots of one degree polynomial in 𝑃 Clearly 𝐴𝑛 ~ ⏟
ℕ × ℕ × − −× ℕ ~ℕ ⇒ 𝐴𝑛 is Countable
By fundamental theorem of algebra, every 𝑛th degree 𝑛 times
polynomial has exactly 𝑛 roots counting with multiplicity. ∴ 𝐴 = ⋃∞ 𝑖=1 𝐴𝑖 is a countable set
Thus, every one degree polynomial has exactly one root in ℂ. Hence, set of all finite subset of natural number is
Since, all one degree polynomials over ℚ is countable set. countable.
Hence, 𝑆1 is countable set. In general,
𝑆2 = {𝑎 ∈ ℂ| 𝛼(𝑎) = 0} Set of all finite subset of countable set is countable.
𝑆2 contains all roots of two degree polynomial in 𝑃 6. 𝑃(ℕ) = {𝑋 ⊆ ℕ| |𝑋| 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒} ∪ {𝑋 ⊆
Clearly, 𝑆2 is countable set. ℕ| |𝑋| is infinite}
In general, = 𝑆1 ⋃𝑆2
𝑆𝑛 = {𝛼 ∈ ℂ| 𝛼𝑛 (𝑥) = 0} Where 𝑃(ℕ) is uncountable set & 𝑆1 is Countable set
𝑆𝑛 contains all roots of 𝑛th degree polynomial in 𝑃 ⇒ 𝑆2 is uncountable set.
As finite Cartesian product of countable set is countable i.e. The set of all infinite subset of Natural numbers is an
Thus, 𝑆𝑛 is countable. uncountable set.
We know that, countable union of countable set is countable. 7. There is no Strictly decreasing function 𝑓: ℕ → ℕ
⇒ 𝑆 = ⋃∞ 𝑛=1 𝑆𝑛 is countable set. 8. The set of all decreasing function from 𝑓: ℕ → ℕ is
countable set.
Note: If 𝑃 = {𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ 𝑎𝑛 𝑥 𝑛 /𝑎𝑖 ∈ 𝑋, 𝑛 ∈ Since, Every decreasing function 𝑓: ℕ → ℕ is an eventually
ℕ U {0}}, then 𝑃 is countable iff 𝑋 is countable. constant function
9. The set of all monotonic function from ℕ 𝑡𝑜 ℕ is
1.6.6 Definition: uncountable 𝑖. 𝑒. = {𝑓: ℕ → ℕ|𝑓 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐 }
1. A complex number is said to be algebraic over ℚ if it is zero 10. If 𝐴 = {〈𝑎𝑛 〉| 𝑎𝑛 is eventually constant in 𝑋}, then 𝐴 is
of non-zero polynomial with rational coefficients. countable iff 𝑋 is countable.
(∵ The set of all algebraic numbers in ℂ is countable set)
the set of all algebraic number in ℝ is countable set) 1.7 Inequality:
2. A real number / Complex number is said to transcendental 1. For any real number 𝑎, 𝑏
if it not algebraic. 1. |𝑎 + 𝑏 | ≤ |𝑎| + |𝑏 |
Clearly, the set of all algebraic number is countable. 2. |𝑎 − 𝑏 | ≥ ||𝑎| − |𝑏 ||
(:ℝ =set of Algebraic numbers ∪ set of transcendental 3. √𝑎2 + 𝑏 2 ≤ |𝑎| + |𝑏 |
numbers) 4. √|𝑎| + |𝑏 | ≤ √|𝑎| + √|𝑏|
If set of transcendental number is countable then ℝ is 5. If 𝑎1 , 𝑎2 , … . , 𝑎𝑛 be any real numbers, then |𝑎1 + 𝑎2 +
countable which a contradiction. ⋯ + 𝑎𝑛 | ≤ +|𝑎2 | + ⋯ + |𝑎𝑛 |
3. Every rational number is an algebraic number 6. If 𝑎1 , 𝑎2 , … , 𝑎𝑛 be any real numbers, then |𝑎1 . 𝑎2 , … .,
4. Every irrational number need not be a transcendental 𝑎𝑛 | = |𝑎1 |. |𝑎2 | … . . |𝑎𝑛 |
number. 7. |𝑥 + 𝑦 | = |𝑥| + |𝑦 | , 𝑖𝑓 𝑥𝑦 ≥ 0
5. All finite subset of set of natural number is countable. 8. |𝑥 + 𝑦 | < |𝑥| + |𝑦 | , 𝑖𝑓 𝑥𝑦 < 0
Reason: |𝑎+𝑏| |𝑎| |𝑏|
9. ≤ +
1+|𝑎+𝑏| 1+|𝑎 | 1+ |𝑏|
Define
𝐴𝑖 = {𝑋 ⊆ ℕ| |𝑋| = 𝑖}
𝐴1 = {𝑋 ⊆ ℕ| |𝑋| = 1}
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2. Cauchy – Schwarz Inequality: 6. Which set is uncountable?
1. If 𝑎1 , 𝑎2 , … . 𝑎𝑛 , 𝑏1 , 𝑏2 , … . 𝑏𝑛 > 0 & (A) The set of positive primes
𝐴 = ∑𝑛𝑖=1 𝑎𝑖2, 𝐵 = ∑𝑛𝑖=1 𝑏𝑖2 & 𝐶 = ∑𝑛𝑖=1 𝑎𝑖 𝑏𝑖 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + (B) The set of integers
⋯ . +𝑎𝑛 𝑏𝑛 (C) The set of rational numbers
Then 𝐶 2 ≤ 𝐴. 𝐵 (D) The set of irrational numbers in [0, 1]
i.e (∑𝑛𝑖=1 𝑎𝑖 𝑏𝑖 )2 ≤ (∑𝑛𝑖=1 𝑎𝑖 2 ) (∑𝑛𝑖=1 𝑏𝑖 2 )
1 1
Let 𝑝 > 1, 𝑃 + 𝑞 = 1 , 𝑎 ≥ 0 , 𝑏 ≥ 0 then 7. Which one of the following is necessarily true?
(A) If 𝑓: ℚ → ℝ is an injective map, then 𝑓(ℚ) ̅̅̅̅̅̅̅ is
𝑎𝑝 𝑏𝑞
𝑎𝑏 ≤ + equality hold iff 𝑎𝑝 = 𝑏 𝑞 uncountable
𝑝 𝑞
(B) There exists no surjective map from ℕ onto ℚ
3. Holder’s Inequality: (C) There exists a surjective map from ℚ × ℝ onto ℂ
If 𝑎𝑖 𝑏𝑖 ≥ 0, ∀ 𝑖 ∈ 𝑆𝑛 = {1 ,2, … . , 𝑛} then (D) None of the above
1⁄ 1⁄ 1 1
∑𝑛𝑖=1 𝑎𝑖 𝑏𝑖 ≤ (∑𝑛𝑖=1 𝑎𝑖 𝑝 ) 𝑃
(∑𝑛𝑖=1 𝑏𝑖 𝑞 ) 𝑞
,𝑃 > 1 & + = 1
𝑃 𝑞 8. Consider the statements:
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
1
[𝑆1 ] {𝑠𝑖𝑛 ( ) | 𝑛 ≥ 1} is uncountable.
4. Minkowskis Inequality: 𝑛
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
1
If 𝑃 ≥ 1 & 𝑎𝑖 , 𝑏𝑖 ≥ 0, ∀ 𝑖 ∈ 𝑆𝑛 then [𝑆2 ] {𝑒 𝑛 | 𝑛 ≥ 1} is uncountable.
1⁄ 1⁄ 1⁄
𝑛 𝑛 𝑛 𝑞
𝑃
𝑝
𝑃
𝑞 [𝑆3 ] ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
{𝜋 𝑛 | 𝑛 ≥ 1} is uncountable.
(∑(𝑎𝑖 + 𝑏𝑖 )𝑝 ) ≤ (∑ 𝑎𝑖 ) + (∑ 𝑏𝑖 )
Which one of the following is the correct option?
𝑖=1 𝑖=1 𝑖=1
(A) None of [𝑆1 ], [𝑆2 ] and [𝑆3 ] is true
Multiple Choice Questions (MCQ) (B) All of 𝑆1 ], [𝑆2 ] and𝑆3 ]] are true
(C) Only [𝑆1 ] is true
1
1. Let 𝑓 ∶ ℝ\{0} → ℝ be defined by 𝑓(𝑥) = 𝑥 + 𝑥 3. On (D) Only 𝑆1 and [𝑆2 ] are true
which of the following interval(s) is 𝑓 one-one?
(A) (−∞, −1) (B) (0, 1) 9. Let 𝑆 = {𝑎 + 𝑏√2 ∶ 𝑎, 𝑏 ∈ ℤ} ⊂ ℝ.
(C) (0, 2) (D) (0, ∞) Which of the following assertions about S is correct?
(A) For any maps 𝑓: ℚ × ℚ → ℝ 𝑎𝑛𝑑 𝑔: ℝ → 𝑆, their
2. Let 𝑓(𝑥) = 2𝑥 3 − 9𝑥 2 + 7. Which of the following is true? composition 𝑔 0 𝑓 never surjective.
(A) f is one-one in the interval [- 1, 1] (B) There exists a surjective map from S onto 𝑅2 /𝑧 2 ,
(B) f is one-one in the interval [2, 4] (C) There exists an injective map from R/Q into S.,
(C) f is NOT one-one in the interval [- 4, 0] (D) For any two maps 𝑓: 𝑆 → ℝ 𝑎𝑛𝑑 𝑔: ℝ → ℝ3 , their
(D) f is NOT one-one in the interval [0, 4] composition of 𝑔𝑜𝑓 has a countable range.
3. Let 𝑋 be a countably infinite subset of ℝ and 𝐴 be a 10. For subsets 𝑇1 and 𝑇2 of ℝ,

countably infinite subset of 𝑋. Then the set 𝑋\𝐴 = we define 𝑇1 + 𝑇2 ∶= {𝑡1 + 𝑡2 |𝑡1 ∈ 𝑇1 , 𝑡2 ∈ 𝑇2 }.
{𝑥 ∈ 𝑋 |𝑥 ∉ 𝐴} Let 𝑆1 ≔ (ℚ ∩ [0,1]) + {√2} and
(A) is empty 𝑆2 ≔ (ℚ ∩ [1,2]) + {√3}. Let 𝑆 = 𝑆1 ∪ 𝑆2 .
(B) is a finite set Which one of the following is true?
(C) can be a countably infinite set (A) There exists a one one map from [0,1] to S
(D) can be an uncountable set (B) There exists an onto map from S to [0,1]
(C) 𝑆1 = 𝑆2
4. Let 𝐴 and 𝐵 be two sets having 𝑚 and 𝑛 many elements (D) None of the above
respectively, where 𝑚 < 𝑛; 𝑚, 𝑛 ∈ ℕ. The number of
injective functions possible from 𝐴 to 𝐵 is: Multiple Select Questions (MSQ)
(A) 𝑚𝑛 (B) 𝑛𝑃𝑚 11. Which of the following sets is in one-to-one
𝑛
(C) 𝐶𝑚 (D) 𝑛𝑚 correspondence with ℕ
1 1 1
(A) {1, , , , … … … }
2 3 4
5. Let 𝑆 be a set with 𝑛 elements. Then the number of (B) {… … … . . , −3, −2, −1, 0, 1, 2, 3, … … … . }
distinct objective mappings from 𝑆 to 𝑆 is: 𝑝
(C) { ∶ 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0}
(A) 𝑛 (B) 𝑛2 𝑞
𝑝
(C) 𝑛! (D) 𝑛𝑛 (D) { ∶ 𝑝, 𝑞 ∈ ℕ}
𝑞
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12. Let 𝑋 = {(𝑥, 𝑦) 𝜖 ℝ2 ∶ 𝑥 2 + 𝑦 2 − 1}, 𝑌 = {(𝑥, 𝑦) 𝜖 ℝ2 ∶ (A) empty (B) nonempty but finite
𝑥 = 𝑦} and 𝑍 = {(𝑥, 𝑦) 𝜖 ℝ2 ∶ 𝑦 = −𝑥}. Consider the (C) countably infinite (D) uncountable
following assertions:
A. X U Y U Z is an equivalence relation on ℝ. 18. What is the number of surjective maps from the set {1,
B. X U Y is a reflexive relation on ℝ but not symmetric. …,10} to the set {1,2}?
C. X U Y is an equivalence relation on ℝ. (A) 90 (B) 1022
D. Y U Z is an equivalence relation on ℝ. (C) 98 (D) 1024
Which of the above assertions are correct?
(A) A and B only (B) A, B and D only 19. Which one of the following set is not countable?
(C) A and D only (D) A, C and D only (A) ℕ𝑟 , where 𝑟 ≥ 1 and ℕ is the set of natural numbers
(B) {0, 1}ℕ , the set of all the sequences which takes
13. Suppose ℛ is a relation defined on ℕ𝑏𝑦 (𝑎, 𝑏) ∈ ℛ iff 𝑎 − values 0 and 1
(C) ℤ, set of integers
𝑏 is a multiple of 4 or a multiple of 6. Consider the
(D) √2ℚ, ℚ is set of rational numbers.
statements:
A. ℛ is symmetric. 20. Let 𝑋 be a countably infinite subset of ℝ and 𝐴 be a
B. ℛ is transitive. countably infinite subset of 𝑋. Then the set 𝑋\𝐴 =
C. There is an infinite subset 𝑈 ⊂ ℕ of ℕ such that ℛ {𝑥 ∈ 𝑋 |𝑥 ∉ 𝐴}
defined on 𝑈 is an equivalence relation. (A) is empty
(B) is a finite set
Then the number of correct statements above is
(C) can be a countably infinite set
(A) 1 (B) 2 (D) can be an uncountable set
(C) 3 (D) 0
Numerical Answer Type (NAT)
14. Consider the subset
21. Consider the set 𝑆 = {𝑥 ∈ ℕ: 𝑥 2 = 2}. Then the number
𝑆 ≔ {(𝑥, 𝑦) ∈ ℝ2 ∶ 𝑦 = 𝑥 or|𝑥| + |𝑦 | = 1} of ℝ2 .
of elements in the power set of S is____.
Consider the following assertions about S.
A. S is reflexive relation on ℝ
22. A set S is described by the roots of the equation 𝑍 𝑛 = 1; Z
B. S is transitive relation on ℝ
is a complex number, then sum of the roots is equal to
C. There exists a surjective map from ℝ onto S
____.
D. There exists an injective map from S into ℚ3
Which of the above assertions is/are correct?
23. A set 𝑆 is described by the roots of the equation 𝑍 𝑛 = 1;
(A) 𝐴 only (B) 𝐴 and 𝐷 only
Z is a complex number, then products of the roots is equal
(C) 𝐴 and 𝐶 only (D) 𝐴 and 𝐵 only
to ____.
15. Let 𝑆 be a collection of subsets of {1, 2, … , 100} such that

24. Let 𝑆 = {[𝑥] ∶ 𝑥 ∈ [−2, 2]}; where [∙] denotes the
the intersection of any two sets in 𝑆 is non-empty. What
greatest integer function. Then order of 𝑆 is _____.
is the maximum possible cardinality |𝑆| of 𝑆?
(A) 100 (B) 2100
99 25. Let 𝑆 = {[𝑥] ∶ 𝑥 ∈ [−2, 2]}; where [∙] denotes the
(C) 2 (D) 298
greatest integer function. Then log(|𝑃(𝑆)|) is ____.
16. Consider the following four sets of maps 𝑓 ∶ ℤ → ℚ:

26. Consider the set S consisting of elements of the form
(i) {𝑓 ∶ ℤ → ℚ | 𝑓 is bijective and increasing}.
𝑛2
(ii) {𝑓 ∶ ℤ → ℚ | 𝑓 is onto and increasing}, [ ] ; 𝑛 ∈ [0, 1]. Cardinality of the set S is _____.
2
(iii) {𝑓 ∶ ℤ → ℚ | 𝑓 is bijective, and satisfies that ∀ 𝑛 ≤
0, 𝑓(𝑛) ≥ 0}, and 27. In a marriage ceremony, 250 peoples like ice-cream, 50
(iv) {𝑓 ∶ ℤ → ℚ | 𝑓 is onto and decreasing}. like cold drink, 20 like milk, 40 like both ice-cream and cold
How many of these sets are empty? drink, 20 like both cold drink and milk, 10 like milk and ice-
(A) 0 (B) 1 cream both and 30 like all. How many peoples are there in
(C) 2 (D) 3 the ceremony____.
17. The set 𝑆 = {𝑥 ∈ ℝ|𝑥 > 0 𝑎𝑛𝑑(1 + 𝑥 2 ) tan(2𝑥) = 𝑥} 28. Let 𝐴 = {𝑥: 𝑥 2 − 3𝑥 + 2 = 0}, 𝐵 = {𝑍 ∈ ℂ ∶ 𝑍 3 = 1}.
is Then the no. of relations from A to B is ______
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29. Let 𝐴 = {𝑥: 𝑥 2 − 3𝑥 + 2 = 0}, 𝐵 = {𝑧 ∈ ℂ ∶ 𝑧 4 = 1}.
Then the no. of relations from A to B is _____
30. If |𝐴| = 5, then number of reflexive relations on 𝐴 are

____
Answer Key
Multiple Choice Questions
1 2 3 4 5 6 7 8 9 10
B D C B C D C A D C
Multiple Select Questions
11 12 13 14 15 16 17 18 19 20
ABCD C B C C D C B B C
Numerical Answer Type
21 22 23 24 25 26 27 28 29 30
1 0 1 5 5log2 1 280 64 256 220
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2
CHAPTER
POINT SET TOPOLOGY
2.1 Archimedean Property: Supremum of 𝑺 / Least upper Bound of 𝑺 (l.u.b. of 𝑺):
1
(1) For any 𝜖 > 0, ∃ 𝑚 ∈ ℕ such that < 𝜖, ∀ 𝑛 ≥ 𝑚 Let 𝛼 ∈ ℝ, 𝑆 ⊆ ℝ. Then 𝛼 is said to be a supremum of 𝑆 if it is
𝑛
(2) ∀ 𝑥, 𝑦 ∈ ℝ, 𝑥 ≠ 0, ∃ 𝑛 ∈ ℤ such that 𝑛𝑥 > 𝑦 the least upper bound of 𝑆 i.e. sup(𝑆) = 𝛼
(3) ∀𝑥, 𝑦 ∈ ℝ, 𝑥 > 0, ∃ 𝑛 ∈ ℕ such that 𝑛𝑥 > 𝑦 Or 𝛼 is a supremum of 𝑆 iff
(i) ∀ 𝑥 ∈ 𝑆, 𝑥 ≤ 𝛼
Order structure: (ii) for any 𝜖 > 0, ∃ 𝑥 ∈ 𝑆 such that 𝑥 > 𝛼 − 𝜖.
A set 𝑆 is said to be an ordered structure if it satisfies the
following postulates Note:
(i) Law of Trichotomy:- (1) Supremum of a set may or may not be a member of the
∀𝑎, 𝑏 ∈ 𝑆 , either 𝑎 < 𝑏 or 𝑎 = 𝑏 or 𝑎 > 𝑏 set.
(ii) Law of Transitivity:- Ex. 𝑆 = (0,1) then sup(𝑆) = 1 ∉ 𝑆
∀𝑎, 𝑏, 𝑐 ∈ 𝑆 , if 𝑎 < 𝑏 & 𝑏 < 𝑐 then 𝑎 < 𝑐 (2) If Sup(𝑆) = 𝛼 & 𝛼 ∈ 𝑆, then 𝛼 is called the greatest
(iii) Order compatibility with respect to addition composition:- element of the set.
∀𝑎, 𝑏, 𝑐 ∈ 𝑆 , if 𝑎 < 𝑏 then 𝑎 + 𝑐 < 𝑏 + 𝑐 (3) sup(𝜙) = −∞
(iv) Order compatibility with respect to multiplication
Bounded Above set:
composition:
A set 𝑆 is said to be bounded above iff 𝑈𝑠 ≠ 𝜙
∀𝑎, 𝑏, 𝑐 ∈ 𝑆 , 𝑐 > 0 if 𝑎 < 𝑏 𝑎𝑐 < 𝑏𝑐
i.e. ∃ 𝛼 ∈ ℝ such that 𝛼 ∈ 𝑈𝑠
Dedekind property of ℝ: i.e. ∃ 𝛼 ∈ ℝ such that ∀ 𝑥 ∈ 𝑆 , 𝑥 ≤ 𝛼
If 𝐴 & 𝐵 are two non-empty subsets of ℝ such that
Unbounded above set:
1. 𝐴 ∪ 𝐵 = ℝ
A set 𝑆 is said to be unbounded above if it has no upper bound.
2. Every member of 𝐴 is less than every member of 𝐵
i.e. 𝑆 is unbounded above set iff 𝑈𝑠 = 𝜙
Then either 𝐴 has maximum or 𝐵 has minimum i.e ∀ 𝑥 ∈
ex. ℤ, ℕ, ℝ, ℚ........
𝐴 & ∀ 𝑦 ∈ 𝐵; 𝑥 < 𝑦, then either 𝐴 has maximum element
or 𝐵 has minimum element.
(2) Lower bound:
Let 𝑆 ⊆ ℝ & 𝛽 ∈ ℝ is said to be lower bound of 𝑆 if ∀ 𝑥 ∈
Absolute value or Modulus of a real Number:
𝑥 𝑥 ≥0 𝑆, 𝛽 ≤𝑥
The absolute value of 𝑥 ∈ ℝ is defined as |𝑥| = { i.e. ∄ 𝑥 ∈ 𝑆 such that 𝛽 > 𝑥
−𝑥, 𝑥 < 0
2.2 Bounds of a Set: Observations:

Upper Bound: (1) A set may or may not have a lower bound ex- ℤ, ℝ, ℚ, ℕ,
Let 𝑆 ⊆ ℝ & 𝛼 ∈ ℝ is said to be upper bound of 𝑆 if ∀ ∀𝑥 ∈ ……
𝑆, 𝑥 ≤ 𝛼 (2) If 𝛽 is a lower bound of 𝑆. Then every real number less
Or, than 𝛽 is also lower bound of 𝑆.
∄ 𝑥 ∈ 𝑆 such that 𝑥 > 𝛼. (3) Let 𝑆 ⊆ ℝ
Define 𝐿𝑆 = {𝑥 ∈ ℝ| 𝑥 is a lower bound of 𝑆}
Observations: = set of all lower bounds of 𝑆
(1) A set may or may not have a upper bound.
Ex. ℤ, ℕ, ℝ, ℚ, ℚ𝑐 ......... Infimum of 𝑺/ Greatest lower Bound of 𝑺 (g.l.b. of 𝑺):
(2) If 𝛼 is an upper bound of 𝑆, then every real number Let 𝑆 ⊆ ℝ & 𝛽 ∈ ℝ, Then 𝛽 is said to be an infimum of 𝑆 if it
greater than 𝛼 is also an upper bound of 𝑆. is the greatest lower bound of 𝑆.
(3) Let 𝑆 ⊆ ℝ Or,
𝑈𝑠 = {𝑥 ∈ ℝ | 𝑥 is an upper bound of 𝑆} (i) ∀𝑥 ∈ 𝑆, 𝛽 ≤ 𝑥
𝑈𝑠 = set of all upper bound of 𝑆. (ii) for any 𝜖 > 0, ∄ 𝑥 ∈ 𝑆 such that 𝑥 < 𝛽 + 𝜖
Chapter – 2 Point Set Topology 17
Note: 2. For any two non-empty set 𝐴, 𝐵 ⊆ ℝ such that 𝑥 ≤
(1) Infimum of the set may or may not belongs to the set 𝐴 = 𝑦, ∀ 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵. Then 𝐴 is bounded above & 𝐵 is
(0,1) bounded below & sup(𝐴) ≤ inf(𝐵)
𝑖𝑛𝑓(𝐴) =0 but 0 ∉ (0,1)
𝐿𝐴 = {𝑥| 𝑥 ≤ 𝑦 , 𝑦 ∈ (0,1)} = (−∞, 0) 3. If 𝐴 & 𝐵 be two non-empty subset of ℝ
(2) If inf (𝑆) = 𝛽 ∈ 𝑆, then 𝛽 is called the smallest element of (i) sup(𝐴 ∪ 𝐵) = max{sup 𝐴, sup 𝐵}
the set 𝑆. (ii) inf(𝐴 ∪ 𝐵) = min{inf 𝐴, inf 𝐵}
(3) inf (𝜙) = ∞
4. sup{𝑥 ∈ ℚ |𝑥 < 𝑎} = 𝑎
Bounded below set: 5. Let 𝐴, 𝐵 ⊆ ℝ be two bounded non-empty subsets, then
A set 𝑆 is said to be bounded below iff 𝐿𝑆 ≠ 𝜙 (i) sup{𝑥 + 𝑦 |𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵} = sup 𝐴 + sup 𝐵
i.e. ∃ 𝛽 ∈ ℝ such that 𝛽 ∈ 𝐿𝑠 (ii) inf{𝑥 + 𝑦 | 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵} = inf 𝐴 + inf 𝐵
i.e. ∃ 𝛽 ∈ ℝ such that 𝛽 ≤ 𝑥, ∀ 𝑥 ∈ 𝑆. (iii) sup{𝑥 − 𝑦|𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵} = sup 𝐴 − inf 𝐵
(iv) inf{𝑥 − 𝑦|𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵} = inf 𝐴 − sup 𝐵.
Unbounded below set:
A set 𝑆 is said to be unbounded below if it has no lower bound Solved Examples:
ex- ℤ, ℝ, … … (1) If 𝑺𝟏 = {𝒙 ∈ ℝ | (𝒙 − 𝟏)(𝒙 − 𝟐)(𝒙 − 𝟑) < 𝟎} then
i) 𝐬𝐮𝐩(𝑺𝟏 )
Bounded set: ii) 𝐢𝐧𝐟(𝑺𝟏 )
Let 𝑆 ⊆ ℝ, We say 𝑆 is bounded iff it is bounded below & Solution: Given
bounded above both i.e both 𝐿𝑠 & 𝑈𝑠 are nonempty sets. 𝑆1 = {𝑥 ∈ ℝ | (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) < 0}
Note:- Suppose, 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3)
(1) If 𝑆 is not bounded above, then sup(𝑆) = ∞ 𝑆1 = {𝑥 ∈ ℝ|𝑓(𝑥) < 0}
(2) If 𝑆 is not bounded below, then inf(𝑆) = −∞ Here 𝑓(𝑥) = 0 for 𝑥 = 1,2,3
Ex. sup(ℝ) = ∞ & inf(ℝ) = −∞ (i) for 𝑥 = 0, 𝑓(0) = (0 − 1)(0 − 2)(0 − 3) = −6 < 0
3 3 3 3 3
(i) Sup(𝑆) & Inf(𝑆) are unique, if they exist (not necessity (ii) for 𝑥 = 2 , 𝑓 (2) = (2 − 1) (2 − 2) (2 − 3) =
equal) 1 −1
( )( )( ) = > 0
−3 3
2 2 2 8
(ii) If 𝑆 ⊆ ℝ be a non-empty set, then inf(𝑆) ≤ sup(𝑆) 5 5 3 1 −1 −3
(iii) for 𝑥 = 2 , 𝑓 (2) = (2) (2) ( 2 ) = <0
8
7 7 5 3 1 15
Completeness property of ℝ: (iv) for𝑥 = , 𝑓 ( ) = ( ) ( ) ( ) = >0
2 2 2 2 2 8
Every non-empty bounded subset of ℝ has a supremum &
𝑥 (−∞, 1) (1, 2) (2,3) (3, ∞)
infimum in ℝ.
1 1 1
sign - + - +
Ex. A= {1 + 1! + 2! + ⋯ … + 𝑛! |𝑛 ∈ ℕ}
⇒ 𝑆1 = (−∞, 1) ∪ (2,3)
1 1 1
If 𝑎𝑛 = 1 + 1! + 2! + ⋯ . + 𝑛! ⇒ sup(𝑆1 ) = 3 , inf (𝑆1 ) = −∞
then for each 𝑛 ∈ ℕ 𝑎𝑛 < 𝑎𝑛+1
1
⇒ inf(𝐴) = 𝑎1 = 1 + 11 = 2 (least element of the set) (2) 𝑺𝟐 = {𝒙 ∈ ℝ | (𝒙 − 𝟏)(𝒙 − 𝟐)(𝒙 − 𝟑) > 𝟎}
Solution: By previous example,𝑆2 = (1,2) ∪ (3, ∞)
sup(𝐴) = 𝑒
Here sup(𝐴) ∉ ℚ & inf(𝐴) ∈ ℚ each element of A is also a sup(𝑆2 ) = ∞, inf(𝑆2 ) = 1
member of ℚ
(3) If 𝑺 = {𝒙 ∈ [𝟎, 𝟒] | 𝐬𝐢𝐧 𝒙 > 𝟎}, then 𝐬𝐮𝐩(𝑺)
⇒ ℚ is not complete.
(i) 4 (ii) 𝜋
𝜋
Properties: (iii) 2
(iv) 0
1. Let 𝐴 & 𝐵 are non-empty bounded subset of ℝ such that Solution:
𝐴 ⊆ 𝐵. Then
(i) sup(𝐴) ≤ sup(𝐵)
(ii) inf(𝐴) ≥ inf(𝐵)
(iii) sup(𝐴) = − inf(−𝐴)
(iv) inf(−𝐴) = − sup(𝐴)
(v) sup(𝑐𝐴) = 𝑐 sup(𝐴) where 𝑐 > 0
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Clearly, for 𝑥 ∈ [0,4] 2.4 Connectedness:
sin 𝑥 > 0 if 𝑥 ∈ (0, 𝜋) Let 𝑆 ⊆ ℝ we say 𝑆 is a connected set iff 𝑆 is an interval.
⇒ 𝑆 = (0, 𝜋) Observations:-
⇒ sup(𝑆) = 𝜋 (1) Every non-empty open interval is an uncountable set.
inf(𝑆) = 0 (2) A closed interval need not be uncountable (ex.{𝛼} =
[𝛼, 𝛼])
(4) If 𝑺 = {𝒙 ∈ [𝟎, 𝟒] | 𝐜𝐨𝐬 𝒙 > 𝟎}, then 𝐬𝐮𝐩(𝑺 ) (3) Non-trivial intervals are uncountable.
𝜋
(a) 4 (b) 2 (4) Any collection of disjoint non-trivial intervals is countable
(c) 0 (d) 𝜋 set.
Solution: (5) Any collection of disjoint open intervals is countable set
(6) A collection of disjoint closed interval need not be
countable set.
Ex. 𝑋 = {{𝑎}| 𝑎 ∈ ℝ} is uncountable.
Nested Interval Theorem:-

If𝐼1 , 𝐼2 , 𝐼3 … . , 𝐼𝑛 … be a sequence of sets such that 𝐼1 ⊇ 𝐼2 ⊇
𝐼3 … . ⊇ 𝐼𝑛 ⊇ 𝐼𝑛+1 … … .. , where 𝐼𝑘 , 𝑘 ∈ ℕ are non-trivial
intervals then ⋂𝑘∈ℕ 𝐼𝑘 is non-empty set.
Clearly, for 𝑥 ∈ [0,4]
cos 𝑥 > 0 , if 𝑥 ∈ (0, 𝜋/2) 2.5 Neighbourhood of a Point (nbd):
⇒ 𝑆 = (0, 𝜋/2) Let 𝑆 ⊆ ℝ & 𝛼 ∈ ℝ we say 𝑆 is a neighbourhood of 𝛼 if ∃ an
⇒ sup 𝑆 = 𝜋/2 open interval say 𝐼 such that 𝛼 ∈ 𝐼 & 𝐼 ⊆ 𝑆.
inf(𝑆) = 0 i.e. 𝑆 is a neighbourhood of 𝛼 iff ∃ 𝛿 > 0 & (𝛼 − 𝛿, 𝛼 + 𝛿) ⊆
𝑆.
2.3 Intervals: Note:
(i) Let 𝐴 ⊆ ℝ Then 𝐴 is said to be an interval if 𝑥, 𝑦 ∈ 𝐴 & 𝑥 ≤
1. for any 𝛿 > 0 define
𝑎 ≤ 𝑦 then 𝑎 ∈ 𝐴.
𝑁𝛿 (𝛼) = (𝛼 − 𝛿, 𝛼 + 𝛿) = {𝑥 ∈ ℝ||𝑥 − 𝛼| < 𝛿}
(ii) Let 𝐴 be a subset of ℝ, we say 𝐴 is an interval if whenever 𝑁𝛿 (𝛼) is called 𝛿 − 𝑛𝑏𝑑 of 𝛼
𝑥, 𝑦 ∈ 𝐴 ⟹ 𝑎𝑥 + 𝑏𝑦 ∈ 𝐴, where 𝑎, 𝑏 ≥ 0 & 𝑎 + 𝑏 = 1. 2. 𝑁𝛿 (𝛼) − {𝛼} is called deleted 𝛿 − 𝑛𝑏𝑑 of 𝛼
Note: Observations:
(1) Singleton sets & empty set are called trivial intervals & the (1) If 𝑆 in nbd of 𝑎 ⇒ 𝑎 ∈ 𝑆(∵ 𝑎 ∈ 𝐼 & 𝐼 ⊆ 𝑆 by definition)
rest are non-trivial intervals. (2) If 𝑆 is nbd of 𝑎 for some 𝛼 ∈ ℝ then 𝑆 is an uncountable
(2) Non-trivial intervals are uncountable (∵ It contains set.
uncountably many irrational numbers) i.e. Uncountability is the necessary condition for the to be
(3) If 𝑆 is an interval & inf(𝑆) = 𝑎, sup(𝐵) = 𝑏 then 𝑆 is one a nbd but not sufficient.
of the following form Example - 𝑆 = ℚ𝑐 is an uncountable set but ℚ𝑐 is not a
(i) 𝑆 = [𝑎, 𝑏] , if 𝑎, 𝑏 ∈ 𝑆 nbd of any 𝑥 ∈ ℝ
(ii) 𝑆 = (𝑎, 𝑏) , if 𝑎, 𝑏 ∉ 𝑆 (3) If 𝑆 is nbd of some 𝛼 ∈ ℝ, then it cannot be free from any
(iii) 𝑆 = [𝑎, 𝑏) , if 𝑎 ∈ 𝑆, 𝑏 ∉ 𝑆 of the four sets ℚ, ℚ𝑐 , Algebraic number set &
(iv) 𝑆 = (𝑎, 𝑏] , if 𝑎 ∉ 𝑆, 𝑏 ∈ 𝑆 Transcendental number set.
(4) ℝ = (−∞, ∞), 𝜙 is an interval (4) Every interval is nbd of all of it’s points excepts the end
points.
Open/ Closed Intervals: (5) Empty set is nbd of each of it’s point
(i) If 𝑆 is an interval such that sup(𝑆) & inf(𝑆) doesn’t (6) Finite intersection of nbds of a point is also a nbd of that
belongs to 𝑆, then 𝑆 is called an open interval. point.
Ex. 𝜙, ℝ, (𝑎, 𝑏) where 𝑎 < 𝑏 (7) Arbitrary intersection of nbd of a point need not be nbd of
(ii) If 𝑆 is an interval such that sup(𝑆) & inf(𝑆) belongs to 𝑆, that point.
then 𝑆 is called a closed interval. 1 1
Ex. 𝑎 = 0 & for each 𝑛 ∈ ℕ, (− 𝑛 , 𝑛 ) is a nbd of 𝛼
Ex. {𝛼} where 𝛼 ∈ ℝ, [𝑎, 𝑏] 𝑎 ≤ 𝑏.
−1 1
Now ⋂∞
𝑛=1 ( , ) = {0} is not nbd of 𝛼 = 0
𝑛 𝑛
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(8) Arbitrary union of family of nbd of a point is also a nbd of (6) Infiniteness is the necessary condition for a set to have a
that point. limit point but not sufficient.
Ex. ℤ, ℕ
2.6 Some Important Types of Points: (7) If 𝑆 is finite set then 𝑆 has no limit point.
Adherent point:- Let 𝑆 ⊆ ℝ & 𝑎 ∈ ℝ. we say 𝑎 is adherent (8) Uncountable set must have limit point.
point of 𝑆. If for any 𝛿 > 0, (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ 𝑆 ≠ 𝜙
i.e. every nbd of 𝑎 contains atleast one point of 𝑆 Perfect set: A set 𝑆 ⊆ ℝ is said to be perfect if 𝑆 = 𝑆′
Notation: − 𝐴𝑑(𝑠) = {𝑥 ∈ ℝ| 𝑥 is adherent point of S} =
Ad(𝑆) Isolated Points:
Let 𝑆 ⊆ ℝ & 𝑎 ∈ 𝑆 − 𝑆 ′ , then 𝑎 is called isolated point of 𝑆
Observations: 𝑖. 𝑒. 𝑎 is an isolated point of 𝑆 iff for some 𝛿 > 0,
(1) Every member of set 𝑆 is adherent point of the set 𝑆, but (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ 𝑆 − {𝑎} = 𝜙
not conversely iff for some 𝛿 > 0, (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ 𝑆 = {𝑎}
Ex. 𝑆 = (0,1) ⇒ 1 ∉ 𝑆
Result:-
But for any 𝛿 > 0, 1 ∈ (1 − 𝛿, 1 + 𝛿)
(1) for any 𝑆 ⊆ ℝ, 𝑆 − 𝑆 ′ is countable set.
Here 𝑆 = (0,1) ⇒ (1— 𝛿, 1) ⊆ 𝑆
(i e. Set of isolated points is countable set)
⇒ (1 − 𝛿, 1) = 𝑆 ∩ (1 − 𝛿, 1 + 𝛿)
(2) If S is an uncountable set then S’ is uncountable.
⇒ 𝑆 ∩ (1 − 𝛿, 1 + 𝛿) ≠ 𝜙
⇒ 1 is an adherent point of a set S. (3) If S’ is countable set then S is countable set.
(2) If 𝑎 ∉ 𝑆 & 𝑎 is adherent point of 𝑆 then (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ (4) If 𝑆′ is finite then 𝑆 is countable set.
𝑆 is an infinite set.
Bolzano Weierstrass Theorem:-
(3) If 𝑆 is finite set, then 𝐴𝑑(𝑆) = 𝑆.
Every infinite Bounded set has a limit point:
Limit Point /Accumulation point:

Compact sets in ℝ:
Let 𝑆 ⊆ ℝ, 𝑎 ∈ ℝ.
Let ⊆ ℝ , then 𝑆 is said to be compact iff 𝑆 is closed &
We say 𝑎 is limit point of 𝑆 if for any 𝛿 > 0
bounded.
(𝑎 − 𝛿 𝑎 + 𝛿) ∩ 𝑆 − {𝑎} ≠ ∅.
𝑖. 𝑒. every nbd of 𝑎 contains at least one point of 𝑆 other than Condensation Point:
𝑎. Let 𝑆 ⊆ ℝ & 𝛼 ∈ ℝ, then 𝛼 is said to be condensation point of
𝑆, if for any 𝛿 > 0, (𝛼 − 𝛿, 𝛼 + 𝛿) ∩ 𝑆 is an uncountable set.
Derived set: The derived set of 𝑆 is the set contains of all limit Notation: Set of condensation points of 𝑆 is denoted as 𝑐𝑑(𝑆).
point of 𝑆 and it is denoted by 𝑆’
𝑆 ′ = {𝛼 ∈ ℝ | 𝛼 is limit point of 𝑆} Observation:
1) If S is countable set & for any ∈ ℝ & 𝛿 > 0 ,(𝛼 −
Results: 𝛿, 𝛼 + 𝛿) ∩ 𝑆 is countable set
If 𝐴 & 𝐵 are two subsets of ℝ, then ⇒ (𝛼 − 𝛿, 𝛼 + 𝛿) ∩ 𝑆 is not uncountable set for any
i) 𝐴 ⊆ 𝐵 ⇒ 𝐴′ ⊆ 𝐵′ countable set 𝑆.
converse need not be true ⇒ If 𝑆 is countable set, then 𝑐𝑑(𝑆) = 𝜙
ex; 𝐴 = [0, 1] ∪ {2}, 𝐵 = [0, 1] ∪ {3} 2) Let 𝑆 be an uncountable subset of ℝ & 𝑇 =
ii) (𝐴 ∪ 𝐵)′ = 𝐴′ ∪ 𝐵′ {𝛼 ∈ ℝ| 𝛼 is a condensation point of 𝑆 }.
iii) (𝐴 ∩ 𝐵)′ ⊆ 𝐴′ ∩ 𝐵′ Then 𝑆 − 𝑇 is countable set
𝐴 = ℚ, 𝐵 = ℚ𝑐 ⇒ 𝐴 ∩ 𝐵 = 𝜙 ⇒ (𝐴 ∩ 𝐵)′ = 𝜙 ′ = 𝜙
𝐴′ = ℝ, 𝐵 ′ = ℝ ⇒ 𝐴′ ∩ 𝐵 ′ = ℝ ∩ ℝ = ℝ Interior Point:
Let 𝑆 ⊆ ℝ, 𝛼 ∈ ℝ then we say 𝛼 is an interior point of 𝑆, if
Observations: ∃ 𝛿 > 0 such that (𝛼 − 𝛿, 𝛼 + 𝛿) ⊆ 𝑆.
(1) Limit point may or may not belongs to the set Notation: Interior of 𝑆 = 𝑆°
(2) 𝑎 is limit point of 𝑆 iff for any 𝛿 > 0, (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ 𝑆 is = {𝑥 ∈ ℝ | 𝑥 is an interior point of 𝑆 }
infinite Ex:
(3) Every limit point of 𝑆 is an adherent point of 𝑆. i) [0, 1]° = (0, 1)
(4) If 𝑎 ∉ 𝑆, then 𝑎 is ad(𝑠) iff 𝑎 ∈ 𝑆 ′ ii) ℤ° = 𝜙
(5) 𝐴𝑑(𝑆) = 𝑆 ∪ 𝑆 ′ iii) If 𝑆 is any countable set, then 𝑆° = 𝜙
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Results: −1 1
⇒ −𝛼 ∉ ( , ) ⇒ −𝛼 ∉ ⋂∞
𝑛=1 𝑆𝑛
𝑛 𝑛
1) 𝑆 is nbd of 𝛼 iff 𝛼 is an interior point of 𝑆
Hence, if 𝛼 ≠ 0 be any real number.
2) Limit point of a set 𝑆 cannot be interior point of 𝑆 𝑐 .
Then 𝛼 ∉ ⋂∞ 𝑛=1 𝑆𝑛
3) Interior point of 𝑆 𝑐 cannot be limit point of 𝑆.
Thus, we have
4) Let 𝑆 be any subset of ℝ, then −1 1
⋂∞ ∞
𝑛=1 𝑆𝑛 = ⋂𝑛=1 ( 𝑛 , 𝑛 ) = {0}, where {0} is not an open set.
𝑆° ⊆ 𝑐𝑑(𝑠) ⊆ 𝑆 ′ ⊆ 𝐴𝑑(𝑠)
Open Set: Closed Set: Let 𝑆 ⊆ ℝ, we say 𝑆 is closed set if 𝑆 𝑐 = ℝ − 𝑆 is

Let 𝑆 ⊆ ℝ, we say 𝑆 is an open set if 𝑆 = 𝑆° an open set.
𝑖. 𝑒. 𝑆 is open set, iff it is neighbourhood of each of its points. Ex. 𝑆 = {1} ⇒ 𝑆 𝑐 = (−∞, 1) ∪ (1, ∞) is an open set ⇒ 𝑆 is
𝑖. 𝑒. 𝑆 is open set iff every element of 𝑆 is an interior point of closed set.
𝑆.
Observation:
𝑖. 𝑒. for any 𝛼 ∈ 𝑆, ∃ 𝛿 > 0 such that (𝛼 − 𝛿, 𝛼 + 𝛿) ⊆ 𝑆.
1) Every finite set is a closed set.
2) 𝜙 & ℝ are closed set.
Observations:
3) The intersection of arbitrary family of closed sets is closed.
(1) Every open interval is an open set (but not conversely)
4) Finite union of closed sets is closed.
Ex: 𝐴 = (2, 3) ∪ (1, 2) is an open set but not open
5) Arbitrary union of closed set need not be closed.
interval.
1 1
(2) ℝ, 𝜙, (𝑎, 𝑏) are open sets. Ex. 𝐴𝑛 = [𝑛 , 𝑎 − 𝑛] where 𝑎 ≥ 1 ⇒ ⋃∞
𝑛=1 𝐴𝑛 = (0, 𝑎)
(3) for any set 𝑆, 𝑆° is an open set (∴ (𝑆°)° = 𝑆° ∀ 𝑆 ⊆ ℝ) which is not closed set.
(4) 𝑆° is the largest open set contained is 𝑆. 6) Derived set of any set is closed set.
(5) 𝑆° is equals to the union of all open subsets of 𝑆.
(6) Arbitrary union of open sets is an open set. Compact Set:
(7) Arbitrary intersection of open sets need not be an open A Set 𝑆 ⊆ ℝ is said to be compact set if 𝑆 is closed and
set. bounded.
(8) Finite intersection of open sets is open. Closure of a set:
(9) Let 𝐴 & 𝐵 are subsets of ℝ, then Let 𝑆 ⊆ ℝ, then the closure of a set S is denoted by 𝑆 and it is
i) 𝐴 ⊆ 𝐵 ⇒ 𝐴° ⊆ 𝐵° the smallest closed set containing 𝑆.
ii) (𝐴 ∩ 𝐵)° = 𝐴° ∩ 𝐵° i.e.
iii) 𝐴° ∪ 𝐵° ⊆ (𝐴 ∪ 𝐵)° i) 𝑆 is closed
(Ex. 𝐴 = ℚ, 𝐵 = ℚ𝑐 ⇒ (𝐴 ∪ 𝐵)° = (ℝ)° = ℝ but 𝐴° ∩ ii) 𝑆⊆𝑆
𝐵° = 𝜙 ∩ 𝜙 = 𝜙) iii) If 𝑇 is a closed set such that 𝑆 ⊆ 𝑇 ⇒ 𝑆 ⊆ 𝑇
Note:
ex. 𝑄 = ℝ, 𝑄𝑐 = ℝ, ℤ = ℤ
i) Between any two rational numbers, there are countably
Observations:
infinite rationals & uncountably many irrational numbers.
1) A set 𝑆 is said to be closed set iff 𝑆 = 𝑆
ii) Between any two irrational numbers, there are countably
2) A closure of a set 𝑆 is a set containing 𝑆 & the set of all
infinite rationals & uncountably many irrational numbers.
limit points of 𝑆 i.e. 𝑆 = 𝑆 ∪ 𝑆′
−1 1
Ex. Let 𝑆𝑛 = ( 𝑛 , 𝑛) then find ⋂∞
𝑛=1 𝑆𝑛
Dense set: A set 𝑆 ⊆ ℝ is said to be dense set in ℝ if 𝑆 = ℝ .
−1 1
Solution: Here for each 𝑛 ∈ ℕ, 0 ∈ ( 𝑛 , 𝑛) = 𝑆𝑛
∞ Exterior point:
0 ∈ ⋂ 𝑆𝑛 = 𝑆 Let 𝑆 ⊆ ℝ & 𝛼 ∈ ℝ, then 𝛼 is said to an exterior point of 𝑆, if
𝑛=1 ∃ 𝛿 > 0 such that (𝛼 − 𝛿, 𝛼 + 𝛿) ∩ 𝑆 = 𝜙
∴ 𝑆 is non-empty
i.e. (𝛼 − 𝛿, 𝛼 + 𝛿) ⊆ 𝑆 𝐶
For any 𝛼 > 0,
⇒ 𝛼 is an interior point of 𝑆 𝐶 .
∴ by Archimedian property, ∃ 𝑚 ∈ ℕ. Such that
1
< 𝛼, ∀ 𝑛 ≥ 𝑚 Frontier point:
𝑛
−1 1 Let 𝑆 ⊆ ℝ & 𝛼 ∈ ℝ, we say 𝛼 is a frontier point of 𝑆 if it is
∴ for this 𝑛 ≥ 𝑚, 𝛼 ∉ ( 𝑛 , 𝑛) = 𝑆𝑛
neither interior point of 𝑆 nor exterior point of 𝑆.
⇒ 𝛼 ∉ ⋂∞ ∞
𝑛≥𝑚 𝑆𝑛 ⇒ 𝛼 ∉ ⋂𝑛=1 𝑆𝑛
1 −1 Fr(𝑆) = {𝑥 ∈ ℝ| 𝑥 is neither interior point of 𝑆 nor exterior
Consider, 𝑛 < 𝛼. ∀ 𝑛 ≥ 𝑚 ⇒ > −𝛼, ∀ 𝑛 ≥ 𝑚
𝑛 point of 𝑆}
IFAS Publications
Boundary Point:
Let 𝛼 ∈ 𝑆, 𝑆 ⊆ ℝ, Then 𝛼 is said to be boundary of 𝑆, if it is
frontier point of 𝑆.
Some Collection of Points:
Necessary 𝑪𝒅(𝑺) 𝑺’ 𝑨𝒅(𝑺) 𝑺𝒐 𝑩𝒅(𝑺) 𝑺=
condition |𝑺| = 𝑪 |𝑺| > ℵ𝟎 𝒂 ∈ 𝑺 ∪ 𝑺′ |𝑺| = 𝑪 𝒂 ∈ 𝑺 ∪ 𝑭𝒓(𝑺) 𝑺 ∪ 𝑺′
1 𝜙 𝜙 𝜙 𝜙 𝜙 𝜙 𝜙
2 ℝ ℝ ℝ ℝ ℝ 𝜙 ℝ
3 {𝑎} 𝜙 𝜙 {𝑎} 𝜙 {𝑎} {𝑎}
4 ℕ 𝜙 𝜙 ℕ 𝜙 ℕ ℕ
5 ℤ 𝜙 𝜙 ℤ 𝜙 ℤ ℤ
6 (𝑎, 𝑏) [𝑎, 𝑏] [𝑎, 𝑏] [𝑎, 𝑏] (𝑎, 𝑏) 𝜙 [𝑎, 𝑏]
7 [𝑎, 𝑏] [𝑎, 𝑏] [𝑎, 𝑏] [𝑎, 𝑏] (𝑎, 𝑏) {0, 1} [0, 1]
8 𝑄 𝜙 ℝ ℝ 𝜙 𝑄 ℝ
9 𝑄𝐶 ℝ ℝ ℝ 𝜙 𝑄𝐶 ℝ
10 1
{ |𝑛 ∈ ℕ} 𝜙 {0} 𝐴 ∪ {0} 𝜙 1 1
𝑛 { /𝑛 ∈ ℕ} { /𝑛 ∈ ℕ} ∪ {0}
𝑛 𝑛
11 1
{ + |𝑚, 𝑛 ∈ ℕ}
1 𝜙 1 1 1 𝜙 1 1 𝐴𝑑(𝑠)
𝑚 𝑛 { /𝑛 ∈ ℕ} ∪ {0} { + |𝑚, 𝑛 ∈ ℕ} { + /𝑛
𝑛 𝑚 𝑛 𝑚 𝑛
1
∪ { |𝑛 ∈ ℕ} ∪ {0} ∈ ℕ}
𝑛
1 1
Note: ⇒ 1 1 1 ≥3
+ +
Let 𝑎1 , 𝑎2 , … , 𝑎𝑛 be positive real numbers, then 𝑛 𝑚 𝑚+𝑛
𝑚𝑛 1
𝐴. 𝑀 =
𝑎1 +𝑎2 +...+𝑎𝑛 ⇒ 𝑚+𝑛+1 ≥ 3 ∀ 𝑚, 𝑛 ∈ ℕ
𝑛 𝑚.𝑛 1 1
1⁄ 𝑚 = 1, 𝑛 = 1 ⇒ 𝑚+𝑛+1 = 1+1+1 = 3
𝐺. 𝑀. = (𝑎1 . 𝑎2 . … . 𝑎𝑛 ) 𝑛
𝑛 1
𝐻. 𝑀. = 1 1 1 Hence inf(𝑆) = 3
+ +...+
𝑎1 𝑎2 𝑎𝑛
𝑚2 𝑚2
& 𝐴. 𝑀 ≥ 𝐺. 𝑀 ≥ 𝐻. 𝑀 Take 𝑚 = 𝑛 ⇒ 𝑚+𝑚+1 = 2𝑚+1 → ∞ as 𝑚 → ∞
⇒ sup(𝑆) does not exist.
Solved Examples: 1
3) 𝑆 = {𝑥 + 𝑥 | 𝑥 > 0}
1) Find supremum & infimum of the following sets. 1
𝑚 4𝑛 Solution: as 𝑥 → ∞, 𝑥 + 𝑥 → ∞ ⇒ sup 𝑆 is does not exist.
i) 𝑆 = { 𝑛 + 𝑚
, 𝑚, 𝑛 ∈ ℕ}, if exist 1⁄
𝑥+1⁄𝑥 1 2
Solution: We know 𝐴. 𝑀 ≥ 𝐺. 𝑀 2
≥ (𝑥. 𝑥)
𝑚 4𝑛 1⁄ 1 1⁄
+ 𝑚 4𝑛 2 ⇒ 𝑥 + 𝑥 ≥ 2(1) 2 ≥2
𝑛 𝑚
≥( × )
2 𝑛 𝑚 1 1
𝑚
+
4𝑛
≥ 2×2 = 4⇒
𝑚
+
4𝑛
≥4 If 𝑥 = 1 ⇒ 𝑥 + 𝑥 = 1 + 1 = 2
𝑛 𝑚 𝑛 𝑚
𝑚 4𝑛 2 4×1 ⇒ inf(𝑆) = 2
𝑚 = 2&𝑛 = 1 ⇒ 𝑛
+ 𝑚
=1+ 2
= 2+2= 4 1⁄
4) 𝑆 = {2𝑥 + 2 𝑥 |𝑥 > 0}
⇒ inf(𝑆) = 4.
𝑚 4𝑛 Solution: We know that
If 𝑎𝑚,𝑛 = 𝑛
+ 𝑚
, then 1⁄ 1⁄
1 1
2𝑥 +2 ⁄𝑥 1 2 𝑥+ 2 1⁄
for 𝑛 = 1, 𝑎𝑚,1 = 𝑚 + 𝑚
4
≥ (2𝑥 . 2 ⁄𝑥 ) = (2 𝑥 ) ≥ (22 ) 2 =2
2
1⁄
Thusas 𝑚 → ∞, 𝑎𝑚,1 → ∞ 1 1⁄ 2 1⁄ 1⁄
(∴ 𝑥 + ≥ 2 ⇒ (2𝑥+ 𝑥) ≥ (22 ) 2 = 22× 2 =2=
𝑥
⇒ sup(𝑆) does not exist.
𝑚𝑛
2)
2) 𝑆 = { | 𝑚, 𝑛 ∈ ℕ}
𝑚+𝑛+1 1⁄
𝑚𝑛 𝑚𝑛⁄
𝑚𝑛 1 ∴ 2𝑥 + 2 𝑥 ≥2×2=4
Solution: 𝑚+𝑛+1 = 𝑚+𝑛+1 =1 1 1 1 1⁄
𝑚𝑛
+ +
𝑛 𝑚 𝑚+𝑛 For 𝑥 = 1, 2 + 2 1 = 2+2= 4
1 1 1
As 𝑚, 𝑛 ∈ ℕ ⇒ 𝑚 , 𝑛 , 𝑚+𝑛 ≤ 1 ⇒ inf(𝑆) = 4
1⁄
1 1
⇒ 𝑛 + 𝑚 + 𝑚+𝑛 ≤ 1 + 1 + 1 = 3
1 As 𝑥 → ∞, 2𝑥 + 2 𝑥 → ∞ ⇒ sup 𝑆 is does not exist.
IFAS Publications
5) 𝑆 = {𝑥 ∈ ℝ |(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)(𝑥 − 𝑑) < 0, 𝑎 < (a) Both 𝐸 & 𝐹 are closed
𝑏 < 𝑐 < 𝑑} (b) 𝐸 is closed but 𝐹 is not
Solution: The graph of the function 𝑓(𝑥) = (𝑥 − 𝑎)(𝑥 − (c) 𝐹 is closed but 𝐸 is not
𝑏)(𝑥 − 𝑐)(𝑥 − 𝑑) is as follows (d) None are closed
𝑛
Solution: Here 𝐸 = { | 𝑛 ∈ ℕ}
𝑛+1
Limit point of 𝐸 is 1 but 1 ∉ 𝐸
⇒ 𝐸 ≠ 𝐸 ⇒ 𝐸 is not closed.
𝑥
𝐹 = {1−𝑥 | 𝑥 ∈ [0,1)} = [0, ∞) is a closed set
⇒ 𝐹 is closed but 𝐸 is not closed set.
Intervals (−∞, 𝑎) (𝑎, 𝑏) (𝑏, 𝑐) (𝑐, 𝑑) (𝑑, ∞) Q. Let 𝑨 ⊆ ℝ be a non-empty bounded subset of ℝ & 𝒍 =
𝐬𝐮𝐩 𝑨. Then which of the following is/are true ?
Sign + − + − + (a) 𝑙 is limit point of set 𝐴.
𝑓(𝑥) < 0, ∀ 𝑥 ∈ (𝑎, (𝑐,
𝑏) ∪ 𝑑) (b) 𝑙 ∈ 𝐴
𝐴 = (𝑎, 𝑏) ∪ (𝑐, 𝑑) (c) 𝑙 ∉ 𝐴
sup 𝐴 = 𝑑 (d) either 𝑙 ∈ 𝐴 or 𝑙 is limit point of 𝐴.
inf 𝐴 = 𝑎 Solution:
(a) 𝐴 = {1} ⇒ sup 𝐴 = 1 but 1 ∉ 𝐴′
Q. Let 𝒔 ∈ (𝟎, 𝟏). Then which of the following is/are (b) 𝐴 = (0, 1) ⇒ sup 𝐴 = 1 but 1 ∉ 𝐴
correct? (c) 𝐴 = {1} ⇒ sup 𝐴 = 1 but 1 ∈ 𝐴
𝑚
(1) ∀ 𝑚 ∈ ℕ, ∃ 𝑛 ∈ ℕ such that 𝑠 > (d) If 𝑙 ∈ 𝐴 then option (d) is true
𝑛
𝑚 If 𝑙 ∉ 𝐴 & 𝑙 = sup 𝐴.
(2) ∀ 𝑚 ∈ ℕ, ∃ 𝑛 ∈ ℕ such that 𝑠 <
𝑛 For any 𝜖 > 0, (𝑙 − 𝜖, 𝑙 + 𝜖) is a nbd of 𝑙.
(3) ∀ 𝑚 ∈ ℕ, ∃ 𝑛 ∈ ℕ such that 𝑠 = 𝑚⁄𝑛
∴ (𝑙 − 𝜖, 𝑙 + 𝜖) has atleast one point of set 𝐴
(4) ∀ 𝑚 ∈ ℕ, ∃ 𝑛 ∈ ℕ such that 𝑠 = 𝑚 + 𝑛
If not then 𝑙 − 𝜖 will be a supremum of 𝐴
Solution:
A contradiction to the fact that 𝑙 = sup 𝐴
(1) By Archimedean property
𝑠 1 𝑠
⇒ For any 𝜖 > 0, (𝑙 − 𝜖, 𝑙 + 𝜖) ∩ 𝐴 ≠ 𝜙
For given 𝑚 > 0, ∃ 𝑘 ∈ ℕ such that 𝑛 < 𝑚 ∀ 𝑛 ≥ 𝑘 ⇒ 𝑙 is limit point of 𝐴.
𝑚
i.e. <𝑠
𝑛
option (1) is correct Q. Let 𝑿 be a connected subset of real numbers. If every
(2) ∀ 𝑚 ∈ ℕ, ∃ 𝑛 = 1 such that
𝑚
≥1>𝑠 element of 𝑿 is irrational, then the cardinality of 𝑿 is
𝑛
(a) Infinite (b) Countably infinite
option (2) is correct
𝑚 (c) 2 (d) 1
(3) if 𝑠 = 𝑒 −1 So ∄ any 𝑚, 𝑛 ∈ ℕ s.t. 𝑠 = 𝑛 Explanation: (d)
(4) 𝑚, 𝑛 ≥ 1 ⇒ 𝑚 + 𝑛 ≥ 2 but 𝑠 ∈ (0, 1) We know that in ℝ, A set is connected if and only if it is
⇒ 𝑠 ≠ 𝑚 + 𝑛, ∀ 𝑚, 𝑛 ∈ ℕ an interval.
If 𝑋 is non-trivial interval, then it contain rational as well
Q. Let 𝑨 = ⋂∞ 𝒏=𝟏[𝒏, ∞) is as irrational numbers.
(1) Finite But 𝑋 contains only irrationals.
(2) Empty ∴ 𝑋 is trivial non-empty interval. 𝑖. 𝑒. singleton set.
(3) Countable ∴ |𝑋| = 1
(4) Uncountable
Solution: by archimedean property for any ∈ ℝ , ∃ 𝑛 ∈ ℕ Q. Let 𝑨 be a closed subset of ℝ , 𝑨 ≠ 𝝓 , 𝑨 ≠ ℝ. Then 𝑨 is
such that 𝑛 > 𝑥, (a) the closure of the interior of 𝐴
⇒ 𝑥 ∉ [𝑛, ∞) (b) a countable set
Since ′𝑥’ is arbitrary (c) a compact set
⇒ ⋂∞ 𝑛=1[𝑛, ∞) = 𝜙 (d) Not open
Explanation: (d)
Q. Consider For options (a) and (c),
𝒏 𝒙
𝑬={ | 𝒏 ∈ ℕ} , 𝑭 = { ∈ [𝟎, 𝟏)} Take 𝐴 = ℤ
𝒏+𝟏 𝟏−𝒙
IFAS Publications
∵ (ℤ̅̅̅∘ ) = 𝜙̅ = 𝜙 ≠ 𝐴 𝑈 = 𝑓 −1 ({ })
1
2
So, option (a) is incorrect.
is a closed set as we know under continuous function
Also ℤ is not compact. So, option (c) is incorrect.
inverse image of closed set is closed.
For option (b) and (d),
[2009: 6 Marks]
Take, 𝐴 = [0,1]
4. The set of all limit points of the sequence
As 𝐴 is closed but uncountable, so option (b) is incorrect. 1 1 3 1 3 5 7 1 3 5 7 9
1, 2 , 4 , 4 , 8 , 8 , 8 , 8 , 16 , 16 , 16 , 16 , 16 , … is
In ℝ, the only sets which are open as well as closed are 𝜙
and ℝ but 𝐴 ≠ 𝜙 , 𝐴 ≠ ℝ (a) [0, 1]
And 𝐴 is closed (b) (0, 1]
∴ 𝐴 is not open (c) The set of all rational numbers in [0, 1]
𝑚
Hence, option (d) is correct. (d) The set of all rational numbers in [0, 1] of the form 2𝑛
where m and n are integers.
Previous Year Questions (PYQ) Explanation: (a)
Given,
[2006: 6 Marks] 𝑚
1. Let 𝐺 be the set of all irrational numbers. The interior and 𝐷 = { 𝑛 : 𝑚, 𝑛 ∈ ℕ, 𝑚 ≤ 𝑛}
2
the closure of G are denoted by 𝐺° and 𝐺, respectively. We claim D is dense [0,1]. It is enough to show that given
Then 𝑎, 𝑏 ∈ [0,1] such that 𝑎 < 𝑏 there exists 𝑟 ∈ 𝐷 such that
(a) 𝐺° = 𝜙, 𝐺 = 𝐺 (b) 𝐺° = ℝ, 𝐺 = ℝ 𝑎 < 𝑟 < 𝑏.
𝑚
(c) 𝐺° = 𝜙, 𝐺 = ℝ (d) 𝐺° = 𝐺, 𝐺 = ℝ We work backwards. If 𝑟 = 2𝑛 is such that 𝑎 < 𝑟 < 𝑏, then
Explanation: (c) we have 2𝑛 𝑎 < 𝑚 < 2𝑛 𝑏. Thus this integer m lies
As we know that ∀𝑥 ∈ 𝐺 and for all 𝜀 > 0, (𝑥 − 𝜀, 𝑥 + 𝜀) between 2𝑛 𝑎 and 2𝑛 𝑏. To ensure this, we choose n large
contains rational numbers which do not belongs to 𝐺. enough so that 2𝑛 (𝑏 − 𝑎) > 1. In the interval (2𝑛 𝑎, 2𝑛 𝑏),
𝑚
∴ 𝑥 is not interior point of G. we can then find an integer m so that 𝑎 < 2𝑛 < 𝑏.
∴ 𝐺° = 𝜙 Hence, limit point is [0,1].
Also, ∀𝑥 ∈ ℝ, (𝑥 − 𝜀, 𝑥 + 𝜀) ∩ (𝐺\{𝑥}) ≠ 𝜙, ∀𝜀 > 0 [2010: 6 Marks]
∴ 𝐺′ = ℝ 5. Let 𝑆 be an infinite subset of ℝ such that 𝑆 ∩ ℚ = 𝜙.
∴𝐺 =𝐺∪ℝ=ℝ Which of the following statements is true?
[2008: 6 Marks] (a) S must have a limit point which belongs to ℚ
2. The set of all boundary point of ℚ in ℝ is (b) S must have a limit point which belongs to ℝ/ℚ
(a) ℝ (b) ℝ/ℚ (c) S cannot be a closed set in ℝ
(c) ℚ (d) 𝜙 (d) ℝ/𝑆 must have a limit point which belongs to S
Explanation: (c) Explanation: (d)
A point 𝑥 ∈ 𝑆 is a boundary point of set For (a), (b), (c) Take 𝑆 = {√𝑝 |𝑝 𝑖𝑠 𝑝𝑟𝑖𝑚𝑒} then 𝑆 ′ = 𝜙
𝑆 ⊆ ℝ if ∀𝛿 > 0, (𝑥 − 𝛿, 𝑥 + 𝛿) ∩ 𝑆 ≠ 𝜙
(d) ∵ 𝑆 ⊆ ℚ𝑐 ⇒ ℝ/𝑆 = ℚ ∪ 𝐴
And (𝑥 − 𝛿, 𝑥 + 𝛿) ∩ 𝑆 𝑐 ≠ 𝜙
where 𝐴 is some subset of ℚ𝑐
For any 𝑎 ∈ ℚ,
∴ (ℝ\𝑆)′ = (ℚ ∪ 𝐴)′
(𝑎 − 𝛿, 𝑎 + 𝛿) ∩ ℚ ≠ 𝜙
∴ (ℝ\𝑆)′ = ℝ
and (𝑎 − 𝛿, 𝑎 + 𝛿) ∩ ℚ𝑐 ≠ 𝜙
∴ ℝ\𝑆 must have a limit point in S.
⇒ Every rational number is boundary point of ℚ.
[2011: 6 Marks]
[2008: 6 Marks] 6. Consider the following subsets of ℝ:
1
3. The set 𝑈 = 𝑅 {𝑥 ∈ ℝ| sin 𝑥 = } is 𝑛 1
2
𝐸={ ∶ 𝑛 ∈ ℕ} , 𝐹 = { ∶ 0 ≤ 𝑥 < 1}
(a) Open 𝑛+1 1−𝑥
(b) Closed Then
(c) Both open and closed (a) Both 𝐸 and 𝐹 are closed
(d) Neither open nor closed (b) 𝐸 is closed and 𝐹 is NOT closed
Explanation: (b) (c) 𝐸 is NOT closed and 𝐹 is closed
1 (d) Neither 𝐸 nor 𝐹 is closed
Given 𝑈 = {𝑥 ∈ ℝ| sin 𝑥 = 2}
Explanation: (c)
And Set 𝑓 ∶ ℝ → ℝ by 𝑓(𝑥) = sin 𝑥 𝑛
Given 𝐸 = {𝑛+1 ; 𝑛 ∈ 𝑁}
where 𝑓 is continuous.
& 𝐹 = (1/(1 − 𝑥); 0 < 𝑥 < 1}
IFAS Publications
∵ 1 is Limit point of 𝐸 but doesn't belongs to E. Hence E is Explanation: (a)
not closed. Given, 𝑆 = {𝑥 ∈ ℝ ∶ 𝑥 6 − 𝑥 5 ≤ 100}
and 𝐹 = [1, ∞) And 𝑇 = {𝑥 2 − 2𝑥 ∶ 𝑥 ∈ (0, ∞)}
⇒ F is closed set. Clearly, S is a closed and bounded set
[2012: 6 Marks] And 𝑇 = [−1, ∞)
𝑥 And 𝑆 = [−2.14,2.361]
7. If 𝑌 = { ∶ 𝑥 ∈ ℝ} then the set of all limit points of 𝑌 is
1+|𝑥|
⇒ 𝑆 ∩ 𝑇 is closed and bounded
(a) (−1, 1) (b) (−1, 1]
[2015: 1 Mark]
(c) [0, 1] (d) [−1, 1]
12. Let 𝐴 be a non-empty subset of ℝ. Let 𝐼(𝐴) denote the set
Explanation: (d)
𝑥 of interior points of 𝐴. Then 𝐼(𝐴) can be
Let 𝑓(𝑥) =
1+|𝑥| (a) Empty
Clearly, range 𝑓 = (−1, 1) (b) Singleton
⇒ 𝑌 = (−1, 1) (c) A finite set containing more than one element
⇒ Y’ (set of limit point of 𝑌) = [−1, 1] (d) Countable but not finite
[2013 : 2 Marks] Explanation: (a)
8. Let A and B be subsets of ℝ. Which of the following is NOT Let 𝐴 ≠ 𝜙 ⊆ ℝ
necessarily true? Then 𝐼(𝐴) is open subset of ℝ then (b), (c) & (d) are
(a) (𝐴 ∩ 𝐵)° ⊆ 𝐴° ∩ 𝐵° (b) 𝐴° ∪ 𝐵° ⊆ (𝐴 ∪ 𝐵)° incorrect
(c) 𝐴 ∪ 𝐵 ⊆ 𝐴 ∪ 𝐵 (d) 𝐴 ∩ 𝐵 ⊆ 𝐴 ∩ 𝐵 Take 𝐴 = {0}
Explanation: (d) Then 𝐼(𝐴) = 𝜙
Let 𝐴, 𝐵 ⊆ ℝ ∴ Option (a) is correct.
Take 𝐴 = ℚ, 𝐵 = ℚ𝑐 [2015: 1 Mark]
ℚ ∩ ℚ𝑐 = ℝ 13. Let 𝑆 be a non-empty subset of ℝ. If S is a finite union of
ℚ ∩ ℚ𝑐 = 𝜙 disjoint bounded intervals, then which one of the
So, 𝐴 ∩ 𝐵 ⊆ 𝐴 ∩ 𝐵 is not true following is true?
[2014 : 2 Marks] (a) If S is not compact, then sup 𝑆 ∉ 𝑆 and inf 𝑆 ∉ 𝑆
𝑥2 (b) Even if sup 𝑆 ∉ 𝑆 and inf 𝑆 ∉ 𝑆, S need not be compact
9. The se {1+𝑥 2 ∶ 𝑥 ∈ ℝ} is
(c) If sup 𝑆 ∉ 𝑆 and inf 𝑆 ∉ 𝑆, then S is compact
(a) Connected but NOT compact in ℝ
(d) Even if S is compact, it is not necessary that sup 𝑆 ∉ 𝑆
(b) Compact but NOT connected in ℝ
and inf 𝑆 ∉ 𝑆
(c) Compact and connected in ℝ
Explanation: (b)
(d) Neither compact nor connected in ℝ
(a) Take 𝑆 = [0,3) ∪ (5,8] then (a) not true
Explanation: (a)
(b) True
𝑥2
Set {1+𝑥 2 ∶ 𝑥 ∈ ℝ} = [0,1) (c) False. Take example same as in (a)
Hence, given set is connected but not compact (d) It is not true because if S is compact
[2014: 2 Marks] ⇒ S is closed.
10. The set of all limit points of the set {
2
∶ 𝑥 ∈ (−1,1)} in ∴ It contains all of its limit points
𝑥+1
∴ 𝑠𝑢𝑝 𝑆 ∈ 𝑆 & 𝑖𝑛𝑓 𝑆 ∈ 𝑆
ℝ is
[2015: 2 Marks]
(a) [1, ∞) (b) (1, ∞) 1 1
(c) [−1,1] (d) [−1, ∞) 14. Let 𝑆 = ⋂∞
𝑛=1 ([0, 2𝑛+1] ∪ [2𝑛 , 1])
Explanation: (a) Which one of the following statements is FALSE?

The set {
2
∶ 𝑥 ∈ (−1,1)} = (1, ∞) (a) There exist sequences {𝑎𝑛 } and {𝑏𝑛 } in [0,1] such that
𝑥+1
𝑆 = [0,1]\ ⋂∞ 𝑛=1(𝑎𝑛 , 𝑏𝑛 )
And limits points of the set = [1, ∞)
(b) [0,1]\S is an open set
[2014: 2 Marks]
(c) If A is an infinite subset of S, then A has a limit point
11. Let 𝑆 = ∈ ℝ ∶ 𝑥 − 𝑥 ≤ 100} and 𝑇 = {𝑥 2 − 2𝑥: 𝑥 ∈
{𝑥 6 5
(d) There exists an infinite subset of S having no limits
(0, ∞)}. The set 𝑆 ∩ 𝑇 is
points
(a) Closed and bounded in ℝ
Explanation: (d)
(b) Closed but NOT bounded in ℝ
Here 𝑆 = {0} ∪ [1/2,1]
(c) Bounded but NOT closed in ℝ
(a) Take 〈𝑎𝑛 〉 = 〈0〉, 〈𝑏𝑛 〉 = 〈1/2〉
(d) Neither closed nor bounded in ℝ
Then 𝑆 = [0,1]\ ⋃∞ 𝑛=1(𝑎𝑛 , 𝑏𝑛 )
IFAS Publications
= [0,1]\(0,1/2) Explanation: (c)
= {0} ∪ [1/2,1] 𝑇 = (0,1], 𝑉 = [1,2)
(b) is true 𝜕𝑇 = {0,1}, 𝜕𝑉 = {1,2}
(c) is true because if A is infinite subsets of S. 𝜕𝑇 ∪ 𝜕𝑉 = {0,1,2}
∵ S is bounded ⇒ A is bounded. 𝜕(𝑇 ∪ 𝑉) = 𝜕(0,2) = {0,2}
∴ A has a limit point. ⇒ 𝜕(𝑇 ∪ 𝑉) ≠ 𝜕𝑇 ∪ 𝜕𝑉
(d) It is false by reason given in (c) [2017: 2 Marks]
[2015 : 2 Marks] 18. Let 𝑆 be an infinite subset of ℝ such that 𝑆\{𝛼} is compact
15. Let G and H be non-empty subsets of R, where G is for some 𝛼 ∈ 𝑆.
connected and 𝐺 ∪ 𝐻 is not connected. Which one of the Then which one of the following is TRUE?
following statements is true for all such G and H? (a) 𝑆 is a connected set
(a) If 𝐺 ∩ 𝐻 = 𝜙, then H is connected (b) 𝑆 contains no limit points
(b) If 𝐺 ∩ 𝐻 = 𝜙, then H is not connected (c) 𝑆 is a union of open intervals
(c) If 𝐺 ∩ 𝐻 ≠ 𝜙 then H is connected (d) Every sequence in 𝑆 has a subsequence converging to
(d) If 𝐺 ∩ 𝐻 ≠ 𝜙 then H is not connected an elements in 𝑆
Explanation: (d) Explanation: (d)
Discard by examples, Let 𝑆 = [1, 2] ∪ {3} then (a), (b) and (c) incorrect.
(a) Let 𝐺 = [2,3] & 𝐻 = {4} ∪ {5} [2019 : 1 Mark]
Here 𝐺 ∪ 𝐻 is not connected 𝑛 √2
19. Let 𝑆 be the set of all limit points of the set { + 𝑛
∶𝑛∈
√2
and 𝐺⋂𝐻 = 𝜙 but H is not connected.
ℕ}. Let ℚ+ be the set of all positive rational numbers.
(b) Take 𝐺 = [1,2] & 𝐻 = {3} then (b) is false.
(c) Take 𝐺 = [1,2] and 𝐻 = {2} ∪ {3} then 𝐺 ∩ 𝐻 ≠ 𝜙 & Then
H is not connected. (a) ℚ+ ⊆ 𝑆 (b) 𝑆 ⊆ ℚ+
(d) Now 𝐺 ∪ 𝐻 is not connected and 𝐺 ∩ 𝐻 ≠ 𝜙 (c) 𝑆 ∩ (ℝ/ℚ+ ) ≠ 𝜙 (d) 𝑆 ∩ ℚ+ ≠ 𝜙
Clearly G is an interval. Explanation: (b)
𝑛 √2
Now H have a point which is in G also. Let 𝐴 = { + ∶ 𝑛 ∈ ℕ}
√2 𝑛
Now if H is an interval ⇒ 𝐺 ∪ 𝐻 is an interval. ⇒𝑆=𝐴 =ϕ ′
⇒ 𝐺 ∪ 𝐻 is connected, which is contradiction. ϕ ⊆ ℚ+
∴ H is not an interval. ⇒ 𝑆 ⊆ ℚ+
[2016 : 1 Mark] [2019: 2 Marks]
16. Let S be a closed subset of ℝ, T a compact subset of ℝ such 𝑥
20. The set {1+𝑥 ∶ −1 < 𝑥 < 1}, as a subset of ℝ, is
that 𝑆 ∩ 𝑇 ≠ 𝜙. Then 𝑆 ∩ 𝑇 is
(a) Connected and compact
(a) Closed but not compact
(b) Connected but not compact
(b) Not closed
(c) Not connected but compact
(c) Compact
(d) Neither connected nor compact
(d) Neither closed nor compact
Explanation:(b)
Explanation: (c)
𝑥
𝑔𝑖𝑣𝑒𝑛 𝑆 = [0, ∞) and 𝑇 = [0,2] 𝑆={ ∶ −1 < 𝑥 < 1}
1+𝑥
Then 𝑆 ∩ 𝑇 = [0,2] ≠ 𝜙 −1 < 𝑥 < 1
Here 𝑆 ∩ 𝑇 = [0,2] is compact so options (a), (b) & (d) are ⇒ <
1 1
2 𝑥+1
incorrect. −1 −1
⇒ >
[2016 : 2 Marks] 2 𝑥+1
1 1
17. Let 𝑆 ⊂ ℝ and 𝜕𝑆 denote the set of points x in ℝ such that ⇒ 1− > 1−
2 𝑥+1
every neighborhood of x contains some points of S as well 1
⇒ 2 > 𝑥+1
𝑥
as some points of complement of S. Further, let 𝑆 denote 1

∴ 𝑆 = (−∞, 2)
the closure of S. Then which one of the following is FALSE?
(a) 𝜕ℚ = ℝ S is connected but not compact.
(b) 𝜕(ℝ/𝑇) = 𝜕𝑇, 𝑇 ⊂ ℝ [2019: 2 Marks]
1 1
(c) 𝜕(𝑇 ∪ 𝑉) = 𝜕𝑇 ∪ 𝜕𝑉 , 𝑇, 𝑉 ⊂ ℝ, 𝑇 ∩ 𝑉 ≠ 𝜙 21. The set {𝑚 + 𝑛 : 𝑚, 𝑛 ∈ ℕ} ∪ {0}, as a subset of ℝ is
(d) 𝜕𝑇 = 𝑇 ∩ (ℝ/𝑇), 𝑇 ⊂ ℝ (a) Compact and open
(b) Compact but not open
IFAS Publications
(c) Not compact but open 4. If ℚ ⊂ 𝐴 ⊂ ℝ, which of the following must be true?
(d) Neither compact nor open (A) If 𝐴 is open, then 𝐴 = ℝ
Explanation: (b) (B) If A is closed, then 𝐴 = ℝ
1 1 (C) If A is uncountable, then A is closed
𝑆 = { + ∶ 𝑚, 𝑛 ∈ ℕ} ∪ {0}
𝑚 𝑛
1 (D) If A is countable, then A is closed
𝑆 ′ = {𝑛 ∶ 𝑛 ∈ ℕ} ∪ {0}
𝑆 ′ ⊆ 𝑆 ⇒ 𝑆 is closed. 5. The subset 𝐴 = {𝑥 ∈ ℚ ∶ 𝑥 2 < 4} of ℝ is
1 1
Also, 0 < 𝑚 + 𝑛 ≤ 2 (A) bounded above but not bounded below.
⇒ 𝑆 is bounded (B) bounded above and sup 𝐴 = 2.
⇒ 𝑆 is compact (C) bounded above but does not have a supremum.
Also 𝑆 is countably infinite (D) not bounded above.
⇒ 𝑆 is not open.
6. The subset 𝐴 = {𝑥 ∈ ℚ ∶ −1 < 𝑥 < 0} ∪ ℕ of ℝ is
[2016 : 2 Marks]
(A) bounded, infinite set and has a limit point in ℝ
22. Which of the following statement(s) is (are) TRUE?
(B) unbounded, infinite set and has a limit point in ℝ
(a) There exists a connected set in ℝ which is not compact
(C) unbounded, infinite set and does not have a limit point
(b) Arbitrary union of closed intervals in ℝ need not be
in ℝ
compact
(D) bounded, infinite set and does not have a limit point in
(c) Arbitrary union of closed intervals in ℝ is always closed
(d) Every bounded infinite subset 𝑉 of ℝ has a limit point ℝ
in 𝑉 itself 1 1
Explanation: (a) & (b) 7. LetS = ∏∞
𝑛=1 [− 𝑛 , 1 + 𝑛 ], then S equals
(a) 𝐴 = [0, ∞) is connected but not compact. (A) [0, 1] (B) (0, 1]
(b) ∪ 𝐴𝛼 = ℝ ; 𝛼 ∈ ℝ define 𝐴𝛼 = {𝛼} ; ∀𝛼 ∈ ℝ which is (C) (0, 1) (D) [0, 1)
not compact.
1 1 8. The set 𝑆 = {sin 𝑥 : 𝑥 ∈ ℕ} has the set of limit points:
(c) ∵ ⋃𝑛∈𝑁 [𝑛 , 1 + 𝑛] = (0, 2] which is not closed in ℝ
1
(A) (−1, 1) (B) [−1, 1]
(d) 𝑉 = {𝑛 } bounded & infinite & have limit point {0} but (C) ℕ (D) ℝ
0 ∉ 𝑉.
9. 𝑆 = {(𝑥, 𝑦) ∈ ℝ2 : 𝑥𝑦 < 0} is
Multiple Choice Questions (MCQ) (A) neither connected nor compact subset of ℝ2
(B) not connected but is compact subset of ℝ2
1. Which one of the following subsets of ℝ has a non-empty
(C) is both connected and compact subset of ℝ2
interior?
(D) is not compact subset of ℝ2 but connected
(A) The set{𝑏 ∈ ℝ ∶ 𝑥 2 + 𝑏𝑥 + 1 = 0 has distinct roots}.
(B) The set {𝑎 ∈ ℝ ∶ sin(𝑎) = 1}. 𝑝
10. The limit points of the set {𝑞 : 𝑝, 𝑞 ∈ ℕ} are
(C) The set of all irrational numbers in ℝ
(A) ℝ (B) 𝑄
(D) The set of all rational number in ℝ
(C) (0, ∞) (D) [0, ∞)
𝑓(𝑛)
2. Let 𝑓 ∶ ℕ → ℕ be a bijective map such that ∑∞
𝑛=1 𝑛2
< Multiple Select Questions (MSQ)
+∞ The number of such bijective maps is
(A) infinite 11. Let X = {(𝑥, 𝑦) ∈ ℝ2 : 𝑥 ∈ ℚ, 𝑦 ∈ ℝ\ℚ} where ℚ is the set
(B) exactly one of rationals. Then
(C) zero (A) X is an open and dense subset of ℝ2
(D) finite but more than one. (B) X is an open but not dense subset of ℝ2
(C) X is not an open but a dense subset of ℝ2
3. Let 𝑆 and 𝑇 be subsets of ℝ. Select the incorrect (D) X is neither an open nor a dense subset of ℝ2
statement:
12. Let S = {𝑥 ∈ ℝ ∶ 3 − 𝑥 2 > 0}. Then
(A) (int 𝑆) ∩ (int 𝑇) = int (𝑆 ∩ 𝑇)
(A) S is bounded above and 3 is the least upper bound of
(B) (int 𝑆) ∪ (int 𝑇) ⊂ int (𝑆 ∪ 𝑇)
S.
(C) 𝑆̅ is closed in ℝ
(B) S is bounded above and does not have a least upper
(D) 𝑇̅ is the largest closed set containing 𝑇.
bound in ℝ
IFAS Publications
(C) S is bounded above and does not have a least upper In light of the above statements, choose the correct
bound in ℚ, the set of rational numbers. answer from the options given below.
(D) S is not bounded above. (A) Both Statement I and Statement II are true
(B) Both Statement I and Statement. II are false
13. Let 𝑆 be the set of all rational numbers in (0, 1).Then which (C) Statement I is true but Statement II is false
of the following statements is/are TRUE? (D) Statement I is false but Statement Il is true
(A) S is a closed subset of ℝ
(B) S is not a closed subset of ℝ 19. Let 𝑆 be a subset of ℝ such that 2018 is an interior point
(C) S is an open subset of ℝ of S. Which of the following is (are) TRUE?
(D) Every 𝑥 ∈ (0, 1)\S is a limit point of S (A) S contains an interval
(B) There is sequence in S which does not converge to
𝑛𝜋
14. If 𝑎𝑛 = 𝑛sin( 2 ) then 2018
(A) lim sup 𝑎𝑛 = +∞, lim inf 𝑎𝑛 = −1 (C) There is element 𝑦 ∈ 𝑠, 𝑦 ≠ 2018 such that y is also an
(B) lim sup 𝑎𝑛 = +∞, lim inf 𝑎𝑛 = 0 interior point of S
(C) lim sup 𝑎𝑛 = +∞, lim inf 𝑎𝑛 = −∞ (D) There is a point 𝑧 ∈ 𝑆, such that |𝑧 − 2018| =
(D) lim sup 𝑎𝑛 = 1, lim inf 𝑎𝑛 = −1 0.002018
15. Which of the following subsets of ℝ is (are) connected? 20. Let for each 𝑛 ≥ 1, 𝑆𝑛 be the open disc in 𝑅2 , with center
(A) {𝑥 ∈ ℝ |𝑥 2 + 𝑥 > 4} at a point (𝑛, 0) and radius equal to n. Then 𝑆 = 𝑈𝑛≥1 𝑆𝑛
(B) {𝑥 ∈ ℝ |𝑥 2 + 𝑥 < 4} is
(C) {𝑥 ∈ ℝ | |𝑥| < |𝑥 − 4|} (A) {(𝑥, 𝑦) ∈ 𝑅2 : 𝑥 > 0 𝑎𝑛𝑑 |𝑦 | < 𝑥}
(D) {𝑥 ∈ ℝ | |𝑥| > |𝑥 − 4|} (B) {(𝑥, 𝑦)} ∈ 𝑅2 : 𝑥 > 0}
(C) {(𝑥, 𝑦) ∈ 𝑅2 : 𝑥 < 0 𝑎𝑛𝑑 |𝑦 | < 2𝑥}
16. Let P and Q be two non-empty disjoint subsets of ℝ. (D) {(𝑥, 𝑦) ∈ 𝑅2 : 𝑥 > 0 𝑎𝑛𝑑 |𝑦 | < 3𝑥}
Which of the following is (are) FALSE?
(A) If P and Q are compact, then 𝑃 ∪ 𝑄 is also compact Numerical Answer Type (NAT)
(B) If P and Q are not connected, then 𝑃 ∪ 𝑄 is also not 1
21. The limit point of the set 𝑆 = {𝑛 ; 𝑛 ∈ ℕ} is ___.
connected
(C) If 𝑃 ∪ 𝑄 and P are closed, then Q is closed
1
(D) If 𝑃 ∪ 𝑄 and P are open, then Q is open 22. The cardinality of limit point of the set 𝑆 = { ; 𝑛 ∈ ℕ} is
𝑛
___.
17. Given below are two statements, one is labelled as
Assertion A and the other is labelled as Reason R. 1
23. The cardinality of interior points of the set 𝑆 = {1 + 𝑛 ∶
1
Assertion A: If 𝑆 = {𝑛2 : 𝑛 ∈ ℕ} then inf 𝑆 = 0
𝑛 ∈ ℕ} is ___.
Reason R: If 𝑥 ∈ ℝ then there exists 𝑛𝑥 ∈ ℕ such that 𝑥 <
𝑛𝑥 . 1 1
24. The set 𝑆 = { , 2, , 3, … } has a limit point __.
In light of the above statements, choose the correct 2 3
answer from the options given below.

𝑛
1
(A) Both A and R are true and R is the correct explanation 25. Let 𝑆𝑛 = {(−1)[ 2 ] (1 + ) : 𝑛 ∈ ℕ}.Then lim sup 𝑆𝑛 is ___
𝑛
of A
(B) Both A and Rare true but R is NOT the correct [ ]
𝑛
1
explanation of A 26. Let 𝑆𝑛 = {(−1) 2 (1 + 𝑛) : 𝑛 ∈ ℕ}.Then sup 𝑆𝑛 is ___.
1
(C) A is true but R is false 27. The limit points of the set 𝑆 = {1 + : 𝑛 ∈ ℕ} is __.
𝑛
(D) A is false but R is true 28. The limit inferior of the sequence (𝑥𝑛 ) where 𝑥𝑛 = 1 +
1
(−1)𝑛 + is _____.
18. Given below are two statements. 3𝑛
Statement l: Every compact subset 𝑆 ⊆ ℝ contains a

29. Consider the set
maximum and a minimum element.
1
Statement II: If 𝑆 ⊆ ℝ contains a maximum and a S = {𝑛 ∶ 𝑛 ∈ ℕ and 𝑛 is prime} T = {𝑥 2 ∶ 𝑥 ∈ ℝ}. Then
minimum element, then 𝑆 is compact. sup S is ___.
IFAS Publications
30. Consider the set
1
S = {𝑛 ∶ 𝑛 ∈ ℕ and 𝑛 is prime} T = {𝑥 2 ∶ 𝑥 ∈ ℝ}. Then
inf S is ___.
Answer Key
Multiple Choice Questions
1 2 3 4 5 6 7 8 9 10
C B D B B B A B A D
Multiple Select Questions
11 12 13 14 15 16 17 18 19 20
C C B,D B B,C,D B,C,D A C A,B,C B
Numerical Answer Type (NAT)
21 22 23 24 25 26 27 28 29 30
0 1 0 0 1 2 1 0 ½ 1
IFAS Publications
3
CHAPTER
SEQUENCES
3.1 Definition: 3.2 Bounded & Monotonic sequence:
Let 𝑋 be a non-empty set. Then a function 𝑓 ∶ ℕ → 𝑋 is called 3.2.1 Bounded Sequence:
sequence of the elements of 𝑋. A sequence is bounded if and only if it’s range set is bounded.
Here, i.e. 〈𝑎𝑛 〉 is bounded iff there exists 𝑚, 𝑀 ∈ ℝ such that
If 𝑓 ∶ ℕ → 𝑋, then the arrangement 𝑚 ≤ 𝑎𝑛 ≤ 𝑀, ∀ 𝑛 ∈ ℕ
𝑓(1), 𝑓(2), 𝑓(3), … , 𝑓(𝑛), 𝑓(𝑛 + 1) …. is called sequence of
elements in 𝑋 Example: 〈𝑎𝑛 〉 = 〈sin(𝑛)〉 is bounded sequence
If 𝑋 = ℝ ⇒ sequence of real numbers i.e. 𝑓 ∶ ℕ → ℝ 〈𝑏𝑛 〉 = 〈𝑛〉 is not bonded sequence.
If 𝑋 = ℚ ⇒ sequence of rational numbers i.e. 𝑓 ∶ ℕ → ℚ
If 𝑋 = ℂ ⇒ sequence of complex numbers i.e. 𝑓 ∶ ℕ → ℂ Note:
If 𝑋 = ℕ ⇒ sequence of natural numbers i.e. 𝑓 ∶ ℕ → ℕ 1) If 𝑓 ∶ 𝐴 → 𝐵 such that 𝐴 is infinite & 𝑓(𝐴) is finite
⇒ 𝑓(𝐴) has smallest & greatest element in 𝐵.
Note: Let 𝛼 = inf(𝑓(𝐴)) & 𝛽 = sup(𝑓(𝐴))
(i) We define 𝑓(𝑛) = 𝑎𝑛 & denote the sequence as 〈𝑎𝑛 〉 = ⇒ 𝛼 ≤ 𝑓(𝑥) ≤ 𝛽, ∀ 𝑥 ∈ 𝐴 ⇒ 𝑓 is bounded
{𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , … } Moreover,
(ii) 𝑎𝑛 is called 𝑛𝑡ℎ term of the sequence 〈𝑎𝑛 〉. If range of a sequence is finite then there exist an element in
(iii) The set containing values of all terms is called as range of 𝑋 which appears in the sequence infinitely many times.
the sequence.
(iv) Range set may finite or infinite but it is always countable. 3.2.2 Monotonic Sequence:
A Sequence 〈𝑎𝑛 〉 is said to be monotonic if it satisfy any of the
Examples: following condition
1. 〈𝑎𝑛 〉 = 〈𝑛〉 1. If 𝑎𝑛 ≤ 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ , Then 〈𝑎𝑛 〉 is non-decreasing/
2. 〈𝑎𝑛 〉 = 〈−𝑛〉 monotonically increasing sequence.
3. 〈𝑎𝑛 〉 = 〈𝑛2 〉 2. 𝑎𝑛 ≥ 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ , Then 〈𝑎𝑛 〉 is Non-increasing/
1
4. 〈𝑎𝑛 〉 = 〈𝑛〉 monotonically decreasing sequence.
5. 〈𝑎𝑛 〉 = sin(𝑛) 3. 𝑎𝑛 < 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ , Then 〈𝑎𝑛 〉 is Increasing/strictly
6. 〈𝑎𝑛 〉 = cos(𝑛) increasing sequence.
3
7. 𝑎1 = , 𝑎𝑛+1 = √2 + 𝑎𝑛 4. 𝑎𝑛 > 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ , Then 〈𝑎𝑛 〉 is Decreasing/strictly
2
1 1 1 decreasing sequence.
8. 𝑎𝑛 = sin(1) − sin (2) + sin (3) … … + (−1)𝑛+1 sin (𝑛)
𝝅
Eventually Monotonic sequence:
Q. Let 𝒂𝒏 = 𝐬𝐢𝐧 𝒏. For the sequence 𝒂𝟏 , 𝒂𝟐 , … the supremum A sequence 〈𝑎𝑛 〉 is said to be eventually monotonic if ∃ 𝑘 ∈ ℕ
is such that 〈𝑏𝑛 〉 is monotonic where 𝑏𝑛 = 𝑎𝑛+𝑘
(a) 0 and it is attained
(b) 0 and it is not attained Constant sequence:
(c) 1 and it is attained A sequence 〈𝑎𝑛 〉 is said to be constant sequence if 𝑎𝑛 =
(d) 1 and it is not attained 𝛼, ∀ 𝑛 ∈ ℕ, where 𝛼 ∈ ℝ
𝜋
Solution: We know that |sin 𝑛| ≤ 1 ∀ 𝑛 ∈ ℕ
𝜋 Eventually constant sequence:
For 𝑛 = 2, 𝑎2 = sin 2 = 1
A sequence 〈𝑎𝑛 〉 is said to be eventually constant sequence, if
∴ for 𝑛 = 2, 1 is the supremum & it is attained
∃ 𝑚 ∈ ℕ such that 𝑎𝑛 = 𝛼, ∀ 𝑛 ≥ 𝑚, for some 𝛼 ∈ ℝ
Clearly, constant sequence is monotonic & bounded

(i) If 𝑎𝑛 = 𝛼, ∀ 𝑛 ∈ ℕ Assume that 𝑎𝑘 ≤ 𝑎
⇒ 𝛼 ≤ 𝑎𝑛 ≤ 𝛼, ∀ 𝑛 ∈ ℕ ⇒ 𝑎𝑘 . 𝑎 ≤ 𝑎. 𝑎
⇒ 〈𝑎𝑛 〉 is bounded sequence. ⇒ √𝑎𝑘 . 𝑎 ≤ √𝑎. 𝑎 = 𝑎
(ii) 𝛼 = 𝑎𝑛 ≤ 𝑎𝑛+1 = 𝛼 , ∀ 𝑛 ∈ ℕ ⇒ 𝑎𝑘+1 ≤ 𝑎
⇒ 〈𝑎𝑛 〉 is monotonic sequence. Thus by mathematical induction
𝑎𝑛 ≤ 𝑎 ∀ 𝑛 ∈ ℕ
Examples: Also 𝑎𝑛 ≥ 0, ∀ 𝑛 ∈ ℕ
Check whether the following sequence is ⇒ 0 ≤ 𝑎𝑛 ≤ 𝑎, ∀ 𝑛 ∈ ℕ
(i) bounded ⇒ 〈𝑎𝑛 〉 is bounded sequence
(ii) monotonic 𝑎1 = √𝑎, 𝑎2 = √𝑎 √𝑎 ≥ √𝑎. 1 = 𝑎1 (∴ 𝑎 ≥ 1)
𝑎2 ≥ 𝑎1
𝟏
Q. 𝒂𝒏 = 〈 〉 Similarly, 𝑎3 ≥ 𝑎2
𝒏
1
Solution: Here for any 𝑛 ∈ ℕ , 0 ≤ 𝑛 ≤ 1 Assume that 𝑎𝑘+1 ≥ 𝑎𝑘
1 𝑎. 𝑎𝑘+1 ≥ 𝑎. 𝑎𝑘
⇒ 〈 〉 is bounded Sequence.
𝑛 ⇒ √𝑎. 𝑎𝑘+1 ≥ √𝑎. 𝑎𝑘
1 1
𝑎𝑛 = 𝑛 , 𝑎𝑛+1 = 𝑛+1 ⇒ 𝑎𝑘+2 ≥ 𝑎𝑘+1
1 1 Thus by mathematical induction
We know that 𝑛 < 𝑛 + 1 ⇒ 𝑛 > 𝑛+1 , ∀ 𝑛 ∈ ℕ
𝑎𝑛 ≤ 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ
⇒ 𝑎𝑛 > 𝑎𝑛+1 ∀ 𝑛 ∈ ℕ
〈𝑎𝑛 〉 is increasing sequence
⇒ 〈𝑎𝑛 〉 is monotonically decreasing sequence.
∴ 〈𝑎𝑛 〉 is monotonically increasing & bounded sequence.
Q. Given 𝒂𝟏 = √𝒂 & 𝒂𝒏+𝟏 = √𝒂 + 𝒂𝒏 , ∀ 𝒏 ≥ 𝟏, 𝒂 ≥ 𝟐 𝟑 𝟏 𝟐

Q. 𝒂𝟏 = 𝟐 & 𝒂𝒏+𝟏 = 𝟐 (𝒂𝒏 + 𝒂 ) , 𝒏 ≥ 𝟏
Solution: Here 𝑎1 = √𝑎 ≤ 𝑎 𝒏
3 1 2
𝑎2 = √𝑎 + 𝑎1 = √𝑎 + √𝑎 ≤ 𝑎 Solution: Given 𝑎1 = 2 & 𝑎𝑛+1 = 2 (𝑎𝑛 + 𝑎 ) , ∀ 𝑛 ≥ 1
𝑛
Suppose, it is true for 𝑛 = 𝑘 i.e. 𝑎𝑘 ≤ 𝑎 Using 𝐴. 𝑀. & 𝐺. 𝑀. inequality
2 1
⇒ 𝑎 + 𝑎𝑘 ≤ 𝑎 + 𝑎 = 2𝑎 (We know that 𝑎𝑘+1 = √𝑎 + 𝑎𝑘 ) 𝑎𝑛 +
𝑎𝑛 2 2
⇒ ≥ (𝑎𝑛 . 𝑎 )
2
⇒ √𝑎 + 𝑎𝑘 ≤ √2𝑎 ≤ 𝑎 2
𝑛
𝑎𝑛 +
⇒ 𝑎𝑘+1 ≤ 𝑎 ⇒ 𝑎𝑛+1 =
𝑎𝑛
≥ √2
2
∴ If 𝑎𝑘 ≤ 𝑎 ⇒ 𝑎𝑘+1 ≤ 𝑎, ∀ 𝑘 ∈ ℕ
⇒ 𝑎𝑛+1 ≥ √2, ∀ 𝑛 ∈ ℕ
∴ by mathematical induction
⇒ 𝑎𝑛 ≥ √2, ∀ 𝑛 ∈ ℕ
𝑎𝑛 ≤ 𝑎, ∀ 𝑛 ∈ ℕ
⇒ 〈𝑎𝑛 〉 is bounded below.
Also 𝑎𝑛 ≥ 0, ∀ 𝑛 ∈ ℕ ⇒ 0 ≤ 𝑎𝑛 ≤ 𝑎, ∀ 𝑛 ∈ ℕ
Consider
⇒ 〈𝑎𝑛 〉 is bounded sequence. 1 2 𝑎𝑛 1 𝑎𝑛 1
𝑎𝑛 − 𝑎𝑛+1 = 𝑎𝑛 − 2 (𝑎𝑛 + 𝑎 ) = 𝑎𝑛 − −𝑎 = −𝑎
Here 𝑎1 = √𝑎, 𝑎2 = √𝑎 + √𝑎 > √𝑎 + 0 = √𝑎 = 𝑎1 𝑛 2 𝑛 2 𝑛
2 −2
1 𝑎𝑛
𝑎1 < 𝑎2 𝑎𝑛 − 𝑎𝑛+1 = 2 ( )
𝑎𝑛
Similarly, 𝑎2 < 𝑎3 (∵ 𝑎2 = √𝑎 + √𝑎, 𝑎3 = √𝑎 + 𝑎2 = Here 𝑎𝑛 ≥ 0, ∀ 𝑛 ∈ ℕ & 𝑎𝑛 ≥ √2 ∀ 𝑛 ∈ ℕ

⇒ 𝑎𝑛2 ≥ 2, ∀ 𝑛 ∈ ℕ
√𝑎 + √𝑎 + √𝑎 > √𝑎 + √𝑎 + 0 = 𝑎2 ) ⇒ 𝑎𝑛2 − 2 ≥ 0, ∀ 𝑛 ∈ ℕ
2 −2
1 𝑎𝑛
⇒ 𝑎𝑛 − 𝑎𝑛+1 = 2 ( ) ≥ 0∀𝑛 ∈ ℕ
Assume that 𝑎𝑘 ≤ 𝑎𝑘+1 𝑎𝑛
(claim: 𝑎𝑘+1 ≤ 𝑎𝑘+2) ⇒ 𝑎𝑛 − 𝑎𝑛+1 ≥ 0, ∀ 𝑛 ∈ ℕ

𝑎𝑘 + 𝑎 ≤ 𝑎𝑘+1 + 𝑎 ⇒ 𝑎𝑛 ≥ 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ
⇒ 〈𝑎𝑛 〉 is monotonically decreasing sequence
√𝑎𝑘 + 𝑎 ≤ √𝑎𝑘+1 + 𝑎
⇒ 𝑎1 ≥ 𝑎2 ≥ 𝑎3 ≥ 𝑎4 ≥. . . ≥ 𝑎𝑛 ≥ 𝑎𝑛+1 ≥. …
⇒ 𝑎𝑘+1 ≤ 𝑎𝑘+2
⇒ Sequence 〈𝑎𝑛 〉 is bounded above by it’s first term 𝑎1
Hence 𝑎𝑛 ≤ 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ 3
Thus, 〈𝑎𝑛 〉 is monotonically increasing & bounded sequence. ⇒ 𝑎𝑛 ≤ 𝑎1 = 2
3
Hence √2 ≤ 𝑎𝑛 ≤ 2 , ∀ 𝑛 ∈ ℕ
Q. 𝒂𝟏 = √𝒂 & 𝒂𝒏+𝟏 = √𝒂. 𝒂𝒏 , ∀ 𝒏 ≥ 𝟏, 𝒂 ≥ 𝟏 ⇒ 〈𝑎𝑛 〉 is bounded & monotonic sequence.
Solution: Here 𝑎1 = √𝑎 ≤ 𝑎
𝑎2 = √𝑎. √𝑎 ≤ √𝑎. 𝑎 = 𝑎
IFAS Publications
Chapter – 3 Sequences 31
𝟏 𝟏 𝟏 1 𝑛−1
Q. 𝒂𝒏 = 𝒏+𝟏 + 𝒏+𝟐 +. . . + 𝒏+𝒏 1−( )
2
𝑎𝑛 ≤ 1 + ( 1 )
1 1 1 1−
Solution: Given 𝑎𝑛 = + +. . . + 2
𝑛+1 𝑛+2 𝑛+𝑛 1 𝑛−1
1 1 1 1 (1−( ) ) 1 2
2
𝑎𝑛+1 = (𝑛+1)+1 + (𝑛+1)+2 +. . . + (𝑛+1)+(𝑛−1) + (𝑛+1)+(𝑛) + 𝑎𝑛 ≤ 1 + 1 = 1 + 2 (1 − 2𝑛−1) = 1 + 2 − 2𝑛−1
1 2
1
(𝑛+1)+(𝑛+1) ⇒ 𝑎𝑛 ≤ 3 − 2𝑛−2 < 3 ∀ 𝑛 ∈ ℕ
1 1 1 1 1
= + +. . . + + + ⇒ 𝑎𝑛 < 3, ∀ 𝑛 ∈ ℕ
𝑛+2 𝑛+3 2𝑛 2𝑛+1 2𝑛+2
1 1 1 1 1 1 Also 0 ≤ 𝑎𝑛 , ∀ 𝑛 ∈ ℕ
𝑎𝑛+1 − 𝑎𝑛 = (𝑛+2 + 𝑛+3 +. . . + 2𝑛 + 2𝑛+1 + 2𝑛+2) − (𝑛+1 +
1 1
⇒ 〈𝑎𝑛 〉 is bounded sequence.
𝑛+2
+. . . + 𝑛+𝑛) Hence, 〈𝑎𝑛 〉 is bounded & monotonically increasing sequence.
1 1 1
𝑎𝑛+1 − 𝑎𝑛 = − 𝑛+1 + 2𝑛+1 + 2𝑛+2
−2 1 1 Exercise:
⇒ 𝑎𝑛+1 − 𝑎𝑛 = + +
2(𝑛 + 1) 2𝑛 + 1 2𝑛 + 2 i) 𝑎1 = √2, 𝑎𝑛+1 = √2 + 𝑎𝑛
−2 1 1
⇒ 𝑎𝑛+1 − 𝑎𝑛 = 2𝑛+2 + 2𝑛+1 + 2𝑛+2 ii) 𝑎1 = √3, 𝑎𝑛+1 = √3 + 𝑎𝑛
1 1
⇒ 𝑎𝑛+1 − 𝑎𝑛 = − iii) 𝑎1 = √7, 𝑎𝑛+1 = √7 ∙ 𝑎𝑛
2𝑛+1 2𝑛+2
Here, 2𝑛 + 1 < 2𝑛 + 2, ∀ 𝑛 ∈ ℕ show that all sequences are monotonically increasing &
1 1 bounded.
⇒ 2𝑛+1 > 2𝑛+2 , ∀ 𝑛 ∈ ℕ
1 1
⇒ 2𝑛+1 − 2𝑛+2 > 0, ∀ 𝑛 ∈ ℕ 3.3 Subsequence of a Sequence:
⇒ 𝑎𝑛+1 − 𝑎𝑛 > 0, ∀ 𝑛 ∈ ℕ If 〈𝑎𝑛 〉 is a sequence and if 〈𝑛1 , 𝑛2 , 𝑛3 , … . . , 𝑛𝑘 , 𝑛𝑘+1 … . . 〉 is a
⇒ 〈𝑎𝑛 〉 is monotonically increasing sequence. strictly increasing sequence of natural numbers. Then the
𝑛 terms
sequence 〈𝑎𝑛𝑘 〉 is called a subsequence of 〈𝑎𝑛 〉.
1 ⏞
1 11 1 1 1
Consider 𝑎𝑛 = 𝑛+1
⏟ + 𝑛+2 +. . . + 𝑛+𝑛 < 𝑛 + 𝑛 + 𝑛 +. . . + 𝑛 =
𝑛 terms Examples:
𝑛.1
=1 (i) Let 〈𝑛𝑘 〉 = 〈2𝑘〉 =< 2, 4, 6, 8, … … > ,then for any 〈𝑎𝑛 〉
𝑛
1 1
(∴ < 𝑛 , 𝑎 > 0, ∀ 𝑛 ∈ ℕ) we have corresponding subsequence 〈𝑎𝑛 𝑘 〉 =<
𝑛+𝑎
⇒ 𝑎𝑛 < 1, ∀ 𝑛 ∈ ℕ 𝑎2 , 𝑎4 , 𝑎6 , … . . >
⇒ 𝑎𝑛 < 1, ∀ 𝑛 ∈ ℕ (ii) 〈𝑛𝑘 〉 = 〈2𝑘 − 1〉 =< 1, 3, 5, 7, 8 … … > then for any 〈𝑎𝑛 〉
Also 0 ≤ 𝑎𝑛 , ∀ 𝑛 ∈ ℕ we have corresponding subsequence
⇒ 0 ≤ 𝑎𝑛 ≤ 1, ∀ 𝑛 ∈ ℕ 〈𝑎𝑛 𝑘 〉 =< 𝑎1 , 𝑎3 , 𝑎5 , … … >
⇒ 𝑎𝑛 is bounded sequence.
Thus, 〈𝑎𝑛 〉 is bounded & monotonically increasing sequence. Complementary Pair of subsequences (CPS) :-
If the sets of suffixes of any two subsequence of 〈𝑎𝑛 〉 defines a
Q. 𝒂𝟏 = 𝟏, 𝒂𝒏 = 𝟏 + 𝟏! + 𝟐! +. . . + (𝒏−𝟏)! ,
𝟏 𝟏 𝟏
𝒏≥𝟐 partition of ℕ, then such pair of subsequences form a CPS of
1 1 1
sequence 〈𝑎𝑛 〉 .
Solution: Given 𝑎𝑛 = 1 + + +. . . + (𝑛−1)! , 𝑛≥2 i.e. If 〈𝑎𝑛 𝑘 〉&〈𝑎𝑚𝑘 〉 are subsequence of the sequence 〈𝑎𝑛 〉 &
1! 2!
1 1 1 1
∴ 𝑎𝑛+1 = 1 + 1! + 2! +. . . + (𝑛−1)! + 𝑛! 𝐴 = {𝑛𝑘 | 𝑘 ∈ ℕ} & 𝐵 = {𝑚𝑘 |𝑘 ∈ ℕ} we say 〈𝑎𝑚 𝑘 〉 & 〈𝑎𝑛𝑘 〉
1 1 1 1 1 form a CPS, if 𝐴 ∪ 𝐵 = ℕ & 𝐴 ∩ 𝐵 = 𝜙
𝑎𝑛+1 − 𝑎𝑛 = (1 + 1! + 2! +. . . + (𝑛−1!) + 𝑛!) − (1 + 1! + 1 1 1
〈𝑎𝑛 〉 = 〈 〉 then 〈𝑎2𝑛 〉 = 〈 〉 and 〈𝑎2𝑛−1 〉 = 〈 〉 are
1 1 𝑛 2𝑛 2𝑛−1
+. . . + (𝑛−1)!)
2! complimentary pair of subsequence of 〈𝑎𝑛 〉.
1
𝑎𝑛+1 − 𝑎𝑛 = ≥ 0, ∀ 𝑛 ∈ ℕ
𝑛!
⇒ 𝑎𝑛+1 ≥ 𝑎𝑛 , ∀ 𝑛 ∈ ℕ Note:
Hence, 〈𝑎𝑛 〉 is monotonically increasing sequence. 1. A sequence is bounded iff it has bounded CPS.
1 1
𝑎𝑛 = 1 + 1! + 2! +. . . + (𝑛−1)!
1 2. If 𝛼 ∈ ℝ & for some 𝛿 > 0 such that (𝛼 − 𝛿, 𝛼 + 𝛿)
contains infinite terms of 〈𝑎𝑛 〉, then we say this nbd of 𝛼
2𝑛−1 ≤ 𝑛!, ∀ 𝑛 ≥ 2
{⇒ 1 1 } contains a subsequence of 〈𝑎𝑛 〉.
≥ ,∀𝑛 ≥ 2 3. If 𝑎𝑛 𝑘 is a subsequence in (𝛼 − 𝛿, 𝛼 + 𝛿)
2𝑛−1 𝑛!
1 1 1 1
⇒ 𝑎𝑛 ≤ 1 + (1 + + 2 + 3 +. . . + 𝑛−2) ⇒ 𝑎𝑛 𝑘 ∈ (𝛼 − 𝛿, 𝛼 + 𝛿) for each 𝑘 ∈ ℕ
2 2 2 2
𝑖 ⇒ 𝛼 − 𝛿 ≤ 𝑎𝑛 𝑘 ≤ 𝛼 + 𝛿, ∀ 𝑘 ∈ ℕ
1
⇒ 𝑎𝑛 ≤ 1 + [∑𝑛−2
𝑖=0 (2) ] ⇒ This subsequence is bounded.
IFAS Publications
4. If 〈𝑎𝑛 〉, 𝑎𝑛 ∈ ℝ is unbounded above then for any 𝑀 ∈ 2. If the range of the sequence is finite, then the sequence
ℝ ∃ 𝑘 ∈ ℕ such that 𝑎𝑘 > 𝑀 ∀𝑛 ≥ 𝑘 must have limit point but converse need not be true.
1 𝑛 is odd
Ex. 𝑎𝑛 = {
Theorem: 𝑛 𝑛 is even
If 𝑆 be an unbounded subset of ℝ, then there is an unbounded
3. If 𝑆 is the range set of 〈𝑎𝑛 〉 & 𝛼 ∈ 𝑆′, then 𝛼 is a limit point
strictly increasing sequence in 𝑆.
of 〈𝑎𝑛 〉
Limit Point if a Sequence:
i.e. every limit point of the range set of 〈𝑎𝑛 〉 is the limit
Let 〈𝑎𝑛 〉 be a sequence of real numbers & 𝛼 ∈ ℝ. Then 𝛼 is said
point of 〈𝑎𝑛 〉 but not conversely
to be a limit point of 〈𝑎𝑛 〉 if for any 𝛿 > 0, 𝑎𝑛 ∈ (𝛼 − 𝛿, 𝛼 + 𝛿)
Example. 〈𝑎𝑛 〉 = 〈(−1)𝑛 〉
for infinitely many 𝑛.
Here range set = {−1,1} which is finite.
i.e. every neighbourhood of 𝛼 contains infinite terms of
∴ 〈𝑎𝑛 〉 has limit point 1 but range set does not have any
sequence 〈𝑎𝑛 〉.
limit point.
Example:
Q. If 〈𝒂𝒏 〉 = 〈(−𝟏)𝒏 〉, find the limit points of 〈𝒂𝒏 〉 . 4. Limit point of set of limit point of sequence is again limit
Solution: of sequence.
Consider (i.e. The set of limit points of sequence is always closed)
〈𝑎𝑛 〉 =< −1, 1, −1, 1, −1, 1, … . >
1 if 𝑛 is even 5. A real number is a limit point of 〈𝑎𝑛 〉 iff it is either limit
𝑎𝑛 = {
−1 if 𝑛 is odd point of range of the sequence or it repeat itself in the
Clearly, for 𝛼 = 1 & any 𝛿 > 0 sequence infinitely many times.
1 ∈ (1 − 𝛿, 1 + 𝛿)
& 𝑎𝑛 = 1, for every even natural number. 6. Bolzano Weierstrass theorem for sequence:
⇒ 𝑎𝑛 ∈ (1 − 𝛿, 1 + 𝛿), for every even natural numbers Every bounded sequence has a limit point.
There are infinitely many even natural number
⇒ 𝑎𝑛 ∈ (1 − 𝛿, 1 + 𝛿) for infinitely many natural number 𝑛 7. If a sequence is monotonic, then it can have at most one
∴ by definition of limit point of a sequence limit point.
1 is a limit point of 〈𝑎𝑛 〉
−1 ∈ (−1 − 𝛿, −1 + 𝛿), ∀ 𝛿 > 0 8. Every monotonic & bounded sequence has unique limit
& 𝑎𝑛 = −1 for every odd natural numbers point.
⇒ −1 is a limit point of a sequence 〈𝑎𝑛 〉
Thus, the limit point of sequence 〈𝑎𝑛 〉 are {1, −1} 9. If 〈𝑎𝑛 𝑘 〉 & 〈𝑎𝑚𝑘 〉 forms a CPS of 〈𝑎𝑛 〉 & 𝐿1 , 𝐿2 are the set
𝟏 of limit points of 〈𝑎𝑛 𝑘 〉 & 〈𝑎𝑚 𝑘 〉 respectively . Then 𝐿 =
𝐐. 𝐈𝐟 〈𝒂𝒏 〉 = 〈𝒏〉 , Then find limit points of 〈𝒂𝒏 〉.
𝐿1 𝑈 𝐿2 is the set of all limit points of 〈𝑎𝑛 〉.
Solution:
1
Given 〈𝑎𝑛 〉 = 〈𝑛〉 10. If ∃ ∝ ∈ ℝ & ∃ 𝛿 > 0 such that (∝ −𝛿, ∝ +𝛿) excludes
By Archimedean property, for given 𝛿 > 0, ∃ 𝑚 ∈ ℕ only finite terms of the sequence 〈𝑎𝑛 〉, then 〈𝑎𝑛 〉 is
1 bounded sequence.
Such that 𝑛 < 𝛿, ∀ 𝑛 ≥ 𝑚
1
i.e. 0 < 𝑛 < 0 + 𝛿, ∀ 𝑛 ≥ 𝑚
Q. There exist a sequence having exactly one limit point.
1
⇒ 0 − 𝛿 < 𝑛 < 0 + 𝛿, ∀𝑛 ≥ 𝑚 [T/F]
⇒ −𝛿 <
1
< 𝛿, ∀ 𝑛 ≥ 𝑚 Solution: 〈𝑎𝑛 〉 = 〈1〉
𝑛
1 ⇒ 1 ∈ (1 − 𝛿, 1 + 𝛿), for any 𝛿 > 0 & ∀𝑛 ∈ ℕ
Thus, ∈ (−𝛿, 𝛿), ∀ 𝑛 ≥ 𝑚
𝑛 ⇒ 1 is the only limit point of 〈𝑎𝑛 〉
⇒ 𝑎𝑛 ∈ (−𝛿, 𝛿), ∀ 𝑛 ≥ 𝑚
Thus, ‘0’ is the limit point of a sequence 〈𝑎𝑛 〉. 𝟏 𝟏
Q. The set {𝒏 𝐬𝐢𝐧 𝒏 : 𝒏 ∈ ℕ} has
Observation: 1. One limit point and it is 0
1. If a real number appears in a sequence infinitely many 2. One limit point and it is 1
times, then it is also limit point of that sequence but 3. One limit point and it is – 1
converse need not be true. 4. Three limit points and these are – 1, 0 and 1.
1 1
Example: 〈𝑛〉 Solution: Since, |sin ( )| ≤ 1, ∀𝑛 ∈ ℕ
𝑛
IFAS Publications
1 1 1 1 The range of Sequence 〈𝑆𝑛 〉 contains only finitely many
⇒− ≤ sin ( ) ≤
𝑛 𝑛 𝑛 𝑛 members {
1
,
−1
, 1, −1, 0}
∴ for given any 𝛿 > 0, ∃ 𝑚 ∈ ℕ √2 √2
1 & it attains infinitely many times all this numbers.
Such that 𝑛 < 𝛿, ∀ 𝑛 ≥ 𝑚 1 −1
−1 1 1 1 ⇒ The set of limit points of the sequence is {1, −1 ,0, , }
√2 √2
⇒ −𝛿 < 𝑛
≤ 𝑛 sin (𝑛) ≤ 𝑛 < 𝛿, ∀ 𝑛 ≥ 𝑚
3.4 Limits of a Sequence:
1 1 Let 〈𝑎𝑛 〉 be a sequence of real number and ℓ ∈ ℝ. We say ℓ is
⇒ −𝛿 < 𝑛 sin (𝑛) < 𝛿, ∀ 𝑛 ≥ 𝑚
the limit of a sequence 〈𝑎𝑛 〉 & denoted by lim 𝑎𝑛 = ℓ, if for
⇒ 𝑎𝑛 ∈ (−𝛿, 𝛿), ∀ 𝑛 ≥ 𝑚 𝑛→∞
Thus, ‘0’ is the limit point of a sequence 〈𝑎𝑛 〉. any 𝜀 > 0, there exist 𝑚 ∈ ℕ such that
|𝑎𝑛 − ℓ| < 𝜀 , ∀ 𝑛 ≥ 𝑚 .
Q. There exist a sequence having exactly ‘𝒌’ limit points,
where ∈ ℕ . [T/F] Here, we need to comprehend that If lim 𝑎𝑛 = ℓ ⇔ for given
𝑛→∞
Solution: 𝜀 > 0, ∃ 𝑚 ∈ ℕ such that |𝑎𝑛 − ℓ| < 𝜀 ∀ 𝑛 ≥ 𝑚.
〈𝑎𝑛 〉 =< 1,2,3,4, … . . , 𝑘, 1,2,3, … . , 𝑘, 1,2,3, … , 𝑘, , … … . > ⇔ 𝑎𝑛 ∈ (ℓ − 𝜀, ℓ + 𝜀), ∀ 𝑛 ≥ 𝑚
Here, the elements of the sequence 〈𝑎𝑛 〉 occurs infinitely many −𝜀 < 𝑎𝑛 − ℓ < 𝜀
times ℓ − 𝜀 < 𝑎𝑛 < ℓ + 𝜀
⇒ 〈𝑎𝑛 〉 has exactly 𝑘 limit points 𝑎𝑛 ∈ (ℓ − 𝜀, ℓ + 𝜀)
The set of limit of 〈𝑎𝑛 〉 is {1,2,3,4, … … . . , 𝑘}. i.e. ℓ is the limit of 〈𝑎𝑛 〉 iff every nbd of ℓ excludes only finitely
many terms of 〈𝑎𝑛 〉.
Q. There exists a sequence which has countably infinite limit Limit of a sequence is a limit point of that sequence.
points. [T/F]
1 1 Convergent Sequence:
Solution: Define 〈𝑎𝑚,𝑛 〉 = {𝑚 + 𝑛 , 𝑛 ∈ ℕ}
A sequence is said to be convergent if it has limit.
Consider, for fix 𝑚, we have subsequence
Moreover, if lim 𝑎𝑛 = ℓ, we say 〈𝑎𝑛 〉 is convergent &
1 𝑛→∞
〈𝑎1,𝑛 〉 = {1 + , 𝑛 ∈ ℕ} has limit point 1
𝑛 converges to ℓ.
1 1
〈𝑎2,𝑛 〉 = { + , 𝑛 ∈ ℕ} has limit point 1⁄2
2 𝑛
. Example:
1
. Find lim 𝑎𝑛 where 〈𝑎𝑛 〉 = 〈𝑛〉
𝑛→∞
. Solution:
1 1
〈𝑎𝑚,𝑛 〉 = { + , 𝑛 ∈ ℕ} has limit point 1⁄𝑚 Consider for 𝜀 > 0
𝑚 𝑛
In general, this sequence 〈𝑎𝑚,𝑛 〉 has limit point set of |𝑎𝑛 − 0| < 𝜀
1 1
{ | 𝑚 ∈ ℕ} ∪ {0} . ⇔ |𝑛 − 0| < 𝜀
𝑚
1
⇔ |𝑛| < 𝜀
Q. There exists a sequence which has uncountable limit ⇔𝑛>
1
𝜀
point. [T\F] 1
Solution: As |ℕ| = |ℚ| Choose 𝑚 = [ ] + 1
𝜀
1
∴ There exist a bijective function𝑓: ℕ → ℚ. For this 𝑚 (= [𝜀 ] + 1)
As 𝑅(𝑓) = ℚ & ℚ′ = ℝ 1
We have |𝑛 − 0| < 𝜀, ∀ 𝑛 ≥ 𝑚
We know that, every limit point of range set of sequence is also
1
limit point of sequence. ∴ By definition of limit of a sequence lim =0
𝑛→∞ 𝑛
∴ any bijective function(sequence) 𝑓: ℕ → ℚ has limit point set Clearly 𝑚 is depend on 𝜀
ℝ (i) If 𝜀 = 10−2 ⇒ 𝜀 = 10−2 = 100
1 1
Thus, there exist a sequence which has uncountable limit 1 1

points. Then |𝑛 − 0| < 𝜀, ∀ 𝑛 ≥ [𝜀 ] + 1 = 100 + 1
So, we get 𝑚 = 101
𝒏𝝅 1 1
Q. Find limit points of the Sequence 〈𝒔𝒊𝒏 ( )〉 (ii) If 𝜀 = 10−10 , then 𝜀 = 10−10 = 1010
𝟒
1 1
Solution: Consider, the range of the sequence. Then |𝑛 − 0| < 𝜀, ∀ 𝑛 ≥ [𝜀 ] + 1 = 1010 + 1
𝑛𝜋 1 1 −1 1
〈𝑆𝑛 〉 = 〈sin ( )〉 = { , 1, , 0, , −1, , 0, … . . } ⇒ for different values of 𝜀, we get different values of 𝑚.
4 √2 √2 √2 √2
IFAS Publications
Results: And,
5
(1) Every convergent sequence is bounded. 𝑎1 = 1 < 2 , 𝑎2 = 3 < 2
(2) A bounded sequence is convergent iff it has unique limit 3+2𝑎𝑛 4+2𝑎𝑛 −1 2(2+𝑎𝑛 ) 1
𝑎𝑛+1 = = = − 2+𝑎
point. 2+𝑎𝑛 2+𝑎𝑛 2+𝑎𝑛 𝑛
1 1
(3) A monotonic sequence is convergent iff it is bounded. 𝑎𝑛+1 = 2 − 2+𝑎 < 2, ∀ 𝑛 ∈ ℕ (∵ 2+𝑎 > 0)
𝑛 𝑛
(4) If a sequence has no limit point, then it’s every ⇒ 𝑎𝑛+1 < 2, ∀ 𝑛 ∈ ℕ
subsequence is unbounded. ⇒ 〈𝑎𝑛 〉 is bounded sequence.
(5) If sequence 〈𝑎𝑛 〉 has a limit point ℓ, then there exist a Hence, 〈𝑎𝑛 〉 is convergent sequence & it has unique limit.
subsequence of 〈𝑎𝑛 〉 that converges to 𝑙 Suppose,
(6) If there exist a CPS of sequence 〈𝑎𝑛 〉 such that both the lim 𝑎𝑛 = ℓ for some ℓ ∈ ℝ
𝑛→∞
subsequence converges to the same limit, then the
⇒ lim 𝑎𝑛+1 = ℓ
sequence 〈𝑎𝑛 〉 converges to the same limit. 𝑛→∞
3+2𝑎𝑛 3+2𝑎𝑛
(7) Let 〈𝑎𝑛 〉 be a bounded sequence & ℓ ∈ ℝ such that every We have 𝑎𝑛+1 = 2+𝑎𝑛
⇒ lim 𝑎𝑛+1 = lim
𝑛→∞ 𝑛→∞ 2+𝑎𝑛
convergent subsequence of 〈𝑎𝑛 〉 converges to ℓ. Then the 3+2ℓ
ℓ=
2+ℓ
sequence 〈𝑎𝑛 〉 converges to ℓ.
⇒ 2ℓ + ℓ2 = 3 + 2ℓ
(8) If lim 𝑎𝑛 = ℓ & 𝑎𝑛 ≥ 𝑎, ∀ 𝑛 ≥ 𝑚, where 𝑚 ∈ ℕ then
𝑛→∞ ⇒ ℓ2 = 3
ℓ≥𝑎 ⇒ ℓ = ±√3
𝟏
𝟏 − 𝒏 𝒏 𝐢𝐬 𝐨𝐝𝐝 But 𝑎𝑛 > 0, ∀ 𝑛 ∈ ℕ
Example: Let 〈𝒂𝒏 〉 = { 𝟏
𝟏 + 𝒏 𝒏 𝐢𝐬 𝐞𝐯𝐞𝐧 ⇒ ℓ > 0 ⇒ ℓ = √3
(Note: If 𝑎𝑛 > 𝑎, ∀𝑛 ∈ ℕ, then lim 𝑎𝑛 ≥ 𝑎)
Solution: Consider the CPS of 〈𝑎𝑛 〉 𝑛→∞
1 1
〈𝑎2𝑛 〉 = 1 + & lim 𝑎2𝑛 = lim (1 + ) = 1
2𝑛 𝑛→∞ 𝑛 𝑛→∞ Q. 𝒂𝟏 = √𝟕 & 𝒂𝒏+𝟏 = √𝒂𝒏 + 𝟕 ∀ 𝒏 ≥ 𝟏
1 1
〈𝑎2𝑛−1 〉 = 1 − & lim 𝑎2𝑛−1 = lim (1 − 2𝑛−1) = 1 Solution: We know that 〈𝑎𝑛 〉 is monotonically increasing &
2𝑛−1 𝑛→∞ 𝑛→∞
Here, the CPS has the same limit i.e. ℓ = 1 bounded sequence.
⇒ lim 𝑎𝑛 = 1 . Therefore, 〈𝑎𝑛 〉 is convergent sequence & it has unique limit.
𝑛→∞
Suppose,
Q. Check the convergence of the following sequences lim 𝑎𝑛 = ℓ for some ℓ ∈ ℝ
𝑛→∞
1.
1 1
𝑎𝑛 = 𝑛+1 + 𝑛+2 + 𝑛+3 + ⋯ + 𝑛+𝑛
1 1 ⇒ lim 𝑎𝑛+1 = ℓ
𝑛→∞
Solution: We know that, 〈𝑎𝑛 〉 is monotonic & bounded ⇒ lim 𝑎𝑛+1 = lim √𝑎𝑛 + 7
𝑛→∞ 𝑛→∞
⇒ 〈𝑎𝑛 〉 is convergent sequence. ⇒ ℓ2 = ℓ + 7
1 1 1
2. 𝑎1 = 1, 𝑎𝑛 = 1 + 1! + 2! +. . . + (𝑛−1)! , 𝑛≥2 ⇒ ℓ2 − ℓ − 7 = 0
−(−1)±√1−4×1(−7) 1±√1+28
Solution: We know that, 〈𝑎𝑛 〉 is monotonic & bounded ℓ= =
2×1 2
⇒ 〈𝑎𝑛 〉 is convergent sequence. 1±√29
ℓ= 2
𝟑+𝟐𝒂𝒏 (∵ 𝑎𝑛 > 0, ∀𝑛 ⇒ ℓ ≥ 0)
Q. Define 𝒂𝟏 = 𝟏, 𝒂𝒏+𝟏 + , 𝒏≥𝟏
𝟐+𝒂𝒏 1+√29
3+2 5 ⇒ℓ=
2
Solution: Consider 𝑎1 = 1, 𝑎2 = =
2+1 3
𝑎1 < 𝑎2 Q. 𝒂𝟏 = 𝟏 & 𝒂𝒏+𝟏 = √𝟑𝒂𝒏 ∀ 𝒏 ≥ 𝟏
Assume that 𝑎𝑘 < 𝑎𝑘+1 =⇒ 𝑎𝑘+1 − 𝑎𝑘 > 0
Solution: we know that 〈𝑎𝑛 〉 is monotonically increasing &
Consider,
3+2𝑎𝑘+1 3+2𝑎𝑘
bounded sequence.
𝑎𝑘+2 − 𝑎𝑘+1 = − Therefore, 〈𝑎𝑛 〉 is convergent sequence & it has unique limit.
2+𝑎𝑘+1 2+𝑎𝑘
Suppose,
6+3𝑎𝑘 +4𝑎𝑘+1 +2𝑎𝑘 𝑎𝑘+1 −6−3𝑎𝑘+1 −4𝑎𝑘 −2𝑎𝑘 𝑎𝑘+1 lim 𝑎𝑛 = ℓ for some ℓ ∈ ℝ
= (2+𝑎𝑘1 )(2+𝑎𝑘 ) 𝑛→∞
3𝑎𝑘 −4𝑎𝑘 +4𝑎𝑘+1 −3𝑎𝑘+1 𝑎𝑘+1 −𝑎𝑘 ⇒ lim 𝑎𝑛+1 = ℓ
𝑎𝑘+2 − 𝑎𝑘−1 = (2+𝑎𝑘+1 )(2+𝑎𝑘 )
= (2+𝑎 >0 𝑛→∞
𝑘+1 )(2+𝑎𝑘 )
⇒ lim 𝑎𝑛+1 = lim √3𝑎𝑛
⇒ 𝑎𝑘+2 − 𝑎𝑘+1 > 0 𝑛→∞ 𝑛→∞
⇒ 𝑎𝑘+2 > 𝑎𝑘+1 ⇒ ℓ2 = 3ℓ ⇒ ℓ2 − 3ℓ = 0
∴By Mathematical induction, 𝑎𝑛 < 𝑎𝑛+1 , ∀ 𝑛 ∈ ℕ ⇒ ℓ = 0 𝑜𝑟 ℓ = 3
⇒ 〈𝑎𝑛 〉 is monotonically increasing sequence. But 𝑎𝑛 > 1, ∀𝑛 ∈ ℕ ⇒ lim 𝑎𝑛 ≥ 1
𝑛→∞
IFAS Publications
⇒ lim 𝑎𝑛 = 3 For this 𝑚. We have
𝑛→∞
1
Results: | − 0| < 𝜀, where 𝑛 ≥ 𝑚
𝑛𝑝
1. If lim 𝑎𝑛 = ℓ then lim |𝑎𝑛 | = |ℓ| ∴ lim
1
= 0, for 𝑝 > 0
𝑛→∞ 𝑛→∞
𝑛→∞ 𝑛 𝑝
Converse need not be true.
Example: 𝑎𝑛 = < (−1)𝑛 > Q. Show that the sequence < 𝒓𝒏 > converges to 0, If |𝝃| < 𝟏
2. lim |𝑎𝑛 − ℓ| = 0 iff lim 𝑎𝑛 = ℓ. Solution: If |𝑟| < 1 ⇒ |𝑟| = 1+ℎ for some ℎ > 0
1
𝑛→∞ 𝑛→∞
3. If |𝑎𝑛 | ≤ |𝑏𝑛 |, ∀ 𝑛 ≥ 𝑚 & lim |𝑏𝑛 | = 0 ⇒ lim |𝑎𝑛 | = 𝑛(𝑛−1)
𝑛→∞ 𝑛→∞ Consider (1 + ℎ)𝑛 = 1 + 𝑛ℎ + 2
ℎ2 + ⋯ + ℎ𝑛
0 (1 + ℎ)𝑛 ≥ 1 + 𝑛ℎ, ∀ 𝑛 ∈ ℕ, & ℎ > 0
4. Every sequence has a monotonic subsequence. 1 𝑛 1
5. A real number ℓ is a limit point of a sequence 𝑎𝑛 > iff ⇒( ) ≤ ,∀ 𝑛 ∈ ℕ
1+ℎ 1+𝑛ℎ
1 𝑛 1
there exist a subsequence < 𝑎𝑛𝑘 > of < 𝑎𝑛 > converges Now, |𝑟 𝑛 − 0| = |𝑟 𝑛 | = |𝑟|𝑛 = ( ) ≤ ,∀𝑛 ∈ ℕ
1+ℎ 1+𝑛ℎ
to ℓ. For given 𝜀 > 0.
1
|𝑟 𝑛 − 0| ≤ < 𝜀,
Results on Monotonic Sequence & Its convergence : 1+𝑛ℎ
1 1 1 1
1. A bounded & monotonically increasing sequence ⇒ 1 + 𝑛ℎ > 𝜀 ⇒ 𝑛ℎ > 𝜀 − 1 ⇒ 𝑛 > ℎ (𝜀 − 1)
converges to its supremum. 1 1
Choose, 𝑚 = [ ( − 1)] + 1
𝑛 𝜀
2. A bounded & monotonically decreasing sequence
For this 𝑚, we have
converges to its infimum.
|𝑟 𝑛 − 0| < 𝜀, whenever 𝑛 ≥ 𝑚
3. An eventually monotonic sequence is convergent iff it is
⇒ lim |𝑟|𝑛 = 0 , if |𝑟| < 1.
bounded. 𝑛→∞
𝟑𝒏+𝟐
Examples: Q. 𝐥𝐢𝐦 =𝟑
𝒏→∞ 𝒏+𝟏
Q. If 〈𝒂𝒏 〉 = √𝒏 + 𝟏 − √𝒏, then find 𝐥𝐢𝐦 𝒂𝒏 Solution: For given 𝜀 > 0
𝒏→∞
3𝑛+2 3𝑛+2−3𝑛−3 −1 1
Solution: Given 𝑎𝑛 = √𝑛 + 1 − √𝑛 |𝑎𝑛 − 3| = | − 3| = | |=| |=| |< 𝜀
𝑛+1 𝑛+1 𝑛+1 𝑛+1
1 1
By using rationalization, ⇒ 𝑛+1 >𝜀 ⇒ 𝑛 > −1
𝜀
Consider, 1
2 2 Choose 𝑚 = [𝜀 − 1] + 1
(√𝑛+1+√𝑛) (√𝑛+1) −(√𝑛)
𝑎𝑛 = (√𝑛 + 1 − √𝑛) (√𝑛+1+√𝑛) = For this 𝑚, we have
√𝑛+1+√𝑛
(𝑛+1)−𝑛 1 3𝑛+2
𝑎𝑛 = = | − 3| < 𝜀 , Whenever 𝑛 ≥ 𝑚
√𝑛+1+√𝑛 √𝑛+1+√𝑛 𝑛+1
3𝑛+2
Now, for given 𝜀 > 0, Consider Thus, lim =3
𝑛→∞ 𝑛+1
1 1
|𝑎𝑛 − 0| = |(√𝑛 + 1 − √𝑛) − 0| = | |<| |=
√𝑛+1+√𝑛 √𝑛+√𝑛
𝟏+𝟑+𝟓+𝟕+⋯+(𝟐𝒏−𝟏)
|
1
|< 𝜀 Q. 𝐥𝐢𝐦 =𝟏
𝒏→∞ 𝒏𝟐
2√𝑛
1 1 1 Solution: Consider
∴ 2√𝑛 > 𝜀 ⇒ √𝑛 > 2𝜀 ⇒ 𝑛 > 4𝜀 2
1
1 + 3 + 5 + 7 + ⋯ + (2𝑛 − 1) = 𝑘
Choose 𝑚 = [4𝜀 2] + 1 ∴ (1 + 3 + 5 + 7 + ⋯ + (2𝑛 − 1)) + (2 + 4 + 6 + 8 + ⋯ +
⇒ for this 𝑚 , We have 2𝑛) = 𝑘 + (2 + 4 + 6 + ⋯ + 2𝑛)
|𝑎𝑛 − 0| < 𝜀, wherever 𝑛 ≥ 𝑚. ⇒
2𝑛(2𝑛+1)
= 𝑘 + 2[1 + 2 + 3 + ⋯ + 𝑛]
2
⇒ lim (√𝑛 + 1 − √𝑛) = 0 𝑛(𝑛+1)
𝑛→∞ ⇒ 𝑛(2𝑛 + 1) = 𝑘 + 2 [ 2
]
𝟏 ⇒ 𝑘 = 𝑛(2𝑛 + 1) − 𝑛(𝑛 + 1) = 2𝑛2 + 𝑛 − 𝑛2 − 𝑛 = 𝑛2

Q. If 𝒑 > 𝟎 then 𝐥𝐢𝐦 𝒏𝒑 = 𝟎 ⇒ 𝑘 = 𝑛2
𝒏→∞
1
Solution: Given 𝑎𝑛 = ,𝑝 > 0 ⇒ 1 + 3 + 5 + 7 + ⋯ + (2𝑛 − 1) = 𝑛2
𝑛𝑃
1+3+5+7+⋯+(2𝑛−1) 𝑛2
Now for given 𝜀 > 0, We have ⇒ = =1
𝑛2 𝑛2
1 1
|𝑎𝑛 − 0| = | − 0| = | 𝑝| < 𝜀 ⇒ lim 1 = 1
𝑛𝑝 𝑛 𝑛→∞
1
1 1 ⁄𝑝
⇒ 𝑛 𝑝 > 𝜀 ⇒ 𝑛 > (𝜀 ) Exercise
1 𝑛 2 +1
1 ⁄𝑝 Q. lim
Choose 𝑚 = = [(𝜀 ) ]+1 𝑛→∞ 2𝑛 2 +5
1
Answer 𝑙 =
2
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Iit Jam - Real Analysis

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INDEX

S. No. Contents Page No.

1. SET THEORY & COUNTABILITY 1

1.4 Classification of Functions………………………………………………………………………………….. 4

1.5 Similar/Equivalent Sets………………………………………………………………………………………. 7

1.6 Countable & Uncountable Sets…………………………………………………………………………… 8

2. POINT SET TOPOLOGY 16

2.2 Bounds of A Set…………………………………………………………………………………………………… 16

2.5 Neighbourhood of A Point……………………………………………………………………………………18

2.6 Some Important Type of Points & Its Properties…………………………………….. …………..19

4.2 Positive Term Series…………………………………………………………………………………………… 65

4.3 Some Important Series & Their Convergence…………………………………………………….. 67

4.4 Important Test for Convergence………………………………………………………………………… 70

4.5 Alternating Term Series……………………………………………………………………………………… 77

4.6 Series of Arbitrary Terms……………………………………………………………………………………. 82

4.7 Sum of Series……………………………………………………………………………………………………… 84

5.2 Wavy Curve Method…………………………………………………………………………………………… 98

5.3 Some Well Known Functions………………………………………………………………………………. 99

5.4 Trigonometric Functions…………………………………………………………………………………….. 100

5.5 Inverse Trignometric Function.…………………………………………………………………………… 101

5.6 Graphical Transformations ………………………………………………………………………………… 102

5.7 Algebraic & Transcendental Function…………………………………………………………………. 105

5.8 Bounded Function………………………………………………………………………………………………. 106

6. LIMIT & CONTINUITY 109

6.2 Left Hand Limit & Right Hand Limit……………………………………………………………………… 111

6.3 Continuity…………………………………………………………………………………………………………… 114

6.4 Discontinuity………………………………………………………………………………………………………. 115

6.5 Cauchy Functional Equation………………………………………………………………………………… 117

6.6 Properties of Continuous Functions……………………………………………………………………. 118

6.7 Roots of A Function…………………………………………………………………………………………….. 121

6.8 Lipschitz Function……………………………………………………………………………………………….. 124

6.9 Uniform Continuity……………………………………………………………………………………………… 124

7.2 Monotonic Functions & Differentiability……………………………………………………………… 143

7.3 Fixed Point………………………………………………………………………………………………………….. 144

7.4 Local Extrema of a Function…..……………………………………………………………………………. 146

7.5 Application of Differentiability……………………………………………………………………………. 147

8. RIEMANN INTEGRATION & BOUNDED VARIATION 159

8.2 Definite Integral as Limit of Sum…………………………………………………………………………. 161

9. IMPROPER INTEGRAL 166

10. FUNCTION OF SEVERAL VARIABLE 171

10.2 Limit of A Function……………………………………………………………………………………………… 172

10.3 Continuity…………………………………………………………………………………………………………… 176

10.4 Directional Derivatives………………………………………………………………………………………… 178

10.5 Differentiability…………………………………………………………………………………………………… 180

10.6 Partial Derivatives of Higher Order …………………………………………………………………….. 183

10.7 Extreme Values…………………………………………………………………………………………………… 184

10.8 Function of 𝑛 Variables……………………………………………………………………………………….. 186

10.9 Vector Valued Function………………………………………………………………………………………. 186

SET THEORY Then 𝐴 is called subset of 𝐵.

1.2.4 Partial order Relation :

(d) Clearly 𝑅 is not an equivalence relation

(f) Clearly, 𝑅 is not a partial order relation. Example :

Note: A function that is not one-one is called to be many-one

Counting number of many to one function:

Hence 𝑓 is bijective map. ⇒ 𝑓 is not an onto function.

1.4.6 Cantor’s Theorem:

Note: Result :- ′ ∼ ′ is an equivalence relation

2) 𝑺𝟐 = {𝒇 |𝒇: ℕ → {𝟏, 𝟐} is a function}

3] 𝑺𝟑 = {𝒇|𝒇: {𝟏, 𝟐} → ℕ 𝐢𝐬 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 & 𝒇(𝟏) ≤ 𝒇(𝟐)}