Introduction to Real Analysis I to Real Analysis I Lecture notes by: Khim R Shrestha, Ph. D. Assistant Professor of Mathematics University of Great Falls Great Falls, Montana

Introduction to Real Analysis I

Lecture notes by:Khim R Shrestha, Ph. D.

Assistant Professor of MathematicsUniversity of Great Falls

Great Falls, Montana

Contents

1 The Real Number System 11.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Ordered Field Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Completeness Axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Inverse Functions and Images . . . . . . . . . . . . . . . . . . . . . . 51.6 Countable and Uncountable Sets . . . . . . . . . . . . . . . . . . . . 9

2 Sequences of Real Numbers 122.1 Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . 172.4 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Limits Supremum and Infimum . . . . . . . . . . . . . . . . . . . . . 21

3 Functions on R 243.1 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 One-Sided Limits and Limits Involving Infinity . . . . . . . . . . . . . 263.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

ii

Chapter 1

The Real Number System

1.1 Review

1.2 Ordered Field Axioms

1

CHAPTER 1. THE REAL NUMBER SYSTEM 2

1.3 Completeness Axiom

Theorem 1.1 (Archimedean Principle). Given a, b ∈ R with a > 0, there is aninteger n ∈ N such that b < na.

Proof. if b < a then take n = 1. So we have b < na. If a < b then the set E = {k ∈N : ka ≤ b} 6= ∅ because 1 ∈ E. Since for every k ∈ E, ka ≤ b =⇒ k ≤ b/a, E isbounded. By completeness axiom, E has a finite supremum, say supE = s. Since Eis a set of integers, s ∈ E. Set n = s+ 1. Notice that n /∈ E =⇒ b < na.

Example 1.1. The supremum of {1, 12, 14, 18, · · · } is 1.

Example 1.2. The supremum of {12, 23, 34, · · · } is 1.

Proof. Let M be an upper bound of E. We want to show that M ≥ 1. If M < 1 then1 −M > 0 and 1

1−M > 0. Then there exists n ∈ N such that 11−M < n =⇒ M <

1− 1n

= n−1n∈ E. This contradicts the assumption that M is an upper-bound.

Theorem 1.2 (Density of Rationals). Let a, b ∈ R and a < b. Then there exists aq ∈ Q such that a < q < b.

Proof. Suppose that a > 0. By Archimedean Principle, choose n ∈ N satisfying

max

{1

a,

1

b− a

}< n.

Observe that 1/n < a and 1/n < (b− a).Consider the set E = {k ∈ N : k/n ≤ a}. 1 ∈ E 6= ∅ and E is bounded above by

na. =⇒ supE = s exists and s ∈ E. Take q = s+1n

. Since s+ 1 /∈ E, q = s+1n> a.

On the other hand,

b = a+ (b− a) >s

n+

1

n= q.

If a ≤ 0, by Archimedean Principle, there exists k ∈ N such that −a < k. Then0 < a+ k < b+ k. By the previous case we have r ∈ Q such that a+ k < r < b+ k.Therefore r − k ∈ Q and a < r − k < b.

How are supremum and infimum of a set are related? For E ⊂ R, we define −Eto be the set

−E = {−x : x ∈ E}.

Theorem 1.3 (Reflection Theorem). Let E ⊆ R be a nonempty set.

(i) E has a supremum ⇐⇒ −E has an infimum. Moreover, − supE = inf(−E).

(ii) E has an infimum ⇐⇒ −E has a supremum. Moreover, − inf E = sup(−E).


Proof. We prove only (i). Suppose supE = s. Then a ≤ s for all a ∈ E =⇒−s ≤ −a for all a ∈ E =⇒ −s is a lower bound of −E. We want to show that−s = inf(−E). Let m be any lower bound of −E. Then m ≤ −a for all a ∈ E. =⇒a ≤ −m for all a ∈ E =⇒ −m is an upper bound of E =⇒ s ≤ −m =⇒ −s ≥ m.Thus

− supE = −s = inf(−E).

Theorem 1.4 (Monotone Property). Suppose that A ⊆ B are nonempty subset of R.

(i) If B has a supremum, then supA ≤ supB.

(ii) If A has an infimum, then inf A ≥ inf B.

Proof. Since supB is an upper bound of B, it is an upper bound of A. Hence, bydefinition of supremum, supA ≤ supB.

1.4 Mathematical Induction

Theorem 1.5 (Well-Ordering Principle). If E is a nonempty subset of N, then Ehas a least element, that is, inf E ∈ E.

Proof. The set −E is bounded above by −1. Thus sup(−E) exists (by Completenessaxiom) and sup(−E) ∈ −E. This implies inf E = − sup(−E) ∈ −(−E) = E.

Theorem 1.6 (Principle of Mathematical Induction). Suppose for each n ∈ N thatA(n) is a proposition which satisfies the following two properties:

1. A(1) is true.

2. For every n ∈ N, if A(n) is true then A(n+ 1) is true.

Then A(n) is true for all n ∈ N.

Proof. Suppose that there are some n ∈ N for which A(n) is not true. Then the setE = {n ∈ N : A(n) is false} 6= ∅. Then by the Well-Ordering Principle, E has aleast element, say inf E = x and x ∈ E. Since 1 /∈ E, x > 1. So x − 1 > 0 and1 ≤ x− 1 ∈ N.

Since x− 1 < x and x is the least element of E, the statement A(x− 1) must betrue. Then hypothesis (2) implies A(x) is true and x /∈ E, a contradiction.

Example 1.3. Prove that

n∑k=1

= 1 + 2 + 3 + · · ·+ n =n(n+ 1)

2.


We can use Pascal’s triangle to expand (a+ b)n.Some notations: 0! = 1, n! = 1 · 2 . . . (n− 1) · n, n ∈ N. The binomial coefficient n

over k: (n

k

):=

n!

(n− k)!k!

Lemma 1.7. If n, k ∈ N and 1 ≤ k ≤ n, then(n

k − 1

)+

(n

k

)=

(n+ 1

k

).

Proof. (n

k − 1

)+

(n

k

)=

n!k

(n− k + 1)!k!+n!(n− k + 1)

(n− k + 1)!k!

=n!(n+ 1)

(n− k + 1)!k!=

(n+ 1

k

).

Theorem 1.8 (Binomial Formula). If a, b ∈ R, n ∈ N, then

(a+ b)n =n∑k=0

(n

k

)an−kbk.

Proof.

(a+ b)n+1 = (a+ b)(a+ b)n

= (a+ b)

(n∑k=0

(n

k

)an−kbk

)

=n∑k=0

(n

k

)an−k+1bk +

n∑k=0

(n

k

)an−kbk+1

= an+1 +n∑k=1

(n

k

)an−k+1bk +

n−1∑k=0

(n

k

)an−kbk+1 + bn+1

= an+1 +n∑k=1

(n

k

)an−k+1bk +

n∑k=1

(n

k − 1

)an−k+1bk + bn+1

= an+1 +n∑k=1

((n

k

)+

(n

k − 1

))an−k+1bk + bn+1

=n+1∑k=0

(n+ 1

k

)an+1−kbk


Remark 1.1. If x > 1 and x /∈ N, then there is an n ∈ N such that n < x < n+ 1.

Proof. By Archimedean Principle, the set E = {m ∈ N : x < m} 6= ∅. By Well-Ordering Principle, E has a least element, say m0. Then m0 − 1 < x < m0. Taken = m0 − 1 to get n < x < n+ 1. Observe that n ≥ 1 and hence n ∈ N.

Remark 1.2. If n ∈ N is not a perfect square, that is, n 6= m2 for any m ∈ N, then√n is irrational.

Proof. Suppose that there is an n ∈ N which is not a perfect square but√n ∈ Q.

Write√n = p/q for some p, q ∈ N. By the previous remark there exists m0 ∈ N such

that m0 <√n < m0 + 1.

The set E := {k ∈ N : k√n ∈ Z} 6= ∅ because q

√n = p ∈ E. By the Well-

Ordering Principle, E has a least element, say n0.Now since 0 <

√n−m0 < 1, n0(

√n−m0) < n0. This implies n0(

√n−m0) /∈ E.

But,n0(√n−m0)

√n = n0n− n0m0

√n ∈ Z

implies n0(√n−m0) ∈ E, a contradiction.

Exercises

1. Let x, y, a ∈ R. Prove that x > y − ε for all ε > 0 if and only if x ≥ y.

2. Let E ⊂ R be a nonempty set and inf E exists. Prove that for any ε > 0 thereexists a point a ∈ E such that inf E ≤ a < inf E + ε.

3. Let E ⊂ R be a nonempty set. Assume that inf E exists. Prove that − inf E =sup(−E).

4. 1.4.2 (a), 1.4.4 (b)

1.5 Inverse Functions and Images

Definition 1.1. A function f : X → Y has an inverse function if and only if range f =Y and each y ∈ Y has a unique preimage x ∈ X, that is, f(x) = y.

Definition 1.2. Let X and Y be sets and f : X → Y .

i. f is said to be one-to-one ( 1 - 1 or an injection) if and only if

x, x′ ∈ X and f(x) = f(x′) =⇒ x = x′

ii. f is said to be onto (or surjection) if and only if for each y ∈ Y there is an x ∈ Xsuch that f(x) = y.


iii. f is called bijection if and only if it is one-to-one and onto.

If f : X → Y has an inverse, we will write f−1 : Y → X. In this case f(x) =y ⇐⇒ x = f−1(y).

Theorem 1.9. Let f : X → Y . Then the following statements are equivalent.

1. f has an inverse.

2. f one-to-one from X onto Y .

3. There is a function g : Y → X such that g(f(x)) = x for all x ∈ X andf(g(y)) = y for all y ∈ Y .

Proof. (1) =⇒ (2). Assume that f has an inverse. Then range f = Y . This impliesf is onto. Also every y ∈ Y has a unique preimage in X. This implies f is one-to-one.

(2) =⇒ (3). f is onto implies range f = Y . This implies every y ∈ Y has apreimage in X. Since f is one-to-one, the preimage is unique. Thus f has an inversef−1 : Y → X. Take g = f−1.

(3) =⇒ (1). Let y ∈ Y . Take x = g(y). Then x ∈ X and f(x) = f(g(y)) = y .This implies that every y ∈ Y has an preimage in X, that is, range f = Y .

If x and x′ are two preimages of y ∈ Y , then f(x) = y = f(x′). Then it followsthat x = g(f(x)) = g(f(x′)) = x′. Hence the preimage is unique. Thus f has aninverse.

Theorem 1.10. Let f : X → Y has an inverse function. Then the inverse functionis unique.

Proof. Suppose that g : Y → X and h : Y → X are two inverse functions of f . Wewant to show that g(y) = h(y) for every y ∈ Y . Take a y ∈ Y . Then there is anx ∈ X such that f(x) = y. Now, g(y) = g(f(x)) = x = h(f(x)) = h(y). Thusg = h.

Remark 1.3. Let I ⊂ R be an interval and f : I → R. If f ′(x) > 0, then f isinjective on I.

Proof. Recall that f is called strictly increasing on I if for x, x′ ∈ I, x < x′ =⇒f(x) < f(x′). Since f ′(x) > 0 on I, f is strictly increasing on I. Let x, x′ ∈ I. Ifx 6= x′ then either x < x′ or x > x′. Since f is strictly increasing, either f(x) < f(x′)or f(x) > f(x′). Hence x 6= x′ implies f(x) 6= f(x′). Thus f is one-to-one.

If f : X → Y has an inverse function f−1 : Y → X then y = f(x) can be solvedfor x.


Definition 1.3. Let f : X → Y . The image of a set E ⊆ X under f is the set

f(E) := {y ∈ Y : y = f(x) for some x ∈ E}.

The inverse image of a set E ⊆ Y under f is the set

f−1(E) := {x ∈ X : f(x) = y for some y ∈ Y }.

Example 1.4. Let E = (0, 1) and f(x) = x2. Find f(E) and f−1(E).

Example 1.5. Let E = (0,∞) and f : E → R be defined by f(x) = e1/x. Find f(E)and f−1 on f(E).

Definition 1.4. Let E = {Eα}α∈A be a collection of sets. Then⋃α∈A

Eα := {x : x ∈ Eα for some α ∈ A}

and ⋂α∈A

Eα := {x : x ∈ Eα for all α ∈ A}

We have,⋃

x∈(0,1]

= [0, 1) and⋂

x∈(0,1]

[0, x) = 0.

Theorem 1.11 (DeMorgan’s Laws). Let X be a set and {Eα}α∈A be a collection ofsubsets of X. Then (⋃

α∈A

Eα

)c

=⋂α∈A

Ecα

and (⋂α∈A

Eα

)c

=⋃α∈A

Ecα

Note that Ec = X \ E.

Proof. Let x ∈(⋃

α∈AEα)c

.

=⇒ x ∈ X and x /∈⋃α∈A

Eα.

=⇒ x /∈ Eα for all α ∈ A.=⇒ x ∈ Ec

α for all α ∈ A.

=⇒ x ∈⋂α∈A

Ecα.


Theorem 1.12. Let X and Y be sets and f : X → Y .

i. If {Eα}α ∈ A is a collection of subsets of X, then

f

(⋃α∈A

Eα

)=⋃α∈A

f(Eα) and f

(⋂α∈A

Eα

)=⋂α∈A

f(Eα).

ii. If B and C are subsets of X, then f(C \B) ⊇ f(C) \ f(B).

iii. If {Eα}α ∈ A is a collection of subsets of Y , then

f−1

(⋃α∈A

Eα

)=⋃α∈A

f−1(Eα) and f−1

(⋂α∈A

Eα

)=⋂α∈A

f−1(Eα).

iv. If B and C are subsets of Y , then f−1(C \B) = f−1(C) \ f−1(B).

v. If E ⊆ f(X), then f(f−1(E)) = E, but if E ⊆ X, then E ⊆ f−1(f(E)).

Proof. (i).

y ∈ f

(⋃α∈A

Eα

)⇐⇒ y = f(x) for some x ∈

⋃α∈A

Eα.

⇐⇒ y = f(x) for some x ∈ Eα and α ∈ A.⇐⇒ y ∈ f(Eα) for some α ∈ A.

⇐⇒ y ∈⋃α∈A

f(Eα).

(ii). Let y ∈ f(C) \ f(B).

=⇒ y ∈ f(C) and y /∈ f(B).

=⇒ y = f(c) for some c ∈ C but y 6= f(b) for any b ∈ B.=⇒ y = f(c) for some c ∈ C \B.=⇒ y ∈ f(C \B).

(v). Let y ∈ f(f−1(E)). Then y = f(x) for some x ∈ f−1(E). By definition,x = f−1(y′) for some y′ ∈ E. But y = f(x) = f(f−1(y′)) = y′ implies y′ ∈ E. Thusf(f−1(E)) ⊆ E.

For the other way around, let y ∈ E. Then y = f(f−1(y)) ∈ f(f−1(E)). ThusE ⊆ f(f−1(E)). Therefore we have f(f−1(E)) = E where E ⊆ f(X).

For the second part, let y ∈ E. Then y = f−1(f(y)) ∈ f−1(f(E)). Hence,E ⊆ f−1(f(E)).


1.6 Countable and Uncountable Sets

Definition 1.5.A set E is said to be finite if and only if either E = ∅ or there exists a bijective

function from {1, 2, · · · , n} to E for some n ∈ N.A set E is said to be countable if and only if there exists a bijective function from

N to E. E is said to be at most countable if it is finite or countable.A set E is said to be uncountable if and only if E neither finite nor countable.

Remark 1.4 (Cantor’s Diagonalization Argument). The interval (0, 1) is uncount-able.

Proof. Suppose that (0, 1) is countable. Then there is a bijection f : N→ (0, 1), thatis, f(N) = (0, 1). Then by using decimal expansion, we can write

f(1) = 0.a11a12a13 · · ·f(2) = 0.a21a22a23 · · ·f(3) = 0.a31a32a33 · · ·f(4) = 0.a41a42a43 · · ·· · · = · · ·

where none of these expansions terminate in 9’s. For example 0.123999 · · · = 0.124.

Now the number x = 0.b1b2b3 · · · where bk :=

{akk + 1 if akk ≤ 5

akk − 1 if akk > 5.Clearly x ∈

(0, 1). Since f(N) = (0, 1) there is an n ∈ N such that f(n) = x, that is,

0.b1b2 · · · bn · · · = 0.an1an2 · · · ann · · ·

Since the decimals on either side don’t terminate in 9’s, we have to have

ann = bn = ann ± 1

which is a contradiction.

Lemma 1.13. A non-empty set is at most countable if and only if there is a surjectivefunction f : N→ E.

Proof. Reading exercise.

Theorem 1.14. Suppose that A and B are sets and A ⊆ B.

i. If B is at most countable, then A is at most countable.

ii. If A is uncountable, then B is uncountable.


Proof. (i). Let B is at most countable. Then there is a surjective function f : N→ B.If A = B then f is automatically a surjective function from N to A. If A ( B, thendefine g : N→ A by

g(n) =

{f(n) if f(n) ∈ Aa0 for some a0 ∈ A if f(n) /∈ A.

Then g : N→ A is surjective. So A is at most countable.(ii). If B is at most countable then by (i), A is at most countable, which is a

contradiction. So B is uncountable if A is.

Since (0, 1) is uncountable and (0, 1) ⊂ R, R is uncountable.

Theorem 1.15.

1. N× N is countable. In particular, if A and B are at most countable sets, thenA×B is at most countable.

2. If {Ai}∞i=1 is a collection of at most countable sets, then E =⋃∞i=1Ai is at most

countable.

Proof. 1. Let us arrange the points of N× N as following,

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) · · ·(2, 1) (2, 2) (2, 3) (2, 4) · · ·(3, 1) (3, 2) (3, 3) · · ·(4, 1) (4, 2) · · ·(5, 1) · · ·· · ·

Let us arrange the natural numbers similarly,

1 3 6 10 15 21 · · ·2 5 9 14 20 · · ·4 8 13 19 · · ·7 12 18 · · ·11 17 · · ·16 · · ·· · ·

Now define a function g : N→ N×N that sends a number n ∈ N to the correspondingpoint in N× N in these two arrangements.

To show that A×B is at most countable define f : N× N→ A×B by

f(m,n) = (φ(m), ψ(n))


where φ : N→ A and ψ : N→ B are surjective functions. Hence f ◦ g : N→ A× Bis surjective and therefore A×B is at most countable.

2. To show that E =⋃i∈NAi is at most countable, define f : N × N → E by

f(i, j) = φi(j) where φi : N → Ai is a surjective function for all i ∈ N. Then fis surjective (why?). And, now f ◦ g : N → E is surjective. Hence E is at mostcountable.

Remark 1.5. The sets Z and Q are countable. The set of irrationals is uncountable.

Proof. Since Z = N∪−N∪{0}, Z is countable. Similarly Q = ∪∞n=1{pn

: p ∈ Z} impliesQ is countable. If the set of irrational were countable then R = Q ∪ {irrationals}would be countable. So the irrationals are uncountable.

Exercises

1. Let I ⊂ R be an interval and f : I → R. Prove that if f ′(x) < 0, then f isinjective on I.

2. Let f(x) = x2 + 1. Finid f(E) and f−1(E), where E = [1, 2].

3. 1.5.3 (c, d)

4. If {Eα}α ∈ A is a collection of subsets of X, then prove that

f

(⋂α∈A

Eα

)=⋂α∈A

f(Eα).

5. Let f : X → Y and E ⊆ X. Construct an example where E 6= f−1(f(E)).

6. Let f : A → B and g : B → C. Then the composition g ◦ f : A → C isdefined by g ◦ f(x) = g(f(x)). Prove that if f and g are one-to-one, then g ◦ fis one-to-one.

7. 1.6.7 is a good Project.

Chapter 2

Sequences of Real Numbers

A sequence is a function defined on N. If f : N → R is a sequence, then we writef(n) = xn and call xn the n-th term of the sequence. Instead of using a function, wewill rather use the terms to denote a sequence. So we will use one of the followingnotations to denote a sequence:

{x1, x2, x3, · · · } or {xn}n∈N or {xn}∞n=1 or {xn}.

Example 2.1. The function f : N → R defined by f(n) = 1/n is the sequence{1, 1/2, 1/3, · · · }.

2.1 Limit of a Sequence

Definition 2.1. A sequence of real numbers {xn} is said to converge to a real numberl ∈ R if for every ε > 0 there exists an n0 ∈ N such that

for every n ≥ n0 we have |xn − l| < ε.

We will call l the limit of the sequence {xn}.

Note that in the above definition the choice of n0 usually depends on ε, thatmeans n0 changes with the different choice of ε. When a sequence {xn} converges toa number l we will write it in one of the following ways:

1. {xn} converges to l.

2. xn converges to l.

3. xn → l as n→∞.

4. limn→∞

xn = l

The following statements follow immediately from the definition:

12

CHAPTER 2. SEQUENCES OF REAL NUMBERS 13

1. A sequence {xn} converges to l if and only if {xn − l} converges to 0.

2. xn → 0 as n→∞ if and only if |xn| → 0 as n→∞.

Example 2.2. The sequence { 1n} converges to 0.

Proof. Let ε > 0 be given. By Archimedean Principle, choose n0 ∈ N such thatn0 >

1ε. Then for every n ≥ n0,∣∣∣∣ 1n − 0

∣∣∣∣ =1

n≤ 1

n0

< ε.

Example 2.3. The sequence {(−1)n} has no limit.

Proof. Suppose that (−1)n → l. Then for given ε = 1, there exists n0 ∈ N such that

|(−1)n − l| < 1 whenever n ≥ n0.

For n odd, | − 1− l| < 1, and for n even |1− l| < 1. Hence,

2 = |1 + 1| = |(1− l) + (1 + l)| ≤ |1− l|+ |1 + l| < 1 + 1 = 2


Theorem 2.1. If a sequence {xn} has a limit, it is unique.

Proof. Suppose that xn → a and xn → b as n → ∞. Then we want to show thata = b. Let ε > 0 be given. Then

xn → a =⇒ there is an n0 ∈ N such that |xn − a| < ε/2 whenever n ≥ n0.

xn → b =⇒ there is an n1 ∈ N such that |xn − b| < ε/2 whenever n ≥ n1.

For n ≥ max {n0, n1},

|a− b| ≤ |xn − a|+ |xn − b| < ε/2 + ε/2 = ε.

That is, |a− b| < ε for all ε > 0. Hence a− b = 0 and a = b.

Definition 2.2. A subsequence of a sequence {xn}n∈N is a sequence of the form{xnk}k∈N, where nk ∈ N and n1 < n2 < · · · . That is, a subsequence of a sequence

{xn} is a sequence which is a subset of {xn}.

Theorem 2.2. If a sequence {xn} converges, then every subsequence of {xn} con-verges to the same limit.


Proof. Let xn → a as n → ∞ and {xnk} be any subsequence of {xn}. We want to

show that xnk→ a as k →∞. Let ε > 0 and n0 ∈ N such that |xn− a| < ε whenever

n ≥ n0. Notice that k ≥ n0 implies nk ≥ n0. Hence it follows that |xnk− a| < ε

whenever k ≥ n0. Thus xnk→ a as k →∞.

Definition 2.3. A sequence {xn} is said to be bounded above if and only if theunderlying set {xn : n ∈ N} is bounded above and it is said to be bounded belowif the underlying set is bounded below. A sequence is called bounded if it is bothbounded above and bounded below.

A sequence {xn} is bounded if and only if there is an M ∈ R such that xn ≤ Mfor all n ∈ N. Similarly, {xn} is bounded below if and only if there is an m ∈ R suchthat m ≤ xn for all n ∈ N. And {xn} is bounded if and only if there is a numberM > 0 such that |xn| < M for all n ∈ N.

Theorem 2.3. If a sequence {xn} is convergent, then it is bounded.

Proof. Let ε = 1 and xn → l as n → ∞. Then by definition, there exists an n0 ∈ Nsuch that

|xn − l| < 1 whenever n ≥ n0.

=⇒ |xn| < 1 + |l| whenever n ≥ n0.

Take M = max{x1, x2, · · · , xn0 , 1 + |l|}. Hence |xn| < M for all n ∈ N.

2.2 Limit Theorems

Sometimes if we know the convergence or divergence of a sequence we can use it todetermine the convergence or divergence of another sequence.

Theorem 2.4 (Squeeze Theorem). Suppose that the sequences {xn} and {yn} con-verge to the same limit l. If {wn} is a sequence such that

xn ≤ wn ≤ yn for n ≥ n0

for some n0 ∈ N then{wn} converges to l.

Proof. Let ε > 0. We want to show that wn → l as n→∞.

xn → l as n→∞ =⇒ there exists n1 ∈ N such that |xn − l| < ε whenever n ≥ n1.

=⇒ l − ε < xn < l + ε whenever n ≥ n1.

yn → l as n→∞ =⇒ there exists n2 ∈ N such that |yn − l| < ε whenever n ≥ n2.

=⇒ l − ε < yn < l + ε whenever n ≥ n2.


Thus we have

l − ε < xn < wn < yn < l + ε whenever n ≥ max {n0, n1, n2}.=⇒ − ε < wn − l < ε whenever n ≥ max {n0, n1, n2}.=⇒ wn → l as n→∞.

Theorem 2.5. If xn → 0 as n→∞ and {yn} is bounded, then xnyn → 0 as n→∞.

Proof. Since {yn} is bounded, there exists an M ∈ R such that |yn| ≤ M for alln ∈ N. We have

0 ≤ |xnyn| = |xn||yn| ≤M |xn|for all n ∈ N. Since xn → 0 as n → ∞, |xn| → 0 as n → ∞ and hence M |xn| → 0as n → ∞. Hence by the Squeeze Theorem |xnyn| → 0 and hence xnyn → 0 asn→∞.

Theorem 2.6. Let E ⊂ R has a finite supremum. Then there is a sequence {xn} inE such that xn → supE as n→∞. Similarly, If E has a finite infimum, then thereis a sequence {yn} in E such that yn → inf E as n→∞.

Proof. For every n ∈ N, by the Approximation Property of supremum, there existsxn ∈ E such that

supE − 1

n< xn ≤ supE.

By the Squeeze Theorem xn → supE as n→∞.

Theorem 2.7. Suppose that {xn} and {yn} are convergent sequences and α ∈ R.Then

1. limn→∞

(xn + yn) = limn→∞

xn + limn→∞

yn

2. limn→∞

(αxn) = α limn→∞

xn

3. limn→∞

(xnyn) = limn→∞

xn limn→∞

yn

4. limn→∞

xnyn

=limn→∞ xnlimn→∞ yn

whenever yn 6= 0 and limn→∞ 6= 0.

Proof. Let limn→∞ xn = a and limn→∞ yn = b. Let ε > 0.

1. We want to prove that limn→∞(xn + yn) = a+ b.

xn → a =⇒ there exists n1 ∈ N such that |xn − a| < ε whenever n ≥ n1.

yn → b =⇒ there exists n2 ∈ N such that |yn − a| < ε whenever n ≥ n2.

Now,|(xn + yn)− (a+ b)| ≤ |xn − a|+ |yn − b| < ε/2 + ε/2 = ε

whenever n ≥ max {n1, n2}.


2. We want to prove that limn→∞(αxn) = αa. Let ε > 0. Then

xn → a =⇒ there exists n0 ∈ N such that |xn − a| < ε/|α| whenever n ≥ n0.

Now|αxn − αa| = |α||xn − a| < |α|

ε

|α|= ε

whenever n ≥ n0.

3. Since the convergent sequence is bounded, there is an M ∈ R such that |yn| ≤Mfor all n ∈ N. Let ε > 0. There exists n1, n2 ∈ N such that |xn − a| < ε/2Mwhenever n ≥ n1 and |yn − b| < ε/2|a| whenever n ≥ n2.

|xnyn − ab| = |(xnyn − ayn) + (ayn − ab)|≤ |xn − a||yn|+ |a||yn − b|

<ε

2MM + |a| ε

2|a|= ε

4. Exercise.

Example 2.4. Evaluate limn→∞

2n3 − 5n+ 8

3n3 + 7.

Definition 2.4. Let {xn} be sequence. We say that {xn} diverges to ∞, writtenlimn→∞ xn = ∞ if and only if for each M ∈ R there is n0 ∈ N such that xn > Mwhenever n ≥ n0. And, we say that {xn} diverges to −∞, written limn→∞ xn = −∞if and only if for each M ∈ R there is n0 ∈ N such that xn < M whenever n ≥ n0.

Theorem 2.8 (Comparison Theorem). Suppose that xn and yn are convergent se-quence. If there is an n0 ∈ N such that xn ≤ yn for n ≥ n0, then limn→∞ xn ≤limn→∞ yn.

Proof. Let xn → a and yn → b as n → ∞. We want to show that a ≤ b. If a > b,then take ε = a − b. Take ε = a−b

2. Then there is an n1 ∈ N such that |xn − a| < ε

and |yn − b| < ε whenever n ≥ n1. Then for n ≥ n1

xn > a− ε = a− a− b2

= b+a− b

2= b+ ε > yn.

This is a contradiction to the hypothesis.

Remark 2.1. If {xn} and {yn} are convergent then

xn < yn, for n ≥ n0 =⇒ limn→∞

xn ≤ limn→∞

yn

but notlimn→∞

xn < limn→∞

yn.

For example, think of { 1n2} and { 1

n}.


Exercises

1. Give an example to show that every bounded sequence may not converge.

2. 2.1.1 (a).

3. Consider the set E = (1, 2) ⊂ R. Then inf E = 1. Construct a sequencexn ∈ (1, 2) such that xn → 1 as n→∞.

4. Suppose that {xn} is a sequence of non-negative real numbers and xn → x asn→∞. Prove that

√xn →

√x as n→∞.

5. Let r ∈ R. Prove that there exists a sequence of rational numbers {rn} suchthat rn → r as n→∞.

6. Let xn → l and yn → l as n → ∞. If xn ≤ wn ≤ yn for all n ∈ N, prove thatwn → l as n→∞.

2.3 Bolzano-Weierstrass Theorem

Definition 2.5. We say that a sequence {xn} is

i. increasing if xn ≤ xn+1 for all n ∈ N.

ii. strictly increasing if xn < xn+1 for all n ∈ N.

iii. decreasing if xn ≥ xn+1 for all n ∈ N.

iv. strictly decreasing if xn > xn+1 for all n ∈ N.

v. monotone if and only if it is either increasing or decreasing.

Theorem 2.9 (Monotone Convergence Theorem).

1. If {xn} is increasing and bounded above, then {xn} converges to a finite limit.

2. If {xn} is decreasing and bounded below, then {xn} converges to a finite limit.

Proof.

1. Suppose that {xn} is increasing and bounded above. Then by the CompletenessAxiom, supremum of {xn : n ∈ N} exists, say a = sup{xn : n ∈ N}. Let ε > 0.Then by the Approximation Property of supremum, there exists n0 ∈ N suchthat

a− ε < xn0 ≤ a.

Since {xn} is increasinga− ε < xn ≤ a

whenever n ≥ n0.


2. Exercise.

Example 2.5. If |a| < 1, then an → 0 as n→∞.

Proof. Since |a| < 1, |a|n+1 < |a|n for all n ∈ N. This implies that the sequence{|a|n} is decreasing. Observe that it is bounded below by 0. Hence the MonotoneConvergence Theorem implies that {|a|n} converges. Let limn→∞ |a|n = l. We wantto show that l = 0. Write

|a|n+1 = |a| · |a|n

=⇒ limn→∞

|a|n+1 = limn→∞

|a| · |a|n

=⇒ l = |a| · l=⇒ l = 0.

Example 2.6. If a > 0, then a1/n → 1 as n→∞.

Proof. Reading Exercise.

Definition 2.6. A sequence of sets {In} is said to be nested if and only if In ⊇ In+1

for all n ∈ N, that is,I1 ⊇ I2 ⊇ I3 ⊇ · · · .

Theorem 2.10 (Nested Interval Property). If {In} is a nested sequence of nonemptyclosed bounded intervals, then E = ∩∞n=1In is nonempty. Moreover, if the length ofthese intervals satisfy |In| → 0 as n→∞, then E is a single point.

Proof. Let In = [an, bn]. Since {In} is nested, we have

a1 ≤ a2 ≤ a3 ≤ · · · ≤ b3 ≤ b2 ≤ b1.

Thus the sequence of left end points{an} is a bounded increasing sequence and thesequence of right end points {bn} is a bounded decreasing sequence. By the MonotoneConvergence Theorem {an} and {bn} are convergent. Let an → a and bn → b. Sincean ≤ bn for all n ∈ N, a ≤ b and hence an ≤ a ≤ b ≤ bn for all n ∈ N. This implies[a, b] ⊆ [an, bn] for all n ∈ N. Hence ∩∞n=1In ⊇ [a, b] 6= ∅.

If |In| → 0 then bn − an → 0, that is, limn→∞(bn − an) = 0 =⇒ b − a =limn→∞ bn − limn→∞ an = 0. Thus a = b and [a, b] = {a}.

Remark 2.2.

1. If the intervals in the nested sequence {In} are not closed, then the conclusionof the above might not hold.


2. If the intervals in the nested sequence {In} are not bounded, then the conclusionof the above might not hold.

Proof. Here are the examples.

1. Take In = (0, 1/n). Then ∩∞n=1In = ∩∞n=1(0, 1/n) = ∅ (why?).

2. Take In = [1,∞). Then ∩∞n=1In = ∩∞n=1[n,∞) = ∅ (why?).

Theorem 2.11 (Bolzano-Weierstrass Theorem). Every bounded sequence of real num-bers has a convergent subsequence.

Proof. Let {xn} bounded sequence. Let a be a lower bound and b be an upper boundof {xn}. Then xn ∈ [a, b] for all n ∈ N. Split [a, b] in to two halves [a, a+b

2] and

[a+b2, b]. At least one of these intervals contains infinitely many terms of {xn}. Call

it I1 and choose n1 such that xn1 ∈ I1. Observe that length of I1 is b−a2

. Split I1 intotwo halves to get I2 where I2 is the interval containing infinitely many terms of {xn}.Choose n2 > n1 such that xn2 ∈ I2. Observe that the length of I2 is b−a

22.

Continue this process to get Ik for every k ∈ N such that Ik contains infinitelymany terms of xn. Then choose nk such that nk > nk−1 and xnk

∈ Ik. Observe thatthe length of Ik is b−a

2k.

Thus we have a nested sequence of nonempty closed and bounded intervals

I1 ⊇ I2 ⊇ I3 ⊇ · · ·

where |Ik| = b−a2k→ 0 as k → ∞. This implies ∩∞k=1Ik contains a single point. Let

x ∈ ∩∞k=1Ik. Now,

0 ≤ |xnk− x| ≤ |Ik| =

b− a2k

.

Since b−a2k→ 0 as k → ∞, by the Squeeze Theorem |xnk

− x| → 0. Thus we have aconvergent subsequence {xnk

}.

2.4 Cauchy Sequences

Definition 2.7. A sequence of real numbers {xn} is called a Cauchy sequence if andonly if for every ε > 0 there is an n0 ∈ N such that

|xn − xm| < ε whenever n,m ≥ n0.

Theorem 2.12. Every convergent sequence is Cauchy.


Proof. Let {xn} be a convergent sequence. Let xn → x as n → ∞. Then for ε > 0,there exists an n0 ∈ N such that

|xn − x| < ε/2 whenever n ≥ n0.

Hence for n,m ≥ n0,

|xn − xm| ≤ |xn − x|+ |xm − x| < ε/2 + ε/2 = ε.

This implies {xn} is Cauchy.

Theorem 2.13. Every Cauchy sequence is bounded.

Proof. Let {xn} be a Cauchy sequence. We want to show that {xn} is bounded. Letε = 1. Then there exists an n0 ∈ N such that

|xn − xm| < 1 whenever n,m ≥ n0.

In particular,|xn − xn0| < 1 whenever n ≥ n0.

This implies|xn| ≤ 1 + |xn0 | whenever n ≥ n0.

Thus|xn| ≤M for all n ∈ N

whereM = max{|x1|, |x2|, · · · , |xn0|, |xn0|+ 1}.

Theorem 2.14. A sequence of real numbers {xn} is Cauchy if and only if {xn} isconvergent.

Proof. Let {xn} be a Cauchy sequence. We want to show that {xn} is convergent.Since {xn} is Cauchy, it is bounded. By the Bolzano-Weierstrass Theorem, {xn}

has a convergent subsequence {xnk}, say xnk

→ a as k →∞. We want to show thatxn → a as n→∞.

Let ε > 0 . Then {xn} is Cauchy, there exists n1 ∈ N such that

|xn − xm| < ε/2 whenever n,m ≥ n1.

Since xnk→ a as k →∞, there exists an n2 ∈ N such that

|xnk− a| < ε/2 whenever k ≥ n2.

Choose k so big so that nk ≥ n1. Now,

|xn − a| ≤ |xn − xnk|+ |xnk

− a| < ε/2 + ε/2 = ε

whenever n ≥ n1. Thus we have xn → a as n→∞.The converse was proven in an earlier theorem.


Example 2.7. Prove that any real sequence {xn} that satisfies

|xn − xn+1| ≤1

2n, n ∈ N,

is convergent.

Proof. Let us show that {xn} is Cauchy. If m > n,

|xn − xm| = |xn − xn+1 + xn+1 − xn+1 + · · ·+ xm−1 − xm|≤ |xn − xn+1|+ |xn+1 − xn+1|+ · · ·+ |xm−1 − xm|

≤ 1

2n+

1

2n+1+

1

2n+2+ · · ·+ 1

2m−1

=1

2n

(1 +

1

2+

1

22+ · · ·+ 1

2m−n−1

)=

1

2n−1

(1− 1

2m−n

)<

1

2n−1.

Nor for ε > 0. Choose n0 ∈ N so that

1

2n−1< ε

for n ≥ n0. Thus {xn} is Cauchy hence convergent.

Remark 2.3. A sequence that satisfies xn+1−xn → 0 is not necessarily Cauchy. Forinstance, {log n} satisfies | log(n+1)− log n| = log

(n+1n

)→ log 1 = 0 but log n→∞.

This shows that {log n} is not convergent and hence not Cauchy.

2.5 Limits Supremum and Infimum

Consider a real sequence {xn}. Then think of all possible convergent subsequences of{xn} including the ones converging to ±∞. Now the set

L = {x : x is a limit of some subsequence of {xn}}.

The infimum of the set L is called the limit inferior of {xn} and the supremum of Lis called the limit superior of {xn}.Definition 2.8. Let {xn} be a real sequence. Then the limit supremum of {xn} isdefined to be

lim supn→∞

xn = limn→∞

(supk≥n

xk

).

Similarly, the limit infimum is defined to be

lim infn→∞

xn = limn→∞

(infk≥n

xk

).


Note that the lim sup and lim inf of a sequence {xn} can very well be the extendedreal numbers ±∞.

Example 2.8. Find the lim sup and lim inf of { 1n}.

Example 2.9. Find the lim sup and lim inf of {(−1)n}.

Theorem 2.15. Let {xn} be a sequence of real numbers. Then

lim infn→∞

xn ≤ lim supn→∞

xn.

Proof. Since infk≥n xk ≤ supk≥n xk, we have

limn→∞

infk≥n

xk ≤ limn→∞

supk≥n

xk.

Let {xn} be a real sequence and ε > 0. By the definition of the limit of a sequence,if xn → x, then the interval (x − ε, x + ε) contains all but finitely many terms ofthe sequence. How does this compare to the lim sup and lim inf of a sequence? Ift = lim inf xn and s = lim supxn, then the interval (t− ε, s + ε) contains all but thefinitely many terms of the sequence {xn}.

We state a few interesting theorems about limit supremum and limit infimumwithout proof as it might be overwhelming at this level. You can read the proof offthe textbook if you are interested.

Theorem 2.16.

1. Let {xn} be a sequence of real numbers and lim supn→∞ xn = s. Then thereexists a subsequence {xnk

} of {xn} such that xnk→ s as k →∞.

2. Let {xn} be a sequence of real numbers and lim infn→∞ xn = t. Then there existsa subsequence {xlj} of {xn} such that xlj → t as j →∞.


Theorem 2.17. Let {xn} be a sequence of real numbers and x be an extended realnumber. Then

limn→∞

xn = x if and only if lim infn→∞

xn = lim supn→∞

xn = x.


Theorem 2.18. Let {xn} be a sequence and x be a limit of a subsequence. Thenlim infn→∞ xn ≤ x ≤ lim supn→∞ xn.has a subsequence

Theorem 2.19. If xn ≤ yn for n large, then

lim supn→∞

xn ≤ lim supn→∞

yn and lim infn→∞

xn ≤ lim infn→∞

yn.


Exercise

1. Prove that if a sequence {xn} is decreasing and bounded below then it convergesto a finite limit.

2. Prove that the sequence {2 1n} converges to 1, that is, 2

1n → 1 as n → ∞.

(Mimic the proof in Example 2.21 Case 2.)

3. Exercise 2.3.2.

4. If {xn} and {yn} are Cauchy sequences prove that {xnyn} is Cauchy.

5. Find lim supn→∞ xn and limn→∞ xn where xn = sin(nπ/2).

Chapter 3

Functions on R

3.1 Limit of a Function

Definition 3.1. Let I ⊆ R be an open interval and a ∈ I. Let f be a functiondefined on I \ {a}. Then f(x) is said to have a limit L as x approaches to a if andonly if for every ε > 0 there exists a δ > 0 such that

|f(x)− L| < ε whenever 0 < |x− δ| < ε.

If f(x) has a limit L at a we write

limx→a

f(x) = L.

Example 3.1. If f(x) = c is a constant function defined on R, then for any a ∈R, limx→a f(x) = c. To see this, for given ε > 0 one can choose any δ > 0 for which|f(x)− c| = |c− c| = 0 < ε holds automatically for all x satisfying 0 < |x− a| < δ.

Example 3.2. For a linear function f(x) = mx+ c, where m, c ∈ R,

limx→a

f(x) = f(a)

for all a ∈ R.

Proof. The case m = 0 is discussed in previous example. So we assume that m 6= 0.Let ε > 0. Choose δ = ε/m. Now,

|f(x)− f(a)| = |mx+ c−ma− c| = |m||x− a| < |m|δ = ε

whenever 0 < |x− a| < δ.

Example 3.3. Based on what we learned in Calculus I, we know that

limx→2

x2 = 4.

24

CHAPTER 3. FUNCTIONS ON R 25

Let us prove this using our ε− δ definition.Let ε > 0. Here,

|f(x)− L| = |x2 − 4| = |x+ 2||x− 2|.

We want to find a δ > 0 such that |f(x) − L| < ε whenever 0 < |x − 2| < δ. If|x − 2| < 1 we get |x| < |x − 2| + |2| < 3. Hence, |x + 2| ≤ |x| + 2 < 5. Takeδ = min{1, ε/5}. Then,

|x2 − 4| = |x+ 2||x− 2| < 5δ ≤ ε

whenever 0 < |x− 2| < δ.

Theorem 3.1. Let I ⊆ R be an open interval and a ∈ I. Let f, g be real functionsdefined on I \ {a}. Let lim

x→af(x) and lim

x→ag(x) exist. Then we have

1. limx→a

(f + g)(x) = limx→a

f(x) + limx→a

g(x).

2. limx→a

(αf)(x) = α limx→a

f(x).

3. limx→a

(fg)(x) = limx→a

f(x) limx→a

g(x)

4. limx→a

(f

g

)(x) =

limx→a f(x)

limx→a g(x)when lim

x→ag(x) 6= 0.

Theorem 3.2 (Squeeze Theorem). Let I ⊆ R be an open interval and a ∈ I. Letf, g, h be real functions defined on I \ {a}. If f(x) ≤ h(x) ≤ g(x) for all x ∈ I \ {a}and

limx→a

f(x) = L = limx→a

g(x),

thenlimx→a

h(x) = L.

Theorem 3.3 (Comparison Theorem). Let I ⊆ R be an open interval and a ∈ I.Let f, g are real functions defined on I \ {a}. If lim

x→af(x) and lim

x→ag(x) exist and

f(x) ≤ g(x) for all x ∈ I \ {a}, then

limx→a

f(x) ≤ limx→a

g(x).

Theorem 3.4. Let I ⊆ R be an open interval and a ∈ I. Let f be a real functiondefined on I \ {a}. Then lim

x→af(x) = L exists if and only if lim

n→∞f(xn) = L for every

sequence {xn}, xn ∈ I \ {a} which converges to a.


3.2 One-Sided Limits and Limits Involving Infinity

Definition 3.2. Let f be a real function and a, L ∈ R.

1. We say that f(x) converges to L as x approaches a from left if and only if forgiven ε > 0 there exists a δ > 0 such that f is defined on (a− δ, a) and

|f(x)− L| < ε whenever a− δ < x < a.

In this case we write limx→a−

f(x) = L.

2. We say that f(x) converges to L as x approaches a from right if and only if forgiven ε > 0 there exists a δ > 0 such that f is defined on (a, a+ δ) and

|f(x)− L| < ε whenever a < x < a+ δ.

In this case we write limx→a+

f(x) = L.

3. We say that f(x) converges to L as x→∞ if and only if for given ε > 0 thereexists an M ∈ R such that f is defined on (M,∞) and

|f(x)− L| < ε whenever M < x <∞.

In this case we write limx→∞

f(x) = L.

4. We say that f(x) converges to L as x→ −∞ if and only if for given ε > 0 thereexists an M ∈ R such that f is defined on (−∞,M) and

|f(x)− L| < ε whenever −∞ < x < M.

In this case we write limx→−∞

f(x) = L.

5. We say that f(x) converges to ∞ as x→ a if and only if for given M > 0 thereexists a δ > 0 such that f is defined on {x : 0 < |x− a| < δ} and

f(x) > M whenever 0 < |x− a| < δ.

In this case we write limx→a

f(x) =∞.

6. We say that f(x) converges to −∞ as x → a if and only if for given M < 0there exists a δ > 0 such that f is defined on {x : 0 < |x− a| < δ} and

f(x) < M whenever 0 < |x− a| < δ.

In this case we write limx→a

f(x) = −∞.


Theorem 3.5. Let f be a real function. Then

limx→a

f(x) = L

if and only iflimx→a−

f(x) = L = limx→a+

f(x).

Example 3.4. Let f(x) =

{x+ 1 if x ≥ 0

x− 1 if x < 0.Prove that lim

x→0−f(x) = 1.

Proof. Given ε > 0. Take δ = ε. Then

|f(x)− 1| = |x| < ε whenever − δ < x < 0.

Example 3.5. Prove that limx→0+

√x = 0.

Proof. Given ε > 0. Take δ = ε2. Then

|f(x)− L| = |√x− 0| =

√x <√δ = ε whenever 0 < x < δ.

Example 3.6. Prove that limx→∞

1

x= 0.

Proof. Given ε > 0. Take M = 1ε. Then

|f(x)− L| =∣∣∣∣1x − 0

∣∣∣∣ =1

|x|<

1

M= ε

whenever M < x <∞.

Exercises

1. 3.1.1 (a)

2. 3.1.6 (a)

3. Prove that limx→0+

1

x=∞.


3.3 Continuity

In Calculus we learned that a function f is called continuous at a point a if

limx→a−

f(x) = limx→a+

f(x) = f(a)

which is simplylimx→a

f(x) = f(a).

Definition 3.3. Let E be a nonempty subset of R and f : E → R. Then f is saidto be continuous at a point a ∈ E if and only if given ε > 0 there exists a δ > 0 suchthat

|f(x)− f(a)| < ε whenever |x− a| < δ and x ∈ E.The functions f is said to be continuous on E if and only if f is continuous at

every point of E.

Remark 3.1. Let I ⊆ R be an open interval and a ∈ I. Then a function f : I → Ris continuous at a if and only if

limx→a

f(x) = f(a).

Proof. limx→a f(x) = f(a) if and only if given ε > 0 there is a δ > 0 such that

|f(x)− f(a)| < ε whenever |x− a| < δ

if and only if f is continuous at a. Notice that we have to choose δ so small that(a− δ, a+ δ) ⊂ I. Also notice that we did not exclude a from (a− δ, a+ δ) becausef(a) is well defined.

Example 3.7. The function f(x) = sinxx

is not continuous at 0 because the functionis not defined at 0.

Example 3.8. The function f : R→ R defined by

f(x) =

{1 if x is rational

−1 if x is irrational

is discontinuous everywhere.

Proof. Take ε = 1. Let a ∈ R be any number. Then for every δ > 0, the interval(a− δ, a+ δ) contains both rational and irrational numbers. If a is rational, then

|f(x)− f(a)| = | − 1− 1| = 2 > ε

for all irrational x. Similarly, if a is irrational, then

|f(x)− f(a)| = |1 + 1| = 2 > ε

for all rational x. Thus f is discontinuous at every a ∈ R.


Example 3.9. The function f : Z→ R defined by f(x) = x is continuous on Z.

Proof. Given ε > 0. Let a ∈ Z be any point. Choose 0 < δ < 1. Then (a− δ, a+ δ) ={a}. Then it automatically follows that,

|f(x)− f(a)| = |f(a)− f(a)| = 0 < ε whenever |x− a| < δ.

Notice that the only x satisfying |x− a| < δ is a itself.

Theorem 3.6. Suppose that E is a nonempty subset of R, a ∈ E and f : E → R.Then the following statements are equivalent.

i. f is continuous at a ∈ E.

ii. If xn ∈ E and xn → a, then f(xn)→ f(a) as n→∞.

Theorem 3.7. Let E be a nonempty subset of R and f, g : E → R. If f, g arecontinuous at a point a ∈ E, then f+g, gf , and αf, α ∈ R are continuous. Moreover,f/g is continuous at a when g(a) 6= 0.

Definition 3.4. Suppose that A and B are subset of R. Let f : A→ R, g : B → Rand f(A) ⊆ B. Then the composition of g with f is the function g ◦ f : A → Rdefined by

g ◦ f(x) = g(f(x)), x ∈ A.

Theorem 3.8. Suppose that A and B are subsets of R, that f : A→ R and g : B → Rand f(A) ⊆ B.

1. If f is continuous at a ∈ A and g is continuous at f(a) ∈ B, then g ◦ f iscontinuous at a ∈ A.

2. If

(a) A = I \ {a}, where I is a nondegenerate interval which either contains aor has a one of its endpoints

(b) limx→a

f(x) = L and L ∈ B, and

(c) g is continuous at L,

thenlimx→a

g(f(x)) = g(

limx→a

f(x)).

Definition 3.5. Let E be a nonempty subset of R. A function f : E → R is saidto be bounded on E if and only if there is an M ∈ R such that |f(x)| ≤ M for allx ∈ E.

Theorem 3.9 (Extreme Value Theorem). If


(a) I ⊂ R is a closed and bounded interval and

(b) f : I → R is continuous on I,

then f is bounded on I. Moreover, there exists xM , xm ∈ I such that

f(xM) = supx∈I

f(x) and f(xm) = infx∈I

f(x).

Proof. Let I ⊂ R be a closed and bounded interval and f : I → R be continuous onI. We want to show that f is bounded on I.

If possible assume that f is not bounded. Then for every n ∈ N there exists xn ∈ Isuch that

|f(xn)| > n.

This implies that |f(xn)| → ∞ as n→∞.The sequence {xn} is bounded. By Bolzano–Weierstrass Theorem {xn} has a

convergent subsequence, say xnk→ a as k → ∞. Since I is closed, a ∈ I. This

implies that f(a) ∈ R. So |f(a)| <∞. But

|f(a)| = limk→∞|f(xnk

)| =∞

which is a contradiction. Hence f is bounded.Let

M = supx∈I

f(x) and m = infx∈I

f(x).

We want to show that there is xM , xm ∈ I such that

f(xM) = M and f(xm) = m.

Assume on the contrary that f(x) < M for all x ∈ I. Then the function1

M − f(x)is continuous on I, hence bounded on I. Let∣∣∣∣ 1

M − f(x)

∣∣∣∣ ≤ C

for all x ∈ I. This implies f(x) ≤M − 1

Cfor all x ∈ I. This implies that

M = supx∈I

f(x) ≤M − 1

C


Theorem 3.10 (Intermediate Value Theorem). Suppose that a < b and f : [a, b]→ Ris continuous. If y0 is a number between f(a) and f(b), then there exists an x0 ∈ (a, b)such that f(x0) = y0.


Proof. Without loss of generality, let us assume that f(a) < f(b). Define

E = {x ∈ [a, b] : f(x) < y0}.

Since f(a) < y0 < f(b), a ∈ E. So E is nonempty and bounded. This implies thatsupE exists, say supE = x0. Observe that f(x0) ≤ y0 and x0 ∈ [a, b).(why ?)

We want to show that f(x0) = y0. If f(x0) < y0, then the function

g(x) = y0 − f(x)

is continuous on [a, b] and g(x0) = y0 − f(x0) > 0. Let ε = g(x0)/2. Since g iscontinuous at x0, there exists δ > 0 such that {x : |x− x0| < δ} ⊂ [a, b] and

|g(x)− g(x0)| <g(x0)

2

whenever |x− x0| < δ. This implies

g(x0)

2< g(x)

whenever |x− x0| < δ. Then for any x1 ∈ (x0, x0 + δ),

0 <g(x0)

2< g(x1) = y0 − f(x1) =⇒ f(x1) < y0.

Thus we have x1 > x0 and x1 ∈ E. This implies that x0 6= supE, a contradiction.Hence f(x0) = y0. Also it follows that x0 6= a and hence x0 ∈ (a, b).

Documents

Introduction to Real Analysis I to Real Analysis I Lecture notes by: Khim R Shrestha, Ph. D. Assistant Professor of Mathematics University of Great Falls Great Falls, Montana