Contentsweb.math.ucsb.edu/~zhou/Math117_2017f/Math 117_lecture...MATH 117 LECTURE NOTES XIN ZHOU...

MATH 117 LECTURE NOTES

XIN ZHOU

Abstract. This is the set of lecture notes for Math 117 during Fall quarter

of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1].

Contents

1. The set N of natural numbers 2

2. The set Q of rational numbers 4

3. Ordered fields 7

4. Real numbers R and the Completeness Axiom 11

5. +∞ and −∞ 15

6. Limits of sequences 16

7. Proofs 18

8. Problem session 21

9. Limit theorems for sequences 23

10. Monotone sequences and Cauchy sequences 29

11. Subsequences 35

12. lim sup and lim inf revisited 40

13. Series 42

14. Alternating series and Integral Tests 47

15. Continuous functions 50

16. Properties of continuous functions 55

References 59

Date: December 5, 2017.1

2 XIN ZHOU

1. The set N of natural numbers

We introduce basic properties for the set of natural numbers.

• N = {1, 2, 3, · · · } denotes the set of all natural numbers, or all positive

integers;

• each n ∈ N1 has a successor n + 1. For example, the successor of 5 is

6.

Proposition 1.1. The set N satisfies the following properties:

N1. 1 ∈ N;

N2. if n ∈ N, then its successor n+ 1 ∈ N;

N3. 1 is not the successor of any element in N;

N4. if n and m have the same successor, then n = m;

N5. suppose S is a subset of N satisfying: 1 ∈ N and if n ∈ S then n+1 ∈ S,

then S = N.

Property (N5) is called the Peano Axiom or Peano Postulate.

Property (N5) is the basis for mathematical induction:

• Let P1, P2, · · · be a list of statements of propositions that may or may

not be true;

• The Principle of Mathematical Induction asserts that:

All of P1, P2, · · · are true provided:

– P1 is true; (this is called the basis of induction)

– if Pn is true, then Pn+1 is true for all n ∈ N. (this is called

induction step)

Example 1.2. All numbers of form 7n − 2n (n ∈ N) is divisible by 5.

Proof. The n-th proposition is:

Pn: “7n − 2n is divisible by 5”.

• Basis of induction P1: 71 − 21 = 5 is divisible by 5, so P1 is true;

• Induction step: Suppose Pn is true, then 7n−2n = 5m for some m ∈ N.

To verify Pn+1, we have

7n+1 − 2n+1 = 7n+1 − 2 · 7n + 2 · 7n − 2n+1

= 7n(7− 2) + 2(7n − 2n)

= 5 · 7n + 2(5m) = 5(7n + 2m).

1The symbol n ∈ N means “n is an element of the set N”

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2. The set Q of rational numbers

We introduce basic properties for the set of integers and rational numbers.

• Z = {0,−1, 1,−2, 2, · · · } denotes the set of all integers;

• Q = {mn

: m,n ∈ Z, n 6= 0} denotes the set of all rational numbers.

– Q is a good algebraic system: one can do basic operations like

additions, multiplication, subtraction, division over Q;

– Q is inadequate when solving algebraic equations like: x2− 2 = 0.

Actually, the solutions of x2 − 2 = 0 makes sense:

(1) by the Pythagorean Theorem, if d is the length of the diagonal of a

square of edge length 1, then d2 = 11+12 = 2, and hence d is a solution;

(2) the graph of the function y = x2 − 2 will pass through the x-axis at

two points which are both solutions of x2 − 2 = 0.

This means that:

there are gaps in Q, as√

2 6= Q.

Moreover, there are more exotic numbers like “π”, and “e”:

• “π” appears when studying the perimeters of circles and spheres;

• “e” appears when studying the infinite sums:∑

n∈N1n!

.

Definition 2.1. A number is called an algebraic number if it satisfies a poly-

nomial equation:

anxn + an−1x

n−1 + · · ·+ a1x+ a0 = 0,

where ai ∈ Z for all i = 0, · · · , n, and an 6= 0.

Example 2.2. We know that:

• all rational numbers are algebraic numbers:

if r = mn

, m,n ∈ Z, n 6= 0, then it satisfies nx−m = 0;

• square root√·, cubic root 3

√·, etc, are algebraic numbers;

• ordinary algebraic operations (including addition, multiplication, sub-

traction, division, taking roots) of algebraic numbers are also algebraic

numbers.

Example 2.3. 171/3, (2+51/3)1/2,[(4−2·31/2)/7

]1/2are all algebraic numbers.

Proof. (1) 171/3 is a solution of x3 − 17 = 0;

(2) Let a = (2 + 51/3)1/2, =⇒ a2 = 2 + 51/3, =⇒ (a2 − 2)3 = 5, =⇒a6 − 2a4 + 4a2 − 8 = 5, =⇒ a6 − 2a4 + 4a2 − 13 = 0;


(3) Let b =[(4−2·31/2)/7

]1/2, =⇒ 7b2 = 4−2·31/2, =⇒ 7b2−4 = −2·31/2,

=⇒ (7b2−4)2 = 12, =⇒ 49b4−56b2 +16 = 12, =⇒ 49b4−56b2 +4 = 0.

�

Recall that

• an integer k is a factor of another integer m, or k divides m, if m/k is

also an integer;

• an integer p ≥ 2 is a prime provided that the only positive factors of p

is 1 and p.

Proposition 2.4.√

2 is not a rational number.

Proof. We prove by contradiction argument. Suppose the statement is not

true, that is√

2 ∈ Q, =⇒√

2 = pq, where p, q are both integers with no

common factor and q 6= 0. Then

p2

q2= 2, =⇒ p2 = 2q2.

Lemma 2.5. If p2 is an even number (p ∈ Z), then p is also even.

Proof. Using contradiction argument again, suppose not, then p is odd, =⇒p = 2m+ 1 for some m ∈ Z. Then

p2 = (2m+ 1)2 = 4m2 + 4m+ 1 = 2(2m2 + 2) + 1,

so p2 is odd, and this is a contradiction. �

Let us go back to the proof of Proposition 2.4. Since p2 is even, by the above

lemma p is also even, =⇒ p = 2m, hence

p2 = 4m2 = 2q2, =⇒ q2 = 2m2.

So q2 is even, and hence q is even by the above lemma. But this is a contra-

diction to the assumption that p, q has no common factor. �

Theorem 2.6 (Rational Zero Theorem). Suppose a0, a1, · · · , an are all inte-

gers, and r is a rational number satisfying the polynomial equation:

anxn + · · ·+ a1x+ a0 = 0,

where n ≥ 1, an 6= 0, a0 6= 0. Write

r =p

q,

where p, q ∈ Z have no common factor and q 6= 0. Then

q divides an, and p divides a0.

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Proof. Plug in r = pq

into the equation:

an(p

q)n + an−1(

p

q)n−1 + · · ·+ a1(

p

q) + a0 = 0.

Multiply by qn, =⇒

anpn + an−1p

n−1q + · · ·+ a1pqn−1 + a0q

n = 0.

Re-arrange the equation:

anpn = −q(an−1pn−1 + · · ·+ a1pq

n−2 + a0qn−1),

=⇒ q divides anpn. Since p, q has no common factor, q must divide an.

We can prove p divides a0 in a similar manner. In particular,

a0qn = −p(anpn−1 + an−1p

n−2 + · · ·+ a1qn−1),

=⇒ p divides a0qn, and hence p must divide a0. �

Example 2.7. Use the above theorem to prove that√

2 is not a rational num-

ber.

Proof. Suppose the equation x2 − 2 = 0 has a rational root pq, where p, q ∈ Z

has no common factor and q 6= 0. Then by Theorem 2.6,

q divides 1, and p divides −2.

There p can only be ±1 or ±2, and q can only be ±1, =⇒ pq

must be of the

forms ±1, ±2. By substituting back to x2 − 2 = 0, we can see that none of

them is a solution. This is a contradiction. �

Example 2.8. Prove that a = (2 + 51/3)1/2 is not a rational number.

Proof. By the above example, we know that a satisfies:

a6 − 2a4 + 4a2 − 13 = 0.

If a is a rational number, then write a = pq, where p, q ∈ Z has no common

factor and q 6= 0. By the Rational Zero Theorem,

q divides 1, and p divides −13.

There p can only be ±1 or ±13, and q can only be ±1, =⇒ pq

must be of

the forms ±1, ±13. By substituting back, we can see that none of them is a

solution. This is a contradiction. �


3. Ordered fields

The basic algebraic operations on Q are “+” and “·”. The following prop-

erties hold in this algebraic system (Q,+, ·):

Properties for “+”:

A1. a+ (b+ c) = (a+ b) + c, ∀a, b, c ∈ Q2 – associative law;

A2. a+ b = b+ a, ∀a, b ∈ Q – commutative law;

A3. a+ 0 = a, ∀a ∈ Q – existence of 0;

A4. For each a ∈ Q, there exists an element −a ∈ Q, such that a+(−a) = 0

– existence of reverse;

properties for “·”:

M1. a · (b · c) = (a · b) · c, ∀a, b, c ∈ Q – associative law;

M2. a · b = b · a, ∀a, b ∈ Q – commutative law;

M3. a · 1 = a, ∀a ∈ Q – existence of 1;

M4. For each a ∈ Q, a 6= 0, there exists an element a−1 ∈ Q, such that

a · (a−1) = 1 – existence of reciprocal;

a property relating “·” and “·”:

DL. a · (b+ c) = a · b+ a · c, ∀a, b, c ∈ Q – distributive law.

Definition 3.1. A set F is a field if it has two operations “+” and “·”, and

elements 0 and 1 satisfying all the properties (A1)...(A4) (M1)...(M4) (DL)

(with Q changed to F ).

Example 3.2. There are two examples which are not fields:

• N (the natural numbers ) fails properties (A4) and (M4);

• Z (the integers) fails (M4).

Q also has an order structure “≤” satisfying the properties:

O1. given a, b ∈ Q, either a ≤ b or b ≤ a;

O2. if a ≤ b and b ≤ a, then a = b;

O3. if a ≤ b and b ≤ c, then a ≤ c – transitive law;

O4. if a ≤ b, then a+ c ≤ b+ c;

O5. if a ≤ b and 0 ≤ c, then a · c ≤ b · c.

Definition 3.3. A field F is called an ordered field if it has an order structure

satisfying properties (O1)...(O5).

2Here the symbol ∀ means “for all”.

8 XIN ZHOU

Remark 3.4. • a < b means that a ≤ b and a 6= b.

Properties of ordered fields: based on the defining properties, we can

deduce more properties for a field or an ordered field.

Theorem 3.5. Suppose (F,+, ·) is a field with elements 0, 1, and a, b, c denote

arbitrary elements in F , then:

(i) a+ c = b+ c, =⇒ a = b;

(ii) a · 0 = a, ∀a;

(iii) (−a) · b = −(a · b);

(iv) (−a) · (−b) = a · b;(v) a · c = b · c, and c 6= 0, =⇒ a = b;

(vi) a · b = 0, =⇒ either a = 0 or b = 0.

Proof.

(i)

(a+ c) + (−c) = a+ (c+ (−c)) = a+ 0 = a

= (b+ c) + (−c) = b+ (c+ (−c)) = b+ 0 = b.

(ii) a · 0 = a · (0 + 0) = a · 0 + a · 0, =⇒ (by (i)) a · 0 = 0.

(iii) (−a) · b+ a · b = (−a+ a) · b = 0 · b = 0 (by (ii)), so (−a) · b = −(a · b).(iv) By (iii), (−a) · (−b) = −a · (−b) = −(−a · b) = ab.

(v) Homework.

(vi) Homework.

�

Theorem 3.6. Suppose F is an ordered field with order structure “≤”, and

a, b, c denotes arbitrary elements in F , then:

(i) if a ≤ b, then −b ≤ −a;

(ii) if a ≤ b and c ≤ 0, then bc ≤ ac;

(iii) if 0 ≤ a and 0 ≤ b, then 0 ≤ ab;

(iv) 0 ≤ a2(= a · a), ∀a;

(v) 0 < 1;

(vi) if 0 < a, then 0 < a−1;

(vii) if 0 < a < b, then 0 < b−1 < a−1.

Proof.

(i) By (O4), a + (−b) ≤ b + (−b), =⇒ a + (−b) ≤ 0; by (O4), −a + [a +

(−b)] ≤ −a+ 0 = −a; by (A1), −b ≤ −a.


(ii) By (i), 0 ≤ −c, so by (O5), a · (−c) ≤ b · (−c); by Theorem 3.5(iii),

−ac ≤ −bc; by (i), bc ≤ ac.

(iii) By(ii).

(iv) If 0 ≤ a, then it follows from (iii); if a ≤ 0, then by (i), 0 ≤ (−a); by

(iii), 0 ≤ (−a) · (−a) = a2 (by Theorem 3.5)(iv).

(v) Since 12 = 1, by (v), 0 ≤ 1; also as 0 6= 1, we have 0 < 1.

(vi) Homework.

(vii) Homework.

�

Definition 3.7. Let F be an ordered field. Given an element a ∈ F , we can

define the absolute value |a| of a as:

|a| := a if a ≥ 0, and |a| := −a if a < 0.

Here a ≥ 0 is an equivalent way to write 0 ≤ a.

The absolute value gives rise to the notion of distance.

Definition 3.8. Given a, b ∈ F , the distance between them is defined as:

dist(a, b) := |a− b|.

Theorem 3.9. Let F be an ordered field. For all a, b ∈ F ,

(i) |a| ≥ 0;

(ii) |a · b| = |a| · |b|;(iii) |a+ b| ≤ |a|+ |b|.

Proof. (i) If a ≥ 0, then by definition |a| = a ≥ 0; if a < 0, then |a| =

−a ≥ 0 by Theorem 3.6(i).

(ii) We can compare both sides in four cases: (1) a ≥ 0, b ≥ 0, (2) a ≥0, b < 0, (3) a < 0, b ≥ 0, (4) a < 0, b < 0.

For case (1): a · b ≥ 0 by Theorem 3.6(iii), so |ab| = ab; on the other

hand, |a| = a, |b| = b, so |a| · |b| = ab.

We will leave other cases as exercises.

(iii) By definition, since either a = |a| or a = −|a|, we have

−|a| ≤ a ≤ |a|, and − |b| ≤ b ≤ |b|.

Using (O4),

−|a|+ (−|b|) ≤ −|a|+ b ≤ a+ b ≤ |a|+ b ≤ |a|+ |b|.

10 XIN ZHOU

Therefore

−(|a|+ |b|) ≤ a+ b ≤ |a|+ |b|−(|a|+ |b|) ≤ −(a+ b) ≤ |a|+ |b| } =⇒

|a+ b| ≤ |a|+ |b|.�

Corollary 3.10. We have the triangle inequality:

dist(a, c) ≤ dist(a, b) + dist(b, c).

Proof.

dist(a, c) = |a− c| = |a− b+ b− c| ≤ |a− b|+ |b− c| = dist(a, b) + dist(b, c).

�


4. Real numbers R and the Completeness Axiom

Heuristically, the set of real numbers R is the order field F containing Qwith “no gaps”. We will make rigorous what does “no gap” mean.

Definition 4.1. Let S ⊂ F be a subset of an ordered field F , and S 6= ∅.(a) If S contains a largest element s0, [that is to say, s0 ∈ S, and s ≤ s0,

∀s ∈ S], we call s0 the maximum of S, and write s0 = maxS;

(b) If S contains a smallest element s′0, [that is to say, s′0 ∈ S, and s′0 ≤ s,

∀s ∈ S], we call s′0 the minimum of S, and write s0 = minS.

Example 4.2. (1) Every finite subset of Q has a maximum and a mini-

mum, i.e.

max{1, 2, 2.6, 3, 4} = 4, and min{1, 2, 2.6, 3, 4} = 1.

(2) Given a, b ∈ F , with a < b, we denote by

[a, b] = {x ∈ F : a ≤ x ≤ b} − −closed interval,

(a, b) = {x ∈ F : a < x < b} − −open interval,

[a, b) = {x ∈ F : a ≤ x < b} − −half-open interval,

(a, b] = {x ∈ F : a < x ≤ b} − −half-open interval.

Among them:

max[a, b] = b, min[a, b] = a,

(a, b) has no maximum and minimum in general.

(3) The set {r ∈ Q : 0 ≤ r ≤√

2} has no maximum.

These examples say that maximums and minimums may not always exist.

Nevertheless, we can generate the concepts to upper and lower bounds.

Definition 4.3. Let S ⊂ F be a subset and S 6= ∅.(a) If M ∈ F and s ≤ M ∀s ∈ S, then M is called an upper bound of S,

and S is said to be bounded from above;

(b) If m ∈ F and m ≤ s ∀s ∈ S, then m is called an lower bound of S,

and S is said to be bounded from below;

(c) S is said to be bounded if it is bounded above and below, and this is

equivalent to say S ⊂ [m,M ] for some m,M ∈ F .

Example 4.4. (1) If S has the maximum maxS, then maxS is an upper

bound of S; similarly minS is a lower bound of S if it exists.

(2) b is an upper bound of the intervals [a, b], (a, b), (a, b] and [a, b).

12 XIN ZHOU

(3) 2 is an upper bound of {r ∈ Q : 0 ≤ r ≤√

2};√

2 is the least upper

bound.

Now we summarize the definition for least upper bound and greatest lower

bound.

Definition 4.5. Let S ⊂ F be a subset and S 6= ∅.(a) If S is bounded from above and S has a least upper bound, then we

call it the supreme of S and denote it by supS.

(b) If S is bounded from below and S has a greatest lower bound, then we

call it the infimum of S and denote it by inf S.

We have the following criterion for least upper bound: M = supS if and

only if

(i)s ≤M,∀s ∈ S,(ii) if M1 < M, then ∃s1 ∈ S, such that s1 > M.

Example 4.6. (a) If S has a maximum, then maxS = supS, and simi-

larly for the minimum and infimum of S.

(b) sup[a, b] = sup[a, b) = sup(a, b] = sup(a, b) = b.

(c) Let A = {r ∈ Q : 0 ≤ r ≤√

2}, then supA =√

2 /∈ A.

Let us elaborate a bit about this example using decimal expansions:

first it is obvious that√

2 is an upper bound, and we will explain why

it is the least one; the decimal expansion for√

2 is 1.4142135623.....; if

M1 is any number less than√

2, even if it is very close to√

2, there

should exist a first decimal of M1 that is less than the corresponding

one for√

2; for instance, M1 = 1.4141, or M1 = 1.414212, or M1 =

1.41421355; and for these numbers we can find s1 = 1.4142, s1 =

1.414213, s1 = 1.41421356 that is less than√

2 but greater than M1.

Remark 4.7. By case (c) above, the least upper bound may not belong to S.

This is a way to say that there exist gaps in F .

Completeness Axiom:

every nonempty subset S ⊂ R that is bounded above has a least upper

bound, ⇐⇒ supS exists and is a real number.

Definition 4.8. The set of real numbers R is an ordered field containing Qthat satisfies the “Completeness Axiom”.


Remark 4.9. The example A = {r ∈ Q : 0 ≤ r ≤√

2} shows that Q does not

satisfy the “Completeness Axiom”.

Corollary 4.10. Every nonempty subset S ⊂ R which is bounded from below

has a greatest upper bound.

Proof. Consider the subset, denoted by −S = {−s : s ∈ S}. Since S is

bounded from below, ∃m ∈ R, such that m ≤ s, ∀s ∈ S, =⇒ −s ≤ −m,

∀ − s ∈ −S. Thus −S is bounded from above.

By the “Completeness Axiom”, the supreme of −S exists and sup(−S) ∈ R.

Let s0 = sup(−S), and we claim that −s0 = inf S. In particular,

(1) −s ≤ s0, ∀ − s ∈ −S, =⇒ −s0 ≤ s, ∀s ∈ S, and hence −s0 is a lower

bound of S;

(2) To show that −s0 is the greatest lower bound, consider any other lower

bound t ∈ R, such that t ≤ s, ∀s ∈ S, =⇒ −s ≤ −t, ∀ − s ∈ −S; thus

−t is an upper bound of −S, so s0 ≤ −t, =⇒ t ≤ −s0; therefore −s0is the greatest lower bound.

�

There are two important corollaries of the completeness axiom.

Theorem 4.11 (Archimedean Property). Given two real numbers a, b ∈ R, if

a > 0 and b > 0, then for some positive integer n ∈ N , we have na > b.

Proof. We use the contradiction argument. Suppose the conclusion is not true,

which means that na ≤ b, ∀n ∈ N . Denote the set S as

S = {na : n ∈ N}.

Then b is an upper bound of S. By the completeness axiom: the supreme of

S exists and s0 = supS ∈ R.

Since a > 0, then s0 < s0 + a, or equivalently s0− a < s0. By the definition

of supreme, we know that s0 − a is no longer an upper bound of S, so there

exists an element n0a, for some n0 ∈ N, such that

s0 − a < n0a.

=⇒ s0 < n0a+ a = (n0 + 1)a ∈ S.There is a contradiction to the fact that s0 is an upper bound of S. �

Corollary 4.12. (1) if a > 0, then 1/n(= n−1) < a for some positive

integer n ∈ N;

(2) if b > 0, then b < n for some positive integer n ∈ N.

14 XIN ZHOU

Proof. (1) follows from the Archimedean Property by letting b = 1, and (2)

by letting a = 1. �

Theorem 4.13 (Denseness of Q). Given two real numbers a, b ∈ R, if a < b,

then there exists a rational number r ∈ Q, such that a < r < b.

Proof. We need to find some integers m,n ∈ Z, n > 0, such that a < mn< b,

and this is equivalent to

na < m < nb.

Since a < b =⇒ (b− a) > 0, by the Archimedean property, there exists n ∈ N,

such that

n(b− a) > 1.

By the Archimedean property again, there exists k ∈ N, such that

k > max{|na|, |nb|}.

Therefore,

−k < na < nb < k.

Consider the set {j ∈ Z : −k < j ≤ k, and na < j}, and it is a finite and

nonempty set (since k belongs to it). As it is finite, we pick the minimum

= min{j ∈ Z : −k < j ≤ k, and na < j}.

Then na < m, and moreover, m− 1 ≤ na. Therefore

na < m = (m− 1) + 1 ≤ na+ 1 < na+ (nb− na) = nb.

So the pair (m,n) satisfies the requirement. �


5. +∞ and −∞

We add in two elements +∞ and −∞ into R so that R∪{+∞,−∞} is still

a good ordered system, and it makes taking supreme and infimum easier in

many cases.

• +∞ and −∞ denote plus infinity and minus infinity;

• we will write +∞ as ∞;

• we can add the ordering to R∪{+∞,−∞} so that ∀a ∈ R∪{+∞,−∞},

−∞ ≤ a ≤ +∞;

• this ordering system satisfies O1, O2, O3;

• +∞,−∞ do not represent any real numbers, and there is no algebraic

structure like “+” and “·” on R ∪ {+∞,−∞};• we can introduce new notations:

[a,∞) = {x ∈ R : a ≤ x} − −unbounded closed interval,

(a,∞) = {x ∈ R : a < x <} − −unbounded open interval,

(−∞, b] = {x ∈ R : x ≤ b} − −unbounded closed interval,

(−∞, b) = {x ∈ R : x < b} − −unbounded open interval,

(−∞,+∞) = R;

• given an nonempty subset S ⊂ R, denote

supS = +∞, if S is not bounded from above,

inf S = −∞, if S is not bounded from below;

• therefore, if S ⊂ R and S 6= ∅, then

supS and inf S always make sense in R ∪ {+∞,−∞}.

16 XIN ZHOU

6. Limits of sequences

We introduce the concept of sequences and their limits in this section.

• A sequence is a function, with domain a set of natural numbers of the

form {n ∈ Z : n ≥ m}, where m is usually chosen to be 1 or 0, with

ranges in R.

• Denote a sequence by (sn)∞n=m, sn ∈ R, or (sm, sm+1, sm+2, · · · ).• If m = 2, write (sn)n∈N.

Example 6.1. a) Let sn = 1n2 , then the sequence is (1, 1

22, 132, 142, · · · ) =

(1, 14, 19, 116, · · · ). The set of values is {1, 1

4, 19, 116, · · · }.

b) Let an = (−1)n for n ≥ 0, then the sequence is (1,−1, 1,−1, · · · ), and

the set of values is {1,−1}.

Remark 6.2. Note the difference between a sequence and its set of

values: we use (·, ·, · · · ) to denote a sequence, and {·, ·, · · · } to denote

the set of values. In particular, the sequence (an = (−1)n) has infinite

number of terms, while its set of values {(−1)n} consists of only two

numbers.

c) Consider the sequence(cos(nπ

3))n∈N. It can be written as

(12,−1

2,−1,−1

2, 12, 1, 1

2, · · · ). The set of values is {1,−1, 1

2, 12}.

d) Let sn = n1n for n ∈ N, then the sequence is (1,

√2, 31/3, · · · ). We will

show that n1/n is close to 1 for very large n.

e) Let(sn = (1 + 1

n)n)n∈N.

The limit of a sequence (sn) is intuitively a real number for which the values

sn accumulate to. To be more precise,

Definition 6.3. A sequence (sn) of real numbers is said to converge to the

real number s, provided that

(6.1)for each ε > 0, there exists a number N , such that

n > N implies |sn − s| < ε.

If (sn) converges to s, we will write limn→∞ sn = s, or sn → s. s is called

the limit of the sequence (sn).

A sequence that does not converge to some real number is said to diverge.

Remark 6.4. (1) ε, δ represent small numbers;

(2) (6.1) is an infinite number of statements. We sometime denote N =

N(ε) to show the dependence of N on ε, and in principle N(ε) is large

when ε is small;


(3) N can be any large real or rational numbers by the Archimedean prop-

erty.

Example 6.5.

a) limn→∞1n2 = 0, and this will be proved soon.

b) The sequence (an = (−1)n) diverges:

Proof. Assume by contradiction that the sequence converges to some real num-

ber s ∈ R, lim(−1)n = s. Since (6.1) is assumed to be true for any ε > 0, we

choose ε = 12. By (6.1), there exists a number N > 0, such that when n > N ,

we have

|(−1)n − s| < 1

2=⇒ s− 1

2< (−1)n < s+

1

2.

In particular, we can let n = 2N or n = 2N + 1 which are both greater than

n, and then{When n = 2N, s− 1

2< 1 < s+ 1

2=⇒ 1

2< s < 3

2,

When n = 2N + 1, s− 12< −1 < s+ 1

2=⇒ −3

2< s < −1

2.

This is a contradiction, so we finish the proof. �

c) The sequence(cos(nπ

3))n∈N diverges, and this is an exercise.

d) limn1n = 1, and this will be proved later.

e) lim sn = (1 + 1n)n = e, and this will be proved later.

Using the definition, we can show that the limit, if it exists, is uniquely

determined.

Proposition 6.6. Let (sn) be a sequence, and assume that lim sn = s and

lim sn = t for some real numbers s, t ∈ R, then s = t.

Proof. By (6.1) and the assumptions, for any ε > 0, there exists N1 > 0, such

that

n > N1 =⇒ |sn − s| <ε

2;

and there exists N2 > 0, such that

n > N2 =⇒ |sn − t| <ε

2.

So for n > max{N1, N2},

|s− t| ≤ |s− sn|+ |sn − t| < ε.

Therefore s = t. (If this were not true, then take ε = 12|s − t| to get a

contradiction.) �

18 XIN ZHOU

7. Proofs

The main purpose of this section is to introduce how to write rigorous proofs

for limits.

Example 7.1. limn→∞1n2 = 0.

Discussion: given ε > 0, we want to find N > 0, such that if n > N , then∣∣∣∣ 1

n2− 0

∣∣∣∣ < ε.

Algebra:∣∣ 1n2 − 0

∣∣ < ε ⇐⇒ 1n2 < ε ⇐⇒ n2 > 1

ε, or n > 1√

ε. Take N = 1√

ε,

then it should serve our purpose.

Proof. For any ε > 0, let N = 1√ε. If n > N , then n > 1√

ε, so n2 > 1

ε, hence

ε > 1n2 .

Thus n > N implies∣∣ 1n2 − 0

∣∣ < ε, and this proves limn→∞1n2 = 0. �

Example 7.2. limn→∞3n+17n−4 = 3

7.

Discussion: given ε > 0, we want to find N > 0, such that if n > N , then∣∣∣∣3n+ 1

7n− 4− 3

7

∣∣∣∣ < ε.

Algebra:∣∣3n+17n−4 −

37

∣∣ =∣∣∣ 197(7n−4)

∣∣∣.∣∣3n+17n−4 −

37

∣∣ < ε ⇐⇒∣∣∣ 197(7n−4)

∣∣∣ < ε ⇐⇒ 197(7n−4) < ε ⇐⇒ 19

7ε< 7n − 4 ⇐⇒

7n > 197ε

+ 4 ⇐⇒ n > 1949ε

+ 47. So put N = 19

49ε+ 4

7.

Proof. For any ε > 0, let N = 1949ε

+ 47. If n > N , then we can reverse all steps

above to show that∣∣3n+17n−4 −

37

∣∣ < ε. �

Example 7.3. limn→∞4n3+3nn3−6 = 4.

Discussion: given ε > 0, we want to find N > 0, such that if n > N , then

(7.1)

∣∣∣∣4n3 + 3n

n3 − 6− 4

∣∣∣∣ < ε,⇐⇒∣∣∣∣3n+ 24

n3 − 6

∣∣∣∣ < ε.

There is no way to easily solve this inequality for n by ε, so we need to do

some rough estimates:

(1) 3n+ 24 ≤ 3n+ 24n ≤ 27n;

(2) n3− 6 > n3/2 provide n is large, actually this is equivalent to n3 > 12,

which holds true when n > 2.


By what we have above, we have

(7.1)⇐=27n

n3/2< ε, and n > 2;

⇐⇒ 54

n2< ε, and n > 2;

⇐⇒ n >

√54

ε, and n > 2.

Proof. For any ε > 0, let N = max{√

54ε, 2}. If n > N , then we can reverse

all steps above to (7.1). �

Example 7.4. Let (sn) be a sequence of nonnegative numbers, i.e. sn ≥ 0.

Suppose that limn→∞ sn = s. (Show that s ≥ 0 as an exercise). Prove that

lim√sn =

√s.

Discussion: given ε > 0, we want to find N > 0, such that if n > N , then

|√sn −

√s| < ε.

Note that√sn −

√s =

sn − s√sn +

√s.

We will divide the discussion into two cases:

(1) if s > 0, then√sn +

√s ≥√s, hence

|√sn −

√s| ≤ sn − s√

s.

By the assumption limn→∞ sn = s, we can select N > 0, such that

n > N =⇒ |sn − s| <√sε, and this will serve our purpose.

(2) The case when s = 0 will be left as homework.

Proof. We only discuss the case when s > 0 here. For any ε > 0, since

limn→∞ sn = s, there exists N > 0, such that n > N =⇒ |sn− s| <√sε. This

implies that

|√sn −

√s| ≤ sn − s√

s< ε,

and this finishes the proof. �

Example 7.5. Let (sn)n∈N be a sequence of real numbers such that sn 6= 0 for

all n ∈ N and lim sn = s 6= 0. Prove that

inf{|sn| : n ∈ N} > 0.

20 XIN ZHOU

Discussion: Let ε = |s|/2 in the definition (6.1). For N = N(ε) in (6.1), if

n > N , then we have

|sn − s| < ε,=⇒ sn ∈ (s− ε, s+ ε).

Proof. Let ε = |s|/2. Since lim sn = s, there exists N > 0, such that

n > N,=⇒ |sn − s| < ε =|s|2.

So n > N implies that

|sn| = |s+ sn − s| ≥ |s| −|s|2

=|s|2.

Now set

m = min{|s|2, |s1|, |s2|, · · · , |sN |},

then |sn| ≥ m, for all nN, and this finishes the proof. �


8. Problem session

(1) (4.15) Let a, b ∈ R. Show if a ≤ b+ 1n

for all n ∈ N, then a ≤ b.

Proof. Suppose not. Then a > b, or a − b > 0. By the Archimedean

Property, there exists n ∈ N, such that

1

n< a− b.

And this is equivalent to a > b+ 1n, but this is a contradiction. �

(2) Show that the Archimedean Property of the real numbers holds if and only

if the set of natural numbers N is not bounded above.

Proof. First assume that the Archimedean Property holds. Then for any

positive number M > 0, there exists a n ∈ N, such that M < n. This

means that N is not bounded above.

Now assume that N is not bounded above, and this means that for any

N > 0, there exists n ∈ N, such that N < n. Now given any two numbers

a, b ∈ R, a > 0, b > 0. Consider the number N = b/a, so there exists n ∈ N,

such that N = b/a < n, which is equivalent to

na > b.

�

(3) (HW3.2) Show that if S ⊂ Z is bounded above then S has a maximum, i.e.,

supS ∈ S.

(4) Let x ∈ R. We define the Greatest Integer function of x as the largest

integer m ∈ Z such that m ≤ x; it is denoted by [x] and satisfies the

inequality

[x] ≤ x < [x] + 1.

For instance, [0.1] = 0, [1.1] = 1, [−2.1] = −3, [5] = 5, [−4] = −4, etc.

Show that for any real number x ∈ R, [x] exists. Hint: Consider the set

S = {n ∈ Z : n ≤ x}.

Then consider two cases depending on whether x ≥ 0 or x < 0. In the case

when x < 0, use the Archimedean property of R to show that S is nonempty.

Then apply the completeness axiom and finally show that supS = [x], i.e.,

show that supS is an integer (this follows from Problem 3) and it satisfies

supS ≤ x < supS + 1.

22 XIN ZHOU

Proof. Let us only consider the case when x ≥ 0, then 0 ∈ S, hence S 6= ∅.Since S is bounded from above, (x is an upper bound), by the Completeness

Axiom, supS exists. By Problem 3, supS ∈ S, and supS = maxS. Then

the only thing we need to prove is

x < supS + 1.

Assume by the contrary that x ≥ supS+ 1. Then since supS is an integer,

and hence supS + 1 is an integer, so by the definition of S, we have

supS + 1 ∈ S.

This is a contradiction to the fact that supS = maxS. �


9. Limit theorems for sequences

Definition 9.1. A sequence (sn) of real numbers is said to be bounded if the

set of values {sn : n ∈ N} is bounded. This is equivalent to saying that:

∃M ∈ R, such that |sn| ≤M for all n ∈ N.

Theorem 9.2. Convergent sequences are bounded.

Proof. Let (sn) be a convergent sequence, and let s be the limit, or equivalently

s = lim sn.

Apply the definition of convergence with ε = 1, then we get some large number

N ∈ N, such that

if n > N , then |sn − s| < 1.

By the triangle inequality, |sn − s| < 1 =⇒

|sn| ≤ |s+ sn − s| ≤ |s|+ |sn − s| ≤ |s|+ 1.

Define

M = max{|s1|, |s2|, · · · , |sN |, |s|+ 1}.Then we have

|sn| ≤{

M if n ≤ N

|s|+ 1 ≤M if n > N.

Therefore |sn| ≤M for all n ∈ N, and so {sn} is bounded. �

Remark 9.3. In the proof above, we only apply the defining condition (6.1) for

a single ε = 1. Actually, we can take ε to be any finite positive number in the

proof, and this is left as an exercise.

Theorem 9.4. If s sequence (sn) converges to s, and k ∈ R, then the sequence

(ksn) converges to ks. That is,

lim(ksn) = k lim sn.

Proof. If k = 0, the proof is trivial as ksn ≡ 0, and lim(0 · sn) = 0 = 0 · lim sn.

Assume k 6= 0 now. Let ε > 0 be any positive real number, and we need to

show that

|ksn − ks| < ε, for large n ∈ N.Note that the above inequality is equivalent to saying that

|sn − s| <ε

|k|, for large n ∈ N.

24 XIN ZHOU

Since lim sn = s, for ε|k| > 0, by (6.1) there exists N > 0, such that

if n ≥ N , then |sn − s| <ε

|k|.

Therefore for n > N , we have |ksn − ks| < ε. �

Theorem 9.5. If (sn) converges to s and (tn) converges to t, then (sn + tn)

converges to s+ t. That is to say,

lim(sn + tn) = lim sn + lim tn.

Proof. Let ε > 0 be any positive real number, and we want to show that

|(sn + tn)− (s+ t)| < ε, for all large n ∈ N.

Note that

|(sn + tn)− (s+ t)| ≤ |sn − s|+ |tn − t|.(Hint: so we can try to show |sn − s| < ε/2 and |tn − t| < ε/2.)

Since lim sn = s, ∃N1 ∈ N, such that

if n ≥ N1, then |sn − s| < ε/2.

Similarly by lim tn = t, ∃N2 ∈ N, such that

if n ≥ N2, then |tn − t| < ε/2.

Let

N = max{N1, N2}.Then if n > N , we have

|(sn + tn)− (s+ t)| ≤ |sn − s|+ |tn − t| < ε.

�

Theorem 9.6. If (sn) converges to s and (tn) converges to t, then (sn · tn)

converges to s · t. That is to say,

lim(sn · tn) = (lim sn) · (lim tn).

Discussion:

|sntn − st| = |sntn − snt+ snt− st|≤ |sn||tn − t|+ |t||sn − s|≤M |tn − t|+ |t||sn − s|.

• In the above inequality, |tn− t| and |sn− s| can be very small for large

n;

• |sn| is bounded by Theorem 9.2, i.e. |sn| ≤M .


Proof. Let ε > 0 be any positive real number. By Theorem 9.2, ∃M > 0, such

that |sn| ≤M , for all n ∈ N.

Since lim tn = t, ∃N1 ∈ N, such that

if n ≥ N1, then |tn − t| < ε/(2M).

Similarly by lim sn = s, ∃N2 ∈ N, such that

if n ≥ N2, then |sn − s| <ε

2(|t|+ 1).

Let

N = max{N1, N2}.Then if n > N , we have

|sntn − st| ≤ |sn||tn − t|+ |t||sn − s|

≤M · ε

2M+ |t| · ε

2(|t|+ 1)< ε.

�

Lemma 9.7. If (sn) converges to s and if sn 6= 0 for all n ∈ N, and if s 6= 0,

then

lim1

sn=

1

s.

Discussion: ∣∣∣∣ 1

sn− 1

s

∣∣∣∣ =

∣∣∣∣sn − ssns

∣∣∣∣≤ |sn − s|m|s|.

We know that |sn − s| is very small for n very large. To show the right hand

side is small, we need the denominator |sns| to be bounded away from 0. By

Example 7.5, we do have that

m = inf{|sn| : n ∈ N} > 0.

Proof. Let ε > 0 be any positive real number, and m be defined as above.

Since lim sn = s, there exists N > 0, such that

if n ≥ N , then |sn − s| < ε ·m · |s|.

Therefore if n ≥ N , then ∣∣∣∣ 1

sn− 1

s

∣∣∣∣ ≤ ε ·m · |s||sn| · |s|

< ε.

�

26 XIN ZHOU

Theorem 9.8. Suppose lim sn = s and lim tn = t. If s 6= 0 and sn 6= 0 for all

n ∈ N, then

lim

(tnsn

)=t

s.

Proof. By the previous lemma, lim 1sn

= 1s. Then

limtnsn

= (lim tn)(lim1

sn) =

t

s.

�

A few basic examples:

a) lim 1np = 0 for p > 0 (assume p ∈ N at this moment);

b) lim an = 0 if |a| < 1;

c) limn1/n = 1;

d) lim a1/n = 1 for a > 0.

Proof. a). Let ε > 0 be any positive real number. (We want to show 1np < ε,

which is equivalent to n >(1ε

)1/p.)

Let N =(1ε

)1/p,

(Exercise: use the completeness axiom to show that: “for any n ∈ N, and

a > 0, there exists x > 0, such that xn = a, and we write x = a1/n”.)

then if n > N , we have 1np < ε. Therefore∣∣∣∣ 1

np− 0

∣∣∣∣ =1

np< ε.

b). If a = 0, then an = 0, and the proof is trivial.

If a 6= 0 and |a| < 1, then |a| = 11+b

for some b > 0. Using the binomial

expansion,

(1 + b)n ≥ 1 + nb > nb.

So

|an = 0| = |a|n =1

(1 + b)n<

1

nb.

Let ε > 0 be any positive real number, and let N = 1εb

. Then if n > N , we

have

|an = 0| < 1

nb< ε.

c). Let sn = n1/n−1. Note that sn ≥ 0. It suffices to show lim sn = 0. Note

that

1 + sn = n1/n ⇐⇒ (1 + sn)n = n.


By the binomial expansion:

n = (1 + sn)n = 1 + n · sn +1

2n(n+ 1)s2n > frac12n(n+ 1)s2n.

So

s2n <2

n− 1,=⇒ sn <

√2

n− 1.

Exercise: use the above formula to show lim sn = 0.

d). First assume that a ≥ 1. Then for n ≥ a, we have

1 ≤ a1/n ≤ n1/n.

Exercise: use the fact limn1/n = 1 to prove lim a1/n = 1.

(Hint: let ε > 0 be any positive real number, there exists N > 0, such that

n > N =⇒ n1/n − 1 < ε =⇒ a1/n − 1 < ε.)

Suppose 0 < a < 1, then 1a> 1. So lim

(1a

)1/n= 1. By Lemma 9.7, we can

get lim a1/n = 1. �

Definition 9.9. A sequence (sn) is said to diverges to +∞, provided:

(9.1) for each M > 0, ∃N > 0, such that n > N , =⇒ sn > M .

We write lim sn = +∞.

Similarly we write lim sn = −∞, provided

(9.2) for each M < 0, ∃N > 0, such that n > N , =⇒ sn < M .

Example 9.10. Prove that lim(√n+ 5) = +∞.

Discussion: consider an arbitrarily large M > 0, we need to find N > 0, such

that

n > N =⇒√n+ 5 > M ⇐⇒ n > (M − 5)2.

Proof. Given M > 0, let N = (M − 5)2, then n > N implies√n+ 5 > M . �

Theorem 9.11. Assume that lim sn = +∞ and lim tn > 0, then

lim(tn · sn) = +∞.

Discussion: consider an arbitrarily large M > 0, we want (sn · tn) > M .

• Since lim sn = +∞, sn can be as large as we want when n is large, so

we need to show that tn’s are bounded away from 0;

• Choose 0 < m < lim tn, and observe that tn > m for n large;

• Then we only need sn > M/m for large n.

28 XIN ZHOU

Proof. Let M > 0 be an arbitrarily large number. Select m > 0, such that

0 < m < lim tn.

By Problem 8.10 in textbook, ∃Nn > 0, such that

n > N1,=⇒ tn > m.

Since lim sn = +∞, ∃N2 > 0, such that

n > N2,=⇒ sn >M

m.

Let N = max{N1, N2}, then

n > N,=⇒ sn · tn >M

m·m = M.

�

Theorem 9.12. Let (sn) be a sequence of positive numbers, then

lim sn = +∞,⇐⇒ lim

(1

sn

)= 0.

Proof. (1) Assume lim sn = +∞. Given any ε > 0, let M = 1ε> 0, then

∃N1, such that

n > N1,=⇒ sn > M =1

ε,=⇒ 1

sn< ε.

(2) Assume lim(

1sn

)= 0. Given M > 0, let ε = 1

M, then ∃N2, such that

n > N2,=⇒1

sn> M =

1

ε,=⇒ sn > M.

�


10. Monotone sequences and Cauchy sequences

Definition 10.1. Let (sn) be a sequence.

• (sn) is called a non-decreasing sequence if sn ≤ sn+1 for all n ∈ N.

• (sn) is called a non-increasing sequence if sn ≥ sn+1 for all n ∈ N.

Such sequences are called monotone sequences.

Example 10.2. • an = 1− 1n

, bn = n3, cn =(1 + 1

n

)nare non-decreasing

sequences;

• dn = 1n2 is a non-increasing sequence;

• sn = (−1)n, tn = cos(nπ3

), un = (−1)n ·n, vn = (−1)n

nare not monotone

sequences;

• xn = n1/n is not monotone.

Theorem 10.3. All bounded monotone sequences converge.

Proof. Let (sn) be a bounded non-decreasing sequence. Consider the set of

values:

S = {sn : n ∈ N}.Since (sn) is bounded, the set S is also bounded. By the Completeness Axiom,

we can take

u = supS.

Claim: lim sn = u.

Given ε > 0, since u is the supremum, there exists N ∈ N, such that

sN > u− ε.

Since (sn) is non-decreasing,

∀n > N, sn ≥ sN > u− ε.

On the other hand, since u is an upper bound for S, we have

∀n ∈ N, sn ≤ u.

Combining the two inequalities above, ∀n > N ,

u− ε < sn ≤ u,=⇒ |sn − u| < ε.

If (sn) is a bounded non-increasing sequence, the proof is similar and will

be left as an exercise. (Hint: let u = inf S.) �

Decimals:

30 XIN ZHOU

• Different decimal expansions can represent the same real number;

• Focus on non-negative decimals expansions and non-negative real num-

bers.

Given a decimal:

k.d1d2d3d4 · · · , k ∈ Z+, dj ∈ {0, 1, 2, · · · , 9}.

Let

sn = k +d110

+d2102

+ · · ·+ dn10n

,

then (sn) is a non-decreasing sequence, and (sn) is bounded from above by

k + 1. By Theorem 10.3, (sn) converges and we denote the limit by

s = k.d1d2d3d4 · · · .

Theorem 10.4.

(1) If (sn) is an unbounded non-decreasing sequence, then lim sn = +∞;

(2) If (sn) is an unbounded non-increasing sequence, then lim sn = −∞.

Proof.

(1) Since (sn) is unbounded, and sn ≥ s1, (sn) is unbounded from above.

This implies that:

for any M > 0, ∃, N > 0, such that sN > M.

Since (sn) is non-decreasing,

∀n > N, sn ≥ sN > M.

(2) Exercise.

�

Corollary 10.5. If (sn) is a monotone sequence, then it either converges,

diverges to +∞, or diverges to −∞. Therefore lim sn always is meaningful.

lim sup and lim inf:Let (sn) be a bounded sequence. It may or may not converge. The limiting

behavior of (sn) depends only on the terms {sn : n > N} for N large. Let

N ∈ N be an arbitrary natural number:

Claim 1: if lim sn exists, then lim sn ∈ [uN , vN ], where

uN = inf{sn : n > N}, vN = sup{sn : s > N}.


Claim 2: we have:

u1 ≤ u2 ≤ u3 ≤ · · · ;

v1 ≥ v2 ≥ v3 ≥ · · · .

Note that as N increases, the sets SN = {sn : n > N} get smaller:

S1 ⊃ S2 ⊃ S3 ⊃ · · · .

The above inequalities follow by Problem 4.7 in textbook.

By Theorem 10.3, the following limits exist:

u = limuN , v = lim vN , and u ≤ v.

Moreover,

if lim sn exists,=⇒ u ≤ lim sn ≤ v.

Definition 10.6.

lim sup sn := limn→∞

(sup{sn : n > N}

);

lim inf sn := limn→∞

(inf{sn : n > N}

).

Remark 10.7. In the above definition, the sequence (sn) is not restricted to be

bounded.

• If sup{sn : n > N} = +∞, then lim sup sn = +∞;

• If inf{sn : n > N} = −∞, then lim inf sn = −∞.

Theorem 10.8. Let (sn) be a sequence.

(i) If lim sn exists (in R, or is +∞, or −∞), then

lim inf sn = lim sn = lim sup sn.

(ii) If lim inf sn = lim sup sn, then lim sn exists, and

lim sn = lim inf sn = lim sup sn.

Proof. As in the definition, denote

uN = inf{sn : n > N}, vN = sup{sn : s > N},

and u = lim inf sn = limuN , v = lim sup sn = lim vN .

32 XIN ZHOU

(i) Case 1: lim sn = +∞. Then ∀M > 0, ∃N > 0, such that

n > N,=⇒ sn > M.

Then uN ≥ M , =⇒ if m > N , then um ≥ uN ≥ M , so limuN = +∞.

Therefore

lim sup sn = lim vN ≥ limuN = +∞.

Case 2: lim sn = −∞ can be proven similarly.

Case 3: assume lim sn = s. Then ∀ ε > 0, ∃N > 0, such that

|sn − s| < ε, for n > N.

We have =⇒ sn < s+ ε, for n > N ;

=⇒ vN = sup{sn : n > N} ≤ s+ ε;

if m > N,=⇒ vm ≤ vN ≤ s+ ε;

=⇒ lim sup sn = lim vm ≤ s.

Similarly, we have=⇒ sn > s− ε, for n > N ;

=⇒ uN = inf{sn : n > N} ≤ s− ε;if m > N,=⇒ um ≥ uN ≥ s− ε;=⇒ lim inf sn = limum ≥ s.

Since we know lim inf sn ≤ lim sup sn, they must all equal to s.

(ii) The proof when lim sup sn = lim inf sn = ±∞ is left as exercise.

Assume lim sup sn = lim inf sn = s. Given ε > 0, since lim vN = s,

there exists N1 > 0, such that if n > N1

|vN − s| = |sup{sn : n > N} − s| < ε.

Note that {vN} is monotone non-increasing, so we have vN ≥ s, and

hence

sup{sn : n > N} − s < ε,

=⇒ sn < s+ ε, if n > N1.

Similarly, since since limuN = s, there exists N2 > 0, such that if

n > N2

|uN − s| = |inf{sn : n > N} − s| < ε.

Note that {uN} is monotone non-decreasing, so we have uN ≤ s, and

hence

s− inf{sn : n > N} < ε,


=⇒ sn > s− ε, if n > N2.

Let N = max{N1, N2}, then if n > N ,

s− ε < sn < s+ ε.

This implies that lim sn = s.

�

Definition 10.9. A sequence (sn) is called a Cauchy sequence if

(10.1) ∀ ε > 0,∃N > 0, such that m,n > M,=⇒ |sm − sn| < ε.

Lemma 10.10. Convergent sequences are Cauchy sequences.

Proof. Assume that lim sn = s. Let ε > 0, then there exists N > 0, such that

n > N,=⇒ |sn − s| <ε

2.

Therefore, if n,m > N ,

|sn − sm| ≤ |sn − s|+ |s− sm| <ε

2+ε

2= ε.

�

Lemma 10.11. Cauchy sequences are bounded.

Proof. Apply (10.1) with ε = 1. Then there exists N > 0, such that

m,n > N,=⇒ |sn − sm| < 1.

In particular, |sn − sN+1| < 1, and so

|sn| < |sN+1|+ 1.

Let

M = max{|sN+1|+ 1, |s1|, · · · , |sN |}.Then |sn| ≤M . �

Theorem 10.12. A sequence converges if and only if it is a Cauchy sequence.

Proof. According Lemma 10.10, we only need to show that Cauchy sequence

must converge. Assume (sn) is Cauchy. By Lemma 10.11, (sn) is bounded, so

by Theorem 10.8 we only need to show that

lim inf sn = lim sup sn.

Given ε > 0, there exists N > 0, such that

if m,n > N,=⇒ |sm − sn| < ε.

34 XIN ZHOU

=⇒ sn < sm + ε;

=⇒ vN = sup sn : n > N ≤ sm + ε,∀m > N ;

=⇒ vN − ε ≤ sm, ,∀m > N ;

=⇒ vN − ε ≤ inf{sm : m > N} = uN ;

=⇒ lim sup sn ≤ vN ≤ uN + ε ≤ lim inf sN + ε.

�


11. Subsequences

Definition 11.1. Let (sn) be a sequence. A subsequence of (sn) is a sequence

(tk)k∈N, such that tk = snk, with

n1 < n2 < n3 < · · · < nk < nk+1 < · · · .

Example 11.2. Consider sn = n2(−1)n. The positive terms form a subse-

quence:

(tk) = (4, 16, 36, · · · ).

Here tk = s2k.

Example 11.3. Consider an = sin(nπ3

). The nonnegative terms are

(

√3

2,

√3

2, 0, 0,

√3

2,

√3

2, · · · )

Example 11.4. Q can be listed as s sequence (rn).

Proposition 11.5. Given any real number a, there exists a subsequence (rnk)

of (rn) that converges to a.

Discussion: we want to construct rnk, such that

|rnk− a| < 1

k, for all k ∈ N.

Proof. We use “Definition/construction by induction”.

(1) Select n1, such that |rn1 − a| < 1, and this follows from the Denseness

of Q;

(2) Suppose n1, · · · , nk have been selected such that

n1 < n2 < · · · < nk,

and |rnj− a| < 1

jfor j = 1, 2, · · · , k.

We want to find rnk+1, with nk+1 > nk and |rnk+1

− a| < 1k+1

, which

is equivalent to rnk+1∈(a− 1

k+1, a+ 1

k+1

).

By Problem 4.11 in the textbook, there are infinitely many rationals

in(a− 1

k+1, a+ 1

k+1

). Therefore, there exists nk+1 > nk, such that

rnk+1∈(a− 1

k+1, a+ 1

k+1

).

By Mathematical Induction, there exists a subsequence (rnk), with lim rnk

=

a. �

36 XIN ZHOU

Example 11.6. Let (sn) be a sequence of positive numbers, i.e. sn > 0, such

that

inf{sn : n ∈ N} = 0.

Prove that a subsequence of (sn) converges monotonically to 0.

Discussion: we want to construct snk, such that

s1 > s2 > · · · > sk · · · ; and

sk <1

k, for all k ∈ N.

Proof. We use ‘Definition/construction by induction”.

(1) Since inf{sn : n ∈ N} = 0, there exists n1 ∈ N, such that s1 < 1;


n1 < n2 < · · · < nk,

and

snj+1< min{snj

,1

j + 1}, for j = 1, 2, · · · , k − 1.

We want to find snk+1, such that snk+1

< min{snk, 1k+1}. Note that

min{sn : 1 ≤ n ≤ nk} > 0,=⇒ inf{sn : n > nk} = 0.

Therefore, there exists nk+1 > nk, such that

snk+1< min{snk

,1

k + 1}.

By Mathematical Induction, there exists a decreasing subsequence (snk),

with lim snk= 0. �

Theorem 11.7. If a sequence (sn) converges, then every subsequence con-

verges to the same limit.

Proof. Let lim sn = s, and (snk) be a subsequence of (sn). Note that nk ≥ k

for all k.

Given ε > 0, there exists N > 0, such that n > N , =⇒ |sn − s| < ε.

If k > N , then nk > N , so |snk− s| < ε. This implies that lim snk

= s. �

Theorem 11.8. Every sequence (sn) has a monotone subsequence.

Proof. Say that “the n-term sn is dominant if sn > sm for all m > n.”

We divide the proof into two cases:


Case 1: there exists infinitely many dominant terms (snk), then snk+1

< snk, so

(snk) is a decreasing sequence.

Case 2: suppose there are only finitely many dominant terms, and we only focus

on this case in the following.

Select n1, such that sn1 is larger than all dominant terms. Then

(11.1) For any N ≥ n1, there exists m > N , such that sm ≥ sN .

We are going to use “inductive construction” to find a monotone sequence.

(1) Applying (11.1) with N = n1 gives n2 > n1, such that sn2 ≥ sn1 ;


n1 < n2 < · · · < nk,

and

sn1 ≤ sn2 ≤ · · · ≤ snk.

Applying (11.1) with N = nk gives nk+1 > nk, such that snk+1≥ snk

.

By Mathematical Induction, (snk) is a nondecreasing subsequence. �

Corollary 11.9. Let (sn) be a sequence, then there exists

• a monotone subsequence with limit lim sup sn, and

• a monotone subsequence with limit lim inf sn.

Proof. We only prove the first statement for lim sup sn.

Let vN = sup{sn : s > N}, N ∈ N, then v = lim vN = lim sup sn.

If v = −∞, then lim sn = −∞, so any monotone sequence will converge to

lim sup sn, and we finished.

If v 6= −∞, select an arbitrary monotone increasing sequence (tN), such

that lim tN = v:

• if v is finite, let tN = v − 1N

;

• if v = +∞, let tN = N .

Now we discuss the two cases as in the proof of the above theorem.

Case 1: by assumption, snk= sup{sm : m ≥ nk} = vnk−1, so {snk

} is decreasing

and

limk→∞

snk= lim vN = lim sup sn.

Case 2: Given N > n1 as above, by (11.1), there exists m1 > N , such that

sm1 ≥ sN . Since

tN < v ≤ vN = sup{sm : m > N},

38 XIN ZHOU

there exists m2 > N , such that sm2 > tN .{either sm1 ≥ sm2 ,=⇒ sm1 ≥ sN and sm1 > tN ;

or sm2 ≥ sm1 ,=⇒ sm2 ≥ sN and sm2 > tN .

Then we have

(11.2) Given N ≥ n1, there exists m > N , such that sm ≥ sN and sm > tN .

We are going to use “inductive construction” to find the monotone subse-

quence.

(1) Applying (11.2) with N = n1 gives n2 > n1, such that sn2 ≥ sn1 and

sn2 > tn1 ;


n1 < n2 < · · · < nk,

sn1 ≤ sn2 ≤ · · · ≤ snk,

and

snj+1> tnj

, for all 1 ≤ j ≤ k − 1.

Applying (11.2) with N = nk gives nk+1 > nk, such that snk+1≥ snk

and snk+1> tnk

. Moreover, snk1≤ sup{sm : m > nk} = vnk

, so

tnk< snk+1

≤ vnk.

By Mathematical Induction, (snk) is a nondecreasing sequence and lim snk

=

lim sup sn. �

Definition 11.10. Let (sn) be a sequence. A subsequential limit is any real

number, or +∞, or −∞, that is the limit of some subsequence.

Example 11.11. If lim sn = s, then the set of subsequential limit is {s}.

Example 11.12. Let (rn) be the list of rational numbers, then the set of

subsequential limits are R ∪ {+∞,−∞}.

Theorem 11.13. Let (sn) be a sequence, and S be the set of subsequential

limits of (sn). Then

(i) S 6= ∅;(ii) supS = lim sup sn, inf S = lim inf sn;

(iii) lim sn exists if and only if S consists of only one element, i.e. S = {s}.


Proof. (i) follows from the above corollary since lim sup sn, lim inf sn ∈ S, and

(iii) follows from (ii).

To prove (ii), let t = lim snkfor some subsequence (snk

), then

t = lim sup snk= lim inf snk

.

Note that

{snk: k > N} ⊂ {sn : n > N}.

So

sup{snk: k > N} ≤ sup{sn : n > N};

inf{snk: k > N} ≥ inf{sn : n > N}.

Therefore,

lim inf sn ≤ lim inf snk= t = lim sup snk

≤ lim sup sn.

So

lim inf sn ≤ inf S ≤ supS ≤ lim sup sn.

Also lim sup sn, lim inf sn ∈ S, so we finish the proof. �

Theorem 11.14. Let S be the set of subsequential limits of (sn). If tn ∈ S∩Rand lim tn = t, then t ∈ S.

Proof. We use “construction by induction”.

(1) Since t1 = lim snkfor some subsequence (snk

), there exists n1, such

that

|sn1 − t1| < 1;

(2) Assume n1 < · · · < nk have been selected with

|snj− tj| <

1

j, for j = 1, · · · , k.

Since tk+1 = lim snifor some other subsequence (sni

), there exists

nk+1 > nk, such that

|snk+1− tk+1| <

1

k + 1.

By induction, we found a subsequence (snk) such that |snk

− tk| < 1k. It is then

an easy exercise to show that t = lim snk. �

Definition 11.15. A subset S ⊂ R is called closed, if for any convergent

sequence (tn) where tn ∈ S, then the limit t = lim tn ∈ S.

40 XIN ZHOU

12. lim sup and lim inf revisited

Theorem 12.1. If (sn) is a sequence with lim sn = s and s > 0, and (tn) is

an arbitrary sequence, then

lim sup(sn · tn) = s · lim sup tn.

Proof. First we will show lim sup(sn ·tn) ≥ s·lim sup tn. Denote β = lim sup tn.

Case 1: β ∈ R. There exists a subsequence (tnk) of (tn), such that lim tnk

=

lim sup tn = β. Then

lim(snktnk

) = sβ ≤ lim sup(sntn).

Case 2: β = +∞. There exists a subsequence (tnk) of (tn), such that lim tnk

=

+∞. Since lim snk= s > 0, we have lim(snk

tnk) = +∞ by Theorem

9.11. Hence

lim sup(sntn) = +∞.

Case 3: β = −∞. Exercise.

To show the reverse direction: assume sn 6= 0 by ignore the first few terms.

Then

lim sup tn = lim sup

[(

1

sn) · (sntn)

]≥ lim

(1

sn

)·lim sup(sntn) =

1

2lim sup(sntn).

This finishes the proof. �

Theorem 12.2. Let (sn) be a sequence with sn 6= 0. Then

lim inf

∣∣∣∣sn+1

sn

∣∣∣∣ ≤ lim inf |sn|1/n ≤ lim sup |sn|1/n ≤ lim sup

∣∣∣∣sn+1

sn

∣∣∣∣ .Proof. Let α = lim sup |sn|1/n, and L = lim sup

∣∣∣ sn+1

sn

∣∣∣. If L = +∞, then we

are done.

Assume L < +∞. It suffices to show α ≤ L1 for any L1 > L. Since

L = limN→∞

(sup

{∣∣∣∣sn+1

sn

∣∣∣∣ : n > N

})< L1,

there exists N > 0, such that

sup

{∣∣∣∣sn+1

sn

∣∣∣∣ : n ≥ N

}< L1.


Then by iteration, for n > N ,

|sn| =∣∣∣∣ snsn−1

∣∣∣∣ · ∣∣∣∣sn−1sn−2

∣∣∣∣ · · · ∣∣∣∣sN+1

sN

∣∣∣∣ · |sN |< Ln−N1 |sN | = Ln1 ·

|sN |LN1

.

Let a = L−N1 |sN | > 0, then

|sn| < a · Ln1 =⇒ |sn|1/n < L1a1/n.

As lim a1/n = 1, we have

α = lim sup |sn|1/n ≤ L1.

The other part lim inf∣∣∣ sn+1

sn

∣∣∣ ≤ lim inf |sn|1/n is left as an exercise. �

42 XIN ZHOU

13. Series

Denoten∑

k=m

ak = am + am+1 + · · ·+ an.

We are interested in infinite series∑∞

n=m an. Let

sn = am + am+1 + · · ·+ an =n∑

k=m

ak.

Definition 13.1.

(1) The infinite series∑∞

n=m an is said to converge provided the sequence

of partial sums (sn) converges to a real number s ∈ R, and we define

∞∑n=m

an = s.

(2) A series that does not converge is said to diverge.

(3) Say∑∞

n=m an = diverges to +∞, and write∑∞

n=m an = +∞, provided

lim sn = +∞. Similarly we can define∑∞

n=m an = −∞.

(4) If we do not care the initial index m, just write∑an.

(5) If an ≥ 0, then (sn) is nondecreasing, so∑an either converges or

diverges to +∞. Hence for any series∑an, the sum of absolute values∑

|an| is meaningful.

(6)∑an is said to converge absolutely or be absolutely convergent if

∑|an|

converges.

Example 13.2. A series∑∞

n=0 arn for constants a and r is called a geometric

series. For r 6= 1,

sn =n∑k=0

ark = a1− rn+1

1− r.

For |r| < 1, limn→∞ rn+1 = 0, we have

limn→∞

sn =a

1− r.

And∞∑n=0

arn =a

1− r.

If a 6= 0, and |r| ≥ 1, the series does not converge.


Example 13.3. The series∑∞

n=11np converges if and only if p > 1. (This will

be proved later.)

Definition 13.4. We say a series∑an satisfies the “Cauchy criterion” if the

sequence of partial sums (sn) is a Cauchy sequence:

(13.1) for each ε > 0, ∃N > 0, such that if n,m > N , =⇒ |sn − sm| < ε.

And this is equivalent to

(13.2) for each ε > 0, ∃N > 0, such that n ≥ m > N , =⇒ |sn − sm−1| < ε.

This is equivalent to

(13.3) for each ε > 0, ∃N > 0, such that n ≥ m > N , =⇒

∣∣∣∣∣n∑

k=m

ak

∣∣∣∣∣ < ε.

As a direct corollary of Theorem 10.12, we have,

Theorem 13.5. A series converges if and only if it satisfies the Cauchy cri-

terion.

Corollary 13.6. If∑an converges, then lim an = 0.

Proof. For any ε > 0, by the Cauchy criterion, there exists N > 0, such that if

n ≥ m > N , then |∑n

k=m ak| < ε. We can take n = m, then we have |am| < ε.

This implies that lim am = 0. �

Remark 13.7. The converse of the above result is not true. For instance,∑1n

= +∞, although lim 1n

= 0.

Theorem 13.8 (Comparison Test). Consider a series∑an, with an ≥ 0 for

all n ∈ N.

(i) If∑an converges, and |bn| ≤ an, then

∑bn converges;

(ii) If∑an = +∞, and bn ≥ an, then

∑bn = +∞.

Proof.

(i) For any n ≥ m, by the triangle inequality,∣∣∣∣∣n∑

k=m

bk

∣∣∣∣∣ ≤n∑

k=m

|bk| ≤n∑

k=m

ak.

Therefore∑an satisfies the Cauchy criterion implies that

∑bk also

satisfies the Cauchy criterion.

(ii) Denote tn =∑n

k=0 bk and sn =∑n

k=0 ak. Then tn ≥ sn. So lim sn =

+∞ implies lim tn = +∞.

44 XIN ZHOU

�

Corollary 13.9. Absolutely convergent series are convergent.

Proof. This is an easy corollary of the comparison test. �

Theorem 13.10 (Ratio Test). Consider a series∑an of nonzero terms (an 6=

0 for all n ∈ N).

(i)∑an converges absolutely, if lim sup

∣∣∣an+1

an

∣∣∣ < 1;

(ii)∑an diverges If lim inf

∣∣∣an+1

an

∣∣∣ > 1;

(iii) otherwise if lim inf∣∣∣an+1

an

∣∣∣ ≤ 1 ≤ lim sup∣∣∣an+1

an

∣∣∣, then the test gives no

information.

We will use the following root test to prove the ratio test.

Theorem 13.11 (Root Test). Let∑an be a series, and α = lim sup |an|1/n.

The the series∑an

(i) converges absolutely, if α < 1;

(ii) diverges if α > 1;

(iii) otherwise if α = 1, then the test gives no information.

Proof.

(i) Suppose α < 1, and select β such that α < β < 1.

By the definition of lim sup, there exists N > 0, such that

sup{|an|1/n : n > N} < β.

In particular, |an|1/n < β for all n > N , and hence

|an| < βn, for all n > N.

Since 0 < β < 1, the series∑βn converges. By the comparison test,∑

|an| converges.

(ii) If α > 1, then a subsequence of |an|1/n will converge to α > 1, hence

|an| > 1 for infinitely many choices of n.

By Corollary 13.6,∑an cannot converge.

(iii) For∑

1n

and∑

1n2 , α turns out to be 1, but

∑1n

diverges, and∑

1n2

converges.

�

Proof of Ratio Test. Let α = lim sup |an|1/n. By Theorem 12.2:


• if lim sup∣∣∣an+1

an

∣∣∣ < 1, then α < 1, so∑an converges absolutely by Root

Test;

• if lim inf∣∣∣an+1

an

∣∣∣ > 1, then α > 1, so∑an diverges by Root Test;

• if lim inf∣∣∣an+1

an

∣∣∣ ≤ 1 ≤ lim sup∣∣∣an+1

an

∣∣∣, nothing is known by checking the

examples:∑

1n

and∑

1n2 .

�

Example 13.12. Given∑∞

n=2

(−1

3

)n= 1

9

∑∞n=0

(−1

3

)n, does it converge or

not? If converge, calculate the limit.

Solution. Since this is a geometric series with a = 19

and r = 13, so

1

9

∞∑n=0

(−1

3

)n=

1/9

1− (−13)

=1

12.

�


n=0n

n2+3, does it converge or not?

Solution.

• If denote an = nn2+3

, then

an+1

an=n+ 1

n· n2 + 3

n2 + 2n+ 4,

so lim∣∣∣an+1

an

∣∣∣ = 1. Neither Ratio nor Root Test works.

• As an approaches 1n

for n large, we have

an =n

n2 + 3≥ n

n2 + 3n2=

1

4n.

Since∑

1n

diverges,∑

14n

also diverges. By Comparison Test,∑∞

n=0n

n2+3

diverges.

�

Example 13.14. Given∑

1n2+1

, does it converge or not?

Solution.

• Neither Ratio nor Root Test works;

• Since 1n2+1

≤ 1n2 and

∑1n2 converges, so by Comparison Test,

∑1

n2+1

converges.

�

46 XIN ZHOU

Example 13.15. Given∑

n3n

, does it converge or not?

Solution. Let an = n3n

.

(1) an+1

an= n+1

3n, so lim

∣∣∣an+1

an

∣∣∣ = 13< 1. Then

∑n3n

converges by Ratio Test;

(2) lim |an|1/n = lim n1/n

3= 1

3< 1, so Root Test also works.

�

Example 13.16. Given∑an where an =

[2

(−1)n − 3]n

, does it converge or

not?

Solution.

• When n is even, an = (−1)n = 1, so |an|1/n = 1; when n is odd,

an =(−1

2

)n, so |an|1/n = 1

2. (As an exercise):

α = lim sup |an|1/n = 1.

So Root Test does not work.

• Since an = 1 for all even n, (an) cannot converge to 0, so∑an diverges

by Corollary 13.6.

�


n=0 2(−1)n−n, does it converge or not?

Solution. Let an = 2(−1)n−n.

• an ≤ 12n−1 , so

∑an converges by Comparison Test;

• an+1

an= 1

8for even n, and an+1

an= 2 for odd n. Ratio Test fails since

1

8= lim inf

∣∣∣∣an+1

an

∣∣∣∣ < 1 < lim sup

∣∣∣∣an+1

an

∣∣∣∣ = 2;

• |an|1/n = 21n−1 for n even, and |an|1/n = 2−

1n−1 for n odd, so

lim |an|1/n =1

2,

and Root Test implies that∑an converges.

�


14. Alternating series and Integral Tests

Example 14.1. Show that∑∞

n=11n

= +∞.

Solution. Consider f(x) = 1x:

n+1∑k=1

1

k= sum of the area of the rectangles with base [i, i+ 1] and height

1

i

for 1 ≤ i ≤ n

≥ area under the curve f(x) =1

xbetween 1 and n+ 1

=

∫ n+1

1

1

x= log(n+ 1).

Since log(n+ 1)→ +∞ as n→∞, we have∑∞

n=11n

= +∞. �

Example 14.2. Show that∑∞

n=11n2 = converges.

Solution. Consider f(x) = 1x2

:

n+1∑k=1

1

k2= sum of the area of the rectangles with base [i− 1, i] and height

1

i2

for 1 ≤ i ≤ n

≤ 1 + area under the curve f(x) =1

x2between 1 and n

= 1 +

∫ n

1

1

x2= 2− 1

n≤ 2.

Therefore the sequence of partial sums (sn) is non-decreasing and bounded

from above by 2, so∑∞

n=11n2 converges. �

Theorem 14.3.∑∞

n=11np , p > 0, converges if and only if p > 1.

Proof. If p > 1, then by similar comparison argument with areas under the

graph f(x) = 1xp

as Example 14.2, we have

n+1∑k=1

1

kp≤ 1 +

∫ n

1

1

xp= 1 +

1

p− 1

(1− 1

np−1

)< 1 +

1

p− 1=

p

p− 1.

Therefore, the sequence of partial sums (sn) is non-decreasing and bounded

from above by pp−1 , so

∑∞n=1

1np converges.

48 XIN ZHOU

If p ≤ 1, then 1np ≥ 1

nfor all n ∈ N. Since

∑1n

diverges to +∞ by Example

14.1, we have∑∞

n=11np diverges to +∞ by the Comparison Test (Theorem

13.8). �

Theorem 14.4 (Alternating Series Theorem). If (an) is a non-increasing se-

quence of nonnegative numbers, i.e.

a1 ≥ a2 ≥ · · · ≥ an ≥ · · · ≥ 0,

and if lim an = 0, then the alternating series∞∑n=1

(−1)nan converges.

Proof. First we introduce the following observation: given n ≥ m, denote

A = am − am+1 + am+1 − am+1 · · · ± an,

where the sign of an depends on whether n−m is even or odd. In particular,

we haven∑

k=m

(−1)kak = (−1)kA.

(1) If n−m is odd, then the last term in A is −an, so

A = [am − am+1] + [am+2 − am+3] + · · ·+ [an−1 − an] ≥ 0,

because each [·]-term is nonnegative by the monotonicity; moreover

A = am − [am+1 − am+2]− · · · − [an−2 − an−1]− an ≤ am,

because each [·]-term is nonnegative and an is nonnegative.

(2) If n−m is even, then the last term in A is an, so

A = [am − am+1] + · · ·+ [an−2 − an−1] + an ≥ 0,

because each [·]-term is nonnegative and an is nonnegative; and

A = am − [am+1 − am+2]− · · · − [an−1 − an] ≤ am,

because each [·]-term is nonnegative.

Therefore, we proved:

0 ≤ A ≤ am.

Using the fact that lim am = 0, for any ε > 0, there exists N > 0, such that

m > N implies am < ε. Therefore if n ≥ m > N , we have∣∣∣∣∣n∑

k=m

(−1)kak

∣∣∣∣∣ = |A| = A ≤ am < ε.


The Cauchy Criterion (Definition 13.4) implies that∑∞

n=1(−1)nan converges.

�

Example 14.5. Prove that∑∞

n=1(−1)n√

nconverges.

Proof.

• Since lim√

nn+1

= 1, the Ratio test does not apply. Similarly since

lim∣∣ 1n1/2n

∣∣ = 1, the Root test does not apply.

• Since∑

1√n

= +∞, the Comparison test does not apply.

• This is a standard situation to apply the Alternating Series Theorem.

We can let an = 1√n, and it is easy to check that this series satisfies the

requirement of Theorem 14.4.

�

50 XIN ZHOU

15. Continuous functions

Let f be a real-valued function. The defining domain of f , denoted as

dom(f) = the subset (in R) where f is defined.

Example 15.1. Let f(x) =√

4− x2, then dom(f) = [−2, 2].

Definition 15.2. Let f be a function.

(i) Given x0 ∈ dom(f), f is said to be continuous at x0 if

for any sequence (xn) in dom(f) converging to x0, we have

limn→∞

f(xn) = f(x0).(15.1)

(ii) If f is continuous at each point of a subset S ⊂ dom(f), then we say

f is continuous on S.

(iii) f is said to be continuous if f is continuous on dom(f).

Theorem 15.3. Let f be a real valued function whose domain is a subset of

R. Then f is continuous at x0 ∈ dom(f) if and only if

(15.2)for each ε > 0, there exists a number δ > 0, such that

if x ∈ dom(f), and |x− x0| < δ, then |f(x)− f(x0)| < ε.

Proof.

Step 1: Suppose (15.2) holds true. Consider a sequence (xn), xn ∈ dom(f),

such that lim xn = x0. We want to show lim f(xn) = f(x0).

Let ε > 0, then there exists δ > 0 satisfying the requirement of (15.2). Since

limxn = x0, there exists N > 0, such that if n > N , then |xn − x0| < δ.

According to (15.2), we have |f(xn) − f(x0)| < ε. This implies lim f(xn) =

f(x0).

Step 2: Assume f is continuous at x0. If (15.2) fails, then there exists ε >

0, such that for any δ > 0, there exists x ∈ dom(f), |x − x0| < δ, but

|f(x) − f(x0)| ≥ ε. In particular, let δ = 1n, then there exists xn ∈ dom(f),

|xn − x0| < 1n, but |f(xn)− f(x0)| ≥ ε.

However, we thus get a sequence (xn), xn ∈ dom(f), such that limxn = x0,

but we cannot have lim f(xn) = f(x0), hence a contradiction. �

Example 15.4. Let f(x) = 2x2 + 1, x ∈ R. Prove that f is continuous on R,

by

(a) using the definition;

(b) using the ε− δ property.


Proof.

(a) Suppose (xn) is a sequence and limxn = x0s. Then we have

lim f(xn) = lim[2x2n + 1] = 2(limxn)2 + 1 = 2x20 + 1 = f(x0).

Therefore f is continuous at x0.

(b) Given x0 ∈ R, let ε > 0, and we want to show

|f(x)− f(x0)| < ε, provided |x− x0| < δ.

We have

– |f(x)− f(x0)| = |2x2 − 2x20| = 2|x+ x0| · |x− x0|;– Need to bound |x+ x0| from above;

– If |x − x0| < 1, then |x| < |x0| + 1, and |x + x0| ≤ |x| + |x0| <2|x0|+ 1.

Thus:

|f(x)− f(x0)| < 2∣∣2|x0|+ 1

∣∣ · |x− x0|, provided |x− x0| < 1.

Let

δ = min

{1,

ε

2(2|x0|+ 1)

}.

Then |x− x0| < δ implies |f(x)− f(x0)| < ε.

�

Example 15.5. Let f(x) = x2 sin(1x

)for x 6= 0 and f(0) = 0. Prove that f

is continuous at 0.

Proof. Let ε > 0. We have

|f(x)− f(0)| = x2∣∣∣∣sin(

1

x)

∣∣∣∣ ≤ x2.

Set δ =√ε. Then if |x− 0| < δ, =⇒ x2 < δ2 = ε, so

|f(x)− f(0)| ≤ x2 < ε.

�

Example 15.6. Let f(x) = 1x

sin( 1x2

) for x 6= 0, and f(0) = 0. Show that f is

discontinuous at 0.

Proof. If suffices to find a sequence (xn) such that limxn = 0, but

(f(xn)) does not converge to f(0) = 0.

52 XIN ZHOU

Since f(xn) = 1xn

sin( 1x2n

), we can find xn → 0, such that sin( 1x2n

) = 1. That is

we can let1

x2n=π

2+ 2nπ, that is: xn =

1√π2

+ 2nπ.

Then limxn = 0, but

lim f(xn) = lim1

xn= +∞.

�

Algebraic operations on functions.Let f be a real-valued function. For any k ∈ R, we can define

(kf)(x) = kf(x), for x ∈ dom(f);

|f |(x) = |f(x)|, for x ∈ dom(f).

Example 15.7. Let f(x) =√x − 4 for x ≥ 0, then 3f = 3

√x − 12, and

|f | = |√x− 4|.

Given f, g two real-valued functions: we can define

(1) (f + g)(x) = f(x) + g(x);

(2) (fg)(x) = f(x)g(x);

(3) (f/g)(x) = f(x)/g(x);

(4) g ◦ f(x) = g(f(x)

);

(5) max{f, g}(x) = max{f(x), g(x)};(6) min{f, g}(x) = min{f(x), g(x)}.

The defining domains are:

• For (1)(2)(5)(6), the defining domain is dom(f) ∩ dom(g);

• For (3), dom(f/g) = dom(f) ∩ {x ∈ dom(g) : g(x) 6= 0};• For (4), dom(g ◦ f) = {x ∈ dom(f) : f(x) ∈ dom(g)}.

Theorem 15.8. Let f be a real-valued function. If f is continuous at x0 ∈dom(f), then |f |, kf (for k ∈ R) are continuous at x0.

Proof. Consider a sequence (xn), xn ∈ dom(f), such that limxn = x0, then we

have

lim f(xn) = f(x0).

Therefore


• Consider the function kf :

lim((kf)(xn)

)= lim k

(f(xn)

)= k lim f(xn) = kf(x0) = (kf)(x0).

This proved that kf is continuous at x0.

• To show |f | is continuous at x0, we need to prove

lim |f(xn)| = |f(x0)|.

By triangle inequality,∣∣|f(xn)| − |f(x0)|∣∣ ≤ |f(xn)− f(x0)|.

For any ε > 0, since lim f(xn) = f(x0), then there exists N > 0, such

that if n > N , we have |f(xn)− f(x0)| < ε, so∣∣|f(xn)| − |f(x0)|∣∣ < ε.

This implies lim |f(xn)| = |f(x0)|.�

Theorem 15.9. Let f, g be real-valued functions that are continuous at x0 ∈dom(f) ∩ dom(g). Then

(i) f + g is continuous at x0;

(ii) fg is continuous at x0;

(iii) f/g is continuous at x0 if g(x0) 6= 0.

Proof. These follow by Limit Theorems (Section 9). �

Theorem 15.10. If f is continuous at x0 ∈ dom(f), and g is continuous at

f(x0) ∈ dom(g), then g ◦ f is continuous at x0.

Proof. Let (xn) be a sequence, such that xn ∈ dom(f), f(xn) ∈ dom(g), and

limxn = x0.

Since f is continuous at x0, we have lim f(xn) = f(x0), and as g is continuous

at f(x0), we have

lim g ◦ f(xn) = lim g(f(xn)

)= g(f(x0)

)= g ◦ f(x0).

This implies that g ◦ f is continuous at x0. �

Example 15.11. Let f, g be continuous at x0 ∈ dom(f)∩dom(g). Prove that

max{f, g} is continuous at x0.

54 XIN ZHOU

Proof. We have the following equation:

max{f, g} =1

2(f + g) +

1

2|f − g|,

since max{a, b} = 12(a+ b) + 1

2|a− b|.

We know that f+g, f−g are both continuous at x0, so is |f−g|. Therefore12(f + g), 1

2|f − g| are continuous at x0. The conclusion then follows. �


16. Properties of continuous functions

Definition 16.1. Let f be a real-valued function. f is said to be bounded if

{f(x) : x ∈ dom(f)} is a bounded set: that is to say, there exists M ∈ R, such

that |f(x)| ≤M for all x ∈ dom(f).

Theorem 16.2. Let f be a continuous function on a closed interval [a, b], then

f is bounded.

Proof. Assume by contradiction that f is not bounded on [a, b]. Then to each

n ∈ N, the set of values {|f(x)| : x ∈ [a, b]} is not bounded by n: that is to

say, there exists an element in [a, b], denoted as xn, such that |f(xn)| > n.

For considering all n ∈ N, we obtain a sequence (xn), xn ∈ [a, b], such that

|f(xn)| > n.

By the Bolzano-Weierstrass Theorem, (xn) has a subsequence (xnk), such that

limxnk= x0. Since a ≤ xn ≤ b, we have x0 ∈ [a, b]. By the continuity of f on

[a, b], we have

limk→∞

f(xnk) = f(x0).

This is a contradiction to limk→∞ |f(xnk)| = +∞. �

Theorem 16.3. Let f be a continuous function on a closed interval [a, b].

Then f assumes it maximum and minimum on [a, b]: that is to say, there

exist x0, y0 ∈ [a, b], such that f(x0) ≤ f(x) ≤ f(y0) for all x ∈ [a, b].

Proof. Since the set of values {f(x) : x ∈ [a, b]} is bounded, the supremum

exists by the Completeness Axiom:

M = sup{f(x) : x ∈ [a, b]}.

For each n ∈ N, by the definition of supremum, there exists an element in

[a, b], denoted as yn, such that

M − 1

n< f(yn) ≤M.

We obtain a sequence (yn), yn ∈ [a, b], by considering all n ∈ N, and

limn→∞

f(yn) = M.

By the Bolzano-Weierstrass Theorem, there exists a subsequence (ynk) of (yn),

such that

limk→∞

yk = y0.

56 XIN ZHOU

Since a ≤ yn ≤ b, we have y0 ∈ [a, b]. By the continuity of f on [a, b], we have

limk→∞

f(ynk) = f(y0) = M.

So M is a maximum of {f(x) : x ∈ [a, b]}.The case for inf{f(x) : x ∈ [a, b]} is left as an exercise. �

Remark 16.4. The theorems fail if we change the defining domain [a, b] to an

open interval (a, b). A counter-example is:

f(x) =1

x: x ∈ (0, 1).

f(x) is continuous on (0, 1), but f is not bounded on (0, 1).

Theorem 16.5 (Intermediate Value Theorem). If f is continuous on on in-

terval I, and when a, b ∈ I, a < b, and

f(a) < y < f(b), or f(b) < y < f(a),

there exists at least one x ∈ (a, b), such that f(x) = y.

Proof. Assume that f(a) < y < f(b), (the other case is left as an exercise).

Let

S = {x ∈ [a, b] : f(x) < y}.Since f(a) < y, a ∈ S and S 6= ∅.

Let x0 = supS, then x0 ∈ [a, b].

For each n ∈ N, there exists an element in S, denoted as sn, such that

x0 −1

n< sn ≤ x0.

Therefore, lim sn = x0.

Since sn ∈ S, we have f(sn) < y, and hence by continuity of f

f(x0) = lim f(sn) ≤ y.

On the other hand, let

tn = min{b, x0 +1

n},

then x0 ≤ tn ≤ x0 + 1n, hence

lim tn = x0.

However, since b /∈ S (as f(b) > y), and x0 + 1n/∈ S, we have tn /∈ S, and hence

f(tn) ≥ y, for all n ∈ N.


Using continuity of f again,

f(x0) = lim f(tn) ≥ y.

Therefore,

f(x0) = y.

Since y 6= f(a), f(b), we deduce that x0 6= a, b, and hence x0 ∈ (a, b). �

Corollary 16.6. Let f be a continuous function on a closed interval I = [a, b],

then the image

f(I) = {f(x) : x ∈ I} is a closed interval or a point.

Proof. Using the Intermediate Value Theorem, we are easily show that J =

f(I) has the following property:

if y0, y1 ∈ J, y0 < y < y1, then y ∈ J.

By Theorem 16.3, J has maximum max J and minimum min J , hence

J ⊂ [min J,max J ].

moreover, there exists y0, y1 ∈ I, such that

max J = f(y0), min J = f(y1).

If inf J < sup J , then J must be an interval. In fact, if inf J < y < sup J ,

then f(y0) < y < f(y1), and by the Intermediate Value Theorem, there exists

x ∈ (y0, y1) or x ∈ (y1, y0), such that f(x) = y ∈ J , hence

(inf J, sup J) ⊂ J.

And hence

J = [min J,max J ].

�

Example 16.7. Let f : [0, 1] → [0, 1] be a continuous function defined on

[0, 1]. Prove that there exists x0 ∈ [0, 1], such that f(x0) = x0.

Proof. Consider the function:

g(x) = f(x)− x.

It is continuous on [0, 1] by Theorem 15.9.

Notice that

g(0) = f(0)− 0 ≥ 0;

g(1) = f(1)− 1 ≤ 0.

58 XIN ZHOU

Actually if either g(0) = 0 or g(1) = 0, then we can let x0 = 0, or x0 = 1.

Otherwise

g(0) = f(0)− 0 > 0;

g(1) = f(1)− 1 < 0.

By the Intermediate Value Theorem, there exists x0 ∈ (0, 1), such that g(x0) =

0, and this is

f(x0) = x0.

�

Example 16.8. If y > 0 is a positive real number, and m ∈ N is an integer,

then y has a positive m-th root.

Proof. Consider the continuous function f(x) = xm defined on R.

First we claim that there exists b > 0, such that y < bm. In fact

• if y ≤ 1, let b = 2;

• if y > 1, let b = y.

Then we hanve that

f(0) = 0 < y < bm = f(b).

By the Intermediate Value Theorem, there exists x ∈ (0, b), such that

f(x) = xm = y.

�


References

[1] Kenneth A. Ross. Elementary Analysis: The Theory of Calculus, (Undergraduate Texts

in Mathematics),1980 or 2013 edition, Springer, New York, ISBN: 978-1461462705.

http://link.springer.com/book/10.1007/978-1-4614-6271-2

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