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MATH 11521

Further Topics in Analysis2013/14

Lecturer: Dr. Julia Wolf

Notes by Dr. Vitaly Moroz

School of MathematicsUniversity of Bristol

Bristol BS8 1TW, UK

Course webpage:

www.juliawolf.org/teaching/MATH11521.shtml

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Contents

I Relations and functions 11 Relations 1

2 Functions 4

II Elements of set theory 6

3 Finite, innite and countable sets 6

4 Uncountable sets and the notion of continuum 12

5 Cardinality 16

6 The power set and the hierarchy of cardinalities 18

III Convergence and continuity 20

7 Subsequences and accumulation points 20

8 The Bolzano-Weierstrass theorem 24

9 Limit superior and limit inferior 26

10 Cauchy sequences 29

11 Uniformly continuous functions 31

12 Pointwise and Uniform convergence 35

IV The Riemann integral 38

13 Denition of the integral 38

14 Criterion of integrability 4215 Classes of integrable functions 43

16 Inequalities and the mean-value property of the integral 45

17 Further properties of the integral 46

18 Integration as the inverse to differentiation 49

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Notation

N the set of natural numbers {0, 1, 2, 3, . . . }N+ the set of positive natural numbers {1, 2, 3, . . . }Z the set of integer numbers {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }Q the set of rational numbers {r = p

q | p ∈Z , q ∈N+ , hcf( p, q ) = 1}R the set of real numbers

R + the set of non-negative real numbers {x ∈R | x ≥ 0}In the remainder of the text ‘Theorem A 1.2.3 ’ etc. refers to ‘Theorem 1.2.3 ’ in Lecture Notes in Analysis 1 (MATH 11006) by Vitali Liskevich.

This material is copyright of the University of Bristol unless explicitly stated otherwise. It is provided exclusively for educational purposes at the University of Bristol and is to be down-loaded or copied for your private study only.

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1 RELATIONS

Part I

Relations and functions

Recommended texts:

1. I. Stewart, D. Tall . The Foundations of Mathematics , Oxford University Press, 1977.2. S. Krantz . Real Analysis and Foundations , Second Edition. Chapman and Hall/CRCPress, 2005.

1 Relations

Denition 1.1 Let X, Y be sets. A set R ⊆ X ×Y is called a relation from X to Y .

If (x, y) ∈ R, we say that x is in relation R to y. We also write xRy .

Example 1.2 1. Let A = {1, 2, 3}, B = {3, 4, 5}. The set R = {(1, 3), (1, 5), (3, 3)} is a

relation from A to B since R ⊆ A ×B .2. G = {(x, y ) ∈R ×R |x > y } is a relation from R to R .

Denition 1.3 Let R be a relation from X to Y . The domain of R is the set

D(R) = {x ∈ X |∃y ∈ Y [(x, y) ∈ R]}.

The range of R is the set

Ran (R) = {y ∈ Y |∃x ∈ X [(x, y) ∈ R]}.

The inverse of R is the relation R−1

from Y to X dened as followsR−1 = {(y, x ) ∈ Y ×X |(x, y) ∈ R}.

Denition 1.4 Let R be a relation from X to Y , S be a relation from Y to Z . The compo-sition of S and R is a relation from X to Z dened as follows

S ◦R = {(x, z ) ∈ X ×Z |∃y ∈ Y [(x, y) ∈ R]∧[(y, z ) ∈ S ]}.

Theorem 1.5 Let R be a relation from X to Y , S be a relation from Y to Z , T be a relation from Z to V . Then

1. (R−1)−1 = R.

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1 RELATIONS

2. D(R−1) = Ran (R).

3. Ran (R−1) = D(R).

4. T ◦(S ◦R) = ( T ◦S ) ◦R.

5. (S ◦R)−1 = R−1 ◦S −1.

Proof. Exercise, or see [Stewart and Tall (1977)].

Next we take a look at some particular types of relations. Let us consider relations from X to X , i.e. subsets of X ×X . In this case, we talk about a relation on X . A simple exampleof such a relation is the identity relation on X which is dened by

iX = {(x, y) ∈ X ×X |x = y}.

Denition 1.6 1. A relation R on X is said to be reexive if

(∀x ∈ X ) (x, x ) ∈ R.

2. R is said to be symmetric if

(∀x ∈ X )(∀y ∈ X ) {[(x, y) ∈ R] ⇒ [(y, x ) ∈ R]}.

3. R is said to be transitive if

(

∀

x

∈ X )(

∀

y

∈ X )(

∀

z

∈ X )

{[((x, y)

∈ R)

∧

((y, z )

∈ R)]

⇒ [(x, z )

∈ R]

}.

Equivalence relations

A particularly important class of relations are equivalence relations.

Denition 1.7 A relation R on X is said to be an equivalence relation if it is reexive,symmetric and transitive.

Example 1.8 1. Let X be the set of students. Dene a relation on X by

R = {(x, y ) ∈ X ×X : x and y are friends}.

Then R is reexive (assuming that each student is a friend to him or herself). It is alsosymmetric. But it is not transitive.

2. Let X = R , a be some positive number. Dene R ⊆ X ×X as

R = {(x, y) | |x −y| ≤ a}.

R is reexive, symmetric, but not transitive.

3. Let X = Z and m ∈Z . Dene

R := {(x, y) ∈Z ×Z : (∃k ∈Z )[x −y = km ]}.

Then R is an equivalence relation on Z .

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1 RELATIONS

Denition 1.9 Let R be an equivalence relation on X . Let x ∈ X . The equivalence classof x with respect to R is the set

[x]R = {y ∈ X |(y, x ) ∈ R}.

Let us take a look at several properties of equivalence classes.

Proposition 1.10 Let R be an equivalence relation on X . Then 1. (∀x ∈ X ) x ∈ [x]R .2. (∀x ∈ X )(∀y ∈ X ) [(y ∈ [x]R ) ⇔ ([y]R = [x]R )].

Proof. 1. Since R is reexive, (x, x ) ∈ R, hence x ∈ [x]R .2. First, let y ∈ [x]R , so that ( y, x ) ∈ R. Suppose that z ∈ [y]R . Then ( z, y) ∈ R. Bytransitivity, ( z, x ) ∈ R; hence, z ∈ [x]R . This shows that [ y]R ⊆ [x]R . The reverse inclusion

[x]R ⊆ [y]R can be shown similarly. Therefore, [ x]R = [y]R .The implication ([ y]R = [x]R ) ⇒ (y ∈ [x]R ) is obvious.

From the above proposition it follows that equivalence classes [ x]R and [y]R either coincide orare disjoint, in other words,

[x]R = [y]R or [x]R ∩[y]R = ∅.Also, every element of the set X belongs to an equivalence class. This may be expressed as

x∈X

[x]R = X.

Example 1.11 Let R be the equivalence class in Example 3 above. Then, for x ∈Z ,

[x] = {x + km : k ∈Z}.

This is called the equivalence class of integers congruent to x modulo m. Often wewrite y ≡ x to mean y ∈ [x].

Remark 1.12 For more on this topic, see [Stewart and Tall (1977)].

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2 FUNCTIONS

2 Functions

The notion of a function is of fundamental importance in all branches of mathematics. You

met functions in your previous study of mathematics, but without a precise denition. Herewe give a denition and make connections to examples you came across before.

Denition 2.1 Let X and Y be sets. Let F be a relation from X to Y . Then F is called afunction from X to Y , if the following properties are satised:

(i) (∀x ∈ X )(∃y ∈ Y ) [(x, y) ∈ F ].

(ii) (∀x ∈ X )(∀y ∈ Y )(∀z ∈ Y ){([(x, y) ∈ F ]∧[(x, z ) ∈ F ]) ⇒ (y = z)}.

It is customary to write y = F (x), the image of x under F . X is called the domain of F

and Y is called codomain .

Let us consider several examples.

Example 2.2 (i) Let X = {1, 2, 3}, Y = {4, 5, 6}. Dene F ⊆ X ×Y as

F = {(1, 4), (2, 5), (3, 5)}.

Then F is a function. In contrast to this, dene G ⊆ X ×Y via

G = {(1, 4), (1, 5), (2, 6), (3, 6)}.

Then G is not a function.(ii) Let X = R and Y = R . Dene F ⊆ X ×Y to be

F = {(x, y) ∈R ×R |y = x2}.

Then F is a function from R to R . In contrast to this, dene G ⊆ X ×Y to be

G = {(x, y) ∈R ×R |x2 + y2 = 1}.

Then G is not a function.

We often use the notationF : X → Y

for a function F from X to Y .

Note that in order to dene a function F from X to Y we have to dene X , Y and a subsetof X ×Y satisfying (i) and (ii) of the denition.

Note: one can assign a value y to each value x ∈ X by means of a “rule” or “formula”. Butthis notion is rather vague. The present denition is more precise.

Theorem 2.3 Let X, Y be sets and F, G functions from X to Y . Then

[(∀x ∈ X )(F (x) = G(x))] ⇔ (F = G).

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2 FUNCTIONS

Proof. 1. (⇒). Assume the LHS above. We rst show that F ⊆ G. Let (x, y ) ∈ F . Theny = F (x). So y = G(x); or, ( x, y) ∈ G. The proof of the inclusion G ⊆ F is similar. Theimplication (⇐) is trivial.

Thus, two functions are equal if they share the same domain and codomain and elements inthe domain have equal images.

Example 2.4 Let f : R → R , g : R → R , h : R → R+ . Here, R+ denotes the set of non-negative real numbers. Suppose that for all x ∈R ,

f (x) = ( x + 1) 2, g(x) = x2 + 2 x + 1 , h(x) = ( x + 1) 2.

Then f and g are equal, but f and h are not since they have different codomains.

The denition of composition of relations can be also applied to functions. If f : X

→ Y and

g : Y → Z theng ◦ f = {(x, z ) ∈ X ×Z |(∃y ∈ Y )[[(x, y ) ∈ f ]∧[(y, z ) ∈ g]]}.

Theorem 2.5 Let f : X → Y and g : Y → Z . Then g ◦ f : X → Z and

(∀x ∈ X ) [(g ◦ f )(x) = g(f (x))].

Proof. We know that g ◦ f is a relation. So we must prove that for every x ∈ X there existsa unique element z ∈ Z such that ( x, z ) ∈ g ◦ f .

Existence: Let x ∈ X be arbitrary. As f is a function, ∃y ∈ Y such that ( x, y) ∈ f . As g is afunction,

∃z

∈ Z such that ( y, z )

∈ g. So

(∃y ∈ Y )[(x, y) ∈ f ∧ (y, z ) ∈ g]

and ( x, z ) ∈ g ◦ f . In short,

(∀x ∈ X )(∃z ∈ Z )[(x, z ) ∈ g ◦ f ].

Uniqueness: Let x ∈ X . Suppose that ( x, z 1) ∈ g ◦f and (x, z 2) ∈ g ◦f . As (x, z1) ∈ g ◦f ,(∃y1 ∈ Y )[(x, y1) ∈ f ∧(y1, z1) ∈ g]. As (x, z2) ∈ g ◦f , (∃y2 ∈ Y )[(x, y2) ∈ f ∧(y2, z2) ∈ g].As f is a function, y1 = y2. As g is a function, z1 = z2.

Example 2.6 Let f :R

→R

, g :R

→R

,f (x) =

1x2 + 2

, g(x) = 2 x −1.

Find ( f ◦g)(x) and ( g ◦ f )(x).

Solution .

(f ◦g)(x) = f (g(x)) = 1

[g(x)]2 + 2 =

1(2x −1)2 + 2

,

(g ◦f )(x) = g(f (x)) = 2 f (x) −1 = 2

x2 + 2 −1.

Warning: As is clear from the above, f ◦g = g ◦f in general.

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3 FINITE, INFINITE AND COUNTABLE SETS

Part II

Elements of set theory

Recommended texts:

1. I. Stewart, D. Tall. The Foundations of Mathematics , Oxford University Press, 1977.2. S. Krantz. Real Analysis and Foundations , Second Edition. Chapman and Hall/CRCPress, 2005.

In this part of the course we study the notion of cardinality . The aim is to formulate anotion of size of an innite set, or to count its elements. It is possible to ask questions suchas how many rational numbers are there? or is the set of real numbers bigger then the set of natural numbers? To do so, however, it is helpful to be able to compare the number of elements in a given set with those in another. We say that sets X and Y have the samecardinality (intuitively, number of elements) if there is a bijection f : X → Y . This idea wasrst conceived by the German mathematician Georg Cantor (1845 – 1918) at the end of the19th century. His “Naive Set Theory” gives rise to various paradoxes (e.g. Russell’s Paradox,see p. 19), which are resolved in the subsequent Axiomatic Set Theory . For our purposes, wetake the “naive” notion of set as a collection of elements .

3 Finite, innite and countable sets

Let X , Y be sets and f : X → Y a function from X to Y . Recall that the function f is aninjection if

(∀x1 ∈ X )(∀x2 ∈ X )[(f (x1) = f (x2)) ⇒ (x1 = x2)].

This means that f is a one-to-one correspondence between X and the range Ran (f ). Thefunction f is a surjection if

(∀y ∈ Y )(∃x ∈ X )[f (x) = y].This means that Ran (f ) = Y . Finally, f is a bijection if f is an injection and surjection. Inthis case, f establishes a one-to-one correspondence between X and Y .

How do we decide how many elements there are in a set? One answer is to count them. Butwhat does this mean? This can be described as follows. We take an element of the set, andthink one . We then look at a different element and think two, and so on. We need to becareful to make sure that all elements have been counted, and none has been counted twice.If this process terminates, we say that the set is nite.

Denition 3.1 A set X is said to be nite if either X = ∅ or for some n ∈ N+ there existsa bijection

f : {1, 2, . . . , n } → X.

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In the former case, we say that X has zero elements; in the latter, that X has n elements. X is said to be innite if it is not nite.

A natural way to extend this process of counting to innite sets is as follows.Denition 3.2 A set X is said to be countable if there exists a bijection

f : N+ → X.

Remark 3.3 By Theorem A1.7.6 , a function f : N+ → X is a bijection if and only if theinverse function f −1 : X → N+ exists. In this case, the inverse f −1 is a bijection too. Inparticular, X is countable if and only if there exists a bijection f : X →N+ .

Example 3.4 N = {0, 1, 2, . . . } is countable.

Proof. Dene f : N+ →N by f (n) = n −1.

1 2 3 4 5 . . . n . . .

↓ ↓ ↓ ↓ ↓ ↓0 1 2 3 4 . . . n −1 . . .

Then f is a bijection.

In this example, we observe the rst remarkable property of innite sets. N+ is a subsetof N, so intuitively it should have fewer elements. Yet N+ and N are both countable. Thenext example, constructed by Galileo in 1638, was considered as paradox for more than twocenturies.

Example 3.5 The set of perfect squares S := {12, 22, 32, . . . , n 2, . . . } is countable.

Proof. Dene f : N+ → S by f (n) = n2.

1 2 3 4 5 . . . n . . .

↓ ↓ ↓ ↓ ↓ ↓1 4 9 16 25 . . . n2 . . .

It is clear that f is a bijection.

The reason Galileo’s example might be considered a paradox is that the set S of perfect squaresseems smaller then N+ . But

S is still countable.

Example 3.6 The set of integers Z is countable.

Proof. Dene f : N+ →Z by

f (n) = n

2 , if n is even,1−n

2 , if n is odd.

The following diagram illustrates f :

1 2 3 4 5 . . . 2n 2n + 1 . . .

↓ ↓ ↓ ↓ ↓ ↓ ↓0 1 −1 2 −2 . . . n −n . . .

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The inverse of f is the function f −1 : Z →N+ given by

f −1(m) = 2m, if m ≥ 1,

1 −2m, if m ≤ 0.By Theorem A1.7.6 we conclude that f is a bijection.

Remark 3.7 Notice that although f in the last example is a bijection, it does not preserveorder, in the sense that m < n does not imply f (m) < f (n).

Example 3.8 The set N+ ×N+ is countable.

Proof. We write elements of N+ ×N+ in a rectangular array (gure on the left). We thenread them off along the cross diagonals (gure on the right): rst (1 , 1), next (1 , 2), (2, 1), then(1, 3), (2, 2), (3, 1) and so on.

(n, m ) 1 2 3 4 . . .1 (1,1) (1,2) (1,3) (1,4) . . .2 (2,1) (2,2) (2,3) . . .3 (3,1) (3,2) . . .4 (4,1) . . .

. . . · · ·

f (n,m )

−−−−→1 2 4 7 . . .3 5 8 . . .6 9 . . .

10 . . .

· · ·This correspondence can also be written

(1, 1) (1, 2) (2, 1) (1, 3) (2, 2) (3, 1) . . .

↓ ↓ ↓ ↓ ↓ ↓1 2 3 4 5 6 . . .

In this way we establish a bijection between N+ ×N+ and N+ . This bijection f : N+ ×N+ →N+

can be represented analytically by the formula

f (n, m ) = (n + m −2)(n + m −1)

2 + n.

Therefore N+ ×N+ is countable.

Remark 3.9 Let X be a countable set and f : N+ → X a bijection. Dene x1 := f (1),

x2 := f (2), x3 := f (3), . . . , xn := f (n), . . . .1 2 3 4 5 . . . n . . .

↓ ↓ ↓ ↓ ↓ ↓x1 x2 x3 x4 x5 . . . xn . . .

Thus all elements of X can be listed in a sequence:

X = {x1, x2, x3, x4, x5, . . . , x n , . . . }.

We now establish several important properties of countable sets.

Lemma 3.10 If a set X is countable then any subset A ⊆ X is nite or countable.

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Proof. Assume A is innite. We need to show that A is countable. Since X is countable thereexists a bijection f : N+ → X . Dene g : N+ → A inductively as follows. Set

g(1) := f (min{k ∈N+

: f (k) ∈ A});g(2) := f (min{k ∈N+ : f (k) ∈ A \ {g(1)}});· · · · · · · · ·

g(n + 1) := f (min{k ∈N+ : f (k) ∈ A \ {g(1) , . . . , g (n)}}).

It is clear that g is a bijection.

Remark 3.11 We say that a set X is at most countable if X is nite or countable.

Remark 3.12 Lemma 3.10 is useful for proving that a given set is countable. For example,let

P be the set of all prime numbers. Then

P ⊆N+ . And it is known that the set of primes

is innite. We may then conclude that P is countable.

Lemma 3.13 Any innite set X has a countable subset A ⊆ X .

Proof. First, X = ∅ as X is innite. So we may choose x1 ∈ X . Dene

g1 : {1} → X ; 1 → x1.

Then g1 is not onto, for otherwise X would be nite. So we may choose x2 ∈ X \{x1}. Dene

g2 : {1, 2} → X ; j → x j .

Then g2 is not onto, for otherwise X would be nite. So we may choose x3 ∈ X \ {x1, x2}.Continuing in this way, we obtain a sequence ( xn )n∈N+ of distinct points in X . The mapping

g : N+ → A := {xn ∈ X |n ∈N+ }; n → xn

is then a bijection from N+ to A ⊆ X . Thus A is a countable subset of X .

Denition 3.14 A set Y is said to be a proper subset of a set X if Y ⊆ X and Y = X .

Proposition 3.15 If a set X is innite then there exists a proper subset Y ⊆ X and a bijection f : X → Y .

Proof. Let A := {x1, x2, . . . , x n , . . . } be a countable subset of X , constructed as in Lemma3.13. Let B := {x1} ⊆ A. Set Y := X \B . Dene f : X → Y by

f (xn ) = xn +1 for xn ∈ Aand

f (x) = x for x ∈ X \A.


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Remark 3.16 We proved that if X is innite and B = {x1} ⊆ X then Y := X \ B is inniteand there is a bijection f : X → Y . Thus removing one element from an innite set doesnot change its ”size”. In a similar way, it is possible to select a countable subset B ⊆ X , set

Y := X \ B and yet construct a bijection f : X → Y . To do so, select a countable subsetA := {x1, x2, . . . , x n , . . . } of X , then let B := {x2k−1|k ∈N+ } ⊆ A, set Y := X \ B and denef : X → Y by

f (xn ) = x2n for xn ∈ Aand

f (x) = x for x ∈ X \A.


Lemma 3.17 Let X and Y be countable sets. Then the Cartesian product X ×Y is countable.

Proof. Let f : X → N+

and g : Y → N+

be bijections. Dene the function h : X ×Y →N+ ×N+ viah(x, y) := ( f (x), g(y)) .

To see that h is an injection suppose that h(x1, y1) = h(x2, y2). This means that

(f (x1), g(y1)) = ( f (x2), g(y2)) .

So f (x1) = f (x2) and g(y1) = g(y2). Since f and g are injections it follows that x1 = x2 andy1 = y2, that is ( x1, y1) = ( x2, y2). To check that h is a surjection, suppose ( n, m ) ∈N+ ×N+ .Then since f and g are both surjections, we can choose x ∈ X and y ∈ Y such that f (x) = nand g(y) = m. Therefore, h(x, y) = ( n, m ), as required. Hence h : X ×Y → N+ ×N+ is a

bijection. By Example 3.8, there is a bijection ϕ : N+

×N+

→ N+

. It is easy to see that thecomposition ϕ ◦h : X ×Y →N+ is also a bijection. Therefore X ×Y is countable.

Lemma 3.18 Let {X k}k∈N+ be a countable collection of pairwise disjoint countable sets. Then the union

∪∞k=1 X k

is countable.

Proof. Since every set X k is countable we can write the elements of X k in a list:

X 1 = {x11 , x12 , x13 , . . . , x 1n , . . . },

X 2 = {x21 , x22 , x23 , . . . , x 2n , . . . },

X 3 = {x31 , x32 , x33 , . . . , x 3n , . . . },

. . . . . .

X k = {xk1, xk2, xk3, . . . , x kn , . . . },

. . . . . .

By denition of the union

∪∞k=1 X k = {xkn |k ∈N+ , n ∈N+ }.

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4 UNCOUNTABLE SETS AND THE NOTION OF CONTINUUM

Denition 4.4 A set X is said to be a continuum if there exists a bijection

f : (0, 1) → X.

Example 4.5 Let a, b ∈R and a < b. Then the open interval ( a, b) is a continuum.

Proof. The function f : (0, 1) → (a, b) dened by the formula

f (x) := ( b−a)x + a

is a bijection with inverse f −1 : (a, b) → (0, 1) given by

f −1(y) = y −ab−a

.

Example 4.6 The interval [0 , 1) is a continuum (as is (0 , 1] and [0, 1]).

Proof. (Compare this proof with the proof of Proposition 3.15! )The function f : [0, 1) → (0, 1) dened by the formula

f (x) := 1 − 1n +1 , if x = 1 − 1

n , n ∈N+ ,x, if x = 1 − 1

n , n ∈N+ .

is a bijection with inverse

f −1(y) := 1− 1

n , if y = 1 − 1n +1 , n ∈N+ ,

y, if y = 1 − 1n +1 , n ∈N+ .

The intervals (0 , 1] and [0, 1] can be considered in a similar way.

Example 4.7 The real line R is a continuum.

Proof. The function g : (−1, 1) →R dened by

g(x) := x

1 − |x|is a bijection with inverse g−1 : R → (−1, 1) given by

g−1(y) = y

1 + |y|.

By Example 4.5 there is a bijection f : (0, 1) → (−1, 1). Then the composition g◦f : (0, 1) →Ris a bijection.

Example 4.8 [0, 1) ×[0, 1) is a continuum.

Idea of proof. Any x ∈ [0, 1) has a decimal expansion

x = 0 . a 1 a2 a3 · · · .We assume that the decimal expansion does not contain a string of consecutive 9’s with innitelength. The expansion is then unique. The expansion may be rewritten in blocks

x = 0 . α 1 α 2 α 3 · · ·13

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where each block α j has the formα j = ∗

where

∗ stands for one of the digits 0 , 1, 2, 3, 4, 5, 6, 7, 8; or

α j = 9 9 · · ·9∗(a nite block of 9’s followed by a non-nine digit).

Dene f : [0, 1) ×[0, 1) → [0, 1) via

f (x, y ) := 0 . α 1 β 1 α 2 β 2 α 3 β 3 . . . .

Here, we writex = 0 . a 1 a2 a3 · · · = 0 . α 1 α 2 α3 · · · ;y = 0 . b1 b2 b3

· · · = 0 . β 1 β 2 β 3

· · · ;

rst in decimal form, then in block form, as described above. Then f is a bijection.

To conclude, compose f with the bijection in Example 4.6.

The interval (0, 1) is uncountable – another proof. 2 Our proof of Theorem 4.2 relies onthe decimal expansion of the real numbers. Cantor’s original proof of Theorem 4.2, published in 1874,was different. It was based on the following “nested intervals” theorem, which is similar to TheoremA 3.2.3 from Analysis.

Theorem 4.9 Let ([bk , ck ])k ∈ N + be a sequence of closed nonempty bounded intervals such that

(

∀

k

∈N+ )([bk +1 , ck +1 ]

⊂ [bk , ck ]).

Then ∩k ∈ N + [bk , ck ] = ∅.Exercise 4.10 Prove Theorem 4.9.

Hint. Modify the proof of Theorem A 3.2.3 from Analysis I.

Cantor’s original proof of Theorem 4.2. Let E ⊂ (0, 1) be a countable subset of (0 , 1), so E canbe represented as

E = {x1 , x 2 , x 3 , . . . }.

Observe, that given a point x ∈ (0, 1) there exists a nonempty interval [ b, c] ⊂ (0, 1) such that x ∈ [b, c].Hence, we can choose a nonempty interval [ b1 , c1 ]

∈ (0, 1) such that x1

∈ [b1 , c1 ]. Now we proceed

inductively. Let n ∈ N+ . Having chosen a nonempty interval ([ bk , ck ])nk =1 such that xk ∈ [bk , ck ] for

k = 1 , . . . , n , and such that [ bk +1 , ck +1 ] ⊂ (bk , ck ) for k = 1 , . . . , n −1, we can choose nonempty interval[bn +1 , cn +1 ] such that xn +1 ∈ [bn +1 , cn +1 ] and such that [ bn +1 , cn +1 ] ⊂ (bn , cn ).

In such a way we inductively dened a “nested” sequence of closed nonempty bounded intervals([bk , ck ])k ∈ N + so that

(∀k ∈N+ )[([bk +1 , ck +1 ] ⊂ [bk , ck ])∧(xk ∈ [bk , ck ])].

According to Theorem 4.9, there exists a point x∗ ∈ ∩k ∈ N + [bk , ck ]. But since xn ∈ ∩n ∈ N + [bk , ck ] forevery n ∈ N + , we conclude that x∗ ∈ E . Therefore E = (0 , 1), which means that the interval (0 , 1) isuncountable.

2 The content of this paragraph will be excluded from the examination.

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Decimal expansions. 3 Let ( a k )k ∈ N + be a sequence of numbers chosen from the set {0, 1, 2, . . . , 9}.Then the series

∞

k =1

ak

10k

converges by comparison with a geometric progression to a sum α ∈ [0, 1]. We write

α = 0 . a 1 a2 a3 a4 a5 . . . ,

and say that the right–hand side is a decimal expansion of the real number α . For example,

13

= 0.333333 . . . , 1

7 = 142857

142857

142857

. . . ,

π4

= 0 .785398163397448 . . . . . . .

Proposition 4.11 Every real number α ∈ [0, 1] has a unique decimal expansion unless it is a rational number of the form α = m

10 n for some m ∈ N+ and n ∈ N+ in which case α has precisely two decimal expansions.

Sketch of the proof. Let α ∈ [0, 1]. Suppose that {a1 , a 2 , . . . , a k } have been chosen from {0, 1, 2, . . . , 9}in such a way thatk := α −

a1

10 +

a2

102 + · · ·+ ak

10k

satises0 ≤ k <

110k .

Let ak +1 be chosen from {0, 1, 2, . . . , 9} so that a k +1 is maximized subject to the constraint k +1 ≥ 0.If ak +1 < 9 then

k +1 < 110k +1 ,

because otherwise we could replace a k +1 by a k +1 + 1. If a k +1 = 9 then

k +1 = k − 910k +1 <

110k −

910k +1 =

110k +1 .

By the “sandwich rule” ( A3.1.6 ) we conclude that lim k →∞ k = 0. Therefore

α =∞

k =1

a k

10k ,

which gives the required decimal expansion of α .We do not consider the issue of uniqueness of the decimal expansion except for the observation

that, where two distinct expansions of α exist then α = m10 n for some m ∈N+ and n ∈N+ and one of

the expansions consists of zeros from some point and the other consists of nines from some point on.In this case we say that a decimal expansion of α terminates . For example,

12

= 510

= 0 .5000 · · · = 0 .4999 . . . .

If we agree that when decimal expansion terminates we will write it ending with a sequence of zeros,not a sequence of nines, then every real number α ∈ [0, 1] has a unique decimal expansion.

3 The content of this paragraph will not be included in the examination paper.

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5 CARDINALITY

5 Cardinality

In this section, we discuss a method for comparing the size of arbitrary sets.

Denition 5.1 Let X , Y be sets. We say that X has the same cardinality as Y if thereexists a bijection

f : X → Y.

In this case, we write X ∼ Y or card (X ) = card (Y ).

Remark 5.2 Intuitively, X ∼ Y or card (X ) = card (Y ) means that the sets X and Y havethe same size or the same number of elements.

Example 5.3 We already know that N+∼

Z ∼ Q ∼ N+ ×N+ and (0 , 1) ∼ [0, 1] ∼ R ∼(0, 1) ×(0, 1).

Proposition 5.4 For any sets X , Y , Z :(a) X ∼ X (reexivity);(b) if X ∼ Y then Y ∼ X (symmetry);(c) if X ∼ Y and Y ∼ Z then X ∼ Z (transitivity).

Remark 5.5 Properties (a), (b), (c) entail that “ ∼” is an equivalence relation on the collec-tion of all sets. We will see however that the collection of all sets cannot be a set. This meansthat “∼” is not a relation in the sense dened before, as its domain is not a set. Even so, wecan still consider the equivalence class [ X ] of a given set X : the collection of sets which areequivalent to X . For example, the equivalence class [ N+ ] of the set of natural numbers N+ isthe collection of all countable sets.

Proof. (a) The identity function iX : X → X is a bijection.(b) Suppose X ∼ Y . Let f : X → Y be a bijection. Hence by Theorem A1.7.6 , the inversefunction f −1 : Y → X exists. But now note that ( f −1)−1 = f : X → Y is an inverse to f −1.So by Theorem A1.7.6 again, f −1 : Y → X is a bijection. Therefore Y ∼ X .(c) Suppose X ∼ Y and Y ∼ Z . Let f : X → Y and g : Y → Z be bijections. Hence byTheorem A1.7.16 there exist inverse functions f −1 : Y → X and g−1 : Z → Y . Consider thecomposition g ◦f : X → Z . It is easy to check that ( g ◦f )−1 = f −1 ◦g−1 : Z → X is theinverse function of g ◦f : X → Z . Hence g ◦ f : X → Z is a bijection. Therefore X ∼ Z .

Exercise 5.6 Suppose A ∼ B and C ∼ D . Prove that:

i) (A ×C ) ∼ (B ×D );ii) if A and C are disjoint and B and D are disjoint, then ( A∪C ) ∼ (B ∪D ).

Proof. See Exercises 3, Q 5.

Denition 5.7 Let X , Y be sets. We say that the cardinality of Y is greater than orequal to the cardinality of X (Y dominates X ), if there exists an injection

f : X → Y.

In this case, we write X Y or card (X ) ≤ card (Y ). We say that the cardinality of Y isstrictly greater then the cardinality of X , if there exists an injection

f : X → Y,

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5 CARDINALITY

but there is no injection (and hence no bijection) from Y to X . In this case, we write X ≺ Y or card (X ) < card (Y ).

Remark 5.8 (a) Intuitively, X Y (or card (X ) ≤ card (Y )) means that the set X has fewerelements than the set Y .(b) If X ∼ Y then X Y and Y X . To see this, consider a bijection f : X → Y . Thenf : X → Y is an injection and f −1 : Y → X is an injection too.(c) If A ⊆ X then the identity map iA : A → X is an injection. Hence if A ⊆ X then A X .(d) If X ≺ Y then X Y and X ∼ Y .

Example 5.9 ∅ ≺ {1} ≺ {1, 2} {3, 4} ≺ {1, 2, . . . , n } ≺ N+ Q ≺ R (0, 1). However

{1, 2} ∼ {3, 4}, N+∼

Q and R∼ (0, 1).

Exercise 5.10 Prove that for any sets X , Y , Z :(a) X X (reexivity);(b) if X Y and Y Z then X Z (transitivity).

Proof. Similar to the proof of Proposition 5.4.

Exercise 5.11 Suppose A B and C D. Prove that:i) (A ×C ) (B ×D );ii) If A and C are disjoint and B and D are disjoint, then ( A∪C ) (B ∪D ).

Proof. Similar to Exercises 3, Q5.

Remark 5.12 Properties (a) and (b) of Exercise 5.10 mean that “ ” is a reexive and tran-sitive relation on the collection of all sets (see Remark 5.5). However “ ” is not an orderrelation, because it is not antisymmetric. This means that X Y and Y X does not implythat X = Y . For example, N+

∼Q , so N+ Q and Q N+ , but N+ = Q .

Theorem 5.13 (Cantor-Schr¨ oder-Bernstein theorem) Let X and Y be sets. If X Y and Y X then X ∼ Y .

Proof. See Stewart and Tall , Theorem 6 on p.238.

The Cantor–Schr¨ oder–Bernstein theorem is extremely useful when it is difficult to nd abijection between two sets. Instead, it is enough to construct two injections.

Example 5.14 [0, 1] ∼ (0, 1).

A simple proof via the Cantor–Schr¨ oder–Bernstein theorem . Dene f : (0, 1) → [0, 1] andg : [0, 1] → (0, 1) by

f (x) := x, g(x) := 12

x + 14

.

It is clear that both f and g are injections.

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6 THE POWER SET AND THE HIERARCHY OF CARDINALITIES

6 The power set and the hierarchy of cardinalities

Denition 6.1 Let X be a set. The power set of X , P (X ) or 2X , is the set of all subsets

of X .

Example 6.2 Let X = {0, 1}. Then 2 X = {∅, {0}, {1}, {0, 1} }. Notice that 2 ∅ = {∅}, so 2∅has one element.

Exercise 6.3 Let X = {1, 2, . . . , N }. Then the set 2 X has 2N elements.

Theorem 6.4 Let X be a set. Then X ≺ 2X .

Proof. It is clear that the function f : X → 2X given by f (x) := {x} is an injection, soX 2X . We now show X ∼ 2X .

Assume that f : X → 2X is a bijection. Put

Z := {x ∈ X |x ∈ f (x)}.

Then Z ∈ 2X . As f is a surjection, f (z) = Z for some z ∈ X . Now,

z ∈ Z ⇒ z ∈ f (z) ⇒ z ∈ Z,

z ∈ Z ⇒ z ∈ f (z) ⇒ z ∈ Z,

a contradiction. So there is no bijection f : X → 2X .

Applied to the set N+ , the theorem says that 2 N+is uncountable. Actually it may be shown

that 2 N+

is a continuum.

Proposition 6.5 2N+

∼R .

Proof. See Stewart and Tall , pp.240-241.

Exercise 6.6 Let F = {f : [0, 1] → R} be the set of all functions from the interval [0 , 1] toR . Prove that [0 , 1] ≺ F .

Hint. Consider the set F 0 = {f A : A ⊂ [0, 1]} of all characteristic functions f A : [0, 1] →{0, 1} of the form

f A(x) = 1 if x ∈ A,0 if x ∈ A.

Here, A ∈ 2[0,1] is a subset of [0, 1]. Establish a bijection between the set F 0 and the set 2 [0,1]

of all subsets of the interval [0 , 1].

Hierarchy of cardinalities. Theorem 6.4 leads us to a hierarchy of cardinalities. We beginwith card (N+ ). Then card (2N+

) = card (R ) is strictly bigger. Then follows card (2R ), and soon:

∅ ≺ {1} ≺ {1, 2} ≺ · · · ≺ {1, 2, . . . , N } ≺ · · · ≺N+≺ 2N+

∼R≺ 2R

≺ 22R

≺ . . . .

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6 THE POWER SET AND THE HIERARCHY OF CARDINALITIES

Continuum Hypothesis. In 1878, Cantor asked whether there exists a set X such thatN+≺ X ≺ R . He conjectured that the answer was no, but was unable to prove it. This

conjecture is known as the

Continuum Hypothesis. Let X be a set. If N+≺ X and X R then X ∼R .

The Continuum Hypothesis was resolved by G¨ odel and Cohen in the middle of the 20thcentury. They proved that in the framework of Axiomatic Set Theory it is impossible to provethe continuum hypothesis and it is also impossible to disprove it: that is, the continuumhypothesis is independent of the axioms. This means that it is possible to construct axiomaticmodels of set theory in which this hypothesis is true, and models in which it is false. Its statustherefore is similar to that of the parallel postulate in Euclidean geometry.

Paradoxes of Naive Set Theory. In 1899 Cantor discovered a paradox which arises if oneconsiders the set of all sets. What is the cardinal number of the set of all sets? Clearly it must

be the greatest possible cardinal, yet the cardinal of the set of all subsets of a set always hasa greater cardinal than the set itself.In 1902, Russell discovered another paradox now known as Russell’s paradox. Suppose thatwe assume the existence of the set y consisting of all sets x which do not belong to themselves,i.e.

y = {x : x ∈ x}.

If y ∈ y then y must satisfy the dening property of y, i.e. y ∈ y. On the other hand, if y ∈ y,then y satises the dening property of y and then y ∈ y. In either case, a contradiction isobtained.A popular version of this paradox concerns a certain barber who shaves everyone in his town

who does not shave himself. The question is: who shaves the barber? Each answer leads to acontradiction. The conclusion is that there is no such barber.Similarly, it is meaningless to speak of the set all sets which do not belong to themselves. Itis also meaningless to speak of the set all sets. There is no such set.These paradoxes can be resolved in the framework of modern Axiomatic Set Theory.

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7 SUBSEQUENCES AND ACCUMULATION POINTS

Part III

Convergence and continuity

Recommended texts:

1. J. Howie. Real Analysis , Springer-Verlag, 2003.2. S. Krantz. Real Analysis and Foundations , Second Edition. Chapman and Hall/CRCPress, 2005.3. G. Wanner. Analysis by its History , Springer-Verlag, 1996.

7 Subsequences and accumulation points

Denition 7.1 Let (an )n∈N+ be a sequence. A subsequence of (an )n∈N+ is a sequence

(am (k) )k∈N+ = ( am (1) , a m (2) , am (3) , . . . , a m (k) , . . . ),

where m : N+ →N+ is a strictly increasing function, so that

1 ≤ m(1) < m (2) < m (3) < · · · < m (k) < . . . .

Remark 7.2 (a) The sequence ( an )n∈N+ is a subsequence of itself (take m(k) = k).(b) The monotonicity of the function m implies that (∀k ∈N+ )(m(k) ≥ k). So,

m(k) → ∞ as k → ∞.(c) Any sequence has innitely many subsequences. Take, for example,

m1(k) = k, m2(k) = 2 k, m3(k) = 3 k, . . .

for k ∈N+ .

Example 7.3 Consider the sequence ( an )n∈N+ dened by the formula an = ( −1)n . Then

(a2k+1 )k∈N+ = (−1, −1, −1, −1 . . . ),(a2k )k∈N+ = (1 , 1, 1, 1 . . . ),(a3k )k∈N+ = (−1, 1, −1, 1 . . . )

are subsequences of ( an )n∈N+ . Note that a2k+1 → −1 and a2k → 1 as k → ∞, while (a3k )k∈N+

diverges.

Example 7.4 Let X = {x1, x2, . . . , x s} ⊆R be a nite set of distinct real numbers. Considerthe sequence

(an )n∈N

+

= ( x1, x2, . . . , x s

, x1, x2, . . . , x s

, x1, x2, . . . , x s

, . . . ).

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This sequence may be described by

an =

x1 if n ≡ 1 (mod s),

x2 if n ≡ 2 (mod s),... ...

xs if n ≡ 0 (mod s).

Then a sn +1 → x1, a sn +2 → x2, . . . , and a sn → xs as n → ∞. In short, the sequence ( an )n∈N+

possesses subsequences that tend to each element in X .

Example 7.5 Consider the sequence

(an )n∈N+ = ( 1

, 1, 2

, 1, 2, 3

, 1, 2, 3, 4

, 1, 2, 3, 4, 5

, . . . ).

Then for any m ∈N+

there is a subsequence

(m,m,m,m,. . . )

of (an )n∈N+ which tends to m as n → ∞.Lemma 7.6 Suppose that the sequence (an )n∈N+ converges to a. Then any subsequence (am (k))k∈N+ also converges to a.

Proof. Let ε > 0. As an → a as n → ∞, we can nd N ∈N+ such that

|an −a| < ε for n > N.

Now, for k > N , m(k) > m (N ) ≥ N . Thus

(∀k ∈N+ ) (k > N ) ⇒ (|am (k) −a| < ε ) .

That is, am (k) → a as k → ∞.

Lemma 7.7 Let (an )n∈N+ be a bounded sequence. Then any subsequence is bounded.

Proof. Let (am (k) )k∈N+ be a subsequence of ( an )n∈N+ . As (an )n∈N+ is bounded, there exists0 ≤ M < ∞ such that

|an

| ≤ M for n

∈N+ .

But then(∀k ∈N+ ) |am (k)| ≤ M ,

and so ( am (k) )k∈N+ is bounded.

Denition 7.8 Let (an )n∈N+ be a sequence in R . We say that a ∈ R is an accumulationpoint of (an )n∈N+ if there exists a subsequence ( am (k) )k∈N+ which converges to a .

Example 7.9 (a) Consider the sequence

(

−1, 1,

−1, 1,

−1, 1,

−1, 1, . . . ).

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from Example 7.3. Then the set of accumulation points of ( an )n∈N+ is

{−1, 1}.

(b) The set of accumulation points of the sequence

(an )n∈N+ = ( x1, x2, . . . , x s

, x1, x2, . . . , x s

, x1, x2, . . . , x s

, . . . )

from Example 7.4 is the setX = {x1, x2, . . . , x s}.

(c) Consider the sequence

(an )n∈N+ = ( 1

, 1, 2

, 1, 2, 3

, 1, 2, 3, 4

, 1, 2, 3, 4, 5

, . . . )

from Example 7.5. The set of accumulation points of ( an )n

∈

N+ is the set of natural numbers

N+ = {1, 2, 3, 4, 5, . . . }.

Exercise 7.10 Let (an )n∈N+ be a sequence that converges to a. Then the set of accumulationpoints of ( an )n∈N+ comprises one element, namely {a}.

Hint. Apply Lemma 7.6.

Exercise 7.11 Let (an )n∈N+ be a bounded sequence. Then the set of accumulation points of (an )n∈N+ is bounded.

Hint. Apply Lemma 7.7.

Characterisation of the set of all accumulation points of a sequence.

Proposition 7.12 A number a ∈R is an accumulation point of the sequence (an )n∈N+ if and only if

(1) for any ε > 0, the set {n ∈N+ : a −ε < a n < a + ε} is innite;

equivalently,

(2) (∀ε > 0)(∀N ∈N+ )(∃n ∈N+ )[(n ≥ N )∧(|an −a| < ε )].

We also formulate the negation of Proposition 7.12.Negation of 7.12 A number a ∈ R is not an accumulation point of the sequence (an )n∈N+

if and only if there is an ε > 0 such that the set {n ∈N+ : a −ε < a n < a + ε} is nite.

Proof. (⇒) Assume that a ∈ R is an accumulation point of ( an )n∈N+ . We prove that (2)holds. Since a ∈ R is an accumulation point, there is a subsequence ( am (k) )k∈N+ such thatlimk→∞am (k) = a. Let ε > 0 and N ∈N+ be given. By denition of the limit

(∃K > 0)(∀k ∈N+ )[(k ≥ K ) ⇒ (|am (k) −a| < ε )].

Now choose k ∈N+ such that k ≥ max{K, N } and put n := m(k) (≥ k). Then

[(n ≥ N )∧(|an −a| < ε )].

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(⇐) Now assume that (2) holds. Set ε = 1 and N = 1. By (2),

(∃n ∈N+ )[(n ≥ N )∧(|an −a| < 1).

Put m(1) = n.Set ε = 1 / 2 and N = m(1) + 1. By (2),

(∃n ∈N+ )[(n ≥ N )∧(|an −a| < 1/ 2).

Put m(2) = n.Set ε = 1 / 3 and N = m(2) + 1. By (2),

(∃n ∈N+ )[(n ≥ N )∧(|an −a| < 1/ 3).

Put m(3) = n.Continuing in this way we construct a subsequence ( am (k) )k∈N+ such that

(∀k ∈N+ )( |am (k) −a| < 1/k ).

In particular, lim k→∞am (k) = a, and a is an accumulation point of ( an )n∈N+ .

Example 7.13 The set of accumulation points of the sequence ( an )n∈N+ dened by

an = ( −1)n + 1n

comprises

{−1, 1

}.

Proof. Asa2k−1 → −1 and a2k → 1 as k → ∞,

−1 and 1 are accumulation points of ( an )n∈N+ .

Claim: Each b ∈R \ {−1, 1} is not an accumulation point of ( an )n∈N+ .That is,

(∃ε > 0)(∃N ∈N+ )(∀n ∈N+ ) [(n ≥ N ) ⇒ (|an −b| > ε |] .

Put ε = (1 / 2)min{|1 −b|, |1 + b|}. Then

|an

−b

| =

|(

−1)n + 1 /n

−b

|≥ |(−1)n −b| −1/n

≥ min{|1 −b|, |1 + b|} −1/n= 2 ε −1/n

using the fact that

|x −y| ≥ ||x| − |y|| for any x, y ∈R .

Choose N ∈N+ such that N > 1/ by the AP. Then for any n ≥ N ,

|an −b| ≥ 2ε −1/n ≥ 2ε −1/N > 2ε −ε = ε.

So b is not an accumulation point.

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8 THE BOLZANO-WEIERSTRASS THEOREM

8 The Bolzano-Weierstrass theorem

Theorem 8.1 (Bolzano-Weierstrass theorem) Any bounded sequence has a convergent

subsequence.

Proof. Let (an )n∈N+ be a bounded sequence; say,

|an | ≤ M for all n ∈N+ ,

where 0 ≤ M < ∞.

Step 1 Set [b1, c1] := [−M, M ] c1 −b1 = 2M Step 2 Divide [b1, c1] into two equal intervals.

At least one interval contains innitely many elements of ( an )n∈N+ . [b2, c2] ⊂ [b1, c1]Denote this by [ b2, c2]. c2

−b2 = M

Step 3 Divide [b2, c2] into two equal intervals.At least one interval contains innitely many elements of ( an )n∈N+ . [b3, c3] ⊆ [b2, c2]Denote this by [ b3, c3]. c3 −b3 = 2−1 M . . . . . . . . .

Step k Divide [bk−1, ck−1] into two equal intervals.At least one interval contains innitely many elements of ( an )n∈N+ . [bk , ck ] ⊆ [bk−1, ck−1]Denote this by [ bk , ck ]. ck −bk = 2 2−k M . . . . . . . . .

In summary, we have constructed a sequence of intervals ([ bk , ck])k

∈

N+ such that:

(i) (∀k ∈N+ )([bk+1 , ck+1 ] ⊂ [bk , ck ]);

(ii) limk→∞(ck −bk ) = lim k→∞22−k M = 0.

By the nested intervals theorem (Theorem A 3.2.3), there exists a unique point α ∈ ∩k∈N+ [bk , ck ];moreover,

(3) limk→∞

bk = limk→∞

ck = α.

We now construct a subsequence ( am (k) )k∈N+ of the sequence ( an )n∈N+ .

k = 1: Set am (1) = a1.k = 2: Choose a j ∈ [b2, c2] with j > m (1)

(possible, as [b2, c2] contains ∞-ly many elements of ( an )n∈N+ ).Set am (2) = a j .

k = 3: Choose a j ∈ [b3, c3] with j > m (2)(possible, as [b3, c3] contains ∞-ly many elements of ( an )n∈N+ ).Set am (3) = a j .. . . . . .

Note that(

∀

k

∈N+ )[bk

≤ am (k)

≤ ck ].

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8 THE BOLZANO-WEIERSTRASS THEOREM

Since limk→∞bk = lim k→∞ck = α , by the sandwich rule, we conclude that

limk

→∞

am (k) = α.

Thus ( am (k) )k∈N+ is a convergent subsequence of ( an )n∈N+ .

Remark 8.2 The following examples show that the Bolzano-Weierstrass theorem does nothold for unbounded sequences:(a) the sequence an = n diverges to + ∞ and has no convergent subsequences;(b) the sequence an = (−n)n diverges and has no convergent subsequences;(c) the sequence an = n + ( −1)n n diverges but has a convergent subsequence a2k+1 = 0.

Example 8.3 Consider the sequence ( an )n∈N+ with an = cos( n). This sequence is bounded,as |cos(n)| ≤ 1 for all n ∈ N+ . The Bolzano-Weierstrass theorem guarantees that ( an )n∈N+

contains a convergent subsequence. It is by no means straightforward, however, to producesuch a subsequence.

Exercise 8.4 Prove that any unbounded sequence has a subsequence that diverges to + ∞or to −∞.

Hint. Imitate the proof of the Bolzano-Weierstrass theorem by selecting a sequence of un-bounded intervals of the form ( −∞, n ] or [n, + ∞) that contain innitely many members of the sequence.

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9 LIMIT SUPERIOR AND LIMIT INFERIOR

9 Limit superior and limit inferior

Denition 9.1 Let (an )n∈N+ be a bounded sequence. Let A denote the set of accumulation

points of ( an )n∈N+ . The limit superior limsup n→∞an of (an )n∈

N+ is dened by

lim supn→∞

an := sup A.

The limit inferior liminf n→∞an of (an )n∈N+ is dened by

lim inf n→∞

an := inf A.

Remark 9.2 The limit superior and the limit inferior of a bounded sequence always exist.Indeed, let A be the set of accumulation points of a bounded sequence ( an )n∈N+ . By theBolzano-Weierstrass theorem the set Ais nonempty, and by Exercise 7.11 the set Ais bounded.

Therefore from the Completeness Axiom and Theorem A 2.4.1 we conclude that both sup Aand inf A exist.

Example 9.3 a) Let an = 1n . This sequence converges to zero. Hence the set of all accumu-

lation points is A = {0}. Therefore

lim supn→∞

an = liminf n→∞

an = 0 .

b) Let an = (−1)n . The set of all accumulation points of this sequence is A = {−1, 1}. Hence

lim supn

→∞

an = 1 , liminf n→∞

an = −1.

c) Let an = (−1)n + 1n . The set of all accumulation points of this sequence is A = {−1, 1}(see Example 7.13). Hence

lim supn→∞

an = 1 , liminf n→∞

an = −1.

d) Let an = n. This sequence is unbounded. The limit superior and the limit inferior isundened for an unbounded sequence.

Exercise 9.4 Let (an )n∈N+ be a convergent sequence and lim n→∞an = a. Prove that

lim supn→∞an = liminf n→∞an = a.

Hint. Use Exercise 7.10.

Theorem 9.5 Let (an )n∈N+ be a bounded sequence and set

α := lim supn→∞

an and β := lim inf n→∞

an .

Then α and β are accumulation points of (an )n∈N+ .

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Proof. Let A denote the set of all accumulation points of ( an )n∈N+ ; then α = sup A. Weprove that α ∈ A. Fix ε > 0 and N ∈N+ . Since α = sup A,

(∃a ∈ A)(a > α − ε2).

As a ∈ A, using Proposition 7.12,

(∃k ∈N+ )[(k ≥ N )∧(|ak −a| < ε2

)].

By the triangle inequality,

|α −ak| = |(α −a) + ( a −ak )| ≤ (α −a)

<ε/ 2

+ |a −ak| <ε/ 2

< ε.

Proposition 7.12 now implies that α is an accumulation point of ( an

)n∈

N+ .

Characterisation of the limit superior and limit inferior

Proposition 9.6 Let (an )n∈N+ be a bounded sequence. Then α = lim sup n→∞an if and only if for any ε > 0

(i) the set {n ∈N+ : an > α + ε} is nite, and

(ii) the set {n ∈N+ : an > α −ε} is innite.

Likewise, β = liminf n→∞

an if and only if for any ε > 0

(iii) the set {n ∈N+ : an < β −ε} is nite, and

(iv) the set {n ∈N+ : an < β + ε} is innite.

Proof. We show equivalence with (i) and (ii) .

(⇒) Let A be the set of accumulation points of ( an )n∈N+ so that α = limsup n→∞an = sup A.Fix ε > 0. Now α ∈ A by Theorem 9.5. So there exists a subsequence ( am (k))k∈N+ such thatlimk→∞am (k) = α . By denition of the limit this means that

(∃K ∈N+ )(∀k ∈N+ )[(k > K ) ⇒ (|am (k) −α | < ε ).

In particular,(∀k > K )(am (k) > α −ε).

So the set {n ∈N+ : an > α −ε} is innite and condition (ii) holds.

Assume, for a contradiction, that (i) does not hold. This means that there exists ε > 0 and asubsequence ( am (k))k∈N+ such that

(∀k ∈N+ )(am (k) > α + ε).

Since (am (k))k∈N+ is bounded, by the Bolzano-Weierstrass theorem and Theorem 7.12 weconclude that ( am (k) )k

∈

N+ has an accumulation point γ

≥ α + ε. This contradicts the denition

of α .

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(⇐) Now let α ∈ R be such that for any ε > 0 properties ( i) and ( ii ) hold. We prove thatα = limsup n→∞an . First, we show that α ∈ A. Indeed, from ( i) and ( ii ), it follows that forany ε > 0 the set {n ∈ N+ : α −ε < a n < α + ε} is innite. By Theorem 7.12 we conclude

that α ∈ A.We now prove that α = sup A. Assume for a contradiction that there exists γ ∈ A such thatγ > α . Set ε := (1 / 2)(γ −α ) > 0. As γ ∈ A, by Proposition 7.12, the set

{n ∈N+ : γ −ε < a n < γ + ε}is innite. But

γ −ε = γ −(1/ 2)(γ −α) = α + (1 / 2)(γ −α) = α + ε.

This contradicts ( i). We conclude that α = sup A.

Theorem 9.7 (Formulae for limit superior and limit inferior) Let (an )n

∈

N+ be a bounded sequence. Then

(4) lim supn→∞

an = limn→∞

supk≥n

ak ,

(5) lim inf n→∞

an = limn→∞

inf k≥n

ak .

Remark 9.8 For each n ∈N+ , put cn := sup k≥n ak . Then the sequence ( cn )n∈N+ is decreas-ing. In fact, for any n ∈N+ ,

cn = sup An where An := {ak : k ≥ n}.

Now, An +1 ⊆ An . Therefore,

cn +1 = sup An+1 ≤ sup An = cn .

The sequence ( cn )n∈N+ is also bounded below. For, as ( an )n∈N+ is bounded,

−M ≤ an (≤ M ) for all n ∈N+

for some 0 ≤ M < ∞. So

cn = sup

{ak : k

≥ n

} ≥ an

≥ −M for any n

∈N+ .

Therefore, by Theorem A 3.2.2, the limit lim n→∞cn = lim n→∞sup k≥n ak exists in R .

Likewise, the sequence ( bn )n∈N+ with bn := inf k≥n ak is increasing and bounded above. ByTheorem A 3.2.1, the limit lim n→∞bn = lim n→∞(inf k≥n ak ) exists in R .

Proof. Put α := lim n→∞supk≥n ak . This means that

(6) (∀ε > 0)(∃N ∈N+ )(∀n ∈N+ )[(n ≥ N ) ⇒ (|α −supk≥n

ak|) < ε )],

or, equivalently,

(7) (∀ε > 0)(∃N ∈N+

)(∀n ∈N+

)[(n ≥ N ) ⇒ (α −ε < supk≥n ak < α + ε)].

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11 UNIFORMLY CONTINUOUS FUNCTIONS

Since (an )n∈N+ is bounded, by the Bolzano-Weierstrass theorem, ( an )n∈N+ has a convergentsubsequence ( am (k))k∈N+ with limit a, say. We prove that ( an )n∈N+ also converges to a. Fixε > 0. Then

(∃N ∈N+ )(∀m, n ∈N+ )[(m ≥ N )∧(n ≥ N ) ⇒ (|am −an | < ε2

)],

and(∃K ∈N+ )(∀k ∈N+ )[(k ≥ K ) ⇒ (|am (k) −a| <

ε2

)].

Choose l ∈N+ such that m(l) ≥ l ≥ max{N, K }. Let N+ n ≥ N . Then

|an −a| ≤ |an −am (l) | <ε/ 2

+ |am (l) −a| <ε/ 2

< ε.

In short, lim n

→∞an = a.

Remark 10.6 Combining Theorems 10.4 and 10.5 we see that

a sequence (an )n∈N+ converges if and only if (an )n∈N+ is a Cauchy sequence.

This convergence criterion is intrinsic in the sense that it characterises convergence in termsof a property of the elements of the sequence.

Remark 10.7 Theorem 10.5 may be useful when we need to show that a sequence diverges.Indeed, combining Theorems 10.4 and 10.5 we see that

a sequence (an )n∈N+ diverges if and only if (an )n∈

N+ is not a Cauchy sequence.

Example 10.8 Prove that the sequence an = (−1)n diverges.

Solution. Let N ∈N+ . Then for N+ m ≥ N we have

|am −am +1 | = 2 .

So, if we x ε := 1 (say), and choose n := m + 1, then

|am −an | > ε.

So (an )n∈N+ is not a Cauchy sequence and therefore ( an )n∈

N+ diverges.

11 Uniformly continuous functions

Denition 11.1 Let A ⊆ R and f : A → R a function. Then f is said to be uniformlycontinuous on A if

(8) (∀ε > 0) (∃δ > 0) (∀x, y ∈ A)[(|x −y| < δ ) ⇒ (|f (x) −f (y)| < ε )].

In the above denition, δ = δ (ε) depends on ε but not on x, nor y.

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Lemma 11.2 Let f : A → R be a uniformly continuous function on A ⊆ R . Then f is continuous on A.

Proof. We need to prove that f is continuous at every point in A. Fix a ∈ A. From (8),(∀ε > 0)(∃δ > 0)(∀x, y ∈ A)[(|x −y| < δ ) ⇒ (|f (x) −f (y)| < ε )].

In particular,

(∀ε > 0)(∃δ > 0)(∀y ∈ A)[(|x −a| < δ ) ⇒ (|f (x) −f (a)| < ε )].

This means that f is continuous at a. As this argument works for any a ∈ A, f is continuouson A.

Example 11.3 Let

−∞ < a < b < +

∞. Then the function f (x) = x is uniformly continuous

on [a, b].

Solution. Let ε > 0. Choose δ = ε. Then for any x, y ∈ [a, b],

(|x −y| < δ ) ⇒ (|f (x) −f (y)| = |x −y| < δ = ε).

This means that f is uniformly continuous on [0 , 1].

Example 11.4 Let −∞ < a < b < + ∞. Then the function f (x) = x2 is uniformly continuouson [a, b].

Solution. Set c := max

{|a

|,

|b

|}. Note that

|x + y| ≤ |x|+ |y| ≤ 2 c

for any x, y ∈ [a, b]. Hence,

|x2 −y2| = |x + y||x −y| ≤ 2 c |x −y|for any x, y ∈ [a, b]. Let ε > 0. Choose δ = ε/ 2 c. Then, for any x, y ∈ [a, b],

(|x −y| < δ ) ⇒ (|f (x) −f (y)| = |x2 −y2| = |x + y|

≤2c

|x −y|

<δ

< 2 c δ = ε).

This means that f is uniformly continuous on [ a, b].

From (11.1), the function f : A →R is not uniformly continuous on A if

(9) (∃ε > 0)(∀δ > 0)(∃x, y ∈ A)[(|x −y| < δ )∧(|f (x) −f (y)| ≥ ε)].

Alternatively,

Negation of Denition 11.1. A function f : A →R is not uniformly continuous on A if

(10) (∃ε > 0)(∀n ∈N+ )(∃xn , yn ∈ A)[(|xn −yn | < 1/n )∧(|f (xn ) −f (yn )| ≥ ε)].

Exercise 11.5 Prove that (10) is equivalent to (9).

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Example 11.6 The function f (x) = 1 /x is continuous on (0 , 1) but is not uniformly contin-uous on (0 , 1).

Solution. The fact that f (x) = 1 /x is continuous on (0 , 1) follows from Theorem A 4.2.1 .We prove that f is not uniformly continuous on (0 , 1).

Fix ε = 1. For each n ∈N+ , choose xn = 1 / 2 n and yn = 1 / 4 n. Then, for any n ∈N+ ,

|xn −yn | = 1 / 4 n < 1/n,

and

|f (xn ) −f (yn )| =1

xn − 1yn

= |2n −4n| = 2n > ε = 1 .

By (10), f is not uniformly continuous on (0 , 1).

Example 11.7 The function f (x) = x2 is not uniformly continuous on (0 , ∞).

Solution. Fix ε = 1. Then for each n ∈ N+ , choose xn = n and yn = n + 12n . Then

|xn −yn | = 1 / 2 n < 1/n and

|f (xn ) −f (yn )| = n2 − n + 12n

2

= 1 + 14n2 > 1 = ε.

By (10), f is not uniformly continuous on (0 , ∞).

The following is a fundamental theorem in Analysis.

Theorem 11.8 (Cantor’s theorem on uniform continuity) Let −∞ < a < b < + ∞.Let f : [a, b] →R be a continuous function on [a, b]. Then f is uniformly continuous on [a, b].

Proof. Assume for a contradiction that f is not uniformly continuous on [ a, b]. Then, accordingto (10),

(11) (∃ε > 0)(∀n ∈N+ )(∃xn , yn ∈ [a, b])[(|xn −yn | < 1/n )∧(|f (xn ) −f (yn )| ≥ ε)].

Observe that the sequence ( xn )n∈N+ ⊆ [a, b] is bounded. By the Bolzano-Weierstrass theorem,we may extract a convergent subsequence ( xm (k) )k∈N+ with limit x0 (say) in [a, b]. Now,

|x

m (k) −y

m (k) | < 1/m (k)

→ 0 as k

→ ∞.

Also, by the triangle inequality,

|ym (k) −x0| ≤ |ym (k) −xm (k)| →0

+ |xm (k) −x0| →0

→ 0 as k → ∞.

In short,ym (k) → x0 as k → ∞.

Now, by hypothesis, f is continuous at x0. Consequently,

f (xm (k))

→ f (x0) and f (ym (k) )

→ f (x0) as k

→ ∞.

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Moreover, by the triangle inequality,

|f (xm (k) ) −f (ym (k))| ≤ |f (xm (k) ) −f (x0)|

→0

+ |f (x0) −f (ym (k) )|

→0→ 0 as k → ∞.

On the other hand, (11) tells us that

|f (xm (k) ) −f (ym (k) )| ≥ ε

for each k ∈ N + . In this way, we arrive at a contradiction. We conclude that f is uniformlycontinuous on [ a, b].

Example 11.9 The function f (x) = √ x is uniformly continuous on [0 , 1].

Solution. f is continuous on [0 , 1]. Now use Theorem 11.8.

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12 POINTWISE AND UNIFORM CONVERGENCE

12 Pointwise and uniform convergence

Denition 12.1 Let A ⊆ R . Let f and ( f n )n∈N+ be R-valued functions on A. We say that

(f n )n∈N+ converges pointwise to f on A if

(12) (∀x ∈ A)(∀ε > 0)(∃N ∈N+ )(∀n ∈N+ )[(n ≥ N ) ⇒ (|f n (x) −f (x)| < ε )];

in other words,(∀x ∈ A)[ lim

n→∞f n (x) = f (x)].

Remark 12.2 In this denition, N = N (ε, x ) depends on both ε and x.

Denition 12.3 Let A ⊆ R . Let f and ( f n )n∈N+ be R-valued functions on A. We say that(f n )n∈N+ converges uniformly to f on A if

(13) (∀ε > 0)(∃N ∈N+

)(∀x ∈ A)(∀n ∈N+

)[(n ≥ N ) ⇒ (|f n (x) −f (x)| < ε )].Remark 12.4 In this denition, N depends on ε only, and not on x.

Remark 12.5 If (f n )n∈N+ converges uniformly to f on A then ( f n )n∈N+ converges pointwiseto f on A. But the converse is not true in general (see Example 12.6 below).

Negation of Denition 12.3 A sequence of functions ( f n )n∈N+ does not converge uniformlyto f on A if

(14) (∃ε > 0)(∀N ∈N+ )(∃x ∈ A)(∃n ∈N+ )[(n ≥ N )∧(|f n (x) −f (x)| ≥ ε)].

Example 12.6 Letf n : [0, 1] →R ; x → xn (n ∈N+ ).

Then ( f n )n∈N+ converges pointwise on [0 , 1] to f : [0, 1] →R given by

f (x) = 0 if x ∈ [0, 1),1 if x = 1 .

Note that f is discontinuous. Moreover, for any a ∈ (0, 1), (f n )n∈N+ converges uniformly to f on [0, a ]. But ( f n )n∈N+ does not converge uniformly to f on [0, 1].

Solution. First, note that

limn→∞

xn = 0 if x ∈ [0, 1),

1 if x = 1 .This means that ( f n )n∈N+ converges pointwise to f on [0, 1] .

Let a ∈ (0, 1). We now show that ( f n )n∈N+ converges uniformly to f on [0, a ]. Fix ε > 0.Choose N ∈N+ such that aN < ε . Then, for any x ∈ [0, a ] and n ≥ N we have

|f n (x) −f (x)| = |xn −0| ≤ an ≤ aN < ε.

This means that the sequence ( f n )n∈N+ converges uniformly to f on [0, a].

Lastly, we show that ( f n )n∈N+ does not converge uniformly to f on [0, 1]. According to (14),we need to verify that

(∃ε > 0)(∀N ∈N+ )(∃x ∈ [0, 1])(∃n ∈N+ )[(n ≥ N )∧(|f n (x) −f (x)| ≥ ε)].

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Let ε = 1 / 2. Fix N ∈N+ . Setx := 2−1/N

and N+ n := N . Note that x

∈ [0, 1]. Then n

≥ N and

|f n (x) −f (x)| = |xn −0| = 1 / 2 = ε ≥ ε.

Thus ( f n )n∈N+ does not converges uniformly to f on [0, 1] .

Theorem 12.7 (Weierstrass’s theorem on uniform convergence) Let A ⊆ R . Let (f n )n∈N+ be a sequence of continuous R-valued functions on A. Let f be an R-valued function on A. Assume that (f n )n∈N+ converges uniformly to f on A. Then f is continuous on A.

Proof. Fix a ∈ A. We show that f is continuous at a. Let ε > 0. From the denition of uniform convergence, there exists N ∈N+ such that

(15) (∀x ∈ A)(∀n ∈N+ )[(n ≥ N ) ⇒ (|f n (x) −f (x)| < ε/ 3)].

Now, f N is continuous at a . So there exists δ > 0 with the property that

(16) (∀x ∈ A)[(|x −a| < δ ) ⇒ (|f N (x) −f N (a)| < ε/ 3)].

Suppose that x ∈ A with |x −a| < δ . By the triangle inequality,

|f (x) −f (a)| ≤ |f (x) −f N (x)|

<ε/ 3 by (15)

+ |f N (x) −f N (a)|

<ε/ 3 by (16)

+ |f N (a) −f (a)|

<ε/ 3 by (15)

< ε.

This shows that f is continuous at a .

Example 12.8 (Example 12.6 revisited) Let

f n : [0, 1] →R ; x → xn (n ∈N+ ).

Then ( f n )n∈N+ does not converge uniformly to f on [0, 1] (with f as before).

Solution. (f n )n∈N+ constitutes a sequence of continuous R -valued functions on [0 , 1]. Assumethat ( f n )n∈N+ converges uniformly to f on [0, 1]. By Theorem 12.7, the limit function f mustbe continuous. But f is not continuous. It follows that ( f n )n∈N+ cannot converge uniformly

to f on [0, 1].

Example 12.9 Let

f n : [0, + ∞) →R ; x → 2nx1 + n2x2 (n ∈N+ ).

Then ( f n )n∈N+ converges pointwise to 0 on [0 , + ∞) but does not converge uniformly to 0there.

Solution. If x = 0 then f n (x) = 0 for each n ∈N+ . For x ∈ (0, ∞),

f n (x) = 2x/n1/n 2 + x2 (n ∈N

+).

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Therefore, f n (x) → 0 as n → ∞. In sum, ( f n )n∈N+ converges pointwise to 0 on [0 , + ∞).

Now,

f n (x) = 2n (1

−n2x2)

(1 + n2x2)2 (x ∈ [0, ∞)) ( n ∈N+ ).

This indicates that f n has a maximum at x = 1 /n . In fact,

f n (1/n ) = 1 ( n ∈N+ ).

This means that the convergence cannot be uniform. For, in (14), choose

ε := 1 / 2, x := 1 /N and n := N.

Series of functions. Let A ⊆ R and ( f n )n∈N+ a sequence of R-valued functions on A.Consider the partial sum

sn (x) :=n

k=0

f k (x) (x ∈ A) (n ∈N+ ).

We say that the series of functions∞

k=0

f k(x)

is pointwise convergent on A if (sn )n

∈

N+ converges pointwise on A to a function s : A

→R .

The function s is called the sum of the series. The series is said to be uniformly convergentif the sequence of partial sums ( sn )n∈N+ converges uniformly on A.

Example 12.10 The geometric series

∞

k=0

xk

converges pointwise tos : (−1, 1) →R ; x →

11

−x

on (−1, 1). For any r ∈ (0, 1), the series converges uniformly to s on [−r, r ].

Example 12.11 The exponential series

∞

k=0

xk

k!

converges pointwise on R . Its sum s is called the exponential function and denoted byexp( x) or ex . For any r > 0, the series converges uniformly on [ −r, r ].

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13 DEFINITION OF THE INTEGRAL

Part IV

The Riemann integral

Recommended texts:

1. J. Howie “Real Analysis”, Springer–Verlag, 2003.2. S. Krantz “Real Analysis and Foundations”, Second Edition. Chapman and Hall/CRCPress, 2005.3. G. Wanner “Analysis by its History”, Springer–Verlag, 1996.

13 Denition of the integral

Denition 13.1 Let −∞ < a < b < ∞. A partition P of [a, b] is a nite collection of points

P = {x0, x1, . . . , x n} ⊆ [a, b]

such thata = x0 < x 1 < · · · < x n = b.

The length of the longest subinterval in P is called the norm of P and denoted

P . Thus,

P = max1≤i≤n

(x i −x i−1).

Example 13.2 Let n ∈N+ . Dene a partition P n of [0, 1] via

P n := {0, 1/n, 2/n , . . . , (n −1)/n, 1}.

P n is said to be a uniform partition of [0, 1]. This is because all subintervals of the partitionhave equal length. Note that P n = 1 /n .

Denition 13.3 Let f : [a, b] → R be a bounded function. Let P = {x0, . . . , x n} be apartition of [ a, b]. Set

m i = inf {f (x) : xi−1 ≤ x ≤ xi}, M i = sup {f (x) : xi−1 ≤ x ≤ xi}.

The lower sum L(f, P ) of f with respect to the partition P is dened by

L(f, P ) :=n

i=1

m i(x i −x i−1).

The upper sum U (f, P ) of f with respect to the partition P is dened by

U (f, P ) :=n

i=1

M i (x i

−x i

−1).

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Remark 13.4 Putm := inf

x∈[a,b ]f (x), M := sup

x∈[a,b]f (x).

Then m ≤ m i ≤ M i ≤ M for each i ∈ {1, 2, . . . , n }. Therefore,

m (b−a) =n

i=1

m(x i −x i−1) ≤n

i=1

m i (x i −x i−1)

≤n

i=1

M i (x i −x i−1) ≤n

i=1

M (x i −x i−1) = M (b−a).

This implies that for any partition P of [a, b] the following inequality holds:

m (b−a) ≤ L(f, P ) ≤ U (f, P ) ≤ M (b−a).

In other words, the sets

{L(f, P ) : P is a partition of [ a, b]},

{U (f, P ) : P is a partition of [ a, b]}in R are bounded (and non-empty). This means that their inmum and supremum exist.

Denition 13.5 The upper integral of f over [a, b] is dened by

J := inf P

U (f, P ),

and the lower integral of f over [a, b] is dened by j := sup

P L(f, P ),

where the inmum and supremum are taken over all partitions P of the interval [ a, b].

Denition 13.6 A function f : [a, b] →R is said to be Riemann integrable over [a, b] if

J = j.

In this case, the common value is called the Riemann integral of f over [a, b] and denotedby

b

a f (x) dx. Thus,

b

af (x) dx = J = j.

Example 13.7 Let f : [a, b] →R ; x → c be a constant function ( c ∈R ). For any partition P of [a, b] we have

L(f, P ) = U (f, P ) = c (b−a).

Thus, J = j = c (b−a). We conclude that f is Riemann integrable over [ a, b] and

b

af (x) dx = c (b−a).

Remark 13.8 If j = J we say that f is not Riemann integrable.

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Example 13.9 (Example of a function which is not Riemann integrable) The Dirich-let function D : [0, 1] →R is dened by

D (x) = 1 if x is rational ,0 if x is irrational .

It is everywhere discontinuous. For any partition P of [0, 1], we have

L(D, P ) = 0 and U (D, P ) = 1 .

Thus j = 0 and J = 1. Hence the Dirichlet function D is not Riemann integrable.

Example 13.10 The function f (x) = x2 is Riemann integrable over [0 , 1] and

1

0f (x)dx = 1 / 3.

Solution. Let P n be the uniform partition of [0 , 1],

0 < 1n

< 2n

< · · · < n −1

n < 1.

Now f (x) = x2 is a strictly increasing function on [0 , 1]. So,

m i = inf [x i − 1 ,x i ]

f (x) = f (x i−1) = (i −1)2

n2

andM i = sup

[x i − 1 ,x i ]f (x) = f (x i) = i2

n2 .

Therefore,

L(f, P n ) =n

i=1

(i −1)2

n21n

= 1n3

n−1

i=0i2

and

U (f, P n ) =n

i=1

i2

n21n

= 1n3

n

i=1

i2.

Recall thatn

k=1

k2 = n(n + 1)(2 n + 1)6

.

(This may be proved by induction.) We then calculate

L(f, P n ) = n(n −1)(2n −1)

6n3 = 13 −

12n

+ 16n2 ,

U (f, P n ) = n(n + 1)(2 n + 1)

6n3 = 13

+ 12n

+ 16n2 .

Therefore

(17) limn→∞L(f, P n ) = 13, limn→∞U (f, P n ) = 13

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and

(18) limn→∞

[U (f, P n ) −L(f, P n ))] = 0 .

Note. The above computations take into account only uniform partitions P n of [0, 1]. Thedenition of the Riemann integral, however, involves all partitions of [0 , 1]. But, by Corollary14.2 below, (18) and (17) together imply that the function f (x) = x2 is Riemann integrableover [0, 1] with

j = limn→∞

L(f, P n ) = limn→∞

U (f, P n ) = J,

and

1

0x2 dx = j = J =

13

.

This completes the argument.

Properties of upper and lower sums.

Proposition 13.11 Let f : [a, b] →R be a bounded function and P, Q partitions of [a, b] such that P ⊆ Q. Then

L(f, P ) ≤ L(f, Q ), U (f, P ) ≥ U (f, Q ).

Remark 13.12 If P ⊆ Q, we say that Q is a renement of P .

Proof. First, let P be a partition formed from P by adding one extra point, say c ∈ [xk−1, xk ].Let

mk = inf x∈[x k − 1 ,c] f (x), mk = inf x∈[c,x k ] f (x).Then mk ≥ mk , m k ≥ mk , and

L(f, P ) =k−1

i=1

m i(x i −x i−1) + mk (c −xk−1) + mk (xk −c) +n

i= k+1

m i (x i −x i−1)

≥k−1

i=1

m i(x i −x i−1) + mk (xk −xk−1) +n

i= k+1

m i (x i −x i−1) = L(f, P ).

Similarly one obtainsU (f, P )

≤ U (f, P ).

To prove the assertion, add a nite number of points to P to form Q, and apply the aboveresult repeatedly.

Proposition 13.13 Let f : [a, b] → R be a bounded function, and let P and Q be arbitrary partitions of [a, b]. Then

L(f, P ) ≤ U (f, Q ).

Proof. Consider the partition P ∪Q. By Proposition 13.11 we obtain

L(f, P ) ≤ L(f, P ∪Q) ≤ U (f, P ∪Q) ≤ U (f, Q ).

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Proposition 13.14 Let f : [a, b] →R be a bounded function and let J and j be the upper and lower integrals of f over [a, b]. Then

j ≤ J.

Proof. Fix a partition Q. By Proposition 13.13, for any partition P of [a, b] we have

L(f, P ) ≤ U (f, Q ).

Therefore j = sup

P L(f, P ) ≤ U (f, Q ).

In other words, for any partition P ,

j ≤ U (f, P ).

Hence j ≤ inf P

U (f, P ) = J.

14 Criterion of integrability

Theorem 14.1 (Criterion of integrability) A bounded function f : [a, b] →R is Riemann integrable if and only if

(19) (

∀

ε > 0) (

∃

a partition P of [a, b]) [U (f, P )

−L(f, P ) < ε ].

Proof. First, assume that f is integrable; that is, J = j . By denition of the supremum, thereexists a partition P 1 such that

L(f, P 1) > j −ε/ 2.

Also, by denition of the inmum, there exists a partition P 2 such that

U (f, P 2) < J + ε/ 2.

Set Q = P 1∪P 2. Then Q is a renement of both P 1 and P 2. By Propositions 13.11 and 13.13we obtain

j

−ε/ 2 < L (f, P 1)

≤ L(f, Q )

≤ U (f, Q )

≤ U (f, P 2) < J + ε/ 2.

Since J = j by assumption, we conclude that

U (f, Q ) −L(f, Q ) < ε.

That is, (19) holds.

Conversely, assume that (19) holds. Fix ε > 0. Let P be a partition such that

U (f, P ) −L(f, P ) < ε.

Note that J ≤ U (f, P ) and j ≥ L(f, P ) by denition of j and J . So

J − j ≤ U (f, P ) −L(f, P ) < ε.

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15 CLASSES OF INTEGRABLE FUNCTIONS

Let P = {x0, . . . , x n} be a partition with norm P < δ . Recall that a continuous functionon a closed bounded interval attains its extrema. Therefore, for each i = 1 , . . . , n ,

M i = supx∈[x i − 1 ,x i ] f (x) = f (yi )

andm i = inf

x∈[x i − 1 ,x i ]f (x) = f (zi )

for some yi , zi ∈ [x i−1, x i ]. So

M i −m i = f (yi ) −f (zi) < η

as |yi −zi | < δ . We then have

U (f, P ) −L(f, P ) =n

i=1(M i −m i)(x i −x i−1) < η

n

i=1(x i −x i−1) = η (b−a) = ε.

By the Criterion of Integrability, we conclude that f is integrable.

There are, however, Riemann integrable functions that are neither monotone nor continuous.

Example 15.3 Dene f : [0, 1] →R by

f (x) = 0 if x ∈Q ,1/q if x ∈Q and x = p/q in lowest terms .

Note that f is continuous on [0 , 1] \Q

and discontinuous on Q

∩[0, 1] and is not monotone.Fix ε > 0. PutS := {x ∈ [0, 1] : f (x) > ε/ 2}.

Then S contains only nitely many points. Say

S = {y1, . . . , ym }.

Choose a partition P = {x0, . . . , x n} of [0, 1] with P < η := ε/ 4m. If the interval [ x i−1, xi ]contains no points in S , then f (x) ≤ ε for all x ∈ [x i−1, xi]. Also, at most 2 m intervals[x i−1, xi] contain points in S . And, in addition, f (x) ≤ 1 for any x ∈ [0, 1]. Therefore,

U (f, P ) =n

i=1M i (x i −x i−1) < ε/ 2 + 2 m η = ε.

It is also the case that L(f, P ) = 0. So f is Riemann integrable and

1

0f (x) dx = 0 .

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16 INEQUALITIES AND THE MEAN-VALUE PROPERTY OF THE INTEGRAL

16 Inequalities and the mean-value property of the integral

The following inequality is useful for estimating integrals.

Theorem 16.1 Let f, g be bounded and integrable on [a, b]. Suppose that

(∀x ∈ [a, b]) (f (x) ≤ g(x)) .

Then

b

af (x) dx ≤ b

ag(x) dx.

Proof. Let P be a partition of [ a, b]. Then

L(f, P )

≤ L(g, P )

≤ b

ag(x) dx.

Taking the supremum over all partitions P of [a, b] yields the result.

Exercise 16.2 Suppose that f is bounded and integrable on [ a, b]. Assume that

(∀x ∈ [a, b]) (m ≤ f (x) ≤ M )

for some m, M ∈R . Then

m(b−a) ≤

b

af (x)dx ≤ M (b−a).

Hint. Apply Theorem 16.1.

Corollary 16.3 (Mean-value property of the Riemann integral) Let f be continuous on [a, b]. Then there exists θ ∈ [a, b] such that

b

af (x)dx = f (θ)(b−a).

Proof. Setm := min

[a,b]f (x), M := max

[a,b]f (x).

By Exercise 16.2,

m ≤ 1b−a b

af (x) dx ≤ M,

Applying the intermediate value theorem (Theorem A 4.3.1 ) to the function f we see thatthere exists θ ∈ [a, b] such that

f (θ) = 1b−a b

af (x)dx.

This completes the proof.

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17 FURTHER PROPERTIES OF THE INTEGRAL

17 Further properties of the integral

In this section we present some elementary properties of integral which are familiar from

Calculus.

Theorem 17.1 Let a < c < b . Let f be integrable on [a, b]. Then f is integrable on [a, c] and on [c, b] and

b

af (x)dx = c

af (x)dx + b

cf (x)dx.

Conversely, if f is integrable on [a, c] and on [c, b] then it is integrable on [a, b].

Proof. Suppose that f is integrable on [ a, b]. Fix ε > 0. Then there exists a partitionP = {x0, . . . , x n} of [a, b] such that

U (f, P ) −L(f, P ) < ε.We can assume that c ∈ P so that c = x j for some j ∈ {0, 1, . . . , n } (otherwise considerthe renement of P adding the point c). Then P 1 = {x0, . . . , x j} is a partition of [ a, c] andP 2 = {x j , . . . , x n} is a partition of [ c, b]. Moreover,

L(f, P ) = L(f, P 1) + L(f, P 2), U (f, P ) = U (f, P 1) + U (f, P 2).

Therefore we have

[U (f, P 1) −L(f, P 1)] + [U (f, P 2) −L(f, P 2)] = U (f, P ) −L(f, P ) < ε.

Since each of the terms on the left hand side is non-negative, each one is less than ε, whichproves that f is integrable on [ a, c] and on [c, b]. Note also that

L(f, P 1) ≤ c

af (x)dx ≤ U (f, P 1), L(f, P 2) ≤ b

cf (x)dx ≤ U (f, P 2),

so that

L(f, P ) ≤ c

af (x)dx + b

cf (x)dx ≤ U (f, P ).

This is true for any partition of [ a, b]. Therefore

b

af (x)dx =

c

af (x)dx +

b

cf (x)dx.

Now assume that f is integrable on [ a, c] and [c, b]. Fix ε > 0. Then there exists a partitionP 1 of [a, c] and a partition P 2 of [c, b] such that

U (f, P 1) −L(f, P 1) < ε/ 2, U (f, P 2) −L(f, P 2) < ε/ 2.

Let P = P 1 ∪P 2. Then

U (f, P ) −L(f, P ) = [U (f, P 1) −L(f, P 1)] + [U (f, P 2) −L(f, P 2)] < ε.

By the Criterion of Integrability we conclude that f is integrable on [ a, b].

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Remark 17.2 The integral ba f (x)dx was dened only for a < b. We add by denition that

a

a

f (x)dx = 0 and

b

a

f (x)dx =

− a

b

f (x)dx if a > b.

With this convention we always have that

b

af (x)dx = c

af (x)dx + b

cf (x)dx.

Theorem 17.3 Let f and g be integrable on [a, b]. Then f + g is also integrable on [a, b] and

b

a[f (x) + g(x)]dx = b

af (x)dx + b

ag(x)dx.

Proof. Let P = {x0, . . . , x n} be a partition of [ a, b]. Let

m i = inf {f (x) : xi−1 ≤ x ≤ xi},m i = inf {g(x) : xi−1 ≤ x ≤ xi},

m i = inf {f (x) + g(x) : xi−1 ≤ x ≤ xi}.

Dene M i , M i , M i similarly. The following inequalities hold

m i ≥ m i + m i , M i ≤ M i + M i .

Therefore we have

L(f, P ) + L(g, P ) ≤ L(f + g, P ), U (f + g, P ) ≤ U (f, P ) + U (g, P ).

Hence for any partition P

L(f, P ) + L(g, P ) ≤ L(f + g, P ) ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ),

or otherwise

U (f + g, P ) −L(f + g, P ) ≤ [U (f, P ) −L(f, P )] + [U (g, P ) −L(g, P )].

Fix ε > 0. Since f and g are integrable there are partitions P 1 and P 2 such that

U (f, P 1) −L(f, P 1) < ε/ 2, U (g, P 2) −L(g, P 2) < ε/ 2.

Thus for the partition P = P 1 ∪P 2 we obtain that

U (f + g, P ) −L(f + g, P ) < ε.

This proves that f + g is integrable on [ a, b]. Moreover,

L(f, P ) + L(g, P ) ≤ L(f + g, P ) ≤ b

a[f (x) + g(x)]dx ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ),

and

L(f, P ) + L(g, P ) ≤ b

af (x)dx + b

ag(x)dx ≤ U (f, P ) + U (g, P ).

Therefore it follows that

b

a[f (x) + g(x)]dx = b

af (x)dx + b

ag(x)dx.

The proof is complete.

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Theorem 17.4 Let f be integrable on [a, b]. Then, for any c ∈ R , cf is also integrable on [a, b] and

b

acf (x)dx = c

b

af (x)dx.

Proof. The proof is left as an exercise. Consider separately two cases: c ≥ 0 and c ≤ 0.

Theorem 17.5 Let f be integrable on [a, b]. Then |f | is integrable on [a, b] and

b

af (x)dx ≤ b

a |f (x)|dx.

Proof. Note that for any interval [ α, β ] we have

(20) sup[α,β ] |f (x)| − inf

[α,β ] |f (x)| ≤ sup[α,β ]

f (x) − inf [α,β ]

f (x).

Indeed,(∀x, y ∈ [α, β ]) f (x) −f (y) ≤ sup

[α,β ]f (x) − inf

[α,β ]f (x) ,

so that(∀x, y ∈ [α, β ]) |f (x)| − |f (y)| ≤ sup

[α,β ]f (x) − inf

[α,β ]f (x) ,

which proves (20) by passing to the supremum in x and y. It follows from (20) that for anypartition of [ a, b]

U (|f |, P ) −L(|f |, P ) ≤ U (f, P ) −L(f, P ),

which proves the integrability of |f | by the Criterion of Integrability. The last assertion followsfrom Theorem 16.1.

Theorem 17.6 Let f : [a, b] →R be an integrable function such that

(∀x ∈ [a, b])(m ≤ f (x) ≤ M ).

Let g : [m, M ] → R be a continuous function. Then the composite function h : [a, b] → Rdened by h(x) = g(f (x)) is integrable.

Proof. Fix ε > 0. Since g is uniformly continuous on [ m, M ], there exists δ > 0 such thatδ < ε and

(∀t, s ∈ [m, M ]) (|t −s| < δ ) ⇒ (|g(t) −g(s)| < ε ) .By integrability of f there exists a partition P = {x0, . . . , x n} of [a, b] such that

(21) U (f, P ) −L(f, P ) < δ 2.

Let m i = inf [x i − 1 ,x i ] f (x), M i = sup [x i − 1 ,x i ] f (x) and m∗i = inf [x i − 1 ,x i ] h(x), M ∗i = sup [x i − 1 ,x i ] h(x).Decompose the set {1, . . . , n } into two subset : ( i ∈ A) ⇔ (M i − m i < δ ) and ( i ∈ B) ⇔(M i −m i ≥ δ ).

For i ∈ A by the choice of δ we have that M ∗i −m∗i < ε .For i ∈ B we have that M ∗i −m∗i ≤ 2K where K = sup t∈[m,M ] |g(t)|. By (21) we have

δ i∈B (x i −x i−1) ≤i∈B (M i −m i)(x i −x i−1) < δ 2

,

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18 INTEGRATION AS THE INVERSE TO DIFFERENTIATION

so that i∈B (x i −x i−1) < δ . Therefore

U (h, P ) −L(h, P ) =i∈A

(M ∗i −m∗i )(x i −x i−1) +i∈B

(M ∗i −m∗i )(x i −x i−1)

< ε (b−a) + 2 Kδ < ε [(b−a) + 2 K ],

which proves the assertion since ε > 0 is arbitrary.

Corollary 17.7 Let f, g be integrable on [a, b]. Then the product fg is integrable on [a, b].

Proof. Since f + g and f −g are integrable on [ a, b], (f + g)2 and ( f −g)2 are integrable on[a, b] by the previous theorem. Therefore

fg = 14

[(f + g)2 −(f −g)2]

is integrable on [ a, b].

18 Integration as the inverse to differentiation

Denition 18.1 Let f : [a, b] → R be a given function. Suppose that F : [a, b] → R iscontinuous and differentiable on ( a, b) with the property that

F (x) = f (x) for all x ∈ (a, b).

Then F is said to be a primitive or antiderivative of f on [a, b].

Remark 18.2 Primitives are not uniquely dened. If F is a primitive of f then so is F + C where C is an arbitrary constant. Also, with C := −F (a), the primitive G(x) := F (x) + C satises G(a) = 0.

Example 18.3 Let n ∈N+ . For each C ∈R , the function F (x) = xn +1

n +1 + C is a primitive of f (x) = xn .

Let f be a bounded integrable function on some interval [ a, b]. We now use the Riemannintegral to construct a primitive of f on [a, b]. Firstly,

Theorem 18.4 Let f be bounded and integrable on [a, b]. Dene F : [a, b] →R by

(22) F (x) := x

af (t) dt (x ∈ [a, b]).

Then F is continuous on [a, b].

Proof. Put M := sup [a,b] |f (x)|. Let x, y ∈ [a, b] with x < y . Making use of Theorem 17.1, weobtain

F (y) −F (x) = y

xf (t) dt ≤ M |x −y|.

It follows from this estimate that F is continuous on [ a, b] (in fact, it is uniformly continuous

there [cf. FTA]).

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18 INTEGRATION AS THE INVERSE TO DIFFERENTIATION

Theorem 18.7 (fundamental theorem of calculus for integrable functions) Let f be bounded and integrable on [a, b] and G a primitive of f on [a, b]. Then

b

af (x) dx = G(b) −G(a).

Proof. Let P = {x0, . . . , x n} be a partition of [ a, b]. Note that G is continuous on [ x i−1, x i ]and differentiable on ( x i−1, x i) by denition of the primitive. By the mean value theorem,there exists t i ∈ (x i−1, x i ) such that

G(x i ) −G(x i−1) = G (t i)(x i −x i−1) = f (t i )(x i −x i−1).

Putm i := inf

[x i − 1 ,x i ]f (x), M i := sup

[x i − 1 ,x i ]f (x).

Thenm i (x i −x i−1) ≤ f (t i ) (x i −x i−1) ≤ M i (x i −x i−1),

in other words,m i (x i −x i−1) ≤ G(x i ) −G(x i−1) ≤ M i (x i −x i−1).

Adding these inequalities for i = 1 , . . . , n we obtain

n

i=1

m i (x i −x i−1) ≤ G(b) −G(a) ≤n

i=1

M i (x i −x i−1).

So, for any partition P , we have

L(f, P ) ≤ G(b) −G(a) ≤ U (f, P ).

As f is integrable, this entails that

G(b) −G(a) = b

af (x) dx.

This material is copyright of the University of Bristol unless explicitly stated otherwise It is

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