Schinazi R.B. From Calculus to Analysis

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x Contents

6.2 Properties of the Integral .......................189

6.3 UniformContinuity .........................201

7 Convergence of Functions ........................207

8 Decimal Representation of Numbers ..................221

9 Countable and Uncountable Sets ....................235

Further Reading ................................245

List of Mathematicians Cited in the Text ...................247

Index ......................................249

Chapter 1

Number Systems

1.1 The Algebra of the Reals

We start with some notation and symbols. We denote the naturals by N =

{1, 2, 3,...}, the integers by Z ={...,−2, −1, 0, 1, 2,...}, the rationals (repre-

sented by fractions of integers) by Q, and the reals (which may be described by

their decimal representation, see Chap. 8)byR. The symbol ∈ means ‘belongs to’.

For instance, 1 ∈N means ‘1 is a natural’. The notation



x ∈R :x



designates the set of reals whose square is 2. The set

{x ∈R :−1 <x≤2}

is the set of real numbers strictly larger than −1 and smaller than or equal to 2.

A shorter notation for such a set is (−1, 2].Theset



:x ∈[−1, 1]



is the set of squares of numbers in [−1, 1]. The notation

{−1, 2, 4}

designates a set with three elements: −1, 2, and 4. The empty set will be denoted

by ∅.

At the beginning of any mathematics (geometry, probability, ...) there is a set

of rules called axioms. These rules are assumed and not proved. Everything else is

proved using these axioms. For instance, all of Euclidean geometry is based on 5

axioms.

The Peano axioms may be used to deﬁne the set of naturals N, the addition op-

eration, and the order relation <. Then one can construct the integers, the rationals,

and ﬁnally the reals. However, this is a rather involved program and is outside the

scope of this text. We will not construct the number systems. The reader may consult

Goodfriend (2005) for a discussion of the Peano axioms and their consequences and

Krantz (1991) for the construction of integers, rationals, and reals, see the references

at the end of the book.

R.B. Schinazi, From Calculus to Analysis,

DOI 10.1007/978-0-8176-8289-7_1, © Springer Science+Business Media, LLC 2012

2 1 Number Systems

We will not recall the familiar properties of the reals regarding addition, multi-

plication, distributivity, and so on. We will however recall the familiar operations

regarding inequalities. First, we give a set of rules.

Ordering the reals

The binary relation < has the following properties.

R1. For x and y in R, there are three possibilities: x = y, x<y,ory<x.

R2. If x<yand y<z, then x<z.

R3. If x<yand z is any real, then x +z<y+z.

R4. If 0 <x and 0 <y, then 0 <xy.

We may use y>xfor x<y. The notation x ≤y means that x<yor x =y.

We now prove some useful consequences.

C1. If x<yand z>0, then xz < yz.

By R3 we have x −x<y−x. Hence, 0 <y−x.ByR4wehave

(y −x)z > 0.

That is,

yz −xz > 0.

Thus, by R3

yz > xz.

This proves C1.

C2. If x<yand z<0, then xz > yz.

The proof is very similar to the proof of C1 and is left as an exercise.

C3. If 0 <x<y, then x

Since x<yand x>0, by C1 xx < xy. Similarly, since x<yand y>0, by C1

xy < yy.ByR2,x

C4. If x>0, then 1/x > 0.

Let x



= 1/x. If we had x



< 0, then by C1 we would have x



x<0x = 0. But



x =1 > 0. Thus, x



> 0.

C5. If x>y>0, then 1/x < 1/y.

By R4 xy > 0, and so by C4

> 0. By C1,

That is, 1/y > 1/x.

1.1 The Algebra of the Reals 3

We now turn to the function absolute value. We deﬁne

|x|=x if x ≥0,

|x|=−x if x ≤0.

The following simple lemma is very useful.

Lemma Let a>0 be a real number. We have |x|≤a if and only if −a ≤x ≤a.

We prove the lemma. It is an ‘If and only if’ statement, which means that we

have two implications to prove.

For the direct implication, assume that |x|≤a. There are two cases. If x ≥ 0,

|x|=x and x ≤ a. Since x ≥ 0, we must have −a ≤ x. Thus, −a ≤ x ≤ a.Onthe

other hand, if x ≤ 0, |x|=−x and −x ≤a, that is, x ≥−a. Since x ≤ 0, we have

x ≤a. In both cases

−a ≤x ≤a, and

the direct implication is proved.

For the converse, assume now that −a ≤x ≤a.If|x|=x, then |x|≤a.If|x|=

−x, since −a ≤ x,wehavea ≥−x and a ≥|x|. In both cases |x|≤a, and the

lemma is proved.

Triangle inequality

For any real numbers a and b,wehave

|a +b|≤|a|+|b|.

For any reals a and b,wehave

−|a|≤a ≤|a| (why?),

−|b|≤b ≤|b|.

By adding the inequalities we get

−



|a|+|b|



≤a +b ≤|a|+|b|.

According to the lemma above, this implies that

|a +b|≤|a|+|b|.

This completes the proof of the triangle inequality.

We now turn to an important algebraic identity. We deﬁne the nth power of a real

a by

and

n+1

a for all n ≥1.

We also set a

=1.

4 1 Number Systems

An algebraic identity

For any natural number n and any real numbers a and b,wehavethat

−b

=(a −b)

n−1



k=0

n−1−k

For n =2 and n =3, these are the well-known identities

−b

=(a −b)



k=0

1−k

=(a −b)(a +b)

and

−b

=(a −b)



k=0

2−k

=(a −b)



+ab +b



For n ≥4, the identity is

−b

=(a −b)



n−1

n−2

b +···+ab

n−2

n−1



We now prove it. We have

(a −b)

n−1



k=0

n−1−k

n−1



k=0

n−1−k

−b

n−1



k=0

n−1−k

Thus,

(a −b)

n−1



k=0

n−1−k

n−1



k=0

n−k

−

n−1



k=0

n−1−k

k+1

The difference between the two sums is

n−1

b +···+a

n−2

+ab

n−1

−



n−1

b +a

n−2

+···+ab

n−1



All the terms cancel except a

−b

. This proves the identity above.

Exercises

1. (a) Describe in words the set



x ∈Q :x

< 3



(b) Describe with mathematical symbols the set of reals whose inverses are

between 1 and 2.

2. Let A ={

:n ∈N}.

(a) Pick an element in A.

(b) Describe in words the set A.

1.1 The Algebra of the Reals 5

3. Prove that if x<yand z<0, then xz > yz.

4. (a) Assume that x<y<0. State an inequality between x

and y

and prove it.

(b) Assume that x<y<0. State an inequality between 1/x and 1/y and prove

it.

5. (a) Show that if 0 <a<1 and x ≥0, then ax ≤x.

(b) Show that if 0 <a<1 and x ≥0, then x/a ≥x.

x +1

≤x.

6. Assume that a>1 and x<0.

(a) Compare ax and x.

(b) Compare x/a and x.

7. Prove that for any real number a,wehave−|a|≤a ≤|a|.

8. When is the triangle inequality |a +b|≤|a|+|b| an equality?

9. (a) Show that

|a|−|b|≤|a −b|.

(b) Show that

|b|−|a|≤|b −a|=|a −b|.



|a|−|b|



≤|a −b|.

10. Assume that |a −b|<. Show that

(a) b −<a<b+.

(b) |

a|< |b|+.

11. Suppose

that a ≤x ≤b and a ≤y ≤b. Show that

|x −y|≤b −a.

12. Assume that −1 ≤x ≤ 2 and 3 ≤y ≤ 4. Find a lower and an upper bound for

x/y.

13. Assume that 0 ≤x<1.

(a) Show that

1 −x

≥1 +x +x

(b) Generalize (a).

14. Show that if a =1 and n is a natural, then

1 +a +a

+···+a

1 −a

n+1

1 −a

15. Let a>0 be a real number. Find d such that the set (a − d,a + d) has only

strictly positive numbers.

6 1 Number Systems

16. (a) Show that for 1/4 ≤x ≤3/4, we have

1 −x

≥

(b) Find a constant c so that for 1/4 ≤x ≤3/4, we have

1 −x

≤cx.

1.2 Natural Numbers and Integers

We ﬁrst need the notion of minimum.

Minimum of a set

A natural number n

is said to be the minimum of a nonempty subset A of the

naturals if the following two conditions hold:

(i) n

belongs to A.

(ii) If n belongs to A, then n ≥n

We denote the minimum of A (if it exists) by min A. We now state three easy

consequences of this deﬁnition.

C1. If a minimum exists, it is unique. Thus, we may talk about the minimum of A.

Assume that A has two minima n

and n

. Since n

is a minimum and n

in A, we have by deﬁnition that n

≥n

. By exchanging n

and n

in the preceding

sentence we get n

≥n

. Thus, n

. That is, there is only one minimum.

C2. If A is a singleton, then it has a minimum, and that minimum is the only element

in A.

Assume that A ={p}. Then p is in A, and if n is in A, then n is necessarily p,

and therefore n ≥p. That is, p satisﬁes the two requirements of the deﬁnition above,

and we get min A =p.

C3. If A ⊂B and they both have a minimum, then min A ≥min B.

This is so because min A ∈A ⊂B, and so min A is in B.ButminB is the smallest

of the elements in B. Thus, min A ≥min B.

With the ﬁve Peano’s axioms one can construct the naturals, deﬁne the addition,

and the order relation <. Here we only state one of the axioms: the well-ordering

principle.

Well-ordering principle

If A is a nonempty subset of naturals, then it has a minimum.

1.2 Natural Numbers and Integers 7

Example 1.1 Let r be a rational such that 0 <r<1. Show that r can be written in

irreducible form. That is, r may be written as the ratio of two naturals that have no

common divisors but 1.

Let

A ={n ∈N :nr ∈N}.

Since r is a rational, there are naturals a and b such that

r =

Therefore, rb = a, and since a is a natural, b is in A. Hence, A is a nonempty subset

of naturals. By the well-ordering principle, A has a minimum m.

Let k = mr. Since m is in A, k is a natural. We now need to show that k and

m are relatively prime. We do a proof by contradiction. Assume that d>1 divides

both k and m. Therefore, there are naturals k



and m



such that

k =k



d and m =m



Since k = mr, we get k



d =m



dr. Thus, k



= m



r. In particular, m



must be in A,

but m



<m, the minimum of A. Hence, we have a contradiction: the naturals k and

m have no common divisor, and r =k/m is irreducible.

Principle of induction

Let P be a subset of N. Assume the following two properties for P :

(i) 1 belongs to P .

(ii) If n belongs to P , then n +1 belongs to P .

Then P is all of N.

We now prove the principle of induction. Let T be the set of all naturals that

are not in P . We want to show that T is empty, so that P = N.Wedoaproofby

contradiction. That is, we assume that T is nonempty, and we use our axioms to

show that this leads to a statement that is not true. This forces to conclude that T

must be empty.

If T is nonempty, by the well-ordering principle, T has a minimum that we de-

note by a. Since 1 belongs to P , a cannot be 1. We must have a>1. The number

a − 1 is a natural (since a>1) and cannot be in T since a is the minimum of T .

Thus, a − 1isinP . By (ii), a − 1 + 1 = a is also in P . That is, a is in P .Buta

is also in T , that is, not in P . We have a contradiction: a is in P and is not in P .

Hence, T is empty, and the principle of induction is proved.

We will now see an application of the induction principle.

Example 1.2 Show that the sum of the ﬁrst n natural numbers is

n(n+1)

8 1 Number Systems

Let P be the set of naturals whose sum of the ﬁrst n integers is

n(n+1)

.More

mathematically, let



i=1

i and a

n(n +1)

Then

P ={n ∈N :S

That is, P is the set of naturals n for which S

.WehaveS

=1 and a

=1, so

1 belongs to P .

Now assume that n belongs to P so that S

. Consider

n+1

=1 +2 +···+n +(n +1) =S

+(n +1) =a

+n +1,

where the last equality follows from the assumption that n is in P . Since

+n +1 =

n(n +1)

+n +1 =

(n +1)(n +2)

n+1

we have

n+1

This proves that n + 1 belongs to P . Thus, by the principle of induction, P = N.

Therefore, for every natural number n, the sum of the ﬁrst n natural numbers is

n(n+1)

Example 1.3 Deﬁne factorials by

n!=1 ×2 ×···×n.

Prove that for every natural n,

n!≥2

n−1

Let

P =



n ∈N :n!≥2

n−1



For n =1, we have n!=1!=1 and 2

n−1

=1. Hence, 1 is in P . Assume now

that n is in P . Then

(n +1)!=n!(n +1) ≥2

n−1

(n +1).

Since n ≥1, we have n +1 ≥2 and

(n +1)!≥2

n−1

(n +1) ≥2

n−1

2 =2

n+1−1

That is, n +1 belongs to P .

We now state an alternative form of the induction principle that is useful.

1.2 Natural Numbers and Integers 9

Alternative form of the principle of induction

Let S(n) be a statement for natural n. Assume that

(i) S(1) is true;

(ii) If S(n) is true, then S(n +1) is also true.

Then S(n) is true for all n in N.

To prove this form of the induction principle, we deﬁne

P =



n ∈N :S(n) is true



That is, P is the set of naturals for which S(n) is true. Since S(1) is true, we have

1inP . Assume that n is in P . By deﬁnition of P , S(n) is true. By property (ii),

we have that S(n +1) is true, and so n + 1isinP . By the principle of induction,

all naturals are in P . That is, S(n) is true for all naturals n.Wehaveprovedthe

alternative form of the principle of induction.

We use the alternative form of the principle of induction to prove the following

inequality.

Bernoulli’s inequality

For any real a ≥0 and any natural n,wehave

(1 +a)

≥1 +na.

We now prove this inequality. Deﬁne the sequences

=(1 +a)

and b

=1 +na.

For n ≥1, let S(n) be the statement: a

≥b

.Taken =1. Then

=(1 +a)

=1 +a =b

That is, S(1) is true. Assume now that S(n) is true. Since a

≥b

,wehave

(1 +a) ≥b

(1 +a),

and therefore,

n+1

=(1 +a)

n+1

=(1 +a)

(1 +a) =a

(1 +a) ≥b

(1 +a).

By the deﬁnition of b

we have

(1 +a) =(1 +na)(1 +a) =1 +a +na +na

≥1 +a +na =1 +(n +1)a =b

n+1