Chapter 3 THE REAL NUMBERS In this chapter we present in some detail many of the important properties of the set

R of real numbers. Our approach will be axiomatic, but not constructive. That

is, we shall not construct the set of real numbers from some simpler set, such as the rational numbers∗ . Instead, we shall assume the existence of , say that it constitutes

R

a “complete ordered field”and postulate the properties that characterize it. For that reason in the next section we present the axioms of an ordered field and then we give the completeness axiom.

3.1 Fields Definition 89 A field is a set F endowed with two operations called addition (+) and multiplication

(·),

which satisfy the following so-called “field axioms” (A),(M), and

(D): (A) Axioms for addition

1. ∀

x, y ∈ F , x + y ∈ F .

2. ∀

x, y ∈ F, x + y = y + x.

3. ∀

x, y, z ∈ F, (x + y ) + z = x + (y + z ) .

4. ∃

0

∈

F : x + 0 = x, ∀x ∈ F.

x ∈ F,

5. ∀

∃

(−x)

∈

F

: (−x) + x = 0

(M) Axioms for multiplication

1. ∀

x, y ∈ F, x · y ∈ F.

2. ∀

x, y ∈ F, x · y = y · x.

3. ∀

x, y, z ∈ F, (x · y ) · z = x · (y · z) .

This is the approach taken by Rudin in the appendix of Chapter 1. The proof is rather tedious and beyond the scope of these notes, and therefore ommited. ∗

The Real Numbers

36

4. ∃ 1 ∈ F

: x · 1 = x, ∀x ∈ F.

5. ∀x ∈ F , x = 0, ∃ 1/x ∈ F : x · (1/x) = 1. (D) The distributive law ∀ x, y, z ∈ F, (x + y) · z = x · z + y · z . Axioms A1 and M1 are known as closure axioms. They can be read as “F is closed under addition and multiplication”. Axioms A2 and M2 are called commutative laws and axioms A3 and M3 are the associative laws. Axiom D is the distributive law, that shows how addition and multiplication relate to each other. When writing multiplication we often omit the raised dot and write xy instead of x y.

Notation

·

Given that the relation “ ” is a complete order on R, we have that R is an ordered field ≤

An ordered field is a field F which is also an ordered set with an order ≤, such that: i) ∀ x, y, z ∈ F, if x ≤ y, then x + z ≤ y + z, ii) ∀ x, y, z ∈ F, if x ≤ y and z ≥ 0 then x z ≤ y z. Definition 90

Exercise 91 Prove that the set of natural numbers is not an ordered field. Exercise 92 Prove the following theorem: Let x,y, and z be real numbers. Then a) If x + z = y + z , then x = y . b) x 0 = 0. c) (−1) x = −x. d) xy = 0 iff x = 0 or y = 0. e) x ≤ y iff −x ≥ −y f) If x ≤ y and z ≤ 0 then x z ≥ y z.

3.2 The completeness axiom (least-upper-bound property) and the real field R

The last property that helps characterize the set of real numbers is that of completeness, or also called the least upped bound. For that purpose, we need some preliminary definitions.

The completeness axiom (least-upper-bound property) and the real field R

37

3.2.1 Upper bounds and Suprema Definition 93 Let S be an ordered set and let E ⊆ S. If there exists β ∈ S such that x
β ∀x ∈ E, then β is called an upper bound for E, and we say that E is bounded above.

lower bound

If there exists of

E,

γ ∈S

such that

γ


x ∀x ∈ E, then γ

and E is bounded below. The set E is said to be

is called a

bounded

if it

is bounded above and below.

Definition 94 If an upper bound b for E is a member of E, then b is called the

maximum (or largest element) of E, and we write b

= max E

Similarly, if a lower bound of E is an element of E, then it is called the

(or least element) of E, denoted by min E.

minimum

A set may have upper or lower bounds, or it may have neither. Note that a set may have many upper and lower bounds, but if it has a maximum or a minimum, those values are unique. Thus we speak of an upper bound and the maximum. Exercise 95 Let S be an ordered set and let E ⊆ S. Prove that if E has a maximum

it is unique.

Exercise 96

Prove that any finite subset E

⊆N

is bounded. Moreover, prove that

E has both a maximum and a minimum element.

Exercise 97

Is

∅⊆N

a bounded set? Does it have a maximum or a minimum?

Since any number larger than an upper bound is also an upper bound, it is useful to identify the smallest (or least) upper bound in the following way

Definition 98

Let

S

be an ordered set and let

E

be a nonempty subset of

S. Suppose

α ∈ S with the following properties: α is an upper bound of E ii) ∀γ ∈ S, γ < α ⇒ γ is not an upper bound of E. Then α is called the least upper bound of E or the supremum of E, and we write α = sup E. Similarly, if there exists β ∈ S such that i) β is a lower bound of E ii) ∀γ ∈ S, γ > β ⇒ γ is not a lower bound of E. Then β is called the greatest lower bound of E or the infimum of E, and we write β = inf E.

there is

i)

As you have probably noticed, the difference between the maximum and the least upper bound is that the least upper bound is not necessarily an element of set

E. The same is true between the minimum and the greatest lower bound.

The Real Numbers

38

3.2.2 The Completeness axiom Now we are in condition to introduce the main distinctive characteristic of

as the completeness axiom

Definition 99 An ordered set S is said to satisfy the

least -upper-bound property

(or the completeness axiom) if for every nonempty subset above has

a least upper bound. That is

sup

R, known

E exists in S.

E ⊆ S that is bounded

While the completeness axiom refers only to sets that are bounded above, the corresponding property for sets bounded below follows readily.

The following

theorem tells you that every ordered set with the least-upper-bound property also has the

greatest-lower-bound property

Theorem 100 Suppose S is an ordered set with the least-upper-bound property. Let B be a nonempty and bounded below subset of S . Let L be the set of all lower bounds of B. Then ∃ α ∈ S : α = sup L = inf B.

Proof. Since B is bounded from below, L is nonempty. S : y
x for all x ∈ B } ⊆ S implies L is bounded from above.

L = {y ∈ Finally, since S has the least upper bound property, then ∃ α ∈ S : α = sup L. Now, if γ < α then γ is not an upper bound of L; hence γ ∈ / B and it follows (by contrapositive) that α
x for all x ∈ B . Thus α ∈ L. Moreover, if α < β then β ∈ / L since α = sup L. We have shown that α ∈ L but β ∈ / L if β > α which means that α is a lower bound of B and β is not if β > α. Hence α = inf B . Moreover

3.2.3 The set of real numbers

Having defined the least upper bound property allows to introduce the set of real numbers

Definition 101 The real number system is an ordered field with a complete order
which has the least-upper-bound property. It is denoted by R. The members of R are called real numbers.

It is known that

R is the only

ordered field with the least-upper-bound property.

x
0 ( x  0 ). We say that a real number x is strictly positive (strictly negative) if x > 0 ( x < 0 ). The set of positive (negative) numbers is denoted by R+ (R− ), and that the set of strictly positive ( strictly negative ) numbers is denoted by R++ (R− − ).We can now We say that a real number

x is positive ( negative

) if

redefine the sets of natural (N) and rational numbers (Q) as subsets of

R

.

Definition 102 Let A be the collection of all subsets A ⊆ R such that 1) 1 ∈ A and

2) ∀x ∈ A, x + 1 ∈ A. Then the set of natural

numbers N is the intersection of all A ∈ A.

The completeness axiom (least-upper-bound property) and the real field R

39

You may wonder why the above was a definition and not a theorem, since we had already defined the set of natural numbers using the Peano axioms. This would have been obtained as a theorem if we had established at some point that N ⊆ R. Some authors , like Rudin, construct the set of real numbers from the set of rational numbers — for which the set of integers is a prerequisite — therefore implying that N ⊆ R. Under this approach, the above definition can be stated as a theorem. Since our approach to R was axiomatic, we cannot guarantee that N ⊆ R, having to provide the above as a definition and not as a theorem. To see why this assumption is necessary, try to prove the above as a theorem: Exercise 103 Assume N ⊆ R. Let A be the collection of all subsets A ⊆ R such that 1) 1 ∈ A and 2) ∀x ∈ A, x + 1 ∈ A. Then the set of natural numbers N is the intersection of all A ∈ A.
(Hint. Call M = A, and prove that a) N ⊆ M — try induction — and b) A ∈A M ⊆ N — N ∈ A —) Definition 104 The set of positive integers is N ∪ {0} and it is denoted by Z+ ,

.

The negative of the elements of Z+ is the set of negative integers, and denoted by Z− The set of integers Z is Z− ∪ Z+ Definition 105 The subset Q of R, called rational numbers is Q = {x ∈ R : ∃ m, n ∈ Z such that n = 0 and x = m/n}. .

.

Having defined set of natural numbers, integers and rational numbers as subsets of the real numbers, helps in finding the differences between them Exercise 106 Prove that 1) N is not a field, 2) Z is not a field either, 3) Q is an ordered field.

Therefore the main difference between Q and R has to lie in the completeness axiom. Before going into this, lets prove another important property of Q Theorem 107 Q is dense with respect to
, that is ∀ p, q ∈ Q, p < q ∃ r ∈ Q, s.t. ;

p < r < q.

Proof. We prove this by construction: Consider r = p+2 q , p < r < q, and r ∈ Q, since Q is a field and addition and multiplication are closed in Q. That is, 1 1 since p, q ∈ Q, p + q ∈ Q, and since 2 ∈ Q, 2 (p + q ) ∈ Q. Being Q dense means that rational numbers are so close to each other that q and every positive distance ε, there is another rational p − ε ≤ s ≤ p + ε. But although Q is dense, it has gaps. To 2 equations x − p = 0 have no solutions in Q when p is a prime

for every rational number number

s

such that

give an example, the number.

40

Theorem 108 Proof.

The Real Numbers

√

Let p be a prime number. Then p is not a rational number.

√p

√p

√p = m , where m and n are integers with no common

We will assume that

is rational and obtain a contradiction. If

n factors. Then n2 p = m2 , so m2 must be a multiple of p. Since p is a prime number, then m must also be a multiple of p. That is m = k p, for some integer k. But then n2p = m2 = k 2 p2, implying n2 = k 2 p. This implies that n is also a multiple of p, contradicting the fact that m and n had no common factors.

is rational, then we can write

This is equivalent to saying that the sets S =

{x ∈ Q : x2 − p ≤ 0} do not

meet the least upper bound property if p is a prime number. The need to fill these gaps was the main reason to ”create” the set of real numbers. Since the set of real numbers satisfies by definition the completeness axiom, it does not have any holes in it. The set of holes in the rational number system is known as the set of irrational numbers and is defined as

Definition 109 The set of irrational numbers is R − Q

.

Any real number can be approximated as closely as we want by rational numbers, but to show you that we need first to prove the Archimedean property. 3.2.4 The Archimedean Property

One of the important consequences of the completeness axiom is called the Archimedean property. It basically states that the natural numbers are not bounded above in R. Although this property may seem obvious, it depends on the completeness axiom. In fact, there are other ordered fields in which it does not hold.

Theorem 110 (Archimedean Property) The set of natural numbers N is not bounded above in R.

This theorem says that in R you can have arbitrarily large integers. We can prove it by contradiction. Suppose N is bounded above. Since N is a non empty subset of R, then by the completeness axiom it must follows that ∃ α ∈ R : α = sup N. Since α is the least upper bound, α − 1 is not an upper bound and thus there exists n0 ∈ N : α − 1 < n0 . But then n0 + 1 ∈ N and α < n0 + 1, contradicting that α was an upper bound for N. Proof.

,

There are several equivalent forms of the Archimedean property that are useful in different contexts. We establish their equivalence in the following theorem

Theorem 111 Each of the following is equivalent to the Archimedean property: (a) For each z ∈ R, there exists an n ∈ N such that n > z. (b) For each x ∈ R, y ∈ R, x > 0 there exists an n ∈ N such that nx > y. (c) For each x ∈ R, x > 0 there exists an n ∈ N such that 0 < 1/n < x.

The completeness axiom (least-upper-bound property) and the real field R

41

Proof. We shall prove that the Archimedean property ⇒ (a) ⇒ (b) ⇒ (c) ⇒ Archimedean property, thereby establishing their equivalence. If (a) were not true, then there would exist z0 ∈ R such that n < z0∀n ∈ N. But then z0 would be an upper bound for N, contradicting the Archimedean property. To see that (a) ⇒ (b), let z = y/x. Then there exists n ∈ N such that n > y/x, implying that nx > y. (c) follows from (b) by making y = 1. Then nx > 1, so that 1/n < x. Since n ∈ N, n > 0 and also 1/n > 0. Finally, suppose that N were bounded above by some real number, say m. That is, n < m for all n ∈ N. But then 1/n > 1/m for all n ∈ N, and this contradicts (c) for x = 1/m. Thus (c) implies the Archimedean property. This theorem says that in R,there is an arbitrarily small positive rational x > 1 number n1 ε y n 0. Then there exists n ∈ N such that n − 1 ≤ y < n. Moreover, n is unique. Proof. Consider the set S = {m ∈ N : m > y} . By the Archimedean property we know S is non empty and by the well ordering property of N we also know that there is a unique n ∈ S such that for every m ∈ S, m ≥ n. Since n ∈ S, we have n > y. Now consider two possibilities, n = 1 or n > 1. If n = 1 then n − 1 = 0 and by assumption 0 < y, then 0 < y < 1. If n > 1, then n − 1 ∈ N but by construction n − 1 ∈/ S. So n − 1 ≤ y, implying the desired result. Theorem 113 (Q is dense in R) If x ∈ R, y ∈ R, and x < y, then there exists a p ∈ Q such that x < p < y. Proof. Begin by supposing x > 0. Since y − x ∈ R and y − x > 0, using the Archimedean property ∃n ∈ N such that n (y − x) > 1, or equivalently nx + 1 < ny. Since nx > 0, by the previous lemma we know there is m ∈ N such that m − 1 ≤ nx < m. But then m ≤ nx + 1 < ny, so that nx < m < ny. It follows that the rational number p = m/n satisfies x < p < y. Finally, if x ≤ 0, choose an integer k such that k > −x and apply the above argument to the positive numbers x+ k and y +k. If q is the rational number satisfying x + k < q < y + k, then p = q − k is the rational number satisfying x < p < y. Based on the above theorem we can prove very easily the statement we posed in the previous section:

Corollary 114 Any real number can be approximated as closely as we want by rational numbers. That is, ∀x ∈ R, ∀ε > 0 ∃ q ∈ Q : x − ε < q < x + ε.

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42

The proof is left as an exercise. Using the density property of Q in R it is also easy to show that between any two real numbers there is an irrational number. The proof is left as an exercise, but the hints are as follow: ,

Exercise 115 Let x ∈ R, y ∈ R, and x < y. Then there exists a w ∈ R − Q such

that x < w < y. (Hint: first show the lemma that if x is a nonzero rational number and y is irrational, then xy is irrational. Then use that lemma and the fact that if x < y then √x2 < √y2 to prove the result)

Finally, we close this section presenting the theorem that proves that R does not have the type of holes through which we initially proved that Q did not satisfy the completeness axiom

Theorem √ 116 written as

∀x ∈ R++ and ∀n ∈ N, ∃! y ∈ R++ n x or x1/n .

such that

yn

= x. This number is

Proof. The steps to prove this theorem are the following. First show the

uniqueness of y if it exists. Second, show that the set Y = {t : tn < x} is nonempty and bounded implying the existence of y = sup Y . Third, show that yn = x by contradiction (assuming yn < x or yn > x and find z ∈ Y such that z > y or find z < y which is an upper bound of Y ). The details are in Rudin, pag. 10.

3.3 The extended real number system It is convenient to introduce upper and lower bounds for subsets of

R which are not

bounded.

The extended real number system R∗ consists of the real field R and two symbols, +∞ and −∞. In R∗we preserve the original order defined for R , and define −∞ < x < +∞ for every x ∈ R. Definition 117

∞

It is then the case that +

is an upper bound of every subset of the extended

real number system, and that every nonempty subset (not necessarily bounded) has a least upper bound.

∞

∞ −∞ y∈R

It is important to note that the symbols +

and

are not really numbers.

x = + means that there is no such that y
x, or that the subset containing x is not bounded. The extended real number system does not fulfill

The statement

the field axioms. It is customary to make the following conventions: x (a) ∀x ∈ R, x + ∞ = +∞, x − ∞ = −∞, +x∞ = −∞ = 0.

(b) ∀x ∈ R++

x · (+∞) = +∞, x · (−∞) = −∞. (c) ∀x ∈ R−− x · ∞ −∞, x · −∞ ∞.

, (+ ) = ( )=+ When it is desired to make the distinction between real numbers and the symbols + and , the former are called finite. ∞

,

−∞

The absolute value function

43

Notation For interval subsets of real numbers we shall use the following notation [a, b] = {x ∈ R : a ≤ x ≤ b} (a, b) = {x ∈ R : a < x < b} [a, b) = {x ∈ R : a ≤ x < b} (a, b] = {x ∈ R : a < x ≤ b}

The set [a, b] is called a closed interval, the set (a, b) is called an open interval and the sets [a,b) and (a,b] are called half-open ( or half-closed ) intervals. We shall also have occasion to refer to the unbounded intervals: [a, ∞) = {x ∈ R : a ≤ x} (a, ∞) = {x ∈ R : a < x} (−∞, b] = {x ∈ R : x ≤ b} (−∞, b) = {x ∈ R : x < b} 3.4

The absolute value function

As we have seen so far, many of the proofs involve manipulating inequalities, and one useful tool in working with inequalities is the concept of absolute value. Definition 118

If

x ∈ R,

then the

x

| |

absolute value of x, denote |x| , is defined by




= −x,x,

if

x≥0 x