8 The real numbers

The whole theory of (real) Analysis rests on a few axioms about the real numbers and a few concepts and tools, like distance, signum and absolute value, which are interwoven with each other. In this context, we usually refer to the numbers as points.

A number x, thought of as a point on the number line (also on the closed real line), is characterised by two properties: its distance from the zero point and its algebraic sign.

• The distance from zero is denoted by |x|, read "the absolute value of x", (Fig. 8.1).
• The algebraic sign is given by sign(x), (Fig. 8.2).

8.1 The Absolute Value Function

The absolute value function |·| maps positive numbers onto themselves, and negative numbers onto their additive inverses, i. e.,
if x ≥ 0, then f(x) = |x| = x = id(x).
If x < 0, then f(x) = |x| = -x.
E. g., |-7| = -(-7) = 7.

8.2 The Distance Function

If a > b, then a - b = c, with c > 0, and b - a = -c < 0. The distance between a and b should be a positive number, therefore, the distance between a and b is defined to be
d(a, b) = d(b, a) = |a - b| = |b - a|.

Properties of the Absolute Value

We always have |a| ≥ 0. We can change the sign within the absolute value bars.
We can keep this in mind thinking of the mirror symmetry
|a| = |-a|, (Fig. 8.1), or of |a - b| = |b - a|.
If a and b are both positive or both negative then their product
a · b > 0 and we have |a · b| = |a| · |b|.
If a and b have different signs, the a · b < 0 and again we have
|a · b| = |a| · |b|, also if a or b or both are zero, we always have
|a · b| = |a| · |b| and
|a/b| = |a|/|b|, if b ≠ 0.

1.) It is always true that a ≤ |a| and -a ≤ |a|.
Proof. If a ≥ 0 then |a| = a and -a ≤ 0, and thus is -a ≤ a ≤ |a|. If a ≤ 0 then |a| = -a and -a ≥ 0, and thus is a ≤ -a ≤ |a|. ☐

2.) For all x, ε ∈ R is |x| ≤ ε equivalent to -ε ≤ x ≤ ε.
Proof. (i) ⇒: From |x| ≤ ε and 1.) we have that x ≤ |x| ≤ ε and -x ≤ |x| ≤ ε. From the last inequality it follows that -ε ≤ x and thus -ε ≤ x ≤ ε.
(ii) ⇐: Supposing -ε ≤ x ≤ ε. Mutiplying by -1 we get also ε ≥ -x ≥ -ε. Is x ≥ 0 then is |x| = x und thus |x| ≤ ε. If x ≤ 0 then is |x| = -x and it follows likewise that |x| ≤ ε. ☐

3.) The triangle inequality: For all x, y ∈ R, |x + y| ≤ |x| + |y|.
Proof. Because of 1.) we have x ≤ |x| and y ≤ |y|. By Addition we obtain x + y ≤ |x| + |y|. In the same manner we obtain -x - y ≤ |-x| + |-y|, i. e., -(x + y) ≤ |x| + |y|. Multiplying the last inequality by -1, we get also x + y ≥ -(|x| + |y|). So we have -(|x| + |y|) ≤ x + y ≤ |x| + |y| and making use of 2.) it follows |x + y| ≤ |x| + |y|. ☐

4.) The reverse triangle inequality: |x - y| ≥ ||x| - |y||.
Proof. In the triangle inequality we plug in the number x - y intstead of x and obtain |x - y + y| ≤ |x - y| + |y| or |x| - |y| ≤ |x - y|. By interchanging x and y in the last inequality, we obtain |y| - |x| ≤ |y - x|, i. e., -(|x| - |y|) ≤ |x - y|, since |y - x| = |x - y|. This is equivalent to |x| - |y| ≥ -|x - y|. Thus we have because of -|x - y| ≤ |x| - |y| ≤ |x - y| and 2.), that |ΙxΙ - ΙyΙ| ≤ |x - y| or |x - y| ≥ |ΙxΙ - ΙyΙ|. ☐

Remark: Alternative method to show 3.): From 1.) we conclude -|x| ≤ x ≤ |x| and likewise -|y| ≤ y ≤ |y|. Adding these two inequalities leads us to -(|x| + |y|) ≤ x + y ≤ |x| + |y| and thus according to 2.), to |x + y| ≤ |x| + |y|. ☐

8.3 The Signum Function

The signum function is defined by:
sign(x) = 1 if x > 0,
sign(x) = -1 if x < 0
sign(x) = 0 if x = 0.

For every number x we have:
x = |x| · sign(x) and also
|x| = x ⋅ sign(x).

8.4 Producing the Reals Out of Thin Air

To indulge in real analysis is to juggle with real numbers. This material, the structure (R, +, ⋅) of the real numbers, is defined by means of a few axioms. Everything that you are going to construct thereafter, will be derived from these axioms.

The field axioms

Let a, b, c  be real numbers.
A1. a + b = b + a   and   a⋅ b = b⋅ a.   -   (Commutativity)
A2. a + (b + c) = (a + b) + c   and   a⋅ (b⋅ c) = (a⋅ b)⋅ c.   -   (Associativity)
A3. a⋅ (b + c) = a⋅ b + a⋅ c.   -   (Multiplication is distributive with regard to addition)
A4. ꓱ 0, 1 ∈ R, 0 ≠ 1 | ∀ a ∈ R   a + 0 = a   and   a ⋅ 1 = a.   -   (Identity or identity element)
A5. ∀ a ∈ R ∃ -a ∈ R | a + (-a) = 0   and   ∀ a ∈ R, a ≠ 0, ∃ a^-1 ∈ R | a ⋅ a^-1 = 1.   -   (Inverse or inverse element)

The order axioms

A6. a < b   or   a = b   or   a > b.   -   (Trichotomy)
A7. a < b   and   b < c ⇒ a < c.   -   (Transitivity)
A8. a < b ⇒ a + c < b + c ∀ c   and   a⋅ c < b⋅ c ∀ c > 0.   -   (Monotony)

The axioms A1 to A5 define a field. Along with the axioms A6 to A8 the field becomes an ordered field.

Alternative Order Axiom

An ordered field is a field F (A1 to A5), along with the following axiom:

There exists a subset P ⊂ F, such that
(a) a, b ∈ P implies a + b ∈ P and a · b ∈ P,
(b) If a ≠ 0 and a ∈ F, then either a ∈ P or -a ∈ P.
The elements of P are called the positive elements.

The thus postulated field corresponds to the set of rational numbers Q, which we use for calculations of all sorts. However, this set of all fractions, (periodic or finite decimal numbers) does not fill the number line. Admittedly, for every real number x and for every arbitrarily small (permissible) deviation ε there exists a rational number y with |y – x| < ε. But continuous functions? Nope! The set Q is nowhere continuous, it has gaps everywhere. For example, the circumference and the diameter of any circle are incommensurable, which means that pi  is an irrational number, π is even a transcendental number. There are infinitely many prime numbers and the square root of each is irrational, there exist triangles with legs of irrational lenght, etc.
Real analysis necessitates the closed real line, which has no gaps. The process of assigning every point of the number line a number is called completing Q.
For the wording of a corresponding axiom, we need a couple of new concepts.

8.4.1 DEFINITION: Intervals, open cover, covering

Let a, b ∈ R, a < b.

The set {x ∈ R | a < x < b} =: (a, b) is the open interval, and {x ∈ R | a ≤ x ≤ b} =: [a, b] is the closed interval with boundary points a and b.

x₀ := (a + b)/2 is the center, (b – a)/2 is the radius and b – a is the length of the interval.

The intervals with center x₀ and radius r, U_r(x₀) and U_r[x₀] are called (open / closed) r-neighborhoods of x₀.

A sequence of closed intervals I_n := [a_n, b_n] is called nested intervals if a_n → y and y ← b_n.

One-sided and two-sided infinite intervals are (-∞, b), (a, +∞) and (-∞, +∞).

A family Ω = {J_α}_α∈S of sets J_α is called a cover of a set A, if A ⊆ ⋃_α∈S J_α. If each J_α is open, then {J_α}_α∈S is an open cover of A.

In particular, a set Ω of open intervals is called an open cover of the closed interval [a, b] if every x ∈ [a, b] is contained in at least one of the open intervals in the set Ω.

The Completeness Axiom

A9b. For every open cover Ω of a closed interval [a, b] there exists a finite subset Ω' ⊂ Ω, which already covers [a, b]. (Remark: Ω' is then called a finite sub-cover).

Oh, Goodness! You must be joking! This can't be an axiom!

With this characterisation A9b of the real numbers, it's actually the Borel-Lebesgue covering theorem, we were confronted in the first week at the Universiteit van Suid-Afrika by Hanno Rund, who was also leading in writing [60] (Inleiding tot die Algebra en Analise vir voorgraadse Studente) wherein completeness is introduced in the same way. This wiskunde (mathematics) book for undergraduates was published in 1970. It was the first book written in Afrikaans (in the past also called Cape Dutch or Colonial Dutch), in the new spirit of 20^th century mathematics. Although Rund was a German, he was influenced by french mathematicians. Maybe it was because of this, that he amused himself, in giving us students a fright with Borel-Lebesgue. We thought that this approach was fanciful and asked ourselves, where it shall end up. Therefore, let's give a genuine axiom, immediately, although we will have to jump a little ahead with terms and concepts.
Here is a little motivation for this new try:
Consider the following three sets of rational numbers:
G := {x ∈ Q | x² < 2},
F := {x ∈ Q | x² ≤ 2} and
O := {x ∈ Q | x² > 2}.
First of all we observe, that G = F. Why is this? Gerhard and Franz are one and the same set, because the number x, such that x² = 2, is not contained in Q; remember: Q is not complete, it has many gaps.
All elements of O are upper bounds for G, but G has no least upper bound (supremum). All elements of G are lower bounds for O, but O has no greatest lower bound (infimum). This will be different in the set of rational numbers R, which contains the number x, such that x² = 2.

A9a. Every nonempty set A ⊂ R which is bounded above (below) has a supremum (an infimum) in R.

---

We are going to show that A9a and A9b are equivalent. This proof will be very instructive and will more than compensate for the fright with this covering business.

Here are some more definitions for the completeness principle. Every statement can be derived from every other statement:

⋅ A9c. Every Dedekind cut has one and only one dividing number.
⋅ A9d The principle of nested intervals.
⋅ A9e. Every Cauchy sequence converges.

In order to prove that those five characterisations of completeness are equivalent, it suffices to the prove a closed chain, for example, A9b ⇒ A9a ⇒ A9c ⇒ A9d ⇒ A9e ⇒ A9b.

Remark: Whereas the rational numbers are enumerable (countably infinite), the real numbers are non-denumerable (uncountably infinite). According to the Continuum Hypothesis we have |R| = 2^|Q| = c.

In order to demonstrate the equivalence of A9a and A9b, we need some new tools and definitions.

8.4.2 DEFINITION: Bounds, minima, suprema, etc.

Let X ⊂ R.

If there is a real number a, such that a ≤ x for all x in X, then we say that a is a lower bound for X.

If X has a lower bound, then X is said to be bounded from below.

If there is a real number b, such that x ≤ b for all x ∈ X, then we say that b is an upper bound for X

If X has an upper bound, then X is said to be bounded from above.

X is said to be bounded, if X is bounded both from above and from below.

If X is bounded from below, its greatest lower bound is called the infimum of X (inf X).

If X is bounded from above, its least upper bound is called the supremum of X (sup X).

If X has a minimum, then inf X = min X.

If X has a maximum, then sup X = max X.

We must thus resort to the infimum if a set has no minimum and
we must resort to the supremum if a set has no maximum.
(This concerns open and half-open intervals.)

8.4.3 Methods of Proof

A mathematician needs imagination, pencil and paper as well as ideas and methods to prove or disprove other ideas which jumped off his imagination or which are asserted by others.

Direct proof:

A ⇒ B: (A is sufficient for B).
Starting from the true statement A we show by algebraic rearrangement and inference that statement B follows from statement A.

Indirect proof or proof by contradiction:

An assertion is proved to be false by inferring from it a contradiction to an already recognized thesis. This method of proof is also called Reductio ad absurdum.

Example: in A4 we have ∃ 1 ∈ R | ∀ a ∈ R a ⋅ 1 = a. We want to prove that 1 is unique and assume that it is not, that there is

a number x with the same property. So x = x ⋅ 1 = 1 ⋅ x = 1. Contradiction! ↯ Absurd!   ♫
A special case of an indirect proof is by

Contraposition:

Instead of A ⇒ B we show ¬B ⇒ ¬A. (B is necessary for A).

Induction:

(i) The assertion A(k) is true;
(ii) if A(n) is true, then A(n+1) is true.
We conclude that A(n) is true for all integers n ≥ k.

Counterexample:

Give an example that contradicts the assumption.

Before we prove the equivalence of A9b and A9a, we need an easily manageable definition for the Supremum.

8.4.4 THEOREM

Let X ⊂ R. Then sup X = γ ϵ R ⇔
    (i) x ≤ γ for all x ϵ X,
    (ii) for every real number ε > 0 there exists an x ϵ X, such that
γ − ε < x.

Proof: Put A ≔ (γ = sup X) and B ≔ (i) ∧ (ii).

A ⇒  B(i): If γ = sup X holds then by definition 8.4.2 γ is an upper bound of X and thus (i) holds.
A ⇒ B(ii): Suppose ¬B(ii) holds, i. e. there exists ε' > 0 such that for every x ϵ X, x ≤ γ − ε' holds. But that would mean, γ − ε' is a smaller upper bound than the supremum γ, i. e. ¬(γ = sup X), i. e. ¬A. ↯

B(i) ⇒ A: ∀ x ∈ X, x ≤ γ ⇒ γ is an upper bound for X by definition 8.4.2.
B(ii) ⇒ A: we start out from ¬A and cause a contradiction by deeming (ii) to be true: Suppose γ', (γ' < γ), is the least upper bound. We choose ε = γ – γ' > 0. Thus, by B(ii), ∃ x ∈ X | γ – ε = γ - (γ – γ') = γ' < x. This would mean that γ', contrary to our assumption, is no upper bound for X. ↯ ❑

8.4.5 THEOREM

The properties A9b and A9a are equivalent.
Reminder:
A9b. For every open cover Ω of a closed interval [a, b] there exists a finite subset Ω' ⊂ Ω, which already covers [a, b].
A9a. Every non-empty set of real numbers which is bounded from above has a least upper bound. (The proof for the greatest lower bound works in the same way.)

Proof: A9b ⇒ A9a
It may not immediately be visible, what it's all about. So let's say a word or two on the situation. A non-empty set X contains at least one element x, that is a single number. If this is the case, x is both the maximim and supremum (also the minimum and the infimum) of X. In the case of a discrete set there exists also a maximum, and the same is true for an interval which is closed above. In any of these instances there is a least upper bound. So what we have to prove is actually, what happens when I demarcate a set X from the set of real numbers, without specifying a greatest element for X, like an interval which is open above. What happens at the boundary between the two ares? Is there a least upper bound for X? How can the seemingly dubious characteristic A9b be applied to prove that X has a supremum?
Let's begin by assuming ¬A9a, that there is no least upper bound and try to derive a contradiction. Since X is bounded from above, it has an upper bound b. Then we choose any a ∈ X, thus a < b. We consider the closed interval [a, b] (A9b requires a closed interval) and split it up in two subsets P and Q, where P contains the elements that belong to X and Q contains the upper bounds of X:
P = {y | y ∈ [a, b] and y is no upper bound of X}
Q = {y | y ∈ [a, b] and y is an upper bound of X}.
We have P ∪ Q = [a, b] and P ∩ Q = ∅. We construct an open cover of [a, b] by building an open interval about each y of [a, b]. For the extreme boundary poinst we take any a₀ < a and any b₀ > b.

(i) For y ∈ P we form the interval I(y) = (a₀, p(y)), where p(y) ∈ X and p(y) > y. Such a choice is always possible since y is no upper bound for X. We observe that y ∈ (a₀, p(y)) and that (a₀, p(y)) ∩ Q = ∅, since p(y) ∈ X.
(ii) For y ∈ Q we form the interval J(y) = (q(y), b₀), where q(y) ∈ Q and q(y) < y. Such a choice is always possible since y is an upper bound for X and according to our assumption that X has no least upper bound. We observe that y ∈ (q(y), b₀) and that (q(y), b₀) ∩ P = ∅, since q(y) ∈ Q.

The set of the open intervals I(y) and J(y) forms a cover of [a, b]. According to A9b there exists a finite subcover, say
{I(y₁), I(y₂), … , I(y_m), J(y_m+1), … , J(y_n)}.
In view of J(y_r) ∩ P = ∅ for every J(y_r), the intervals I(y₁), I(y₂), … , I(y_m) form a cover of P. Let p(y_k) be the greatest of the finitely many p(y₁), p(y₂), … , p(y_m). Considering p(y_k) ∈ P, we have p(y_k) ∈ I(y_d) for some d. But this means that p(y_k) < p(y_d) contrary to the fact that p(y_k) is the greatest of the numbers p(y₁), p(y₂), … , p(y_m).  ↯ Absurd!  ☺ From this contradiction we conclude that X has a supremum. 😀

Remark: The supremum of X, call this number c ∈ Q, is the minimum of Q. For c you cannot find a p(c) < c ∈ Q, so it is absurd to choose an interval (q(c), b₀) for c. The crux of the matter is however, that for every element x ∈ X we can find an element y ∈ X, such that dass x < y.   ♬

Proof: A9a ⇒ A9b
Suppose we have an infinite set Ω of open intervals that covers [a, b]. For the purpose of this proof we shall call a subset Ω' ⊂ Ω "admissible", if it meets the following two requirements:
(a) Ω' consists of only finitely many intervals I₁, I₂, … , I_n
(b) ⋃_j=1ⁿ I_j has no "gaps", i. e. x ∈ ⋃_j=1ⁿ I_j and a ≤ y ≤ x implies y ∈ ⋃_j=1ⁿ I_j.
There exist admissible subsets of Ω; For example, every interval I ∈ Ω for which a ∈ I, is admissible. If now Ω' = {I₁, I₂, … , I_n} with I_i = (a_i, b_i) is admissible, we put c(Ω') := max{b_i}. It is obvious, that if for some Ω' we have b < c(Ω'), then Ω' is a finite cover of [a, b].
We assume now, that c(Ω') ≤ b for every admissible Ω' and prove thereupon, that this assumption leads to a contradiction.
If c(Ω') ≤ b for every admissible Ω', then also r := sup {c(Ω')} ≤ b. According to our assumption A9a this supremum exists. Since r ∈ [a, b], there exists an interval ω ∈ Ω such that r ∈ ω. We choose now an admissible Ω'' such that c(Ω'') ∈ ω. This is always possible because of theorm 8.4.4 (ii). But the set Ω''' which consists of the intervals of Ω'' together with the interval ω, is also admissible and therefore holds that c(Ω''') > r, because r ∈ ω. This is in conflict with the assumption that c(Ω') ≤ r for every admissible Ω'.  ↯
This completes the prove for A9b ⇔ A9a.  ☐  ♫

8.4.6 EXAMPLE

The Borel-Lebesgue theorem does not hold for the set Q of rational numbers. To see this, we construct an infinite open cover of rational numbers for the closed interval [1, 2] of rational numbers, of which no finite subset covers [1, 2].
For every natural number n we form the open interval I_n = (0, a_n). Here, a_n is the greatest n-digit decimal number with a_n² < 2.
For every natural number n we form the open interval J_n = (b_n, 3). Here, b_n is the smallest n-digit decimal number with b_n² > 2.
Now we prove that the two sets of intervals
{I_n} = {(0, 1), (0, 1.4), (0, 1.41), (0, 1.414), …} and
{J_n} = {(2, 3), (1.5, 3), (1.42, 3), (1.415, 3), …} together cover [1, 2].
We observe that a₁ ≤ a₂ ≤ a₃ ≤ a₄ ≤ … and that b₁ ≥ b₂ ≥ b₃ ≥ b₄ ≥ …
Suppose now, that α ∈ [1, 2] is not covered, i. e. a_n < α < b_n, for every n and thus
a_n² < α² < b_n²     …    (1)
and
α² - 2 < b_n² – 2.     …    (2)
Since we also have 2 - a_n² > 0 for every n, it follows from (2) that
α² - 2 < (b_n² – 2) + (2 – a_n²) = b_n² – a_n²     …    (3)
Now is b_n² – a_n² = (b_n + a_n)(b_n – a_n) ≤ 3 · 10^-n+1 and it follows from (3) that α² - 2 < 3 · 10^-n+1 for every n.

This is only possible if α² - 2 ≤ 0 or
α² ≤ 2     …    (4)

From (1) follows furthermore that 2 - α² < 2 – a_n².     …    (5)
Since b_n² – 2 > 0, it follows from (5) that
2 - α² < (2 – a_n²) + (b_n² – 2) = b_n² – a_n².
We have thus that 2 - α² ≤ 0 and therefore
α² ≥ 2.     …    (6)
From (4) and (6) it follows that
α² = 2.
This is in conflict with the fact, that there is no rational number α which satisfies this equation. Thus, contrary to the assumption, the sets of intervals {I_n} and {J_n} form a cover for [1, 2].  ☐  ♫

On the other hand, however, no finite subset of these intervals forms a cover for [1, 2]. Chose the finite subset
{I_r1, I_r2, … ,I_rn, J_s1, J_s2, … ,J_sm} where
r₁ < r₂ < … < r_n < and s₁ < s₂ < … < s_m, then none of the rational numbers {x | a_rn ≤ x ≤ b_sm} is covered by these intervals.  ☐  ♫

REMARK
We must keep in mind, that in the theorem of Borel-Lebesgue, the intervals which form the cover are open and the interval which is being covered, is closed. For example, the following infinite set of open intervals
(0, 1/2), (0,2/3), (0, 3/4), ... , (0, (n-1)/n), ... ,
covers the open interval (0, 1).
However, there is no finite subset of this infinite set, which covers (0, 1).

Similarly, there are closed covers of closed intervals, but no finite subset of these forms a cover. Thus, the closed interval [-1, 1] is covered by the following family of closed intervals: [-1, 0], [1/2, 1], [1/3, 1], ... , [1/n, 1], ... No finite subfamily of this covering forms a cover for [-1, 1].

Archimedes' Priciple

Eureka! Archimedes exclaimed, when it dawned to him: A body partly (or completely) submerged in liquid is buoyed up (buoyancy) by a force equal to the weight of the liquid (or gas) displaced.
There is another principle, less widely known, which is also named after Archimedes. Let's begin with the following

Proposition

If a, b ∈ R and a > 0, then there is an n ∈ N, such that n · a > b.
By dividing over the a, we aim to prove that there is some n ∈ N, such that n > b/a. But b/a is just some real number, say x. Thus, if a, b ∈ R and a > 0, then there is an n ∈ N, such that n > x. This latter form is called

The Archimedean Principle

Every real number is surpassed by a natural number. Equivalently, the set of natural numbers N is not bounded above.

Proof. Suppose for a contradiction that N is bounded above. That would mean, that there exists γ := sup N, and thus there would exist some n ∈ N such that γ - 1 < n, i. e. γ < n + 1, which interferes with the meaning of the supremum.  ↯ ☐
Another formulation of this principle gives us the following

8.4.7 THEOREM (Eudoxos)

For every positive ε there exists a natural number n such that 1/n < ε.