One must have exercise. You need adventures in your head.

In Analysis . . .

we study relations in the orderstructure of the set R of real numbers. Important tools and concepts for this are the absolute value of a number and the distance between two numbers. The absolute value function |·| maps positive numbers onto themselves, 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. The distance between two numbers a, b ∈ R is given by d(a,b) = |a - b|. If a - b ≥ 0 then a ≥ b.
We always have |a| ≥ 0. We can change the sign "within" the absolute value. Often we make use of the symmetry |a| = |-a| and therefore also 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.

0 -a a x y f(x) = |x| = -x f(x) = |x| = x = id(x) The Absolute Value Function

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|. ☐

Only on account of having a negative sign, an algebraic expression need not be negative; (-(-x) = x). Sometimes the signum function comes in useful: sign(x) = 1 if x > 0, sign(x) = -1 if x < 0 und sign(x) = 0 if x = 0.
For every number x we have: x = |x| · sign(x).

The Signum Function f(x) = sign(x) 0 x y 1 -1

8 The real numbers R or R1

In chapter 7 we found the structure (R, +, ⋅) to be a field. Both operations (R x R) → R (addition and multiplication) are both commutative and associative, multiplication is distributive with regard to addition, for each operation there exists an identity element, for every element in R there exists an additive inverse element and for each element except zero there exists a multiplicative inverse element. Reminder:
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 [element])
A5. ∀ a ∈ R ∃ -a ∈ R | a + (-a) = 0   and   ∀ a ∈ R, a ≠ 0, ∃ a-1R | a ⋅ a-1 = 1.   -   (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 field axioms A1 to A5 are also satisfied by the complex numbers. C, however, is not an ordered set. With A1 to A8, the set of the rational numbers Q is a well-ordered field and for most practical purposes, Q is just fine. 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, it is even a transcendental number. There are infinitely many prime numbers and the square root of each is irrational, etc.
With a ninth axiom, the field Q will now be enriched to become the continuum, that we call the real numbers R. We need first a few new concepts.

8.1. DEFINITION: Intervals, open cover or cover (also covering is in use)
Let a, b ∈ R.
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, Ur(x0) and Ur[x0] are called (open resp. closed) r-neighborhoods of x0.
A sequence of closed intervals In := [an, bn] is called nested intervals if an → y and y ← bn. One-sided and two-sided infinite intervals are (-∞, b), (a, +∞) and (-∞, +∞).
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
A9. 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).

This characterisation A9 of the real numbers (it's actually the Borel-Lebesgue theorem about compactness) we were taught at the Universiteit van Suid-Afrika by Hanno Rund, who was also leading in writing [60], 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 20th century mathematics. Although Rund was a German, he was influenced by french mathematicians. I think it is because of this, that he had no scruples, to give students a fright with Borel-Lebesgue. We thought that this was over-ambitious and it was by no means motivating. We are going to show further down that our A9 is equivalent to the more plausible supremum principle. This proof will then be instructive and will compensate for the fright with this covering business.

Here are some more propositions which are equivalent to the completeness axiom A9:
⋅ A9a. The supremum principle: Every non-empty set of real numbers which is bounded from above has a least upper bound.
⋅ A9b. Every Dedekind cut has one and only one dividing number.
⋅ A9c. The principle of nested intervals.
⋅ A9d. Every Cauchy sequence converges.
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.
We want to demonstrate the equivalence of A9 and the supremum principle A9a. For doing so, we need again some new tools and definitions.

8.2. DEFINITION: Bounds, minima, suprema, etc.
Let XR.
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 and X is said to be bounded from below.
If there is a real number b, such that x ≤ b for all xX, then we say that b is an upper bound for X and 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 resort to infimum and/or supremum if a set has no minimum and/or no maximum. (Open or half-open intervals.)

8.3. 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:
In order to prove the assertion A, we assume A to be false and try to reach a contradiction. 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
Instead of A ⇒ B we show ¬B ⇒ ¬A. (B is necessary for A).
(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.
Give an example that contradicts the assumption.

Before we prove the equivalence of A9 and A9a, we need a formal characterisation of the notion supremum.
8.4. THEOREM: Let XR. 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 8.2 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 ist eine obere Schranke von X gemäß Definition 8.2.
 B(ii) ⇒ A: wir gehen von ¬A aus und führen einen Widerspruch herbei, indem wir B(ii) als wahr erachten: Angenommen γ', (γ' < γ), ist die kleinste obere Schranke. Wir wählen ε = γ – γ' > 0. Dann, gemäß B(ii), ∃ x ∈ X | γ – ε = γ - (γ – γ') = γ' < x. Das würde bedeuten, dass γ', im Gegensatz zu unserer Annahme, keine obere Schranke für X darstellt. ❑

B(i) ⇒ A: ∀ x ∈ X, x ≤ γ ⇒ γ is an upper bound for X by definition 8.2.
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. ❑

8.5 THEOREM: The properties A9 and A9a are equivalent.
A9. 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.
Proof: A9 ⇒ A9a:
This is the crucial point where analysis begins, and it is perhaps not immediately visible, what it's all about. So let's drop some more words 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 axiom A9 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. At a distance > 0 on the upper side of X we choose an upper bound b of X. Then we choose any a ∈ X, thus a < b. We consider the closed interval [a, b] (A9 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 A9 there exists a finite subcover, say
{I(y1), I(y2), … , I(ym), J(ym+1), … , J(yn)}.
In view of J(yr) ∩ P = ∅ for every J(yr), the intervals I(y1), I(y2), … , I(ym) form a cover of P. Let p(yk) be the greatest of the finitely many p(y1), p(y2), … , p(ym). Considering p(yk) ∈ P, we have p(yk) ∈ I(yd) for some d. But this means that p(yk) < p(yd) contrary to the fact that p(yk) is the greatest of the numbers p(y1), p(y2), … , p(ym). 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. ♬
A9a ⇒ A9:
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, … , In
(b) ⋃j=1n Ij has no "gaps", i. e. x ∈ ⋃j=1n Ij and a ≤ y ≤ x implies y ∈ ⋃j=1n Ij.
There exist admissible subsets of Ω; For example, every interval I ∈ Ω for which a ∈ I, is admissible. If now Ω' = {I1, I2, … , In} with Ii = (ai, bi) is admissible, we put c(Ω') := max{bi}. 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 (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 A9 ⇔ A9a. ♫

We are now going to construct an infinite cover of 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 In = (0, an). Here, an is the greatest n-digit decimal number with (an)2 < 2.
For every natural number n we form the open interval Jn = (bn, 3). Here, bn is the smallest n-digit decimal number with (bn)2 > 2.
Now we prove that the two sets of intervals
{In} = {(0, 1), (0, 1.4), (0, 1.41), (0, 1.414), … } and
{Jn} = {(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. an < α < bn, for every n and thus
(an)2 < α2 < (bn)2     …    (1)
α2 - 2 < (bn)2 – 2.     …    (2)
Since we also have 2 - (an)2 > 0 for every n, it follows from (2) that
α2 - 2 < [(bn)2 – 2] + [2 – (an)2] = (bn)2 – (an)2     …    (3)
Now is (bn)2 – (an)2 = (bn + an)(bn – an) ≤ 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 – (an)2.     …    (5)
Since (bn)2 – 2 > 0, it follows from (5) that
2 - α2 < [2 – (an)2] + [(bn)2 – 2] = (bn)2 – (an)2.
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, the sets of intervals {In} and {Jn} 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
{Ir1, Ir2, … ,Irn, Js1, Js2, … ,Jsm} where
r1 < r2 < … < rn < and s1 < s2 < … < sm, then none of the rational numbers {x | arn ≤ x ≤ bsm} is covered by these intervals. ♫
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