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 an multiplication) are commutative and associative, multiplication is distributive regarding 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.
The field axioms
Let a, b, c be real numbers.
A1. a + b = b + a and ab = ba (commutativity)
A2. a + (b + c) = (a + b) + c and a(bc) = (ab)c (associativity)
A3. a(b + c) = ab + ac (distributivity)
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, a = b, a > b (trichotomy)
A7. a < b ∧ b < c → a < c (transitivity)
A8. a < b → a + c < b + c ∀ c and ac < bc ∀ c > 0 (monotony)
The field axioms A1 to A5 are satisfied by the complex numbers. C, however, cannot be arranged. With A1 to A8, the set of the rational numbers Q is a well-ordered field and for all 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. However, when we take an interest in the nature of the numbers themselves, we must observe, that Q has infinitely many gaps. 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.
Intervals
Let a, b ∈ R.
The set {x ∈ R | a < x < b} =: (a, b) is the open, 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 or 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 (-∞, +∞).
Open cover or covering
A set G of open intervals is called an open cover or covering of the closed interval [a, b] if every x ∈ [a, b] is contained in at least one of the open intervals in the set G.
The completeness axiom
A9. For every open covering G of a closed interval [a, b] there exists a finite subset G' ⊂ G, which already covers [a, b] (finite sub-cover).
Whereas the rational numbers are enumerable, the real numbers are nondenumerable. According to the Continuum Hypothesis we have |R| = 2|Q| = c.
Bounds, minima, suprema, etc.
Let X ⊂ R.
If ∃ a ∈ R | a ≤ x ∀ x ∈ X, then we say that a is a lower bound for X and X is said to be bounded from below.
If ∃ b ∈ R | x ≤ b ∀ x ∈ X, 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 only resort to infimum and/or supremum if a set has no minimum and/or no maximum. (Open or half-open intervals.)
There are several propositions that are equivalent to A9. One proposition can be deduced from the other:
⋅ Every Dedekind cut has one and only one dividing number.
⋅ The nested intervals principle.
⋅ Every Cauchy sequence (an) converges.
⋅ The supremum principle
Let's demonstrate the equivalence of A9 and the supremum principle
The supremum principle:
A9a. Every non-empty set of real numbers which is bounded from above has a least upper bound.

Methods of Proof
At this point we should pause for a moment to give thought to an essential element that constitutes our subject: how to go about to prove an assertion.
Direct proof: A → B:
Starting from the true statement A we show by rearrangement and inference that statement B follows.
Contraposition:
Instead of A → B we show ¬B → ¬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.
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.
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.
There is a number x with the same property. We then have x = x ⋅ 1 = 1. Contradiction! ▮

We are going to prove the equivalence of A9 and A9a presently. Let's first dissect the following theorem.
STELLING: As X ⊂ R dan is sup X = γ ϵ R as en slegs as
    (i) x ≤ γ vir alle x ϵ X,
    (ii) vir elke reële getal ε > 0 ’n x ϵ X bestaan sodanig dat γ − ε < x.
This was the first analysis theorem I encountered. In fact in Rund [60]; it made me feel uneasy. Was this the mathematics I wanted to study? Even today, the proof of this theorem gives me a queasy feeling, because normally this is how you define the supremum to be. For these very reasons we shall do it here, and because it's instructive.
We put
    A := (γ = sup X)
   (i) := ∀ x ∈ X, x ≤ γ
  (ii) := ∀ ε > 0 ∃ x ∈ X | γ – ε < x.
We must prove: A ↔ (i) ∧ (ii)
A → (i): γ = sup X → ∀ x ∈ X, x ≤ γ
A → (ii): we show ¬(ii) → ¬A:
¬(ii) = ∃ ε' > 0 | ∀ x ∈ X, x ≤ γ – ε' → ¬(γ = sup X)
(i) → A: ∀ x ∈ X, x ≤ γ → γ is an upper bound for X
(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 (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. ▮

Now we are going to prove that properties A9 and A9a are equivalent:
We suppose that A9 holds. Let us prove that a set X which is bounded from above has a least upper bound. Since X is bounded from above it has an upper bound b. Let's assume that X has no least upper bound and then, with the aid of A9, derive a contradiction. Take any a ∈ X then, by our assumption, a < b. (a cannot be an upper bound, b is an upper bound). We split up [a, b] as follows:
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 = ∅. Also P ≠ ∅ since a ∈ P and Q ≠ ∅ since b ∈ Q. We construct an open covering 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 the 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) form a covering 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 covering 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). From this contradiction we conclude that X has a supremum. ▮

Now we prove that A9a implies A9. Suppose we have an infinite set G of open intervals that covers [a, b]. For the purpose of this proof we shall call a subset G' ⊂ G admissible, if it meets the following two requirements:
(a) G' consists of only finitely many intervals I1, I2, … , In
(b) Every x of [a, b] is covered by the union of the intervals Ij.
There exist admissible subsets of G; For example, every interval I ∈ G for which a ∈ I, is admissible. If now G' = {I1, I2, … , In} with Ii = (ai, bi) is admissible, we put max(bi) = c(G'). It is obvious, that if for some G' we have b < c(G'), then G' is a finite covering of [a, b].
We assume now, that c(G') ≤ b for all admissible G' and prove thereupon, that this assumption leads to a contradiction.
If c(G') ≤ b, then also sup c(G') = r ≤ b. According to our assumption (A9a) this supremum exists. Since r ∈ [a, b], there exists an interval g ∈ G such that r ∈ g. We choose now an admissible G'' such that c(G'') ∈ g (stelling). But the set G''' which consists of the intervals of G'' together with the interval g, is also admissible and therefore holds that c(G''') > r, because r ∈ g. This is in conflict with the assumption that c(G') ≤ r for every admissible G'. 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)
and
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. ▮



This page is under construction    As from today, April 18, 2020 (Covid-19): a page of mathematical goodies is under construction.

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Beispiele für Aussagen und deren Negation: a = b, a ≠ b. a ∈ M, a ∉ M. A ⊂ M, A ⊄ M. a < b, a ≮ b. a < b, a ≥ b.

1.3. DEFINISIE: Gestel X ⊂ R. As daar ’n getal a bestaan so dat so dat vir elke x ϵ X geld dat x ≤ a dan heet a ’n bogrens van X, ...

1.4. DEFINISIE: As X ⊂ R van bo begrens is dan heet die kleinste van die bogrense van X die supremum of die kleinste bogrens van X en word geskryf as sup X of k.b.g. X. ...

Als ich 21 Jahre alt war, hörte ich im Radio von einem Städter, der sich in eine ländliche Gegend zurückgezogen hatte. Auf die Frage, was er in seiner freien Zeit mache, antwortete er, dass er sich unter anderem mit Mathematik beschäftige. So etwas hatte ich noch nie gehört. Da kamen mir kurz darauf die Bücher Du und die Natur [34] und Du und der Zauber der Zahlen [35], unter.

Paul Karlson
A geschnitten mit B B Teilmenge von A A und B sind disjunkt B ohne A symmetrisch Differenz A vereinigt mit B A x B Potenzmenge

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