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c ≤ x) for all x ∈ X.The number c in this case is called an upper bound (resp. lower bound) of theset X. It is also called a majorant (resp. minorant) of X.Definition 3 A set that is bounded both above and below is called bounded.Definition 4 An element a ∈ X is called the largest or maximal (resp.
smallest orminimal) element of X if x ≤ a (resp. a ≤ x) for all x ∈ X.We now introduce some notation and at the same time give a formal expressionto the definition of maximal and minimal elements:(a = max X) := a ∈ X ∧ ∀x ∈ X (x ≤ a) ,(a = min X) := a ∈ X ∧ ∀x ∈ X (a ≤ x) .Along with the notation max X (read “the maximum of X”) and min X (read “theminimum of X”) we also use the respective expressions maxx∈X x and minx∈X x.It follows immediately from the order Axiom 1≤ that if there is a maximal (resp.minimal) element in a set of numbers, it is the only one.However, not every set, not even every bounded set, has a maximal or minimalelement.For example, the set X = {x ∈ R | 0 ≤ x < 1} has a minimal element. But, as onecan easily verify, it has no maximal element.Definition 5 The smallest number that bounds a set X ⊂ R from above is called theleast upper bound (or the exact upper bound) of X and denoted sup X (read “thesupremum of X”) or supx∈X x.This is the basic concept of the present subsection.
Thus(s = sup X) := ∀x ∈ X (x ≤ s) ∧ ∀s < s ∃x ∈ X s < x .The expression in the first set of parentheses on the right-hand side here says thats is an upper bound for X; the expression in the second set says that s is the smallest2.1 Axioms and Properties of Real Numbers45number having this property. More precisely, the expression in the second set ofparentheses asserts that any number smaller than s is not an upper bound of X.The concept of the greatest lower bound (or exact lower bound) of a set X isintroduced similarly as the largest of the lower bounds of X.Definition 6(i = inf X) := ∀x ∈ X (i ≤ x) ∧ ∀i > i ∃x ∈ X x < i .Along with the notation inf X (read “the infimum of X”) one also uses the notation infx∈X x for the greatest lower bound of X.Thus we have given the following definitions:sup X := min c ∈ R | ∀x ∈ X (x ≤ c) ,inf X := max c ∈ R | ∀x ∈ X (c ≤ x) .But we said above that not every set has a minimal or maximal element.
Therefore the definitions we have adopted for the least upper bound and greatest lowerbound require an argument, provided by the following lemma.Lemma (The least upper bound principle) Every nonempty set of real numbers thatis bounded from above has a unique least upper bound.Proof Since we already know that the minimal element of a set of numbers isunique, we need only verify that the least upper bound exists.Let X ⊂ R be a given set and Y = {y ∈ R | ∀x ∈ X (x ≤ y)}. By hypothesis,X = ∅ and Y = ∅. Then, by the completeness axiom there exists c ∈ R such that∀x ∈ X ∀y ∈ Y (x ≤ c ≤ y).
The number c is therefore both a majorant of X and aminorant of Y . Being a majorant of X, c is an element of Y . But then, as a minorantof Y , it must be the minimal element of Y . Thus c = min Y = sup X.Naturally the existence and uniqueness of the greatest lower bound of a nonemptyset of numbers that is bounded from below is analogous, that is, the following proposition holds.Lemma (X nonempty and bounded below) ⇒ (∃! inf X).We shall not take time to give the proof.We now return to the set X = {x ∈ R | 0 ≤ x < 1}. By the lemma just proved itmust have a least upper bound.
By the very definition of the set X and the definitionof the least upper bound, it is obvious that sup X ≤ 1.To prove that sup X = 1 it is thus necessary to verify that for any number q < 1there exists x ∈ X such that q < x; simply put, this means merely that there arenumbers between q and 1. This of course, is also easy to prove independently (forexample, by showing that q < 2−1 (q + 1) < 1), but we shall not do so at this point,462The Real Numberssince such questions will be discussed systematically and in detail in the next section.As for the greatest lower bound, it always coincides with the minimal elementof a set, if such an element exists. Thus, from this consideration alone we haveinf X = 0 in the present example.Other, more substantive examples of the use of the concepts introduced here willbe encountered in the next section.2.2 The Most Important Classes of Real Numbersand Computational Aspects of Operations with RealNumbers2.2.1 The Natural Numbers and the Principle of MathematicalInductiona.
Definition of the Set of Natural NumbersThe numbers of the form 1, 1 + 1, (1 + 1) + 1, and so forth are denoted respectivelyby 1, 2, 3, . . . and so forth and are called natural numbers.Such a definition will be meaningful only to one who already has a completepicture of the natural numbers, including the notation for them, for example in thedecimal system of computation.The continuation of such a process is by no means always unique, so that theubiquitous “and so forth” actually requires a clarification provided by the fundamental principle of mathematical induction.Definition 1 A set X ⊂ R is inductive if for each number x ∈ X, it also containsx + 1.For example, R is an inductiveset; the set of positive numbers is also inductive.The intersection X = α∈A Xα of any family of inductive sets Xα , if not empty,is an inductive set.Indeed,x∈X=Xα ⇒ ∀α ∈ A (x ∈ Xα ) ⇒α∈A⇒ ∀α ∈ A (x + 1) ∈ Xα ⇒ (x + 1) ∈Xα = X .α∈AWe now adopt the following definition.Definition 2 The set of natural numbers is the smallest inductive set containing 1,that is, the intersection of all inductive sets that contain 1.2.2 Classes of Real Numbers and Computations47The set of natural numbers is denoted N; its elements are called natural numbers.From the set-theoretic point of view it might be more rational to begin the naturalnumbers with 0, that is, to introduce the set of natural numbers as the smallestinductive set containing 0; however, it is more convenient for us to begin numberingwith 1.The following fundamental and widely used principle is a direct corollary of thedefinition of the set of natural numbers.b.
The Principle of Mathematical InductionIf a subset E of the set of natural numbers N is such that 1 ∈ E and together witheach number x ∈ E, the number x + 1 also belongs to E, then E = N.Thus,(E ⊂ N) ∧ (1 ∈ E) ∧ x ∈ E ⇒ (x + 1) ∈ E ⇒ E = N.Let us illustrate this principle in action by using it to prove several useful properties of the natural numbers that we will be using constantly from now on.10 .
The sum and product of natural numbers are natural numbers.Proof Let m, n ∈ N; we shall show that (m + n) ∈ N. We denote by E the set ofnatural numbers n for which (m + n) ∈ N for all m ∈ N. Then 1 ∈ E since (m ∈N) ⇒ ((m + 1) ∈ N) for any m ∈ N. If n ∈ E, that is, (m + n) ∈ N, then (n + 1) ∈ Ealso, since (m + (n + 1)) = ((m + n) + 1) ∈ N. By the principle of induction, E = N,and we have proved that addition does not lead outside of N.Similarly, taking E to be the set of natural numbers n for which (m · n) ∈ N forall m ∈ N, we find that 1 ∈ E, since m · 1 = m, and if n ∈ E, that is, m · n ∈ N, thenm · (n + 1) = mn + m is the sum of two natural numbers, which belongs to N bywhat was just proved above.
Thus (n ∈ E) ⇒ ((n + 1) ∈ E), and so by the principleof induction E = N.20 . (n ∈ N) ∧ (n = 1) ⇒ ((n − 1) ∈ N).Proof Consider the set E consisting of all real numbers of the form n − 1, wheren is a natural number different from 1; we shall show that E = N. Since 1 ∈ N, itfollows that 2 := (1 + 1) ∈ N and hence 1 = (2 − 1) ∈ E.If m ∈ E, then m = n − 1, where n ∈ N; then m + 1 = (n + 1) − 1, and sincen + 1 ∈ N, we have (m + 1) ∈ E. By the principle of induction we conclude thatE = N.30 . For any n ∈ N the set {x ∈ N | n < x} contains a minimal element, namelymin{x ∈ N | n < x} = n + 1.482The Real NumbersProof We shall show that the set E of n ∈ N for which the assertion holds coincideswith N.We first verify that 1 ∈ E, that is,min{x ∈ N | 1 < x} = 2.We shall also verify this assertion by the principle of induction.
LetM = x ∈ N | (x = 1) ∨ (2 ≤ x) .By definition of M we have 1 ∈ M. Then if x ∈ M, either x = 1, in which casex + 1 = 2 ∈ M, or else 2 ≤ x, and then 2 ≤ (x + 1), and once again (x + 1) ∈ M.Thus M = N, and hence if (x = 1) ∧ (x ∈ N), then 2 ≤ x, that is, indeed min{x ∈N | 1 < x} = 2. Hence 1 ∈ E.We now show that if n ∈ E, then (n + 1) ∈ E.We begin by remarking that if x ∈ {x ∈ N | n + 1 < x}, then(x − 1) = y ∈ {y ∈ N | n < y}.For, by what has already been proved, every natural number is at least as large as 1;therefore (n + 1 < x) ⇒ (1 < x) ⇒ (x = 1), and then by the assertion in 20 we have(x − 1) = y ∈ N.Now let n ∈ E, that is, min{y ∈ N | n < y} = n + 1.
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