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Foundations of Mathematical Analysis Foundations of Mathematical Analysis Fabio Bagagiolo Dipartimento di Matematica, Universit`adi Trento email:[email protected] Contents 1 Introduction 2 2 Basic concepts in mathematical analysis and some complements 5 2.1 Limit, supremum, and monotonicity . 5 2.2 The Bolzano-Weierstrass theorem and the Cauchy criterium . 8 2.3 Superior and inferior limit and semicontinuous functions . 11 2.4 Infinite number series . 16 2.5 Rearrangements . 20 2.6 Sequences of functions . 25 2.7 Series of functions . 30 2.8 Power series . 31 2.9 Fourier series . 35 2.10 Historical notes . 38 3 Real numbers and ordered fields 43 3.1 Ordering and Archimedean properties . 44 3.2 Isomorphisms and complete ordered fields . 50 3.3 Choose your axiom, but choose one! . 57 3.4 Historical notes and complements . 60 1 3 Real numbers and ordered fields In this section we are going to point out the properties of the set R of the real numbers, both from an algebraic and from an analytical point of view. At this stage, we suppose that the reader already knows what the real numbers are. The first properties of the real numbers are of course the algebraic properties. Let us start from these ones. Definition 3.1 1) Given a nonempty set A, an operation on it is a function from the cartesian product A × A to A. 2) A nonempty set G is said to be a commutative (or abelian) group if the there is an operation ' :(a; b) 7! '(a; b) =: a + b on it which satisfies the following properties: 2i) (associative property) a + (b + c) = (a + b) + c 8 a; b; c 2 G; 2ii) (neutral element) 9 G 3 b =: 0 such that a + 0 = a 8 a 2 G; 2iii) (opposite element) 8 a 2 G 9 G 3 b =: (−a) such that 0 = a + (−a) =: a − a; 2iv) (commutative property) a + b = b + a 8 a; b; 2 G: (3.1) 3) A nonempty set F is said a field if there are two operations on it: '1(a; b) =: a + b, 59 '2(a; b) =: ab such that F is an abelian group with respect to '1, F n f0g is an abelian group with respect to '2, and if the following compatibility condition between the two operations is satisfied: (distributive property) c(a + b) = ca + cb 8 a; b; c 2 F (3.2) By the associative and commutative properties it follows that the neutral element of a group is unique as well as the opposite element. Indeed if 0; 00 are two neutral elements, we get 0 = 0 + 00 = 00 + 0 = 00; and for any a; b; c: a + c = b + c =) a + c − c = b + c − c =) a = b; which implies the uniqueness of the opposite element since, if b; b0 are two opposite ele- ments of a, we have a + b = 0 = a + b0 =) b = b0: 59 Here, 0 is the neutral element of '1. The group F n f0g has a multiplicative representation, that is we indicate by 1 its unique neutral element and by a−1 or even by 1=a the opposite element. Also note that, by definition, it follows that the restriction of '2 to (F n f0g) × (F n f0g) is an operation on F n f0g; in particular, if a; b 2 F n f0g then ab 6= 0 and a−1 6= 0. This immediately implies that ab = 0 =) a = 0 or b = 0. 43 Also note that, in the case of a field F , it is, for any a 2 F , 0 · a = a · 0 = 0: Indeed a · 0 = a(0 + 0) = a · 0 + a · 0 =) a · 0 = 0 and similarly for 0 · a. Proposition 3.2 If F is a field, then, for every a; b 2 F the following holds: i) (−1)a = −a60; ii) (−a)−1 = −a−1; Proof. i) a + (−1)a = a(1 − 1) = a · 0 = 0; ii) (−a)(−a−1) = (−1)(−1)(aa−1) = 1. ut With the usual operations of sum and multiplication, the set of the real numbers R is a field. Other well-known fields are the set of rational numbers Q61 and the set of complex numbers C62. The set of integers Z is not a field63, but the quotient sets Z=pZ, with p 2 N a prime number, are all fields64. There is a main difference between Q; R; C from one side and Z=pZ from the other side: the first ones are infinite fields, that is with infinitely many elements, the second one are finite fields, that is with finitely many elements: just p elements. Mathematical analysis is mainly devoted to the study of the infinite fields. As we are going to see in this notes, there are also big differences between Q; R and C: the first is ordered but not complete, the second is ordered and complete, the third is complete65 but not ordered. The (real) mathematical analysis is mainly devoted to the study of R. 3.1 Ordering and Archimedean properties As anticipated at the end of the previous subsection, R is ordered. Definition 3.3 An order relation (or an ordering) on a nonempty set A (which is then said to be an ordered set) is a relation, denoted by \≤", between its elements66 such that, for every a; b; c 2 A, (transitive property) a ≤ b; b ≤ c =) a ≤ c; (reflexive property) a ≤ a; (anti-symmetric property) a ≤ b; b ≤ a =) a = b: The order relation is said to be total if, for every couple of elements a; b 2 A, it is always true that at least one of the following relations hold: a ≤ b or b ≤ a. If the ordering 60From which (−1)(−1) = 1. 61With the same operations as in R. 62With the known operations which extend the ones in R. 63Z n f0g is not a group with respect to the multiplication: there is no opposite element. 64With natural extension of sum and multiplication from elements of Z to the elements of Z=pZ, which are equivalence classes. 65In the sense of metric space. 66A \relation" is often defined as a subset R of the cartesian product A × A, so that a is in relation with b if and only if (a; b) 2 R. 44 is total, then A is said to be totally ordered, if instead the ordering is not total67 then A is said partially ordered. Given an order relation on A we can always define a strict order relation on A, denoted by \<", as a < b () a ≤ b and a 6= b: Such a strict order relation satisfies the transitive property only. If F is a field and it is also a totally ordered set, we say that F is an ordered field if the following compatibility conditions between ordering and operations hold i) a ≤ b =) a + c ≤ b + c 8c 2 F; (3.3) ii) a ≤ b =) ac ≤ bc 8 0 ≤ c 2 F: Remark 3.4 Instead of writing a ≤ b, we will often say \a is smaller than or equal to b", as well as "b is larger than or equal to a", and also, in a more ambiguous manner, we will sometimes say \a is smaller than b" as well as \b is larger than a" for indicating both a ≤ b and a < b. Finally, a ≥ b (as well as a > b) will mean b ≤ a (as well as b < a). Example 3.5 We give a simple (but important for the sequel of these notes) example of a partially, but not totally, ordered set. Let x0 2 R and define the set n o A = A ⊆ R 9 r > 0 such that ]x0 − r; x0 + r[⊆ A ; 68 so that the elements of A are subsets of R containing open intervals centered in x0 . We define the following relation in A: 8 A1;A2 2 A;A1 ≤ A2 () A2 ⊆ A1 : (3.4) that is the \inverse inclusion order". The reader is invited to prove that (3.4) defines a partial, but not total, ordering on A. With the usual ordering, both Q and R are ordered fields. A first natural total ordering in the complex field C is the lexicographical one: given z1 = a1 + ib1; z2 = a2 + ib2 2 C z1 ≤ z2 () a1 < a2 or (a1 = a2 and b1 ≤ b2) : It is immediate to see that this is a total ordering on C, but also that, with such an ordering, C is not an ordered field: the property ii) of (3.3) does not hold69. 67That is there exist two elements a; b 2 A such neither a ≤ b nor b ≤ a holds true. 68 Neighborhoods of x0? 690 ≤ i, but i2 = i · i = −1 < 0 = 0 · i. 45 Proposition 3.6 If F is an ordered field, then, the implications in (3.3) are indeed equivalence, and moreover, for every a; b 2 F , we have: i) a ≤ b () −b ≤ −a70, ii) a2 := aa ≥ 071, iii) a > 0 () a−1 > 0, iv) 0 < a < b () 0 < b−1 < a−1, v) a ≤ b () a − b ≤ 0, vi) a < 0; b > 0 =) ab < 0. Proof. For the first sentence, just take in (3.3-i).ii)) c = 0 and c = 1 respectively. i) a ≤ b =) a + (−a − b) ≤ b + (−a − b) =) −b ≤ −a; the opposite implication similarly comes starting from −b ≤ −a; ii) If a ≥ 0 then it is obvious by point ii) of (3.3), if instead a < 0 then −a > 0 and so, being a = −(−a), −a2 = (−1)(−1)(−1)(−a)2 = −(−a)2 ≤ 0 =) a2 ≥ 0; iii) if a−1 < 0 then 1 = a(a−1) < 0 which is absurd; the opposite implication similarly comes from the equality (a−1)−1 = a; iv) if by absurd hypothesis we have 0 < a−1 < b−1, then multiplying by ab > 0 we obtain a contradiction, v) a ≤ b =) 0 = a − a ≤ b − a, vice versa 0 ≤ b − a =) a = 0 + a ≤ (b − a) + a = b; vi) if it was ab > 0 then a = abb−1 > 0, which is absurd.
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