 By Steven G. Krantz

The new version of this well known textual content is revised to fulfill the feedback of clients of the former version. A readable but rigorous method of an important a part of mathematical pondering, this article bridges the distance among vintage theoretical texts and no more rigorous ones, delivering a delicate transition from common sense and proofs to actual research. in addition to the fundamental fabric, the textual content covers Riemann-Stieltjes integrals, Fourier research, metric areas and functions, and differential equations.

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Additional resources for Real Analysis and Foundations

Sample text

Then c′ ∈ −C and z = c + c′ . Hence 0 ⊆ C + (−C). We conclude that C + (−C) = 0. Having verified the axioms for addition, we turn now to multiplication. 17 If C and D are cuts then we define the product C ·D as follows: • If C, D > 0 then C · D = {q ∈ Q : q < c · d for some c ∈ C, d ∈ D with c > 0, d > 0 } • If C > 0, D < 0 then C · D = − (C · (−D)) • If C < 0, D > 0 then C · D = − ((−C) · D) • If C, D < 0 then C · D = (−C) · (−D) • If either C = 0 or D = 0 then C · D = 0. Notice that, for convenience, we have defined multiplication of negative numbers just as we did in high school.

Since addition of rational numbers is commutative, it follows immediately that addition of cuts is commutative. Associativity follows in a similar fashion. Now we show that if C is a cut then C + 0 = C. For if c ∈ C and z ∈ 0 then c + z < c + 0 = c hence C + 0 ⊆ C. Also, if c′ ∈ C then choose a d′ ∈ C such that c′ < d′ . Then c′ − d′ < 0 so c′ − d′ ∈ 0. And c′ = d′ + (c′ − d′ ). Hence C ⊆ C + 0. We conclude that C + 0 = C. Finally, for Axiom A5, we let C be a cut and set −C to be equal to {d ∈ Q : c + d < 0 for all c ∈ C}.

The other field axioms for C are easy to check. We conclude that the number system C forms a field. You will prove in the exercises that it is not possible to order this field. If α is a real number then we associate α with the complex number (α, 0). Thus we have the natural “embedding” R ∋ α −→ (α, 0) ∈ C . In this way, we can think of the real numbers as a subset of the complex numbers. In fact, the real field R is a subfield of the complex field C. This means that if α, β ∈ R and (α, 0), (β, 0) are the corresponding elements in C then α + β corresponds to (α + β, 0) and α · β corresponds to (α · β, 0).