Introduction to Analysis in One Variable. Michael E. Taylor

Transcription

1 Introduction to Analysis in One Variable Michael E. Taylor 1

2 2 Contents Chapter I. Numbers 1. Peano arithmetic 2. The integers 3. Prime factorization and the fundamental theorem of arithmetic 4. The rational numbers 5. Sequences 6. The real numbers 7. Irrational numbers 8. Cardinal numbers 9. Metric properties of R 10. Complex numbers Chapter II. Spaces 1. Euclidean spaces 2. Metric spaces 3. Compactness A. The Baire category theorem Chapter III. Functions 1. Continuous functions 2. Sequences and series of functions 3. Power series 4. Spaces of functions A. Absolutely convergent series Chapter IV. Calculus 1. The derivative 2. The integral 3. Power series 4. Curves and arc length 5. Exponential and trigonometric functions 6. Unbounded integrable functions A. The fundamental theorem of algebra B. π 2 is irrational C. More on (1 x) b D. Archimedes approximation of π E. Computing π using arctangents F. Power series of tan x G. Abel s power series theorem

3 3 Chapter V. Further topics in analysis 1. Convolutions and bump functions 2. The Weierstrass approximation theorem 3. The Stone-Weierstrass theorem 4. Fourier series 5. Newton s method A. Inner product spaces

4 4 Introduction This is a text for students who have had a three course calculus sequence, and who are ready for a course that explores the logical structure of this area of mathematics, which forms the backbone of analysis. This is intended for a one semester course. An accompanying text, Introduction to Analysis in Several Variables [T2], can be used in the second semester of a one year sequence. The main goal of Chapter 1 is to develop the real number system. We start with a treatment of the natural numbers N, obtaining its structure from a short list of axioms, the primary one being the principle of induction. Then we construct the set Z of all integers, which has a richer algebraic structure, and proceed to construct the set Q of rational numbers, which are quotients of integers (with a nonzero denominator). After discussing infinite sequences of rational numbers, including the notions of convergent sequences and Cauchy sequences, we construct the set R of real numbers, as ideal limits of Cauchy sequences of rational numbers. At the heart of this chapter is the proof that R is complete, i.e., Cauchy sequences of real numbers always converge to a limit in R. This provides the key to studying other metric properties of R, such as the compactness of (nonempty) closed, bounded subsets. We end Chapter 1 with a section on the set C of complex numbers. Many introductions to analysis shy away from the use of complex numbers. My feeling is that this forecloses the study of way too many beautiful results that can be appreciated at this level. This is not a course in complex analysis. That is for another course, and with another text (such as [T3]). However, I think the use of complex numbers in this text both simplifies the treatment of a number of key concepts, and extends their scope in natural and useful ways. In fact, the structure of analysis is revealed more clearly by moving beyond R and C, and we undertake this in Chapter 2. We start with a treatment of n-dimensional Euclidean space, R n. There is a notion of Euclidean distance between two points in R n, leading to notions of convergence and of Cauchy sequences. The spaces R n are all complete, and again closed bounded sets are compact. Going through this sets one up to appreciate a further generalization, the notion of a metric space, introduced in 2. This is followed by 3, exploring the notion of compactness in a metric space setting. Chapter 3 deals with functions. It starts in a general setting, of functions from one metric space to another. We then treat infinite sequences of functions, and study the notion of convergence, particularly of uniform convergence of a sequence of functions. We move on to infinite series. In such a case, we take the target space to be R n, so we can add functions. Section 3 treats power series. Here, we study series of the form (1) a k (z z 0 ) k, k=0 with a k C and z running over a disk in C. For results obtained in this section, regarding the radius of convergence R and the continuity of the sum on D R (z 0 ) = {z C : z z 0 <

5 R}, there is no extra difficulty in allowing a k and z to be complex, rather than insisting they be real, and the extra level of generality will pay big dividends in Chapter 4. A final section in Chapter 3 is devoted to spaces of functions, illustrating the utility of studying spaces beyond the case of R n. Chapter 4 gets to the heart of the matter, a rigorous development of differential and integral calculus. We define the derivative in 1, and prove the Mean Value Theorem, making essential use of compactness of a closed, bounded interval and its consequences, established in earlier chapters. This result has many important consequences, such as the Inverse Function Theorem, and especially the Fundamental Theorem of Calculus, established in 2, after the Riemann integral is introduced. In 3, we return to power series, this time of the form 5 (2) a k (t t 0 ) k. k=0 We require t and t 0 to be in R, but still allow a k C. Results on radius of convergence R and continuity of the sum f(t) on (t 0 R, t 0 + R) follow from material in Chapter 3. The essential new result in 3 of Chapter 4 is that one can obtain the derivative f (t) by differentiating the power series for f(t) term by term. In 4 we consider curves in R n, and obtain a formula for arc length for a smooth curve. We show that a smooth curve with nonvanishing velocity can be parametrized by arc length. When this is applied to the unit circle in R 2 centered at the origin, one is looking at the standard definition of the trigonometric functions, (3) C(t) = (cos t, sin t). We provide a demonstration that (4) C (t) = ( sin t, cos t) that is much shorter than what is usually presented in calculus texts. In 5 we move on to exponential functions. We derive the power series for the function e t, introduced to solve the differential equation dx/dt = x. We then observe that with no extra work we get an analogous power series for e at, with derivative ae at, and that this works for complex a as well as for real a. It is a short step to realize that e it is a unit speed curve tracing out the unit circle in C R 2, so comparison with (3) gives Euler s formula (5) e it = cos t + i sin t. That the derivative of e it is ie it provides a second proof of (4). Thus we have a unified treatment of the exponential and trigonometric functions, carried out further in 5, with details developed in numerous exercises. Section 6 extends the scope of the Riemann integral to a class of unbounded functions. Chapter 4 has several appendices, one proving the fundamental theorem of algebra, one showing that π is irrational, one exploring in

6 6 more detail than in 3 the power series for (1 x) b, and one describing an approximation to π pioneered by Archimedes. Chapter 5 treats further topics in analysis. If time permits, the instructor might cover one or more of these at the end of the course. The topics center around approximating functions, via various infinite sequences or series. Topics include approximating continuous functions by polynomials, Fourier series, and Newton s method for approximating the inverse of a given function.

7 7 Chapter I Numbers Introduction One foundation for a course in analysis is a solid understanding of the real number system. Texts vary on just how to achieve this. Some take an axiomatic approach. In such an approach, the set of real numbers is hypothesized to have a number of properties, including various algebraic properties satisfied by addition and multiplication, order axioms, and, crucially, the completeness property, sometimes expressed as the supremum property. This is not the approach we will take. Rather, we will start with a small list of axioms for the natural numbers (i.e., the positive integers), and then build the rest of the edifice logically, obtaining the basic properties of the real number system, particularly the completeness property, as theorems. Sections 1 3 deal with the integers, starting in 1 with the set N of natural numbers. The development proceeds from axioms of G. Peano. The main one is the principle of mathematical induction. We deduce basic results about integer arithmetic from these axioms. A high point is the fundamental theorem of arithmetic, presented in 3. Section 4 discusses the set Q of rational numbers, deriving the basic algebraic properties of these numbers from the results of 1 3. Section 5 provides a bridge between 4 and 6. It deals with infinite sequences, including convergent sequences and Cauchy sequences. This prepares the way for 6, the main section of this chapter. Here we construct the set R of real numbers, as ideal limits of rational numbers. We extend basic algebraic results from Q to R. Furthermore, we establish the result that R is complete, i.e., Cauchy sequences always have limits in R. Section 7 provides examples of irrational numbers, such as 2, 3, 5,... Section 8 deals with cardinal numbers, an extension of the natural numbers N, that can be used to count elements of a set, not necessarily finite. For example, N is a countably infinite set, and so is Q. We show that R uncountable, and hence much larger than N or Q. Section 9 returns to the real number line R, and establishes further metric properties of R and various subsets, with an emphasisis on the notion of compactness. The completeness property established in 6 plays a crucial role here. Section 10 introduces the set C of complex numbers and establishes basic algebraic and metric properties of C. While some introductory treatments of analysis avoid complex numbers, we embrace them, and consider their use in basic analysis too precious to omit. Sections 9 and 10 also have material on continuous functions, defined on a subset of R or C, respectively. These results give a taste of further results to be developed in Chapter 3, which will be essential to material in Chapters 4 and 5.

8 8 1. Peano arithmetic In Peano arithmetic, we assume we have a set N (the natural numbers). We assume given 0 / N, and form Ñ = N {0}. We assume there is a map (1.1) s : Ñ N, which is bijective. That is to say, for each k N, there is a j Ñ such that s(j) = k, so s is surjective; and furthermore, if s(j) = s(j ) then j = j, so s is injective. The map s plays the role of addition by 1, as we will see below. The only other axiom of Peano arithmetic is that the principle of mathematical induction holds. In other words, if S Ñ is a set with the properties (1.2) 0 S, k S s(k) S, then S = Ñ. Actually, applying the induction principle to S = {0} s(ñ), we see that it suffices to assume that s in (1.1) is injective; the induction principle ensures that it is surjective. We define addition x + y, for x, y Ñ, inductively on y, by (1.3) x + 0 = x, x + s(y) = s(x + y). Next, we define multiplication x y, inductively on y, by (1.4) x 0 = 0, x s(y) = x y + x. We also define (1.5) 1 = s(0). We now establish the basic laws of arithmetic. Proposition 1.1. x + 1 = s(x). Proof. x + s(0) = s(x + 0). Proposition x = x. Proof. Use induction on x. First, = 0. Now, assuming 0 + x = x, we have 0 + s(x) = s(0 + x) = s(x).

9 9 Proposition 1.3. s(y + x) = s(y) + x. Proof. Use induction on x. First, s(y + 0) = s(y) = s(y) + 0. Next, we have s(y + s(x)) = ss(y + x), s(y) + s(x) = s(s(y) + x). If s(y + x) = s(y) + x, the two right sides are equal, so the two left sides are equal, completing the induction. Proposition 1.4. x + y = y + x. Proof. Use induction on y. The case y = 0 follows from Proposition 1.2. Now, assuming x + y = y + x, for all x Ñ, we must show s(y) has the same property. In fact, x + s(y) = s(x + y) = s(y + x), and by Proposition 1.3 the last quantity is equal to s(y) + x. Proposition 1.5. (x + y) + z = x + (y + z). Proof. Use induction on z. First, (x + y) + 0 = x + y = x + (y + 0). Now, assuming (x + y) + z = x + (y + z), for all x, y Ñ, we must show s(z) has the same property. In fact, (x + y) + s(z) = s((x + y) + z), and we perceive the desired identity. x + (y + s(z)) = x + s(y + z) = s(x + (y + z)), Remark. Propositions 1.4 and 1.5 state the commutative and associative laws for addition. We now establish some laws for multiplication. Proposition 1.6. x 1 = x. Proof. We have x s(0) = x 0 + x = 0 + x = x, the last identity by Proposition 1.2. Proposition y = 0. Proof. Use induction on y. First, 0 0 = 0. Next, assuming 0 y = 0, we have 0 s(y) = 0 y + 0 = = 0.

10 10 Proposition 1.8. s(x) y = x y + y. Proof. Use induction on y. First, s(x) 0 = 0, while x = = 0. Next, assuming s(x) y = x y + y, for all x, we must show that s(y) has this property. In fact, s(x) s(y) = s(x) y + s(x) = (x y + y) + (x + 1), x s(y) + s(y) = (x y + x) + (y + 1), and identity then follows via the commutative and associative laws of addition, Propositions 1.4 and 1.5. Proposition 1.9. x y = y x. Proof. Use induction on y. First, x 0 = 0 = 0 x, the latter identity by Proposition 1.7. Next, assuming x y = y x for all x Ñ, we must show that s(y) has the same property. In fact, x s(y) = x y + x = y x + x, the last identity by Proposition 1.8. s(y) x = y x + x, Proposition (x + y) z = x z + y z. Proof. Use induction on z. First, the identity clearly holds for z = 0. Next, assuming it holds for z (for all x, y Ñ), we must show it holds for s(z). In fact, (x + y) s(z) = (x + y) z + (x + y) = (x z + y z) + (x + y), x s(z) + y s(z) = (x z + x) + (y z + y), and the desired identity follows from the commutative and associative laws of addition. Proposition (x y) z = x (y z). Proof. Use induction on z. First, the identity clearly holds for z = 0. Next, assuming it holds for z (for all x, y Ñ), we have while (x y) s(z) = (x y) z + x y, x (y s(z)) = x (y z + y) = x (y z) + x y, the last identity by Proposition 1.10 (and 1.9). These observations yield the desired identity. Remark. Propositions 1.9 and 1.11 state the commutative and associative laws for multiplication. Proposition 1.10 is the distributive law. Combined with Proposition 1.9, it also yields z (x + y) = z x + z y, used above. We next demonstrate the cancellation law of addition:

11 11 Proposition Given x, y, z Ñ, (1.6) x + y = z + y = x = z. Proof. Use induction on y. If y = 0, (1.6) obviously holds. Assuming (1.6) holds for y, we must show that (1.7) x + s(y) = z + s(y) implies x = z. In fact, (1.7) is equivalent to s(x+y) = s(z +y). Since the map s is assumed to be one-to-one, this implies that x + y = z + y, so we are done. We next define an order relation on Ñ. Given x, y Ñ, we say (1.8) x < y y = x + u, for some u N. Similarly there is a definition of x y. We have x y if and only if y R x, where (1.9) R x = {x + u : u Ñ}. Other notation is y > x x < y, y x x y. Proposition If x y and y x then x = y. Proof. The hypotheses imply (1.10) y = x + u, x = y + v, u, v Ñ. Hence x = x + u + v, so, by Proposition 1.12, u + v = 0. Now, if v 0, then v = s(w), so u + v = s(u + w) N. Thus v = 0, and u = 0. Proposition Given x, y Ñ, either (1.11) x < y, or x = y, or y < x, and no two can hold. Proof. That no two of (1.11) can hold follows from Proposition It remains to show that one must hold. Take y Ñ. We will establish (1.11) by induction on x. Clearly (1.11) holds for x = 0. We need to show that if (1.11) holds for a given x Ñ, then either (1.12) s(x) < y, or s(x) = y, or y < s(x). Consider the three possibilities in (1.11). If either y = x or y < x, then clearly y < s(x) = x + 1. On the other hand, if x < y, we can use the implication (1.12A) x < y = s(x) y to complete the proof of (1.12). See Lemma 1.17 for a proof of (1.12A). We can now establish the cancellation law for multiplication.

12 12 Proposition Given x, y, z Ñ, (1.13) x y = x z, x 0 = y = z. Proof. If y z, then either y < z or z < y. Suppose y < z, i.e., z = y + u, u N. Then the hypotheses of (1.13) imply hence, by Proposition 1.12, x y = x y + x u, x 0, (1.14) x u = 0, x 0. We thus need to show that (1.14) implies u = 0. In fact, if not, then we can write u = s(w), and x = s(a), with w, a Ñ, and we have (1.15) x u = x w + s(a) = s(x w + a) N. This contradicts (1.14), so we are done. Remark. Note that (1.15) implies (1.16) x, y N = x y N. We next establish the following variant of the principle of induction, called the wellordering property of Ñ. Proposition If T Ñ is nonempty, then T contains a smallest element. Proof. Suppose T contains no smallest element. Then 0 / T. Let (1.17) S = {x Ñ : x < y, y T }. Then 0 S. We claim that (1.18) x S = s(x) S. Indeed, suppose x S, so x < y for all y T. If s(x) / S, we have s(x) y 0 for some y 0 T. On the other hand (see Lemma 1.17 below), (1.19) x < y 0 = s(x) y 0. Thus, by Proposition 1.13, (1.20) s(x) = y 0. It follows that y 0 must be the smallest element of T. Thus, if T has no smallest element, (1.18) must hold. The induction principle then implies that S = Ñ, which implies T is empty. Here is the result behind (1.12A) and (1.19).

13 13 Lemma Given x, y Ñ, (1.21) x < y = s(x) y. Proof. Indeed, x < y y = x + u with u N, hence u = s(v), so hence s(x) y. y = x + s(v) = s(x + v) = s(x) + v, Remark. Proposition 1.16 has a converse, namely, the assertion (1.22) T Ñ nonempty = T contains a smallest element implies the principle of induction: ) (1.23) (0 S Ñ, k S s(k) S = S = Ñ. To see this, suppose S satisfies the hypotheses of (1.23), and let T = Ñ \ S. If S Ñ, then T is nonempty, so (1.22) implies T has a smallest element, say x 1. Since 0 S, x 1 N, so x 1 = s(x 0 ), and we must have (1.24) x 0 S, s(x 0 ) T = Ñ \ S, contradicting the hypotheses of (1.23). Exercises Given n N, we define n k=1 a k inductively, as follows. (1.25) 1 a k = a 1, k=1 n+1 ( n ) a k = a k + a n+1. k=1 k=1 Use the principle of induction to establish the following identities. n (1) 2 k = n(n + 1). k=1 n (2) 6 k 2 = n(n + 1)(2n + 1). k=1

14 14 (3) (a 1) In (3), we define a n inductively by n a k = a n+1 a, if a 1. k=1 (1.26) a 1 = a, a n+1 = a n a. We also set a 0 = 1 if a N, and n k=0 a k = a 0 + n k=1 a k. Verify that (4) (a 1) n a k = a n+1 1, if a 1. k=0 5. Given k N, show that 2 k 2k, with strict inequality for k > Show that, for x, x, y, y Ñ, x < x, y y = x + y < x + y, and x y < x y, if also y > Show that the following variant of the principle of induction holds: ( ) 1 S N, k S s(k) S = S = N. Hint. Consider {0} S Ñ. More generally, with R x as in (1.9), show that, for x N, Hint. Use induction on x. ( x S R x, ) k S s(k) S = S = R x. 8. With a n defined inductively as in (1.26) for a Ñ, n N, show that if also m N, Hint. Use induction on n. a m a n = a m+n, (a m ) n = a mn.

15 15 2. The integers An integer is thought of as having the form x a, with x, a Ñ. To be more formal, we will define an element of Z as an equivalence class of ordered pairs (x, a), x, a Ñ, where we define (2.1) (x, a) (y, b) x + b = y + a. We claim (2.1) is an equivalence relation. In general, an equivalence relation on a set S is a specification s t for certain s, t S, which satisfies the following three conditions. (a) Reflexive. s s, s S. (b) Symmetric. s t t s. (c) Transitive. s t, t u = s u. We will encounter various equivalence relations in this and subsequent sections. Generally, (a) and (b) are quite easy to verify, and we will be content with verifying (c). Proposition 2.1. The relation (2.1) is an equivalence relation. Proof. We need to check that (2.2) (x, a) (y, b), (y, b) (z, c) = (x, a) (z, c), i.e., that, for x, y, z, a, b, c Ñ, (2.3) x + b = y + a, y + c = z + b = x + c = z + a. In fact, the hypotheses of (2.3), and the results of 1, imply (x + c) + (y + b) = (z + a) + (y + b), and the conclusion of (2.3) then follows from the cancellation property, Proposition Let us denote the equivalence class containing (x, a) by [(x, a)]. We then define addition and multiplication in Z to satisfy (2.4) [(x, a)] + [(y, b)] = [(x, a) + (y, b)], [(x, a)] [(y, b)] = [(x, a) (y, b)], (x, a) + (y, b) = (x + y, a + b), (x, a) (y, b) = (xy + ab, ay + xb). To see that these operations are well defined, we need:

16 16 Proposition 2.2. If (x, a) (x, a ) and (y, b) (y, b ), then (2.5) (x, a) + (y, b) (x, a ) + (y, b ), and (2.6) (x, a) (y, b) (x, a ) (y, b ). Proof. The hypotheses say (2.7) x + a = x + a, y + b = y + b. The conclusions follow from results of 1. In more detail, adding the two identities in (2.7) gives x + a + y + b = x + a + y + b, and rearranging, using the commutative and associative laws of addition, yields (x + y) + (a + b ) = (x + y ) + (a + b), implying (2.5). The task of proving (2.6) is simplified by going through the intermediate step (2.8) (x, a) (y, b) (x, a ) (y, b). If x > x, so x = x + u, u N, then also a = a + u, and our task is to prove (xy + ab, ay + xb) (xy + uy + ab + ub, ay + uy + xb + ub), which is readily done. Having (2.8), we apply similar reasoning to get and then (2.6) follows by transitivity. (x, a ) (y, b) (x, a ) (y, b ), Similarly, it is routine to verify the basic commutative, associative, etc. laws incorporated in the next proposition. To formulate the results, set (2.9) m = [(x, a)], n = [(y, b)], k = [(z, c)] Z. Also, define (2.10) 0 = [(0, 0)], 1 = [(1, 0)], and (2.11) m = [(a, x)].

17 17 Proposition 2.3. We have (2.12) m + n = n + m, (m + n) + k = m + (n + k), m + 0 = m, m + ( m) = 0, mn = nm, m(nk) = (mn)k, m 1 = m, m 0 = 0, m ( 1) = m, m (n + k) = m n + m k. To give an example of a demonstration of these results, the identity mn = nm is equivalent to (xy + ab, ay + xb) (yx + ba, bx + ya). In fact, commutative laws for addition and multiplication in Ñ imply xy + ab = yx + ba and ay + xb = bx + ya. Verification of the other identities in (2.12) is left to the reader. We next establish the cancellation law for addition in Z. Proposition 2.4. Given m, n, k Z, (2.13) m + n = k + n = m = k. Proof. We give two proofs. For one, we can add n to both sides and use the results of Proposition 2.3. Alternatively, we can write the hypotheses of (2.13) as x + y + c + b = z + y + a + b and use Proposition 1.12 to deduce that x + c = z + a. Note that it is reasonable to set (2.14) m n = m + ( n). This defines subtraction on Z. There is a natural injection (2.15) N Z, x [(x, 0)], whose image we identify with N. Note that the map (2.10) preserves addition and multiplication. There is also an injection x [(0, x)], whose image we identify with N.

18 18 Proposition 2.5. We have a disjoint union: (2.16) Z = N {0} ( N). Proof. Suppose m Z; write m = [(x, a)]. By Proposition 1.14, either a < x, or x = a, or x < a. In these three cases, x = a + u, u N, or x = a, or a = x + v, v N. Then, either (x, a) (u, 0), or (x, a) (0, 0), or (x, a) (0, v). We define an order on Z by: (2.17) m < n n m N. We then have: Corollary 2.6. Given m, n Z, then either (2.18) m < n, or m = n, or n < m, and no two can hold. The map (2.15) is seen to preserve order relations. Another consequence of (2.16) is the following. Proposition 2.7. If m, n Z and m n = 0, then either m = 0 or n = 0. Proof. Suppose m 0 and n 0. We have four cases: m > 0, n > 0 = mn > 0, m < 0, n < 0 = mn = ( m)( n) > 0, m > 0, n < 0 = mn = m( n) < 0, m < 0, n > 0 = mn = ( m)n < 0, the first by (1.16), and the rest with the help of Exercise 3 below. This finishes the proof. Using Proposition 2.7, we have the following cancellation law for multiplication in Z.

19 19 Proposition 2.8. Given m, n, k Z, (2.19) mk = nk, k 0 = m = n. Proof. First, mk = nk mk nk = 0. Now mk nk = (m n)k. See Exercise 3 below. Hence mk = nk = (m n)k = 0. Given k 0, Proposition 2.7 implies m n = 0. Hence m = n. Exercises 1. Verify Proposition We define n k=1 a k as in (1.25), this time with a k Z. We also define a k inductively as in Exercise (3) of 1, with a 0 = 1 if a 0. Use the principle of induction to establish the identity n ( 1) k 1 k = m if n = 2m, k=1 3. Show that, if m, n, k Z, m + 1 if n = 2m + 1. (nk) = ( n)k, and mk nk = (m n)k. Hint. For the first part, use Proposition 2.3 to show that nk + ( n)k = 0. Alternatively, compare (a, x) (y, b) with (x, a) (y, b). 4. Deduce the following from Proposition Let S Z be nonempty and assume there exists m Z such that m < n for all n S. Then S has a smallest element. Hint. Given such m, let S = {( m) + n : n S}. Show that S N and deduce that S has a smallest element. 5. Show that Z has no smallest element.

20 20 3. Prime factorization and the fundamental theorem of arithmetic Let x N. We say x is composite if one can write (3.1) x = ab, a, b N, with neither a nor b equal to 1. If x 1 is not composite, it is said to be prime. If (3.1) holds, we say a x (and that b x), or that a is a divisor of x. Given x N, x > 1, set (3.2) D x = {a N : a x, a > 1}. Thus x D x, so D x is non-empty. By Proposition 1.16, D x contains a smallest element, say p 1. Clearly p 1 is a prime. Set (3.3) x = p 1 x 1, x 1 N, x 1 < x. The same construction applies to x 1, which is > 1 unless x = p 1. Hence we have either x = p 1 or (3.4) x 1 = p 2 x 2, p 2 prime, x 2 < x 1. Continue this process, passing from x j to x j+1 as long as x j is not prime. The set S of such x j N has a smallest element, say x µ 1 = p µ, and we have (3.5) x = p 1 p 2 p µ, p j prime. This is part of the Fundamental Theorem of Arithmetic: Theorem 3.1. Given x N, x 1, there is a unique product expansion (3.6) x = p 1 p µ, where p 1 p µ are primes. Only uniqueness remains to be established. This follows from: Proposition 3.2. Assume a, b N, and p N is prime. Then (3.7) p ab = p a or p b. We will deduce this from:

21 Proposition 3.3. If p N is prime and a N, is not a multiple of p, or more generally if p, a N have no common divisors > 1, then there exist m, n Z such that (3.8) ma + np = Proof of Proposition 3.2. Assume p is a prime which does not divide a. Pick m, n such that (3.8) holds. Now, multiply (3.8) by b, to get Thus, if p ab, i.e., ab = pk, we have so p b, as desired. To prove Proposition 3.3, let us set mab + npb = b. p(mk + nb) = b, (3.9) Γ = {ma + np : m, n Z}. Clearly Γ satisfies the following criterion: Definition. A nonempty subset Γ Z is a subgroup of Z provided (3.10) a, b Γ = a + b, a b Γ. Proposition 3.4. If Γ Z is a subgroup, then either Γ = {0}, or there exists x N such that (3.11) Γ = {mx : m Z}. Proof. Note that n Γ n Γ, so, with Σ = Γ N, we have a disjoint union Γ = Σ {0} ( Σ). If Σ, let x be its smallest element. Then we want to establish (3.11), so set Γ 0 = {mx : m Z}. Clearly Γ 0 Γ. Similarly, set Σ 0 = {mx : m N} = Γ 0 N. We want to show that Σ 0 = Σ. If y Σ \ Σ 0, then we can pick m 0 N such that and hence m 0 x < y < (m 0 + 1)x, y m 0 x Σ is smaller than x. This contradiction proves Proposition 3.4.

22 22 Proof of Proposition 3.3. Taking Γ as in (3.9), pick x N such that (3.11) holds. Since a Γ and p Γ, we have a = m 0 x, p = m 1 x for some m j Z. The assumption that a and p have no common divisor > 1 implies x = 1. We conclude that 1 Γ, so (3.8) holds. Exercises 1. Prove that there are infinitely many primes. Hint. If {p 1,..., p m } is a complete list of primes, consider What are its prime factors? x = p 1 p m Referring to (3.10), show that a nonempty subset Γ Z is a subgroup of Z provided (3.12) a, b Γ = a b Γ. Hint. a Γ 0 = a a Γ a = 0 a Γ, given (3.12). 3. Let n N be a 12 digit integer. Show that if n is not prime, then it must be divisible by a prime p < Determine whether the following number is prime: (3.13) Hint. This is for the student who can use a computer. 5. Find the smallest prime larger than the number in (3.13). Hint. Same as above.

23 23 4. The rational numbers A rational number is thought of as having the form m/n, with m, n Z, n 0. Thus, we will define an element of Q as an equivalence class of ordered pairs m/n, m Z, n Z\{0}, where we define (4.1) m/n a/b mb = an. Proposition 4.1. This is an equivalence relation. Proof. We need to check that (4.2) m/n a/b, a/b c/d = m/n c/d, i.e., that, for m, a, c Z, n, b, d Z \ {0}, (4.3) mb = an, ad = cb = md = cn. Now the hypotheses of (4.3) imply (mb)(ad) = (an)(cb), hence (md)(ab) = (cn)(ab). We are assuming b 0. If also a 0, then ab 0, and the conclusion of (4.3) follows from the cancellation property, Proposition 2.8. On the other hand, if a = 0, then m/n a/b mb = 0 m = 0 (since b 0), and similarly a/b c/d cb = 0 c = 0, so the desired implication also holds in that case. We will (temporarily) denote the equivalence class containing m/n by [m/n]. We then define addition and multiplication in Q to satisfy (4.4) [m/n] + [a/b] = [(m/n) + (a/b)], (m/n) + (a/b) = (mb + na)/(nb), [m/n] [a/b] = [(m/n) (a/b)], (m/n) (a/b) = (ma)/(nb). To see that these operations are well defined, we need: Proposition 4.2. If m/n m /n and a/b a /b, then (4.4A) (m/n) + (a/b) (m /n ) + (a /b ), and (4.4B) (m/n) (a/b) (m /n ) (a /b ).

24 24 Proof. The hypotheses say (4.4C) mn = m n, ab = a b. The conclusions follow from the results of 2. In more detail, multiplying the two identities in (4.4C) yields man b = m a nb, which implies (4.4B). To prove (4.4A), it is convenient to establish the intermediate step (4.4D) This is equivalent to hence to or to (m/n) + (a/b) (m /n ) + (a/b). (mb + na)/nb (m b + n a)/(n b), (mb + na)n b = (m b + n a)nb, mn bb + nn ab = m nbb + n nab. This in turn follows readily from (4.4C). Having (4.4D), we can use a similar argument to establish that (m /n ) + (a/b) (m /n ) + (a /b ), and then (4.4A) follows by transitivity of. From now on, we drop the brackets, simply denoting the equivalence class of m/n by m/n, and writing (4.1) as m/n = a/b. We also may denote an element of Q by a single letter, e.g., x = m/n. There is an injection (4.5) Z Q, m m/1, whose image we identify with Z. This map preserves addition and multiplication. define We (4.6) (m/n) = ( m)/n, and, if x = m/n 0, (i.e., m 0 as well as n 0), we define (4.7) x 1 = n/m. The results stated in the following proposition are routine consequences of the results of 2.

25 25 Proposition 4.3. Given x, y, z Q, we have x + y = y + x, (x + y) + z = x + (y + z), x + 0 = x, x + ( x) = 0, x y = y x, (x y) z = x (y z), x 1 = x, x 0 = 0, x ( 1) = x, x (y + z) = x y + x z. Furthermore, x 0 = x x 1 = 1. For example, if x = m/n, y = a/b with m, n, a, b Z, n, b 0, the identity x y = y x is equivalent to (ma)/(nb) (am)/(bn). In fact, the identities ma = am and nb = bn follow from Proposition 2.3. We leave the rest of Proposition 4.3 to the reader. We also have cancellation laws: Proposition 4.4. Given x, y, z Q, (4.8) x + y = z + y = x = z. Also, (4.9) xy = zy, y 0 = x = z. Proof. To get (4.8), add y to both sides of x+y = z +y and use the results of Proposition 4.3. To get (4.9), multiply both sides of x y = z y by y 1. It is natural to define (4.10) x y = x + ( y), and, if y 0, (4.11) x/y = x y 1. We now define the order relation on Q. Set (4.12) Q + = {m/n : mn > 0},

26 26 where, in (4.12), we use the order relation on Z, discussed in 2. This is well defined. In fact, if m/n = m /n, then mn = m n, hence (mn)(m n ) = (mn ) 2, and therefore mn > 0 m n > 0. Results of 2 imply that (4.13) Q = Q + {0} ( Q + ) is a disjoint union, where Q + = { x : x Q + }. Also, clearly (4.14) x, y Q + = x + y, xy, x y Q+. We define (4.15) x < y y x Q +, and we have, for any x, y Q, either (4.16) x < y, or x = y, or y < x, and no two can hold. The map (4.5) is seen to preserve the order relations. In light of (4.14), we see that (4.17) given x, y > 0, x < y x y < 1 1 y < 1 x. As usual, we say x y provided either x < y or x = y. Similarly there are natural definitions of x > y and of x y. The following result implies that Q has the Archimedean property. Proposition 4.5. Given x Q, there exists k Z such that (4.18) k 1 < x k. Proof. It suffices to prove (4.18) assuming x Q + ; otherwise, work with x (and make a few minor adjustments). Say x = m/n, m, n N. Then S = {l N : l x} contains m, hence is nonempty. By Proposition 1.16, S has a smallest element; call it k. Then k x. We cannot have k 1 x, for then k 1 would belong to S. Hence (4.18) holds. Exercises 1. Verify Proposition 4.3.

27 27 2. Look at the exercise set for 1, and verify (3) and (4) for a Q, a 1, n N. 3. Here is another route to (4) of 1, i.e., (4.19) n k=0 a k = an+1 1 a 1, a 1. Denote the left side of (4.19) by S n (a). Multiply by a and show that as n (a) = S n (a) + a n Given a Q, n N, define a n as in Exercise 3 of 1. If a 0, set a 0 = 1 and a n = (a 1 ) n, with a 1 defined as in (4.7). Show that, if a, b Q \ 0, a j+k = a j a k, a jk = (a j ) k, (ab) j = a j b j, j, k Z. 5. Prove the following variant of Proposition 4.5. Proposition 4.5A. Given ε Q, ε > 0, there exists n N such that ε > 1 n. 6. Work through the proof of the following. Assertion If x = m/n Q, then x 2 2. Hint. We can arrange that m and n have no common factors. Then ( m n ) 2 = 2 m 2 = 2n 2 m even (m = 2k) 4k 2 = 2n 2 n 2 = 2k 2 n even. Contradiction? (See Proposition 7.2 for a more general result.) 7. Given x j, y j Q, show that x 1 < x 2, y 1 y 2 = x 1 + y 1 < x 2 + y 2. Show that 0 < x 1 < x 2, 0 < y 1 y 2 = x 1 y 1 < x 2 y 2.

28 28 5. Sequences In this section, we discuss infinite sequences. For now, we deal with sequences of rational numbers, but we will not explicitly state this restriction below. In fact, once the set of real numbers is constructed in 6, the results of this section will be seen to hold also for sequences of real numbers. Definition. A sequence (a j ) is said to converge to a limit a provided that, for any n N, there exists K(n) such that (5.1) j K(n) = a j a < 1 n. We write a j a, or a = lim a j, or perhaps a = lim j a j. Here, we define the absolute value x of x by (5.2) x = x if x 0, x if x < 0. The absolute value function has various simple properties, such as xy = x y, which follow readily from the definition. One basic property is the triangle inequality: (5.3) x + y x + y. In fact, if either x and y are both positive or they are both negative, one has equality in (5.3). If x and y have opposite signs, then x + y max( x, y ), which in turn is dominated by the right side of (5.3). Proposition 5.1. If a j a and b j b, then (5.4) a j + b j a + b, and (5.5) a j b j ab. If furthermore, b j 0 for all j and b = 0, then (5.6) a j /b j a/b. Proof. To see (5.4), we have, by (5.3), (5.7) (a j + b j ) (a + b) a j a + b j b.

29 29 To get (5.5), we have (5.8) a j b j ab = (a j b j ab j ) + (ab j ab) b j a j a + a b b j. The hypotheses imply b j B, for some B, and hence the criterion for convergence is readily verified. To get (5.6), we have (5.9) a j a 1 { b a aj + a b b j }. b j b b b j The hypotheses imply 1/ b j M for some M, so we also verify the criterion for convergence in this case. We next define the concept of a Cauchy sequence. Definition. A sequence (a j ) is a Cauchy sequence provided that, for any n N, there exists K(n) such that (5.10) j, k K(n) = a j a k 1 n. It is clear that any convergent sequence is Cauchy. On the other hand, we have: Proposition 5.2. Each Cauchy sequence is bounded. Proof. Take n = 1 in the definition above. Thus, if (a j ) is Cauchy, there is a K such that j, k K a j a k 1. Hence, j K a j a K + 1, so, for all j, a j M, M = max ( a 1,..., a K 1, a K + 1 ). Now, the arguments proving Proposition 5.1 also establish: Proposition 5.3. If (a j ) and (b j ) are Cauchy sequences, so are (a j + b j ) and (a j b j ). Furthermore, if, for all j, b j c for some c > 0, then (a j /b j ) is Cauchy. The following proposition is a bit deeper than the first three. Proposition 5.4. If (a j ) is bounded, i.e., a j M for all j, then it has a Cauchy subsequence. Proof. We may as well assume M N. Now, either a j [0, M] for infinitely many j or a j [ M, 0] for infinitely many j. Let I 1 be any one of these two intervals containing a j for infinitely many j, and pick j(1) such that a j(1) I 1. Write I 1 as the union of two closed intervals, of equal length, sharing only the midpoint of I 1. Let I 2 be any one of them with the property that a j I 2 for infinitely many j, and pick j(2) > j(1) such that a j(2) I 2. Continue, picking I ν I ν 1 I 1, of length M/2 ν 1, containing a j for infinitely many j, and picking j(ν) > j(ν 1) > > j(1) such that a j(ν) I ν. Setting b ν = a j(ν), we see that (b ν ) is a Cauchy subsequence of (a j ), since, for all k N, b ν+k b ν M/2 ν 1.

30 30 Proposition 5.5. Each bounded monotone sequence (a j ) is Cauchy. Proof. To say (a j ) is monotone is to say that either (a j ) is increasing, i.e., a j a j+1 for all j, or that (a j ) is decreasing, i.e., a j a j+1 for all j. For the sake of argument, assume (a j ) is increasing. By Proposition 5.4, there is a subsequence (b ν ) = (a j(ν) ) which is Cauchy. Thus, given n N, there exists K(n) such that (5.11) µ, ν K(n) = a j(ν) a j(µ) < 1 n. Now, if ν 0 K(n) and k j j(ν 0 ), pick ν 1 such that j(ν 1 ) k. Then a j(ν0 ) a j a k a j(ν1 ), so (5.12) k j j(ν 0 ) = a j a k < 1 n. We give a few simple but basic examples of convergent sequences. Proposition 5.6. If a < 1, then a j 0. Proof. Set b = a ; it suffices to show that b j 0. Consider c = 1/b > 1, hence c = 1 + y, y > 0. We claim that c j = (1 + y) j 1 + jy, for all j 1. In fact, this clearly holds for j = 1, and if it holds for j = k, then c k+1 (1 + y)(1 + ky) > 1 + (k + 1)y. Hence, by induction, the estimate is established. Consequently, b j < 1 jy, so the appropriate analogue of (5.1) holds, with K(n) = Kn, for any integer K > 1/y. Proposition 5.6 enables us to establish the following result on geometric series. Proposition 5.7. If x < 1 and a j = 1 + x + + x j, then a j 1 1 x.

31 31 Proof. Note that xa j = x + x x j+1, so (1 x)a j = 1 x j+1, i.e., The conclusion follows from Proposition 5.6. Note in particular that a j = 1 xj+1 1 x. (5.13) 0 < x < 1 = 1 + x + + x j < 1 1 x. It is an important mathematical fact that not every Cauchy sequence of rational numbers has a rational number as limit. We give one example here. Consider the sequence (5.14) a j = Then (a j ) is increasing, and a n+j a n = n+j l=n+1 j l=0 1 l!. 1 l! < 1 ( 1 n! n (n + 1) ) (n + 1) j, since (n + 1)(n + 2) (n + j) > (n + 1) j. Using (5.13), we have (5.15) a n+j a n < 1 (n + 1)! n+1 = 1 n! 1 n. Hence (a j ) is Cauchy. Taking n = 2, we see that (5.16) j > 2 = < a j < Proposition 5.8. The sequence (5.14) cannot converge to a rational number. Proof. Assume a j m/n with m, n N. By (5.16), we must have n > 2. Now, write (5.17) m n = n l=0 1 + r, r = lim l! (a n+j a n ). j Multiplying both sides of (5.17) by n! gives (5.18) m(n 1)! = A + r n! where (5.19) A = n l=0 n! l! N.

32 32 Thus the identity (5.17) forces r n! N, while (5.15) implies (5.20) 0 < r n! 1/n. This contradiction proves the proposition. Exercises 1. Show that and more generally for each n N, Hint. See Exercise Show that and more generally for each n N, lim k lim k lim k k 2 k = 0, k n 2 k = 0. 2 k k! = 0, lim k 2 nk k! = 0. The following two exercises discuss continued fractions. We assume (5.21) a j Q, a j 1, j = 1, 2, 3,..., and set (5.22) f 1 = a 1, f 2 = a a 2, f 3 = a a a 3,.... Having f j, we obtain f j+1 by replacing a j by a j + 1/a j+1. In other words, with (5.23) f j = φ j (a 1,..., a j ), given explicitly by (5.22) for j = 1, 2, 3, we have (5.24) f j+1 = φ j+1 (a 1,..., a j+1 ) = φ j (a 1,..., a j 1, a j + 1/a j+1 ). 3. Show that f 1 f j, j 2, and f 2 f j, j 3.

33 33 Going further, show that (5.25) f 1 f 3 f 5 f 6 f 4 f If also ã j+1 Q, ã j+1 1, show that (5.26) with (5.27) φ j+1 (a 1,..., a j, a j+1 ) φ j+1 (a 1,..., a j, ã j+1 ) = φ j (a 1,..., a j 1, b j ) φ j (a 1,..., a j 1, b j ), b j = a j + 1 a j+1, bj = a j + 1 ã j+1, b j b j = 1 a j+1 1 ã j+1 = ãj+1 a j+1 ã j+1 a j+1. Note that b j, b j > 1. Iterating this, show that (5.28) f 2j f 2j+1 0, as j. Deduce that (f j ) is a Cauchy sequence. 5. Suppose a sequence (a j ) has the property that there exist r < 1, K N such that j K = a j+1 a j r. Show that a j 0 as j. How does this result apply to Exercises 1 and 2? 6. If (a j ) satisfies the hypotheses of Exercise 5, show that there exists M < such that k a j M, k. j=1 Remark. This yields the ratio test for infinite series. 7. Show that you get the same criterion for convergence if (5.1) is replaced by j K(n) = a j a < 5 n. Generalize, and note the relevance for the proof of Proposition 5.1. observation to the criterion (5.10) for (a j ) to be Cauchy. Apply the same

34 34 6. The real numbers We think of a real number as a quantity that can be specified by a process of approximation arbitrarily closely by rational numbers. Thus, we define an element of R as an equivalence class of Cauchy sequences of rational numbers, where we define (6.1) (a j ) (b j ) a j b j 0. Proposition 6.1. This is an equivalence relation. Proof. This is a straightforward consequence of Proposition 5.1. In particular, to see that (6.2) (a j ) (b j ), (b j ) (c j ) = (a j ) (c j ), just use (5.4) of Proposition 5.1 to write a j b j 0, b j c j 0 = a j c j 0. We denote the equivalence class containing a Cauchy sequence (a j ) by [(a j )]. We then define addition and multiplication on R to satisfy (6.3) [(a j )] + [(b j )] = [(a j + b j )], [(a j )] [(b j )] = [(a j b j )]. Proposition 5.3 states that the sequences (a j + b j ) and (a j b j ) are Cauchy if (a j ) and (b j ) are. To conclude that the operations in (6.3) are well defined, we need: Proposition 6.2. If Cauchy sequences of rational numbers are given which satisfy (a j ) (a j ) and (b j) (b j ), then (6.4) (a j + b j ) (a j + b j), and (6.5) (a j b j ) (a jb j). The proof is a straightforward variant of the proof of parts (5.4)-(5.5) in Proposition 5.1, with due account taken of Proposition 5.2. For example, a j b j a j b j = a jb j a j b j + a j b j a j b j, and there are uniform bounds a j A, b j B, so a j b j a jb j a j b j b j + a j a j b j A b j b j + B a j a j.

35 35 There is a natural injection (6.6) Q R, a [(a, a, a,... )], whose image we identify with Q. This map preserves addition and multiplication. If x = [(a j )], we define (6.7) x = [( a j )]. For x 0, we define x 1 as follows. First, to say x 0 is to say there exists n N such that a j 1/n for infinitely many j. Since (a j ) is Cauchy, this implies that there exists K such that a j 1/2n for all j K. Now, if we set α j = a K+j, we have (α j ) (a j ); we propose to set (6.8) x 1 = [(α 1 j )]. We claim that this is well defined. First, by Proposition 5.3, (α 1 j ) is Cauchy. Furthermore, if for such x we also have x = [(b j )], and we pick K so large that also b j 1/2n for all j K, and set β j = b K+j, we claim that (6.9) (α 1 j ) (β 1 j ). Indeed, we have (6.10) α 1 j β 1 j = β j α j α j β j 4n2 β j α j, so (6.9) holds. It is now a straightforward exercise to verify the basic algebraic properties of addition and multiplication in R. We state the result. Proposition 6.3. Given x, y, z R, all the algebraic properties stated in Proposition 4.3 hold. For example, if x = [(a j )] and y = [(b j )], the identity xy = yx is equivalent to (a j b j ) (b j a j ). In fact, the identity a j b j = b j a j for a j, b j Q, follows from Proposition 4.3. The rest of Proposition 6.3 is left to the reader. As in (4.10)-(4.11), we define x y = x + ( y) and, if y 0, x/y = x y 1. We now define an order relation on R. Take x R, x = [(a j )]. From the discussion above of x 1, we see that, if x 0, then one and only one of the following holds. Either, for some n, K N, (6.11) j K = a j 1 2n,

36 36 or, for some n, K N, (6.12) j K = a j 1 2n. If (a j ) (b j ) and (6.11) holds for a j, it also holds for b j (perhaps with different n and K), and ditto for (6.12). If (6.11) holds, we say x R + (and we say x > 0), and if (6.12) holds we say x R (and we say x < 0). Clearly x > 0 if and only if x < 0. It is also clear that the map Q R in (6.6) preserves the order relation. Thus we have the disjoint union (6.13) R = R + {0} R, R = R +. Also, clearly (6.14) x, y R + = x + y, xy R +. As in (4.15), we define (6.15) x < y y x R +. If x = [(a j )] and y = [(b j )], we see from (6.11) (6.12) that (6.15A) x < y for some n, K N, j K b j a j 1 n ( i.e., a j b j 1 ). n The relation (6.15) can also be written y > x. Similarly we define x y and y x, in the obvious fashions. The following results are straightforward. Proposition 6.4. For elements of R, we have (6.16) x 1 < y 1, x 2 < y 2 = x 1 + x 2 < y 1 + y 2, (6.17) x < y y < x, (6.18) 0 < x < y, a > 0 = 0 < ax < ay, (6.19) 0 < x < y = 0 < y 1 < x 1. Proof. The results (6.16) and (6.18) follow from (6.14); consider, for example, a(y x). The result (6.17) follows from (6.13). To prove (6.19), first we see that x > 0 implies

37 37 x 1 > 0, as follows: if x 1 > 0, the identity x ( x 1 ) = 1 contradicts (6.14). As for the rest of (6.19), the hypotheses imply xy > 0, and multiplying both sides of x < y by a = (xy) 1 gives the result, by (6.18). As in (5.2), define x by (6.20) Note that x = x if x 0, x if x < 0. (6.20A) x = [(a j )] = x = [( a j )]. It is straightforward to verify (6.21) xy = x y, x + y x + y. We now show that R has the Archimedean property. Proposition 6.5. Given x R, there exists k Z such that (6.22) k 1 < x k. Proof. It suffices to prove (6.22) assuming x R +. Otherwise, work with x. Say x = [(a j )] where (a j ) is a Cauchy sequence of rational numbers. By Proposition 5.2, there exists M Q such that a j M for all j. By Proposition 4.5, we have M l for some l N. Hence the set S = {l N : l x} is nonempty. As in the proof of Proposition 4.5, taking k to be the smallest element of S gives (6.22). Proposition 6.6. Given any real ε > 0, there exists n N such that ε > 1/n. Proof. Using Proposition 6.5, pick n > 1/ε and apply (6.19). Alternatively, use the reasoning given above (6.8). We are now ready to consider sequences of elements of R. Definition. A sequence (x j ) converges to x if and only if, for any n N, there exists K(n) such that (6.23) j K(n) = x j x < 1 n. In this case, we write x j x, or x = lim x j. The sequence (x j ) is Cauchy if and only if, for any n N, there exists K(n) such that (6.24) j, k K(n) = x j x k < 1 n. We note that it is typical to phrase the definition above in terms of picking any real ε > 0 and demanding that, e.g., x j x < ε, for large j. The equivalence of the two definitions follows from Proposition 6.6. As in Proposition 5.2, we have that every Cauchy sequence is bounded. It is clear that, if each x j Q, then the notion that (x j ) is Cauchy given above coincides with that in 5. If also x Q, the notion that x j x also coincides with that given in 5. Here is another natural but useful observation.

38 38 Proposition 6.6A. If each a j Q, and x R, then (6.25) a j x x = [(a j )]. Proof. First assume x = [(a j )]. In particular, (a j ) is Cauchy. Now, given m, we have from (6.15A) that (6.26) x a k < 1 m K, n such that j K a j a k < 1 m 1 n = K such that j K a j a k < 1 2m. On the other hand, since (a j ) is Cauchy, for each m N, K(m) such that j, k K(m) a j a k < 1 2m. Hence k K(m) = x a k < 1 m. This shows that x = [(a j )] a j x. For the converse, if a j x, then (a j ) is Cauchy, so we have [(a j )] = y R. The previous argument implies a j y. But so x = y. Thus a j x x = [(a j )]. x y x a j + a j y, j, Next, the proof of Proposition 5.1 extends to the present case, yielding: Proposition 6.7. If x j x and y j y, then (6.27) x j + y j x + y, and (6.28) x j y j xy. If furthermore y j 0 for all j and y 0, then (6.28) x j /y j x/y. So far, statements made about R have emphasized similarities of its properties with corresponding properties of Q. The crucial difference between these two sets of numbers is given by the following result, known as the completeness property.

39 Theorem 6.8. If (x j ) is a Cauchy sequence of real numbers, then there exists x R such that x j x. Proof. Take x j = [(a jl : l N)] with a jl Q. Using (6.25), take a j,l(j) = b j Q such that (6.29) x j b j 2 j. Then (b j ) is Cauchy, since b j b k x j x k + 2 j + 2 k. Now, let (6.30) x = [(b j )]. It follows that (6.31) x j x x j b j + x b j 2 j + x b j, and hence x j x. If we combine Theorem 6.8 with the argument behind Proposition 5.4, we obtain the following important result, known as the Bolzano-Weierstrass Theorem. Theorem 6.9. Each bounded sequence of real numbers has a convergent subsequence. Proof. If x j M, the proof of Proposition 5.4 applies without change to show that (x j ) has a Cauchy subsequence. By Theorem 6.8, that Cauchy subsequence converges. Similarly, adding Theorem 6.8 to the argument behind Proposition 5.5 yields: Proposition Each bounded monotone sequence (x j ) of real numbers converges. A related property of R can be described in terms of the notion of the supremum of a set. Definition. If S R, one says that x R is an upper bound for S provided x s for all s S, and one says (6.32) x = sup S provided x is an upper bound for S and further x x whenever x is an upper bound for S. For some sets, such as S = Z, there is no x R satisfying (6.32). However, there is the following result, known as the supremum property. Proposition If S is a nonempty subset of R that has an upper bound, then there is a real x = sup S. Proof. We use an argument similar to the one in the proof of Proposition 5.4. Let x 0 be an upper bound for S, pick s 0 in S, and consider I 0 = [s 0, x 0 ] = {y R : s 0 y x 0 }. 39

40 40 If x 0 = s 0, then already x 0 = sup S. Otherwise, I 0 is an interval of nonzero length, L = x 0 s 0. In that case, divide I 0 into two equal intervals, having in common only the midpoint; say I 0 = I l 0 I r 0, where I r 0 lies to the right of I l 0. Let I 1 = I r 0 if S I r 0, and otherwise let I 1 = I l 0. Note that S I 1. Let x 1 be the right endpoint of I 1, and pick s 1 S I 1. Note that x 1 is also an upper bound for S. Continue, constructing I ν I ν 1 I 0, where I ν has length 2 ν L, such that the right endpoint x ν of I ν satisfies (6.33) x ν s, s S, and such that S I ν, so there exist s ν S such that (6.34) x ν s ν 2 ν L. The sequence (x ν ) is bounded and monotone (decreasing) so, by Proposition 6.10, it converges; x ν x. By (6.33), we have x s for all s S, and by (6.34) we have x s ν 2 ν L. Hence x satisfies (6.32). We turn to infinite series k=0 a k, with a k R. We say this series converges if and only if the sequence of partial sums (6.35) S n = converges: n k=0 a k (6.36) a k = A S n A as n. k=0 The following is a useful condition guaranteeing convergence. Proposition The infinite series k=0 a k converges provided (6.37) a k <, k=0 i.e., there exists B < such that n k=0 a k B for all n. Proof. The triangle inequality (the second part of (6.21)) gives, for l 1, (6.38) S n+l S n = n+l k=n+1 n+l k=n+1 a k a k,