DR.RUPNATHJI( DR.RUPAK NATH ) - PDF Free Download

Contents 1 Sets 1 2 The Real Numbers 9 3 Sequences 29 4 Series 59 5 Functions 81 6 Power Series 105 7 The elementary functions 111

Chapter 1 Sets It is very convenient to introduce some notation and terminology from set theory. A set is just a collection of objects which will usually be certain mathematical objects, such as numbers, points in the plane, functions or some such. If A denotes some given set and x denotes an object belonging to A, then this fact is indicated by the expression x A to be read as x belongs to A, or x is a member of A, or x is an element of A. If x denotes some object which does not belong to the set A, then this is indicated by the symbolism x / A and is read as x does not belong to A, or x is not a member of A, or x is not an element of A. To say that the sets A and B are equal is to say that they have the same elements. In other words, to say that A = B is to say both that if x A then also x B and if y B then also y A. We can write this as { x A = x B A = B is the same as y B = y A. The verification that given sets A and B are equal is made up of two parts. The first is the verification that every element of A is also an element of B and the second part is the verification that every element of B is also an element of A. We list a few examples of sets and also introduce some notation. 1

2 Chapter 1 Examples 1.1. 1. The set consisting of the three integers 2, 3, 4. We write this as { 2, 3, 4 }. 2. The set of natural numbers { 1, 2, 3, 4, 5, 6,... } (i.e., all strictly positive integers). This set is denoted by N. Notice that 0 / N. 3. The set of all real numbers, denoted by R. For example, 8, 11, 0, 5, 1 2, 1 3, π are elements of R. 4. The set of complex numbers is denoted by C. 5. The set of all integers (positive, negative and including zero) is denoted by Z. 6. The set of all rational numbers (all real numbers of the form m n for integers m, n with n 0) is denoted by Q. For example, the real numbers 3 4, 17 9, 0, 78, 3 belong to Q, but 2 / Q. 7. The set of even natural numbers { 2, 4, 6, 8,... }. This could also be written as { n N : n = 2m for some m N }. (The colon : stands for such that (or with the property that ), so this can be read as the set of all n in N such that n = 2m for some m in N.) 8. The set { x R : x > 1 } is the set of all those real numbers strictly greater than 1. 9. The set { z C : z = 1 } is the set of complex numbers with absolute value equal to 1. This is the unit circle in C ( the circle with centre at the origin and with radius equal to 1). Certain sets of real numbers, so-called intervals, are given a special notation with the use of round and square brackets. Let a R and b R and suppose that a < b. { x R : a x b } is denoted [ a, b] (closed interval) { x R : a < x < b } is denoted (a, b) (open interval) { x R : a x < b } is denoted [ a, b) (closed-open interval) { x R : a < x b } is denoted (a, b] (open-closed interval) { x R : x a } is denoted (, a] { x R : x < a } is denoted (, a) { x R : a x } is denoted [ a, ) { x R : a < x } is denoted (a, ). Department of Mathematics

Sets 3 It is important to realize that all this is just notation a useful visual short-hand. In particular, the symbol is used in four of the cases. This in no way is meant to imply that represents a real number it positively, absolutely, certainly is not. is not a real number. There is no such real number as. Given sets A and B, we say that A is a subset of B if every element of A is also an element of B, i.e., x A = x B. If this is the case, we write A B read A is a subset of B. By virtue of our earlier discussion of the equality A = B, we can say that A = B is equivalent to if and only if We have N R, Q R, N Z. both A B and B A. Definition 1.2. Suppose that A and B are given sets. The union of A and B, denoted by A B, is the set with elements which belong to either A or B (or both); A B = { x : x A or x B } read A union B equals.... Note that the usage of the word or allows both. In non-mathematical language, the union A B is obtained by bundling together everything in A and everything in B. Clearly, by construction, A A B and also B A B. Example 1.3. Suppose that A = { 1, 2, 3 } and B = { 3, 6, 8 }. Then we find that A B = { 1, 2, 3, 6, 8 }. Definition 1.4. The intersection of A and B, denoted by A B, is the set with elements which belong to both A and B; read A intersect B equals.... A B = { x : x A and x B } In non-mathematical language, the intersection A B is got by selecting everything which belongs to both A and B. Clearly, by construction, we see that A B A and also A B B. King s College London

4 Chapter 1 Example 1.5. With A = { 1, 2, 3 } and B = { 3, 6, 8 }, as in the example above, we see that A B = { 3 }. If A and B have no elements in common then their intersection A B has no elements at all. It is convenient to provide a symbol for this situation. We let denote the set with no elements. is called the empty set. Then A B = if A and B have no common elements. In such a situation, we say that A and B are disjoint. Example 1.6. Let A and B be the intervals in R given as A = (1, 4] and B = (4, 6). Then A B = and A B = (1, 6). Remark 1.7. Let A and B be given sets and consider the truth, or otherwise, of the statement A B. This fails to be true precisely when A possesses an element which is not a member of B. Now suppose that A =. The statement B is false provided that there is some nuisance element of which is not an element of B. However, has no elements at all, so there can be no such nuisance element. In other words, the statement B cannot be false and consequently must be true; obeys B for any set B. This might seem a bit odd, but is just a logical consequence of the formalism. Theorem 1.8. For sets A, B and C, we have (1) A (B C) = (A B) (A C). (2) A (B C) = (A B) (A C). Proof. (1) We must show that lhs rhs and that rhs lhs. First, we shall show that lhs rhs. If lhs =, then we are done, because is a subset of any set. So now suppose that lhs and let x lhs = A (B C). Then x A or x (B C) (or both). (i) Suppose x A. Then x A B and also x A C and therefore x (A B) (A C), that is, x rhs. (ii) Suppose that x (B C). Then x B and x C and so x A B and also x A C. Therefore x (A B) (A C), that is, x rhs. So in either case (i) or (ii) (and at least one these must be true), we find that x rhs. Since x lhs is arbitrary, we deduce that every element of the left hand side is also an element of the right hand side, that is, lhs rhs. Now we shall show that rhs lhs. If rhs =, then there is no more to prove. So suppose that rhs. Let x rhs. Then x (A B) and x (A C). Case (i): suppose x A. Then certainly x A (B C) and so x lhs. Case (ii): suppose x / A. Then since x (A B), it follows that x B. Also x (A C) and so it follows that x C. Hence x B C and so x A (B C) which tells us that x lhs. Department of Mathematics

Sets 5 We have seen that every element of the right hand side also belongs to the left hand side, that is, rhs lhs. Combining these two parts, we have lhs rhs and also rhs lhs and so it follows that lhs = rhs, as required. (2) This is left as an exercise. The notions of union and intersection extend to the situation with more than just two sets. For example, A 1 A 2 A 3 = { x : x A 1 or x A 2 or x A 3 } = { x : x belongs to at least one of the sets A 1, A 2, A 3 } = { x : x A i for some i = 1 or 2 or 3 } = { x : x A i for some i { 1, 2, 3 } }. More generally, for n sets A 1, A 2,..., A n, we have A 1 A 2 A n = { x : x A i for some i { 1, 2,..., n } }. This union is often denoted by n A i which is somewhat more concise than i=1 the alternative A 1 A 2 A n. Let Λ denote the index set { 1, 2,..., n }. This is just the set of labels for the collection of sets we are considering. Then the above can be conveniently written as A i = { x : x A i for some i Λ }. i Λ This all makes sense for any non-empty index set. Indeed, suppose that we have some collection of sets indexed (that is, labelled,) by a set Λ. Suppose the set with label λ Λ is denoted by A λ. The union of all the A λ s is defined to be A λ = { x : x A λ for some λ Λ }. λ Λ If Λ = N, one often writes Examples 1.9. i=1 A i for λ Λ 1. Suppose that Λ = { 1, 2, 3,..., 57, 58 } and A j = [j, j + 1] for each j Λ. (So, for example, with j = 7, A 7 = [7, 7 + 1] = [7, 8].) Then 58 j=1 A λ. A j = [1, 59]. King s College London

6 Chapter 1 2. Suppose that Λ = N and A j = [1, j + 1] for j N. Then A j = [1, ). j=1 To see this, suppose that x j=1. Then x is an element of at least one of the A j s, that is, there is some j 0, say, in N such that x A j0. This means that x [1, j 0 + 1], that is, 1 x j 0 + 1 and so certainly x [1, ). It follows that lhs rhs. Now suppose that x [1, ). Then, in particular, x 1. Let N be any natural number satisfying N > x. Then certainly x satisfies 1 x N + 1 which means that x A N and so x j=1 A j. Hence rhs lhs and the equality lhs = rhs follows. 3. Suppose that Λ is the interval (0, 1) and, for each λ (0, 1), A λ is given by A λ = { (x, y) R 2 : x = λ }. In other words, A λ is the vertical line x = λ in the plane R 2. Then A λ = { (x, y) R 2 : 0 < x < 1 } λ Λ which is the vertical strip in R 2 with boundary edges given by the lines with x = 0 and x = 1, respectively. Note that these lines (boundary edges) are not part of the union of the A λ s. 4. Let Λ be the interval [3, 5] and for each λ [3, 5] let A λ = { λ }. In other words, A λ consists of just one point, the real number λ. Then A λ = [3, 5] λ Λ which just says that the interval [3, 5] is the union of all its points (as it should be). A similar discussion can be made regarding intersections. A 1 A 2 A 3 = { x : x A 1 and x A 2 and x A 3 } = { x : x belongs to every one of the sets A 1, A 2, A 3 } = { x : x A i for all i = 1 or 2 or 3 } = { x : x A i for all i { 1, 2, 3 } }. In general, if { A λ } Λ is any collection of sets indexed by the (non-empty) set Λ, then the intersection of the A λ s is A λ = { x : x A λ for all λ Λ }. λ Λ Department of Mathematics

Sets 7 If Λ = { 1, 2,..., n }, we usually write we usually write Examples 1.10. i=1 A i for λ Λ A λ. n i=1 A i for λ Λ A λ and if Λ = N, then 1. Suppose that Λ = N and for each j Λ = N, let A j = [0, j]. Then A j = [0, 1]. j N 2. Let Λ = N and set A j = [j, j + 1] for j N. Then A j =. j=1 3. Let Λ = N and set A j = [j, ) for j N. Then A j =. j=1 To see this, note that x j=1 provided that x belongs to every A j. This means that x satisfies j x j + 1 for all j N. But clearly this fails whenever j is a natural number strictly greater than x. In other words, there are no real numbers which satisfy this criterion. 4. Suppose that Λ = N and for each k N let A k be the interval given by A k = [0, 1/k). Then, in this case, A k = { 0 }. k=1 This follows because the only non-negative real number which is smaller than every 1/k (where k N) is zero. 5. Let Λ = N and let A k = [0, 1 + 1/k] for k N. Then A k = [0, 1]. k=1 Indeed, [0, 1] A k for every k and if x / [0, 1] then x must fail to belong to some A k. King s College London

8 Chapter 1 Theorem 1.11. Suppose that A and B λ, for λ Λ, are given sets. Then (1) A ( ) B λ = (A B λ ). λ Λ λ Λ (2) A ( ) B λ = (A B λ ). λ Λ λ Λ Proof. (1) Suppose that x A ( λ Λ B λ). If x A, then x A Bλ for all λ and so x λ Λ (A B λ). If x / A, then it must be the case that x λ Λ B λ, in which case x B λ for all λ and so x λ Λ (A B λ). We have shown that A ( λ Λ B ) λ λ Λ (A B λ). To establish the reverse inclusion, suppose that x λ Λ (A B λ). Then x A B λ for every λ Λ. If x A, the certainly x A ( λ Λ B λ). If x / A, then we must have that x B λ for every λ, that is, x λ Λ B λ. But then it follows that x A ( λ Λ B λ). Hence λ Λ (A B λ) A ( λ Λ B λ) and so the equality follows. A ( ) B λ = (A B λ ) λ Λ λ Λ (2) The proof of this proceeds along similar lines to part (1). Department of Mathematics

Chapter 2 The Real Numbers In this chapter, we will discuss the properties of R, the real number system. It might well be appropriate to ask exactly what a real number is? It is the job of mathematics to set out clear descriptions of the objects within its scope, so it is not at all unreasonable to expect an answer to this. One must start somewhere. For example, in geometry, one might take the concept of point as a basic undefined object. Lines are then specified by pairs of points the line passing through them. Beginning with the natural numbers, N, one can construct Z and from Z one constructs the rationals, Q. Finally from Q it is possible to construct the real numbers R. We will not do this here, but rather we will take a close look at the structure and special properties of R. Of course, everybody knows that numbers can be added and multiplied and even subtracted and it makes sense to divide one number by another (as long as the latter, the denominator, is not zero). We can also compare two numbers and discuss which is the larger. It is precisely these properties (or axioms) that we wish to isolate and highlight. Arithmetic To each pair of real numbers a, b R, there corresponds a third, denoted a + b. This pairing, denoted + and called addition obeys (A1) a + (b + c) = (a + b) + c, for all a, b, c R. (A2) a + b = b + a, for all a, b R. (A3) There is a unique element, denoted 0, in R such that a + 0 = a, for any a R. (A4) For any a R, there is a unique element (denoted a) in R such that a + ( a) = 0. The properties (A1) (A4) say that R is an abelian group with respect to the binary operation +. 9

10 Chapter 2 Next, we consider multiplication. To each pair a, b R, there is a third, denoted a.b, the product of a and b. The operation., called multiplication, obeys (A5) a.(b.c) = (a.b).c, for any a, b, c R. (A6) a.b = b.a, for any a, b R. (A7) There is a unique element, denoted 1, in R, with 1 0 and such that a.1 = a, for any a R. (A8) For any a R with a 0, there is a unique element in R, written a 1 or 1 a, such that a.a 1 = 1. The element a 1 is called the (multiplicative) inverse, or reciprocal, of a. (A9) a.(b + c) = a.b + a.c, for all a, b, c R. Remarks 2.1. 1. 0 1 is not defined. The element 0 has no reciprocal. Such an object simply does not exist in R. 1/0 has no meaning. 2. Subtraction is given by a b = a + ( b), for a, b R. 3. Division is defined via a b = a.(b 1 ) ( = a. 1 b = a/b) provided b 0. If it should happen that b = 0, then the expression a/b has no meaning. 4. It is usual to omit the dot and write just ab for the product a.b. There is almost never any confusion from this. All the familiar arithmetic results are consequences of the above properties (A1) (A9). Examples 2.2. 1. For any x R, x.0 = 0. Proof. By (A3), 0 + 0 = 0 and so x.(0 + 0) = x.0. x.0 + x.0 = x.0. Adding (x.0) to both sides gives Hence, by (A9), (x.0 + x.0) + ( (x.0)) = x.0 + ( (x.0)) = 0, by (A4). Hence, by (A1), x.0 + (x.0 + ( (x.0)) = 0 and so, using property (A4) again, we get x.0 + 0 = 0. However, by (A3), x.0 + 0 = x.0 and so by equating these last two expressions for x.0 + 0 we obtain x.0 = 0, as required. Department of Mathematics

The Real Numbers 11 2. For any x, y R, x.( y) = (x.y). Proof. (x.y) + x.( y) = x.(y + ( y)), = x.0, by (A4), by (A9), = 0, by the previous result. By (A4) (uniqueness), (x.y) must be the same as x.( y). 3. For any x R, ( x) = x. Proof. We have as required. x = x + 0, by (A3), = x + (( x) + ( ( x))), by (A4), = (x + ( x)) + ( ( x)), by (A1), = 0 + ( ( x)), by (A4), = ( x), by (A3) 4. For any x, y R, x.y = ( x).( y). Proof. By example 2, above, α.( β) = (α, β) for any α, β R. If we now choose α = x and β = y, we get and we are done. ( x).( y) = (( x).y) = (y.( x)), by (A6), = ( (y.x)), by example 2, above, = ( (x.y)), by (A6), = x.y by example 3, above, King s College London

12 Chapter 2 Order properties Here we formalize the idea of one number being greater than another. We can order two numbers by thinking of the larger as being the higher in order. More precisely, there is a relation < (read less than ) between elements of R satisfying the following: (A10) For any a, b R, exactly one of the following is true: a < b, b < a or a = b (trichotomy). The notation u > a (read u is greater than a ) means that a < u. (A11) If a 0, then aγ < bγ. Notation We write a b to signify that either a b. The notation x w is used to mean that w x and as already noted above, x > w is used to mean w < x. By (A10), if x 0, then either x > 0 or else x < 0. If x > 0, then x is said to be (strictly) positive and if x < 0, we say that x is (strictly) negative. Thus, if x is not zero, then it is either positive or else it is negative. It is quite common to call a number x positive if it obeys x 0 or negative if it obeys x 0. Should it be necessary to indicate that x is not zero, then one adds the adjective strictly. Examples 2.3. 1. For any x R, we have x > 0 ( x) < 0. Proof. Using (A12), we have 0 < x = 0 + ( x) < x + ( x) (adding ( x) to both sides), = ( x) < 0 since rhs = 0, by (A4). Conversely, again from (A12), ( x) < 0 = ( x) + x < 0 + x and the result follows. = 0 < x by (A2), (A3) and (A4) (adding x to both sides), Department of Mathematics

The Real Numbers 13 2. For any x 0, we have x 2 > 0. Proof. Since x 0, we must have either x > 0 or x < 0, by (A10). If x > 0, then by (A13) we have x 2 > 0 (take a = 0, b = γ = x). On the other hand, if x < 0, then x > 0 by the example above. Hence, by (A13) (with a = 0, b = γ = ( x)), it follows that ( x)( x) > 0. But we know (from the arithmetic properties) that ( α)( β) = α β, for any α, β R and so we have as required. x 2 = x x = ( x)( x) > 0 The number 1 was introduced in (A7). If we set x = 1 here, then we see that 1 = 1 2 > 0, i.e., 1 > 0. We have deduced that the number 1 is positive. Nobody would doubt this, but we see explicitly that this is a consequence of our set-up. Note that it follows from this, by (A12), that a < a + 1, for any a R. 3. If a, b R with a b, then a b. Proof. If a = b then certainly a = b, so we need only consider the case when a < b. a < b = a + ( a) < b + ( a) and the result follows. = 0 < b + ( a) by (A12), = b < ( b) + b + ( a) by (A12) and (A4), = b < a by (A1), (A2) and (A4) From now on, we will work with real numbers and inequalities just as we normally would and will not follow through a succession of steps invoking the various listed properties as required as we go. Suffice it to say that we could do so if we wished. Next, we introduce a very important function, the modulus or absolute value. King s College London

14 Chapter 2 Definition 2.4. For any x R, the modulus (or absolute value) of x is the number x defined according to the rule { x, if x 0, x = x, if x < 0. For example, 5 = 5, 0 = 0, 3 = 3 and 1 2 = 1 2. Note that x is never negative. We also see that x = max{ x, x }. Let f(x) = x and g(x) = x. Then x = f(x) when x 0 and x = g(x) when x < 0. Now, we know what the graphs of y = f(x) = x and y = g(x) = x look like and so we can sketch the graph of the function x. It is made up of two straight lines, meeting at the origin. x y = x y = x x 0 Figure 2.1: The absolute value function x. The basic properties of the absolute value are contained in the following two propositions. They are used time and time again in analysis and it is absolutely essential to be fluent in their use. Proposition 2.5. (i) For any a, b R, we have ab = a b. (ii) For any a R and r > 0, the inequality a < r is equivalent to the pair of inequalities r < a < r. Proof. (i) We just consider the various possibilities. If either a or b is zero, then so is the product ab. Hence ab = 0 and at least one of a or b is also zero. Therefore ab = 0 = a b. If both a > 0 and b > 0, then ab > 0 and we have ab = ab, a = a and b = b and so ab = a b in this case. Now, if a > 0 but b < 0, then ab < 0 so we have a = a, b = b and ab = ab = a b. The case a < 0 and b > 0 is similar. Finally, suppose that both a < 0 and b < 0. Then ab > 0 and we have ab = ab, a = a and b = b. Hence, ab = ab = ( a)( b) = a b. Department of Mathematics

The Real Numbers 15 (ii) Suppose that a < r. Then max{ a, a } < r and so both a < r and a < r. In other words, a < r and r < a which can be written as r < a < a. On the other hand, if r < a < r, then both a < r and a < r so that max{ a, a } < r. That is, a < r, as required. Remark 2.6. Putting b = 1 in (i), above, and using the fact that 1 = 1, we see that a = a. Proposition 2.7. For any real numbers a and b, (i) a + b a + b. (ii) a b a + b. (iii) a b a b. Proof. (i) We have a + b a + b and (a + b) = a b a + b. Hence a + b = max{ a + b, (a + b) } a + b. (ii) Let c = b and apply (i) to the real numbers a and c to get the inequality a + c a + c. But then this means that a b a + b. (iii) We have a = (a b) + b a b + b by part (i) (with (a b) replacing a). This implies that a b a b. Swapping around a and b, we have ( a b ) = b a b a = a b and therefore as required. a b = max{ a b, ( a b ) } a b If a and b are real numbers, how far apart are they? For example, if a = 7 and b = 11 then we might say that the distance between a and b is 4. If, on the other hand, a = 10 and b = 6, then we would say that the distance between them is 16. In either case, we notice that the distance is given by a b. It is extremely useful to view a b as the distance between the numbers a and b. For example, to say that a b is very small is to say that a and b are close to each other. King s College London

16 Chapter 2 Proposition 2.8. Let a, b R be given and suppose that for any given ε > 0, a and b obey the inequality a 0, then x 0. Proof. We know that either a b or else a > b. Suppose the latter were true, namely, a > b. Set ε 1 = a b. Then ε 1 > 0 and a = b + ε 1. Taking ε = ε 1, we see that this conflicts with the hypothesis that a 0 (it fails for the choice ε = ε 1 ). We conclude that a > b must be false and so a b. For the last part, simply set a = x and b = 0 to get the desired conclusion. We have listed a number of properties obeyed by the real numbers: (A1)... (A9) arithmetic (A10)... (A13) order. Is this it? Are there any more to be included? We notice that all of these properties are satisfied by the rational numbers, Q. Are all real numbers rational, i.e., is it true that Q = R? Or do we need to consider yet further properties which distinguish between Q and R? Consider an apparently unrelated question. Do all numbers have square roots? Since a 2 is positive for any a R, it is clear that no negative number can have a square root in R. (Indeed, it is the consideration of C, the complex numbers, which allows for square roots of negative numbers.) So we ask, does every positive real number have a square root? Does every natural number n have a square root in R? In particular, is there such a real number as the square root of 2? It would be nice to think that there is such a real number. In fact, according to Pythagoras Theorem, this should be the length of the diagonal of a square whose sides have unit length. The following proposition tells us that there is certainly no such rational number. Proposition 2.9. There are no integers m, n N satisfying m 2 = 2n 2. In particular, 2 is not a rational number. Proof. To say that 2 is rational is to say that there are integers m and n (with n 0) such that m/n = 2. This means that m 2 /n 2 = 2 and so m 2 = 2n 2 for m, n N. (By replacing m or n by m or n, if necessary, we may assume that m and n in this last equality are both positive.) So the fact that 2 / Q is a consequence of the first part of the proposition. Consider the equality m 2 = 2n 2 ( ) To show that m 2 = 2n 2 is impossible for any m, n N, suppose the contrary, namely that there are numbers m and n in N obeying ( ). We will show that this leads to a contradiction. Department of Mathematics

The Real Numbers 17 Indeed, if m 2 = 2n 2, then m 2 is even. The square of an odd number is odd and so it follows that m must also be even. This means that we can express m as m = 2k for some suitable k N. But then (2k) 2 = m 2 = 2n 2 which means that 2k 2 = n 2 and so n 2 is even. Arguing as above, we deduce that n can be expressed as n = 2j for some j N. Substituting, we see that k and j also obey ( ), namely, k 2 = 2j 2. This tells us that m/2 and n/2 are integers also obeying ( ). Repeating this whole argument with m = m/2 and n = n/2, we find that both m /2 and n /2 belong to N and also satisfy ( ). In other words, m/2 2 and n/2 2 belong to N and obey ( ). We can keep repeating this argument to deduce that m/2 j and n/2 j are integers obeying ( ). In particular, m/2 j N implies that m/2 j 1 and so m 2 j. But this holds for any j N and we can take j as large as we wish. We can take j so large that 2 j > m. This leads to a contradiction, as we wanted it to. We finally conclude that there are no natural numbers m and n obeying ( ) and as a consequence, there is no element of Q whose square is equal to 2, that is, 2 is not a rational number. Remark 2.10. A somewhat similar argument can be used to show that many other numbers do not have square roots in Q. For example, 3 / Q. In fact, one can show that if n N, then either n N, that is, n is a perfect square, or else n / Q. For example, 16 = 4 N but 17 / Q. Returning to the discussion of the defining properties of R, we still have to pinpoint the extra property that R has which is not shared by Q. First we need some terminology. Definition 2.11. A non-empty subset S of R is said to be bounded from above if there is some M R such that a M for all a S. Any such number M is called an upper bound for the set S. Evidently, if M is an upper bound for S, then so is any number greater than M. We say that a non-empty subset S of R is bounded from below if there is some m R such that m a for all a S. Any such number m is called a lower bound for the set S. If m is a lower bound for S, then so is any number less than m. If S is both bounded from above and from below, then S is said to be bounded. King s College London

18 Chapter 2 Example 2.12. Consider the set A = ( 6, 4]. Then A is bounded because any x A obeys 6 x 4. (In fact, any x A obeys the inequalities 6 < x 4.) Any real number greater than or equal to 4 is an upper bound for A and any real number less than or equal to 6 is a lower bound for A. The set A has a maximal element, namely 4, but A does not have a minimal element. Let B = (1, 3 2 ) (2, 5 2 ) (3, 7 2 ) = (k, k + 1 2 ). Then the set B is bounded from below (the number 1 is clearly a lower bound for B). However, B contains k + 1 4 for every k N and so B is not bounded from above (so B is not bounded). We also see that B does not have a minimal element. Remark 2.13. What does it mean to say that a set S is not bounded from above? Consider the inequality x M. Now, given S and some particular real number M, the inequality ( ) may hold for some elements x in S but may fail for other elements of S. To say that S is bounded from above is to say that there is some M such that ( ) holds for all elements x S. If S is not bounded from above, then it must be the case that whatever M we try, there will always be some x in S for which ( ) fails, that is, for any given M there will be some x S such that x > M. In particular, if we try M = 1, then there will be some element (many, in fact) in S greater than 1. Let us pick any such element and label it as x 1. Then we have x 1 S and x 1 > 1. We can now try M = 2. Again, ( ) must fail for at least one element in S and it could even happen that x 1 > 2. To ensure that we get a new element from S, let M = max{ 2, x 1 }. Then there must be at least one element of S greater than this M. Let x 2 denote any such element. Then we have x 2 S and x 2 > 2 and x 2 x 1. Now setting M = max{ 3, x 1, x 2 }, we may say that there is some element in S, which we choose to denote by x 3, such that x 3 > 3 and x 3 x 1 and x 3 x 2. We can continue to do this and so we see that if S is not bounded from above, then there exist elements x 1, x 2, x 3,..., x n,... (which are all different) such that x n > n for each n N. k=1 The following concepts play an essential rôle. ( ) Department of Mathematics

The Real Numbers 19 Definition 2.14. Suppose that S is a non-empty subset of R which is bounded from above. The number M is the least upper bound (lub) of S if (i) a M for all a S (i.e., M is an upper bound for S). (ii) If M is any upper bound for S, then M M. If S is a non-empty subset of R which is bounded from below, then the number m is the greatest lower bound (glb) of S if (i) m a for all a S (i.e., m is a lower bound for S). (ii) If m is any lower bound for S, then m m. Note that the least upper bound and the greatest lower bound of a set S need not themselves belong to S. They may or they may not. The least upper bound is also called the supremum (sup) and the greatest lower bound is also called the infimum (inf). The ideas are illustrated by some examples. Examples 2.15. 1. Let S be the following set consisting of 4 elements, S = { 3, 1, 2, 5 }. Then clearly S is bounded from above and from below. The least upper bound is 5 and the greatest lower bound is 3. 2. Let S be the interval S = ( 6, 4]. Then lub S = 4 and glb S = 6. Note that 4 S whereas 6 / S. 3. Let S = (1, ). S is not bounded from above and so has no least upper bound. S is bounded from below and we see that glb S = 1. Note that glb S / S in this case. Remark 2.16. Suppose that M is the lub for a set S. Let δ > 0. Then M δ < M and since any upper bound M for S has to obey M M, we see that M δ cannot be an upper bound for S. But this means that it is false that a M δ for all a S. In other words, there must be some a S which satisfies M δ < a. Since M is an upper bound for S, we also have a M and so a obeys M δ < a M. So no matter how small δ may be, there will always be some element a S (possibly depending on δ and there may be many) such that M δ < a M, where M = lub S. For any δ > 0, there is a S such that lub S δ < a lub S. King s College London

20 Chapter 2 Now suppose that m = glb S. Then for any δ > 0 (however small), we note that m < m + δ and so m + δ cannot be a lower bound for S (because all lower bounds for S must be less than or equal to m). Hence, there is some a S such that a < m + δ, which means that m a < m + δ. For any δ > 0, there is a S such that glb S a < glb S + δ. Remark 2.17. As already noted above, lub S and glb S may or may not belong to the set S. If it should happen that lub S S, then in this case lub S (or sup S) is the maximum element of S, denoted max S. If glb S S, then glb S (or inf S) is the minimum element of S, denoted min S. For example, the interval S = ( 2, 5] is bounded and, by inspection, we see that sup S = 5 and inf S = 2. Since sup S = 5 S, the set S does indeed have a maximum element, namely, 5 = sup S. However, inf S / S and so S has no minimum element. We are now in a position to discuss the final property satisfied by R and it is precisely this last property which distinguishes R from Q. (A14) (The completeness property of R) Any non-empty subset of R which is bounded from above possesses a least upper bound. Any non-empty subset of R which is bounded from below possesses a greatest lower bound. These statements might appear self-evident, but as we will see, they have far-reaching consequences. We note here that these two statements are not independent, in fact, each implies the other, that is, they are equivalent. Remark 2.18. It is very convenient to think of R as the set of points on a line (the real line). Indeed, this is standard procedure when sketching graphs of functions where the coordinate axes represent the real numbers. Department of Mathematics

The Real Numbers 21 Imagine now the following situation. } {{ } A )( R }{{} B Figure 2.2: The real line has no gaps. The set A consists of all points on the line (real numbers) to the left of the arrow and B comprises all those points to the right. Numbers are bigger the more they are to the right. The arrow points to the least upper bound of A (which is also the greatest lower bound of B). The completeness property (A14) ensures the existence of the real number in R that the arrow supposedly points to. There are no gaps or missing points on the real line. We can think of the integers Z or even the rationals Q as collections of dots on a line, but it is property (A14) which allows us to visualize R as the whole unbroken line itself. The next result is so obvious that it seems hardly worth noting. However, it is very important and follows from property (A14). Theorem 2.19 (Archimedean Property). For any given x R, there is some n N such that n > x. Proof. Let x R be given. We use the method of proof by contradiction so suppose that there is no n N obeying n > x. This means that n x for all n N, that is, x is an upper bound for N in R. By the completeness property, (A14), N has a least upper bound, α, say. Then α is an upper bound for N so that n α ( ) for all n N. Since α is the least upper bound, α 1 cannot be an upper bound for N and so there must be some k N such that α 1 < k. But we can rewrite this as α < k + 1 which contradicts ( ) since k + 1 N. We conclude that there is some n N obeying n > x, as claimed. Corollary 2.20. (i) For any given δ > 0, there is some n N such that 1 n < δ. (ii) For any α > 0, β > 0, there is n N such α n < β. Proof. (i) Let δ > 0 be given. By the Archimedean Property, there is some n N such that n > 1/δ. But then this gives 1/n < δ, as required. (ii) For given α > 0 and β > 0, set δ = β/α. By (i), there is n N such that 1/n < δ = β/α and so α/n < β. King s College London

22 Chapter 2 The next result is no surprise either. Theorem 2.21. For any a R, there is a unique integer n Z such that n a < n + 1. Proof. Let S = { k Z : k > a }. By theorem 2.19, S is not empty and is bounded below (by a). Hence, by the completeness property (A14), S has a greatest lower bound α, say, in R. We have a α k for all k S. (The inequality a α follows because a is a lower bound and α is the greatest lower bound and the inequality α k follows because α is a lower bound of S.) Since α is the greatest lower bound, α + 1 cannot be a lower bound of S and so there is some m S such that m < α + 1, that is, m 1 < α. m 1 = n a m = n + 1 Figure 2.3: The integer part of a. Now, α is a lower bound for S and m 1 < α and so m 1 / S. But then, by the defining property of S, this means that it is false that m 1 > a. In other words, we have m 1 a. But m S and so m > a and so m satisfies m 1 a < m. Putting n = m 1, we get n Z and n satisfies the required inequalities n a < n + 1. To show the uniqueness of such n Z, suppose that also n Z obeys n a < n +1. Suppose that n < n. Then n+1 n and so the inequalities n a and a < n + 1 give n a < n + 1 n giving n < n which is impossible. Similarly, the assumption that n < n would lead to the impossible inequality n < n. We conclude that n = n which is to say that n is unique. Remark 2.22. For x R, let n Z be the unique integer obeying the inequalities n x < n+1. Set r = x n. Then we see that 0 x n = r < 1 and so x = n + r with n Z and where 0 r < 1. The unique integer n here is called the integer part of the real number x and is denoted by [x] (or sometimes by x ). R Department of Mathematics

The Real Numbers 23 Theorem 2.23. Between any pair of real numbers a < b, there are infinitelymany rational numbers and also infinitely-many irrational numbers. Proof. First, we shall show that there is at least one such rational, that is, we shall show that for any given a < b in R, there is some q Q such that a < q < b. The idea of the proof is as follows. If there is an integer between a and b, then we are done. In any case, we note that since the integers are spread one unit apart, there should certainly be at least one integer between a and b if the distance between a and b is greater than 1. If the distance between a and b is less than 1, then we can open up the gap between them by multiplying both by a sufficiently large (positive) integer, n, say. The gap between na and nb is n(b a). Clearly, if n is large enough, this value is greater than 1. Then there will be some integer m, say, between na and nb, i.e., na < m < nb. But then we see (since n is positive) that a < m/n 1 so that na + 1 < nb and let m = [na] + 1. Since [na] na < [na] + 1, it follows that [na] na < [na] + 1 na + 1 < nb }{{} m and so na < m < nb and hence a < m/n < b. (Note that n > 0, so this last step is valid.) Setting q = m/n, we have that q Q and q obeys a < q < b, as required. To see that there are infinitely-many rationals between a and b, we just repeat the above argument but with, say, q and b instead of a and b. This tells us that there is a rational, q 2, say, obeying q < q 2 < b. Once again, repeating this argument, there is a rational, q 3, say, obeying q 2 < q 3 < b. Continuing in this way, we see that for any n N, there are n rationals, q, q 2,..., q n obeying a < q < q 2 < q 3 < < q n < b. Hence it follows that there are infinitely-many rationals between a and b. To show that there are infinitely-many irrational numbers between a and b, we use a trick together with the observation that if r is rational, then r/ 2 is irrational. The trick is simply to apply the first part to the numbers a 2 and b 2 to deduce that for any n N there are rational numbers r 1, r 2,..., r n obeying a 2 < r 1 < r 2 < < r n < b 2. Now let µ j = r j / 2 for j = 1, 2,..., n. Then each µ j is irrational and we have a < µ 1 < µ 2 < < µ n < b and the result follows. King s College London

24 Chapter 2 As a further application of the Completeness Property of R, we shall show that any positive real number has a positive n th root. Theorem 2.24. Let x 0 and n N be given. Then there is a unique s 0 such that s n = x. The real number s is called the (positive) n th root of x and is denoted by x 1/n. Proof. If x = 0, then we can take s = 0, so suppose that x > 0. Let A be the set A = { t 0 : t n < x }. Then 0 A and so A is not empty and, by the Archimedean Property, there is some integer K with K > x. But then every t A must obey t < K because otherwise we would have t K and therefore t n K n K > x, which is not possible for any t A. This means that A is bounded from above. By the Completeness Property of R, A has a least upper bound, lub A = s, say. Note now that, since x > 0, by the Archimedean Property there is some m N such that m > 1/x. Hence m n m > 1/x which implies that 1/m n < x so that 1/m n A. This means that s 1/m n. In particular, s > 0. Now, exactly one of the statements s n = x, s n < x or s n > x is true. We claim that s n = x and to show this we shall show that the last two statements must be false. Indeed, suppose that s n < x. For k N, let s k = s(1 + 1 k ). Then evidently s k > s and we will show that s n k < x for suitably large k. Let d = x s n. Then d > 0 and x s n k = x sn + s n s n k = d (sn k sn ) = d s n( (1 + 1 k )n 1 ). Now, writing α = (1 + 1 k ) and noting that 1 < α 2, we estimate Hence (1 + 1 k )n 1 = α n 1 = (α 1)(α n 1 + α n 2 + + 1) (α 1)(2 n 1 + 2 n 2 + + 1) (α 1) n 2 n = 1 k n 2n. s n k sn = s n( (1 + 1 k )n 1 ) sn n 2 n. k For sufficiently large k, the right hand side of this inequality is less than d and so x s n k = x sn + s n s n k = d (sn k sn ) > 0. It follows that if k is large enough, then s k A. But s k > s which means that s cannot be the least upper bound of A and we have a contradiction. Hence it must be false that s n < x. Department of Mathematics

The Real Numbers 25 Suppose now that s n > x and let δ = s n x. For given k N, let t k = s(1 1 k ). Writing β = 1 1 k and noting that 0 β 1, we estimate that It follows that 1 (1 1 k )n = 1 β n = (β n 1) = (β 1)(β n 1 + β n 2 + + 1) = (1 β)(β n 1 + β n 2 + + 1) (1 β) n = 1 k n. s n t n k = sn (1 (1 1 k )n ) 1 k sn n < δ for sufficiently large k. But then this means that t n k x = tn k sn + s n x = δ (s n t n k ) > 0 for large k. However, t k < s and since s = lub A, it follows that t k is not an upper bound for A. In other words, there is some τ A such that τ > t k and therefore τ n x > t n k x > 0. However, τ A means that τ n < x which is a contradiction and so it is false that s n > x. We have now shown that s n < x is false and also that s n > x is false and so we conclude that it must be true that s n = x, as required. We have established the existence of some s 0 such that s n = x and so, finally, we must prove that such an s is unique. If x = 0, then s = 0 obeys s n = 0 = x. No s 0 can obey s n = 0 because s n (1/s) n = 1 0, so s = 0 is the only solution to s n = 0. Now let s > 0 and t > 0. If s > t, then s/t > 1 so that (s/t) n > 1 and we find that s n > t n. Interchanging the rôles of s and t, it follows that if s < t, then s n < t n. We conclude that if s n = x = t n then both s < t and s > t are impossible and so s = t. The proof is complete. King s College London

26 Chapter 2 Principle of induction Suppose that, for each n N, P (n) is a statement about the number n such that (i) P (1) is true. (ii) For any k N, the truth of P (k) implies the truth of P (k + 1). Then P (n) is true for all n. Example 2.25. For any n N, 1 2 + 2 2 + 3 2 + + n 2 = Proof. For n N, let P (n) be the statement that 1 2 + 2 2 + 3 2 + + n 2 = Then P (1) is the statement that n(n + 1)(2n + 1) 6 n(n + 1)(2n + 1) 6 1 2 1(1 + 1)(2 + 1) = 6 which is true. Now suppose that k N and that P (k) is true. We wish to show that P (k + 1) is also true. Since we are assuming that P (k) is true, we see that 1 2 + 2 2 + 3 2 + + k 2 + (k + 1) 2 k(k + 1)(2k + 1) = + (k + 1) 2, 6 using the truth of P (k), k(k + 1)(2k + 1) + (k + 1)(6k + 6) = 6 = (k + 1)(2k2 + k + 6k + 6) 6 (k + 1)(k + 2)(2k + 3) = 6 which is to say that P (k + 1) is true. By the principle of induction, we conclude that P (n) is true for all n N. We can rephrase the principle of induction as follows. Let T be the set given by T = { k N : P (k) is true }, so k T if and only if P (k) is true. In particular, P (1) is true if and only if 1 T. Hence the principle of induction may be rephrased as follows. Let T be a set of natural numbers such that 1 T and such that if T contains k then it also contains k + 1. Then T = N... Department of Mathematics

The Real Numbers 27 Principle of induction (2nd form) Suppose that Q(n) is a statement about the natural number n such that (i) Q(1) is true. (ii) For any k N, the truth of all Q(1), Q(2),..., Q(k) implies the truth of Q(k + 1). Then Q(n) is true for all n. In a nutshell: Suppose that: Q(1) is true and Q(1) true Q(2) true Q(3) true = Q(k +1) true. Q(k) true Conclusion: Q(n) is true for all n N. This follows from the usual form of the principle. To see this, let S = { m N : Q(m) is true }. We shall use the usual form of induction to show that the hypotheses above imply that S = N. For any n N, let P (n) be the statement { 1, 2,..., n } S. Now, by hypothesis, Q(1) is true and so 1 S. Hence { 1 } S which is to say that P (1) is true. Next, suppose that the truth of P (k) implies that of P (k+1) and assume that P (k) is true. This means that { 1, 2,..., k } S, that is, each of Q(1), Q(2),..., Q(k) is true. But then by the 2nd part of the hypothesis above, Q(k + 1) is true, that is to say, k + 1 S. Hence { 1, 2,... k, k + 1 } S. But this just tells us that P (k + 1) is true. By induction (usual form), it follows that P (n) is true for all n N. This means that { 1, 2,..., n } S for all n. In particular, n S for every n N, that is, Q(n) is true for all n N which is the content of the 2nd form of the principle. King s College London

28 Chapter 2 Department of Mathematics

Chapter 3 Sequences A sequence of real numbers is just a listing a 1, a 2, a 3,... of real numbers labelled by N, the set of natural numbers. Thus, to each n N, there corresponds a real number a n. Not surprisingly, a n is called the n th term of the sequence. a 1, a 2, a 3,..., a k, a k+1,... labelled by N k th term Figure 3.1: The sequence (a n ) n N. Whilst it may seem a trivial comment, it is important to note that the essential thing about a sequence is that it has a notion of direction it makes sense to talk about one term being further down the sequence than another. For example, a 101 is further down the sequence than, say, a 45. It is convenient to denote the above sequence by (a n ) n N or even simply by (a n ). Note that there is no requirement that the terms be different. It is quite permissible for a j to be the same as a n for different j and n. Indeed, one could have a n = α, say, for all n. This is just a sequence with constant terms (all equal to α) a somewhat trivial sequence, but a sequence nonetheless. Remark 3.1. On a more formal level, one can think of a sequence of real numbers to be nothing but a function from N into R. Indeed, we can define such a function f : N R by setting f(n) = a n for n N. Conversely, any f : N R will determine a sequence of real numbers, as above, via the assignment a n = f(n). One might wish to consider a finite sequence such as, say, the four term sequence a 1, a 2, a 3, a 4. We will use the word sequence to mean an infinite sequence and simply include the adjective finite when this is meant. 29

30 Chapter 3 Examples 3.2. 1. 1, 4, 9, 16,... Here the general term a n is given by the simple formula a n = n 2. 2. 2, 3/2, 4/3, 5/4, 6/5,... The general term is a n = (n + 1)/n. 3. 2, 0, 2, 0, 2, 0,... { 0, if n is even Here a n = 2, if n is odd. This can also be expressed as a n = 1 ( 1) n. 4. Let a n be defined by the prescription a 1 = a 2 = 1 and a n = a n 1 + a n 2 for n 3. The sequence (a n ) is then 1, 1, 2, 3, 5, 8, 13,... These are known as the Fibonacci numbers. We are usually interested in the long-term behaviour of sequences, that is, what happens as we look further and further down the sequence. What happens to a n when n gets very large? Do the terms settle down or do they get sometimes big, sometimes small,..., or what? In examples 3.2.1 and 3.2.4, the terms just get huge. In example 3.2.2, we see, for example, that a 99 = 100/99, a 10000 = 10001/10000, a 10 20 = (10 20 + 1)/10 20,..., so it looks as though the terms become close to 1. In example 3.2.3, the terms just keep oscillating between the two values 0 and 2. In example 3.2.2, we would like to say that the sequence approaches 1 as we go further and further down it. Indeed, for example, the difference between a 10 10 and 1 is that between (10 10 + 1)/10 10 and 1, that is, 10 10. How can we formulate this idea of convergence of a sequence precisely? We might picture a sequence in two ways, as follows. The first is as the graph of the function n a n. (Notice that we do not join up the dots.) a 1 a 2 1 2 3 4 a 3 a 4... a n n Figure 3.2: A sequence as a graph. N Department of Mathematics

Sequences 31 The second way is just to indicate the values of the sequence on the real line. a4 a2 a1 a3 R Figure 3.3: Plot the values of the sequence on the real line. The example 3.2.3 above, would then be pictured either as a 1 a 3 a 5 a 2 a 4 a 6 1 2 3 4... N Figure 3.4: Graph with values 0 or 2. or as 0 a2 a 4. 2 a1 a 3. R Figure 3.5: The values of a n are either 0 or 2. King s College London

32 Chapter 3 Returning to the general situation now, how should we formulate the idea that a sequence (a n ) converges to α? According to our first pictorial description, we would want the plotted points of the sequence (the graph) to eventually become very close to the line y = α. y = α 1 2 3 4... Figure 3.6: The graph gets close to the line y = α. In terms of the second pictorial description, we would simply demand that the values of the sequence eventually cluster around the value x = α. x = α Figure 3.7: The values of (a n ) cluster around x = α. If we think of the index n as representing time, then we can think of a n as the value of the sequence at the time n. The sequence can be considered to have some property eventually provided we are prepared to wait long enough for it to become established. It is very convenient to use this word eventually, so we shall indicate precisely what we mean by it. We say that a sequence eventually has some particular property if there is some N N such that all the terms a n after a N (i.e., all a n with n > N) have the property under consideration. (The number N can be thought of as some offered time after which we are guaranteed that the property under consideration will hold and will continue to hold.) As an example of this usage, let (a n ) be the sequence given by the prescription a n = 100 n, for n N. Then a 1 = 99, a 2 = 98,... etc. It is clear that a n is negative whenever n is greater than 100. Thus, we can say that this sequence (a n ) is eventually negative. Now we can formulate the notion of convergence of a sequence. The idea is that (a n ) converges to the number α if eventually it is as close to α as desired. That is to say, given some preassigned tolerance ε, no matter how small, we demand that eventually (a n ) is close to within ε of α. In Department of Mathematics R R

Sequences 33 other words, the distance between a n and α (as points on the real line) is eventually smaller than ε. Definition 3.3. We say that the sequence (a n ) n N of real numbers converges to the real number α if for any given ε > 0, there is some natural number N N such that a n α < ε whenever n > N. α is called the limit of the sequence. In such a situation, we write a n α as n or alternatively lim n a n = α. The use of the symbol is just as part of a phrase and it has no meaning in isolation. There is no real number. Remark 3.4. The positive number ε is the assigned tolerance demanded. Typically, the smaller ε is, so the larger we should expect N to have to be. For example, consider the sequence (a n ) where a n = 1/n. We would expect that a n 0 as n. To see this, let ε > 0 be given. (We are not able to choose this. It is given to us and its actual value is beyond our control.) It will be true that a n 0 < ε provided n > 1/ε. So after some contemplation, we proceed as follows. We are unable to influence the choice of ε given to us, but once it is given then we can (and must) base our tactics on it. So let N be any natural number larger than 1/ε. If n > N, then n > N > 1/ε and so 1/n < ε. That is, if n > N, then a n 0 = 1/n < ε and so, according to our definition, we have shown that a n 0 as n. Notice that the smaller ε is, the larger N has to be. Note that the statement can also be written as or also as if n > N then a n α < ε a n α < ε whenever n > N n > N = a n α < ε. Also, we should note that the inequality a n α < ε telling that the distance between the real numbers a n and α is less than ε can also be expressed by the pair of inequalities ε < a n α < ε or equivalently by the pair α ε < a n < α + ε. This simply means that a n lies on the real line somewhere between the two values α ε and α + ε. This must happen eventually if the sequence is to be convergent (to α). King s College London