Introduction to Abstract Mathematics

Introduction to Abstract Mathematics

Analysis (first half of Lent term)

Martin Anthony 2003/4

Introduction Arrangements This part of the course will be lectured by Martin Anthony (Room B409, email [email protected]) This second part of the course will consist of ten lectures and five classes. I will place all handouts and sketch solutions to classwork (after classes have taken place) on the course website, http://www.maths.lse.ac.uk/Courses/MA103/

What is Analysis? Analysis is the theory behind real numbers, sequences, and functions. The word ‘theory’ is important. You might, for example, think that you have a good idea of what we mean by a ‘limit’, but in this part of the course we try to formalise such notions.

Topics The topics I hope to cover are as follows. (We might not cover all of these.)

• Inequalities, bounded sets of real numbers, supremum and infimum. • Sequences of real numbers. • The limit of a sequence: formal definition. 1

• The Sandwich Theorem. • The algebra of limits. Limit of an . Relative growth rates of powers and polynomials. • Bounded sequences. The fact that convergent sequences are bounded. • The fact that monotonic bounded sequences converge. • Subsequences. The fact that every sequence has a monotonic subsequence. The Bolzanno-Weierstrass theorem. • Limits of functions, and the algebra of these limits. • Continuity. • The fact that continuous functions on closed intervals are bounded and attain their bounds • The Intermediate Value Theorem

2

Real Numbers Textbooks for the Analysis part of the course This part of the course does not follow any one particular book, but there are many relevant books available, all with titles like ‘Introduction to Analysis’. Here are some: Mathematical Analysis: A Straightforward Approach, 2nd ed., K. G. Binmore, Cambridge University Press 1982. QA300 B61 The Elements of Real Analysis, 2nd ed., R. G. Bartle, Wiley 1976. QA300 B28 Introduction to Real Analysis, 2nd ed., R. G. Bartle and D. R. Sherbert, Wiley 1992. QA300 B28 Yet Another Introduction to Analysis, V. Bryant, Cambridge University Press 1990. QA300 B91 The Bryant book is written informally and entertainingly, and is the one I recommend most. All books cover the material in this course, as do many other books with the words ‘analysis’, ‘real analysis’ or ‘mathematical analysis’ in their title. An interactive website covering the concepts of real analysis (in essence an online textbook) is http://www.shu.edu/projects/reals/index.html There are many other analysis resources on the web, since this is a topic covered in all undergraduate Mathematics degrees.

3

The real numbers The underlying set for all our considerations is the set R of real numbers (or reals, for short). This by far the most important set used in mathematics. The reals combine our geometric intuition in the form of the real line with the elementary computations known from counting, namely addition and multiplication and their inverse operations, subtraction and division. These operations can also be carried out on the set of rational numbers. But as soon as one wants to ‘do algebra’, that is, solve equations such as x2 = 2, then we know √ that rational numbers are no longer sufficient. Furthermore, irrational numbers like 2 that solve such an equation, do ‘exist’, like in geometry√as the length of the diagonal of a square with a side of length one. A way of defining 2 would be to consider the set S given by S = {x : x2 ≤ 2}

(1)

√ and to let 2 be the ‘largest’ member of that set. But does this largest element of S exist? It does not if x in (1) ranges over the rational numbers only. However, for reals such an element of S does exist, and this is in fact the fundamental property of the reals that distinguishes them from the rationals. That is, R is equipped with the order relation ≤ that guarantees the existence of a ‘largest element’—more precisely a supremum—for any set of real numbers, apart from certain obvious exceptions. This will be our first topic.

Infimum and supremum, completeness of the reals For any subset S of R, • an upper bound of S is a real number x such that for all y in S, y ≤ x; • a lower bound of S is a real numbers x such that for all y in S, x ≤ y; • a least upper bound or supremum of S is an upper bound σ of S such that if z is an upper bound of S, then σ ≤ z; • a greatest lower bound or infimum of S is a lower bound τ of S such that if z is a lower bound of S, then z ≤ τ ; • a maximum of S is an upper bound of S that belongs to S; • a minimum of S is a lower bound of S that belongs to S.

4

We say that S ⊆ R is: • bounded above if it has an upper bound; • bounded below if it has a lower bound; • bounded if it is both bounded above and bounded below. We assume that R satisfies the completeness axiom: Completeness Axiom. Any non-empty subset of R that is bounded above has a supremum; and any non-empty subset of R that is bounded below has an infimum. It is easy to show that any minimum of a set is an infimum, any maximum a supremum, and that infimum and supremum are unique whenever these exist. Note that, by the completeness axiom, any bounded set of real numbers will have a supremum and an infimum, but need not have a maximum or a minimum. Notation. If S ⊆ R, then sup S denotes the supremum of S if it exists (that is, if S is bounded above) and inf S denotes the infimum of S if it exists. Similarly, min S and max S denote minimum and maximum of S, whenever they exist.

Distance The distance between two real numbers x and y is defined as the absolute value |x − y| of their difference. In an abstract way, we can define this distance as d(x, y) = |x − y|, which has the following properties (all easy to prove) d(x, y) ≥ 0, d(x, y) = 0 ⇐⇒ x = y, d(x, y) = d(y, x), d(x, z) ≤ d(x, y) + d(y, z). These properties capture what is essential about ‘distance’ and what it means to say that a point x on the real line is close to another point y: namely, that d(x, y) is small. This abstraction of the idea of distance is useful since it applies to other contexts as well. For example, for vectors x and y in R2 the distance d(x, y) can be defined as the length of the vector x − y. With this measure of distance on R2 , the same properties 5

of d listed above hold. The inequality d(x, z) ≤ d(x, y) + d(y, z) is, generally, called the ‘triangle inequality’ and it is easy to see why in the context of R2 . Furthermore, this distance measure on R2 is also the concept of distance used for complex numbers, with the usual depiction of C as R2 with a special way of multiplying these vectors in R2 . Nevertheless, we shall not treat ‘distance’ (or ‘metric’) as an abstract concept in this part of the course. We merely emphasize again that the distance of two objects (reals, vectors, complex numbers) is a real number, so one clearly should get to know the reals first before going any further. (In the course MA203, distance is studied more abstractly, in the context of metric spaces.)

Intervals For real numbers a and b, we define the intervals [a, b] = {x ∈ R : a ≤ x ≤ b} , (a, b] = {x ∈ R : a < x ≤ b} , (a, b) = {x ∈ R : a < x < b} , [a, b) = {x ∈ R : a ≤ x < b} , [a, ∞) = {x ∈ R : x ≥ a} , (a, ∞) = {x ∈ R : x > a} , (−∞, b] = {x ∈ R : x ≤ b} , (−∞, b) = {x ∈ R : x < b} . Note that (a, b) may not denote an interval but instead stand for the ordered pair with components a and b; which meaning is intended will be clear from the context. In the above notation for intervals, a parenthesis ‘(’ or ‘)’ means that the respective endpoint is not included (the interval is said to be open at that side), and a square bracket ‘[’ or ‘]’ that the endpoint is included (the interval is closed at that endpoint). Infinity, ∞, and negative infinity, −∞, are never included since they are not real numbers.

Properties of supremum and infimum We make some easy observations on sup and inf. The first is that if an upper bound of a set S belongs to that set, it is a supremum. (The corresponding result for infimum also holds.) A more common term is then ‘maximum’ since it is the largest element of that set. 6

Theorem 1 Let S be a set of reals and let z be an upper bound of S with z ∈ S. Then z = sup S.

Proof. Since z is an upper bound of S, all we have to show is that it is a least upper bound. So let x be any other upper bound of S, which means that for all y in S, y ≤ x. In particular, since z is a member of S, we have z ≤ x. So z is indeed at least as small as any other upper bound of S, that is, z = sup S. Secondly, suprema, if they exist, are unique (which justifies the use of the notation ‘sup S’ to denote a single real number that is the supremum of S): Theorem 2 Let S be a set of reals and let σ and σ 0 be least upper bounds of S. Then σ = σ0. Proof. By definition, σ 0 is an upper bound of S and hence at least as large as the least upper bound σ, so we have σ 0 ≥ σ. Exchanging σ 0 and σ, the same reasoning shows σ ≥ σ 0 . But these inequalities imply σ 0 = σ. The next observation is a little less trivial and gives a condition that characterizes a supremum. The theorem is most interesting if that supremum is not already a maximum, i.e. does not belong to the set in question. It shows that no matter how small the value of  is, a supremum cannot be decreased by  without losing the property that it is an upper bound.

Theorem 3 Let S be a nonempty set of reals and let σ be an upper bound of S. Then σ = sup S

⇐⇒

∀ > 0 ∃y ∈ S

y > σ − .

Proof. The proof is almost there once it is understood what the condition on the right hand side means. It says ‘for all positive  there is an element y of S so that y > σ − ’. So let us prove that this condition holds if σ is a supremum, i.e., show the implication ‘=⇒’ of the equivalence that we want to prove. Suppose otherwise; that is, σ is a supremum but ‘∀ > 0 ∃y ∈ S y > σ − ’ is false, that is, there is some  > 0 so that ∃y ∈ S y > σ − ’ is false, that is, for all y in S we have that ‘y > σ − ’ is false, that is, for all y in S we have y ≤ σ − . But this means that σ −  is an upper bound for S, which is less than σ since  > 0, contradicting our assumption that σ is the least upper bound. So ‘=⇒’ indeed holds.

7

Conversely, assume that σ is an upper bound for S and that for all positive  there is an element y of S so that y > σ − . Then σ must be the least upper bound of S since if there was an upper bound σ 0 of S with σ 0 < σ, taking  = σ − σ 0 (which is > 0) would give an  > 0 with ∀y ∈ S y ≤ σ − , contradicting the assumption. Hence σ = sup S (since σ is an upper bound of S by assumption), showing the implication ‘⇐=’ as well. In fact, the preceding theorem is rather easy, but we have argued very carefully using the logic of quantifiers and the negations of quantified formulae to get you used to that. Remember that ‘∀y P (y)’ means ‘for all y property P (y) holds’, and that its negation ‘¬∀y P (y)’ is equivalent to ‘∃y ¬P (y)’, meaning that P (y) is not true for all y if it is false for at least one y (a rather obvious statement). Similarly, the statement ‘∃y P (y)’ means ‘there is a y so that P (y) holds’, and its negation ‘there is no y such that P (y) holds’, that is, ‘¬∃y P (y)’, is equivalent to ‘∀y ¬P (y)’ and means that for all y, P (y) is false (again, rather obviously). When quantifiers start to pile up, it is useful to go through these negation processes step by step, as illustrated in the preceding proof.

8

Sequences and limits Examples of sequences A sequence is an infinite ordered list a1 , a2 , a3 , . . . of real numbers. An example is given by an = 1/n for all natural numbers (positive integers) n; that is, the sequence 1 1 1 1 1, , , , , . . . . 2 3 4 5 What happens to the numbers 1/n as n gets larger and larger? Clearly they ‘approach’ 0. But how do we formalise this? Before considering limits of sequences in a formal way, we consider another important sequence of numbers (which will show us that things are not always so simple as they are for the sequence just considered). For a fixed x > 0, let the numbers a1 , a2 , a3 , . . . be defined by a1 = 1, a2 = 1 + x, a3 = 1 + x +

x2 x3 x2 , a4 = 1 + x + + 2 2 6

and, in general, n−1

an = 1 + x +

X xi xn−1 x2 + ··· + = . 2 (n − 1)! i! i=0

Although that sequence (obviously) increases with each step, it turns out that, as n increases, the numbers approach a fixed number. That is, the sequence has a limit, which we will denote by limn→∞ an . This is far from clear. The value of that limit depends on x and defines a very important function in mathematics, which has its own name, exp(x). That is, n−1 i X x exp(x) = lim , n→∞ i! i=0 which we write as exp(x) =

∞ X xi i=0

9

i!

.

(Be aware that the infinite sum notation used here is just a shorthand for the limiting value of the finite sum of the an ; we don’t really ‘add up’ an infinite sequence of numbers. You will find out more about such matters if you take MA203, in which series are studied.) Considered as a function of x, this is the so-called exponential function, a function of central importance in analysis and calculus.

Sequences and limits Sequences: formal definition Let N = {1, 2, 3, . . .} be the set of natural numbers (positive integers). Formally, a sequence is a function from N to R. Only the notation is unusual: Instead of writing f (n) for the value of the function f , say, for the natural number n, that value is written an . The entire sequence is then written in one of the following ways, all legitimate: (an )n∈N ,

(an )∞ n=1 ,

(an )n≥1 ,

(an ),

where the last expression (an ), the one we will use most often, is an abbreviation that should only be used when it is clear that the index n runs through the natural numbers. The nth element (or nth term) an of a sequence may be defined explicitly by a formula involving n, as in the two examples given above. Alternatively, a sequence might be defined recursively. For example, we might have a1 = 1 and an+1 = an /2 for n ≥ 1. The sequence may also be given abstractly, without assuming a specific form.

The definition of a limit The single most important idea in this part of MA103 is the notion of the limit of a sequence. This can sometimes appear daunting, and many students struggle with it when they first meet it. But it is of such importance in this course that you are urged to grapple with it until it makes sense to you. Let’s take things easy to start with. We intuitively know what we mean by saying that the sequence (1/n) tends to the limit 0 as n tends to infinity. In symbols, we may write 1/n → 0 as n → ∞. Well, I say we know what it means, but do we, really? Before we go further, try for yourself to write down, precisely, what it means to say that a sequence (an ) of numbers tends to 0 as n → ∞. It’s not that easy.

10

You might suggest something like ‘the terms get closer and closer to 0’, but although this is indeed a property of the sequence (1/n), it is not a sufficient definition of tending towards 0. For, consider an = 1/2 + 1/n. This does not tend to 0 (—it tends to 1/2, as you can convince yourself by calculating some of the numbers), although it is the case that the terms get closer and closer to 0. (The point is, they don’t get ‘close enough’.) Consider now the sequence bn given as follows 1 1 1 1 1 1 1 , 1, , , , , , , . . . . 2 3 2 4 3 5 4 So, b2k = 1/k and b2k−1 = 1/(k +1). This sequence tends to the limit 0, but it is not the case that the terms get closer and closer to 0, because we have, for any k, b2k > b2k−1 . Anyway, all this is just to try to convince you that we really do need a formal, precise, definition of what it means for a sequence to tend to 0 if we are to have a notion of limit that works well. And here it is. . .. Definition 4 The sequence (an ) is said to tend to 0 (as n tends to infinity) if for all positive real numbers  there is a number N = N () such that for any natural number n > N , the distance of the element an from 0 is at most : ∀ > 0 ∃N ∀n > N

|an | < .

(∗)

If this holds, we write an → 0

as

n→∞

and also lim an = 0.

n→∞

The definition may also be understood as saying that for any  > 0, there is N such that if n > N then the ‘size’ |an | of the nth term is smaller than . Another interpretation is: given any ‘error margin’ , the terms of the sequence are ultimately equal to 0 within that error margin. It should be understood in (*) that  is an arbitrary positive real number, whereas n is a natural number so that an makes sense. Also, N depends on , something I have stressed by writing N = N (). As an example, consider the sequence (an ) defined by an = 1/n for n ∈ N. The limit of this sequence is zero, which can be seen, formally, as follows: To show (*) holds, consider any  > 0. Then we look for an N such that whenever n > N , then |an | < , that is, 1/n <  or, equivalently, 1/n < , that is, n > 1/. So a suitable value for N in that case is N = 1/ (which may not be an integer, but this poses no problem since n in (*) is always an integer). 11

In general, (*) states that no matter how close one wants to be to the limit 0 (that is, no matter how small  is chosen), the elements an are eventually all closer to 0 than distance . ‘Eventually all’ means ‘all except for a finite number’, where these finite exceptions must be among the elements an with 1 ≤ n ≤ N . In general, and fairly obviously, the smaller  is, the larger N must be. We see this in the above example, in which N = N () = 1/. What about a sequence tending to some limit other than 0? The following definition shows how we can easily adapt the one just given. Definition 5 The sequence (an ) is said to have the limit a (as n tends to infinity) if and only if an − a tends to 0. Explicitly, we therefore have: Definition 6 The sequence (an ) is said to have the limit a (as n tends to infinity) if for all positive real numbers  there is a number N such that for any natural number n > N , the distance of the element an from a is at most : ∀ > 0 ∃N = N () ∀n > N

|an − a| < .

(∗∗)

If this holds, one also says that the sequence tends to a as n tends to infinity, written as an → a as n → ∞ and also as a = lim an . n→∞

A sequence that has a limit is called convergent, otherwise divergent. In general, (**) states that no matter how close one wants to be to the limit a (that is, no matter how small  is chosen), the elements an are eventually all closer to a than distance . Definition (**) is crucial and should be understood thoroughly. Note that the following conditions are equivalent: |an − a| < , a −  < an < a + , an ∈ (a − , a + ) . The interval (a − , a + ) is also called the -neighbourhood of a. 12

Note also, for example, that the reference to N in (**) cannot be omitted: The (rather useless) statement ∀ > 0 ∃n |an − a| <  would be trivially true if a was one of the elements an of the sequence. Furthermore, it is also not sufficient that infinitely many elements of the sequence are close to a: The sequence (an ) defined by an = (−1)n is given by −1, 1, −1, 1, −1, . . . and alternates between 1 and −1. Either of these two numbers has an infinite number of elements of the sequence close to it, but not eventually all of them (simply take  = 1 in (**) where there is no N with the property stated there). This is an example of a divergent sequence. Certain sequences diverge because their members become arbitrarily large. For these sequences, a useful definition can be given.

Definition 7 The sequence (an ) tends to infinity, written an → ∞ as n → ∞, if ∀K ∃N ∀n > N

an > K.

This is also written as lim an = ∞ .

n→∞

Similarly, the sequence tends to minus infinity if (−an ) tends to infinity.

Note: If a sequence tends to infinity, it is divergent (not convergent). Note also that a sequence may diverge because it tends to ∞ or −∞, or because, while remaining bounded like an = (−1)n , it simply fails to tend towards any fixed real number. Do not confuse divergence with tending to infinity or minus infinity: divergence is more general than this. The key thing is to be clear than convergence means tending towards a (finite) real number.

Therapy The definition just given is the most important thing in this part of the course. It’s even more important than any of the Theorems. Definitions matter. We need them because we need to know precisely what we mean by saying, for example, that a sequence converges to a number. You cannot even properly begin to prove things about the limits of sequences until you know the formal definition. This should be clear: how can you prove that a sequence converges to 0 if you don’t even know what ‘converges to 0’ means? Try to make sure, now, that you understand this definition. Its meaning will become clearer as we work further through the course and use it. If you cannot understand it, then do something about it now, before it’s too late: talk to your class teacher or me. The rest of the analysis material won’t make much sense 13

if you cannot get your head around the definition of a limit, and the concept of a limit is so central in mathematics that it will be required for subsequent courses. I know it might be difficult: this is well-known to be one of the hardest conceptual hurdles in university mathematics. Bryant (on the back cover of his book) claims that ‘A first course in Analysis at college is always regarded as one of the hardest in the curriculum’. Academic articles have been written about the difficulty of the limit concept (e.g., ‘The Notion of Limit: Some Seemingly Unavoidable Misconception Stages’, Journal of Mathematical Behavior, 5, 1986, 281–303). Anybody who has taught–or even studied– university-level mathematics is very aware that it can initially be difficult to handle the concept of limit. So, what I’m saying is (to quote Bill Clinton) ‘I feel your pain’, but nonetheless you must do your best to grapple with this important concept.

Some standard results on and properties of limits The notation limn→∞ an suggests that the limit of a sequence is unique. Indeed, this is the case:

Theorem 8 A sequence has at most one limit.

Proof. Consider a sequence (an ) and assume, to the contrary, that it has two limits a and b with a 6= b. We will arrive at a contradiction by choosing  in the definition of limit small enough. Intuitively, if  is less than half of the distance between a and b, then an element that is less than  away from a must be farther than  away from b. So it can’t be the case that eventually all elements of the sequence can be that close to both a and b. To formalize this, let  = |a − b|/2, which is positive since a 6= b. Then for some N1 and all n > N1 , since a is a limit of the sequence, |an − a| < , and similarly for some N2 and all n > N2 , since b is a limit of the sequence, |an − b| < . But then for n > N = max{N1 , N2 }, 2 = |a − b| ≤ |a − an | + |an − b| = |an − a| + |an − b| <  +  = 2, which is a contradiction. Hence it is not possible that the sequence has more than one limit. A slightly different way to argue this, which is essentially the same as that just given, is as follows. As above, let  > 0 be any given positive number. Then, with N1 and N2 as above, we deduce that for any n > N = max{N1 , N2 }, |a − b| < 2. Now, this 14

must be true for any  > 0. The only way that a non-negative number X can satisfy X < 2 for all  > 0 is if X = 0. (For, if X > 0, then the inequality X < 2 will not hold if  = X/2. A sequence is said to be bounded if all its elements belong to an interval of the form [a, b]. (That is, if the set of terms of the sequence is a bounded set of real numbers.) Then, a is any lower bound and b any upper bound of the set of elements of the sequence. The case that only lower or upper bounds exist is also of interest. Definition 9 Let (an ) be a sequence and S = {an : n ∈ N}. Then the sequence (an ) is said to be bounded below if S has a lower bound, bounded above if S has an upper bound, and bounded if it is bounded above and below. Theorem 10 Any convergent sequence is bounded. Proof. Let (an ) be a convergent sequence with limit a. Then the sequence is bounded since for sufficiently large n, the elements an of the sequence are close to a, whereas there is only a finite number of them that are not close to a, and those finitely many elements are also bounded. To make this precise, we use the definition of a limit with  = 1, so that for some N and all n > N we have a − 1 < an < a + 1 (any other positive number  instead of 1 would do as well). Then we define as lower and upper bounds L and U L = min ({an : 1 ≤ n ≤ N } ∪ {a − 1}) ,

U = max ({an : 1 ≤ n ≤ N } ∪ {a + 1}) ,

where these are minima and maxima of finite sets and therefore exist. Then clearly, L ≤ an ≤ U holds for all n in N, so the sequence is indeed bounded. The next definition concerns sequences a1 , a2 , a3 , . . . that satisfy a1 ≤ a2 ≤ a3 ≤ · · · or a1 ≥ a2 ≥ a3 ≥ · · · which are called increasing and decreasing, respectively. Note that some elements in such a sequence can be equal to their predecessors, so the sequence does not have to go strictly up or down. Definition 11 A sequence (an ) is increasing if an ≤ an+1 for all n ∈ N, decreasing if an ≥ an+1 for all n ∈ N, and monotonic if it is increasing or decreasing. 15

The following result is very useful.

Theorem 12 An increasing sequence that is bounded above has a limit.

Proof. Let (an ) be an increasing sequence that is bounded above and consider the set of its elements S = {an : n ∈ N}. There is a natural candidate for the limit of (an ), namely σ = sup S, which exists since S is nonempty and has an upper bound by assumption. Because σ = sup S, if  > 0 there is some y ∈ S such that y > σ − . But, because S is simply the set of an , this means there is some N ∈ N such that aN > σ −. Now, the sequence is increasing, so for all n > N , we must have an ≥ aN > σ − . Also, since σ is an upper bound on S, an ≤ σ for all n. Thus, for all n > N , σ −  < an ≤ σ and, in particular, |an − σ| < . This shows that an → σ as n → ∞. We could also, alternatively, prove the result by contradiction. Suppose, to the contrary, that σ is not the limit of (an ). Then ∃ > 0 ∀N ∃n > N

|an − σ| ≥ .

So consider such an  > 0 and take any N ∈ N, so that, according to this condition, there is some n > N with |an − σ| ≥ . Since the sequence is increasing and since σ is an upper bound for S, this means σ − aN ≥ σ − an = |an − σ| ≥ . Because N is an arbitrary natural number here, aN ≤ σ −  for all N ∈ N. But this means that σ −  is an upper bound for S that is strictly less than the supremum σ of that set, and this is a contradiction. So σ is indeed the limit of the sequence. We have not only shown that an increasing sequence that is bounded above has a limit, but that this limit is the supremum of its elements. Similarly, one can see the following:

Theorem 13 A decreasing sequence that is bounded below converges to the infimum of its elements.

16

Algebra of limits When computing the limits of a sequence, it is useful to apply arithmetic rules if the terms are the sum, product, etc., of terms of sequences that have a known limit behaviour. For example, one can prove using the formal definition of a limit that the sequence (an ) defined by 4n2 + 9 an = 2 3n + 7n + 11 converges to 4/3. However, it is simpler to observe that an =

4 + 9/n2 3 + 7/n + 11/n2

where the terms 9/n2 , 7/n and 11/n2 all have limit zero and can be replaced by their limit to obtain that
lim 4 + 9/n2 n→∞
lim an = n→∞ lim 3 + 7/n + 11/n2 n→∞

lim (4) + lim (9/n2 )

=

n→∞ n→∞

=

n→∞

lim (3) + lim (7/n) + lim (11/n2 ) n→∞

n→∞

4 4+0 = . 3+0+0 3

Such an ‘algebra of limits’ is possible because of the following observation.

Theorem 14 Let (an ) and (bn ) be convergent sequences with limit a and b, respectively. Let C be a real number and let k be a positive integer. Then as n → ∞, (a)

Can → Ca,

(b)

|an | → |a|,

(c)

an + bn → a + b,

(d)

an bn → ab,

(e)

akn → ak ,

(f )

if bn 6= 0 for all n and b 6= 0, then 1/bn → 1/b as n → ∞.

17

Proof. We prove (d) as an example to illustrate the proof method in one of the more difficult cases. Note that (d) implies (e) by induction on k, taking (bn ) to be the sequence defined by bn = ak−1 n . To show that an bn → ab as n → ∞, consider any  > 0. The goal is to show that |an bn − ab| <  for all sufficiently large n. Here we need a trick in order to exploit that an → a and bn → b as n → ∞, which is all we know at this point. The trick is to insert the term −abn + abn , which is zero, into the sum to obtain |an bn − ab| = |an bn − abn + abn − ab| ≤ |an bn − abn | + |abn − ab| = |(an − a) bn | + |a (bn − b)| = |an − a| · |bn | + |a| · |bn − b| where it suffices to make the last two summands smaller than /2. Since the sequence (bn ) converges, it is bounded, as proved above, so that for some real number B we have |bn | < B for all n, where B > 0 (we will need this shortly since we divide by B). Then |an − a| · |bn | < |an − a| · B, so in order to have the left hand side smaller than /2 it suffices to have |an − a|
N1 , say, since (an ) converges to a, so that then |an − a| · |bn | < |an − a| · B
N2 so that |bn − b|
N , |an bn − ab| < , as desired. This ‘re verse construction’ of N1 and N2 using a modified term like 2B instead of  in the definition of limit is very typical in convergence proofs. Be careful not to misinterpret (e). This works only for a fixed number k. So, for example, since 1 + 1/n → 1 as n → ∞, we have (1 + 1/n)k → 1k = 1 for any k ∈ N. However, (1 + 1/n)n → e as n → ∞.

18

The sandwich theorem A useful result is the sandwich theorem (also known as the squeeze theorem).

Theorem 15 Let (an ), (bn ) and (cn ) be sequences such that an ≤ bn ≤ cn for all n ∈ N and lim an = b = lim cn . n→∞

n→∞

Then lim bn = b.

n→∞

Proof. Let  > 0. By assumption, there are N1 and N2 so that |an − b| <  if n > N1 and |cn − b| <  if n > N2 . Letting N = max{N1 , N2 }, we observe that for n > N this implies b −  < an ≤ bn ≤ cn < b +  which proves that |bn − b| <  as required. As an example, consider the sequence (an )n∈N where an = Clearly an ≤ n ·

n2

1 1 1 + + . . . + . n2 + 1 n2 + 2 n2 + n

n 1 1 = . Furthermore, an ≥ 0. Thus, since 1 ≤ +1 n n + n2 1 , n→∞ n

lim 0 = 0 = lim

n→∞

we have by the sandwich theorem that   1 1 1 lim + + ... + 2 = 0. n→∞ n2 + 1 n2 + 2 n +n Another application of the sandwich theorem is to prove that xn tends to zero if |x| < 1. The standard approach to proving this goes as follows: Let  > 0. We are looking for a number N so that n > N implies |xn | < , which is equivalent to |x|n < . Here we can without loss of generality assume x > 0, since otherwise we can replace x by −x and if x = 0 the claim holds trivially. Now observe the following equivalences, where the

19

logarithm is taken for any basis and log x < 0 since x < 1, so the inequality reverses when dividing both sides by log x: xn <  log(xn ) < log  n log x < log  log  n> . log x That is, if N = log / log x, then n > N implies xn <  as desired. The problem with this derivation of N is that it uses the definition of the logarithm, which if done formally requires a number of other concepts in analysis that are not yet proven at this point. In short, we want a more elementary proof that xn converges to zero if 0 < x < 1. To achieve this, observe that then x can be written as 1/(1 + h) for some h > 0. By induction on n, one can easily prove (1 + h)n ≥ 1 + hn which implies xn =

1 1 1 ≤ < . (1 + h)n 1 + hn hn

Then taking an = 0, bn = xn and cn = 1/hn and applying the sandwich theorem shows that xn → 0 since cn → 0 as n → ∞, as desired. This is not the only way to show that xn → 0 if 0 < x < 1. Here’s another way: let an = xn . Then, since an+1 = xan and 0 < x < 1, we can see that (an ) is a decreasing sequence. But it’s bounded below by 0. So, as a decreasing sequence which is bounded below, it converges (to its infimum). We now have to show the limit is 0. Suppose that an → a. We’ve already noted that an+1 = xan . So, by the algebra of limits, the sequence (an+1 ) converges to xa. This sequence, with nth term equal to an+1 is essentially the same as (an ), except that the numbers have all been shifted one place to the left (with a1 ) omitted. Now we can use a fairly obvious fact (which you should convince yourself of): if an → a as n → ∞, then an+1 → a as n → ∞. So the sequence (an+1 ) must converge to a. But we’ve just proved it converges to xa. So, since it can have only one limit, a = xa. Given that 0 < x < 1, this means a = 0. So an → 0 as n → ∞, as required.

20

Subsequences Consider the sequence 1 1 1 1 1 1, , , , , , . . . 2 3 4 5 6 If we were to cross out every other term our sequence would become 1 1 1 , , ,... 2 4 6 The resulting sequence is called a subsequence of the original sequence. The first term of the new sequence is the second term of the original sequence; the second term of the new sequence is the fourth term of the original sequence, and so on. Thus the elements of the new subsequence can be found by looking at the appropriate position in the original sequence. More precisely, the nth term of the subsequence is simply the (2n)th term of the original sequence. The formal definition of a subsequence is as follows.

Definition 16 Let (an )n∈N be a sequence and consider strictly increasing natural numbers k1 , k2 , k3 , . . ., that is, k1 < k2 < k3 < k4 < · · ·. Then the sequence (akn )n∈N is called a subsequence of the sequence (an )n∈N .

In the above example, we have obtained a subsequence given by the indices k1 = 2, k2 = 4, etc., in general kn = 2n. The indices kn tell us precisely which terms to keep from the original sequence. An alternative definition of a subsequence changes the set of indices N to a subset S of the natural numbers. Namely, let S be an infinite subset of N. Then a subsequence of (an )n∈N is simply given by (am )m∈S with the understanding that the index m runs through the elements of S in increasing order. Namely, here m = kn and the index set S is the set S = {k1 , k2 , k3 , . . .} = {kn : n ∈ N} . The set S can always be written in this way because it is infinite.

Theorem 17 Let (an ) be a sequence which tends to a limit a. Then any subsequence also tends to the limit a.

21

Proof. The proof is almost immediate: Let an → a as n → ∞, and let (akn ) be a subsequence. Then by the definition of limits, ∀ > 0 ∃N ∀n > N

|an − a| < .

Now, because the sequence k1 , k2 , . . . is increasing, we have kn ≥ n for all n. So, for all n > N , since kn > N , we have |akn − a| < . This means that the subsequence (ank ) converges to a. Here are two examples. The sequences (1/2n )n∈N and (1/n!)n∈N are subsequences of (1/n)n∈N . Secondly, ((−1)2n )n∈N and ((−1)2n−1 )n∈N are subsequences of ((−1)n )n∈N . The latter are two constant sequences with constant terms 1 and −1, respectively. From this we can see that ((−1)n ) has no limit by using the result that each subsequence of a convergent sequence has the same limit. This is because ((−1)2n ) tends to 1 as n → ∞, whereas ((−1)2n−1 ) tends to −1.

Theorem 18 Every sequence has a monotonic subsequence.

Proof. We first give a nice illustration of this proof taken from the book by V. Bryant, Yet Another Introduction to Analysis. Assume that (an ) is the given sequence, and that an is the height of a hotel with number n, which is followed by hotel n + 1, and so on, along an infinite line where at infinity there is the sea. A hotel is said to have the seaview property if it is higher than all hotels following it. That is, a hotel without seaview is followed sooner or later by a hotel of at least that height. Now there are only two possibilities: Either there are infinitely many hotels with seaview. Then they form a decreasing (in fact strictly decreasing) subsequence. Or there is only a finite number of hotels with seaview, so that after the last hotel with seaview, one can start with any hotel and then always find one later that is at least as high, which is taken as the next hotel, then considering yet another that is at least as high as that one, and so on. Then the subsequence of hotels generated in this way is increasing, although not necessarily strictly. Formally, let S = {m ∈ N : ∀n > m am > an } (which is the set of numbers of hotels with seaview). Consider the following two possible cases. Case 1. S is infinite. Then clearly, (am )m∈S is a decreasing subsequence of (an )n∈N . Case 2. S is finite. Then if S is empty, let k1 = 1, otherwise let k1 = max(S) + 1. This means that for all n ≥ k1 , the sequence element an does not belong to S (hotel n does not have seaview). Then define inductively for n = 1, 2, 3, . . . kn+1 = min{m ∈ N : m > kn and am ≥ akn } 22

which is possible since the set on the right hand side is not empty (otherwise kn would belong to S). In other words, kn+1 — which we intend to be the next index in the subsequence — is the smallest number greater than kn so that akn+1 ≥ akn . This implies that k1 , k2 , k3 , . . . are the indices of a subsequence of (an )n∈N so that ak1 ≤ ak2 ≤ ak3 ≤ · · ·. That is, (akn )n∈N is an increasing subsequence of (an )n∈N . We obtain from this as a corollary the following famous result.

Theorem 19 (The Bolzano-Weierstrass Theorem) Every bounded sequence of real numbers has a convergent subsequence.

Proof. Let (an ) be a bounded sequence. By the preceding theorem, it has a monotonic subsequence (akn ). This subsequence is then also bounded and we have seen that bounded monotonic sequences are convergent.

23

Limits of Functions and Continuity Limit of a function Definition of limit 6 6

L

L

a

-

u

a

-

Figure 1

We start by defining what we mean by the limit of a function f (x) as the argument x approaches some value a ∈ R. For example, for the first function illustrated in Figure 1, we would say that the limit of the function as x approaches a, is the value L. In fact, we are not concerned about the value of the function at the point a (indeed we don’t even care if f is defined at a); for example, for the second function illustrated in Figure 1, we would still say that the limit is L, even though f (a) is defined to be some value strictly greater than L. Thus the notion of limit applies only to the behaviour of the function f in a neighbourhood near to a. The formal definition now follows.

Definition 20 Let f : R → R be a function. We say that L is the limit of f (x) as x approaches a, denoted by limx→a f (x) = L if for each  > 0, there exists δ > 0 such that 0 < |x − a| < δ =⇒ |f (x) − L| < .

24

The definition states that if someone gives us any arbitrarily small , then there is some neighbourhood of a, (a − δ, a + δ), such that any x in this neighbourhood - other than a itself - will have f (x) in the -neighbourhood (L − , L + ) of L.

Examples Example: Suppose that f (x) = 3x − 1. We can then show directly from the definition of a limit that limx→1 f (x) = 2. To prove this, we must argue that for any  we can bound the value |f (x) − 2| = |(3x − 1) − 2| <  in some neighbourhood of 1. But we easily see that |f (x) − 2| = 3|x − 1|. In other words, the distance of f (x) from 2 is three times the distance of x from 1. Thus if we choose x to be within distance /3 from 1, then f (x) is within distance of 2. A formal proof would go as follows. Let  > 0. Then, let δ = /3. For any x s.t. 0 < |x − 1| < δ we have: |f (x) − 2| = 3|x − 1| < 3 ·

 < . 3

Therefore f (x) → 2 as x → 1. Example: As a second example, consider f (x) = x2 + x. We show that limx→2 f (x) = 6. To see this, suppose that  > 0 is given. If x is such that 0 < |x − 2| < δ, then |f (x) − 6| = |x2 + x − 6| = |x + 3||x − 2| < |x + 3|δ. Now, if |x − 2| < δ then |x + 3| ≤ 5 + |x − 2| < 5 + δ, by the triangle inequality. So to make |f (x) − 6| less than , it suffices to have (5 + δ)δ ≤ . There is no need to solve a quadratic equation in δ. If we just put δ = min{1, /6}, the smaller of 1 and /6, then we have 5 + δ ≤ 6 and δ ≤ /6, so (5 + δ)δ ≤ 6(/6) = , and we’re done.

Algebra of limits Let f, g : R → R be two functions and c be any real number, then we derive other functions by applying an algebra on the set of functions. For example, a new function 25

(f + g) is obtained by defining for each x, (f + g)(x) = f (x) + g(x). We say that (f + g) is derived point-wise since the value of (f + g) at x is defined by the normal arithmetic sum of the two real numbers f (x) and g(x). Similarly, we may define the functions |f |, (cf ), (f − g), (f + g), (f · g) and (f /g), provided g(x) 6= 0.

Theorem 21 Let f, g : R → R be two functions and c be any real number. Suppose that limx→a f (x) = L and limx→a g(x) = M . Then 1. limx→a (cf )(x) = cL 2. limx→a (|f |)(x) = |L| 3. limx→a (f + g)(x) = L + M 4. limx→a (f − g)(x) = L − M 5. limx→a (f g)(x) = LM 6. limx→a (f /g)(x) = L/M provided g(x) 6= 0 for each x in some neighbourhood of a.

More on limits Sometimes, we may have a configuration as in Figure 2 where the limit from one side is different from that from the other. We adapt the definition as follows.

Definition 22 Let f : R → R be a function. We say that L is the limit of f (x) as x approaches a from the left (or from below), denoted by limx→a− f (x) = L if for each  > 0, there exists δ > 0 such that a − δ < x < a ⇒ |f (x) − L|.

A similar definition applies to limits from the right (or from above), denoted limx→a+ f (x) = L. In Figure 2, the left-hand and right-hand limits are L, M respectively.

26

6

M

u

L

-

a Figure 2

So far we’ve discussed limits of a function as x approaches some value a. But we can also make the following definition (similar to the definition of the limit of a sequence).

Definition 23 Let f : R → R be a function. We say that L is the limit of f (x) as x approaches ∞, denoted by limx→∞ f (x) = L if for each  > 0, there exists M > 0 such that x ≥ M ⇒ |f (x) − L| < .

Continuity Loosely speaking, a function is continuous if its graph can be drawn without lifting pen from paper. For example, the first function in Figure 1 is continuous, but the second is not. For a formal definition of this notion, we first define continuity at a point.

Definition 24 A function is continuous at the point a if f (a) is defined and is equal to limx→a f (x). Definition 25 A function is continuous if it is continuous at each point a.

Definition 26 A function is continuous on the interval [a, b] if it is continuous at each point in (a, b) and (i) f (a) = limx→a+ f (x) and (ii) f (b) = limx→b− f (x). 27

As an example of a discontinuous function (that is, one that is not continuous), consider the ‘sign’ function ( 1 if x > 0, f (x) = 0 if x = 0, −1 if x < 0. This function f (x) makes a “jump” when x = 0. This represents a discontinuity since when we approach zero from the left (that is, with negative values for x), the function has always value −1, no matter how close we are to zero, whereas at zero it has a different value. That is, the function values when approaching zero “tend to” something else than what the value of the function is when the limit is taken first. (The same happens when zero is approached from the right, but already a strange behaviour like that when coming from one direction is enough to make the function discontinuous.) It follows from the results on the algebra of limits that: Theorem 27 Let f, g : R → R be functions which are continuous at a ∈ R and c be any real number. Then |f |, (cf ), (f − g), (f + g), (f · g) are all continuous at a, and (f /g) is continuous provided g(x) 6= 0 for any x in some neighbourhood of a. P As a corollary we see that any polynomial p(x) = ki=0 ai xi is continuous. This can be proved by induction as follows. Clearly any constant function is continuous by the above results. Moreover, the function f (x) = x is continuous. Thus by the product rule, so is f (x) = x2 , and by induction f (x) = xk is continuous for any finite k. Hence so is any function of the form f (x) = ai xi . Finally, by repeated application of the summation rule, the polynomial p(x) is deduced to be continuous. Note that if f, g are functions, then the composition function f ◦ g is defined by (f ◦ g)(x) = f (g(x)) for each x. Theorem 28 If g is a function which is continuous at a, and f is a function which is continuous at g(a). Then (f ◦ g) is continuous at a. Proof. Let’s be clear about what we want to prove. To show that the composite function is continuous at a, we must show that for any  > 0, there is δ > 0 such that |x − a| < δ implies |f (g(x)) − f (g(a))| < . By the continuity of f at g(a), we know that for any given  > 0 there is some δ1 > 0 such that if |y − g(a)| < δ1 then |f (y) − f (g(a))| < . So if we can find δ > 0 such that |x − a| < δ implies that (taking y = g(x)) |g(x) − g(a)| < δ1 , then we’ll have that |x − a| < δ implies |f (g(x)) − f (g(a))| < . But we can certainly find such a δ, simply from the definition of g being continuous at a. (Note the use of the notation δ1 here, so that we do not confuse ‘intermediate δ-values’ with the δ we need to finally produce.) 28

Continuity and sequences We now give an alternative definition of continuity which ties in the concept of limits for sequences. Theorem 29 A function f is continuous at a if and only if for each sequence (xn ) such that limn→∞ xn = a we have limn→∞ f (xn ) = f (a). We now prove that this is, in fact, equivalent to the previous definition. Let (*) be the statement that for any sequence (xn ) such that limn→∞ xn = a, limn→∞ f (xn ) = f (a). Suppose first that f is continuous at a. We prove that this implies (*). Let (xn ) be a sequence of reals converging to a. We want to show that f (xn ) → f (a) as n → ∞, that is, ∀ > 0 ∃N ∀n ≥ N |f (xn ) − f (a)| < . (∗∗) To prove this, let  > 0. Choose, according to the definition of continuity, a δ > 0 so that for all x, whenever |x − a| < δ, then |f (x) − f (a)| < . Since xn → a as n → ∞, there is an N so that n ≥ N implies |xn − a| < δ, which in turn implies |f (xn ) − f (a)| < . This shows (**) as desired. Conversely, assume that property (*) holds. In order to show continuity, we assume, to the contrary, that the function is discontinuous at a. This means that there is an  > 0 so that for all δ > 0 there is an x with |x − a| < δ but |f (x) − f (a)| ≥ . In particular, for every natural number n, letting δ = 1/n, there is a real number x, call it xn , with |xn − a| < 1/n but |f (xn ) − f (a)| ≥ . But then clearly xn → a as n → ∞, but we do not have f (xn ) → f (a) as n → ∞, a contradiction to (*). We find this a very useful alternative for the sake of manipulation of limits since it states that for any sequence xn → a, the limit of the f (xn )’s is simply f applied to the limit of the xn ’s:   lim f (xn ) = f lim xn . n→∞

n→∞

Continuous functions on closed intervals Definition 30 For X subset of the domain of a function f , we say that f is bounded on X if there exists M such that |f (x)| ≤ M for each x ∈ M . Definition 31 We define the supremum (or maximum) of f on X as sup{f (x) : x ∈ X} (or max{f (x) : x ∈ X} if it exists). 29

The following result is very important and useful.

Theorem 32 Let f be continuous on [a, b]. Then f is bounded on [a, b] and it achieves its maximum; that is, the supremum is equal to the maximum.

Proof. Suppose first that f is unbounded above. For each n ∈ N, let xn be a point in [a, b] such that f (xn ) > n. The sequence (xn ) is bounded, so has a convergent subsequence (xkn ), tending to some limit c. Necessarily c ∈ [a, b]. Since f is continuous at c, f (xkn ) → f (c) as n → ∞. But this contradicts the construction of the sequence (xn ), since f (xkn ) > n → ∞. So f is bounded above. Let M = sup{f (x) : x ∈ [a, b]}. For each n ∈ N, let xn be a point in [a, b] such that f (xn ) > M − n1 . Again take a convergent subsequence (xkn ) of (xn ), tending to some limit c ∈ [a, b]. Arguing as before, we see f (c) = M . The same result holds with max/sup replaced by min/inf.

The Intermediate Value Theorem In this section we prove one of the most fundamental (and obvious!) theorems in real analysis. One useful property of continuous functions f lies in the fact that they have solutions x to equations of the form f (x) = C for any given C where such a solution might reasonably be expected, namely if f takes values below and above C. In other words, a continuous function cannot “hop over” intermediate values as it moves from one value to another. This central property of continuous functions is known as the “intermediate value theorem”.

Theorem 33 (The Imtermediate Value Theorem) Let f be a continuous function on [a, b] and let K be such that f (a) < K < f (b). Then for some c ∈ (a, b), f (c) = K.

We prove a special case of the result, from which the full Theorem follows. Theorem. Let f be a continuous function on [a, b] such that f (a) < 0 and f (b) > 0. Then for some c ∈ (a, b), f (c) = 0. Proof. We construct a sequence of intervals [an , bn ] such that 30

1. f (an ) < 0, f (bn ) > 0 for each n 2. [an+1 , bn+1 ] ⊆ [an , bn ] for each n. We start by letting [a1 , b1 ] = [a, b]. Then for each n ≥ 1, we define [an+1 , bn+1 ] as follows. Let cn = (an + bn )/2, be the midpoint of the previous interval. If f (cn ) = 0, then the theorem is proved and so we need not continue constructing intervals! Otherwise, if f (cn ) < 0, we define an+1 = cn and bn+1 = bn . And if f (cn ) > 0, we define bn+1 = cn and an+1 = an . Note that the condition 1. is satisfied by choosing our intervals in this manner. Moroever, note that the (n + 1)st interval is half the size of the nth interval and so bn+1 − an+1 ≤ (b1 − a1 )/2n . It follows that lim (bn − an ) = 0.

n→∞

(1)

Finally, note that (an ) is increasing and bounded above (by b1 ) and so it has a limit; similarly (bn ) is decreasing and bounded below and so has a limit. Thus by (1) (and algebra of limits) these limits are equal to, say, c. Thus by continuity (using Theorem 29) f (c) = lim f (bn ) ≥ 0 n→∞

where the last inequality follows from the fact that each f (bn ) ≥ 0 (in fact > 0). Similarly, f (c) = lim f (an ) ≤ 0. n→∞

Thus f (c) must be equal to zero, and the proof is complete.

31

Exercises Do not panic: not all of these will be assigned. Specific exercises will be required for classes each week. Solutions to all problems, including those not assigned, will be made available on the website. Exercise 1 The distance between two real numbers x and y is defined as the absolute value |x − y| of their difference. This is a nonnegative real number, and satisfies |x − y| = |y − x| and |x + y| ≤ |x| + |y|. Use these properties to show the following “triangle inequality” |x − z| ≤ |x − y| + |y − z| and the inequality |x − y| ≥ ||x| − |y||. Exercise 2 For each of the following sets, find the sup, inf, max and min whenever these exist. [−1, ∞), {x ∈ R : x2 − 2x − 1 < 0}, {x2 − 2x − 1 : x ∈ R}. Exercise 3 Suppose A is a bounded subset of R. Let B be the set defined by B = {b : b = 2a + 3, a ∈ A}. Prove that sup B = 2 sup A + 3. [Hint: let σA = sup A and start by proving that 2σA + 3 is an upper bound on B.] Exercise 4 A bounded subset A of R has the property that for all x, y in A, |x−y| < 1. Prove by a contradiction argument that (sup A − inf A) ≤ 1. 32

Exercise 5 Let S be a nonempty set of reals and let τ be a lower bound of S. Prove that τ = inf S ⇐⇒ ∀ > 0 ∃y ∈ S y < τ + .

Exercise 6 Suppose that (xn ) is a sequence of real numbers which converges to a limit L > 0. Show that there is N ∈ N such that for all n ≥ N , xn > 0. √ Exercise 7 Suppose that xn = 1/( n + 1). Prove that, for any  > 0, there is N = N () such that if n > N , then xn < . Exercise 8 Suppose that xn = 1 − n12 . Prove that, for any  > 0, there is N = N () such that if n > N , then |xn − 1| < .

Exercise 9 Let (xn ) be the sequence defined by xn =

2n − 3 . n+3

By considering the difference |xn − 2| , show explicitly that xn → 2 as n → ∞. [This means that you must explicitly produce an N = N () such that for all n > N , |xn − 2| < .]

Exercise 10 Let (an ) be the sequence defined by an =

4n2 + 9 . 3n2 + 7n + 11

Using the formal definition of the limit of a sequence, show explicitly that an →

4 as n → ∞. 3

[Like the previous exercise, this means you must produce N ().]

Exercise 11 Prove that limn→∞ xn+1 = limn→∞ xn , assuming either side of the equation exists. 33

Exercise 12 Let (an )n∈N be a sequence, and let (bn )n∈N be the sequence defined by bn = |an | for n ∈ N. Which of the following two statements implies the other? (a) (an ) converges. (b) (bn ) converges. Exercise 13 Prove that a sequence which is decreasing and bounded below converges. Exercise 14 Let (xn ) be a sequence of non-negative real numbers (i.e., xn ≥ 0 for each n). Suppose that the sequence converges to x. Prove that x ≥ 0. [Note, however, that a sequence of positive terms need not have a positive limit; for example, (1/n) is a sequence of positive terms converging to 0.] Exercise 15 Suppose the sequence (xn ) of positive terms converges to 0. Prove that if yn = 1/xn then the sequence (yn ) tends to infinity. Let (xn ) be a sequence of positive terms such that the sequence (1/xn ) tends to infinity. Show that (xn ) converges to 0. Exercise 16 Find the limits as n → ∞ of: 4n − 5 5(32n ) − 1 1 + 2 + · · · + n n2 − 1 , , , 3 . 22n − 7 4(9n ) + 7 n2 n +1 Exercise 17 A sequence (xn ) is defined as follows. Let x1 be any positive real number and, for n ≥ 1, let x2 + K xn+1 = n , 2xn where K is a fixed positive number. 2 2 (i) By using √ the inequality a + b ≥ 2ab for real numbers a, b (or otherwise), prove that xn ≥ K for n ≥ 2. (ii) Prove that xn+1 ≤ xn for all n ≥ 2. (iii) Deduce that (xn ) converges and find its limit. Exercise 18 Find limn→∞ xn in the following cases: xn = xn =

2n3 + 1 , 3n3 + n + 2

(n + 1) 2n n + + ··· + 2. 2 2 n n n 34

Exercise 19 Prove by induction that if n ∈ N and x ≥ −1 then (1 + x)n ≥ 1 + nx. √ Taking x = 1/ n in this, show that 1≤n

1/n

√
2/n ≤ < 1+ n



1 1+ √ n

2 .

Deduce that lim n1/n = 1.

n→∞

Exercise 20 Using the fact that for a > 0, a1/n → 1 as n → ∞, and a sandwiching argument, find lim (1n + 2n + 3n + 4n )1/n . n→∞

Exercise 21 Find  lim

n→∞

1

1 √ +√ + ··· + √ n2 + 1 n2 + 2 n2 + n 1

 .

Exercise 22 For the positive sequence (an ), it is given that there exist numbers N and α such that α < 1 and 0
N. an

Prove that (an ) is eventually decreasing and that an → 0 as n → ∞. Give an example to show that if we relax this condition and only have 0
N, an

then we cannot conclude that an → 0. Exercise 23 Discuss the behaviour as n → ∞ of the following:  n  n 1 2n 2n3 + 1 3 3 (n + n) , , , 2 n3 + n n+1 4 22n + n , n3 3n + 1

√

35

n+1−

√

n.

Exercise 24 Let (an ) be a√sequence of non-negative numbers. Prove that if an → L √ as n → ∞ then an → L as n → ∞. [This is not covered by the ‘algebra of limits’ results of the lectures (—why not?): prove the result explicitly using the formal definition of a limit.] Exercise 25 Prove, from the formal definition of limit, that f defined on R \ {2} by f (x) =

5 (x − 2)2

tends to infinity as x → 2. Exercise 26 Evaluate the following limit: x3 + 5x + 7 . x→1 x4 + 6x2 + 8 lim

Exercise 27 Let bxc denote the largest integer n such that n ≤ x. Determine lim (x − bxc) and

x→1+

lim (x − bxc).

x→1−

Exercise 28 Let f : R → R be defined by ( (x − 1)2 f (x) = 1 3x + 2

if x < 1; if x = 1; if x > 1.

Use the definitions of the lectures to prove that f (x) → 0 as x → 1− and f (x) → 5 as x → 1+. Is the function • continuous • continuous on the right • continuous on the left at the point 1? Exercise 29 The function f is defined on R by n 0, if x is rational; f (x) = 1 if x is irrational. 36

Prove that f is discontinuous at every point of R. [Hint: use the fact (which you might recall from the first term of this course) that there are irrational numbers arbitrarily close to any rational number and that there are rational numbers arbitrarily close to any irrational number.]

Exercise 30 Suppose the real function f is continuous at c and f (c) > 0. Prove, directly from the definition of continuity, that there is δ > 0 such that f (x) is positive for x in the interval (c − δ, c + δ).

Exercise 31 Suppose the real function f is continuous, positive and unbounded on R and that inf{f (x) : x ∈ R} = 0. Use the Intermediate Value Theorem to prove that the range of f is (0, ∞), the set of all positive real numbers. [It might be obvious, but give a watertight proof. Explicitly, prove that for any y > 0 there is some c ∈ R such that y = f (c).]

Exercise 32 Let T (θ), for 0 ≤ θ ≤ 2π, be the surface temperature at the point at θ degrees longitude on the equator. (Note that T (0) = T (2π).) Assuming T is a continuous function of θ, prove that at any given time, there are two points on the equator which have the same temperature and are diametrically opposite. [Regrettably, field trips are not available. Instead, consider the function f (θ) = T (θ) − T (θ + π), and use the Intermediate Value Theorem.]

Exercise 33 Suppose that the real function f is continuous on the closed interval [a, b] and that f maps [a, b] into [a, b]. By considering the function h(x) = f (x) − x, show that there is c ∈ [a, b] with f (c) = c. Suppose the real function g is continuous on R and that g maps [a, b] into [d, e] and maps [d, e] into [a, b], where a < b, d < e. By considering the function k(x) = g(g(x)), prove that there are p, q ∈ R such that g(p) = q, g(q) = p. Hence show that there is c ∈ R such that g(c) = c.

37