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Chapter 6

Mathematical analysis

Sets and functions,

supremum and infimum,

limits of sequences,

limits and continuity of

real functions, open and

closed sets, compact

sets, analysis in Rm

Reading

Simon and Blume contains most of what is needed for this chapter, but for sometopics (such as the sandwich theorem and the intermediate value theorem), youshould consult a specialist text in mathematical analysis (or real analysis). Someexamples are listed below.

Simon, C.P. and Blume, L., Mathematics for Economists Appendix A1, sectionsA1.1 and A1.2; Chapter 12; Chapter 13, section 13.4.

Bartle, R.G. and Sherbert, D.R., Introduction to Real Analysis. Chapter 2,section 2.4; Chapter 3, sections 3.13.4 and 3.6; Chapter 4, sections 4.1 and4.2; Chapter 5, sections 5.15.3; Chapter 10, sections 10.110.3.

Bryant, V., Yet Another Introduction to Analysis. Chapters 1,2, and 3.

Introduction

In this large chapter we introduce the topic of mathematical analysis, often re-ferred to as real analysis or simply analysis. This is a formal approach to theideas underlying calculus. The emphasis in mathematical analysis is on formalisingnotions such as the limit of a sequence, and on proving certain facts using theseformalisations. Students do not find analysis easy when they first meet it, but it is

probably the single most important part of this subject, so you should persevere!

Sets and functions: some terminology

For sets A, B, we define the set

A \ B = {a A : a B} .

If B A, then A \ B is the complement of B in A.

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Suppose that S is some set and I is some non-empty (indexing) set such that for eachi I, we have a set Ai S. Thus, {Ai : i I} is a set, or collection, of sets. Then

iI

Ai = {x : x Ai for all i I}

and

iI

Ai =

{x : x

Ai for at least one i

Ai

}.

We have the De Morgan laws of complementation:

S\I

Ai =I

(S\ Ai), S\I

Ai =I

(S\ Ai).

In what follows, let f : A B be a function or mapping. (The terms functionand mapping will, for our purposes, mean the same thing.) The set A is called thedomain of f and B is called the codomain of f. For C A, we define

f(C) = {f(c) : c C} .

The set f(A) is called the image of f.

The function f is surjective if f(A) = B. It is injective if

f(x) = f(y) = x = y.

It is bijective if it is both injective and surjective, in which case it has an inversefunction f1 : B A.

The cartesian product of A and B is the set

A B = {(a, b) : a A, b B}

of all ordered pairs (a, b). Given sets A1, A2, . . . , An, we define

A1 A2 . . . An = {(a1, a2, . . . , an) : a1 A1, a2 A2, . . . , an An} .

Quantifiers

There are two very useful symbols often used in mathematical analysis.

First, we have the symbol , called the universal quantifier, which means for all.For example, consider the following mathematical statement:

x R, x2 0.

This says, literally, that for all x R, x2 is non-negative; that is, the square of anyreal number is non-negative, and this is something we know is true. On the otherhand, the statement

x [1, 2], x > 1is clearly false, since if we take x = 1, we have x [1, 2] but x is not greater than 1.This serves to demonstrate a useful observation. Suppose that P(x) is a statementdepending on x, which is either true or false for each particular x. (So, for example,P(x) could be x > 1.) A statement of the form x S, P(x) is called an existential

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statement, and it says that P(x) is true for every x S. Such a statement canbe shown to be false (or, equivalently, can be disproved) by exhibiting a particularx0 S such that P(x0) is false.

Next, we have the existential quantifier , meaning there exists (or there is).Consider the statement

x R such that x2 > 5.

This says there is an x R such that x2 > 5, and this is clearly true. (Often, oneabbreviates such that to s. t., or simply omits it.) On the other hand, the statement

x R such that x2 + 1 = 0is false, since x2+1 = 0 has no real solutions; that is, for all x R, x2 + 1 = 0. Againwe have a general principle. Suppose that P(x) is a statement depending on x, whichis either true or false for each particular x. A statement of the form x S, P(x) iscalled a universal statement, and it says that P(x) is true for some x S. Sucha statement can be shown to be false (or, equivalently, can be disproved) by provingthat for every x S, P(x) is false.

Statements can involve both types of quantifier. Consider

M R x R such that x2 > M,or, equivalently (omitting the such that)

M R, x R, x2 > M.This says for every real number M there is some real number x such that x2 > M.Well, this is true: for example (and there are other possibilities!), if M < 0 we maytake x = 0, if 0 < M < 1 we may take x = 1, and if M > 1 then (since M2 > M) wemay take x = M.

The order in which the quantifiers appear can be crucial. Suppose we were to reverse

the order of the universal and existential quantifiers in the statement just considered.Then we obtain

x R, M R, x2 > M.Think about what this means: it asserts that there is some fixed real number x suchthat for every real number M, x2 > M. But this is nonsense. For a fixed x, there isalways M such that x2 M. Indeed, take M = x2 to see this. So the statement wehave now is very different from the previous one.

Activity 6.1 Ensure you understand why the order of the quantifiers is vital in theseexamples.

The triangle inequality

For any real number x, the absolute value of x, denoted by |x|, is x itself if x 0 andx if x < 0. For instance, |5| = 5 and | 3.5| = 3.5.

The basic triangle inequality is very simple, but extremely useful. It states that forany two real numbers x, y, we have

|x + y| |x| + |y|.

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Activity 6.2 Prove the triangle inequality. (Just consider each of four cases: x >0, y < 0, etc.)

This has many useful consequences, which we shall use often. In particular, for anythree real numbers x,y,z, we have

|x

y

| |x

z

|+

|y

z

|(simply because |x y| = |(x z) + (z y)| |x z| + |z y|).

Sometimes useful is the fact that for any two real numbers x, y,

|x y| ||x| |y||.

Properties of real numbers: supremum and infimum

Recall that for real numbers a, b, we define different types of interval as follows:

[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}

.

We shall sometimes use the symbol R+ to denote the set of all positive real numbers.

It is very important (especially in what follows) that you understand that is nota number: it is merely a symbol. You cannot perform arithmetic with it. It is quitesimply nonsense to write 1/ = 0 and so on.

Some of these intervals are bounded above, some bounded below and some boundedabove and below. The formal definitions of boundedness are:

Definition 6.1 S R is bounded above if and only if there is M R such thatfor all x S, x M.

In this case, M is called an upper bound of (or on or for) S.

Definition 6.2 S R is bounded below if and only if there is m R such that

for all x S, x m.

In this case, m is called a lower bound of (or on or for) S.

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A set is said simply to be bounded if it is bounded above and bounded below.

Definition 6.3 An element x S is the maximum of S if it is an upper bound ofS (similarly, we can define the minimum of S).

A set that is bounded above (below) need not have a maximum (minimum); a set has

a maximum element if and only if it has an upper bound that belongs to the set; thiswill not be true in general.

Every subset of the integers has a maximum and a minimum. The correspondingnotions for subsets of the real line are supremum and infimum, defined below.

We assume the following property of real numbers (as an axiom).

The Continuum property: Every non-empty set of real numbers that is boundedabove has a least upper bound and any non-empty set of real numbers that is boundedbelow has a greatest lower bound.

Definition 6.4 For a non-empty subsetS ofR, ifS is bounded above, the least upperbound of S (which exists, by the continuum property) is called the supremum of S,and is denoted sup S. If S is bounded below, the greatest lower bound is called theinfimum of S and is denoted infS.

Note that the supremum and infimum of a set S (when they exist; i.e., when S isbounded above/below) need not belong to S.

Example: Consider the real interval S = (1, 2). This is bounded above: for example,2 is an upper bound, since every member of S is at most 2. In fact, any numbergreater than or equal to 2 is an upper bound, so the set of upper bounds is [2 , ).It is clear that the least upper bound is therefore 2. That is, sup S = 2. But noticethat the set S has no maximum. For, suppose that M is a maximum. Then (twothings!): (i) M S and, (ii) for all x S, x M. Now, because M S, we musthave M < 2. But now consider the number (M + 2)/2, half-way between M and 2.This is still less than 2, so it belongs to S. But it is not true that x M for allx S, because this doesnt hold when we take x = M. So S has a supremum but nomaximum.

Example: On the other hand, the interval S = (1, 2] has a maximum: its maximum(and its supremum) is 2.

Example: Consider the set S = {1/n : n N} = {1, 1/2, 1/3, . . .}. This is clearlybounded below, by 0 for example. In fact, infS = 0. To see this, we observe (as wealready have) that 0 is a lower bound on S. To prove it is the greatest lower bound,we need to show that no number > 0 is a lower bound on S. So lets suppose that is any positive number. To prove it is not a lower bound on S we have to showthat some member of S is less than . Well, this means we need to find a number ofthe form 1/n such that 1/n < . But its clear that this inequality is equivalent ton > 1/. So if we take any natural number n that is greater than 1/, then 1/n Sand 1/n < . So we are done: any > 0 cant be a lower bound, and 0 is, so 0 is thegreatest lower bound, the infimum.

Activity 6.3 Make sure you understand the preceding three examples.

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An alternative description of the supremum is: = sup S if and only if is an upperbound for S and if for any < , there is some x S with x > . (A similarcondition can be given for infimum.)

Activity 6.4 Write down the corresponding description of infimum.

A further characterisation (which follows from the one just given by taking = )is as follows.

Theorem 6.1 For S R, S non-empty, = sup S if and only if is an upperbound for S and

> 0, x S with x > .

Activity 6.5 Convince yourself that this characterisation of the supremum is equiv-alent to the one given in the definition.

There is also a similar characterisation of the infimum.

Theorem 6.2 For S R, S non-empty, = infS if and only if is a lower boundfor S and

> 0, x S with x < + .

Sequences of real numbers

Formally, a sequence is a function f from N to R. We call f(n) the nth

term of thesequence and we often denote the sequence by (f(n))n=1 or simply (f(n)). Informally,and more usually, a sequence is thought of as an infinite list of real numbers, one foreach positive integer: for example,

a1, a2, a3, . . .

We denote it (an)n=1 or (an) (or indeed, (ar), (ai) etc.). Then we call an the nth term

of this sequence.

A sequence may be defined by giving an explicit formula for the nth term. For examplethe formula an = 1/n defines the sequence whose value at the positive integer n is1/n.

A sequence may also be defined inductively. For instance, we might have

a1 = 1, an+1 =an2

+3

2an(n 1).

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Limits of sequences

The definition of limit, and some basic results

As an example, let us calculate the first few terms of the sequence given by

a1 = 1, an+1 =an2 +

3

2an (n 1).We have

a1 = 1, a2 = 2, a3 = 1.75, a4 = 1.73214, a5 = 1.7320508, . . . .

Activity 6.6 Check these calculations of a2, . . . , a5.

Note that the terms get closer and closer to 1.73205 . . . as n increases. In fact, thenumber the terms approach can be shown to be

3 = 1.7320508057 . . .. We say that

the sequence has limit

3, or that it converges to

3.

What do we really mean by saying that a sequence (xn) of numbers tends towardssome fixed limit L? Well, one informal approach is to say that the terms of thesequence are eventually approximately equal to L. But this will not do: we needsomething more formal. First, the notion of approximately equal is very vague.What we really mean is that for any small given > 0, the terms of the sequenceshould be between L and L + ; that is,

L < xn < L + ,or, equivalently, |xn L| < . And by eventually, we mean for large enough n: thatis, for all n > N, where N is some number. The smaller is, the larger N might haveto be, so N will generally depend on . The formal definition of a limit is as follows.

Definition 6.5 (Finite limit of a sequence) The sequence (xn) is said to tendto the (finite) limit L if for all > 0, there is N such that for all n > N we have|xn L| < . That is, (xn) tends to L if

> 0, N s.t. for n > N, |xn L| < .We write

xn L a s n or

limn

xn = L,

and say that xn tends to L as n tends to .

You will need to know this definition thoroughly, but dont just memorise it: under-stand it! Without the understanding, you wont be able to use it. Be very aware thatthe order of the quantifiers is important: comes before N.

Any sequence that tends to a finite limit is said to be convergent.

In the definition, we can assume if we like that N is an integer, or we can replacen > N by n N: neither of these changes makes any difference.

The following result is easy to prove, but very useful.

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Theorem 6.3 A sequence has no more than one limit.

Activity 6.7 Prove this. [Hint: Suppose that L1 andL2 are two limits of a sequence.Show that in fact L1 must equal L2 by proving that for all > 0, |L1L2| < 2, whichcan only be true if |L1 L2| = 0.]

We can also talk of a sequence tending to plus or minus infinity, as in the followingdefinition.

Definition 6.6 (Infinite limit) The sequence (xn) tends to infinity as n if:

K, N s.t. for n > N, xn > K.We write xn or

limn

xn = .Similarly we say that the sequence (xn) tends to minus infinity as n if:

K, N s.t. n > N xn < K.

Bounded sequences

A sequence is bounded above if its terms are bounded above (and similarly for boundedbelow, and bounded). Formally:

Definition 6.7 (Bounded sequences) A sequence is (respectively) bounded above,bounded, or bounded below if the set

S = {xn : n N} Ris (respectively) bounded above, bounded, or bounded below.

Note that a sequence is bounded if and only if it is both bounded below and boundedabove.

The following result is often useful.

Theorem 6.4 Any convergent sequence is bounded.

Proof Suppose (xn) is convergent, with limit L. Then for all > 0 there is N suchthat ifn > N then |xn L| < . Take = 1. Then there is N (which we may assumeis a positive integer) so that for all n > N, |xn L| < 1. Then, for such n,

|xn| = |(xn L) + L| |xn L| + |L| < 1 + L.

So all the terms xN+1, xN+2, . . . are bounded in absolute value by the constant 1+ |L|.The remaining terms x1, x2, . . . , xN are only finite in number (there are N of them)and so they can be bounded in absolute value by

M = max{|x1|, |x2|, . . . , |xN|}.

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Then, for all n, we have either |xn| |L| + 1 or |xn| M, so certainly, |xn| 0, there are N1, N2 such that forn > N1, |an L| < and, for n > N2, |cn L| < . If we set N = max{N1, N2}, thelarger of N1 and N2, then for n > N we have n > N1 and n > N2, so both the aboveconditions hold:

|an

L|

< and|cn

L|

< . These may be written as

L < an < L + , L < cn < L + ,so for n > N,

L < an bn cn < L + ,which says |bn L| < . Thus bn L as n .

Example: 0 f(b) then thesame conclusion holds.

We prove a special case of the result, from which the full theorem follows.

Theorem 6.15 Let f be a continuous function on [a, b] such that f(a) < 0 andf(b) > 0. Then for some c (a, b), f(c) = 0.

Proof We construct a sequence of intervals [an, bn] such that

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] asfollows. 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 iff(cn) > 0, wedefine bn+1 = cn and an+1 = an. Note that condition 1. is satisfied by choosing our

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Theorem 6.16 The union of any collection of open sets is again an open set.

It should not be thought that all open sets are open intervals: although every openinterval is open, there are many other types of open set. For example, the set (1, 2) (3, 4) is open, but is not an open interval.

Closed sets

We also have the notion of closed sets. But before defining what this means, weneed to clear up one source of potential confusion. By analogy with the use of thewords open and closed in everyday language, we might think that a given set ofreal numbers must be either open or closed, and that if it is not open, it is closed.This, unfortunately, will not be the case: as we shall see, sets can be open but notclosed, closed but not open, both open and closed, or neither open nor closed!

Here is the formal definition of a closed set:

Definition 6.18 A set C R is a closed set (or is closed) if whenever (xn) is aconvergent sequence and xn C for all n, then the limit of the sequence, lim xn, isin C.

So a set C is closed if for any convergent sequence of members of C, the limit of thesequence is in C. This is a tricky definition to work with, but as we shall shortly see,there is another way of describing closed sets.

Example: The interval C = [0, 1] is closed. To see this, suppose that (xn) is anysequence in C, converging to a limit L. Then for each n, xn C, so 0 xn 1.Now, it follows from this that 0

L

1 (prove this!), so L

C, and hence C is

closed.

Example: The interval C = (0, 1] is not closed. Consider the sequence (xn) wherexn = 1/n. For all n, xn C. The sequence converges to 0, but 0 is not in C. So Cis not closed.

We mentioned that closed is not the opposite of open, but the following resultlinking open sets and closed sets is very useful.

Theorem 6.17 A set C of real numbers is closed if and only if its complementR \ C is open.

Proof Because this is an if and only if result, there are two things to prove here:first, that if C is closed then R \ C is open; secondly, that if R \ C is open then C isclosed.

Suppose, first, that C is closed and consider its complement U = R \ C. We want toshow U is open. Suppose it isnt. Then there is some y U such that for no > 0do we have (y , y + ) U. In other words, for all > 0, the interval (y , y + )doe not lie entirely within U = R \ C and hence must contain points of C. For anypositive integer n, lets take = 1/n. Then there is some xn (y 1/n,y + 1/n) suchthat xn C. Because |xn y| < 1/n, we have that xn y as n . So here we

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have a sequence (xn) in C such that lim xn = y C. But this cannot happen sinceC is closed. So whats gone wrong? Well, we supposed that R \ C was not open, andthis supposition must therefore be wrong. So R \ C is open.

Next, suppose that R \ C is open. To prove that C is closed, we need to show thatthe limit of any convergent sequence of points of C is in C. So suppose (xn) is aconvergent sequence, with xn C, and L = lim xn. We need to show L C. Sup-pose this isnt so. Then L is in the open set R

\C, so there is some > 0 such that

(L , L + ) R \ C. Now, because xn L, there is some N such that for n > N,|xn L| < , that is xn (L , L + ). But then for n > N, xn R \ C. This is acontradiction to the fact that xn C. So we have gone wrong in assuming that L isnot in C. Therefore it is in C, and C is closed.

This theorem is an extremely useful characterisation of closed sets. In fact, we could,if we had wanted, have taken the definition of a closed set to be a set C whosecomplement R \ C is open, and many texts do this.

Example: Consider again the set C = [0, 1]. We showed this was closed by using

Definition 6.18. But we can also see that it is closed by considering its complement.For,R \ C = R \ [0, 1] = (, 0) (1, ),

and this is open because it is the union of two open sets. So, since the complementof C is open, C is closed.

Example: The set R of all real numbers is both open and closed. The interval (0, 1]is neither open nor closed.

Activity 6.11 Convince yourself of the statements in the example just given.

Analysis on Rm

We now turn our attention to functions defined on Rm. (We use m rather than themore usual n because we have used n in representing a typical member of a sequenceas xn.) All the above may be generalised considerably using the notion of a metricor distance function.

Distance in Rm

The Euclidean distance (or simply distance) between x = (x1, x2, . . . , xm) andy = (y1, y2, . . . , ym) in R

m is defined to be

x y = n

i=1

(xi yi)2.

(The case in which m = 1 corresponds to the distance |x y| between two realnumbers.)

There is a certain mathematical attraction in defining distances on rather more ab-stract or unusual spaces, and this leads to the notion of a metric space. But in this

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subject, we are primarily interested in Rm, and we shall always use the Euclideandistance.

The Euclidean distance between two vectors x, y is simply the norm of x y, wherethe norm is the one arising from the usual inner product on Rm. Consequently, thedistance measure has some nice properties. For example, by the triangle inequalityfor norms, we have that for any x, y, z Rm,

y z = (y x) + (x z) y x + x z.

Lets recall the definition of continuity of a function f : R R. We say that f iscontinuous at a if and only if given any > 0, there is > 0 such that

|x a| < = |f(x) f(a)| < .This appears to use the fact that one can form the difference between any two realnumbers, However, as above, we can interpret |x a| as the distance between the realnumbers x and a, in which case the above condition can be restated as

distance(x, y) < = distance(f(x), f(a)) < .It should be clear from this that the condition doesnt really use any algebraic prop-erties of R, only distance properties. This definition and many of its consequenceswill remain if we have as domain and codomain Rm and Rk for any m, k 1.

Bounded subsets of Rm

A bounded subset of Rm is one in which there is some fixed number bounding thedistance between any two points in the set. Formally:

Definition 6.19 (Bounded subset of Rm

) B Rm

is bounded if there is K > 0such that for all x, y B, x y K.

Note that K is fixed: it does not depend on x, y. (The definition would be meaninglessif that were the case.)

Open balls

For m > 1, the counterpart in Rm to the open interval in R is the open ball.

Definition 6.20 (Open Ball) For x Rm and > 0, the open ball of radius around x is

B(x) = {y : x y < } .This is the set of y whose distance from x is less than .

Example: When m = 1, B(x) is exactly the open interval (x , x + ).

Example: In R2 the open ball B(x) is the region enclosed by a circle of radius centred at x note that the points on this circle do not lie in B(x).

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Continuity in general

We mentioned earlier that with a general concept of distance, we ought to be able togeneralise the definition of continuity to functions having as domain and codomainRm and Rk.

Definition 6.21 (Continuity off : Rm

Rk

) Suppose that f : Rm

Rk

andthat a Rm. Then, we say that f is continuous at a if given any > 0 there exists > 0 such that if x a < then f(x) f(a) < .

(Note that in this definition we use, as is customary, the notation . to denote boththe distance function on Rm and that on Rk.)

This can also be phrased in terms of open balls.

Definition 6.22 The function f : Rm Rk is continuous at a Rm if given anyopen ball B (f(a)), there exists such that

f(B(a)) B (f(a)) .

For a subset X of Rm, we say that f is continuous on X if for all a Rm, f iscontinuous at a. If f is continuous on Rm then we simply say that f is continuous.

The following result is sometimes useful.

Theorem 6.18 Suppose that f : Rm Rk and that f1, f2, . . . , f k : Rm R aresuch that for all x Rm,

f(x) = (f1(x), f2(x), . . . , f k(x))T.

Then f is continuous if and only if f1, f2, . . . , f k are continuous.

Open sets in Rm

We have already investigated in some depth the notion of open sets of real numbers.All the ideas and results extend to Rm, and the proofs are very similar. You shouldtry them, following the proofs given earlier in this chapter.

The definition of open set

Definition 6.23 A subset U of Rm is open if for any y U, there is = (y) > 0such that B(y) U.

Informally, a set is open if, from any point of the set, we can move some positivedistance in any direction without going outside the set.

It can easily be shown that any open ball is an open set, but there are other types ofopen set. Just as for open sets in R, the union of any collection of open subsets ofRm is again open.

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Sequences in Rm

A sequence (xn) in Rm is simply an ordered list x1, x2, . . . of elements of R

m. (Thisis a straightforward extension of the definition of a sequence of real numbers.) Itsfairly simple to extend to Rm ideas about convergence of sequences. The followingdefinition is a straightforward extension of the earlier definition for convergence forsequences of real numbers. (We use x rather than L just for cosmetic reasons.)

Definition 6.24 (Convergence of sequences in Rm) Suppose that (xn) is a se-quence of points of Rm. Then we say that the sequence has limit x Rm if for every > 0 there is N such that if n > N thenxn x < . Such a sequence is said to beconvergent and to converge towards x.

Having defined convergence of sequences in Rm, it is possible to describe an alternativeapproach to the definition of continuity of functions from Rm to Rk. We have alreadyseen for the case of real functions on R that if a function is continuous and xn xthen f(xn) f(x) and we have noted that this can in fact be used as the basis of adefinition of continuity: a function f : R R is continuous at a R if and onlyif for every sequence xn converging to a, f(xn) f(a). This can be generalisedfor functions mapping from Rm into Rk (and this is the approach taken in Simonand Blume). We have the following theorem (which may be taken as a definition ofcontinuity).

Theorem 6.19 Suppose that f : Rm Rk and that a Rm. Thenf is continuousat a if and only if for every sequence (xn) converging to a we have f(xn) f(a).Therefore f is continuous (on the whole of Rm) if for every convergent sequence (xn)in Rm, we have lim f(xn) = f(lim xn).

Closed sets in Rm

Just as for subsets of R, we can define the notion of a closed subset of Rm.

Definition 6.25 A set C Rm is a closed set (or is closed) if whenever (xn) isa convergent sequence and xn C for all n, then the limit of the sequence, lim xn, isin C.

So a set C is closed if for any convergent sequence of members of C, the limit of thesequence is in C.

As for the case m = 1 investigated above, we have the following result, the proof ofwhich is similar to the one given earlier.

Theorem 6.20 A set C Rm is closed if and only if its complement Rm \ C isopen.

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(Note that a function f is strictly increasing if f(n + 1) > f(n) for all positive n.)

In the example given above, we see that we have in fact obtained a subsequencecorresponding to the increasing function defined by f(n) = 2n. Note that this functiontells us precisely which terms to keep from the original sequence.

Often we will use shorthand to denote a sequence. For example, for a sequence

x1, x2, x3, x4, . . .

we may say that we wish to choose a subsequence

xk1 , xk2 , xk3 , xk4 , . . . .

We have written the increasing function f explicitly in terms of its value, by sayingexactly which terms of the original sequence to take; that is, we take terms k1, k2, k3and so on. Note that this will be just for notations sake and in these cases theunderlying function has not disappeared; in fact the increasing function here is givenby f(i) = ki for i = 1, 2, 3, . . .. One result that is easy to show is:

Theorem 6.22 Let (xn) be a sequence which tends to a limit L. Then any subse-quence also tends to the limit L.

Activity 6.12 Prove this.

We have the following result on sequences lying in compact subsets of Rm. It isknown as the Bolzanno-Weierstrass Theorem. (You need not know the proof. 2) 2 The proof can be found,

for the case m = 1, inBartle and Sherbert,

Section 3.4, for example.Theorem 6.23 (Bolzanno-Weierstrass Theorem) LetC be a compact subset ofRm and let (xn) be any sequence of points of C. Then (xn) has a convergent subse-

quence whose limit is in C.

Consider the case m = 1. Since any bounded subset of the real numbers lies in someclosed interval (because if M = sup S and m = infS then S [m, M]), it followsthat any bounded sequence can be assumed to be a sequence of numbers from somecompact set C. The theorem tells us, therefore, that any bounded sequence of realnumbers has a convergent subsequence. This is often useful in proving results in realanalysis.

Compactness and open covers

Another interesting aspect of compact sets is to do with open covers.

Definition 6.28 A cover for a subset B of Rm (or a cover of B) is a collectionCof sets such that

B UC

U.

(That is, the union of all the sets inC contains B.) A subcover ofC is a subcollectionD C such that D itself is a cover for B. A cover is finite if it consists of a finitenumber of sets.

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Definition 6.29 (Open cover) An open cover for B Rm is a cover for Bconsisting of open sets.

The following result holds, and indeed is taken in many texts to be the definitionof a compact set.3 (It might seem weird to take it as the definition of compactness, 3 For a proof of the result,

see, for example, Simon

and Blume, Section 29.5

since the definition we have given is so much more straightforward. But there is abigger picture here, involving more general notions of metric spaces and topological

spaces, in which it is appropriate to use the characterisation given below.)

Theorem 6.24 A subset B of Rm is compact if and only if any open covering ofB has a finite subcover. More explicitly, B is compact if and only if whenever we aregiven a family{UI : i I} of open sets such that B

IUi, then there is some finite

subfamilyUi1 , Ui2 , . . . , U in

such thatB Ui1 Ui2 . . . Uin.

There are some important points to be made on this theorem. First, we only considercoverings by open sets. Secondly and this is crucial! it is important to realisewhat the theorem does not say. It does not say that the set is compact if there isa finite open cover for B. Indeed, that is always true, since we may simply take thecover consisting of the single open set {Rm}. What it does say is something vastlydifferent: given any open cover C ofB, there is a finite number of the sets in C whichare enough to cover B.

Learning outcomes

At the end of this chapter and the relevant reading, you should be able to:

use existential and universal quantifiers explain exactly what is meant by the supremum and infimum and be able to

determine these

explain what is meant by a sequence work with sequences defined by a formula or defined recursively explain precisely what it means to say that a sequence converges or tends to

infinity or minus infinity determine limits and prove results using the formal definition of the limit of a

sequence

explain what is meant by a sequence being bounded, bounded above or boundedbelow

comprehend the links between boundedness and convergence prove and use the fact that bounded monotonic sequences converge calculate limits of sequences using the algebra of limits

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prove and use the Sandwich Theorem explain what is meant formally by the limit of a function at a point, or as x

tends to or , and by one-sided limits determine limits and prove results using the formal definitions of the limit of a

function

calculate limits of functions using the algebra of limits

explain what is meant by continuity prove functions are continuous or discontinuous using the formal definition and

heredity results

explain and be able to use the interaction between continuous functions andconvergent sequences

use the fact that a continuous function has a maximum and a minimum valueon a closed bounded interval

state and use the Intermediate Value Theorem

explain what is meant by open and closed sets and be able to prove that setsare open or closed or neither

use the definitions and results concerning convergence, continuity, open andclosed sets, and compact sets in Rm.

Sample examination questions

Question 6.1 Prove that for any two real numbers x, y, |x y| ||x| |y||.

Question 6.2 For each of the following sets, find the sup, inf, max and min wheneverthese exist.

[1, ),{x R : x2 2x 1 < 0},

{x2 2x 1 : x R}.

Question 6.3 Suppose A is a bounded subset of R. Let B be the set defined by

B = {b : b = 2a + 3, , a A}.

Prove thatsup B = 2 sup A + 3.

Question 6.4 Suppose that (xn) is a sequence of real numbers that converges to alimit L > 0. Show that there is N N such that for all n > N, xn > 0.

Question 6.5 Let (an) be the sequence defined by

an =4n2 + 9

3n2 + 7n + 11.

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By considering the difference an 43 ,

show explicitly that

an 43

as n .

Question 6.6 Show that if a sequence is decreasing and bounded below then itconverges.

Question 6.7 Let (xn) be a sequence of non-negative real numbers (i.e. xn 0 foreach 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.]

Question 6.8 Find the limits as n of:4n

5

22n 7 ,5(32n)

1

4(9n) + 7 ,1 + 2 +

+ n

n2 ,n2

1

n3 + 1 .

Question 6.9 A sequence (xn) is defined as follows. Let x1 be any positive realnumber and, for n 1, let

xn+1 =x2n + k

2xn,

where k is a fixed positive number.(i) By using the inequality a2 + b2 2ab for real numbers a, b (or otherwise), provethat xn

k for n 2.

(ii) Prove that xn+1 xn for all n 2.(iii) Deduce that (xn) converges and find its limit.

Question 6.10 Find limn xn in the following cases:

xn =2n3 + 1

3n3 + n + 2,

xn =n

n2+

(n + 1)

n2+ . . . +

2n

n2.

Question 6.11 Find

limn

1

n2 + 1+ 1

n2 + 2+ . . . + 1

n2 + n

.

Question 6.12 Prove, from the formal definition of limit, that f defined on R \ {2}by

f(x) =5

(x 2)2tends to infinity as x 2.

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=

28n + 179n2 + 21n + 33

45n9n2

=5

n.

This is < for n > N = 5/. [There is no need to solve a horrible quadratic for n:

make the approximation suggested and life is simple.]

Question 6.6 Suppose the sequence is (xn) and let = inf{xn : n N} (whichexists, by boundedness). Let > 0. Since + cannot be a lower bound on the setof xn, theres N so that xN < + . But the sequence is decreasing, so for n > N, xn xN < + and hence |xn | < . This shows the sequence converges to.

Question 6.7 We show x cant be negative. Suppose it was. Then x/2 > 0, sotaking = x/2 in the definition of limit, there is N so that for n > N,

|xn

x|

Na contradiction.

Question 6.8 We use the heredity results (that is, the algebra of limits).

4n 522n 7 =

4n 54n 7 =

1 5/4n1 7/4n

1

1= 1.

5(32n 14(9n) + 7

=5 1/9n4 + 7/9n

54

.

For the next one, the Sandwich theorem is no use (even though you might think so:try it!). Instead:

1 + 2 + + nn2

=n(n + 1)/2

n2=

n2/2 + n/2

n2=

1 + 1/(2n)

1 1

1= 1.

Finally,n2 1n3 + 1

=1/n 1/n3

1 + 1/n3 0

1= 0.

Question 6.9 We have, for n 2,

xn =x2n1 + k

2xn1=

x2n1 + (

k)2

2xn1 2xn1

k

2xn1=

k.

Also, for n 1,xn+1 xn = x

2n + k

2xn xn = k x

2n

2xn 0,

where we have used the fact that xn

k. So xn+1 xn. So the sequence isbounded below (by

k) and it is decreasing. These two facts mean that it converges.

Suppose xn L. Then xn+1 L too. But

xn+1 =x2n + k

2xn L

2 + k

2L,

and so L = (2L2 + k)/2L, or L2 = k. Since xn is positive for all n, L 0, so L =

k,and we have xn

k.

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Question 6.10 First,

2n3 + 1

3n3 + n + 2=

2 + 1/n3

3 + 1/n2 + 2/n3 2

3.

For the second, although it looks like its worth trying, the Sandwich Theorem doesnthelp much. Instead, we use

n + (n + 1) + + 2n = (n + 1)(n + 2n)/2 = 3n2

/2 + 3n/2,

so

xn =3n2/2 + 3n/2

n2=

3/2 + 3/(2n)

1 3/2.

(Straightforward Sandwich theorem application only tells us that xn lies between 1and 2.)

Question 6.11 Let xn denote the quantity we need to find the limit of. Then

n 1n2 + n

xn n 1n2 + 1

.

Now,n

n2 + n=

11 + 1/n

11

= 1,

andn

n2 + 1=

11 + 1/n2

11

= 1.

By the Sandwich theorem, xn 1.

Question 6.12 Let K > 0. We want to find so that if 0 < |x 2| < thenf(x) > K. If |x 2| < then

f(x) =5

(x 2)2 >5

2,

so it suffices to have 5/2 K, so we may take =

5/K.

Question 6.13

limx1

x3 + 5x + 7

x4 + 6x2 + 8=

13 + 5(1) + 7

14 + 6(1)2 + 8=

13

15.

Question 6.14 If 1 < x < 2 then x

x

= x

1, so

limx1+

(x x) = limx1+

(x 1) = 0.

If 0 < x < 1 then x x = x 0 = x, so

limx1

(x x) = limx1

(x) = 1.

Question 6.15 Taking = f(c)/2 > 0 in the formal definition of continuity at c,there is some such that if x (c , c + ) then |f(x) f(c)| < f(c)/2. Thus, forx (c , c + ), f(x) > f(c) f(c)/2 = f(c)/2 > 0, so f is positive on that interval.

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Question 6.16 Let y (0, ). We show that there is some c R such that f(c) = y.This shows that the range is the whole of (0, ). (The fact that it is no larger followsfrom the given fact that f is positive.) Now, f(R) = inf{f(x) : x R} = 0, so, sincey > 0, there must be some y1 f(R) with y1 < y. This means there is some x1 Rsuch that y1 = f(x1) < y. Similarly, because f is unbounded, which means f(R)is unbounded, there must be some y2 f(R) with y2 > y, and there will be somex2 R such that y2 = f(x2) > y. Then y lies between f(x1) and f(x2) and, sincef is continuous, the Intermediate Value Theorem shows that there is some c betweenx1 and x2 with f(c) = y.

Question 6.17 The function h satisfies h(a) = f(a) a 0, since f(a) [a, b]implies f(a) a. Similarly, h(b) 0. So a < b and h(a) 0 h(b). Since fis continuous, as is the function x x, it follows that h is continuous. By theIntermediate Value Theorem, c [a, b] with h(c) = 0; that is, f(c) = c.

Question 6.18 Let be such that 0 < < f(x). Such an can be chosen sincef(x) > 0. Because f is continuous at x, there is > 0 such that

f(B(x)) B(f(x)) = (f(x) , f(x) + ).

So, for x B = B(x), we have f(x) > f(x) > 0.

Question 6.19 Let > 0. Then there is > 0 such that if x a < then|f(x) f(a)| < . If |t a1| < then

(t, a2, . . . , am) a = (t, a2, . . . , am) (a1, a2, . . . , am)=

(t a1)2 + 0 + + 0

= |t a1| < ,so we have

|f(t, a2, . . . , am) f(a1, a2, . . . , am)| < ,which says precisely that

|g(t)

g(a1)

|< . This proves that g is continuous at a1.

Question 6.20 Suppose xn x and let > 0 be given. Then there is N such thatfor n > N, xn x < . But

xn x = m

i=1

(xin xi)2 |xin xi|,

for any i between 1 and m, so for n > N, |xin xi| < and hence xin xi. On theother hand, if, for each i, |xin xi| < , then

xn

x

=

m

i=1

(xin

xi)2

Ni, |xin xi| < /

m. Let N bethe largest of N1, . . . , N m. Then for n > N, |xin xi| < /

m for all i and hence

xn x < . This shows that xn x.

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Question 6.21 There are a number of ways of proving this. We show that thecomplement of B is open. Suppose that y Rm \ B. This complement is the set ofy such that y z > . To prove that Rm \ B is open, we want to show that someopen ball B(y) around y lies entirely in the complement ofB. Leave undeterminedfor now: well choose it later. Let x B(y), so that x y < . Then

y z y x + x z < + x z,

so x z > y z .Now, we know that y z > and we want to choose so that x z > (to showthat x B). Because y z is strictly bigger than , it is possible to find > 0such that y z > . (Clearly we just need 0 < < y z .) With such a, we see that B(y) Rm \ B and we are done.

Question 6.22 This is straightforward if we use the fact (indeed, the definition) thata set is compact if and only if it is closed and bounded. Clearly the intersection ofany collection of bounded sets is bounded (convince yourself!). To show that theintersection of any set of closed sets is closed can either be shown by considering their

open complements or (perhaps more easily) by simply noting that any convergentsequence whose terms are in the intersection must (by closure of the sets) have a limitwhich lies in each of the sets and hence in the intersection. The details are left toyou.

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