real analysis mathematics

Cambridge Books Onlinehttp://ebooks.cambridge.org/

Real Analysis

N. L. Carothers

Book DOI: http://dx.doi.org/10.1017/CBO9780511814228

Online ISBN: 9780511814228

Hardback ISBN: 9780521497497

Paperback ISBN: 9780521497565

Chapter

1 - Calculus Review pp. 3-17

Chapter DOI: http://dx.doi.org/10.1017/CBO9780511814228.002

Cambridge University Press

CHAPTER ONE

Calculus Review

Our goal in this chapter is to provide a quick review of a handful of important ideas fromadvanced calculus (and to encourage a bit of practice on these fundamentals). We willmake no attempt to be thorough. Our purpose is to set the stage for later generalizationsand to collect together in one place some of the notation that should already be moreor less familiar. There are sure to be missing details, unexplained terminology, andincomplete proofs. On the other hand, since much of this material will reappear in laterchapters in a more general setting, you will get to see some of the details more thanonce. In fact, you may find it entertaining to refer to this chapter each time an old nameis spoken in a new voice. If nothing else, there are plenty of keywords here to assistyou in looking up any facts that you have forgotten.

The Real Numbers

First, let's agree to use a standard notation for the various familiar sets of numbers. Rdenotes the set of all real numbers; C denotes the set of all complex numbers (althoughour major concern here is R, we will use complex numbers from time to time); Z standsfor the integers (negative, zero, and positive); N is the set of natural numbers (positiveintegers); and Q is the set of rational numbers. We won't give the set of irrationalnumbers its own symbol; rather we'll settle for writing R\Q (the set-theoretic differenceofRandQ).

We will assume most of the basic algebraic and order properties of these sets, butwe will review a few important ideas. Of greatest importance to us is that the set R ofreal numbers is complete - in more than one sense! First, recall that a subset A of Ris said to be bounded above if there is some x e R such that a < x for all a e A. Anysuch number x is called an upper bound for A. The real numbers are constructed sothat any nonempty set with an upper bound has, in fact, a least upper bound (l.u.b.).We won't give the details of this construction; instead we'll take this property as anaxiom:

The Least Upper Bound Axiom (sometimes called the completeness axiom).Any nonempty set of real numbers with an upper bound has a least upper bound.That is, if A c R is nonempty and bounded above, then there is a number s e R

satisfying: (i) s is an upper bound for A; and (ii) if x is any upper bound for A, then

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4 Calculus Review

s < x. In other words, if y < s, then we must have y < a < s for some a eA. (Why?) We even have a notation for this: In this case we write s = l.u.b. A =sup A (for supremum). If A fails to be bounded above, we set sup A -f oo, and ifA = 0 , we put sup A = 00 since, after all, every real number is an upper boundfor A.

Example 1.1sup(00, 1) = 1 and sup{2 (l/n): n = 1, 2 , . . .} = 2. Notice, please, that sup Ais not necessarily an element of A.

An immediate consequence of the least upper bound axiom is that we also havegreatest lower bounds (g.l.b.), just by turning things around. The details are left asExercise 1.

EXERCISE1. If A is a nonempty subset of R that is bounded below, show that A has a greatestlower bound. That is, show that there is a number m e K. satisfying: (i) m is a lowerbound for A; and (ii) if x is a lower bound for A, then x < m. [Hint: Consider theset A = {a : a A} and show that m = sup(A) works.]

We have a notation for greatest lower bounds, too, of course: We write m = g.l.b. A =inf A (for infimum). It follows from Exercise 1 that inf A = sup (A). Thus, inf A =00 if A isn't bounded below, and inf 0 = +00. In case a set A is both bounded aboveand bounded below, we simply say that A is bounded.

EXERCISES2. Let A be a bounded subset of E containing at least two points. Prove:(a) 00 < inf A < sup A < +00.(b) If B is a nonempty subset of A, then inf A < inf B < sup B < sup A.(c) If B is the set of all upper bounds for A, then B is nonempty, bounded below,

and inf B = sup A.3. Establish the following apparently different (but "fancier") characterization ofthe supremum. Let A be a nonempty subset of M that is bounded above. Prove thats = sup A if and only if (i) s is an upper bound for A, and (ii) for every s > 0, thereis an a A such that a > s s. State and prove the corresponding result for theinfimum of a nonempty subset of R that is bounded below.Recall that a sequence (xn) of real numbers is said to converge to x G ffi. if, for everys > 0, there is a positive integer N such that \xn x\ < s whenever n > N. In thiscase, we call x the limit of the sequence (xn) and write x = lim^oo xn.4. Let A be a nonempty subset of R that is bounded above. Show that thereis a sequence (xn) of elements of A that converges to sup A.



The Real Numbers

5. Suppose that an < b, for all n, and that a = limôo an exists. Show that a < b.Conclude that a < supn an = s\xp{an : n e N}.

> 6. Prove that every convergent sequence of real numbers is bounded. Moreover, if(an) is convergent, show that infn an < limôo an < supn an.

As an application of the least upper bound axiom, we next establish the Archimedeanproperty in E.

Lemma 1.2, Ifx and y are positive real numbers, then there is some positive integern such that nx > y.

PROOF. Suppose that no such n existed; that is, suppose that nx < y for all n eN.Then A = {nx : n e N} is bounded above by y, and so s = sup A is finite. Now,since s x < 5, we must have some element of A in between, that is, 5 x < nx < sforsomen e N.Butthens < (n + l)x. And what's wrong? Well, since (n + l)jc e A,we should instead have (n + 1 )x < s. This contradiction tells us that it is unacceptableto have nx < y for all n, and so we must have nx > y for some n.

This simple observation does a lot of good:

Theorem 1.3. If a and b are real numbers with a < b, then there is a rational r e Qwith a < r < b.

PROOF. Since b a > 0, we may apply Lemma 1.2 to get a positive integer qsuch that q(b a) > 1. But if qa and qb differ by more than 1, there must be someinteger in between. That is, there is some p Z with qa < p < qb. Thus a 0. Since x s < x = supn*n, we must have xN > x s for some N. Butthen, for any n > N, we have x s < xN < xn < x. (Why?) That is, \x xn \ < efor all n > N. Consequently, (xn) converges and x = supn xn = limôo xn.

Finally, if (xn) is decreasing, consider the increasing sequence (~xn). From thefirst part of the proof, (xn) converges to supn(xn) = infn xn. It then followsthat (xn) converges to infw xn. D

In subsequent chapters we will consider certain properties of the real line that may bedefined either in terms of sequences or in terms of subsets of M. To better appreciate theconnection between sequences and sets, we will show how Theorem 1.4 gives a quick proofof the nested interval theorem. Later in this chapter we will use the nested interval theoremto define a strange and beautiful subset of R called the Cantor set.

The Nested Interval Theorem 1.5. If (/) is a sequence of closed, bounded,nonempty intervals in E with I\ D h ^ h D , then H^Li In / 0- tf> inaddition, length (/) - 0, then H^Li h contains precisely one point.

PROOF. Write / = [an, bn ]. Then / 3 h+\ means that an < an+\ < bn+\ ooZ?n = infnbnboth exist (as finite real numbers) and satisfy a < b. (Why?) Thus we must haveP|^= 1 ln [a, b]. Indeed, ifx e / for all n, then an < x < bn for all n, andhence a < x < b. Conversely, if a < x < b, then an < x < bn for all n. That is,x e In for all n. Finally, if bn an = length (/) -> 0, then a b and so H^LiIn = {a). Examples 1.6(a) Please note that it is essential that the intervals used in the nested interval theorem

be both closed and bounded. Indeed, fXLdn> ) = 0 a n d Pl^liC0' 1/"] = 0-(b) Suppose that (/) is a sequence of closed intervals with / D In+\, for all n and with

length (/) -> 0 as n -> oo. If H^Li h = W*tnen anY sequence of points (xn),with xn G / for all w, must converge to x. (Why?)

A sequence of sets (/) with ln D In+\ for all n is often said to be a decreas-ing sequence of sets. Thus, the nested interval theorem might be paraphrased by say-ing that a decreasing sequence of closed, bounded, nonempty intervals "converges" to anonempty set. In this language, the nested interval theorem is at least reminiscent of thefact that a monotone bounded sequence of real numbers is convergent. And with good rea-son: The fact that monotone bounded sequences converge is actually equivalent to theleast upper bound axiom, as is the nested interval theorem. That is, we might just aswell have assumed the conclusion of either Theorem 1.4 or Theorem 1.5 as an axiom



The Real Numbers 1

for R and deduced the existence of least upper bounds as a corollary. As evidence, hereis a proof that the nested interval theorem implies the existence of least upper bounds (thisis similar in spirit to Bolzano's original proof):

Let A be a nonempty subset of M that is bounded above. Specifically, let a\ A and let b\be an upper bound for A. For later reference, set /i = [a\,b\]. Now consider the point x\ =(a\ -\-b\)/2, halfway between a\ and b\. If X\ is an upper bound for A, we set l2 = [ ci\, X\ ];otherwise, there is an element a2 G A with a2 > x\. In this case, set I2 = [ a2, b\ ]. In eitherevent, I2 is a closed subinterval of I\ of the form [a2,b2], where a2 A and b2 is an upperbound for A. Moreover, length(/2) < length(I\)/2. We now start the process all overagain, using I2 in place of /1 , and obtain a closed subinterval 1$ = [#3, 3 ] C /2, wherea3 G A and 3 is an upper bound for A, with length (73) < length (I2)/2 < length (/i)/4.By induction, we get a sequence of nested closed intervals / = [ an,bn ], where an e Aand bn is an upper bound for A, with length (/) < length {I\)/2n~l -> 0 as n -> 00. Thesingle point Z? G H^Li ^ ^s m e ^eas t uPPe r bound for A. (Why?)

EXERCISES10. Let a\ = A/2 and let an+\ = \/2an for n > 1. Show that (an) converges andfind its limit. [Hint: Show that (an) is increasing and bounded.]11. Fix a > 0 and let x\ > v^- F r ^ > 1, define

Show that (jcn) converges and that limn^oo xn =12. Suppose that 1^ > s2 > 0 and let sn+\ |(^n + sn-\) for n > 2. Show that(sn) converges. [Hint: Show that (s2n-\) decreases and (5*2^ ) increases.]

> 13. Let an > 0 for all n, and let sn YH=i ai- Show that (sn) converges if andonly if (sn) is bounded.

Recall that a sequence of real numbers (xn) is said to be Cauchy if, for everys > 0, there is an integer N > 1 such that \xn xm \ < s whenever n, m > N.

> 14. Prove that a convergent sequence is Cauchy, and that any Cauchy sequence isbounded.

> 15. Show that a Cauchy sequence with a convergent subsequence actually converges.16.(a) Why is 0.4999.. . = 0.5? (Try to give more than one reason.)(b) Write 0.234234234... as a fraction.(c) Precisely which real numbers between 0 and 1 have more than one decimal

representation? Explain.

Our second approach to describing the elements of K. as limits of sequences of rationalnumbers is to consider decimals. We might as well do this in some generality.



Calculus Review

Proposition 1.7. Fix an integer p > 2, and let (an) be any sequence of integerssatisfying 0 < an < p 1 for all n. Then, Yl'nLi an/pn converges to a number in[ 0 , 1 ] .

PROOF. Since an > 0, the partial sums J2n=\ an/pn are nonnegative and increasewith N. Thus, to show that the series converges to some number in [ 0, 1 ], we justneed to show that 1 is an upper bound for the sequence of partial sums. But this iseasy:

h pn

(Why? What does this say when p = 10?)

Conversely, each x in [ 0, 1 ] can be so represented:Proposition 1.8. Let p be an integer, p > 2, and let 0 < x < 1. Then there is asequence of integers (an) with 0 < an < p I for all n such thatx = X!li an/pn-

PROOF. Certainly the case x = 0 causes no real strain, so let us suppose that0 < x < 1. We will construct (an) by induction.

Choose a\ to be the largest integer satisfying a\/p < x. (How?) Since x > 0,it follows that a\ > 0; and since x < 1, we have a\ < p. Because a\ is an in-teger, this means that a\ < p 1. Also, since a\ is largest, we must have a\/p 1, a ^ 0, then (1 + a)n > 1 + na for anyinteger n > 1.

The proof of Bernoulli's inequality is left as an exercise. We'll apply it to prove:

Proposition 1.10.(i) (l + ^)n is strictly increasing.

(ii) (l + ~) is strictly decreasing.(iii) 2 < (1 + \Y < (1 + lnT+l < 4(iv) Both sequences converge to the same limit e = limn^oo (1 + (l/n))n, where

2 < e < 4.

PROOF, (i) We need to show that (1 + l/(/i + l))n+1/(l + (I /"))" > 1- For this werewrite and apply Bernoulli's inequality:

> I 1 + - V ( 1 I = 1 (by Bernoulli).V n / \ w + 1 /

(ii) This case is very similar to (i).

+2

( ) I 1 + - I = 1 (by Bernoulli).\n + lj V n)(iii) Since 1 + ( 1 / ) > 1 , we have ( 1 + (1/n))" < ( l + ( l / n ) ) " + 1 . Since

(1 + (l/n))n increases, the left-hand side is at least 2 (the first term); and since(1 + ( l /n))"+ 1 decreases, the right-hand side is at most 4 (the first term).



10 Calculus Review

(iv) Finally, we define e = limôo (1 + (l/n))n, and conclude that

lim 1 + - = lim 1 + - lim 1 + - = e. ->oo y ) y ) y

1 \- lin) ->

The same proof applies to the sequence ( 1 + (x/n))n for any x e l , and we may definee

x = limn_>oo(l + (x/n))n. The full details of this last conclusion are best left for another

day. See Exercise 18(b).

EXERCISES17. Given real numbers a and b, establish the following formulas: \a + b\ 1 + na for any

integer n > 1.(b) Use (a) to show that, for any x > 0, the sequence (1 + (x/n))n increases.(c) If a > 0, show that (1 + a) r > 1 + ra holds for any rational exponent r > 1.

[Hint: If r = p/q, then apply (a) with n = q and (b) with x = ap.](d) Finally, show that (c) holds for any real exponent r > 1.19. If 0 < c < 1, show that cn -> 0; and if c > 0, show that cl/n -> 1. [Hint:Use Bernoulli's inequality for each, once with c = 1/(1 + x), x > 0 and once withc

xln = 1 + xn, where xn > 0.]20. Given a, b > 0, show that \fab < \{a + b) (this is the arithmetic-geometricmean inequality). Generalize this to {a\ a2 an)x/n < (l/n)(a\ +a2-\ h an).[Hint: Induction and Bernoulli's inequality.]21. Let p > 2 be a fixed integer, and let 0 < x < 1. If x has a finite-length base pdecimal expansion, that is, if x = a\/p-\ \- an/pn with an ^ 0, prove that x hasprecisely two base p decimal expansions. Otherwise, show that the base p decimalexpansion for x is unique. Characterize the numbers 0 < x < 1 that have repeatingbase p decimal expansions. How about eventually repeating?

As long as we are on the subject of sequences, this is a good time to outline part ofthe master plan! Virtually everything that we need to know about the real line R andabout functions / : R -> R can be described in terms of convergent sequences. In-deed, a continuous function / : R > R could be defined as a function that "pre-serves" convergent sequences: /(lim^-ôo^) limn_ôo f(xn). If we hope to under-stand continuous functions (and we do!), then it is of great importance to us to knowprecisely which real sequences converge. So far we know that monotone, bounded se-quences converge, and that any convergent sequence is necessarily bounded. (Why?)These two facts together raise the question: Does every bounded sequence converge?Of course not. But just how "far" from convergent is a typical bounded sequence? To



The Real Numbers 11

answer this, we will want to broaden our definition of limit. First a few easy observa-tions.

Let (an) be a bounded sequence of real numbers, and consider the sequences:

tn = mf{an, an+\, an+2, } and Tn = sup{an, an+u an+2, . }

Then (tn) increases, (Tn) decreases, and infâ^ < tn < Tn < supâ^ for all n. (Why?)Thus we may speak of limnôo tn as the "lower limit" and limôo Tn as the "upper limit"of our original sequence (an). And that is exactly what we will do.

Now these same considerations are meaningful even if we start with an unboundedsequence (an), although in that case we will have to allow the values oo for at least someof the tn's or Tn's (possibly both). That is, if we permit comparisons to 00, then the tn'sstill increase and the Tn's still decrease. Of course we will want to use supn tn and infn Tnin place of limôo tn and limôo Tn, since "sup" and "inf" have more or less obviousextensions to subsets of the extended real number system [00, +00] whereas "lim" doesnot. Even so, we are sure to get caught saying something like il{tn) converges to +00." Butwe will pay a stiff penalty for too much rigor here; even a simple fact could have a tediouslylong description. For the remainder of this section you are encouraged to interpret wordssuch as "limit" and "converges" in this looser sense.

Given any sequence of real numbers (an), we define

lim inf an = lim an = sup(inf{an, an+u an+2, })

and

lim sup an = lim an = inf (sup{an, an+u an+2, })

That is, liminf^ôo^ = supn tn (^lim^ôo^ if (an) is bounded from below) andlimsup^^tfn = infw ^(^ l im^ôo Tn if (an) is bounded from above). The name "lim inf"is short for "limit inferior," while "lim sup" is short for "limit superior."

EXERCISES

22. Show that infn an < liminfôo an < lim supnôo an < supn an.23. If (an) is convergent, show that lim inf^ -^ oo an = lim sup^^^ an = limnôo an.

> 24. Show that lim supn_>oo(an) = liminfnôo an.> 25. If lim s u p ^ ^ an = 00, show that (an) diverges to 00. If lim s u p ^ ^ an =

+00, show that (an) has a subsequence that diverges to +00. What happens ifoodn = 00?

If we start with a bounded sequence (an), then

M = lim sup an = lim (supja^ : k > n}) / 00,



12 Calculus Review

and hence:f for every s > 0, there is an integer N > 1 such that[ M s < supjafc : k > n] < M + s for all n > N.

Thus, the number M = lim s u p ^ ^ an is characterized by the following:

f for every s > 0, we have an < M + e for all but finitely[ many n, and M s < an for infinitely many n.

EXERCISES> 26. Prove the characterization of lim sup given above. That is, given a bounded se-

quence (an), show that the number M = lim sup^^^ an satisfies (*) and, conversely,that any number M satisfying (*) must equal l imsup^â , , . State and prove thecorresponding result for m = liminfôo an.

> 27. Prove that every sequence of real numbers (an) has a subsequence (ank) thatconverges to lim s u p , ^ ^ an. [Hint: If M = lim sup^^^ an oo, we must inter-pret the conclusion loosely; this case is handled in Exercise 25. If M / oo, use (*)to choose (ank) satisfying \ank M\ < l/k, for example.] There is necessarily alsoa subsequence that converges to liminfôo an. Why?28. By modifying the argument in the previous exercise, show that every sequenceof real numbers has a monotone subsequence.29. If (ank) is a convergent subsequence of (an), show that liminfyi->ooa;l 0. From our char-acterizations of lim inf and lim sup, there is an N\ > 1 such that a s < an for alln > N\ (since a = liminfôo an), and there is an N2 > 1 such that an < a + for all n > N2 (since a = \imsupn_^0Oan). Thus, for n > max{Afi, N2} we have\a-an\ < s.

You may recall that a sequence of real numbers converges if and only if it is Cauchy.Although one approach to this fact has already been suggested in the exercises, it is suchan important property of the real numbers that it is well worth the effort to give a secondproof!

First recall that if a sequence converges, then it is Cauchy; and if a sequence is Cauchy,then it is also bounded. (See the exercises for more details.) We want to reverse the firstimplication, and so we may assume that we have a bounded sequence to start with. Thishelps, since for a bounded sequence (an) both lim s u p ^ ^ an and lim infôo an are (finite)real numbers. Given a Cauchy sequence, then, we only need to check that these two numbersare equal, which is easier than it might sound.

Theorem 1.16. A sequence of real numbers converges if {and only if) it is Cauchy.

PROOF. Let (an) be Cauchy, and let > 0. Choose N > 1 such that \an am\ < for all m,n > N. Then, in particular, we have aN < an < aN + for all n > N\thus, (an) is bounded. But aN < an for n > N implies that aN < lim inf an,while an < aN + for n > N implies that lim s u p ^ ^ an < aN + . (Why?)Since 00 < lim inf an < limsupnôo an < 00, we may subtract these results andconclude that limsupn_>oo an liminfôo an < 2. Since > 0 is arbitrary, weget that lim s u p ^ ^ an = lim infôo an.

EXERCISES33. Show that (xn) converges to x E if and only if every subsequence (xnk) of(xn) has a further subsequence (jcnjk/) that converges to x.34. Suppose that an > 0 and that J2T=\ an < -(i) Show that lim infnôo nan = 0.

(ii) Give an example showing that lim s u p ^ ^ nan > 0 is possible.



14 Calculus Review

35. (The ratio test): Let an > 0.(i) I f l i m s u p ^ ^ ^ + i / a ^ < 1, show that J2T=\ an < -

(ii) If limin^ôo an+\/an > 1, show that X^Li an diverges.(iii) Find examples of both a convergent and a divergent series having

limwôo^+i/a^ = 1.36. (The root test): Let an > 0.

(i) If l i m s u p ^ ^ ^fcTn < 1, show that J2T=i an < -(ii) If liminf^ôo tya^ > 1, show that X^Li an diverges.

(iii) Find examples of both a convergent and a divergent series havinglinv+oo ?/a~= 1.

t> 37. If (En) is a sequence of subsets of a fixed set 5, we defineo o / o o \ o o / o o \

limsup En = P | I I ) * 1 and liminf = M ( P | * I Show that

/ \ clim inf En C lim sup En and that lim inf ( E c ) I lim sup 2sn I .^ n->oo ^ V ->oo /

38. Show that

lim sup En = {x e S : x En for infinitely many n}

and that

l iminf Zsn = {x e S : x e En for all but finitely many n}.n-*oo

39. How would you define the limit (if it exists) of a sequence of sets? What shouldthe limit be if Ex D E2 D ? If Ex C E2 C ? Compute liminfôo En andlim supn_>oo En in both cases and test your conjecture.

Limits and Continuity

In this section we present a brief refresher course on limits and continuity for real-valuedfunctions. With any luck, much of what we have to say will be very familiar. To begin,let / be a real-valued function defined (at least) for all points in some open interval con-taining the point a e M. except, possibly, at a itself. We will refer to such a set as apunctured neighborhood of a. Given a number L e R, we write linWa f(x) = L tomean:

I for every s > 0, there is some 8 > 0 such that | / (JC) L\ < e[ whenever x satisfies 0 < \x a\ < 8.

Wesaythatlimâ f(x) exists if there is some number L ffi. that satisfies the requirementsspelled out above. The proof of our first result is left as an exercise.



Limits and Continuity 15

Theorem 1.17. Let f be a real-valued function defined in some punctured neighbor-hood of a R. Then, the following are equivalent:

(i) There exists a number L such that lim*-^ f{x) = L (by the s-8 definition).(ii) There exists a number L such that f(xn) -> L whenever xn *> a, where

xn ^z a for all n.(iii) (f(xn)) converges (to something) whenever xn -> a, where xn / a for all n.

The point to item (iii) is that if lim^^oo f(xn) always exists, then it must actually beindependent of the choice of (xn). This is not as mystical as it might sound; indeed, ifxn > a and yn -> a, then the sequence X\, y\9 #2, )>2> also converges to a. (How doesthis help?) This particular phrasing is interesting because it does not refer to L. That is, wecan test for the existence of a limit without knowing its value.

Now suppose that / is defined in a neighborhood of a, this time including the point aitself. We say that / is continuous at a if l im*^ f(x) = f(a). That is, if:

I for every s > 0, there is a 8 > 0 (that depends on / , a, and s)[ such that | / ( JC) f(a)\ < s whenever x satisfies \x a\ < 8.

Notice that we replaced L by f(a) and we dropped the requirement that x ^ a. Theo-rem 1.17 has an obvious extension to this case (and its proof is also left as an exercise).

Theorem 1.18. Let f be a real-valued function defined in some neighborhood ofa R. Then, the following are equivalent:

(i) / is continuous at a (by the s-8 definition)',(ii) f(xn) > f(a) whenever xn - a;

(iii) (f(xn)) converges (to something) whenever xn > a.

Notice that we dropped the requirement that xn ^ a. Thus, if limn_>oo f(xn) always exists,then it must equal f(a). (Why?)

You might also recall that we have a notation for left- and right-hand limits and left andright continuity. For example, if we define

f(a-) = lim f(x) and f(a+) = lim f(x)JC>-a~ x>a+

(provided that these limits exist, of course), then we could add another equivalence toTheorem 1.18:

1.18. (iv) f(a) and f(a+) both exist, and both are equal to f(a).One-sided limits are peculiar to functions defined on R, and they do not generalize very

well (because they are tied to the order in R) . But they are very good at what they do: Theypermit the cataloguing of very refined types of discontinuities. For example, we say that /is right-continuous at a if f(a+) exists and equals / ( a ) , and we say that / has a jumpdiscontinuity at a if f(a) and f(a+) both exist but at least one is different from f(a).A function having only jump discontinuities is not so very bad. In particular, monotonefunctions are rather well behaved:



16 Calculus Review

Proposition 1.19. Let f : (a,b) -^ R be monotone and let a < c < b. Then,f(c) and / ( c + ) both exist. Thus, f can have only jump discontinuities.

PROOF. We might as well suppose that / is increasing (otherwise, consider / ) .In that case, f(c) is an upper bound for {f(t) : a < t < c} and a lower bound for{f(t) : c < t < b}. All that remains is to check that sup{/0) : a < t < c} =l im*^- f(x) andinf{/(/) : c < t < b] = limJC_>c+ f(x). We will sketch the proofof the first of these.

Given > 0, there is some JCO with a < Xo < c such that supr R be continuous and suppose that f(x) = 0 whenever x isrational. Show that f(x) = 0 for every x in[a,b].46. Let / : R -> R be continuous.(a) If / (0 ) > 0, show that f(x) > 0 for all x in some open interval (a, a).(b) If f(x) > 0 for every rational x9 show that f(x) > 0 for all real x. Will this

result hold with ">0" replaced by ">0 "? Explain.47. Let f,g9h, and k be defined on [ 0, 1 ] as follows:

0 if* Q \ \ - x if xh ( x ) \ x iix

[ 1 - J C if= \x if

0 i f x ^ Q

(in lowest terms).Prove that / is not continuous at any point in [0 , 1 ], that g is continuous only atx = 0, that h is continuous only at x = 1/2, and that k is continuous only at theirrational points in [ 0, 1 ].48. Give an example of a one-to-one, onto function / : [ 0 , 1 ] -> [ 0, 1 ] that isnot monotone. Can you find a monotone, one-to-one function that is not onto? Or amonotone, onto function that is not one-to-one?



Notes and Remarks 17

49. L e t / : {a, b) > E be monotone and let a < x < b. Show that / i s continuousat x if and only if / ( * - ) = f(x+).50. Let D denote the set of rationals in [ 0, 1 ] and suppose that / : D -> M.is increasing. Show that there is an increasing function g : [0, 1 ] - E such thatg(x) = / (JC) whenever x is rational. [Hint: For x [0, 1 ], define g(x) = sup{/(0 :0 < t < 1, t Q}.]51. Let / : [a, b] > ]R be increasing and define g : [a, b] > IR by g(jc) =f(x+) for a < x < b and (^Z?) = f(b). Prove that g is increasing and right-continuous.

0

Notes and Remarks

Although we cannot claim to have reviewed every last detail that you might need for anuntroubled reading of these pages, we have managed to at least recall several importantissues. Bartle [1964] and Fulks [1969] are good sources for a review of advanced calculus;Apostol [1975] and Stromberg [1981] are good sources for further details on the topicsdiscussed in this chapter.

Full details of the construction of the real numbers "from scratch" can be found inBirkhoff and MacLane [1965], Goffman [1953a], Hewitt and Stromberg [1965], andSprecher [1970]. For more on the various equivalent notions of completeness for the realnumbers, see the aptly titled article "Completeness of the real numbers" in Goffman [1974].For more on the history of rigorous analysis, see Boyer [1968], Edwards [1979], Grabiner[1983], Grattan-Guinness [1970], Kitcher [1983], Kleiner [1989], and Kline [1972]. Asan interesting tidbit in this vein, Dudley [1989] points out that no proof of the so-calledBolzano-Weierstrass theorem (Corollary 1.11) has ever been found among Bolzano's writ-ings. For a curious observation about real numbers with "ambiguous" decimal representa-tions, see Petkovsek [1990].

Exercise 42 is taken from Apostol [1975].



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