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Supplements to the Exercises in Chapters 1-7 of Walter Rudin’s Principles of Mathematical Analysis, Third Edition

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URL: http://www.math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.ps or .pdf

Principles of Mathematical Analysis, Third Edition

by George M. Bergman

This packet contains both additional exercises relating to the material in Chapters 1-7 of Rudin, and information on Rudin’s exercises for those chapters. For each exercise of either type, I give a title (an idea borrowed from Kelley’s General Topology), an estimate of its difficulty, notes on its dependence on other exercises if any, and sometimes further comments or hints.

Numbering. I have given numbers to the sections in each chapter of Rudin, in general taking each of his capitalized headings to begin a new numbered section, though in a small number of cases I have inserted one or two additional section-divisions between Rudin’s headings. My exercises are referred to by boldfaced symbols showing the chapter and section, followed by a colon and an exercise-number; e.g., under section 1.4 you will find Exercises 1.4:1 , 1.4:2 , etc.. Rudin puts his exercises at the ends of the chapters; in these notes I abbreviate ‘‘Chapter M, Rudin’s Exercise N ’’ to M :RN. However, I list both my exercises and his under the relevant section.

It could be argued that by listing Rudin’s exercises by section I am effectively telling the student where to look for the material to be used in solving the exercise, which the student should really do for his or her self. However, I think that the advantage of this work of classification, in showing student and instructor which exercises are appropriate to attempt or to assign after a given section has been covered, outweighs that disadvantage. Similarly, I hope that the clarifications and comments I make concerning many of Rudin’s exercises will serve more to prevent wasted time than to lessen the challenge of the exercises.

Difficulty-codes. My estimate of the difficulty of each exercise is shown by a code d : 1 to d : 5. Codes d : 1 to d : 3 indicate exercises that it would be appropriate to assign in a non-honors class as ‘‘easier’’,

‘‘typical’’, and ‘‘more difficult’’ problems; d : 2 to d : 4 would have the same roles in an honors course, while d : 5 indicates the sort of exercise that might be used as an extra-credit ‘‘challenge problem’’ in an honors course. If an exercise consists of several parts of notably different difficulties, I may write something like d : 2, 2, 4 to indicate that parts (a) and (b) have difficulty 2, while part (c) has difficulty 4.

However, you shouldn’t put too much faith in my estimates – I have only used a small fraction of these exercises in teaching, and in other cases my guesses as to difficulty are very uncertain. (Even my sense of what level of difficulty should get a given code has probably been inconsistent. I am inclined to rate a problem that looks straightforward to me d : 1; but then I may remember students coming to office hours for hints on a problem that looked similarly straightforward, and change that to d : 2.)

The difficulty of an exercise is not the same as the amount of work it involves – a long series of straightforward manipulations can have a low level of difficulty, but involve a lot of work. I discovered how to quantify the latter some years ago, in an unfortunate semester when I had to do my own grading for the basic graduate algebra course. Before grading each exercise, I listed the steps I would look for if the student gave the expected proof, and assigned each step one point (with particularly simple or complicated steps given 12 or 112 points). Now for years, I had asked students to turn in weekly feedback on the time their study and homework for the course took them; but my success in giving assignments that kept the average time in the appropriate range (about 13 hours per week on top of the 3 hours in class) had been erratic; the time often ended up far too high. That Semester, I found empirically that a 25-point assignment regular kept the time quite close to the desired value.

I would like to similarly assign point-values to each exercise here, from which it should be possible to similarly calibrate assignments. But I don’t have the time to do this at present.

Dependencies. After the title and difficulty-code, I note in some cases that the exercise depends on some other exercise, writing ‘‘>’’ to mean ‘‘must be done after ... ’’.

Comments on Rudin’s exercises. For some of Rudin’s exercises I have given, after the above data, notes clarifying, motivating, or suggesting how to approach the problem. (These always refer the exercise listed immediately above the comment; if other exercises are mentioned, they are referred to by number.)

True/False questions. In most sections (starting with §1.2) the exercises I give begin with one numbered ‘‘0’’, and consisting of one or more True/False questions, with answers shown at the bottom of the next page. Students can use these to check whether they have correctly understood and absorbed the definitions, results, and examples in the section. No difficulty-codes are given for True/False questions. I tried to write them to check for the most elementary things that students typically get confused on, such as the difference between a statement and its converse, and order of quantification, and for the awareness of what Rudin’s various counterexamples show. Hence these questions should, in theory, require no original thought; i.e., they should be ‘‘d : 0’’ relative to the classification described above. But occasionally, either I did not see a good way to give such a question, or I was, for better or worse, inspired with a question that tested the student’s understanding of a result via a not-quite-trivial application of it.

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Terminology and Notation. I have followed Rudin’s notation and terminology very closely, e.g. using R for the field of real numbers, J for the set of positive integers, and ‘‘at most countable’’ to describe a set of cardinality ≤ ℵ0. But on a few points I have diverged from his notation: I distinguish between sequences (si) and sets {si} rather than writing {si} for both, and I use ⊆ rather than ⊂ for inclusion. I also occasionally use the symbols ∀ and ∃, since it seems worthwhile to familiarize the student with them.

Advice to the student. An exercise may only require you to use the definitions in the relevant section of Rudin, or it may require for its proof some results proved there, or an argument using the same method of proof as some result proved there. So in approaching each problem, first see whether the result becomes reasonably straightforward when all the relevant definitions are noted, and also ask yourself whether the statement you are to prove is related to the conclusion of any of the theorems in the section, and if so, whether that theorem can be applied as it stands, or whether a modification of the proof can give the result you need. (Occasionally, a result listed under a given section may require only material from earlier sections, but is placed there because it throws light on the ideas of the section.)

Unless the contrary is stated, solutions to homework problems are expected to contain proofs, even if the problems are not so worded. In particular, if a question asks whether something is (always) true, an affirmative answer requires a proof that it is always true, while a negative answer requires an example of a case where it fails. Likewise, if an exercise or part of an exercise says ‘‘Show that this result fails if such-and-such condition is deleted’’, what you must give is an example which satisfies all the hypotheses of the result except the deleted one, and for which the conclusion of the result fails. (I am not counting the true/false questions under ‘‘homework problems’’ in this remark, since they are not intended to be handed in; but when using these to check yourself on the material in a given section, you should be able to justify with a proof or counterexample every answer that is not simply a statement taken from the book.)

From time to time students in the class ask ‘‘Can we use results from other courses in our homework?’’

The answer is, in general, ‘‘No.’’ Rudin shows how the material of lower division calculus can be developed, essentially from scratch, in a rigorous fashion. Hence to call on material you have seen developed in the loose fashion of your earlier courses would defeat the purpose. Of course, there are certain compromises: As Rudin says, he will assume the basic properties of integers and rational numbers, so you have to do so too. Moreover, once one has developed rigorously the familiar laws of differentiation and integration (a minor aspect of the material of this course), the application of these is not essentially different from what you learned in calculus, so it is probably not essential to state explicitly in homework for later sections which of those laws you are using at every step. When in doubt on such matters, ask your instructor.

Unfinished business. I have a large list of notes on errata to Rudin, unclear points, proofs that could be done more nicely, etc., which I want to write up as a companion to this collection of exercises, when I have time. For an earlier version, see http://www.math.berkeley.edu/~gbergman/ug.hndts/Rudin_notes.ps .

As mentioned in the paragraph in small print on the preceding page, I would like to complement the

‘‘difficulty ratings’’ that I give each exercise with ‘‘amount-of-work ratings’’. I would also like to complement the dependency notes with reverse-dependency notes, marking exercises which later exercises depend on, since this can be relevant to an instructor’s decision on which exercises to assign. This will require a bit of macro-writing, to insure that consistency is maintained as exercises are added and moved around, and hence change their numbering. On a much more minor matter, I want to rewrite the page- header macro so that the top of each page will show the section(s) of Rudin to which the material on the page applies.

I am grateful to Charles Pugh for giving me comments on an early draft of this packet. I would welcome further comments and corrections on any of this material.

George Bergman

Department of Mathematics University of California Berkeley, CA 94720-3840 gbergman @math.berkeley.edu

July 2001, December 2003, May 2006, December 2006

2006 George M. Bergman

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Chapter 1. The Real and Complex Number Systems.

1.1. INTRODUCTION. (pp.1-3)

Relevant exercise in Rudin:

1:R2. There is no rational square root of 12. (d : 1)

Exercise not in Rudin:

1.1:1. Motivating Rudin’s algorithm for approximating 0-2. (d : 1)

On p.2, Rudin pulls out of a hat a formula which, given a rational number p, produces another rational number q such that q2 is closer to 2 than p2 is. This exercise points to a way one could come up with that formula. It is not an exercise in the usual sense of testing one’s grasp of the material in the section, but is given, rather, as an aid to students puzzled as to where Rudin could have gotten that formula. We will assume here familiar computational facts about the real numbers, including the existence of a real number 0-2, though Rudin does not formally introduce the real numbers till several sections later.

(a) By rationalizing denominators, get a non-fractional formula for 1 ⁄ (0-2+ 1). Deduce that if x = 0-2+ 1, then x = (1 ⁄ x) + 2.

(b) Suppose y > 1 is some approximation to x = 0-2+ 1. Give a brief reason why one should expect (1 ⁄ y) + 2 to be a closer approximation to x than y is. (I don’t ask for a proof, because we are only seeking to motivate Rudin’s computation, for which he gives an exact proof.)

(c) Now let p > 0 be an approximation to 0-2 (rather than to 0-2+ 1). Obtain from the result of (b) an expression f ( p) that should give a closer approximation to 0-2 than p is. (Note: To make the input p of your formula an approximation of 0-2, substitute y = p + 1 in the expression discussed in (b); to make the output an approximation of 0-2, subtract 1.)

(d) If p < 0-2, will the value f ( p) found in part (c) be greater or less than 0-2? You will find the result different from what Rudin wants on p.2. There are various ways to correct this. One would be to use f ( f ( p)), but this would give a somewhat more complicated expression. A simpler way is to use 2 ⁄ f ( p). Show that this gives precisely (2p + 2) ⁄ ( p + 2), Rudin’s formula (3).

(e) Why do you think Rudin begins formula (3) by expressing q as p – ( p2–2) ⁄ ( p + 2) ? 1.1:2. Another approach to the rational numbers near 0-2. (d : 2)

Let sets A and B be the sets of rational numbers defined in the middle of p.2. We give below a quicker way to see that A has no largest and B no smallest member. Strictly speaking, this exercise belongs under §1.3, since one needs the tools in that section to do it. (Thus, it should not be assigned to be done before students have read §1.3, and students working it may assume that Q has the properties of an ordered field as described in that section.) But I am listing it here because it simplifies an argument Rudin gives on p.2.

Suppose A has a largest member p.

(a) Show that the rational number p′ = 2 ⁄ p will be a smallest member of B.

(b) Show that p′ > p.

(c) Let q = ( p + p′) ⁄ 2, consider the two possibilities q∈A and q∈B, and in each case obtain a contradiction. (Hint: Either the condition that p is the greatest element of A or that p′ is the smallest element of B will be contradicted.)

This contradiction disproves the assumption that A had a largest element.

(d) Show that if B had a smallest element, then one could find a largest element of A. Deduce from the result of (c) that B cannot have a smallest element.

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1.2. ORDERED SETS. (pp.3-5)

Relevant exercise in Rudin:

1:R4. Lower boundupper bound. (d : 1)

Exercises not in Rudin:

1.2:0. Say whether each of the following statements is true or false.

(a) If x and y are elements of an ordered set, then either xy or y > x.

(b) An ordered set is said to have the ‘‘least upper bound property’’ if the set has a least upper bound.

1.2:1. Finite sets always have suprema. (d : 1)

Let S be an ordered set (not assumed to have the least-upper-bound property).

(a) Show that every two-element subset {x, y}S has a supremum. (Hint: Use part (a) of Definition 1.5.)

(b) Deduce (using induction) that every finite subset of S has a supremum.

1.2:2. If one set lies above another. (d : 1)

Suppose S is a set with the least-upper-bound property and the greatest-lower-bound property, and suppose X and Y are nonempty subsets of S.

(a) If every element of X isevery element of Y, show that sup Xinf Y.

(b) If every element of X is < every element of Y, does it follow that sup X < inf Y ? (Give a proof or a counterexample.)

1.2:3. Least upper bounds of least upper bounds, etc. (d : 2)

Let S be an ordered set with the least upper bound property, and let Ai (i∈I ) be a nonempty family of nonempty subsets of S. (This means that I is a nonempty index set, and for each i∈I, Ai is a nonempty subset of S.)

(a) Suppose each set Ai is bounded above, let αi = sup Ai, and suppose further that {αi ! i∈I } is bounded above. Then show that

i∈I Ai is bounded above, and that sup (

i∈I Ai) = sup {αi ! i∈I } . (b) On the other hand, suppose that either (i) not all of the sets Ai are bounded above, or (ii) they are all bounded above, but writing αi = sup Ai for each i, the set {αi ! i∈I } is unbounded above. Show in each of these cases that

i∈I Ai is unbounded above.

(c) Again suppose each set Ai is bounded above, with αi = sup Ai. Show that

i∈I Ai is also

bounded above. Must it be nonempty? If it is nonempty, what can be said about the relationship between sup (

i∈I Ai) and the numbers αi (i∈I ) ?

1.2:4. Fixed points for increasing functions. (d : 3)

Let S be a nonempty ordered set such that every nonempty subset ES has both a least upper bound and a greatest lower bound. (A closed interval [a, b] in R is an example of such an S.) Suppose f : SS is a monotonically increasing function; i.e., has the property that for all x, y∈S , x ≤ yf (x)f (y).

Show that there exists an x∈S such that f (x) = x.

1.2:5. If everything that is >α is ≥β ... (d : 2)

(a) Let S be an ordered set such that for any two elements p < r in S, there is an element q∈S with p < q < r. Suppose α and β are elements of S such that for every x∈S with x >α, one has x≥β. Show that β≤α.

(b) Show by example that this does not remain true if we drop the assumption that whenever p < r there is a q with p < q < r.

1.2:6. L.u.b.’s can depend on where you take them. (d : 3)

(a) Find subsets ES1S2S3Q such that E has a least upper bound in S1, but does not have any least upper bound in S2, yet does have a least upper bound in S3.

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(b) Prove that for any example with the properties described in (a) (not just the example you have given), the least upper bound of E in S1 must be different from the least upper bound of E in S3.

(c) Can there exist an example with the properties asked for in (a) such that E = S1? (If your answer is yes, you must show this by giving such an example. If your answer is no, you must prove it impossible.) 1.2:7. A simpler formula characterizing l.u.b.’s. (d : 2)

Let S be an ordered set, E a subset of S, and x an element of S.

If one translates the statement ‘‘x is the least upper bound of E ’’ directly into symbols, one gets ((∀y∈E ) x≥ y) ∧ ((∀z∈S ) ((∀y∈E ) z≥ y)zx ).

This leads one to wonder whether there is any simpler way to express this property.

Prove, in fact, that x is the least upper bound of E if and only if (∀y∈S ) (y < x⇔ ((∃ z∈E ) (z > y))).

1.2:8. Some explicit sup’s and inf’s. (d : 2)

(a) Prove that inf {x + y + z! x, y, z∈R, 0 < x < y < z } = 0.

(b) Determine the values of each of the following. If a set is not bounded on the appropriate side, answer

‘‘undefined’’. No proofs need be handed in; but of course you should reason out your answers to your own satisfaction.

a = inf {x + y + z! x, y, z∈R, 1 < x < y < z } . d = sup {x + y + z! x, y, z∈R, 1 < x < y < z } . b = inf {x + y – z ! x, y, z∈R, 1 < x < y < z } . e = sup {x + y – 2z! x, y, z∈R, 1 < x < y < z } . c = inf {x – y + z! x, y, z∈R, 1 < x < y < z } .

1.3. FIELDS. (pp.5-8)

Relevant exercise in Rudin:

1:R3. Prove Proposition 1.15. (d : 1)

Exercise 1:R5 can also be done after reading this section, if one replaces ‘‘real numbers’’ by ‘‘elements of an ordered field with the least upper bound property’’.

Exercises not in Rudin:

1.3:0. Say whether each of the following statements is true or false.

(a) Z (the set of integers, under the usual operations) is a field.

(b) If F is a field and also an ordered set, then it is an ordered field.

(c) If x and y are elements of an ordered field, then x2+ y2 ≥ 0.

(d) In every ordered field, –1 < 0.

1.3:1. sup ({s + y! s∈S }) = (sup S ) + y . (d : 1, 1, 2) Let F be an ordered field.

(a) Suppose S is a subset of F and y an element of F, and let T = {s + y ! s∈S } . Show that if S has a least upper bound, sup S, then T also has a least upper bound, namely (sup S ) + y.

(b) Deduce from (a) that if x is a nonzero element of F and we let S = {n x! n is an integer}, then S has no least upper bound.

(c) Deduce Theorem 1.20(a) from (b) above.

1.4. THE REAL FIELD. (pp.8-11)

Relevant exercises in Rudin:

1:R1. Rational + irrational = irrational. (d : 1)

(‘‘Irrational’’ means belonging to R but not to Q.)

...

Answers to True/False question 1.2:0. (a)T. (b)F.

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1:R5. inf A = – sup ( – A). (d : 1)

1:R6. Rational exponentiation of positive real numbers. (d : 3. > 1:R7(a-c))

Part (a) is more easily done if the first display is changed from ‘‘(bm)1 ⁄ n= (bp)1 ⁄ q’’ to

‘‘(b1 ⁄ n)m= (b1 ⁄ q)p’’, and in the next line ‘‘(bm)1 ⁄ n’’ is changed to ‘‘(b1 ⁄ n)m’’. Part (d ) requires part (c) of the next exercise, so the parts of these two exercises should be done in the appropriate order.

1:R7. Logarithms of positive real numbers. (d : 3)

In part (g), ‘‘is unique’’ should be ‘‘is the unique element satisfying the above equation’’.

It is interesting to compare the statement proved in part (a) of this exercise with the archimedean property of the real numbers. That property says that if one takes a real number > 0 and adds it to itself enough times, one can get above any given real number. From that fact and part (a) of this exercise, we see that if one takes a real number > 1 and multiplies it by itself enough times, one can get above any given real number. One may call the former statement the ‘‘additive archimedean property’’, and this one the ‘‘multiplicative archimedean property’’.

Exercises not in Rudin:

1.4:0. Say whether each of the following statements is true or false.

(a) Every ordered field has the least-upper-bound property.

(b) For every real number x, (x2)12= x.

(c) 0-- >12 12.

(d) If a subset E of the real numbers is bounded above, and x = sup E, then x∈E.

(e) If a subset E of the real numbers has a largest element, x (i.e., if there exists an element x∈E which is greater than every other element of E), then x = sup E.

(f ) If E is a subset of R, and s is a real number such that s > x for all x∈E, then s = sup E.

1.4:1. Some explicit sup’s and inf’s. (d : 2)

(a) Prove that inf {x + y + z! x, y, z∈R, 0 < x < y < z } = 0.

(b) Determine the values of each of the following. If a set is not bounded on the appropriate side, answer

‘‘undefined’’. No proofs need be handed in; but of course you should reason out your answers to your own satisfaction.

a = inf {x + y + z! x, y, z∈R, 1 < x < y < z } . d = sup {x + y + z! x, y, z∈R, 1 < x < y < z } . b = inf {x + y – z ! x, y, z∈R, 1 < x < y < z } . e = sup {x + y – 2z! x, y, z∈R, 1 < x < y < z } . c = inf {x – y + z! x, y, z∈R, 1 < x < y < z } .

1.4:2. Details on decimal expansions of real numbers. (d : 3)

This exercise gives some of the details skipped over in Rudin’s sketch of the decimal expansion of real numbers.

In parts (a) and (b) below, let x be a positive real number, and let n0, n1, ... , nk, ... be constructed as in Rudin’s 1.22 (p.11).

(a) Prove that for all nonnegative integers k, one has 0x – Σki = 0 ni10–i < 10–k, and that for all positive integers k, 0nk < 10.

We would like to conclude from the former inequality that x is the least upper bound of {Σki = 0 ni10–i! k0 } . However, in this course we want to prove our results, and to prove this, we need a fact about the numbers 10–k. This is obtained in the next part:

(b) For any real number c > 1, show that {ck ! k0 } is not bounded above. (Hint: Write c = 1 + h and note that cn1 + n h. Then what?) Deduce that the greatest lower bound of {c–k ! k0 } is 0.

Taking c = 10, show that this together with the result of (a) implies that the least upper bound of {Σki = 0 ni10–i! k0 } is x.

...

Answers to True/False question 1.3:0. (a)F. (b)F. (c) T. (d)T.

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In the remaining two parts, let m0 be any integer, and m1, m2, ... any nonnegative integers < 10.

(c) Show that {Σki = 0 mi10–i ! k0 } is bounded above. (Suggestion: Show m0+ 1 is an upper bound.) Thus this set will have a least upper bound, which we may call x.

(d) Let x be as in part (c), and n0, n1, n2, ... be constructed from this x as in (a) and (b). Show that if there are infinitely many values of k such that nk ≠ 9, then nk = mk for all k. (Why is the restriction on cases where nk = 9 needed?)

The remaining three exercises in this section go beyond the subject of the text, and examine the relationship of R with other ordered fields which do or do not satisfy the archimedean property. As their difficulty-numbers show, these should probably not be assigned in a non-honors course, though students whose curiosity is piqued by these questions might find them interesting to think about.

1.4:3. Uniqueness of the ordered field of real numbers. (d : 5)

On p.21, Rudin mentions, but does not prove, that any two ordered fields with the least-upper-bound property are isomorphic. This exercise will sketch how that fact can be proved. For the benefit of students who have not had a course in Abstract Algebra, I begin with some observations generally included in that course (next paragraph and part (a) below).

If F is any field, let us define an element nF∈F for each integer n as follows: Let 0F and 1F be the elements of F called ‘‘0’’ and ‘‘1’’ in conditions (A4) and (M4) of the definition of a field. (We add the subscript F to avoid confusion with elements 0, 1∈Z .) For n≥1, once nF is defined we recursively define (n + 1)F = nF+ 1F; in this way nF is defined for all nonnegative integers. Finally, for negative integers n we define nF = – ( – n)F. (Note that in that expression, the ‘‘inner’’ minus is applied in Z , the ‘‘outer’’ minus in F.)

(a) Show that under the above definitions, we have (m + n)F = mF+ nF and (m n)F = mFnF for all m, n∈Z .

(b) Show that if F is an ordered field, then we also have mF < nFm < n for all m, n∈Z . Deduce that in this case, the map nnF is one-to-one.

The results of (a) and the first sentence of (b) above are expressed by saying that the map nnF

‘‘respects’’ the operations of addition and multiplication and the order relation ‘‘<’’.

(c) Show that if F is an ordered field, and if for every rational number r = m ⁄ n (m, n∈Z , n ≠ 0) we define rF = mF ⁄ nF∈F, then r → rF is a well-defined one-to-one map QF, which continues to respect addition, multiplication, and ordering (i.e., satisfies (r + s)F = rF+ sF, (r s)F = rFsF, and rF < sFr < s for all r, s∈Q). Thus, Q is isomorphic as an ordered field to a certain subfield of F.

(The statement that the above map is ‘‘well-defined’’ means that the definition is consistent, in the sense that if we write a rational number r in two different ways, r = m ⁄ n = m⁄ n′ (m, m′ n , n′ ∈Z ) then the two candidate values for rF, namely mF ⁄ nF and m′F ⁄ nF, are the same. We have to prove such a ‘‘well-definedness’’ result whenever we give a definition that depends on a choice of how to write something.)

(Remark: We constructed the map ZF of (a) without assuming F ordered. Could we have done the same with the above map QF ? No. The trouble is that for some choices of F, the map ZF would not have been one-to-one, hence starting with a rational number r = m ⁄ n, we might have found that nF = 0F even though n0, and then mF ⁄ nF would not be defined.)

We will call an ordered field F archimedean if for all x, y∈F with x > 0F, there exists a positive integer n such that nFx > y. Note that by the proof of Theorem 1.20(a), every ordered field with the least-upper-bound property is archimedean. If F is an archimedean ordered field, then for every x∈F let us define Cx = {r∈Q! rF < x} . (This set describes how the element x∈F ‘‘cuts’’ Q in two; thus it is called ‘‘the cut in Q induced by the element x∈F ’’.)

(d) Let F be an archimedean ordered field, and K an ordered field with the least-upper-bound property ...

Answers to True/False question 1.4:0. (a)F. (b)F. (c) T. (d)F. (e)T. (f )F.

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(hence also archimedean). Let us define a map f : FK by setting f (x) = sup {rK ! r∈Cx} for each x∈F. Show that f is a well-defined one-to-one map which respects addition, multiplication, and ordering.

(Note that the statements f (r + s) = f (r) + f (s) etc. must now be proved for all r, s in F, not just in Q.) Show, moreover, that f is the only one-to-one map FK respecting addition, multiplication, and ordering. In other words, F is isomorphic as an ordered field, by a unique isomorphism, to a subfield of K.

(e) Deduce that if two ordered fields F and K both have the least-upper-bound property, then they are isomorphic as an ordered fields.

(Hint: For such F and K, step (d) gives maps f : FK and k : KF which respect the field operations and the ordering. Hence the composite maps f k : KK and k f : FF also have these properties. Now the identity maps idK: KK and idF: FF, defined by idK(x) = x (x∈K ) and idF(x) = x (x∈F ) also respect the field operations and the ordering. Apply the uniqueness statement of (d) to each of these cases, and deduce that f and k are inverse to one another.)

Further remarks : It is not hard to write down order-theoretic conditions that a subset C of Q must satisfy to arise as a cut Cx in the above situation. If we define a ‘‘cut in Q ’’ abstractly as a subset CQ satisfying these conditions, then we can show that if F is any ordered field with the least-upper- bound property, the set of elements of F must be in one-to-one correspondence with the set of all cuts in Q. If we know that there exists such a field F, this gives a precise description of its elements. If we do not, it suggests that we could construct such a field by defining it to have one element xC corresponding to each cut CQ, and defining addition, subtraction, and an ordering on the resulting set, {xC ! C is a cut in Q } . After doing so, we might note that the symbol ‘‘x’’ is a superfluous place-holder, so the operations and ordering could just as well be defined on {C ! C is a cut in Q }. This is precisely what Rudin will do, though without the above motivation, in the Appendix to Chapter 1.

1.4:4. Properties of ordered fields properly containing R . (d : 4) Suppose F is an ordered field properly containing R.

(a) Show that for every element α∈F which does not belong to R, either (i) α is greater than all elements of R, (ii) α is less than all elements of R, (iii) there is a greatest element a∈R that is

< α, but no least element of R that is > α, or (iv) there is a least element that is > α, but no greatest element of R that is < α.

(b) Show that there will in fact exist infinitely many elements α of F satisfying (i), infinitely many satisfying (ii), infinitely many as in (iii) for each a∈R, and infinitely many as in (iv) for each a∈R.

(Hint to get you started: There must be at least one element satisfying one of these conditions. Think about how the operation of multiplicative inverse will behave on an element satisfying (i), respectively on an element satisfying (iii) with a = 0.)

In particular, from the existence of elements satisfying (i), we see that F will be non-archimedean.

(c) Show that for every α∈F not lying in R there exists β > α such that no element x∈R satisfies α < x < β. (This shows that R is not dense in F.)

1.4:5. Constructing a non-archimedean ordered field. (d : 4)

We will indicate here how to construct a non-archimedean ordered field F containing the field R of real numbers.

The elements of F will be the rational functions in a variable x, that is, expressions p(x) ⁄ q(x) where p(x) and q(x) are polynomials with coefficients in R, and q is not the zero polynomial.

Unfortunately, though expressions p(x) ⁄ q(x) are called rational ‘‘functions’’, they are not in general functions on the whole real line, since they are undefined at points x where the denominator is zero. We may consider each such expression as a function on the subset of the real line where its denominator is nonzero (consisting of all but finitely many real numbers); but we then encounter another problem: We want to consider rational functions such as (x2 – 1) ⁄ (x – 1) and (x + 1) ⁄ 1 as the same; but they are not, strictly, since they have different domains.

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There are technical ways of handling this, based on defining a rational function to be an ‘‘equivalence class’’ of such partial functions under the relation of agreeing on the intersections of their domains, or as an equivalence classes of pairs ( p(x), q(x)) under an appropriate equivalence relation. Since the subject of equivalence classes is not part of the material in Rudin, I will not go into the technicalities here, but will simply say that we will consider two rational functions to ‘‘be the same’’ if they can be obtained from one another by multiplying and dividing numerator and denominator by equal factors, equivalently, if they agree wherever they are both defined; and will take for granted that the set of these elements form a field (denoted R(x) by algebraists). We can now begin.

(a) Show that if q is a polynomial, then either q is ‘‘eventually positive’’ in the sense that (∃B∈R) (∀r∈R) (r > B ⇒ q(r) > 0)

or q is ‘‘eventually negative’’, i.e.,

(∃B∈R) (∀r∈R) (r > B ⇒ q(r) < 0).

or q = 0. (Hint: look at the sign of the coefficient of the highest power of x in q(x).) (b) Deduce that if f is a rational function, then likewise either

(∃B∈R) (∀r∈R) (r > B ⇒ f (r) > 0), or (∃B∈R) (∀r∈R) (r > B ⇒ f (r) < 0).

or r = 0. Again, let us say in the first two cases that f is ‘‘eventually positive’’, respectively ‘‘eventually negative’’.

Given rational functions f and f′, let us write f < f′ if f– f is eventually positive.

(c) Show that the above relation ‘‘<’’ makes the field of rational functions an ordered field F.

We shall regard the real numbers as forming a subfield of F, consisting of the constant rational functions r ⁄ 1 (r∈R). In particular, the sets of integers and of rational numbers also become subsets of F.

(d) Show that in F, the polynomial x (i.e., the rational function x ⁄ 1, which is the function f given by f (r) = r) is > n for all integers n. Thus, F is not archimedean.

We note a consequence:

(e) Deduce that the element 1 ⁄ x∈F is positive, but is less than all positive rational numbers, hence less than all positive real numbers. (Thus, ‘‘from the point of view of F ’’, the field of real numbers has a

‘‘gap’’ between 0 and the positive real numbers. It similarly has ‘‘gaps’’ between every real number and all the numbers above or below it.)

1.4:6. A smoother approach to the archimedean property. (d : 2) Let F be an ordered field.

(a) Suppose A is a subset of F which has a least upper bound, α∈F, and x is an element of F.

Show that {a + x | a∈A} has a least upper bound, namely α+ x.

(b) Suppose x is an element of F, and we let A = {n x | n∈Z}. Show that {a + x | a∈A} = A.

(c) Combining the results of (a) and (b), deduce that if x is a nonzero element of F, then the set {n x | n∈Z} cannot have a least upper bound in F.

(d) Now suppose x is a positive element, and that F has the least upper bound property. Deduce from (c) that {n x | n∈Z} is not bounded above; deduce from this that {n x | n∈J} is not bounded above, where J denotes the set of positive integers, and deduce from this the statement of Theorem 1.20(a).

1.4:7. An induction-like principle for the real numbers. (d : 1, 2, 2)

Parts (a) and (b) below will show that some first attempts at formulating analogs of the principle of mathematical induction with real numbers in place of integers do not work. Part (c) gives a form of the principle that is valid.

Let symbols x, y, etc. denote nonnegative real numbers. Suppose that for each x, a statement P (x)

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about that number is given, and consider the following four conditions:

(i) P (0) is true.

(ii) For every x, if P (y) is true for all y < x, then P (x) is true.

(iii) For every x such that P (x) is true, there exists y > x such that for all z with xzy, P (z) is true.

(iv) For all x, P (x) is true.

(a) Show that (i) and (ii) do not in general imply (iv). (I.e., that there exist statements P for which (i) and (ii) hold but (iv) fails. Suggestion: Try the statement ‘‘x ≤ 1’’.)

(b) Show, likewise, that (i) and (iii) (without (ii)) do not in general imply (iv).

(c) Show that (i), (ii) and (iii) together do imply (iv).

(Suggestion: If the set of nonnegative real numbers x for which P (x) is false is nonempty, look at the greatest lower bound of that set.)

(Remark: Condition (i) is really a special case of (ii), and so could be dropped from (a) and (c), since with our variables restricted to nonnegative real numbers, ‘‘P (y) holds for all y < 0’’ is vacuously true.

But I include it to avoid requiring you to reason about the vacuous case.) 1.5. THE EXTENDED REAL NUMBER SYSTEM. (pp.11-12)

Relevant exercises in Rudin: None Exercises not in Rudin:

1.5:0. Say whether each of the following statements is true or false.

(a) In the extended real numbers, ( +

) · 0 = 1.

(b) In the extended real numbers, ( –12) · ( –

) = +

.

The next exercise is in the same series as the last two exercises in the preceding section, and like them is tangential to the material in Rudin.

1.5:1. Mapping a non-archimedean ordered field to the extended reals. (d : 2. > 1.4:4) Suppose that F is an ordered field that properly contains R.

(a) Assuming the result of 1.4:4, show that F can be mapped onto the extended real number system by a map f that carries each α∈R to itself, and which ‘‘respects’’ addition and multiplication, in the sense that it satisfies f (α+β) = f (α) + f (β) and f (α β) = f (α) f (β) except in the cases where these operations are not defined for the extended real numbers f (α) and f (β). Briefly discuss the behavior of f in the latter cases.

(b) How does the f you have constructed behave with respect to the order-relation < ? 1.6. THE COMPLEX FIELD. (pp.12-16)

Relevant exercises in Rudin:

1:R8. C cannot be made an ordered field. (d : 1) 1:R9. C can be made an ordered set. (d : 1, 2)

Note that your answer to the final question, about the least-upper-bound property, requires either a proof that the property holds, or an argument showing why some set that is bounded above does not have a least upper bound.

1:R10. Square roots in C . (d : 2)

1:R11. C = ( positive reals) · (unit circle). (d : 1) 1:R12. The n-term triangle inequality. (d : 1) 1:R13. An inequality on absolute values. (d : 2)

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1:R14. An identity on the unit circle. (d : 2)

1:R15. When does equality hold in the Schwarz inequality? (d : 3) Exercise not in Rudin:

1.6:0. Say whether each of the following statements is true or false.

(a) C (the set of complex numbers, under the usual operations) is a field.

(b) For every complex number z, Im (z-) = Im ( – z).

(c) For all complex numbers w and z, Re (w z) = Re (w) Re (z).

1.7. EUCLIDEAN SPACES. (pp.16-17)

Relevant exercises in Rudin:

1:R16. Solutions to | z – x | = | z – y | = r. (d : 2) 1:R17. An identity concerning parallelograms. (d : 2) 1:R18. Vectors satisfying x · y = 0. (d : 1)

1:R19. Solutions to | x – a | = 2 | x – b | . (d : 2)

The middle two lines of the above exercise should be understood to mean ‘‘{ x ! |x – a| = 2 |x – b|} = { x ! |x – c| = r }’’. Rudin actually gives the ‘‘solution’’ to this problem, but you have to prove that his solution has the asserted property.

Exercises not in Rudin:

1.7:0. Say whether the following statement is true or false.

(a) For all x, y∈Rk, | x · y || x | · | y | .

1.7:1. Relations between | x | , | y | and | x + y | . (d : 3)

(a) Show that for any points x and y of Rk one has | x + y || x | – | y | and | x + y || y | – | x |.

(b) Combining the above inequalities with that of Theorem 1.37(e), what is the possible set of values for

| x + y | if | x | = 3 and | y | = 1? If | x | = 1 and | y | = 4? Do there exist pairs of points x , y∈R2 with

| x | = 3, | y | = 1, and | x + y | taking on each value allowed by your answer to the former question?

1.7:2. A nicer proof of the Schwarz inequality for real vectors. (d : 3)

Rudin’s proof of the Schwarz inequality is short, but messy. This exercise will indicate what I hope is a more attractive proof in the case of real numbers, and the next exercise will show how, with a bit of additional work, it can be extended to complex numbers. In the exercise after that, we indicate briefly still another version of these proofs which can be used if we consider the quadratic formula as acceptable background material

Although Rudin proves the Schwarz inequality on the page before he introduces Euclidean space Rk, we will here assume the reverse order, so that we can write our relations in terms of dot products of vectors.

Suppose a1, ... , ak and b1, ... , bk are real numbers, and let us write a = (a1, ... , ak)∈Rk, b = (b1, ... , bk)∈Rk. Then a | b | – | a | b is also a member of Rk. Its dot product with itself, being a sum of squares, is nonnegative. Expand the inequality stating this nonnegativity, using the distributive law for the dot product, and translate occurrences of a · a and b · b to | a |2 and | b |2. Assuming a and b both nonzero, you can now cancel a factor of 2 | a | | b | from the whole formula and obtain an inequality close to the Schwarz inequality, but missing an absolute-value symbol. Putting – a in place of a in this inequality, you get a similar inequality, but with a sign reversed. Verify that these two inequalities are together equivalent to the Schwarz inequality.

The above derivation excluded the cases a = 0 and b = 0. Show, finally, that the Schwarz inequality ...

Answers to True/False question 1.5:0. (a)F. (b)T.

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holds for trivial reasons in those cases.

1.7:3. Extending the above result to complex vectors. (d : 3)

One can regard k-tuples of complex numbers as forming a complex vector space Ck; but it is not very useful to define the dot product a · b of vectors in this space as Σ aibi, because this would not satisfy the important condition that a · a0 for nonzero a . So one instead defines a · b = Σ ai..bi

, and notes that for a nonzero vector a , a · a is a positive real number by Theorem 1.31(e) (p.14). Thus, one can again define | a | = | a · a |12. The above dot product satisfies most of the laws holding in the real case, but note two changes: First, though as before we have (c a ) · b = c ( a · b ), we now have a · (c b ) = c- ( a · b ).

Secondly, the dot product is no longer commutative. Rather, b · a = a · b....

.

With these facts in mind, repeat the calculation of the preceding exercise for vectors a and b of complex numbers. You will get an inequality close to the one you first got in that exercise, except that in place of a · b , you will have the expression 12( a · b + a · b....

) = Re ( a · b ). To get around this problem, verify that for every complex number z there exists a complex number γ with |γ| = 1 such that γz = | z |. Choosing such a γ for z = a · b , verify that on putting γ a in place of a in your inequality, you get the Schwarz inequality. Again, give a separate quick argument for the case where a or b is zero.

1.7:4. The Schwarz inequality via the quadratic formula. (d : 3)

The method of proving the real Schwarz inequality given in 1.7:2 requires us to remember one trick,

‘‘Take the dot product of a | b | – | a | b with itself, and use its nonnegativity.’’ One can also prove the result with a slightly different trick: ‘‘Let t be a real variable, regard the dot product of a + t b with itself as a real quadratic function of the real variable t, and use its nonnegativity.’’

Namely, expand that dot product in the form a t2+ b t + c, note why the coefficients a, b and c are real, and recall that a function of that form has a change of sign if the discriminant b2– 4 a c is positive.

Conclude that the discriminant must here be ≤ 0. Verify that that conclusion yields the real case of the Schwarz inequality (this time without special treatment of the situation where a or b is zero). Again, you can get the complex Schwarz inequality by applying the ideas of 1.7:3 to this argument.

The only difficulty is that since we are developing the properties of R from scratch, we should not assume without proof the above property of the discriminant! So for completeness, you should first prove that property. This is not too difficult. Namely, assuming the discriminant is positive, check by computation that if a ≠ 0, the quadratic formula you learned in High School leads to a factorization of a t2+ b t + c, and that this results in a change in sign. The case a = 0 can be dealt with by hand.

1.8. APPENDIX to Chapter 1. (Constructing R by Dedekind cuts.) (pp.17-21) Relevant exercise in Rudin:

1:R20. What happens if we weaken the definition of cut? (d : 3) Exercises not in Rudin:

1.8:0. Say whether each of the following statements is true or false.

Here α, β denote cuts in Q, and r, s elements of Q.

(a) α+β = {r + s : r∈α, s∈β}.

(b) α β = {r s : r∈α, s∈β}.

(c) –α = { –r : r∈α}.

(d) α < β ⇔ α is a proper subset of β. (e) r * = {s : sr }.

1.8:1. Some details of the proof of the distributive law for real numbers. (d : 3)

Verify the assertion in Step 7 of the proof of Theorem 1.19 (p.20) that multiplication of cuts satisfies ...

Answers to True/False question 1.6:0. (a)T. (b)T. (c) F. Answer to True/False question 1.7:0. (a)T.

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the distributive law for the following list of typical cases

(i) α < 0*, β > 0*, γ > 0*, (iii) α > 0*, β > 0*, β + γ = 0*, (ii) α < 0*, β> 0*, β + γ< 0*, (iv) α = 0*;

or, for variety, the cases

(v) α > 0*, β< 0*, γ< 0*, (vii) α < 0*, β > 0*, β + γ = 0*, (vi) α > 0*, β < 0*, β+ γ > 0*, (vii) β = 0*.

1.8:2. The distributive law for the real numbers: another approach. (d : 4)

This exercise shows an alternative to the separate verification of the 27 cases of the distributive law in Step 7 of the proof of Theorem 1.19 (p.20). We begin with two preparatory steps:

(a) Suppose F is a set with operations of addition and multiplication satisfying axioms (A) and (M) on p.5 of Rudin. Show that F satisfies axiom (D) on p.6, i.e.,

(D) (∀x, y, z∈F ) x (y + z) = x y + x z

if and only if it satisfies

(D′) (∀x, y, z, w∈F ) (y + z + w = 0) ⇒ (xy + xz + xw = 0).

(Suggestion: First show that each of (D) and (D′) implies that x( – y) = – (x y), and then prove with the help of this identity that each of (D) and (D′) implies the other.)

(b) Rudin noted in Step 6 that (D) (which in the case of real numbers we will here write α(β+γ) = α β+α γ) held for positive real numbers. With multiplication extended to all real numbers as in Step 7, verify that (D) still holds when α = 0*, then that it holds when β = 0*, then note that the case γ = 0*

follows from the case β = 0* using commutativity of addition. Thus, we now know that it holds whenever α, β, γ ≥ 0*. Verify also that the definition of multiplication of not-necessarily positive real numbers in Rudin’s Step 7 implies the properties α( –β) = – (α β) = ( –α)β.

By part (a), in order to show that the multiplication we have defined is distributive for all real numbers, it will suffice to show that these satisfy condition (D′), i.e.,

(∀α,β,γ,δ∈R ) (β+γ+δ = 0*) ⇒ (αβ+αγ+αδ = 0*).

We will do this in two parts:

(c) Prove that if the above formula is true for all α ≥ 0*, then it is true for all α∈R.

(d) To prove the above formula in the case α ≥ 0*, note that if β,γ,δ∈R satisfy β+γ+δ = 0*, then either two of β, γ, δ are ≥ 0* and one is ≤ 0*, or two are ≤ 0* and one is ≥ 0*. Since we know from Rudin’s Step 4 that the addition of R is commutative, we can in each of these cases rename and rearrange terms so that the ‘‘two’’ referred to are α and β and the ‘‘one’’ is γ. Show that the result needed in the first case follows quickly from Rudin’s Step 6, while the second follows easily from the first, using the identity α( –β) = – (α β).

1.8:3. A second round of cuts doesn’t change R . (d : 3)

The constructions of this Appendix can be carried out starting with any ordered field F in place of Q;

the result will be a set F′ with an ordering, with two operations of addition and multiplication, and with a map rr* of F into F′.

Show that if we start with F = R, the ordered field of real numbers, then F′ will be isomorphic to R; more precisely, that the map rr* will be an isomorphism (a one-to-one and onto map respecting the operations and the ordering). This shows that in a sense, the field of real numbers has ‘‘no gaps left to fill’’.

In doing this exercise, you may take for granted that the assertions Rudin makes in Step 8 of his construction remain valid in this context of a general ordered field F. (This in fact leaves just one nontrivial statement for you to prove. Make clear what it is before setting out to prove it.)

...

Answers to True/False question 1.8:0. (a)T. (b)F. (c) F. (d)T. (e)F.

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1.8:4. Cuts don’t work well on a nonarchimedean ordered field. (d : 4)

Rudin notes in the middle of p.19 that the archimedean property of Q is used in proving that R satisfies axiom (A5) (additive inverses). Suppose F is an ordered field which does not have the archimedean property, and that (as in the preceding exercise) we carry out the construction of this Appendix, getting an ordered set Fwith operations of addition and multiplication. Show that F′ fails to satisfy axiom (A5).

1.8:5. Proving r * s * = (r s)*. (d : 3)

In parts (a)-(c) below, let r and s be positive rational numbers. In those parts you will prove that these r and s satisfy r * s * = (r s)* i.e., assertion (b) of Step 8 in the construction of the real numbers (p.20). In part (d) you will look at the case where the factors are not both positive.

(a) Verify that r * s * and (r s)* contain the same nonpositive elements, and use the definitions given in Rudin to describe the positive elements each of them contains in terms of operations and inequalities in Q.

Thus, it remains to prove that the sets of positive elements you have described are equal.

(b) Verify the inclusion r * s *(r s)*.

(c) To obtain the inclusion (r s)*r * s *, suppose p is a positive element of (r s)*. From the fact that p < r s, deduce that there are rational numbers t1 and t2, both > 1, such that r s ⁄ p = t1t2. (Suggestion: Show there is a t1 such that 1 < t1 < r s ⁄ p, and then choose t2 in terms of t1.) As the analog of the first display on p.21 of Rudin, write r′ = r ⁄ t1, s= s ⁄ t2, and complete the proof as in Rudin, using multiplication instead of addition.

This completes the proof that r * s * = (r s)* for r and s positive. As with the axioms for multi- plication in R, the remaining cases are deduced from this one. I will just ask you to do one of these:

(d) Deduce from the case proved above that Rudin’s assertion (b) also holds when r < 0 and s > 0.

Chapter 2. Basic Topology.

2.1. FINITE, COUNTABLE, AND UNCOUNTABLE SETS. (pp.24-30) Relevant exercises in Rudin:

2:R1. The empty set is everywhere. (d : 1)

2:R2. The set of algebraic numbers is countable. (d : 3)

For this exercise, the student should take as given the result (generally proved in a course in Abstract Algebra) that a polynomial of degree n over a field has at most n roots in that field.

2:R3. Not all real numbers are algebraic. (d : 1. > 2:R2) 2:R4. How many irrationals are there? (d : 2)

Exercises not in Rudin:

2.1:0. Say whether each of the following statements is true or false.

(a) If f : XY is a mapping, and E is a subset of Y containing exactly one element, then the subset f–1(E ) of X also contains exactly one element.

(b) If Y is a set and {Gα ! α∈A} is a family of subsets of Y, then

α∈A Gα is a subset of Y.

(c) Every proper subset of J (the set of positive integers) is finite.

(d) The range of any sequence is at most countable.

(e) The set of rational numbers r satisfying r2< 2 is countable.

(f ) Every infinite subset of an uncountable set is uncountable.

(g) Every countable set is a subset of the integers.

(h) Every subset of the integers is at most countable.

(i) Every finite set is countable.

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(j) The union of any collection of countable sets is countable.

(k) If A and B are countable sets, then AB is countable.

(l) If A is countable and B is any set, then AB is countable.

(m) If A is countable and B is any set, then AB is at most countable.

2.1:1. An infinite image of a countable set is countable. (d : 2)

Suppose E is a countable set, and f is a function whose domain is E and whose image f (E ) is infinite. Show that f (E ) is countable. (Hint: The proof will be like that of Theorem 2.8, but this time, take n1 = 1, and for each k > 1, assuming n1, ... , nk –1 have been chosen, let nk be the least integer such that xn

k∈{xn

1, ... , xn

k –1} . To do this you must note why there is at least one such nk.) 2.1:2. Functions and cardinalities. (d : 2, 1, 1, 1)

Suppose A and B are sets, and f : AB a function.

(a) Assume A and B infinite. We can divide this situation into four cases, according to whether A is countable or uncountable and whether B is countable or uncountable. Show that if f is one-to-one, then three of these four cases can occur, but one cannot. To do this, you must give examples of three cases, and a proof that the fourth cannot occur. (Hint: Some or all of your examples can be of trivial sorts; e.g., using functions that don’t move anything, but satisfy f (x) = x for all x.) Express your nonexistence result as an implication saying that if f : AB is one-to-one, and a certain one of A or B has a certain property, then the other has a certain property.

(b) If we don’t assume A and B infinite, then each of these sets can be finite, countable, or uncountable, giving 3×3 = 9 rather than 4 combinations. Again, for f one-to-one, certain of these 9 cases are possible and certain impossible. I won’t ask your to prove your assertions (since your understanding of the consequences of finiteness is largely intuitive, and a course in set theory, not Math 104, is where you will learn the theory that will make it precise); but make a 3×3 chart showing which cases can occur, and which cannot. (Label the rows with the properties of A, in the order ‘‘finite; countable; uncountable’’, the columns with the properties of B in the same order, and use 0 and × for ‘‘possible’’ and

‘‘impossible’’. If you don’t know the answer in some case, use ‘‘?’’)

(c) As in (a), assume A and B infinite, so that we have four cases depending on whether each is countable or uncountable; but now suppose f is onto, rather than one-to-one. Again, give examples showing that three cases can occur, but show that the fourth cannot. (Hint: Use 2.1:1 .) Again, express your nonexistence result as an implication.

(d) Analogously to part (b), modify part (c) by not assuming A and B infinite, and make a 3×3 chart showing which cases can occur, and which cannot.

2.1:3. Cardinalities of sets of functions. (d : 3)

Suppose A is a countable set, and B is a set (finite, countable, or infinite), which contains at least two elements. Show that there are uncountably many functions AB.

Suggestion: Consider first the case where A = J and B = {0,1}, and convince yourself that this case is equivalent to a result Rudin has proved for you.

This would probably not be a good problem to assign, because students who had seen some set theory might have a big advantage over those who hadn’t. But it is a good one for students to think about if they haven’t seen these ideas.

2.2. METRIC SPACES. (pp.30-36)

Relevant exercises in Rudin:

2:R5. Find a bounded set with 3 limit points. (d : 2) 2:R6. Properties of {limit points of E } . (d : 2)

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2:R7. Closures of unions versus unions of closures. (d : 2)

In the last sentence of this exercise, ‘‘this inclusion can be proper’’ means that there are some choices of metric space and subsets Ai such that the union of closures shown at the end of (b) is a proper subset of B..

. (If you’re unsure what ‘‘proper subset’’ means, use the index!) 2:R8. Limits points of closed and open sets. (d : 2)

2:R9. Basic properties of the interior of a set. (d : 2) 2:R10. The metric with d( p, q) = 1 for all pq. (d : 2)

(But the last sentence of this exercise refers to the concept of compactness, and so requires §2.3.) 2:R11. ‘‘Which of these five functions are metrics?’’ (d : 2)

In this exercise you must, for each case, either prove that the properties of a metric are satisfied, or give an example showing that one of these properties fails.

Exercises not in Rudin:

2.2:0. Say whether each of the following statements is true or false.

(a) Every unbounded subset of R is infinite.

(b) Every infinite subset of R is unbounded.

(c) If E is a subset of a metric space X, then every interior point of E is a member of E.

(d) If E is a bounded subset of a metric space X, then every subset of E is also bounded.

(e) Q is a dense subset of R.

(f ) R is a dense subset of C.

(g) If E is a subset of a set X, then (Ec)c (the complement of the complement of E in X ) is E.

(h) If E is an open subset of a metric space X, then every subset of E is also an open subset of X.

(i) If E is an open subset of a metric space X, then Ec is a closed subset of X.

(j) If E is a subset of a metric space X, and E is not open, then it is closed.

(k) If Y is a subset of a metric space X, and {Gα} is a family of subsets of Y that are open relative to Y, then

Gα is also open relative to Y.

(l) If E is a subset of a metric space X and p is a limit point of E, then there exists q∈E such that qp and such that q belongs to every neighborhood of p in X .

(m) If E is a subset of a metric space X and p is a limit point of E, then for every neighborhood N of p in X , there exists q∈E∩N – { p }.

(n) The union of any two convex subsets of Rk is convex.

(o) The intersection of any family of convex subsets of Rk is convex.

2.2:1. Possible distances among 3 points. (d : 1. > 1.7:1)

(a) Show that for any points p, q and r of a metric space, one has d( p , r)d( p , q) – d( q , r) and d( p , r)d( q , r) – d( p , q).

(b) Combining the above inequalities with that of Definition 2.15(c), what is the possible set of values for d( p , r) if d( p , q) = 3 and d( q , r) = 1? If d( p , q) = 1 and d( q , r) = 4? For each value d( p , r) = c allowed by your answer to the former question, do there in fact exist points p, q and r in some metric space X such that d( p , q) = 3, d( q , r) = 1, and d( p , r) = c ? Hint: Use 1.7:1.

2.2:2. A characterization of open sets. (d : 1)

Show that a subset E of a metric space X is open if and only if it is the union of a set of neighborhoods.

...

Answers to True/False question 2.1:0. (a) F. (b) T. (c) F. (d) T. (e) T. (f ) F. (g) F. (h) T. (i)F. (j)F. (k) T.

(l)F. (m) T.

(17)

2.2:3. A characterization of the closure of a set. (d : 1) Let E be a subset of a metric space X. Show that E..

= {x∈X ! (∀ε> 0) (∃y∈E ) d(y, x) < ε} . (I prefer this to the definition of closure that Rudin gives, since the splitting of the points of E..

into two sorts, those in E and the limit points, seems to me unnatural.)

2.2:4. A characterization of limit points. (d : 1)

Let X be a metric space and E a subset. Show that a point p∈X is a limit point of E if and only if every open subset GX which contains p contains some point of E other than p. (This differs from part (b) of Definition 2.18 only in the replacement of ‘‘neighborhood of p’’ by ‘‘open subset of X containing p’’.)

2.2:5. Open and closed subsets of open and closed sets. (d : 2)

(a) Suppose E is an open subset of a metric space X, and F is a subset of E. Show that F is open relative to E if and only if it is open as a subset of X.

(b) Suppose E is a closed subset of a metric space X, and F is a subset of E. Show that F is closed relative to E if and only if it is closed as a subset of X.

(c) Suppose E is a open subset of a metric space X, and F is a subset of E which is closed relative to E. Show by an example that F need not, in general, either be open or closed as a subset of X. (Give a specific metric space X and specific subsets E and F .)

(d) Give a similar example showing that for a closed subset E, a relatively open subset F of E need neither be open nor closed in X.

2.2:6. The boundary of a subset of a metric space. (d : 2)

Let X be a metric space, and E a subset of X. One defines the boundary of E to be the setE of all points x∈X such that every neighborhood of x contains at least one point of E and at least one point of Ec. (In saying ‘‘at least one point’’, we do not exclude the point x itself.)

(a) Show that E..

= E∪ ∂E.

(b) Deduce that E is closed if and only ifEE.

(c) Show that E is open if and only ifEEc.

(d) Deduce that E is both open and closed if and only ifE = ∅.

2.2:7. Equivalent formulations of boundedness. (d : 2)

Let E be a nonempty subset of a metric space X. Show that the following conditions are equivalent:

(a) E is bounded. (Definition 2.18(i).)

(b) For every point q∈X there exists a real number M such that for all p∈E, d( p, q) < M.

(c) There exists a real number M and a point q∈E such that for all p∈E, d( p, q) < M.

(d) There exists a real number M such that for all p, q∈E, d( p, q) < M.

(Warning: The ‘‘M ’’s of statements (a)-(d) will not necessarily be the same.) 2.2:8. Another description of the closure of a set. (d : 1)

Let E be a subset of a metric space X. Show that E..

equals the intersection of all closed subsets of X containing E.

2.2:9. The closure of a bounded set is bounded. (d : 1)

Let E be a bounded subset of a metric space X. Show that E..

is also bounded.

2.2:10. Finding bounded perfect subsets of perfect sets. (d : 3. > 2.2:9) Let X be a metric space.

(a) Show that if E is a perfect subset of X and A is an open subset of X, then E...∩ A

is perfect.

(b) Deduce that if X has a nonempty perfect subset, then it has a bounded nonempty perfect subset.

...

Answers to True/False question 2.2:0. (a) T. (b) F. (c) T. (d) T. (e) T. (f ) F. (g) T. (h) F. (i)T. (j)F. (k) T.

(l)F. (m) T. (n)F. (o)T.

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