Math 202 (Introduction to Real Analysis) Fall and Spring 2006-2007Lecture 1

Instructor: Dr. Kin Y. LiOffice: Room 3471 Office Phone: 2358 7420 e-mail address: makyli@ust.hkOffice Hours: Tu. 3:30pm - 4:30pm (or by appointments)

Prerequisite: A-level Math or One Variable Calculus

Website for Lecture Notes and Practice Exercises: http://www.math.ust.hk/makyli/UG.html

Grade Scheme: Homeworks (4%), Tutorial Presentations (6%), 3 Examinations (30% each)

All records of grades will be put on the website http://grading.math.ust.hk/checkgrade/ as soon as they areavailable. At the end of Fall semester, students who have done a reasonable job on homeworks and exams will be givena pp-grade for permission to proceed to the next semester. Those who have not done a reasonable job in the opinion ofthe instructor (for example, received less than 40 marks in the first examination) will get a F-grade and will repeatthe course next year. For those who are permitted to proceed to the second semester, the overall grade will be given atthe end of the second semester. This course is essentially graded by curve with one exception, namely studentswho achieve 40% or less of the overall grade will fail the course.

Students should make copies of homeworks before submitting the originals. In case homeworks are notreceived, students will be required to resubmit copies within a short period of time (may be less than a day).

For tutorial presentations, students will form groups (of 1 to 3 students) and present solutions to assigned problemsin the tutorial sessions. All members of a group must attend the sessions to assist in answering possible questions frompresentations. Marks will be deducted for failure to present solutions or for absence in supporting his/her group.

Course Description: Math 202 is a 2 semester course. It is the first of two required courses on analysis for Mathmajors. It is to be followed by Math 301 (Real Analysis). This course will focus on the proofs of basic theoremsof analysis, as appeared in one variable calculus. Along the way to establish the proofs, many new concepts will beintroduced. These include countability, sequences and series of numbers, of functions, supremum/infimum, Cauchycondition, Riemann integrals and improper integrals, etc. Understanding them and their properties are important forthe development of the present and further courses.

Textbooks:Steven Lay, Analysis With an Introduction to Proof , 4th ed., Prentice Hall, 2005.

References:1. J. A. Fridy, Introductory Analysis; 2nd ed., Academic Press, 2000.2. Manfred Stoll, Introduction to Real Analysis, Addison Wesley, 1997.3. Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.4. Tom Apostol, Mathematical Analysis, 2nd ed., Addison-Wesley, 1974.

*5. Chinese Solution Manual to Tom Apostols Mathematical Analysis, 2nd ed.

Reminders: Students are highly encouraged to come to office hours for consultation. This is a difficult course formany, but not all students. Although there are lecture notes, students should attend all lectures and tutorials aslectures notes are only brief records of materials covered in class, which may contain typographical errors. Of course,questions from students and answers from instructors or other digressions will not be recorded. Students are advisedto take your own notes. All materials presented in lectures and tutorials as well as proper class conduct are thestudents responsibility. The instructor reserves the right to make any changes to the course throughout thesemester. The only way to succeed in this course is to do the work.

1

Objectives of the Course

The objectives of the course are to learn analysis and learn proofs.

Questions: What is analysis? How is it different from other branches of mathematics?

Analysis is the branch of mathematics that studies limit and concepts derived from limit, such as continuity,differentiation and integration, while geometry deals with figures, algebra deals with equations and inequalitiesinvolving addition, subtraction, multiplication, division and number theory studies integers.

When we try to solve problems involving real or complex numbers, such as finding roots of polynomials orsolving differential equations, we may not get the right answers the first time. However, we can get approximationsand the limits of these approximations, we hope, will give us the right answers. At least, we can know solutions existeven though we may not be able to write them explicitly.

Problems involving integers can be much harder since integers are discrete, i.e. there are minimum distanceseparating distinct integers so that one cannot find integers arbitrarily close to another integer.

Try to see if there is a real solution to the equation2x x2

4x4 C 1D 987654321: Then try to see if there is an integer

solution. What are the differences in the way you solve these two problems?

Question: Why should we learn proofs?

A statement is true not because your teacher tells you it is true. A teacher can make mistakes! There were famousmathematicians who made conjectures that were discovered to be wrong years later. How can we be certain the factswe learned are true? How can we judge when more than one proposed solutions are given, which is correct?

Suppose we want to find limx!0

x2 sin 1x

sin x: Since the numerator is between x2 and x2; the numerator has limit 0.

The denominator also has limit 0. So, by lHopitals rule, limx!0

x2 sin 1x

sin xD lim

x!0

2x sin 1x cos 1

x

cos x: However, the new

numerator does not have a limit because cos 1x

has no limit as x ! 0; while the new denominator has limit 1. So thelimit of the original problem does not exist. Is this reasoning correct? No. Where is the mistake?

Sometimes we explain facts by examples or pictures. For instance, the statement that every odd degree polynomialwith real coefficients must have at least one root is often explained by some examples or some pictures. In our lifetime,we can only do finitely many examples and draw finitely many pictures. Should we believe something is true by seeinga few pictures or examples?

Draw the graphs of a few continuous functions on [0; 1]: Do you think every continuous function on [0; 1]is differentiable in at least one point on .0; 1/? Or do you think there exists a continuous function on [0; 1] notdifferentiable at any point of .0; 1/?

Consider the function f .n/ D n2 n C 41: Note f .1/ D 41; f .2/ D 43; f .3/ D 47; f .4/ D 53; f .5/ D61; f .6/ D 71; f .7/ D 83; f .8/ D 97; f .9/ D 113; f .10/ D 131 are prime numbers. Should you believe that f .n/is a prime number for every positive integer n? What is the first n that f .n/ is not prime?

In order to have confident, you have to be able to judge the facts you learned are absolutely correct. Almostcorrect is not good enough in mathematics.

2

Chapter 1. Logic

To reason correctly, we have to follow some rules. These rules of reasoning are what we called logic. We willonly need a few of these rules, mainly to deal with taking opposite of statements and to handle conditional statements.

We will use the symbol (or :) to denote the word not. Also, we will use the symbol 8 to denote for all,for any, for every. Similarly, the symbol 9 will denote there is (at least one), there exists, there are (some)and usually followed by such that. The symbols 8 and 9 are called quantifiers.

Negation. Below we will look at rules of negation (i.e. taking opposite). They are needed when we do indirectproofs (or proofs by contradiction). For any expression p; we have . p/ D p:

Examples. (1)expression :

pz }| {

x > 0 andq

z }| {

x < 1opposite expression : x 0 or x 1

rule : .p and q/ D . p/ or . q/

(2) expression : x < 0 or x > 1opposite expression : x 0 and x 1

rule : .p or q/ D . p/ and . q/

(3)statement : For every x 0; x has a square root. (True)

quantified statement : 8x 0 .x has a square root/:opposite statement : There exists x 0 such that x does not have a square root. (False)

quantified opposite statement : 9x 0 .x has a square root/:

(4) statement : For every x 0; there is y 0 such that y2 D x : (True)quantified statement : 8x 0; 9y 0 .y2 D x/:

opposite statement : There exists x 0 such that for every y 0; y2 6D x : (False)quantified opposite statement : 9x 0; 8y 0 .y2 D x/:

From examples (3) and (4), we see that the rule for negating statements with quantifiers is first switch every8 to 9 and every 9 to 8, then negate the remaining part of the statement.

If-then Statements. If-then statements occur frequently in mathematics. We will need to know some equivalentways of expressing an if-then statement to do proofs. The statement if p; then q may also be stated as p implies q,p only if q, p is sufficient for q, q is necessary for p and is commonly denoted by p H) q. For example,the statement if x D 3 and y D 4; then x2 C y2 D 25 may also be stated as x D 3 and y D 4 are sufficient forx2 C y2 D 25 or x2 C y2 D 25 is necessary for x D 3 and y D 4.

Example. (5) statement : If x 0; then jxj D x : (True)opposite statement : x 0 and jxj 6D x : (False)

rule : .p H) q/ D p and . q/

Remark. Notep H) q D . .p H) q//

D .p and . q//D . p/ or . q/D . p/ or q:

3

For the statement if p; then q .p H) q/; there are two related statements: the converse of the statement is ifq , then p (q H) p) and the contrapositive of the statement is if . q/; then . p/ ( q H) p).

Examples. (6) statement : If x D 3; then x2 D 9: (True)converse : If x2 D 9; then x D 3: (False, as x may be 3.)

contrapositive : If x2 6D 9; then x 6D 3: (True)

(7) statement : x D 3 H) 2x D 6 (True)converse : 2x D 6 H) x D 3 (True)

contrapositive : 2x 6D 6 H) x 6D 3 (True)

(8) statement : If jxj D 3; then x D 3: (False, as x may be 3.)converse : If x D 3; then jxj D 3: (True)

contrapositive : If x 6D 3; then jxj 6D 3: (False, as x may be 3.)

Remarks. Examples (6) and (7) showed that the converse of an if-then statement is not the same as the statementnor the opposite of the statement in general. Examples (6), (7) and (8) showed that an if-then statement and itscontrapositive are either both true or both false. In fact, this is always the case because by the remark on the last page,

. q/ H) . p/ D . q/ or . p/D q or . p/D . p/ or qD p H) q:

So an if-then statement and its contrapositive statement are equivalent.

Finally, we introduce the terminology p if and only if q to mean if p; then q and if q; then p. The statementp if and only if q is the same as p is necessary and sufficient for q. We abbreviate p if and only if q by p ()q. So p () q means p H) q and q H) p. The phrase if and only if is often abbreviated as iff.

Caution! Note 88 D 88 and 99 D 99; but 89 6D 98: For example, every student is assigned anumber is the same as 8 student, 9 number such that the student is assigned the number. This statement impliesdifferent students may be assigned possibly different numbers. However, if we switch the order of the quantifiers, thestatement becomes 9 number such that 8 student, the student is assigned the number. This statement implies thereis a number and every student is assigned that same number!

4

Chapter 2. Sets

To read and write mathematical expressions accurately and concisely, we will introduce the language of sets. Aset is a collection of objects (usually numbers, ordered pairs, functions, etc.) If object x is in a set S, then we say x isan element (or a member) of S and write x 2 S: If x is not an element of S, then we write x 62 S: A set having finitelymany elements is called a finite set, otherwise it is called an infinite set. The empty set is the set having no objects andis denoted by ;:

A set may be shown by listing its elements enclosed in braces (eg. f1; 2; 3g is a set containing the objects 1; 2; 3; thepositive integer N D f1; 2; 3; : : :g, the integerZD f: : : ;2;1; 0; 1; 2; : : :g, the empty set ; D fg) or by descriptionenclosed in braces (eg. the rational numbersQD fm

n: m 2Z; n 2 Ng; the real numbersRD fx : x is a real numberg

and the complex numbers C D fx C iy : x; y 2 Rg:) In describing sets, the usual convention is to put the form of theobjects on the left side of the colon and to state the conditions on the objects on the right side of the colon. The setwill consist of all elements satisifying all the conditions. It is also common to use a vertical bar in place of colon in setdescriptions.Examples. (i) The closed interval with endpoints a; b is [a; b] D fx : x 2 R and a x bg:(ii) The set of square numbers is f1; 4; 9; 16; 25; : : :g D fn2 : n 2 Ng:

(iii) The set of all positive real numbers isRC D fx : x 2 R and x > 0g: (If we want to emphasize this is a subsetofR; we may stress x is real in the form of the objects and writeRC D fx 2 R : x > 0g: If numbers are alwaystaken to mean real numbers, then we may write simplyRC D fx : x > 0g:)

(iv) The set of points (or ordered pairs) on the line `m with equation y D mx is f.x;mx/ : x 2 Rg:

For sets A; B; we say A is a subset of B (or B contains A) iff every element of A is also an element of B: Inthat case, we write A B: (For the case of the empty set, we have ; S for every set S:) Two sets A and B areequal if and only if they have the same elements (i.e. A D B means A B and B A:) So A D B if and only if(x 2 A () x 2 B). If A B and A 6D B; then we say A is a proper subset of B and write A B: (For example,if A D f1; 2g; B D f1; 2; 3g;C D f1; 1; 2; 3g; then A B is true, but B C is false. In fact, B D C: Repeatedelements are counted only one time so that C has 3 elements, not 4 elements.)

For a set S; we can collect all its subsets. This is called the power set of S and is denoted by P.S/ or 2S : Forexamples, P.;/ D f;g; P.f0g/ D f;; f0gg and P.f0; 1g/ D f;; f0g; f1g; f0; 1gg: For a set with n elements, its powerset will have 2n element. This is the reason for the alternative notation 2S for the power set of S: Power set is oneoperation of a set. There are a few other common operations of sets.

Definitions. For sets A1; A2; : : : ; An ;

(i) their union is A1 [ A2 [ [ An D fx : x 2 A1 or x 2 A2 or or x 2 Ang;(ii) their intersection is A1 \ A2 \ \ An D fx : x 2 A1 and x 2 A2 and and x 2 Ang;

(iii) their Cartesian product is

A1 A2 An D f.x1; x2; : : : ; xn/ : x1 2 A1 and x2 2 A2 and and xn 2 Ang;

(iv) the complement of A2 in A1 is A1 n A2 D fx : x 2 A1 and x 62 A2g:

Examples. (i) f1; 2; 3g [ f3; 4g D f1; 2; 3; 4g; f1; 2; 3g \ f2; 3; 4g D f2; 3g; f1; 2; 3g n f2; 3; 4g D f1g:(ii) [2; 4]\N D f1; 2; 3; 4g; [0; 2][ [1; 5][ [4; 6] D [0; 6]:

(iii) .[0; 7] \Z/ n fn2 : n 2 Ng D f0; 1; 2; 3; 4; 5; 6;7g n f1; 4; 9; 16; 25; : : :g D f0; 2; 3; 5; 6; 7g:(iv) RRRD f.x; y; z/ : x; y; z 2 Rg;Q .RnQ/ D f.a; b/ : a is rational and b is irrationalg:Remarks. (i) For the case of the empty set, we have

A [ ; D A D ; [ A; A \ ; D ; D ; \ A; A ; D ; D ; A; A n ; D A and ; n A D ;:

5

(ii) The notions of union, intersection and Cartesian product may be extended to infinitely many sets similarly. Theunion is the set of objects in at least one of the sets. The intersection is the set of objects in every one of the sets.The Cartesian product is the set of ordered tuples such that the i-th coordinate must belong to the i-th set.

(iii) The set A1 [ A2 [ [ An may be written asn[

kD1Ak : If for every positive integer k; there is a set Ak ; then the

notation A1 [ A2 [ A3 [ may be abbreviated as1

[

kD1Ak or

[

k2N

Ak : If for every x 2 S; there is a set Ax ; then

the union of all the sets Ax s for all x 2 S is denoted by[

x2SAx : Similar abbreviations exist for intersection and

Cartesian product.

Examples. (i) .[1; 2][ [2; 3][ [3; 4][ [4; 5][ / \ZD [1;C1/ \ZD N:

(ii)\

n2N

h

0; 1C1n

D [0; 2/ \h

0; 112

\

h

0; 113

\

h

0; 114

\ D [0; 1]:

(iii) For every k 2 N; let Ak D f0; 1g; then

A1 A2 A3 D f.x1; x2; x3; : : :/ : each xk is 0 or 1 for k D 1; 2; 3; : : :g:

(iv) For each m 2 R; let `m be the line with equation y D mx on the plane, then[

m2R

`m D R2n f.0; y/ : y 2 R; y 6D 0g

and\

m2R

`m D f.0; 0/g:

(v) Show that if A B and C D; then A \ C B \ D:Reason. For every x 2 A \C; we have x 2 A and x 2 C: Since A B and C D; we have x 2 B and x 2 D;which imply x 2 B \ D: Thus, we see that every element in A \C is also in B \ D: Therefore, A \C B \ D:

(vi) Show that .A [ B/ n C D .A n C/ [ .B n C/:Reason. For every x 2 .A[ B/nC; we have x 2 A[ B and x 62 C: So either x 2 A or x 2 B: In the former case,x 2 A n C or in the latter case, x 2 B n C: So x 2 .A n C/ [ .B n C/: Hence, .A [ B/ n C .A nC/ [ .B n C/:

Conversely, for every x 2 .A n C/ [ .B n C/; either x 2 A n C or x 2 B n C: In the former case, x 2 Aand x 62 C or in the latter case, x 2 B and x 62 C: In both cases, x 2 A [ B and x 62 C: So x 2 .A [ B/ n C:Hence, .A n C/ [ .B n C/ .A [ B/ n C: Combining with the conclusion of the last paragraph, we have.A [ B/ n C D .A n C/ [ .B n C/:

We shall say that sets are disjoint iff their intersection is the empty set. Also, we say they are mutually disjoint iffthe intersection of every pair of them is the empty set. A relation on a set E is any subset of E E : The following isan important concept that is needed in almost all branches of mathematics. It is a tool to divide (or partition) the set ofobjects we like to study into mutually disjoint subsets.Definition. An equivalence relation R on a set E is a subset R of E E such that

(a) (reflexive property) for every x 2 E; .x; x/ 2 R;(b) (symmetric property) if .x; y/ 2 R; then .y; x/ 2 R;(c) (transitive property) if .x; y/; .y; z/ 2 R; then .x; z/ 2 R:

We write x y if .x; y/ 2 R: For each x 2 E; let [x] D fy : x yg: This is called the equivalence classcontaining x : Note that every x 2 [x] by (a) so that

[

x2E[x] D E : If x y; then [x] D [y] because by (b) and (c),

z 2 [x] () z x () z y () z 2 [y]: If x 6 y; then [x] \ [y] D ; because assuming z 2 [x] \ [y] willlead to x z and z y; which imply x y; a contradiction. So every pair of equivalence classes are either the sameor disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes.

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Examples. (1) (Geometry) For triangles T1 and T2; define T1 T2 if and only if T1 is similar to T2: This is anequivalence relation on the set of all triangles as the three properties above are satisfied. For a triangle T; [T ] is the setof all triangles similar to T :

(2) (Arithmetic) For integers m and n; define m n if and only if m n is even. Again, properties (a), (b), (c) caneasily be verified. So this is also an equivalence relation on Z: There are exactly two equivalence classes, namely[0] D f: : : ;4;2; 0; 2; 4; : : :g (even integers) and [1] D f: : : ;5;3;1; 1; 3; 5; : : :g (odd integers). Two integersin the same equivalence class is said to be of the same parity.(3) Some people think that properties (b) and (c) imply property (a) by using (b), then letting z D x in (c) to conclude.x; x/ 2 R: This is false as shown by the counterexample that E D f0; 1g and R D f.1; 1/g; which satisfies properties(b) and (c), but not property (a). R fails property (a) because 0 2 E; but .0; 0/ 62 R as 0 is not in any ordered pair in R:

A function (or map or mapping) f from a set A to a set B (denoted by f : A ! B) is a method of assigning toevery a 2 A exactly one b 2 B: This b is denoted by f .a/ and is called the value of f at a: Thus, a function must bewell-defined in the sense that if a D a0; then f .a/ D f .a0/: The set A is called the domain of f (denoted by dom f )and the set B is called the codomain of f (denoted by codom f ). We say f is a B-valued function (eg. if B D R; thenwe say f is a real-valued function.) When the codomain B is not emphasized, then we may simply say f is a functionon A: The image or range of f (denoted by f .A/ or im f or ran f ) is the set f f .x/ : x 2 Ag: (To emphasize this is asubset of B; we also write it as f f .x/ 2 B : x 2 Ag:) The set G D fx; f .x/ : x 2 Ag is called the graph of f: Twofunctions are equal if and only if they have the same graphs. In particular, the domains of equal functions are the sameset.

Examples. The function f : Z! R given by f .x/ D x2 has dom f D Z; codom f D R: Also, ran f Df0; 1; 4; 9; 16; : : :g: This is different from the function g : R! Rgiven by g.x/ D x2 because dom g D R 6D dom f:Also, a function may have more than one parts in its definition, eg. the absolute value function h : R! Rdefined byh.x/ D

n

x if x 0x if x < 0 : Be careful in defining functions. The following is bad: let xn D .1/

n and i.xn/ D n: Therule is not well-defined because x1 D 1 D x3; but i.x1/ D 1 6D 3 D i.x3/:

Definitions. (i) The identity function on a set S is IS : S ! S given by IS.x/ D x for all x 2 S:(ii) Let f : A ! B; g : B 0 ! C be functions and f .A/ B 0: The composition of g by f is the function

g f : A ! C defined by .g f /.x/ D g. f .x// for all x 2 A:(iii) Let f : A ! B be a function and C A: The function f jC : C ! B defined by f jC.x/ D f .x/ for every x 2 C

is called the restriction of f to C:(iv) A function f : A ! B is surjective (or onto) iff f .A/ D B:(v) A function f : A ! B is injective (or one-to-one) iff f .x/ D f .y/ implies x D y:

(vi) A function f : A ! B is a bijection (or a one-to-one correspondence) iff it is injective and surjective.(vii) For an injective function f : A ! B; the inverse function of f is the function f 1 : f .A/ ! A defined by

f 1.y/ D x () f .x/ D y:

Remarks. A function f : A ! B is surjective means f .A/ D B; which is the same as saying every b 2 B is an f .a/for at least one a 2 A: In this sense, the values of f do not omit anything in B: We will loosely say f does not omitany element of B for convenience. However, there may possibly be more than one a 2 A that are assigned the sameb 2 B: Hence, the range of f may repeat some elements of B: If A and B are finite sets, then f surjective implies thenumber of elements in A is greater than or equal to the number of elements in B:

Next, a function f : A ! B is injective means, in the contrapositive sense, that x 6D y implies f .x/ 6D f .y/;which we may loosely say f does not repeat any element of B: However, f may omit elements of B as there maypossibly be elements in B that are not in the range of f: So if A and B are finite sets, then f injective implies thenumber of elements in A is less than or equal to the number of elements in B:

Therefore, a bijection from A to B is a function whose values do not omit nor repeat any element of B: If A andB are finite sets, then f bijective implies the number of elements in A and B are the same.

7

Remarks (Exercises). (a) Let f : A ! B be a function. We have f is a bijection if and only if there is a functiong : B ! A such that g f D IA and f g D IB : (In fact, for f bijective, we have g D f 1 is bijective.)(b) If f : A ! B and h : B ! C are bijections, then h f : A ! C is a bijection.(c) Let A; B be subsets ofRand f : A ! B be a function. If for every b 2 B; the horizontal line y D b intersects thegraph of f exactly once, then f is a bijection.

Example. Show that f : [0; 1] ! [3; 4] defined by f .x/ D x3 C 3 is a bijection.Method 1 If f .x/ D f .y/; then x3 C 3 D y3 C 3; which implies x3 D y3: Taking cube roots of both sides, we getx D y: Hence f is injective. Next, for every y 2 [3; 4]; solving the equation x3 C 3 D y for x; we get x D 3py 3:Since y 2 [3; 4] implies y 3 2 [0; 1]; we see x 2 [0; 1]: Then f .x/ D y: Hence f is surjective. Therefore, f isbijective.Method 2. Define g : [3; 4] ! [1; 2] by g.y/ D 3py 3: For every x 2 [0; 1] and y 2 [3; 4]; .g f /.x/ D g.x3C3/ D3p

.x3 C 3/ 3 D x and . f g/.y/ D f . 3py 3/ D . 3py 3/3 C 3 D y: By remark (a) above, f is a bijection.

To deal with the number of elements in a set, we introduce the following concept. For sets S1 and S2; we willdefine S1 S2 and say they have the same cardinality (or the same cardinal number) if and only if there exists abijection from S1 to S2: This is easily checked to be an equivalence relation on the collection of all sets. For a set S; theequivalence class [S] is often called the cardinal number of S and is denoted by card S or jSj: This is a way to assigna symbol for the number of elements in a set. It is common to denote, for a positive integer n; card f1; 2; : : : ; ng D n;card ND @0 (read aleph-naught) and card RD c (often called the cardinality of the continuum).

8

Chapter 3. Countability

Often we compare two sets to see if they are different. In case both are infinite sets, then the concept of countablesets may help to distinguish these infinite sets.

Definitions. A set S is countably infinite iff there exists a bijection f : N! S (i.e. N and S have the same cardinalnumber @0:) A set is countable iff it is a finite or countably infinite set. A set is uncountable iff it is not countable.

Remarks. Suppose f :N! S is a bijection. Then f is injective means f .1/; f .2/; f .3/; : : : are all distinct and fis surjective means f f .1/; f .2/; f .3/; : : :g D S: So n 2 N$ f .n/ 2 S is a one-to-one correspondence between Nand S: Therefore, the elements of S can be listed in an orderly way (as f .1/; f .2/; f .3/; : : :) without repetition oromission. Conversely, if the elements of S can be listed as s1; s2; : : : without repetition or omission, then f : N! Sdefined by f .n/ D sn will be a bijection as no repetition implies injectivity and no omission implies surjectivity.

Bijection Theorem. Let g: S ! T be a bijection. S is countable if and only if T is countable.(Reasons. The finite set case is clear. For infinite sets, it is true because S countable implies there is a bijective functionf : N! S; which implies h D g f : N! T is bijective, i.e. T is countable. For the converse, h is bijective impliesf D g1 h is bijective.)Remarks. Similarly, taking contrapositive, S is uncountable if and only if T is uncountable.

Basic Examples. (1) N is countably infinite (because the identity function IN

.n/ D n is a bijection).(2) Zis countably infinite because the following function is a bijection (one-to-one correspondence):

N D f 1; 2; 3; 4; 5; 6; 7; 8; 9; : : : gf # # # # # # # # # # : : :Z D f 0; 1; 1; 2; 2; 3; 3; 4; 4; : : : g:

The function f : N!Zis given by f .n/ D n

2 if n is even.

n12 / if n is odd

and its inverse function g : Z! N is given

by g.m/ D

2m if m > 01 2m if m 0 : Just check g f D IN and f g D IZ:

(3)NN D f.m; n/: m; n 2 Ng is countably infinite.

(Diagonal Counting Scheme) Using the diagram on the right, de-fine f :N! N N by f .1/ D .1; 1/, f .2/ D .2; 1/, f .3/ D .1; 2/,f .4/ D .3; 1/, f .5/ D .2; 2/, f .6/ D .1; 3/, : : :, then f is injectivebecause no ordered pair is repeated. Also, f is surjective because

.m; n/ D f

mCn2X

kD0k C n

!

D f

.m C n 2/.m C n 1/2

C n

:

.1; 1/ .1; 2/ .1; 3/ .1; 4/

.2; 1/ .2; 2/ .2; 3/:

:

:

:

:

:

.3; 1/ .3; 2/:

:

:

:

:

:

:

:

:

:

:

:

.4; 1/:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

(4) The open interval .0; 1/ D fx : x 2 R and 0 < x < 1g is uncountable. Also,R is uncountable.f .1/ D 0:a11a12a13a14 : : :f .2/ D 0:a21a22a23a24 : : :f .3/ D 0:a31a32a33a34 : : :f .4/ D 0:a41a42a43a44 : : :

:

:

:

:

:

:

Suppose .0; 1/ is countably infinite and f :N ! .0; 1/ is a bijection asshown on the left. Consider the number x whose decimal representation is

0:b1b2b3b4 : : :, where bn D

2 if ann D 11 if ann 6D 1

. Then 0 < x < 1 and x 6D f .n/for all n because bn 6D ann . So f cannot be surjective, a contradiction. NextRis uncountable because tan.x 12 / provides a bijection from .0; 1/ ontoR:

To determine the countability of more complicated sets, we will need the theorems below.

9

Countable Subset Theorem. Let A B: If B is countable, then A is countable. (Taking contrapositive, if A isuncountable, then B is uncountable.)Countable Union Theorem. If An is countable for every n 2 N; then

[

n2N

An is countable. In general, if S is countable

(say f : N! S is a bijection) and As is countable for every s 2 S; then[

s2SAs D

[

n2N

A f .n/ is countable. (Briefly,countable union of countable sets is countable.)Product Theorem. If A, B are countable, then AB D f.a; b/: a 2 A; b 2 Bg is countable. In fact, if A1; A2; : : : ; Anare countable, then A1 A2 An is countable (by mathematical induction).

(Sketch of Reasons. For the countable subset theorem, if B is countable, then we can list the elements of B and tocount the elements of A, we can skip over those elements of B that are not in A: For the countable union theorem, ifwe list the elements of A1 in the first row, the elements of A2 in the second row, : : : ; then we can count all the elementsby using the diagonal counting scheme. As for the product theorem, we can imitate the example of N N and alsouse the diagonal counting scheme.)

Examples. (5)QD 1[nD1

Sn;where Sn Dnm

n: m 2Z

o

:For every n 2 N; the function fn : Z! Sn given by fn.m/ D mn

is a bijection (with f 1n

mn

D m), so Sn is countable by the bijection theorem. Therefore, Q is countable by thecountable union theorem. (Then subsets ofQ likeZn f0g;N[ f0g;Q\ .0; 1/ are also countable.)

(6) Rn Q is uncountable. (In fact, if A is uncountable and B is countable, then A n B is uncountable as A n Bcountable implies .A \ B/ [ .A n B/ D A countable by the countable union theorem, which is a contradiction).

(7) C D fx C iy : x; y 2 Rg containsRandR is uncountable, so by the countable subset theorem, C is uncountable.(8) Show that the set A D frpm : m 2 N; r 2 .0; 1/g is uncountable, but the set B D frpm : m 2 N; r 2 Q\ .0; 1/g

is countable.

Solution. Taking m D 1; we see that .0; 1/ A: Since .0; 1/ is uncountable, A is uncountable. Next we willobserve that B D

[

m2N

Bm; where Bm D frp

m : r 2 Q \ .0; 1/g D[

r2Q\.0;1/

frp

mg for each m 2 N: Since

Q\ .0; 1/ is countable and frp

mg has 1 element for every r 2 Q\ .0; 1/; Bm is countable by the countable uniontheorem. Finally, since N is countable and Bm is countable for every m 2 N; B is countable by the countableunion theorem.

(9) Show that the set L of all lines with equation y D mx C b; where m; b 2 Q; is countable.Solution. Note that for each pair m; b of rational numbers, there is a unique line y D mx C b in the set L : So thefunction f : QQ! L defined by letting f .m; b/ be the line y D mx C b (with f 1 sending the line back to.m; b/) is a bijection. SinceQQ is countable by the product theorem, so the set L is countable by the bijectiontheorem.

(10) Show that if An D f0; 1g for every n 2 N; then A1 A2 A3 is uncountable. (In particular, this shows thatthe product theorem is not true for infintely many countable sets.)Solution. Assume A1 A2 A3 D f.a1; a2; a3; : : :/ : each ai D 0 or 1g is countable and f : N !A1 A2 A3 is a bijection. Following example (4), we can change the n-th coordinate of f .n/ (from 0 to1 or from 1 to 0) to produce an element of A1 A2 A3 not equal to any f .n/; which is a contradiction.So it must be uncountable.

(11) Show that the power set P.N/ of all subsets of N is uncountable.Solution. As in example (10), let An D f0; 1g for every n 2 N: Define g : P.N/ ! A1 A2 A3 byg.S/ D .a1; a2; a3; : : :/; where am D

1 if m 2 S0 if m 62 S : (For example, g.f1; 3; 5; : : :g/ D .1; 0; 1; 0; 1; : : :/:) Note

g has the inverse function g1

.a1; a2; a3; : : :/

D fm : am D 1g: Hence g is a bijection. Since A1 A2 A3 is uncountable, so P.N/ is uncountable by the bijection theorem.

10

(12) Show that the set S of all nonconstant polynomials with integer coefficients is countable.Solution. For n 2 N; the set of Sn of all polynomials of degree n with integer coefficients is countable becausethe function f : Sn ! .Zn f0g/ Z Zdefined by f .anxn C an1xn1 C C a0/ D .an; an1; : : : ; a0/is a bijection and .Zn f0g/ Z Zis countable by the product theorem. So, S D

[

n2N

Sn is countable by

the countable union theorem.(13) Show that there exists a real number, which is not a root of any nonconstant polynomial with integer coefficients.

Solution. For every nonconstant polynomial f with integer coefficients, let R f denotes the set of roots of f: ThenR f has at most (deg f ) elements, hence R f is countable. Let S be the set of all nonconstant polynomials withinteger coefficients, which is countable by the last example. Then

[

f 2SR f is the set of all roots of nonconstant

polynomials with integer coefficients. It is countable by the countable union theorem. Since R is uncountable,Rn

[

f 2SR f is uncountable by the fact in example (6). So there exist uncountably many real numbers, which are

not roots of any nonconstant polynomial with integer coefficients.

Remarks. Any number which is a root of a nonconstant polynomial with integer coefficients is called an algebraicnumber. A number which is not a root of any nonconstant polynomial with integer coefficients is called a transcendentalnumber. transcendental numbers? If so, are there finitely many or countably many such numbers? Since every rationalnumber ab is the root of the polynomial bx a; every rational number is algebraic. There are irrational numbers like

p

2; which are algebraic because they are the roots of x2 2: Using the identity cos 3 D 4 cos3 3 cos ; theirrational number cos 20 is easily seen to be algebraic as it is a root of 8x3 6x 1: Example (13) showed that thereare only countably many algebraic numbers and there are uncountably many transcendental real numbers. In a numbertheory course, it will be shown that and e are transcendental.Theorem.(1) (Injection Theorem) Let f : A ! B be injective. If B is countable, then A is countable. (Taking contrapositive,

if A is uncountable, then B is uncountable.)(2) (Surjection Theorem) Let g : A ! B be surjective. If A is countable, then B is countable. (Taking contrapositive,

if B is uncountable, then A is uncountable.)(Reasons. For the first statement, observe that the function h : A ! f .A/ defined by h.x/ D f .x/ is injective(because f is injective) and surjective (because h.A/ D f .A/). So h is a bijection. If B is countable, then f .A/ iscountable by the countable subset theorem, which implies A is countable by the bijection theorem.

For the second statement, observe that B D g.A/ D[

x2Afg.x/g: If A is countable, then it is a countable union of

countable sets. By the countable union theorem, B is countable.)Examples. (14) ShowQ is countable by using the injection theorem.

Solution. Define f : Q!Z N by f .x/ D .m; n/; where m=n is the reduced fraction form of x : Then f isinjective because f .x/ D f .x 0/ D .m; n/ implies x D m=n D x 0: SinceZN is countable by product theorem,soQ is countable by the injection theorem.

(15) Let A1 be uncountable and A2; : : : ; An be nonempty sets. Show that A1 A2 An is uncountable.Solution. Define g : A1 A2 An ! A1 by g.x1; x2; : : : ; xn/ D x1: Since A2; : : : ; An are nonempty, leta2 2 A2; : : : ; an 2 An : Then for every a1 2 A1; we have g.a1; a2; : : : ; an/ D a1 so that g is surjective. Since A1is uncountable, by the surjection theorem, A1 A2 An is uncountable.The following is a famous statement in mathematics.

Continuum Hypothesis. If S is uncountable, then there exists at least one injective function f : R! S; i.e. everyuncountable set has at least as many elements as the real numbers.

In 1940, Kurt Godel showed that the opposite statement would not lead to any contradiction. In 1966, Paul Cohenwon the Fields Medal for showing the statement also would not lead to any contradiction. So proof by contradictionmay not be applied to every statement.

11

Chapter 4. Series

Definitions. A series is the summation of a countable set of numbers in a specific order. If there are finitely manynumbers, then the series is a finite series, otherwise it is an infinite series. The numbers are called terms. The sum ofthe first n terms is called the n-th partial sum of the series.

An infinite series is of the form a1|{z}

1st term

C a2|{z}

2nd term

C a3|{z}

3rd term

C : : : or we may write it as1

X

kD1ak:

The first partial sum is S1 D a1. The second partial sum is S2 D a1Ca2. The nth partial sum is Sn D a1Ca2C: : :Can .

Series are used frequently in science and engineering to solve problems or approximate solutions. (E.g. trigono-metric or logarithm tables were computed using series in the old days.)

Examples.(1) 1C 12C 14C 18C 116C: : : D? (Sn D 1C 12C 14C: : :C 12n D 2 12n ; 1C 12C 14C 18C 116C: : : D limn!1

212n

D 2.)We say the series converges to 2, which is called the sum of the series.

(2) 1C 1C 1C 1C 1C 1C : : : D 1 (Sn D 1C 1C : : : C 1| {z }

n

D n, limn!1

Sn D 1.) We say the series diverges (to1).

(3) 1 1C 1 1C 1 1C 1 1C : : :. (Sn Dn1 if n is odd

0 if n is even , limn!1 Sn doesnt exist.) We say the series diverges.

Definitions. A series1

X

kD1ak D a1Ca2Ca3C : : : converges to a number S iff lim

n!1.a1Ca2C : : :Can/ D lim

n!1Sn D S:

In that case, we may write1

X

kD1ak D S and say S is the sum of the series. A series diverges to1 iff the partial sum Sn

tends to infinity as n tends to infinity. A series diverges iff it does not converge to any number.

Remarks. (1) For every series1

X

kD1ak , there is a sequence (of partial sums) fSng. Conversely, if the partial sum sequence

fSng is given, we can find the terms an as follows: a1 D S1, a2 D S2 S1, : : :, ak D Sk Sk1 for k > 1. Then

a1C : : :C an D S1C .S2 S1/C : : :C .Sn Sn1/ D Sn. So fSng is the partial sum sequence of1

X

kD1ak . Conceptually,

series and sequences are equivalent. So to study series, we can use facts about sequences.

(2) Let N be a positive integer.1

X

kD1ak converges to A if and only if

1

X

kDNak converges to B D A .a1 C C aN1/

because

B D limn!1

.aN C C an/ D limn!1

.a1 C a2 C C an/ .a1 C C aN1/ D A .a1 C C aN1/:

So to see if a series converges, we may ignore finitely many terms.

Theorem. If1

X

kD1ak converges to A and

1

X

kD1bk converges to B; then

1

X

kD1.ak C bk/ D AC B D

1

X

kD1ak C

1

X

kD1bk;

1

X

kD1.ak bk/ D A B D

1

X

kD1ak

1

X

kD1bk;

1

X

kD1cak D cA D c

1

X

kD1ak

for any constant c:

For simple series such as geometric or telescoping series, we can find their sums.

12

Theorem (Geometric Series Test). We have

1

X

kD0rk D lim

n!1.1C r C r2 C : : : C rn/ D lim

n!1

1 rnC1

1 rD

( 11 r

if jrj < 1doesnt exist otherwise

:

Example. 0:999 D 910

C

9100

C

91000

C D

910

.

11 110

/ D 1 D 1:000 : So 1 has two decimal representa-

tions!

Theorem (Telescoping Series Test). We have1

X

kD1.bk bkC1/ D lim

n!1

.b1 b2/ C .b2 b3/ C C .bn bnC1/

D limn!1

.b1 bnC1/ D b1 limn!1

bnC1 converges if and only if limn!1

bn is a number.

Examples. (1)1

X

kD1

1k.k C 1/

D

1

X

kD1.

1k

1k C 1

/ D

112

C

12

13

C

13

14

C D 1 limn!1

1n C 1

D 1:

(2)1

X

kD1.51=k 51=.kC1// D .5

p

5/C .p

5 3p

5/ C D 5 limk!1

51=.kC1/ D 5 50 D 4:

If a series is not geometric or telescoping, we can only determine if it converges or diverges. This can be donemost of the time by applying some standard tests. If the series converges, it may be extremely difficult to find the sum!

Theorem (Term Test). If1

X

kD1ak converges, then lim

k!1ak D 0. (If lim

k!1ak 6D 0, then the series

1

X

kD1ak diverges.) If

limk!1

ak D 0, the series1

X

kD1ak may or may not converge.

(Reason. Suppose1

X

kD1ak converges to S: Then lim

n!1Sn D S and lim

k!1ak D lim

k!1.Sk Sk1/ D S S D 0.)

Term test is only good for series that are suspected to be divergent!

Examples. (1) 1C 1C 1C 1C : : :. Here ak D 1 for all k, so limk!1 ak D 1. Series diverges.

(2)1

X

kD1cos.

1k/ D cos 1C cos

12C cos

13C : : : diverges because lim

k!1cos.

1k/ D cos 0 D 1 6D 0.

(3)1

X

kD1cos k D cos 1 C cos 2 C cos 3 C : : : diverges because lim

k!1cos k 6D 0: (Otherwise, lim

k!1cos k D 0: Then

limk!1

j sin kj D limk!1

p

1 cos2 k D 1 and 0 D limk!1

j cos.k C 1/j D limk!1

j cos k cos 1 sin k sin 1j D sin 1 6D 0; acontradiction.)

(4) 1 12 C 14 18 C : : :. Here ak D . 12 /k1 for all k, so limk!1 ak D 0: (Term test doesnt apply!) Series convergesby the geometric series test.

(5) 1 C 12C

12

| {z }

2 times

C

14C

14C

14C

14

| {z }

4 times

C

18C : : : C

18

| {z }

8 times

C : : :. We have limk!1

ak D 0: (Term test doesnt apply.) Series

diverges to1 because S1 S2 S3 and S2n1 D n has limit1:

13

For a nonnegative series1

X

kD1ak (i.e. ak 0 for every k), we have S1 S2 S3 : : : and lim

n!1Sn must exist as a

number or equal toC1. So either1

X

kD1ak converges to a number or

1

X

kD1ak diverges toC1. (In short, either

1

X

kD1ak D S

or1

X

kD1ak D C1.) For nonnegative series, we have the following tests.

Theorem (Integral Test). Let f : [1;C1/! Rdecrease to 0 as x !C1. Then1

X

kD1f .k/ converges if and only if

Z

1

1f .x/ dx 1;Z

1

1

1x p

dx D .ln x/j11 D 1 if p D 1 andZ

1

1

1x p

dx D xpC1

p C 1

1

1D 1 if p < 1, the integral test gives the conclusion.)

Remarks. For even positive integer p; the value of .p/ was computed by Euler back in 1736. He got

.2/ D

2

6 ; .4/ D

4

90 ; : : : ; .2n/ D .1/nC1 .2/2n B2n

2.2n/!; : : : ;

14

where B0 D 1 and .k C 1/Bk D k1X

mD0

k C 1m

Bm for k 1: The values of .3/; .5/; : : : are unknown. Only in the

1980s, R. Apery was able to show .3/ was irrational.

Theorem (Comparison Test). Given vk uk 0 for every k. If1

X

kD1vk converges, then

1

X

kD1uk converges. If

1

X

kD1uk

diverges, then1

X

kD1vk diverges.

(Reason. vk uk 0 )1

X

kD1vk

1

X

kD1uk 0. If

1

X

kD1vk is a number, then

1

X

kD1uk is a number. If

1

X

kD1uk D C1, then

1

X

kD1vk D C1.)

Theorem (Limit Comparison Test). Given uk , vk > 0 for every k: If limk!1vk

ukis a positive number L ; then either (both

1

X

kD1uk and

1

X

kD1vk converge) or (both diverge to C1). If lim

k!1

vk

ukD 0; then

1

X

kD1uk converges )

1

X

kD1vk converges. If

limk!1

vk

ukD 1; then

1

X

kD1uk diverges )

1

X

kD1vk diverges.

(Sketch of Reason. For k large, vkuk

L . For L > 0;P

vk P

Luk D LP

uk. If one series converges, then theother also converges. If one diverges (toC1), so does the other. For L D 0; vk < uk eventually. For L D 1; vk > ukeventually. So the last two statements follow from the comparison test.)

Examples. Consider the convergence or divergence of the following series:

(1)1

X

kD1

1k2

cos1

k (2)

1

X

kD2

3k

k2 1(3)

1

X

kD1

p

k C 1k2 C 5k

(4)1

X

kD1sin1

k

.

Solutions. (1) Since 0 < 1k2

cos1

k

>

>

>

>

>

>

>

>

>

>

>

>

>

>

:

< 1 )1

X

kD1ak converges absolutely

D 1 )1

X

kD1ak may converge

e:g:1

X

kD1

1k2

!

or diverge

e:g:1

X

kD1

1k

!

> 1 )1

X

kD1ak diverges

:

16

(Sketch of reason. Let r D limk!1

akC1

ak

, then for k large,

akC1

ak

,

akC2

akC1

, : : :,

akCn

akCn1

r, so jakCnj jak jrn and

jakj C jakC1j C jakC2j C : : : jakj.1 C r C r2 C r3 C : : :/ which converges if r < 1 by the geometric series test andP

akCn P

akrn diverges if r > 1 by the term test.)

Theorem (Root Test). If limk!1

kp

jakj exists, then

limk!1

kpjak j

8

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

:

< 1 )1

X

kD1ak converges absolutely

D 1 )1

X

kD1ak may converge

e:g:1

X

kD1

1k2

!

or diverge

e:g:1

X

kD1

1k

!

> 1 )1

X

kD1ak diverges

:

(Sketch of reason. Let r D limk!1

kpjakj, then for k large, k

p

jakj r: So jak j rk ,P

jak j P

rk .)

Examples. Consider the convergence or divergence of the following series:

(1)1

X

kD1

13k 2k

(2)1

X

kD1

k!kk

.

Solutions. (1) Since limk!1

13kC12kC1

13k2k

D limk!1

3k 2k

3kC1 2kC1D lim

k!1

13 .

23/

k 13

1 . 23 /kC1D

13< 1; by the ratio test,

1

X

kD1

13k 2k

converges. Alternatively, since limk!1

kr

13k 2k

D limk!1

1kp3k 2k

D limk!1

1

3 kq

1 . 23 /kD

13< 1, by the root

test,1

X

kD1

13k 2k

converges.

(2) Since limk!1

.k C 1/!.k C 1/kC1

kk

k!D lim

k!1

1.1 C 1k /k

D

1e< 1; by the ratio test,

1

X

kD1

k!kk

converges.

Remarks. You may have observed that in example (1), the limit you got for applying the root test was the same as thelimit you got for applying the ratio test. This was not an accident!

Theorem. If ak > 0 for all k and limk!1

akC1

akD r 2 R; then lim

k!1kpak D r: (This implies that the root test can be

applied to more series than the ratio test.)

Examples. (1) Let ak D k; then limk!1

akC1

akD lim

k!1

k C 1k

D 1: So, limk!1

kpk D 1:

(2) Let ak D k!kk ; then limk!1akC1

akD

1e

as above. So limk!1

kpak D limk!1

kpk!k

D

1e; i.e. when k is large, k!

ke

k;

which is a simple version of what is called Stirlings formula. It is useful in estimating n! when n is large. Forexample, since log10

100e

1:566; so 100e

101:566; then we get 100! 10156:6; which has about 157 digits.

Theorem (Summation by Parts). Let Sj DjX

kD1ak D a1 C a2 C : : : C aj and 1bk D

bkC1 bk.k C 1/ k

D bkC1 bk; then

nX

kD1akbk D Snbn

n1X

kD1Sk1bk :

17

(Reason. Note a1 D S1 and ak D Sk Sk1 for k > 1. So,nX

kD1akbk D S1b1 C .S2 S1/b2 C : : : C .Sn Sn1/bn

D Snbn S1.b2 b1/ : : : Sn1.bn bn1/:/

Example. Show that1

X

kD1

sin kk

converges.

Let ak D sin k and bk D1k: Using the identity sin m sin

12D

12

cos.m 12/ cos.m C

12/

; we have

Sk D sin 1C sin 2C C sin k Dcos 12 cos.k C

12 /

2 sin 12:

This implies jSkj 1

sin.1=2/for every k: Applying summation by parts and noting that lim

n!1

SnnD 0; we get

1

X

kD1

sin kk

D limn!1

nX

kD1

sin kk

D limn!1

Snn

n1X

kD1Sk

1k C 1

1k

!

D

1

X

kD1Sk

1k

1k C 1

:

Now1

X

kD1

Sk

1k

1k C 1

1sin.1=2/

1

X

kD1

1k

1k C 1

D

1sin.1=2/

by the telescoping series test. So by the

absolute convergence test,1

X

kD1

sin kk

D

1

X

kD1Sk

1k

1k 1

converges.

Inserting Parentheses and Rearrangements of Series.

Definition. We say1

X

kD1bk is obtained from

1

X

kD1ak by inserting parentheses iff there is a strictly increasing function

p : N[f0g ! N[f0g such that p.0/ D 0; b1 D a1C Cap.1/; b2 D ap.1/C1C Cap.2/; b3 D ap.2/C1C Cap.3/; : : : :(Note bn is the sum of kn D p.n/ p.n 1/ terms.)

Grouping Theorem. Let1

X

kD1bk be obtained from

1

X

kD1ak by inserting parentheses. If

1

X

kD1ak converges to s; then

1

X

kD1bk

will converge to s: Next, if limn!1

an D 0; kn is bounded and1

X

kD1bk converges to s; then

1

X

kD1ak will converge to s:

(Reason. Let sn DnX

kD1ak and tn D

nX

kD1bk: For the first part,

1

X

kD1bk D lim

n!1tn D lim

n!1

p.n/X

kD1ak D s: For the second part,

let p.n/ p.n 1/ be bounded by M: For a positive integer j; let p.i/ j < p.i C 1/: For r D 1; 2; : : : ; M; definecr; j D

ap.i/Cr if p.i/ C r j0 if p.i/ C r > j :Then

1

X

kD1ak D limj!1 sj D limi!1 ti C limj!1.c1; j C C cM; j / D s C 0C C 0 D s:)

Examples. (1) Since1

X

kD1

12kD

12C

14C

18C

116 C converges to 1, so by the theorem,

12C

14C

18

C

116

C

132C

164

C

1128

C

1256

C

1512

C

11024

C D 1:

18

(2) .11/C .11/C converges to 0, but 11C11C diverges by term test. So limn!1

an D 0 is important.

Also, .1 1/C1

2C

12

12

12

C

13C

13C

13

13

13

13

C converges to 0. However, the serieswithout parentheses diverges (as Sn2 D 1 and Sn2Cn D 0) even though the terms have limit 0. So kn bounded isimportant.

(3) Since

112

C

13

14

C D

1

X

jD1

12 j 1

12 j

D

1

X

jD1

12 j .2 j 1/ converges (by the limit compari-

son test with1

X

jD1

1j 2 ), so by the theorem, 1

12C

13

14C D

1

X

kD1

.1/kC1

kconverges to the same sum.

Definition.1

X

kD1bk is a rearrangement of

1

X

kD1ak iff there is a bijection : N! N such that bk D a.k/:

Example. Given ln 2 D 112C

13

14C

15

16 C : : : (which converges conditionally). Consider the rearrangement

1C13

| {z }

2C

12

|{z}

1

C

15 C

17

| {z }

2C

14

|{z}

1

C

19 C

111

| {z }

2C

16

|{z}

1

C : : :. Observe that

.1 12 / C .13

14 / C .

15

16 / C .

17

18/ C : : : D ln 2

C

12

14 C

16

18 : : : D

12 ln 2

1 C . 13 12 / C

15 C .

17

14/ C : : : D

32 ln 2:

Riemanns Rearrangement Theorem. Let ak 2 Rand1

X

kD1ak converge conditionally. For any x 2 Ror x D 1,

there is a rearrangement1

X

kD1a.k/ of

1

X

kD1ak such that

1

X

kD1a.k/ D x :

(Sketch of reason. Let pk D

ak if ak 00 if ak < 0

and qk D

0 if ak 0jak j if ak < 0

: Then ak D pk qk and jakj D pk C qk :

Now both1

X

kD1pk;

1

X

kD1qk must diverge toC1: (If both converges, then their sum

1

X

kD1jakjwill be finite, a contradiction.

If one converges and the other diverges to C1; then1

X

kD1ak D

1

X

kD1pk

1

X

kD1qk will diverges to 1; a contradiction

also.) Let un; vn be sequences of real numbers having limits x and un < vn; v1 > 0: Now let P1; P2; : : : be thenonnegative terms of

1

X

kD1ak in the order they occur and Q1; Q2; : : : be the absolute value of the negative terms in

the order they occur. Since1

X

kD1Pk;

1

X

kD1Qk differ from

1

X

kD1pk;

1

X

kD1qk only by zero terms, they also diverges to C1:

Let m1; k1 be the smallest integers such that P1 C C Pm1 > v1 and P1 C C Pm1 Q1 Qk1 < u1:Let m2; k2 be the smallest integers such that P1 C C Pm1 Q1 Qk1 C Pm1C1 C C Pm2 > v2 andP1C C Pm1 Q1 Qk1 C Pm1C1C C Pm2 Qk1C1 Qk2 < u2 and continue this way. This is possiblesince the sums of Pk and Qk areC1: Now if sn; tn are the partial sums of this series P1C CPm1Q1 Qk1C whose last terms are Pmn ; Qkn ; respectively, then jsn vnj Pmn and jtn unj Qkn by the choices of mn; kn : SincePn; Qn have limit 0, so sn; tn must have limit x : As all other partial sums are squeezed by sn and tn; the series weconstructed must have limit x :)

Dirichlets Rearrangement Theorem. If ak 2 Rand1

X

kD1ak converges absolutely, then every rearrangement

1

X

kD1a.k/

converges to the same sum as1

X

kD1ak.

19

(Reason. Define pk; qk as in the last proof. Since pk; qk jak j;1

X

kD1pk;

1

X

kD1qk converge, say to p and q; respectively.

Since a.k/ D p.k/q.k/; we may view

1

X

kD1p.k/ as a rearrangement of the nonnegative terms of

1

X

kD1ak and inserting

zeros where a.k/ < 0: For any positive integer m; the partial sum sm D

mX

kD1p.k/

1

X

kD1pk D p: Since pk 0; the

partial sum sm is also increasing, hence1

X

kD1p.k/ converges. Now, for every positive integer n;

nX

kD1pk

1

X

kD1p.k/ p:

As n !1; we get1

X

kD1p.k/ D p: Similarly,

1

X

kD1q.k/ D q: Then

1

X

kD1a.k/ D p q D

1

X

kD1ak :)

Example.1

X

kD1.

12/

kD

12C

122

123C

124

125C : : : converges (absolutely) to

12

1 . 12 /D

13

.

12C

122C

124

123

| {z }

2 terms

C

128

127C

126

125

| {z }

4 terms

C

1216

1215

C

1214

1213

C

1212

1211

C

1210

129

| {z }

8 terms

C : : :

is a rearrangement of1

X

kD1.

12/

k, so it also converges to

13

.

Remarks. As a consequence of the rearrangement theorem, the sum of a nonnegative series is the same no matter howthe terms are rearranged.

Complex Series

Complex numbers S1; S2; S3; : : : with Sn D un C ivn are said to have limit limn!1

Sn D u C iv iff limn!1

un D u

and limn!1

vn D v: A complex series is a series where the terms are complex numbers. The definitions of convergent,absolutely convergent and conditional convergent are the same. The remarks and the basic properties following thedefinitions of convergent and divergent series are also true for complex series.

The geometric series test, telescoping series test, term test, absolute convergence test, ratio test and root test are

also true for complex series. For zk D xk C iyk; we have1

X

kD1zk converges to z D x C iy if and only if

1

X

kD1xk converges

to x and1

X

kD1yk converges to y: So complex series can be reduced to real series for study if necessary.

Examples. (1) Note limn!1

in 6D 0 (otherwise 0 D limn!1

jin j D limn!1

1 is a contradiction). So1

X

kD1ik diverges by term test.

(2) If jzj 1; then

zk

k2

1k2

and1

X

kD1

1k2

converges by p-test implies1

X

kD1

zk

k2converges absolutely. However, if

jzj > 1; then limk!1

zkC1

.k C 1/2k2

zk

D limk!1

k2

.k C 1/2jzj D jzj > 1 implies

1

X

kD1

zk

k2diverges by the ratio test.

20

Chapter 5. Real Numbers

Decimal representations and points on a line are possible ways of introducing real numbers, but they are not tooconvenient for proving many theorems. Instead we will introduce real numbers by its important properties.

Axiomatic Formulation. There exists a setR (called real numbers) satisfying the following four axioms:(1) (Field Axiom) R is a field (i.e. Rhas two operationsC and such that for any a, b, c 2 R,

(i) aCb, a b 2 R, (ii) aCb D bCa, a b D b a, (iii) .aCb/Cc D aC .bCc/, .a b/ c D a .b c/,(iv) there are unique elements 0, 1 2 Rwith 1 6D 0 such that a C 0 D a, a 1 D a,(v) there is a unique element a 2 Rsuch that aC .a/ D 0; if a 6D 0; then there is a unique element a1 such

that a .a1/ D 1:(vi) a .b C c/ D a b C a c.)

(This axiom allows us to do algebra with equations. Define a b to mean a C .b/I ab to mean a bI ab tomean a .b1/: Also, define 2 D 1C 1; 3 D 2C 1; : : : :)

(2) (Order Axiom) Rhas an (ordering) relation < such that for any a, b 2 R(i) exactly one of the following a < b, a D b, b < a is true,

(ii) if a < b, b < c, then a < c,(iii) if a < b, then a C c < b C c,(iv) if a < b and 0 < c, then ac < bc.

(This axiom allows us to work with inequalities. For example, using (ii) and (iii), we can see that if a < band c < d; then aC c < bC d because aC c < bC c < bC d: Also, we can get 0 < 1 (for otherwise 1 < 0would imply by (iii) that 0 D 1 C .1/ < 0C .1/ D 1; which implies by (iv) that 0 < .1/.1/ D 1;a contradiction). Now define a > b to mean b < aI a b to mean a < b or a D bI etc. Also, defineclosed interval [a; b] D fx : a x bgI open interval .a; b/ D fx : a < x < bgI etc. Part (i) of theorder axiom implies any two real numbers can be compared. We define max.a1; : : : ; an/ to be the maximumof a1; : : : ; an and similarly for minimum. Also, define jxj D max.x;x/: Then x jxj and x jxj;i.e. jxj x jxj: Next jxj a if and only if x a and x a; i.e. a x a: Finally, addingjxj x jxj andjyj y jyj; we getjxj jyj x C y jxj C jyj; which is the triangle inequalityjx C yj jxj C jyj:)

(3) (Well-ordering Axiom) N D f1; 2; 3; : : :g is well-ordered (i.e. for any nonempty subset S of N, there is m 2 Ssuch that m x for all x 2 S. This m is the least element (or the minimum) of S).

(This axiom allows us to formulate the principle of mathematical induction later.)

Definitions. For a nonempty subset S of R, S is bounded above iff there is some M 2 R such that x M for allx 2 S. Such an M is called an upper bound of S. The supremum or least upper bound of S (denoted by sup S or lub S)is an upper bound QM of S such that QM M for all upper bounds M of S.

(4) (Completeness Axiom) Every nonempty subset ofRwhich is bounded above has a supremum inR.(This axiom allows us to prove results that have to do with the existence of certain numbers with specificproperties, as in the intermediate value theorem.)

Examples. (1) For S D 1n

: n 2 N

D

1; 12 ;13 ; : : :

; the upper bounds of S are all M 1: So sup S D 1 2 S:(2) For S D fx 2 R: x < 0g, the upper bounds of S are all M 0: So sup S D 0 62 S:

Definitions. N D f1; 2; 3; 4; : : :g is the natural numbers (or positive integers),ZD f: : : ;3;2;1; 0; 1; 2; 3; : : :gis the integers, Q D fm

n: m 2 Zand n 2 Ng is the rational numbers and Rn Q D fx 2 R: x 62 Qg is the irrational

numbers.

Remarks (Exercises). The first three axioms are also true if R is replaced by Q: However, the completeness axiomis false for Q: For example, S D fx : x 2 Q; x > 0; x2 < 2g is bounded above by 3 in Q; but it does not have asupremum inQ:

21

As above, we define S to be bounded below if there is some m 2 R such that m x for all x 2 S. Such an m iscalled a lower bound of S. The infimum or greatest lower bound (denoted by inf S or glb S) of S is a lower bound Qmof S such that m Qm for all lower bounds m of S.

lower boundsare here

upper boundsare hereS

inf S sup S

Remarks (Exercises). (1) Let B D fx : x 2 Bg: (This is the reflection of B about 0.) If B is bounded below,thenB is bounded above and inf B D sup.B/: Similarly, if B is bounded above, thenB is bounded below andsup B D inf.B/: From these and the completeness axiom, we get the following statement.

(Completeness Axiom for Infimum) Every nonempty subset ofRwhich is bounded below has an infimum inR.

( )-B B

0inf(-B) sup(-B) supBinfA supAAinfB

(2) For a set B, if it is bounded above and c 0; then let cB D fcx : x 2 Bg: (This is the scaling of B by a factor ofc:) We have sup cB D c sup B: If ; 6D A B, then inf B inf A whenever B is bounded below and sup A sup Bwhenever B is bounded above.

infB supB sup(c+B)inf(c+B)

c units

B c+B

(3) For c 2 R; let cC B D fc C x : x 2 Bg: (This is a translation of B by c units.) It follows that B has a supremumif and only if c C B has a supremum, in which case sup.c C B/ D c C sup B: The infimum statement is similar, i.e.inf.c C B/ D cC inf B: More generally, if A and B are bounded, then letting A C B D fx C y : x 2 A; y 2 Bg; wehave sup.A C B/ D sup A C sup B and inf.A C B/ D inf AC inf B:

If S is bounded above and below, then S is bounded. Note sup S, inf S may or may not be in S. Also, if Sis bounded, then for all x 2 S; jxj D max.x;x/ max.sup S; inf S/ (because x sup S and x inf S:)Conversely, if there is c 2 R such that for all x 2 S; jxj c; then c x c so that S is bounded (above by c andbelow by c:)

Simple Consequences of the Axioms.

Theorem (Infinitesimal Principle). For x, y 2 R, x < yC " for all " > 0 if and only if x y: (Similarly, y " < xfor all " > 0 if and only if y x.)Proof. If x y, then for all " > 0, x y D y C 0 < y C " by (iv) of the field axiom and (iii) of the order axiom.

Conversely, if x < y C " for all " > 0; then assuming x > y; we get x y > 0 by (iii) of the order axiom. Let"0 D x y; then x D y C "0: Since "0 > 0; we also have x < y C "0: These contradict (i) of the order axiom. Sox y: The other statement follows from the first statement since y " < x is the same as y < x C ":

Remarks. Taking y D 0, we see that jxj < " for all " > 0 if and only if x D 0. This is used when it is difficult toshow two expressions a; b are equal, but it may be easier to show ja bj < " for every " > 0:

Theorem (Mathematical Induction). For every n 2 N, A.n/ is a (true or false) statement such that A.1/ is true andfor every k 2 N, A.k/ is true implies A.k C 1/ is also true. Then A.n/ is true for all n 2 N.

22

Proof. Suppose A.n/ is false for some n 2 N. Then S D fn 2 N: A.n/ is falseg is a nonempty subset of N. By thewell-ordering axiom, there is a least element m in S: Then A.m/ is false. Also, if A.n/ is false, then m n. Takingcontrapositive, this means that if n < m; then A.n/ is true.

Now A.1/ is true, so m 6D 1 and m 2 N imply m 2. So m 1 1: Let k D m 1 2 N; then k D m 1 < mimplies A.k/ is true. By hypothesis, A.k C 1/ D A.m/ is true, a contradiction.

Theorem (Supremum Property). If a set S has a supremum in R and " > 0; then there is x 2 S such thatsup S " < x sup S.Proof. Since sup S " < sup S, sup S " is not an upper bound of S. Then there is x 2 S such that sup S " < x .Since sup S is an upper bound of S, x sup S. Therefore sup S " < x sup S.

Theorem (Infimum Property). If a set S has an infimum inRand " > 0; then there is x 2 S such that inf S C " >x inf S.Proof. Since inf SC " > inf S, inf SC " is not a lower bound of S. Then there is x 2 S such that inf SC " > x . Sinceinf S is a lower bound of S, x inf S. Therefore inf S C " > x inf S.

Theorem (Archimedean Principle). For any x 2 R, there is n 2 N such that n > x .Proof. Assume there exists x 2 R such that for all n 2 N; we have n x : Then N D fn: n 2 Ng has an upperbound x . By the completeness axiom, N has a supremum in R. By the supremum property, there is n 2 N such thatsupN 1 < n, which yields the contradiction supN < n C 1 2 N.

Question. How isQ contained inR? How isRnQ contained inR?Below we will show thatQ is dense inRin the sense that between any two distinct real numbers x; y, no matter

how close, there is a rational number. Similarly,RnQ is dense inR: First we need a lemma.

Lemma. For every x 2 R; there exists a least integer greater than or equal to x : (In computer science, this is calledthe ceiling of x and is denoted by dxe:) Similarly, there exists a greatest integer less than or equal to x : (This is denotedby [x]: In computer science, this is also called the floor of x and is denoted by bxc:)Proof. By the Archimedean principle, there is n 2 N such that n > jxj: Thenn < x < n: By (iii) of the order axiom,0 < x C n < 2n: The set S D fk 2 N : k x C ng is a nonempty subset of N because 2n 2 S: By the well-orderingaxiom, there is a least positive integer m x C n: Then m n is the least integer greater than or equal to x : So theceiling of every real number always exist.

Next, to find the floor of x; let k be the least integer greater than or equal to x; then k is the greatest integerless than or equal to x :

Theorem (Density of Rational Numbers). If x < y, then there is mn2 Q such that x < m

n< y.

Proof. By the Archimedean principle, there is n 2 N such that n > 1=.y x/: So nynx > 1 and hence nxC1 < ny:Let m D [nx]C 1; then m 1 D [nx] nx < [nx]C 1 D m: So nx < m nx C 1 < ny; i.e. x < m

n< y:

Theorem (Density of Irrational Numbers). If x < y, then there is w 2 RnQ such that x < w < y.Proof. Let w0 2 Rn Q. By the density of rational numbers, there is mn 2 Q such that

xjw0 j

0: By the Archimedeanprinciple, there is n 2 N such that n > 1=t : Then t > 1=n 2 S; a contradiction to t being a lower bound of S: So0 is the greatest lower bound of S:

23

(3) Let S D [2; 6/ \Q: Since 2 x < 6 for every x 2 S; S has 2 as a lower bound and 6 as an upper bound. Wewill show inf S D 2 and sup S D 6: (Note 2 2 S and 6 62 S.) Since 2 2 S; so every lower bound t satisfy t 2:Therefore inf S D 2: For supremum, assume there is an upper bound u < 6: Since 2 2 S; so 2 u: By thedensity of rational numbers, there is a r 2 Q such that u < r < 6: Then r 2 [2; 6/\QD S: As u < r contradictsu being an upper bound of S; so every upper bound u 6: Therefore, sup S D 6:

24

Chapter 6. Limits

Limit is the most important concept in analysis. We will first discuss limits of sequences, then limits of functions.

Definitions. An (infinite) sequence in a set S (e.g. S D Ror S D [0; 1]) is a list x1; x2; x3; : : : of elements of S in aspecific order. Briefly it is denoted by fxng: (Mathematically it may be viewed as a function x :N! S with x.n/ D xnfor n 2 N:) We say the sequence fxng is bounded above iff the set fx1; x2; x3; : : :g is bounded above. (Bounded belowand bounded sequences are defined similarly.) We will also write supfxng for the supremum of the set fx1; x2; x3; : : :gand inffxng for the infimum of the set fx1; x2; x3; : : :g:

CAUTION: Since we seldom talk about a set with one element from now on, so notations like fxng will denotesequences unless explicitly stated otherwise.

For x; y 2 R; the distance between x and y is commonly denoted by d.x; y/; which equals jx yj: Below we willneed a quantitative measure of what it means to be close for a discussion of the concept of limit. For " > 0; the openinterval .c "; cC "/ is called the "-neighborhood of c: Note x 2 .c "; cC "/ if and only if d.x; c/ D jx cj < ";i.e. every number in .c "; cC "/ has distance less than " from c:

Limit of a sequence fxng is often explained by saying it is the number the xns are closer and closer to as n getslarger and larger. There are two bad points about this explanations.

(1) Being close or large is a feeling! It is not a fact. It cannot be proved by a logical argument.(2) The effect of being close can accumulate to yield large separation! If two numbers having a distance less than or

equal to 1 are considered close, then 0 is close to 1 and 1 is close to 2 and 2 is close to 3, : : : ; 99 is close to 100,but 0 is quite far from 100.

So what is the meaning of close? How can limit be defined so it can be checked? Intuitively, a sequence fxng getsclose to a number x if and only if the distance d.xn; x/ goes to 0. This happens if and only if for every positive "; thedistance d.xn; x/ eventually becomes less than ": The following example will try to make this more precise.

Example. As n gets large, intuitively we may think xn D2n2 1n2 C 1

gets close to 2. For " D 0:1; how soon (that is, forwhat n) will the distance d.xn; 2/ be less than "? (What if " D 0:01? What if " D 0:001? What if " is an arbitrarypositive number?)

Solution. Consider d.xn; 2/ D

2n2 1n2 C 1

2

D

3n2 C 1

< ": Solving for n; we get n2 > .3="/ 1: If " D 0:1; then

n >p

29: So as soon as n 6; the distance between xn and 2 will be less than " D 0:1:(If " D 0:01; then n > p299: So n 18 will do. If " D 0:001; then n > p2999: So n 55 will do.

If 0 < " 3; then n [p.3="/ 1] C 1 will do. If " > 3; then since 3n2C1 < 3 < " for every n 2 N; so

n 1 will do. So for every " > 0; there is a K 2 N so that as soon as n K ; the distance d.xn; 2/ will be lessthan ":) Note the value of K depends on the value of "I the smaller " is, the larger K will be. (Some people write K

"

to indicate K depends on ":)

Definition. A sequence fxng converges to a number x (or has limit x) iff for every " > 0, there is K 2 N such that forevery n K ; it implies d.xn; x/ D jxn xj < " (which means xK ; xKC1; xKC2; : : : 2 .x "; x C "/:/

Remarks. (i) From the definition, we see that fxng converges to x; fxn xg converges to 0 and fjxn xjg convergesto 0 are equivalent because in the definition, jxn xj is the same as j.xn x/ 0j D jjxn xj 0j:

(ii) To show fxng converges to x means for every " > 0; we have to find a K as in the definition or show such a Kexists. On the other hand, if we are given that fxng converges to x , then for every " > 0; (which we can evenchoose for our convenience,) there is a K as in the definition for us to use.

Let us now do a few more examples to illustrate how to show a sequence converges by checking the definition.Later, we will prove some theorems that will help in establishing convergence of sequences.

25

Examples. (1) Let vn D c: For every " > 0, let K D 1, then n K implies jvn cj D 0 < ". So fvng converges to c:

(2) Let wn D c 1n: For every " > 0, there exists an integer K >

1"

(by the Archimedean principle). Then n K

implies jwn cj D1n

1K

< ". So fwng converges to c:

(3) Let xn D n.cos n/ n

: Show that fxng converges to1 by checking the definition.

Solution. For every " > 0; there exists an integer K > 1C 1"

by the Archimedian principle. Then n K implies

n

.cos n/ n .1/

D

cos n

.cos n/ n

1n 1

1K 1

< ": So fxng converges to 1:

(4) Let yn D .1/n : Show that fyng does not converge.Solution. Assume fyng converges, say to y: Let " D 0:1: Then there exists K 2 N such that n K impliesj.1/n yj < " D 0:1: Taking an odd integer n k; we get j 1 yj < 0:1; which implies y 2 .1:1;0:9/:Taking a even integer n K ; we get j1 yj < 0:1; which implies y 2 .0:9; 1:1/: Since no y is in both .1:1;0:9/and .0:9; 1:1/; we have a contradiction.

(5) Let zn D n1=n: Show that fzng converges to 1 by checking the definition.Solution. (Let un D jzn 1j D zn 1: By the binomial theorem,

n D znn D .1C un/n D 1C nun Cn.n 1/

2u2n C C u

nn

n.n 1/2

u2n

so that un r

2n 1

:) For every " > 0; there exists integer K > 1C 2"

2 (by the Archimedean principle). Then n K

implies jzn 1j D un r

2n 1

r

2K 1

< ": So fzng converges to 1.

Theorem (Uniqueness of Limit). If fxng converges to x and y, then x D y (and so we may write limn!1

xn D x).Proof. For every " > 0, we will show jxyj < ". (By the infinitesimal principle,we will get x D y.) Let "0 D "=2 > 0:By the definitionof convergence, there are K1, K2 2 N such that n K1 ) jxnxj < "0 and n K2 ) jxn yj < "0.Let K D max.K1; K2/. By the triangle inequality, jx yj D j.xxK /C.xK y/j jxxK jCjxK yj < "0C"0 D ".

Boundedness Theorem. If fxng converges, then fxng is bounded.Proof. Let lim

n!1xn D x . For " D 1, there is K 2 N such that n K ) jxnxj < 1 ) jxnj D j.xnx/Cxj < 1Cjxj.

Let M D max.jx1j, : : :, jxK1j, 1C jxj/; then for every n 2 N; jxnj M (i.e. xn 2 [M; M]).

Remarks. The converse is false. The sequence f.1/n g is bounded, but not convergent by example (4). In general,bounded sequences may or may not converge.

Theorem (Computation Formulas for Limits). If fxng converges to x and fyng converges to y, then(i) fxn yng converges to x y, respectively, i.e. lim

n!1.xn yn/ D lim

n!1xn lim

n!1yn,

(ii) fxn yng converges to x y, i.e. limn!1

.xn yn/ D

limn!1

xn

limn!1

yn

,

(iii) fxn=yng converges to x=y, provided yn 6D 0 for all n and y 6D 0.

26

Proof. (i) For every " > 0, there are K1; K2 2 N such that n K1 ) jxn xj < "=2 and n K2 ) jyn yj < "=2.Let K D max.K1; K2/: Then n K implies n K1 and n K2: So for these ns,

j.xn yn/ .x y/j D j.xn x/ .yn y/j jxn xj C jyn yj < "=2C "=2 D ":

(ii) We prove a lemma first.Lemma. If fang is bounded and lim

n!1bn D 0, then lim

n!1anbn D 0.

Proof. Since fang is bounded, there is M such that janj < M for all n. For every " > 0, since "=M > 0 andfbng converges to 0, there is K 2 N such that n K ) jbn 0j < "=M ) janbn 0j Mjbnj < ".

To prove (ii), we write xn yn x y D xn yn xn y C xn y x y D xn.yn y/ C y.xn x/. Since fxng converges,fxng is bounded by the boundedness theorem. So by (i) and the lemma,

limn!1

xn yn D limn!1

.xn yn x y/C limn!1

x y D limn!1

xn.yn y/ C limn!1

y.xn x/ C x y D 0C 0C x y D x y:

(iii) Note 12 jyj > 0: Since fyng converges to y; there is K0 2 N such that n K0 implies jyn yj < 12 jyj: Bythe triangle inequality, jyj jynj jyn yj < 12 jyj )

12 jyj < jynj for n K0: Then for every n 2 N;

jynj m D min.jy1j; : : : ; jyK01j; 12 jyj/ > 0:

Next we will show limn!1

1ynD

1y

(then by (ii), limn!1

xn

ynD lim

n!1.xn

1yn/ D x

1yD

x

y). For every " > 0; let

"0 D mjyj" > 0: Since limn!1

yn D y 6D 0, there is K 2 N such that n K ) jyn yj < "0: Then

n K )

1yn

1y

D

jy ynjjynjjyj

n j j implies n jC1 j C1:

Subsequence Theorem. If limn!1

xn D x, then limj!1 xnj D x for every subsequence fxnj g of fxng. (The converse istrivially true because every sequence is a subsequence of itself.)Proof. For every " > 0, there is K 2 N such that n K ) jxn xj < ". Then j K ) n j K ) jxnj xj < ".

Question. How can we tell if a sequence converges without knowing the limit (especially if the sequence is given bya recurrence relation)?

For certain types of sequences, the question has an easy answer.

Definitions. fxng is

8

>

:

increasingdecreasing

strictly increasingstrictly decreasing

9

>

=

>

;

iff

8

>

:

x1 x2 x3 : : :x1 x2 x3 : : :x1 < x2 < x3 < : : :x1 > x2 > x3 > : : :

9

>

=

>

;

; respectively. fxng is

monotonestrictly monotone

iff

fxng is

increasing or decreasingstrictly increasing or decreasing

; respectively.

Monotone Sequence Theorem. If fxng is increasing and bounded above, then limn!1

xn D supfxng. (Similarly, if fxngis decreasing and bounded below, then lim

n!1xn D inffxng.)

28

Proof. Let M D supfxng, which exists by the completeness axiom. By the supremum property, for any " > 0, there isxK such that M " < xK M . Then j K ) M " < xK x j M ) jx j Mj D M x j < ":Remark. Note the completeness axiom was used to show the limit of xn exists (without giving the value).Examples. (1) Let 0 < c < 1 and xn D c1=n: Then xn < 1 and cnC1 < cn ) xn D c1=n < c1=.nC1/ D xnC1: So bythe monotone sequence theorem, fxng has a limit x : Now x22n D .c1=2n/2 D c1=n D xn: Taking limits and using thesubsequence theorem, we get x2 D x : So x D 0 or 1. Since 0 < c D x1 x; the limit x is 1. Similarly, if c 1; thenc1=n will decrease to the limit 1.

(2) Doesr

2Cq

2Cp

2C represent a real number?Here we have a nested radical defined by x1 D

p

2 and xnC1 Dp

2C xn: The question is whether fxngconverges to a real number x . (Computing a few terms, we suspect that fxng is increasing. To find an upper bound,observe that if lim

n!1xn D x; then x D

p

2C x implies x D 2:) Now by mathematical induction, we can show thatxn < xnC1 < 2: (If xn < xnC1 < 2; then 2C xn < 2C xnC1 < 4; so taking square roots, we get xnC1 < xnC2 < 2:)By the monotone sequence theorem, fxng has a limit x : We have x2 D lim

n!1x2nC1 D lim

n!12C xn D 2C x : Then

x D 1 or 2. Sincep

2 D x1 x; so x D 2:

Another common type of sequences is obtained by mixing a decreasing sequence and an increasing sequence intoone of the form a1; b1; a2; b2; a3; b3; : : : : In the next example, we will have such a situation and we need two theoremsto handle these kind of sequences.

Nested Interval Theorem. If In D [an; bn] is such that I1 I2 I3 : : :, then1

\

nD1In D [a; b]; where

a D limn!1

an limn!1

bn D b: If limn!1

.bn an/ D 0, then1

\

nD1In contains exactly one number.

[ ][ [ ] ]a1 a2 a3 b3 b2 b1...

Proof. I1 I2 I3 : : : implies fang is increasing and bounded above by b1 and fbng is decreasing and boundedbelow by a1. By the monotone sequence theorem, fang converges to a D supfang and fbng converges to b D inffbng.Since an bn for every n 2 N, taking limits, we have an a b bn . Consequently, x 2 [an; bn] (i.e. an x bn)for all n if and only if lim

n!1an D a x b D lim

n!1bn: So

1

\

nD1In D [a; b]. If 0 D lim

n!1.bn an/ D b a; then a D b

and1

\

nD1In D fag:

Remarks. Note in the proof, the monotone sequence theorem was used. So the nested interval theorem also implicitlydepended on the completeness axiom.

Intertwining Sequence Theorem. If fx2mg and fx2m1g converge to x; then fxng also converges to x :Proof. For every " > 0; since fx2mg converges to x; there is K0 2 N such that m K0 ) jx2m xj < ": Since fx2m1galso converges to x; there is K1 2 N such that m K1 ) jx2m1 xj < ": Now if n K D max.2K0; 2K1 1/;then either n D 2m 2K0 ) jxn xj D jx2m xj < " or n D 2m 1 2K1 1 ) jxn xj D jx2m1 xj < ":

Example. Does1

1C1

1C

represent a number?

Here we have a continued fraction defined by x1 D 1 and xnC1 D 1=.1 C xn/: We have x1 D 1; x2 D 1=2; x3 D2=3; x4 D 3=5; : : : : Plotting these on the real line suggests 1=2 x2n < x2nC2 < x2nC1 < x2n1 1 for all n 2 N:This can be easily established by mathematical induction. (If 1=2 x2n < x2nC2 < x2nC1 < x2n1 1; then1 C x2n < 1 C x2nC2 < 1 C x2nC1 < 1 C x2n1: Taking reciprocal and applying the recurrence relation, we havex2nC1 > x2nC3 > x2nC2 > x2n : Repeating these steps once more, we get x2nC2 < x2nC4 < x2nC3 < x2nC1:)

Let In D [x2n; x2n1]; then I1 I2 I3 : Now

jxm xmC1j D

11C xm1

11C xm

D

jxm1 xm j

.1C xm1/.1 C xm/ 0; since lim

n!1xn D x; there is K 2 N such that j K ) jx j xj < "=2. For m, n K , we

have jxm xnj jxm xj C jx xnj < "=2C "=2 D ". So fxng is a Cauchy sequence.

The converse of the previous theorem is true, but it takes some work to prove that. The difficulty lies primarilyon how to come up with a limit of the sequence. The strategy of showing every Cauchy sequence inRmust convergeis first to find a subsequence that converges, then show that the original sequence also converge to the same limit.

Theorem. If fxng is a Cauchy sequence, then fxng is bounded.Proof. Let " D 1: Since fxng is a Cauchy sequence, there is K 2 N such that m; n K ) jxm xnj < " D 1:In particular, for n K ; jxK xnj < 1 ) jxnj D j.xn xK / C xK j jxn xK j C jxK j < 1 C jxK j: LetM D max.jx1j; : : : ; jxK1j; 1C jxK j/; then for all n 2 N; jxnj M (i.e. xn 2 [M; M]/:

Bolzano-Weierstrass Theorem. If fxng is bounded, then fxng has a subsequence fxnj g that converges.Proof. (Bisection Method) Let a1 D inffxng, b1 D supfxng and I1 D [a1; b1]. Let m1 be the midpoint of I1. If thereare infinitely many terms of fxng in [a;m1], then let a2 D a1, b2 D m1 and I2 D [a2; b2]. Otherwise, there will beinfinitely many terms of fxng in [m1; b1], then let a2 D m1, b2 D b1 and I2 D [a2; b2]. For k D 2, 3, 4, : : :, repeatthis bisection on Ik to get IkC1. We have Ij D [aj; bj] and I1 I2 I3 : : :. By the nested interval theorem, sincelimj!1

.bj aj/ D limj!1b1 a1

2 j1D 0,

1

\

nD1In contains exactly one number x .

Take n1 D 1; then xn1 D x1 2 I1. Suppose n j is chosen with xnj 2 Ij . Since there are infinitely many terms xn inIjC1, choose n jC1 > n j and xnjC1 2 IjC1. Then limj!1 jxnj xj limj!1.bj aj/ D 0. Therefore, limj!1 xnj D x .

30

Remarks. In the proof, the nested interval theorem was used, so the Bolzano-Weierstrass theorem depended on thecompleteness axiom.

Alternate Proof. We will show every sequence fxng has a monotone subsequence. (If fxng is bounded, then themonotone sequence theorem will imply the subsequence converges.)

Call xm a peak of fxng if xm xk for all k > m: If fxng has infinitely many peaks, then we order the peaks bystrictly increasing subscripts m1 < m2 < m3 < : By the definition of a peak, xm1 xm2 xm3 : So fxm j gis a decreasing subsequence of fxng: On the other hand, if fxng has only finitely many peaks xm1; ; xmk; then letn1 D maxfm1; ;mkg C 1: Since xn1 is not a peak, there is n2 > n1 such that xn2 > xn1 : Inductively, if xnj is not apeak, there is n jC1 > n j with xnjC1 > xnj : So fxnj g is a strictly increasing subsequence of fxng:

Remarks. This alternate proof used the monotone sequence theorem, so it also depended on the completeness axiom.

Cauchys Theorem. fxng converges if and only if fxng is a Cauchy sequence.Proof. The only if part was proved. For the if part, since fxng is a Cauchy sequence, fxng is bounded. By theBolzano-Weierstrass theorem, fxng has a subsequence fxnj g that converges inR, say limj!1 xnj D x .

We will show limn!1

xn D x . For every " > 0, since fxng is a Cauchy sequence, there is K1 2 N such that m,n K1 ) jxm xnj < "=2. Since limj!1 xnj D x , there is K2 2 N such that j K2 ) jxnj xj < "=2. Ifn J D max.K1; K2/, then n J J K1, J K2 and jxn xj jxn xnJ j C jxnJ xj < "=2C "=2 D ".

Example. Does the sequence fxng converge, where x1 D sin 1 and xk D xk1 Csin kk2

for k D 2; 3; 4; : : :?

We will check the Cauchy condition. For m > n; xm xn DmX

kDnC1.xk xk1/ D

mX

kDnC1

sin kk2

and

jxmxnj 1

.n C 1/2C C

1m2

0, there is > 0 such that for every x 2 S; 0 < jx x0j < implies j f .x/ L j < ". This is denoted bylim

x!x0x2S

f .x/ D L (or limx!x0

f .x/ D L in short.)

In the definition, depends on " and x0: For different " (or different x0), will be different. If a limit value exists,then it is unique. The proof is similar to the sequential case and is left as an exercise for the readers.

Examples. (1) For g : [0;1/ ! Rdefined by g.x/ D px; show that limx!0

g.x/ D 0 and limx!4

g.x/ D 2 by checkingthe "- definition.

Solution. For every " > 0; let D "2: Then for every x 2 [0;1/; 0 < jx0j < implies jg.x/0j D px 0; let D 2": Then for every

x 2 [0;1/; 0 < jx 4j < implies jg.x/ 2j jx 4j2

0 such that for every > 0; there is x 2 S with 0 < jx x0j <

and j f .x/ L j ". Now, setting D 1n; there is xn 2 S with 0 < jxn x0j < D 1n and j f .xn/ L j ". By the

sandwich theorem, xn ! x0 in S n fx0g: So limn!1

f .xn/ D L : Then 0 D limn!1

j f .xn/ L j ", a contradiction.Remarks. If lim

n!1f .xn/ exists for every xn ! x0 in S nfx0g; then all the limit values are the same. To see this, suppose

xn ! x0 and wn ! x0 in S n fx0g: Then the intertwining sequence fzng D fx1; w1; x2; w2; x3; w3; : : :g converges to x0in S n fx0g: Since f f .xn/g and f f .wn/g are subsequences of the convergent sequence f f .zn/g, they have the same limit.

Consequences of Sequential Limit Theorem.

(1) (Computation Formulas) If f; g : S ! Rare functions, limx!x0x2S

f .x/ D L1 and limx!x0x2S

g.x/ D L2; then

limx!x0x2S

f .x/

8

>

:

C

=

9

>

=

>

;

g.x/

D L1

8

>

:

C

=

9

>

=

>

;

L2 D limx!x0x2S

f .x/

8

>

:

C

=

9

>

=

>

;

limx!x0x2S

g.x/

respectively (in the case of division, provided g.x/ 6D 0 for every x 2 S and L2 6D 0).Proof. Since f .x/ and g.x/ have limits L1 and L2; respectively, as x tends to x0 in S; by the sequential limittheorem, f .xn/ and g.xn/ will have limits L1 and L2; respectively, for every xn ! x0 in S n fx0g: By thecomputation formulas for sequences, the limit of f .xn/C g.xn/ is L1 C L2 for every xn ! x0 in S n fx0g: By thesequential limit theorem, f .x/ C g.x/ has limit L1 C L2 as x tends to x0 in S: ReplacingC by; ; =;we get theproofs for the other parts.

Alte