ADVANCED CALCULUS
1996–1997
Gilbert Weinstein
Office: CH 493B Tel: (205) 934-3724
(205) 934-2154 FAX: (205) 934-9025
Email: [email protected]
About the Course
Welcome to Advanced Calculus! In this course, you will learn Analysis, that is the theory of differentiation and integration of functions. However, you will also learn something more fundamental than Analysis. You will learn what is a mathematical proof. You may have seen some proofs earlier, but here, you will learn how to write your own proofs. You will also learn how to understand someone else’s proof, find a flaw in a proof, fix a deficient proof if possible and discard it if not. In other words you will learn the trade of mathematical exploration.
Mathematical reasoning takes time. In Calculus, you expected to read a prob-lem and immediately know how to proceed. Here you may expect some frustration and you should plan to spend a lot of time thinking before you write down anything. Analysis was not discovered overnight. It took centuries for the correct approach to emerge. You will have to go through an accelerated process of rediscovery. Twenty hours of work a week outside class is not an unusual average for this course.
The course is run in the following way. In these notes, you will findDefinitions,
Theorems, andExamples. I will explain the definitions. At home,on your own, you will try to prove the theorems and the statements in the examples. You will use no books and no help from anyone. It will be just you, the pencil and the paper. Every statement you make must be justified. In your arguments, you may use any result which precedes in the notes the item you are proving. You may use these results even if they have not yet been proven. However, you may not use results from other sources. Then, in class, I will call for volunteers to present their solutions at the board. Every correct proof is worth one point. If more than one person volunteer for an item, the one with the fewest points is called to the board, ties to be broken by lot. Your grade will be determined by the number of points you have accumulated during the term.
Introduction
1. Mathematical Proof
What is a proof? To explain, let us consider an example.
Theorem 1.1. There is no rational numberr which is a square root of2.
This theorem was already known to the ancient Greeks. It was very important to them since they were particularly interested in geometry, and, as follows from the Theorem of Pythagoras, a segment of length √2 can be constructed as the hypotenuse of a right triangle with both sides of length 1.
Before we prove this theorem, in fact before we prove any theorem, we must understand its statement. To understand its statement, we must understand each of the terms used. For instance: what is a rational number? For this we need a
definition.
Definition 1.1. A numberrisrational if it can be represented as the ratio of two integers:
r= n m (1.1)
wherem6= 0.
Of course, in this definition, we are using other terms that need to be defined, such asnumber,ratio,integer. We will not dwell on this point, and instead assume for now that these have been defined previously. However, already one point is clear. If we wish to be absolutely rigorous, we must begin from some given assumptions. We will call these axioms. They do not require proof. We will discuss this point further later. For the time being, let us assume that we have a system of numbers where the usual operations of arithmetic are defined.
Next we need to define what we mean by asquare root of 2.
Definition 1.2. Let y be a number. The number x is a square root of y if x2=y.
Again, we assume that the meaning of x2 is understood. Note that we have
said a square root, and not the square root. Indeed if x6= 0 is a square root of y, then −x is another one. Note that then, one of the two numbers xand −xis positive. Now, we may give the proof of Theorem 1.1.
Proof. The proof is by contradiction. Suppose thatxis a square root of 2, and that x is rational. Clearly, x6= 0, hence we may assume thatx > 0. Then, x2= 2, and there are integersn,m
6
= 0, such that
x= n m. (1.2)
Of course there are many such pairsn, andm. In fact, ifnandmis any such pair, then 2n and 2m is another pair. Also, there is one pair in which n >0. Among all these pairs, with n > 0, pick one for which n is the smallest positive integer possible, i.e.,x=n/m,n >0, and ifx=k/l thenn✻k. We have:
n
m
2
= 2, (1.3)
or equivalently
n2= 2m2. (1.4)
Thus, 2 divides n2 = n
·n. It follows that 2 divides n, i.e. n is even. We may therefore writen= 2k and thus
n2= 4k2= 2m2, (1.5)
or equivalently
2k2=m2. (1.6)
Now, 2 dividesm2, hencem is even. Writem= 2l. We obtain
x= n m =
2k 2l =
k l. (1.7)
But k is positive and clearlyk < n, a contradiction. Thus no such xexists, and the theorem is proved.
A close examination of this proof will be instructive. The first observation is that the proof is by contradiction. We assumed that the statement to be proved is false, and we reached an absurdity. Here the statement to be proved was that
there is no rational x for which x2 = 2, so we assumed there is one such x. The
absurdity was that we could certainly take x=n/m, withn >0 and as small as possible, but we deduced x= k/l with 0 < k < n. Next, we see that each step follows from the previous one, and possibly some additional information. Take for example, the argument immediately following (1.4). If 2 dividesn2 then 2 divides
n. This seemingly obvious fact requires justification. We will not do this here; it is done in algebra, and relies on the unique factorization by primes of the integers. It is extremely important to identify the information which you import into your proof from outside. Usually, this is done by quoting a known theorem. Remember that before you quote a theorem, you must check its hypotheses.
Read this proof again and again during the term. Try to find its weak points, those points which could use more justification. Try to improve it. Try to imagine how it was discovered.
2. Set Theory and Notation
which mean that the two statements are equivalent. We will abbreviateif and only if by iff. For example, ifAandB are two sets, thenA=B iffA⊂B andB⊂A. This is usually the way one checks that two setsAandB are equal: A⊂Band B⊂A. Again, we are learning early an important lesson: break a proof into smaller parts. In these notes, you will find that I have tried to break the development of the material into the proof of a great many small facts. However, in the more difficult problems, you might want to continue this process further on your own, i.e., decompose the harder problems into a number of smaller problems. Try to take the proof of Theorem 1.1 and break it into the proof of several facts. Think about how you could have guessed that these were intermediate steps in proving Theorem 1.1.
We will also use the symbol ∀ to meanfor every, and the symbol ∃ to mean
there is. Finally we will use∅to denote theempty set, the set with no elements, LetAandBbe sets, theirintersection, which is denoted byA∩B, is the set of all elements which belong to bothAandB. Thus,x∈A∩B iffx∈Aandx∈B. If A∩B = ∅ we say that A and B are disjoint. Similarly, their union, denoted A∪B, is the set of all elements which belong to either Aor B (or both), so that x∈A∪B iffx∈A orx∈B.
Theorem 1.2. Let A,B, andC be sets. Then
(A∩B)∪C= (A∪C)∩(B∪C). (1.8)
Theorem 1.3. Let A,B, andC be sets. Then
(A∪B)∩C= (A∩C)∪(B∩C). (1.9)
IfA is a subset ofX, then thecomplement ofA in X is the setX\Awhich consists of all the elements ofX which do not belong toA, i.e.x∈X\Aiffx∈X and x 6∈ A. If X is understood, then the complement is also denoted Ac. The
notationX\Ais occasionally used also whenAis not a subset ofX. However, in this caseX\A is not called the complement ofAinX.
LetX andY be sets. The set of all pairs (x, y) withx∈X andy∈Y is called the Cartesian product ofX and Y, and is written X×Y. A subset f of X ×Y is afunction, if for eachx∈X there is a unique y∈Y such that (x, y)∈f. The last statement includes two conditions,existence and uniqueness. These are often treated separately. Thusf ⊂X×Y is a function iff:
(i) ∀x∈X,∃y∈Y, such that (x, y)∈f; (ii) If (x, y1)∈f, and (x, y2)∈f, theny1=y2.
To simplify the notation, we will writey=f(x) instead of (x, y)∈f. Iff ⊂X×Y is a function, we write f: X → Y. We call X the domain of f, and say that f
mapsX toY. The set ofy∈Y such that∃x∈X for whichy=f(x) is called the
range off, and will be written as Ran(f).
Let f:X → Y. If A ⊂ X, we define the function f|A:A → Y, called the
restriction off toAby setting (f|A)(x) =f(x) for everyx∈A. We define the set f(A)⊂Y by
f(A) ={y∈Y: ∃x∈A, f(x) =y} ={f(x) : x∈A}.
(1.10)
Theorem 1.4. Let f:X→Y, andA, B⊂X, then
f(A∪B) =f(A)∪f(B). (1.11)
Theorem 1.5. Let f:X→Y, andA, B⊂X, then
f(A∩B)⊂f(A)∩f(B). (1.12)
Example1.1. Equality does not always hold in (1.12).
IfA⊂Y, we define thepre-image of Aunder f, denotedf−1(A)
⊂X, by: f−1(A) =
{x∈X: f(x)∈A}. (1.13)
Theorem 1.6. Let f:X→Y, andA, B⊂Y, then
f−1(A
∪B) =f−1(A)
∪f−1(B).
(1.14)
Theorem 1.7. Let f:X→Y, andA, B⊂Y, then
f−1(A
∩B) =f−1(A)
∩f−1(B).
(1.15)
A function f: X → Y is onto (or surjective) if Ran(f) = Y. A function f:X →Y isone-to-one (or injective) iff(x1) =f(x2) implies that x1=x2. Iff
is both one-to-one and onto, we say thatf isbijective. Then there exists a unique functiong such thatg(f(x)) =xfor allx∈X, andf(g(y)) =y for ally∈Y. The functiongis denotedf−1, and is called theinverse off.
Example1.2. Letf:X →Y be bijective, and let f−1 be its inverse. Then ∀y∈Y:
f−1(
{y}) ={f−1(y) }
(1.16)
Caution: part of the problem here is to explain the notation. In particular, the notation is abused in the sense that f−1 on the left-hand side has a different
meaning than on the right-hand side.
3. Induction
Induction is an essential tool if we wish to write rigorous proof using repetitions of an argument an unknown number of times. In a more elementary course, a combination of words such as ‘and so on’ could probably be used. Here, this will not be acceptable. We will denote by◆the set of natural numbers, i.e. the counting numbers 1,2,3, . . ., and by❩the integers, i.e., the natural numbers, their negatives, and zero.
Axiom 1 (Well-Ordering Axiom). Let∅ 6=S⊂◆. ThenS contains a smallest element.
As the name indicates, we will take this as an axiom. No proof need be given. However, we will prove thePrinciple of Mathematical Induction.
Theorem 1.8 (The Principle of Mathematical Induction). Suppose that S ⊂ ◆satisfies the following two conditions:
(i) 1∈S.
(ii) If n∈S then n+ 1∈S.
We will denote by P(n) a statement about integers. You may want to think ofP(n) as a function from ❩to{0,1}(0 represents false, 1 represents true.) P(n) might be the statement: nis odd; or the statement: the number of primes less than or equal tonis less then n/log(n).
Theorem 1.9. Let P(n) be a statement depending on an integer n, and let n0∈❩. Suppose that
(i) P(n0)is true;
(ii) If n❃n0 andP(n)is true, then P(n+ 1) is true. ThenP(n) is true for alln∈❩such thatn❃n0.
This last result is what we usually refer to as Mathematical Induction. Here is an application.
Example1.3. Letn∈◆, then
n
X
k=1
(2k−1) =n2 (1.17)
See Theorem 2.13, and Equation (2.6) for a precise definition of the summa-tion’sPnotation.
4. The Real Number System
In a course on the foundations of mathematics, one would construct first the natural numbers ◆, then the integers ❩, then the rational numbers ◗, and then finally the real numbers ❘. However, this is not a course on the foundations of mathematics, and we will not labor on these constructions. Instead, we will assume that we are given the set of real numbers❘with all the properties we need. These will be stated as axioms. Of course, you probably already have some intuition as to what real numbers are, and these axioms are not meant to substitute for that intuition. However, when writing your proofs, you should make sure that all your statements follow from these axioms, and their consequences proved so far only.
The set of real numbers❘ is characterized as being a complete ordered field. Thus, the axioms for the real numbers are divided into three sets. The first set of axioms, the field axioms, are purely algebraic. The second set of axioms are the order axioms. It is extremely important to note that ◗ satisfies the axioms in the first two sets. Thus ◗ is an ordered field. Nevertheless, if one wishes to do analysis, the rational numbers are totally inadequate. Another ordered field is given in the appendix. The last axiom required in order to study analysis is the Least Upper Bound Axiom. An ordered field which satisfies this last axiom is said to be complete. I suggest that at first, you try to understand, or rather recognize, the first two sets of axioms.
Axiom 2 (Field Axioms). For each pairx, y∈❘, there is an element denoted x+y∈❘, called thesumofxandy, such that the following properties are satisfied:
(i) (x+y) +z=x+ (y+z) for all x, y, z∈❘. (ii) x+y=y+xfor allx, y∈❘.
Furthermore, for each pairx, y∈❘, there is an element denotedxy∈❘, and called theproduct ofxandy, such that the following properties are satisfied:
(v) (xy)z=x(yz) for all x, y, z∈❘. (vi) xy=yx for allx, y∈❘.
(vii) There exists a non-zero element 1∈❘such that 1·x=xfor allx∈❘. (viii) For all non-zero x ∈ ❘, there is an element denoted x−1
∈ ❘ such that xx−1= 1.
(ix) For allx, y, z∈❘, (x+y)z=xz+yz.
Any set which satisfies the above field axioms is called afield. Using these, it is possible to prove that all the usual rules of arithmetic hold.
Example1.4. LetX ={x}be a set containing one elementx, and define the operations + and· in the only possible way. IsX a field?
Example1.5. Let X = {0,1}, and define the addition and multiplication operations according to the following tables:
+ 0 1 0 0 1 1 1 0
· 0 1 0 0 0 1 0 1
Prove thatX is a field.
Axiom 3 (Ordering Axioms). There is a subset P ⊂❘, called the set of
posi-tive numbers, such that:
(x) 06∈P.
(xi) Let 06=x∈❘. Ifx6∈P then−x∈P, and ifx∈P then−x6∈P. (xii) Ifx, y∈P, thenxy, x+y∈P.
A field with a subsetP satisfying the ordering axioms is called anordered field. Let N =−P ={x∈❘: −x∈P} denote the negative numbers. Then (x) and (xi) simply say that❘=P∪ {0} ∪N, and these three sets are disjoint. We define x < yto meany−x∈P. ThusP ={x∈❘: 0< x}. The usual rules for handling inequalities follow.
Theorem 1.10. Letx, y, z∈❘. Ifx < y andy < zthen x < z. Theorem 1.11. Letx, y, z∈❘. Ifx < y and0< z thenxz < yz.
We will also usex > yto meany < x.
Theorem 1.12.
0<1 (1.18)
Theorem 1.13. Letx, y, z∈❘, then (i) If x < ythen x+z < y+z.
(ii) If x >0 thenx−1>0.
(iii) If x, y >0and x < ytheny−1< x−1.
We also definex✻y to mean that eitherx < yor x=y.
Theorem 1.15. Suppose that x, y ∈ ❘ and x < y, then there is z ∈ ❘ such
that x < z < y.
Note that since we will only use the axioms for an ordered field, this holds also in the rational numbers◗.
Definition 1.3. LetS⊂❘. We say that S isbounded above if there exists a numberb∈❘such thatx✻bfor allx∈S. The numberb is then called anupper bound forS. A numberc is called aleast upper bound forS if it has the following two properties:
(i) cis an upper bound forS;
(ii) ifbis an upper bound forS, thenc✻b.
Theorem 1.16. If a least upper bound forS exists then it is unique, i.e., if c1
andc2 are both least upper bounds forS thenc1=c2.
Thus we can speak ofthe least upper bound of a setS.
Definition 1.4. A setS is bounded below if there exists a numberb∈❘such that b ✻ x for all x∈ S. The number b is then called a lower bound for S. A numberc is called agreatest lower bound forS if it has the following properties:
(i) cis a lower bound forS;
(ii) ifbis a lower bound forS, thenb✻c.
Theorem 1.17. S is bounded above iff −S = {x∈❘: −x∈S} is bounded
below.
Theorem 1.18. c is the least upper bound of S iff −c is the greatest lower
bound of −S.
We now state the Least Upper Bound Axiom.
Axiom 4 (Least Upper Bound Axiom). Let ∅ 6= S ⊂ ❘ be bounded above. ThenS has a least upper bound.
We list a few consequences of this axiom.
Theorem 1.19. Let∅ 6=S ⊂❘be bounded below. ThenS has a greatest lower
bound.
Theorem 1.20. Letx∈❘, then there exists an integer n∈❩such thatn > x. Example1.6. Letx∈❘satisfy
0✻x < 1 n, (1.19)
for all integersn >0. Thenx= 0.
Theorem 1.21. Let y ❃0. Then there exists a real number x❃0 such that x2=y.
In particular, in contrast with Theorem 1.1, the number 2 has a real square root.
Theorem 1.22. Show that the set ◗ of rational numbers is an ordered field,
Note that ify❃0, andxis a square root ofy, then also−xis a square root of y, since
(−x)2= (
−1)2x2=y.
(1.20)
Thus, ify❃0 it has exactly one non-negative square rootx.
Definition 1.5. Let y ❃ 0, we define the square root of y to be the non-negative numberx❃0 such thatx2=y, and we denote it byx=√y.
This is somewhat in disagreement with our previous definition of square roots, but we will no longer use definition 1.2, hence no problems should arise from this.
Example1.7. Leta❃0, andb❃0, then√ab=√a√b. Definition1.6. Letx∈❘. We define itsabsolute value|x|by:
|x|=
(
x ifx❃0,
−x ifx <0. (1.21)
Theorem 1.23. Letx∈❘, then√x2=|x|.
Theorem 1.24. Letx, y∈❘, then|xy|=|x||y|. Theorem 1.25. Letx, y∈❘then
|x+y|✻|x|+|y|. (1.22)
This last inequality is calledthe triangle inequality. It is used widely.
Theorem 1.26. Letx, y∈❘then
|x−y|❃|x| − |y|. (1.23)
Theorem 1.27. Letx, y∈❘then
|x−y|❃
|x| − |y| .
(1.24)
Appendix
In this appendix, we sketch the construction of an ordered field ❋which does not satisfy the Least Upper Bound Axiom, and in which Theorem 1.20 does not hold.
We will use the set of polynomials with real coefficients:
❘[t] =
(
p:❘→❘: p(t) =
n
X
k=0
aktk;ak∈❘
)
. (1.25)
Note that a functionf:❘→❘is zero iff it assigns zero to all realt∈❘. Thus, a polynomial is zero iff all its coefficients are zero. Ifp∈❘[t], andp6= 0, we call the coefficient of the highest power ofttheleading coefficient.
Let❋be the set of rational functions with real coefficients. More precisely, let ❋={p/q: p, q∈❘[t];q6= 0},
(1.26)
r =p/q. We define addition and multiplication in❋ as usual for functions. It is easy to see that the sum and the product of two functions in❋ is again in❋. It is also not difficult to check that❋is a field. We will not carry out all these steps here. They are, although tedious, quite straightforward. We only note that❋contains a copy of❩, in fact a copy of❘, namely the constant functions.
Define P ⊂ ❋ to be the set of non-zero rational functions p/q such that the leading coefficient of pq is positive. The set P is the set of all non-zero rational functions which can be written as a ratiop/qwhere the leading coefficients of both pandq are positive. Now it is clear that 06∈P. Also, one can check that all the axioms forP are satisfied. Thus❋is an ordered field.
However, Theorem 1.20 does not hold in ❋. In fact, consider the function t ∈ ❘[t]. This is the function which assigns to each real number x ∈ ❘ the real number x∈ ❘. Now, if n ∈❩, then t−n has leading coefficient 1, hence lies in P. Thusn < t. Since this is true for everyn∈❩, we have found an elementt∈❋ which is larger than every integern.
Ordered fields in which Theorem 1.20 does not hold are callednon-Archimedean ordered fields. In such a field, there are ‘infinitely large’ elements. We have con-structed here a non-Archimedean ordered field❋. In❋, the elementst,t2, etc, are
Sequences
1. Limits of Sequences
Definition 2.1. Asequence of real numbers, is a functionx:◆→❘.
When no confusion can arise, we will usually abbreviate this simply as a se-quence. If x: ◆ → ❘ is a sequence, we write, in keeping with tradition, xn
in-stead of x(n), and we will use the notation {xn}
∞
n=1 for the sequence x, where
xn=x(n)∈❘. Note that there is a distinction between the sequence{xn}
∞
n=1and
the subset{xn: n∈◆} ⊂❘.
Definition2.2. Let{xn}∞
n=1 be a sequence, and letL∈❘. We say that the
sequence{xn}
∞
n=1 converges toL, if for everyε >0, there is an integerN ∈◆such
that for all integersn❃N, there holds:
|xn−L|< ε.
(2.1)
Theorem 2.1. Suppose that {xn}∞
n=1 converges to L1, and also converges to
L2. ThenL1=L2.
Thus the numberL in Definition 2.2 is unique, and we may speak ofthe limit
Lof the sequence, if it exists.
Example2.1. Let c ∈ ❘, and for each n ∈ ◆ let xn = c. Then {xn}∞
n=1
converges toc.
Example2.2. Letxn= 1/nfor eachn∈◆. Then{xn}∞
n=1 converges to 0.
Definition 2.3. Let{xn}∞
n=1 be a sequence. We say that {xn}
∞
n=1 converges
if there is an L ∈ ❘ such that {xn}
∞
n=1 converges to L. If a sequence does not
converge, we say that itdiverges.
Example2.3. Letxn= (−1)n. Then{xn}∞
n=1 diverges.
Theorem 2.2. Let {xn}∞
n=1 be a sequence which converges. Then {xn: n ∈
◆} ⊂❘is bounded above and below.
In order to compute, we need theorems which allow us to perform simple arith-metic operations with limits. Let {xn}
∞
n=1 and {yn}
∞
n=1 be sequences. We define
their sum as the sequence {xn+yn}n∞=1, i.e. the function (x+y) :◆→❘which
assigns to each n∈◆the number xn+yn ∈❘. Similarly, we define their product
as the sequence{xnyn}∞n=1.
Theorem 2.3. Suppose that {xn}∞
n=1 converges to L, and {yn}
∞
n=1 converges toM. Then {xn+yn}∞n=1 converges to L+M.
Theorem 2.4. Suppose that {xn}∞ Theorem 2.10. Ifk, n∈◆0, we define thebinomial coefficients by
n
The function f given by Theorem 2.12 is called the power function, and we denote it asf(n) =an.
Theorem 2.13 (Summation). Letx:◆m→❘, and letk❃m, then there is a
We denote:
wheref is the function given in Theorem 2.13.
Theorem 2.14. Leta, b∈❘, and letn∈◆. Then
Then we say thatf isincreasing. Similarly, iff has the following property:
ifa, b∈S, anda < b, then f(a)> f(b),
then we say thatf isdecreasing.
Definition 2.6. Let{xn}∞
n=1be a sequence, and letg:◆→◆be an increasing
sequence of natural numbers. Then the sequence{xg(n)}∞n=1is called asubsequence
of the sequence{xn}
∞
n=1.
We often write g(j) = nj, and thus, we also often write the subsequence
{xg(n)}∞n=1as {xnj}
When the sequence{xn}
∞
n=1 converges toL, we will write:
lim
bounded below. A sequence isbounded if it is bounded above and bounded below.
Theorem 2.16. Let {xn}∞
n=1 be an non-decreasing sequence which is bounded above. LetL∈❘ be the least upper bound of the set{xn: n∈◆}. Then{xn}∞n=1
converges to L.
Theorem 2.17. Let {xn}∞
n=1 be a non-increasing sequence which is bounded below. Let L ∈ ❘ be the greatest lower bound of {xn: n ∈ ◆}. Then {xn} sequence if the following holds:
For eachε >0, there isN ∈◆, such that for all integersn, m❃N,
The natural question is then: Does every Cauchy sequence converge? The rest of this section is devoted to the proof of this fact.
Theorem 2.19. Let {xn}∞
LetS⊂❘be bounded above. We will denote its least upper bound by supS. LetS be bounded below. We will denote its greatest lower bound by infS. Note that ifS⊂❘is bounded above, andS′
⊂S, thenS′
is also bounded above. Thus if
{xn}
n=1 is bounded. Define
yn = sup{xk: k❃n}.
(2.15)
Then the sequence {yn}
∞
Definition 2.10. The limit of the sequence{yn}∞
n=1 defined in Theorem 2.22
is called thelimit superior of{xn}
∞
n=1 and is denoted lim supn→∞xn. Thus, if{xn}
∞
n=1is bounded,
lim sup
n→∞
xn= lim
n→∞(sup{xk: k❃n}). (2.16)
Theorem 2.23. Suppose{xn}∞
n=1 is bounded. Define
yn = inf{xk: k❃n}.
(2.17)
Then the sequence {yn}
∞
n=1 converges.
Definition 2.11. The limit of the sequence{yn}∞
n=1 defined in Theorem 2.23
is called thelimit inferior of {xn}
∞
n=1and is denoted lim infn→∞xn.
Thus, thelimit inferior of a bounded sequence{xn}
∞
n=1 is:
lim inf
n→∞ xn= limn→∞(inf{xk: k❃n}). (2.18)
Theorem 2.24. Let {xn}∞
n=1 be bounded, and let L = lim supn→∞xn. Then
there exists a subsequence{xnj}
∞
j=1 which converges toL.
Theorem 2.25. If{xn}∞
n=1 is bounded, it has a convergent subsequence.
Theorem 2.26. Let {xn}∞
n=1 be a Cauchy sequence, and let L be its limit su-perior. Then{xn}
∞
n=1 converges to L.
Theorem 2.27. A sequence{xn}∞
n=1 converges iff it is a Cauchy sequence.
Theorem 2.28. Let {xn}∞
n=1 be a Cauchy sequence, and let L be its limit in-ferior. Then {xn}
∞
n=1 converges to L.
Theorem 2.29. Let{xn}∞
n=1 be a bounded sequence. Then{xn}
∞
n=1 converges if and only if
lim sup
n→∞
Series
1. Infinite Series
Let {xn}
∞
n=1 be a sequence of real numbers. We will not define what is the series:
Definition 3.1. Let{xn}∞
n=1be a sequence, and consider the series (3.1). We
define thesequence of partial sums {sm}
∞
form ∈◆. We say that the series (3.1) converges if the sequence of partial sums
{sm}
∞
m=1converges. In this case, we define thesum of the series as the limit of the
sequence{sm}
If the series (3.1) does not converge, we say that it diverges. Similarly, ifm ∈❩, andx:Nm→❘, then we define the partial sums of the seriesP
and define its sum by:
∞
provided the series converges, i.e., provided the limit on the right-hand side of (3.10) exists.
Theorem 3.6. Suppose that both seriesP∞
n=1xnand
Theorem 3.7. Suppose that the series P∞
n=1xn converges, and let c ∈ ❘.
Example3.3. Show that the seriesP∞
Theorem 3.9. Let m∈◆. Then, the seriesP∞
n=1xn converges iff the series
P∞
n=m+1xn converges. Furthermore, in this case, we have:
∞
n=1xn converges iff the following criterion is satisfied:
Theorem 3.11. Suppose that the series P∞
n=1xn converges, and let ym =
P∞
n=mxn. Thenlimm→∞ym= 0.
2. Series with Nonnegative Terms
Theorem 3.12. LetN ∈◆and suppose thatxn❃0 for alln❃N. Then, the
seriesP∞
n=1xn converges iff the sequence of partial sums is bounded.
Theorem 3.13. Let N ∈◆, and suppose that 0 ✻xn ✻yn for all n❃N. If
the seriesP∞
n=1yn converges, then the series P
∞
Theorem 3.18. Letxn❃0, and suppose that:
n=1 is a non-increasing sequence of nonnegative real numbers, and define
yn = 2nx2n.
(3.22)
If the series P∞
n=1xn converges, then the series P
∞
n=1yn converges.
Example3.5. Prove that the seriesP∞
n=11/n diverges.
3. Absolute Convergence
Definition 3.2. The series P∞
n=1xn is said to converge absolutely if the
se-riesP∞
n=1|xn| converges. If the series
P∞
n=1xn converges, but does not converge
absolutely, then we say that itconverges conditionally.
Theorem 3.20. If the seriesP∞
n=1xn converges absolutely, then it converges.
Example3.6. Show that the series
∞
n=1 is a non-increasing sequence of nonneg-ative real numbers which converges to 0, and suppose that the sequence of partial sums of{xn}
∞
n=1 is bounded. Then, the series
P∞
n=1xnyn converges.
Theorem 3.23 (Leibnitz). Let {xn}∞
n=1 be a increasing sequence of non-negative real numbers, and suppose thatlimn→∞xn= 0. Then, then series
∞
Theorem 3.24. Suppose that P∞
n=1xn converges absolutely, and let P
∞
n=1yn
be a rearrangement of P∞
n=1xn. ThenP
∞
n=1yn converges, and
∞
X
n=1
yn =
∞
X
n=1
xn.
(3.27)
Example3.8. Suppose the seriesP∞
n=1xnconverges conditionally. Then there
is a rearrangementP∞
n=1yn of
P∞
n=1xn such that
P∞
n=1yn diverges.
Example3.9. Suppose that the series P∞
n=1xn converges conditionally, and
lets∈❘. Then, there is a rearrangementP∞
n=1yn ofP
∞
n=1xn such that
∞
X
n=1
yn=s.
(3.28)
Theorem 3.25. Suppose that the series P∞
n=0xn converges absolutely, and
that the series P∞
n=0yn converges. Define for eachn∈◆:
zn= n
X
k=0
xkyn−k.
(3.29)
Then, the series P∞
n=0zn converges, and:
∞
X
n=0
zn=
∞
X
n=0
xn
! ∞
X
n=0
yn
!
Functions and Continuity
1. Open and Closed Sets of ❘
Definition 4.1. Let D ⊂ ❘, and a ∈ ❘. We say that a is a limit point of D if there exists a sequence {xn}
∞
n=1 such that xn ∈ D for every n ∈ ◆, and
limn→∞xn=a. The set of all limit points ofD is denotedD.
Example4.1. LetD⊂❘, thenD⊂D. However, it may be thatD6=D. Definition4.2. A setD⊂❘isclosed ifD=D. A setD⊂❘isopen if❘\D is closed.
Theorem 4.1. Let D⊂❘. Then D is closed.
Definition 4.3. Leta < b, I ={x∈❘: a < x < b}, andJ ={x∈❘: a✻ x ✻ b}. I is called an open interval, and is denoted (a, b). J is called a closed interval, and is denoted [a, b].
Theorem 4.2. Let D⊂❘. ThenD is open if and only if the following
state-ment holds: for every x∈D, there exists an ε >0 such that (x−ε, x+ε)⊂D.
Example4.2. Every open interval is open. Every closed interval is closed.
Example4.3. A set may be neither open nor closed.
Theorem 4.3. If a set D ⊂❘ is both open and closed, then eitherD =❘or D=∅.
Definition 4.4. LetD⊂❘. We define the boundary ofD by ∂D=D∩(❘\D).
(4.1)
LetD⊂❘. We define the interior ofD by: intD=D\∂D. (4.2)
Theorem 4.4. Let D ⊂❘, then x∈intD if and only if there is ε > 0 such
that (x−ε, x+ε)⊂D.
Theorem 4.5. Let D⊂❘, then D is open if and only ifD= intD.
Definition 4.5. Let D ⊂ ❘. A set U ⊂ D is relatively open in D, if there exists an open setV ⊂❘such thatV ∩D=U.
Example4.4. A subset U ⊂ D may be relatively open in D but not open. However, ifD is open andU is relatively open inD, then U is open.
Let{Uα: α∈A} be a family of sets. The set A is the index set, and it can
be finite or infinite. IfA=◆for instance, we have a sequence of sets, but we may
want to consider even more general situations. We define the unionS
α∈AUαto be
the set of all elements xwhich belong to at least one Uα. Thus, x∈Sα∈AUα iff
∃α∈Asuch thatx∈Uα. Similarly, we define the intersectionTα∈AUαto be the
set of all elementsxwhich belong to all theUα. Thus,x∈Tα∈AUαiff∀α∈Awe
havex∈Uα.
Theorem 4.6. Let {Uj: 0 ✻j ✻n} be a finite family of open subsets of ❘.
ThenTn
j=0Uj is open.
Example4.5. The intersection of a collection of open sets may not be open.
Theorem 4.7. Let {Uα: α ∈ A} be a family of open subsets of ❘. Then
S
α∈AUα is open.
Theorem 4.8. The union of a finite number of closed sets is closed. The
in-tersection of an arbitrary collection of closed sets is closed.
Definition4.6. A setK⊂❘iscompact if every sequence inK has a subse-quence which converges to an element ofK.
Theorem 4.9. Let K⊂❘be compact. ThenK is closed and bounded.
Theorem 4.10. LetK⊂❘be closed and bounded. ThenK is compact.
2. Limit and Continuity
Definition4.7. LetD⊂❘,f:D→❘,a∈D, andL∈❘. We say that Lis thelimit off ataif the following condition holds:
Ifxn ∈D, and lim
n→∞xn=a, then limn→∞f(xn) =L.
Remark. In other words, we require that for any sequence{xn}
∞
n=1of elements
in D which converges to a, the sequence{f(xn)}∞n=0 converges to L. When L is
the limit of the functionf ata, we write
lim
x→af(x) =L.
(4.3)
Theorem 4.11. Let D ⊂❘,f:D →❘, a∈D, and L, M ∈❘. Suppose that
both limx→af(x) =L, andlimx→af(x) =M. Then L=M.
Hence, iff: D→❘has a limit ata∈D, then that limit is unique. Thus, we may speak ofthelimit of the functionf ata. Next, we give an equivalent condition forLto be the limit off ata.
Example4.6. Let a, c ∈ ❘, and let f:❘ → ❘ be the function which sends x∈❘toc. Then
lim
x→af(x) =c.
(4.4)
Example4.7. Letf:❘→❘be the function which sendsx∈❘tox, and let a∈❘. Then
lim
x→af(x) =a.
Theorem 4.12. Let D ⊂ ❘, f, g: D → ❘, a ∈ D, L, M ∈ ❘. Suppose that limx→af(x) =L, andlimx→ag(x) =M.
lim
x→a f(x) +g(x)
=L+M, (4.6)
lim
x→a f(x)g(x)
=LM. (4.7)
Example4.8. Let D ⊂ ❘, a ∈ D, c ∈ ❘, and f:D → ❘. Suppose that limx→af(x) =L. Then limx→acf(x) =cL.
Theorem 4.13. Let D ⊂ ❘, a ∈ D, f, g: D → ❘, L, M ∈ ❘. Suppose limx→af(x) = L,limx→ag(x) = M,M 6= 0, and that for every x∈D, g(x)6= 0.
Then
lim
x→a
f(x) g(x) =
L M. (4.8)
Definition 4.8. Letak ∈❘, fork= 0,1, . . . , n. A function p:❘→❘which can be written as
f(x) =
n
X
k=0
akxk
(4.9)
is called apolynomial. The set of all polynomials is denoted by❘[x]. Letp, q∈❘[x], q 6= 0, and let D ⊂❘ be the set of all x∈ ❘ such that q(x) 6= 0. The function r:D→❘, given by
r(x) = p(x) q(x) (4.10)
is called arational function.
Theorem 4.14. Letp∈❘[x], and leta∈❘. Then lim
x→ap(x) =p(a).
(4.11)
Theorem 4.15. Letr:D→❘ be a rational function, and leta∈D. Then lim
x→ar(x) =r(a).
(4.12)
Definition 4.9. Let D ⊂ ❘, f: D → ❘, and a ∈ D. We say that f is
continuous ataif
lim
x→af(x) =f(a).
(4.13)
We say thatf is continuous onD if for eacha∈D,f is continuous ata.
Example4.9. Letr:D →❘be a rational function, then r is continuous on D.
Theorem 4.16. Let D ⊂❘, and let f, g: D →❘. Suppose thatf, g are
con-tinuous ata∈D. Thenf +g andf g are continuous ata.
Theorem 4.17. Let D ⊂❘, f: D →❘, a∈D, and L ∈❘. The function f
has the limitL ataif and only if the following condition holds:
For everyε >0, there is aδ >0 such that ifx∈D and|x−a|< δ, then|f(x)−L|< ε.
Theorem 4.18. Let D ⊂❘,f:D →❘, a∈D, and L ∈❘. Thenf has the
For every open set W ⊂❘, such that L ∈ W, there is an open set
U ∈❘such that a∈U andf(U∩D)⊂W.
Theorem 4.19. Let D ⊂ ❘, f: D →❘. Then f is continuous on D if and
only if the following condition holds:
For every open setV ⊂❘, the setf−1(V) is relatively open inD.
Theorem 4.20. Let f: D→❘be a continuous function, and suppose that D
is compact. Then f(D)is compact.
Definition 4.10. A functionf:D→❘is said to bebounded above iff(D) is bounded above. For such a function, we will write:
sup
D
f = supf(D). (4.14)
The functionf is said to bebounded below iff(D) is bounded below. For such a function, we will write:
inf
D f = inff(D).
(4.15)
It is said to bebounded if it is bounded above and below.
Example4.10. There is a functionf:D→❘bounded above for which there is nox∈Dsuch thatf(x) = supDf.
Theorem 4.21. LetDbe compact, and letf:D→❘be a continuous function.
Then there isx∈D such that f(x) = supDf.
Theorem 4.22. LetDbe compact, and letf:D→❘be a continuous function.
Then there isx∈D such that f(x) = infDf.
Theorem 4.23. Let a, b ∈ ❘, a < b, let f: [a, b] → ❘ be continuous, and
suppose that
f(a)✻0✻f(b). (4.16)
Then there isx∈[a, b] such thatf(x) = 0.
3. Uniform Continuity and Uniform Convergence
Definition 4.11. LetD ⊂❘, and letf:D→❘. We say thatf isuniformly
continuous onD is for every ε >0, there exists aδ >0 such that for all x, y∈D which satisfy|x−y|< δ there holds|f(x)−f(y)|< ε.
Theorem 4.24. Let D ⊂❘be bounded, and let f:D →❘ be uniformly
con-tinuous. Thenf is bounded.
Example4.11. Give a domain D ⊂ ❘ and a function f: D → ❘ which is continuous onDbut not uniformly continuous.
Theorem 4.25. Let D be compact, and let f:D →❘ be continuous. Thenf
is uniformly continuous.
Definition 4.12. LetD⊂❘. Asequence of functions{fn}∞
n=1onD is a map
from◆into the set of functions❘D={f:D→❘}. If for eachn∈◆the function
fn is continuous on D, we say that{fn}
∞
n=1 is asequence of continuous functions
onD. Let{fn}
∞
n=1 be a sequence of functions onD, and let f:D →❘. We say
that{fn}∞n=1converges tof if for eachx∈Dthe sequence of numbers{fn(x)}∞n=1
Example4.12. There exists D ⊂❘, and a sequence of continuous functions
{fn}∞n=1 onD which converges to a functionf:D→❘which is not continuous.
Definition 4.13. Let D ⊂❘, let {fn}∞
n=1 be a sequence of functions on D,
and letf:D→❘. We say that{fn}
∞
n=1converges uniformlytof onDif for every
ε >0, there existsN ∈◆such that for all n∈◆which satisfyn❃N, and for all x∈D, there holds|fn(x)−f(x)|< ε.
Theorem 4.26. LetD⊂❘, and let{fn}∞
n=1be a sequence of continuous func-tions onD which converges uniformly tof. Then f is continuous on D.
LetD⊂❘, and let{fn}
∞
n=1be a sequence of functions defined onD. As with
numerical series, we define the sequence of partial sums of the series P∞
n=1fn by
sm = Pmn=1fn, and we say that the seriesP
∞
n=1fn converges if the sequence of
partial sums{sm}
∞
m=1converges.
Theorem 4.27. LetD⊂❘, and let{fn}∞
n=1be a sequence of continuous func-tions on D. Suppose that there are numbers cn ∈ ❘ such that |fn| ✻ cn, and
such that P∞
n=1cn converges. Then the series
P∞
Integration and Differentiation
1. The Lower and Upper Intergals
We will use the notation{xj}nj=0for afinite sequence, i.e., a functionx:N0n→
❘, wherenis some positive integer, and Nn
0 ={j ∈❩: 0✻j✻n}. As before, a
finite sequence{xj}nj=0 isincreasing ifxj−1< xj for all 1✻j ✻n.
Definition 5.1. Letn∈◆. An increasing finite sequence P ={xj}n j=0 with
x0=a, andxn =bis called apartition of [a, b]. We denote the set of all partitions
of [a, b] by P[a, b]. If P = {xj}nj=1 ∈ P[a, b], we define ∆j(P) = xj −xj−1. If
P1, P2∈P[a, b], we say thatP1is arefinementofP2, writtenP1⊃P2, if the image
ofP2as a function is contained in the image of P1.
Theorem 5.1. LetP1, P2∈P[a, b]. Then there isP ∈P[a, b]such thatP⊃P1,
andP ⊃P2.
A partition P satisfying P ⊃ P1, and P ⊃ P2, as given by this theorem, is
called acommon refinement ofP1 andP2.
Definition 5.2. Let f be a bounded function defined on [a, b], and letP ∈ P[a, b]. We define for each 1✻j✻n:
L(f, P) =
n
X
j=1
mj(f, P)∆j(P), U(f, P) = n
X
j=1
Mj(f, P)∆j(P),
where
mj(f, P) = inf
[xj−1,xj]
f, Mj(f, P) = sup
[xj−1,xj]
f.
We define thelower integral and theupper integral off over [a, b] by:
Z b
a
f(x)dx= sup
L(f, P) : P ∈P[a, b] ,
Z b
a
f(x)dx= inf
U(f, P) : P ∈P[a, b] .
Theorem 5.2. Let f be a bounded function on[a, b], and letP ∈P[a, b], then L(f, P) =−U(−f, P).
Theorem 5.3. Let f be a bounded function on [a, b], and let P1, P2 ∈ P[a, b]
satisfy P1⊃P2. Then,L(f, P2)✻L(f, P1), andU(f, P1)✻U(f, P2).
Theorem 5.4. Let f be a bounded function on [a, b], and let P1, P2 ∈P[a, b],
thenL(f, P1)✻U(f, P2).
Theorem 5.5. Let f be a bounded function on [a, b], then
Z b
a
f dx✻
Z b
a
f dx.
2. The Riemann Integral and its Properties
Definition5.3. Let f be a bounded function on [a, b]. We say that f is
Riemann integrable, writtenf ∈❘[a, b], if the lower and upper integrals off over [a, b] coincide. In this case, we denote their common value by:
Z b
a
f(x)dx.
Theorem 5.6. Let f ∈❘[a, b]. Then, for everyε >0 there isP ∈P[a, b]such
that
U(f, P)−L(f, P)< ε.
Theorem 5.7. Let f be a bounded function on [a, b]. Suppose that for every ε >0 there isP ∈P[a, b]such that
U(f, P)−L(f, P)< ε.
Thenf ∈❘[a, b].
Theorem 5.8. Suppose that f is continuous on [a, b]. Thenf ∈❘[a, b]. Theorem 5.9. Suppose that f ∈❘[a, b], and letc∈❘. Then cf∈❘[a, b] and
Z b
a
cf(x)dx=c
Z b
a
f(x)dx.
Definition5.4. A function f:D → ❘ is non-decreasing on D if for every x, y∈D such thatx✻y, there holdsf(x)✻f(y). We say thatf isnon-increasing
on D is −f is non-decreasing. A function f: D → ❘is monotonic if f is either non-decreasing or non-increasing on [a, b].
Theorem 5.10. Suppose thatf is monotonic on[a, b]. Thenf ∈❘[a, b]. Theorem 5.11. Suppose thatf, g∈❘[a, b]. Thenf+g∈❘[a, b] and
Z b
a
f(x) +g(x)
dx=
Z b
a
f(x)dx+
Z b
c
g(x)dx.
Theorem 5.12. Let f ∈❘[a, b] and a < c < b. Then, we have f ∈ ❘[a, c]∩ ❘[c, b], and:
Z b
a
f(x)dx=
Z c
a
f(x)dx+
Z b
c
f(x)dx.
Theorem 5.13. Suppose thatf ∈❘[a, b], andf ❃0. Then
Z b
a
f(x)dx❃0.
Theorem 5.14. Suppose thatf, g∈❘[a, b], andf ✻g. Then
Z b
a
f(x)dx✻
Z b
a
Theorem 5.15. Suppose thatf ∈❘[a, b]. Then |f| ∈❘[a, b], and is differentiable at eacha∈D, then we say thatf is differentiable onD.
Theorem 5.19. Let f: D →❘ be differentiable at a∈intD. Then f is
con-function f+g is differentiable ata, and
(f+g)′
function f g is differentiable ata, and
(f g)′
(a) =f′
Theorem 5.23. For each 1 ✻ k ✻ n, let ck ∈ R, and let p:❘ → ❘ be the
polynomial given by:
p(x) =
n
X
k=0
ckxk.
(5.6)
Let a∈❘, thenpis differentiable ata, and
p′ (a) =
n
X
k=1
k ckak−1.
(5.7)
4. The Mean Value Theorem
Theorem 5.24. Letf: [a, b]→❘be continuous on[a, b], and differentiable on (a, b). Suppose that there isc∈(a, b), such that f(c) = sup
(a,b)
f. Thenf′
(c) = 0.
Theorem 5.25. Letf: [a, b]→❘be continuous on[a, b], and differentiable on (a, b). Suppose also thatf(a) =f(b) = 0. Then there isc∈(a, b)such that
f′ (c) = 0. (5.8)
Theorem 5.26. Letf: [a, b]→❘be continuous on[a, b], and differentiable on (a, b). Then there isc∈(a, b)such that
f′
(c) = f(b)−f(a) b−a . (5.9)
Theorem 5.27. Letf: [a, b]→❘be continuous on[a, b], and differentiable on (a, b). Suppose also that
f′
(c) = 0, (5.10)
for allc∈(a, b). Then, there isd∈❘, such thatf(x) =dfor all x∈[a, b].
Theorem 5.28. Letf: [a, b]→❘be continuous on[a, b], and differentiable on (a, b). Suppose also that
f′
(c)❃0, (5.11)
for eachc∈(a, b). Thenf is non-decreasing on [a, b].
5. Integration and Differentiation
Theorem 5.29. Letf ∈❘[a, b]. DefineF: [a, b]→❘by:
F(x) =
Z x
a
f(t)dt,
forx∈[a, b]. Iff is continuous at c∈[a, b], thenF is differentiable atc, and
F′
(c) =f(c).
Theorem 5.30. Let f ∈ ❘[a, b], and suppose there is a continuous function F: [a, b]→❘such that F′(x) =f(x)for each x
∈(a, b). Then
Z b
a
