solutions manual apostol.pdf

7/22/2019 Solutions manual Apostol.pdf

1/77

Mathematical AnalysisT. M. Apostol


2/77

Chapter 3


3/77

Mathematical Analysis by Tom M. Apostol

Chapter 3: Elements of Point Set Theory: Notes

Let E1 denote the set of all real numbers (the real line). Let E2 denote the set of all

complexnumbers (the complex plane). An open interval (a, b) is defined as (a, b) = {x | a < x 0, and let x be a given point. The open interval (x-h, x+h) is called a

neighbourhood with x as centreand of radiush. We denote the neighbourhood by N(x; h), or

simply by N(x) if the radius is unimportant.

Let S be a set in E1and assume that x S. Then x is called an interior point of S if thereis some neighbourhood N(x) all of whose points belong to S.

Let S be a set in E1. S is called an open set if every point of S is an interior point of S.

Thus an open set is such that each of its pointscan be enclosed in a neighbourhood which is

completely contained in the set. The simplest kind of open set is an open interval. The empty

set is also open, as is the real line E1.

The union of any collection of open sets is an open set. The intersection of a finite

collection of open sets is open. Note that arbitrary intersectionswill not always lead to open

sets. The union of a countable collection of disjoint open intervals is an open set and,

remarkably enough, every open set on the real line can be obtained in this way.

The structure of open sets in E1. A set of points in E1 is said to be bounded if it is a

subset of some finite interval. Let S be an openset in E1 and let (a, b) be an open interval

which is contained in S, but whose endpoints are notin S. Then (a, b) is called a component

interval of S.

If S is a bounded open set, then (i) each point of S belongs to auniquely determined

component interval of S; and (ii) the component intervals of S form a countable collection of

disjoint sets whose union is S. From this, it follows that an open interval in E1 cannot be

expressed as the union of two disjoint open setswhen neither set is empty.

Everyopen set in E1is the union of a countable collection of disjoint open intervals.

Accumulation points and the Bolzano-Weierstrass theorem in E1. Let S be a setin E1and

x a pointin E1, x not necessarilyin S. Then x is called an accumulation pointof S, provided

every neighbourhood of x contains at least one point of S distinct from x. If x is an

accumulation point of S, then everyneighbourhood N(x) contains infinitely many points of S.


4/77

Theorem 3-13 (Bolzano-Weierstrass): If a bounded set S in E1contains infinitelymany

points, then there is at least one point in E1which is an accumulation pointof S. Examples: (1)

The set of numbersof the form 1/n,n = 1, 2, 3, ... has 0 as an accumulationpoint. (2) The set of

rational numbers has every real number as an accumulation point. (3) Every point of theclosed

interval[a, b] is an accumulation point of the set of numbers in the open interval (a, b).

Closed sets in E1. A set is called closedif it contains all its accumulation points. Thus aclosed interval is a closed set. An open interval, however, is not a closed set because it does

notcontain its endpoints, both of which are accumulation points of the set. A set which has no

accumulation points is automatically closed and, in particular, every finite set is closed. The

empty set is also closed, as is the whole real line E1. (These last two sets are alsoopen!) A set

which is not closed need not be open, as, for example, the half-open interval (a, b], which is

neitheropen nor closed.

If S is open, then the complement E1S is closed. Conversely, if S is closed, then

E1S is open. It follows from the above that the union of a finite collection of closed setsis

closed and that the intersection of an arbitrary collection of closed sets is closed.

Generalisation: If S is closed, then the complementof S (relative to any open set containing S)

is open. If S is open, then the complement of S (relative to any closed set containing S) is

closed.

Extensions to higher dimensions. The definition of a neighbourhood, |x-x0| < h, still

makes sense if x and x0belong to higher dimensional spaces.

Let n > 0 be an integer. An ordered setof n real numbers (x1, x2, ..., xn) is called an

n-dimensionalpointor a vectorwith n components. Points or vectors will be denoted by singleboldfacedletters, e.g. x= (x1, x2, ..., xn), where xkis the k

th co-ordinate of the point or vector x.

The set of all n-dimensional pointsis called the n-dimensional Euclidean space and is denoted

by En. The usual vector rules apply. Note: Unit coordinate vectors: uk= (k,1, k,2, ..., k,n) (k =

1, 2, ..., n), where k,jis the Kroneker delta, defined by k,j= 0 if k j, and k,k= 1.

Let x= (x1, ..., xn) and y= (y1, ..., yn) be in En. The absolute value, or length, or normof x

is given by |x| = . The distancebetween xand yis given by |x-y| = .x12 + ... +xn

2Si=1

n (xi yi)2

|x| > 0, and |x| = 0 iff x= 0. |x-y| = |y-x|. |x+y| |x| + |y|.

By an open sphere of radiusr > 0 and centreat the point x0in En, we mean the set of all

xin Ensuch that |x-x0| < r. The set of xsuch that |x-x0| r form a closed sphere, the boundaryof the sphere being the set of xwith |x-x0| = r. An open spherewith centre at x0 is called a

neighbourhood of x0 and is denoted by N(x0) or by N(x0; r) if r is the radius. The

corresponding closedsphere is denoted by (x0). The open sphere with its centre removed isN

called a deleted neighbourhood of x0and is denoted by N(x0).

Let a= (a1, ..., an) and b= (b1, ..., bn) be two distinctpoints in En, such that akbk for

each k = 1, 2, ..., n. The n-dimensional closed interval[a, b] is defined to be the set [a, b] ={(x1, ..., xn) | akxkbk, k = 1, 2, ..., n}. If ak< bkfor every k, the n-dimensional open interval(a, b) is the set (a, b) = {(x1, ..., xn) | ak< xk< bk, k = 1, 2, ..., n}.


5/77

Thus, for example, (a, b) can be considered as the cartesian product (a, b) = (a1, b1)

(a2, b2) ... (an, bn) of the none-dimensional open intervals (ak, bk). An open interval in Encould then be called a neighbourhoodof any of its points, and in what follows, it makes no

difference whether a neighbourhood is taken to be a sphere or an interval (we will use

spheres).

Let S be a set of points in En, and assume that xS. Then xis called an interior point ofS if there exists a neighbourhoodN(x) S. The set S is said to be openif each of its points isan interior point. The interior of S is the collectionof its interior points.

Examples of open sets in the plane are: the interior of a disk; the first quadrant; the

wholespace. Any n-dimensional open sphere and any n-dimensional open interval is an open

setin En. Caution: an open interval (a, b) in E1is no longer an open set when it is considered as

a subset of the plane. In fact, nosubset of E1(except the empty set) can be open in E2, because

such a set can contain no two-dimensional neighbourhoods.

The union of an arbitrary collection of open sets in En is an open set in En, and the

intersectionof a finite collection of open sets in Enis an openset in En.

Assume that S En, x En. Then x is called an accumulation point of S if every

neighbourhood N(x) contains atleastonepoint of S distinct from x, that is, if N(x) S is notempty. If x is an accumulation point of S, then every neighbourhood N(x) contains infinitely

many points of S.

A set S in Enis said to be bounded if S lies entirelywithin some sphere; that is, for somer > 0, we have S N(0; r). Theorem 3-29 (Bolzano-Weierstrass): If a bounded set S in Encontains infinitely many points, then there exists at least one point in En which is an

accumulationpoint of S.

A set S in Enis said to be closedif it contains all its accumulationpoints. Examples: A

closed sphere in En is a closed set. An n-dimensional closed interval is a closed set. A set

which is closed in E1is also closed in Enfor n > 1.

A set S in Enis closed iff EnS is open. The union of afinite collection of closed sets inEnis closed and the intersection of an arbitrary collection of closed sets in Enis closed.

Assume that S A En. If A is open and if S is closed, then AS is open. If A isclosedand if S is open, then AS is closed.

The Heine-Borel covering theorem. We begin by defining a covering of a set. A

collection F of sets is said to be a covering of a given set S if S AFA. The collection F isalso said to cover S. If F is a collection of open sets, then F is called an open coveringof S.


6/77

Examples: (1) The collection of all intervalsof the form 1/n< x 0, then we have x(+) = +and x(-) = -. (3) If x < 0, then we have x(+) = -and

x(-) = +. (4) (+)+(+) = (+)(+) = (-)(-) = +, (-)+(-) = (+)(-) = -. (5) If x

E1, then we have -< x < +.

Every open interval (a, +) is called a neighbourhoodof +, and every open interval

(-, a) is called a neighbourhoodof -. E1= (-, ); E1* = [-, ]. Every set in E1* has a sup

it is finiteif the set is bounded above, and it is +if the set is not bounded above. Note thatE1* does not satisfy all the axioms for the real number system.

By the extended complex number system E2* we shall mean the unionof the complex

plane E2with a symbol which satisfies the following properties: (1) If z E2, then z+=

z-= , z/= 0. (2) If z E2, but z 0, then z() = , and z/0= . (3) += ()() = .

Note that -is not needed herebecause no orderingis involved with complex numbers. Every

open set in E2of the form {z| |z| > r > 0} is called a neighbourhoodof .


7/77

Chapter 3: Selected Exercises

3-1. Prove that an open intervalin E1is an open set and that a closed intervalis a closed

set.

Answer: Let us first look at some definitions. Definition 1: An open interval (a, b) is

defined by (a, b) = {x | a < x < b}. Definition 2: Let S be a set in E 1and assume that x S.Then x is called an interiorpoint of S if there is some neighbourhood N(x) all of whose points

belong to S. Definition 3: If S is a set in E1, S is an open set if every point of S is an interior

point of S.

To prove that an open intervalin E1is an openset, we need to prove that every point in

an open interval is an interior point. To do this, we must prove that every point in an open

interval has an associated neighbourhood, all of whose points belong to the open interval.

Consider an arbitrary open interval (a, b). Let x be anypoint in this interval, x (a, b).A neighbourhood of x is given by the interval (x-h, x+h), where h > 0. What we have to show

is that oneof these neighbourhoods is entirely contained within the interval (a, b), i.e. we have

to show that (x-h, x+h) (a, b) for every x and for some h > 0. To do this, we have to show

that for every y(x-h, x+h), we alsohave y (a, b).

Let us chooseh to be given by half the value of the

minimum distance from x to each of the two endpoints, so

that h = min(|x-a|, |x-b|). Because we know that x > a

and that x < b, we can rewrite h as h = min(x-a, b-x).

From this definition, we see that (i) h (x-a), and that (ii) h (x-b).

Let us now analyse what we know about the point y. First of all, we know that y (by

definition) sits in the interval (x-h, x+h), so that x-h < y < x+h. In order to show that a < y < b,

which is what we want to prove, all we have to show is that (1) a < x-h, and that (2) x+h < b.

From (i) above, we can do the following manipulation:

h (x-a)

x-h x-(x-a)

x-h(x+a)

x-h > (a+a) (because a < x)

x-h > a provingpart (1)

From (ii) above, we can do the following manipulation:

h (x-b)

x+h x+(x-b)

x+h (3x-b) x+h < (3b-b) (because x < b)

x+h < b provingpart (2)

( )

a b

( )x-h x x+h


8/77

Because we have shown that y (x-h, x+h) y (a, b), then we have proved that anopen interval is an open set. To show that a closed intervalis a closedset, we use the following

trick: the complementof an open set is a closed set.

Let us consider an arbitrary closed interval [a, b]. The complement of this interval is

given by E1[a, b] = (-, a)(b, ), the union of two open intervals, which we know are open

sets by the above. Further, the unionof two open sets is also an open set (Theorem 3.5), so thatE1[a, b] is an open set. So, as the complement of a closed interval is an open set, then a

closed interval must therefore be a closed set. QED.

3-2. Determine all the accumulation points of the following sets in E1 and decide

whether the sets are openor closed(or neither). (a) All integers. (b) The interval (a, b]. (c) All

numbers of the form 1/n, (n = 1, 2, 3, ...). (d) All rational numbers. (e) All numbers of the form

2-n+ 5-m, (m, n = 1, 2, ...). (f) All numbers of the form (-1)n+ (1/m), (m, n = 1, 2, ...). (g) All

numbers of the form (1/n) + (1/m), (m, n = 1, 2, ...). (h) All numbers of the form (-1)

n/[1+(1/n)], (n

= 1, 2, ...).

Answer: Let us first remind ourselves of the definitionof an accumulation point: Let S

be a set in E1and x a point in E1, x not necessarily in S. Then x is called an accumulation point

of S provided every neighbourhood of x contains at least one point of S distinct from x.

(a) Let S = {x | x Z}. S contains noaccumulation points because every point x has aneighbourhood (x-, x+) which contains no other integers, so that x is not an accumulation

point. Because S contains allits accumulation points (of which there are none!), it follows that

S is a closedset, and hence not an open set (because of exercise 3-5).

(b) Let S = (a, b]. Consider an arbitrary point x S. Taking an arbitrary neighbourhoodN(x) of x, N(x) = (x-h, x+h), with h > 0, then it is clear that the point max{x-a/2, x-

h/2} will

always be in S and in N(x) so that everyneighbourhood of a point in S will always contain at

least one other point from S. We therefore conclude that all the points in S are accumulation

points. Further, a is an accumulation point of S so that theset of accumulation pointsof S is

given by the set {x | a x b}.

S is not closed because it doesnt contain the accumulation point a. Further, S is notopen because b is notan interior point (in the neighbourhood (b-h, b+h), with any h > 0, the

points in (b, b+h) (b-h, b+h) do not belong to S but do belong to the neighbourhood. Itfollows that b is not an interior point).

(e) Let S = 2-n+5-m(n, m = 1, 2, ...). Claim: all numbers of the form x = 2-n(n = 1, 2, ...)

and of the form y = 5-m(m = 1, 2, ...) are accumulation points of S. Justification: to show that

each point x = 2-n is an accumulation point of S (n = 1, 2, ...), consider an arbitrary

neighbourhood N(x) of x, N(x) = (x-h, x+h), with h > 0. If we now choose an integer lso that

5-lh/2(and there will alwaysbe such an l, given by l= ceiling(-log(n/2)/log(5))), then the pointy = 2-n+5-lwill be in N(x) and in S.


9/77

It therefore follows that all points of the form x = 2-n (n = 1, 2, ...) are accumulation

points of S. A similar argument can be used to show that all points of the form y = 5-m(m = 1,

2, ...) are accumulation points of S. There is one other accumulation point of S: zero (every

neighbourhood of zero will contain a point from S if n and m are large enough), and because 0

S, then S is nota closed set.

The question remains as to whether S is an open set. To prove that it is not, it issufficient to find a single point in S which does not have a neighbourhood all of whose points

belong to S. Take the point in S given by setting n = 1 and m = 1, namely the point 7/10. Clearly

this is the largest element of the set S, and so every neighbourhood of this point (i.e. the

interval (7/10-h,7/10+h) for some h > 0) will contain elements which are larger than

7/10,

elements which cannot be expressed in the form 2-n+5-m, and so the point 7/10 S is not aninterior point so that S is not an openset.

(h) Let S = (n = 1, 2, ...). Writing S out as asequence, S = {-1/2,2/3, -

3/4,4/5, -

5/6, ...}.(1)n

1+ 1n

From this series representation, we see that S is an alternating series, one half of the seriestending to -1, and the other halftending to 1. Therefore, -1 and 1 are accumulation pointsof

the set S, because the set has elements which are arbitrarily close to -1 and 1. There are no

other accumulation points in the set.

Because S does not containits two accumulation points (-1 S and 1 S), then S is nota closed set. By the same argument as in part (e), no neighbourhoodof the largest element of S,2/3, will contain elementsjustfrom S, so that

2/3is not an interiorpoint of S, and so S is not an

openset.

3-6. Show that every closed set in E1is the intersectionof a countable collection of open

sets.

Answer: Let us first consider some examplesof closed sets and the intersection of open

sets that is equivalent to the closed set in question. To start with, consider the set S consisting

of a single point x E1. This set is closedbecause it has no accumulation points, and it can beexpressed as the intersection of a countable collection of open sets as follows:

S = n=1

(x-

1

/n, x+

1

/n).

Now consider a closed interval[a, b], which we have shown in a previous exercise to be

a closed set, which we shall denote by T. This closed set can be expressed as the intersection of

a countable collection of open sets as follows:

T = n=1(a-1/n, b+

1/n).

We therefore have some evidence that the statement that we are trying to prove is

correct. We now want to show that everyclosed set C can be expressed as the intersection of

some countable collection of open sets, i.e. we want to show that C can be written as

C = n=1On, where every Onis an openset.


10/77

To start with, let us define D to be given by D i, j= (di-1/j, di+

1/j), where diC, and j R>0. In other words, D is the 1/jinterval of a point from C. We see immediately that Di, jis an

open set because it consists of an open interval, which we know (from exercise 3-1) to be an

open set. Now let us define Ej to be the following expression: Ej= iDi, j, the union over allpoints of C of 1/jintervals of all those points. Each Ejis an open set because it is an arbitrary

union of open sets, which (by theorem 3-5) is an open set. Finally, let us define F to be the set

given by F = j=1

Ej. It follows that F is the intersection of a countable collection of open sets.

The question now arises as to whether we have the conclusion F = C. We have already

shown that F is the intersection of an arbitrary collection of open sets, and so it is in the right

form to be considered as thesolutionto this exercise. We will in fact show that F = , i.e. showC

that F is the closure of the set C. If we can do this, then we will also show that F = C because

for a closed set C, we have (looking at part (f) of exercise 3-12) C = . Therefore, if we canC

show that F = , then we reach the required conclusion. To do this, as = CC (see exerciseC C3-12 for this definition), then we first need to show that F contains all of Cs accumulation

pointsand all of thepointsfrom C.

Claim 1: F contains all of Cs accumulation points, whether they are in C or not. Proof

of Claim: Consider an arbitrary accumulation point of the set C, say the point x. If x C, then

we refer to the proof of claim 2 below to show that x F. If the accumulation point x is notinC, then we must take a little more care.

Following the definition of an accumulation point, every

neighbourhood N(x) of x will contain a point y C. But for every

such y, there exists a corresponding interval Dy,jwhich in turn is alsoin every Ej. When we take the intersection of all the Ejs, we close

in on the point x (see the diagram on the right), the crucial feature

being that every interval Ejwill containthe point x. Therefore, F will also contain the point x.

End of Proof.

Claim 2: F contains all of the points from C, i.e. C F. Proof of Claim. Consider a point

y C. If y C, then there is a set Dy, j= (y-1/j, y+1/j), where j R>0. It follows that y belongs to

eachEj, where Ej= iDi, j, and so y belongs to F = j=1Ej, so that y F. Because y C y

F, then C F. End of Proof.

Conclusion: Because we have shown that C F and that C F, then CC F. To

complete the proof, we need to show that F CC.

If z F, then z belongs to all of the sets Ej, where Ej= iD i, j. If z belongs to all of the

sets Ej, then there must be a set Dx, j= (x-1/j, x+

1/j), where j R>0. It follows (by the definition

of Di, jand by the proof of claims 1 and 2 above) that z CC, and so z F z CC, so

that F CC. QED. Conclusion: we have expressed an arbitrary closed set C as theintersection of a countable collection of open sets, namely the set F. QED.

(x

E1

E2

E3

etc.


11/77

3-9. Show that the interior of a set S in Enis an open set.

Answer: Let I be the set of all interior pointsfrom S, i.e. (by definition) I is the interior

of S. For a set A to be an openset, all the points in A must be interior points. Because all the

points in our set I are interior points of S, it follows that I is an open set, and therefore the

interior of a set S in Enis an openset. QED.

3-12. If S is a set in En, let S denote the set of accumulation pointsof S, and let =S

SS. (The set S is called the derivedsetof S and is called the closureof S.) Show that (a)S

S is a closedset, that is, (S) S. (b) If S T, then S T. (c) (ST) = ST. (d) ( ) =SS. (e) is a closedset. (f) S is closedif, and only if, S = .S S

Answer: (a) If x Enand x (S), then x is an accumulation point of the set S. If x is

an accumulation point of the set S, then every neighbourhood N(x) of x contains at least one

other point y S (y x), i.e. another accumulation point of S. It follows that all

neighbourhoods N(y) of y will contain a point z S. Pick a

neighbourhood N(y) such that N(y) N(x) and xN(y). Note that

in order to do this, we must have < max(dist(x,y), -dist(x,y)). So as

every neighbourhood N(x) of x contains a point z S (with z x), then

it follows that x must be an accumulation pointof the set S, so that x

(S) x S, and so (S) S as required. QED.

(b) If S T, then x S x T. Now if y S, then y is an accumulation point of the

set S. But if y is an accumulation point of S, then eithery S, which implies that y T, and so

y is also an accumulation point of the set T; or y S, but every neighbourhood of y contains apoint from z from S, which is also (because S T) a point from T, so that everyneighbourhood of y contains a point z from T, and so y is an accumulation point of the set T.

Bringing the two cases together, we see that if y S, then y T, and so S T. QED.

(c) To prove that (ST) = ST, we must prove that (i) (ST) ST, and that (ii)

ST (ST). (i) If x (ST), then x is an accumulation point of the set ST, whichimplies that either(1) x is an accumulation point of S, or(2) an accumulation point of T, or (3)

an accumulation point of both S and T; which implies that either (1) x S ST, or (2) x

T ST, or (3) x STST. In all cases, x (ST) x ST, so that(ST) ST. (ii) If x ST, then either x S (ST), or x T (ST), so that x

ST x (ST), and so ST (ST). QED.

(d) ( ) = (SS) = (from (c)) = S(S) = ST (where T S) = S. QED.S

(e) To be a closed set, must contain all its accumulation points. From (d), we knowS

that the set of the accumulation points of is the set S. But S as = (SS). So asS S S Scontains all its accumulation points, we conclude that isa closed set. QED.S

x N(x)

yN(y)


12/77

(f) If. If S = , then either S = , i.e. S contains no accumulation points, and so S isS

closed by definition; or S consists of pointsfromS, i.e. S S, so that again S contains all itsaccumulation points, and is again a closed set by definition. Only If. If S is closed, then it

containsall its accumulation points. Therefore, S will consist of elementsfromS, so that S

S, and so = (SS) = (using S S) = S in this case. QED.S

3-15. The collection F of intervals of the form1

/n< x ng. But G (by construction) has noelements of this type, so that the set Gcovers no part of the interval (0, 1/ng]. We therefore conclude that the set G does not fully

cover the interval (0, 1). QED. (Note: a quickerproof is to show that 1/ng(0, 1) but 1/ng

G.)

3-19. Assume that S En. A point xin Enis said to be a condensation pointof S if every

neighbourhood N(x) has the property that N(x)S is not countable. Show that if S is notcountable, then there exists a point xinS such that xis a condensation point of S.

Answer: Divide the set S up into n (possibly partially overlapping) sets Si,

where n is afinitenumber; i = 1, 2, ..., n; and iSi= S. If the set S is notcountable, then at least one of these regionswill have to be uncountable as

well, or else we will have a finite collection of countable sets and thus S

will be countable and we will have a contradiction.

Choose one of the sets Siwhich is uncountable, and repeat the above with this set (i.e.

divide it up into afinite collection of smaller sets, at least oneof which will be uncountable). If

we continue to do this, then this process will converge onto a point zS which will be thecondensation point we are looking for. Justification: with everysubdivision, we find a smaller

and smaller region which is uncountable. In the limit of this process, every neighbourhood of

the point zwe converge onto will contain one of the regions which we know to be uncountable

(i.e. some region SjN(z)). Therefore, every neighbourhood of this point zhas the property

that N(z)S is not countable, i.e. z is a condensationpoint. Conclusion: given an uncountable

set S, we can always find a condensation point zS. QED.

S1 S2 S3 Sn

S


13/77

3-20. Assume that S Enand assume that S is not countable. Let T denote the set ofcondensationpointsof S. Show that

(a) ST is countable, (b) ST is notcountable,(c) T is a closed set, (d) T contains noisolated points.

Note that Exercise 3-19 is a veryspecialcase of (b).

Answer: (a) If ST was not countable, then by 3-19 it would containa condensation

point. But if T is the set of condensation points of S, then ST contains no condensation

points. It follows (by contradiction) that ST is countable. QED

(b) The set ST is the set of condensation points which are in S. We must thereforeprove that theset of condensation pointsin S is not countable. To do this, we use three pieces

of information: (1) From set theory, we know that ST = STc; (2) We know that S is not

countable; and (3) We know that ST is countable. If ST is countable, and if S is notcountable, then STcmust be non countable. Using (1), it follows that ST must also be noncountable. QED.

(c) By definition, T is closed if it contains all its accumulation points. We know that

every point xT is a condensation point, so that every neighbourhood N(x) of a point xT is

not countable (more precisely, N(x)S is not countable). This implies (by Exercise 3-19) that

the neighbourhood contains another condensation point y T (N(x)S uncountable a

condensation point yN(x)S), so that the point xis an accumulation point of the set T. This

appliesfor allcondensation points, so that every point xT is an accumulation point.

It remains to show that T has no accumulation points that are notin T. Assume that zis

an accumulation point of the set T with z T. Because z is not a condensation point, itfollows that not all neighbourhoods N(z) are uncountable. Pick one such neighbourhood, say

N1(z). By the definitionof an accumulation point, N1(z) contains an element z1T. But theneighbourhoods of z1 are uncountable (by the definition of a condensation point). Pick a

neighbourhood of z1 which is contained in N1(z), say the

neighbourhood N2(z1) (We can always do this see the answer toexercise 3-12, part (a)). Because N2(z1) is uncountable, N1(z) cannot

possibly be countable, contradicting our assumption that N1(z) was

countable. It follows that allthe neighbourhoods of the point zmust be

uncountable, and so we must have zT, and so T has no accumulationpoints that are not in T. QED

(d) Because we concluded in (c) that every point in T is an accumulation point, it

follows that nopoint in T is an isolatedpoint, i.e. notan accumulation point. QED.

z N 1(z)

z1N2(z1)


14/77

3-21. A set S is said to beperfectif S = S, that is, if S is a closed set which contains no

isolated points. (Note: S is the set of accumulation points of S). Show that if F is an

uncountable closed set in En, then F can be expressed in the form F = AB, where A is aperfect setand B is a countable set (Cantor-Bendixon theorem). [Hint: Use Exercise 3-20.].

Answer: Let T be the set of condensation pointsof S. It follows that TS is the set of

condensation points inS, and that ST is the set of points in S which are not condensationpoints. We can therefore write S = (TS)(ST), i.e. S is the union of all the condensationpoints of S together with the points of S which are not condensation points. From 3-20, we

know that all the points in TS are accumulation points, i.e. we have TS = (TS). It follows

that TS is a perfect set. Also from 3-20, we know that ST is countable. Therefore, S =

(TS)(ST) is the required form, where A = TS is a perfect set, and B = ST iscountable. QED.


15/77

Possible Further Work / Evaluation

Exercise 3-19: I could make a slight adjustment in the answer due to the comment in

Draft 2 to divide the set S up into a countable collection of sets instead of into a finite

collection of sets.Adjustedanswer:

Answer: Divide the set S up into a countable collection of (possibly partiallyoverlapping) sets Si, so that iSi= S. If the set S is not countable, then at least one of theseregionswill have to be uncountable as well, or else we will have a countable collection of

countable sets and thus S will be countable and we will have a contradiction.

Choose one of the sets Siwhich is uncountable, and repeat the above with this set (i.e.

divide it up into a countable collection of smaller sets, at least one of which will be

uncountable). If we continue to do this, then this process will converge onto a point z Swhich will be the condensation point we are looking for. Justification: with every subdivision,

we find a smaller and smaller region which is uncountable. In the limit of this process, everyneighbourhood of the point zwe converge onto will contain one of the regions which we know

to be uncountable (i.e. some region SjN(z)). Therefore, every neighbourhood of this point z

has the property that N(z)S is not countable, i.e. z is a condensationpoint. Conclusion: given

an uncountable set S, we can always find a condensation point zS. QED.


16/77

Chapter 4


17/77

Chapter 4: The Limit Concept and Continuity: Notes

Let us begin with the equation xn = A, which is meant to convey the idea thatlimxdx0when x is sufficientlynear to x0, then f(x) will be as near to A asdesired. The statements x

sufficiently near to x0 and f(x) will be as near to A as desired are made mathematically

precise by the following type of definition: the symbolism means that for everyxdx0lim f(x) =A

number > 0, there is another number > 0 such that whenever 0 < |x-x0| < , then |f(x)-A| < .

We write xn = A to mean that for every > 0, there is an integer N such thatlim

nd

whenever n > N, then |xn-A| < .

If we use neighbourhood terminology, we see that both above definitions involve the

sameprinciple. To say that 0 < |x-x0| < means that x is in a neighbourhood of x0, but x x0.That is, x is in a deletedneighbourhood N(x0). Also, to say that n > N is the same as saying

that n is in a deletedneighbourhood of +.

The parts of the definitions concerned with can also be rephrased in neighbourhood

terminology. For example, the inequality |f(x)-A| < states that f(x) N(A; ) and, similarly,

|xn-A| < means that xnN(A; ). The basicprincipleinvolved in both definitions of limit is

that for every neighbourhood N(A) there must exist a neighbourhood N(x0) such that x

N(x0) implies f(x) N(A).

If f is a real-valued functiondefined at a point x= (x1, ..., xm) in Em, we use either f(x1,

..., xm) or f(x) to denote the value of f at that point. If we have several real-valued functions,

say f1, ..., fk defined on a common subsetS of Em, it is extremelyconvenient to introduce avector-valued function f, defined by the equation f(x) = (f1(x), ..., fk(x)), if xS.

Our definition of limit now assumes thefollowingform: Let fbe a function defined on a

set S in Emand let the range of fbe a subset T of Ek. If ais an accumulation pointof S and if b

Ek, then the symbolism f(x) = b is defined to mean the following: For everylimxda

neighbourhood N(b) Ek, there exists a neighbourhood N(a) Em such that x N(a)S

implies f(x) N(b).

Note 1: we writexN(a)S rather thanxN(a) in order to make certain that xis inthe domainof f. Also, we require that abe an accumulationpointof S in order to make certain

that the intersection N(a)S will never be the emptyset. Note 2: If limxaf(x) exists (finiteorinfinite), its value is uniquelydetermined.

Strictly speaking, we should somehow indicate the fact that the limit just defined

depends on the set S through which xis allowed to range. This will usually be clearfrom the

context but, if necessary, we will write f(x) = b to emphasise the fact more explicitly. Anlimxda

x S

important special case of this occurs when S is an interval in E1having aas its left endpoint.

We then write f(x) = f(x) = b, and b is called the right-hand limitat a. (left-handlimitslimxdax S

limxda+

aresimilarlydefined).


18/77

Theorem 4-4. Let a be an accumulation point of a set S in Em. Then there exists an

infinite sequence{xn} whose terms are distinctpoints of S, such that limnxn= a.

Note: Suppose a sequence {xn} has a limit, say a= limnxn, and let S = {x1, x2, ...}

denote the rangeof the sequence. If S is infinite, it follows at once from the definition of limit

that a is an accumulation point of S. The above theorem tells us that S can have no further

accumulation points. Therefore, a sequence {xn} whose range S is infinitehas a limit if, andonly if, S has exactly one accumulation point, in which case the accumulation point is also the

limit of the sequence.

Theorem 4-5. Let fbe defined on a set S in Emwith function values in Ek, and let abe an

accumulation point of S. Let {xn} be an infinite sequence whose terms are points of S, such

that eachterm xnabut such that xn= a. Then we have (i) If limxaf(x) = b, then limnlim

nd

f(xn) = b; (ii) Conversely, if for eachsuch sequencewe know that limnf(xn) exists, then all

these sequences have thesamelimit (call it y) and also limxaf(x) exists and equalsy.

In the definition of thestatementlimxaf(x) = b, it was assumed that the limitingvalueb

was given. The following theorem gives us a condition (called the Cauchycondition) which

enables us to determine, without knowing its value in advance, whether such a point bexists.

Theorem 4-6 (Cauchy condition for sequences). Let {xn} be an infinitesequence whose

terms are points in Ek. There exists a point yin Ek such that limnxn= yif, and only if, the

following condition is satisfied: For every > 0, there exists an integerN such that n > N and

m > N implies |xn-xm| < .

Theorem 4-7 (Cauchy condition for functions). Let f be defined on a set S in Em, the

function valuesbeing in Ek. Let abe an accumulation point of S. There exists a point bin Ek

such that limxaf(x) = bif, and only if, thefollowingcondition holds: For every > 0, there is

a neighbourhood N(a) such that xand yin N(a)S implies |f(x)-f(y)| < .

Let f and g be twofunctions, each defined on a set S in En, with function values in E1or

in E2. Let abe an accumulation point of S and assume we have f(x) = A, g(x) = B. Thenlimxda

limxda

we also have (i) [f(x)g(x)] = AB, (ii) f(x)g(x) = AB, (iii) f(x)/g(x) = A/B if B 0.limxdalimxda

limxda

Continuity. Definition 4-9. Let fbe defined on a set S in Enwith function values in Em,

and let abe an accumulation point of S. We say that fis continuous at the point aprovidedthat

(i) fis defined at a, (ii) limxaf(x) = f(a). If a is notan accumulation point of S, we say f is

continuous at aprovided only (i) holds. If f is continuous at every point of S, we say f is

continuous on the setS.

It is convenient to note that whenever fis continuousat a, we can write part (ii) of the

above definition as follows: f(x) = f( x). Thus, when we deal with continuousfunctions,limxda

limxda

the limitsymbol may be interchanged with the functionsymbol. Observe also that continuityat ameans that for every neighbourhoodN(f(a)), there exists a neighbourhood N(a) such that

f[N(a)S] N[f(a)].


19/77

Theorem 4-10. Let f and g be continuous at the point a in Enand assume that f and g

have function values in E1or E2. Then f+g, f-g, and fg are each continuousat a. The quotient

f/g is also continuous at a, provided that g(a) 0. Note that the product fg should not beconfused with the compositionfg, defined by fg(x) = f[g(x)].

Theorem 4-11. Let fbe a function definedon a set S in Enand assume f(S) Em. Let g

be defined on f(S) with values in Ek, and let gfbe the compositefunctiondefined on S by theequation gf(x) = g[f(x)]. Suppose that we have (i) The point a is anaccumulation point of S

and fis continuous at a; (ii) The point f(a) is an accumulation point of f(S) and gis continuous

at f(a). Then the composite function gfis also continuousat a; that is, g[f(x)] = g[f(a)].limxda

Examples of continuous functions. In E2, constant functions (f(x) = c), the identity

function (f(x) = x) and hence all polynomialfunctions are continuous. A rationalfunction g/f

is continuous whenever the denominator does notvanish. The familiar real-valued functions

of elementary calculus such as exponential, trigonometric and logarithmic functions, are

continuous whenever they are defined.

Functions continuous on open or closed sets. Definition 4-12. Let f be a function

whose domainis S and whose rangeis T, and let Y be asubsetof T. By the inverseimage of Y

under f, denoted by f-1(Y), we shall mean the largestsubset of S which fmaps onto Y, that is

to say, f-1(Y) = {x| xS, f(x) Y}.

Note: If fhas an inverse functionf-1, the inverseimageof Y under fis the sameas the

image of Y under f-1, and in the case there is no ambiguityin the notation f-1(Y).

Theorem 4-13. Let fbe a function with domainS and rangeT.AssumeX S and Y

T. Then we have (i) If X = f-1(Y), then Y = f(X), (ii) If Y = f(X), then X f-1(Y). It should beobserved that, in general, we cannotconclude that Y = f(X) implies X = f-1(Y). Note that the

statements in this theorem can also be expressedas follows: Y = f[f-1(Y)], X f-1[f(X)].

Theorem 4-14. Let fbe a function which is continuouson a closed set S in Em, and let

the range of fbe a set T in Ek. Then if Y is a closedsubsetof T, the inverse image f-1(Y) will be

a closed subsetof S.

Theorem 4-15. Let fbe a function which is continuouson an open set S in Em, and let the

rangeof fbe a set T in Ek. Then if Y is an opensubset of T, the inverseimage f-1(Y) will be an

opensubset of S.

We have already seen that the imageof an open set under a continuousmappingis not

alwaysopen. The example in E1, defined by the equation f(x) = tan-1(x), shows that the image

of a closed set under a continuous mapping need notbe closed, since f maps E1onto the open

interval (-/2,/2). However, the image of a compactset under a continuousmappingis always

compact. This will be proved in the nexttheorem.


20/77

Functions continuous on compact sets. Theorem 4-16. Let f be a function which is

continuouson a compact set S in Em, and assume that f(S) Ek. Then f(S) is a compactset.

Theorem 4-17. Let fbe a function which is continuouson a compact set S in Em, and

assume that f(S) Ek. Suppose further that fis one-to-oneon S so that the inversefunctionf-1

exists. Then f-1is continuouson f(S).

Topological mappings. Definition 4-18. Let fbe continuous on S and assume that fis

one-to-oneon S so that the inversefunction f-1exists. Then fis called a topological mappingor

a homeomorphism if, in addition, f-1 is continuous on f(S). Note: A property of a set which

remains invariantunder everytopological mapping is called a topological property.

Properties of real-valued continuous functions. Definition 4-19. Let f be a real-valued

functiondefined on a set S in En. Then f is said to have an absolute maximumon the set S if

there exists a point a in S such that f(x) f(a), for all x in S. If a S and if there is a

neighbourhood N(a) such that f(x) f(a), for all xin N(a)S, then f is said to have a relativemaximum at the point a.Absoluteminimumand relative minimumare similarly defined, using

f(x) > f(a).

Theorem 4-20. Let f be a real-valued functionwhich is continuous on a compact set S in

En. Then f has an absolute maximumand an absolute minimumon S.

Theorem 4-21 (Bolzano). Let f be real-valued and continuouson a closed interval [a, b]

in E1, and suppose that f(a) and f(b) have different signs; that is, assume f(a)f(b) < 0. Then

there is at least onepoint x in the open interval (a, b) such that f(x) = 0.

An immediate consequence of the above theorem is the intermediate value theoremfor

continuous functions (Theorem 4-22): If f is real-valuedand continuouson a closed interval

S in E1, then f assumes everyvalue between its maximum, sup f(S), and its minimum, inf f(S).

Uniform Continuity. Suppose we have a function fcontinuouson a set S, so that for

each accumulation point aof S, we have limxa f(x) = f(a). In the and terminology, this

means that, given the accumulationpointaand given , there is a number > 0 (dependingon

aand on ) such that 0 < |x-a| < implies |f(x)-f(a)| < . In general, we cannot expect that for a

fixed the samevalue of will serve equally well for everysuch point a. This mighthappen,however; when it does, the function is called uniformly continuouson the set S.

Definition 4-23. Let fbe defined on a set S in Enwith function values in Em. Then f is

said to be uniformly continuous on S if the following statement holds: For every > 0, there

exists a > 0 (depending onlyon ) such that if xS and yS and |x-y| < , then |f(x)-f(y)| 0 be givenand let be a secondnumber ( > 0) which

will be made to dependon in a way to be described later. Choose a neighbourhoodN(a) such

that if xN(a)S, then we have both |f(x)-A| < and |g(x)-B| < .

Now |f(x)+g(x) - (A+B)| = |f(x)-A + g(x)-B|

|f(x)-A| + |g(x)-B| (by the trianglerule)

< + = 2.

And |f(x)-g(x) - (A-B)| = |f(x)-A - (g(x)-B)|

|f(x)-A| + |g(x)-B| (using |-| ||+||)

< + = 2.

Conclusion: |f(x)g(x) - (AB)| < 2, and choosing = /2, we see that|f(x)g(x) - (AB)| < as required. QED.

(iii) Let an arbitrary > 0 begivenand let be a secondnumber ( > 0) which will be

made to dependon in a way to be described later. Choose a neighbourhoodN(a) such that if

xN(a)S, then we have both |f(x)-A| < and |g(x)-B| < . In this question, the conclusion

we want is to show that |f(x)/g(x)-A/B| < whenever xN(a)S.

Now |f(x)

g(x) AB |= |

f(x)BAB+ABAg(x)+ABAB

Bg(x) |= |B(f(x)A)+ABA(g(x)B)AB

Bg(x) |= |B(f(x)A)A(g(x)B)

Bg(x) |

.[ |B(f(x)A)

Bg(x) | + |A(g(x)B)

Bg(x) |< | BeBg(x) | + |

AeBg(x) |

Now |g(x)| = |B+g(x)-B| < |B|+ < |B|+1, so that

.| Be

Bg(x) | + |Ae

Bg(x) |=|Be|

|B||g(x)|+|Ae|

|B||g(x)| 0, there exists a > 0 such that if |h| < , then

|f(x+h)-f(x)| < /2(---(1)). Similarly, if we replace h by -h (which holds since |-h| = |h|), then if

|-h| < (which is equivalent to the condition |h| < ), we have |f(x-h)-f(x)| < /2(---(2)).

Adding equations (1)and (2)together, we get

|f(x+h)-f(x)| + |f(x-h)-f(x)| < ,

|f(x+h)-f(x)+f(x-h)-f(x)| < , (using the triangle inequality, |A+B| |A|+|B|)|f(x+h)+f(x-h)| < .

Therefore, given an arbitrary > 0, there exists a > 0 such that if |h| < , then

|f(x+h)+f(x-h)| < , i.e. |f(x+h)-f(x-h)| = 0 as required. QED.limhd0

(b) Let us define a function f to be given by thefollowing: f(x) = x if x 0, and f(x) = 1

if x = 0 (with x E1). At x = 0, |f(x+h)-f(x-h)| = |0-0| = 0, but |f(x+h)-f(x)| = |0-1| = 1.limhd0

limhd0

4-5. If x is anyreal number in [0, 1], show that the followinglimit exists:

.limmd [lim

nd cos2n(m!ox)]

and that its valueis 0 or 1, according to whether x is irrationalor rational.

Answer: When n in the inner limit, we will have an infinite power of

cos(m!x). For any m > 2, m!will always be an evennumber of s, so that it is of the form

2kfor some integer k > 0.Recallthat cos(2k) = 1 for all k > 0. If x is rational, say x = a/b,

where a, b R, then m!x = m!a/b. If m is large enough, then b will be a factor of m! (i.e. we

will have m! = 12...b...m), and thus the bs cancel in the numerator and the denominator,leaving us with A = a, where = m!/b. A will still consist of an evennumber of s (a largem! will still have an even factor lots of them in fact!), so that cos(A) = 1 for large m.

Therefore, when x is rational and when m is large enough, our expression consists of an

infinite power of 1s which will still be 1, so that we have = 1 provided thatlim

nd cos2n(m!ox)

m is large enough. In other words, = 1 when x is rational. Conclusion:mmd [m

nd cos2n(m!ox)]

the limit in question exists when x is rational, and it has the value 1.

If x is irrational, then m!x cannot possibly be an even or an odd number (no irrationalnumber is ever even or odd). It follows that we must have |cos(m!x)| 1 in this case (for any

m), and so it follows that cos(m!x)(-1, 1).


25/77

Claim: if y (-1, 1), then yn= 0. Proof of Claim. We need to find an N such that if nlimnd> N, then for an arbitrary > 0 and a particular y (-1, 1) we have |y|n < (We need the

modulussigns in because if we dont put them in, then for any > 0 we will always have yn<

for any y (-1, 0) and any oddn. We will therefore prove astrongercondition than the one we

need). But |y|n< happens exactly when nlog|y| < log(), or when n > log()/log|y|. Therefore,

if we choose N = log()/log|y|, then we will always have |y|n< for all n > N. End of Proof.

Therefore, because for any y (-1, 1) we have yn = 0, it follows that for anylimndcos(m!x) (-1, 1) we have cos2n(m!x) = 0. This will hold for anym and so we havelimnd

= 0 when x is irrational. Conclusion: the limit in question existswhen xlim

md [lim

nd cos2n(m!ox)]

is irrational, and it is given by zero. QED.

4-7. Let f be continuouson [a, b] and let f(x) = 0 when x is rational. Show that f(x) = 0

for everyx in [a, b].

Answer: If f is continuous on [a, b], then f is continuous at every point in [a, b]. We

know that f(x) = 0 when x is rational. But every irrationalnumber y is an accumulation point

of the set of rational numbers, so by definition 4-9, and by knowing that f is continuous at the

point y, it follows that (i) f is definedat y, and (ii) limxyf(x) = f(y).

We now know that f(y) = limxyf(x) for any irrational point y, but what is this limit? It

must be zero, as in every neighbourhood N(y), because y is an accumulation point, there is a

rational number z such that f(z) = 0 and so f(x) N(0), thus satisfying the definition for a

limit. Conclusion: f(x) = 0 for all x [a, b].

4-8. Let f be continuousat the point a= (a1, a2, ..., an) in En. Keep a2, a3, ..., anfixed and

define a newfunction g of one real variable by the equationg(x) = f(x, a2, ..., an). Show that g

is continuousat the point x = a1. (This is sometimes stated as follows: A continuous function

of n variables is continuous in each variable separately.)

Answer: Let us define a new function s: E1Enby thefollowingdefinition:

s(x) = (x, a2, a3, ..., an) for an arbitrary x E1and fixed a2, a3, ..., anE1.

Looking at the above definitionand at the question itself, we see that we can now write

g(x) as follows: g(x) = f(s(x)) for an arbitrary x E1. Looking at Theorem 4-11, we see that inorder to prove that g(x) is continuous at the point a= (a1, a2, ..., an) in En, we need to show that

(i) the arbitrary point x E1is an accumulation pointof E1, (ii) the function s(x) is continuousat x, (iii) The point s(x) is an accumulation pointof En, and (iv) f is continuous at the point

s(x). If we can show that the fourconditions above hold, then we can say that g is continuous

at the point x = a.

(i) We know that any real number is an accumulation point of the real line E1, so thatcondition (i) is satisfied for an arbitrary real number x = a1.


26/77

(ii) To prove that the function s(x) is continuousat an arbitrary point x E1, we need toshow that the function is defined at x and that . The first part follows

limadxs(a) =s(x)

immediately because the function s(x) is defined for allrealnumbersx E1by its definitionon the previous page. It remainsto show that we have .limadxs(a) =s(x)

But = = s(x).limadxs(a)

limadx (a, a2, a3, ...,an) = (

limadx (a),

limadx (a2),...,

limadx (an)) = (x, a2,..., an)

We can therefore say that condition (ii) holds. For condition (iii) to hold, we must show thatthe point s(x) is an accumulation pointof the set En. But by the same argument as in condition

(i), as s(x) is a point in En, and as all points in Enare accumulation points, then s(x) = (x, a2, ...,

an) is an accumulation point of the set En, satisfying condition (iii).

Finally, we must show that f is continuous at the point s(x). But if we let x = a1, then

s(a1) = (a1, a2, ..., an) = a, and we know by the assumption made in the question that f is

continuousat the point ain En. Therefore, condition (iv) is satisfied, and thus we have shown

that allfourconditionshold, and can subsequently say that the function g is continuousat the

point x = a1. QED.

4-11. For each x in [0, 1], let f(x) = x if x is rational, and let f(x) = 1-x if x is irrational.

Show that (a) f is continuousonly at the point x = ; (b) f assumes everyvalue between 0 and

1.

Answer: Let us first prove that f is continuousat the point x = . If x = , then x is

rational and so f(x) = . To show that f is continuous at this point, then for every arbitrary >

0, we need to find a > 0 such that if |x-y| < , with y [0, 1], then |f(x)-f(y)| < . Now for any

arbitrary > 0, the deleted neighbourhood N()[0, 1] will always contain irrational andrational numbers from [0, 1]. If y is a rationalnumber in N()[0, 1], then |f(x)-f(y)| = |-y|.

If we choose = , then we want to show that if |x-y| < = , then |f(x)-f(y)| < . But in this

case, |x-y| = |-y| = |f(x)-f(y)|, so if |x-y| < = , then we will alwayshave |f(x)-f(y)| 0, we need to find a > 0 such that if |x-y| < , with y [0, 1] x, then |f(x)-f(y)|

< . In order to prove that the above definition cannot be satisfied, it is sufficient to find a

single 1> 0 such that there is noneighbourhood N(x)[0, 1] in which allthe points within

of x in [0, 1] satisfy |f(x)-f(y)| < 1.

Consider that we fix x to be a number in [0, 1] which is not a half. As stated above,every neighbourhood of this point x will contain both rational andirrational numbers from [0,

1]. Consider the case when x is rational and we therefore have f(x) = x.


27/77

Consider also that we define a number 1 = |2x-1| > 0. Given this 1 (which will be

greater than zero if x is not a half), we need to find a > 0 such that if |x-y| < for y [0, 1],

then we will alwayshave |f(x)-f(y)| < 1. Assume that such a exists, and let us pick out an

element y from the neighbourhood N(x)[0, 1] which is an irrational number.

In this case, we have f(x) = x and f(y) = 1-y, with |x-y| < .

It follows that |f(x)-f(y)| = |x-(1-y)| = |x+y-1|.

Now if |x-y| < , then y is bounded from above by x+, i.e. y < x+.

Therefore, we will have |x+y-1| < |x+(x+)-1| = |2x-1+|.

But if f is continuousat x, then we need |2x-1+| < 2, i.e.

|2x-1+| < |2x-1|.

There is no possible > 0 that we can pick that will satisfy the above inequation.

Therefore, we conclude that when x is rational and not equal to a half, if we pick 1= |2x-1| >

0, then there is no > 0 that will suffice so that if |x-y| < , then |f(x)-f(y)| < 1 for ALLirrational y in this neighbourhood. Conclusion: f(x) is not continuous when x is a rational

number in [0, 1] and x .

The other case to consider is when x is a fixed irrationalnumber in [0, 1]. Consider now

that for this fixed irrational number x, we define a number 2= 1. Given this 2, we need to find

a > 0 such that if |x-y| < for y [0, 1] x, then we will alwayshave |f(x)-f(y)| < 2.Assume

that such a exists, and let us pick out an element y from the neighbourhood N(x)[0, 1]

which is a rational number.

In this case, we have f(x) = 1-x and f(y) = y, with |x-y| < .It follows that |f(x)-f(y)| = |1-x+y| = |-x+y+1|.

Now if |x-y| < , then y is again bounded from above by x+, i.e. y < x+.

Therefore, we will have |-x+y+1| < |-x+(x+)+1| = |+1|.

But if f is continuousat x, then we need |+1| < 2, i.e.

|+1| < 1.

But there is no possible> 0 that we can pick that will satisfy the above inequation.

Therefore, we conclude that when x is irrational, if we pick 2= 1, then there is no> 0 that

will suffice so that if |x-y| < , then |f(x)-f(y)| < 1for ALL rational y in this neighbourhood.Conclusion: f(x) is notcontinuous when x is an irrational number in [0, 1].

We have now considered allcases for x in [0, 1]: x = , so that f(x) is continuous; and x

, so that f(x) is not continuous. QED.


28/77

(b) It is a trivial matter to show that f assumes the value of every rationalnumber in the

interval [0, 1], as f(x) = x for all rational numbers in [0, 1]. Our only problem occurs in

showing that f assumes the value of every irrationalnumber in the interval [0, 1], i.e. if we

have an arbitrary irrational number y [0, 1], then we need to find a number z [0, 1] suchthat f(z) = y.

If y is our arbitrary irrational number in the interval [0, 1], then 1-y will also be anarbitrary irrational number in this interval. Thus f(1-y) = 1-(1-y) = y (because 1-y is irrational),

and thus z = 1-y is the number we are looking for so that f(z) = y for all irrational numbers y [0, 1]. Conclusion: f assumes every value between 0 and 1. QED.

4-14. Let f be definedand boundedon the closed interval S = [a, b]. (That is, assume that

f(S) is a boundedset.) If T is a subsetof S, the number f(T) = sup{f(x)-f(y) | xT, y T} is

called the oscillationof f on T. If xS, the oscillation of f at x is defined to be the number

f(x) = f(N(x; h)S). Show that this limit alwaysexists and that f(x) = 0 if, and only if,lim

hd0+

f is continuousat x.

Answer:By definition, f(T) is the biggestvalue of f(x)-f(y) in T (x, y T). It follows

that f(x) is the biggest value of f(x)-f(y) in N(x; h)S (x, y N(x; h)S), i.e. in the

neighbourhood of x inS. The value f(x) is thus determined by twopoints (say x1and x2) in

N(x; h)S. Because f(S) is bounded(i.e. f(S) is a subset of somefiniteinterval), then the twopoints in question will havefinitevalues, and so their difference (i.e. f(x1)-f(x2)) will also be

finite. Therefore, for a particular h, f(N(x; h)S) will exist and is a finite value.

Does the limit f(N(x; h)S) exist? To show that it does exist, all we need do is tolimhd0+show that if h2> h1, then f(N(x; h1)S) f(N(x; h2)S). If this is true, then the sequence of

numbers f(N(x; h)S) given by decreasing h will be a bounded monotonic sequence of finitepositive numbers (bounded above by, say, the entry where h = 1), and thus we can apply

Theorem 12-6 to say that this sequence convergesto a value, say A, and so we can write

A = f(N(x; h)S), where 0 A f(N(x; 1)S), say.lim

hd0+

Claim: If h2> h1, then f(N(x; h1)S) f(N(x; h2)S). Proof of Claim. To start with,let us look at what exactlyis sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)}? Well, it is the

maximum value of f(x1)-f(x2) in the neighbourhood N(x; h)S. But if this is so, then f(x1)

must be the highest value of f(x) in the neighbourhood N(x; h)S, and f(x2) must be the

smallestvalue of f(x) in the neighbourhood N(x; h)S. Therefore, we have

sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)} =

sup{f(x1) | x1((x; h)S)} - inf{f(x2) | x2((x; h)S}.

The proof centreson the fact that if h2> h1, then sup{f(x) | x N(x; h1)S} sup{f(x) |xN(x; h2)S}, and inf{f(x) | x N(x; h1)S} > inf{f(x) | x N(x; h2)S}


29/77

I will not prove these statements (intuitively, if we are considering a smaller

neighbourhood, there are less points to consider, and so the maximum value in a smaller

neighbourhood will be smaller and the smallest value in a smaller neighbourhood will be

bigger).Assumingthat these statements are correct, we have

sup{f(x1) | x1((x; h1)S)} - inf{f(x2) | x2((x; h1)S}

sup{f(x1) | x1((x; h2)S)} - inf{f(x2) | x2((x; h2)S},or sup{f(x1)-f(x2) | x1(N(x; h1)S), x2((x; h1)S)}

sup{f(x1)-f(x2) | x1(N(x; h2)S), x2((x; h2)S)},

or f(N(x; h1)S) f(N(x; h2)S) as required. QED.

We now need to show that f(x) = 0 f is continuous at x. If. If f is continuous at x,

then given an arbitrary > 0, there exists a h > 0 such that if |x-y| < h, then |f(x)-f(y)| < , where

y N(x; h)S. Now if x1, x2N(x; h)S, then the distance between x1and x2is at most 2h,

so that for a given > 0, |x1-x2| < 2h |f(x1)-f(x2)| < 2.

But if h 0+, then |f(x1)-f(x2)| 0+ so that

sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)} 0 when h 0+,or, in other words,

sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)} = 0 as required. QED.lim

hd0+

Only If. If f(x) = 0, then we need to show that f is continuousat x. To do this, we need

to show that f is definedat x and that limyxf(y) = f(x). The first bit follows immediately from

the fact that f is definedand boundedon the interval [a, b], so that f is defined for all x [a,b]. For the second bit, we note that limyx f(y) = f(x) can be equivalently written as the two

following equations: and ;limhd0+ (x + h) =f(x)lim

hd0+ (x h) =f(x)or and .limhd0+ (x + h) f(x) = 0

limhd0+f(x) (x h) = 0

For a given h, the neighbourhoodgiven by (x; h)S will contain the point x and theN

points (x+h) and (x-h). Now if f(x) = 0, then

sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)} = 0.lim

hd0+

From ourpreviousanalysis of sup{f(x1)-f(x2) | x1 (N(x; h)S), x2((x; h)S)}, weknow that f(x1) must be the highestvalue of f(x) in the neighbourhood N(x; h)S, and that

f(x2) must be the smallestvalue of f(x) in the neighbourhood N(x; h)S. Knowing this, then

for a point x3in the neighbourhood N(x; h)S, then we must have f(x3)-f(x) sup{f(x1)-f(x2) |

x1(N(x; h)S), x2((x; h)S)}. Note also that we must have0 f(y)-f(z) sup{f(x1)-f(x2) |

x1(N(x; h)S), x2((x; h)S)} if y, z N(x; h)S, and if y > z.


30/77

If we let x3= x h, then we see that we have

(i) f(x + h) - f(x) sup{f(x1)-f(x2) | x1(N(x; h)S), x2(N(x; h)S)}.Therefore,

f(x + h) - f(x) sup{f(x1)-f(x2) | x1(N(x; h)S), x2(N(x; h)S)}.lim

hd0+lim

hd0+

But we have assumed that sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)} = 0,lim

hd0+

so that f(x + h) - f(x) 0, i.e. we have f(x + h) - f(x) = 0 as required (because we mustlimhd0+lim

hd0+

have f(x+h)-f(x) > 0).

(ii) f(x) - f(x-h) sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)}.Therefore,

f(x) - f(x-h) sup{f(x1)-f(x2) | x1(N(x; h)S), x2((x; h)S)}.lim

hd0+lim

hd0+

But again we have assumed that sup{f(x1)-f(x2) | x1 (N(x; h)S), x2 (N(x;lim

hd0+

h)S)} = 0, so that f(x) - f(x-h) 0, i.e. we have f(x) - f(x-h) = 0 as required (becauselimhd0+lim

hd0+

we must have f(x)-f(x-h) > 0). QED. Technical Note: There is a slight mistake in the above in

that f(x h) N(x; h)S, but I have assumed that in taking the limit this is of no consequenceto the proof.

4-17. Let f be definedand continuouson a closed set S in En. Let A = {x| xS, f(x) =0}. Show that A is a closedset.

Answer: Assume that the rangeof f is a set T in E1(i.e. T = f(S)). Let Y be the subset of

T consisting only of thezeropoint, i.e. Y = {0}. Because Y consists of just a single point, then

Y contains all its accumulation points (of which there are none of), and thus Y is a closed

subsetof T.

By Theorem 4-14, the inverseimage f-1(Y) is a closed subset of S. But f-1(Y) is the set A,

i.e. the elements of S which map onto 0, and thus A must be a closed subset of S. QED. Note:

in the case that 0 f(S), then A = {}, and is a closed set by definition.

4-19. Let f be continuous on a closed interval [a, b]. Suppose that f has a local maximum

at x1and a local maximum at x2. Show that there must be a thirdpoint between x1and x2where

f has a local minimum.

Answer: Consider the interval [x1, x2] [a, b]. Because f is continuous on the closedinterval [a, b], then it will be continuous on thesubinterval[x1, x2] as well. Further, the interval

[x1, x2] is a closed interval. By Exercise 3-1, this interval is a closed set, and so by the note

following Definition 3-39 (and by knowing that [x1, x2] is a bounded set (it is a subset of [a,

b]), we can say that [x1, x2] is a compactset. Therefore, we can apply Theorem 4-20 to say that

f has an absolute maximumand an absolute minimumon [x1, x2].


31/77

If f has an absolute minimumon [x1, x2], then there exists a point c [x1, x2] such that

f(x) > f(c) for all x [x1, x2]. If we now take any neighbourhoodaround the point c such that

N(c) [x1, x2], then we can say that f has a local minimumat the point c in question.

The only detail left to mop up is to make sure that c is not either x1or x2(we need a third

point betweenx1and x2). But, using the above, if we find that we arrive at the conclusion c =

x1or c = x2, then we proceed as follows: for the case c = x1, because x1is a local maximum,then there is a neighbourhood N(x1) such that for all x N(x1)[a, b], we have f(x) f(x1).

But because c is the absolute minimumof f on [x1, x2], then we must have f(x) = f(x1)

for x [x1, x1+h), where h is the radiusof the neighbourhood N(x1), or else c would have beenincorrectly found. Therefore, f(x1) = f(x1+h), and so we can equally well define c to be given by

c = x1+h to represent our absolute/local minimum on [x1, x2]. Note: if x1+h > x2, then just take a

point half way between x1 and x2 to represent c in this situation f must be a constant

function between x1 and x2.

The case c = x2goes through in much thesameway as above (i.e. we redefine c to be

given by c = x2-h or by c = (x1+x2)/2 as appropriate), so that we have found a point c between

x1and x2which is alocal minimumin the interval [x1, x2]. QED.

4-21. Show that a function which is uniformly continuous on a set S is also continuous

on S.

Answer: Let f be defined on a set S in En with function values in Em. f is said to be

uniformly continuous on S if the following statement holds: For every > 0, there exists a >0 (depending onlyon ) such that if xS and yS and |x-y| < , then |f(x)-f(y)| < . fis said

to be continuous on S provided that at each accumulation point a, and given , there is a

number (depending on aand on ) such that 0 < |x-a| < |f(x)-f(a)| < .

Let adenote any arbitrary accumulation pointin S, and let > 0 be an arbitrary number.

If f is uniformly continuous, then for every there exists a > 0 (depending onlyon ) such

that if xS, and |x-a| < , then |f(x)-f(a)| < . But this is exactly the condition we want so that

the function fis continuous on S (it doesnt matterthat does not depend on a), and thereforewe have reached the conclusion we want, in that uniform continuityon S does imply continuityon S. QED.

4-22. Show that the function f defined by f(x) = x is not uniformly continuous on E1.

Answer: Consider that we are given two arbitrary pointsx and y in E1. We need to show

that if the distancebetween x and y is less than , then the distance between f(x) and f(y) is

less than , where depends only on . In other words, given , if |x-y| < , then we need to

show that |f(x)-f(y)| < .


32/77

Now |f(x)-f(y)| = |x-y| = |(x-y)(x+y)| = |(x-y)||(x+y)|. If |x-y| < , then |f(x)-f(y)| < |x+y|.In the example in the book where we considered points in the interval (0, 1], the value |x+y|

was bounded above by 2, and so as |x+y| 2, then |f(x)-f(y)| < 2, so that if we choose = /2

for a given arbitrary > 0, then we get the required result: |f(x)-f(y)| < if |x-y| < .

However, if x, y E1, then |x+y| is notbounded above, so that we cannot choose a value

n so that |f(x)-f(y)| < n, and therefore cannot define = /n for a given arbitrary > 0 so that|f(x)-f(y)| < . Proof by Contradiction: Suppose that there wasa positive number n so that if

we defined = /n for a given arbitrary > 0, then we would get |f(x)-f(y)| < for every |x-y| |(c-d)||(n+n)| = 2n(/2) = n= . So we have a contradiction, and thus

cannot define in terms of onlyso that the function f(x) = x in uniformlycontinuouson E1.End of Proof, and QED.

4-25. Let f be a function definedon a set S in Enand assume that f(S) Em. Let gbedefined on f(S) with values in Ek, and let gfdenote the compositefunction defined by gf(x) =

g[f(x)], if xS. If f is uniformly continuouson S and if g is uniformly continuous on f(S),show that gfis uniformly continuouson S.

Answer: (1) If f is uniformly continuouson S, then given an 1> 0, we can always find a1> 0 (dependent only on 1) such that |x-y| < 1 |f(x)-f(y)| < 1. (x, y En). (2) If g is

uniformly continuouson f(S), then given an 2> 0, we can always find a 2> 0 (dependent only

on 2) such that |p-q| < 2|g(p)-g(q)| < 2. (p, qEm).

Now let a= f(x) and let b= f(y). We know that if |x-y| < 1, then |a-b| < 1. Assume for

the moment that 2= 1, so that |a-b| < 1|a-b| < 2. But if |a-b| < 2, then |g(a)-g(b)| < 2, i.e.

|g(f(x))-g(f(y))| < 2. Therefore, given an 2> 0, by (2) we can always find a 2> 0 (dependent

onlyon 2) such that |f(x)-f(y)| < 2 |g(f(x))-g(f(y))| < 2. If we let2= 1, then given that

particular1(which is dependent onlyon 2), we can (by (1)) alwaysfind a 1> 0 (which nowis dependent onlyon 2) such that |x-y| < 1 |f(x)-f(y)| < 1.

Putting the above all together, we concludethat given an 2> 0, we can alwaysfind a 1> 0 (dependent onlyon 2) such that |x-y| < 1|f(x)-f(y)| < 1= 2|g(f(x))-g(f(y))| < 2, i.e.

|x-y| < 1|g(f(x))-g(f(y))| < 2(x, yEn); and so gfis uniformly continuous on S. QED.


33/77

4-27. Locate and classify the discontinuities of the function f defined on E1 by the

following equations:

(a) f(x) = (sin x)/x if x0, f(0) = 0.

(b) f(x) = e1/x if x 0, f(0) = 0.

(c) f(x) = e1/x+ sin(1/x) if x 0, f(0) = 0.

(d) f(x) = 1/(1-e1/x) if x 0, f(0) = 0.

Answer: (a) f(x) is

continuous at every point in E1except at x = 0, where f(x) has a

removable discontinuity. As

limx0-= 1 and limx0+= 1, we can

make the function continuous by

redefining f(0) to be f(0) = 1. (b)

Again the function is continuousat

every point in E1 except at x = 0,where f(x) has an infinite jump

discontinuity: f(0-) = 0, f(0) = 0,

and f(0+) = . As the function has an infinite right-hand jump, then it is discontinuousat x = 0(and cannotbe made continuous at this point).

(c) In this case, as before, f(x) is continuous everywhereexcept at x = 0. This time, f(x)

has an infinite right-hand jump discontinuity as in part (b), and f(0-) does not exist. We

therefore conclude that f(x) is discontinuousat x = 0. (d) In this final example, we have f(0-) =

1, f(0+) = 0, and f(0) = 0. As f(x) has a left-handjump at x = 0, then f(x) has a left-hand jumpdiscontinuity at x = 0. Apart from this, f(x) is continuous at everyother point in E1.

4-29. Let f be defined in the openinterval (a, b) and assume that for each interior point x

of (a, b) there exists a neighbourhood N(x) in which f is increasing. Show that f is an

increasingfunction throughout (a, b).

Answer: We saw in exercise 3-1 that an open interval (a, b) is an open set so

therefore all points x (a, b) are interior points. This implies that for allpoints x (a, b), there

exists a neighbourhood N(x) in which f is increasing. Let us now prove that f is increasing bycontradiction.

If f is not increasing on (a, b), then there is a subinterval (say (c, d)) of (a, b) in which f

is not increasing, i.e. f is strictly decreasing in this subinterval. Consider a point z (c, d). If z

(c, d), then because (c, d) (a, b), we have z (a, b), and thus there is some neighbourhoodof z in which f is increasing. Because of this, then we must throw away this neighbourhood

from our subinterval (c, d) in which we are claiming that f is strictly decreasing. But we can

apply the above argument with any point in the subinterval (c, d), and thus there cannot

possibly be anysuch subinterval (c, d) in which f is strictly decreasing because for any point in

such an interval, f will be increasing in a neighbourhood of that point. We therefore conclude

that f must be increasingon the interval (a, b). QED.

x

f(x)(a)

x

f(x)(b)

x

f(x)(c)

x

f(x)(d)

1

1


34/77

Here is a second proof for the above question. Suppose that we pick two arbitrary

numbers x1and x2from the interval (a, b) so that a < x1< x2< b. To show that f is an increasing

function throughout (a, b), all we must show is that f(x1) f(x2) for our arbitrary numbers x1and x2.

The interval [x1, x2] is a closed interval. By Exercise 3-1, this interval is a closedset, and

so by the note following Definition 3-39 (and by knowing that [x1, x2] is a bounded set (it is asubset of (a, b)) we can say that [x1, x2] is a compact set. Now for every point x [x1, x2],there will be a corresponding neighbourhood N(x) in which f is increasing. Consider that we

define a covering of the compact set [x1, x2] given by the union of all such neighbourhoods of

points in [x1, x2] in which f is increasing.In other words, we can write

[x1, x2] = .4

all x c[x1,x2]N(x)

Now because [x1, x2] is a compact set, a finite number of these neighbourhoods will

cover [x1, x2], say a collection of nneighbourhoods. Therefore, we can write

[x1, x2] = , with x [x1, x2].4 i=1n Ni(x)

Because f is increasing throughout allof these neighbourhoods, and because this finite

collection of neighbourhoods covers[x1, x2], then f will be increasing throughout the interval

[x1, x2], enabling us to say that f(x1) f(x2) as required. QED.

4-31. If f is one-to-oneand continuouson [a, b], show that f must be strictlymonotonic

on [a, b]. That is, prove that every topological mappingof [a, b] onto an interval [c, d] must bestrictly monotonic.

Answer: If f is one-to-oneon [a, b], then for all points y in the range, there exists one

and only onepoint x in the domain such that f(x) = y. If f is continuous on [a, b], then we

know that f is definedfor all points in the interval [a, b]. Because of this, we know that f(a)

will have a value: say f(a) = y1; and that f(b) will have a value: say f(b) = y2.

Because f is one-to-one, then we cannot have y1= y2, and thus we must have either y1>

y2or y1< y2. Consider the case where y1< y2(we will only consider this case the othercasecan be derived from the following by replacing all and vice-versa). Because of the

continuity of f on the closed interval [a, b], then the intermediate value theorem says that f

assumes all values between y1and y2.


35/77

Consider an arbitrary point c (a, b). What is f(c)? First ofall, f(c) cannot be equal to either y1 or y2 because f is a

one-to-one function. Further (as seen in Diagram 1), we cannot

have f(c) > y2as then (according to the IVT) there will be a point

p in the interval (a, c) where f(p) = y2, and this cannot be the

case if f is one-to-one (because then we will have f(p) = f(b) =

y2). Similarly(as seen in Diagram 2), we cannot have f(c) < y1asthen (again according to the IVT) there will be a point q in the

interval (c, b) where f(q) = y1, and this cannotbe the case if f is

one-to-one (because then we will have f(q) = f(a) = y1).

Therefore, we conclude that we musthave y1< f(c) < y2.

If we apply thesame sort of argumentfor the point d (c,b), we would reach the following conclusion: y1< f(c) < f(d) < y2. In other words, given any

pairof points c and d in [0, 1], with c < d, then we must have f(c) < f(d), and thus f is strictly

increasing on [a, b].

With the other case, where y1 > y2, we would reach the conclusion that f would be

strictly decreasingon [a, b]. Putting all of this together, we conclude that f is eithera strictly

monotonically increasing function or a strictly monotonically decreasing function, i.e. f is a

monotonicfunction. QED.

f(a)f(b)

f(q)

f(c)

f(a)

f(b)

f(c)

f(p)

Diagram 1

Diagram 2


36/77

Possible Further Work / Evaluation

Exercise 4-7: Clarification in the wording, or possibly an alternativesolution, e.g. if we

consider an arbitrary irrational number in [a, b], and because we know that there exists a

sequence of rational numbers xnconverging to y, the result follows because of the continuity of

f?

Exercise 4-11: A small gap left unfilledin the first part of the answer. In the second

part, I made a slip up with a negative sign. New version:

...The other case to consider is when x is a fixed irrationalnumber in [0, 1]. Consider

now that for this fixed irrational number x, we define a number 2= 1. Given this 2, we need

to find a > 0 such that if |x-y| < for y [0, 1] x, then we will alwayshave |f(x)-f(y)| < 2.

Assume that such a exists, and let us pick out an element y from the neighbourhood

N(x)[0, 1] which is a rational number.

In this case, we have f(x) = 1-x and f(y) = y, with |x-y| < .It follows that |f(x)-f(y)| = |1-x-y| = |-x-y+1| = |x+y-1| (as |A-B| = |B-A|). But this is what

we had in the first section of the answer so the result follows by the validity of this first

section. QED.

Exercise 4-14: I was mistaken in thinking that sup was the same thing as maximum.

I therefore looked at the following definitions in Chapter 1:

Definition 1.6: Let A be a set of real numbers. If there is a real number x such that a A

implies a x, then x is called an upper bound for the set A and we say that A is boundedabove.

Definition 1.7: Let A be a set of real numbers bounded above. Suppose there is a real

number x satisfying the following two conditions:

(i) x is an upper boundfor A, and

(ii) if y is any upper bound for A, then x

y.

Such a number x is called a least upper bound, or a supremum, of the set A.

Exercise 4-22: Need to prove thegeneralcase not just for a particular formula for

in terms of .

Exercise 4-29: Need to correct the indexingin the penultimate paragraph:

....Now because [x1, x2] is a compactset, a finitenumber of these neighbourhoods willcover [x1, x2], say a collection of nneighbourhoods N(y1), ...., N(yn). Therefore, we can write

[x1, x2] , with yi[x1, x2].....4 i=1n N(yi)


37/77

Chapter 13


38/77

Chapter 13: Sequences of Functions

In this chapter, we will be dealing with complex-valuedfunctions defined on certain

subsets of E2 (but if results hold only for real-valuedfunctionsdefined on subsets of E1, this

will be explicitlystated).

Given a sequence {fn}, each term of which is a function defined on a set S, for each x inS we can form anothersequence {fn(x)} whose terms are the corresponding function values.

Let T denote the set of these points x in S for which the second sequence converges. The

function f defined by f(x) = fn(x), if x T, will be referred to as the limitfunctionof thelimndsequence {fn} and we will say that {fn} convergespointwiseon the set T.

If each termof the sequence {fn} has a certain property, then to what extent does the

function f alsopossess this property? In general, we need a study of stronger methods that

do preserve these properties. The most important of these is the notion of uniform

convergence.

When we ask whether continuityat each fnat x0implies continuity of the limit functionf

at x0, we are really asking whether the equation fn(x) = fn(x0) implies the equation f(x)lim

xdx0lim

xdx0

= f(x0). The last equation can also be written as follows: .lim

xdx0lim

ndfn(x) =lim

ndlim

xdx0fn(x)Therefore, our question about continuityamounts to this: Can we changethe limit symbols in

the above?

In general, we shall see that we cannotinterchange the symbols. First of all, the limit in

the equation ( f n(x) = fn(x0)) may not exist; or, even if it doesexist, it may not be equal tolimxdx0f(x0). In chapter 12, it is stated that m=1n=1 f(m, n) is not necessarily equal to n=1m=1

f(m, n). For example, consider the following function:

f(m, n) = 1 if m = n+1, n = 1, 2, ...;

f(m, n) = -1 if m = n-1, n = 1, 2, ...;

and f(m, n) = 0 otherwise.

Then m=1n=1f(m, n) = -1, but n=1m=1f(m, n) = 1.

The question arises frequently as to whether we can change the order of the limit

processes. We shall find that uniformconvergenceis a far-reaching sufficient condition for

the validity of interchangingcertain limits, but it does not provide the completeanswer to the

question. Examples can be found in which the order of the two limits can be interchanged

although the sequence is notuniformly convergent.

An exampleof a sequence of real-valued functions: Let fn(x) = nx(1-x)nif x E1, n = 1,

2, ... Here, limnfn(x) exists if 0 x 1, and the limit function has the value0 at each point in[0, 1]. fnhas a local maximum at x = 1/(n+1), but fn(1/(n+1)) +as n .


39/77

Let {fn} be a sequence of functions which converges pointwise on a set T to a limit

function f. Going back to the basic definitionof limit, this means that for each pointx in T and

for each > 0, there exists an N (depending on at most bothx and ) such that n > N implies|fn(x)-f(x)| < . If the sameN works equally well for everypoint in T, the convergence is saidto be uniformon T. That is, we have the following definition:

Asequence of functions {fn} is said to converge uniformlyto f on a set T if, for every > 0, there exists an N (depending onlyon ) such that n > N implies |fn(x)-f(x)| < , for every xin T. We denote thissymbolicallyby writing fnf uniformlyon T.

When each term of the sequence {fn} is real-valued, there is a useful geometric

interpretation of uniform convergence. The inequality |fn(x)-f(x)| < is then equivalent to thetwo inequalities f(x)-< fn(x) < f(x)+. If this is to hold foralln > N and for allx in T, this means that the entiregraph

of fn(that is, the set {(x, y) | y = fn(x), x T}) lies within a

band of height 2 situatedsymmetricallyabout the graphof f (see the diagramon the right).

A sequence {fn} is said to be uniformly boundedon T if there exists a constant M > 0

such that |fn(x)| M for all x in T and for all n = 1, 2, ... The number M is called a uniformboundfor {fn}. If each individual function is boundedand if fnf uniformlyon T, then it iseasy to prove that {fn} is uniformly bounded on T. This observation often enables us to

conclude that a sequence is notuniformly convergent.

Theorem 13-2. Let f be a doublesequence and let P denote the set ofpositive integers.

For each n = 1, 2, ..., define a function gnon P as follows: gn(m) = f(m, n), if m P. Assumethat gng uniformlyon P, where g(m) = limnf(m, n). If the iterated limit limm (limnf(m, n)) exists, then the doublelimit limm,nf(m, n) alsoexists and has thesamevalue.

Uniform Convergence and Continuity. Theorem 13-3: Assume that fn f uniformlyon T. If each fnis continuousat a point x0of T, then the limit function f is alsocontinuous at

x0. Note 1: If x0 is an accumulation point of T, the conclusion implies that

. Note 2: Uniform convergence of {fn} is sufficient but notm

xdx0m

ndfn(x) =m

ndm

xdx0fn(x)necessaryto transmit continuity from the individualtermsto the limitfunction.

The Cauchy condition for uniform convergence. Theorem 13-4: Let {fn} be a

sequence of functions definedon a set T. There exists a function f such that fnf uniformlyon T if, and only if, the following condition (called the Cauchy condition) is satisfied: For

every > 0 there exists an N such that m > N and n > N implies |fm(x)-fn(x)| < , for every x inT.

Uniform Convergence on infinite series. Definition 13-5: Given a sequence {fn} of

functions defined on a set T. For eachx in T, let sn(x) = k=1nfk(x) (n = 1, 2, ...). If there exists a

functionf such that snf uniformly on T, we say that the series fn(x) converges uniformlyonT and we write n=1fn(x) = f(x) (uniformlyon T).

y = fn(x) y = f(x)+

y = f(x)-

y = f(x)


40/77

Theorem 13-6: (Cauchy condition for uniform convergence of series). The series fn(x)converges uniformlyon T if, and only if, for every > 0 there is an N such that n > N implies

fk(x)| < , for each p = 1, 2, ..., and everyx in T.|Sk=n+1n+p

Theorem 13-7: (WeierstrassM-test). Let {Mn} be a sequence of non-negativenumbers

such that 0 |fn(x)| Mn, for n = 1, 2, ..., and for every x in T. Then fn(x) converges uniformly

on T if Mnconverges.

Theorem 13-8: Assume that fn(x) = f(x) (uniformlyon T). If each fnis continuousat apoint x0 of T, then f is also continuous at x0. Note: If x0 is an accumulation point of T, this

theorem permits us to write fn(x) = fn(x).lim

xdx0 Sn=1 Sn=1

limxdx0

A space-filling curve. We can apply the abovetheorem to construct an example of what

is known as a space-fillingcurve. This is a continuouscurve in E2 that passes through every

point of the unit square [0, 1][0, 1]. Consider thefollowingexample: Let be defined on theinterval[0, 2] by the following formulas:

(t) = 0, if 0 t 1/3, or if 5/3t 2; (t) = 3t-1, if 1/3t 2/3; (t) = 1, if 2/3t 4/3; and(t) = -3t+5, if 4/3t 5/3.Extendthe definition of to all of E1by the equation (t+2) = (t).This makes periodicwith period 2 (as shown below). Now define twofunctions 1and 2bythe following equations:

1(t) = , 2(t) = .Sn=1 v(3

2n2t)2n Sn=1

v(32n1t)2n

Both series converge absolutely for each real t and they converge uniformly on E1. In

fact, since |(t)| 1 for all t, the Weierstrass M-test is applicable with Mn = 2-n. Since f iscontinuouson E1, Theorem 13-8 tells us that 1and 2are alsocontinuous on E1. Let = (1,2), and let denote the image of the unit interval[0, 1] under . We will show that fillsthe unit square, i.e. that = [0, 1][0, 1].

First, it is clear that 0 1(t)1 and that 0 2(t) 1 for each t, since n=12-n= 1.Hence, is asubsetof the unit square. Next, we must show that (a, b) whenever (a, b) [0, 1][0, 1]. For this purpose, we write a and b in the binarysystem. That is, we write

a = , b = ,Sn=1 an

2n Sn=1 bn

2n

where each anand each bnis either0 or 1. Now let c = , where c2n-1= anand c2n=2Sn=1 cn

3n

bn, n = 1, 2, ... Clearly, 0 c 1 since 2n=13-n= 1. We will show that 1(c) = a and that 2(c)= b. If we canprovethat (3kc) = ck+1, for each k = 0, 1, 2, ... (---(1)), then we will h

solutions manual apostol.pdf

Documents