compactness - weebly

6COMPACTNESS

6.1 Compact Spaces

17B. Compact Subsets

I Exercise 108. A subset E of X is compact iff every cover of E by open subsets

of X has a finite subcover.

Remark (Lee 2011, p. 94). To say that a subset of a topological space is com-

pact is to say that it is a compact space when endowed with the subspace

topology. In this situation, it is often useful to extend our terminology in the

following way. If X is a topological space and A � X , a collection of subsets of

X whose union contains A is also called a cover of A; if the subsets are open

in X we sometimes call it an open cover of A. We try to make clear in each

specific situation which kind of open cover of A is meant: a collection of open

subsets of A whose union is A, or a collection of open subsets of X whose

union contains A.

Proof. The “only if” part is trivial. So we focus on the “if” part. Let U be an

open cover of E, i.e., U DSfU W U 2 Ug. For every U 2 U, there exists an open

set VU in X such that U D VU \ E. Then fVU W U 2 Ug is an open cover of E,

i.e., U �SfVU W U 2 Ug. Then there exists a finite subcover, say VU1 ; : : : ; VUn

of fVU W U 2 Ug, such that E �SniD1 VUi . Hence, E D

SniD1.VUi \E/; that is, E

is compact. ut

I Exercise 109. The union of a finite collection of compact subsets of X is

compact.

Proof. Let A and B be compact, and U be a family of open subsets of X

which covers A [ B . Then U covers A and there is a finite subcover, say,

UA1 ; : : : ; UAm of A; similarly, there is a finite subcover, say, UB1 ; : : : ; U

Bn of B .

But then fUA1 ; : : : ; UAm ; U

B1 ; : : : ; U

Bn g is an open subcover of A [ B , so A [ B is

compact. ut

43

References

[1] Adamson, Iain T. (1996) A General Topology Workbook, Boston:

Birkhäuser. [33]

[2] Ash, Robert B. (2009) Real Variables with Basic Metric Space Topology,

New York: Dover Publications, Inc. [22]

[3] Dugundji, James (1966) Topology, Boston: Allyn and Bacon, Inc. [39]

[4] Lee, John M. (2011) Introduction to Topological Manifolds, 202 of Grad-

uate Texts in Mathematics, New York: Springer-Verlag, 2nd edition. [25,

43]

[5] McCleary, John (2006) A First Course in Topology: Continuity and Di-

mension, 31 of Student Mathematical Library, Providence, Rhode Island:

American Mathematical Society. [29]

[6] Royden, Halsey and Patrick Fitzpatrick (2010) Real Analysis, New

Jersey: Prentice Hall, 4th edition. [41]

[7] Willard, Stephen (2004) General Topology, New York: Dover Publica-

tions, Inc. [i]

45

Calculus on Manifolds

A Solution Manual for Spivak (1965)

Jianfei Shen

School of Economics, The University of New South Wales

Sydney, Australia 2010

Contents

1 Functions on Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Norm and Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Functions and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.6 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Measure Zero and Content Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Integration on Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

iii

1FUNCTIONS ON EUCLIDEAN SPACE

1.1 Norm and Inner Product

I Exercise 1 (1-1�). Prove that kxk 6PniD1

ˇ̌̌xiˇ̌̌.

Proof. Let x D�x1; : : : ; xn

�. Then0@ nX

iD1

ˇ̌̌xiˇ̌̌1A2 D nX

iD1

�xi�2C

Xi¤j

ˇ̌̌xixj

ˇ̌̌>

nXiD1

�xi�2D kxk

2 :

Taking the square root of both sides gives the result. ut

I Exercise 2 (1-2). When does equality hold in Theorem 1-1 (3)�kx C yk 6 kxk C kyk

�?

Proof. We reprove thatˇ̌hx;yi

ˇ̌6 kxk � kyk for every x;y 2 Rn. Obviously, if

x D 0 or y D 0, then hx;yi D kxk � kyk D 0. So we assume that x ¤ 0 and y ¤ 0.

We first find some w 2 Rn and ˛ 2 R such that hw; ˛yi D 0. Write w D x � ˛y .

Then

0 D hw; ˛yi D hx � ˛y; ˛yi D ˛ hx;yi � ˛2 kyk2

implies that

˛ D hx;yiıkyk

2 :

Then

kxk2D kwk

2C k˛yk

2 > k˛yk2D

�hx;yi

kyk

�2:

Hence,ˇ̌hx;yi

ˇ̌6 kxk � kyk. Particularly, the above display holds with equality if

and only if kwk D 0, if and only if w D 0, if and only if x � ˛y D 0, if and only

if x D ˛y .

Since

kx C yk2D hx C y;x C yi D kxk

2C kyk

2C 2 hx;yi 6 kxk2 C kyk2 C 2 kxk � kyk

D�kxk C kyk

�2;

1

2 CHAPTER 1 FUNCTIONS ON EUCLIDEAN SPACE

equality holds precisely when hx;yi D kxk�jjyjj, i.e., when one is a nonnegative

multiple of the other. ut

I Exercise 3 (1-3). Prove that kx � yk 6 kxk C kyk. When does equality hold?

Proof. By Theorem 1-1 (3) we have kx � yk D kx C .�y/k 6 kxk C k�yk D

kxkC kyk. The equality holds precisely when one vector is a non-positive mul-

tiple of the other. ut

I Exercise 4 (1-4). Prove thatˇ̌kxk � kyk

ˇ̌6 kx � yk.

Proof. We have kx � yk2DPniD1

�xi � yi

�2D kxk

2C kyk

2� 2

PniD1 xiyi >

kxk2C kyk

2� 2 kxk kyk D

�kxk � kyk

�2. Taking the square root of both sides

gives the result. ut

I Exercise 5 (1-5). The quantity ky � xk is called the distance between x and y .

Prove and interpret geometrically the “triangle inequality”: kz � xk 6 kz � ykC

ky � xk.

Proof. The inequality follows from Theorem 1-1 (3):

kz � xk D k.z � y/C .y � x/k 6 kz � yk C ky � xk :

Geometrically, if x, y , and z are the vertices of a triangle, then the inequality

says that the length of a side is no larger than the sum of the lengths of the

other two sides. ut

I Exercise 6 (1-6). If f and g be integrable on Œa; b�.

a. Prove thatˇ̌̌R baf � g

ˇ̌̌6�R baf 2� 1

2

�

�R bag2� 1

2

.

b. If equality holds, must f D �g for some � 2 R? What if f and g are continu-

ous?

c. Show that Theorem 1-1 (2) is a special case of (a).

Proof.

a. Theorem 1-1 (2) implies the inequality of Riemann sums:ˇ̌̌̌ˇ̌Xi

f .xi / g .xi /�xi

ˇ̌̌̌ˇ̌ 6

0@Xi

f .xi /2�xi

1A1=20@Xi

g .xi /2�xi

1A1=2 :Taking the limit as the mesh approaches 0, one gets the desired inequality.

b. No. We could, for example, vary f at discrete points without changing

the values of the integrals. If f and g are continuous, then the assertion

is true. In fact, suppose that for each � 2 R, there is an x 2 Œa; b� with

SECTION 1.1 NORM AND INNER PRODUCT 3�f .x/ � �g .x/

�2> 0. Then the inequality holds true in an open neighbor-

hood of x since f and g are continuous. SoR ba

�f � �g

�2> 0 since the in-

tegrand is always non-negative and is positive on some subinterval of Œa; b�.

Expanding out givesR baf 2 � 2�

R baf � g C �2

R bag2 > 0 for all �. Since the

quadratic has no solutions, it must be that its discriminant is negative.

c. Let a D 0, b D n, f .x/ D xi and g .x/ D yi for all x 2 Œi � 1; i/ for i D 1; : : : ; n.

Then part (a) gives the inequality of Theorem 1-1 (2). Note, however, that

the equality condition does not follow from (a). ut

I Exercise 7 (1-7). A linear transformation M W Rn ! Rn is called norm pre-

serving if kM xk D kxk, and inner product preserving if hM x;M yi D hx;yi.

a. Prove that M is norm preserving if and only if M is inner product preserving.

b. Prove that such a linear transformation M is 1-1 and M�1 is of the same sort.

Proof.

(a) If M is norm preserving, then the polarization identity together with the

linearity of M give:

hM x;M yi DkM x CM yk

2� kM x �M yk

2

4

DkM .x C y/k2 � kM .x � y/k2

4

Dkx C yk

2� kx � yk

2

4

D hx;yi :

If M is inner product preserving, then one has by Theorem 1-1 (4):

kM xk DphM x;M xi D

phx;xi D kxk :

(b) Take any M x;M y 2 Rn with M x D M y . Then M x �M y D 0 and so

0 D hM x �M y;M x �M yi D hx � y;x � yi I

but the above equality forces x D y ; that is, M is 1-1.

Since M 2 L.Rn/ and M is injective, it is invertible; see Axler (1997, Theorem

3.21). Hence, M�1 2 L.Rn/ exists. For every x;y 2 Rn, we have

kM�1 xk D

M �M�1 x

� D kxk ;and D

M�1 x;M�1 yED

�M�M�1 x

�;M

�M�1 y

��D hx;yi :

Therefore, M�1 is also norm preserving and inner product preserving. ut


I Exercise 8 (1-8). If x;y 2 Rn are non-zero, the angle between x and y ,

denoted † .x;y/, is defined as arccos�hx;yi

ıkxk � kyk

�, which makes sense by

Theorem 1-1 (2). The linear transformation T is angle preserving if T is 1-1, and

for x;y ¤ 0 we have † .Tx;Ty/ D † .x;y/.

a. Prove that if T is norm preserving, then T is angle preserving.

b. If there is a basis .x1; : : : ;xn/ of Rn and numbers �1; : : : ; �n such that Txi D

�ixi , prove that T is angle preserving if and only if all j�i j are equal.

c. What are all angle preserving T W Rn ! Rn?

Proof.

(a) If T is norm preserving, then T is inner product preserving by the previous

exercise. Hence, for x;y ¤ 0,

† .Tx;Ty/ D arccos

�hTx;Tyi

kTxk � kTyk

�D arccos

�hx;yi

kxk � kyk

�D † .x;y/ :

(b) We first suppose that T is angle preserving. Since .x1; : : : ;xn/ is a basis of

Rn, all xi ’s are nonzero. Since

†�Txi ;Txj

�D arccos

˝Txi ;Txj

˛kTxik �

Txj !D arccos

˝�ixi ; �jxj

˛k�ixik �

�jxj !

D arccos

�i�j

˝xi ;xj

˛j�i j �

ˇ̌�jˇ̌� kxik � kxj k

!D †

�xi ;xj

�;

it must be the case that

�i�j Dj�i j �ˇ̌�jˇ̌:

Then �i and �j have the same signs. ut

I Exercise 9 (1-9). If 0 6 � < � , let T W R2 ! R2 have the matrix

A D

cos � sin �

� sin � cos �

!:

Show that T is angle preserving and if x ¤ 0, then † .x;Tx/ D � .

Proof. For every�x; y

�2 R2, we have

T�x; y

�D

cos � sin �

� sin � cos �

! x

y

!D

x cos � C y sin �

�x sin � C y cos �

!:

Therefore, T �x; y� 2 D x2 C y2 D �x; y� 2 Ithat is, T is norm preserving. Then it is angle preserving by Exercise 8 (a).

SECTION 1.1 NORM AND INNER PRODUCT 5

Let x D .a; b/ ¤ 0. We first have

hx;Txi D a .a cos � C b sin �/C b .�a sin � C b cos �/ D�a2 C b2

�cos �:

Hence,

† .x;Tx/ D arccos

�hx;Txi

kxk � kTxk

�D arccos

�a2 C b2

�cos �

a2 C b2

!D �: ut

I Exercise 10 (1-10�). If M W Rm ! Rn is a linear transformation, show that

there is a number M such that kM hk 6M khk for h 2 Rm.

Proof. Let M’s matrix be

A D

�a11 � � � a1m:::

: : ::::

an1 � � � anm

˘

´

�a1

:::

an

˘

:

Then

M h D Ah D

� ˝a1;h

˛:::

han;hi

˘

;

and so

kM hk2D

nXiD1

Dai ;h

E26

nXiD1

�kaik � khk

�2D

0@ nXiD1

kaik2

1A � khk2 ;that is,

kM hk 6

0B@p

nXiD1

kaik

1CA � khk :

Let M D

pnXiD1

kaik and we get the result. ut

I Exercise 11 (1-11). If x;y 2 Rn and z;w 2 Rm, show that h.x; z/ ; .y;w/i D

hx;yi C hz;wi and k.x; z/k Dqkxk

2C kzk

2.

Proof. We have .x; z/ ; .y;w/ 2 RnCm. Then

h.x; z/ ; .y;w/i D

nXiD1

xiyi C

mXjD1

zjwj D hx;yi C hz;wi ;

and

k.x; z/k2 D h.x; z/ ; .x; z/i D hx;xi C hz; zi D kxk2 C kzk2 : ut


I Exercise 12 (1-12�). Let .Rn/� denote the dual space of the vector space Rn. If

x 2 Rn, define 'x 2 .Rn/� by 'x .y/ D hx;yi. Define M W Rn ! .Rn/� by M x D 'x .

Show that M is a 1-1 linear transformation and conclude that every ' 2 .Rn/� is

'x for a unique x 2 Rn.

Proof. We first show M is linear. Take any x;y 2 Rn and a; b 2 R. Then

M .ax C by/ D 'axCby D a'x C b'y D aM x C bM y;

where the second equality holds since for every z 2 Rn,

'axCby .z/ D hax C by; zi D a hx; zi C b hy; zi D a'x .z/C b'y .z/ :

To see M is 1-1, we need only to show that ıM D f0g, where ıM is the null set

of M. But this is clear and so M is 1-1. Since dim .Rn/� D dim Rn, M is also onto.

This proves the last claim. ut

I Exercise 13 (1-13�). If x;y 2 Rn, then x and y are called perpendicular (or

orthogonal) if hx;yi D 0. If x and y are perpendicular, prove that kx C yk2D

kxk2C kyk

2.

Proof. If hx;yi D 0, we have

kx C yk2D hx C y;x C yi D kxk

2C 2 hx;yi C kyk2 D kxk2 C kyk2 : ut

1.2 Subsets of Euclidean Space

I Exercise 14 (1-14�). Simple. Omitted.

I Exercise 15 (1-15). Prove that˚x 2 Rn W kx � ak < r

is open.

Proof. For any y 2˚x 2 Rn W kx � ak < r

µ B.aI r/, let " D r�ka;yk. We show

that B.yI "/ � B.aI r/. Take any z 2 B.yI "/. Then

ka; zk 6 ka;yk C ky; zk < ka;yk C " D r: ut

I Exercise 16 (1-16). Simple. Omitted.

I Exercise 17 (1-17). Omitted.

SECTION 1.2 SUBSETS OF EUCLIDEAN SPACE 7

I Exercise 18 (1-18). If A � Œ0; 1� is the union of open intervals .ai ; bi / such

that each rational number in .0; 1/ is contained in some .ai ; bi /, show that @A D

Œ0; 1� X A.

Proof. Let X ´ Œ0; 1�. Obviously, A is open since A DSi .ai ; bi /. Then X X A

is closed in X and so X X A D X X A. Since @A D xA \ X X A D xA \ .X X A/, it

suffices to show that

X X A � xA: (1.1)

But (1.1) holds if and only if xA D X . Now take any x 2 X and any open nhood

U of x in X . Since Q is dense, there exists y 2 U . Since there exists some i such

that y 2 .ai ; bi /, we know that U \ .ai ; bi / ¤ ¿, which means that U \ A ¤ ¿,

which means that x 2 xA. Hence, X D xA, i.e., A is dense in X . ut

I Exercise 19 (1-19�). If A is a closed set that contains every rational number

r 2 Œ0; 1�, show that Œ0; 1� � A.

Proof. Take any r 2 .0; 1/ and any open interval r 2 I � .0; 1/. Then there

exists q 2 Q\ .0; 1/ such that q 2 I . Since q 2 A, we know that r 2 xA D A. Since

0; 1 2 A, the claim holds. ut

I Exercise 20 (1-20). Prove the converse of Corollary 1-7: A compact subset of

Rn is closed and bounded.

Proof. To show A is closed, we prove that Ac is open. Assume that x … A,

and let Gm D˚y 2 Rn W kx � yk > 1=m

, m D 1; 2; : : :. If y 2 A, then x ¤ y ;

hence, kx � yk > 1=m for some m; therefore y 2 Gm (see Figure 1.1). Thus,

A �S1mD1Gm, and by compactness we have a finite subcovering. Now observe

that the Gm for an increasing sequence of sets: G1 � G2 � � � � ; therefore, a

finite union of some of the Gm is equal to the set with the highest index. Thus,

K � Gs for some s, and it follows that B.xI 1=s/ � Ac . Therefore, Ac is open.

1=m

xA

Figure 1.1. A compact set is closed

Let A be compact. We first show that A is bounded. Let


O D˚.�i; i/n W i 2 N

be an open cover of A. Then there is a finite subcover

˚.�i1; i1/

n ; : : : ; .�im; im/n

of A. Let i 0 D max fi1; : : : ; img. Hence, A ��i 0; i 0

�; that is, A is bounded. ut

I Exercise 21 (1-21�).

a. If A is closed and x … A, prove that there is a number d > 0 such that

ky � xk > d for all y 2 A.

b. If A is closed, B is compact, and A \ B D ¿, prove that there is d > 0 such

that ky � xk > d for all y 2 A and x 2 B .

c. Give a counterexample in R2 if A and B are closed but neither is compact.

Proof.

(a) A is closed implies that Ac is open. Since x 2 Ac , there exists an open ball

B.xI d/ with d > 0 such that x 2 B.xI d/ � Ac . Then ky � xk > d for all y 2 A.

(b) For every x 2 B , there exists dx > 0 such that x 2 B.xI dx=2/ � Ac and

ky � xk > dx for all y 2 A. Then the family fB.xI dx=2/ W x 2 Bg is an open

cover of B . Since B is compact, there is a finite set fx1; : : : ; xng such that˚B.x1I dx1

=2/; : : : ;B.xnI dxn=2/

covers B as well. Now let

d D min˚dx1

=2; : : : ; dxn=2.2:

Then for any x 2 B , there is an open ball B.xi I xi=2/ containing x and ky � xik >di . Hence,

ky � xk > ky � xik � kxi � xk > di � di=2 D di=2 > d:

(c) See Figure 1.2.

0

Figure 1.2.ut

SECTION 1.3 FUNCTIONS AND CONTINUITY 9

I Exercise 22 (1-22�). If U is open and C � U is compact, show that there is a

compact set D such that C � DB and D � U .

Proof. ut

1.3 Functions and Continuity

I Exercise 23 (1-23). If f W A ! Rm and a 2 A, show that limx!a f .x/ D b if

and only if limx!a fi .x/ D bi for i D 1; : : : ; m.

Proof. Let f W A! Rm and a 2 A.

If: Assume that limx!a fi .x/ D bi for i D 1; : : : ; m. Then for every "=

pm > 0,

there is a number ıi > 0 such that f i .x/ � bi < "=

pm for all x 2 A which

satisfy 0 < kx � ak < ıi , for every i D 1; : : : ; m. Put

ı D min fı1; : : : ; ımg :

Then for all x 2 A satisfying 0 < kx � ak < ı, f i .x/ � bi < "pm; i D 1; : : : ; m:

Therefore, for every x 2 A which satisfy 0 < kx � ak < ı,

kf .x/ � bk D

pmXiD1

�f i .x/ � bi

�2<

pmXiD1

�"2=m

�D "I

that is, limx!a f .x/ D b.

Only if: Now suppose that limx!a f .x/ D b. Then for every number " > 0

there is a number ı > 0 such that kf .x/ � bk < " for all x 2 A which satisfy

0 < kx � ak < ı. But then for every i D 1; : : : ; m, f i .x/ � bi 6 kf .x/ � bk < ";i.e. limx!a f

i .x/ D bi . ut

I Exercise 24 (1-24). Prove that f W A ! Rm is continuous at a if and only if

each f i is.

Proof. By definition, f is continuous at a if and only if limx!a f .x/ D

f .a/; it follows from Exercise 23 that limx!a f .x/ D f .a/ if and only if

limx!a fi .x/ D f i .a/ for every i D 1; : : : ; m; that is, if and only if f i is contin-

uous at a for each i D 1; : : : ; m. ut

I Exercise 25 (1-25). Prove that a linear transformation T W Rn ! Rm is con-

tinuous.


Proof. Take any a 2 Rn. Then, by Exercise 10 (1-10), there exists M > 0 such

that

Tx � Ta D T .x � a/ 6M kx � ak :

Hence, for every " > 0, let ı D "=M . Then Tx � Ta < " when x 2 Rn and

0 < kx � ak < ı D "=M ; that is, limx!a Tx D Ta, and so T is continuous. ut

I Exercise 26 (1-26). Let A Dn�x; y

�2 R2 W x > 0 and 0 < y < x2

o.

a. Show that every straight line through .0; 0/ contains an interval around .0; 0/

which is in R2 X A.

b. Define f W R2 ! R by f .x/ D 0 if x … A and f .x/ D 1 if x 2 A. For h 2 R2

define gh W R! R by gh .t/ D f .th/. Show that each gh is continuous at 0, but

f is not continuous at .0; 0/.

Proof.

(a) Let the line through .0; 0/ be y D ax. If a 6 0, then the whole line is in

R2 XA. If a > 0, then ax intersects x2 at�a; a2

�and .0; 0/ and nowhere else; see

Figure 1.3.

0 x

y

yDx2

yDax

A

Figure 1.3.

(b) We first show that f is not continuous at 0. Clearly, f .0/ D 0 since 0 … A.

For every ı > 0, there exists x 2 A satisfying 0 < kxk < ı, butjf .x/ � f .0/j D 1.

We next show gh .t/ D f .th/ is continuous at 0 for every h 2 R2. If h D 0,

then g0 .t/ D f .0/ D 0 and so is continuous. So we now assume that h ¤ 0. It

is clear that

gh .0/ D f .0/ D 0:

The result is now from (a) immediately. ut

SECTION 1.3 FUNCTIONS AND CONTINUITY 11

I Exercise 27 (1-27). Prove that˚x 2 Rn W kx � ak < r

is open by considering

the function f W Rn ! R with f .x/ D kx � ak.

Proof. We first show that f is continuous. Take a point b 2 Rn. For any " > 0,

let ı D ". Then for every x satisfying kx � bk < ı, we have

jf .x/ � f .b/j Dˇ̌kx � ak � kb � ak

ˇ̌6 kx � ak � kb � ak 6 kx � bk < ı D ":

Hence,˚x 2 Rn W kx � ak < r

D f �1 .�1; r/ is open in Rn. ut

I Exercise 28 (1-28). If A � Rn is not closed, show that there is a continuous

function f W A! R which is unbounded.

Proof. Take any x 2 @A. Let f .y/ D 1= ky � xk for all y 2 A. ut

I Exercise 29 (1-29). Simple. Omitted.

I Exercise 30 (1-30). Let f W Œa; b�! R be an increasing function. If x1; : : : ; xn 2

Œa; b� are distinct, show thatPniD1 o

�f; xi

�< f .b/ � f .a/.

Proof. ut

2DIFFERENTIATION

2.1 Basic Definitions

I Exercise 31 (2-1�). Prove that if f W Rn ! Rm is differentiable at a 2 Rn, then

it is continuous at a.

Proof. Let f be differentiable at a 2 Rn; then there exists a linear map � W Rn !

Rm such that

limh!0

kf .aC h/ � f .a/ � �.h/k

khkD 0;

or equivalently,

f .aC h/ � f .a/ D �.h/C r.h/; (2.1)

where the remainder r.h/ satisfies

limh!0kr.h/k

ıkhk D 0: (2.2)

Let h! 0 in (2.1). The error term r.h/! 0 by (2.2); the linear term �.h/ aslo

tends to 0 because if h DPniD1 hiei , where .e1; : : : ; en/ is the standard basis of

Rn, then by linearity we have �.h/ DPniD1 hi�.ei /, and each term on the right

tends to 0 as h! 0. Hence,

limh!0

�f .aC h/ � f .a/

�D 0I

that is, limh!0 f .aC h/ D f .a/. Thus, f is continuous at a. ut

I Exercise 32 (2-2). A function f W R2 ! R is independent of the second vari-

able if for each x 2 R we have f .x; y1/ D f .x; y2/ for all y1; y2 2 R. Show that f

is independent of the second variable if and only if there is a function g W R! R

such that f .x; y/ D g.x/. What is f 0.a; b/ in terms of g0?

Proof. The first assertion is trivial: if f is independent of the second variable,

we can let g be defined by g.x/ D f .x; 0/. Conversely, if f .x; y/ D g.x/, then

f .x; y1/ D g.x/ D f .x; y2/.

If f is independent of the second variable, then

13

14 CHAPTER 2 DIFFERENTIATION

lim.h;k/!0

ˇ̌f .aC h; b C k/ � f .a; b/ � g0.a/h

ˇ̌k.h; k/k

D lim.h;k/!0

ˇ̌g.aC h/ � g.a/ � g0.a/h

ˇ̌k.h; k/k

6 limh!0

ˇ̌g.aC h/ � g.a/ � g0.a/h

ˇ̌jhj

D 0I

hence, f 0.a; b/ D .g0.a/; 0/. ut

I Exercise 33 (2-3). Define when a function f W R2 ! R is independent of the

first variable and find f 0.a; b/ for such f . Which functions are independent of

the first variable and also of the second variable?

Proof. We have f 0.a; b/ D .0; g0.b// with a similar argument as in Exercise 32.

If f is independent of the first and second variable, then for any .x1; y1/,

.x2; y2/ 2 R2, we have f .x1; y1/ D f .x2; y1/ D f .x2; y2/; that is, f is con-

stant. ut

I Exercise 34 (2-4). Let g be a continuous real-valued function on the unit

circle˚x 2 R2 W kxk D 1

such that g.0; 1/ D g.1; 0/ D 0 and g.�x/ D �g.x/.

Define f W R2 ! R by

f .x/ D

˚kxk � g

�xıkxk

�if x ¤ 0;

0 if x D 0:

a. If x 2 R2 and h W R ! R is defined by h.t/ D f .tx/, show that h is differen-

tiable.

b. Show that f is not differentiable at .0; 0/ unless g D 0.

Proof. (a) If x D 0 or t D 0, then h.t/ D f .0/ D 0; if x ¤ 0 and t > 0,

h.t/ D f .tx/ D t kxk � g

�tx

t kxk

�D

hkxk � g

�x= kxk

�i� t D f .x/t I

finally, if x ¤ 0 and t < 0,

h.t/ D f .tx/ D �t kxk � g

�tx

�t kxk

�D �t kxk � g

��x= kxk

�D

hkxk � g

�x= kxk

�i� t

D f .x/t:

Therefore, h.t/ D f .x/t for every given x 2 R2, and so is differentiable: Dh D h.

(b) Since g.1; 0/ D 0 and g.�x/ D �g.x/, we have g.�1; 0/ D g.�.1; 0// D

�g.1; 0/ D 0. If f is differentiable at .0; 0/, there exists a matrix .a; b/ such that

Df .0; 0/.h; k/ D ahC bk. First consider any sequence .h; 0/! .0; 0/. Then

SECTION 2.1 BASIC DEFINITIONS 15

0 D limh!0

jf .h; 0/ � f .0; 0/ � ahj

jhjD limh!0

ˇ̌̌jhj � g

�h=jhj ; 0

�� ah

ˇ̌̌jhj

D limh!0

ˇ̌jhj � g .˙1; 0/ � ah

ˇ̌jhj

Djaj

implies that a D 0. Next let us consider .0; k/! .0; 0/. Then

0 D limk!0

jf .0; k/ � f .0; 0/ � bkj

jkjD limk!0

ˇ̌̌jkj � g

�0; k=jkj

�� bk

ˇ̌̌jkj

Djbj

forces that b D 0. Therefore, f 0.0; 0/ D .0; 0/ and Df .0; 0/.x; y/ D 0. If g.x/ ¤ 0,

then

limx!0

jf .x/ � f .0/ � 0j

kxkD lim

x!0

ˇ̌̌kxk � g

�x= kxk

�ˇ̌̌kxk

D limx!0

ˇ̌̌g�x= kxk

�ˇ̌̌¤ 0;

and so f is not differentiable.

Of course, if g.x/ D 0, then f .x/ D 0 and is differentiable. ut

I Exercise 35 (2-5). Let f W R2 ! R be defined by

f .x; y/ D

�x jyj

�qx2 C y2 if .x; y/ ¤ 0;

0 if .x; y/ D 0:

Show that f is a function of the kind considered in Exercise 34, so that f is not

differentiable at .0; 0/.

Proof. If .x; y/ ¤ 0, we can rewrite f .x; y/ as

f .x; y/ Dx �jyjqx2 C y2

Dx �jyj

k.x; y/kD k.x; y/k �

�x

k.x; y/k�jyj

k.x; y/k

�: (2.3)

If we let g W˚x 2 R2 W kxk D 1

! R be defined as g.x; y/ D x �jyj, then (2.3) can

be rewritten as

f .x; y/ D k.x; y/k � g..x; y/= k.x; y/k/:

It is easy to see that

g.0; 1/ D g.1; 0/ D 0; and g.�x;�y/ D �xj�yj D �xjyj D �f .x; y/I

that is, g satisfies all of the properties listed in Exercise 34. Since g.x/ ¤ 0

unless x D 0 or y D 0, we know that f is not differentiable at 0. A direct proof

can be found in Berkovitz (2002, Section 1.11). ut

I Exercise 36 (2-6). Let f W R2 ! R be defined by f .x; y/ Dpjxyj. Show that

f is not differentiable at .0; 0/.


Proof. It is clear that

limh!0

jf .h; 0/j

jhjD 0 D lim

k!0

jf .0; k/j

jkjI

hence, if f is differentiable at .0; 0/, it must be that Df .0; 0/.x; y/ D 0 since

derivative is unique if it exists. However, if we let h D k > 0, and take a

sequence f.h; h/g ! .0; 0/, we have

lim.h;h/!.0;0/

jf .h; h/ � f .0; 0/ � 0j

k.h; h/kD lim.h;h/!.0;0/

ph2

k.h; h/kD

1p2¤ 0:

Therefore, f is not differentiable. ut

I Exercise 37 (2-7). Let f W R2 ! R be a function such that jf .x/j 6 kxk2. Show

that f is differentiable at 0.

Proof. jf .0/j 6 k0k2 D 0 implies that f .0/ D 0. Since

limx!0

jf .x/ � f .0/j

kxkD lim

x!0

jf .x/j

kxk6 lim

x!0kxk D 0;

Df .0/.x; y/ D 0. ut

I Exercise 38 (2-8). Let f W R ! R2. Prove that f is differentiable at a 2 R if

and only if f 1 and f 2 are, and that in this case

f 0.a/ D

.f 1/0.a/

.f 2/0.a/

!:

Proof. Suppose that f is differentiable at a with f 0.a/ D

c1

c2

!. Then for i D

1; 2,

0 6 limh!0

ˇ̌̌f i .aC h/ � f i .a/ � ci � h

ˇ̌̌jhj

6 limh!0

kf .aC h/ � f .a/ � Df .a/.h/k

jhjD 0

implies that f i is differentiable at a with .f i /0.a/ D ci .

Now suppose that both f 1 and f 2 are differentiable at a, then by Exercise 1,

0 6kf .aC h/ � f .a/ � Df .a/.h/k

jhj6

2XiD1

ˇ̌̌f i .aC h/ � f i .a/ � .f i /0.a/ � h

ˇ̌̌jhj

implies that f is differentiable at a with f 0.a/ D

.f 1/0.a/

.f 2/0.a/

!. ut

I Exercise 39 (2-9). Two functions f; g W R! R are equal up to n-th order at a

if


limh!0

f .aC h/ � g.aC h/

hnD 0:

a. Show that f is differentiable at a if and only if there is a function g of the

form g.x/ D a0C a1.x � a/ such that f and g are equal up to first order at a.

b. If f 0.a/; : : : ; f .n/.a/ exist, show that f and the function g defined by

g.x/ D

nXiD0

f .i/.a/

i Š.x � a/i

are equal up to n-th order at a.

Proof. (a) If f is differentiable at a, then by definition,

limh!0

f .aC h/ ��f .a/C f 0.a/ � h

�h

D 0;

so we can let g.x/ D f .a/C f 0.a/ � .x � a/.

On the other hand, if there exists a function g.x/ D a0 C a1.x � a/ such that

limh!0

f .aC h/ � g.aC h/

hD limh!0

f .aC h/ � a0 � a1h

hD 0;

then a0 D f .a/, and so f is differentiable at a with f 0.a/ D a1.

(b) By Taylor’s Theorem1 we rewrite f as

f .x/ D

n�1XiD0

f .i/.a/

i Š.x � a/i C

f .n/.y/

nŠ.x � a/n;

where y is between a and x. Thus,

limx!a

f .x/ � g.x/

.x � a/nD limx!a

f .n/.y/nŠ

.x � a/n � f .n/.a/nŠ

.x � a/n

.x � a/n

D limx!a

f .n/.y/ � f .n/.x/

nŠ

D 0: ut

1 (Rudin, 1976, Theorem 5.15) Suppose f is a real function on Œa; b�, n is a positive integer,f .n�1/ is continuous on Œa; b�, f .n/ exists for every t 2 .a; b/. Let ˛, ˇ be distinct points ofŒa; b�, and define

P.t/ D

n�1XkD0

f .k/.˛/

kŠ.t � ˛/k :

Then there exists a point x between ˛ and ˇ such that

f .ˇ/ D P.ˇ/Cf .n/.x/

nŠ.ˇ � ˛/n:


2.2 Basic Theorems

I Exercise 40 (2-10). Use the theorems of this section to find f 0 for the follow-

ing:

a. f .x; y; z/ D xy .

b. f .x; y; z/ D .xy ; z/.

c. f .x; y/ D sin.x siny/.

d. f .x; y; z/ D sin�x sin.y sin z/

�.

e. f .x; y; z/ D xyz.

f. f .x; y; z/ D xyCz .

g. f .x; y; z/ D .x C y/z .

h. f .x; y/ D sin.xy/.

i. f .x; y/ D�sin.xy/

�cos3.

j. f .x; y/ D�sin.xy/; sin

�x siny

�; xy

�.

Solution. Compare this with Exercise 47.

(a) We have f .x; y; z/ D xy D elnxyD ey lnx D exp B.�2 � ln�1/.x; y; z/. It

follows from the Chain Rule that

f 0.a; b; c/ D exp0h.�2 ln�1/.a; b; c/

i�

��2 ln�1

�0.a; b; c/

D exp.b ln a/ �h.ln�1/.�2/0 C �2.ln�1/0

i.a; b; c/

D ab �h�0; ln a; 0

�C�b=a; 0; 0

�iD

�ab�1b ab ln a 0

�:

(b) By (a) and Theorem 2-3(3), we have

f 0.a; b; c/ D

ab�1b ab ln a 0

0 0 1

!:

(c) We have f .x; y/ D sin B.�1 sin.�2//. Then, by the chain rule,

f 0.a; b/ D sin0h.�1 sin.�2//.a; b/

i�

h�1 sin.�2/

i0.a; b/

D cos .a sin b/ �h.sin�2/.�1/0 C �1.sin�2/0

i.a; b/

D cos .a sin b/ ��sin b .1; 0/C a .0; cos b/

�D

�cos .a sin b/ � sin b a � cos .a sin b/ � cos b

�:

(d) Let g.y; z/ D sin.y sin z/. Then

SECTION 2.2 BASIC THEOREMS 19

f .x; y; z/ D sin�x � g.y; z/

�D sin.�1 � g.�2; �3//:

Hence,

f 0.a; b; c/ D sin0�ag .b; c/

�� .�1g.�2; �3//0.a; b; c/

D cos�ag .b; c/

��

hg .b; c/ .�1/0 C ag0.�2; �3/

i.a; b; c/

D cos�ag .b; c/

��

h�g .b; c/ ; 0; 0

�C ag0.�2; �3/.a; b; c/

i:

It follows from (c) that

g0.�2; �3/.a; b; c/ D�0 cos .b sin c/ � sin c b � cos .b sin c/ � cos c

�:

Therefore,

f 0.a; b; c/

D cos�a sin .b sin c/

� �sin .b sin c/ a cos .b sin c/ sin c ab cos .b sin c/ cos c

�:

(e) Let g.x; y/ D xy . Then

f .x; y; z/ D xg.y;z/ D g�x; g.y; z/

�D g

��1; g.�2; �3/

�:

Then

Df .a; b; c/ D Dg�a; g .b; c/

�B

hD�1;Dg.�2; �3/

i.a; b; c/:

By (a),

Dg�a; g .b; c/

�.x; y; z/ D

�ag.b;c/g .b; c/ =a ag.b;c/ ln a 0

�0B@xyz

1CADab

cbc

ax C

�ab

c

ln a�y;

D�1.a; b; c/.x; y; z/ D x;

and

Dg.�2; �3/.a; b; c/.x; y; z/ D Dg .b; c/ B�D�2;D�3

�.a; b; c/.x; y; z/

Dbcc

by C

�bc ln b

�z:

Hence,

Df .a; b; c/.x; y; z/ Dab

cbc

ax C

�ab

c

ln a� �bcc

by C

�bc ln b

�z

�;

and


f 0.a; b; c/ D�ab

c

bcıa ab

c

bcc ln aıb ab

cbc ln a ln b

�:

(f) Let g.x; y/ D xy . Then f .x; y; z/ D xyCz D g�x; y C z

�D g

��1; �2 C �3

�.

Hence,

Df .a; b; c/.x; y; z/ D Dg .a; b C c/ B�D�1;D�2 C D�3

�.a; b; c/.x; y; z/

D Dg .a; b C c/ B�x; y C z

�DabCc .b C c/

ax C

�abCc ln a

� �y C z

�;

and

f 0.a; b; c/ D�abCc.bCc/

aabCc ln a abCc ln a

�:

(g) Let g.x; y/ D xy . Then

f .x; y; z/ D .x C y/z D g�x C y; z

�D g

��1 C �2; �3

�:

Hence,

Df .a; b; c/.x; y; z/ D Dg .aC b; c/ BhD�1 C D�2;D�3

i.a; b; c/.x; y; z/

D Dg .aC b; c/ B�x C y; z

�D.aC b/c c

.aC b/.x C y/C

�.aC b/c ln .aC b/

�z;

and

f 0.a; b; c/ D�.aCb/cc.aCb/

.aCb/cc.aCb/

.aC b/c ln .aC b/�:

(h) We have f .x; y/ D sin.xy/ D sin B��1�2

�. Hence,

f 0.a; b/ D .sin/0 .ab/ �hb.�1/0.a; b/C a.�2/0.a; b/

iD cos .ab/ �

�b .1; 0/C a.0; 1/

�D cos .ab/ � .b; a/

D

�b � cos .ab/ a � cos .ab/

�:

(i) Straightforward.

(j) By Theorem 2-3 (3), we have

f 0.a; b; c/ D

0BB@�sin.xy/

�0.a; b; c/h

sin�x siny

�i0.a; b; c/

Œxy �0 .a; b; c/

1CCAD

0B@ b � cos .ab/ a � cos .ab/

cos .a sin b/ � sin b a � cos .a sin b/ � cos b

ab�1b ab ln a

1CA : ut

I Exercise 41 (2-11). Find f 0 for the following (where g W R! R is continuous):


a. f .x; y/ DR xCya

g.

b. f .x; y/ DR xyag.

c. f .x; y; z/ DR sin

�x sin.y sin z/

�xy g.

Solution. (a) Let h.t/ DR tag. Then f .x; y/ D

�h B .�1 C �2/

�.x; y/, and so

f 0.a; b/ D h0 .aC b/ �h.�1 C �2/0.a; b/

iD g .aC b/ � .1; 1/

D

�g .aC b/ g .aC b/

�:

(b) Let h.t/ DR tag. Then f .x; y/ D

R xyag D h.xy/ D

hh B

��1 � �2

�i.x; y/.

Hence,

f 0.a; b/ D h0 .ab/ �hb � .�1/0.a; b/C a � .�2/0.a; b/

iD g .ab/ � .b; a/

D

�b � g .ab/ a � g .ab/

�:

(c) We can rewrite f .x; y; z/ as

f .x; y; z/ D

Z a

xy

g C

Z sin.x sin.y sin z//

a

g D

Z sin.x sin.y sin z//

a

g �

Z xy

a

g:

Let .x; y; z/ D sin�x sin.y sin z/

�, k.x; y; z/ D

R .x;y;z/a

g, and h.x; y; z/ DR xyag.

Then f .x; y; z/ D k.x; y; z/ � h.x; y; z/, and so

f 0.a; b; c/ D k0.a; b; c/ � h0.a; b; c/:

It follows from Exercise 40 (d) that

k0.a; b; c/ D k0� .a; b; c/

�� 0.a; b; c/:

The other parts are easy. ut

I Exercise 42 (2-12). A function f W Rn � Rm ! Rp is bilinear if for x;x1;x2 2

Rn, y;y1;y2 2 Rm, and a 2 R we have

f .ax;y/ D af .x;y/ D f .x; ay/ ;

f .x1 C x2;y/ D f .x1;y/C f .x2;y/ ;

f .x;y1 C y2/ D f .x;y1/C f .x;y2/ :

a. Prove that if f is bilinear, then

lim.h;k/!0

kf .h;k/k

k.h;k/kD 0:


b. Prove that Df .a;b/.x;y/ D f .a;y/C f .x;b/.

c. Show that the formula for Dp.a;b/ in Theorem 2-3 is a special case of (b).

Proof. (a) Let�en1 ; : : : ; e

nn

�and

�em1 ; : : : ; e

mm

�be the stand bases for Rn and Rm,

respectively. Then for any x 2 Rn and y 2 Rm, we have

x D

nXiD1

xiein; and y D

mXjD1

yj ejm:

Therefore,

f .x;y/ D f

0@ nXiD1

xiein;

mXjD1

yj ejm

1A D nXiD1

f

0@xiein; mXjD1

yj ejm

1AD

nXiD1

mXjD1

f�xiein; y

j ejm

�

D

nXiD1

mXjD1

xiyjf�ein; e

jm

�:

Then, by letting M DPi;j

f �ein; ejm

� , we have

kf .x;y/k D

Xi;j

xiyjf�ein; e

jm

� 6Xi;j

ˇ̌̌xiyj

ˇ̌̌ f �ein; ejm

� 6M

"maxi

�ˇ̌̌xiˇ̌̌�

maxj

�ˇ̌̌yjˇ̌̌�#

6M kxk kyk :

Hence,

lim.h;k/!0

kf .h;k/k

k.h;k/k6 lim.h;k/!0

M khk kkk

k.h;k/k

D lim.h;k/!0

M khk kkkpXi;j

��hi�2C

�kj�2�

D lim.h;k/!0

M khk kkkqkhk

2C kkk

2

:

Now

khk kkk 6

˚khk

2 if kkk 6 khkkkk

2 if khk 6 kkk :

Hence khk kkk 6 khk2 C kkk2, and so


lim.h;k/!0

M khk kkkqkhk

2C kkk

2

6 lim.h;k/!0

M

qkhk

2C kkk

2D 0:

(b) We have

lim.h;k/!0

kf .aC h;bC k/ � f .a;b/ � f .a;k/ � f .h;b/k

k.h;k/k

D lim.h;k/!0

kf .a;b/C f .a;k/C f .h;b/C f .h;k/ � f .a;b/ � f .a;k/ � f .h;b/k

k.h;k/k

D lim.h;k/!0

kf .h;k/k

k.h;k/k

D 0

by (a); hence, Df .a;b/.x;y/ D f .a;y/C f .x;b/.

(c) It is easy to check that p W R2 ! R defined by p.x; y/ D xy is bilinear.

Hence, by (b), we have

Dp.a; b/.x; y/ D p�a; y

�C p .x; b/ D ay C xb: ut

I Exercise 43 (2-13). Define IP W Rn � Rn ! R by IP.x;y/ D hx;yi.

a. Find D .IP/ .a;b/ and .IP/0 .a;b/.

b. If f; g W R ! Rn are differentiable and h W R ! R is defined by h.t/ D

hf .t/; g.t/i, show that

h0.a/ DDf 0.a/T; g.a/

EC

Df .a/; g0.a/T

E:

c. If f W R! Rn is differentiable and kf .t/k D 1 for all t , show thatDf 0.t/T; f .t/

ED

0.

d. Exhibit a differentiable function f W R ! R such that the function jf j defined

by jf j .t/ Djf .t/j is not differentiable.

Proof. (a) It is evident that IP is bilinear; hence, by Exercise 42 (b), we have

D .IP/ .a;b/.x;y/ D IP .a;y/C IP .x;b/

D ha;yi C hx;bi

D hb;xi C ha;yi ;

and so .IP/0 .a;b/ D .b; a/.

(b) Since h.t/ D IP B�f; g

�.t/, by the chain rule, we have

Dh.a/ .x/ D D .IP/�f .a/; g.a/

�B�Df .a/ .x/ ;Dg.a/ .x/

�D hg.a/;Df .a/ .x/i C hf .a/;Dg.a/ .x/i

D˝g.a/; f 0.a/

˛x C

˝f .a/; g0.a/

˛x:

(c) Let h.t/ D hf .t/; f .t/i with kf .t/k D 1 for all t 2 R. Then


h.t/ D kf .t/k2 D 1

is constant, and so h0.a/ D 0; that is,

0 DDf 0.a/T; f .a/

EC

Df .a/; f 0.a/T

ED 2

Df 0.a/T; f .a/

E;

and soDf 0.a/T; f .a/

ED 0.

(d) Let f .t/ D t . Then f is linear and so is differentiable: Df D t . However,

limt!0C

jt j

tD 1; lim

t!0�

jt j

tD �1I

that is, jf j is not differentiable at 0. ut

I Exercise 44 (2-14). Let Ei , i D 1; : : : ; k be Euclidean spaces of various

dimensions. A function f W E1 � � � � � Ek ! Rp is called multilinear if for

each choice of xj 2 Ej , j ¤ i the function g W Ei ! Rp defined by g.x/ D

f .x1; : : : ;xi�1;x;xiC1; : : : ;xk/ is a linear transformation.

a. If f is multilinear and i ¤ j , show that for h D .h1; : : : ;hk/, with h` 2 E`, we

have

limh!0

f �a1; : : : ;hi ; : : : ;hj ; : : : ; ak� khk

D 0:

b. Prove that

Df .a1; : : : ; ak/ .x1; : : : ;xk/ DkXiD1

f .a1; : : : ; ai�1;xi ; aiC1; : : : ; ak/ :

Proof.

(a) To light notation, define

a�i�j ´�a1; : : : ; ai�1; aiC1; : : : ; aj�1; ajC1; : : : ; ak

�:

Let g W Ei�Ej ! Rp be defined as g�xi ;xj

�D f

�a�i�j ;xi ;xj

�. Then g is bilinear

and so

limh!0

g �a�i�j ;hi ;hj � khk

6 limh!0

g �a�i�j ;hi ;hj � �hi ;hj � D 0

by Exercise 42 (a).

(b) It follows from Exercise 42 (b) immediately. ut

I Exercise 45 (2-15). Regard an n � n matrix as a point in the n-fold product

Rn � � � � � Rn by considering each row as a member of Rn.

a. Prove that det W Rn � � � � � Rn ! R is differentiable and


D�det

�.a1; : : : ; an/ .x1; : : : ;xn/ D

nXiD1

det

0BBBBBBBB@

a1:::

xi:::

an

1CCCCCCCCA:

b. If aij W R! R are differentiable and f .t/ D det�aij .t/

�, show that

f 0.t/ D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � a1n.t/:::

: : ::::

a0j1.t/ � � � a0jn.t/:::

: : ::::

an1.t/ � � � ann.t/

1CCCCCCCCA:

c. If det�aij .t/

�¤ 0 for all t and b1; : : : ; bn W R ! R are differentiable, let

s1; : : : ; sn W R ! R be the functions such that s1.t/; : : : ; sn.t/ are the solutions

of the equationsnX

jD1

aj i .t/sj .t/ D bi .t/; i D 1; : : : ; n:

Show that si is differentiable and find s0i .t/.

Proof.

(a) It is easy to see that det W Rn � � � � � Rn ! R is multilinear; hence, the con-

clusion follows from Exercise 44.

(b) By (a) and the chain rule,

f 0.t/ D�det

�0 �aij .t/

�B�a01.t/; : : : ; a

0n.t/

�

D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � a1n.t/:::

: : ::::

a0j1.t/ � � � a0jn.t/:::

: : ::::

an1.t/ � � � ann.t/

1CCCCCCCCA:

(c) Let

A D

[email protected]/ � � � an1.t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCA ; s D

[email protected]/:::

sn.t/

1CCA ; and b D

[email protected]/:::

bn.t/

1CCA :Then

As D b;


and so

si .t/ Ddet .Bi /

det .A/;

where Bi is obtained from A by replacing the i -th column with the b. It follows

from (b) that si .t/ is differentiable. Define f .t/ D det .A/ and gi .t/ D det .Bi /.

Then

f 0.t/ D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � an1.t/:::

: : ::::

a01j .t/ � � � a0nj .t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCCCCCCCA;

and

g0i .t/ D

nXjD1

0BBBBBBBB@

a11.t/ � � � ai�1;1.t/ b1.t/ aiC1;1.t/ � � � an1.t/:::

: : ::::

::::::

: : ::::

a01j .t/ � � � a0i�1;j .t/ b0j .t/ a0iC1;j .t/ � � � a0nj .t/:::

: : ::::

::::::

: : ::::

a1n.t/ � � � ai�1;n.t/ bn.t/ aiC1;n.t/ � � � ann.t/

1CCCCCCCCA:

Therefore,

s0i .t/ Df 0.t/g0i .t/ � f .t/g

0i .t/

f 2.t/: ut

I Exercise 46 (2-16). Suppose f W Rn ! Rn is differentiable and has a differen-

tiable inverse f �1 W Rn ! Rn. Show that�f �1

�0.a/ D

hf 0�f �1.a/

�i�1.

Proof. We have f B f �1.x/ D x. On the one hand D�f B f �1

�.a/ .x/ D x since

f B f �1 is linear; on the other hand,

D�f B f �1

�.a/ .x/ D

�Df

�f �1 .a/

�B Df �1 .a/

�.x/:

Therefore, Df �1 .a/ DhDf

�f �1 .a/

�i�1. ut

2.3 Partial Derivatives

I Exercise 47 (2-17). Find the partial derivatives of the following functions:

a. f .x; y; z/ D xy .

b. f .x; y; z/ D z.

c. f .x; y/ D sin.x siny/.

d. f .x; y; z/ D sin�x sin.y sin z/

�.

SECTION 2.3 PARTIAL DERIVATIVES 27

e. f .x; y; z/ D xyz.

f. f .x; y; z/ D xyCz .

g. f .x; y; z/ D .x C y/z .

h. f .x; y/ D sin.xy/.

i. f .x; y/ D�sin.xy/

�cos3.

Solution. Compare this with Exercise 40.

(a) D1f .x; y; z/ D yxy�1, D2f .x; y; z/ D xy ln x, and D3f .x; y; z/ D 0.

(b) D1f .x; y; z/ D D2f .x; y; z/ D 0, and D3f .x; y; z/ D 1.

(c) D1f .x; y/ D�siny

�cos

�x siny

�, and D2f .x; y/ D x cosy cos

�x siny

�.

(d) D1f .x; y; z/ D sin.y sin z/ cos�x sin.y sin z/

�,

D2f .x; y; z/ D cos�x sin.y sin z/

�x cos.y sin z/ sin z, and

D3f .x; y; z/ D cos�x sin.y sin z/

�x cos.y sin z/y cos z.

(e) D1f .x; y; z/ D yzxyz�1, D2f .x; y; z/ D xy

zzyz�1 ln x, and D3f .x; y; z/ D

yz lny�xy

zln x

�.

(f) D1f .x; y; z/ D�y C z

�xyCz�1, and D2f .x; y; z/D3f .x; y; z/ D xyCz ln x.

(g) D1f .x; y; z/ D D2f .x; y; z/ D z.x C y/z�1, and

D3f .x; y; z/ D .x C y/z ln.x C y/.

(h) D1f .x; y/ D y cos.xy/, and D2f .x; y/ D x cos.xy/.

(i) D1f .x; y/ D cos 3�sin.xy/

�cos3�1y cos.xy/, and

D2f .x; y/ D cos 3�sin.xy/

�cos3�1x cos.xy/. ut

I Exercise 48 (2-18). Find the partial derivatives of the following functions

(where g W R! R is continuous):

a. f .x; y/ DR xCya

g.

b. f .x; y/ DR xyg.

c. f .x; y/ DR xyag.

d. f .x; y/ DR .R y

bg/

ag.

Solution.

(a) D1f .x; y/ D D2f .x; y/ D g.x C y/.

(b) D1f .x; y/ D g.x/, and D2f .x; y/ D �g�y�.

(c) D1f .x; y/ D yg.xy/, and D2f .x; y/ D xg.xy/.


(d) D1f .x; y/ D 0, and D2f .x; y/ D g�y�� g�R ybg�. ut

I Exercise 49 (2-19). If

f .x; y/ D xxxxy

C�ln x

�0@arctan

arctan

�arctan

�sin

�cos xy

�� ln.x C y/

��!1Afind D2f

�1; y

�.

Solution. Putting x D 1 into f .x; y/, we get f�1; y

�D 1. Then D2f

�1; y

�D

0. ut

I Exercise 50 (2-20). Find the partial derivatives of f in terms of the deriva-

tives of g and h if

a. f .x; y/ D g.x/h�y�.

b. f .x; y/ D g.x/h.y/.

c. f .x; y/ D g.x/.

d. f .x; y/ D g�y�.

e. f .x; y/ D g.x C y/.

Solution.

(a) D1f .x; y/ D g0 .x/ h�y�, and D2f .x; y/ D g.x/h0

�y�.

(b) D1f .x; y/ D h�y�g.x/h.y/�1g0 .x/, and D2f .x; y/ D h0

�y�g.x/h.y/ lng.x/.

(c) D1f .x; y/ D g0 .x/, and D2f .x; y/ D 0.

(d) D1f .x; y/ D 0, and D2f .x; y/ D g0�y�.

(e) D1f .x; y/ D D2f .x; y/ D g0.x C y/. ut

I Exercise 51 (2-21�). Let g1; g2 W R2 ! R be continuous. Define f W R2 ! R by

f .x; y/ D

Z x

0

g1 .t; 0/ dt C

Z y

0

g2 .x; t/ dt:

a. Show that D2f .x; y/ D g2.x; y/.

b. How should f be defined so that D1f .x; y/ D g1.x; y/?

c. Find a function f W R2 ! R such that D1f .x; y/ D x and D2f .x; y/ D y. Find

one such that D1f .x; y/ D y and D2f .x; y/ D x.

Proof.


(a) D2f .x; y/ D 0C g2.x; y/ D g2.x; y/.

(b) We should let

f .x; y/ D

Z x

0

g1�t; y

�dt C

Z y

0

g2 .a; t/ dt;

where t 2 R is a constant.

(c) Let

� f .x; y/ D�x2 C y2

�=2.

� f .x; y/ D xy. ut

I Exercise 52 (2-22�). If f W R2 ! R and D2f D 0, show that f is independent

of the second variable. If D1f D D2f D 0, show that f is constant.

Proof. Fix any x 2 R. By the mean-value theorem, for any y1; y2 2 R, there

exists a point y� 2�y1; y2

�such that

f�x; y2

�� f

�x; y1

�D D2f

�x; y�

� �y2 � y1

�D 0:

Hence, f�x; y1

�D f

�x; y2

�; that is, f is independent of y.

Similarly, if D1f D 0, then f is independent of x. The second claim is then

proved immediately. ut

I Exercise 53 (2-23�). Let A D˚.x; y/ 2 R2 W x < 0, or x > 0 and y ¤ 0

.

a. If f W A! R and D1f D D2f D 0, show that f is constant.

b. Find a function f W A! R such that D2f D 0 but f is not independent of the

second variable.

Proof.

(a) As in Figure 2.1, for any .a; b/; .c; d/ 2 R2, we have

f .a; b/ D f .�1; b/ D f .�1; d/ D f .c; d/ :

(b) For example, we can let

f .x; y/ D

˚0 if x < 0 or y < 0

x otherwise.ut

I Exercise 54 (2-24). Define f W R2 ! R by

f .x; y/ D

˚xy x

2�y2

x2Cy2 .x; y/ ¤ 0;

0 .x; y/ D 0:

a. Show that D2f .x; 0/ D x for all x and D1f�0; y

�D �y for all y.


.a; b/ .�1; b/

.�1; d/ .c; d/

x

y

Figure 2.1. f is constant

b. Show that D1;2f .0; 0/ ¤ D2;1f .0; 0/.

Proof.

(a) We have

D2f .x; y/ D

�x.x4�y4�4x2y2/

.x2Cy2/2 .x; y/ ¤ 0;

0 .x; y/ D 0;

and

D1f .x; y/ D

��y.y4�x4�4x2y2/

.x2Cy2/2 .x; y/ ¤ 0;

0 .x; y/ D 0:

Hence, D2f .x; 0/ D x and D1f�0; y

�D �y.

(b) By (a), we have D1;2f .0; 0/ D D2�D1f

�0; y

��.0/ D �1; but D2;1f .0; 0/ D

D1�D2 .x; 0/

�.0/ D 1. ut

I Exercise 55 (2-25�). Define f W R! R by

f .x/ D

˚e�x

�2x ¤ 0

0 x D 0:

Show that f is a C1 function, and f .i/ .0/ D 0 for all i .

Proof. Figure 2.2 depicts f .x/. We first show that f 2 C1.

Let pn�y�

be a polynomial with degree n with respect to y. For x ¤ 0 and

k 2 N, we show that f .k/ .x/ D p3k�x�1

�e�x

�2. We do this by induction.

Step 1 Clearly, f 0 .x/ D 2x�3e�x�2

.

Step 2 Suppose that f .k/ .x/ D p3k�x�1

�e�x

�2.

Step 3 Then by the chain rule,


0 x

y

−2 −1 1 2

Figure 2.2.

f .kC1/ .x/ Dhf .k/ .x/

i0D p03k

�x�1

��

��x�2

�� e�x

�2

C p3k

�x�1

�� 2x�3 � e�x

�2

D

�p03k

�x�1

��

��x�2

�C p3k

�x�1

�� 2x�3

�� e�x

�2

D

�q3kC1

�x�1

�C q3kC3

�x�1

�� e�x

�2

D p3.kC1/

�x�1

�� e�x

�2

;

where q3kC1 and q3kC3 are polynomials.

Therefore, f .x/ 2 C1 for all x ¤ 0. It remains to show that f .k/ .x/ is

defined and continuous at x D 0 for all k.

Step 1 Obviously,

f 0 .0/ D limx!0

f .x/ � f .0/

xD limx!0

e�x�2

xD limx!0

2x�3e�x�2

D 0

by L’Hôpital’s rule.

Step 2 Suppose that f .k/ .0/ D 0.

Step 3 Then,

f .kC1/ .0/ D limx!0

f .k/ .x/ � f .k/ .0/

x

D limx!0

p3kC1

�x�1

�e�x

�2

D limx!0

p3kC1�x�1

�ex�2

:

Hence, if we use L’Hôpital’s rule 3k C 1 times, we get f .kC1/ .0/ D 0.

A similar computation shows that f .k/ .x/ is continuous at x D 0. ut

I Exercise 56 (2-26�). Let

f .x/ D

˚e�.x�1/

�2

� e�.xC1/�2

x 2 .�1; 1/ ;

0 x … .�1; 1/ :


a. Show that f W R ! R is a C1 function which is positive on .�1; 1/ and 0

elsewhere.

b. Show that there is a C1 function g W R ! Œ0; 1� such that g.x/ D 0 for x 6 0and g.x/ D 1 for x > ".

c. If a 2 Rn, define g W Rn ! R by

g.x/ D f

x1 � a1

"

!� � � f

�xn � an

"

�:

Show that g is a C1 function which is positive on�a1 � "; a1 C "

��

�an � "; an C "

�and zero elsewhere.

d. If A � Rn is open and C � A is compact, show that there is a non-negative

C1 function f W A ! R such that f .x/ > 0 for x 2 C and f D 0 outside of

some closed set contained in A.

e. Show that we can choose such an f so that f W A ! Œ0; 1� and f .x/ D 1 for

x 2 C .

Proof.

(a) If x 2 .�1; 1/, then x � 1 ¤ 0 and x C 1 ¤ 0. It follows from Exercise 55 that

e�.x�1/�2

2 C1 and e�.xC1/�2

2 C1. Then it is straightforward to check that

f 2 C1. See Figure 2.3

0 x

y

−1 1

Figure 2.3.

(b) By letting z D xC 1, we derive a new function j W R! R from f as follows:

j .z/ D

˚e�.z�2/

�2

� e�z�2

z 2 .0; 2/ ;

0 z … .0; 2/ :

By letting w D "z=2, we derive a function k W R! R from j as follows:

k .w/ D

˚e�.2w="�2/

�2

� e�.2w="/�2

w 2 .0; "/ ;

0 w … .0; "/ :


0 x

y

1 2

j

0 x

yk

Figure 2.4.

It is easy to see that k 2 C1, which is positive on .0; "/ and 0 elsewhere. Now

let

g.x/ D

�Z x

0

k .x/

�,�Z "

0

k .x/

�:

Then g 2 C1; it is 0 for x 6 0, increasing on .0; "/, and 1 for x > ".

(c) It follows from (a) immediately.

(d) For every x 2 C , let Rx ´ .�"; "/n be a rectangle containing x, and Rx is

contained in A (we can pick such a rectangle since A is open and C � A).

Then fRx W x 2 C g is an open cover of C . Since C is compact, there exists

fx1; : : : ; xmg � C such that˚Rx1

; : : : ; Rxm

covers C . For every xi , i D 1; : : : ; n,

we define a function gi W Rxi! R as

gi .x/ D f

x1i � a

1i

"

!� � � f

�xni � a

ni

"

�;

where�a1i ; : : : ; a

ni

�2 Rn is the middle point of Rxi

.

Finally, we define g W Rx1[ � � � [Rxm

! R as follows:

g.x/ D

mXiD1

gi .x/ :

Then g 2 C1; it is positive on C , and 0 outside Rx1[ � � � [Rxm

.

(e) Follows the hints. ut

I Exercise 57 (2-27). Define g; h W˚x 2 R2 W kxk 6 1

! R3 by

g.x; y/ D

�x; y;

q1 � x2 � y2

�;

h.x; y/ D

�x; y;�

q1 � x2 � y2

�:

Show that the maximum of f on˚x 2 R3 W kxk D 1

is either the maximum of

f B g or the maximum of f B h on˚x 2 R2 W kxk 6 1

.

Proof. Let A ´˚x 2 R2 W kxk 6 1

and B ´

˚x 2 R3 W kxk D 1

. Then B D

g .A/ [ h .A/. ut


2.4 Derivatives

I Exercise 58 (2-28). Find expressions for the partial derivatives of the follow-

ing functions:

a. F.x; y/ D f�g.x/k

�y�; g.x/C h

�y��

.

b. F.x; y; z/ D f�g.x C y/; h

�y C z

��.

c. F.x; y; z/ D f�xy ; yz ; zx

�.

d. F.x; y/ D f�x; g.x/; h.x; y/

�.

Proof.

(a) Letting a´ g.x/k�y�; g.x/C h

�y�, we have

D1F.x; y/ D D1f .a/ � g0 .x/ � k

�y�C D2f .a/ � g

0 .x/ ;

D2F.x; y/ D D1f .a/ � g.x/ � k0�y�C D1f .a/ � h

0�y�:

(b) Letting a´ g.x C y/; h�y C z

�, we have

D1F.x; y; z/ D D1f .a/ � g0.x C y/;

D2F.x; y; z/ D D1f .a/ � g0.x C y/C D2f .a/ � h

0�y C z

�;

D3F.x; y; z/ D D2f .a/ � h0�y C z

�:

(c) Letting a´ xy ; yz ; zx , we have

D1F.x; y; z/ D D1f .a/ � yxy�1C D3f .a/ � z

x ln z;

D2F.x; y; z/ D D1f .a/ � xy ln x C D2f .a/ � zy

z�1;

D3F.x; y; z/ D D2f .a/ � yz lny C D3f .a/ � xz

x�1:

(d) Letting a´ x; g.x/; h.x; y/, we have

D1F.x; y/ D D1f .a/C D2f .a/ � g0 .x/C D3f .a/ � D1h.x; y/

D2F.x; y/ D D3f .a/ � D2h.x; y/: ut

I Exercise 59 (2-29). Let f W Rn ! R. For x 2 Rn, the limit

limt!0

f .aC tx/ � f .a/

t;

if it exists, is denoted Dxf .a/, and called the directional derivative of f at a, in

the direction x.

a. Show that Deif .a/ D Dif .a/.

b. Show that Dtxf .a/ D tDxf .a/.

SECTION 2.4 DERIVATIVES 35

c. If f is differentiable at a, show that Dxf .a/ D Df .a/.x/ and therefore

DxCyf .a/ D Dxf .˛/C Dyf .a/.

Proof.

(a) For ei D .0; : : : ; 0; 1; 0; : : : ; 0/, we have

Deif .a/ D lim

t!0

f .aC tei / � f .a/

t

D limt!0

f .a1; : : : ; ai�1; ai C t; aiC1; : : : ; an/ � f .a/

t

D Dif .a/

by definition.

(b) We have

Dtxf .a/ D lims!0

f .aC stx/ � f .a/

sD limst!0

tf .aC stx/ � f .a/

stD tDxf .a/:

(c) If f is differentiable at a, then for any x ¤ 0 we have

0 D limt!0

jf .aC tx/ � f .a/ � Df .a/.tx/j

ktxk

D limt!0

jf .aC tx/ � f .a/ � t � Df .a/.x/j

jt j�1

kxk

D limt!0

ˇ̌̌̌f .aC tx/ � f .a/

t� Df .a/.x/

ˇ̌̌̌�1

kxk;

and so

Dxf .a/ D limt!0

f .aC tx/ � f .a/

tD Df .a/.x/:

The case of x D 0 is trivial. Therefore,

DxCyf .a/ D Df .a/ .x C y/

D Df .a/.x/C Df .a/ .y/

D Dxf .a/C Dyf .a/: ut

I Exercise 60 (2-30). Let f be defined as in Exercise 34. Show that Dxf .0; 0/

exists for all x, but if g ¤ 0, then DxCyf .0; 0/ ¤ Dxf .0; 0/ C Dyf .0; 0/ for all

x;y .

Proof. Take any x 2 R2.

limt!0

f .tx/ � f .0; 0/

tD lim

t!0

jt j � kxk � g

�tx.�jt j � kxk

��t

:

Therefore, Dxf .0; 0/ exists for any x.

Now let g ¤ 0; then, D.0;1/f .0; 0/ D D.1;0/f .0; 0/ D 0, but D.1;0/C.0;1/f .0; 0/ D

D.1;1/f .0; 0/ ¤ 0. ut


I Exercise 61 (2-31). Let f W R2 ! R be defined as in Exercise 26. Show that

Dxf .0; 0/ exists for all x, although f is not even continuous at .0; 0/.

Proof. For any x 2 R2, we have

limt!0

f .tx/ � f .0/

tD lim

t!0

f .tx/

tD 0

by Exercise 26 (a). ut

I Exercise 62 (2-32).

a. Let f W R! R be defined by

f .x/ D

˚x2 sin 1

xx ¤ 0

0 x D 0:

Show that f is differentiable at 0 but f 0 is not continuous at 0.

b. Let f W R2 ! R be defined by

f .x; y/ D

‚ �x2 C y2

�sin 1q

x2 C y2.x; y/ ¤ 0

0 .x; y/ D 0:

Show that f is differentiable at .0; 0/ but Dif is not continuous at .0; 0/.

Proof.

(a) We have

limx!0

f .x/ � f .0/

xD limx!0

x2 sin 1x

xD limx!0

x sin1

xD 0:

Hence, f 0 .0/ D 0. Further, for any x ¤ 0, we have

f 0 .x/ D 2x sin1

x� cos

1

x:

It is clear that limx!0 f0 .x/ does not exist. Therefore, f 0 is not continuous at

0.

(b) Since

lim.x;y/!.0;0/

�x2 C y2

�sin 1q

x2 C y2qx2 C y2

D lim.x;y/!.0;0/

qx2 C y2 sin

1qx2 C y2

D 0;

we know that f 0.0; 0/ D .0; 0/. Now take any .x; y/ ¤ .0; 0/. Then

D1f .x; y/ D 2x sin1q

x2 C y2� 2x cos

1qx2 C y2

:

SECTION 2.4 DERIVATIVES 37

0

Figure 2.5.

As in (a), limx!0 D1f .x; 0/ does not exist. Similarly for D2f . ut

I Exercise 63 (2-33). Show that the continuity of D1f j at a may be eliminated

from the hypothesis of Theorem 2-8.

Proof. It suffices to see that for the first term in the sum, we have, by letting�a2; : : : ; an

�µ a�1,

limh!0

ˇ̌̌f�a1 C h1; a�1

�� f .a/ � D1f .a/ � h1

ˇ̌̌khk

6 limh1!0

ˇ̌̌f�a1 C h1; a�1

�� f .a/ � D1f .a/ � h1

ˇ̌̌ˇ̌h1ˇ̌ D 0:

See aslo Apostol (1974, Theorem 12.11). ut

I Exercise 64 (2-34). A function f W Rn ! R is homogeneous of degree m if

f .tx/ D tmf .x/ for all x. If f is also differentiable, show that

nXiD1

xiDif .x/ D mf .x/:

Proof. Let g.t/ D f .tx/. Then, by Theorem 2-9,

g0.t/ D

nXiD1

Dif .tx/ � xi : (2.4)

On the other hand, g.t/ D f .tx/ D tmf .x/; then

g0.t/ D mtm�1f .x/: (2.5)

Combining (2.4) and (2.5), and letting t D 1, we then get the result. ut

I Exercise 65 (2-35). If f W Rn ! R is differentiable and f .0/ D 0, prove that

there exist gi W Rn ! R such that


f .x/ D

nXiD1

xigi .x/:

Proof. Let hx.t/ D f .tx/. ThenZ 1

0

h0x.t/dt D hx .1/ � hx .0/ D f .x/ � f .0/ D f .x/:

Hence,

f .x/ D

Z 1

0

h0x.t/dt D

Z 1

0

f 0.tx/dt D

Z 1

0

24 nXiD1

xiDif .tx/

35 dt

D

nXiD1

xiZ 1

0

Dif .tx/dt

D

nXiD1

xigi .x/;

where gi .x/ DR 10

Dif .tx/dt . ut

2.5 Inverse Functions

For this section, Rudin (1976, Section 9.3 and 9.4) is a good reference.

I Exercise 66 (2-36�). Let A � Rn be an open set and f W A! Rn a continuously

differentiable 1-1 function such that det�f 0.x/

�¤ 0 for all x. Show that f .A/ is

an open set and f �1 W f .A/ ! A is differentiable. Show also that f .B/ is open

for any open set B � A.

Proof. For every y 2 f .A/, there exists x 2 A such that f .x/ D y . Since f 2

C 0 .A/ and det�f 0.x/

�¤ 0, it follows from the Inverse Function Theorem that

there is an open set V � A containing x and an open set W � Rn containing y

such that W D f .V /. This proves that f .A/ is open.

Since f W V ! W has a continuous inverse f �1 W W ! V which is differen-

tiable, it follows that f �1 is differentiable at y ; since y is chosen arbitrary, it

follows that f �1 W f .A/! A is differentiable.

Take any open set B � A. Since f�B 2 C 0 .B/ and det��f�B

�0.x/

�¤ 0 for

all x 2 B � A, it follows that f .B/ is open. ut


a. Let f W R2 ! R be a continuously differentiable function. Show that f is not

1-1.

SECTION 2.5 INVERSE FUNCTIONS 39

b. Generalize this result to the case of a continuously differentiable function

f W Rn ! Rm with m < n.

Proof.

(a) Let f 2 C 0. Then both D1f and D2f are continuous. Assume that f is 1-1;

then both D1f and D2f cannot not be constant and equal to 0. So suppose that

there is�x0; y0

�2 R2 such that D1f

�x0; f0

�¤ 0. The continuity of D1f implies

that there is an open set A � R2 containing�x0; y0

�such that D1f .x/ ¤ 0 for

all x 2 A.

Define a function g W A! R2 with

g.x; y/ D�f .x; y/; y

�:

Then for all .x; y/ 2 A,

g0.x; y/ D

D1f .x; y/ D2f .x; y/

0 1

!;

and so det�g0.x; y/

�D D1f .x; y/ ¤ 0; furthermore, g 2 C 0 .A/ and g is 1-1. Then

by Exercise 66, we know that g .A/ is open. We now show that g .A/ cannot be

open actually.

Take a point�f�x0; y0

�; zy�2 g .A/ with y ¤ y0. Then for any .x; y/ 2 A, we

must have

g.x; y/ D�f .x; y/; y

�D

�f�x0; y0

�; zy�H) .x; y/ D

�x0; y0

�I

that is, there is no .x; y/ 2 A such that g.x; y/ D�f�x0; y0

�; zy�. This proves

that f cannot be 1-1.

(b) We can write f W Rn ! Rm as f D�f 1; : : : ; f m

�, where f i W Rn ! R for every

i D 1; : : : ; m. As in (a), there is a mapping, say, f 1, a point a 2 Rn, and an open

set A containing a such that D1f 1.x/ ¤ 0 for all x 2 A. Define g W A! Rm as

g�x1;x�1

�D

�f .x/;x�1

�;

where x�1´�x2; : : : ; xn

�. Then as in (a), it follows that f cannot be 1-1. ut


a. If f W R! R satisfies f 0.a/ ¤ 0 for all a 2 R, show that f is 1-1 (on all of R).

b. Define f W R2 ! R2 by f .x; y/ D�ex cosy; ex siny

�. Show that det

�f 0.x; y/

�¤

0 for all .x; y/ but f is not 1-1.

Proof.


(a) Suppose that f is not 1-1. Then there exist a; b 2 R with a < b such that

f .a/ D f .b/. It follows from the mean-value theorem that there exists c 2 .a; b/

such that

0 D f .b/ � f .a/ D f 0 .c/ .b � a/ ;

which implies that f 0 .c/ D 0. A contradiction.

(b) We have

f 0.x; y/ D

Dxex cosy Dyex cosy

Dxex siny Dyex siny

!

D

ex cosy �ex siny

ex siny ex cosy

!:

Then

det�f 0.x; y/

�D e2x

�cos2 y C sin2 y

�D e2x ¤ 0:

However, f .x; y/ is not 1-1 since f .x; y/ D f�x; y C 2k�

�for all .x; y/ 2 R2 and

k 2 N.

This exercise shows that the non-singularity of Df on A implies that f is

locally 1-1 at each point of A, but it does not imply that f is 1-1 on all of A.

See Munkres (1991, p. 69). ut

I Exercise 69 (2-39). Use the function f W R! R defined by

f .x/ D

˚x2C x2 sin 1

xx ¤ 0

0 x D 0

to show that continuity of the derivative cannot be eliminated from the hypoth-

esis of Theorem 2-11.

Proof. If x ¤ 0, then

f 0 .x/ D1

2C 2x sin

1

x� cos

1

xI

if x D 0, then

f 0 .0/ D limh!0

h=2C h2 sin�1=h

�h

D1

2:

Hence, f 0 .x/ is not continuous at 0. It is easy to see that f is not injective for

any neighborhood of 0 (see Figure 2.6).

2.6 Implicit Functions

I Exercise 70 (2-40). Use the implicit function theorem to re-do Exercise 45 (c).

Proof. Define f W R � Rn ! Rn by

SECTION 2.6 IMPLICIT FUNCTIONS 41

0

Figure 2.6.ut

f i .t; s/ D

nXjD1

aj i .t/sj� bi .t/;

for i D 1; : : : ; n. Then

M´

0BB@D2f 1 .t; s/ � � � D1Cnf 1 .t; s/

:::: : :

:::

D2f n .t; s/ � � � D1Cnf n .t; s/

1CCA [email protected]/ � � � an1.t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCA ;and so det .M/ ¤ 0.

It follows from the Implicit Function Theorem that for each t 2 R, there is a

unique s.t/ 2 Rn such that f�t; s.t/

�D 0, and s is differentiable. ut

I Exercise 71 (2-41). Let f W R�R! R be differentiable. For each x 2 R define

gx W R! R by gx�y�D f .x; y/. Suppose that for each x there is a unique y with

g0x�y�D 0; let c .x/ be this y.

a. If D2;2f .x; y/ ¤ 0 for all .x; y/, show that c is differentiable and

c0 .x/ D �D2;1f

�x; c .x/

�D2;2f

�x; c .x/

� :b. Show that if c0 .x/ D 0, then for some y we have

D2;1f .x; y/ D 0;

D2f .x; y/ D 0:

c. Let f .x; y/ D x�y logy � y

�� y log x. Find


max1=26x62

"min

1=36y61f .x; y/

#:

Proof.

(a) For every x, we have g0x�y�D D2f .x; y/. Since for every x there is a unique

y D c .x/ such that D2f�x; c .x/

�D 0, the solution c .x/ is the same as ob-

tained from the Implicit Function Theorem; hence, c .x/ is differentiable, and

by differentiating D2f�x; c .x/

�D 0 with respect to x, we have

D2;1f�x; c .x/

�C D2;2f

�x; c .x/

�� c0 .x/ D 0I

that is,

c0 .x/ D �D2;1f

�x; c .x/

�D2;2f

�x; c .x/

� :(b) It follows from (a) that if c0 .x/ D 0, then D2;1f

�x; c .x/

�D 0. Hence, there

exists some y D c .x/ such that D2;1f .x; y/ D 0. Furthermore, by definition,

D2�x; c .x/

�D D2f .x; y/ D 0.

(c) We have

D2f .x; y/ D x lny � ln x:

Let D2f .x; y/ D 0 we have y D c .x/ D x1=x . Also, D2;2f .x; y/ D x=y > 0 since

x; y > 0. Hence, for every fixed x 2�1=2; 2

�,

minyf .x; y/ D f

�x; c .x/

�:

0

x

y

0.5 1 1.5 2

0.5

1

1.5 c .x/

Figure 2.7.

It is easy to see that c0 .x/ > 0 on�1=2; 2

�, c .1/ D 1, and c.a/ D 1=3 for some

a > 1=2 (see Figure 2.7). Therefore,

min1=36y61

f�x; y

�D f

�x; y� .x/

�;

where (see Figure 2.8)

SECTION 2.6 IMPLICIT FUNCTIONS 43

y� .x/ D

„1=3 if 1=2 6 x 6 ac .x/ D x1=x if a < x 6 11 if 1 < x 6 2:

0

x

y

0.5 1 1.5 2

0.5

1

1.5

y� .x/

Figure 2.8.

1=2 6 x 6 a In this case, our problem is

max1=26x6a

f�x; 1=3

�D �

�1C ln 3

3

�x �

1

3ln x:

It is easy to see that x� D 1=2, and so f�x�; 1=3

�D ln

�4=3e

�=6.

a < x 6 1 In this case, our problem is

maxa<x61

f�x; x1=x

�D �x1C1=x :

It is easy to see that the maximum of f occurs at x� D a and y��x��D 1=3.

1 < x 6 2 In this case, our problem is

max1<x62

f .x; 1/ D �x � ln x:

The maximum of f occurs at x� D 1.

Now, as depicted in Figure 2.9, we have x� D 1=2, y� D 1=3, and f�x�; y�

�D

ln�4=3e

�=6. ut


0

xf�x; y� .x/

�0.5 1 1.5 2

−2.5

−2

−1.5

−1

−0.5

Figure 2.9.

3INTEGRATION

3.1 Basic Definitions

I Exercise 72 (3-1). Let f W Œ0; 1� � Œ0; 1�! R be defined by

f�x; y

�D

˚0 if 0 6 x < 1=21 if 1=2 6 x 6 1:

Show that f is integrable andRŒ0;1��Œ0;1�

f D 1=2.

Proof. Consider a partition P D .P1; P2/ with P1 D P2 D�0; 1=2; 1

�. Then

L�f; P

�D U

�f; P

�D 1=2. It follows from Theorem 3-3 (the Riemann condition)

that f is integrable andRŒ0;1��Œ0;1�

f D 1=2. ut

I Exercise 73 (3-2). Let f W A! R be integrable and let g D f except at finitely

many points. Show that g is integrable andRAf D

RAg.

Proof. Fix an " > 0. It follows from the Riemann condition that there is a

partition P of A such that

U�f; P

�� L

�f; P

�<"

2:

Let P 0 be a refinement of P such that:

� for every x 2 A with g .x/ ¤ f .x/, it belongs to 2n subrectangles of P 0, i.e.,

x is a corner of each subrectangle.

� for every subrectangle S of P 0,

v .S/ <"

2nC1d .u � `/;

where

45

46 CHAPTER 3 INTEGRATION

d Dˇ̌̌˚

x W f .x/ ¤ g .x/ˇ̌̌;

u D supx2A

fg .x/g � infx2Aff .x/g ;

` D infx2Afg .x/g � sup

x2A

ff .x/g :

x

Figure 3.1.

With such a choice of partition of A, we have

U�g; P 0

�� U

�f; P 0

�D

dXiD1

24 2nXjD1

hMSij

�g��MSij

�f�iv�Sij�35

6 d2nuv;

where v ´ supS2P 0 fv .S/g is the least upper bound of the volumes of the

subrectangles of P 0. Similarly,

L�g; P 0

�� L

�f; P 0

�D

dXiD1

24 2nXjD1

hmSij

�g��mSij

�f�iv�Sij�35

> d2n`v:

Therefore,

U�g; P 0

�� L

�g; P 0

�6hU�f; P 0

�C d2nuv

i�

hL�f; P 0

�C d2n`v

i6"

2C d2n .u � `/ v

D"

2C d2n .u � `/

"

2nC1d .u � `/

D "I

that is, g is integrable. It is easy to see now thatRAg D

RAf . ut

I Exercise 74 (3-3). Let f; g W A! R be integrable.

a. For any partition P of A and subrectangle S , show that mS�f�C mS

�g�6

mS�f C g

�and MS

�f C g

�6 MS

�f�C MS

�g�

and therefore L�f; P

�C

L�g; P

�6 L

�f C g; P

�and U

�f C g; P

�6 U

�f; P

�C U

�g; P

�.


b. Show that f C g is integrable andRA

�f C g

�DRAf C

RAg.

c. For any constant c, show thatRAcf D c

RAf .

Proof.

(a) We show that mS�f�C mS

�g�

is a lower bound ofn�f C g

�.x/ W x 2 S

o. It

is clear that mS�f�6 f .x/ and mS

�g�6 g .x/ for any x 2 S . Then for every

x 2 S we have

mS�f�CmS

�g�6 f .x/C g .x/ D

�f C g

�.x/ :

Hence, mS�f�CmS

�g�6 mS

�f C g

�.

Similarly, for every x 2 S we have MS

�f�> f .x/ and MS

�g�> g .x/;

hence,�f C g

�.x/ D f .x/ C g .x/ 6 MS

�f�C MS

�g�

and so MS

�f C g

�6

MS

�f�CMS

�g�.

Now for any partition P of A we have

L�f; P

�C L

�g; P

�D

XS2P

mS�f�v .S/C

XS2P

mS�g�

D

XS2P

hmS

�f�CmS

�g�iv .S/

6XS2P

mS�f C g

�v .S/

D L�f C g; P

�;

(3.1)

and

U�f; P

�C U

�g; P

�D

XS2P

MS

�f�v .S/C

XS2P

MS

�g�v .S/

D

XS2P

hMS

�f�CMS

�g�iv .S/

>XS2P

MS

�f C g

�v .S/

D U�f C g; P

�:

(3.2)

(b) It follows from (3.1) and (3.2) that for any partition P ,

U�f C g; P

�� L

�f C g; P

�6hU�f; P

�C U

�g; P

�i�

hL�f; P

�C L

�g; P

�iD

hU�f; P

�� L

�f; P

�iC

hU�g; P

�� L

�g; P

�i:

Since f and g are integrable, there exist P 0 and P 00 such that for any " > 0,

we have U�f; P 0

��L

�f; P 0

�< "=2 and U

�g; P 00

��L

�g; P 00

�< "=2. Let xP refine

both P 0 and P 00. Then

U�f; xP

�� L

�f; xP

�<"

2and U

�g; xP

�� L

�g; xP

�<"

2:

Hence,


U�f C g; xP

�� L

�f C g; xP

�< ";

and so f C g is integrable.

Now, by definition, for any " > 0, there exists a partition P (by using a

common refinement partition if necessary) such thatRAf < L

�f; P

�C "=2,R

Ag < L

�g; P

�C"=2, U

�f; P

�<RAf C"=2, and U

�g; P

�<RAgC"=2. Therefore,Z

A

f C

ZA

g � " < L�f; P

�C L

�g; P

�6 L

�f C g; P

�6ZA

�f C g

�6 U

�f C g; P

�6 U

�f; P

�C U

�g; P

�<

ZA

f C

ZA

g C ":

Hence,RA

�f C g

�DRAf C

RAg.

(c) First, suppose that c > 0. Then for any partition P and any subrectangle

S , we have mS�cf�D cmS

�f�

and MS

�cf�D cMS

�f�. But then L

�cf; P

�D

cL�f; P

�and U

�cf; P

�D cU

�f; P

�. Since f is integrable, for any " > 0 there

exists a partition P such that U�f; P

�� L

�f; P

�< "=c. Therefore,

U�cf; P

�� L

�cf; P

�D c

hU�f; P

�� L

�f; P

�i< "I

that is, cf is integrable. Further,

c

ZA

f �"

c< cL

�f; P

�D L

�cf; P

�6ZA

cf 6 U�cf; P

�D cU

�f; P

�< c

ZA

f C"

c;

i.e.,RAcf D c

RAf .

Now let c < 0. Then for any partition P of A, we have mS�cf�D cMS

�f�

and MS

�cf�D cmS

�f�. Hence L

�cf; P

�D cU

�f; P

�and U

�cf; P

�D cL

�f; P

�.

Since f is integrable, for every " > 0, choose P such that U�f; P

�� L

�f; P

�<

�"=c. Then

U�cf; P

�� L

�cf; P

�D �c

hU�f; P

�� L

�f; P

�i< "I

that is, cf is integrable. Furthermore,

�c

ZA

f C"

c< �cL

�f; P

�D �U

�cf; P

�6 �

ZA

cf 6 �L�cf; P

�D �cL

�f; P

�< �c

ZA

f �"

c;

i.e.,RAcf D c

RAf . ut


I Exercise 75 (3-4). Let f W A! R and let P be a partition of A. Show that f is

integrable if and only if for each subrectangle S the function f�S is integrable,

and that in this caseRAf D

PS

RSf�S .

Proof. Let P be a partition of A, and S be a subrectangle with respect to P .

Only if: Suppose that f is integrable. Then there exists a partition P1 of A

such that U�f; P1

�� L

�f; P1

�< " for any given " > 0. Let P2 be a common

refinement of P and P1. Then

U�f; P2

�� L

�f; P2

�6 U

�f; P1

�� L

�f; P1

�< ";

and there are rectangles˚S12 ; : : : ; S

n2

µ �2 .S/ with respect to P2, such that

S DSniD1 S

i2. Therefore,

U�f; P2

�� L

�f; P2

�D

XS2

hMS2

�f��mS2

�f�iv .S2/

>X

S22�2.S/

hMS2

�f��mS2

�f�iv .S2/

D U�f�S;P2

�� L

�f�S;P2

�I

that is, f�S is integrable.

0

x

y

a b

c

d

Figure 3.2.

If: Now suppose that f �S is integrable for each S . For each partition P 0, letˇ̌P 0ˇ̌

be the number of subrectangles induced by P 0. Let PS be a partition such

that

U�f�S;PS

�� L

�f�S;PS

�<

"

2jP j:

Let P 0 be the partition of A obtained by taking the union of all the sub-

sequences defining the partitions of the PS ; see Figure 3.2. Then there are


refinements P 0S of PS whose rectangles are the set of all subrectangles of P 0

which are contained in S . Hence,XS

ZS

f�S � " <XS

L�f�S;PS

�6XS

L�f�S;P 0S

�D L

�f; P 0

�6 U

�f; P 0

�D

XS

U�f�S;P 0S

�6XS

U�f�S;PS

�<XS

ZS

f�S C ":

Therefore, f is integrable, andRAf D

PS

RSf�S . ut

I Exercise 76 (3-5). Let f; g W A ! R be integrable and suppose f 6 g. Show

thatRAf 6

RAg.

Proof. Since f is integrable, the function �f is integrable by Exercise 74 (c);

then g � f is integrable by Exercise 74 (b). It is easy to seeRA

�g � f

�> 0

since g > f . It follows from Exercise 74 thatRA

�g � f

�DRA

�g C

��f

��DR

Ag C

RA

��f

�DRAg �

RAf ; hence,

RAf 6

RAg. ut

I Exercise 77 (3-6). If f W A! R is integrable, show that jf j is integrable andˇ̌RAfˇ̌6RAjf j.

Proof. Let f C D max ff; 0g and f � D max f�f; 0g. Then

f D f C � f � and jf j D f C C f �:

It is evident that for any partition P of A, both U�f C; P

�� L

�f C; P

�6

U�f; P

��L

�f; P

�and U

�f �; P

��L

�f �; P

�6 U

�f; P

��L

�f; P

�; hence, both

f C and f � are integrable if f is. Further,ˇ̌̌̌ZA

f

ˇ̌̌̌D

ˇ̌̌̌ZA

�f C � f �

�ˇ̌̌̌D

ˇ̌̌̌ZA

f C �

ZA

f �ˇ̌̌̌

6ZA

f C C

ZA

f �

D

ZA

�f C C f �

�D

ZA

jf j : ut

I Exercise 78 (3-7). Let f W Œ0; 1� � Œ0; 1�! R be defined by

SECTION 3.3 FUBINI’S THEOREM 51

f�x; y

�D

„0 x irrational

0 x rational, y irrational

1=q x rational, y D p=q is lowest terms.

Show that f is integrable andRŒ0;1��Œ0;1�

f D 0.

Proof. ut

3.2 Measure Zero and Content Zero

I Exercise 79 (3-8). Prove that Œa1; b1�� Œan; bn� does not have content 0 if

ai < bi for each i .

Proof. Similar to the Œa; b� case. ut


a. Show that an unbounded set cannot have content 0.

b. Give an example of a closed set of measure 0 which does not have content 0.

Proof.

(a) Finite union of bounded sets is bounded.

(b) Z or N. ut


a. If C is a set of content 0, show that the boundary of C has content 0.

b. Give an example of a bounded set C of measure 0 such that the boundary of

C does not have measure 0.

Proof. ut

3.3 Fubini’s Theorem

I Exercise 82 (3-27). If f W Œa; b� � Œa; b�! R is continuous, show thatZ b

a

Z y

a

f�x; y

�dx dy D

Z b

a

Z b

x

f�x; y

�dy dx:

Proof. As illustrated in Figure 3.3,


C Dn�x; y

�2 Œa; b�2 W a 6 x 6 y and a 6 y 6 b

oD

n�x; y

�2 Œa; b�2 W a 6 x 6 b and x 6 y 6 b

o:

0

x

y

0

x

y

a b

a

byDx

C

y first

x first

Figure 3.3. Fubini’s Theoremut

I Exercise 83 (3-30). Let C be the set in Exercise 17. Show thatZŒ0;1�

ZŒ0;1�

1C�x; y

�dx

!dy D

ZŒ0;1�

ZŒ0;1�

1C�x; y

�dy

!dx D 0:

Proof. There must be typos. ut

I Exercise 84 (3-31). If A D Œa1; b1�� Œan; bn� and f W A! R is continuous,

define F W A! R by

F .x/ D

ZŒa1;x1��Œan;xn�

f:

What is DiF .x/, for x 2 int.A/?

Solution. Let c 2 int.A/. Then

SECTION 3.3 FUBINI’S THEOREM 53

DiF .c/ D limh!0

F�c�i ; ci C h

�� F .c/

h

D limh!0

RŒa1;c1��Œai ;c

iCh��Œan;cn� f � F .c/

h

D limh!0

R ciCh

ai

�RŒa1;c1��Œai�1;x

i�1��ŒaiC1;ciC1��Œan;cn� f

�dxi � F .c/

h

D limh!0

R ciCh

ci

�RŒa1;c1��Œai�1;c

i�1��ŒaiC1;ciC1��Œan;cn� f

�dxi

h

D

ZŒa1;c1��Œai�1;c

i�1��ŒaiC1;ciC1��Œan;cn�

f�x�i ; ci

�: ut

I Exercise 85 (3-32�). Let f W Œa; b� � Œc; d � ! R be continuous and suppose

D2f is continuous. Define F�y�DR baf�x; y

�dx. Prove Leibnitz’s rule: F 0

�y�DR b

aD2f

�x; y

�dx.

Proof. We have

F 0�y�D limh!0

F�y C h

�� F

�y�

h

D limh!0

R baf�x; y C h

�dx �

R baf�x; y

�dx

h

D limh!0

Z b

a

f�x; y C h

�� f

�x; y

�h

dx:

By DCT, we have

F 0�y�D

Z b

a

"limh!0

f�x; y C h

�� f

�x; y

�h

#dx

D

Z b

a

D2f�x; y

�dx: ut

I Exercise 86 (3-33). If f W Œa; b� � Œc; d �! R is continuous and D2f is contin-

uous, define F�x; y

�DR xaf�t; y

�dt .

a. Find D1F and D2F .

b. If G .x/ DR g.x/a

f .t; x/ dt , find G0 .x/.

Solution.

(a) D1F�x; y

�D f

�x; y

�, and D2F D

R xa

D2f�t; y

�dt .

(b) It follows that G .x/ D F�g .x/ ; x

�. Then

G0 .x/ D g0 .x/D1F�g .x/ ; x

�C D2F

�g .x/ ; x

�D g0 .x/ f

�g .x/ ; x

�C

Z g.x/

a

D2f .t; x/ dt: ut

4INTEGRATION ON CHAINS

4.1 Algebraic Preliminaries

I Exercise 87 (4-1�). Let e1; : : : ; en be the usual basis of Rn and let '1; : : : ; 'nbe the dual basis.

a. Show that 'i1 ^ � � � ^ 'ik .ei1 ; : : : ; eik / D 1. What would the right side be if the

factor .k C `/Š=kŠ`Š did not appear in the definition of ^?

b. Show that 'i1 ^ � � � ^ 'ik .v1; : : : ; vk/ is the determinant of the k � k minor of�v1:::

vk

˘

obtained by selecting columns i1; : : : ; ik .

Proof.

(a) Since 'ij 2 T .Rn/, for every j D 1; : : : ; k, we have

'i1 ^ � � � ^ 'ik .ei1 ; : : : ; eik / DkŠ

1Š � � � 1ŠAlt

�'i1 ˝ � � � ˝ 'ik

�.ei1 ; : : : ; eik /

D

X�2Sk

.sgn.�//'i1.e�.i1// � � �'ik .e�.ik//

D 1:

If the factor .k C `/Š=kŠ`Š did not appear in the definition of ^, then the

solution would be 1=kŠ.

(b) ut

I Exercise 88 (4-9�). Deduce the following properties of the cross product in

R3.

a.e1 � e1 D 0 e2 � e1 D �e3 e3 � e1 D e2

e1 � e2 D e3 e2 � e2 D 0 e3 � e2 D �e1

e1 � e3 D �e2 e2 � e3 D e1 e3 � e3 D 0

Proof.

55

56 CHAPTER 4 INTEGRATION ON CHAINS

(a) We just do the first line.

hw; zi D

e1

e1

w

D 0 H) z D e1 � e1 D 0;

hw; zi D

e2

e1

w

D �w3 H) e2 � e1 D �e3;

hw; zi D

e3

e1

w

D w2 H) e3 � e1 D e2:

ut

References

[1] Apostol, Tom M. (1974) Mathematical Analysis: Pearson Education, 2nd

edition. [37]

[2] Axler, Sheldon (1997) Linear Algebra Done Right, Undergraduate Texts

in Mathematics, New York: Springer-Verlag, 2nd edition. []

[3] Berkovitz, Leonard D. (2002) Convexity and Optimization in Rn, Pure

and Applied Mathematics: A Wiley-Interscience Series of Texts, Mono-

graphs and Tracts, New York: Wiley-Interscience. [15]

[4] Munkres, James R. (1991) Analysis on Manifolds, Boulder, Colorado:

Westview Press. [40]

[5] Rudin, Walter (1976) Principles of Mathematical Analysis, New York:

McGraw-Hill Companies, Inc. 3rd edition. [17, 38]

[6] Spivak, Michael (1965) Calculus on Manifolds: A Modern Approach to

Classical Theorems of Advanced Calculus, Boulder, Colorado: Westview

Press. [i]

57

Index

Directional derivative, 24

59

Convexity and Optimization in Rn

A Solution Manual for Berkovitz (2002)

Jianfei Shen


Sydney, Australia 2011

Contents

1 Topics in Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Vectors in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Convex Sets in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1 Lines and Hyperplanes in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Properties of Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Separation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.4 Supporting Hyperplanes: Extreme Points . . . . . . . . . . . . . . . . . . . . . . . 17

2.5 Systems of Linear Inequalities: Theorems of the Alternative . . . . . 17

2.6 Affine Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1 Definition and Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

iii

1TOPICS IN REAL ANALYSIS

1.1 Introduction

No exercises.

1.2 Vectors in Rn

1

2CONVEX SETS IN RN

2.1 Lines and Hyperplanes in Rn

I Exercise 1 (2.1.1).

a. Draw a sketch in R2 and R3 that illustrates Lemma 1.1.

b. In R2 sketch the hyperplane x C 2y D 1 and the parallel subspace.

Proof. See Figure 2.1 for the R2 case. See Rockafellar (1970, Sec. 1) for a de-

tailed discussion.

x0

x

x � x0

u

uC x0

0

H1.1;2/

H0.1;2/

Figure 2.1. H0.1;2/

and H1.1;2/

ut

3

4 CHAPTER 2 CONVEX SETS IN RN


a. In R4 find an equation of the hyperplane through the points .1; 1; 1; 1/,

.2; 0; 1; 0/, .0; 2; 0; 1/, and .1; 1;�1; 0/.

b. Let x1; : : : ;xn be n points in Rn. Find a sufficient condition for the existence of

a unique hyperplane containing these points.

Proof.

(a) If there is a hyperplane Hˇa through those four points, then

ˇ1 1 1 1

2 0 1 0

0 2 0 1

1 1 �1 0

ˇa1

a2

a3

a4

D

ˇˇ

ˇ

ˇ

ˇ

:

Solving this equation system, we get a D�ˇ=2; ˇ=2; 0; 0

�. Then the hyperplane

is determined by

�ˇ2

ˇ2

0 0�ˇx1

x2

x3

x4

D ˇI

that is,

Hˇa D

nx 2 R4 W x1 C x2 D 2

o:

(b) A hyperplane in Rn is an .n � 1/-dimensional affine set. Let

H D aff .x1; : : : ;xn/ ;

the affine hull of .x1; : : : ;xn/. If .x1; : : : ;xn/ is affinely independent, that is, if

.x2 � x1; : : : ;xn � x1/ is linearly independent, then .x2 � x1; : : : ;xn � x1/ deter-

mines a .n � 1/-dimensional subspace, L´ span.x2 � x1; : : : ;xn � x1/, and

aff .x1; : : : ;xn/ D LC x1

is the unique .n � 1/-dimensional affine set. In particular, since dimL D n � 1,

there exists a nonzero vector a such that a ? L, i.e., fag is the orthogonal

complement of L. Since hy; ai D 0 for any y 2 L, we have

hxi � x1; ai D hxi ; ai � hx1; ai D 0; 8 i D 2; : : : ; n;

i.e., hxi ; ai D hx1; ai D ˇ, for any i D 2; : : : ; n. Now any x 2 aff .x1; : : : ;xn/ can

be represented uniquely as

x D

nXiD1

�ixi ; where �i > 0 andnXiD1

�i D 1;

and so

SECTION 2.1 LINES AND HYPERPLANES IN RN 5

hx; ai D

*nXiD1

�ixi ; a

+D

nXiD1

�i hxi ; ai D ˇ:

Hence, if .x1; : : : ;xn/ is affinely independent, there is a unique hyperplane Hˇa ,

where a ? span.x2 � x1; : : : ;xn � x2/ and ˇ D hx1; ai. ut

I Exercise 3 (2.1.3). Show that a hyperplane is a closed set.

Proof. Let Hˇa be a hyperplane, and y be a limit point of H

ˇa . Then for any

" > 0, there is x 2 Hˇa such that ky � xk < ". By Cauchy-Schwarz Inequality,ˇ̌hy � x; ai

ˇ̌6 ky � xk � kak < " kak :

But then hy � x; ai D 0, that is, hy; ai D hx; ai. Hence, y 2 Hˇa and so H

ˇa is

closed.

x

y � x

y

Hˇa

0

a

Figure 2.2. Hˇa is closed

ut

I Exercise 4 (2.1.4). Given a hyperplane H˛a show that a is indeed normal to

H˛a in the sense that if x1 and x2 are any two points in H˛

a , then a is orthogonal

to x2 � x1.

Proof. If x1;x2 2 H˛a , then hx1; ai D hx2; ai D ˛; hence

hx2 � x1; ai D hx2; ai � hx1; ai D ˛ � ˛ D 0: ut

I Exercise 5 (2.1.5). Let V be an .n � 1/-dimensional subspace of Rn. Let y be

a vector in Rn not in V . Show that every x in Rn has a unique representation

x D vC ˛y where v 2 V and ˛ 2 R.

Proof. Let .v1; : : : ; vn�1/ be a basis of V . Since y 2 Rn X V , the list

.v1; : : : ; vn�1;y/


is linear independent; then .v1; : : : ; vn�1;y/ is a basis of Rn, and so every x 2 Rn

can be represented uniquely as

x D

n�1XiD1

˛ivi C ˛y D vC ˛y;

where v´Pn�1iD1 ˛ivi 2 V . ut

2.2 Properties of Convex Sets

Lemma 2.1 (2.2.3). Let A1 � Rn1 , A2 � Rn2 , : : :, Ak � Rnk and let A1; : : : ; Akbe convex. Then A1 � � � �Ak is a convex set in Rn1C��nk .

Proof. It suffices to prove the lemma when k D 2. Take any .a1; a2/ ; .b1; b2/ 2

A1 �A2 and let .c1; c2/ D ˛ .a1; a2/C ˇ .b1; b2/, where ˛; ˇ 2 Œ0; 1� and ˛C ˇ D 1.

We can rewrite .c1; c2/ as

.c1; c2/ D�˛a1 C ˇb1; ˛a2 C ˇb2

�;

and it belongs to A1 � A2 since both A1 and A2 are convex. ut

I Exercise 6 (2.2.1). Sketch the following sets in R2 and determine from figure

which sets are convex and which are not:

a.n�x; y

�W x2 C y2 6 1

o,

b.n�x; y

�W 0 < x2 C y2 6 1

o,

c.n�x; y

�W y > x2

o,

d.n�x; y

�W jxj Cjyj 6 1

o, and

e.n�x; y

�W y > 1

ı1C x2

o.

Solution. It follows from Figure 2.3 that (b) and (e) are not convex.

I Exercise 7 (2.2.2). Show that

SH˛C

a D˚x W ha;xi > ˛

; SH˛�

a D˚x W ha;xi 6 ˛

:

Proof. By definition,

SH˛C

a D H˛Ca D

˚x W ha;xi > ˛

D˚x W ha;xi > ˛

:

Similarly, SH˛C

a D˚x W ha;xi 6 ˛

. ut

SECTION 2.2 PROPERTIES OF CONVEX SETS 7

0 0

x

y0

00 0

xy

Figure 2.3.ut

I Exercise 8 (2.2.3). Using the definition of a convex set, show that (a) the non-

negative orthant in Rn, fx W x D .x1; : : : ; xn/ ; xi > 0; i D 1; : : : ; ng is convex and

(b) a hyperplane H˛a is convex.

Proof. Take any points x;y 2 RnC. Then for any � 2 Œ0; 1�,

.1 � �/x C �y D�.1 � �/ x1 C �y1; : : : ; .1 � �/ xn C �yn

�2 RnC

since .1 � �/ xi C �yi > 0 for all i D 1; : : : ; n.

Now take any points x;y 2 H˛a . Then for any � 2 Œ0; 1�,

h.1 � �/x C �y; ai D .1 � �/ hx; ai C � hy; ai D ˛I

that is, .1 � �/x C �y 2 H˛a . ut

I Exercise 9 (2.2.4). Let Sx D Tx C b be an affine transformation. Show that

if C is a convex set in Rn, then S .C / D fy W y D Tx C b;x 2 C g is a convex set in

Rn.

Proof. Let y1;y2 2 S .C /. Then there exist x1;x2 2 C such that y1 D Tx1 C b

and y2 D Tx2 C b. For any ˛ 2 Œ0; 1�, we have

˛y1 C .1 � ˛/y2 D ˛ .Tx1 C b/C .1 � ˛/ .Tx2 C b/

D ˛Tx1 C .1 � ˛/Tx2 C b

D T�˛x1 C .1 � ˛/x2

�C b

2 S .C /

since ˛x1 C .1 � ˛/x2 2 C . ut

I Exercise 10 (2.2.5). Show that the sets Pn are compact and convex.


Proof. Let Pn Dnp D

�p1; : : : ; pn

�W pi > 0;

PniD1 pi D 1

o. Then Pn is bounded,

closed and so is compact by the Heine-Borel Theorem. To see Pn is convex,

take any points pn; qn 2 Pn, and � 2 Œ0; 1�. Then

.1 � �/p C �q D�.1 � �/ p1 C �q1; : : : ; .1 � �/ pn C �qn

�2 Pn

since .1 � �/ pi C �qi > 0 for any i D 1; : : : ; n and

nXiD1

�.1 � �/ pi C �qi

�D .1 � �/

nXiD1

pi C �

nXiD1

qi D 1: ut

I Exercise 11 (2.2.6). Show that for any set A the set K .A/ of all convex com-

binations of points in A is convex.

Proof. Let x;y 2 K .A/. Then there exist x1; : : : ;xm 2 A and y1; : : : ;yn 2 A

such that

x D

mXiD1

pixi ; y D

nXjD1

qjyj ;

where�p1; : : : ; pm

�2 Pm and

�q1; : : : ; qn

�2 Pn. Take any ˛ 2 Œ0; 1�. Then

˛x C .1 � ˛/y D ˛

0@ mXiD1

pixi

1AC .1 � ˛/0@ nXjD1

qjyj

1AD

mXiD1

�˛pi

�xi C

nXjD1

�.1 � ˛/ qj

�yj :

SincePmiD1 ˛pi C

PnjD1 .1 � ˛/ qj D ˛ C .1 � ˛/ D 1, we have�

˛p1; : : : ; ˛pm; .1 � ˛/ q1; : : : ; .1 � ˛/ qn�2 PmCnI

therefore, ˛x C .1 � ˛/y 2 K .A/. ut

I Exercise 12 (2.2.7). Show that the open ball B.0I r/ is convex.

Proof. Take any x;y 2 B.0I r/ and � 2 Œ0; 1�. Then

k.1 � �/x C �yk 6 .1 � �/ kxk C � kyk < r: ut

I Exercise 13 (2.2.8). Consider the linear programming (LP) problem: Mini-

mize hc;xi subject to Ax D b, x > 0. Let S D˚x W x is a solution of the problem LP

.

Show that if S is not empty, then it is convex.

Proof. Suppose S ¤ ¿ and take any x;y 2 S and � 2 Œ0; 1�. Let z D .1 � �/x C

�y . We show that z is also a solution to the LP. It is easy to see that Az D b and

z > 0. Suppose that there exists some z0 with Az0 D b and z0 > 0 so that˝c; z0

˛< hc; zi D .1 � �/ hc;xi C � hc;yi I


then either hc;xi >˝c; z0

˛or hc;yi >

˝c; z0

˛or both, which contradicts to the

assumption that both x and y are solutions to the LP. ut

I Exercise 14 (2.2.9). Let C � Rn. Show that C is convex if and only if �C C

�C D��C �

�C for all � > 0, � > 0.

Proof.

If: Let �C C�C D��C �

�C for all � > 0 and � > 0. Let x;y 2 C and ˛ 2 Œ0; 1�.

Then

.1 � ˛/x C ˛y 2�.1 � ˛/C ˛

�C D C I

hence, C is convex.

Only if: Now let C be convex. If � D 0 or � D 0, the proof is trivial; thus we

assume that � > 0 and � > 0. If x 2��C �

�C , there exists c 2 C such that

x D��C �

�c D �c C �c 2 �C C �C . To establish the reverse inclusion, let

x D �c1 C �c2 2 �C C �C , where c1; c2 2 C . Since C is convex,��

�C �

�c1 C

��

�C �

�c2µ c3 2 C:

Therefore,

�c1 C �c2 D��C �

�c3 2

��C �

�C: ut

I Exercise 15 (2.2.10). A set C is said to be a cone with vertex at the origin,

or simply a cone, if whenever x 2 C , all vectors �x 2 C for all � > 0. If C is also

convex, C is said to be a convex cone.

a. Give an example of a cone that is not convex.

b. Give an example of a cone that is convex.

c. Let C be a nonempty set in Rn. Show that C is a convex cone if and only if x1

and x2 2 C implies that �1x1 C �2x2 2 C for all �1 > 0, �2 > 0.

Proof. See Figure 2.4 for the examples.

0 0

Figure 2.4. A cone and a convex cone


It is clear that if �1x1C�2x2 2 C whenever x1;x2 2 C and �1; �2 > 0, then C

is a convex cone. To see the reverse implication, let C ¤ ¿ be a convex cone;

we also assume that both �1 > 0 and �2 > 0; otherwise the proof is trivial. For

any x1;x2 2 C and �1; � > 0, we have �1x1 2 C and �2x2 2 C since C is a cone.

Then ��1

�1 C �2

�x1 C

��2

�1 C �2

�x2µ x 2 C

since C is convex. But then �1x1C�2x2 D .�1 C �2/x 2 C since C is a cone. ut

I Exercise 16 (2.2.11). Show that if C1 and C2 are convex cones, then so is

C1 C C2 and that C1 C C2 D conv .C1 [ C2/.

Proof. It is easy to see that C1 C C2 is convex. To see C1 C C2 is a cone, take

any c1 C c2 2 C1 C C2 and � > 0. Then

� .c1 C c2/ D �c1 C �c2 2 C1 C C2:

We first show that C1 C C2 � conv .C1 [ C2/. For any c1 C c2 2 C1 C C2, we

have 2c1 2 C1 and 2c2 2 C2; hence 2c1 2 C1 [ C2 and 2c2 2 C1 [ C2. But then

c1 C c2 D12.2c1/C

12.2c2/ 2 conv .C1 [ C2/.

To establish the other inclusion relation, let c 2 conv .C1 [ C2/. Then

c D

mXiD1

pici ;

where�p1; : : : ; pm

�2 Pm and c1; : : : ; cm 2 C1 [ C2. Without loss of generality,

suppose that c1; : : : ; cm1 2 C1 and cm1C1; : : : ; cm 2 C2. In this case,

p1c1 C � � � C pm1cm1 2 C1;

and

pm1C1cm1C1 C � � � C pmcm 2 C2;

since both C1 and C2 are convex cone (by Exercise 15 (c)). Therefore, c 2 C1 C

C2. ut

I Exercise 17 (2.2.12). Show that if A is a bounded set in Rn, then so is conv .A/.

Proof. Take any x 2 conv .A/. Then by the Carathéodory Theorem, x DPnC1iD1 pixi , where x1; : : : ;xnC1 2 A and

�p1; : : : ; pnC1

�2 PnC1. Therefore,

kxk D

nC1XiD1

pixi

6nC1XiD1

pi kxik < C1: ut

I Exercise 18 (2.2.13). Let X be a convex set in Rn and let y be any vector not

in X . Let Œy; X� denote the union of all line segments Œy;x� with x 2 X .

a. Sketch a set Œy; X� in R3.


b. Show that Œy; X� is convex.

Proof.

(a) See Figure 2.5.

(b) Take any a;b 2 Œy; X� DS

x2X Œy;x�. Then there exist w; z 2 X and ˛; ˇ 2

Œ0; 1�, such that a D .1 � ˛/yC˛w and b D�1 � ˇ

�yCˇz. Then for any � 2 Œ0; 1�,

.1 � �/ aC �b D .1 � �/ .1 � ˛/y C .1 � �/ ˛wC ��1 � ˇ

�y C �ˇz

D

h1 �

�˛ � �˛ C �ˇ

�iy C .˛ � �˛/wC �ˇz

is a convex combination of y;w; z. Let

p1´ 1 ��˛ � �˛ C �ˇ

�; p2´ ˛ � �˛; and p3´ �ˇ:

Then�p1; p2; p3

�2 P3. We can rewrite x as follows:

x D p1y C p2wC p3z

D�1 � p2 � p3

�y C

�p2 C p3

� � p2

p2 C p3wC

p3

p2 C p3z

�:

Clearly, p2w.�p2 C p3

�C p3z

.�p2 C p3

�µ v 2 X since X is convex. Then

x D�1 � p2 � p3

�C�p2 C p3

�vI

that is, x 2 Œy; v� � Œy; X�.

I Exercise 19 (2.2.14). Let C be a proper subset of Rn. For each " > 0 define a

set �" .C /, the "-neighborhood of C , by �" .C / DS

x2C B.xI "/. Show that

�" .C / D˚y W kx � yk < " for some x 2 C

D

[x2C

�B.0I "/C x

�:

Show that �" .C / is open. Show that if C is convex, then so is �" .C /.

Proof. Since �" .C / is a union of open balls, it is open. We first show that

B.xI "/ D B.0I "/C x for each x 2 C . We have y 2 B.0I "/ iff y D zC x for some

z 2 B.0I "/ iff y � x D z with kzk < " iff ky � xk < " iff y 2 B.xI "/. The other

equality is just by definition. �" .C / is open since it is a union of open sets.

Take two points x and y in �" .C /. Then there exists B.xI "/ and B.yI "/ in-

cluding x and y , separately. For any ˛; ˇ 2 Œ0; 1� with ˛ C ˇ D 1, we have

z ´ ˛x C ˇy 2 C since C is convex. Then, by definition, there exists B.zI "/

including z and so z 2 �" .C /, that is, �" .C / is convex if C is. ut

I Exercise 20 (2.2.15). Show that if a vector x 2 Rn has distinct representations

as a convex combination of a set of vectors x0;x1; : : : ;xr , then the vectors x1 �

x0; : : : ;xr � x0 are linear dependent.


y

Figure 2.5.�y;X

�ut

Proof. By assumption, there exists a ´ .a0; a1; : : : ; ar / 2 PrC1 and b ´

.b0; b1; : : : ; br / 2 PrC1 with a ¤ b such that

x D a0x0 C � � � C arxr ; (2.1)

and

x D b0x0 C � � � C brxr : (2.2)

We can rewrite (2.1) as

x D x0 C

0@ rXiD0

aixi � x0

1A D x0 C

rXiD1

ai .xi � x0/ :

Similarly, we can rewrite (2.2) as

x D x0 C

rXiD1

bi .xi � x0/ :

Hence,

0 D

rXiD1

.ai � bi / .xi � x0/ :

Since a ¤ b, not all .ai � bi / are zeros, that is, .x1 � x0; : : : ;xr � x0/ is linear

dependent. ut

SECTION 2.3 SEPARATION THEOREMS 13

I Exercise 21 (2.2.16). Let X be a set contained in the closed negative half

space determined by the hyperplane H˛a . Show that if x 2 int.X/, then ha;xi <

˛.

Proof. Since x 2 int.X/, there is a ball B.xI r/ � X � H˛�

a . Then x C z 2 X for

all z 2 B.0I r/; see ??. But then

hx C z; ai D hx; ai C hz; ai 6 ˛:

Since there exists z 2 B.xI r/ such that hz; ai > 0, e.g., z D "a, where " is small

enough but strictly positive, we thus have hx; ai 6 ˛ � hz; ai < ˛.

ut

I Exercise 22 (2.2.17). Let X be a convex set and let H˛a be a hyperplane such

that X \ H˛a D ¿. Show that X is contained in one of the open half spaces

determined by H˛a .

Proof. Take any x 2 X and suppose hx; ai < ˛. If there is y 2 X with hy; ai >

˛, then

f .t/ D htx C .1 � t /y; ai D t hx; ai C .1 � t / hy; ai ;

where t 2 .0; 1/, is continuous and there is t� 2 .0; 1/ such that f�t��D 0, that

is,

t�x C�1 � t�

�y 2 X \H˛

a : ut

2.3 Separation Theorems

Theorem 2.2 (2.3.2). Let C be a convex set and let y … C . Then there exists

a hyperplane H˛a such that for all x 2 C , ha;xi 6 ˛ and ha;yi D ˛. If

int.C / ¤ ¿, then for all x 2 int.C /, ha;xi < ˛.

Proof. See Figure 2.6.

I Exercise 23 (2.3.1). Sketch the pairs of sets .X1; Y1/, .X2; Y2/,�X�2 ; Y

�2

�,

.X3; Y3/, and�X4�; Y4�

�.

Solution. See Figure 2.7.

I Exercise 24 (2.3.2). Let V be a linear subspace and let y … V . Show that

x� 2 V is the closest point in V to y if and only if y � x� is orthogonal to V ; that

is, for every w 2 V , y � x� is orthogonal to w.

Proof. Let x� D PV y , the projection of y on V . Then for any w 2 V ,


0

x

z

�xkxk

x

z

x0

H0a

x

z

x0

H0a

Figure 2.6. The hyperplane for y D 0

ut

0 x1

x2

X1

Y10 x1

x2

X1

Y1

separated

0 x1

x2

X2

Y20 x1

x2

X2

Y2

properly

0 x1

x2

X3

Y30 x1

x2

X3

Y3

strictly

0 x1

x2

X4�

Y4�0 x1

x2

X4�

Y4�

strongly

Figure 2.7.ut

ky � PV yk2h1i6 ky � PV yk

2C kPV y �wk

2

h2iD k.y � PV y/C .PV y �w/k2

D ky �wk2 ;

where h2i comes from the Pythagorean theorem, which applies because y �

PV y 2 V ? and PV y � w 2 V . Taking square roots gives the desired inequal-

ity. The above inequality is an equality if and only if h1i is an equality, which

happens if and only if kPV y �wk D 0, which happens if and only if w D PV y .

ut

I Exercise 25 (2.3.3). Let C be a closed convex set and let y … C . Show that

x� 2 C is closed to y if and only if hx � y;x� � yi > kx� � yk2 for all x 2 C .

Proof. As in ??, we have x � y D .x� � y/C .x � x�/; then

SECTION 2.3 SEPARATION THEOREMS 15

V

PV y D x�

y

0

y � PV y

PV y �w

y �w

w

Figure 2.8. x� D PV y

hx � y;x� � yi D h.x� � y/C .x � x�/ ; .x� � y/i

D kx� � yk2C hx � x�;x� � yi I

that is,

hx � y;x� � yi � kx� � yk2D hx � x�;x� � yi : (2.3)

It follows from (2.3) that

hx � y;x� � yi > kx� � yk2() hx � x�;x� � yi > 0

() hy � x�;x � x�i 6 0:

But the later inequality holds if and only if x� is the closet point to y by Lemma

2.3.4. ut

I Exercise 26 (2.3.4). In reference to Theorem 2.3.5, show that if K is assumed

to be convex and closed, then it may not be possible to separate K and C strictly,

let along strongly.

Proof. Let

K Dn.x1; x2/ 2 R2 W x1 6 0

o;

C Dn.x1; x2/ 2 R2 W x1 > 0; x2 > 0; x1x2 > 1

o:

See Bertsekas (2003, Proposition 2.4.3, p.111) for a series of sufficient con-

ditions for strict separation. ut

I Exercise 27 (2.3.5). Show that if F is closed and K is compact, then K C F

is closed. Give an example in which F and K are closed yet F CK is not.

Proof. Let z be a limit point of F CK. Then there exist sequence fxng � F and

fyng � K such that xn C yn ! z. Since K is compact, there is a subsequence˚ynk

of fyng such that ynk ! y 2 K. By passing to this subsequence, we can

assume that yn ! y 2 K. But then fxng must also be convergent, say, xn ! x.


0

CK

Figure 2.9. K and C cannot be strictly seperated

Since F is closed, we have x 2 F . Therefore, xn C yn ! x C y D z with x 2 F

and y 2 K, and so F CK is closed.

To see that F CK is not closed when K is not, let

F Dn�x; y

�2 R2 W xy > 1

o;

K Dn�x; y

�2 R2 W xy 6 �1

o:

In this case, both F and K are closed, but F C K Dn�x; y

�2 R2 W x > 0

ois

open. ut

I Exercise 28 (2.3.6). Let A and B be two compact sets. Show that A and B

can be strongly separated if and only if conv .A/ \ conv .B/ D ¿.

Proof. Let both A and B are compact. Then conv .A/ and conv .B/ are compact

by Lemma 2.7.

If: Assume that conv .A/ \ conv .B/ D ¿. It follows from Theorem 3.5 that

conv .A/ and conv .B/ can be strongly separated. Then A can B can be strongly

separated since A � conv .A/ and B � conv .B/.

Only if: Now assume that A and B can be strongly separated. Then there

exist a hyperplane H˛a and � > 0 such that ha;xi < ˛ � � for all x 2 A, and

ha;yi < ˛ C � for all y 2 B . By Carathéodory Theorem, for every x0 2 conv .A/,

there exist�p1; : : : ; pnC1

�2 PnC1 and x1; : : : ;xnC1 2 A, such that

˝a;x0

˛D

*a;

nC1XiD1

pixi

+D

nC1XiD1

pi ha;xi i < ˛ � �:

Similarly,˝a;y 0

˛> ˛ C � for every y 0 2 conv .B/. Hence, conv .A/ and conv .B/

are strongly separated by H˛a . ut

SECTION 2.6 AFFINE GEOMETRY 17

I Exercise 29 (2.3.7). Prove Theorem 2.3.5 using Theorem 2.3.1 and Exer-

cise 27 (2.3.5).

Proof. Let K be a compact convex set and let C be a closed convex set such

that K \ C D ¿. Then K C C is closed (by Exercise 27) and convex (by Lemma

2.2.2).

ut

I Exercise 30 (2.3.9). Let A be a closed convex set such that Ac is convex. Show

that A is a closed half space.

Proof. Since A is closed and convex, we can write A as

A D\i2I

Hi ;

where Hi is a closed (positive) half space for each i 2 I . Then Ac DSi2I H c

i .

For simplicity, define Ki D Hi for each i 2 I ; then each Ki is an open (negative)

half space. Thus, A DSi2I Ki . We can show that Ki DKj for all i; j 2 I since

Ac is convex. Hence, Hi D Hj and so A is a closed half space. ut

I Exercise 31 (2.3.10). Let C1 and C2 be two convex subsets of Rn. Show that

there exists a hyperplane that strongly separates C1 and C2 if and only if

inf˚kx � yk W x 2 C1;y 2 C2

> 0: (2.4)

Proof. Let C1 and C2 be two convex subsets of Rn.

If: Let (2.4) hold. ut

2.4 Supporting Hyperplanes: Extreme Points

2.5 Systems of Linear Inequalities: Theorems of the

Alternative

2.6 Affine Geometry

Remark (Corollary 1). If M1 and M2 are linear manifolds, then so is M1C

M2´ fx W x D m1 Cm2;m1 2M1;m2 2M2g.

Proof. Let x;y 2M1 CM2. Then

.1 � �/x C �y D .1 � �/�mx1 Cmx

2

�C �

�m

y1 Cm

y2

�D�.1 � �/mx

1 C �my1

�C�.1 � �/mx

2 C �my2

�2M1 CM2: ut


Remark (Corollary 3). A set S is a linear manifold if and only if every

affine combination of points in S is in S .

Proof. The if part is clear, so we only prove the only if part by induction

on k, the number of points in the affine combination. If S is affine, then the

definition of affinity says that every affine combination of two points is in S .

Suppose that we have shown that if S is affine, then every affine combination

of k points is in S . We shall show that every affine combination of kC 1 points

is in S . Let

x D

kC1XiD1

qixi ;

kC1XiD1

qi D 1:

Let q < 1, thenPkiD1 qi > 0. Hence,

x D

0@ kXiD1

qi

1A" q1PkiD1 qi

x1 C � � � CqkPkiD1 qi

xk

#C qkC1xkC1 2 S: ut

I Exercise 32 (2.6.1). Let k be a positive integer, 1 6 k 6 n� 1. Show that for a

given subspace Vn�k of Rn, there exist k linearly independent vectors a1; : : : ; ak

such that

Vn�k D˚x 2 Rn W hai ;xi D 0; i D 1; : : : ; k

:

Proof. We have Rn D Vn�k ˚ Vk , where Vk D V ?n�k

and dimVk D k. Take a

basis .a1; : : : ; an/ of Vk , and we get the result. Thus, Vn�k is the intersection of

k hyperplanes through the origin. ut

I Exercise 33 (2.6.2). Show that a set of points x1; : : : ;xk is affinely dependent

if and only if one of the points is an affine combination of the other points.

Proof. .x1; : : : ;xk/ is affinely dependent if and only if .x2 � x1; : : : ;xk � x1/ is

linearly dependent.

If: Let j be the largest index such that xj can be written as an affine combina-

tion of the other points:

xj DXi¤j

qixi ;Xi¤j

qi D 1I

that is,

q1x1 C � � � C qj�1xj�1 �

0@Xi¤j

qi

1Axj C qjC1xjC1 C � � � C qkxk D 0:

Equivalently,

q1�x1 � xj

�C � � � C qj�1

�xj�1 � xj

�C qjC1xjC1 C � � � C qk

�xk � xj

�D 0:

SECTION 2.6 AFFINE GEOMETRY 19

Hence,�x1 � xj ; : : : ;xj�1 � xj ;xjC1 � xj ; : : : ;xk � xj

�is linear dependent, and

so .x1; : : : ;xk/ is affinely dependent.

Only if: Let .x1; : : : ;xk/ be affinely dependent; then .x2 � x1; : : : ;xk � x1/ is

linearly dependent. Then there exists�q2; : : : ; qk

�¤ 0 such that

kXiD2

qi .xi � x1/ D 0I

that is,kXiD2

qixi �

0@ kXiD2

qi

1Ax1 D 0:

Let j be the largest index with qj ¤ 0. Then

xj D

PkiD2 qi

qjx1 �

q2

qjx2 � � � � �

qj�1

qjxj�1 �

qjC1

qjxjC1 � � � � �

qk

qjxk ;

and PkiD2 qi

qj�q2

qj� � � � �

qj�1

qj�qjC1

qj� � � � �

qk

qjDqj

qjD 1: ut

I Exercise 34 (2.6.3). In R3 consider the plane 2x C y C z D 5.

a. Find an affine basis for this plane that does not include the point x D

.�1;�2; 9/.

b. Express x D .�1;�2; 9/ in barycentric coordinates relative to your basis.

Proof.

(a) Let

x0 D .1; 1; 2/ ; x1 D .2; 1; 0/ ; and x3 D .0; 0; 5/ :

(b) Let

.�1;�2; 9/ D�1 � q1 � q2

�x0 C q1x1 C x2:

then ��1

�2

9

�

D

�1C q1 � q2

1 � q2

2 � 2q1 C 3q2

�

:

hence, q1 D 1, q2 D 3, and so .�1;�2; 9/ D �3x0 C x1 C 3x2. ut

I Exercise 35 (2.6.4). Let x0;x1; : : : ;xr be affinely independent. Show that any

r C 2 points y0;y1; : : : ;yr ;yrC1 in conv .x0;x1; : : : ;xr / are affinely dependent.

Proof. Let .x0;x1; : : : ;xr / be affinely independent. Let .y0;y1; : : : ;yr ;yrC1/ be

r C 2 points in conv .x0;x1; : : : ;xr /, the simplex spanned by .x0;x1; : : : ;xr /.

Then each yi , i D 0; 1; : : : ; r; r C 1, can be written uniquely as


yi D

rXjD0

pijxj ;

where�pi0; : : : ; pir

�2 PrC1. Now consider the list

.y1 � y0; : : : ;yrC1 � y0/ ;

which can be expressed as0@ rXjD0

�p1j � p0j

�xj ; : : : ;

rXjD0

�prC1;j � p0j

�xj

1A :ut

3CONVEX FUNCTIONS

3.1 Definition and Elementary Properties

I Exercise 36 (3.1.1). Show that if f and g are convex functions defined on a

convex set C , then for any � > 0, � > 0, the function �f C �g is convex.

Proof. Let x;y 2 C and�˛; ˇ

�2 P2. Then�

�f C �g� �˛x C ˇy

�D �f

�˛x C ˇy

�C �g

�˛x C ˇy

�6 � f̨ .x/C � f̌ .y/C �˛g .x/C �ˇg .y/

D ˛��f C �g

�.x/C ˇ

��f C �g

�.y/ : ut

I Exercise 37 (3.1.2). Let ffng be a sequence of convex functions defined on a

convex set C . Show that if f .x/ D limn!1 fn .x/ exists for all x 2 C , then f is

convex.


�2 P2. Then

f�˛x C ˇy

�D limn!1

fn�˛x C ˇy

�6 limn!1

�f̨n .x/C f̌ .y/

�D ˛ lim

n!1fn .x/C ˇ lim

n!1fn .y/

D f̨ .x/C f̌ .y/ : ut

I Exercise 38 (3.1.3). If f is a convex function defined on a convex set C in

Rn, then for any real the set A´ fx W f .x/ 6 g is either empty or convex.

Proof. Suppose that A ¤ ¿. Let x;y 2 A and�˛; ˇ

�2 P2. Then f .x/ 6 ,

f .y/ 6 , and

f�˛x C ˇy

�6 f̨ .x/C f̌ .y/ 6 ˛ C ˇ D : ut

21

22 CHAPTER 3 CONVEX FUNCTIONS

I Exercise 39 (3.1.4). Let f be a convex function defined on a convex set C in

Rn. Let g be a nondecreasing convex function defined on an interval I in R. Let

f .C / � I . Then the composite function g B f is convex on C .


�2 P2. Then�

g B f� �˛x C ˇy

�D g

�f�˛x C ˇy

��6 g

�f̨ .x/C f̌ .y/

�6 ˛

�g B f

�.x/C ˇ

�g B f

�.y/ : ut

I Exercise 40 (3.1.5). Let I D Œa; b� be a real interval and let C be a convex set

in Rn. Let f W .t;x/ 7! f .t;x/ be a real-valued function defined on I � C that is

continuous on I for each fixed x 2 C and is convex on C for each fixed t 2 I .

Show that

F .x/ D

Z b

a

f .t;x/ dt

is defined for each x 2 C and that F is convex on C .

Proof. F .x/ is well-defined since f .t;x/ is continuous for each fixed x 2 C .

Now let x;y 2 C and�˛; ˇ

�2 P2. Then

F�˛x C ˇy

�D

Z b

a

f�t; ˛x C ˇy

�dt

6Z b

a

�f̨ .t;x/C f̌ .t;y/

�dt

D ˛F .x/C ˇF .y/ : ut


a. Show that any norm � on Rn is convex on Rn.

b. Let S be a nonempty convex set, and let k�k denote the euclidean norm in Rn.

Let f be the distance function to S defined by

f .y/ D inf˚ky � xk W x 2 S

:

Show that f is convex on Rn.

c. Show that f is Lipschitz continuous on compact subsets of Rn.

Proof. Let x;y 2 Rn and�˛; ˇ

�2 P2.

(a) � is convex on Rn since

��˛x C ˇy

�6 � .˛x/C �

�ˇy�D ˛� .x/C ˇ� .y/ :

(b) See Rockafellar (1970, p.34). ut

References

[1] Berkovitz, Leonard D. (2002) Convexity and Optimization in Rn, Pure

and Applied Mathematics: A Wiley-Interscience Series of Texts, Mono-

graphs and Tracts, New York: Wiley-Interscience. [i]

[2] Bertsekas, Dimitri P. (2003) Convex Analysis and Optimization, Bel-

mont, Massachusetts: Athena Scientific. [15]

[3] Rockafellar, R. Tyrrell (1970) Convex Analysis, Princeton Mathemati-

cal Series, New Jersey: Princeton University Press, 2nd edition. [3, 22]

23

Introduction to Topology

Introduction to Topology

A Solution Manual for Gamelin and Greene (1999)

Jianfei Shen

SCHOOL OF ECONOMICS

THE UNIVERSITY OF NEW SOUTH WALES

SYDNEY 2052, AUSTRALIA

I hear, I forget;I see, I remember;I do, I understand.

Old Proverb

Contents

Preface vii

1. Metric Spaces 1

1.1 Open and Closed Sets 1

1.2 Completeness 12

1.3 The Real Line 18

1.4 Products of Metric Spaces 22

1.5 Compactness 24

1.6 Continuous Functions 27

1.7 Normed Linear Spaces 30

2. Topological Spaces 33

2.1 Topological Spaces 33

2.2 Subspaces 39

2.3 Continuous Functions 41

2.4 Base for a Topology 43

2.5 Separation Axioms 47

2.6 Compactness 49

2.7 Locally Compact Spaces 52

3. 55

4. Higher Dimensional Homotopy 57

4.1 Simplexes and Barycentric Subdivision 57

Preface

Chapter 1

Metric Spaces

1.1 Open and Closed Sets

Exercise 1 (1.1.1). Let U; V , and W be subsets of some set. Recall thatU XW consists of all points in U that do not belong to V .

(a) Prove that .U [ V / XW D .U XW / [ .V XW /.

(b) Prove that .U \ V / XW D .U XW / \ .V XW /.

(c) Does U X.V XW / coincide with .U XV /XW ? Justify your answerby proof or counterexample.

(d) Prove the two set-theoretic identities used in the proof of Theo-rem 1.1.9.

Proof.

(a)

x 2 .U [ V / XW ) .x 2 U or x 2 V / and x … W

) .x 2 U and x … W / or .x 2 V and x … W /

) x 2 .U XW / [ .V XW /:

x 2 .U XW / [ .V XW /) .x 2 U and x … W / or .x 2 V and x … W /

) .x 2 U [W and x … W / or .x 2 U [ V and x … W /

) x 2 .U [W / XW:

Or:

.U [ V / XW D .U [ V / \W c

D�U \W c

�[�V \W c

�D .U XW / [ .V XW / :

2 CHAPTER 1

(b)

x 2 .U \ V / XW () x 2 U \ V and x … W

() x 2 U and x 2 V and x … W

()�x 2 U and x … W

�and

�x 2 V and x … W

�() x 2 .U XW / \ .V XW / :

Or:

.U \ V / XW D .U \ V / \W c

D�U \W c

�\�V \W c

�D .U XW / \ .V XW / :

(c)

U X .V XW / D U \�V \W c

�c;

but

.U X V / XW D�U \ V c

�\W c

D U \�V c \W c

�D U \ .V \W /c :

(d) De Morgan Laws.

Exercise 2 (1.1.2). Show that

d.x; y/ D

8<:1; if x ¤ y

0; if x D y;

defines a metric on X . Show that every subset of the resulting metricspace is both open and closed.

Proof. To see d is a metric on X , we need only to prove the triangle inequal-ity. Let x; y; z 2 X . There are two cases: (i) x D z. In this case, d.x; z/ D 0

and there is nothing to prove. (ii) x ¤ z. In this case, we must have x ¤ y

or y ¤ z. Therefore, d.x; z/ D 1 6 d.x; y/C d.y; z/.By this metric, for any 0 < r < 1 we have B .xI r/ D fxg for all x 2 X ; that

is, any singleton is open since a ball is open. Then for any subset Y of X ,

Y D[x2Y

B .xI r/ D[x2Y

fxg

METRIC SPACES 3

is open since it is the union of a class of open balls. On the other hand,

X X Y D[

x2XXY

B .xI r/

is open; thus Y D X X .X X Y / is closed.

Exercise 3 (1.1.3).

(a) Show that if a; b; c 2 R are such that for all � 2 R, a�2Cb�Cc > 0,then b2 � 4ac 6 0.

(b) Show that for any .x1; : : : ; xn/ ;�y1; : : : ; yn

�2 Rn,

nXiD1

xiyi 6

0@ nXiD1

x2i

1A 120@ nXiD1

y2i

1A 12

:

[This is a version of the important Cauchy-Schwarz inequality.]

(c) Using the Cauchy-Schwarz inequality, show that the function ddefined by

d.x;y/ D

24 nXjD1

�xi � yi

�235 12

; x;y 2 Rn

satisfies the triangle inequality.

Proof.

(a) �a�2 C b�C c

�0D 2a�C b D 0) �� D �

b

2a;�

a�2 C b�C c�00D 2a > 0) a > 0I

hence,

min�a�2 C b�C c

�D4ac � b2

4a:

Therefore, a�2 C b�C c > 0 means that

4ac � b2

4a> 0) b2 � 4ac 6 0:

(b) (Axler, 1997). Notice that

x D ay C .x � ay/ ;

4 CHAPTER 1

where x´ .x1; : : : ; xn/, y ´�y1; : : : ; yn

�, and a 2 R. We want to find

a such that y is orthogonal to .x � ay/. In other words, we want

0 D hy;x � ayi D

nXiD1

yi ��xi � ayi

�D

nXiD1

xiyi � a

nXiD1

y2i D hx;yi � akyk2 ;

that is

a Dhx;yi

kyk2:

Making the choice of a, we can write

x Dhx;yi

kyk2

y C

x �hx;yi

kyk2

y

!´hx;yi

kyk2

y Cw:

By the Pythagorean theorem,

kxk2D

hx;yikyk2

y

2

Ckwk2D

ˇ̌hx;yi

ˇ̌2kyk

2Ckwk

2 >ˇ̌hx;yi

ˇ̌2kyk

2:

Therefore, ˇ̌hx;yi

ˇ̌6kxk �kyk ;

or

nXiD1

xiyi 6

0@ nXiD1

x2i

1A 120@ nXiD1

y2i

1A 12

:

(c) We need to check the triangle inequality, i.e., if z D .z1; : : : ; zn/, wemust show that d .x;y/ 6 d .x; z/C d .z;y/.For i D 1; : : : ; n, set

ai D xi � zi ; bi D zi � yi :

Then

d .x; z/ D

0@ nXiD1

a2i

1A1=2 ; and d .z;y/ D

0@ nXiD1

b2i

1A1=2 ;and

d .x;y/ D

24 nXiD1

�xi � yi

�2351=2

D

8<:nXiD1

h.xi � zi /C

�zi � yi

�i29=;1=2

D

24 nXiD1

.ai C bi /2

351=2 :

METRIC SPACES 5

We must show that24 nXiD1

.ai C bi /2

351=2 60@ nXiD1

a2i

1A1=2 C0@ nXiD1

b2i

1A1=2 : (1.1)

Squaring both sides of (1.1), and we see that (1.1) is equivalent to

nXiD1

.ai C bi /2 6

nXiD1

a2i C

nXiD1

b2i C 2

0@ nXiD1

a2i

1A1=20@ nXiD1

b2i

1A1=2

nXiD1

aibi 6

0@ nXiD1

a2i

1A1=20@ nXiD1

b2i

1A1=2 ;which is just the Cauchy-Schwarz inequality.

Remark 1.1.1 (Hölder’s Inequality, Shirali and Vasudeva, 2006). Letxi > 0 and yi > 0 for i D 1; 2; : : : ; n, and suppose that p > 1 and q > 1are such that 1

pC

1qD 1. Then

nXiD1

xiyi 6

0@ nXiD1

xpi

1A 1p0@ nXiD1

yqi

1A 1q

: (1.2)

In the special case when p D q D 2, the above inequality reduces tothe Cauchy-Schwarz inequality.

Proof. We need consider only the case whenPniD1 x

pi ¤ 0 ¤

PniD1 y

qi . We

first have the following inequality: if x > 0, y > 0, p > 1 and 1pC

1qD 1,

then

xiyi 61

pxpi C

1

qyqi ; 8 i D 1; : : : ; n: (1.3)

(1.3) holds because we have the AM-GM Inequality: If a > 0 and b > 0 and if0 < � < 1 is fixed, then

a�b1�� 6 �aC .1 � �/ b: (AM-GM)

Then let � D 1p

, a� D xi and b1�� D yi . In this case,

xiyi 61

px1=�i C

�1 �

1

p

�y1=.1��/i

D1

pxpi C

1

qyqi :

6 CHAPTER 1

Let

x0i Dxi�Pn

iD1 xpi

�1=p ; y0i Dyi�Pn

iD1 yqi

�1=q ;then (1.3) becomes

x0iy0i 6

1

px0pi C

1

qy0qi

D1

p

xpiPn

iD1 xpi

!C

1

q

yqiPn

iD1 yqi

!:

(1.30)

Now add up the inequalities of (1.30) and we get

nXiD1

x0iy0i 6

1

pC1

qD 1; (1.4)

that is,

nXiD1

"xi�Pn

iD1 xpi

�1=p #" yi�PniD1 y

qi

�1=q # 6 1;which is just (1.2).

Exercise 4 (1.1.4). Show that the semiopen interval .0; 1� is neither opennor closed in R.

Proof. 0 is adherent to .0; 1�, but 0 … .0; 1�; hence .0; 1� is not closed; 1 2.0; 1�, but 1 … int

�.0; 1�

�; hence .0; 1� is not open.

Exercise 5 (1.1.5). Show that

d�f; g

�D sup j̊f .s/ � g.s/j

ˇ̌s 2 S

defines a metric on the space B.S/ of bounded real-valued functionson a set S .

Proof. (1.1) and (1.3) are obvious. To see (1.2), let d�f; g

�D 0, then f .s/ D

g.s/ for all s 2 S , so f D g by definition. The other direction is trivial.To see the triangle inequality, let f; g; h 2 B.S/. Then for each s 2 S ,

jf .s/ � h.s/j 6 jf .s/ � g.s/j Cjg.s/ � h.s/j6 sup j̊f .s/ � g.s/j

ˇ̌s 2 S

C sup j̊g.s/ � h.s/j

ˇ̌s 2 S

D d.f; g/C d.g; h/I

METRIC SPACES 7

hence,

d.f; h/ D sup j̊f .s/ � g.s/jˇ̌s 2 S

6 d.f; g/C d.g; h/:

This is because that d.f; g/C d.g; h/ is an upper bound of

j̊f .s/ � h .s/jˇ̌s 2 S

;

but d�f; h

�is the minimal element of the upper bounds.

Exercise 6 (1.1.6). Prove that the interior of a subset Y of X coincideswith the union of all open subsets of X that are contained in Y . [Thusthe interior of Y is the largest open set contained in Y .]

Proof. Let Y � X and U ´Si2I Ui , where Ui �

openX and Ui � Y for any

i 2 I ; hence, U �open

X and U � Y . We have to prove that

int .Y / D U: (1.5)

To see that U � int .Y /, pick any point x 2 U . Since U �open

X , there

exists an open ball B .xI r/ such that B .xI r/ � U � Y , which means thatx 2 int .Y /.

To see that int .Y / � U , note that int .Y / �open

X and int .Y / � Y , so

there exists an index i 2 I such that int .Y / D Ui , and which means thatint .Y / � U .

Exercise 7 (1.1.7). Prove that the closure of a subset Y of X coincideswith the intersection of all closed subsets of X that contain Y . [Thusthe closure of Y is the smallest closed set containing Y .]

Proof. Let Y � X and V DTi2I Vi , where Vi �

closedX and Y � Yi , for any

i 2 I ; hence, Y �closed

X and Y � V . We have to prove that

Y D V: (1.6)

To see that Y � V , pice any x 2 Y ; then for any open ball B .xI r/ wehave B .xI r/\ Y ¤ ¿, but which means that B .xI r/\ V ¤ ¿ since Y � V ;hence, x 2 V D V since V �

closedX .

To see that V � Y , notice that Y �closed

X and Y � Y , that is, there is an

index i 2 I such that Y D Vi , which implies that V � Y .

8 CHAPTER 1

Exercise 8 (1.1.8). A set of the form

�y 2 X

ˇ̌̌d�x; y

�6 r

�is called

a closed ball. Show that a closed ball is a closed set. Is the closed

ball

�y 2 X

ˇ̌̌d�x; y

�6 r

�always the closure of the open ball B.xI r/?

What if X D Rn? Prove your answers.

Proof. Let B 0 .xI r/´

�y 2 X

ˇ̌̌d�x; y

�6 r

�.

(a) Choose any point y 2 X X B 0 .xI r/, then d�x; y

�> r . We now show

that y 2 int�.X/ X B 0 .xI r/

�. Let " D d.x; y/ � r > 0, and consider

the ball B�yI "

�. Pick any point z 2 B

�yI "

�. Then d

�z; y

�< ". By the

triangle inequality,

d .x; z/ > d�x; y

�� d

�y; z

�> d

�x; y

��

hd�x; y

�� "

iD r;

that is, z 2 X XB 0 .x; I r/. We thus proved that B�yI "

�� X XB 0 .xI r/,

so X X B 0 .xI r/ is open, and consequently, B 0 .xI r/ is closed.

(b) To see if B 0 .xI r/ D B .xI r/, consider the discrete metric in Exercise 2and let r D 1. In this case

B 0 .xI 1/ D X; but B .xI 1/ D fxg ) B .xI 1/ D fxg :

Thus, this statement fails in general metric space.

(c) This statement, however, holds in Rn with the metric defined in Exer-cise 3.

Exercise 9 (1.1.9). Let Y be a subspace of X and let S be a subset of Y .Show that the closure of S in Y coincides with S \ Y , where S is theclosure of S in X .

Proof. Let S � Y � X , SY be the closure of S in Y , and S be the closure ofS in X . We need to show that

SY D S \ Y: (1.7)

METRIC SPACES 9

To see that SY � S \ Y , choose any point s 2 SY . Then s 2 Y and thereexists a sequence fsng

1nD1 � S such that sn ! s. In this way, s 2 X and

sn ! s, i.e., s 2 S . Hence, s 2 S \ Y .To see that S \ Y � SY , choose any point s 2 S \ Y . In this case, s is

adherent to S and s 2 Y . By Theorem 1.11, there is a sequence fsng1nD1 � S

such that sn ! s, that is, s 2 SY .

Exercise 10 (1.1.10). A point x 2 X is a limit point of a subset S of Xif every ball B .xI r/ contains infinitely many points of S . Show that xis a limit point of S if and only if there is a sequence

˚xj1jD1

in S suchthat xj ! x and xj ¤ x for all j . Shwo that the set of limit points of Sis closed.

Proof. Notice that a limit point is also defined as in the following way: x isa limit point of S if �

B .xI r/ X fxg�\ S ¤ ¿; 8 r > 0:

See Rudin (1976). These two definitions are equivalent.

(a) Suppose that there is a sequence˚xj1jD1

in S such that xj ! x andxj ¤ x for all j . We want to show that x is a limit point of S . Considerany B .xI r/. Since xj ! x, there exists Nr 2 N such that

d .xn; x/ < r 8 n > Nr ;

that is, B .xI r/ contains infinitely many points of S . Hence, x is a limitpoint of S .

To prove the other direction, let x be a limit point of S . For any 1n

,where n 2 N, pice a point

xn 2 B

�xI1

n

�\ S; xn ¤ x:

In this case, fxng1nD1 � S , xn ! x, and xn ¤ x 8 n. [Since there are

infinitely many points in S \ B�xI 1

n

�for every n, we can find such a

sequence.]

(b) Let S 0 be the set of limit points of S . Pick y 2 X X S 0. Then there is aball B

�yI r

�which contains only finitely many points of S . In this case,

every point in B�yI r

�cannot be a limit point of S [Suppose not. Then

there is z 2 B�yI r

�such that there is an open ball B .zI "/ � B

�yI r

�and there are infinite points in S \ B .zI "/, but which means thereare infinite points in S \ B

�yI r

�. A contradiction.]; hence, B

�yI r

��

X X S 0, that is, X X S 0 is open, and consequently, S 0 is closed.

10 CHAPTER 1

Exercise 11 (1.1.11). A point x 2 S is an isolated point of S if thereexists r > 0 such that B .xI r/ \ S D fxg. Show that the closure ofa subset S of X is the disjoint union of the limit points of S and theisolated points of S .

Proof. Let S be the closure of S , S 0 be the set of limit points of S , and S 00

be the set of isolated points of S . It is easy to see that S 0\S 00 D ¿. We nowshow that

S D S 0 t S 00:

(a) Let x 2 S . Then for any ball B .xI r/ we have B .xI r/ \ S ¤ ¿. Thereare two cases:

� jB .xI r/ \ S j D 1. In this case, x 2 S 0, or

� jB .xI r/ \ S j <1. In particular, let�B .xI r/ X fxg

�\S D fx1; : : : ; xng.

Now let

r� D min fd .x1; x/ ; : : : ; d .xn; x/g :

Since jB .xI r/ \ S j is finite, we know that r� is well-defined andr� > 0. First note that x 2 S ; otherwise, B

�xI r�

�\ S D ¿,

that is, x … S . A contradiction. Since x 2 S , we then haveB�xI r�

�\ S D fxg, that is, x 2 S 00.

In sum, S � S 0 t S 00.

(b) Let x 2 S 0 t S 00. There are two cases:

� x 2 S 00. In this case, x 2 S by definition; therefore, x 2 S sinceS � S .

� x 2 S 0. In this case, x is a limit point of S , so B .xI r/ \ S ¤ ¿for any r > 0; hence, x 2 S .

We thus proved that S 0 t S 00 � S .

Exercise 12 (1.1.12). Two metrics onX are equivalent if they determinethe same open subsets. Show that two metrics d , � on X are equivalentif and only if the convergent sequences in .X; d/ are the same as theconvergent sequences in .X; �/.

METRIC SPACES 11

Proof.

(a) A sequence in X converges to x if and only if for any open set Ucontaining x, there exists N 2 N such that xn 2 U for all n > N . Thisproves that if two metrics determine the same open sets, then the twometric spaces have the same convergent sequences.

(b) From Theorem 1.11 we see that a set in a metric space is closed if andonly if it contains the limit of any sequence in the set that converges.Thus, if two metrics have the same convergent sequences, then theyhave the same closed sets, consequently they have the same open sets.

Exercise 13 (1.1.13). Define � on X �X by

��x; y

�D min

n1; d

�x; y

�o; x; y 2 X:

Show that � is a metric that is equivalent to d . [Hence every metric isequivalent to a bounded metric.]

Proof. It is easy to see that

��x; y

�! 0 () d

�x; y

�! 0:

So they determine the same convergent sequences. By Exercise 12, the twometrics are equivalent.

Exercise 14 (1.1.14). The boundary @E of a set E is defined to be theset of points adherent to both E and the complement of E,

@E D E \ .X XE/:

Show that E is open if and only if E \ @E is empty. Show that E isclosed if and only if @E � E.

Proof.

(a) (a) Let E �open

X . Then E D int .E/, and x 2 E implies that x …

X XE, so E \ .X XE/ D ¿. Since E � E,

E \ @E D E \hE \ .X XE/

iD

�E \E

�\ .X XE/

D E \ .X XE/

D ¿:

12 CHAPTER 1

(b) Suppose that E \ @E D ¿. Then from the above equation weknow that E \ .X XE/ D ¿, that is, no point of E is adherent toX XE. Hence, if x 2 .X XE/, then x 2 .X XE/, that is,

.X XE/ � X XEI

therefore, X XE is closed, and E is open.

(b) (a) Let E �closed

X . Then X XE �open

X , and so .X XE/\ @ .X XE/ D

¿ by part (1.a). Since @ .X XE/ D .X XE/ \ E D @E, we have.X XE/ \ @E D ¿, that is, @E � E.

(b) If @E � E, then @E\.X XE/ D @ .X XE/\.X XE/ D ¿; hence,X XE �

openX by part (1.b). Then E �

closedX .

1.2 Completeness

Remark 1.2.1 (Baire Category Theorem). Let X is open, and x 2 X .Then there is a ball B .xI r/ such that B .xI r/ � X .

Proof. Since X is open, there is a ball B�xI r 0

�� X . Let r D r 0 � ", where

" 2�0; r 0

�and consider B .xI r/. Let y 2 B .xI r/. Then there exists z 2 X

such that

z 2 B .xI r/ \ B�yI "

�:

We now prove that y 2 B�xI r 0

�. This is because

d�xIy

�6 d .xI z/C

�zIy

�< r C "

D r 0:

This proves that y 2 X since B�xI r 0

�� X .

Exercise 15 (1.2.1). A sequence fxkg1kD1 in a metric space .X; d/ is a

fast Cauchy sequence if

1XkD1

d�xk ; xkC1

�<1:

Show that a fast Cauchy sequence is a Cauchy sequence.

METRIC SPACES 13

Proof. Let fxkg1kD1 be a fast Cauchy sequence. Let n > m, then by the trian-

gle inequality,

d .xm; xn/ 6 d .xm; xmC1/C d .xmC1; xn/6 d .xm; xmC1/C

�d .xmC1; xmC2/C d .xmC2; xn/

�6 d .xm; xmC1/C � � � C d .xn�1; xn/ :

SinceP1kD1 d

�xk ; xkC1

�converges, there exists N 2 N such that

d .xm; xmC1/C � � � C d .xn�1; xn/ 6 "; 8 " > 0;

when n > m > N . (See Rudin 1976, Theorem 3.22, or Zorich 2004, Section3.1.4, p. 95). Therefore,

d .xm; xn/ 6 ";8 " > 0;

that is, fxkg1kD1 is a Cauchy sequence.

Exercise 16 (1.2.2). Prove that every Cauchy sequence has a subse-quence that is a fast Cauchy sequence.

Proof. Let fxng1nD1 be a Cauchy sequence. For "1 D

12, there exists n1 such

that

d�xn1

; xm�<1

2; 8 m > n1:

Now let " D 122 , and pick n2 > n1 such that

d�xn2

; xm�<1

22; 8 m > n2:

In this way, we can find a subsequence such that

d .xn; xnC1/ 61

2n;

and so1XkD1

d�xnk

; xnkC1

�<1

2C

1

22C � � � D 1 <1:

Exercise 17 (1.2.3). Prove that the set of isolated points of a countablecomplete metric space X forms a dense subset of X .

14 CHAPTER 1

Proof. Refer Exercise 11 for the definition of an isolated point.1 Let X becomplete and jX j D @0. Let X0 be the set of isolated points of X . We needto prove that X0 D X .

Let X 0 D X XX0, the set of points of X that are not isolated points of X .For each x 2 X 0, let Yx D X X fxg. Since X0 D X XX 0, we have

X0 D X X

0@ [x2X 0

fxg

1A D X \0@ [x2X 0

fxg

1Ac

D X \

0@ \x2X 0

fxgc

1AD

\x2X 0

�X X fxg

�D

\x2X 0

Yx :

Note that Yx D X Xfxg �open

X since fxg is closed in a metric space. It is also

easy to see that Yx is dense, that is, X X fxg D X . To prove this, we needonly to prove that x 2 X X fxg. Since x is not an isolated point of X , thenfor any ball B .xI r/ we must have a point y ¤ x such that y 2 B .xI r/ \X ,but this implies that B .xI r/\

�X X fxg

�¤ ¿, that is, x 2 X X fxg. Then, by

the Baire Category Theorem, X0 is dense in X .

Exercise 18 (1.2.4). Suppose that F is a subset of the first categoryin a metric space X and E is a subset of F . Prove that E is of thefirst category in X . Show by an example that E may not be of the firstcategory in the metric space F .

Proof. Since F is of the first category in a metric space X , we can write Fas

F D

1[nD1

Fn;

where the Fn’s are nowhere dense in X , that is, int�Fn

�D ¿, 8 n 2 N.

Since E � F , we have

E D E \ F D E \

0@ 1[nD1

Fn

1A D 1[nD1

.E \ Fn/ :

Thus, to prove E is of the first category in X , we need only to prove thatthe .E \ Fn/’s are nowhere dense in X . This is true because .E \ Fn/ � Fn

1x 2 X is an isolated point of X if there is an open ball B .xI r/ such that�B .xI r/X fxg

�\X D ¿.

METRIC SPACES 15

and Fn is nowhere dense. Particularly,

.E \ Fn/ � Fn) .E \ Fn/ � Fn) int�E \ Fn

�� int

�Fn

�:

For the example, note that R is nowhere dense in R2, but R is dense inR.

Exercise 19 (1.2.5). Prove that any countable union of sets of the firstcategory in X is again of the first category in X .

Proof. A countable union of countable unions is still a countable union.

Exercise 20 (1.2.6). Let S be a nonempty set, let .X; d/ be a metricspace, and let F be the set of functions from S to X . For f; g 2 F ,define

��f; g

�D sup

s2S

minn1; d

�f .s/; g.s/

�o:

Show that � is a metric in F . Show that a sequence ffng converges tof in the metric space

�F ; �

�if and only if ffng converges uniformly to

f on X . Show that�F ; �

�is complete if and only if .X; d/ is complete.

Proof.

(a) We need only to show the triangle inequality. For any s 2 S , and anyf; g; h 2 F ,

minn1; d

�f .s/; h.s/

�o6 min

n1; d

�f .s/; g.s/

�C d

�g.s/; h.s/

�o6 min

n1; d

�f .s/; g.s/

�oCmin

n1; d

�g.s/; h.s/

�o6 sup

s2S

minn1; d

�f .s/; g.s/

�oC sup

s2S

minn1; d

�g.s/; h.s/

�oD �

�f; g

�C �

�g; h

�:

Hence, the triangle inequality is proved after take supremum over S .

(b) Notice that when " < 1,

��fn; f

�6 " () min

s2Sd�fn .s/ ; f .s/

�6 "

() d�fn .s/ ; f .s/

�6 " 8 s 2 S:

16 CHAPTER 1

(c) First suppose that .X; d/ is complete. Let ffng be a Cauchy sequence inF DS X . The definition of � shows that ff .s/g is a Cauchy sequencein X . Since X is complete, we have

fn .s/! f .s/ 2 X; 8 s 2 S:

So now we need to prove that fn ! f 2S X [in this case, F is com-plete]. Suppose " 2 .0; 1/. Choose N 2 N such that

��fn; fm

�< "; 8 n;m > N:

Then

d�fn .s/ ; fm .s/

�< "; 8 s 2 S:

Let m " 1, and we obtain

d�fn .s/ ; f .s/

�6 "; 8 s 2 S:

This proves that

��fn; f

�6 "; 8 n > N;

that is, fn ! f 2 F . Thus,�F ; �

�is complete.

Now suppose that .X; d/ is not complete. We want to prove�F ; �

�is

not complete, too. Let fxng is a Cauchy sequence in X which is notconvergent. Then the constant functions

fn .s/ D xn; 8 n 2 N

form a Cauchy sequence in F that does not converge.

METRIC SPACES 17

Exercise 21 (1.2.7). Let .X; d/ be a metric space and let S be the set ofCauchy sequences in X . Define a relation � in S by declaring “fskg �ftkg” to mean that d .sk ; tk/! 0 as k !1.

(a) Show that the relation “�” is an equivalence relation.

(b) Let eX denote the set of equivalence classes of S and letes denotethe equivalence class of s D fskg

1kD1. Show that the function

��zs; zt�D limk!1

d .sk ; tk/ ; zs; zt 2 eXdefines a metric on eX .

(c) Show that�eX; �� is complete.

(d) For x 2 X , define ex to be the equivalence class of the constantsequence fx; x; : : :g. Show that the function x ,ex is an isometryof X onto a dense subset of eX . (By an isometry, we mean thatd�x; y

�D �

�ex;ey�, x; y 2 X .)

(e) Show that when Y is a completion of X , then the inclusion mapX ! Y extends to an isometry of eX onto Y .

Proof.

(a) � is called an equivalence on S if it is reflexive, symmetric, and transi-tive in S . See Hrbacek and Jech (1999). � is

� Reflexive: This is because d .sk ; sk/ D 0 for any fskg 2 S .

� Symmetric: This is just follows from the symmetricity of d .�; �/.

� Transitive: Let fskg ; ftkg ; fukg 2 S , d .sk ; tk/! 0, and d .tk ; uk/!0. Then d .sk ; uk/! 0 by the triangle-inequality of d .�; �/.

(b) With the language of set theory, eX D X= � andes D Œs��.

Exercise 22 (1.2.8). The diameter of a nonempty subset E of a metricspace .X; d/ is defined to be

dd (1.8)

Equation 1.8

18 CHAPTER 1

1.3 The Real Line

Exercise 23 (1.3.1). Use the Least Upper Bound Axiom, together withthe usual manipulations of algebraic identities and inequalities, toprove the following:

(a) The set Z of integers is not bounded above.

(b) For each " > 0, there exists a rational number r 2 .0; "/.

(c) If a; b 2 R satisfy a < b, then there exists a rational numbers 2 .a; b/.

(d) The set Q of rational numbers is dense in R.

Proof.

(a) If Z is bounded above, then by the Least Upper Bound Axiom,

sup Z D b

exists. Then there must exist z 2 Z such that

z C 1 > b;

but since zC 1 2 Z, we conclude that b is not an upper bound of Z. Acontradiction.

(b) Given " > 0, we can find z" 2 ZC such that

z" � " > 1I

otherwise, 1"

is an upper bound of Z. Hence,

1

z"2 .0; "/ \Q:

(c) [Rudin Rudin (1976) provides a detailed, but not very elegant proof.]We can choose a large integer N such that N > �a. Then

0 < N C a < N C b:

It suffices to find

s 2 .N C a;N C b/ \Q

since then

s �N 2 .a; b/ \Q:

METRIC SPACES 19

We can thus assume that 0 < a < b. In this case, b � a > 0, and by (b),we know that there exists m 2 ZC such that

1

m2 .0; b � a/ :

Let

k D max˚k0 2 Z

ˇ̌k0 < bm

:

[This k must exist; otherwise bm would be an upper bound of Z.] Set

s Dk

m:

Then s < b, and since k C 1 > bm, we have

s Dk C 1 � 1

m

Dk C 1

m�1

m

> b �1

m

> b � .b � a/

D a:

(d) We want to show Q D R. Pick any x 2 R and any " > 0. Consider theopen ball

B .xI "/ D .x � "; x C "/ :

(c) implies that there exists q 2 Q such that q 2 B .xI "/, that is,

B .xI "/ \Q ¤ ¿; 8 x 2 R and 8 " > 0:

Thus, x 2 Q, and Q is dense in R.

Exercise 24 (1.3.2). Prove that the set of irrational numbers is dense inR.

Proof. Suppose that R XQ is not dense in R. Then 9x 2 R and " > 0 suchthat

B .xI "/ \ .R XQ/ D ¿;

but which means that B .xI "/ � Q. This is impossible because card .Q/ D@0, but card

�B .xI "/

�D c.

20 CHAPTER 1

Exercise 25 (1.3.3). Regard the rational numbers Q as a subset of R.Does the metric space Q have any isolated points? Why does this notcontradict Exercise 17?

Proof. Q has no isolated points. This does not contradict Exercise 17 be-cause Q is not complete.

Exercise 26 (1.3.4). Prove that every open subset of R is a union ofdisjoint open intervals (finite, semi-infinite, or infinite).

Proof. Let U � R be open. For any x 2 U , let Ix be the union of openintervals containing x that are contained in U . It is easy to see that Ix isopen, Ix � U , and [

x2U

Ix D U:

Moreover, if Ix ¤ Iy , then Ix \ Iy D ¿.

Exercise 27 (1.3.5). Prove that the set of irrational numbers cannot beexpressed as the union of a sequence of closed subsets of R.

Proof. Suppose R XQ DS1nD1En, where fEng

1nD1 is a sequence of closed

subsets of R. Then we have

R D .R XQ/ [Q

D

0@ 1[nD1

En

1A [0@[r2Q

frg

1A ;that is, R is the countable union of a closed sequence. We thus know thatthere at least exists one set in the sequence which has nonempty interior.This is because, by the Baire Category Theorem, if the interior of every setin the sequence is empty, that is, if every set in the sequence is nowheredense, then their countable union is nowhere dense, too. [Note that for aclosed set E, E D E.]

It is clear that frgı D ¿, 8 r 2 Q, so it must be the case that there existsEm such that Eım ¤ ¿. But if it is the case, then Em ¤ ¿, and there mustexist a rational number in Em, that is,

R XQ ¤1[nD1

En:

This contradiction proves the claim.

METRIC SPACES 21

Exercise 28 (1.3.6). Prove that the set of rational numbers cannot beexpressed as the intersection of a sequence of open subsets of R.

Proof. Suppose

Q D1\nD1

Fn;

where fFng1nD1 � R is a sequence of open sets. Then

R XQ D

0@ 1\nD1

Fn

1Ac

D

1[nD1

F cn ;

that is, RXQ can be expressed as the union of a sequence of closed subsetsof R. We have proved in Exercise 27 that it is impossible.

Exercise 29 (1.3.7). Let E0 be the closed unit interval Œ0; 1�. Let E1 bethe closed subset of E0 obtained by removing the open, middle third

of E0, so that E1 Dh0; 13

i[

h23; 1i. Let E2 be the closed subset of

E1 obtained by removing the open middle thirds of each of the twointervals in E1, so that E2 consists of four closed intervals, each oflength 1

32 . Proceeding in this manner, we construct En to consist of2n closed intervals, and we obtain EnC1 by removing the open middlethird from each interval of En. The Cantor set is defined to be theintersection of the En’s. Prove the following:

(a) The Cantor set is a closed subset of R with empty interior.

(b) The Cantor set has no isolated points.

(c) The Cantor set is uncountable.

Proof. Let C be the Cantor set, that is,

C D

1\nD0

En:

(a) C is closed since for all n, En is closed.

(b) Suppose that C has an isolated point x. Then x 2 C and there exists" > 0 such that

B .xI "/ \ C D ¿:

22 CHAPTER 1

1.4 Products of Metric Spaces

Exercise 30 (1.4.1). Prove that the functions on X �X given by

d .x;y/ Dhd1�x1; y1

�2C d2

�x2; y2

�2i 12

; (1.9)

d .x;y/ D maxnd1�x1; y1

�; d2

�x2; y2

�o; (1.10)

d .x;y/ D d1�x1; y1

�C d2

�x2; y2

�; (1.11)

are metrics.

Proof. In all of the cases, I just prove that there satisfy the triangle inequal-ity, and suppose x;y; z 2 X �X .

d .x; z/2 D d1 .x1; z1/2C d2 .x2; z2/

2

6hd1�x1; y1

�C d1

�y1; z1

�i2C

hd2�x2; y2

�C d2

�y2; z2

�i2D d .x;y/2 C d .y; z/2 C 2 �

hd1�x1; y1

�d1�y1; z1

�C d2

�x2; y2

�d2�y2; z2

�iD d .x;y/2 C d .y; z/2 C 2 �

��d1�x1; y1

�; d2

�x2; y2

��;�d1�y1; z1

�; d2

�y2; z2

��6 d .x;y/2 C d .y; z/2 C 2 �

�d1 �x1; y1� ; d2 �x2; y2�� d1 �y1; z1� ; d2 �y2; z2�� D d .x;y/2 C d .y; z/2 C 2 �

hd1�x1; y1

�2C d2

�x2; y2

�2i 12

�

hd1�y1; z1

�2C d2

�y2; z2

�2i 12

D d .x;y/2 C d .y; z/2 C 2 � d .x;y/ � d .y; z/

D�d .x;y/C d .y; z/

�2I

hence,

d .x; z/ 6 d .x;y/C d .y; z/ :

d .x; z/ D max fd1 .x1; z1/ ; d2 .x2; z2/g

6 max

�hd1�x1; y1

�C d1

�y1; z1

�i;hd2�x2; y2

�C d2

�y2; z2

�i�6 max

nd1�x1; y1

�; d2

�x2; y2

�oCmax

nd1�y1; z1

�; d2

�y2; z2

�oD d .x;y/C d .y; z/ :

METRIC SPACES 23

d .x; z/ D d1 .x1; z1/C d2 .x2; z2/

6hd1�x1; y1

�C d1

�y1; z1

�iC

hd2�x2; y2

�C d2

�y2; z2

�iD

hd1�x1; y1

�C d2

�x2; y2

�iC

hd1�y1; z1

�C d2

�y2; z2

�iD d .x;y/C d .y; z/ :

Exercise 31 (1.4.2). For each of the metrics (1.9), (1.10), and (1.11) onthe plane R2, sketch the open ball B .0; 1/. What inclusion relation holdbetween these balls?

Proof. See Figure 1.1.

Figure 1.1 B1, B2, and B3

Exercise 32 (1.4.3). Prove that each of the metrics (1.9), (1.10), and(1.11) has property (4.4).

Proof. Note that each of (1.9), (1.10), and (1.11) has property (4.5) [See Gamelinand Greene (Gamelin and Greene, 1999, p. 17)], that is,

dk�xk ; yk

�6 d .x;y/ ; x;y 2 X; 1 6 k 6 n:

24 CHAPTER 1

Hence, x.j/ ! x implies that

x.j/k! xk ; 8 1 6 k 6 n:

On the other hand, note that, for example,

d .x;y/ D

24 nXiD1

di�xi ; yi

�235 12

6

"n �

�max

nd1�x1; y1

�; : : : ; dn

�xn; yn

�o�2# 12

Dpn �max

nd1�x1; y1

�; : : : ; dn

�xn; yn

�o:

Hence, the other direction of (4.4) also holds.

1.5 Compactness

Exercise 33 (1.5.1). Give an example of a totally bounded metric spacewhich is not compact.

Solution. Consider the interval .a; b/ � R, where�1 < a; b <1. By Lemma5.4, .a; b/ is totally bounded since it is bounded; however, .a; b/ is not com-pact since it is not closed by the Heine-Borel Theorem.

Exercise 34 (1.5.2). Give an example of a complete metric space whichis not compact.

Solution. Consider R, which is complete, but is not bounded, and so is notcompact by the Heine-Borel Theorem.

Exercise 35 (1.5.3). Show directly that a compact metric space is totallybounded.

Proof. Let X be a compact metric space. For any " > 0, consider the follow-ing collection:

C D˚B .xI "/ j x 2 X

:

Then C is an open cover of X . By definition, there is a finite subcoverB .x1I "/ ; : : : ; B .xnI "/.

METRIC SPACES 25

Exercise 36 (1.5.4). Let fU˛g˛2ƒ be a finite open cover of a compactmetric space X .

(a) Show that there exists " > 0 such that for each x 2 X , the openball B .xI "/ is contained in one of the U˛ ’s. [Such an " is called aLebesgue number of the cover.]

(b) Show that if at least one of the U˛ ’s is a proper subset of X , thenthere is a largest Lebesgue number for the cover.

Proof. An extension of this exercise can be found in Lee (2000, Lemma 4.21,p. 76). I feel the solution provided by the authors is incorrect.

(a) Let X be a compact metric space and fU˛g˛2ƒ be a finite open coverof X . For any x 2 X , there is an open subset U˛.x/ such that x 2 U˛.x/;since U˛.x/ �

openX , there exists an " .x/ > 0 such that B

�xI 2" .x/

��

U˛.x/. Further, nB�xI " .x/

�j x 2 X

oforms an open cover of X , and so there is a finite subcover, say

B�x1I " .x1/

�; : : : ; B

�xnI " .xn/

�:

We now show that

" D min f" .x1/ ; : : : ; " .xn/g

is a Lebesgue number for fU˛g˛2ƒ. For any y 2 X , there is an openball B

�xi I " .xi /

�, where i D 1; : : : ; n, such that y 2 B

�xi I " .xi /

�. We

then can prove that

B�yI "

�� B

�xi I 2" .xi /

�� U˛.xi/:

[See Figure 1.2.] By the triangle inequality, for any z 2 B�yI "

�,

d .z; xi / 6 d�z; y

�C d

�y; xi

�< "C " .xi /

6 2" .xi / ;which proves the claim.

(b)

Exercise 37 (1.5.5). Prove that the product of a finite number of com-pact metric spaces is compact.

26 CHAPTER 1

"

" .xi /

2" .xi /

zy

xi

U˛.xi/

Figure 1.2 A Lebesgue number

Proof. For the general Tychonoff’s Theorem, see Theorem 2.10.4.

Exercise 38 (1.5.6). A metric space is a Polish space if it is separableand if there is an equivalent metric for which it is complete. Prove thatany closed subspace of a Polish space is Polish.

Proof. By Theorem 1.5.7.

Exercise 39 (1.5.7). Let B .S/ be the metric space of bounded real-valued functions on the set S , with the metric of uniform convergenceon S . Show that B .S/ is separable if and only if S is finite.

Proof.

METRIC SPACES 27

1.6 Continuous Functions

Exercise 40 (1.6.3). Let E be the subspace of R2 obtained from thecircle centered at

�0; 1=2

�of radius 1=2 by deleting the point .0; 1/.

Define a function h from R to E so that h .s/ is the point at which theline segment from .s; 0/ to .0; 1/ meets E.

(a) Show that h is a bicontinuous function from R to E.

(b) Show that

� .s; t/ Djh .s/ � h .t/j ; s; t 2 R;

defines a metric � on R that is equivalent to the usual metric onR.

(c) Show that�R; �

�is totally bounded but not complete.

(d) Identify the completion of�R; �

�.

h .s/

.s; 0/0

.0; 1/

Figure 1.3 h is a homeomorphism from R to E

Proof.

(a) The segment through .0; 1/ to .s; 0/ can be parametrized as

t , .ts; 1 � t / ; t 2 Œ0; 1� :

The intersection of this segment and the circle thus is

.ts/2 C

�1 � t �

1

2

�2D1

4) t D 0 or t D

1

1C s2:

Hence, the function h W R! E can be written as

h .s/ D

s

1C s2;

s2

1C s2

!: (1.12)

28 CHAPTER 1

0−1 −0.5 0.5 1

0.5

1

s D 1

s D 2s D 3

s D �12

0−1 −0.5 0.5 1

0.5

1

s D 1

s D 2s D 3

s D �12

(x.s/ D s

1Cs2

y.s/ D s2

1Cs2

Figure 1.4 R � E

0−10 −8 −6 −4 −2 2 4 6 8 10

−0.5

0.5

1

s1Cs2

s2

1Cs2

Figure 1.5 s1Cs2 and s2

1Cs2

METRIC SPACES 29

Since both s1Cs2 and s2

1Cs2 are continuous, h .s/ is continuous. It isevident that h .s/ is bijective.

Now any point�x; y

�on E can be represented as

y

xD s;

and which gives the continuous inverse of h with the definition thath�1 .0; 0/ D 0.

(b)

Exercise 41 (1.6.4). Let f be a continuous function from the metricspace X to the metric space Y and let E be a subset of X . Show thatthe restriction of f to E is a continuous function from E to Y .

Proof. Let f W X ! Y be continuous and E � X . For any open subsetV � Y , f �1 .V / �

openX since f is continuous. Because�f � E

��1.V / D f �1 .V / \E �

openE;

we know that f � E is continuous.

Exercise 42 (1.6.5). Let X0, X1, and X2 be metric spaces, let f be afunction from X0 to X1, and let g be a function from X1 to X2. Showthat if f is continuous at x0 2 X0 and if g is continuous at f .x0/, thenthe composition g ı f is continuous at x0.

Proof. Consider any sequence xn ! x0. Since f is continuous at x0, we havef .xn/ ! f .x0/; since g is continuous at f .x0/, we have

�g ı f

�.xn/ !�

g ı f�.x0/. This proves that g ı f is continuous at x0 by definition.

Exercise 43 (1.6.6). Let f be a continuous function from the closedunit interval Œ0; 1� to itself. Show that there exists t 2 Œ0; 1� such thatf .t/ D t .

Proof. Let S D˚s 2 Œ0; 1� j f .s/ > s

. S ¤ ¿ since 0 2 S . We first show

that S is closed. For any convergent sequence in S , sn ! s, we know thatf .sn/ ! f .s/ since f is continuous. Because f .sn/ > sn, it must be thatf .s/ > s, that is, s 2 S ; this proves that S is closed.

30 CHAPTER 1

Let t D supS . Since S is closed, t 2 S . If t < 1, then f .t C "/ < t C ".Let " ! 0, we get f .t C "/ ! f .t/ 6 t , and so f .t/ D t . If t D 1, thenf .t/ D f .1/ 6 1 D t , and so f .t/ D t .

Exercise 44 (1.6.7). Show that if f is a continuous function from acompact metric space X to a metric space Y , then the image of f isa compact subset of Y . In particular, show that f is bounded, that is,that the range of f is a bounded subset of Y .

Proof. Let f W X ! Y be continuous, and let X be a compact metric space.Let fV˛g˛2ƒ be an open cover of f .X/ in Y . Then

˚f �1 .V˛/

˛2ƒ

is an opencover of X since

X D f �1

0@[˛2ƒ

V˛

1A D [˛2ƒ

f �1 .V˛/ :

SinceX is compact, there exists a finite subcover, say,˚f �1 .V1/ ; : : : ; f

�1 .Vn/,

that is, X DSniD1 f

�1 .Vi / D f�1�Sn

iD1 Vi�. Applying f to both sides and

we get

f .X/ D

n[iD1

Vi ;

that is, f .X/ is compact.

1.7 Normed Linear Spaces

Exercise 45 (1.7.1). Sketch the unit ball of R2 when it is endowed witheach of the following norms: k�k1, k�k2, k�k4, and k�k1. Sketch the ballof radius 1 and center .2; 1/ in the metric determined by each of thesenorms.

Solution. The unit balls determined by the norms are depicted in Figure 1.6.The other part is straightforward.

METRIC SPACES 31

0−2 −1 1

−1

1

x1

x2

k�k4k�k2

k�k1

k�k1

Figure 1.6 The unit balls

Exercise 46 (1.7.2). Show the following for x 2 Rn:

(a) kxk1 6kxkp , 1 6 p <1,

(b) kxkp 6 n1=pkxk1, 1 6 p <1,

(c) limp!1kxkp Dkxk1.

Proof.

32 CHAPTER 1

(a) Suppose that max j̊x1j ; : : : ;jxnjDjxkj. Then

kxk1 D max j̊x1j ; : : : ;jxnj

Djxkj

D�jxkj

p�1=p

6

0@ nXiD1

jxi jp

1A1=pDkxkp :

(b) We still suppose that max j̊x1j ; : : : ;jxnjDjxkj. Then

kxkp D

0@ nXiD1

jxi jp

1A1=p

6�n �jxkj

p�1=p

D n1=pkxk1 :

(c) Let p !1. Then from (a) and (b):

kxk1 6kxkp 6 n1=pkxk1p!1) kxkp !kxk1 :

Exercise 47 (1.7.3). Two normsk�k and jjj � jjj on a vector space X areequivalent if there exists c > 0 such that

1

c� jjjxjjj 6kxk 6 c � jjjxjjj; x 2 X:

(a) Show that the normsk�k and jjj � jjj are equivalent if and only if theidentity map of X is bicontinuous between thek�k-topology of X

and the jjj � jjj-topology of X.

(b)

Proof.

Chapter 2

Topological Spaces

2.1 Topological Spaces

Exercise 48 (2.1.1). Show that the intersection of a family of topologiesfor X is a topology for X .

Proof. Consider a family of topologies for X : fT˛g˛2A.

� .X;¿/ 2 T˛ , 8 ˛, so

.X;¿/ 2\˛2A

T˛:

� Let˚Uˇˇ2B�T˛2A T˛ , then˚Uˇˇ2B2 T˛ )

[ˇ2B

Uˇ 2 T˛; 8 ˛I

hence, [ˇ2B

Uˇ 2\˛2A

T˛:

� With the same logic, we can show thatT˛2A T˛ is closed under finite

intersection.

Exercise 49 (2.1.2). Let X be a set and let T be the family of subsetsU of X such that X X U is finite, together with the empty set ¿. Showthat T is a topology. (T is called the cofinite topology of X .)

Proof. Let

T D˚U � X j card .X X U/ <1

[¿:

� ¿ 2 T by hypothesis, and X 2 T since X XX D ¿ and ¿ is finite.

34 CHAPTER 2

� Let fU˛g˛2A � Ti . Consider U DS˛2A U˛ . Note that

X X U D X X

0@[˛2A

U˛

1AD

\˛2A

.X X U˛/

��X X U 0˛

�;

where ˛0 2 A. Hence, card .X X U/ <1, and U 2 T .

� Let fUigniD1 � T , and U D

TniD1 Ui . Then

X X U D X X

0@ n\iD1

Ui

1AD

n[iD1

.X X Ui /

is finite. Hence, U 2 T .

Exercise 50 (2.1.3). Let T be the cofinite topology on the set Z of in-tegers. Show that the sequence f1; 2; 3; : : :g converges in .Z; T / to eachpoint of Z. Describe the convergent sequences in .Z; T /.

Proof. Let

T D˚U � Z j card .Z X U/ <1

[¿:

Then each every U 2 T must satisfy

maxU c D NU 2 ZI

hence, n 2 U for any n > NU . This proves that f1; 2; 3; : : :g converges in.Z; T /.

Exercise 51 (2.1.4). Let S be a subset of a set X . Describe the closureof S when

(i) X has the discrete topology,

(ii) X has the indiscrete topology, and

(iii) X has the cofinite topology.

TOPOLOGICAL SPACES 35

Proof. Note that S is the smallest closed set including S [Aliprantis andBorder (Aliprantis and Border, 2006, Sec. 2.2)], Exercise 54, and Exercise 55;therefore,

(i) S D S since S is both open and closed in a discrete topology;

(ii) In the indiscrete topology, T D f¿; Xg, so

S D

8<:¿; if S D ¿X; otherwise.

(iii) In the cofinite topology, the closed sets are ¿; X , and all finite set.Hence,

S D

8<:S; if S is finite

X; otherwise.

Exercise 52 (2.1.5). Let S be a subset of a topological space X in whichsets consisting of one point are closed. A point x 2 X is a limit point ofS if every neighborhood of x contains a point of S other than x itself.A point s 2 S is an isolated point of S if there is a neighborhood U ofs such that U \ S D fsg.

(a) Show that the set of limit points of S is closed.

(b) Show that S is the disjoint union of the set of limit points of Sand the isolated points of S .

Proof. Let

S 0 D

�x 2 X

ˇ̌̌S \

�U X fxg

�¤ ¿ for all U 2 Nx

�;

where Nx is the neighborhood system of x; hence, S 0 is the set of limitpoints of S . We also let S0 be the set of isolated points of S .

(a) Let x 2 S 0c . Then there exists U 2 Nx such that S \�U X fxg

�D ¿.

There are two cases:

(a) x 2 S . In this case, x 2 S0. Then there is an open neighborhoodUx of x such that S \

�Ux X fxg

�D ¿. Since Ux is open, we have

S \�Ux X fug

�D ¿; 8 u 2 Ux ;

that is, Ux � S0c .

36 CHAPTER 2

(b) x 2 Sc . In this case, there exists an open neighborhood Ux of xsuch that S \ Ux D ¿, so Ux � S

0c .

(a) and (b) show that for any x 2 S 0c , there is an open neighborhoodUx of x such that Ux � S

0c ; therefore,

S 0c D[x2S 0c

Ux

is open, and S 0 is closed.

(b) To prove

S D S 0 t S0;

first notice that a point x 2 X cannot be both an isolated point and alimit point of S .

(a) By definition, S0 � S � S , and S 0 � S ; hence, S 0 t S0 � S .

(b) Let x 2 S . Then there are two cases:

i. S \�U X fxg

�¤ ¿, for any U 2 Nx . In this case, x 2 S 0, or

ii. S \�U X fxg

�D ¿, for some U 2 Nx . In this case, x 2 S0.

Thus, S � S 0 t S0.

Exercise 53 (2.1.6).

(a) Show that if X is a metrizable topological space and if p and qare distinct points of X , then there are open sets U and V suchthat p 2 U , q 2 V , and U \ V D ¿.

(b) Let X be an infinite set and let T be the cofinite topology on X(Exercise 49). Prove that the property described in part (a) doesnot hold for the open sets in X and hence that X with the cofinitetopology is not metrizable.

Proof.

(a) Suppose that the topology forX is the metric topology associated withthe metric d . Since p ¤ q, we have

d�p; q

�> 0:

Let

U D B

pId�p; q

�2

!; V D B

qId�p; q

�2

!:

Then both U and V are open sets, p 2 U , q 2 V , and U \ V D ¿.


(b) Let card .X/ be infinite. Then every open set in the cofinite topologyT is infinite; otherwise, if U 2 T and card .U / <1, then card .U c/ isinfinite. Now suppose that U; V 2 T , p 2 U , q 2 V , and U \ V D ¿.In this case, V � U c , so V is finite. A contradiction.

Exercise 54 (2.1.7). Show that if S is a subset of a topological space X ,then Sı is the union of all open sets contained in S .

Proof. Let .X; T / be a topological space, and

U D˚Ui 2 T j Ui � S

:

We need to prove that

Sı D[Ui2U

Ui ;

that is, Sı is the largest open set included in S .Let x 2

SUi2U Ui . Then there exists Ux 2 U such that x 2 Ux � S , that

is, x 2 Sı. This proves that [Ui2U

Ui � Sı:

Let x 2 Sı. Then there exists an open set Ux such that x 2 Ux � S ; hence,Ux 2 U, that is,

Sı �[Ui2U

Ui :

Exercise 55 (2.1.8). Show that if S is a subset of a topological space X ,then S is the intersection of all closed sets containing S .

Proof. Let .X; T / be a topological space. Let

V D˚Vi 2 T j S � Vi

:

We want to prove that

S D\Vi2V

Vi ;

that is, S is the smallest closed set including S [see Exercise 51].Notice that S � S , and S 2 V [Theorem 2.1.3]; thus,

TVi2V Vi � S .

Let x 2 S , Vi 2 V . There are tow cases:

38 CHAPTER 2

(a) x 2 S . In this case, x 2 S � Vi , that is, x 2 Vi , for all Vi 2 V .

(b) x … S . Suppose x … Vi . Since S � Vi , we have

V ci � Sc ; and x 2 V ci :

Since x 2 S , and V ci is open, we have

V ci \ S ¤ ¿:

A contradiction [because V ci � Sc ) V ci \ S D ¿]. Hence, x 2 Vi for

all Vi 2 V .

Exercise 56 (2.1.9). Show that if S is a subset of a topological space X ,then

(a) X X S D X X Sı,

(b) .X X S/ı D X X S .

Proof.

(a) Set arithmetic means that

Sı � S ) X X S � X X Sı;

and X X Sı is closed [since Sı is open]. By Exercise 55, we need onlyto prove that X X Sı is the smallest closed set including X X S . Thisis true because Sı is the largest open set included in S [Exercise 54].Precisely, suppose X X U is a closed set including X X S , then

X X S � X X U ) U � S;

and U is open. Hence, U � Sı, and

X X Sı � X X U:

(b) By definition,

S � S ) X X S � X X S:

To prove X X S D .X X S/ı, we need only to prove that X X S is thelargest open set included in X X S by Exercise 54.

Let V be a closed set, and X X V � X X S . Then S � V and S � Vsince S is the smallest closed set including S . Hence,

X X S � X X V:


Exercise 57 (2.1.10). If U is open, then prove that U�0�0� D U�, where

the bar means closure and the prime means complement.

Proof. We first prove that for any subset S of a topological space X ,�Sc�cD Sı: (2.1)

Clearly,

Sı � S ) Sc ��Sı�c

) Sc � .Sı/c D�Sı�c

) Sı ��Sc�c:

Also, Sc � Sc )�Sc�c� S . Since

�Sc�c

is an open set and Sı is the

largest open set included in S , we see that Sı D�Sc�c

.

Now let S D U�. Then by (2.1),

.U�/0�0

D .U�/ı

h1iD U;

where h1i holds because U is open, and U � U�. [Note that U� D U ı t

@U D U t @U when U is open.] Therefore,

U�0�0�D U�:

Exercise 58 (2.1.11). Prove the following result, originally noted by Ku-ratowski: If S is a subset of a topological space X , then there are atmost 14 subsets of X that can be obtained from S by successively tak-ing either complements or closures. Find a subset S or R such thatexactly fourteen subsets of R can be obtained from S in this manner.

2.2 Subspaces

Exercise 59 (2.2.1). Let X be a topological space, let S be a subspaceof X , and let E be a subset of S . Show that the relative topology thatE inherits from S coincides with the relative topology that E inheritsform X .

40 CHAPTER 2

Proof. Let .E;EX / be a subspace of .X; T /, and let .E;ES / be a subspace of.S; SS/. We need to prove that EX D ES . This is because

ES D˚U \E j U 2 SS

D˚.V \ S/ \E j V 2 T

D˚V \ .S \E/ j V 2 T

D˚V \E j V 2 T

D EX :

Exercise 60 (2.2.2). Prove that if A and S are subsets of a topologicalspace X , then the closure of A \ S in S in the relative topology for Sis a subset of the intersection A \ S , where A is the closure of A in X .Give an example where the relative closure of A \ S is a proper subsetof A \ S .

Proof. Note that A \ S � S ; hence, by Theorem 2.2.2, the closure of A \ Sin S is �

A \ S�\ S;

where A \ S is the closure of A \ S in X. Now it is straightforward to seethat �

A \ S�\ S � A \ S:

As an example, let X D R, A D .�1; 0/, S D Œ0; 1�, and with the standardmetric. Then A \ S D ¿, and so ¿ \ Œ0; 1� D ¿, but

.�1; 0/ \ Œ0; 1� D f0g :

Exercise 61 (2.2.3). Let S be a subset of a topological space X . Showthat a sequence fxig in S converges to x0 2 S in the relative topology ifand only if, considered as a sequence inX , the sequence fxig convergesto x0.

Proof. Straightforward. Just like Theorem 2.2.2.


2.3 Continuous Functions

Exercise 62 (2.3.1). Let f be a function from a metric space X to a met-ric space Y and let x 2 X . Show that the definition of continuity of fat x given in this section (for each open neighborhood V of f .x/, thereexists an open neighborhood U of x such that f .U / � V ) coincideswith the definition given in Section 1.6 (whenever fxng is a sequence inX such that xn ! x, then f .xn/! f .x/).

Proof. Let f be continuous at x by the first definition. Let xn ! x. LetB�f .x/I "

�be an open ball in Y . Then there exists an open ball B .xI ı/ in

X such that f�B .xI ı/

�� B

�f .x/I "

�. Since xn ! x, there is N 2 N such

that xn 2 B .xI ı/ for all n > N , that is, f .xn/ 2 B�f .x/I "

�for all n > N .

This proves that f .xn/! f .x/.The inverse direction follows Theorem 1.6.2.

Exercise 63 (2.3.2). Show that a function f W X ! Y is continuous ifand only if f �1 .E/ is closed subset of X for every closed subset E ofY .

Proof. Let f W X ! Y be continuous. Let E � Y be closed. Then Y X E isopen, and

f �1 .Y XE/ D f �1 .Y / X f �1 .E/ D X X f �1 .E/

is open. Therefore, f �1 .E/ is closed.For the inverse direction, let U � Y be open. Then Y X U is closed, and

so

f �1 .Y X U/ D X X f �1 .U /

is closed; therefore, f �1 .U / is open, and so f is continuous.

42 CHAPTER 2

Exercise 64 (2.3.3). Prove the following statements about continuousfunctions and discrete and indiscrete topological spaces.

(a) If X is discrete, then every function f from X to a topologicalspace Y is continuous.

(b) If X is not discrete, then there is a topological space Y and afunction f W X ! Y that is not continuous.

(c) If Y is an indiscrete topological space, then every function f froma topological space X to Y is continuous.

(d) If Y is not indiscrete, then there is a topological space X and afunction f W X ! Y that is not continuous.

Proof.

(a) If X is discrete, then every subset f �1 .U / is open.

(b) Let Y D X with the discrete topology. Consider IdY .

(c) If Y is an indiscrete topological space, then there are just two opensets in Y : Y and ¿; thus,

f �1 .Y / D X; and f �1 .¿/ D ¿:

(d) Let X D Y , and X is indiscrete. Consider IdX .

Exercise 65 (2.3.4). Prove that all open intervals in R (finite, semi-infinite, or infinite) are homeomorphic.

Proof. See Dudley (Dudley, 2002, p. 40). A finite, non-empty open interval.a; b/ is homeomorphic to .0; 1/ by a linear transformation: let

f .x/´ aC .b � a/ x:

A bit more surprisingly, .�1; 1/ is homeomorphic to all of R, letting

f .x/´ tan

��x

2

�:


Exercise 66 (2.3.5). Prove that all semiopen intervals in R (finite orsemi-infinite) are homeomorphic.

Proof. Similar to Exercise 65.

Exercise 67 (2.3.6). Show that the unit balln�x; y

�2 R2

ˇ̌x2 C y2 < 1

ois homeomorphic to the open square

n�x; y

�j 0 < x < 1; 0 < y < 1

o.

Proof.

Exercise 68 (2.3.13). Prove that if f W X ! Y is continuous and if S isa subspace of X , then the restriction f � S W S ! Y is continuous.

Proof. Since f is continuous, f �1 .U / �open

X for any U �open

Y . Because�f � S

��1.U / D S \ f �1 .U / �

openS;

f � S W S ! Y is continuous.

Exercise 69 (2.3.14).

2.4 Base for a Topology

Exercise 70 (2.4.1). Let X be a topological space with the cofinitetopology (Exercise 49). Show that X is separable. When is X second-countable?

Proof. Let fxng1nD1 � X be a sequence. Suppose fxng ¤ X , that is, X is not

the smallest closed set including fxng. Then, X X fxng ¤ ¿ and X X fxng isopen. We thus have

X X�X X fxng

�D fxng

is finite.If card .X/ D @0, then the family of finite sets is countable, and conse-

quently the set of open sets is countable. This proves a countable set issecond-countable.

On the other hand, let X is uncountable. Let fUng be a countable familyof opens sets in X .

44 CHAPTER 2

Exercise 71 (2.4.2). LetX be a topological space with the discrete topol-ogy. Find a base B of open sets for X such that B is included in anyother base of open sets for X .

Solution. Let B D˚fxg j x 2 X

.

Exercise 72 (2.4.3). Let X and Y be topological spaces and let B be abase of open sets for Y . Show that a function f W X ! Y is continuousif and only if f �1 .U / is an open subset of X for every U 2 B.

Proof. It is evident that if f is continuous then the result holds. For theinverse direction, suppose that f �1 .U / �

openX for every U 2 B. Choose

any V �open

Y , then Y DSBi for some Bi 2 B. Hence,

f �1 .V / D f �1�[

Bi

�D

[f �1 .Bi /

�open

X:

Exercise 73 (2.4.4). A topological space X satisfies the first axiom ofcountability, or is first countable, if for each x 2 X , there exists a se-quence of open neighborhoods fUng

1nD1 of x such that each neighbor-

hood of x includes one of the Un’s. Prove the following assertions:

(a) Any metric space is first-countable.

(b) Any second-countable space is first-countable.

(c) In a first-countable space, any point adherent to a set S is a limitof a sequence in S .

Proof.

(a) Take Un D B�xI 1

n

�.

(b) Take the Un’s to be the sets in a countable base that contain x.

(c) If x is adherent to S , then S\Un ¤ ¿ for each Un. Choose xn 2 S\Un.We now show that xn ! x. Take any open neighborhood U , then thereexist a UN such that UN � U , and so xN 2 U . Further, there exists aUn such that Un � UN since UN is a neighborhood of x.


Exercise 74 (2.4.5). Let X be a set and let � be a family of subsets ofX .

(a) Show that there exists a unique smallest topology T on X suchthat � � T . The topology T is called the topology generated by� , and � is called a subbase for the topology T .

(b) Let B be the family of subsets of X consisting of X , ¿, and allfinite intersections of sets in � . Show that B is a base of opensets for the topology generated by � .

(c) Let X have the topology generated by � , let Y be a topologicalspace, and let f W Y ! X be a function. Show that f is contin-uous if and only if f �1 .S/ is an open subset of Y for every setS 2 � .

Proof.

(a) Let fT˛g˛2A be the set of topologies containing � . This set is not emptysince the discrete topology satisfies this criterion. Then let

T D\˛2A

T˛:

It is evident that � � T . We now show that T is a topology [and so T

is the smallest topology].

� ¿ 2 T and X 2 T since ¿ 2 T˛ and X 2 T˛ for each ˛ 2 A;

� IfU1; : : : ; Un 2 T , thenU1; : : : ; Un 2 T˛ , 8 ˛ 2 A; hence,TniD1 Ui 2

T˛ , 8 ˛ 2 A, implies thatTniD1 Ui 2 T , that is, T is closed under

finite intersection;

� Similarly, we can show that T is closed under arbitrary union.

(b) Clear.

(c) Clear.

46 CHAPTER 2

Exercise 75 (2.4.6). Let B be the family of subsets of R of the formŒa; b/, where

�1 < a < b <1:

(a) Show that B is a base of open sets for a topology T of R. Thetopology determined by B is the half-open interval topology.

(b) Show that every open subset of R (in the metric topology) is T -open.

(c) Show that each interval Œa; b/ is T -closed.

Proof.

(a) To show that B is a base for a topology T of R, we need to show thatB satisfies the two conditions of Gamelin and Greene (1999, Theorem2.4.2).

� For any x 2 R, consider the interval Œx; x C "/, where " > 0. It isevident that x 2 Œx; x C "/ 2 B;

� Let Œa; b/ ; Œc; d/ 2 B. Then Œa; b/ \ Œcd/ is an empty set or anelement of B.1

Therefore, B is the base of T .

(b) Let U �open

R in the metric topology. Then U is a union of open inter-

vals .a; b/. However,

.a; b/ D[ŒaC "; b/ :

(c)

Exercise 76 (2.4.7). Prove that a subspace of a second-countable spaceis second-countable.

Proof. Let A be a subspace of X . Since X is second-countable, it admits acountable base, B. We show that

BA D˚A \ B j B 2 B

is a [countable] base of the subspace A through showing that every opensubset of A satisfies the basis criterion with respect to BA (Lee, 2000, Ch.

1I suspect that B should include ¿.


3.1). Choose any U �open

A and let q 2 U . Then U D A\V for some V �open

X .

By the basis criterion in X , there is a B 2 B such that q 2 B � V . Hence,

q 2 A \ B � A \ V D U:

2.5 Separation Axioms

Exercise 77 (2.5.1). Show that a sequence in a Hausdorff space cannotconverge to more than one point.

Proof. Let X be a Hausdorff space and fxng1nD1 be a sequence in X . Suppose

that xn ! x and xn ! x0, where x ¤ x0. By the Hausdorff property,there exist disjoint neighborhoods U of x and U 0 of x0. By definition ofconvergence, there exist N;N 0 such i > N implies xi 2 U and i > N 0

implies xi 2 U0. But since U \ U 0 D ¿, this is a contradiction when i >

max˚N;N 0

.

Exercise 78 (2.5.2). Let X be a topological space and let X0 be thetopological space that is the set X with the cofinite topology. Showthat the identity map of X to X0 is continuous if and only if X is aT1-space.

Proof. Let X be a T1-space. If U �open

X0, then Id�1 .U / D U � X . To see

U �open

X , note that X XU is finite by definition of the cofinite topology, and

since X is T1, every singleton set is closed in X ; hence,

X X U D

n[iD1

fxig �closed

X:

Thus, U �open

X , and Id W X ! X is continuous.

For the converse direction, let Id W X ! X be continuous. Then Id�1�fxg

�D

fxg �closed

X since Id sends closed subset of X0 to X and fxg �closed

X0.

Exercise 79 (2.5.3). A property of a topological space is hereditary if,whenever a topological space X has that property, then every subspaceof X has the property. Show that the properties of being a T1-space,Hausdorff space, and regular space are hereditary.

48 CHAPTER 2

Proof. Let X is a topological space and A � X is a subspace.

(a) Let X be a T1-space. Choose any two distinct points x; y 2 A. Thenthere exists U �

openX containing y such that x … U . In particular,

y 2 A \ U �open

A and x … A \ U . Thus, A is a T1-space.

(b) Let X be Hausdorff. Choose any two distinct points x; y 2 A. Thenthere exist disjoint open sets U and V such that x 2 U and y 2 V . Inthis case, A\U;A\ V �

openA, .A \ U /\ .A \ V / D ¿, and x 2 A\U ,

y 2 A \ V , i.e., A is Hausdorff.

(c) Let X be regular. Choose any x 2 A and E �closed

A such that x … E.

Since E �closed

A, there exist F �closed

X such that E D A \ F ; since

x 2 A and x … E, we know that x … F . Then there exist disjoint opensets U and V in X such that F � U and x 2 V . Now consider thefollowing two open sets: A\U and A\V . First, they are disjoint opensubsets of A; next,

E D A \ F � A \ U; and x 2 A \ V:

Therefore, A is regular, too.

Exercise 80 (2.5.4). Prove that a topological space X is normal if andonly if the conclusion of Urysohn’s Lemma is valid for X . Prove thatthis occurs if and only if the conclusion of the Tietze Extension Theo-rem is valid for X .

Proof.

(a) It is obvious that if X is normal, then Urysohn’s Lemma holds byLemma 2.5.2. For the other direction, suppose the conclusion of Urysohn’sLemma is valid forX , that is, ifE;F �

closedX andE\F D ¿, then there

exists a continuous function f W X ! Œ0; 1� such that f .E/ D f0g andf .F / D f1g. Since Œ0; "/ and ."; 1�, where " 2 .0; 1/, are open subsetsof Œ0; 1�, and f is continuous, we conclude that

E � f �1�Œ0; "/

��

openX; and F � f �1

�."; 1�

��

openX;

and

f �1�Œ0; "/

�\ f �1

�."; 1�

�D ¿:

Thus, X is normal by definition.


(b) Suppose that Tietze Extension Theorem is valid for X . Suppose E andF are disjoint closed sets in X . Then E [ F �

closedX . Then define

h W E [ F ! Œ0; 1� be h .E/ D f0g and h .F / D f1g. Evidently h iscontinuous on E [ F . Let f W X ! Œ0; 1� be a continuous extension of

h to X . Again f �1�h0; 12

��and f �1

��12; 1i�

are disjoint open sets

containing E and F respectively, and X is normal.

2.6 Compactness

Exercise 81 (2.6.1). Prove in detail that compactness is a topologicalproperty.

Proof. Let X Š Y and f be a homeomorphism. Suppose X is compact. LetfV˛g˛2A be an open cover of Y . Then

˚f �1 .V˛/

˛2A

is an open cover of X .Since X is compact, there is a finite sub open cover of X :n

f �1�V˛1

�; : : : ; f �1

�V˛n

�o:

Hence,˚V˛1

; : : : ; V˛n

is a finite sub open cover of Y , that is, Y is compact.

Exercise 82 (2.6.2). Prove that a topological space X is compact if andonly if it has the following property: if fE˛g˛2A is any family of closedsubsets of X such that any finite intersection of the E˛ ’s is nonempty,then

T˛2AE˛ is nonempty.

Proof. First suppose that X is compact and fE˛g˛2A is any family of closedsubsets of X such that any finite intersection of the E˛ ’s is nonempty. Sup-pose

T˛2AE˛ D ¿. Consider the family of open subsets of X derived from

fE˛g˛2A:

fX XE˛g˛2A :

50 CHAPTER 2

Since [˛2A

.X XE˛/ D[˛2A

�X \Ec˛

�D X \

0@[˛2A

Ec˛

1AD X \

0@\˛2A

E˛

1Ac

D X X

0@\˛2A

E˛

1AD X;

it can be concluded that fX XE˛g˛2A forms an open cover of X . Since X iscompact, there exists a finite subcover of X :˚

X XE˛1; : : : ; X XE˛n

:

In this case,

X D�X XE˛1

�[� � �[

�X XE˛n

�D XX

�E˛1\ � � � \E˛n

�) E˛1

\� � �\E˛nD ¿:

A contradiction.For the other direction, suppose the hypothesis holds. Let fU˛g˛2A is an

open cover of X , and let E˛ D X X U˛ . ThenT˛2AE˛ D ¿. Note that the

statement in the problem is equivalent to the following one:\˛2A

E˛ D ¿) 9 ˛1; : : : ; ˛n such that E˛1\ � � � \E˛n

D ¿: (2.2)

Then by (2.2),˚U˛1

; : : : ; U˛n

is a subcover of X , i.e., X is compact.

Exercise 83 (2.6.3). Show that any space with the cofinite topology iscompact.

Proof. Let fU˛g˛2A be an open cover of X . Choose any U˛1¤ ¿ from the

U˛ ’s. Then X X U˛1D fx2; : : : ; xng is finite. For any xi , i D 2; : : : ; n, choose

a U˛icontaining xi . Thus,

˚U˛1

; : : : ; U˛n

is a finite subcover of X .

Exercise 84 (2.6.4). Show that a discrete topological space is compactif and only if it is finite.


Proof. Let�X; T D 2X

�is a discrete topological space. If X is finite, then

2X is, and consequently X is compact since there are only finite many opensets in T .

Now suppose X is compact. It is obvious that˚fxg j x 2 X

is an open

cover of X . Since it has a finite subcover, we know that X is finite.

Exercise 85 (2.6.5). Show that a continuous real-valued function on acompact space attains its maximum value and its minimum value. Inparticular, show that a continuous real-valued function on a compactspace is bounded.

Proof. If f W X ! R is continuous, and X is compact, then by Theorem2.6.6, f .X/ is a compact subset of R. By the Heine-Borel Theorem, f .X/is a closed bounded subset of R. Hence f .X/ contains its supremum andinfimum, and consequently f attains its maximum and minimum values.

Exercise 86 (2.6.6). Prove that if X is a compact Hausdorff space and ifx; y 2 X satisfy x ¤ y, then there is a continuous real-valued functionf on X such that f .x/ ¤ f

�y�. (In other words, the continuous real-

valued functions on X “separate the points” of X .)

Proof. By Theorem 2.6.5, a compact Hausdorff space is normal; in a Haus-dorff space, fxg ; fyg �

closedX . By the Urysohn’s Lemma, there exists a con-

tinuous function f W X ! Œ0; 1� such that f .x/ D 0 ¤ 1 D f�y�.

Exercise 87 (2.6.7). Let X be a compact Hausdorff space and letfU˛g˛2A be an open cover of X . Show that there exist a finite numberof continuous real-valued functions h1; : : : ; hm on X with the followingproperties:

(a) 0 6 hj 6 1, 1 6 j 6 m,

(b)Phj D 1,

(c) For each 1 6 j 6 m, there is an index j̨ such that the closure ofthe set

˚xˇ̌hj .x/ > 0

is contained in U

j̨.

Proof. As in Exercise 86, X is normal and fxg �closed

X for any x 2 X . For

each x 2 X , take an open set U˛.x/ containing x from the U˛ ’s. We then use

52 CHAPTER 2

Urysohn’s Lemma to the closed sets x and X X U˛.x/: there exists a contin-uous function gx W X ! Œ0; 1� such that f .x/ D 1 and f

�X X U˛.x/

�D f0g.

We then show that the set Vx D˚x j gx > 0

satisfies V x � U˛.x/. Sup-

pose it were not the case. Then there exists a point y satisfies y 2 V x buty … U˛.x/; hence f

�y�D 0

Unsolved

2.7 Locally Compact Spaces

Exercise 88 (2.7.1). Prove that a Hausdorff topological space X is lo-cally compact if and only if, for each p 2 X and each open neighbor-hood U of p, there is an open neighborhood V of p such that V � Uand V is compact.

Proof. Let X be a locally compact Hausdorff topological space. Choose anyp and open neighborhood U of p. By definition, there is an open set Wsuch that p 2 W and W is compact. Let V D U \ W . Then V �

openX and

V �closed

W . Since W is compact, V is compact.

For the inverse direction, note that V �open

U �open

U implies that V �open

X .

Then the proof is trivial.

Exercise 89 (2.7.2). Prove in detail that local compactness is a topolog-ical property.

Proof. Let X Š Y and f be a homeomorphism. Suppose X is locally com-pact. For any y 2 Y , there exists x 2 X such that x D f �1

�y�. Since X is

locally compact, there is an open set W such that x 2 W and W is compact.

Then f .W / �open

Y , y 2 f .W /, and f�W�D f .W / is compact.

Exercise 90 (2.7.3). Prove that a locally compact Hausdorff space isregular.

Proof. Let X be a locally compact Hausdorff space. Choose any E �closed

X

and x 2 X X E. Since X is locally compact, there is an open set W suchthat x 2 W and W is compact. Replacing W by W X E, we can assume thatW \ E D ¿. Then x … W \ E �

closedX . By Lemma 2.6.3, there exist disjoint


neighborhoods U of x and V of W \ E. Consider the following two opensets:

U \W and V [�X XW

�:

It is easy to see that they are disjoint, x 2 U \W , and E � V [�X XW

�.

Remark 2.7.1. A compact Hausdorff space is normal.

Exercise 91 (2.7.4). Prove that if every point ofX has an open neighbor-hood U such that U is a compact Hausdorff space, thenX is Hausdorff.

Proof.

Chapter 3

Chapter 4

Higher Dimensional Homotopy

4.1 Simplexes and Barycentric Subdivision

Exercise 92 (4.3.6). A subset S of a vector space V is convex if S in-cludes the line segment between any two of its points, that is, if when-ever x; y 2 S and 0 < t < 1, then tx C .1 � t / y 2 S . A convex combi-nation of vectors in a set S is a finite sum of the form

Ptjxj , where

xj 2 S , tj > 0, andPtj D 1. Show that S is convex if and only if any

convex combination of vectors in S again belongs to S .

Proof. It is evident that if any convex combination of vectors in S is againbelongs to S , then S is convex. For the nontrival direction, suppose that Sis convex. The claim holds when there are two vectors. Now suppose thatit is hold for n vectors, that is,

PnjD1 xj 2 S for any xj 2 S , tj > 0, andPn

jD1 tj D 1. Consider the case of nC 1 vectors. Without loss of generality,we assume that tnC1 > 0. Note that

v D

nC1XjD1

tjxj

D

nXjD1

tjxj C tnC1xnC1

D

0@ nXjD1

tj

1A24 t1PnjD1 tj

!x1 C � � � C

tnPnjD1 tj

!xn

35C tnC1xnC1:Since

�t1Pn

j D1 tj

�x1C� � �C

�tnPn

j D1 tj

�xn D u 2 S be the induction hypothesis,

and

v D

0@ nXjD1

tj

1AuC tnC1xnC1D .1 � tnC1/ uC tnC1xnC1;

we know that v 2 S .

58 CHAPTER 4

Exercise 93 (4.3.7). Let S be a subset of a vector space V . Show that theset co .S/ of convex combinations of elements of S is a convex subsetof V , which is the smallest convex subset of V containing S . The setco .S/ is called the convex hull of S .

Proof. Choose any v1; v2 2 co .S/. Then

v1 D

n1XjD1

s1j x1j and v2 D

n2XjD1

s2j x2j :

For any t 2 Œ0; 1�,

tv1 C .1 � t / v2 D t

0@ n1XjD1

s1j x1j

1AC .1 � t /0@ n2XjD1

s2j x2j

1Ais again a convex combination of elements of S since

t

0@ n1XjD1

s1j

1AC .1 � t /0@ n2XjD1

s2j

1A D 1:This proves that co .S/ is convex.

Exercise 94 (4.3.8). Let S be a subset of Rn such that every elementof co .S/ has a unique representation as a convex combination of ele-ments of S . Show that S is an affinely independent set and that co .S/is a simplex with vertices S .

Proof. Suppose that S is not an affinely independent set. Then there existx1; : : : ; xn and t1; : : : ; tn 2 R such that

nXiD1

tixi D 0;

nXiD1

ti D 0; and .t1; : : : ; tn/ ¤ .0; : : : ; 0/ :

Let

T D˚ti j ti > 0

; and T 0 D

˚ti j ti < 0

:

Hence,

0 D

nXiD1

ti DXti2T

ti CXti2T

0

ti

implies thatPti2T

ti DPti2T

0 .�ti /. Now define si DtiP

ti 2T tiif ti 2 T and

s0i D �tiP

ti 2T 0 tiif ti 2 T

0. Finally,PniD1 tixi D 0 implies thatX

ti2T

tixi DXti2T

0

.�ti / xi )X

sixi DX

s0ixi ;

that is, there is an element of co .S/ which has two representations. A con-tradiction.

References

[1] Aliprantis, C. D. and K. C. Border (2006), Infinite Dimensional Anal-ysis: A Hitchhiker’s Guide, Berlin: Springer-Verlag, 3rd edition. [35]

[2] Axler, S. (1997), Linear Algebra Done Right, Undergraduate Texts inMathematics, New York: Springer-Verlag, 2nd edition. [3]

[3] Dudley, R. M. (2002), Real Analysis and Probability, Vol. 74 of Cam-bridge Studies in Advanced Mathematics, New York: Cambridge Uni-versity Press, 2nd edition. [42]

[4] Gamelin, T. W. and R. E. Greene (1999), Introduction to Topology, NewYork: Dover Publications, Inc. 2nd edition. [iii, 23, 46]

[5] Hrbacek, K. and T. Jech (1999), Introduction to Set Theory, Vol. 220of Pure and Applied Mathematics: A Series of Monographs and Text-books, New York: Taylor & Francis Group, LLC, 3rd edition. [17]

[6] Lee, J. M. (2000), Introduction to Topological Manifolds, Vol. 202 ofGraduate Texts in Mathematics, New York: Springer-Verlag. [25, 46]

[7] Rudin, W. (1976), Principles of Mathematical Analysis, New York:McGraw-Hill Companies, Inc. 3rd edition. [9, 13, 18]

[8] Shirali, S. and H. L. Vasudeva (2006), Metric Spaces, London:Springer-Verlag. [5]

[9] Zorich, V. A. (2004), Mathematical Analysis I, Universitext, Berlin:Springer-Verlag. [13]

Real Analysis

Solution Manual for Royden and Fitzpatrick (2010)

Jianfei Shen


Sydney, Australia October 15, 2011

Jianfei Shen

School of Economics

The University of New South Wales

Sydney 2052

Australia

[email protected]

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix

Part I Lebesgue Integration for Functions of A Single Real Variable

1 The Real Numbers: Sets, Sequences, and Functions . . . . . . . . . . . . . . . . 3

1.1 The Field, Positivity, and Completeness Axioms . . . . . . . . . . . . . . . . 3

1.2 The Natural and Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Lebesgue Outer Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 The � -algebra of Lebesgue Measurable Sets . . . . . . . . . . . . . . . . . . . . . 7

2.4 Outer and Inner Approximation of Lebesgue Measurable Sets . . . 8

2.5 Countable Additivity, Continuity, and the Borel-Cantelli Lemma . 12

2.6 Nonmeasurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.7 The Cantor Set and the Cantor-Lebesgue Function . . . . . . . . . . . . . . 14

3 Lebesgue Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.1 Sums, Products, and Compositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.1 Continuity of Monotone Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 The Lp Spaces: Completeness and Approximation . . . . . . . . . . . . . . . . . 21

5.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.2 The Inequalities of Young, Hölder, and Minkowski . . . . . . . . . . . . . . 22

6 The Lp Spaces: Duality and Weak Convergence . . . . . . . . . . . . . . . . . . . . 23

6.1 The Riesz Representation for the Dual of Lp , 1 6 p <1 . . . . . . . . 23

Part II Abstract Spaces: Metric, Topological, Banach, and Hilbert Spaces

v

vi CONTENTS

7 Metric Spaces: General Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7.1 Examples of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7.2 Open Sets, Closed Sets, and Convergent Sequences . . . . . . . . . . . . . 30

7.3 Continuous Mappings between Metric Spaces . . . . . . . . . . . . . . . . . . . 30

7.4 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7.5 Compact Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

7.6 Separable Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

8 Metric Spaces: Three Fundamental Theorems . . . . . . . . . . . . . . . . . . . . . . 37

8.1 The Arzelà-Ascoli Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

9 Topological Spaces: Three Fundamental Theorems . . . . . . . . . . . . . . . . 39

9.1 Urysohn’s Lemma and the Tietze Extension Theorem . . . . . . . . . . . 39

10 Continuous Linear Operators between Banach Spaces . . . . . . . . . . . . . . 41

10.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

10.2 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Preface

Sydney, Jianfei Shen

October 15, 2011

vii

Acknowledgements

ix

Part I

Lebesgue Integration for Functions of ASingle Real Variable

1THE REAL NUMBERS: SETS, SEQUENCES, ANDFUNCTIONS

1.1 The Field, Positivity, and Completeness Axioms

1.2 The Natural and Rational Numbers

I Exercise 1 (1.8). Use an induction argument to show that for each natural

number n, the interval .n; nC 1/ fails to contain any natural number.

Proof. ut

3

2LEBESGUE MEASURE

2.1 Introduction

In the first three problems, let m be a set function defined for all sets in a � -

algebra A with values in Œ0;1�. Assume m is countably additive over countable

disjoint collections of sets in A.

I Exercise 2 (2.1). Prove that if A and B are two sets in A with A � B , then

m.A/ 6 m.B/.

Proof. If m.B/ D1, then the conclusion is trivial. So suppose that m.B/ <1,

and write B D A t .B X A/. Then m.B/ D m.A/Cm.B X A/ > m.A/. (Countable

additivity implies finite additivity; see the following exercise.) ut

I Exercise 3 (2.2). Prove that if there is a set A in the collection A for which

m.A/ <1, then m.¿/ D 0.

Proof. We have m.A/ D m.A[¿[¿[ � � � / D m.A/CPm.¿/. Since m.A/ <1,

we obtainPm.¿/ D 0, i.e., m.¿/ D 0 (since m.¿/ > 0). ut

I Exercise 4 (2.3). Let fEkg1kD1 be a countable collection of sets in A. Prove

that m.S1kD1Ek/ 6

P1kD1m.Ek/.

Proof. Define F1 D E1 and Fn D En X .Sn�1iD1 Ei / for n > 2. Then

S1kD1En DF1

kD1 Fk . It follows from countable additivity and monotonicity that

m

0@ 1[kD1

Ek

1A D m0@ 1GkD1

Fk

1A D 1XkD1

m.Fn/ 61XkD1

m.En/: ut

I Exercise 5 (2.4). A set function c, defined on all subsets of R, is defined as

follows. Define c.E/ to be 1 if E has infinitely many members and c.E/ to be

equal to the number of elements in E if E is finite; define c.¿/ D 0. Show that c

is a countably additive and translation invariant set function.

5

6 CHAPTER 2 LEBESGUE MEASURE

Proof. Let fEkg1kD1 � 2R be a disjoint sequence. If j

F1kD1Ekj D 1, then either

there exists a set En with jEnj D 1 or there are infinite number of nonempty

sets. In both cases, the countable additivity follow immediately. If jF1kD1Ekj <

1, then each Ek is finite and so the countable additivity holds again. To see c is

translation invariant, observe that a translation does not change the cardinality

of a set. ut

2.2 Lebesgue Outer Measure

I Exercise 6 (2.5). By using properties of outer measure, prove that the interval

Œ0; 1� is not countable.

Proof. Suppose that Œ0; 1� is countable. Thenm�.Œ0; 1�/ D 0. However,m�.Œ0; 1�/ D

`.Œ0; 1�/ D 1. A contradiction. ut

I Exercise 7 (2.6). Let A be the set of irrational numbers in the interval Œ0; 1�.

Prove that m�.A/ D 1.

Proof. Let A0 D Œ0; 1� \Q. Then A D Œ0; 1� X A0. Since A0 is countable, we have

m�.A0/ D 0. The finite additivity yields m�.A/ D m�.Œ0; 1�/ �m�.A0/ D `.Œ0; 1�/ D

1. ut

I Exercise 8 (2.7). A set of real numbers is said to be a Gı set provided it is the

intersection of a countable collection of open sets. Show that for any bounded

set E, there is a Gı set G for which

E � G and m�.G/ D m�.E/:

Proof. For each n 2 N, there exists a sequence fInkg1kD1 of nonempty open,

bounded intervals such that E �S1kD1 Ink and

P1kD1 `.Ink/ < m

�.E/C 1=n. Let

On´S1kD1 Ink for each n 2 N. Then On is open, E � On, and

m�.E/ 6 m�.On/ 61XkD1

`.Ink/ < m�.E/C 1=n;

for all n 2 N. Let G DT1nD1 On. Then G is a Gı set, E � G, and for every n 2 N

we have

m�.E/ 6 m�.G/ 6 m�.On/ < m�.E/C 1=n:

Hence, m�.E/ D m�.G/. ut

I Exercise 9 (2.8). Let B be the set of rational numbers in the interval Œ0; 1�,

and let fIkgnkD1 be a finite collection of open intervals that covers B . Prove thatPnkD1m

�.Ik/ > 1.

SECTION 2.3 THE � -ALGEBRA OF LEBESGUE MEASURABLE SETS 7

Proof. If fIkgnkD1 covers Œ0; 1�, then m�.Œ0; 1�/ D `.Œ0; 1�/ D 1 6 m�.SnkD1 Ik/ 6Pn

kD1m�.Ik/. Thus, if

PnkD1m

�.Ik/ < 1, then fIkgnkD1 fails to cover Œ0; 1�. But

then fIkgnkD1 fails to cover B , too. ut

I Exercise 10 (2.9). Prove that if m�.A/ D 0, then m�.A [ B/ D m�.B/.

Proof. Monotonicity implies that m�.B/ 6 m�.A [ B/. To see the inverse in-

clusion relation, observe that m�.A [ B/ 6 m�.A/Cm�.B/ D m�.B/. ut

I Exercise 11 (2.10). Let A and B be bounded sets for which there is an ˛ > 0

such that ja�bj > ˛ for all a 2 A, b 2 B . Prove that m�.A[B/ D m�.A/Cm�.B/.

Proof. It is clear that A \ B D ¿. Then the conclusion follows from finite

additivity. ut

2.3 The � -algebra of Lebesgue Measurable Sets

I Exercise 12 (2.11). Prove that if a � -algebra of subsets of R contains intervals

of the form .a;1/, then it contains all intervals.

Proof. Denote this � -algebra as A. We have

� .�1; a� D .a;1/c 2 A;

� .�1; a/ DS1kD1.�1; a � 1=k� 2 A;

� Œa;1/ D .�1; a/c 2 A;

� .a; b� D�.�1; a� [ .b;1/

�c2 A;

� Œa; b� D�.�1; a/ [ .b;1/

�c2 A;

� Œa; b/ D�.�1; a/ [ Œb;1/

�c2 A;

� .a; b/ D�.�1; a� [ Œb;1/

�c2 A. ut

I Exercise 13 (2.12). Show that every interval is a Borel set.

Proof. Since .a;1/ is open, for all a 2 R, it follows that .a;1/ 2 B. Then by

the previous exercise we know that every interval is a Borel set. ut

I Exercise 14 (2.13). Show that (i) the translate of an F� set is also F� , (ii) the

translate of a Gı set is also Gı , and (iii) the translate of a set of measure zero

also has measure zero.

Proof. (i) Let F be a F� set; that is, F DS1kD1 Fk , where each Fk is closed.

We show that .S1kD1 Fk/C y D

S1kD1.Fk C y/ and each Fk C y is closed (this is

clear), for all y 2 R. If x 2 .S1kD1 Fk/C y, then there exists k 2 N and xk 2 Fk


such that x D xk C y; but then x 2 Fk C y �S1kD1.Fk C y/. Conversely, if

x 2S1kD1.Fk C y/, then there exists k 2 N such that x 2 Fk C y, i.e., x D xk C y

for some xk 2 Fk . But then xk 2S1kD1 Fk and so x 2 .

S1kD1 Fk/C y.

(ii) Similarly.

(iii) Let E be measurable with m.E/ D 0. Since m.E/ D m�.E/ D 0, for every

" > 0, there exists a countable collection of nonempty open, bounded intervals

fIkg1kD1

such thatP1kD1 `.Ik/ < ". Then for every y 2 R we have E C y �S1

kD1.Ik C y/, and `.Ik C y/ D `.Ik/ for all k 2 N. Thus,

m�.E C y/ 6 m�

0@ 1[kD1

ŒIk C y�

1A 61XkD1

`.Ik C y/ D

1XkD1

`.Ik/ < ";

i.e., m�.E C y/ D 0. Then E C y is measurable and so m.E C y/ D 0. ut

I Exercise 15 (2.14). Show that if a set E has positive outer measure, then

there is a bounded subset of E that also has positive outer measure.

Proof. Take an arbitrary sequence of nonempty open, bounded intervals

fIkg1kD1

covering E. Then

0 < m�.E/ D m�

0B@E \24 1[kD1

Ik

351CA 6

1XkD1

m�.E \ Ik/:

But then there must exist k 2 N such that m�.E \ Ik/ > 0. Evidently, E \ Ik is

bounded and E \ Ik � E. ut

I Exercise 16 (2.15). Show that if E has finite measure and " > 0, then E is the

disjoint union of a finite number of measurable sets, each of which has measure

at most ".

Proof. ut

2.4 Outer and Inner Approximation of Lebesgue

Measurable Sets

I Exercise 17 (2.16). Complete the proof of Theorem 11 by showing that mea-

surability is equivalent to (iii) and also equivalent to (iv).

Proof. (E is measurable H) (iii)) If E is measurable, then Ec is measurable.

It follows from (i) that for each " > 0, there exists an open set O containing Ec

for which m�.O XEc/ < ". Then Oc � E, and

m�.E XOc/ D m�.O XEc/ < ":

SECTION 2.4 OUTER AND INNER APPROXIMATION OF LEBESGUE MEASURABLE SETS 9

Let F D Oc . We then get (iii).

((iii) H) (iv)) Assume property (iii) holds for E. For each natural number k,

choose a closed set Fk that is contained in E and for which m�.E X Fk/ < 1=k.

Define F DS1kD1 Fk . Then F is a F� set that contained in E. Moreover, since

E X F � E X Fk for each k, by the monotonicity of outer measure,

m�.E X F / 6 m�.E X Fk/ < 1=k:

Therefore m�.E X F / D 0 and so (iv) holds.

((iv) H) E is measurable) Since F and E X F are both measurable, so

E D F [ .E X F /

is measurable, too. ut

I Exercise 18 (2.17). Show that a set E is measurable if and only if for each

" > 0, there is a closed set F and open set O for which F � E � O and

m�.O X F / < ".

Proof. For the only if part, let E be measurable. It follows from Theorem 11

that there exist an open set O containing E and a closed set F contained in E,

such that m�.O XE/ < "=2 and m�.E X F / < "=2. Then

m�.O X F / D m�.ŒO XE� [ ŒE X F �/ 6 m�.O XE/Cm�.E X F / < ":

We next show the if part. Since for each " > 0 there exists a closed set F and

O for which F � E � O and m�.O X F / < ", monotonicity implies that

m�.E X F / 6 m�.O X F / < ":

Then Theorem 11 means that E is measurable. ut

I Exercise 19 (2.18). Let E have finite outer measure. Show that there is a

Gı set G for which E � G and m�.E/ D m�.G/. Use this to show that E is

measurable if and only if there is an F� set F for which F � E and m.F / D

m�.E/.

Proof. Let E � R with m�.E/ < 1. Then for each n 2 N there exists a se-

quence fInkg1kD1 of open intervals covering E and for which

1XkD1

`.Ink/ < m�.E/C

1

n:

Let On´S1kD1 Ink . Then for every n 2 N, we have

m�.On/ 61XkD1

`.Ink/ < m�.E/C

1

n:


Define G DT1nD1 On. Then G is a Gı set, E � G, and G � On for all n, so

m�.G/ 6 m�.On/ < m�.E/C

1

nI

that is, m�.G/ D m�.E/. The remainder follows from Theorem 11. ut

I Exercise 20 (2.19). Let E have finite outer measure. Show that if E is not

measurable, then there is an open set O containing E that has finite outer mea-

sure and for which m�.O XE/ > m�.O/ �m�.E/.

Proof. It follows from Theorem 11 that E is measurable if and only if for each

" > 0, there is an open set O containing E for which m�.O XE/ < ". Hence,

E is not measurable () .9" > 0/.8 open set O � E/.m�.O XE/ > "/:

Fix such an " > 0. Sincem�.E/ < ", there is a sequence of open intervals fIkg1kD1covering E such that

P1kD1 `.Ik/ < m

�.E/C ". Let O DS1kD1 Ik . Then

m�.O/ 61XkD1

`.Ik/ < m�.E/C "I

that is, m�.O/ <1 and m�.O/ �m�.E/ < ". But then m�.O X E/ > " > m�.O/ �

m�.E/. ut

I Exercise 21 (2.20, Lebesgue). Let E have finite outer measure. Show that E

is measurable if and only if for each open, bounded interval .a; b/,

b � a D m�..a; b/ \E/Cm�..a; b/ XE/:

Proof. The only if part is trivial. So we focus on the if part. Since m�.E/ <1,

for every " > 0 there exists a sequence fIkg1kD1 of open, bounded intervals

which covers E andP1kD1 `.Ik/ < m�.E/C ". Let O D

S1kD1 Ik . Then O X E DS1

kD1.Ik XE/, and so

m�.O XE/ 61XkD1

m�.Ik XE/ D

1XkD1

�`.Ik/ �m

�.Ik \E/�

D

1XkD1

`.Ik/ �

1XkD1

m�.Ik \E/

61XkD1

`.Ik/ �m�

0B@24 1[kD1

Ik

35 \E1CA

D

1XkD1

`.Ik/ �m�.E/

< ":

SECTION 2.4 OUTER AND INNER APPROXIMATION OF LEBESGUE MEASURABLE SETS 11

It follows from Theorem 11 that E is measurable. ut

I Exercise 22 (2.21). Use property (ii) of Theorem 11 as the primitive defini-

tion of a measurable set and prove that the union of two measurable sets is

measurable. Then do the same for property (iv).

Proof. Let E1 and E2 be measurable. Then there exist Gı sets G1 and G2 with

E1 � G1 and E2 � G2, such that

m�.G1 XE1/ D m�.G2 XE2/ D 0:

We first show that G1 [ G2 is Gı . Write G1 DT1kD1 O1k and G2 D

T1nD1 O2n,

where O1k and O2n are open sets for all k; n 2 N. Then

G1 [G2 D

0@ 1\kD1

O1k

1A [0@ 1\nD1

O2n

1A D\k;n

.O1k [O2n/

is a countable intersection of open sets and so is Gı .

We next show that E1 [ E2 is measurable by showing that m�.ŒG1 [ G2� X

ŒE1 [E2�/ D 0. We have

m�.ŒG1 [G2� X ŒE1 [E2�/ D m�.ŒG1 X .E1 [E2/� [ ŒG2 X .E1 [E2/�/

6 m�.G1 X ŒE1 [E2�/Cm�.G2 X ŒE1 [E2�/

6 m�.G1 XE1/Cm�.G2 XE2/

D 0:

The case for property (iv) is similarly. ut

I Exercise 23 (2.22). For any set A, define m��.A/ 2 Œ0;1� by

m��.A/ D inffm�.O/ W O � A;O open.g

How is this set function m�� related to outer measure m�?

Proof. m�� D m� since the union of any covering sequence of open intervals

is an open superset of A. ut

I Exercise 24 (2.23). For any set A, define m��.A/ 2 Œ0;1� by

m��.A/ D supfm�.F / W F � A;F closed.g

How is this set function m�� related to outer measure m�?

Proof. If A is measurable, then

m�.A/ D supfm.F / W F � A;F bounded and closedg:


Conversely, if this equation holds and m�.A/ < 1, then A is measurable. See

Oxtoby (1980, Theorem 3.18). ut

2.5 Countable Additivity, Continuity, and the

Borel-Cantelli Lemma

I Exercise 25 (2.24). Show that if E1 and E2 are measurable, then m.E1 [

E2/Cm.E1 \E2/ D m.E1/Cm.E2/.

Proof. If m.E1/ D 1 or m.E2/ D 1 or both, then m.E1 [ E2/ D 1 and so the

equality holds. Hence we assume that both m.E1/ and m.E2/ are finite. Then

m.E1 [E2/ D m.E1 [ ŒE2 X .E1 \E2/�/ D m.E1/Cm.E2 X ŒE1 \E2�/

D m.E1/Cm.E2/ �m.E1 \E2/:

Since m.E1 \E2/ 6 m.E1/ <1, we get the equality. ut

I Exercise 26 (2.25). Show that the assumption that m.B1/ < 1 is necessary

in part (ii) of the theorem regarding continuity of measure.

Proof. Consider the sequence f.n;1/g1nD1. This is a decreasing sequence with

limit ¿; however, m..n;1// D 1 for all n, and so limn!1m..n;1// D 1 ¤

m.T1nD1.n;1// D 0. ut

I Exercise 27 (2.26). Let fEkg1kD1 be a countable disjoint collection of measur-

able sets. Prove that for any set A,

m�

0@A \ 1[kD1

Ek

1A D 1XkD1

m�.A \Ek/:

Proof. Countable subadditivity implies that

m�

0B@A \24 1[kD1

Ek

351CA 6

1XkD1

m�.A \Ek/:

It follows from Proposition 6 (p. 36) that for every n 2 N we have

m�

0B@A \24 n[kD1

Ek

351CA D nX

kD1

m�.A \Ek/:

Therefore,

SECTION 2.6 NONMEASURABLE SETS 13

m�

0B@A \24 1[kD1

Ek

351CA > m�

0B@A \24 n[kD1

Ek

351CA D nX

kD1

m�.A \Ek/

holds for all n; that is,

m�

0B@A \24 1[kD1

Ek

351CA >

1XkD1

m�.A \Ek/:

We thus obtain the desired result. ut

I Exercise 28 (2.27). Easy.

I Exercise 29 (2.28). Show that continuity of measure together with finite

additivity of measure implies countable additivity of measure.

Proof. Let fEkg1kD1 be a disjoint sequence. Define Fn DSnkD1Ek . Then fFng is

increasing, andS1nD1 Fn D

S1kD1. Hence

m

0@ 1[kD1

Ek

1A D m0@ 1[nD1

Fn

1A D limn!1

m.Fn/ D limn!1

m

0@ n[kD1

Ek

1AD limn!1

nXkD1

m.Ek/

D

1XkD1

m.Ek/: ut

2.6 Nonmeasurable Sets

I Exercise 30 (2.29). a. Show that rational equivalence defines an equivalence

relation on any set.

b. Explicitly find a choice set for the rational equivalence relation on Q.

c. Define two numbers to be irrationally equivalent provided their difference is

irrational. Is this an equivalence relation on R? Is this an equivalence relation

on Q?

Proof. (a) Take an arbitrary set E � R. Denote the rational equivalence �.

Then � is reflexive: for every x 2 E we have x � x since x � x D 0 2 Q; �

is symmetric: if x � y, i.e., x � y 2 Q, then y � x D �.x � y/ 2 Q, and so

y � x. Finally, � is transitive: if x � y � z, then x � y 2 Q and y � z 2 Q; then

x � z D .x � y/C .y � z/ 2 Q, i.e., x � z.

(b) Take an arbitrary q 2 Q. For every y 2 Q, we have q � y 2 Q, i.e., q � y.

Thus, Q= �D Œq� for any q 2 Q, and so CQ D fqg.


(c) Observe that for an arbitrary r 2 R we have r � r D 0 2 Q; hence, the

irrational equivalence is not an equivalence relation on R. Similarly, it is not an

equivalence relation on Q. ut

I Exercise 31 (2.30). Show that any choice set for the rational equivalence

relation on a set of positive outer measure must be uncountably infinite.

Proof. If note, then CE is measurable. Contradicts to Vitali’s Theorem. ut

I Exercise 32 (2.31). Justify the assertion in the proof of Vitali’s Theorem that

it suffices to consider the case that E is bounded.

Proof. Let E be unbounded. Then E can be write as a disjoint countable union

of bounded sets fEkg1kD1. Sincem�.E/ > 0, countable subadditivity implies that

0 < m�.E/ 61XkD1

m�.Ek/I

that is, there exists Ek such that m�.Ek/ > 0. For this Ek , we apply Vitali’s

Theorem. ut

I Exercise 33 (2.32). Does Lemma 16 remain true if � is allowed to be finite or

to be uncountably infinite? Does it remain true if � is allowed to be unbounded?

Proof. Yes, No, and No. ut

I Exercise 34 (2.33). Let E be a nonmeasurable set of finite outer measure.

Show that there is a Gı set G that contains E for which m�.E/ D m�.G/, while

m�.G XE/ > 0.

Proof. It follows from Theorem 2.11 that E is measurable if and only if there

exists a Gı set G containing E for which m�.G X E/ D 0. Therefor, E is non-

measurable if and only if for all Gı set G � E, we have m�.G XE/ > 0.

Since m�.E/ < 1, it follows from Exercise 19 that there is a Gı set G con-

taining and m�.E/ D m�.G/. The argument in the previous paragraph also

means that m�.G XE/ > 0. ut

2.7 The Cantor Set and the Cantor-Lebesgue Function

I Exercise 35 (2.34). Show that there is a continuous, strictly increasing func-

tion on the interval Œ0; 1� that maps a set of positive measure onto a set of mea-

sure zero.

Proof. Define z.x/ D .x/=2 D Œ'.x/Cx�=2 for x 2 Œ0; 1�. Then z W Œ0; 1�! Œ0; 1�

is continuous, strictly increasing, and maps the Cantor set C onto a measurable

set W of positive measure. Let

SECTION 2.7 THE CANTOR SET AND THE CANTOR-LEBESGUE FUNCTION 15

D z�1:

Then is continuous, strictly increasing, and maps W onto C. ut

I Exercise 36 (2.35). Let f be an increasing function on the open interval I .

For x0 2 I show that f is continuous at x0 if and only if there are sequences fang

and fbng in I such that for each n, an < x0 < bn, and limn!1Œf .bn/�f .an/� D 0.

Proof. The only if part is clear, so we focus on the if part. Suppose that there

are sequences fang and fbng in I such that an < x0 < bn for each n, and

limn!1Œf .bn/ � f .an/� D 0. If f is not continuous at x0, then f .xC0 / > f .x�0 /

(since f is increasing); that is, we can find such sequences fang and fbng. A

contradiction. ut

I Exercise 37 (3.36). Show that if f is any increasing function on Œ0; 1� that

agrees with the Cantor-Lebesgue function ' on the complement of the Cantor

set, then f D ' on all of Œ0; 1�.

Proof. Suppose that f .x/ D '.x/ for all x 2 O. Take an arbitrary x 2 C. As in

the proof of Proposition 2.20, for each k 2 N there exists ak ; bk 2 Ok such that

ak < x < bk and '.bk/ � '.ak/ < 1=2k :

But then f .x/ D '.x/. ut

I Exercise 38 (2.45). Show that a strictly increasing function that is defined on

an interval has a continuous inverse.

Proof. Let f be strictly increasing. Then f is injective, and so its inverse

function f �1 exists. Then the proof is straightforward. ut

I Exercise 39 (2.46). Let f be a continuous function and B be a Borel set. Show

that f �1.B/ is a Borel set.

Proof. ut

3LEBESGUE MEASURABLE FUNCTIONS

3.1 Sums, Products, and Compositions

I Exercise 40 (1.8). Use an induction argument to show that for each natural

number n, the interval .n; nC 1/ fails to contain any natural number.

Proof. ut

17

4DIFFERENTIATION AND INTEGRATION

4.1 Continuity of Monotone Functions

I Exercise 41 (6.1). Let C be a countable subset of the nondegenerate closed,

bounded interval Œa; b�. Show that there is an increasing function on Œa; b� that

is continuous only at points in Œa; b� X C .

Proof. Consider Proposition 6.2. If a; b 2 C , then there is nothing to prove. If

a … C , let f .c/ D 0; if b … C , let f .d/ D 1. ut

I Exercise 42 (6.2). Show that there is a strictly increasing function on Œ0; 1�

that is continuous only at the irrational numbers in Œ0; 1�.

Proof. Let C ´ Œ0; 1� \Q. Enumerate C as fqng1nD1. Define f W Œ0; 1�! Œ0; 1� by

letting

f .x/ DX

fnWqn6xg

1

2nfor all x 2 Œ0; 1�:

Define g W Œ0; 1�! Œ0; 1� by g.x/ D f .x/C x. ut

I Exercise 43 (6.3). Let f be a monotone function on a subset E of R. Show

that f is continuous except possibly at a countable number of points in E.

Proof. This holds even if E D R; see Chung (2001). ut

19

5THE LP SPACES: COMPLETENESS ANDAPPROXIMATION

5.1 Normed Linear Spaces

I Exercise 44 (7.1). For f in C Œa; b�, define kf k1 DR bajf j. Show that this is a

norm on C Œa; b�. Also show that there is no number c > 0 for which

kf kmax 6 ckf k1 for all f in C Œa; b�;

but there is a c > 0 for which

kf k1 6 ckf kmax for all f in C Œa; b�:

Proof. kf k1 is a norm on C Œa; b� since C Œa; b� � L1Œa; b�. We first show that for

every c > 0, there exists f 2 C Œa; b� such that kf kmax > ckf k1. Pick d 2 .a; b/.

Define f W Œa; b�! Œ0; b � a� by letting

fd .x/ D

˚b�dd�a

x if x 2 Œa; d �

.b � d/C d�ab�a

x if x 2 .d; b�:

(See Figure 5.1.) This function is continuous on Œa; b� and bounded. Observe

that when d ! b, we have

c

Z b

a

jfd j ! 0:

To see that there exists c > 0 such that ckf kmax > kf k1 for all f 2 C Œa; b�,

notice that for all f 2 C Œa; b�, we haveZ b

a

kf kmax >Z b

a

jf j () .b � a/kf kmax > kf k1I

hence, by letting c D b � a, we obtain the desired result. ut

I Exercise 45 (7.2). Let X be the family of all polynomials with real coefficients

defined on R. Show that this is a linear space. For a polynomial p, define kpk to

be the sum of the absolute values of the coefficients of p. Is this a norm?

21

22 CHAPTER 5 THE LP SPACES: COMPLETENESS AND APPROXIMATION

0a bd

Figure 5.1. fd .

Proof. X is a linear space, and kpk is a norm. ut

I Exercise 46 (7.3). For f in L1Œa; b�, define kf k DR bax2jf .x/jdx. Show that

this is a norm on L1Œa; b�.

Proof. It suffices to show the triangle inequality. For f; g 2 L1Œa; b�,

kf C gk D

Z b

a

x2jf .x/C g.x/jdx 6Z b

a

x2�jf .x/j C jg.x/j

�D kf k C kgk: ut

I Exercise 47 (7.4). For f 2 L1Œa; b�, show that

kf k1 D inf˚M W mfx 2 Œa; b� W jf .x/j > M g D 0

and if, furthermore, f is continuous on Œa; b�, then kf k1 D kf kmax.

Proof. Trivial. ut

I Exercise 48 (7.5). Show that `1 and `1 are normed linear spaces.

Proof. Trivial. ut

5.2 The Inequalities of Young, Hölder, and Minkowski

6THE LP SPACES: DUALITY AND WEAK CONVERGENCE

6.1 The Riesz Representation for the Dual of Lp,

1 6 p < 1

Remark (Proposition 2, p. 157). For p D 1, argue by contradiction. If kgk1 >

kTk�, there is a set E0 of finite positive measure on which jgj > kTk�, and one

gets a contradiction by choosing f to be Œ1=m.E0/� � sgn.g/ � 1E0:Z

E

jf j D

ZE

ˇ̌̌̌1

m.E0/� sgn.g/ � 1E0

ˇ̌̌̌D

1

m.E0/

ZE

1E0D 1I

that is, f 2 L1.E/. But

T.f / D

ZE

g �1

m.E0/� sgn.g/ � 1E0

D1

m.E0/

ZE

jgj1E0

>1

m.E0/

ZE

kTk�1E0

D kTk�:

I Exercise 49 (8.1). Verify (8):

kTk� D sup˚jT.f /j W f 2 X; kf k 6 1

: (8)

Proof. By definition,

kTk� D inf˚0 6 M <1 W jT.f /j 6 Mkf k for all f 2 X

:

We first show that

kTk� D inf˚0 6 M <1 W jT.f /j 6 M for all f 2 X with kf k D 1

(6.1)

D sup˚jT.f /j W f 2 X with kf k D 1

: (6.2)

Define

23

24 CHAPTER 6 THE LP SPACES: DUALITY AND WEAK CONVERGENCE

M1 D f0 6 M <1 W jT.f /j 6 Mkf k for all f 2 Xg;

M2 D f0 6 M <1 W jT.f /j 6 M for all f 2 X with kf k D 1g:

It is evident that if M 2 M1, then M 2 M2. So suppose that M 2 M2. Pick

f 2 X . If kf k D 0, then f D 0 and so T .f / D 0, hence we trivially have jT.f /j 6Mkf k. If kf k > 0, then, writing g D f=kf k, we have kgk D kf=kf kk D 1 and

hence jT.g/j 6 M . Using linearity, we have

jT.g/j 6 M ()

ˇ̌̌̌T

�f

kf k

�ˇ̌̌̌6 M ()

ˇ̌̌̌1

kf kT.f /

ˇ̌̌̌6 M

() jT.f /j 6 Mkf kI

that is, M 2 M1. We thus proved that M1 D M2 and so (6.1) holds. The equiv-

alence of (6.1) and (6.2) is clear.

Now define

M3 D˚jT.f / W f 2 X; kf k 6 1

;

M4 D˚jT.f / W f 2 X; kf k D 1

:

Since M4 �M3, we have sup M4 6 sup M3. Next, for all f 2 X with 0 < kf k 61, we have (by (6.2))

kTk� >ˇ̌̌̌T

�f

kf k

�ˇ̌̌̌DjT.f /j

kf k> jT.f /j:

(We trivially have kTk� > jT.f /j when kf k D 0.) It follows that

kTk� > sup M3:

Hence, sup M3 D sup M4, and so (8) holds. ut

I Exercise 50 (8.2). Let X be a normed linear space. Then the collection of

bounded linear functionals on X is a linear space on which k � k� is a norm.

Proof. Denote this space as L.X/. Let S;T 2 L.X/. Then

kSC Tk� D sup˚j.SC T/j.f / W f 2 X; kf k D 1

D sup

˚jS.f /C T.f /j W f 2 X; kf k D 1

6 sup

˚jS.f /j C jT.f /j W f 2 X; kf k D 1

6 sup

˚jS.f /j W f 2 X; kf k D 1

C sup

˚jT.f /j W f 2 X; kf k D 1

D kSk� C kTk�:

Others are easy. ut

Part II

Abstract Spaces: Metric, Topological,Banach, and Hilbert Spaces

7METRIC SPACES: GENERAL PROPERTIES

7.1 Examples of Metric Spaces

I Exercise 51 (9.1). Show that tow metrics � and � on the same set X are

equivalent if and only if there is c > 0 such that for all u; v 2 X ,

1

c�.u; v/ 6 �.u; v/ 6 c�.u; v/:

Proof. The “if” part is clear: let c1 D 1=c and c2 D c. So assume that � and �

are equivalent; that is, there are positive numbers c1 and c2 such that for all

u; v 2 X ,

c1�.u; v/ 6 � 6 c2�.u; v/:

First suppose that c1 > 1. Then 1=c2 6 1=c1 6 c1; hence,

1

c2�.u; v/ 6 c1�.u; v/ 6 �.u; v/ 6 c2�.u; v/:

Let c D c2 and we are done. Next suppose that 0 < c2 6 1. Then c2 6 1=c2 61=c1; hence

c1�.u; v/ 6 �.u; v/ 6 c2�.u; v/ 61

c1�.u; v/:

Let c D 1=c1 and the proof is complete. ut

I Exercise 52 (9.2). Show that the following define equivalent metrics on Rn:

��.x;y/ D jx1 � y1j C � � � C jxn � ynjI

�C.x;y/ D max˚jx1 � y1j; : : : ; jxn � ynj

:

Proof. For all x;y 2 Rn, we have ��.x;y/ 6 n�C.x;y/. By letting n be large

enough, we obtain1

n�C.x;y/ 6 ��.x;y/ 6 n�C.x;y/:

It then follows from Exercise 51 that �� and �C are equivalent on Rn. ut

27

28 CHAPTER 7 METRIC SPACES: GENERAL PROPERTIES

I Exercise 53 (9.3). Find a metric on Rn that fails to be equivalent to either of

those defined in the preceding problem.

Proof. Consider the discrete metric. ut

I Exercise 54 (9.4). For a closed, bounded interval Œa; b�, consider the set

X D C Œa; b� of continuous real-valued functions on Œa; b�. Show that the met-

ric induced by the maximum norm and that induced by the L1Œa; b� norm are

not equivalent.

Proof. It follows from Exercise 44 that there is no c > 0 for which

kf kmax 6 ckf k1 for all f 2 C Œa; b�:

This completes the proof. ut

I Exercise 55 (9.5, The Nikodym Metric). Let E be a Lebesgue measurable set

of real numbers of finite measure, X the set of Lebesgue measurable subsets

of E, and m Lebesgue measure. For A;B 2 X , define �.A;B/ D m.A�B/. Show

that this is a pseudometric on X . Define two measurable sets to be equivalent

provided their symmetric difference has measure zero. Show that � induces a

metric on the collection of equivalence classes. Finally, show that for A;B 2 X ,

�.A;B/ D

ZE

j1A � 1B j:

Proof. It suffices to show the triangle inequality. Let A;B;C 2 X . ut

I Exercise 56 (9.6). Show that for a; b; c > 0,

if a 6 b C c, thena

1C a6

b

1C bC

c

1C c:

Proof. If a D 0 or b D 0 or c D 0, then the inequality holds trivially. So we

assume that a > 0, b > 0 and c > 0. First suppose that b > c. Then a 6 b C c 62b, i.e., b > a=2. We have

b

1C bC

c

1C cD

1

1C 1=bC

1

1C 1=c>

2

1C 1=b>

2

1C 2=aD

a

1C a=2>

a

1C a:

Similarly for b 6 c. ut

I Exercise 57 (9.7). Let E be a Lebesgue measurable set of real numbers that

has finite measure and X the set of Lebesgue measurable real-valued functions

on E. For f; g 2 X , define

�.f; g/ D

ZE

jf � gj

1C jf � gj:

SECTION 7.1 EXAMPLES OF METRIC SPACES 29

Use the preceding problem to show that this is a pseudometric on X . Define two

measurable functions to be equivalent provided they are equal a.e. on E. Show

that � induces a metric on the collection of equivalence classes.

Proof. It suffices to show the triangle inequality. Let f; g; h 2 X . Then jf �hj 6jf � gj C jg � hj. Hence,

�.f; h/ D

ZE

jf � hj

1C jf � hj6ZE

�jf � gj

1C jf � gjCjg � hj

1C jg � hj

�D �.f; g/C �.g; h/:

Now define an equivalence relation � on X by letting f � g iff f D g a.e. on

E. Define z� on X= � as follows:

z��Œf �; Œg�

�D �.f; g/:

It is evident that z� is a metric on X= � (where the triangle inequality follows

from �). ut

I Exercise 58 (9.8). For 0 < p < 1, show that .aCb/p 6 apCbp for all a; b > 0.

Proof. Without loss of generality, we assume that a; b > 0. If a D b, we have

.aC b/p D 2pap < 2ap D ap C bp:

Next we assume that a < b. Since 0 < p < 1, the function xp defined on .0;1/

is concave. Write b as a convex combination of a and aC b as follows:

b Da

baC

b � a

b.aC b/:

Then

bp D

�a

baC

b � a

b.aC b/

�p6a

bap C

b � a

b.aC b/pI

that is,

.aC b/p 6bpC1 � apC1

b � a6.b � a/.ap C bp/

b � aD ap C bp;

where the second inequality holds since

.b � a/.ap C bp/ D bpC1 � apC1 C ab.ap�1 � bp�1/ > bpC1 � apC1: ut


I Exercise 60 (9.10). Let f.Xn; �n/g1nD1 be a countable collection of metric

spaces. Use Exercise 56 to show that �� defines a metric on the Cartesian product

�1nD1Xn, where for points x D .xn/ and y D .yn/ in�1nD1Xn,

��.x;y/ D

1XnD1

1

2n�

�n.xn; yn/

1C �n.xn; yn/:

Proof. First observe that �� is well-defined since for all n 2 N,

30 CHAPTER 7 METRIC SPACES: GENERAL PROPERTIES

1

2n�

�n.xn; yn/

1C �n.yn; zn/<1

2n;

so the series defining �� converges. We only prove the triangle inequality. Let

x;y; z 2�1nD1Xn. For every n 2 N, we have �n.xn; zn/ 6 �n.xn; yn/C �n.yn; zn/.

Hence,

��.x; z/ D

1XnD1

1

2n�

�n.xn; zn/

1C �n.xn; zn/

61XnD1

1

2n�

��n.xn; yn/

1C �n.yn; zn/C

�n.yn; zn/

1C �n.yn; zn/

�D ��.x;y/C ��.y; z/: ut


I Exercise 62 (9.12). Show that the triangle inequality for Euclidean space Rn

follows from the triangle inequality for L2Œ0; 1�.

Proof. ut

7.2 Open Sets, Closed Sets, and Convergent Sequences

7.3 Continuous Mappings between Metric Spaces

I Exercise 63 (9.36). Let X D C Œa; b�. Define the function W X ! R by

.f / D

Z b

a

f .x/dx for each f 2 X:

Show that is Lipschitz on the metric space X , where X has the metric induced

by the maximum norm.

Proof. Take arbitrary f; g 2 X . Then

j .f / � .g/j D

ˇ̌̌̌ˇZ b

a

�f .x/ � g.x/

�dx

ˇ̌̌̌ˇ

6Z b

a

jf .x/ � g.x/jdx

6Z b

a

maxx2Œa;b�

˚jf .x/ � g.x/j

dx

D .b � a/kf � gk:

Hence, is Lipschitz. ut

compactness - weebly

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