reading notes of “real analysis” 3rd edition by h. l....

Reading Notes of Real Analysis 3rd Edition

by H. L. Royden

Zigang Pan

October 22, 2017

Preface

This is a reading note of the book Royden (1988) and the MATH 441& 442 notes by Prof. Peter Leob of University of Illinois at Urbana-Champaign. In Chapter 6, I have also included some material from thebook Maunder (1996). The Chapters 610 include a significant amountof material from the book Luenberger (1969). Chapter 9 also referencedBartle (1976). Chapter 11 includes significant amount of self-developed ma-terial due to lack of reference on this subject. The proof of Radon-NikodymTheorem 11.169 was adapted from the MATH 442 notes by Prof. Peck ofUniversity of Illinois at Urbana-Champaign. The book Royden (1988) doesoffer some clues as to how to invent the wheel and the book Bartle (1976)is sometimes used to validate the result. The book Spivak (1965) providesthe goal of the Fundamental Theorem of Calculus that Chapter 12 is toprove. But the result in Spivak (1965) may not be correct to begin with.Now, Chapter 12 is complete.A,B, C,D, E ,F ,G,H, I,J ,K,L,M,N ,O,P ,Q,R,S, T ,U ,V ,W ,X ,Y,ZA,B,C,D,E,F,G,H, I, J,K,L,M,N,O,P,Q,R, S,T,U,V,W,X,Y,ZA,B,C,D,E,F,G,H, I, J,K,L,M,N,O,P,Q,R, S,T,U,V,W,X,Y,Z

3

Contents

Preface 3

1 Notations 9

2 Set Theory 192.1 Axiomatic Foundations of Set Theory . . . . . . . . . . . . 192.2 Relations and Equivalence . . . . . . . . . . . . . . . . . . . 202.3 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . 222.5 Algebra of Sets . . . . . . . . . . . . . . . . . . . . . . . . . 232.6 Partial Ordering and Total Ordering . . . . . . . . . . . . . 242.7 Basic Principles . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Topological Spaces 353.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 353.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Basis and Countability . . . . . . . . . . . . . . . . . . . . . 413.4 Products of Topological Spaces . . . . . . . . . . . . . . . . 423.5 The Separation Axioms . . . . . . . . . . . . . . . . . . . . 473.6 Category Theory . . . . . . . . . . . . . . . . . . . . . . . . 483.7 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 503.8 Continuous Real-Valued Functions . . . . . . . . . . . . . . 553.9 Nets and Convergence . . . . . . . . . . . . . . . . . . . . . 59

4 Metric Spaces 734.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 734.2 Convergence and Completeness . . . . . . . . . . . . . . . . 754.3 Uniform Continuity and Uniformity . . . . . . . . . . . . . 784.4 Product Metric Spaces . . . . . . . . . . . . . . . . . . . . . 804.5 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.6 Baire Category . . . . . . . . . . . . . . . . . . . . . . . . . 864.7 Completion of Metric Spaces . . . . . . . . . . . . . . . . . 874.8 Metrization of Topological Spaces . . . . . . . . . . . . . . . 924.9 Interchange Limits . . . . . . . . . . . . . . . . . . . . . . . 93

5

6 CONTENTS

5 Compact and Locally Compact Spaces 97

5.1 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . 975.2 Countable and Sequential Compactness . . . . . . . . . . . 103

5.3 Real-Valued Functions and Compactness . . . . . . . . . . . 1055.4 Compactness in Metric Spaces . . . . . . . . . . . . . . . . 107

5.5 The Ascoli-Arzela Theorem . . . . . . . . . . . . . . . . . . 1095.6 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.7 Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . 1145.7.1 Fundamental notion . . . . . . . . . . . . . . . . . . 114

5.7.2 Partition of unity . . . . . . . . . . . . . . . . . . . . 117

5.7.3 The Alexandroff one-point compactification . . . . . 1205.7.4 Proper functions . . . . . . . . . . . . . . . . . . . . 121

5.8 -Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . 1235.9 Paracompact Spaces . . . . . . . . . . . . . . . . . . . . . . 124

5.10 The Stone-Cech Compactification . . . . . . . . . . . . . . . 128

6 Vector Spaces 131

6.1 Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1316.2 Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

6.3 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1346.4 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 134

6.5 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1366.6 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

6.7 Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.8 Linear Independence and Dimensions . . . . . . . . . . . . . 144

7 Banach Spaces 1477.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . 147

7.2 The Natural Metric . . . . . . . . . . . . . . . . . . . . . . . 1537.3 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 156

7.4 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1587.5 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . 163

7.6 Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . 1657.7 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . 167

7.8 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . 175

7.9 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 1797.9.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . 179

7.9.2 Duals of some common Banach spaces . . . . . . . . 1807.9.3 Extension form of Hahn-Banach Theorem . . . . . . 184

7.9.4 Second dual space . . . . . . . . . . . . . . . . . . . 1927.9.5 Alignment and orthogonal complements . . . . . . . 194

7.10 The Open Mapping Theorem . . . . . . . . . . . . . . . . . 1997.11 The Adjoints of Linear Operators . . . . . . . . . . . . . . . 203

7.12 Weak Topology . . . . . . . . . . . . . . . . . . . . . . . . . 206

CONTENTS 7

8 Global Theory of Optimization 217

8.1 Hyperplanes and Convex Sets . . . . . . . . . . . . . . . . . 217

8.2 Geometric Form of Hahn-Banach Theorem . . . . . . . . . 220

8.3 Duality in Minimum Norm Problems . . . . . . . . . . . . . 222

8.4 Convex and Concave Functionals . . . . . . . . . . . . . . . 225

8.5 Conjugate Convex Functionals . . . . . . . . . . . . . . . . 229

8.6 Fenchel Duality Theorem . . . . . . . . . . . . . . . . . . . 236

8.7 Positive Cones and Convex Mappings . . . . . . . . . . . . 243

8.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . 245

9 Differentiation in Banach Spaces 253

9.1 Fundamental Notion . . . . . . . . . . . . . . . . . . . . . . 253

9.2 The Derivatives of Some Common Functions . . . . . . . . 257

9.3 Chain Rule and Mean Value Theorem . . . . . . . . . . . . 261

9.4 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . 267

9.4.1 Basic concept . . . . . . . . . . . . . . . . . . . . . . 267

9.4.2 Interchange order of differentiation . . . . . . . . . . 273

9.4.3 High order derivatives of some common functions . . 278

9.4.4 Properties of high order derivatives . . . . . . . . . . 282

9.5 Mapping Theorems . . . . . . . . . . . . . . . . . . . . . . . 288

9.6 Global Inverse Function Theorem . . . . . . . . . . . . . . . 300

9.7 Interchange Differentiation and Limit . . . . . . . . . . . . 307

9.8 Tensor Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 311

9.9 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . 313

10 Local Theory of Optimization 325

10.1 Basic Notion . . . . . . . . . . . . . . . . . . . . . . . . . . 325

10.2 Unconstrained Optimization . . . . . . . . . . . . . . . . . . 331

10.3 Optimization with Equality Constraints . . . . . . . . . . . 334

10.4 Inequality Constraints . . . . . . . . . . . . . . . . . . . . . 339

11 General Measure and Integration 349

11.1 Measure Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 349

11.2 Outer Measure and the Extension Theorem . . . . . . . . . 355

11.3 Measurable Functions . . . . . . . . . . . . . . . . . . . . . 368

11.4 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

11.5 General Convergence Theorems . . . . . . . . . . . . . . . . 398

11.6 Banach Space Valued Measures . . . . . . . . . . . . . . . . 414

11.7 Calculation With Measures . . . . . . . . . . . . . . . . . . 446

11.8 The Radon-Nikodym Theorem . . . . . . . . . . . . . . . . 477

11.9 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493

11.10Dual of C(X ,Y) . . . . . . . . . . . . . . . . . . . . . . . . . 509

8 CONTENTS

12 Differentiation and Integration 53312.1 Caratheodory Extension Theorem . . . . . . . . . . . . . . 53312.2 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . 53912.3 Product Measure . . . . . . . . . . . . . . . . . . . . . . . . 54412.4 Functions of Bounded Variation . . . . . . . . . . . . . . . . 57112.5 Absolute and Lipschitz Continuity . . . . . . . . . . . . . . 59512.6 Fundamental Theorem of Calculus . . . . . . . . . . . . . . 61212.7 Representation of (Ck(,Y)) . . . . . . . . . . . . . . . . . 641

13 Hilbert Spaces 64313.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 64313.2 Projection Theorems . . . . . . . . . . . . . . . . . . . . . . 64713.3 Dual of Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . 64913.4 Hermitian Adjoints . . . . . . . . . . . . . . . . . . . . . . . 65213.5 Approximation in Hilbert Spaces . . . . . . . . . . . . . . . 65313.6 Other Minimum Norm Problems . . . . . . . . . . . . . . . 663

14 Probability Theory 66914.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 669

15 Numerical Methods 67115.1 Newtons Method . . . . . . . . . . . . . . . . . . . . . . . . 671

A Elements in Calculus 675A.1 Some Formulas . . . . . . . . . . . . . . . . . . . . . . . . . 675A.2 Convergence of Infinite Sequences . . . . . . . . . . . . . . . 675

Bibliography 677

Index 678

Chapter 1

Notations

IN, Z, and Q the sets of natural number, integers, and rationalnumbers, respectively

IR and C the sets of real numbers and complex numbers, re-spectively

IK either IR or CZ+, Z IN {0}, Z \ IN, respectivelyIR+, IR, C+, C (0,) IR, (, 0) IR, the open right half of

the complex plane, the open left half of the complexplane, respectively

belong to6 not belong to contained in contains strict subset of strict super set of for all exists! exists a unique because therefore such that(xn )

n=1 the sequence x1, x2, . . .

(x ) the ordered collectionidA the identity map on a set A| | the absolute value of a real or complex number the complex conjugate of a complex number Re () the real part of a complex number Im () the imaginary part of a complex number a b the maximum of two real numbers a and ba b the minimum of two real numbers a and b

9

10 CHAPTER 1. NOTATIONS

the empty set; See Page 19.{x, y} an unordered pair; See Page 19.X2 the collection of all subsets of X ; See Page 19. the set union; See Page 19.(x, y) an ordered pair; See Page 20.X Y the Cartesian or direct product of sets X and Y ;

See Page 20.A = {x B | P (x)} Definition of a set A; See Page 20.x y x and y are related in a relation; See Page 20.X/ the quotient of the set X with respect to an equiv-

alence relation ; See Page 20.f : X Y a function ofX to Y ; {(x, f(x)) XY | x X }

is the graph of f ; See Page 21.graph (f ) the graph of a function f ; See Page 21.dom(f ) the domain of f ; See Page 21.f(A) the image of A X under f ; See Page 21.range (f ) the range of f , equals to f(X); See Page 21.f inv(A) the preimage of A Y under f ; See Page 21.onto, surjective f(X) = Y ; See Page 21.1-1, injective f(x1) 6= f(x2) if x1, x2 X and x1 6= x2; See Page

21.bijective both surjective and injective; See Page 21.f inv the inverse function of f ; See Page 21.g f the composition of g : Y Z with f : X Y ; See

Page 21.f |A the restriction of f to A; See Page 21.Y X the set of all functions of X to Y ; See Page 21. the set intersection; See Page 22.A the compliment of a set A, where the whole set is

clear from context; See Page 22.\ set minus; See Page 22.A B the symmetric difference of A and B, equals to (A \

B) (B \A); See Page 22.card (X) the number of elements in the finite set X ; See Page

23.X the Cartesian or direct product of Xs; See Page

33.(x) the projection of an element in a Cartesian product

space to one of the coordinates; See Page 33.A the closure of a set A, where the whole set is clear

from context; See Page 35.A the interior of a set A, where the whole set is clear

from context; See Page 35.A the boundary of a set A, where the whole set is clear

from context; See Page 35.

11

(X,O) the product topological space; See Page 42.

T1 Tychonoff space; See Page 47.T2 Hausdorff space; See Page 47.T3 regular space; See Page 47.T4 normal space; See Page 47.T3 12 completely regular space; See Page 58.

(x )A a net; See Page 59.limA x the limit of a net; See Page 59.limxx0 f(x) the limit of f(x) as x x0; See Page 63.IRe the set of extended real numbers, which equals to

IR {,+} ; See Page 66.lim supA x the limit superior of a real-valued net; See Page 67.lim infA x the limit inferior of a real-valued net; See Page 67.lim supxx0 f(x) the limit superior of f(x) as x x0; See Page 68.lim infxx0 f(x) the limit inferior of f(x) as x x0; See Page 68.BX (x0, r ) the open ball centered at x0 with radius r; See Page

73.BX (x0, r ) the closed ball centered at x0 with radius r; See

Page 73.dist(x0, S) the distance from a point x0 to a set S in a metric

space; See Page 75.(X, X) (Y, Y ) the finite product metric space; See Page 80.(i=1Xi, ) the countably infinite product metric space; See

Page 83.supp f the support of a function; See Page 117.(X ) the Stone-Cech compatification of a completely reg-

ular topological space X ; See Page 128.(M(A,Y),F) the vector space of Y-valued functions of a set A

over the field F ; See Page 136.X the null vector of a vector space X ; See Page 138.N (A ) the null space of a linear operator A; See Page 138.R (A ) the range space of a linear operatorA; See Page 138.S the scalar multiplication by of a set S in a vector

space; See Page 139.S + T the sum of two sets S and T in a vector space; See

Page 139.span (A ) the subspace generated by the set A; See Page 140.v (P ) the linear variety generated by a nonempty set P ;

See Page 140.co (S ) the convex hull generated by S in a vector space;

See Page 141.x the norm of a vector x; See Page 147.|x | the Euclidean norm of a vector x; See Page 147.


C1([a, b]) the normed linear space of continuously differen-tiable real-valued functions on the interval [a, b]; SeePage 148.

lp The normed linear space of real-valued sequenceswith finite p-norm, 1 p +; See Page 149.

lp(X) the normed linear space of X-valued sequences withfinite p-norm, 1 p +; See Page 153.

V (P ) the closed linear variety generated by a nonemptyset P ; See Page 156.

P the relative interior of a set P ; See Page 156.X Y the finite Cartesian product normed linear space;

See Page 156.C([a, b]) the Banach space of continuous real-valued func-

tions on the interval [a, b]; See Page 160.C(K,X) the normed linear space of continuous X-valued

functions on a compact space K; See Page 160.XIR the real normed linear space induced by a complex

normed linear space X; See Page 164.[x ] the coset of a vector x in a quotient space; See Page

165.X/M the quotient space of a vector space X modulo a

subspace M ; See Page 165.X/M the quotient space of a normed linear space X mod-

ulo a closed subspace M ; See Page 166.Cv(X ,Y) the vector space of continuous Y-valued functions on

X ; See Page 168.B (X,Y) the set of bounded linear operators of X to Y; See

Page 175.X the dual of X; See Page 179.x a vector in the dual; See Page 179.x, x the evaluation of a bounded linear functional x at

the vector x, that is x(x); See Page 179.c0(X) the subspace of l(X) consisting of X-valued se-

quences with limit X; See Page 183.X the second dual of X; See Page 192.S the orthogonal complement of the set S; See Page

194.S the pre-orthogonal complement of the set S; See

Page 194.A the adjoint of a linear operator A; See Page 203.A the adjoint of the adjoint of a linear operator A; See

Page 204.Oweak (X ) the weak topology on a normed linear space X; See

Page 206.

13

Xweak the weak topological space associated with a normedlinear space X; See Page 207.

Oweak (X ) the weak topology on the dual of a normed linearspace X; See Page 209.

Xweak the weak topological space associated with a

normed linear space X; See Page 210.Ksupp the support of a convex set K; See Page 222.[f, C ] the epigraph of a convex function f : C IR; See

Page 225.Cconj the conjugate convex set; See Page 229.f conj the conjugate convex functional; See Page 229.[f, C ] conj the epigraph of the conjugate convex functional; See

Page 230.

conj the pre-conjugate convex set; See Page 232.

conj the pre-conjugate convex functional; See Page 232.

conj[,] the epigraph of the pre-conjugate convex functional;See Page 232.

= greater than or equal to (with respect to the positive

cone); See Page 243.= less than or equal to (with respect to the positive

cone); See Page 243. greater than (with respect to the positive cone); See

Page 243. less than (with respect to the positive cone); See

Page 243.S the positive conjugate cone of a set S; See Page 243.S the negative conjugate cone of a set S; See Page

243.AD (x0 ) the set of admissible deviations in D at x0; See Page

253.

f (1)(x0), Df(x0) the Frechet derivative of f at x0; See Page 254.Df(x0;u) the directional derivative of f at x0 along u; See

Page 254.fy (x0, y0) partial derivative of f with respect to y at (x0, y0);

See Page 255.ro(A)(B) right operate: ro(A)(B) = BA; See Page 260.Bk (X,Y) the set of bounded multi-linear Y-valued functions

on Xk; See Page 267.BSk (X,Y) the set of symmetric bounded multi-linear Y-valued

functions on Xk; See Page 267.

Dkf(x0), f(k)(x0) the kth order Frechet derivative of f at x0; See Page

267.Ck, C k-times and infinite-times continuously differen-

tiable functions, respectively; See Page 268.


kfxik xi1

kth-order partial derivative of f ; See Page 273.

Sm (D ) =IK D; See Page 273.

Ck(,Y) the normed linear space of k-times continuouslydifferentiable Y-valued functions on a compact set X; See Page 309.

Cb (X ,Y) the normed linear space of bounded continuous Y-valued functions on a topological space X ; See Page309.

Cbk(,Y) the normed linear space of k-times bounded con-tinuously differentiable Y-valued functions on a set X; See Page 310.

ATn1,...,nm the transpose of an mth order tensor A with thepermutation (n1, . . . , nm); See Page 311.

AB the outer product of two tensors; See Page 312.0m1mn an nth order IK-valued tensor in

B (IKmn , . . . ,B (IKm1 , IK) ) with all elementsequal to 0; See Page 313.

1m1mn an nth order IK-valued tensor inB (IKmn , . . . ,B (IKm1 , IK) ) with all elementsequal to 1; See Page 313.

S+ X, Spsd X sets of positive definite and positive semi-definiteoperators over the real normed linear space X, re-spectively; See Page 326.

SX, Snsd X sets of negative definite and negative semi-definiteoperators over the real normed linear space X, re-spectively; See Page 326.

SX BS 2 (X, IR); See Page 326.(IR,BL, L) Lebesgue measure space; See Page 360.Lo Lebesgue outer measure; See Page 361.BB (X ) Borel sets; See Page 361.A (X ) the algebra generated by the topology of a topolog-

ical space X ; See Page 361.B the Borel measure on IR; See Page 362.R ((IR, IR, | |),BB ( IR) , B); See Page 366.P a.e. in X P holds almost everywhere in X ; See Page 370.P (x) a.e. x X P holds almost everywhere in X ; See Page 370.P f P f : X [0,) IR defined by P f(x) =

f(x), x X ; See Page 380.R (X ) the collection of all representation of X; See Page

383.I (X ) the integration system on X; See Page 383.Xf d the integral of a function f on a set X with respect

to measure ; See Page 384.X f(x) d(x) the integral of a function f on a set X with respect

to measure ; See Page 384.

15

P the total variation of a Banach space valued pre-measure ; See Page 414.

P the total variation of a Banach space valued measure; See Page 419.

1 + 2 the Y-valued measure that equals to the sum of twoY-valued measures on the same measurable space;See Page 449.

the Y-valued measure that equals to the scalar prod-uct of IK and Y-valued measure ; See Page 450.

y the Y-valued measure that equals to scalar productof a IK-valued measure and y Y; See Page 450.

A the Z-valued measure that equals to product of anbounded linear operator A and a Y-valued measure; See Page 450.

M(X,B,Y) the vector space of -finite Y-valued measures onthe measurable space (X,B); See Page 456.

Mf(X,B,Y) the normed linear space of finite Y-valued measureson the measurable space (X,B); See Page 457.

limnIN n = a sequence of -finite (Y-valued) measures (n )n=1

converges to a -finite (Y-valued) measure ; SeePage 460.

1 2 the measures 1 and 2 on the measurable space(X,B) can be compared if 1(E) 2(E), E B;See Page 460.

1,1 1,m

......

n,1 n,m

the vector measure; See Page 463.

1 2 the measure 1 is absolutely continuous with respectto the measure 2; See Page 473.

1 2 the measures 1 and 2 are mutually singular; SeePage 473.

dd the Radon-Nikodym derivative of the -finite Y-

valued measure with respect to the -finite IK-valued measure ; See Page 483.

Pp f the function f()p; See Page 493.ess sup the essential supremum; See Page 495.limnIN zn

.= z in Lp the sequence (zn )

n=1 Lp converges to z Lp in

Lp pseudo-norm; See Page 497.Mft(X ,Y) the normed linear space of finite Y-valued topologi-

cal measures on X ; See Page 514.Mt(X ,Y) the vector space of -finite Y-valued topological

measures on X ; See Page 514.M(X,B) the set of -finite measures on the measurable space

(X,B); See Page 515.


Mf(X,B) the set of finite measures on the measurable space(X,B); See Page 515.

Mt(X ) the set of -finite topological measures on the topo-logical space X ; See Page 515.

Mft(X ) the set of finite topological measures on the topo-logical space X ; See Page 515.m

j=1 j the product measure of 1, . . . , m; See Page 547.mi=1 Xi product topological measure space of X1, . . . ,Xm,

where m IN; See Page 564.rx1,x2 the semi-open rectangle in IR

m with corners x1 andx2; See Page 572.

rx1,x2 the closed rectangle in IRm with corners x1 and x2;

See Page 573.rx1,x2 the open rectangle in IR

m with corners x1 and x2;See Page 573.

P() the principal of a region ; See Page 573.VRecti,x1,x2 the set of vertexes of rx1,x2 with i coordinates equal

to that of x1; See Page 575.F (rx1,x2) the increment of F on rx1,x2 ; See Page 575.TF (rx1,x2 ) the total variation of a function F on the semi-open

rectangle rx1,x2 ; See Page 575.(IRm,BLm, Lm), Lmom-dimensional Lebesgue measure space and the m-

dimensional Lebesgue outer measure; See Page 594.U g(x) dF (x) the integral of function g with respect to Y-valued

measure space (P(),BB (P()) , ) whose distribu-tion function is F : Y over the set U P();See Page 595.

bi0 : IRm IRm1 bi0(x) = (1(x), . . . , i01(x), i0+1(x), . . . , m(x)),

x IRm; See Page 595. baf(x) dx the integral of function f from a IR to b IR with

respect to B; See Page 606.dia(S) the diameter of a subset of a metric space; See Page

613.em,i the ith unit vector in IR

m; See Page 616.x, y the inner product of vectors x and y; See Page 643.x y the vectors x and y are orthogonal in a pre-Hilbert

space; See Page 645.x S the vector x is orthogonal to the set S in a pre-

Hilbert space; See Page 645.x x X that satisfies x, y = y, x, y X;

See Page 649.M N the direct sum of M and N ; See Page 651.A the Hermitian adjoint of A; See Page 652.Gram(y1, . . . , yn) the Gram matrix of y1, . . . , yn; See Page 655.gram(y1, . . . , yn) the Gram determinant of y1, . . . , yn; See Page 655.

17

E(x) expectation of x; See Page 669.

E(x|B) conditional expectation of x given B; See Page 669.

Chapter 2

Set Theory

2.1 Axiomatic Foundations of Set Theory

We will list the nine axioms of ZFC axiom system. The ninth axiom, whichis the Axiom of Choice, will be introduced in Section 2.7. Let A and B besets and x and y be objects (which is another name for sets).

Axiom 1 (Axiom of Extensionality) A = B if x A, we have x B;and x B, we have x A.

Axiom 2 (Axiom of Empty Set) There exists an empty set , whichdoes not contain any element.

Axiom 3 (Axiom of Pairing) For any objects x and y, there exists a set{x, y}, which contains only x and y.

Axiom 4 (Axiom of Regularity) Any nonempty set A 6= , there existsa A, such that b A, we have b 6 a.

Axiom 5 (Axiom of Replacement) x A, let there be one and onlyone y to form an ordered pair (x, y). Then, the collection of all such ys isa set B.

Axiom 6 (Axiom of Power Set) The collection of all subsets of A is aset denoted by A2.

Axiom 7 (Axiom of Union) For any collection of sets (A ), where is a set, then

A is a well defined set.

Axiom 8 (Axiom of Infinity) There exists a set A such that A andx A, we have {, x} A.

19

20 CHAPTER 2. SET THEORY

By Axiom 2, there exists the empty set , which we may call 0. Now,by Axiom 3, there exists the set {}, which is nonempty and we may call1. Again, by Axiom 3, there exists the set {, {}}, which we will call 2.After we define n, we may define n + 1 := {0, n}, which exists by Axiomof Pairing. This allows us to define all natural numbers. By Axiom 8,these natural numbers can form the set, IN := {1, 2, . . .}, which is the setof natural numbers. Furthermore, by Axiom 6, we may define the set of allreal numbers, IR.

For any x A and y B, we may apply Axiom 3 to define the orderedpair (x, y) := {{{{x}}, 1}, {{{y}}, 2}}. Then, the set A B is defined byxA

yB{(x, y)}, which is a valid set by Axiom 7. By Axiom 5, any

portion As of a well-defined set A is again a set, which is called a subset ofA, we will write As A. Thus, the formula

{x A | p(x) is true.}

defines a set as long as A is a set and p(x) is unambiguous logic expression.

2.2 Relations and Equivalence

Definition 2.1 Let A and B be sets. A relation R from A to B is a subsetof AB. x A, y B, we say x y if (x, y) R. We will say that Ris a relation on A if it is a relation from A to A. We define

dom(R ) := {x A | y B, such that x y }range (R ) := {y B | x A, such that x y }

which are well-defined subsets.

Definition 2.2 Let A be a set and R be a relation on A. x, y, z A,

1. R is reflexive if x x.

2. R is symmetric if x y implies y x.

3. R is transitive if x y and y z implies x z.

4. R is a equivalence relationship if it is reflexive, symmetric, and tran-sitive, and will be denote by .

5. R is antisymmetric if x y and y x implies x = y.

Let be an equivalence relationship on A, then, it partitions A intodisjoint equivalence classes Ax := {y A | x y }, x A. The collectionof all equivalence classes, A/ := {Ax A | x A}, is called the quotientof A with respect to .

2.3. FUNCTION 21

2.3 Function

Definition 2.3 Let X and Y be sets and D X. A function f of D toY , denoted by f : D Y , is a relation from D to Y such that x D,there is exactly one y Y , such that (x, y) f ; we will denote that y asf(x). The graph of f is the set graph (f ) := {(x, f(x)) XY | x D}.The domain of f is dom (f ) = D. A X, the image under f of A isf(A) := {y Y | x AD such that f(x) = y }, which is a subset of Y .The range of f is range (f ) = f(X). B Y , the inverse image under fof B is f inv(B) := {x D | f(x) B }, which is a subset of D. f is saidto be surjective if f(X) = Y ; and f is said to be injective if f(x1) 6= f(x2),x1, x2 D with x1 6= x2; f is said to be bijective if it is both surjective andinjective, in which case it is invertible and the inverse function is denotedby f inv : Y D. We will say that f is a function from X to Y .

Let f : D Y and g : Y Z be functions, we may define a functionh : D Z by h(x) = g(f(x)), then h is called the composition of g with f ,and is denoted by gf . Let A X . We may define a function l : AD Yby l(x) = f(x), x A D. This function is called the restriction of f toA, and denoted by f |A. Let f : D Y , g : Y Z, and h : Z W , wehave (h g) f = h (g f). Let f : D D and k Z+, we will writefk := f f

k

, where f0 := idD.

A function f : X Y is a subset of X Y . Then, f XY2. Thecollection of all functions of X to Y is then a set given by

Y X :={f XY2 | x X, ! y Y (x, y) f }

We have the following result concerning the inverse of a function.

Proposition 2.4 Let : X Y , where X and Y are sets. Then, isbijective if, and only if, i : Y X, i = 1, 2, such that 1 = idY and2 = idX . Furthermore, inv = 1 = 2.

Proof Sufficiency Let i : Y X , i = 1, 2, exist. y Y , 1(y) = idY (y) = y, which implies that y range (), and hence, issurjective. Suppose that is not injective, then x1, x2 X with x1 6= x2such that (x1) = (x2). Then, we have

x1 = idX(x1) = 2((x1)) = 2((x2)) = idX(x2) = x2

which is a contradiction. Hence, is injective. This proves that isbijective.

Necessity Let be bijective. Then, inv : Y X exists. x X ,let y = (x), then x = inv(y), hence inv((x)) = x. Therefore, we haveinv = idX . y Y , let x = inv(y), then y = (x), hence (inv(y)) = y.Therefore, we have inv = idY . Hence, 1 = 2 = inv.


Let 1 and 2 satisfy the assumption of the proposition, and inv bethe inverse function of . Then, we have

1 = idX 1 = (inv ) 1 = inv ( 1) = inv idY = inv2 = 2 idY = 2 ( inv) = (2 ) inv = idX inv = inv

This completes the proof of the proposition. 2For bijective functions f : X Y and g : Y Z, g f is also bijective

and (g f)inv = f inv ginv.

2.4 Set Operations

Let X be a set, X2 is the set consisting of all subsets of X . A,B X , wewill define

A B := {x X | x A or x B }A B := {x X | x A and x B }

A := {x X | x 6 A}A \B := {x A | x 6 B } = A BA B := (A \B) (B \A)

We have the following results.

Proposition 2.5 Let A,B,D,A X2, f : D Y , C,E,C Y2, whereX and Y are sets, , and is an index set. Then, we have

1. A B = B A and A B = B A;

2. A A B and A = A B if, and only if, B A;

3. A = A, A = , A X = X, and A X = A;

4. = X, A = A, A A = X, A A = , and A B if, and only if,B A;

5. The De Morgans Laws:

(

A)

=

A;

(

A)

=

A

6. B (

A)

=

(B A) and B

(

A)

=

(B A);

7. f(

A)

=

f(A) and f

(

A)

f(A);

2.5. ALGEBRA OF SETS 23

8. f inv(

C)

=

f inv(C) and f inv

(

C)

=

f inv(C);

9. f inv(C \ E) = f inv(C) \ f inv(E), f(f inv(C)) = C range (f ), andf inv(f(A)) A dom(f ) = A D.

The proof of the above results are standard and is therefore omitted.

2.5 Algebra of Sets

Definition 2.6 A set X is said to be finite if it is either empty or the rangeof a function of {1, 2, . . . , n}, with n IN. In this case, card (X) denotesthe number of elements in X. It is said to be countable if it is either emptyor the range of a function of IN.

Definition 2.7 Let X be a set and A X2. A is said to be an algebra ofsets on X (or a Boolean algebra on X) if

(i) , X A;

(ii) A,B A, A B A and A A.A is said to be a -algebra on X if it is an algebra on X and countableunions of sets in A is again in A.

Let M X2, where X is a set, then, there exists a smallest algebra onX , A0 X2, containing M, which means that any algebra on X , A1 X2,that contains M, we have A0 A1. This algebra A0 is said to be thealgebra on X generated by M. Also, there exists a smallest -algebra onX , A X2, containing M, which is said to be the -algebra on X generatedby M.

Proposition 2.8 Let X be a set, E be a nonempty collection of subsets ofX, A be the algebra on X generated by E, and

A :={A X

n,m IN, i1, . . . , i2n {1, . . . ,m}, Fi1,...,i2n X

with Fi1,...,i2n E or (Fi1,...,i2n ) E , such that

A =

m

i1=1

m

i2=1

m

i2n=1

Fi1,...,i2n

}

Then, A = A.

Proof E E , let n = 1, m = 1, and F1,1 = E. Then, E =1i1=1

1i2=1

Fi1,i2 A. Hence, we have E A. It is clear that A A. Allwe need to show is that A is an algebra on X . Then, A A and the resultfollows.


Fix E E 6= . E X . Let n = 1, m = 2, F1,1 = E, F1,2 = E,F2,1 = E, and F2,2 = E. Then, =

2i1=1

2i2=1

Fi1,i2 A. Let n = 1,m = 2, F1,1 = E, F1,2 = E, F2,1 = E, and F2,2 = E. Then, X =2i1=1

2i2=1

Fi1,i2 A.A,B A, nA,mA IN, i1, . . . , i2nA {1, . . . ,mA}, F Ai1,...,i2nA X

with FAi1,...,i2nA

E or(F

Ai1,...,i2nA

) E such that A =

mAi1=1

mAi2=1

mAi2nA=1 FAi1,...,i2nA

, and nB,mB IN, i1, . . . , i2nB {1, . . . ,mB}, F Bi1,...,i2nB X with F

Bi1,...,i2nB

E or(F

Bi1,...,i2nB

) E

such that B =mBi1=1

mBi2=1

mBi2nB =1 FBi1,...,i2nB

.

Note that A =mAi1=1

mAi2=1

mAi2nA=1(F

Ai1,...,i2nA

). Let n = nA +1,

m = mA, i1, . . . , i2n {1, . . . ,m}, Gi1,...,i2n =(F

Ai2,...,i2n1

). Then,

A =mi1=1

mi2=1

mi2n=1Gi1,...,i2n A.Without loss of generality, assume nA nB. Let n = nA and m =

mA + mB. Define i = 1 + (i mod mA) and i = 1 + (i mod mB), i IN.i1, . . . , i2n {1, . . . ,m}, let

Gi1,...,i2n =

FAi1 ,i2,...,i2nA

if i1 mAF

Bi1mA ,i2,...,i2nB

if i1 > mA

Then, it is easy to check that AB = mi1=1mi2=1

mi2n=1Gi1,...,i2n A.Hence, A is an algebra on X .

This completes the proof of the proposition. 2

2.6 Partial Ordering and Total Ordering

Definition 2.9 Let A be a set and be a relation on A. will be calleda partial ordering if it is reflexive and transitive. It will be called a totalordering is it is an antisymmetric partial ordering and satisfies x, y Awith x 6= y, we have either x y or y x (not both).

As an example, the set containment is a partial ordering on anycollection of sets; while is a total ordering on any subset of IR.

Definition 2.10 Let A be a set with a partial ordering .

1. a A is said to be minimal if, x A, x a implies a x;

2. a A is said to be the least element if, x A, a x, and x aimplies that x = a.

3. a A is said to be maximal if, x A, a x implies x a;

2.6. PARTIAL ORDERING AND TOTAL ORDERING 25

4. a A is said to be the greatest element if, x A, x a, and a ximplies that x = a.

Definition 2.11 Let A be a set with a partial ordering , and E A.

1. a A is said to be an upper bound of E if x a, x E. It is theleast upper bound of E if it is the least element in the set of all upperbounds of E;

2. a A is said to be a lower bound of E if a x, x E. It is thegreatest lower bound of E if it is the greatest element in the set of alllower bounds of E;

We have the following results.

Proposition 2.12 Let A be a set with a partial ordering . Then, thefollowing holds.

(i) If a A is the least element, then it is minimal.

(ii) There is at most one least element in A.

(iii) Define a relation by x, y A, x y if y x. Then, isa partial ordering on A. Furthermore, is antisymmetric if isantisymmetric.

1. a A is the least element for (E,) if, and only if, it is thegreatest element for (E,).

2. a A is minimal for (E,) if, and only if, it is maximal for(E,).

(iv) If a A is the greatest element, then it is maximal.

(v) There is at most one greatest element in A.

(vi) If is antisymmetric, then a A is minimal if, and only if, theredoes not exist x A such that x a and x 6= a.

(vii) If is antisymmetric, then a A is maximal if, and only if, theredoes not exist x A such that a x and x 6= a.

(viii) If is antisymmetric, then it is a total ordering if, and only if,x1, x2 A, we have x1 x2 or x2 x1.

Proof (i) is straightforward from Definition 2.10.For (ii), let a1 and a2 be least elements of A. By a1 being the least

element, we have a1 a2. By a2 being the least element, we then havea1 = a2. Hence, the least element is unique if it exists.

For (iii), x, y, z A, Since x x implies x x, then is reflexive. Ifx y and y z, we have y x and z y, which implies z x, and hence


x z. This shows that is transitive. Hence, is a partial ordering onA.

When is antisymmetric, x y and y x implies that x y andy x, and therefore x = y. Hence, is also antisymmetric.

For 1, only if let a A be the least element in (E,). a x impliesx a, x A. x A with a x, we have x a, by a being the leastelement in (E,), we have x = a. Hence, a is the greatest element in(E,). The if part is similar to the only if part.

For 2, only if let a A be a minimal element for (E,). Then,x A with a x implies x a and hence a x, which yields x a.Hence, a is a maximal element for (E,). The if part is similar to theonly if part.

(iv) is straightforward from Definition 2.10.For (v), Let a1 and a2 be greatest elements of A. By a1 being the

greatest element, we have a2 a1. By a2 being the greatest element, wethen have a1 = a2. Hence, the greatest element is unique if it exists.

For (vi), if, x A with x a, then, we have a = x, which meansthat a x; hence a is minimal. Only if, suppose that x A such thatx a and x 6= a. Note that a x since a is minimal. Then, a = x, since is antisymmetric, which is a contradiction.

For (vii), if, x A with a x, then, we have a = x, which meansthat x a; hence a is maximal. Only if, suppose that x A such thata x and x 6= a. Note that x a since a is maximal. Then, a = x, since is antisymmetric, which is a contradiction.

For (viii), if, x1, x2 A with x1 6= x2, we must have x1 x2 orx2 x1. They can not hold at the same time since, otherwise, x1 = x2,which is a contradiction. Only if, x1, x2 A, when x1 = x2, thenx1 x2; when x1 6= x2, then x1 x2 or x2 x1; hence, in both cases, wehave x1 x2 or x2 x1.


2.7 Basic Principles

Now, we introduce the last axiom in ZFC axiom system.

Axiom 9 (Axiom of Choice) Let (A ) be a collection of nonemptysets, is a set, (this collection is a set by Axiom 5), then, there exists afunction f : A, such that, , we have f() A.

With Axioms 18 holding, the Axiom of Choice is equivalent to thefollowing three results.

Theorem 2.13 (Hausdorff Maximal Principle) Let be a partial or-dering on a set E. Then, there exists a maximal (with respect to set con-tainment ) subset F E, such that is a total ordering on F .

2.7. BASIC PRINCIPLES 27

Theorem 2.14 (Zorns Lemma) Let be an antisymmetric partial or-dering on a nonempty set E. If every nonempty totally order subset F ofE has an upper bound in E, then, there is a maximal element in E.

Definition 2.15 A well ordering of a set is a total ordering such that everynonempty subset has a least element.

Theorem 2.16 (Well-Ordering Principle) Every set can be well or-dered.

To prove the equivalence we described above, we need the followingresult.

Lemma 2.17 Let E be a nonempty set and is an antisymmetric partialordering on E. Assume that every nonempty subset S of E, on which isa total ordering, has a least upper bound in E. Let f : E E be a mappingsuch that x f(x), x E. Then, f has a fixed point on E, i. e., w E,f(w) = w.

Proof Fix a point a E, since E 6= . We define a collection ofgood sets:

B ={B E

(i) a B (ii) f(B) B (iii) F B,F 6= F is totally ordered with implies that the least upperbound of F belongs to B.

}

Consider the setB0 := {x E | a x}

Clearly, B0 is nonempty since a B0 and

f(B0) = {f(x) E | a x f(x)} B0

since f satisfies x f(x), x E.For any F B0, such that F is totally ordered with and F 6= .

Let e0 be the least upper bound of F in E. Then, x0 F such thata x0 e0, and therefore e0 B0.

This shows that B0 B, and B is nonempty.The following result holds for the collection B.

Claim 2.17.1 Let {B | } be any nonempty subcollection of B, thenB B.

Proof of claim: (i) a B, . This implies a B.

(ii) By Proposition 2.5, we have f(B)

f(B)

B, where the last follows from the fact f(B) B, .(iii) Let F B, which is totally ordered by and F 6= .


For any , F B implies that the least upper bound of F is anelement of B. Therefore, the least upper bound of F is in the intersectionB.

This establishesB B, and completes the proof of the claim.

2

The claim shows that the collection B is closed under arbitrary inter-section, as long as the collection is nonempty. Define A :=

BB B. By

the above claim, we have A B, i. e., A is the smallest set in B.Hence, A B0, i. e., the set A satisfies, in addition to (i) (iii),

(iv) x A, a x.Define the relation on E as x, y E, x y if, and only if, x y

and x 6= y.Define the set P by

P = {x A | y A, y x f(y) x}

Clearly, a P , since there does not exists any y A such that y a,by being antisymmetric. Therefore, P is nonempty.

We claim that

Claim 2.17.2 (v) x P , z A, then z x or f(x) z.

Proof of claim: Fix x P , and let

B := {z A | z x} {z A | f(x) z }

We will show that B B.(i) a A, x P A, by (iv), a x, which further implies that a B.(ii) z B A, then f(z) A since A B. There are three exhaustivescenarios. If z x, since x P and z B A, then f(z) x. Thisimplies that f(z) B. If z = x, then f(x) f(x) = f(z). This impliesthat f(z) B. If f(x) z, then f(x) z f(z). This again impliesthat f(z) B. Hence, in all three scenarios, we have f(z) B. Then,f(B) B by the arbitraryness of z B.(iii) Let F 6= be any totally ordered subset of B and e0 E be the leastupper bound of F . Since F B A and A B, then e0 A. Thereare two exhaustive scenarios. If there exists y F such that f(x) y,then, f(x) y e0. This implies e0 B. If, for any y F , y x, thenF {z A | z x}. This implies that x is an upper bound of F ande0 x, since e0 is the least upper bound of F . Therefore, e0 B. In bothof the cases, we have e0 B.

This establishes that B B. By A being the smallest set in B, we haveA = B. Therefore, the claim is proven. 2

Now, we show that P B.(i) a P and therefore P 6= .(ii) Fix an x P A. Then, f(x) A. y A such that y f(x).We need to show that f(y) f(x), which then implies f(x) P . By (v),


there are two exhaustive scenarios. If y x, then y x. If f(x) y, thenf(x) y f(x) form a contradiction by being antisymmetric. Therefore,we must have y x, which results in the following two exhaustive scenarios.If y x, then f(y) x since x P . This implies that f(y) x f(x). Ify = x, then f(y) = f(x) f(x). In both cases, we have f(y) f(x). Bythe arbitraryness of y, we have f(x) P , which further implies f(P ) Pby the arbitraryness of x P .(iii) Let F 6= be a totally ordered subset in P . Let e0 E be the leastupper bound of F . We have F A implies that e0 A by A B. z Awith z e0, implies that z must not be an upper bound of F . Therefore,x0 F such that x0 6 z. By (v), we have z x0. Hence, by x0 F P ,z A, and z x0, we have f(z) x0. Therefore, f(z) e0 since e0 is anupper bound of F . This further implies that e0 P by the arbitrarynessof z.

This proves that P B.Since P A and A is the smallest set in B, then, P = A.The set A satisfies properties (i) (v).For any x1, x2 A, by (v), there are two exhaustive scenarios. If

x1 x2, then, x1 and x2 are related through . If f(x2) x1, then, x2 f(x2) x1, which implies that x1 and x2 are related through . Therefore,x1 and x2 are related through in both cases. Then, by Proposition 2.12(viii), A is totally ordered by and nonempty. Let w E be the leastupper bound of A. Then, w A, since A B.

Therefore, f(w) A by f(A) A, which implies that f(w) w. Thiscoupled with w f(w) yields f(w) = w, since is antisymmetric.

This completes the proof of the lemma. 2

Theorem 2.18 Under the Axioms 18, the following are equivalent.

1. Axiom of Choice

2. Hausdorff Maximum Principle

3. Zorns Lemma

4. Well-ordering principle

Proof 1. 2. Define

E := {A E | defines a total ordering on A}

Clearly, E , then E 6= . Define a partial ordering on E by , which isset containment. This partial ordering is clearly reflexive, transitive, andantisymmetric.

A E , define a collection

AA :={

{B E | A B } if B E such that A B{A} otherwise


Clearly, AA 6= . By Axiom of Choice, T : E E such that T (A) = B AA, A E .

We will show that T admits a fixed point by Lemma 2.17. Let B Ebe any nonempty subset on which is a total ordering. Let C := BB B.Clearly, C E. We will show is a total ordering on C. Since is apartial ordering on E, then it is a partial ordering on C. x1, x2 C,B1, B2 B such that x1 B1 and x2 B2. Since is a total orderingon B, then, we may without loss of generality assume B1 B2. Then,x1, x2 B2. Since B2 B E , then is a total ordering on B2, whichmeans that we have x1 x2 or x2 x1. Furthermore, if x1 x2 andx2 x1, then x1 = x2 by being antisymmetric on B2. Therefore, byProposition 2.12, is a total ordering on C. Hence, C E . This showsthat B admits least upper bound C in E with respect to . By the definitionof T , it is clear that A T (A), A E . By Lemma 2.17, T has a fixedpoint on E , i. e., A0 E such that T (A0) = A0.

By the definitions of T and AA0 , there does not exist B E such thatA0 B. Hence, by Proposition 2.12 (vii), A0 is maximal in E with respectto .

2. 3. Let E be a nonempty set with an antisymmetric partial ordering. By Hausdorff Maximum Principle, there exists a maximal (with respectto ) totally ordered (with respect to ) subset F E. We must haveF 6= , otherwise, let x0 E (since E 6= ), F {x0} E and {x0}is totally ordered by , which violates the fact that F is maximal (withrespect to ). Then, F has an upper bound e0 E.Claim 2.18.1 e0 F .Proof of claim: Suppose e0 6 F . Define A := F {e0} E. Clearly,F A and F 6= A. We will show that is a total ordering on A. Clearly, is an antisymmetric partial ordering on A since it is an antisymmetricpartial ordering on E. x1, x2 A, we will distinguish 4 exhaustive andmutually exclusive cases: Case 1: x1, x2 F ; Case 2: x1 F , x2 = e0;Case 3: x1 = e0, x2 F ; Case 4: x1 = x2 = e0. In Case 1, we have x1 x2or x2 x1 since is a total ordering on F . In Case 2, we have x1 x2 = e0since e0 is an upper bound of F . In Case 3, we have x2 x1 = e0. In Case4, we have x1 = e0 e0 = x2. Hence, is a total ordering on A. Notethat F A and F 6= A. By Proposition 2.12 (vii), this contradicts withthe fact that F is maximal with respect to . Therefore, we must havee0 F . This completes the proof of the claim. 2

e1 E such that e0 e1. x F , we have x e0 e1. Hence, e1is an upper bound of F . By Claim 2.18.1, we must have e1 F . Then,e1 e0 since e0 is an upper bound of F . This shows that e0 is maximal inE with respect to .

3. 4. Let E be a set. It is clear that E is well-ordered by theempty relation. Define

E := { (A,) | A E, A is well-ordered by }


Then, E 6= . Define an ordering on E by (A1,1), (A2,2) E , we say(A1,1) (A2,2) if the following three conditions hold: (i) A1 A2; (ii)2 = 1 on A1; (iii) x1 A1, x2 A2 \A1, we have x1 2 x2.

Now, we will show that defines an antisymmetric partial ordering onE . (A1,1), (A2,2), (A3,3) E . Clearly, (A1,1) (A1,1). Hence, is reflexive. If (A1,1) (A2,2) and (A2,2) (A3,3), we haveA1 A2 A3, and (i) holds; (ii) 3 = 2 on A2 and 2 = 1 on A1implies that 3 = 1 on A1; (iii) x1 A1, x2 A3 \ A1, we have 2exhaustive senarios: if x2 A2, then x2 A2 \ A1 which implies x1 2 x2and hence x1 3 x2; if x2 A3 \ A2, then we have x1 A2 and x1 3 x2,thus, we have x1 3 x2 in both cases. Therefore, (A1,1) (A3,3) andhence is transitive. If (A1,1) (A2,2) and (A2,2) (A1,1),then A1 A2 A1 A1 = A2 and 2 = 1 on A1. Hence, (A1,1) =(A2,2), which shows that is antisymmetric. Therefore, defines anantisymmetric partial ordering on E .

Let A E be any nonempty subset totally ordered by . Take A ={(A,) | } where 6= is an index set. Define A :=

A.

Define an ordering on A by: x1, x2 A, (A1,1), (A2,2) Asuch that x1 A1 and x2 A2, without loss of generality, assume that(A1,1) (A2,2) since A is totally ordered by , then x1, x2 A2, wewill say that x1 x2 if x1 2 x2. We will now show that this ordering isuniquely defined independent of (A2,2) A. Let (A3,3) A be suchthat x1, x2 A3. Since A is totally ordered by , then there are two ex-haustive cases: Case 1: (A3,3) (A2,2); Case 2: (A2,2) (A3,3).In Case 1, we have A3 A2 and 3 = 2 on A3, which implies that x1 x2 x1 2 x2 x1 3 x2. In Case 2, we have A2 A3 and 3 = 2 on A2,which implies that x1 x2 x1 2 x2 x1 3 x2. Hence, the ordering is well-defined on A.

Next, we will show that is a total ordering on A. x1, x2, x3 A.(Ai,i) A such that xi Ai, i = 1, 2, 3. Since A is totally orderedby , then, without loss of generality, assume that (A1,1) (A2,2) (A3,3). Then, x1, x2, x3 A3. Clearly, x1 x1 since x1 3 x1, whichimplies that is reflective. If x1 x2 and x2 x3, then, x1 3 x2 3 x3,which implies x1 3 x3 since 3 is transitive on A3, and hence, x1 x3.This shows that is transitive. If x1 x2 and x2 x1, then x1 3 x2 andx2 3x1, which implies that x1 = x2 since 3 is antisymmetric on A3. Thisshows that is antisymmetric. Since 3 is a well-ordering on A3, then wemust have x1 3 x2 x1 x2 or x2 3 x1 x2 x1. Hence, defines atotal ordering on A.

Next, we will show that is a well-ordering on A. B A withB 6= . Fix x0 B. Then, (A1,1) A such that x0 A1. Note that 6= B A1 A1. Since A1 is well-ordered by 1, then e B A1,which is the least element of B A1. y B A, (A2,2) A suchthat y A2. We have 2 exhaustive and mutually exclusive cases: Case 1:y A1; Case 2: y A2 \A1. In Case 1, e1 y since e is the least element


of B A1, which implies that e y. In Case 2, since A is totally orderedby , we must have (A1,1) (A2,2), which implies that e2 y, by (iii)in the definition of , and hence e y. In both cases, we have shown thate y. Since is a total ordering on A, then e is the least element of B.Therefore, is a well ordering on A, which implies (A,) E .

(A1,1) A. (i) A1 A. (ii) x1, x2 A1, x1 1 x2 x1 x2;hence = 1 on A1. (iii) x1 A1, x2 A \ A1, (A2,2) A suchthat x2 A2 \ A1; since A is totally ordered by , then, we must have(A1,1) (A2,2); hence x1 2x2 and x1 x2. Therefore, we have shown(A1,1) (A,). Hence, (A,) E is an upper bound of A.

By Zorns Lemma, there is a maximal element (F,F ) E . We claimthat F = E. We will prove this by an argument of contradiction. SupposeF E, then x0 E \F . Let H := F {x0}. Define an ordering H on Hby: x1, x2 H , if x1, x2 F , we say x1 H x2 if x1 F x2; if x1 F andx2 = x0, then we let x1 H x2; if x1 = x2 = x0, we let x1 H x2. Now, wewill show that H is a well ordering on H . x1, x2, x3 H . If x1 F , then,x1 F x1 and x1 H x1; if x1 = x0, then x1 H x1. Hence, H is reflexive.If x1 H x2 and x2 H x3. We have 4 exhaustive and mutually exclusivecases: Case 1: x1, x3 F ; Case 2: x1 F and x3 = x0; Case 3: x3 F andx1 = x0; Case 4: x1 = x3 = x0. In Case 1, we must have x2 F and thenx1 F x2 and x2 F x3, which implies that x1 F x3, and hence x1 H x3.In Case 2, we have x1 H x3. In Case 3, we must have x2 = x0, which leadsto a contradiction x0 H x3, hence, this case is impossible. In Case 4, wehave x1 H x3. In all cases except that is impossible, we have x1 H x3.Hence, H is transitive. If x1 H x2 and x2 H x1. We have 4 exhaustiveand mutually exclusive cases: Case 1: x1, x2 F ; Case 2: x1 F andx2 = x0; Case 3: x2 F and x1 = x0; Case 4: x1 = x2 = x0. In Case 1,we have x1 F x2 and x2 F x1, which implies that x1 = x2 since F isantisymmetric on F . In Case 2, we have x0 H x1, which is a contradiction,and hence this case is impossible. In Case 3, we have x0 H x2, whichis a contradiction, and hence this case is impossible. In Case 4, we havex1 = x2. In all cases except those impossible, we have x1 = x2. Hence,H is antisymmetric. When x1, x2 F , then, we must have x1 F x2 orx2 F x1 since F is a well ordering on F , and hence x1 H x2 or x2 H x1.When x1 F and x2 = x0, then x1 H x2. When x2 F and x1 = x0,then x2 H x1. When x1 = x2 = x0, then x1 H x2. This shows thatH is a total ordering on H . B H with B 6= . We will distinguishtwo exhaustive and mutually exclusive cases: Case 1: B = {x0}; Case 2:B 6= {x0}. In Case 1, x0 is the least element of B. In Case 2, B \ {x0} Fand is nonempty, and hence admits a least element e0 B \ {x0} F withrespect to F . x B, if x B \ {x0}, then e0 F x and hence e0 H x; ifx = x0, then e0H x. Hence, e0 is the least element of B since H is a totalordering on H . Therefore, H is a well ordering on H and (H,H) E .

Clearly, F H , F = H on F , and x1 F and x2 H \F , we havex2 = x0 and x1H x2. This implies that (F,F ) (H,H). Since (F,F )


is maximal in E with respect to , we must have (H,H) (F,F ), andhence, H F . This is a contradiction. Therefore, F = E and E is wellordered by F .

4. 1. Let (A ) be a collection of nonempty sets, and is a set.Let A :=

A. By Well-Ordering Principle, A may be well ordered by

. , A A is nonempty and admits the least element e A.This defines a function f : A by f() = e A, .

This completes the proof of the theorem. 2

Example 2.19 Let be an index set and (A ) be a collection ofsets. We will try to define the Cartesian (direct) product

A. Let

A =A, which is a set by the Axiom of Union. Then, as we discussed

in Section 2.3, A is a set, which consists of all functions of to A. Definethe projection functions : A

A, , by, f A, (f) = f().Then, we may define the set

A :=

{f A

(f) A, }

When all of As are nonempty, then, by Axiom of Choice, the productA is also nonempty.

Chapter 3

Topological Spaces

3.1 Fundamental Notions

Definition 3.1 A topological space (X,O) consists of a set X and a col-lection O of subsets (namely, open subsets) of X such that

(i) , X O;

(ii) O1, O2 O, we have O1 O2 O;

(iii) (O ) O, where is an index set, we haveO O.

The collection O is called a topology for the set X.

Definition 3.2 Let (X,O) be a topological space and F X. The com-plement of F is F := X \ F . F is said to be closed if F O. The closureof F is given by F :=

FBeBO

B, which is clearly a closed set. The interior

of F is given by F :=

BFBO

B, which is clearly an open set. A point of

closure of F is a point in F . An interior point of F is a point in F . Aboundary point of F is a point x X such that O O with x O, wehave O F 6= and O F 6= . The boundary of F , denoted by F , isthe set of all boundary points of F . An exterior point of F is a point inF , where F is called the exterior of F . An accumulation point of F is apoint x X such that O O with x O, we have O (F \ {x}) 6= .

Clearly, and X are both closed and open.

Proposition 3.3 Let (X,O) be a topological space and A,B,E are subsetsof X. Then,

35

36 CHAPTER 3. TOPOLOGICAL SPACES

(i) E E, E = E, E E, (E) = E, and E = E;(ii) x X, x is a point of closure of E if, and only if, O O with

x O, we have O E 6= ;(iii) x X, x is an interior point of E if, and only if, O O with

x O such that O E;(iv) A B = A B, (A B) = A B;(v) E is closed if, and only if, E = E;

(vi) E = E E.

(vii) X equals to the disjoint union E E (E);

Proof (i) Clearly, E E. Then, E E. C E with C O, wehave C E. Then, E =

CEeCO

C

CEeCO

C = E. Hence, we have E = E.

Clearly, E E. Note that

E =

(

BEeBO

B

)=

BEeBO

B =

O eEOO

O = E

Furthermore,

(E) =

(E

)=

( E

)=E =

E =

(E

)= E

(ii) Only if x E, we have x BEeBO

B. O O with x O, let

O1 := O E O. Note that x O1 and E E = . Suppose O E = .Then, E O1 = , which further implies that E O1. Then, E O1 andx O1. This contradicts with x O. Hence, O E 6= .

If x E = E O. Then, O := E O such that E O = .Hence, the result holds.

(iii) Only if x E E, then E O.If x X , O O such that x O E. Then, x O BE

BOB =

E. Hence, the result holds.(iv) Let B := {O O | O A B }, BA := {O O | O A}, and

BB := {O O | O B }. O1 BA and O2 BB, then, O1 O2 B.On the other hand, O B, we have O = O O and O BA and O BB.Then,

(A B) =

OABOO

O =

O1A,O2BO1,O2O

(O1 O2)

=

(

O1AO1O

O1

)(

O2BO2O

O2

)= A B

3.1. FUNDAMENTAL NOTIONS 37

We also have

A B =((

A B))

=((

A B))

=((

A)

(B))

=A B = A B

(v) If E is closed since E = E and E is closed.Only if Since E is closed, then E E. Then, we have E = E. Hence,

the result holds.(vi) This result follows directly from (ii), (iii), and Definition 3.2.

(vii) Note that X = E E = E E E. By (iii) and Definition 3.2,E and E are disjoint. It is obvious that E is disjoint with E E = E.Hence, the result holds. 2

To simplify notation in the theory, we will abuse the notation to writex X when x X and A X when A X for a topological spaceX := (X,O). We will later simply discuss a topological space X withoutfurther reference to components of X , where the topology is understood tobe OX . When it is clear from the context, we will neglect the subscript X .

Proposition 3.4 Let (X,O) be a topological space and A X. A admitsthe subset topology OA := {O A | O O}.

Proof Clearly, OA is a collection of subsets of A. = A OA andA = X A OA. OA1, OA2 OA, O1, O2 O such that OA1 = O1 Aand OA2 = O2 A. Then, O1 O2 O since O is a topology. Then,OA1 OA2 = (O1 O2)A OA. (OA ) OA, where is an indexset, we have, , O O such that OA = OA. Then,

O

O since O is a topology. Therefore, OA =(

O)A OA.

Hence, OA is a topology on A. 2Let (X,O) be a topological space and A X . The property of a set

E A being open or closed is relative with respect to (X,O), that is, thisproperty may change if we consider the subset topology (A,OA).

Proposition 3.5 Let X be a topological space, A X be endowed with thesubset topology OA, and E A. Then,

(1) E is closed in OA if, and only if, E = A F , where F X is closedin OX ;

(2) the closure of E relative to (A,OA) (the closure of E in OA) is equalto E A, where E is the closure of E relative to X .

Proof Here, the set complementation and set closure operation arerelative to X .

(1) If A \ E = A \ (A F ) = A A F = A F . Since F is closedin OX , then F OX . Then, A \ E OA. Hence, E is closed in OA.


Only if A\E OA. Then, O OX such that A\E = AO. Then,E = A \ (A \ E) = A A O = A O. Hence, the result holds.

(2) By (1), E A is closed in OA. Then, the closure of E relative to(A,OA) is contained in E A. On the other hand, by Proposition 3.3, ifx X is a point of closure of E relative to X , then it is a point of closure ofE in OA if x A. Then, E A is contained in in the closure of E relativeto (A,OA). Hence, the result holds.


Definition 3.6 For two topologies over the same set X, O1 and O2, wewill say that O1 is stronger (finer) than O2 if O1 O2, in which case, O2is said to be weaker (coarser) than O1.

Proposition 3.7 Let X be a set and A X2. Then, there exists theweakest topology O on X such that A O. This topology is called thetopology generated by A.

Proof Let M :={X X2 | A X and X is a topology on X } and

O = XM X . Clearly, X2 M and hence O is well-defined. Then,(i) , X X , X M. Hence, , X O.(ii) A1, A2 O, we have A1, A2 X , X M. Then, A1 A2 X ,

X M. Hence, A1 A2 O.(iii) (A ) O, where is an index set, we have, , X M,

A X . Then,A X , X M. Hence, we have

A

O.Therefore, O is a topology on X . Clearly, A O since A X , X M.Therefore, O is the weakest topology containing A. 2

3.2 Continuity

Definition 3.8 Let (X,OX) and (Y,OY ) be topological spaces, D Xwith the subset topology OD, and f : D Y (or f : (D,OD) (Y,OY ) tobe more specific). Then, f is said to be continuous if, OY OY , we havef inv(OY ) OD. f is said to be continuous at x0 D if, OY OY withf(x0) OY , U OX with x0 U such that f(U) OY . f is said to becontinuous on E D if it is continuous at x, x E.

Proposition 3.9 Let X and Y be topological spaces, D X with the subsettopology OD, and f : D Y. f is continuous if, and only if, x0 D, fis continuous at x0.

Proof If OY OY , x f inv(OY ) D. Since f is con-tinuous at x, then Ux OX with x Ux such that f(Ux) OY ,which implies, by Proposition 2.5, that Ux D f inv(OY ). Then,

3.2. CONTINUITY 39

f inv(OY ) =xf inv(OY )(Ux D) = (

xf inv(OY ) Ux) D OD. Hence, f

is continuous.Only if x0 D, OY OY with f(x0) OY , let U = f inv(OY )

OD. By Proposition 3.4, U OX such that U = U D. Then, x0 U .By Proposition 2.5, f(U) = f(U) OY . Hence, f is continuous at x0.


Proposition 3.10 Let X and Y be topological spaces and f : X Y. f iscontinuous if, and only if, B Y with B OY , we have f inv(B) OX ,that is, the inverse image of any closed set in Y is closed in X .

Proof If O OY , we have, by Proposition 2.5, f inv(O) =f inv(O) OX . Hence, f is continuous.

Only if B Y with B OY . Since f is continuous, then, byProposition 2.5, f inv(B) = f inv(B) OX . Hence, the result holds.


Theorem 3.11 Let X and Y be topological spaces, f : X Y, and X =X1 X2, where X1 and X2 are both open or both closed. Let X1 and X2be endowed with subset topologies OX1 and OX2 , respectively. Assume thatf |X1 : X1 Y and f |X2 : X2 Y are continuous. Then, f is continuous.

Proof Consider the case that X1 and X2 are both open. x0 X .Without loss of generality, assume x0 X1. O OY with f(x0) O.Since f |X1 is continuous, then, by Proposition 3.9, U OX1 with x0 Usuch that f |X1 (U) O. Since X1 OX , then U OX . Note thatf(U) = f |X1 (U) O, since U X1. Hence, f is continuous at x0. By thearbitraryness of x0 and Proposition 3.9, f is continuous.

Consider the case that X1 and X2 are both closed. closed subsetB Y, we have f inv(B) X . Then, f inv(B) X1 = (f |X1)inv(B) isclosed in OX1 , by Proposition 3.10 and the continuity of f |X1 . Similarly,f inv(B) X2 = (f |X2)inv(B) is closed in OX2 . Since X1 and X2 are closedsets in OX , then, f inv(B) X1 and f inv(B) X2 are closed in OX , byProposition 3.5. Then, f inv(B) = (f inv(B)X1) (f inv(B)X2) is closedin OX . By Proposition 3.10, f is continuous.

This completes the proof of the theorem. 2

Proposition 3.12 Let X , Y, and Z be topological spaces, f : X Y,g : Y Z, and x0 X . Assume that f is continuous at x0 and g iscontinuous at y0 := f(x0). Then, g f : X Z is continuous at x0.

Proof OZ OZ with g(f(x0)) OZ . Since g is continuous atf(x0), then OY OY with f(x0) OY such that g(OY ) OZ . Since f iscontinuous at x0, then OX OX with x0 OX such that f(OX) OY .Then, g(f(OX)) OZ . Hence, g f is continuous at x0. This completesthe proof of the proposition. 2


Definition 3.13 Let X and Y be topological spaces and f : X Y. fis said to be a homeomorphism between X and Y if it is bijective andcontinuous and f inv : Y X is also continuous. The spaces X and Y aresaid to be homeomorphic if there exists a homeomorphism between them.

Any properties invariant under homeomorphisms are called topologicalproperties.

Homeomorphisms preserve topological properties in topological spaces.Isomorphisms preserve algebraic properties in algebraic systems. Isometriespreserve metric properties in metric spaces.

Definition 3.14 Let X be a topological space, D X with the subsettopology OD, and f : D IR. f is said to be upper semicontinuous ifa IR, f inv((, a)) OD. f is said to be upper semicontinuous atx0 X if (0,) IR, U OX with x0 U such that f(x) a2 implies thata2 (a2, a0) Q such that x0 Oa2 Oa2 . Therefore x0 V . Thisshows that V O with x0 V such that f(V ) U .

Case 2: a0 = 0. Then, we must have a1 < 0 = a0 < a3 < a4. TakeV = Oa3 O. We must have x0 V . x V , 0 f(x) a3. Hence,f(V ) [0, a3] (a1, a4) U . Hence, V O with x0 V such thatf(V ) U .

Case 3: a0 = 1. Then, we must have a1 < a2 < a0 = 1 < a4. Take

V = Oa2 O. Since f(x0) = a0 = 1, then x0 O 1+a22

Oa2 = V . x V ,


f(x) a2. Hence, f(V ) [a2, 1] (a1, a4) U . Hence, V O withx0 V such that f(V ) U .

Therefore, in all cases, V O with x0 V such that f(V ) U . Hence,f is continuous at x0. By the arbitraryness of x0 and Proposition 3.9, f iscontinuous. This completes the proof of the theorem. 2

Proposition 3.56 Let X and Y be topological spaces and Y be Hausdorff.f1 : X Y and f2 : X Y are continuous. Let D X be dense. Assumethat f1|D = f2|D. Then, f1 = f2.

Proof Suppose f1 6= f2. Then, x X such that f1(x) 6= f2(x). SinceY is Hausdorff, then O1, O2 OY such that f1(x) O1, f2(x) O2, andO1O2 = . Since f1 and f2 are continuous, we have U1 := f1inv(O1) OXand U2 := f2inv(O2) OX . Note that x U1 U2 OX and x D, then,by Proposition 3.3, x D U1 U2. Then, f1(x) O1 and f2(x) O2,which implies that f1|D (x) 6= f2|D (x). This is a contradiction. Hence, wemust have f1 = f2.


Theorem 3.57 (Tietzes Extension Theorem) Let (X,O) be a nor-mal topological space, A X be closed, and h : A IR. Let A be endowedwith the subset topology OA. Assume that h is continuous. Then, thereexists a continuous function k : X IR such that k|A = h.

Proof Let f :=h

1 + |h| . Then, |f(x) | < 1, x A, and by Proposi-tion 3.12, f is continuous.

Claim 3.57.1 Let l : A IR be a continuous function such that | l(x) | c1 IR, x A, where c1 > 0. Then, there exists a continuous functiong : X IR such that |g(x) | c1/3, x X, and | l(x) g(x) | 2c1/3,x A.

Proof of claim: Let B := {x A | l(x) c1/3} and C := {x A |l(x) c1/3}. Then, B anc C are closed sets in OA, by the continuityof l and Proposition 3.10. Since A is closed, then B and C are closedin O, by Proposition 3.5. Clearly, B C = . By Urysohns Lemma,there exists a continuous function g : X IR such that |g(x) | c1/3,x X , g(x) = c1/3, x B, and g(x) = c1/3, x C. Hence,| l(x) g(x) | 2c1/3, x A. This completes the proof of the claim. 2

By repeated application of Claim 3.57.1, we may define fi : X IR,i IN, such that fi is continuous, |fi(x) | 2

i1

3i , x X , andf(x)

ik=1 fk(x)

2i3i , x A.Define g : X IR by g(x) = limiIN

ik=1 fk(x), x X . Clearly, g

is well-defined, g|A = f , and |g(x) |

i=12i1

3i = 1, x X . x0 X .

3.8. CONTINUOUS REAL-VALUED FUNCTIONS 57

(0,) IR. N IN such that

i=N+1 fi(x) < /3, x X . By

the continuity of f1, . . . , fN and Proposition 3.9, U O with x0 U suchthat

N

i=1 fi(x) N

i=1 fi(x0) < /3, x U . Then, we have, x U ,

|g(x) g(x0) | g(x)

N

i=1

fi(x)+N

i=1

fi(x) N

i=1

fi(x0)

+N

i=1

fi(x0) g(x0) <

Therefore, g is continuous at x0. Then, g is continuous, by the arbitrarinessof x0 and Proposition 3.9.

Let D := {x X | |g(x) | = 1}. Clearly, D is a closed set, byProposition 3.10. Note that A D = , since g|A = f and |f(x) | < 1,x A. Then, by Urysohns Lemma, there exists a continuous functiong : X [0, 1] such that g|A = 1 and g|D = 0. Define k : X IR byk(x) =

g(x)g(x)

1 g(x) |g(x)| , x X . By Propositions 3.12 and 3.32 and thefact that 1 g(x) |g(x) | 6= 0, x X , we have k is continuous. x A,k(x) =

g(x)

1 |g(x)| = h(x). Hence, k|A = h.This completes the proof of the theorem. 2

Definition 3.58 Let X be a set and F be a collection of real-valued func-tions of X. Then, there is the weakest topology on X such that all functionsin F are continuous. This topology is called the weak topology generatedby F .

Let X be a set, I := [0, 1] IR, and F be a collection of functions ofX to I such that x, y X with x 6= y, f F , we have f(x) 6= f(y).Each f F is a point in IX and F can be identified with a subset of IX .The topology that F inherits as a subspace of IX is called the topology ofpointwise convergence. Now, X can be identified with a subset of IF by,x X , f (x) = f(x), f F . Then, the topology of X as a subset of IFis the weak topology generated by F .

Proposition 3.59 Let X be a topological space, I := [0, 1] IR, and Fbe a collection of continuous functions of X to I such that x, y X withx 6= y, f F , we have f(x) 6= f(y). Let E : X IF be the equivalencemap given by, x X , f (E(x)) = f(x), f F . Then, E is continuous.Furthermore, if closed set F X and x X with x 6 F , f F withf(x) = 1 and f |F = 0, then E : X E(X ) is a homeomorphism.

Proof x0 X . Fix a basis open set O in IF with E(x0) O. ByProposition 3.25, O =

fF Of , where Of OI with OI being the subset

topology on I, f F , and Of = I for all f s except finitely many f s, say


f FN . Let U =fFN f inv(Of ) OX . By E(x0) O, we have x0 U .

x U , we have f (E(x)) Of = I, f F \ FN , and f (E(x)) =f(x) Of , f FN . Hence, E(x) O. Then, E(U) O. Therefore, Eis continuous at x0. By the arbitrariness of x0 and Proposition 3.9, E iscontinuous.

Under the additional assumption on F , we need to show that E is ahomeomorphism between X and E(X ). x, y X with x 6= y, f Fsuch that f (E(x)) = f(x) 6= f(y) = f (E(y)). Then, E(x) 6= E(y).Hence, E : X E(X ) is injective. Clearly, E : X E(X ) is surjective.Then, E : X E(X ) is bijective and admits inverse Einv : E(X ) X .x0 X , we will show that Einv is continuous at E(x0). O OX withx0 O. O is closed and x0 6 O. Then, f0 F such that f0(x0) = 1and f0| eO = 0. Define U =

fF Uf IF by Uf = I, f F \ {f0} and

Uf0 = (1/2, 1] OI . Clearly, U is open in IF . Clearly, E(x0) U . x Xwith E(x) U , we have f0 (E(x)) = f0(x) > 1/2. Then, x 6 O andx O. This shows that Einv(E(X )U) O. Hence, Einv is continuous atE(x0). By the arbitrariness of x0 and Proposition 3.9, Einv : E(X ) X iscontinuous. This implies that E : X E(X ) is a homeomorphism.


Definition 3.60 A topological space X is said to be completely regular(or T3 12 ) if it is Tychonoff and x0 X and closed set F X withx0 6 F , there exists a continuous real-valued function f : X [0, 1] suchthat f(x0) = 1 and f |F = 0.

Proposition 3.61 A normal topological space is completely regular. Acompletely regular topological space is regular.

Proof Let X be a normal topological space. Then, X is Tychonoff.x0 X and closed set F X with x0 6 F , we have {x0} is closed,by Proposition 3.34. By Urysohns Lemma, there exists a continuous real-valued function f : X [0, 1] such that f(x0) = 1 and f |F = 0. Hence, Xis completely regular.

Let X be a completely regular topological space. Then, X is Tychonoff.x0 X and closed set F X with x0 6 F , there exists a continuousreal-valued function f : X [0, 1] such that f(x0) = 1 and f |F = 0.Let O1 := {x X | f(x) > 1/2} and O2 := {x X | f(x) < 1/2}.Then, O1, O2 O by the continuity of f . Clearly, x0 O1, F O2, andO1 O2 = . Hence, X is regular.


Corollary 3.62 Let X be a completely regular topological space, I =[0, 1] IR, and F := {f : X I | f is continuous}. Then, the equiva-lence map: E : X IF defined by f (E(x)) = f(x), x X , f F , isa homeomorphism between X and E(X ) IF .

3.9. NETS AND CONVERGENCE 59

Proof Since X is completely regular, then X is Tychonoff and allsingleton subset of X is closed. Then, it is easy to check that all assumptionsin Proposition 3.59 are satisfied. Then, the result follows. This completesthe proof of the corollary. 2

3.9 Nets and Convergence

Definition 3.63 A directed system is a nonempty set A and a relation onA, , such that

(i) is transitive;

(ii) , A, A such that and .

A net is a mapping of a directed system A := (A,) to a topologicalspace X . A, the image is x. The net is denoted by (x )A, wherewe have abuse the notation to say A when A. It is understoodthat the relation for A is A, where we will ignore the subscript A if noconfusion arises.

A point x X is a limit of the net (x )A if O O with x O,0 A A with 0 , we have x O. We also say that(x )A converges to x.

A point x X is a cluster point of (x )A if O O with x O, A, A with x O.

Clearly, a limit point of a net is a cluster point of the net.In Definition 3.63, we may restrict O to be a basis open set without

changing the meaning of the definition.

Example 3.64 (IN,) is a directed system. A net over (IN,) corre-sponds to a sequence.

Proposition 3.65 Let X be a topological space. Then, the following state-ments holds.

(i) X is Hausdorff if, and only if, for all net (x )A X , there existsat most one limit point for the net. We then write x = limA xwhen the limit exists.

(ii) If X is Hausdorff, any convergent net (x )A X with limit x Xhas exactly one cluster point, which is x.

Proof (i) Only if Suppose there exists a net (x )A X suchthat xA, xB X with xA 6= xB and xA and xB are limit points of the net.Since X is Hausdorff, then O1, O2 O such that xA O1, xB O2, andO1 O2 = . Since xA is the limit of the net, then 1 A, A with1 , we have x O1. Similarly, since xB is the limit of the net, then


2 A, A with 2 , we have x O2. Since A is a directedsystem, 3 A such that 1 3 and 2 3. Then, we have x3 O1and x3 O2, which implies that O1 O2 6= , which is a contradiction.Therefore, every net in X has at most one limit point.

If Suppose X is not Hausdorff. Then, xA, xB X with xA 6= xBsuch that OA, OB O with xA OA and xB OB, we have OAOB 6= .Let := {(OA, OB) | xA OA O, xB OB O}. Clearly, (X ,X ) ,then 6= . Define a relation on by, (OA1, OB1), (OA2, OB2) , wesay (OA1, OB1) (OA2, OB2) if OA1 OA2 and OB1 OB2. Clearly, istransitive on . (OA1, OB1), (OA2, OB2) , we have xA OA3 := OA1OA2 O and xB OB3 := OB1OB2 O. Then, we have (OA3, OB3) ,(OA1, OB1) (OA3, OB3), and (OA2, OB2) (OA3, OB3). Hence, A :=(,) is a directed system. (OA, OB) , OA OB 6= . By Axiomof Choice, we may have a mapping x(OA ,OB) OA OB , (OA, OB) .Then, the net

(x(OA,OB)

)(OA,OB)A X . OA1 O with xA OA1. Fix

OB1 := X O with xB OB1. Then, (OA1, OB1) . (OA2, OB2) with (OA1, OB1) (OA2, OB2), we have x(OA2,OB2) OA2 OB2 OA1 OB1 = OA1. Hence, xA is a limit point of

(x(OA,OB)

)(OA,OB)A.

OB1 O with xB OB1. Fix OA1 := X O with xA OA1. Then,(OA1, OB1) . (OA2, OB2) with (OA1, OB1) (OA2, OB2), we havex(OA2,OB2) OA2 OB2 OA1 OB1 = OB1. Hence, xB is a limit pointof(x(OA,OB)

)(OA,OB)A. This contradicts with the assumption that every

net has at most one limit point. Therefore, X is Hausdorff.(ii) Let X be Hausdorff and net (x )A X satisfy limA x = x

X . Clearly, x is a cluster point of the net (x )A. Let y X with y 6= xbe another cluster point of the net (x )A. Then, there exist O1, O2 Owith x O1 and y O2 and O1 O2 = . Since x is the limit of the net,then 1 A such that A with 1 , we have x O1. Then,x 6 O2. This contradicts the definition that y is cluster point of (x )A.Hence, the net has exactly one cluster point, which is x.


Proposition 3.66 Let X and Y be topological spaces, D X with subsettopology OD, and f : D Y. Then, the following are equivalent.

(i) f is continuous at x0 D;

(ii) net (x )A D with x0 as a limit point, we have that the net(f(x) )A has a limit point f(x0).

(iii) net (x )A D with x0 as a cluster point, we have that the net(f(x) )A Y has a cluster point f(x0).

Proof (i) (ii). Fix a net (x )A D with x0 D as a limit point.OY OY with f(x0) OY . By the continuity of f at x0, OX OX withx0 OX such that f(OX) OY . Since x0 is a limit point of (x )A,


then 0 A such that, A with 0 , we have x OX . Then,f(x) OY . Hence, we have f(x0) is a limit point of (f(x) )A.

(ii) (i). Suppose f is not continuous at x0 D. Then, OY 0 OYwith f(x0) OY 0 such that, OX OX with x0 OX , we have f(OX) 6OY 0. Let M := {O OX | x0 O}. Clearly, X M and M 6= . Definea relation on M by, O1, O2 M, we say O1 O2 if O1 O2. Clearly, is transitive on M. O1, O2 M, let O3 = O1 O2 OX and x0 O3.Then, O3 M, O1 O3, and O2 O3. Hence, A := (M,) is a directedsystem. O M, f(O) \ OY 0 6= . By Axiom of Choice, we may definea net (xO )OA by xO O D with f(xO) 6 OY 0. Clearly, x0 is a limitpoint of (xO )OA. Yet, f(x0) OY 0 OY and f(xO) 6 OY 0, O M.Then, f(x0) is not a limit point of the net (f(xO) )OA. This contradictswith the assumption. Therefore, f is continuous at x0.

(i) (iii). Fix a net (x )A D with x0 as a cluster point. OY OY with f(x0) OY , by the continuity of f at x0, U OX with x0 Usuch that f(U) OY . By Definition 3.63, A, 0 A with 0,x0 U . Then, f(x0) OY . Hence, f(x0) is a cluster point of the net(f(x) )A.

(iii) (i). Suppose f is not continuous at x0. Let M := {O OX |x0 O}. Clearly, A := (M,) is a directed system. OY 0 OY withf(x0) OY 0 such that U M, we have f(U) 6 OY 0. By Axiom ofChoice, we may assign to each U M an xU U D such that f(xU ) OY 0. Consider the net (xU )UA D. Clearly, x0 is a limit point of the net,and therefore is a cluster point of the net. Consider the net (f(xU ) )UA.

For the open set OY 0 f(x0), U A, f(xU ) OY 0. Then, f(x0) isnot a cluster point of (f(xU ) )UA. This contradicts with the assumption.Therefore, f must be continuous at x0.


Proposition 3.67 Let (X,O) be a topological space, , where is an index set. Let (X,O) be the product space (X,O). Let(x )A X be a net. Then, x0 X is a limit point of (x )A if, andonly if, , (x0) X is a limit point of ((x) )A.

Proof Only if , by Proposition 3.27, is continuous. Then, is continuous at x0 X , by Proposition 3.9. By Proposition 3.66, (x0)is a limit point of the net ((x) )A.

If Suppose that x0 X is not a limit point of the net (x )A.Then, a basis open set B O with x0 B such that, 0 A, Awith 0 , we have x 6 B. Then, B =

O, O O, ,

and O = X for all s except finitely many s, say N . Then,0 A, A with 0 , we have x 6 B. This implies that (x) 6 O , for some N . Then, by an argument of contradiction,we may show that 0 N such that, 0 A, A with 0 ,we have 0(x) 6 O0 . Hence, 0(x0) O0 is not the limit of the net


(0(x) )A. This contradicts with the assumption. Hence, we have x0is a limit point of (x )A.


Proposition 3.68 Let (X,O) be a topological space, E X, and x X.x E if, and only if, a net (x )A E such that x is a limit point ofthe net.

Proof Only if Let M := {O O | x O}. Clearly, X M, thenM 6= . Clearly, A := (M,) is a directed system. Since x E, then,by Proposition 3.3, O A, O E 6= . By Axiom of Choice, a net(xO )OA E such that xO O E, O A. O O with x O, thenO A. O1 A with O O1, we have xO1 O1 E O. Hence, x is alimit point of (xO )OA.

If Let (x )A E be the net such that x is a limit point of thenet. O O with x O, 0 A, A with 0 , we havex E O. Since (x )A is a net, then 1 A with 0 1. Then,x1 O E 6= . By Proposition 3.3, x E.


Definition 3.69 Let (X,O) be a topological space, A := (A,) be a di-rected system, and (x )A X be a net. Let As A be a subset withthe same relation as A such that A, s As such that s.Then, As := (As,) is a directed system and (x )As is a net, which iscalled a subnet of (x )A.

Proposition 3.70 Let (X,O) be a topological space and (x )A X bea net. Then, x0 X is a limit point of (x )A if, and only if, any subnet(x )As has a limit point x0.

Proof Only if Since x0 X is a limit point of (x )A, thenO O with x0 O, 0 A such that, A with 0 , we havex O. Let (x )As be a subnet. Then, s0 As such that 0 s0 .s As with s0 s, we have 0 s and xs O. Hence, x0 is alimit point of the subnet.

If Since (x )A is a subnet of itself, then it has limit x0.This completes the proof of the proposition. 2A cluster point of a subnet is clearly a cluster point of the net.

Proposition 3.71 Let (X,O) be a topological space and (x )A X bea net. Then, x0 X is a limit point of (x )A if, and only if, for everysubnet (x )As of (x )A, there exists a subnet (x )Ass that has alimit point x0.

Proof Sufficiency We assume that every subnet (x )As of(x )A, there exists a subnet (x )Ass that has a limit point x0. We


will prove the result using an argument of contradiction. Suppose x0 is nota limit point of (x )A. Then, O0 O with x0 O0, 0 A, Awith 0 such that x O0. Define As := (

{ A

x O0},).

Clearly, (x )As is a subnet of (x )A. Any subnet (x )Ass of

(x )As , ss Ass, we have xss O0. Then, x0 is not a limit of(x )Ass . This contradicts the assumption. Therefore, x0 is a limit pointof (x )A.

Necessity Let x0 be a limit point of (x )A and (x )As be asubnet. By Proposition 3.70, x0 is a limit point of (x )As , which is asubnet of itself. Then, the result holds.


Definition 3.72 Let X := (X,OX) and Y := (Y,OY ) be topological spaces,D X , f : D Y, and x0 X be an accumulation point of D. y0 Yis said to be a limit point of f(x) as x x0 if OY OY with y0 OY ,U OX with x0 U such that f(U \ {x0}) = f((D U) \ {x0}) OY .We will also say that f(x) converges to y0 as x x0.

When basis are available on topological spaces X and Y, in Definition 3.72,we may restrict the open sets OY and U to be basis open sets withoutchanging the meaning of the definition.

Proposition 3.73 Let X := (X,OX) and Y := (Y,OY ) be topologicalspaces, D X , f : D Y, and x0 X be an accumulation point of D. IfY is Hausdorff, then there is at most one limit point of f(x) as x x0. Inthis case, we will write limxx0 f(x) = y0 Y when the limit exists.

Proof Suppose f(x) admits limit points yA, yB Y as x x0 withyA 6= yB. Since Y is Hausdorff, then UA, UB OY such that yA UA,yB UB, and UA UB = . Since yA is a limit point of f(x) as x x0,then VA OX with x0 VA such that f(VA \ {x0}) UA. Since yBis a limit point of f(x) as x x0, then VB OX with x0 VB suchthat f(VB \ {x0}) UB. Then, x0 V := VA VB O. Since x0is an accumulation point of D, then x (D V ) \ {x0}. Then, we havef(x) UA since x (DVA)\{x0} and f(x) UB since x (DVB)\{x0}.Then, f(x) UA UB 6= . This contradicts with UA UB = . Hence,the result holds. This completes the proof of the proposition. 2

Proposition 3.74 Let X := (X,OX) and Y := (Y,OY ) be topologicalspaces, D X with subset topology OD, f : D Y, and x0 D. Then,the following statements are equivalent.

(i) f is continuous at x0.

(ii) If x0 is an accumulation point of D, then f(x0) is a limit point off(x) as x x0.


Proof (i) (ii). This is straightforward.(ii) (i). We will distinguish two exhaustive and mutually exclusive

cases: Case 1: x0 is not an accumulation point of D; Case 2: x0 is anaccumulation point of D. Case 1: x0 is not an accumulation point of D.V OX with x0 V such that V D = {x0}. U OY with f(x0) U ,we have f(V ) = {f(x0)} U . Hence, f is continuous at x0.

Case 2: x0 is an accumulation point of D. U OY with f(x0) U ,V OX with x0 V such that f(V \{x0}) U . Then, we have f(V ) U .Hence, f is continuous at x0.

In both cases, f is continuous at x0.This completes the proof of the proposition. 2

Proposition 3.75 Let X := (X,OX), Y := (Y,OY ), and Z := (Z,OZ) betopological spaces, D X , f : D Y, x0 X be an accumulation point ofD, y0 Y be a limit point of f(x) as x x0, and g : Y Z be continuousat y0. Then, g(y0) Z is a limit point of g(f(x)) as x x0. When Y andZ are Hausdorff, then we may write limxx0 g(f(x)) = g(limxx0 f(x)).

Proof OZ OZ with g(y0) OZ , by the continuity of g at y0,OY OY with y0 OY such that g(OY ) OZ . Since y0 is the limit off(x) as x x0, then OX OX with x0 OX such that f(OX \ {x0}) OY . Then, g(f(OX \ {x0})) OZ . Hence, g(f(x)) converges to g(y0) asx x0. This completes the proof of the proposition. 2

Proposition 3.76 Let X be a topological space, D X , x0 X be anaccumulation point of D, Y and Z be Hausdorff topological spaces, D Y,y0 Y be an accumulation point of D, f : D D, and g : D Z.Assume that

(i) O0 OX with x0 O0 such that f(O0 \ {x0}) D \ {y0};

(ii) limxx0 f(x) = y0 and limyy0 g(y) = z0 Z.Then, limxx0 g(f(x)) = z0.

Proof OZ OZ with z0 OZ , by limyy0 g(y) = z0, OY OYwith y0 OY such that g(OY \ {y0}) = g((OY D) \ {y0}) OZ . Bylimxx0 f(x) = y0, OX OX with x0 OX such that f(OX \ {x0}) =f((OX D) \ {x0}) OY . Let O1 := O0 OX OX . Clearly, x0 O1.Then, x (O1 D) \ {x0}, we have f(x) OY (D \ {y0}) = (OY D) \ {y0} and g(f(x)) OZ . Hence, g(f((O1 D) \ {x0})) OZ . Hence,limxx0 g(f(x)) = z0. This completes the proof of the proposition. 2

Proposition 3.77 Let X , Y, and Z be Hausdorff topological spaces, D X , x0 X be an accumulation point of D, D Y, y0 Y be an accumu-lation point of D, f : D D be bijective, and g : D Z. Assume thatlimxx0 f(x) = y0 and limyy0 f inv(y) = x0. Then, limxx0 g(f(x)) =limyy0 g(y) whenever one of the limits exists in Z.


Proof We will prove the result by distinguishing two exhaustive cases:Case 1: limyy0 g(y) = z0 Z; Case 2: limxx0 g(f(x)) = z0 Z.

Case 1: limyy0 g(y) = z0 Z. We will further distinguish threeexhaustive and mutuall

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