Reading Notes of “Real Analysis” 3rd Edition by H. L. Royden

8/22/2019 Reading Notes of Real Analysis 3rd Edition by H. L. Royden

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Reading Notes of Real Analysis 3rd Edition

by H. L. Royden

Zigang Pan

April 19, 2013


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Preface

This is a reading note of the book Royden (1988) and the MATH 441 & 442notes by Prof. Peter Leob of University of Illinois at Urbana-Champaign.

In Chapter 6, I have also included some material from the book Maun-der (1996). The Chapters 610 include a significant amount of materialfrom the book Luenberger (1969). Chapter 9 also referenced Bartle (1976).Chapter 11 includes significant amount of self-developed material due tolack of reference on this subject. The proof of Radon-Nikodym Theorem11.167 was adapted from the MATH 442 notes by Prof. Peck of Universityof Illinois at Urbana-Champaign. The book Royden (1988) does offer someclues as to how to invent the wheel and the book Bartle (1976) is sometimesused to validate the result.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

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Contents

Preface 3

1 Notations 9

2 Set Theory 172.1 Axiomatic Foundations of Set Theory . . . . . . . . . . . . 172.2 Relations and Equivalence . . . . . . . . . . . . . . . . . . . 182.3 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . 202.5 Algebra of Sets . . . . . . . . . . . . . . . . . . . . . . . . . 212.6 Partial Ordering and Total Ordering . . . . . . . . . . . . . 222.7 Basic Principles . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Topological Spaces 33

3.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 333.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.3 Basis and Countability . . . . . . . . . . . . . . . . . . . . . 393.4 Products of Topological Spaces . . . . . . . . . . . . . . . . 403.5 The Separation Axioms . . . . . . . . . . . . . . . . . . . . 453.6 Category Theory . . . . . . . . . . . . . . . . . . . . . . . . 463.7 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 483.8 Continuous Real-Valued Functions . . . . . . . . . . . . . . 533.9 Nets and Convergence . . . . . . . . . . . . . . . . . . . . . 57

4 Metric Spaces 714.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 714.2 Convergence and Completeness . . . . . . . . . . . . . . . . 73

4.3 Uniform Continuity and Uniformity . . . . . . . . . . . . . 764.4 Product Metric Spaces . . . . . . . . . . . . . . . . . . . . . 784.5 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.6 Baire Category . . . . . . . . . . . . . . . . . . . . . . . . . 844.7 Completion of Metric Spaces . . . . . . . . . . . . . . . . . 854.8 Metrization of Topological Spaces . . . . . . . . . . . . . . . 904.9 Interchange Limits . . . . . . . . . . . . . . . . . . . . . . . 91

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6 CONTENTS

5 Compact and Locally Compact Spaces 955.1 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . 955.2 Countable and Sequential Compactness . . . . . . . . . . . 1015.3 Real-Valued Functions and Compactness . . . . . . . . . . . 1035.4 Compactness in Metric Spaces . . . . . . . . . . . . . . . . 1055.5 The Ascoli-Arzela Theorem . . . . . . . . . . . . . . . . . . 1075.6 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1095.7 Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . 112

5.7.1 Fundamental notion . . . . . . . . . . . . . . . . . . 1125.7.2 Partition of unity . . . . . . . . . . . . . . . . . . . . 1155.7.3 The Alexandroff one-point compactification . . . . . 1185.7.4 Proper functions . . . . . . . . . . . . . . . . . . . . 119

5.8 -Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . 1215.9 Paracompact Spaces . . . . . . . . . . . . . . . . . . . . . . 1225.10 The Stone-Cech Compactification . . . . . . . . . . . . . . . 126

6 Vector Spaces 129

6.1 Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.2 Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1316.3 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1326.4 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 1326.5 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1346.6 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1356.7 Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 1396.8 Linear Independence and Dimensions . . . . . . . . . . . . . 142

7 Banach Spaces 145

7.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . 1457.2 The Natural Metric . . . . . . . . . . . . . . . . . . . . . . . 1517.3 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1547.4 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1567.5 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . 1617.6 Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . 1637.7 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . 1657.8 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . 1737.9 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

7.9.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . 177

7.9.2 Duals of some common Banach spaces . . . . . . . . 1787.9.3 Extension form of Hahn-Banach Theorem . . . . . . 1827.9.4 Second dual space . . . . . . . . . . . . . . . . . . . 1907.9.5 Alignment and orthogonal complements . . . . . . . 192

7.10 The Open Mapping Theorem . . . . . . . . . . . . . . . . . 1977.11 The Adjoints of Linear Operators . . . . . . . . . . . . . . . 2017.12 Weak Topology . . . . . . . . . . . . . . . . . . . . . . . . . 204


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CONTENTS 7

8 Global Theory of Optimization 2138.1 Hyperplanes and Convex Sets . . . . . . . . . . . . . . . . . 213

8.2 Geometric Form of Hahn-Banach Theorem . . . . . . . . . 216

8.3 Duality in Minimum Norm Problems . . . . . . . . . . . . . 218

8.4 Convex and Concave Functionals . . . . . . . . . . . . . . . 221

8.5 Conjugate Convex Functionals . . . . . . . . . . . . . . . . 225

8.6 Fenchel Duality Theorem . . . . . . . . . . . . . . . . . . . 232

8.7 Positive Cones and Convex Mappings . . . . . . . . . . . . 239

8.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . 241

9 Differentiation in Banach Spaces 249

9.1 Fundamental Notion . . . . . . . . . . . . . . . . . . . . . . 249

9.2 The Derivatives of Some Common Functions . . . . . . . . 254

9.3 Chain Rule and Mean Value Theorem . . . . . . . . . . . . 257

9.4 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . 263

9.4.1 Basic concept . . . . . . . . . . . . . . . . . . . . . . 263

9.4.2 Interchange order of differentiation . . . . . . . . . . 269

9.4.3 High order derivatives of some common functions . . 274

9.4.4 Properties of high order derivatives . . . . . . . . . . 278

9.5 Mapping Theorems . . . . . . . . . . . . . . . . . . . . . . . 285

9.6 Global Inverse Function Theorem . . . . . . . . . . . . . . . 296

9.7 Interchange Differentiation and Limit . . . . . . . . . . . . 304

9.8 Tensor Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 307

10 Local Theory of Optimization 311

10.1 Basic Notion . . . . . . . . . . . . . . . . . . . . . . . . . . 311

10.2 Unconstrained Optimization . . . . . . . . . . . . . . . . . . 317

10.3 Optimization with Equality Constraints . . . . . . . . . . . 320

10.4 Inequality Constraints . . . . . . . . . . . . . . . . . . . . . 325

11 General Measure and Integration 333

11.1 Measure Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 333

11.2 Outer Measure and the Extension Theorem . . . . . . . . . 339

11.3 Measurable Functions . . . . . . . . . . . . . . . . . . . . . 352

11.4 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 366

11.5 General Convergence Theorems . . . . . . . . . . . . . . . . 38111.6 Banach Space Valued Measures . . . . . . . . . . . . . . . . 397

11.7 Calculation With Measures . . . . . . . . . . . . . . . . . . 429

11.8 The Radon-Nikodym Theorem . . . . . . . . . . . . . . . . 461

11.9 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477

11.10Dual ofC(X,Y) . . . . . . . . . . . . . . . . . . . . . . . . . 492


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8 CONTENTS

12 Differentiation and Integration 51712.1 Caratheodory Extension Theorem . . . . . . . . . . . . . . 51712.2 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . 52312.3 Product Measure . . . . . . . . . . . . . . . . . . . . . . . . 52812.4 Functions of Bounded Variation . . . . . . . . . . . . . . . . 55512.5 Absolute and Lipschitz Continuity . . . . . . . . . . . . . . 57812.6 Fundamental Theorem of Calculus . . . . . . . . . . . . . . 59012.7 Representation of (Ck(,Y)) . . . . . . . . . . . . . . . . . 599

13 Hilbert Spaces 60313.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 60313.2 Projection Theorems . . . . . . . . . . . . . . . . . . . . . . 60713.3 Dual of Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . 609

13.4 Hermitian Adjoints . . . . . . . . . . . . . . . . . . . . . . . 61113.5 Approximation in Hilbert Spaces . . . . . . . . . . . . . . . 61313.6 Other Minimum Norm Problems . . . . . . . . . . . . . . . 620

14 Probability Theory 62514.1 Fundamental Notions . . . . . . . . . . . . . . . . . . . . . . 625

15 Numerical Methods 62715.1 Newtons Method . . . . . . . . . . . . . . . . . . . . . . . . 627

Bibliography 631

Index 632


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Chapter 1

Notations

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

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

IK either IR orCZ+,Z,C+,C IN {0},Z \ IN, the open right half of the complex

plane, the open left half of the complex plane, re-spectively

belong to 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 sequencex1, x2, . . .

( x ) the ordered collectionidA the identity map on a setA

| | the absolute value of a real or complex number the complex conjugate of a complex numberRe ( ) the real part of a complex numberIm ( ) the imaginary part of a complex numbera b the maximum of two real numbersa and ba b the minimum of two real numbersa and b the empty set; See Page 17.

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10 CHAPTER 1. NOTATIONS

{x, y} an unordered pair; See Page 17.X2 the collection of all subsets of X; See Page 17. the set union; See Page 17.(x, y) an ordered pair; See Page 18.X Y the Cartesian or direct product of sets X and Y;

See Page 18.A= {x B| P(x)} Definition of a set A; See Page 18.x y x and y are related in a relation; See Page 18.X/ the quotient of the setXwith respect to an equiv-

alence relation; See Page 18.f :X Y a function ofXto Y; { (x, f(x)) XY | x X}

is the graph off; See Page 19.graph( f) the graph of a function f; See Page 19.

dom(f) the domain of f; See Page 19.f(A) the image of A Xunder f; See Page 19.range( f) the range of f, equals to f(X); See Page 19.finv(A) the preimage of A Y under f; See Page 19.onto, surjective f(X) = Y; See Page 19.1-1, injective f(x1)=f(x2) ifx1, x2X and x1=x2; See Page

19.bijective both surjective and injective; See Page 19.finv the inverse function off; See Page 19.g f the composition ofg : Y Zwith f :X Y; See

Page 19.f|A the restriction off toA; See Page 19.YX the set of all functions ofX toY ; See Page 19.

the set intersection; See Page 20.A the compliment of a set A, where the whole set isclear from context; See Page 20.

\ set minus; See Page 20.A B the symmetric difference ofA and B, equals to (A \

B) (B \ A); See Page 20.card(X) the number of elements in the finite setX; See Page

21. X the Cartesian or direct product ofXs; See Page

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

space to one of the coordinates; See Page 31.

A the closure of a setA, where the whole set is clearfrom context; See Page 33.A the interior of a set A, where the whole set is clear

from context; See Page 33.A the boundary of a setA, where the whole set is clear

from context; See Page 33.(X, O) the product topological space; See Page 40.


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T1 Tychonoff space; See Page 45.T2 Hausdorff space; See Page 45.T3 regular space; See Page 45.T4 normal space; See Page 45.T3 12 completely regular space; See Page 56.

( x )A a net; See Page 57.limA x the limit of a net; See Page 57.limxx0f(x) the limit of f(x) as x x0; See Page 61.IRe the set of extended real numbers, which equals to

IR {, +}; See Page 63.lim supA x the limit superior of a real-valued net; See Page 65.liminfA x the limit inferior of a real-valued net; See Page 65.lim supxx0f(x) the limit superior off(x) as x

x0; See Page 66.

liminfxx0f(x) the limit inferior of f(x) as x x0; See Page 66.BX( x0, r ) the open ball centered atx0 with radiusr; See Page

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

Page 71.dist(x0, S) the distance from a point x0 to a set S in a metric

space; See Page 73.(X, X ) (Y, Y) the finite product metric space; See Page 78.(

i=1 Xi, ) the countably infinite product metric space; SeePage 81.

supp f the support of a function; See Page 115.(X) the Stone-Cech compatification of a completely reg-

ular topological spaceX

; See Page 126.(M(A, Y), F) the vector space of Y-valued functions of a set A

over the fieldF; See Page 134.X the null vector of a vector spaceX; See Page 136.

N( A ) the null space of a linear operatorA; See Page 136.R ( A ) the range space of a linear operator A; See Page 136.S the scalar multiplication by of a set Sin a vector

space; See Page 137.S+ T the sum of two setsSand T in a vector space; See

Page 137.span( A ) the subspace generated by the setA; See Page 138.v ( P) the linear variety generated by a nonempty set P;

See Page 138.

co ( S) the convex hull generated by S in a vector space;See Page 139.x the norm of a vectorx; See Page 145.|x | the Euclidean norm of a vectorx; See Page 145.C1([a, b]) the normed linear space of continuously differen-

tiable real-valued functions on the interval [a, b]; SeePage 146.


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lp The normed linear space of real-valued sequenceswith finitep-norm, 1 p +; See Page 147.

lp(X) the normed linear space of X-valued sequences withfinitep-norm, 1 p +; See Page 151.

V ( P) the closed linear variety generated by a nonemptyset P; See Page 154.

P the relative interior of a setP; See Page 154.X Y the finite Cartesian product normed linear space;

See Page 154.C([a, b]) the Banach space of continuous real-valued func-

tions on the interval [a, b]; See Page 158.C(K,X) the normed linear space of continuous X-valued

functions on a compact space

K; See Page 158.

XIR the real normed linear space induced by a complexnormed linear space X; See Page 162.

[ x ] the coset of a vectorx in a quotient space; See Page163.

X/M the quotient space of a vector spaceX modulo asubspaceM; See Page 163.

X/M the quotient space of a normed linear space X mod-ulo a closed subspace M; See Page 164.

Cv(X,Y) the vector space of continuous Y-valued functions onX; See Page 166.

B (X,Y) the set of bounded linear operators of X to Y; SeePage 173.

X the dual ofX; See Page 177.

x a vector in the dual; See Page 177.x, x the evaluation of a bounded linear functionalx at

the vectorx, that is x(x); See Page 177.c0(X) the subspace of l(X) consisting of X-valued se-

quences with limit X; See Page 181.X the second dual ofX; See Page 190.S the orthogonal complement of the set S; See Page

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

Page 192.A the adjoint of a linear operator A; See Page 201.A the adjoint of the adjoint of a linear operatorA; See

Page 202.Oweak(X ) the weak topology on a normed linear space X; SeePage 204.

Xweak the weak topological space associated with a normedlinear space X; See Page 205.

Oweak(X ) the weak topology on the dual of a normed linearspace X; See Page 207.


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Xweak the weak

topological space associated with anormed linear space X; See Page 208.

Ksupp the support of a convex setK; See Page 218.[ f, C] the epigraph of a convex function f : C IR; See

Page 221.Cconj the conjugate convex set; See Page 225.fconj the conjugate convex functional; See Page 225.[ f, C] conj the epigraph of the conjugate convex functional; See

Page 226.

conj the pre-conjugate convex set; See Page 228.

conj the pre-conjugate convex functional; See Page 228.

conj[ , ] the epigraph of the pre-conjugate convex functional;See Page 228.

= greater than or equal to (with respect to the positivecone); See Page 239.

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

cone); See Page 239. greater than (with respect to the positive cone); See

Page 239. less than (with respect to the positive cone); See

Page 239.S the positive conjugate cone of a set S; See Page 239.S the negative conjugate cone of a set S; See Page

239.AD( x0 ) the set of admissible deviations inDat x0; See Page

249.f(1)(x0), Df(x0) the Frechet derivative off atx0; See Page 250.Df(x0; u) the directional derivative of f at x0 along u; See

Page 250.fy (x0, y0) partial derivative of fwith respect to y at (x0, y0);

See Page 251.ro(A)(B) right operate: ro(A)(B) =BA; See Page 256.Bk(X,Y) the set of bounded multi-linear Y-valued functions

on Xk; See Page 263.BSk(X,Y) the set of symmetric bounded multi-linear Y-valued

functions on Xk; See Page 263.Dkf(x0), f(k)(x0) thekth order Frechet derivative offatx0; See Page

263.

Ck,C k-times and infinite-times continuously differen-tiable functions, respectively; See Page 264.

kfxik xi1

kth-order partial derivative off; See Page 269.

Sm ( D ) =

IK D; See Page 269.


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Ck(,Y) the normed linear space of k-times continuouslydifferentiable Y-valued functions on a compact set X; See Page 305.

Cb (X,Y) the normed linear space of bounded continuous Y-valued functions on a topological space X; See Page306.

Cb k(,Y) the normed linear space of k-times bounded con-tinuously differentiable Y-valued functions on a set X; See Page 307.

ATn1,...,nm the transpose of an mth order tensor A with thepermutation (n1, . . . , nm); See Page 308.

A B the outer product of two tensors; See Page 309.0m1mn an nth order IK-valued tensor in

B (IKmn , . . . , B ( IKm1 , IK ) ) with all elementsequal to 0; See Page 309.

1m1mn an nth order IK-valued tensor inB (IKmn , . . . , B ( IKm1 , IK ) ) with all elementsequal to 1; See Page 309.

S+X,SpsdX sets of positive definite and positive semi-definiteoperators over the real normed linear space X, re-spectively; See Page 312.

SX,SnsdX sets of negative definite and negative semi-definiteoperators over the real normed linear space X, re-spectively; See Page 312.

SX BS 2(X, IR ); See Page 312.(IR,

BL, L) Lebesgue measure space; See Page 344.

Lo Lebesgue outer measure; See Page 345.BB(X) Borel sets; See Page 345.A (X) the algebra generated by the topology of a topolog-

ical spaceX; See Page 345.B the Borel measure on IR; See Page 346.R ((IR, IR, | |), BB( IR ) , B); See Page 350.Pa.e. inX Pholds almost everywhere inX; See Page 354.P(x) a.e. x X Pholds almost everywhere inX; See Page 354.P f P f : X [0, ) IR defined byP f(x) =

f(x),x X; See Page 364.R (X ) the collection of all representation of X; See Page

366.I (X ) the integration system on X; See Page 366.

Xfd the integral of a functionfon a set Xwith respect

to measure; See Page 367.X

f(x) d(x) the integral of a functionfon a set Xwith respectto measure; See Page 367.

P the total variation of a Banach space valued pre-measure; See Page 397.


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P the total variation of a Banach space valued measure; See Page 402.

1+ 2 theY-valued measure that equals to the sum of twoY-valued measures on the same measurable space;See Page 433.

the Y-valued measure that equals to the scalar prod-uct of IK and Y-valued measure; See Page 433.

y the Y-valued measure that equals to scalar productof a IK-valued measure and y Y; See Page 433.

A the Z-valued measure that equals to product of anbounded linear operatorA and a Y-valued measure; See Page 433.

M(X,

B,Y) the vector space of -finite Y-valued measures on

the measurable space (X, B); See Page 440.Mf(X, B,Y) the normed linear space of finite Y-valued measures

on the measurable space (X, B); See Page 441.limnIN n= a sequence of-finite (Y-valued) measures ( n )

n=1

converges to a -finite (Y-valued) measure ; SeePage 444.

1 2 the measures 1 and 2 on the measurable space(X, B) can be compared if1(E) 2(E),E B;See Page 444.

1,1 1,m...

...n,1 n,m

the vector measure; See Page 447.

1 2 the measure 1is absolutely continuous with respectto the measure 2; See Page 457.

1 2 the measures1 and 2 are mutually singular; SeePage 457.

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

valued measure with respect to the -finite IK-valued measure; See Page 466.

Pp f the functionf()p; See Page 477.ess sup the essential supremum; See Page 479.limnIN zn

.=z in Lp the sequence (zn )

n=1 Lp converges to z Lp in

Lp pseudo-norm; See Page 481.Mft (X,Y) the normed linear space of finite Y-valued topologi-

cal measures on

X; See Page 497.

Mt(X,Y) the vector space of -finite Y-valued topologicalmeasures onX; See Page 498.

M(X, B) the set of -finite measures on the measurable space(X, B); See Page 498.

Mf(X, B) the set of finite measures on the measurable space(X, B); See Page 498.


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Mt(X) the set of -finite topological measures on the topo-logical spaceX; See Page 499.

Mft (X) the set of finite topological measures on the topo-logical spaceX; See Page 499.


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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 andx and y be objects (which is another name for sets).

Axiom 1 (Axiom of Extensionality) A= B ifx A, we havex B;andx B, we havex A.

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

Axiom 3 (Axiom of Pairing) For any objectsx andy, there exists a set{x, y}, which contains onlyx andy .

Axiom 4 (Axiom of Regularity) Any nonempty setA = , there existsa A, such thatb A, we haveb a.

Axiom 5 (Axiom of Replacement)x A, let there be one and onlyoney to form an ordered pair(x, y). Then, the collection of all suchy s isa setB .

Axiom 6 (Axiom of Power Set) The collection of all subsets ofA is a

set denoted by A

2.

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 setA such that A andx A, we have{, x} A.

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18 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 anyx 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 by

xA

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 subsetofA, we will write As

A. Thus, the formula

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

defines a set as long asA is a set andp(x) is unambiguous logic expression.

2.2 Relations and Equivalence

Definition 2.1 LetA andB be sets. A relationR fromA to B is a subsetofA B.x A,y B, we sayx y if(x, y) R. We will say thatRis arelation on A if it is a relation fromA to A. We define

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

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

which are well-defined subsets.

Definition 2.2 LetA be a set andR be a relation onA.x,y,z A,1. R isreflexive ifx x.2. R issymmetric ifx y impliesy x.3. R istransitive ifx y andy z impliesx z.4. Ris aequivalencerelationship if it is reflexive, symmetric, and tran-

sitive, and will be denote by

.

5. R isantisymmetric ifx y andy x impliesx= y.

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


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2.3. FUNCTION 19

2.3 FunctionDefinition 2.3 LetX and Y be sets andD X. A function f of D toY, denoted by f : D Y, is a relation from D to Y such thatx D,there is exactly oney Y, such that (x, y) f; we will denote thaty asf(x). Thegraphoffis the setgraph( f) := {(x, f(x)) XY | x D }.The domain of f isdom( f) = D. A X, the image under f of A isf(A) := {y Y | x AD such thatf(x) =y }, which is a subset ofY.The range off is range( f) =f(X).B Y, theinverse image underfofB isfinv(B) := {x D| f(x) B }, which is a subset ofD . f is saidto besurjective iff(X) = Y; andfis said to beinjectiveiff(x1) =f(x2),x1, x2 Dwithx1=x2;fis said to bebijectiveif it is both surjective andinjective, in which case it isinvertible and the inverse function is denoted

byfinv: Y D. We will say thatfis a function fromX to Y.Let f : DY and g : YZbe functions, we may define a function

h: D Zbyh(x) = g(f(x)), then h is called thecompositionofg withf,and is denoted bygf. LetA X. We may define a function l : AD Ybyl(x) =f(x),xA D. This function is called the restrictionoff toA, and denoted by f|A. Let f : DY, g : YZ, and h : Z W, wehave (h g) f = h (g f). Let f : D D and k Z+, we will writefk :=f f

k

, wheref0 := idD.

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

YX

:= 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 yrange( ), and hence, issurjective. Suppose that is not injective, thenx1, x2X withx1=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: YX exists.xX,let y = (x), then x = inv(y), hence inv((x)) = x. Therefore, we haveinv= idX .y Y, let x= inv(y), theny = (x), hence(inv(y)) = y.Therefore, we have inv= idY. Hence, 1 = 2 = inv.


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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. For bijective functions f :X Y andg : Y Z, g f is also bijective

and (g f)inv= finv ginv.

2.4 Set Operations

LetXbe a set, X

2 is the set consisting of all subsets ofX.A, B X, wewill define

A B := {x X| x Aor x B }A B := {x X| x Aand x B }A := {x X| x A}A \ B := {x A| x B } =A B

A B := (A \ B) (B \ A)

We have the following results.

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

1. A B= B A andA B= B A;2. A A B andA = A B if, and only if, B A;3. A =A, A = , A X=X, andA X=A;

4.= X,A = A, A A = X, A A =, andAB if, and only if,BA;5. The De Morgans Laws:

A

= A; A

= A6. B

A

=

(B A) andB

A

=

(B A);

7. f

A

=

f(A) andf

A

f(A);


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2.5. ALGEBRA OF SETS 21

8. finv

C =

finv(C) andfinv

C =

finv(C);

9. finv(C\E) = finv(C)\finv(E), f(finv(C)) = Crange( f), andfinv(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 setXis said to befiniteif 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 inX. It is said to becountableif it is either empty

or the range of a function ofIN.

Definition 2.7 LetXbe a set andA X2.A is said to be analgebra ofsets onX(or aBoolean algebra onX) if

(i), X A;(ii)A, B A, A B A andA A.

A is said to be a -algebra on X if it is an algebra on X and countableunions of sets inA is again inA.

LetM X2, where Xis 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 the

algebra onX generated byM. Also, there exists a smallest -algebra onX, A X2, containing M, which is said to be the-algebra onXgeneratedbyM.Proposition 2.8 LetXbe a set,Ebe a nonempty collection of subsets ofX,A be the algebra onXgenerated byE, and

A :=

A Xn, m IN, i1, . . . , i2n {1, . . . , m}, Fi1,...,i2n X

withFi1,...,i2n E or (Fi1,...,i2n) E, such that

A=

m

i1=1m

i2=1

m

i2n=1Fi1,...,i2n

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

i1=1

1i2=1

Fi1,i2 A. Hence, we haveE A. It is clear that A A. Allwe need to show is that Ais an algebra onX. Then,A Aand the resultfollows.


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Fix E E =. E X. Let n = 1, m = 2, F1,1 = E, F1,2 =E,F2,1 = E, and F2,2 =E. Then, =2i1=12i2=1 Fi1,i2 A. Let n = 1,m = 2, F1,1 = E, F1,2 = E, F2,1 =E, and F2,2 =E. Then, X =2

i1=1

2i2=1

Fi1,i2 A.A, B A, nA, mA IN, i1, . . . , i2nA {1, . . . , mA}, FAi1,...,i2nA X

with FA

i1,...,i2nA E or

FA

i1,...,i2nA

E such that A =mA

i1=1

mAi2=1

mAi2nA=1 FAi1,...,i2nA , and nB , mB IN, i1, . . . , i2nB {1, . . . , mB},FBi1,...,i2nB X with F

Bi1,...,i2nB

E or

FB

i1,...,i2nB

E

such that B =mB

i1=1

mBi2=1

mBi2nB=1 FBi1,...,i2nB .Note thatA= mAi1=1mAi2=1 mAi2nA=1 FAi1,...,i2nA

. Letn = nA+ 1,

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

FAi2,...,i2n1

. Then,A= mi1=1mi2=1 mi2n=1 Gi1,...,i2n A.Without loss of generality, assume nA nB . Let n = nA and m =

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

Gi1,...,i2n =

FA

i1 ,i2,...,i2nAifi1 mA

FBi1mA ,i2,...,i2nB

ifi1 > mA

Then, it is easy to check thatAB =

mi1=1

mi2=1

mi2n=1

Gi1,...,i2n A.Hence,

Ais an algebra on X.

This completes the proof of the proposition.

2.6 Partial Ordering and Total Ordering

Definition 2.9 LetA be a set and be a relation onA. will be calleda partial ordering if it is reflexive and transitive. It will be called atotalordering is it is an antisymmetric partial ordering and satisfiesx, y Awithx =y, we have eitherx y ory 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 LetA be a set with a partial ordering .1. a A is said to beminimal if,x A, x a impliesa x;2. a A is said to be the least element if,x A, a x, andx a

implies thatx= a.

3. a A is said to bemaximal if,x A, a x impliesx a;


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2.6. PARTIAL ORDERING AND TOTAL ORDERING 23

4. a A is said to be thegreatest element if,x A, x a, anda ximplies thatx= a.

Definition 2.11 LetA be a set with a partial ordering , andE A.1. aA is said to be anupper bound ofE ifxa,xE. It is the

least upper boundofEif it is the least element in the set of all upperbounds ofE;

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

We have the following results.

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

(i) Ifa A is the least element, then it is minimal.(ii) There is at most one least element inA.

(iii) Define a relation byx, 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) Ifa A is the greatest element, then it is maximal.(v) There is at most one greatest element inA.

(vi) If is antisymmetric, then a A is minimal if, and only if, theredoes not existx A such thatx a andx =a.

(vii) If is antisymmetric, thena A is maximal if, and only if, theredoes not existx A such thata x andx =a.

(viii) If is antisymmetric, then it is a total ordering if, and only if,

x1, x2

A, we havex1

x2 orx2

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, Sincex x implies x x, then is reflexive. Ifx y and y z, we havey xand z y, which impliesz x, and hence


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xz . 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 thereforex = 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, bya 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 thegreatest element, we have a2 a1. By a2 being the greatest element, wethen havea1= a2. Hence, the greatest element is unique if it exists.

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

For (vii), if,xA with ax, then, we have a = x, which meansthatx a; hence a is maximal. Only if, suppose thatx A such thatax and x=a. Note that xa sincea is maximal. Then, a= x, since is antisymmetric, which is a contradiction.

For (viii), if,x1, x2 A with x1= 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, thenx1x2; when x1=x2, then x1x2 or x2 x1; hence, in both cases, wehavex1 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 a

functionf :

A, such that, , we havef() A.With Axioms 18 holding, the Axiom of Choice is equivalent to thefollowing three results.

Theorem 2.13 (Hausdorff Maximal Principle) Letbe a partial or-dering on a setE. Then, there exists a maximal (with respect to set con-tainment) subsetF E, such that is a total ordering onF.


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2.7. BASIC PRINCIPLES 25

Theorem 2.14 (Zorns Lemma) Let be an antisymmetric partial or-dering on a nonempty setE. If every nonempty totally order subsetF ofEhas an upper bound inE, then, there is a maximal element inE.

Definition 2.15 Awell orderingof 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 LetEbe a nonempty set and

is an antisymmetric partialordering onE. Assume that every nonempty subsetSofE, on which isa total ordering, has a least upper bound inE. Letf :E Ebe a mappingsuch thatx f(x),x E. Then,fhas a fixed point onE, i. e.,w E,f(w) =w.

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

B = B E (i)a B (ii)f(B) B (iii)F B, F= Fis totally ordered withimplies that the least upperbound ofFbelongs toB .

Consider the set

B0 := {x E| a x}Clearly, B0 is nonempty since a B0 and

f(B0) = {f(x) E| a x f(x)} B0sincef satisfiesx f(x),x E.

For any F B0, such that F is totally ordered with and F=.Let e0 be the least upper bound of F in E. Then,x0 F such thata x0 e0, and therefore e0 B0.

This shows thatB0 B, andB is nonempty.The following result holds for the collectionB.

Claim 2.17.1 Let{B| } be any nonempty subcollection ofB, then B 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 lastfollows from the fact f(B) B, .(iii) Let F B, which is totally ordered byand F= .


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For any , F B implies that the least upper bound ofF is anelement ofB. Therefore, the least upper bound ofF is in the intersection

B.This establishes

B B, and completes the proof of the claim.

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

BBB. By

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

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

andx =y.Define the set P by

P= {x A| y A, y x f(y) x}Clearly, aP, since there does not exists any yA such that y a,

by being antisymmetric. Therefore, Pis nonempty.We claim that

Claim 2.17.2 (v)x P,z A, thenz x orf(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. Ifz = 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 havef(z) B. Then,f(B) B by the arbitraryness ofz B.(iii) LetF=be any totally ordered subset ofB ande0 Ebe the leastupper bound ofF. 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 impliese0 B . If, for any y F, y x, thenF {z A | z x}. This implies thatx is an upper bound ofF ande0 x, sincee0 is the least upper bound ofF. Therefore, e0 B. In bothof the cases, we have e0 B.

This establishes thatB B. ByA being the smallest set inB, we haveA= B. Therefore, the claim is proven.

Now, we show thatP B.(i)a P and therefore P= .(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),


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there are two exhaustive scenarios. Ify x, then y x. Iff(x) y, thenf(x) y f(x) form a contradiction by being antisymmetric. Therefore,we must havey x, which results in the following two exhaustive scenarios.Ify x, thenf(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 ofy , we have f(x)P, which further implies f(P)Pby the arbitraryness ofx P.(iii) Let F= be a totally ordered subset in P. Let e0 E be the leastupper bound ofF. We haveF A implies that e0 A by A B.z Awith ze0, implies that z must not be an upper bound ofF. Therefore,x0 Fsuch thatx0 z. By (v), we havez x0. Hence, byx0 F P,z A, and zx0, we have f(z) x0. Therefore, f(z) e0 sincee0 is anupper bound ofF. This further implies thate0

Pby the arbitraryness

ofz .This proves that P B.SinceP A and A is the smallest set inB, 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 andx2 are related through. Iff(x2) x1, then, x2f(x2) x1, which implies thatx1 and x2 are related through . Therefore,x1 andx2 are related throughin both cases. Then, by Proposition 2.12(viii), A is totally ordered by and nonempty. Let w Ebe the leastupper bound ofA. Then, w A, sinceA B.

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

This completes the proof of the lemma.

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

Clearly, E, thenE =. Define a partial ordering onE by, which isset containment. This partial ordering is clearly reflexive, transitive, andantisymmetric.

A E, define a collection

AA:= {B E | A B } ifB Esuch that A B

{A} otherwise


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Clearly,AA=. By Axiom of Choice,T :E E such that T(A) =BAA,A E.We will show that T admits a fixed point by Lemma 2.17. LetB E

be any nonempty subset on whichis a total ordering. Let C:= BBB.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 orderingonB, 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

Badmits least upper boundCin

Ewith respect to

. By the definition

ofT, it is clear that A T(A),A E. By Lemma 2.17, T has a fixedpoint onE, i.e.,A0 E such thatT(A0) = A0.

By the definitions ofT andAA0 , there does not exist B E such thatA0 B. Hence, by Proposition 2.12 (vii), A0 is maximal inEwith respectto.

2. 3. LetEbe 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=, otherwise, let x0 E (since E=), F {x0} E and{x0}is totally ordered by, which violates the fact that F is maximal (withrespect to). Then, Fhas an upper bound e0 E.Claim 2.18.1 e0 F.Proof of claim: Suppose e0 F. Define A:= F {e0} E. Clearly,F A and F=A. We will show thatis 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 havex1 x2orx2 x1since is a total ordering onF. In Case 2, we havex1 x2 = e0sincee0 is an upper bound ofF. 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= 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.

e1

E such that e0

e1.

x

F, we have x

e0

e1. Hence,e1

is an upper bound ofF. By Claim 2.18.1, we must have e1 F. Then,e1 e0 sincee0 is an upper bound ofF. This shows that e0 is maximal inEwith respect to.

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

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


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Then,E = . Define an ordering on Eby (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 2x2.

Now, we will show thatdefines 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: ifx2A2, then x2A2 \ A1 which implies x1 2x2and hence x1 3x2; ifx2 A3 \ A2, then we have x1 A2 and x1 3x2,thus, we have x1 3x2 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 onE.

LetA Ebe any nonempty subset totally ordered by. TakeA ={(A,)| } where = is an index set. DefineA :=

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) sinceA is totally ordered by, then x1, x2 A2, wewill say that x1 x2 ifx1 2x2. We will now show that this ordering isuniquely defined independent of (A2,2) A. Let (A3,3) A be suchthat x1, x2 A3. SinceA 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 haveA3 A2 and 3= 2on A3, which implies thatx1x2

x1 2x2

x1 3x2. In Case 2, we have A2

A3 and 3 = 2 onA2,

which implies that x1 x2 x1 2 x2 x13x2. Hence, the ordering is well-defined onA.

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. SinceA 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. Ifx1 x2 and x2 x3, then, x1 3x2 3x3,which implies x1 3 x3 since 3 is transitive on A3, and hence, x1 x3.This shows that is transitive. Ifx1 x2 andx2 x1, thenx1 3x2 andx23 x1, which implies that x1 = x2 since 3 is antisymmetric onA3. Thisshows that is antisymmetric. Since 3 is a well-ordering on A3, then wemust havex1 3x2 x1 x2 or x2 3x1 x2 x1. Hence, defines atotal ordering on A.

Next, we will show that is a well-ordering on A. B A withB=. Fix x0 B. Then,(A1,1) A such that x0 A1. Note that = BA1 A1. Since A1 is well-ordered by 1, thene B A1,which is the least element ofB A1. y B A,(A2,2) A suchthaty A2. We have 2 exhaustive and mutually exclusive cases: Case 1:y A1; Case 2: y A2 \ A1. In Case 1,e 1y sincee is the least element


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ofB A1, which implies that e y. In Case 2, sinceA is totally orderedby, we must have (A1,1) (A2,2), which implies thate 2 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 ofB.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; sinceA is totally ordered by, then, we must have(A1,1) (A2,2); hencex12 x2 and x1x2. Therefore, we have shown(A1,1) (A,). Hence, (A,) E is an upper bound ofA.

By Zorns Lemma, there is a maximal element (F,F) E. We claimthatF =E. We will prove this by an argument of contradiction. SupposeF

E, then

x0

E

\F. LetH :=F

{x0

}. Define an ordering HonH

by:x1, x2 H, ifx1, x2 F, we say x1 Hx2 ifx1 Fx2; ifx1 F andx2 = x0, then we let x1 Hx2; ifx1 = x2 = x0, we let x1 Hx2. Now, wewill show that His a well ordering onH.x1, x2, x3 H. Ifx1 F, then,x1 Fx1 andx1 Hx1; ifx1 = x0, thenx1 Hx1. Hence, H is reflexive.Ifx1 Hx2 and x2 Hx3. We have 4 exhaustive and mutually exclusivecases: Case 1: x1, x3 F; Case 2: x1 Fandx3 = x0; Case 3: x3 Fandx1 = x0; Case 4: x1 = x3 = x0. In Case 1, we must havex2 Fand thenx1 Fx2 andx2 Fx3, which implies that x1 Fx3, and hence x1 Hx3.In Case 2, we havex1Hx3. In Case 3, we must havex2 = x0, which leadsto a contradiction x0 Hx3, hence, this case is impossible. In Case 4, wehave x1 Hx3. In all cases except that is impossible, we havex1 Hx3.Hence, H is transitive. Ifx1 Hx2 andx2 Hx1. We have 4 exhaustiveand mutually exclusive cases: Case 1: x

1, x

2 F; Case 2: x

1 F and

x2 =x0; Case 3: x2 F and x1 = x0; Case 4: x1 = x2 = x0. In Case 1,we have x1 F x2 and x2 Fx1, which implies that x1 = x2 since F isantisymmetric onF. In Case 2, we have x0Hx1, which is a contradiction,and hence this case is impossible. In Case 3, we have x0 Hx2, 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. Whenx1, x2 F, then, we must have x1 Fx2 orx2Fx1 since Fis a well ordering on F, and hencex1Hx2 or x2Hx1.When x1 F and x2 = x0, then x1 Hx2. Whenx2 F and x1 = x0,then x2 Hx1. When x1 = x2 = x0, then x1 Hx2. This shows thatH is a total ordering on H. B H with B=. We will distinguishtwo exhaustive and mutually exclusive cases: Case 1: B ={x0}; Case 2:B= {x0}. In Case 1, x0 is the least element ofB. In Case 2, B \ {x0} Fand is nonempty, and hence admits a least element e0 B \ {x0} F withrespect to F.x B, ifx B \ {x0}, thene0Fx and hencee0Hx; ifx= x0, thene0Hx. Hence,e0is the least element ofB since His a totalordering onH. Therefore, His a well ordering on Hand (H,H) E.

Clearly,F H, F = HonF, and x1 Fand x2 H\F, we havex2 = x0 and x1Hx2. This implies that (F,F) (H,H). Since (F,F)


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is maximal inEwith 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.LetA :=

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.

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 discussedin Section 2.3,A is a set, which consists of all functions of toA. Definetheprojectionfunctions : A

A,

, by,

f

A,(f) = f().

Then, we may define the set

A :=

f A (f) A, When all ofAs are nonempty, then, by Axiom of Choice, the product

A is also nonempty.


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Chapter 3

Topological Spaces

3.1 Fundamental Notions

Definition 3.1 A topological space (X, O) consists of a setXand a col-lectionO of subsets (namely, open subsets) ofXsuch that

(i), X O;(ii)O1, O2 O, we haveO1 O2 O;

(iii) (O ) O, where is an index set, we have

O O.

The collectionO is called a topology for the setX.Definition 3.2 Let (X, O) be a topological space andF X. The com-plement ofF isF :=X\ F. F is said to beclosed ifF 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 inF. An interior point ofF is a point inF. Aboundary pointof F is a pointx X such thatO O withx O, wehave O F= and O

F=. The boundary ofF, denoted byF, is

the set of all boundary points of F. An exterior point of F is a point inF, whereF is called theexteriorofF. Anaccumulation point ofF is apointx Xsuch thatO O withx O, we haveO (F\ {x}) = .

Clearly,and Xare both closed and open.

Proposition 3.3 Let(X, O)be a topological space andA, B,E are subsetsofX. Then,

33


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34 CHAPTER 3. TOPOLOGICAL SPACES

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

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

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

(vi) E= E E.(vii) Xequals to the disjoint unionE E

E;

Proof (i) Clearly, E E. Then, E E.C E withC O, wehaveC E. Then, E=

CEeCO

CCEeCO

C= E. Hence, we haveE= E.

Clearly, E E. Note thatE= BEeBO

B

=BEeBO

B = O eEOO

O=EFurthermore,

(E) =

E

=

E

=

E=

E=

E

=E

(ii) Only ifxE, we have xBEeBO B.O O with xO, letO1 :=O E O. Note thatxO1 and EE=. SupposeO E=.Then,E O1 = , which further implies that EO1. Then, EO1 andx O1. This contradicts with x O. Hence, O E= .

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

(iii) Only ifx E E, then E O.Ifx X,O O such that x O E. Then,x OBE

BOB =

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

BB :={O O | OB }.O1 BA andO2 BB , then, O1 O2 B.On the other hand,O B, we haveO = O O and O BA andO BB.Then,

(A B) =

OABOO

O=

O1A,O2BO1,O2O

(O1 O2)

=

O1AO1O

O1

O2BO2O

O2

=A B


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3.1. FUNDAMENTAL NOTIONS 35

We also have

A B =

A B

=A B = A B

= A B= A B

(v) IfEis closed since E= Eand Eis closed.Only if SinceEis closed, thenE E. Then, we haveE= E. Hence,

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

(vii) Note thatX=E E= E EE. By (iii) and Definition 3.2,E and Eare disjoint. It is obvious that

E is disjoint with E E= E.

Hence, the result holds.

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 spaceX withoutfurther reference to components ofX, where the topology is understood tobeOX. When it is clear from the context, we will neglect the subscriptX.Proposition 3.4 Let(X, O) be a topological space andA X. A admitsthe subset topologyOA:= {O A| O O}.Proof Clearly, OAis a collection of subsets ofA. = A OAandA= X A OA.OA1, OA2 OA,O1, O2 O such thatOA1 = O1 Aand OA2 = O2A. Then, O1 O2 O sinceO is a topology. Then,OA1

OA2 = (O1

O2)

A

OA.

( OA )

OA, where is an index

set, we have, , O O such thatOA= O A. Then, OO sinceOis a topology. Therefore, OA= O A OA.

Hence,OA is a topology onA. Let (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 LetXbe a topological space,A Xbe endowed with thesubset topologyOA, andE A. Then,

(1) Eis closed inOA if, and only if, E= A F, whereF X is closedinOX;

(2) theclosure ofErelative to (A, OA) (the closure ofE inOA)is equaltoE A, whereEis the closure ofE relative toX.

Proof Here, the set complementation and set closure operation arerelative toX.

(1) If A \ E= A \ (A F) =A A F = A F. Since F is closedinOX, thenF OX. Then, A \ E OA. Hence, Eis closed inOA.


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Only ifA \ E OA. Then, O OXsuch thatA \ E= A O. Then,E= A \ (A \ E) =A A O= A O. Hence, the result holds.

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


Definition 3.6 For two topologies over the same set X,O1 andO2, wewill say thatO1 isstronger (finer) thanO2 ifO1 O2, in which case,O2is said to beweaker (coarser) thanO1.

Proposition 3.7 Let X be a set andA X

2. Then, there exists theweakest topologyO on X such thatA O. This topology is called thetopology generated byA.Proof Let M:=

X X2| A X andX is a topology on X} andO= XMX. Clearly, X2 Mand henceOis 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 sinceA X,X M.Therefore,O is the weakest topology containingA.

3.2 Continuity

Definition 3.8 Let (X, OX ) and (Y, OY) be topological spaces, D Xwith the subset topologyOD, andf :D Y (orf : (D, OD) (Y, OY) tobe more specific). Then,f is said to be continuous if,OY OY, we havefinv(OY) OD. f is said to becontinuous at x0 D if,OY OY withf(x0)OY,U OX withx0U such thatf(U)OY. f is said to becontinuous onE D if it is continuous atx,x E.

Proposition 3.9 LetX andYbe topological spaces,D Xwith the subsettopologyOD, andf :D Y. f is continuous if, and only if,x0 D, fis continuous atx0.

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


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3.2. CONTINUITY 37

finv(OY) =xfinv(OY)(Ux D) = (xfinv(OY) Ux) D OD. Hence, fis continuous.

Only ifx0 D ,OY OY with f(x0) OY, let U = finv(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 LetX andYbe topological spaces andf : X Y. f iscontinuous if, and only if,B Y withB OY, we have finv(B) OX,that is, the inverse image of any closed set inY is closed inX.Proof IfO OY, we have, by Proposition 2.5, finv(O) =

finv( O) OX. Hence, f is continuous.Only ifB Y withB OY. Since f is continuous, then, byProposition 2.5, finv(B) =finv( B) OX. Hence, the result holds.


Theorem 3.11 LetX andY be topological spaces, f :X Y, andX =X1 X2, where X1 and X2 are both open or both closed. Let X1 andX2be endowed with subset topologiesOX1 andOX2 , respectively. Assume thatf|X1 :X1 Yand f|X2 :X2 Yare continuous. Then, fis continuous.Proof Consider the case thatX1 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 withx0 Usuch that f

|X1

(U)

O. Since X1

OX, then U

OX. Note that

f(U) = f|X1 (U) O, sinceU X1. Hence, f is continuous atx0. By thearbitraryness ofx0 and Proposition 3.9, f is continuous.

Consider the case that X1 and X2 are both closed. closed subsetB Y, we have finv(B) X. Then, finv(B)X1 = ( f|X1)inv(B) isclosed inOX1 , by Proposition 3.10 and the continuity of f|X1 . Similarly,finv(B) X2 = ( f|X2)inv(B) is closed inOX2 . SinceX1 and X2 are closedsets inOX, then, finv(B)X1 and finv(B)X2 are closed inOX, byProposition 3.5. Then, finv(B) = (finv(B) X1) (finv(B) X2) is closedinOX. By Proposition 3.10, f is continuous.

This completes the proof of the theorem.

Proposition 3.12 LetX,Y, andZbe topological spaces, f :X Y,g :

Y Z, and x0

X. Assume that f is continuous at x0 and g is

continuous aty0 := f(x0). Then, g f : X Z is continuous atx0.Proof OZ OZ with g(f(x0)) OZ. Since g is continuous atf(x0), thenOY OYwithf(x0) OY such thatg(OY) OZ. Sincef iscontinuous atx0, thenOX OX with x0 OX such that f(OX )OY.Then, g(f(OX )) OZ. Hence,g fis continuous at x0. This completesthe proof of the proposition.


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Definition 3.13 LetX andY be topological spaces and f :X Y. fis said to be a homeomorphism betweenX andY if it is bijective andcontinuous andfinv:Y X is also continuous. The spacesX andY aresaid to behomeomorphic 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 LetX be a topological space, D X with the subsettopologyOD, and f : D IR. f is said to be upper semicontinuous ifa IR, finv((, a)) OD. f is said to beupper 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 . Thereforex0 V. This

shows that

V O

with x0

V such that f(V)

U.Case 2: a0 = 0. Then, we must have a1 < 0 = a0 < a3 < a4. Take

V = Oa3 O. We must havex0 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. Sincef(x0) = a0 = 1, thenx0 O 1+a22

Oa2 =V.x V,


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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 Vsuch that f(V) U. Hence,f is continuous at x0. By the arbitraryness ofx0 and Proposition 3.9, f iscontinuous. This completes the proof of the theorem.

Proposition 3.56 LetX andY be topological spaces andYbe Hausdorff.f1 : X Y andf2: X Yare continuous. LetD Xbe dense. Assumethat f1|D = f2|D. Then, f1= f2.Proof Supposef1=f2. Then, x Xsuch thatf1(x) =f2(x). SinceYis Hausdorff, thenO1, O2 OY such that f1(x) O1, f2(x)O2, andO1O2 = . Sincef1and f2are continuous, we haveU1 := f1inv(O1) OXandU2 := f2inv(O2) OX. Note that x U1 U2 OX andx D, then,by Proposition 3.3,xD U1 U2. Then, f1(x)O1 and f2(x)O2,which implies that f1|D(x) = f2|D(x). This is a contradiction. Hence, wemust havef1 = f2.


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

Proof Let f := h

1 + |h | . Then,|f(x) | 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: LetB := {x A| l(x) c1/3}and C := {x A|l(x) c1/3}. Then, B anc C are closed sets inOA, by the continuityof l and Proposition 3.10. Since A is closed, then B and C are closedinO, 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.

By repeated application of Claim 3.57.1, we may define fi : X IR,i IN, such that fi is continuous,|fi(x) | 2i13i ,x X, and

f(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,xX.x0 X.


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3.8. CONTINUOUS REAL-VALUED FUNCTIONS 55

(0, )IR.N IN such that i=N+1 fi(x) < /3,xX. Bythe continuity off1, . . . , f Nand Proposition 3.9, U Owithx0 U suchthat

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

g(x) Ni=1

fi(x)+ N

i=1

fi(x) N

i=1

fi(x0)

+ N

i=1

fi(x0) g(x0) <

Therefore,g is continuous atx0. Then,g is continuous, by the arbitrarinessofx0 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) | = 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.

Definition 3.58 LetXbe a set andFbe a collection of real-valued func-tions ofX. Then, there is the weakest topology onXsuch that all functionsin

Fare continuous. This topology is called theweak topology generated

byF.Let X be a set, I := [0, 1] IR, andF be a collection of functions of

X to I such thatx, y X with x= y,f F, we have f(x)= f(y).Each f Fis a point in IX andFcan be identified with a subset ofIX .The topology thatFinherits as a subspace ofIX 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 ofXas a subset ofIFis the weak topology generated byF.Proposition 3.59 LetX be a topological space, I := [0, 1] IR, andFbe a collection of continuous functions ofX to Isuch thatx, y X withx

= y ,

f

F, we havef(x)

=f(y). LetE :

X IF be the equivalence

map given by,x X, f(E(x)) =f(x),f F. Then, E is continuous.Furthermore, if closed setF X andx X withxF,f F withf(x) = 1 and f|F = 0, thenE: X E(X) is a homeomorphism.Proof x0 X. Fix a basis open setO in IF with E(x0) O. ByProposition 3.25,O =

fFOf, whereOf OI withOIbeing the subset

topology onI,f F, andOf =I for all fs except finitely many fs, say


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f FN. Let U= fFN finv(Of) OX. By E(x0) O, we havex0 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 ofx0 and Proposition 3.9, E iscontinuous.

Under the additional assumption onF, we need to show that E is ahomeomorphism betweenX and E(X). x, y X with x= y,f Fsuch that f(E(x)) = f(x)= f(y) = f(E(y)). Then, E(x)= 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 Einvis continuous at E(x0). O OX withx0 O.

O is closed and x0

O. Then,f0 F such that f0(x0) = 1

and f0

| eO = 0. Define U =fFUf IF byUf = I,f F \ {f0} andUf0 = (1/2, 1] OI. Clearly,Uis open in IF. Clearly,E(x0) U.x X

with E(x) U, we have f0(E(x)) = f0(x) > 1/2. Then, xO andx O. This shows that Einv(E(X) U) O. Hence,Einvis continuous atE(x0). By the arbitrariness ofx0 and Proposition 3.9, Einv: E(X) X iscontinuous. This implies that E: X E(X) is a homeomorphism.


Definition 3.60 A topological spaceX is said to becompletely regular(or T3 12 ) if it is Tychonoff andx0 X and closed set F X withx0 F, there exists a continuous real-valued functionf :X [0, 1] suchthatf(x0) = 1 and f|F = 0.

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

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

Let X be a completely regular topological space. Then,Xis Tychonoff.x0 X and closed set F X with x0 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 off. Clearly, x0 O1, F O2, andO1 O2 = . Hence,Xis regular.


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


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3.9. NETS AND CONVERGENCE 57

Proof SinceX is completely regular, thenX is Tychonoff and allsingleton subset ofXis 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.

3.9 Nets and Convergence

Definition 3.63 Adirected system is a nonempty setA and a relation onA,, such that

(i) is transitive;(ii), A, A such that and .

A net is a mapping of a directed systemA := (A, ) to a topologicalspaceX. A, the image isx. The net is denoted by( x )A, wherewe have abuse the notation to say A when A. It is understoodthat the relation forA isA, where we will ignore the subscriptA if noconfusion arises.

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

A point x X is acluster point of ( x )A ifO O withx 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 withoutchanging 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 LetXbe a topological space. Then,X is Hausdorff if,and only if, for all net( x )A X, there exists at most one limit point

for the net. We then writex= limA x when the limit exists.

Proof Only if Suppose there exists a net (x )A X such thatxA, xB X with xA= xB and xA and xB are limit points of the net.SinceXis Hausdorff, thenO1, O2 O such that xA O1, xBO2, andO1 O2 =. Since xA is the limit of the net, then1 A, A with1, we have xO1. Similarly, since xB is the limit of the net, then2 A, A with 2 , we have x O2. SinceA is a directedsystem,3 Asuch that1 3 and2 3. Then, we have x3 O1and x3 O2, which implies that O1 O2=, which is a contradiction.Therefore, every net inXhas at most one limit point.

If SupposeX is not Hausdorff. Then,xA, xB X with xA= xBsuch that OA, OB O withxA OAand xB OB, we haveOAOB= .


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Let := {(OA, OB )| xA OA O, xB OB O}. Clearly, (X, X) ,then = . Define a relationon by,(OA1, OB1), (OA2, OB2) , wesay (OA1, OB1) (OA2, OB2) ifOA1 OA2 andOB1 OB2. Clearly,istransitive on .(OA1, OB1), (OA2, OB2) , we havexA 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) , OAOB=. 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 Owith xAOA1. FixOB1 :=X O with xB OB1. Then, (OA1, OB1) . (OA2, OB2) with (OA1, OB1) (OA2, OB2), we have x(OA2,OB2) OA2 OB2OA1 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.This completes the proof of the proposition.

Proposition 3.66 LetX andY be topological spaces, D Xwith subsettopologyOD, andf :D Y. Then, the following are equivalent.

(i) f is continuous atx0 D;(ii) net ( x )A D with x0 as a limit point, we have that the net

(f(x) )A has a limit pointf(x0).

(iii) net( x )A D withx0 as a cluster point, we have that the net(f(x) )A Yhas a cluster pointf(x0).

Proof (i) (ii). Fix a net ( x )A Dwithx0 Das a limit point.OY OYwithf(x0) OY. By the continuity offatx0, OX OX withx0 OX such that f(OX ) OY. Since x0 is a limit point of ( x )A,then0 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). Supposefis not continuous at x0 D. Then,OY0 OYwith f(x0)OY0 such that,OX OX with x0OX , we havef(OX ) OY0. LetM := {O OX | x0 O }. Clearly,X MandM = . Definea relationonMby,O1, O2 M, we say O1 O2 ifO1 O2. Clearly, is transitive onM.O1, O2 M, letO3 = O1 O2 OX andx0 O3.Then,O3 M, O1 O3, and O2 O3. Hence,A := (M, ) is a directedsystem.O M, f(O) \ OY0=. By Axiom of Choice, we may definea net (xO )OA byxO O D with f(xO) OY0. Clearly, x0 is a limitpoint of ( xO )OA. Yet,f(x0) OY0 OY and f(xO) OY0,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.


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(i) (iii). Fix a net ( x )AD with x0 as a cluster point.OYOY with f(x0)OY, by the continuity off atx0,U OX with x0Usuch that f(U)OY . By Definition 3.63, A,0 Awith 0,x0 U. Then, f(x0) OY. Hence,f(x0) is a cluster point of the net( f(x) )A.

(iii) (i). Supposef is not continuous at x0. LetM :={O OX |x0 O }. Clearly,A := (M, ) is a directed system. OY0 OY withf(x0) OY0 such thatU M, we have f(U) OY0. By Axiom ofChoice, we may assign to each U Man xU U D such that f(xU)OY0. Consider the net ( xU)UA D. Clearly,x0is 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 OY0 f(x0),U A, f(xU)

OY0. Then, f(x0) is

not a cluster point of ( f(xU) )UA

. This contradicts with the assumption.Therefore,fmust 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 )AXbe a net. Then, x0X 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

Owith x0

B such that,

0

A,

Awith 0 , we have x 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 B. This implies that (x ) O , for some N. Then, by an argument of contradiction,we may show that0 N such that,0 A, A with 0 ,we have 0(x )O0 . Hence,0(x0)O0 is not the limit of the net( 0(x ) )A. This contradicts with the assumption. Hence, we havex0is a limit point of ( x )A.


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

Proof Only if LetM := {O O | x O }. Clearly,X M, thenM =. Clearly,A := (M, ) is a directed system. Since x E, then,by Proposition 3.3,O A, O E=. By Axiom of Choice, a net( xO )OA E such that xO O E,O A.O Owith xO, thenO A.O1 Awith OO1, we have xO1 O1 EO. Hence,x is alimit point of (xO )OA.


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If Let ( x )A Ebe 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, then1 Awith 0 1. Then,x1 O E= . 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. LetAs A be a subset withthe same relation asA such that A,s As such that s.Then,As := (As, ) is a directed system and( x )As is a net, which iscalled asubnet of( x )A.

Proposition 3.70 Let(X, O) be a topological space and( x )A X bea net. Then,x0 Xis a limit point of( x )A if, and only if, any subnet( x )As has a limit pointx0.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 0s0 .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. A 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 every

subnet( x )As of (x )A, there exists a subnet ( x )Ass that has alimit pointx0.

Proof Sufficiency We assume that every subnet (x )As of( x )A, there exists a subnet ( x )Ass that has a limit point x0. Wewill 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 xO0. DefineAs := ( A xO0 , ).Clearly, ( x )As is a subnet of ( x )A. Any subnet ( x )Ass of

( x )As ,ss Ass, we have xssO0. 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 LetX := (X, OX )andY:= (Y, OY)be topological spaces,D X, f : D Y, and x0 X be an accumulation point ofD. y0 Y


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is said to be a limit point off(x) as xx0 ifOY OY withy0 OY,U OX withx0 U such thatf(U\ {x0}) = f((D U) \ {x0})OY.We will also say thatf(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 LetX := (X, OX ) andY := (Y, OY) be topologicalspaces, D X, f :D Y, andx0 X be an accumulation point ofD. IfYis Hausdorff, then there is at most one limit point off(x)asx x0. Inthis case, we will write limxx0f(x) =y0 Ywhen the limit exists.

Proof Suppose f(x) admits limit points yA, yB Y as xx0 withyA= yB . SinceY is Hausdorff, thenUA, UB OY such that yA UA,yB UB, and UA UB =. SinceyA is a limit point off(x) as xx0,thenVA OX with x0 VA such that f(VA\ {x0}) UA. Since yBis a limit point of f(x) as x x0, thenVB OX with x0 VB suchthat f(VB\ {x0}) UB. Then, x0 V := VA VB O. Since x0is an accumulation point ofD , thenx(D V) \ {x0}. Then, we havef(x) UAsincex (DVA)\{x0} and f(x) UBsincex (DVB )\{x0}.Then, f(x) UA UB=. This contradicts withUA UB =. Hence,the result holds. This completes the proof of the proposition.

Proposition 3.74 LetX := (X, OX ) andY := (Y, OY) be topologicalspaces, D X with subset topologyOD, f : D Y, andx0 D. Then,the following statements are equivalent.

(i) f is continuous atx0.

(ii) If x0 is an accumulation point of D, then f(x0) is a limit point off(x) asx 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 ofD. Case 1: x0 is not an accumulation point ofD.V OX withx0 V such thatV D= {x0}.U OY withf(x0) U,we havef(V) = {f(x0)} U. Hence, f is continuous at x0.

Case 2: x0 is an accumulation point ofD.

U

OY with f(x0)

U,

V OX with x0 Vsuch 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.

Proposition 3.75 LetX := (X, OX ),Y:= (Y, OY), andZ:= (Z, OZ) betopological spaces,D X, f :D Y, x0 Xbe an accumulation point of


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D, y0 Ybe a limit point off(x)asx x0, andg : Y Zbe continuousaty0. Then,g (y0) Zis a limit point ofg (f(x))asx x0. WhenY andZare Hausdorff, then we may writelimxx0g(f(x)) = g(limxx0f(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) asx x0, thenOX 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.

Proposition 3.76 LetX be a topological space, D X,

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