Undergraduate Texts in Mathematics Editors F. w. Gehring P. R. Halmos Advisory Board C. DePrima I. Herstein Undergraduate Texts in Mathematics Apostol: Introduction to Analytic Number Theory. 1976. xii, 338 pages. 24 illus. Foulds: Optimization Techniques: An Introduction. 1981. xii, 502 pages. 72 illus. Armstrong: Basic Topology. 1983. xii, 260 pages. 132 mus. Franklin: Methods of Mathematical Economics. Linear and Nonlinear Programming. Fixed-Point Theorems. 1980. x, 297 pages. 38 iIlus. Bak/Newman: Complex Analysis. 1982. x, 224 pages. 69 illus. Banchoff/Wermer: Linear Algebra Through Geometry. 1983. x, 257 pages. 81 mus. Childs: A Concrete Introduction to Higher Algebra. 1979. xiv, 338 pages. 8 ilIus. Chung: Elementary Probability Theory with Stochastic Processes. 1975. xvi, 325 pages. 36 ilIus. Croom: Basic Concepts of Algebraic Topology. 1978. x, 177 pages. 46 ilIus. Curtis: Linear Algebra: An Introductory Approach (4th edition) 1984. x, 337 pages. 37 illus. Dixmier: General Topology. 1984. x, 140 pages. 13 illus. Ebbinghaus/FlurnlThomas Mathematical Logic. 1984. xii, 216 pages. I illus. Fischer: Intermediate Real Analysis. 1983. xiv, 770 pages. 100 illus. Fleming: Functions of Several Variables. Second edition. 1977. xi, 411 pages. 96 ilIus. Halmos: Finite-Dimensional Vector Spaces. Second edition. 1974. viii, 200 pages. Halmos: Naive Set Theory. 1974, vii, 104 pages. Iooss/Joseph: Elementary Stability and Bifurcation Theory. 1980. xv, 286 pages. 47 iIlus. lanich: Topology 1984. ix, 180 pages (approx.). 180 illus. Kemeny/Snell: Finite Markov Chains. 1976. ix, 224 pages. II illus. Lang: Undergraduate Analysis 1983. xiii. 545 pages. 52 illus. Lax/Burstein/Lax: Calculus with Applications and Computing, Volume I. Corrected Second Printing. 1984. xi, 513 pages. 170 iIlus. LeCuyer: College Mathematics with A Programming Language. 1978. xii, 420 pages. 144 illus. Macki/Strauss: Introduction to Optimal Control Theory. 1981. xiii, 168 pages. 68 illus. Malitz: Introduction to Mathematical Logic: Set Theory - Computable Functions - Model Theory. 1979. xii, 198 pages. 2 mus. continued after Index Jacques Dixmier General Topology Springer Science+Business Media, LLC Jacques Dixmier Univcrsite de Pari" VI Mathematiques UER 48 9 Quai Saint Bernard 75 Paris France TraT7slatedJrom the French bv SterlIng K Berbenan Department 01 ivl3thematics The Univ3r~ity 01 Texas Austin, TX 75712 UoS.A. Editorial Bourd F. W. Gehring Department of Mathematics Universit) ufMich!gan Ann Arbor, M1 48109 U.S.A. P. R. Haimos Ocpanment of Mathematics Indiana University Bloominpton, IN 47405 U.S.A. AlviS SUhject Classiikation: 54-(JI - - - - - - - - - - - -- _ _ _ _ 0 ___ 0_ I ibioaioy of Congioess Cnaioging in PllblicaTion Oat" Dixnner, Jacques. Gf'nf'ral lOpolo~y (UnJergraduak ti.:xLs in mathematic,) 1 ramlation of. Topologic generale. Includes indexes l. Topology. I. Title l!. Scri.;s. 83-10444 QA611.D48613 1984 514.322 With 11 m1!stmtiun~ Tillt,; of the origiua! fr..;nch editioll: Topoi()gie Gener"l..:, @ Presses Uni\elsitaires de France, Pari~, 191\ l. © 1984 Springer Science+Business Media New York Originally published by Springer-Verlag New York Inc. in 1984 All nrhts reserved No part of thl~ book may b~ tr;ln~laleJ 01 r~prodl!ct,;d in any fcm, v,ithout written pttmission from Springer-Verlag, 175 Fifth Avenue, New York, N"", York 10010, U.S.A. Type~et bv Compusilion HOl!se Ltd., Salisburv, England. 9 8 7 (; 5 4 3 ~ 1 ISBN 978-1-4419-2823-8 DOI 10.1007/978-1-4757-4032-5 ISBN 978-1-4757-4032-5 (eBook) Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .. CHAPTER 1 Topological Spaces 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. IX ......... . Open Sets and Closed Sets in a Metric Space Topological Spaces . . . . Neighborhoods. . . . . . Interior, Exterior, Boundary Closure . . . . . . . . . Separated Topological Spaces. I 4 6 7 9 10 CHAPTER II Limits. Continuity 2.1. 2.2. 2.3. 2.4. 2.5. 2.6. Filters. . . . . Limits Along a Filter Base Mappings Continuous at a Point Continuous Mappings. . . . . Homeomorphisms. . . . . . . Adherence Values Along a Filter Base 12 13 14 17 18 20 22 CHAPTER IJI Constructions of Topological Spaces 3.1. 3.2. 3.3. 3.4. Topological Subspaces. . . . . . . Finite Products of Topological Spaces Infinite Products of Topological Spaces. Quotient Spaces . . . . . . . . . . . 25 25 29 34 34 vi Contents CHAPTER IV Compact Spaces . . . . . . . 4.1. 4.2. 4.3. 4.4. 4.5. Definition of Compact Spaces. Properties of Compact Spaces. Complement: Infinite Products of Compact Spaces. The Extended Real Line . Locally Compact Spaces . 36 36 37 42 44 45 CHAPTER V Metric Spaces 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. .......... . . . . . . Continuity of the Metric . . . . . . . . . . . The Use of Sequences of Points in Metric Spaces. Uniformly Continuous Functions Equicontinuous Sets of Functions . . Complete Metric Spaces . . . . . . Complete Spaces and Compact Spaces The Method of Successive Approximations. 49 49 51 53 54 55 59 60 CHAPTER VI Limits of Functions. 6.1. Uniform Convergence 6.2. Simple Convergence . 6.3. Ascoli's Theorem 62 62 67 69 CHAPTER VII Numerical Functions . . . . . . . . . . . 70 7.1. 7.2. 7.3. 7.4. 7.5. 7.6. 71 Bounds ofa Numerical Function . . . . . Limit of an Increasing Numerical Function . Limit Superior and Limit Inferior of a Numerical Function Semicontinuous Functions . Stone-Weierstrass Theorem. Normal Spaces . . . . . . 73 74 77 80 85 CHAPTER VIII Normed Spaces. . . . . . . . 8.1. 8.2. 8.3. 8.4. 8.5. 8.6. 8.7. 8.8. Definition of Normed Spaces. Continuous Linear Mappings . Bicontinuous Linear Mappings Pre-Hilbert Spaces. . . . . . Separated Pre-Hilbert Spaces. Banach Spaces, Hilbert Spaces Linear Subspaces of a N ormed Space Riesz's Theorem . . . . . . . . . 90 90 93 96 98 10\ 104 106 108 CHAPTER IX Infinite Sums . . . . . . . . . 9.1. Summable Families . . . . . 9.2. Associativity, Commutativity. 113 113 115 Contents vii 9.3. Series . . . . . . . . . . . . . . . . . . . . 9.4. Summable Families of Real or Complex Numbers 9.5. Certain Summable Families in Hilbert Spaces. 119 123 117 CHAPTER X Connected Spaces. . . . . . 10.1. Connected Spaces . . . . 10.2. Arcwise Connected Spaces. 10.3. Connected Components. 125 125 127 128 Exercises . . . . . 129 Index of Notations 137 Index of Terminology . 139 Introduction This book is a course in general topology, intended for students in the first year of the second cycle (in other words, students in their third university year). The course was taught during the first semester of the 1979-80 academic year (three hours a week of lecture, four hours a week of guided work). Topology is the study of the notions of limit and continuity and thus is, in principle, very ancient. However, we shall limit ourselves to the origins of the theory since the nineteenth century. One of the sources of topology is the effort to clarify the theory of real-valued functions of a real variable: uniform continuity, uniform convergence, equicontinuity, Bolzano-Weierstrass theorem (this work is historically inseparable from the attempts to define with precision what the real numbers are). Cauchy was one of the pioneers in this direction, but the errors that slip into his work prove how hard it was to isolate the right concepts. Cantor came along a bit later; his researches into trigonometric series led him to study in detail sets of points of R (whence the concepts of open set and closed set in R, which in his work are intermingled with much subtler concepts). The foregoing alone does not justify the very general framework in which this course is set. The fact is that the concepts mentioned above have shown themselves to be useful for objects other than the real numbers. First of all, since the nineteenth century, for points of Rn. Next, especially in the twentieth century, in a good many other sets: the set of lines in a plane, the set of linear transformations in a real vector space, the group of rotations, the Lorentz group, etc. Then in 'infinite-dimensional' sets: the set of all continuous functions, the set of all vector fields. etc. Topology divides into 'general topology' (of which this course exposes the rudiments) and 'algebraic topology', which is based on general topology x Introduction but makes use of a lot of algebra. We cite some theorems whose most natural proofs appeal to algebraic topology: (1) let B be a closed ball in Rn, f a continuous mapping of B into B; then f has a fixed point; (2) for every x E S2 (the 2-dimensional sphere) let V(x) be a vector tangent to S2 at x; suppose that Vex) depends continuously on x; then there exists an Xo E S2 such that V(xo) = 0; (3) let U and V be homeomorphic subsets of R"; if U is open in R", then V is open in R". These theorems cannot be obtained by the methods of this course, but, having seen their statements, some readers will perhaps want to learn something about algebraic topology. The sign. in the margin pertains to theorems that are especially deep or especially useful. The choice of these statements entails a large measure of arbitrariness: there obviously exist many little remarks, very easy and constantly used, that are not graced by the sign •. The sign * signals a passage that is at the limits of ' the program' (by which I mean what has been more or less traditional to teach at this level for some years). Quite a few of the statements have already been encountered in the First cycle. For clarity and coherence of the text, it seemed preferable to take them up again in detail. The English edition differs from the French by various minor improvements and by the addition of a section on normal spaces (Chapter 7, Section 6). CHAPTER I Topological Spaces After reviewing in §l certain concepts already known concerning metric spaces, we introduce topological spaces in §2, then the simplest concepts associated with them. For example, one has an intuitive notion of what is a boundary point of a set E (a point that is 'at the edge' of E), a point adherent to E (a point that belongs either to E or to its edge), and an interior point of E (a point that belongs to E but is not on the edge). The precise definitions and the corresponding theorems occupy §§4 and 5. Separated topological spaces are introduced in §6; on first reading, the student can suppose in what follows that all of the spaces considered are separated. 1.1. Open Sets and Closed Sets in a Metric Space 1.1.1. Let E be a set. Recall that a metric (or 'distance function') on E is a function d, defined on E x E, with real values ~ 0, satisfying the following conditions: (i) d(x, y) (ii) d(x, y) (iii) d(x, z) = O=-x = y; x) for all x, y in E; d(x, y) + dey, z) for all x, y, z in E (,triangle inequality')' = dey, ~ (On occasion we shall admit the value + 00 for a metric; this changes almost nothing in what follows.) A set equipped with a metric is called a metric space. One knows that the preceding axioms imply (iv) Id(x, z) - d(x, y)1 ~ dey, z) for all x, y, z in E. 2 I. Topological Spaces There is an obvious notion of isomorphism between metric spaces. Let E' be a subset ofE. Take the restriction to E' x E' of the given distance function d on E x E. Then E' becomes a metric space, called a metric subspace of E. 1.1.2. Examples. The ordinary plane and ordinary space, with the usual Euclidean distance, are metric spaces. For x = (Xl' x 2 , . . . , X n) E R" and Y = (Yl' Y2, ... , Yn) E R", set One knows that d is a metric on R", and in this way R" becomes a metric space, as do all of its subsets. In particular R, equipped with the metric (x, y) H Ix - Y I, is a metric space. 1.1.3. Definition. Let E be a metric space (thus equipped with a metric d), A a subset of E. One says that A is open if, for each Xo E A, there exists an e> 0 such that every point x of E satisfying d(xo, x) < f. belongs to A. 1.1.4. Example. Let E be a metric space, a E E, p a number:;:,. 0, A the set of all x E E such that dCa, x) < p. Then A is open. For, let Xo EA. Then dCa, xo) < p. Set e = p - dCa, xo) > O. If x E E is such that d(xo, x) < e, then dCa, x) ::; dCa, xo) + d(xo. x) < dCa, xo) + e = p, thus x E A. The set A is called the open ball with center a and radius p. If p > 0 then a E A; if p = 0 then A = 0. In the ordinary plane, one says 'disc' rather than 'ball'. 1.1.5. In particular, let a, b be real numbers such that a ::; b. The interval (a, b) is nothing more than the open ball in R with center + b) and radius t(b - a). One verifies easily that the intervals ( - 00, a), (a, + 00) are open. This justifies the expression 'open interval' employed in elementary courses. tea 1.1.6. Theorem. Let E be a metric space. (i) The subsets 0 and E ofE are open. (ii) Every union of open subsets ofE is open. (iii) Every finite intersection of open subsets of E is open. The assertion (i) is immediate. Let (AJiEI be a family of open subsets of E, and let A = UAi' iEf A' = nAi' iEI Let us show that A is open. Let Xo E A. There exists i E I such that Xu E Ai' Then there exists e > 0 such that the open ball B with center Xu and radius 3 1.1. Open Sets and Closed Sets in a Metric Space e is contained in Ai' A fortiori, A =:J B. Thus A is open. Assuming I is finite, let us show that A' is open. Let XI EA'. For every i E 1, there exists ej > Osuch that the open ball with center XI and radius Li is contained in Ai' Let f, be the smallest of the ei' Then e > 0, and the open ball with center X I and radius L is contained in each Ai, hence in A'. Thus A' is open. 1.1.7. Let us maintain the preceding notations. If I is infinite, niEI Ai is not always open. For example, in R, the intersection of the open intervals ( ~ lin, lin) for n = 1, 2, 3, ... reduces to {O}, thus is not open. 1.1.8. Definition. Let E be a metric space, B a subset of E. One says that B is closed if the subset E ~ B is open. 1.1.9. Example. Let a E E, P 2': 0, B the set of X E E such that dCa, x) ~ p. Then B is closed. For, let Xo E E ~ B. Then dCa, xo) > p. Set e = dCa, x o) ~ p > 0. If x E E is such that d(xo, x) < e, then dCa, x) 2': dCa, xo) ~ d(xo, x) > dCa, xo) ~ e = p, therefore x E E ~ B. Thus E ~ B is open, consequently B is closed. The set B is called the closed ball with center a and radius p. One has a E B. If p = then B = {a}. In the ordinary plane, one says 'disc' rather than 'ball'. ° 1.1.10. In particular, let a, b be real numbers such that a ~ b. The interval [a, b] is nothing more than the closed ball in R with center l(a + b) and radius l(b ~ a). One verifies easily that the intervals [a, +XJ) and (~XJ, a] are closed. This justifies the expression 'closed interval' employed in elementary courses. One also sees that an interval of the form [a, b) or (a, b], with a < b, is neither open nor closed. 1.1.11. Theorem. Let E be a metric space. (i) The subsets 0 and E ofE are closed. (ii) Every intersection of closed subsets ofE is closed. (iii) Every finite union of closed subsets of E is closed. This follows from 1.1.6 by passage to complements. 1.1.12. Example. Let E be a metric space, a E E, p 2': 0, S the set of all x E E such that dCa, x) = p. Then S is closed. For, let A (resp. B) be the open ball (resp. closed ball) with center a and radius p. Then E ~ A is closed. Since S = B n (E ~ A), S is closed by 1.1.11(ii). The set S is called the sphere with center a and radius p. If P = then S = {a}. In R, a sphere of radius> reduces to a set with 2 points. In the ordinary plane, one says 'circle' rather than 'sphere', ° ° 4 I. Topological Spaces 1.1.13. Let E be a metric space. On comparing 1.1.6 and 1.1.11, one sees that E (and similarly 0) is a subset that is both open and closed. This is exceptional: in the most common examples of metric spaces, it is rare that a subset is both open and closed (cf. Chapter X). On the other hand, although it is easy to exhibit examples of subsets that are either open or closed, it should be understood that a subset of E chosen 'at random' is in general neither open nor closed. For example, the subset Q of R is neither open nor closed. 1.1.14. Theorem. Let E be a set, d and d' metrics on E. Suppose there exist constants c, c' > 0 such that c d(x, y) :s;; d'(x, y) :s;; c' d(x, y) for all x, y E E. The open subsets of E are the same for d and d'. Let A be a subset of E that is open for d. Let Xo E A. There exists an 6 > 0 such that every point x of E satisfying d(xo, x) < 6 belongs to A. If x E E satisfies d'(x O ' x) < CIl, then d(xo, x) < 6, therefore x E A. This proves that A is open for d'. Finally, one can interchange the roles of d and d' in the foregoing. 1.1.15. However (with the preceding notations) the balls and spheres of E are in general different for d and d'. For example, for x = (xt. ... , x.) E Rn or C·, and y = (YI, ... , y.) E Rn or en, set d(x, y) = (lxI - yIl 2 + .. , + Ix. - YnI 2 )1/2, d'(x,y) = IXI - hi + ... + Ix. - Ynl, = sup(lxI - YII,···, Ix. - Ynl)· d"(x, y) One knows that d, d', d" satisfy the conditions of 1.1.14, hence define the same open subsets ofRn. However, for d" the open ball with center (Xl' ... , x n) and radius p is an 'open slab with center (xt. ... , x.)': (Xl - P, Xl + p) X (Xl - P, X2 + p) x ... X (x. - P, Xn + p). 1.2. Topological Spaces 1.2.1. Definition. One calls topological space a set E equipped with a family the open sets ofE) satisfying the following conditions: (!) of subsets ofE (called (i) the subsets 0 and E of E are open; (ii) every union of open subsets of E is open; (iii) every finite intersection of open subsets of E is open. 5 1.2. Topological Spaces One also says that (1) defines a topology on E. 1.2.2. For example, a metric space is automatically a topological space, thanks to 1.1.3 and 1.1.6; this structure of topological space does not change if one replaces the metric d of E by a metric d' related to d by the condition of 1.1.14. In particular, every subset ofthe ordinary plane, or of ordinary space, or of R", is a topological space. For a large part of the course, these are the only interesting examples we shall have at our disposal; but they already exhibit a host of phenomena. 1.2.3. Let E be a set. In general there is more than one way of choosing in E a family (1) of subsets satisfying the conditions 1.2.1. In other words, a set E may be equipped with more than one structure of topological space. For example, if one takes for (1) the family of all subsets ofE, the conditions 1.2.1 are satisfied, thus E becomes a topological space called a discrete space (one also says that the topology of E is discrete). For another example, if one takes for (1) the family consisting only of 0 and E, the conditions 1.2.1 are satisfied, thus E becomes a topological space called a coarse space (or 'indiscrete space'); one also says that E carries the coarsest topology (or 'indiscrete topology'). If ITl and !T2 are topologies on E, IT! is said to be finer than !T2 (and !T2 coarser than ITJ if every open set for !T2 is open for !Tl; this is an order relation among topologies. Every topology on E is finer than the coarsest topology, and coarser than the discrete topology. 1.2.4. For example on R" one can consider, in addition to the topology defined in 1.2.2, the discrete topology and the coarsest topology (and, to be sure, many other topologies). However, it is the topology defined in 1.2.2 that is by far the most interesting. Whenever we speak ofR" as a topological space without further specification, it is always the topology defined in 1.2.2 that is understood. 1.2.5. Definition. Let E be a topological space, A a subset of E. One says that A is closed if the subset E - A is open. 1.2.6. Theorem. Let E be a topological space. (i) The subsets 0 and E of E are closed. (ii) Every intersection of closed subsets ofE is closed. (iii) Every finite union of closed subsets ofE is closed. This follows from 1.2.1 by passage to complements. 6 I. Topological Spaces 1.3. Neighborhoods 1.3.1. Definition. Let E be a topological space and let x E E. A subset V of E is called a neighborhood of x in E if there exists an open subset U of E such that XEU c V. According to this definition, an open neighborhood of x is nothing more than an open subset of E that contains x. 1.3.2. Example. Let E be a metric space, x E E, VeE. The following conditions are equivalent: (i) V is a neighborhood of x; (ii) there exists an open ball with center x and radius > in V. °that is contained (ii) => (i). This is clear since the ball considered in (ii) is open and contains x. (i) => (ii). If V is a neighborhood of x, there exists an open subset U ofEsuch ° that x E U c V. By 1.1.3, there exists e > such that the open ball with center x and radius e is contained in U, thus afortiori contained in V. 1.3.3. Example. In R, consider the subset A = [0, 1]. Let x E R. IfO < x < 1 then A is a neighborhood of x. If x ~ 1 or x :::;; 0, A is not a neighborhood ofx. 1.3.4. Theorem. Let E be a topological space, and let x E E. (i) If V and V' are neighborhoods of x, then V n V' is a neighborhood of x. (ii) If'V is a neighborhood of x, and W is a subset of E containing V, then W is a neighborhood of x. Let V, V' be neighborhoods of x. There exist open subsets U, U' ofE such that x E U C V, X E U' c V'. Then x E U n U' c V n V', and U n U' is open by 1.2.1(iii), therefore V n V' is a neighborhood of x. The assertion (ii) is obvious. 1.4. Interior, Exterior, Boundary 7 .. 1.3.5. Theorem. Let E be a topological space, A a subset of E. The following conditions are equivalent: (i) A is open; (ii) A is a neighborhood of each qf its points. (i) => (ii). Suppose A is open. Let x E A. Then x E A c A, thus A is a neighborhood of x. (ii) => (i). Suppose condition (ii) is satisfied. For every x E A, there exists an open subset Bx of E such that x E Bx c A. Let A' = UxeA Bx. Then A' is open by 1.2.l(ii), A' c A since Bx c A for all x E A, and A' => A since each point x of A belongs to Bx , hence to A'. Thus A is open. 1.3.6. Definition. Let E be a topological space, and let x E E. One calls fundamental system of neighborhoods ofx any family (Vi)ieI of neighborhoods of x, such that every neighborhood of x contains one of the Vi' 1.3.7. Examples. (a) Suppose E is a metric space, x E E. For n = I, 2, 3, ... , let Bn be the open ball with center x and radius lin. Then the sequence (B1' 8 2 , •.• ) is a fundamental system of neighborhoods of x. For, if V is a neighborhood of x, there exists an 8 > 0 such that V contains the open ball B with center x and radius e (1.3.2). Let n be a positive integer such that lin S e. Then Bn c B c V. (b) Let us keep the same notations, and let B~ be the closed ball with center x and radius lin. Then (B~, B~, ... ) is a fundamental system of neighborhoods of x. For, Bn c B~ c Bn -1' thus our assertion follows from (a). (c) Let E be a topological space, and let x E E. The set of all neighborhoods of x is a fundamental system of neighborhoods of x. The set of all open neighborhoods of x is a fundamental system of neighborhoods of x (cf. 1.3.1). 1.3.8. Let E be a topological space, and let x E E. If one knows a fundamental system (Vi)i el of neighborhoods of x, then one knows all the neighborhoods of x. For, let VeE; in order that V be a neighborhood of x, it is necessary and sufficient that V contain one of the Vi (this follows from 1.3.4(ii) and 1.3.6). 1.4. Interior, Exterior, Boundary 1.4.1. Definition. Let E be a topological space, AcE, and x E E. One says that x is interior to A if A is a neighborhood of x in E, in other words if there exists an open subset ofE contained in A and containing x. The set of all points interior to A is called the interior of A and is often denoted A. 1.4.2. If x is interior to A, then of course x E '\. But the converse is not true. For example, if E = R and A = [0, 1], then A = (0, 1); the points 0 and 1 belong to A but are not interior to A. If E = R and A = Z, then A = 0. 8 I. Topological Spaces o 1.4.3. Theorem. Let E be a topological space, A a subset ofE. Then A is the largest open set contained in A. Let U be an open subset of E contained in A. If x E U then A is a neighborhood of x, therefore x EA. Thus U C o ° It is clear that A c A. Let us show that A is open. By 1.3.5, it suffices to prove that if x E A, then Ais a neighborhood of X. Now, there exists an open o subseto B of E such that x E B c A. Then B c A by the first part of the proof, thus A is a neighborhood of x. A- 1.4.4. Theorem. Let E be a topological space, A a subset ofE. Thefollowing conditions are equivalent: (i) A is open; (ii) A = A. (i) = (ii). If A is open, then the largest open set contained in A is A, therefore A by 1.4.3. (ii) (i). If A = then A is open because is open (1.4.3). A= = A, A 1.4.5. Theorem. Let E be a topological space, A and B subsets of E. Then (A n BY = A n B. The set (A n BY is open (1.4.3) and is contained in A n B, hence afortiori in A. Consequently (A n B)O c A 0 by 1.4.3. Similarly, (A n BY c BO, therefore (A 0 B)O CO A °n BO. 0 0 0 One has A c A, B c B, therefore A nBc A n B. But A n B is open (1.2.l(iii», therefore A nBc (A n BY by 1.4.3. ° 1.4.6. However, in general (A u B)O '" A u B. For example, take E = R, A = [0, 1], B = [1, 2]. Then A u B = [0, 2], (A u Bt = (0,2) '" 0 A= (0, 1), B= (1,2), (0, 1) u (1, 2). 1.4.7. Definition. Let E be a topological space, A a subset of E, x a point of E. One says that x is exterior to A if it is interior to E - A, in other words if there exists an open subset of E disjoint from A and containing X. The set of all exterior points of A is called the exterior of A; it is the interior of E - A. Interchanging A and E - A, we see that the exterior of E - A is the interior of A. 1.4.8. Let E be a topological space, AcE, Al the interior of A, A2 the exterior of A. The sets Al and A2 are disjoint. Let A3 = E - (Al U A2)' Then A l , A 2 , A3 form a partition ofE. One says that A3 is the boundary of A.1t is a closed set, since At u A2 is open. If one interchanges A and E - A, then Al 9 1.5. Closure and A2 are interchanged, therefore A3 is unchanged: a set and its complement have the same boundary. 1.4.9. Example. Let A be the subset [0, 1) of R. The interior of A is (0, 1), the exterior of A is ( ~ 00, 0) u (1, + (0), therefore the boundary of A is {O} u {I}. 1.5. Closure 1.5.1. Definition. Let E be a topological space, AcE and x E E. One says that x is adherent to A if every neighborhood of x in E intersects A. The set of all points adherent to A is called the closure (or 'adherence') of A, and is denoted A. 1.5.2. If x E A then of course x is adherent to A; but the converse is not true. For example, if E = R and A = (0, 1), then A = [0, 1]. 1.5.3. Theorem. Let E be a topological space, A a subset ofE. Then A is the complement of the exterior of A. Let x E E. One has the following equivalences: x ~ A<=> there exists a neighborhood of x disjoint from A <=> there exists a neighborhood of x contained in E ~ A <=> x is interior to E ~ A <=> x belongs to the exterior of A, whence the theorem. 1.5.4. Theorem. Let E be a topological space, AcE, BeE. (i) A is the smallest closed subset ofE containing A; (ii) A is closed <=> A = A; (iii) (A u Bf = A u B. In view of 1.5.3, this follows from 1.4.3, 1.4.4, 1.4.5 by passage to complements. For example, let us prove (i) in detail. The exterior of A is (E - At (1.4.7), that is to say, the largest open set contained in E - A (1.4.3). Therefore its complement A (1.5.3) is closed and contains A. If F is a closed subset of E containing A, then E ~ F is open and is contained in E - A, therefore E - F c (E - At = E - A, thus F =:> A. 1.5.5. Taking up again the notations of 1.4.8, Theorem 1.5.3 shows that U A 3 . Therefore the boundary A3 is the intersection of the closures A and (E - Af. A = Al U A 3 , (E - Af = A2 10 I. Topological Spaces 1.5.6. Theorem. Let E be a topological space, A a subset of E. The following conditions are equivalenc: (i) every nonempty open subset of E intersects A; (ii) the exterior of A is empty; (iii) the closure qf A is all qfE. Condition (i) means that the only open set contained in E - A is 0, thus, by 1.4.3, that the interior of E - A is empty. This proves (i) ~ (ii). The equivalence (ii) =- (iii) follows from 1.5.3. 1.5.7. Definition. A subset A of E satisfying the conditions 1.5.6 is said to be dense in E. 1.5.8. Example. Q is dense in R: for, every nonempty open subset of R contains a nonempty open interval, hence contains rational numbers. The complement R - Q of Q in R is also dense in R, because every nonempty open interval contains irrational numbers. 1.5.9. Theorem. Let A be a nonempty subset ofR that is bounded above, x its supremum. Then x is the largest element ofA. Let V be a neighborhood of x in R. There exists an open subset U ofR such that x E U c V. Then there exists e > 0 such that (x - e, x + e) c U. By the definition of supremum (least upper bound), there exists YEA such that x - e < y ~ x. Then y E U c V, therefore V n A i= 0. Thus x is adherent to A. Let x' E A. If x' > x, set e = x' - x > O. Then (x' - e, x' + e) is a neighborhood of x', therefore intersects A. Let YEA n (x' - e, x' + e). Since y > x' - L = x, x is not an upper bound for A, which is absurd. Thus x' ~ x. This proves that x is the largest element of A. l.6. Separated Topological Spaces 1.6.1. Definition. A topological space E is said to be separated (or to be a Hausdorff space) if any two distinct points ofE admit disjoint neighborhoods. 1.6.2. Examples. (a) A metric space E is separated. For, let x, y E E with x i= y. Set e = d(x, y) > O. Then the open balls V, W with centers x, y and radius e/2 are disjoint (because if Z E V n W then d(x, z) < e/2 and d(y, z) < e/2, therefore d(x, y) < e, which is absurd). (b) A discrete topological space E is separated. For, if x, Y E E and x i= y, then {x} and {y} are disjoint open neighborhoods of x, y. 11 1.6. Separated Topological Spaces (c) A coarse topological space E is not separated (if it contains more than one point). For, let x, y E E with x # y. Let V, W be neighborhoods of x, y. Then V contains an open subset U of E containing x, whence U = E and afortiori V = E. Similarly W = E. Thus V n W # 0. 1.6.3. Theorem. Let E be a separated topological space, x E E. Then {x} is closed. Let y E E - {x}. Then y # x, therefore there exist neighborhoods V, W of x, y that are disjoint. In particular, WeE - {x}, therefore E - {x} is a neighborhood of y. Thus E - {x} is a neighborhood of each of its points, consequently is open (1.3.5). Therefore {x} is closed. CHAPTER II Limits. Continuity As mentioned in the introduction, the limit concept is one of those at the origin of topology. The student already knows several aspects of this concept: limit of a sequence of points in a metric space, limit of a function at a point, etc. To avoid a proliferation of statements later on, we present in §2 a framework (limit along a 'filter base') that encompasses all of the useful aspects oflimits.1t doesn't hurt to understand this general definition, but it is much more important to be familiar with a host of special cases. The definition of limit of course brings with it that of continuity of functions: see §§3 and 4. Two topological spaces are said to be homeomorphic (§5) ifthere exists a bijection of one onto the other which, along with the inverse mapping, is continuous; two such spaces have the same topological properties, and one could almost consider them to be the same topological space. (For example, a circle and a square are homeomorphic; a circle and a line are not homeomorphic; and, what is perhaps more surprising, a line and a circle with a point omitted are homeomorphic.) A reasonable goal of topology would be to classify all topological spaces up to homeomorphism, but this seems to be out of reach at the present time. One knows very well that a sequence does not always have a limit. As a substitute for limit, we introduce in §6 the concept of adherence value. 13 2.1. Filters 2.1. Filters 2.1.1. Definition. Let X be a set. Afilteron Xis a set IF of non empty subsets of X satisfying the following conditions: (i) if A E IF and BE IF, then A n BE IF (in particular, A n B # 0); (ii) if A E IF and if A' is a subset of X containing A, then A' E IF. One calls filter base on X a set !?4 of nonempty subsets of X satisfying the following condition: (i') if A E !?4 and B E 14. there exists C E !?4 such that C cAn B (in particular, An B # 0). A filter is a filter base, but the converse is not true. If !?4 is a filter base on X, one sees immediately that the set of subsets of X that contain an element of !?4 is a filter. 2.1.2. Example. Let X be a topological space, Xo E X. The set "f/' of neighborhoods of Xo is a filter on X (1.3.4). If jf" is a fundamental system of neighborhoods of xo, then jf" is a filter base on X. 2.1.3. Example. Let Xo E R. The set of intervals (xo - 6, Xo + 6), where 6 > 0, is a filter base on R. This is, moreover, a special case of 2.1.2. But here are some examples of filter bases on R that are not special cases of 2.1.2: the set of [xo, Xo the set of (xo, Xo + 6), + 6), where e > 0; where e> 0; the set of (xo - e, x o], where e> 0; the set of (xo - e, xo), where e > 0; (xo, Xo + e), the set of (xo - e, xo) U 2.1.4. Example. The set of intervals [a, R. Similarly for the set of (- 00, a]. + where e > O. 00), where a E R, is a filter base on 2.1.5. Example. On N, the set of subsets {n, n n E N. is a filter base. + 1, n + 2, ... }, where 2.1.6. Example. Let X be a topological space, Y c X, and Xo E Y. The set of subsets of Y of the form Y n V, where V is a neighborhood of Xo in X, is a filter on Y (notably Y n V # 0 because Xo E V). If Y = X, one recovers 2.1.2. 14 II. Limits. Continuity 2.2. Limits Along a Filter Base 2.2.1. Definition. Let X be a set equipped with a filter base 81, E a topological space, f a mapping of X into E, I a point of E. One says that f tends to I along 81 if the following condition is satisfied: for every neighborhood V of I in E, there exists B E 81 such that feB) c V. If one knows a fundamental system (V;) of neighborhoods of I in E, it suffices to verify this condition for the Vi (for, every neighborhood of I contains a Va. 2.2.2. Example. Suppose X = N, with the filter base 81 considered in 2.1.5. A mapping of N into E is nothing more than a sequence (a o , aI' a 2 , ••• ) of points of E. To say that this sequence tends to I along 81 means: for every neighborhood V of I in E, there exists a positive integer N such that n ;:::: N => an E V. One then writes limn~ 00 an = I. If, for example, E is a metric space, this condition can be interpreted as follows: for every e > 0, there exists N such that n ;:::: N => dean, l) :::; e. 15 2.2. Limits Along a Filter Base One recognizes the classical definition of the limit of a sequence of points in a metric space (for example, of a sequence of real numbers). 2.2.3. Example. Let X and E be topological spaces, f a mapping of X into E, Xo E X, lEE. Take for fJI the filter of neighborhoods of Xo in X (2.1.2). To say that ftends to I along fJI means: for every neighborhood V of [in E, there exists a neighborhood W of Xo in X such that x E W ~ f(x) E v. One then writes limx_ xo f(x) = [. If X = E = R one recovers the definition of the limit of a real-valued function of a real variable at a point. 2.2.4. Examples. One knows that the concept of limit of a real-valued function of a real variable admits many variants. These variants fit into the general framework of 2.2.1. For example, if X = E = R and if one takes the filter bases considered in 2.1.3, 2.1.4, one recovers the following known concepts: lim f(x) x--+xo,X~XO lim f(x) x-xo,:C>Xo lim f(x) X-Xo, x.sXo lim f(x) x ..... xO,x<xQ lim f(x) x-.xQ,x*'XQ lim f(x) %-+00 lim f(x). 2.2.5. Example. Let X and E be topological spaces, Y c X, Xo E Y, fa mapping ofY into E, lEE. Take for fJI the filter defined in 2.1.6. To say that f tends to I along PJ means: for every neighborhood V of 1in E, there exists a neighborhood W of Xo in X such that x E Y (1 W ~ f(x) E v. One then writes limx_xo,xeY f(x) = I. This example generalizes 2.2.3. On taking X = E = R, and for Y various subsets of R, one recovers the first five examples of 2.2.4. 16 II. Limits. Continuity 2.2.6. Theorem. Let X, E be topological spaces, f a mapping of X into E, E X, lEE, (Wi)ie! afundamental system of neighborhoods ofx o in X, (Vj)jeJ a fundamental system of neighborhoods of I in E. The following conditions are equivalent: Xo (i) lim" .... f(x) = I; (ii) for every j E J, there exists i E I such that feW;) c Vj' "Q Suppose condition (i) is satisfied. Letj E J. There exists a neighborhood W of Xo in X such that few) c Vj' Next, there exists i E I such that Wi c W. Then f(W j ) c Vj' Suppose condition (ii) is satisfied. Let V be a neighborhood of I in E. There exists j E J such that Vj c V. Then there exists i E I such that f(Wi) c Vj' therefore f(W j ) c V. 2.2.7. Corollary. Let X, E be metric spaces, f a mapping of X into E, Xo E X, lEE. Thefollowing conditions are equivalent: (i) lim" .... f(x) = I; (ii) for every B > 0, there exists '1 > 0 such that "Q X E X, d(x, X o) ::; '1 => d(f(x), l) ::; B. For, the closed balls with center Xo (resp. /) and radius> 0 in X (resp. E) form a fundamental system of neighborhoods of Xo (resp. l) in X (resp. E). 2.2.8. Theorem. Let X be a set equipped with a filter base 111, E a separated topological space, f a mapping of X into E. Iff admits a limit along 111, this limit is unique. Let I, I' be distinct limits of f along f?J. Since E is separated, there exist disjoint neighborhoods V, V' of 1, l' in E. There exist B, B' E 111 such that feB) c V, f(B /) c V'. Next, there exists B" E f?J such that B" c B ! l B/. Then f(B") c V !l V'. Since B" oF 0, one has f(B") oF 0, therefore V ! l V' oF 0, which is absurd. 2.2.9. However, if E is not separated, f may admit more than one limit along 111. For example, if E is a coarse space, one verifies easily that every point ofE is a limit of f along 111. The use ofthe limit concept in non-separated spaces is risky; in this course, we shall scarcely speak of limits except for separated spaces. Under these qualifications, Theorem 2.2.8 permits one to speak of the limit (if it exists! the reader already knows many examples of mappings that have no limit at all). 17 2.3. Mappings Continuous at a Point 2.2.10. Theorem (Local Character of the Limit). Let X be a set equipped with a filter base fJI, E a topological space, f a mapping of X into E, lEE. Let X E fJI and let I' be the restriction off to x. The sets B n X, where BE fJI, form a filter base fJI' on X'. The following conditions are equivalent: (i) f tends to I along fJI; (ii) I' tends to I along f!4'. Suppose that f tends to I along f!4. Let V be a neighborhood of I in E. There exists B E f!4 such that feB) c V. Then I'(B n X') c V and B n X' E f!4', thus I' tends to I along f!4'. Suppose that I' tends to I along PJ'. Let V be a neighborhood of I in E. There exists B' E f!4' such that I'(B') c V. But B' is of the form B n X' with BE fJI. Since X' E f!4, there exists Bl E f!4 such that BI c B n X'. Then f(B 1 ) c I'(B') c V, thus f tends to I along fJI. 2.3. Mappings Continuous at a Point 2.3.1. Definition. Let X and Y be topological spaces,f a mapping of X into Y, and Xo EX. One says that f is continuous at Xo if lim x _ xQ f(x) = f(xo), in other words (2.2.3) if the following condition is satisfied: f for every neighborhood W of f(xo) in Y, there exists a neighborhood V of Xo in X such that fey) c w. 2.3.2. Example. Let X and Y be metric spaces, f a mapping of X into Y, and Xo EX. By 2.2.7, to say that f is continuous at Xo means: for every c > 0, there exists '1 > 0 such that x E X and d(x, x o) S '1 => d(f(x), f(x o» S c. We recognize here a classical definition. For example, if X = Y = R, one recovers the continuity of a real-valued function of a real variable at a point. 18 II. Limits. Continuity 2.3.3. Theorem. Let T be a set equipped with ajilter base 81. X and Y topological spaces, I E X, f a mapping of T into X that tends to I along 81, and g a mapping of X into Y that is continuous at l. Then g f tends to gel) along 81. 0 Let W be a neighborhood of gel) in Y. There exists a neighborhood V of I in X such that g(V) c W. Next, there exists aBE 81 such thatf(B) c V. Then (g nCB) c g(V) c W, whence the theorem. 0 2.3.4. Corollary. Let T, X, Y be topological spaces, f: T -+ X and g: X -+ Y mappings, and to E T. Iff is continuous at to, and g is continuous at fCto), then 9 f is continuous at to· 0 One applies 2.3.3 on taking for 81 the filter of neighborhoods of to in T, and I = f(t o). Then go f tends to g(f(t o)) along 81, that is, go f is continuous at to. 2.4. Continuous Mappings 2.4.1. Definition. Let X and Y be topological spaces,! a mapping of X into Y. One says that f is continuous on X if f is continuous at every point of X. The set of continuous mappings of X into Y is denoted ~(X, Y). 2.4.2. Examples. This notion of continuity is well known for real-valued functions of a real variable, and more generally for mappings of one metric space into another. 2.4.3. Theorem. Let X, Y, Z be topological spaces, f Z). Then 9 f E ~(X, Z). ~(Y, 0 This follows at once from 2.3.4. E ~(X, Y) and g E 2.4. Continuous Mappings 19 .. 2.4.4. Theorem. Let X and Y be topological spaces, f a mapping of X into Y. The following conditions are equivalent: (i) (ii) (iii) (iv) f is continuous; the inverse image under f of every open subset of Y is an open subset of X; the inverse image under f of every closed subset of Y is a closed subset of X; for every subset A of X, f(A) c f(A). (i) => (iv). Suppose f is continuous. Let A c X and Xo EA. Let W be a neighborhood of f(xo) in Y. Since f is continuous at xo, there exists a neighborhood V of Xo in X such that f(V) c W. Since Xo E A, V (\ A i= 0· Since f(V (\ A) c W (\ f(A), one sees that W (\ f(A) i= 0. This being true for every neighborhood W of f(xo), one has f(xo) E f(A). Thus f(A) c f(A). (iv) => (iii). Suppose condition (iv) is satisfied. Let Y be a closed subset ofY and let X' = f-l(y). Then feX') c Y', therefore f(X') c Y' (1.5.4). If x E X' then f(x) E feX') by condition (iv), therefore f(x) E Y and so x E X'. Thus X' = X', which proves that X' is closed. (iii) => (ii). Suppose condition (iii) is satisfied. Let Y be an open subset ofY. Then Y - Y' is closed, therefore f-I(y - V') is closed. But f-I(y - Y) = X - f-I(y). Therefore f-I(y,) is open. (ii) => (i). Suppose condition (ii) is satisfied. Let Xo EX; let us prove that f is continuous at Xo' Let W be a neighborhood of f(x o) in Y. There exists an open subset Y' of Y such that f(xo) E Y' c W. Let X' = f - I(Y'). Then X' is open by condition (ii), and Xo E X', thus X' is a neighborhood of Xo. Since f(X /) e Y e W, this proves the continuity of fat Xo. 2.4.5. Example. Let a, b be numbers> O. One knows that the mapping + y2/b z - I of R Z into R is continuous. On the other hand, [0, + (0) is a closed subset of R. Therefore the set of (x, y) E R Z such that x 2 /a 2 + yZ/b 2 - 1 ~ 0 is closed in R2. Similarly, the set of (x, y)ER Z such that x 2/a 2 + y2/b 2 - 1 = 0 (an ellipse) is closed in R2, etc. (x, y) f---+ x 2/a 2 2.4.6. Mistake to Avoid. There is a risk of confusing conditions (ii) and (iii) of 2.4.4 with the following conditions: (ii') the direct image of every open subset of X is an open subset of Y; (iii') the direct image of every closed subset of X is a closed subset of Y. Mappi!1gs satisfying (ii') (resp. (iii'» are called open mappings (resp. closed mappings). There exist continuous mappings that are neither open nor closed, open mappings that are neither continuous nor closed, and closed mappings that are neither continuous nor open. 2.4.7. Let :YI,:YZ be two topologies on a set E. Denote by E 1, E2 the set E equipped with the topologies:Y1 , :Yz . To say that :YI is finer than:Yz means, by 2.4.4(ii), that the identity mapping of EI into E z is continuous. 20 II. Limits. Continuity 2.5. Homeomorphisms 2.5.1. Theorem. Let X and Y be topological spaces, / a bijective mapping o/X onto Y. The/ollowing conditions are equivalent: (i) / and /-1 are continuous; (ii) in order that a subset 0/ X be open, it is necessary and sufficient that its image in Y be open; (iii) in order that a subset o/X be closed, it is necessary and sufficient that its image in Y be closed. . This follows at once from 2.4.4. 2.5.2. Definition. A mapping / of X into Y that satisfies the conditions of 2.5.1 is called a bicontinuous mapping of X onto Y, or a homeomorphism of X onto Y. (By 2.5.1(ii), this is the natural concept of isomorphism for the structure of topological space.) 2.5.3. It is clear that the inverse mapping of a homeomorphism is a homeomorphism. By 2.4.3, the composite of two homeomorphisms is a homeomorphism. 2.5.4. Let X and Y be topological spaces. If there exists a homeomorphism of X onto Y, then X and Yare said to be homeomorphic. By virtue of 2.5.3, this is an equivalence relation among topological spaces. If X and Yare homeomorphic, the open set structure is the same in X and Y; since all topological properties are defined on the basis of open sets, X and Y will have the same topological properties; X and Yare almost the 'same' topological space. One ofthe goals of topology (not the only one, far from it) consists in recognizing whether or not two given spaces are homeomorphic, and classifying topological spaces up to homeomorphism; this goal is far from being attained at the present time. 2.5.5. Examples. All nonempty open intervals in R are homeomorphic. For, if the open intervals Ii and 12 are bounded, there exists a homothety or a translation / that transforms Ii into 12 , and / is obviously bicontinuous. Similarly ifl! and 12 are of the form (a, + (0) or (- 00, a). Thus, it remains to compare the intervals (0, 1), (0, + (0) and (- 00, + (0). Now, the mapping x f-+ tan(1t/2)x of (0, 1) into (0, + 00) is bijective and continuous, and the inverse mapping x f-+ (2/1t)Arctan x is continuous; therefore (0, 1) and (0, + (0) are homeomorphic. Similarly, the mapping x f-+ tan(1t/2)x of( -1, 1) into ( - 00, + (0) is a homeomorphism. The intervals (0, 1) and [0, 1] are not homeomorphic (cf. 4.2.8). 2.5.6. Example. A circle and a square in R2 are homeomorphic (via a translation followed by a central projection). 21 2.5. Homeomorphisms 2.5.7. Example: Stereographic Projection. Let S. be the set of x such that xi = (XI' X2' •.. , X n + l)eRn+l + . . . + x; + 1 = 1 ('n-dimensional sphere'). a = (0, 0, ... , 0, 1) e S •. Let Let us identify Rn with the set Of(Xl, ... ,Xn,0)ERn+ 1. We are going to define a homeomorphism ofSn - {a} onto Rn. Let x = (X., ... , Xn + I) e Sn - {a}. The line D in R·+ I joining a and x is the set of points of the form (Ax l , ... , Ax., 1 + .Ie(Xn+l - 1» with .Ie e R. This point is in R· when 1 + .Ie(Xn+ 1 - 1) = 0, that is, when .Ie = (1 - Xn+1)-1 (Xn+l"# 1 because x"# a). Thus D ( l R n reduces to the point f(x) with coordinates , X2 (1) = 1 X2 - x. , X n+ 1 ' ... , x. = 1 - X.+ I ' , X.+ I We have thus defined a mapping f of Sn - {a} into Rn. Given x' (x'i> ... , x~, 0) in Rn, there exists one and only one point x = (Xl' ... ' X.+ 1 )ES. - {a} such that f(x) = x'. For, the solution of (1) yields the conditions Xi = xi(1 - X.+ 1) n 2:: x?(1 i= 1 - X n + 1)2 for 1::0:; + X;+ 1 i ::0:; n, = 1, then, after dividing out the nonzero factor 1 - Xn + 1, (x,? + ... + x~2)(1 - X n +l) - 1- X.+ 1 = 0, 0 = . = n. 22 Limits. Continuity whence X (2) { _ n+1 - X~:-;;-2_+_._._.+_X____~2;_--1 _ x'? + ... + X~2 + l' 2x; (1 :::; i :::; n). Thus f is a bijection of Sn - {a} onto Rn. The formulas (1) and (2) prove, moreover, that f and f - 1 are continuous. The homeomorphism f is called stereographic projection of Sn - {a} onto Rn. 2.6. Adherence Values Along a Filter Base 2.6.1. Definition. Let X be a set equipped with a filter base fA, E a topological space, f a mapping of X into E, and I a point of E. One says that I is an adherence value off along fA if the following condition is satisfied: for every neighborhood V of I in E and for every BE fA, feB) intersects V. If one knows a fundamental system of neighborhoods (V;) of I in E, it suffices to verify this condition for the Vi' 2.6.2. Example. Suppose in 2.6.1 that X = N, with the filter base 2.1.5. We are thus considering a sequence (aQ, ab ... ) of points of E. To say that I is an adherence value of the sequence means: for every neighborhood V of I in E, and every positive integer N, there exists n ~ N such that an E V. If, for example, E is a metric space, this condition may be rewritten as follows: for every c; > 0, and every positive integer N, there exists n such that dean, l) :::; B. For example, taking E 1- ~ N = R, consider the sequence of numbers t, t, 1- t, i, 1- i ..... The verification that the adherence values of this sequence are as an exercise. ° and 1 is left 23 2.6. Adherence Values Along a Filter Base 2.6.3. Example. Let X and E be topological spaces, Y c X, Xo E Y, f a mapping ofY into E, and lEE. Take for fJI the filter 2.1.6. To say that I is an adherence value of f along fJI means: for every neighborhood V of I in E, and every neighborhood W of Xo in X, f(W 11 Y) intersects V. One then says that I is an adherence value of f as x tends to Xo while remaining in Y. -1 For example, take X = E = R, Y = R - {OJ, Xo = 0, and f(x) = sin(1/x) for x E R - {OJ. It is left as an exercise to show that the adherence values of f, as x tends to 0 through values =F 0, are all the numbers in [-1,1]. From the examples 2.6.2, 2.6.3, one recognizes that the concept of adherence value is a kind of substitute for the limit concept. This point will now be elaborated. 2.6.4. Theorem. Let X be a set equipped with a filter base fA, E a separated topological space, f a mapping of X into E, and I a point ofE. Iff tends to I along fJI, then I is the unique adherence value off along fA. Let V be a neighborhood of I in E and let B E fJI. There exists B' E fA such that f(B') c V. Then B 11 B' =1= 0, therefore f(B 11 B') =1= 0, and f(B 11 B') c f(B) 11 V. Therefore f(B) intersects V. Thus I is an adherence value of f alongfJI. Let l' be an adherence value of f along fJI, and suppose l' =1= I. There exist neighborhoods V, V' of I, l' that are disjoint. Next, there exists B E fJI such that f(B) c V. Then f(B) 11 V' = 0, which contradicts the fact that l' is an adherence value. 2.6.5. Thus, when f admits a limit along fA, the concept of adherence value brings nothing new with it and is thus without interest. But what can happen 24 II. Limits. Continuity if f does not have a limit along f!J? (a) it can happen that f has no adherence value; example: the sequence (0, 1, 2, 3, ... ) in R has no adherence value; however, cf. 4.2.1; (b) it can happen that f has a unique adherence value; example: the sequence (0, 1,0,2,0,3, ... ) in R has no limit, and its only adherence value is 0; however, cf. 4.2.4; (c) it can happen that f has more than one adherence value (cf. 2.6.2, 2.6.3). To sum up, in relation to the limit concept, one loses uniqueness but one gains as regards existence. 2.6.6. Theorem. Let X be a set equipped with a filter base f!J, E a topological space, f a mapping of X into E. The set of adherence values off along f!J is the intersection of the feB) as B runs over f!J. Let I be an adherence value of f along f!J. Let BE f!J. Every neighborhood of I intersects feB), therefore IE feB). Thus IE n feB). BEBiI Suppose mE nBE9iI feB). Let V be a neighborhood of m and let BE f!J. Since mE feB), feB) intersects V. Therefore m is an adherence value of f. (There is thus a connection between the concept of adherence value and that of adherent point. However, the two concepts should not be confused.) CHAPTER III Constructions of Topological Spaces The study of any structure often leads to the study of substructures, product structures and quotient structures. For example, the student has already seen this in the study of vector space structure. The same is true for topological spaces, This yields important new spaces (for example, the tori T\ cf. also the projective spaces, in the exercises for Chapter IV). 3.1. Topological Subspaces 3.1.1. Theorem. Let E be a topological space, F a subset ofE. Let 011 be the set ofopen subsets of E, Let Y be the set of subsets of F oftheform V n F, where V E UIJ. Then Y satisfies the axioms (i), (ii), (iii) of 1.2.1. (i) One has 0 E UIJ and E E UIJ, therefore 0 = 0 n FE1" and F = En FEY. (ii) Let (Vi)iEI be a family of subsets belonging to Y. For every i E I, there exists Vi E UIJ such that Vi = Vi n F. Then UiEI Vi E UIJ, therefore UVi = ieI U (Vi n ieI n F) = (U Vi) n ieI FE 1/'. (iii) With notations as in (ii), suppose moreover that I is finite. Then Vi E UIJ, therefore iE I n n(Vi ieI Vi = ieI n F) = (n Vi) iel n F E "f~. 26 HI. Constructions of Topological Spaces 3.1.2. By 3.1.1, j~ is the set of open sets of a topology on F, called the topology induced on F by the given topology of E. Equipped with this topology, F is called a topological subspace (or simply a subspace) ofE. The open sets of Fare thus, by definition, the intersections with F of the open sets of E. 3.1.3. Remark. Let E be a topological space, F a subspace of E. If A is a subset of F, the properties of A relative to F and relative to E may differ. For example, if A is open in E then A is open in F (because A = A n F), but the converse is in general not true (for example, F is an open subset of F but in general is not an open subset of E). However, if F is an open subset of E and if A is open in F. then A is open in E (because A = B n F with B open in E, and the intersection of two open subsets of E is an open subset of E). 3,1.4. Theorem. Let E bea topological space, F a subspace orE, A a subset ofF. The following conditions are equivalent: (i) A is closed in F; (ii) A is the intersection with F of a closed subset of E. (i) => Oi). Suppose A is closed in F. Then F - A is open in F, therefore there exists an open subset U of E such that F - A = Un F. Since A = (E - U) n F and since E - U is closed in E, we see that condition (ii) is satisfied. (it) => (i). Suppose A = X n F, where X is a closed subset of E. Then F - A = (E - X) n F, and E - X is open in E, therefore F - A is open in F, thus A is closed in F. 3.1.5. Remark. Let us maintain the notations of 3.1.4. If A is closed in E then A is closed in F, but the converse is in general not true (for example, F is closed in F but in general not in E). However, if F is closed in E and if A is closed in F, then A is closed in E (because A = X n F with X closed in E, and the intersection of two closed subsets of E is a closed subset of E). 3.1.6. Theorem. Let E be a topological space, F a subspace of E, and x Let W c F. The following conditions are equivalent: E F. (i) W is a neighborhood ofx in F; Oi) W is the intersection with F of a neighborhood of x in E. (i) => (ii). Suppose W is a neighborhood of x in F. There exists an open subset B of F such that x E B c W. Then there exists an open subset A of E such that B = F n A. Let V = A u W. Then x E A c: V, thus V is a neighborhood of x in E. On the other hand, F n V = (F n A) u (F n W) = B u W = W. 27 3.1. Topological Subspaces (ii) => (i). Suppose W = F n V, where V is a neighborhood of x in E. There exists an open subset A of E such that x E A c V. Then x E F n A c F n V = W, and F n A is open in F, thus W is a neighborhood of x in F. 3.1.7. Theorem. Let E be a topological space, F a subspace of E. If E is separated, then F is separated. Let x, y be distinct points of F. There exist disjoint neighborhoods V, W of x, y in E. Then F n V, F n Ware neighborhoods of x, y in F (3.1.6) and they are disjoint. Thus F is separated. 3.1.8. Definition. Let E be a topological space, AcE and x E A. One says that x is an isolated point of A if there exists a neighborhood V of x in E such that V n A = {x}. 3.1.9. Theorem. Let E be a topological space and let FeE. The following conditions are equivalent: (i) the topological space F is discrete; (ii) every point ofF is isolated. (i) => (ii). Suppose F is discrete. Let x E F. Then {x} is an open subset of F, therefore there exists an open subset U of E such that {x} = U n F. Since U is a neighborhood of x in E, we see that x is an isolated point of F. (ii) => (i). Suppose that every point of F is isolated. Let x E F. There exists a neighborhood V of x in E such that V n F = {x}. Passing to a subset of V, we can suppose that V is open in E. Then {x} is open in F. Since every subset of F is the union of one-element subsets, every subset of F is open in F. Thus the topological space F is discrete. 3.1.10. Example. In R, consider the subset Z. If n {n} E Z then = Z n (n - t, n + !), therefore n is isolated in Z. Thus the topological subspace Z ofR is discrete. 3.1.11. Theorem (Transitivity of Subspaces). Let E, E', E" be sets such that E ::J E' ::J E". Let :T be a topology on E, :Y' the topology induced by :T on E'. :T" the topology induced by :T' on E". Then .'T" is the topology induced by :T on En. Let :T'1 be the topology induced by :T on En. Let un c En be an open set for :T". There exists a subset U' ofE', open for :T', such that U' n E" = U". Next, there exists a subset U of E. open for:T, such that U n E' = U'. Then U n En = un, thus un is open for :T'1' 28 III. Constructions of Topological Spaces Let V" c E" be an open set for .r~. There exists a subset V ofE, open for .r, such that V n E" = V". Set V' = V n E'. Then V' is open for .r', and V" = V' n E", thus V" is open for .rH • 3.1.12. Theorem. Let E be a metric space, E' a metric subspace of E(1.1.1). Let .r, .r' be the topologies of E, E' defined by their metrics (1.2.2). Then .r' is nothing more than the topology induced by.r on E'. Let .r'1 be the topology induced by .r on E'. Let U' c E' be an open set for .r'. For every x E V', there exists 6 x > 0 such that the open ball B~ in E' with center x and radius 6 x is contained in V'. Let Bx be the open ball in E with center x and radius 6x . Then Bx is a neighborhood of x for.r, therefore B~ is a neighborhood of x for.r 1(3.1.6). Thus V' is a neighborhood of x for .r1. This being true for every x E 0', we see that V' is open in E' for .r1 (1.3.5). Let V' c E' be an open set for .rI . There exists an open subset V of E such that V' = V n E'. For every x E V', there exists '1x > 0 such that the open ball C x in E with center x and radius '1x is contained in V. Let C~ be the open ball in E' with center x and radius '1x' Then C~ = C x n E' c V n E' = V'. Therefore V' is open for .r'. 3.1.13. For example, if a subset A ofRn is regarded as a topological subspace of Rn, the topology of A is none other than the one considered in 1.2.2. 3.1.14. Theorem. Let X be a set equipped with a filter base 91, E a topological space, E' a subspace ofE, fa mapping of X into E', I a point ofE'. The following conditions are equivalent: (i) f tends to I along 91 relative to E'; (ii) f tends to I along 91 relative to E. Suppose condition (i) is satisfied. Let V be a neighborhood of I in E. Then V n E' is a neighborhood of I in E' (3.1.6). There exists BE!JI such that f(B) c V n E'. Afortiori,J(B) c V. Thus ftends to I along 91 relative to E. Suppose condition (ii) is satisfied. Let V' be a neighborhood of I in E'. There exists a neighborhood V of I in E such that V n E' = V' (3.1.6). Then there exists BE!JI such that feB) c V. Since f(X) c E', one has f(B) c V n E' = V'. Thus f tends to I along 9B relative to E/. ~ 3.1.15. Theorem. Let T, E be topological spaces, E' a subspace of E, f a mapping of T into E'. The following conditions are equivalent: (i) f is continuous; (ii) f, regarded as a mapping of T into E, is continuous. 29 3.2. Finite Products of Topological Spaces Indeed, for every to E T, the condition lim, .... ,Q f(t) = f(to) has the same meaning, according to 3.1.14, whether one considers f to have values in E' or to have values in E. 3.1.16. Corollary. Let E be a topological space, E' a subspace of E. The identity mapping of E' into E is continuous. For, the identity mapping of E' into E' is obviously continuous. One then applies 3.1.15. 3.2. Finite Products of Topological Spaces 3.2.1. Let E l' E 2 , . . . , En be topological spaces. We are going to define a natural topology on E = El X E2 X ... x En. Let us call elementary open set in E a subset of the form V 1 X V 2 X .•. X Un, where Vi is an open subset of E j • Let us call open set in E any union of elementary open sets. To justify this terminology, we are going to show that this family of subsets of E satisfies the axioms (i), (ii), (iii) of 1.2.1. First of all, E = El X E2 X ... x En 0 and = 0 X E2 x En X ... are open sets, even elementary open sets. Axiom (ii) is obvious. Finally, let A, B be open subsets of E and let us show that A n B is an open subset of E. One has A = U A l , B = U B/L' where the Al and B/L are elementary open sets. Then A n B is the union of the A.. n B/L and it suffices to prove that, for fixed A and Il, Al n B/L is an elementary open set. Now, A;. = VI x ... x V", B/L = VI X ... X Vn , where V j , Vi are open subsets of E i • It follows that A). n BI' =(U 1 nV I ) x ... x (VnnV n); since U i n Vi is an open subset of Ei , this completes the proof. 3.2.2. We have thus defined a topology on E, called the product topology of the given topologies on E I , ••• , En. We also say that E is the product topological space of the topological spaces Ej, E 2 , ..• , En. 3.2.3. Example (Product of Metric Spaces). Let E 1 , E 2 • . • . spaces. Set E = EI X ... x En. , En be metric 30 III. Constructions of Topological Spaces One verifies as in the case of R n that d is a metric on E. Thus, a product of metric spaces is automatically a metric space. Let !f be the topology on E defined by this metric (1.2.2). In addition, E l , ... , En are topological spaces (1.2.2), thus by 3.2.2 there is defined a product topology!f' on E. Let us show that !f = !f'. Let V be a subset ofE that is open for !f'. For every x = (Xl> ... , xn) E V, there exist open neighborhoods V 1> ••• , Vn of Xl' ... , Xn in E l , ... , En such that VI X ... x Vn C U. There exists 6j > 0 such that the open ball in Ei with center Xi and radius 6i is contained in Vi. Let B be the open ball in E with center X and radius 6 = inf(6l •...• 6 n). If Y = (Yl, ...• Y.) E B then d(x, y) < 6, therefore d(x" y;) < 6 ~ 6i for all i, thus Yi E Vi' and so y E U. Thus B c V. We have thus proved that V is open for !f. Let V be a subset of E that is open for !f. For every x = (Xl' ... , xn) E V, there exists 6 > 0 such that V contains the open ball with center X and radius 6. Let Bi be the open ball in Ei with center Xi and radius eln. If then therefore y E V. Thus, the elementary open set B 1 X ..• x Bn is contained in V and contains x. Consequently, V is the union of elementary open sets, therefore is open for !f'. In particular, if Rn is equipped with the topology defined by its usual metric (1.1.2), R n appears as the product topological space R x R x ... x R. 3.2.4. Theorem. Let E = El X •.. x En be a product of topological spaces. Let X = (Xl> ... , x n) E E. The sets of the form VI X .•• X V n , where Vi is a neighborhood ofxj in E i , constitute afundamental system of neighborhoods ofx inE. For i = 1, ... , n, let Vi be a neighborhood of Xi in E i . There exists an open subset Vi of Ei such that Xi E Vi C Vi. Then and U 1 x ... inE. X Vn is open in E, thus VI x ... X Vn is a neighborhood of X 31 3.2. Finite Products of Topological Spaces Let V be a neighborhood of x in E. There exists an open subset V of E such that x EVe V. The set V is the union of elementary open sets, therefore x belongs to one of these sets, say VI X ••• X Vn' where Vi is an open subset of E i . Then XiEVi, thus Vi is a neighborhood of Xi in E i , and VI x ... X Vn cV. 3.2.5. Theorem. Let E = EI X ... x En be a product of topological spaces. If each Ei is separated, then E is separated. Let x = (Xl' ... , xn) and Y = (YI, ... , Yn) be two distinct points of E. One has Xi #- Yi for at least one i, for example XI #- YI' There exist disjoint neighborhoods V, W of XI> Y\ in E\. Then V x E2 X .•• x En and W x E2 X .•. x En are neighborhoods of X, Y in E (3.2.4) and they are disjoint. 3.2.6. Theorem. Let X be a set equipped with a filter hase f!I, E = El X E2 X ... x En a product of topological spaces, I = (/\, ... , In) E E. Let f be a mapping of X into E, thus of the form x 1-4 (fl(X), ... , fix», where /; is a mapping of X into E i . Then the following conditions are equivalent: (i) f tends to I along f!I; (ii) for i = 1, 2, ... , n, /; tends to Ii along f!I. E, If':I~~::""~-=-=-=-.-_-"-1 -,-(X-)---' Ez /' I I /, I E I I I " , ()() Suppose that f tends to I along f!I. Let us show, for example, that fl tends to 1\ along f!I. Let V I be a neighborhood of II in E 1 . Then V I x E2 X ... x En is a neighborhood of I in E (3.2.4). Therefore there exists BE f!I such that feB) c V 1 X E2 X ... x En. Then f\ (B) c VI' thus fl tends to 11 along f!I. Suppose condition (ii) is satisfied. Let V be a neighborhood of I in E. There exist neighborhoods VI' ... , Vn of iI' ... , In in E 1 , ••• , En such that V \ x ... x Vn C V (3.2.4). Then there exist B 1, ... , Bn E.OJ such that fl (B 1 ) c V \, ... , fn(Bn) c Vn' Next, there exists BE f!I such that B c BIn ... n Bn. Then thus f tends to I along f!I. 32 III. Constructions of Topological Spaces 3.2.7. Theorem. Let E = EI X ••• x En be a product of topological spaces, and T a topological space. Let f be a mapping of T into E, thus of the form t H (ft(t), ... , j,,(t», where /; is a mapping of T into E i . The following conditions are equivalent: ~ (i) f is continuous; (ii) fl' ... ,f,. are continuous. Indeed, for every to E T, the conditions lim f(t) = f(to), lim j;(t) = j;(t o) for i = 1, ... , n are equivalent by 3.2.6. 3.2.8. Corollary. Let E = El X .•. x En be a product of topological spaces. The canonical projections of E onto E l , ... , En are continuous. Let fbe the identity mapping ofE. It is continuous. Now, it is the mapping x H (f1(X), ... , fn{x», where ft, ... , fn are the canonical projections of E onto E t , ... , En. It then suffices to apply 3.2.7. 3.2.9. Theorem. Let X, Y, Z be topological spaces. The mapping (x, y, z) H «x, y), z) of X x Y x Z onto (X x Y) x Z is a homeomorphism. This mapping is obviously bijective. The mappings (x, y, z) H x and (x, y, z) H y of X x Y x Z into X and Y are continuous (3.2.8), therefore the mapping (x, y, z) H (x, y) of X x Y x Z into X x Y is continuous (3.2.7). Similarly, the mapping (x, y, z) H z of X x Y x Z into Z is continuous (3.2.8), therefore the mapping (x, y, z) H «x, y), z) of X x Y x Z into (X x Y) x Z is continuous (3.2.7). Similarly, one proves successively the continuity of the following mappings: «x, y), z) H (x, y), «x, y), z) H x, «x, y), z) H y, «x, y), z) H z, «x, y), z) H (x, y, z). 3.2.10. On account of3.2.9, one identifies the topological spaces (X x Y) x Z and X x Y x Z. This reduces step by step the study of finite products of topological spaces to the study of a product of two spaces. For 1 :5: P < n, one identifies R n with RP x Rn- P, etc. 3.2.11. Theorem. Let X, Y be topological spaces, Yo a fixed point ofY, A the subspace X x {Yo} of X x Y. The mapping x H (x, Yo) of X onto A is a homeomorphism. 33 3.2. Finite Products of Topological Spaces This mapping is obviously bijective. It is continuous from X into X x Y by 3.2.7, therefore it is continuous from X into A :~ ______ t--_ _ _ (x...,..':_O)----1 X X K Y x by 3.1.15. The inverse mapping is the composite ofthecanonical injection of A into X x Y, which is continuous (3.1.16), and of the canonical projection of X x Y onto X, which is also continuous (3.2.8). 3.2.12. For example, one can identify R with the subspace R x {O} ofR 2 , etc. 3.2.13. Theorem. Let X be a separated topological space, ~ the diagonal of X x X (that is, the set of all (x, x), where x runs over X). Then ~ is closed in X xX. Let us show that (X x X) - ~ is open in X x X, that is, is a neighborhood of each of its points. Let (x, y) E X X X. If (x, y) rio ~ then x # y. Since X is separated, there exist disjoint neighborhoods V, W of x, y. Then V x W is a neighborhood of (x, y) in X x X (3.2.4), and V x W is disjoint from~, that is to say contained in (X x X) - ~. Thus (X x X) - ~ is a neighborhood of (x, y). 3.2.14. Corollary. Let E be a topological space, F a separated topological space, f and g continuous mappings of E into F. The set of x E E such that f(x) = g(x) is closed in E. For, the mapping x 1-+ hex) = (f(x), g(x» of E into F x F is continuous (3.2.7). Let ~ be the diagonal of F x F, which is closed in F x F (3.2.13). The set studied in the corollary is nothing more than h - iCM, therefore is closed (2.4.4). 3.2.15. Corollary. Let E, F, f, g be as in 3.2.14. Iff and g are equal on a dense subset ofE, then f = g. For, the set of 3.2.14 is here both dense and closed, therefore equal to E. 34 III. Constructions of Topological Spaces 3.3. Infinite Products of Topological Spaces 3.3.1. Let (EJiEI be a family of topological spaces. Let E = ITE' E i • There is an obvious way of extending the definitions 3.2.1, 3.2.2 to this situation, but this does not lead to a useful concept. One calls elementary open set in E a subset of the form DiEI Vi' where Vi is an open subset of E i , and where Vi = Ei for almost all i E I (which, in this context, will mean that Vi = Ei for all but a finite number of indices). For I finite, one recovers the definition 3.2.1. One again calls open set of E any union of elementary open sets. One verifies as in 3.2.1 that this defines a topology on E, called the product topology of the topologies of the E j • 3.3.2. Most of the arguments of 3.2 may be extended with minimal complications. We state the results: (a) Let x = (Xi)iEI E E, where Xi E Ei for all i E I. The sets of the form DiEI Vi' where Vi is a neighborhood of Xi in Ei and where Vi = Ei for almost all i, constitute a fundamental system of neighborhoods of X in E. (b) If every E; is separated, then E is separated. (c) Let X be a set equipped with a filter base Pl, f a mapping of X into E (thus of the form x f---> (};(X))iEI, where Ii is a mapping of X into Ei)' and 1= (li)iEI E E. The following conditions are equivalent: (i) f tends to I along /14; (ii) for every i E I, Ii tends to Ii along Pl. (d) Let T be a topological space, f a mapping ofT into E (thus of the form t f---> (};(t»iEI, where Ii is a mapping ofT into EJ The following conditions are equivalent: (i) f is continuous; (ii) every Ii is continuous. (e) The canonical projections of E onto the Ei are continuous. (f) If I is the union of disjoint subsets I), where Aruns over a set A, then the topological space DiEI Ei may be identified with the topological space DAE i\ <DiEI;. EJ (,associativity of the topological product'). 3.4. Quotient Spaces 3.4.1. Theorem. Let E be a topological space, R an equivalence relation on E, F the quotient set EjR, rr the canonical mapping of E onto F. Let (9 be the set of subsets A ofF such that rr-I(A) is open in E. Then (9 satisfies the axioms (i), (ii), (iii) of 1.2.l. This is immediate. 35 3.4. Quotient Spaces 3.4.2. Thus, lP is the set of open sets of a topology on F, called the quotient topology of the topology of E by R. Equipped with this topology, F is called the quotient space of E by R. The mapping 'It ofE onto F is continuous (for, if A is open in F, then 'It-I(A) is open in E). 3.4.3. Example. The set T is defined to be the quotient ofR by the equivalence relation x - Y E Z. By 3.4.2, T is equipped with a topology that makes it a quotient space of R. .. 3.4.4. Theorem. Let F = E/R be the quotient space of a topological space E by an equivalence relation R, 'It the canonical mapping of E onto F, Y a topological space, and f a mapping of F into Y. The following conditions are equivalent: (i) f is continuous; (ii) the mapping f a 'It ofE into Y is continuous. [email protected] -------~- E/R f-1 (U) Suppose f is continuous. Since 'It is continuous (3.4.2), f 'It is continuous. Suppose f a 'It is continuous. Let U be an open subset of Y. Then 0 'It- 1 (f-I(U)) = (f a n)-I(U) is open in E (2.4.4), therefore f-I(U) is open in F (3.4.2). Thusfis continuous (2.4.4). 3.4.5. Example. Denote by U the set of complex numbers of absolute value 1. One knows that the mapping x f--+ g(x) = exp(2nix) of R into U is surjective, and that g(x) = g(x') <=> X - x' E Z. Thus, if p denotes the canonical mapping of R onto T, then 9 defines, by passage to the quotient, a bijection f of Tonto U such that fop = g. Since 9 is continuous, f is continuous (3.4.4). We shall see later on (4.2.16) that f is a homeomorphism. Let us show that T is separated. Let x, y be distinct points ofT. Then f(x) f= fey). Since U is separated, there exist disjoint open neighborhoods V, W of f(x), fey) in U. Then f-I(V), f -I(W) are disjoint open neighborhoods of x, y in T. CHAPTER IV Compact Spaces This chapter is probably the most important of the course. Although the definition of compact spaces (§1) suggests no intuitive image, it is a very fruitful definition (see the properties of compact spaces in §§2 and 3, and the applications in nearly all the rest of the course). In §4, we adjoin to the real line R a point + ex; and a point - CIJ so as to obtain a compact space, the 'extended real line' R; the student has made use of this for a long time, even though the terminology may appear to be new. Locally compact spaces are introduced in §5; for this course, they are much less important than the compact spaces. 4.1. Definition of Compact Spaces 4.1.1. Theorem. Let E be a topological space. The following conditions are equivalent: (i) If a family of open subsets of E covers E, one can extract from it a finite subfamily that again covers E. (ii) If a family of closed subsets oi E has empty intersection, one can extract from it a finite subfamily whos; intersection is again empty. This is immediate by passage to complements. 4.1.2. Definition. A separated space that satisfies the equivalent conditions of 4.1.1 is called a compact space. Let E be a topological space. A subset A of E such that the topological space A (3.1.2) is compact is of course called a compact subset of E. 37 4.2. Properties of Compact Spaces 4.1.3. Theorem. Let E be a topological space, A a separated subspace of E. Thefollowing conditions are equivalent: (i) A is compact; (ii) if a family of open subsets of E covers A, one can extract from it a .finite subfamily that again covers A. Suppose A is compact. Let (Vi)iel be a family of open subsets ofE covering A. The Vi ( l A are open in A (3.1.2) and cover A, therefore there exists a finite subset J of I such that the family (Vi ( l A)ieJ covers A. A fortiori, the family (Vi)ieJ covers A. Suppose conditions (ii) is satisfied. Let (Vi)iel be a family of open sets of A covering A. For every i E I, there exists an open set Wi of E such that Vi = Wi ( l A (3.1.2). Then (Wi)iel covers A, thus there exists a finite subset J of I such that (Wi)ieJ covers A. Therefore (Vi)ieJ covers A. 4.1.4. Theorem (Borel-Lebesgue). Let a, b E R with a [a, b] is compact. ~ b. Then the interval It is clear that [a, b] is separated. Let (Vi)iel be a family of open subsets of R covering [a, b]. Let A be the set of x E [a, b] such that [a, x] is covered by a finite number of the sets Vi' The set A is nonempty because a E A. It is contained in [a, b], therefore it is bounded above. Let m be its supremum. Then a ~ m ~ b. There exists j E I such that m E V j ' Since V j is open in R, there exists E > 0 such that [m - E, m + e] c Vj. Since m is the supremum of A, there exists x E A such that m - E < X ~ m. Then [a, x] is covered by a finite number of the Vi' and [x, m + e] c Vj, therefore [a, m + e] is covered by a finite number of the Vi' Ifm < b then, after reducing e if necessary so that m + E E [a, b], one sees that m + e E A, which contradicts the definition of supremum. Therefore m = b, and [a, b] is covered by a finite number of the Vi' It then suffices to apply 4.1.3. 4.2. Properties of Compact Spaces We are going to show: (1) that compact spaces have useful properties; (2) that there exist interesting examples of compact spaces (besides those of 4.1.4). The logical order of the proofs unfortunately obliges us to mix the two objectives. ~ 4.2.1. Theorem. Let X be a set equipped with a.filter base {fl, E a compact space, f a mapping of X into E. Then f admits at least one adherence value along rJI. 38 IV. Compact Spaces Consider the family of subsetsf(B) of E, where B runs over (fl. These are closed sets. Let A = nBeat feB). If A = 0, there exist B!, ... , B. E (fl such that feB!) n ... n f(B.) = 0 (because E is compact). Now, there exists Bo E (fl such that Bo c Bl n ... n B., whence f(Bo) c f(B l ) n ... n f(B.) and consequently f(Bd n ... n f(B.) '" 0. This contradiction proves that A '" 0. In view of 2.6.6, this proves the theorem. 4.2.2. Corollary. In a compact space, every sequence of points admits at least one adherence value. 4.2.3. Theorem. Let X, (fl, E, f be as in 4.2.1. Let A be the set (nonempty) of adherence values of f along (fl. Let U be an open subset of E containing A. There exists B E (fl such that feB) c U (and even feB) c U). One has (E - U) n A = 0, thus (E - U) n n Beat feB) = 0· Since E - U and the feB) are closed we infer, by the compactness of E, that there exist B l' ... , B. E (fl such that (E - U) nf(Bt) n ... nf(B.) Next, there exists Bo E (fl = 0· such that Bo c Bl n ... n B•. Then (E - U) n f(Bo) = 0, that is, f(Bo) c U. • 4.2.4. Corollary. Let X, (fl, E, f be as in 4.2.1. herence value I along (fl, then f tends to I along (fl. !f f admits only one ad- With the preceding notations, we have A = {I}, and U can be taken to be any open neighborhood of I. 4.2.5. Corollary. In a compact space, herence value I, then it tends to I. if a sequence of points has only one ad- • 4.2.6. Theorem. Let E be a compact space, F a closed subspace of E. Then F is compact. Since E is separated, F is separated. Let (Fi)ieI be a family of closed subsets of F with empty intersection. Since F is closed in E, the Fi are closed in E (3.1.5). Since E is compact, there exists a finite subfamily (Fi)ieJ with empty in tersection. 4.2. Properties of Compact Spaces 39 4.2.7. The converse of 4.2.6 is true. Better yet: Theorem. Let E be a separated space, F a compact subspace of E. Then F is closed in E. ~ We are first going to prove the following: (*) Let Xo E E - F. There exist disjoint open sets U, V of E such that Xo E U and Fe V. Since E is separated, for every y E F there exist open neighborhoods U y , Vy of xo, y in E that are disjoint. The Vy, as y runs over F, cover F. Since F is compact, there exist (4.1.3) points Y1,' .. , Yn of F such that F c V YI U ... U V Yn' Let U = U YI n ... n U Yn' which is an open neighborhood of Xo in E. Set V = V Y1 U ... U V yn , which is an open subset of E containing F. Then U and V are disjoint, and we have proved (*). Since U c E - F, it follows in particularfrom (*) that E - F is a neighborhood of Xo in E. This being true for every Xo E E - F, E - F is open in E, therefore F is closed in E. 4.2.8. Corollary. In R, the compact subsets are the closed bounded subsets. Let A be a compact subset of R. Then A is closed in R (4.2.7). On the other hand, it is clear that A c (x - 1, x + 1); by 4.1.3, A is covered by a finite number of intervals (Xi - 1, Xi + 1), hence is bounded. Let B be a closed, bounded subset of R. There exists an interval [a, b] such that B c [a, b]. Then [a, b] is compact (4.1.4), and B is closed in [a, b] (3.1.5) hence is compact (4.2.6). UxeA 4.2.9. Theorem. Let E be a separated space. (i) If A, B are compact subsets of E, then A u B is compact. (ii) If (Ai);E! is a nonempty family of compact subsets of E, then n;E! Ai is compact. Let (U;)iE! be a covering of A u B by open subsets of E. There exist finite subsets J 1 , J 2 of I such that (U;);eJI covers A and (U;)ie], covers B. Then (U;)ieJ, u J2 covers A u B, which proves that A u B is compact (4.1.3). The Ai are closed in E (4.2.7), therefore niEI Ai is closed in E, hence in each A; (3.1.5). Since the Ai are compact, niEi A; is compact (4.2.6). 4.2.10. However, an infinite union of compact subsets is not in general compact. For example, the intervals [-1, 1], [ - 2, 2], [- 3, 3], ... of R are compact, but their union, which is R, is not compact (4.2.8). 40 IV. Compact Spaces 4.2.11. Theorem. (i) Let E be a separated space, A and B disjoint compact suhsets ofE. There exist disjoint open sets V, V of E such that A c V and B c V. (ii) In a compact space, every point has a fundamental system of compact neighborhoods. (i) For every x E A, there exist disjoint open subsets W x' W~ of E such that x E W x' B c W~ (cf. the assertion (*) in the proof of 4.2.7). Since A is compact, there exist x I, ... , xp E A such that A c W Xl U ... U W Xp' Set V = W Xl U' .. U Wand V = W'Xl n··· n W'Xp . Then V and V are Xp open subsets ofE, A c V, Be V and V n V = 0. (ii) Suppose E is compact. Let x E E and let Y be an open neighborhood of x in E. Then {x} and E - Yare disjoint compact subsets of E. By (i), there exist disjoint open sets V, V of E such that x E V and E - Y c V. Then is a compact neighborhood of x. We have VeE - V, therefore 0 c E - V since E - V is closed, whence 0 c Y. o 4.2.12. Theorem. Let E be a compact space, F a separated space,f a continuous mapping ofE into F. Then feE) is compact. ~ First, feE) is separated since F is separated. Let (Vi)ieI be a family of open subsets of F covering feE). Since f is continuous, the f-I(V;) are open subsets of E (2.4.4). Since the Vi cover feE), the f-l(V i ) cover E. Since E is compact, there exists a finite subset J of I such that U-I(Vi»iEJ covers E. Then (V;)ieJ covers feE). Therefore feE) is compact (4.1.3). 4.2.13. Corollary. Let E be a nonempty compact space, f a continuous realvalued function on E. Then f is bounded and attains its infimum and supremum. By 4.2.l2,f(E) is a compact subset of R, hence is a closed, bounded subset of R (4.2.8). Since feE) is bounded, f is bounded. Since, moreover, fCE) is closed, feE) has a smallest and a largest element (1.5.9). If, for example, f(xo) is the largest element of f(E), then f attains its supremum at Xo' 4.2.14. Corollary. Let E be a compact space, f a continuous real-valued function on E with values >0. There exists rt. > 0 such that f(x) ;;:: rt. for all XE E. Let rt. be the infimum of f on E. By 4.2.13, there exists Xo E E such that f(x o) = rt.. Therefore rt. > O. It is clear that f(x) ;;:: rt. for all x E E. 41 4.2. Properties of Compact Spaces 4.2.15. Corollary. Let E be a compact space, F a separated space, f a continuous bUective mapping ofE onto F. Then f - 1 is continuous (in other words, f is a homeomorphism of E onto F). Let 9 = f - I. If A is a closed subset of E, then A is compact (4.2.6), therefore f(A) is compact (4.2.12), therefore f(A) is closed in F (4.2.7), in other words g-'(A) is closed in F. This proves that 9 is continuous (2.4.4). 4.2.16. Example. Let p be the canonical mapping of R onto T. It is continuous (3.4.2). Since T is separated (3.4.5) and [0, I] is compact (4.1.4), p([O, 1]) is compact (4.2.12). But p([O, I]) = T. Thus the space T is compact. In 3.4.5 we defined a continuous bijectionf ofT onto U. Now, T is compact and U is separated. Therefore f is a homeomorphism (4.2.15). Thus, the spaces T and U are homeomorphic. Since U 2 may be identified with the surface of the space commonly called a 'torus', one says that T2 is the 2-dimensional torus, and more generally that Tn is the n-dimensional torus. In particular, T is called the I-dimensional torus . • 4.2.17. Theorem. The product of a .finite number of compact spaces is compact. It suffices to show that if X and Yare compact, then X x Y is compact. First, X x Y is separated (3.2.5). Let (Uj);EI be an open covering of X x Y. For every m = (x, Y) E X X Y, choose an i(m) E I such that mE U;(m)' By 3.2.4, there exist an open neighborhood Vm of x in X and an open neighborhood W m of y in Y such that Vm x Wm c U;(m)' Set Pm = Vm x W m . Provisionally, fix Xo EX. The subset {xo} x Y of X x Y is homeomorphic to Y (3.2.11) hence is compact. The subsets Pm' where m runs over {xo} x Y, are open in X x Y and cover {xo} x Y, therefore {xo} x Y is contained in a finite union P ml U ... u P mn (4.1.3). The intersection Vml n ... n Vmn is y x - - .- - .. - _. .- - - .--+--+-+--. --.-----.--.~~=W , I I r I , 1Xol x x y 42 IV. Compact Spaces an open neighborhood Axo of k such that (x o , y) E Pmk = Vmk in X. If (x, y) E Axo x Y, there exists a W mk , whence y E W mk and Xo X (x, y) E Axo x W mk C V mk X W mk = Pm.· Thus Axo x Y is covered by a finite number of the sets V j • Ifnow Xo runs over X, the Axo form an open covering of X, from which one can extract a finite covering X=A XI u,,·uA Xp . Each Ax. x Y is covered by a finite number of the sets V j , therefore X x Y is covered by a finite number of the sets Vi' to 4.2.18. Corollary. In R", the compact sets are the closed bounded sets. (A subset of R" is said to be bounded if its n canonical projections onto R are bounded.) Let A be a compact subset ofR". Then A is closed in R" (4.2.7). Its canonical projections onto R are compact (3.2.8 and 4.2.12), hence bounded (4.2.8), thus A is bounded. Let B be a closed, bounded subset ofR". Since B is bounded. one has Be [al. b l ] x [az. b2] x ... x [a., b"] = C. Now, C is compact (4.1.4 and 4.2.17), and B is closed in C (3.1.5) hence is compact (4.2.6). 4.2.19. Examples. The space R· is not compact. The sphere Sn (2.5.7) is bounded in R"+ I, and closed in R"+ I (1.1.12), hence is compact (4.2.18). With the notations of 2.5.7, the space S" - {a} is homeomorphic to R", hence is not compact. 4.3. Complement: Infinite Products of Compact Spaces * 4.3.1. Let X be a set. If ffl and ffz are filters on X, the relation '~I c ffz is meaningful since ffl and ffz are subsets of .?l(X) (the set of all subsets of X). Thus, the set of filters on X is ordered by inclusion. One calls ultrafilter on X a filter on X that is maximal for this order relation (that is to say, a filter ff such that if'fj is a filter on X containing .?, then 'fj = ff). * 4.3.2. Let (ffA)A~A be a totally ordered family of filters on X (thus, for any A and /1, either '~A c ~ or ~ c ~). Then ff = UAEA ffA is a filter 4.3. Complement: Infinite Products of Compact Spaces 43 on X. For, let Y E:F and let Z c X be such that Z :::> Y; then Y E :IF;. for some A, therefore Z E :FA' therefore Z E :F. On the other hand, let Y l, Y 2 E :F; there exist A, f1 E A such that Y 1 E :FA' Y 2 E y;;,; if, for example, .'FA c .fFu , then Y lEy;;" therefore Y 1 n Y 2 E:Fu c :F. This proves our assertion. It then follows from Zorn's theorem (see, for example, Bourbaki, Theory of Sets, Ch. III, §2, Cor. 1 of Th. 2) that every filter on X is contained in an ultrafilter. * 4.3.3. Theorem. Let :F be afilter on X. The following conditions are equiva- lent: (i) :F is an ultrafilter; (ii) for every subset Y of X, either Y E:F or X - Y E:F. Suppose that :F is not an ultrafilter. Let :F' be a filter on X strictly containing :F. There exists Y E :F' such that Y $ .fF, Then X - Y $ :F' (because Y n (X - Y) = 0) and afortiori X - Y $:F. Suppose there exists a subset Y of X such that Y $ :F and X - Y $ :F. Let q; be the set of all subsets of X containing a set of the form F n Y with F E.'F, Let us show that q; is a filter. For every FE:F, one has F ¢ X - Y (otherwise, X - Y E :F), therefore F n Y :f= 0; thus, every element of q; is nonempty. It is clear that every subset of X containing an element of 'If belongs to q;. Finally,letGl,G z beelementsofq;;thenG l :::> Fl n Y,G 1 :::> F2 n Ywith Flo Fl E:F, therefore G l n G 2 :::> (Fl n F 2) n Y and Fl n F2 E:F; therefore G l n G 2 E q;. We have thus shown that q; is indeed a filter. It is clear that q; :::> .'F and that Y E 'If, therefore ~§ "# .fF and :F is not an ultrafilter. * 4.3.4. Theorem. Let X and X' be sets,f a mapping of X into X', :F an ultrafilter on X. Let .fF' be the set of all subsets of x' that contain a set of the form f(F), where FE:F. Then :F' is an ultrafilter on X'. It is clear that:F' is a filter on X'. Let Y' c X'. Set Y = f-I(Y'). Then Y E :F or X - Y E:F (4.3.3). If Y E:F, then Y' E:F' because fey) c Y'. If X - Y E:F, then X' - Y' E .fF' because f(X - Y) c X' - Y'. Therefore :F' is an ultrafilter (4.3.3). * 4.3.5. Theorem. Let E be a separated topological space. The following conditions are equivalent: (i) E is compact; (ii) if UIf is an ultrafilter on a set X, and iff is a mapping of X into E, then f has a limit along UIf. (a) Suppose E is compact. Let X, UIf, f be as in (ii). Let Xo be an adherence value of f along ott (4.2.1). The sets of the form feU) n V, where U E UIf and V is a neighborhood of Xo in E. are nonempty 44 IV. Compact Spaces (2.6.1). They constitute a filter base 81 on E, because if U 1, U 2 V 2 are neighborhoods of x o , then E <fI and VI' Let ~ be the set of subsets of E that contain an element of 91. Then ~ is a filter on E (2.1.1). Let ~ be the set of subsets of E that contain a set of the form f(U), where U E <fl. Then ~ is an ultrafilter (4.3.4). It is clear that ~ c ff. Therefore ~ = ff. Now let V be a neighborhood of Xo' Then V E !F, therefore V E ~ therefore there exists U E <fI such that f(U) c V. Thus f tends to Xo along <fl. (b) Suppose condition (ii) is satisfied. Let (Fi);eJ be a family of closed subsets of E. Suppose that for every finite subset J of I, the set F J = nieJ Fi is nonempty. We are to show that n;eJ F; "# 0. Now, the F J form a filter base. Let <fI be an ultrafilter on E containing all the F J (4.3.2). Let us apply condition (ii) to the identity mapping of E into E: we see that there exists an Xo E E such that every neighborhood of Xo contains an element of <fl. Fix i E I. Let V be a neighborhood of xo' Let U E <fI be such that U c V. Then F; n U "# 0 since F; E <fI and U E OU. Therefore F; n V "# 0. This being true for all V, we have Xo E Fi = F j • This being true for all i, we see that Xo E nieI F i· * 4.3.6. Theorem. Let (Ei)ieJ be a family of compact spaces, and let E D;eJ E j • Then E is compact. = Let X be a set, OU an ultrafilter on X, f a mapping of X into E, hence of the form x H (Ji(X»ieJ, where Ji is a mapping of X into E i . By 4.3.5, Ji has a limit Ii E Ei along OU. Let I = (li)ieI' By 3.3.2(c),ftends to I alongOU. Therefore E is compact (4.3.5). 4.4. The Extended Real Line 4.4.1. Let R be the set obtained by adjoining to R two elements, denoted + 00 and - 00, not belonging to R. If x, y E R, the relation x ::;; y is defined in the following way: (1) if x, y E R, then x ::;; y has the usual meaning; (2) for all x E R, one sets x < + 00 and - 00 < X; (3) one sets - 00 < + 00. It is easily verified that one obtains in this way a total order on R. The ordered set R is called the extended real line. 45 4.5. Locally Compact Spaces 4.4.2. Consider the mapping f of [ - n12, nl2J into f(x) = tan x 'f I R defined as follows: n n --<x<- 2 2' Then f is bijective and increasing, as is f - 1, and is therefore an isomorphism of ordered sets. Every property of the ordered set [ - n12, nl2J is therefore also true in the ordered set R. In particular, every nonempty subset of R admits a supremum and an infimum. 4.4.3. Topology of R. Let us call open subset of R the image under f of an open subset of [ - n12, n12]. One thus defines a topology on R, and f is a homeomorphism of [ - n12, nl2J onto R. It follows that R is compact and that, in R, every nonempty closed subset has a smallest and a largest element. Since the restriction off to ( - n12, n12) is a homeomorphism of ( - n12, n12) onto R (2.5.5), the topology ofR induces on R the usual topology. The intervals [b, n12J, where - nl2 < b < n12, form a fundamental system of neighborhoods of nl2 in [ - n12, n12]. Therefore the intervals [a, + 00 J, where a E R, form a fundamental system of neighborhoods of + 00 in R. Similarly, the intervals [ - 00, aJ, where a E R, form a fundamental system of neighborhoods of -00 in R. It follows that if(x 1 ,x 2 , ••. ) is a sequence of real numbers, to say that Xn --+ + 00 in R means that Xn --+ + 00 in the usual sense. 4.4.4. Theorem (Passage to the Limit in Inequalities). Let X be a set equipped with afilter base go, f and r mappings of X into R, admitting limits I, t along go. Suppose that f(x) ~ rex) for all x EX. Then I ~ r. Suppose I> t. Let a E R be such that I> a > l'. Then (a, + ooJ and a) are neighborhoods of I and l' in R. Therefore there exist B Ego and B' E go such that [ - 00, X E If x E B = f(x) > a, X E B' = rex) < a. B ('\ B', one sees that f(x) > rex), which is absurd. 4.5. Locally Compact Spaces 4.5.1. Theorem. Let E be a topological space. The following conditions are equivalent: (i) every point of E admits a compact neighborhood; (ii) every point of E admits a fundamental system of compact neighborhoods. 46 IV. Compact Spaces Obviously (ii) => (i). Let x E E and let V be a compact neighborhood of x. By 4.2.11(ii), x admits in V a fundamental system (Vi) of compact neighborhoods. One verifies easily that (Vi) is a fundamental system of neighborhoods of x in E (cf. 3.1.6). 4.5.2. Definition. A topological space is said to be locally compact if it is separated and satisfies the equivalent conditions of 4.5.1. 4.5.3. Examples. (a) Every compact space is locally compact. (b) R n is locally compact (without being compact). For, R n is separated, and every point ofRn admits as neighborhood a closed ball, which is compact (4.2.18). (c) Let us show that Q is not locally compact. Suppose that the point 0 of Q possesses in Q a compact neighborhood V. There exists a neighborhood W of 0 in R such that V = W n Q (3.1.6). Then there exists IX > 0 such that ( -IX, IX) c: W, whence (-IX, IX) n Q c: V. Moreover, since V is compact, V is closed in R (4.2.7). Now, every real number in (-IX, x) is adherent to ( -IX, IX) n Q, whence ( -IX, IX) c: V, which is absurd since V c: Q. 4.5.4. Theorem. Let X be a locally compact space, Y an open or closed subset of X. Then the space Y is locally compact. - First, Y is separated. On the other hand, let y E Y. There exists a compact neighborhood V of yin X. Then V n Y is a neighborhood of y in Y (3.1.6). Ify is closed in X, then V n Y is closed in V (3.1.4), hence is compact (4.2.6). IfY is open in X, one can suppose V c: Y (4.5.1(ii» and then V n Y = V. 4.5.5. Theorem. Let Xl' X2 , •.• , Xn be locally compact spaces and let X XI X ••• X Xn. Then X is locally compact. = First, X is separated. On the other hand, let x = (Xlo •.• , xn) E X. For every i, there exists a compact neighborhood Vi of Xi in Xi' Then V I x ... x Vn is a neighborhood of x in X (3.2.4) and is compact (4.2.17). * 4.5.6. Remark. Let E be a compact space, OJ a point of E. By 4.5.4, the space X = E - {OJ} is locally compact. * 4.5.7. Let us show, under the conditions of 4.5.6, how one can reconstruct the topology of E starting from that of X. We are going to prove that the open sets of E are: (1) the open sets of X; (2) the sets of the form (X - C) u {OJ}, where C is a compact subset ofX. First, an open set of X is open in E (3.1.3). Next, ifC is a compact subset of X, then C is closed in E (4.2.7), therefore E - C is open in E; now, E - C = (X - C) u {OJ}. 47 4.5. Locally Compact Spaces Finally, let U be an open set in E. If w 1= U then U (3.1.3); if W E U, then U = (X - C) c: X, and U is open in X u {w}, where C is the complement of U in E; this complement is closed in E, hence is compact (4.2.6). * 4.5.8. The interest of Remark 4.5.6 is that it yields all of the locally compact spaces, as we shall now see. Let X be a locally compact space, X' a set obtained by adjoining to X a point w not belonging to X. The construction we shall give is inspired by 4.5.7. Let us say that a subset U of X' is open in the following two cases: (a) U is an open set of X; (13) V is of the form (X - C) u {w}, where C is a compact subset of X. Let us show that the axioms (i), (ii), (iii) of 1.2.1 are satisfied. This is clear for (i). Let (Vi)iEI be a family of open sets of X', and V the union of the Vi' Then I = J u K, where: (1) for i E J, Vi is open in X; (2) for i E K, Vi = (X - C j ) u {w} with C j compact in X. If K = 0 then V is an open subset ofX. Suppose K #- 0. Then W E V and X' - V = n (X' - V;) = tEl (n Ci ) n lEK (n (X - V.». IEJ Now, njEK C j is compact in X (4.2.9(ii» and niEJ (X - Vj) is closed in X, therefore X' - VisacompactsubsetCofX(4.2.6),and V = (X - C) u {w}. We have thus verified axiom (ii). Now let V I, U 2 be open sets of X' and let us show that V I n U z is an open set of X'. This is clear if V I, V 2 are open in X. If VI is open in X and V 2 = (X - C) u {w} with C compact in X, then VI n V 2 = VI n (X - C), and X - C is open in X (4.2.7), therefore VI n V 2 is open in X. If VI = (X - C I ) u {w} and V z = (X - C z ) u {w} with C I , C z compact in X, then and C I U C 2 is compact in X (4.2.9(i». We have thus defined a topology on X'. The subset X of X' is open in X'. The intersections with X of the open sets of X' are the open sets of X. In other words, the topology induced on X by that of X' is the given topology on X. Let us show that X' is separated. Let x, y be distinct points of X', and let us show that x and y admit disjoint neighborhoods in X'. This is clear if x, y E X. Suppose x = wand y E X. Let W be a compact neighborhood of y in X. This is also a neighborhood of y in X' (because X is open in X'). Set v= (X - W) u {w}. This is an open neighborhood of w in X', and V n W = 0. 48 IV. Compact Spaces Let us show that X' is compact. Let (Vi)i",! be an open covering of X'. There exists io E I such that W E V io' Then Vio = (X - C) u {w}, where C is compact. The Vi cover C, therefore there exists a finite subset J of I such that (Vi)iEI covers C (4.1.3). Then X' = Vio U (V ,eI Vi)' * 4.5.9. One often says that W is the point at irifinity of X', and that X' results from X by the adjunction oj a point at infinity. One also says that X' is the Alexandroff compactification oj X. * 4.5.10. Let X be a locally compact space, El and E2 compact spaces such that X is a subspace of EI and E 2 , and such that Ei - X reduces to a point Wi (for i = 1 and 2). Then the unique bijection J of EI onto E2 that reduces to the identity on X and transforms WI into W 2 is a homeomorphism ofE I onto E 2; for, by 4.5.7, Jtransforms the open sets ofEI into the open sets ofE z · This proves, in a certain sense, the uniqueness of the Alexandroff compactification. * 4.5.11. Example. We defined in 2.5.7 a homeomorphism of Rn onto the complement Sn - {a} of {a} in the sphere Sn' Now, R" is locally compact and Sn is compact. In view of 4.5.10, the Alexandroff compactification oJRn may be identified with Sn' For example, the Alexandroff compactification of R may be identified with the .circle SI, hence with V (hence with the 1-dimensional torus T, by 4.2.16). CHAPTER V Metric Spaces This chapter, in which we reconsider metric spaces in greater detail, is heterogeneous. In §1, we introduce some concepts that are quite geometrical: diameter of a set, distance between two sets. In §2, we note that some of the earlier definitions take on a much more intuitive aspect in metric spaces. For example, a point adherent to a set A is nothing more than the limit of a sequence of points of A. The student already knows what is meant by uniform continuity for a real-valued function of a real variable. This concept is studied in the setting of arbitrary metric spaces in §3, and the somewhat analogous concept of equicontinuity in §4. The very important concept of complete metric space is studied in §§5, 6, 7. Among the numerous useful theorems concerning such spaces, we cite Baire's theorem (5.5.12). 5.1. Continuity of the Metric 5.1.1. Theorem. Let E be a metric space, d its metric. The mapping (x, Y) 1-+ d(x, Y) ofE x E into R is continuous. Let (xo, Yo) E E x E and let e > O. Let Y, W be the closed balls with centers Xo. Yo and radius e/2. Then V x W is a neighborhood of (xo, Yo) v. Metric Spaces 50 in E x E. If (x, y) E V X W, then + d(xo, Yo) + d(yo, y) d(x, y) :s; d(x, xo) :s; G G "2 + d(xo, Yo) + "2 = d(xo, Yo) + + d(x, d(x o , Yo) :s; d(x o , x) :s; G :2 + d(x, therefore Id(x, y) - d(xo, Yo) I :s; 8. y) G, + dey, Yo) y) G + "2 = d(x, y) + G, This proves the continuity of d at (xo, Yo). 5.1.2. Definition. Let E be a metric space, A a nonempty subset of E. One cans diameter of A the supremum, finite or infinite, of the set of numbers d(x, y), where x and y run over A. 5.1.3. Theorem. The sets A and A have the same diameter. Let D c R Crespo D' c R) be the set of d(x, y) where x, y run over A (resp. A). Then D c: D'. Since every point of Ax A is adherent to A x A (3.2.4), we have D' cis (2.4.4Civ) and 5.1.1). Therefore is = 0'. If 0 is bounded, the diameter of A (resp. A) is the largest element of is (resp. 0') (1.5.9). If 0 is unbounded, then 0 and D' both have supremum + rtJ. 5.1.4. Definition. Let E be a metric space, A and B nonempty subsets of E. One calls distC'nce from A to B the infimum of the numbers d(x, y), where x runs over A and y runs over B. It is denoted dCA, B). One has dCA, B) = d(B, A). We remark that if dCA, B) = 0, A and B need not be equal and may even be disjoint. If Z E E and AcE, we set d(z, A) = d( {z}, A) = infxEA d(z, x). 5.1.5. Theorem. Let E be a metric space, A and B non empty subsets of E. Then dCA, B) = dCA, B). The proof is analogous to that of 5.1.3. 5.1.6. Theorem. Let E be a metric space, A a nonempty subset of E, and x, y E E. Then Id(x, A) - dey, A) I :s; d(x, y). Let I: > o. There exists z E A such that d(x, dey, A) :s; dey, z) :s; dey, x) + d(x, z) z) :s; d(x, A) :s; d(x, y) + E. Then + d(x, A) + E. 51 5.2. The Use of Sequences of Points in Metric Spaces Thus dey, A) - d(x, A) :::;; d(x, y) + s. Interchanging the roles of x and y, one sees that d(x, A) - d(y, A) s d(x, y) + s. Therefore Id(x, A) - dey, A) I s d(x, y) + s. This being true for every s > 0, one obtains Id(x, A) - d(y, A) I s d(x, y). 5.2. The Use of Sequences of Points in Metric Spaces 5.2.1. Theorem. Let E be a metric space, AcE, x conditions are equivalent: ~ E E. The following (i) xEA; (ii) there exists a sequence (Xl' x 2 , ... ) (?fpoints of A that tends to x. If condition (ii) is satisfied, every neighborhood of X contains an x.' therefore intersects A; consequently x EA. If x E A then, for every integer n ~ 1, there exists a point x. of A that belongs to the closed ball with center x and radius lin. Then Xn tends to x by 2.2.2. 5.2.2. Theorem. Let E be a metric space, (Xl' X2,"') a sequence of points ofE, and x E E. Thefollowing conditions are equivalent: ~ (i) x is an adherence value of (x.); (ii) there exists a subsequence (x." x. 2 ' to x. ••• ), where nl < n2 < .. " that tends Suppose condition (ii) is satisfied. Then x is an adherence value of (x." x. 2 ' ••• ) and afortiori of (Xl' x 2 , ... ). Suppose condition (i) is satisfied. There exists n l such that d(x.I' x) :::;; 1. Then there exists n 2 > n l such that d(x. 2 , x) :::;; Then there exists n3 > n2 such that d(x.J' x) :::;; t, etc. The sequence (x." x. 2 ' ••• ) then tends to x. t. 5.2.3. Theorem. Let X, Y be metric spaces, A a subset of X, f a mapping of A into Y, a E A, and y E Y. The following conditions are equivalent: (i) y is an adherence value off as x tends to a while remaining in A (2.6.3); (ii) there exists a sequence (x n ) of points of A such that Xn -+ a and f(x.) -+ y. The proof is analogous to that of 5.2.2. 52 V. Metric Spaces 5.2.4. Theorem. Let X, Y be metric spaces, f a mapping of X into Y, and x E X. The following conditions are equivalent: (i) f is continuous at x; (ii) for every sequence (Xn) of points of X tending to x, the sequence (f(xn» tends to f(x). (i) ~ (ii). This is true for any topological spaces (2.3.3). Not (i) ~ not (ii). Suppose that f is not continuous at x. There exists s> such that, for every '1 > 0, there is ayE X with d(x, y) :-s; '1 yet d(f(x), fey»~ ~ s. Successively take '1 = 1,~, t, .... One obtains points YI' Yz, YJ, ... of X such that d(x, Yn) :-s; lin and d(f(x),f(Yn» ~ s. Then Yn --> X but f(Yn) does not tend to f(x) . ° .. 5.2.5. Theorem. Let X be a metric space. The following conditions are equivalent: (i) X is compact; (ii) every sequence of points in X admits at least one adherence value. (i) ~ (ii). This is true for every compact space (4.2.2). (ii) ~ (i). Suppose condition (ii) is satisfied. Let (Ui)iE) be an open covering of X, and let us show that X can be covered by a finite number of the Vi' We denote by B(x, p) the open ball with center x and radius p. (a) Let us show the existence of an a > 0 such that every ball B(x, a) is contained in some Vi' Suppose that no such a exists. Then, for n = 1, 2, ... , there exists an Xn E X such that B(x n , lin) is not contained in any V j • Let x be an adherence value of (Xl' X2'" .). Then X E Vi for some i, and so B(x, liN) c: Vi for some N. Next, there exists n ~ 2N such that Xn E B(x, 1/2N). Then B(Xn'~) c: B(Xn,2~) C B(X, 2~ + 2~) cUi' which is absurd. (b) It now suffices to prove that X can be covered by a finite number of balls B(x, a). Let Xl E X. If B(XI' a) = X, the proof is over. Otherwise, let X2 EX - B(x l , a). If B(x l , a) v B(X2' a) = X, the proof is over. Otherwise, let X3 E X - (B(XI' a) v B(X2' a»; etc. If the process stops, the theorem is established. Otherwise, there exists a sequence (Xl> X2,"') of points of X such that for every n. The mutual distances of the Xi are ~ a. Let x E X be an adherence value of (Xl' X 2 , ..• ). There exists an n such that Xn E B(x, a/2). Next, there exists an n' > n such that X n ' E B(x, a/2). Then d(xn' x n') < a, which is absurd. 53 5.3. Uniformly Continuous Functions 5.2.6. Corollary. Let X be a metric space, A a subset of X. The following conditions are equivalent: (i) A is compact; (ii) from every sequence of points of A, one can extract a subsequence that has a limit in X. Suppose A is compact. Let (x.) be a sequence of points of A, hence of A. It has an adherence value x E A. By 5.2.2, some subsequence tends to x. Suppose condition (ii) is satisfied. Let (Yb Y2,"') be a sequence of points of A. Let Xi E A be such that d(Yi' Xi) :::;; 1/i. Some subsequence (x. " x. 2 ' ••• ) tends to a point x E X. Then x E A (5.2.1). On the other hand, d(y." x) :::;; d(y." x.) + d(x." 1 x) :::;; ;. + d(x n" x) ~ 0, I therefore Y., tends to x. Thus, the sequence (Yb Y2' ... ) has an adherence value in A. Consequently, A is compact (5.2.5). 5.3. Uniformly Continuous Functions 5.3.1. Definition. Let X and Y be metric spaces, f a mapping of X into Y. We say that f is uniformly continuous if, for every I: > 0, there exists an '1 > 0 such that 5.3.2. It is clear that a uniformly continuous mapping of X into Y is continuous at every point, therefore is continuous. But the converse is not true. For example, the mapping x 1-+ x 2 of R into R is continuous, but it is not uniformly continuous. For, suppose it were. Taking I: = 1 in Definition 5.3.1, there would exist '1 > 0 such that Xl, Xl E Now, let x, = 1/'1, X2 R and lx, - x 2 1 :::;; '1 => Ixt - x~1 :::;; 1. = 1/'7 + '1. Then lx,2 - x 22 = 11'12 1 IXI - x21 :::;; '1 and 1 - 2 - '1 21 = 2 - '12 + '1 2 > 1. 5.3.3. The example 5.3.2 lends weight to the following theorem: Theorem. Let X and Y be metric spaces, f a continuous mapping of X into Y. Assume X is compact. Then f is uniformly continuous. ~ 54 V. Metric Spaces Fix 8 > O. We are to construct an '1 > 0 with the property of 5.3.1. For every x E X, there exists an '1x > 0 such that x' E X, d(x', x) ~ '1x =:> d(f(x'),f(x» ~ 8 2' Let Bx be the open ball with center x and radius t'1x. The Bx, as x runs over X, form an open covering of X. Since X is compact, X is covered by a finite number of such balls Bx, ' BX2 ' ___ , Bx n. Set '1 = inf(t'1x" t'1x2"'" t'1xJ > O. Now let x', XU E X be such that d(x', x' E B x " in other words, XU) ~ '1. There exists an i such that d(x;, x') < t'1x,. (1) Then thus d(x;, x") < '1x,. (2) The inequalities (1) and (2) imply d(f(x'), f(x;) :::;; }' whence d(f(x'), f(x"» :::;; d(f(x U ), f(x;) :::;; 8 2' 8. 5.4. Equicontinuous Sets of Functions 5.4.1. Let X and Y be metric spaces,! a mapping of X into Y. We recall: f is continuous at Xo if, for every 8 > 0, there exists an 17 > 0 such that d(x, x o) ~ '1 =:> d(f(x), f(xo» ~ 8; (b) f is continuous on X if f is continuous at every point of X; (c) f is uniformly continuous on X if, for every 8 > 0, there exists an '1 > such that d(x, x') :::;; '1 =:> d(f(x), f(x'» ~ e. (a) ° Now let (f.) be a family of mappings of X into Y. (a) The family (J.) is said to be equicontinuous at there exists an 17 > 0 such that for all a. Xo if, for every E > 0, 55 5.5. Complete Metric Spaces (b) The family (h) is said to be equicontinuous on X if it is equicontinuous at every point of X. (c) The family (f~) is said to be uniformly equicontinuous on X if, for every e > 0, there exists an I] > 0 such that d(x, x') :::;; '1 => d(h(X), h(x'» :::;; e for all ex. 5.4.2. Example. Take X = Y = R. Let (h) be the family of all differentiable real-valued functions on R whose derivative is bounded by 1 in absolute value. Then (h) is uniformly equicontinuous. For, let e > 0; if x, x' E R are such that Ix - x'i :::;; e, then If~(x) - hex') I :::;; e for all ex by the mean value theorem. 5.4.3. Remark. If a family (h) is equicontinuous, then each h is continuous. However, the converse is not true. For example, take X = Y = R and let (h) be the family of all linear functions. Each h is continuous, but the family (f~) is not equicontinuous at any point of R. For, suppose (f~) were equicontinuous at Xo. Take e = 1 in Definition 5.4.1. There would exist an '1 > 0 such that Ix - xol :::;; '1 => Ih(x) - h(xo) I :::;; 1 for all ex. Now, the function x H (2/I])x is linear, and I~ (xo + 1]) - ~ Xo I = 2 > 1. 5.4.4. Theorem. Let X and Y be metric spaces, (h) an equicontinuous family of mappings of X into Y. Assume that X is compact. Then the family (f~) is uniformly equicontinuous. The proof is nearly the same as in 5.3.3. 5.5. Complete Metric Spaces 5.5.1. Definition. Let X be a metric space, (ah a2' ...) a sequence of points of X. Recall that the sequence is called a Cauchy sequence if d(a ... , an) -+ 0 as m and n -+ 00, in other words, if for every e > 0, there exists an N such that m, n d(a m , an) :::;; e. ~ N => 56 V. Metric Spaces 5.5.2. Theorem. Let X be a metric space, (a 1, a2, ... ) a sequence of points of X If the sequence has a limit in X, then it is a Cauchy sequence. Suppose that (an) tends to a. Let £ > O. There exists a positive integer N such that n ~ N => d(a", a) :s; £/2. Then m, n ~ N => d(a m, a) :s; £ e :2 and d(a", a) :s; :2 => d(an" an) :s; £. 5.5.3. It is well known that the converse of 5.5.2 is in general not true (for example in X = Q). We therefore make the following definition: Definition. A metric space X is said to be complete if every Cauchy sequence has a limit in X 5.5.4. Example. One knows that the metric space R is complete. 5.5.5. Theorem. Let X be a metric space, (Xl' x 2 , ••• ) a Cauchy sequence in X, (x n" x n2 ' ... ) a subsequence. If the sequence (xn.) has a limit I, then the sequence (xn) also tends to l. n Let £ > O. There exists an N such that m, n N. For ni ~ N, we have ~ N => d(xm' xn) ::; £. Fix ~ (1) As i --- x, we have X n , --- I, therefore d(x"" x,,) --- d(l, xn) (5.1.1). By 4.4.4, the inequality (1) implies in the limit that d(l, XII) :s; I;. This being true for all n ~ N, we have Xn --- I. 5.5.6. Theorem. Let X be a complete metric space, Y a closed subspace of X Then Y is complete. ~ Let (Yl, Y2,"') be a Cauchy sequence in Y. It is also a Cauchy sequence in X Therefore it has a limit a in X. Since Yi E Y for all i, we have a E Y = Y, thus (ya has a limit in Y. 5.5.7. The converse of 5.5.6 is true. Better yet: ~ Theorem. Let X be a metric space, Y a complete subspace. Then Y is closed in X Let a E Y. There exists a sequence (y 1, Y2, ... ) in Y that tends to a (5.2.1). The sequence (yJ is a Cauchy sequence (5.5.2). It thus has a limit b in Y since Y is complete. In X, (Yi) tends to a and to b, therefore a = bEY. Thus Y = Y. 57 5.5. Complete Metric Spaces 5.5.S. Theorem. Let XI"'" Xp be complete metric spaces, X Xl X •.• x Xp the product metric space. Then X is complete. ~ = Let (YI, Y2, ... ) be a Cauchy sequence in X. Each point Yi is of the form (Yib Yi2, ... , Yip), where Yil E Xb ... , Yip E Xp. We have d(yml> Y.I) ~ d(Ym' Yn) -+ 0 as m, n -+ 00, therefore (Yll, Y21, Y3l> ... ) is a Cauchy sequence in XI' Since XI is complete, this sequence has a limit II in XI' Similarly, (YI2, h2' Y32,"') has a limit 12 in X2 , · · · , (YIP' Y2p, Y3p' ... ) has a limit Ip in Xp. Therefore (YI' Yz, ... ) tends in X to the point 1= (/1"'" Ip) (3.2.6). 5.5.9. Examples. By 5.5.4 and 5.5.8, Rn is complete. By 5.5.6, every closed subset of R· is a complete space. 5.5.10. Theorem. Let X be a complete metric space. Let (F 1> F 2, ... ) be a decreasing sequence of non empty closed subsets of X with diameters bI' b 2 , •••• Assume that bi -+ 0 as i -+ exactly one point. 00. Then the intersection of the Fi consists of Let F = F I ( l F 2 ( l F 3 ( l . . . . If F is nonempty then its diameter is ~ (ji for all i, hence is zero. There are thus two possibilities: either F is empty or it reduces to one point. Let us show that F is nonempty. Let aiEF i. Let us show that (a b a2,''') is a Cauchy sequence. Let e > O. There exists an N such that bN ~ e. If m, n ~ N then am, a" E F N , therefore d(a rn , a.) ~ e. Since X is complete, the sequence (ai) tends to a limit a in X. Since ai E F. for i ~ n, we have a E F. = F •. This being true for every n, we have a E F. 5.5.11. Theorem. Let X be a complete metric space, T a set equipped with a filter base 11, f a mapping of T into X. Assume that for every e > 0, there exists BE !JI such that feB) has diameter ~ e. Then f has a limit along 11. Denote by b(A) the diameter of a subset A of X. There exists BI E!JI such that b(f(Btl) ~ 1. Next, there exists B2 E!JI such that (j(f(B 2 » ~ t; replacing B2 by an element of 11 contained in BI ( l B2 , we can suppose that B2 c: BI' Next, there exists B3 E!JI such that (j(f(B3» ~ t and B3 c B2 , etc. By 5.1.3, (j(f(Bi» :$; l/i. By 5.5.10, the intersection of the feB,) consists of a point I. Let us show that f tends to I along 11. Let e > O. There exists a positive integer n such that l/n :$; E. If x E Bn then f(x) E f(B.) and 1E f(B.), therefore d(f(x), /) ~ - (j(f(B.» 1 :$; - :$; n e. Thus f(Bn) is contained in the closed ball with center I and radius e. 58 V. Metric Spaces .. 5.5.12. Theorem (Baire). Let X be a complete metric space, V 1 , V 2, ... a sequence of dense open subsets of X. Then V 1 (") V 2 " •.. is dense in X. Set V = V 1 " V 2 " .. '. Weare to prove that for every open ball B with radius> 0, V" B is nonempty. Denote by B(x, p) (resp. B'(x, p» the open ball (resp. closed ball) with center x and radius p. The set V 1 "B is open and non empty, therefore it contains a ball Bl = B'(Xl' Pl) such that 0 < Pl S 1. The set V 2 " B(Xl' Pi) is open and nonempty, therefore it contains a ball B2 = B'(X2' P2) such that 0 < P2 S l The set V 3 (") B(X2, P2) is open and nonempty, therefore it contains a ball B3 = B'(X3' P3) such that 0 < P3 S t, etc. By construction, B :::l Bl :::l B2 :::l B3 :::l • " . The B; are closed and their diameter tends to zero. By 5.5.10, there exists a point a that belongs to all of the B;. Then a E B. Moreover, a E Bn c Vn for all n; therefore a E U. 5.5.13. Theorem. Let X be a metric space, X' a dense subspace of X, Y a complete metric space, r a uniformly continuous mapping of x' into Y. (i) There exists one and only one continuous mapping f of X into Y that extends f'. (ii) f is uniformly continuous. The uniqueness of f in (i) follows from 3.2.15. Let us prove the existence of f. For each x E X, choose a sequence (xn) in X' that tends to x. Let e > O. There exists an '1 > 0 such that Zl, Z2 E X' and d(zl> Z2) s '1 => d(f'(Zl),r(Z2» s e. Now, (xn) is a Cauchy sequence, therefore there exists an N such that m, n Then m, n ~ N => d(x m , xn) S 1'/. ~ N => d(f'(xm)' r(x n» S Il. Thus (f'(xn» is a Cauchy sequence in Y, consequently has a limit in Y which we denote f(x). We have thus defined a mapping f of X into Y. If x E X' then r(xn) tends to rex), therefore f(x) = rex); in other words, f extends f'. Let u, v E X be such that d(u, v) S 1'//2. Let (u 1 , u 2, ... ) and (Vl' v 2, ... ) be the chosen sequences in X' tending to u and v. Then d(u n , vn) -+ d(u, v) (5.1.1), therefore there exists an N such that n ~ N => d(u n , vn) s'1 => d(f'(u n), r(vn» S e. Letting n tend to infinity, one obtains d(f(u), f(v» S e. Thus u, v E X and d(u, v) S ~ => d(f(u), f(v» S e, which proves that f is uniformly continuous, and afortiori continuous. 59 5.6. Complete Spaces and Compact Spaces 5.6. Complete Spaces and Compact Spaces ~ 5.6.1. Theorem. Let X be a metric space. The folloWing conditions are equivalent: (i) X is compact; (ii) X is complete and, jor every e > 0, there exists a jinite covering of X by balls of radius e. Suppose X is compact. Let (Xl, xz, ... ) be a Cauchy sequence in X. One can extract a subsequence that has a limit in X (5.2.6). Therefore (x I. x 2 , ••• ) has a limit in X (5.5.5). Thus X is complete. Let B > O. The open balls with radius c form a covering of X; since X is compact, a finite number of such balls suffices for covering X. Suppose condition (ii) is satisfied. Let (x 1. X 2 • .•• ) be a sequence of points of X. Cover X by a finite number of balls of radius one of these balls contains Xi for infinitely many i. One can therefore extract from (x;) a subsequence of points whose mutual distances are ::s; 1. Let us start anew with replaced by t, t, i, .... We obtain an infinity of sequences ±; ± yi. yS. YL··· yi, y~, y~, ... each of which is a subsequence of the preceding one, and such that dey?, y'}) ::S; lin for all i and j. The 'diagonal' sequence (yL yL yt ... ) is a subsequence of (x;), and dey;;:, y~) ::S; 11m for m ::S; n. Therefore (yD is a Cauchy sequence; consequently, it has a limit. Then X is compact by 5.2.6 applied with A = X. One can obviously replace condition (ii) by the following: X is complete and, for every IJ > 0, there exists a finite covering of X by sets of diameter ::S; IJ. 5.6.2. Theorem. Let X be a complete metric space, A a suhset ~lX. Thefollowing conditions are equivalef1 t · (i) A is compact; (ii) for every c > 0, one can cover A by a finite number of halls of X with radius 8. = (i) (ii). This is obvious. (ii) = (i). Suppose condition (ii) is satisfied. Let e > O. There exist closed balls B 1 , ..• , Bn in X with radius e that cover A. Then A c: Bl u··· u B n. On the other hand, A is complete (5.5.6). Therefore A is compact (5.6.1). 60 V. Metric Spaces 5.7. The Method of Successive Approximations 5.7.1. Theorem. Let X be a complete metric space,! a mapping of X into X. Assume that there exists a A E [0, 1) such that d(f(x), f(x')) ~ U(x, x') for all x, x' E X. (i) There exists one and only one a E X such that f(a) = a. (ii) For any Xo E X, the sequence of points tends to a. Let a, bE X be such that a dCa, b) = f(a), b = feb). = d(f(a), feb»~ ~ Then Ad(a, b), thus (1 - A)d(a, b) ~ 0. Since 1 - A > 0 we infer that d(a. b) ~ O. whence dCa, b) = 0 and a = b. Let Xo E X. Set Xl = f(xo), X2 = f(X1), .... We are going to prove that (x.) tends to a limit a and that f(a) = a. The theorem will thus be established. Let us show that d(x., x.+ d ~ A·d(xo, x 1)' This is clear for n = O. If it is true for n, then d(xn+ 1, X.+ 2) = d(f(x.),!(X.+ 1» ~ U(X., X.+ 1) ~ AAnd(xo, Xl) = .1."+ Id(xo, X1), whence our assertion by induction. From this, one deduces that if nand p are integers ~ 0, then d(x., x.+ p) ~ ~ + d(Xn+l' X.+2) + ... + d(x n +p - l , x.+ p ) (A." + .1."+1 + ... + A·+ P - 1)d(xo, Xl) d(x., ~ .1."(1 X.+l) + A+ .1. 2 + .. .)d(xo, Xl) = A. d~~ ~l). Since 0 ~ A < 1, A" -+ 0 as n -+ 00. We thus see that the sequence (x.) is a Cauchy sequence, consequently tends to a limit a. We have d(x., a) -+ 0, therefore d(f(x.), f(a» -+ 0, that is, d(xn+ h f(a» -+ O. Thus, the sequence (x.) tends also to f(a), whence f(a) = a. 5.7.2. Example. Let I be a closed interval of R, f a function defined on I, with values in I, such that sUPxedf'(x)! < 1. By the mean value theorem, f satisfies the condition of 5.7.1. Consequently, the equation X = f(x) has a unique solution in I which can be obtained, starting with any point X o of I, by the 'successive approximations' Xl = f(xo), X 2 = f(XI)" ... 5.7. The Method of Successive Approximations 61 * 5.7.3. Remark. Let X be a complete metric space, B the closed ball in X with center Xo and radius p, f a mapping olB into X such that d(f(x}, I(x'}) :::;; A.d(x, x') for all x, x' E B (where 0:::;; A. < 1). Assume, moreover, that d(f(xo), xo) :::;; (1 - A.)p. Then there exists one and only one a E B such that I(a) = a. Uniqueness is proved as in 5.7.1. One again forms Xl = I(xo), X2 = I(x l ), •.. , but, for these points to be defined, one must prove that they do not exit B. Indeed, let us suppose that xp E Band d(xp, Xp+1) :::;; A.Pd(xo. Xl) for p = 0, 1, ... , n. Then therefore x n+IE B, and d(xn+ 1, Xn+2) :::;; A. n +1d(xo, Xl) may be proved as in 5.7.1. This established. (x n ) is again a Cauchy sequence and tends to a limit a E B; I(a) = a is proved as in 5.7.1. CHAPTER VI Limits of Functions For real-valued functions of a real variable, the student already knows what it means for a sequence fl' f2' ... of functions to tend uniformly, or to tend simply, to a function f. In this chapter we study these concepts in the general setting of metric spaces. We obtain in this way certain of the 'infinite-dimensional' spaces alluded to in the Introduction, and, thanks to Ascoli's theorem, the compact subsets of these spaces. 6.1. Uniform Convergence 6.1.1. Let X and Y be two sets. The mappings of X into Y form a set which will henceforth be denoted jO(X, V). 6.1.2. Let X be a set, Y a metric space. For f, 9 E jO(X, V), we set d(f, g) = sup d(f(x), g(x» E [0, + 00]. x.x Let us show that d is a metric (with possibly infinite values) on jO(X, V). If d(f, g) = 0 then, for every x E X, d(f(x), g(x» therefore f(x) = g(x); thus f = g. hE jO(X, Y) then, for every x E X, d(f(x), hex»~ ~ d(f(x), g(x» = 0, It is clear that d(f, g) = d(g, f). Finally, if + d(g(x), hex»~ ~ this being true for all x E X, we infer that d(f, h) ~ dC!, g) + d(g, h). d(f, g) + d(g, h); 63 6.1. Uniform Convergence This metric is called the metric of uniform convergence on ff(X, V). The corresponding topology is called the topology of uniform convergence. 6.1.3. Let f, fl, f2' f3, ... E ff(X, V). To say that (f,,) tends to f for this topology means that SUPXEX dCf(x), f.(x» ~ 0, in other words: for every e > 0 there exists an N such that n ~ N => d(f,,(x), f(x» ::; e for all x EX. We then also say that the sequence (f,,) tends uniformly to f. 6.1.4. Let A be a set equipped with a filter base 81. For every A. E A, let fA E ff(X, V). Let f E ff(X, V). To say that fA. tends to f along 81 for the topology of uniform convergence means: for every e > 0, there exists B E 81 such that A. E B => dCf;.(x), f(x» ::; e for all x E X. We then also say that fA. tends to f uniformly along 81. 6.1.5. Example. Take X = Y = A = R. For filter base 81 on A, take the set of intervals [a, + 00). For A. E R and x E R, set fix) = e-.l.(x 2 + 1). Then fA. tends to 0 uniformly as A ~ + 00 (that is, along 81). For, let e > O. There exists a E R such that A. ~ a=> e-). ::; e. Then (provided a ~ 0): A. ~ a => le-;'(X 2 +1) 01 = e-).(x 2 +1)::; e-A.::; e forallxER. - 6.1.6. Theorem. Let X be a set, Y a complete metric space. Then the metric space ff(X, Y) is complete. Let Cf.) be a Cauchy sequence in ff(X, V). Let x E X. Then dCfm(x), f,,(x» ::; dCfm' f,,) ~ 0 as m, n ~ 00, thus Cf.(x» is a Cauchy sequence in Y, consequently has a limit in Y which we denote f(x). We have thus defined a mapping f of X into Y. Let e > O. There exists an N such that m, n ~ N => => dCfm' f,,) ::; e dCfm(x), f,,(x» ::; e for all x E X. We provisionally fix x E X and m yields in the limit ~ d(j~(x), N. As n ~ 00, the preceding inequality f(x» ::; e. This being true for all x E X, we have dCfm, f) ::; e. Thus, m ~ N => dCfm,f) ::; e. In other words, Cfm) tends to f in ff(X, V). 64 VI. Limits of Functions 6.1.7. Thus, to verify that a sequence of mappings of X into Y tends uniformly to a limit, it suffices to verify a 'Cauchy criterion' for the sequence. 6.1.8. Let Uh U2, ••• e ff"(X, C). Let s e ff"(X, C). We say that the series with general term Un converges uniformly and has sum s if the sequence of finite partial sums Ul + Uz + ... + Un tends uniformly to s as n -+ 00, in other words if, for every e > 0, there exists an N such that n~ N => lu1(x) + uix) + '" + uix) - s(x) I ~ e for all x e X. For the series with general term Un to converge uniformly, it is necessary and sufficient, by 6.1.6, that the following condition be satisfied: for every e > 0, there exists an N such that 6.1.9. Let Uh U2' ••• e ff"(X, C). We say that the series with general term Un converges normally if there exists a sequence aI' a2"" of numbers ~ Osuch that an < + 00 and such that Iu.(x) I ~ a. for all n = 1,2, ... and all xeX. Lf' 6.1.10. Theorem. A normally convergent series is uniformly convergent, LetUl'UZ,'" eff"(X, C). Letahaz, .. , be numbers ~ OsuchthatLf' an < an for all nand x. Let B > O. There exists an N such that + 00 and Iu.(x) I ~ n ~ m ~ N => IXm + IXm+ 1 + ... + IX. ~ B. Then n~ m~ N => Ium(x) + '" + U.(x) I ~ Ium(x) I + ... + Iun(x) I ~ am + ... + IXn ~ e for all x e X. By 6.1.8, the series with general term Un is uniformly convergent. • 6.1.11. Theorem. Let X be a topological space, Y a metric space, A a set equipped with a filter base fll. For every A. e A, let fl e rt(X, Y). Assume thatfl tends tof E 9"(X, Y) uniformly along fll. Thenfe ~(X, V). Let Xo e X and e > O. There exists A. e A such that d(fix), f(x» ~ B 3 for all x e X. Next, there exists a neighborhood V of Xo in X such that X E V => d(fix), !;(x o)) ~ B 3' 65 6.1. Uniform Convergence Then, for all x E V, we have thus f is continuous at Xo. 6.1.12. Corollary. Let X be a topological space, Y a metric space. (i) ~(X, Y) is closed in ~(X, V). (ii) If Y is complete, then ~(X, Y) is complete. The assertion (i) follows from 6.1.11 and 5.2.1. Assertion (ii) follows from (i), 6.1.6 and 5.5.6. 6.1.13. Theorem. Let T and X be topological spaces, Y a metric space, and g E ~(T X X, V). For every t E T, set I,(x) = get, x) (where x E X), so that the I, are mappings of X into Y. Let to E T. Assume that X is compact. Then I, tends uniformly to I,o as t tends to to. Let e > O. For each x E X, g is continuous at (to, x), therefore there exist an open neighborhood Vx of to in T and an open neighborhood W x of x in X such that (1) t E Vx and x' E Wx => d(g(t, x'), g(to, x» :::;; ~. As x runs over X, the W x form an open covering of X, therefore there exist ,XnEX such that X=Wx,v .. ·vWXn ' Set V=Vx,n .. ·nVXn ; this is a neighborhood of to in T. Let t E V, Z EX. There exists an i such that Z E WXi' Moreover, t E VXi' Therefore d(g(t, z), g(to, Xi» :::;; e/2 by (1). Similarly, (1) implies that d(g(to, z), g(to, x;) :::;; e/2. Then Xl> ... d(g(t, z), g(to, z» :::;; e. This being true for every Z E X, we have d(I" I,o) :::;; e. Thus t E V=> d(I" I,o) :::;; e, so that I, tends to I,o uniformly as t tends to to. 6.1.14. Theorem (Interchange of Order of Limits). Let Sand T be sets equipped with filter bases [Jl, ~. Let Y be a complete metric space, g a mapping of S x T into Y. For every s E S, set !set) = g(s, t) (where t E T), so that !s E §(T, V). We make the following assumptions: each!s has a limit I. along ~; !s tends uniformly along [JJ to an f E §(T, V). 66 VI. Limits of Functions Then: (i) f has a limit I along ~; (ii) s H Is has a limit I' along fJI; (iii) I = 1'. Let Then (1) e > O. There exists BE fJI such that s E B => d(/s, f) :s;: 8/3. Fix So E B. 8 :S;::3 d(!saCt), f(t» for all t E T. There exists e E ~ such that t, t' E C => d(!so(t), !soCt'» :s;: <:/3. Then if t, t' E e, we have d(f(t), f(t'» :s;: d(fCt), j~o(t» 8 <: 8 :S;::3 +:3 +:3 + dCfs/t), .fso(t'» + dUso(t'), f(t') = e. Since Y is complete, we deduce from this (5.5.11) that f has a limit I along~. Again let So E B. Along ~, the pair (j~o(t), f(t» E Y X Y tends to (is", I). The inequality (1) yields in the limit d(lso' I) :s;: e/3. This being true for every So E B, we conclude that Is tends to I along fJI. * 6.1.15. Remark. Let X be a topological space, Y a metric space, (fn) a sequence of mappings of X into Y, f a mapping of X into Y. We say that (f.) tends to f uniformly on every compact set if, for every compact subset e of X, the sequence of restrictions 1.1 e tends uniformly to f 1C. There exists a 'topology of compact convergence' such that the preceding concept is precisely the concept of a sequence tending to a limit for this topology; however, we shall not define it. If (fn) tends to f uniformly on X, then of course (fn) tends to f uniformly on every compact set. However, the converse is not true. For example, take X = Y = Randfn(x) = e-(x-n)2 forn = 1,2, ... . Leta > OandA > O.Then x E [-a, a] and n 2 A +a => (x - n)2 2 A2 => e-(x-n)2 :s;: e- A2 ; therefore (f.) tends uniformly to 0 on [ - a, a], and this for every a. Therefore (f.) tends to 0 uniformly on each compact subset of R. However, (fn) does not tend to 0 uniformly on R, because j~(n) = 1 and so SUpxER 1 fn(x) 1 2 1. * 6.1.16. Nevertheless, we have the following result: Theorem. Let X be a locally compact space, Ya metric space, Un) a sequence of continuous mappings of X into Y, f a mapping of X into Y. Assume that (f.) tends to f uniformly on every compact set. Then f is continuous. 67 6.2. Simple Convergence Let Xo E X. There exists a compact neighborhood V of Xo in X. The sequence Un IV) tends uniformly to f IV, therefore f IVis continuous (6.1.11). By 2.2.10, f is continuous at Xo. Thus f is continuous. 6.2. Simple Convergence 6.2.1. Let X and Y be sets. Consider the family of sets (Y X>xeX, where Y x = Y for all x E X. An element of[Jxex Y x is the giving, for each x E X. of an element of Y; in other words. it is a mapping of X into Y. Thus: nY xeX x = ~(X, Y). Now suppose Y is a topological space. Then [JXEX Y x, in other words Y), carries a product topology (3.3.1), called the topology of simple convergence (or the 'topology of pointwise convergence'). ~(X. 6.2.2. Theorem. Let X be a set, Y a topological space, and f xl> ... , Xn E X and let Wi be a neighborhood of f(x;) in Y. Let E ~(X, Y). Let V(x) •... , X n , Wi .... ' W n ) be the 8et of all g E ff(X, Y) such that g(Xj)EW I , g(X 2 )EW 2 , ... , g(X n ) EWn • Then the V(XI,"" X n , WI, ... , W n ) constitute a fundamental system of neighborhoods off in ff(X, Y) for the topology of simple convergence. This follows from 3.3.2(a). 6.2.3. Theorem. If Y is separated, then ff(X, Y) is separated for the topology of simple convergence. This follows from 3.3.2(b). 6.2.4. Theorem. Let A be a set equipped with a filter base .gu. For every 1 EA. let f~ E ff (X, Y). Let f E ff (X, Y). The following conditions are equivalent: (i) f~ tends to f along :J4 for the topology of simple convergence; (ii) for every x E X, fix) tends to f(x) along .gu. This follows from 3.3.2(c). 6.2.5. In particular, if f, fl. f2' f3, ... are mappings of X into Y, to say that the sequence Un) tends to ffor the topology of simple convergence means that, 68 VI. Limits of Functions for every x E X, f.(x) tends to f(x). We then also say that (J,.) tends simply to f. 6.2.6. Theorem. Let Xo E X. The mapping f f---> f(x o) of 3F(X, Y) into Y is continuous for the topology of simple convergence. This follows from 3.3.2(e). 6.2.7. Let X be a set, Y a metric space. On 3F(X, V), there is the topology :Yl of uniform convergence and the topology !!l;, of simple convergence. Then 5; isfiner than :Yz. By 2.4.7, it suffices to show that the identity mapping of 3F(X, Y) equipped with g;- into 3F(X, Y) equipped with:Yz is continuous. For this, it suffices by 6.2.4 to show that, for any fixed Xo in X, the mapping ff---> f(x o) of 3F(X, Y) equipped with ~ into Y is continuous. Now, this is clear since d(f(xo), g(xo» :::;; d(f, g) for f, 9 E 3F(X, V). 6.2.8. Let X be a set, Y a metric space. It follows from 6.2.7 that if a sequence (fn) of elements of 3F(X, Y) tends uniformly to an element f of 3F(X, V), then (fn) tends simply to f. The converse is in general not true (see the example in 6.1.15). 6.2.9. Nevertheless, we have the following result: ~ Theorem (Dini). Let X be a compact space. Let Assume that fl :::;; f2 :::;; f3 :::;; ... and that (fn) tends simply to tends uniformly to f. f. Then (fn) Wehavefn(x):::;; f(x) for all x EX.Setg n = f - fn.Thegnarecontinuous, tend simply to 0, and Let I: > 0. Let Xn be the set of x E X such that g.(x) ~ 1:. Then Xl => X 2 => X3 => ... and the Xn are closed (cf. 2.4.5). If x En Xn then gn{x) ~ I: for all n, which is absurd. Therefore n Xn = 0. Since X is compact, the intersection of a finite number of the Xn is empty. Since the Xn decrease, this intersection is one of the X n • Thus Xno = 0 for some no. Then, for n ~ 110, we have g.(x) :::;; e for all x E X, thus I fix) - f(x) I :::;; e for all x E X. °: :; 6.2.10. Changing fn to - fn, we see that Dini's theorem remains valid for decreasing sequences. 69 6.3. Ascoli's Theorem 6.3. Ascoli's Theorem 6.3.1. Theorem (Ascoli). Let X be a compact metric space, Y a complete metric space, A an equicontinuous subset of YJ'(X, Y). Assume that for each x E X the set of f(x), where f runs over A, has compact closure in Y. Then A has compact closure in the metric space ~(X, Y). ~ The metric space YJ'(X, Y) is complete (6.1.12). By 5.6.2, it suffices to prove the following: given any 8 > 0, A can be covered by a finite number of balls of radius 8. By 5.4.4, there exists an 'I > 0 such that x, x' E X, d(x, x') ::; 1'/, f E A ~ d(f(x), f(x'» ::; 8 4' We can cover X by a finite number of open balls with centers Xl> ... , X. and radius 'I. The set of values of the elements of A at Xl, ..• , Xn has compact closure in Y (4.2.9(i»; we cover it by a finite number of open balls with centers Yl' ... , Yp and radius 8/4. Let r be the set of all mappings of {I, 2, ... , n} into {t, 2, ... , p} ; this is a finite set. For each Y E r let Ay be the set of f E A such that By construction, the Ay cover A. It remains only to show that for any fixed y, A y is contained in some ball of radius 8. Now, let f, gEAr' Let X E X. There exists an Xi such that d(x, Xi) < 1'/. Therefore d(f(x), f(x i» e ::; 4' d(g(x), g(x i » ::; 8 4' Moreover, Therefore d(f(x), g(x» ::; 8. This being true for all X E X, we have d(f, g) ::; 8. 6.3.2. Example. Take X = [0, 1], Y = R. Let A be the set of differentiable real-valued functions on [0, 1] such that I f(x) I ::; 1 and I f'(x) I ::; 1 for all X E [0,1]. As in 5.4.2, A is equicontinuous. By 6.3.1, A has compact closure in ~([O, 1], R). In particular (5.2.6), every sequence of functions belonging to A has a uniformly convergent subsequence. CHAPTER VII Numerical Functions This chapter, devoted to real-valued functions, is heterogeneous. In §§l and 2 we take up again some familiar concepts. perhaps in a little more general setting. Let (UI, U2, U3,' .. ) be a sequence of real numbers. The sequence does not always have a limit, but it does have adherence values in R; among these, two play an important role: they are called (perhaps inappropriately) the limit superior and the limit inferior of the sequence. A more general definition is presented in §3. In §4 we define semicontinuous functions, which generalize (for real-valued functions) the continuous functions. In connection with Theorem 7.4.15 (which is a corollary of Baire's theorem) we point out that even if we limited ourselves to continuous functions, the proof would naturally introduce semicontinuous functions. The student is already familiar with various theorems on the approximation of real-valued functions of a real variable: by ordinary polynomials, or by trigonometric polynomials (cf. the theory of Fourier series). In §5 we give a very general result that encompasses these earlier results. It is applied in §6 (devoted to 'normal' spaces) to the approximation of continuous functions on product spaces. The mappings of a set X into R are called numerical functions. If the mapping has values in R we sometimes say, more precisely, finite numerical function. 71 7.1. Bounds of a Numerical Function 7.1. Bounds of a Numerical Function 7.1.1. Let f be a numerical function on X. Recall that the supremum of f on X, denoted SUPXEX f(x), is the supremum of the set f(X). This is the element a ofR characterized by the following two properties: (1) f(x) :::;; a for all x E X; (2) for any b < a, there exists an x E X such that f(x) > b. The infimum of f on X, denoted infxEx f(x), is the infimum of the set f(X). One has inf f(x) = -sup( -f(x», XEX XEX which reduces the properties of the infimum to the properties of the supremum. 7.1.2. Recall that f is said to be bounded above if SUpxEX f(x) < + 00, bounded below if infxEx f(x) > - 00, and bounded if it is both bounded above and bounded below. A bounded function is finite, but a finite function is not necessarily bounded. 7.1.3. One calls oscillation off over X the number sup f(x) - inf f(x) xeX XEX (provided that f is neither constantly difference is defined). + 00 nor constantly - 00, so that the 7.1.4. Theorem. Let X be a set, f a numerical function on X, (X;)iEI afamily of subsets of X covering X. Then: ~~~ f(x) si~f (~~i f(X»). = Set a = SUPXEX f(x), ai = SUPXEX.!(X). It is clear that ai :::;; a for all i E 1. Let b < a. There exists x E X such that f(x) > b. Next, there exists i E I such that x E Xi' Then b < ai' Thus a = SUPiEI ai' 7.1.5. Corollary. Let X, Y be sets,f a numericalfunction on X x Y. Then: sup f(x, y) = sup (suPf(X, y») (x. y)e X x Y XEX = YEY sup (sup f(x, y»). yeY xeX 72 VII. Numerical Functions The set X x Y is the union, as x runs over X, of the sets {x} x Y. The first equality thus follows from 7.1.4. 1.1.6. Theorem. Let X be a set,f and 9 numerical/unctions on X. (i) SUPXEX (f(x) + g(x» ::; SUPxeX f(x) (ii) 1/ f ~ 0 and g ~ 0, then + SUPxeX g(x). SUp (f(x)g(x» ::; (suPf(X»)(SUpg(X»). xeX xeX xeX Set a = SUPXEX f(x), b = SUPXEX g(x). For every x E X, we havef(x) ::; a, g(x) ::; b, therefore f(x) + g(x) ::; a + b. Consequently SUPXEX (f(x)+g(x» + b. Assertion (ii) is proved in an analogous manner. (The statement presumes that the numbers f(x) + g(x), for example, are defined; this would not be the case if, at some point Xo of X, one had for example f(x o) = + 00 and g(xo) = - 00. Also excluded are expressions such as 0 . + 00. Here, and in what follows, it is implicitly understood that we are avoiding such indeterminate expressions.) ::; a 1.1.7. Theorem. Let X be a set, f a numerical function on X, and k E R. (i) SUPXEX (f(x) + k) = (supxExf(x» + k. (ii) If k ~ 0, then SUPXEX (kf(x» = k(SUPXEX f(x». Set a = sUPxExf(x). By 7.1.6, sup (f(x) xeX + k) ::; a + k. + oc, equality clearly holds; suppose k < + 00. Now let b < a + k. We have b = c + k with c < a. There exists Xo E X with f(xo) ~ c, whence f(xo) + k ~ c + k = b; therefore sUPxeX (f(x) + k) ~ b. This proves (i). One reasons analogously for (ii). If k = 1.1.8. Corollary. Let X, Y be sets, / a numerical/unction on X, and 9 a numerical/unction on Y. (i) SUPXEX.YEY (f(x) + g(y» (ii) If f ~ 0 and g ~ 0, then = SUPXEX f(x) + SUPYEY g(y). sup (f(x)g(y» = (sup f(X») . (SUpg(y»). xeX.yeY yeY xeX Set a = c = supf(x), XEX b = supg(y), YEY sup (f(x) + g(y». xeX,yeY 73 7.2. Limit of an Increasing Numerical Function By 7.1.5, c = sup [sup (f(x) + g(y»J. yeY xeX Inside the brackets, g(y) is a constant. By 7.1.7, c = sup(a + g(y}). yeY Then, again by 7.1.7, c = a + b. One reasons similarly for (ii). 7.2. Limit of an Increasing Numerical Function 7.2.1. Let X be an ordered set. We say that X is increasingly filtering (or 'directed upward') if, for every x E X and x' E X, there exists an x" E X such that x" ;::: x and x" ;::: x'. Decreasingly filtering ordered sets are defined in an analogous way. 7.2.2. Examples. (a) A totally ordered set is both increasingly filtering and decreasingly filtering. (b) Let I be a set, X the set of all finite subsets of I. Order X by inclusion. Then X is both increasingly filtering and decreasingly filtering. (c) Let fJB be a filter base on a set. Order fJB by inclusion. Then fJB is decreasingly filtering. 7.2.3. Let X be an increasingly filtering ordered set. For every x E X, let Bx be the set of majorants of x in X (that is, the set of elements of X that are ;::: x). Then, the Bx form a filter base fJI on X. For. x E Bx, thus Bx #- 0. On the other hand, if x, x' E X, there exists a majorant x" of x and x', and one has Bx" c Bx n Bx" When f is a mapping of X into a topological space, the limit of f along fJI-if it exists-is called the limit of f along the increasingly filtering set X and is denoted limx f or limx f(x}. There are analogous definitions for decreasingly filtering sets. 7.2.4. Theorem. Let X be an increasingly filtering ordered set, f an increasing mapping of X into R, and I the supremum of f. Then the limit of f along X exists and is equal to l. We can suppose I > - 00. Let V be a neighborhood of I in R; it contains a neighborhood of the form [a, b], where a < I ~ b. There exists an x E X such that f(x} ;::: a. Then, for all y ;::: x in X, we have fey) ;::: f(x) ;::: a, whereas fey) ~ I, therefore f(y) E [a, b] c V. 74 VII. Numerical Functions 7.2.5. Changing f to - f, we see that if f is decreasing and l' is its infimum, then the limit of f along X exists and is equal to 1'. 7.2.6. Suppose X is decreasingly filtering. Iff is increasing (resp. decreasing), then the limit off along X exists and is equal to the infimum (resp. supremum) off Indeed, for the opposite order on X, X is then increasingly filtering and fis decreasing (resp. increasing). 7.3. Limit Superior and Limit Inferior of a Numerical Function 7.3.1. Definition. Let X be a set equipped with a filter base IJI, f a mapping of X into R, A the set of adherence values of f along IJI. By 2.6.6 and 4.2.1, A is closed in R and is nonempty, hence admits a smallest and a largest element (4.4.3). These elements are called the limit inferior of f along IJI and the limit superior off along IJI. They are denoted lim infBIJ f and lim sUpBIJ f (or lim infBIJf(x), lim supBIJf(x». This definition admits many special cases: (a) If (un) is a sequence of real numbers, one can speak of lim SUPn_oo Un and lim infn _ oo Un (these are elements of R and always exist, whereas limn_ 00 Un does not always exist). (b) If f is a mapping of a topological space X into R and if a E X, one can speak of lim supx_. f(x) and lim infx_ a f(x). Etc. to- 7.3.2. Theorem. Let X be a set equipped with a filter base IJI, f a mapping of X into R. (i) lim sUpBIJ f(x) 2 lim infBIJ f(x). (ii) For f to have a limit along IJI, it is necessary and sufficient that lim sup f(x) = lim inf f(x), BIJ BIJ and the common value is then the limit off. (i) This is obvious. (ii) The space R is compact (4.4.3). Therefore, in order that f have a limit along IJI, it is necessary and sufficient that the set of adherence values of f along IJI reduce to a single point, which is then the limit (2.6.4 and 4.2.4). This implies (ii) at once. 75 7.3. Limit Superior and Limit Inferior of a Numerical Function 7.3.3. Theorem. Let X be a set equipped with a filter base !!J, f a mapping of X into R, and m, n E R such that m > lim sup f(x) Then there exists B E n < lim inf f(x). and !!J such that xEB => n <f(x) < m. For, (n, m) is an open interval of R that contains the set of all adherence values of f along /11, and it suffices to apply 4.2.3. 7.3.4. Theorem. Let X be a set equipped with a filter base !!J, f a mapping of X into R. For every B E !!J, let UB = sup f(x), XEB inf f(x). XEB = VB Then (cf 7.2.2(c) and 7.2.3): lim sup f(x) = inf UB = lim ilfI BEilfI ilfI lim inf f(x) = sup fj/) = VB lim UB, VB' ilfI B"fj/) Since f can be replaced by - f, it suffices to prove the first group of formulas. Set a = lim sup f(x), fj/) b = inf UB' BEfj/) If B, B' E !!J and B ::) B', then UB :2 UB '; by 7.2.4, limfj/) UB exists and is equal to b. Since a is an adherence value of f along !!J, we have a E feB) for all B E !!J (2.6.6); since UB is the largest element of feB) (1.5.9 and 4.4.3), we have a :-;:; UB; this being true for every B E !!J, we have a :-;:; b. Suppose a < b. Let I/. E (a, b). As in 7.3.3, there exists B E f!J such that x E B => f(x) < 1/.; then UB :-;:; I/. and afortiori b :-;:; 1/., which is absurd. 7.3.5. Example. Let (un) be a sequence of real numbers. Then: lim sup n~oo Un = inf(SuP Un) p n~p = lim (sup Un), p-oo n~p lim inf Un = SUP(inf Un) = lim (inf n-oo p n?:p p-a) n?:p un). 76 VII. Numerical Functions 7.3.6. Theorem. Let X be a set equipped with a filter base Pl, f and 9 mappings of X into R such that f(x) ::;; g(x) for all x E X. Then: ~ lim sup f(x) ::;; lim sup g(x), iJ/i iJ/i lim inf f(x) ::;; lim inf g(x). iJ/J iJ/J Let BE Pl. Then f(x) ::;; SUPXEB g(x) for all x E B, therefore SUpxEB f(x) ::;; SUpxEB g(x). Passing to the limit in this inequality, and taking account of 7.3.4. one obtains lim SUpiJ/i f(x) ::;; lim SUpiJ/J g(x). One sees similarly that lim infiJ/i f(x) ::;; lim infiJ/i g(x). 7.3.7. Theorem. Let X be a set equipped with afilter base Pl,j and 9 mappings of X into R. Then + g(x» lim sup (f(x) iJ/i and, if f + lim sup g(x), ::; lim sup f(x) iJ/i i1lJ :2: 0, 9 :2: 0, then limiJ/Jsup (f(x)g(x» ::; (limiJ/JSUP f(X») . (limiJ/iSUP g(X»). If one of the functions f, 9 has a limit along Pl, these inequalities become equalitie s. Let B E Pl. Then sup (f(x) XEB + g(x» ::; sup f(x) xeB + sup g(x) XEB by 7.1.6; passing to the limit along the ordered set Pl, and using 7.14, we deduce that lim sup (f(x) iJ/i (1) Set VB = + g(x» infxEB f(x). For all x VB ::; lim sup f(x) iJ/i E + g(x) + lim sup g(x). iJ/J B, ::; f(x) + g(x), therefore, in view of 7.1.7, VB + sup g(x) xeB ::; sup (f(x) + g(x»; xeD passing to the limit along the ordered set Pl, we deduce that (2) lim inf f(x) iJ/J + lim sup g(x) iJ/J ::;; lim sup (f(x) iJ/i + g(x». 77 7.4. Semicontinuous Functions If f has a limit along 86, then lim sUPml f(x) (1) and (2), we see that lim sup (f(x) mI + g(x» = lim f(x) mI = lim inf,~ f(x). Comparing + lim sup g(x). mI One reasons in an analogous way if 9 has a limit, and in the case of products. 7.4. Semicontinuous Functions 7.4.1. Definition. Let X be a topological space, Xo E X. and f E F(X, R), We say that f is lower semicontinuous at Xo if, for every A. < f(xo), there exists a neighborhood V of Xo in X such that x E V = f(x) ;=:: A.. We say that f is upper semicontinuous at Xo if, for every Jl. > f(xo), there exists a neighborhood V of Xo in X such that x E V = f(x) ~ Jl.. 7.4.2. To say that f is lower semicontinuous at Xo amounts to saying that - f is upper semicontinuous at Xo. It therefore suffices, in principle, to study the properties of lower semicontinuous functions. 7.4.3. Example. Let X be a topological space, Xo E X, and f E F(X, R). Then: f is continuous at Xo ¢ > f is both lower and upper semicontinuous at Xo' 7.4.4. Example. For x E R, set f(x) = x if x "I- 0, and f(O) = 1. Then f is continuous at every point of R - {OJ, and f is upper semicontinuous at 0 but not lower semicontinuous there. 7.4.5. Definition. Let X be a set, (Diei a family of numerical functions on X. We denote by SUPiei /;" infieI /;, the functions x r--;. SUPie'/;{X), X r--;. infieI/;{x) 78 VII. Numerical Functions on X. These are called the upper envelope and lower envelope of the family (h)ieI' In particular, if f and 9 are two numerical functions on X, one can speak of sup(f, g) and inf(j, g). 7.4.6. Theorem. Let X be a topological space, Xo EX, f and g numerical functions on X that are lower semicontinuous at Xo' Then sup(j, g), inf(j, g) and f + 9 are lower semicontinuous at Xo· The same is true of fg if f ~ 0 and 9 ~ O. We give the proof for f + 9 (the other cases may be treated in an analogous way). If f(xo) = - 00 or g(xo) = - 00, the result is evident. Otherwise, let A. < f(xo) + g(xo). There exist {t, v E R such that {t + v = A., {t < f(xo), v < g(xo). Next, there exist neighborhoods V, W of Xo in X such that X E V => f(x) ~ {t, X E W => g(x) ~ v. Then X E V n W => f(x) + g(x) ~ {t + v = A.. 7.4.7. The assertion 7.4.6 may be extended, one step at a time, to finite families of numerical functions. 7.4.8. The case of infinite families requires some precautions, as the following example shows. For n = 1,2, ... , let In be the numerical function on R defined by fix) = e- nx2 • Each fn is continuous. However, inf(ln) is the function f on R such that f(x) = 0 for x =j; 0, f(O) = 1; and f is not lower semicontinuous at O. 7.4.9. Nevertheless, we have the following result: .. Theorem. Let X be a topological space, Xo EX, COieI a family of numerical functions on X, and f = sUPie'Ji. If the /; are lower semicontinuous at Xo <for example, continuous at xo), then f is lower semicontinuous at Xo· Let A.' < f(xo). We have f(xo) = SUPie'/;(XO)' Therefore there exists an I such that A. < fi(xo}. Next, there exists a neighborhood V of Xo such that x E V => /;(x) ~ A.. Then x E V => f(x) ~ A.. i E 7.4.10. Definition. Let X be a topological space and f E $>(X, R). We say that f is lower (resp. upper) semicontinuous on X iff is lower (resp. upper) semicontinuous at every point of X. 79 7.4. Semi continuous Functions .. 7.4.11. Theorem. Let X be a topological space, and fE ff(X, following conditions are equivalent: R:). The (i) f is lower semicontinuous on X; (ii) for every AE R, the set of x E X such that f(x) :S; Ais closed; (iii) for every AE R, the set ofx E X such that f(x) > Ais open. Since f-1([ -00, A]) and f- 1«A, +00]) are complementary 10 X, conditions (ii) and (iii) are equivalent. (i) => (iii). Suppose f is lower semicontinuous on X. Let A = f- 1 «A, + 00]). If Xo E A then f(x o) > A; since f is lower semicontinuous at x o , there exists a neighborhood V of Xo such that x E V => f(x) > A. Therefore V c A. Thus, A is a neighborhood of each of its points, consequently is open. (iii) => (i). Suppose condition (iii) is satisfied. Let Xo E X. Let A < f(xo). The set A = f- \(..1., + cx;]) is open, and Xo E A, thus A is a neighborhood of Xo' Since f(x) > A for all x E A, we see that f is lower semicontinuous at Xo. 7.4.12. There are analogous characterizations of upper semi continuous functions. 7.4.13. Corollary. Let X be a topological space, Y a subset of X, cp the characteristic function of Y in X. For Y to be open (resp. closed), it is necessary and sufficient that cp be lower (resp. upper) semicontinuous. Let X,1 be the set of x E X such that cp(x) > A. If A < 0, then X,1 = X. If O:S; A < l, then X" = Y. If A ~ 1, then X" = 0. The sets X and 0 are open in X. By 7.4.11, we thus have cp lower semicontinuous ¢> Y open. From this, one deduces the characterization of the closed sets. 7.4.14. Theorem. Let X be a compact space, f a lower semicontinuous Junction = infxEx f(x). There exists an Xo E X such that f(xo} = m. on X, and m For every A > m, let X,1 be the set of x E X such that f(x) :S; A. This set is closed (7.4.ll), and it is nonempty by the definition of the infimum m. Every finite intersection is nonempty (because if Ai is the smallest of the numbers AI, ... ,An' then X"' n ... n X,1n = X,,). Since X is compact, X" is nonempty. Let Xo be a point of this intersection. For every A > m, we have Xo E X"' that is, f(xo) :S; A. Therefore f(xo) :::;; m. Also f(xo) ~ m, thus f(xo) = m. n,1>m 80 VII. Numerical Functions ~ 7.4.15. Theorem. Let X be a complete metric space, (1)iEI a family of lower semicontinuous functions on X such that SUPiEI J;(x) < + 00 for each x E X. Then there exist a non empty open subset U of X and a finite number M such that J;(x) :s; M for all i E I and x E U. Let f = SUPiEI.[;. Then f(x) < + 00 for all x E X by hypothesis, and f is lower semicontinuous by 7.4.9. For n = 1,2, ... , let Un be the set of x E X such that f(x) > n. This is an open subset of X (7.4.11). If all of the Un were dense in X, there would exist a point Xo belonging to all of the Un (5.5.12). One would then have f(xo) > n for all n, therefore f(xo) = + 00, which is absurd. Thus for some integer no, Uno is not dense; in other words, there exists a non empty open subset U of X disjoint from Uno' For all x E U, we have x rfo Uno' therefore f(x) :s; no, consequently J;(x) :s; no for all i E I. 7.5. Stone-Weierstrass Theorem 7.5.1. Lemma. Let X he a compact space, Yf a subset ofCC(X, R) having the following properties: (i) ifu E Yf and VEX, then sup(u, v) E Yf and inf(u, v) E .n'; (ii) ifx, yare points of X and ifIY., f3 E R (with IY. = f3 ifx = y), then there exists u E Yf such that u(x) = IY., u(y) = f3. Then every function in CC(X, R) is the uniform limit of a sequence of functions in JIf. Let f 9 :s; E CC(X, R) and f + e. E > O. We are to construct 9 E Yf such that f - e :s; (a) Let Xo E X. Let us show that there exists a function u E Yf such that u(xo) = f(xo) and u z f - e. For every y E X, there exists a function uy E Yf such that uixo) = f(xo) and u/y) = fey). The set Vy of all x E X such that uy(x) > f(x) - e is open (cf. 2.4.5). We have y E V y , thus the V y , as y runs over X, form an open covering of X. Since X is compact, it is covered by sets Vy, , " " V Yn ' Let u = sup(u yl , un' ... , uyJ E JIf. We have uy,(xo) = f(xo) for all i, therefore u(xo) = f(xo}. Let x E X. We have x E V Yi for some i. Then u(x) z uy,(x) > f(x) - e, and u satisfies the stated conditions. (b) The function u constructed in (a) depends on xo. For every x E X, let us define similarly a function Vx E Yf such that vix) = f(x) and Vx z f - e. The set W x of all z E X such that viz) < fez) + e is open. We have x E W X' Since X is compact, it is covered by sets Wx,, ... , W Xp' Let 9 = inf(vx" ... ,vx ) E JIf. 81 7.5. Stone-Weierstrass Theorem We have VXi ~ f - B for all i, therefore 9 ~ f - B. Let x for some j. Then g(x) ~ VXj(x) < f(x) + B. Thus 9 ~ f E X. We have x E W Xj + B. Jt 7.5.2. Lemma. The function on [0,1] is the uniform limit of a sequence of polynomials in t with real coefficients. We define functions poet), PI(t), P2(t), . .. , for t the following way: poet) Pn+ let) = 0, = Pn(t) + !(t E [0, 1], recursively, in - Pn(t)2). Let us show by induction that poet), PI(t), . .. , pit) are polynomials in t, and °~ poet) ~ PI(t) ~ ... ~ Pn(t) ~ Jt on [0, 1]. This is certainly the case for n = 0. Let us admit the preceding statements and let us prove the corresponding results for n + 1. It is first of all immediate that Pn+ let) is a polynomial in t. Next, for t E [0, 1J, we have t ~ Pn(t)2, therefore Pn+ let) ~ Pn(t). Finally, Pn+ let) - Jt = Pn(t) - Jt + !(t - Jt 2Jt, = (Pn(t) - Pn(t)2) JtX1 - !<Pn(t) + Jt». Jt) Jt ° Now, Pn(t) + ~ therefore 1 - !<Pn(t) + ~ 1~ in [0, IJ, and pit) - .)t ~ 0, therefore Pn+ 1(t) ~ .)t. For every t E [0, 1], the sequence (Pn(t» is increasing and bounded above by therefore has a finite limit f(t) ~ that satisfies ° Jt, f(t) = f(t) whencef(t) + !(t - f(t)2), = Jt. Finally, the Pn tend to funiformly on [0, 1J by 6.2.9. • 7.5.3. Theorem (Stone-Weierstrass). Let X be a compact space, .:It' a subset of CC(X, R) having the following properties: (i) the constant functions belong to .:It'; (ii) if u, v E JF, then u + v E .:It' and uv E JIl'; (iii) if x, yare distinct points of X, there exists U E .:It' such that u(x) #- u(y). Then every function in CC(X, R) is the uniform limit ofa sequence offunctions in JIf. Let .:It' be the closure of .:It' in CC(X, R) for the topology of uniform convergence. We are going to show that .:It' possesses the properties (i) and (ii) 82 VII. Numerical Functions of 7.S.l. Then, by 7.5.1, the closure of Jf will be equal to <t'(X, R), whence Jf = ce(X, R), which will imply the theorem. (a) If u E Jf and v E JIf, then u + v E Yf. For, there exist sequences (u,,), (v n ) of functions in Jf such that Un -> U, Vn -> v uniformly; then u" + Vn -> U + v uniformly, whence u + v E Yf. Similarly, uv E Jf; and if A E R, then AU E £. Thus, every polynomial in u, that is, every function of the form ..1.0 + A1U + A2u 2 + '" + Anu", where . 1. 0 , ... , An E R, belongs to .j{: (b) Let x, y be distinct points of X, and a, such that vex) #- v(y). Set fi E R. There exists v E Jf I v' = v() x - v(y) (v - v(y». Then v' E .1f, v'(x) = 1, v'(y) = O. Let /3 + (a - fJ)v'. Then u E Jf and u(x) = a, u(y) = /3. (c) Let u E Jf and let us show that Iu I E Yf. The function u, being continuous u = on X, is bounded (4.2.13). On multiplying u by a suitable constant, we are thus reduced to the case that - 1 ~ U ~ 1. Then 0 ~ u 2 ~ 1. Let 8 > O. By 7.5.2, there exists a polynomial pet) with real coefficients, such that Ip(t) ~ 8 for all t E [0, 1]. Then Ip(U(X)2) - ylu(.x? I ~ e for all x E X, that is, Ip(u 2 ) - Iu II ~ E. Now, p(u 2 ) E Jf by (a). Thus Iu I is adherent to .Yf, consequently Iu I E Yf. (d) Let u, v E Yf. In view of (a) and (c), we have 01 + v + lu i(u + v - lu - sup(u, v) = i(u inf(u, v) = vi) E.Yf, vI) E Yf. 7.5.4. Corollary. Let X be a compact space, Jf a set of continuous, complexvalued jimctions on X, having the jollowing properties: (i) the complex-valued constant functions belongs to Jf; (ii) if u, v E .1f, then u + v E.1f, UV E Jf and Ii E Jf; (iii) if x, y, are distinct points of X, there exists u E Jf such that u(x) #- u(y). Then every function in <t'(X, C) is the uniform limit of a sequence of functions in Yf. Let Jf' be the set of functions belonging to Jf that are real-valued. Then Jf' satisfies the conditions (i) and (ii) of 7.5.3. If x, yare distinct points of X, there exists u E Jf such that u(x) #- u(y). Then either Re u(x) #- Re u(y) or 1m u(x) #- 1m u(y). Now, Re u = i(u + Ii) E Jf' and 1m u = 1 2i (u - Ii) E Jf', 83 7.5. Stone-Weierstrass Theorem therefore Jf' also satisfies condition (iii) of 7.5.3. Let 9 E rt'(X, C). Then = 91 + i9 2 with gl' gz E "6'(X, R). By 7.5.3, gl and g2 are uniform limits of functions in Jf', therefore 9 is the uniform limit of functions in £: 9 7.5.5. Corollary. Let X be a compact subset of Rn, and f E "6'(X, C). Then f is the uniform limit on X of a sequence of polynomials in n variables with com- plex coefficients. Consider the polynomials in n variables with complex coefficients. These are functions on R" whose restrictions to X form a subset Jf of "6'(X, C). It is clear that Jf satisfies conditions (i), (ii), (iii) of 7.5.4, whence the corollary. 7.5.6. Corollary. Let f be a continuous, complex-valued periodic function on R of period 1. Then f is the uniform limit on R of a sequence of trigonometric polynomials (that is, functions of the form n t ~ "L.. r= -n a r e Z7tir ! , where the ar are complex constants). Let p be the canonical mapping of R onto T (3.4.3). Since f has period 1, there exists a complex-valued function 9 on T such that f(x) = g(p(x» for all x E R. By 3.4.4, 9 is continuous. By 4.2.16, there exists a homeomorphism of U onto T such that 0-1(p(X» = e Z7tix for all x E R. Let h = 9 E "6'(U, C). For all x E R, f(x) = g(p(x» = h(e- 1 (p(x») = h(e Z7tiX ). Now, U is a compact subset of R2 = C. Let E > O. There exists a polynomial am.xPny' in x and y, with complex coefficients, such that e 0 e Lm," Ihex + iy) - ~" amnxPny' I ::5: for every point x E + iy ofU (7.5.5). Consequently, for every t E R, we have I h(eZ"i!) - ~n am.(cos 2m)m(sin 2m)" I ::5: E, that is, Since cos 2m function = t(e 27tit + e- 2 "it) and sin 2nt = (l/2i)(e 2nit m,n is a trigonometric polynomial. - e- 2xit ), the 84 VII. Numerical Functions * 7.5.7. Let X be a noncompact locally compact space. The sets of the form X - C, where C is compact in X, form a filter base f!J on X (4.2.9(i». Let X' = X v {w} be the Alexandroff compactification of X (4.5.9). By 4.5.8, the sets (X - C) v {w}, where C is compact in X, are the open neighborhoods of w in X'. Consequently, if f is a complex-valued function on X, the following conditions are equivalent: (i) f tends to 0 along f!J; (ii) if f' is the function on X' that extends limx_w f'(x) = O. f and vanishes at w, then When these conditions are satisfied, we say that f tends to 0 at infinity on X. We remark that if, in addition,fis continuous on X, thenf' is continuous on X'. * 7.5.S. Corollary. Let X be a noncompact locally compact space, !'(lo the set of continuous complex-valued functions on X that tend to 0 at infinity, and :ff a subset of!'(lo having the following properties: (i) if u, V E :ff and A E C, then u + v E Yf, UV E Yf, ii E :ff and AU E :ff ; (ii) ifx, yare distinct points of X, there exists u E:ff such that u(x) '# u(y); (iii) ifx E X, there exists u E .J'{' such that u(x) '# O. Then every function in !'(lois the uniform limit of a sequence of functions in .Yt. Let us keep the notations of 7.5.7. Let :ff' be the set of functions on X' of the form f' + A, where f E :ff and A E C. Then :ff' c !'(leX', C). Obviously :ff' satisfies condition (i) of 7.5.4. Let u, v E :ff'. We have u = f' + A, v = g' + J1 with f, g E :ff and A, J1 E C. Then: u + v = f' + g' + A + J1 = (.f + g)' + (A + J1) =:ff', uv = f'g' + }..g' + J1f' + AJ1 = (.fg + Ag + J1f), + AJ1 E :ff', and ii = t + X= ely + XE :ff'. Finally, if x, yare distinct points of X', there exists u E:ff' such that u(x) '# u(y) (if x, y E X, this follows from the hypothesis (ii); if x = w or y = w, it follows from the hypothesis (iii». Now let hE 16'0 and e> O. By 7.5.4, there exist f E :ff and A E C such that Ih'(x) - f'(x) - A.I f. ~:2 85 7.6. Normal Spaces for all x E X'. In particular, el2 for all x E X. ~ Ih'(w) - f'(w) - AI = 11.1. Therefore * 7.6. Normal Spaces 7.6.1. Theorem. Let X be a topological space. The following conditions are equivalent: (i) For any disjoint closed subsets A and B of X, there exist disjoint open sets U and V of X such that A c: U, B c: V. (ii) For every closed subset A of X and every open set W of X such that A c: W, there exists an open set W' of X such that A c: W' c: W' c: W. (iii) For any disjoint closed subsets A and B of X, there exists a continuous mapping of X into [0, 1] equal to 0 at every point of A and to 1 at every point ofB. (iv) For every closed subset A of X and every numerical function f defined and continuous on A, there exists a numerical function defined and continuous on X that extends f. (iv) => (i). Suppose that condition (iv) is satisfied. Let A and B be disjoint closed subsets of X. Then C = A u B is a closed subset of X. Set f(x) = for x E A and f(x) = 1 for x E B. Then f is continuous on C (because A and B are open in C). By (iv), there exists a continuous mapping 9 of X into R that extends f. Let ° U=g-l«-oo,t», V =g-l«t, +00». Then U and V are disjoint open sets in X, and A c: U, B c: V. (i) => (ii). Suppose that condition (i) is satisfied. Let A (resp. W) be a closed (resp. open) set in X with A c: W. Set B = X - W; this is a closed set in X disjoint from A. By (i), there exist disjoint open sets U and V of X such that A c: U, B c: V. Then U c: X - V and X - V is closed, therefore c: X - V c: X - B = W. Thus A c: V c: 0 c: W. (ii) => (iii). Suppose that condition (ii) is satisfied. Let A and B be disjoint closed subsets of X. We are to construct a continuous mapping of X into [0, 1] equal to 0 on A and to 1 on B. Let D be the set of 'dyadic' numbers belonging to [0, 1], that is, the set of numbers of the form kl2n, where n = 0, 1, 2, ... and k = 0, 1, 2, ... , 2n. This set is dense in [0, 1]. For every a E D, we are going to construct an open subset V(d) of X in such a way that o (1) d < d' => V(d) c: V(d'). 86 We set U(1) VII. Numerical Functions =X- B and we choose an open set U(O) such that (2) A c U(O) c U(O) c U{l) (this is possible by (ii». Suppose the U(d) already chosen for d 2/2", ... ,2"/2" = 1, in such a way that U(k/2") c U«k = 0, 1/2", + 1)/2") for 0 ~ k < 2". Let us define U(k/2" + 1) for k =0,1, ... ,2"+1. For keven, U(k/2" + 1) has already been chosen. For k odd, thus of the form 2h + 1, we choose an open set U «2h + 1)/2" + 1) such that U(2h/2" + 1) c U«2h + 1)/2"+1) c U«2h + 1)/2"+1) c U«2h + 2)/2"+1), which is possible by (ii). By induction, the open sets U(d) are thus defined for all dE D, and property (1) certainly holds. If x E B we set f(x) = 1. If x ¢ B, then x E U(I); let f(x) be the infimum in R of the d E D such that x E U(d). We have thus defined a mapping f of X into [0,1], equal to 1 at every point of B. If x E A then x E U(O) by (2), therefore f(x) = O. Finally, let us show that fis continuous at every point x ofX. Let a = f(x) and e > O. If 0 < a < 1, then there exist d, d', d" E D such that a- e ~ d < d' < a < d" ~ a + e. If one had x ¢ U(d"), it would follow that f(x) ~ d", which is absurd; thus x E U(d"). On the other hand, f(x) > d', therefore x ¢ U(d'), therefore x ¢ U(d) by (1). Consequently, if one sets V = U(d") n (X - U(d», V is an open neighborhood of x. Let y E V. Then y E U(d"), therefore fey) ~ d"; and fey) ~ d, since otherwise one would have y E U(d). Thus, Y E V ::;. I fey) - f(x) I ~ e, which proves our assertion when 0 < a < 1. If a = 1, there exist d, d' E D such that a - e ~ d < d' < a = 1. We see as above that V = X - U(d) is an open neighborhood of x and that, for every y E V, one has fey) ~ d, therefore I fey) - f(x) I ~ B. If a = 0, there exists d" E D such that 0 = a < d" ~ a + B. One sees as above that V = U(d") is an open neighborhood of x and that, for every y E V, one hasf(y) ~ d", therefore If(y) - f(x) I ~ B. (iii) ::;. (iv). Suppose that condition (iii) is satisfied. Let A be a closed set in X and let fbe a continuous mapping of A into R. Let us define a continuous mapping of X into R that extends j: Since Rand [ -1, 1] are homeomorphic, we can suppose that f takes its values in [ - 1, 1]; we will define a continuous mapping of X into [ -1, 1] that extends f. 87 7.6. Normal Spaces We shall first prove the following intermediary result: (*) i If u is a continuous mapping of A into [ -1, 1], there exists a continuousmappingvofXinto [ -1, iJ such that lu(x) - v(x) 1 :s; ! for all x E A. For, let H (resp. K) be the set of x E A such that -1 $ u(x) :s; -i (resp. :s; u(x) :s; 1). The sets Hand K are closed in A, therefore in X, and they are disjoint. By (iii), there exists a continuous mapping v of X into [ -i,·n equal to - i at every point of H and to i at every point of K. It is clear that lu(x) - v(x) 1 :s; 1for all x E A. This established, we are going to recursively construct continuous numerical functions go, g 1, g2, ... on X such that (3) { -1 + mn + 1 :s; gn(x):s; If(x) - 9n(X) 1 :s; 1 - (~)n+l (!r+ I for all x E X for all x E A. The existence of 90 results from applying (*) with u = gl,···, gn already constructed. Define u(x) = @n+ 1(f(X) - 9n(X» for x E f. Suppose go, A. Then u is a continuous mapping of A into [ -1, 1]. By (*), there exists a continuous mapping v of X into [-i, iJ such that lu(x) - v(x) 1 $1 for all x E A. Then, for all x E A, If(x) - gn(x) - (1r+ On the other hand, for all x Ign(x) + (1)"+ Iv(x)1 E 1 v(x)1 = (~-)"+llu(x) - v(x) 1 :s; mn+Z. X, :s; 1 - (~)n+ 1 Setting gn+ I(X) = 9n(X) + mn+ 1. i = 1 - mn+Z. + mn+ IV(X) for every x E X, the construction of the gn is complete by induction. We observe, moreover, that Ign+ 1 (x) - gn(x) I $ ¥~)"+ 1 for all x E X. Since the series with general term i(i)n+ 1 is convergent, the series with general term gn+ 1 - gn is normally convergent, therefore uniformly convergent (6.1.10); in other words, gn has a uniform limit 9 on X. This limit is continuous (6.1.11). By (3), we have -1 :s; g(x) :s; 1 for all x E X andf(x) = g(x) for all x E A. 7.6.2. Definition. One calls normal space a separated space that satisfies the equivalent conditions of 7.6.1. 7.6.3. Examples. (a) Every metric space is normal. For let X be a metric space, d its metric, A and B disjoint, nonempty closed subsets of X. Since the 88 VII. Numerical Functions functions x t--+ d(x, A) and x t--+ d(x, B) are continuous on X (5.1.6), the set U (resp. V) of x E X such that d(x, A) < d(x, B) (resp. d(x, B) < d(x, A» is openinX.ItisclearthatU n V = 0.Ifx E A then d(x, A) = Oandd(x,B»O (because the relation d(x, B) = would imply x E B, therefore x E B). Thus A c U, and similarly B c V. (b) Every compact space is normal. This follows from 4.2. 11 (i). (c) If I is an uncountable set, it can be shown that RI is not normal. One can also construct locally compact spaces that are not normal. ° 7.6.4. Remark. Let X be a normal space, A a closed su bset of X, f a numerical function defined and continuous on A. We know that there exists a numerical function g, defined and continuous on X, that extends f. Suppose, moreover, that f is finite. We shall see that g can then be chosen to be finite. Suppose first that f ;;:.: on A. A fortiori, f takes its values in [0, + 00], therefore we can suppose that g takes its values in [0, + 00] (which is homeomorphic to R). Let B = g - 1( { + 00 }). Then B is closed and is disjoint from A. The function h on the closed set A u B that is equal to f on A and to on B is therefore continuous. Let 0' be a continuous extension of h to X taking its values in [0, + 00]. Replacing 0 by inf(o, 0'), we obtain a continuous exextension of f to X that is finite at every point of X. In the general case, let fl = sup(f, 0), f2 = sup( - f, 0). Then f = fl - f2, andfl'/2 are finite, ;;:.: and continuous. It now suffices to apply the preceding paragraph to fl and f2 . ° ° ° 7.6.5. Theorem. Let X be a compact space. There exists a set I such that X is homeomorphic to a closed subset of [0, 1l Let (Diel be a family of continuous mappings of X into [0, 1]. By 3.3.2(d), the mapping f: x t--+ (/;(X»iel of X into [0, 1]1 is continuous. The set f(X) is a compact, hence closed, subset of [0, 1]1 (4.2.12, 4.2.7). Iff is injective, then f is a homeomorphism of X onto f (X) (4.2.15). Now, if one takes for (/;)iel the family of all continuous mappings of X into [0, 1], then f is injective. For, let a and b be distinct points of X. Since X is normal (7.6.3(b», there exists a continuous mapping 0 of X into [0, l]suchthatg(a) = O,g(b) = 1. Since 0 is oneofthe/;,wehavef(a) #f(b). 7.6.6. Theorem 7.6.5 is sometimes expressed by saying that every compact space may be embedded in a 'generalized cube'. 7.6.7. Theorem. Let X and Y be compact spaces. Let .Yf be the set of functions on X x Y oftheform (x, y) t--+ fl(X)Ol(Y) + f2(X)02(Y) + ... + J..(x)On(Y), where fl,'" ,J.. E CC(X, R), 01"" ,g. E CC(Y, R), n = 1,2, .... Then .Yf is dense in CC(X x Y, R) for the topology of uniform convergence. 7.6. Normal Spaces 89 To apply the Stone-Weierstrass theorem, it suffices to check that :K satisfies the conditions (i), (ii), (iii) of 7.5.3 relative to the compact space X x Y. This is clear for conditions (i) and (ii). Let (a, b) and (a', b') be two distinct points of X x Y. Suppose for example that a '# a'. Since X is normal, there exists f E ~(X, R) such that f(a) = 0, f(a') = 1. Set g(y) = 1 for all y E Y. Then the function (x, y) 1--+ f(x)g(y) on X x Y takes on different values at (a, b) and Ca', b'). CHAPTER VIII Normed Spaces We take up again the theory of normed spaces and pre-Hilbert spaces. §§1 to 5 are already familiar, excepting possibly Theorem 8.3.4 on equivalent norms. In §§6, 7, 8 we make the connection between this theory and that of complete spaces; some of these results (BanachSteinhaus theorem, Riesz's theorem) are very fruitful, but the reader can hardly be convinced of this unless (s)he studies 'functional analysis' later on. 8.1. Definition of Normed Spaces 8.1.1. Definition. Let E be a vector space over R or C. A seminorm on E is a function x I---> Ilx II defined on E, with finite values 2: 0, such that (a) IIAxl1 = IAlllxl1 forallxEEandallscalard; (b) Ilx + yll :s; Ilxll + Ilyll for all x E E and y E E (triangle inequality). ° It follows from (a) that x = 0 => Ilxll = 0. If, conversely, Ilxll = => x = 0, the seminorm is called a norm. When a seminorm (resp. norm) is given on E, we say that E is a seminormed (resp. normed) vector space. Conditions (a) and (b) imply at once: II-xii = Ilxll Ilx - yll :s; Ilxll forallxEE, + Ilyll forall xEEandYEE. There is an obvious notion of isomorphism between normed or seminormed spaces. 91 8.1. Definition of Normed Spaces 8.1.2. Example. In R n or norms: en one can define, for example, the following + ... + IXnIZ)l/Z, + .,. + Ix.l, II(xl .... , xn)ll = (Ixllz II(xI.···. xn)11 = IXII II(x l , ... , x.)11 = sup(lxll,··· .lx.I)· 8.1.3. Example. Let X be a set, E the vector space of all bounded real-valued (or complex-valued) functions defined on X. For every lEE, set II!II = sup xeX I/(x)l. One verifies immediately that I t--+ II I II is a norm on E. Now let A be a subset of X. This time, for every lEE set II!II = sup I/(x)l. XEA One verifies that I t--+ "I" is a semi norm on E. 8.1.4. Example. Let E be the vector space of all sequences (At, Az, ... ) of real (or complex) numbers. The bounded sequences form a linear subspace of E, denoted I'R or ICC (or simply 100 ). For every s = (x t , xz, ... ) E 100 , set Iisil = sup(lxtl, IX21,.· .). Then 100 becomes a normed space. This is the special case of 8.1.3 where one takes X = {I, 2, 3, ... }. 8.1.5. Example. We denote by l~ or Ii (or simply It) the set of sequences S = (X1' x 2 , •.• ) of complex or real numbers such that L:'=1 Ix.1 < +00. This is a linear subspace of 100. For, if s = (x t , x 2 , • •• ) E [t and t = (Yt, Y2,"') E [I. then 00 00 00 L Ix. + Ynl:::; n=l L Ix.1 + I .=1 (1) n=1 IYnl < +00. thus S + t E P. It is clear that AS E /t for every scalar A. For s = (Xt, X2,"') E [I, set IIsll = Then s t--+ IIsll 00 I Ix.l. is a norm on Jl (the triangle inequality results from (1». 8.1.6. The Metric Deduced From a Norm. Let E be a normed vector space. For x, Y E E, set d(x, y) = Ilx - YII. One verifies without difficulty that d is a metric on E. (Thus. a normed space is automatically a metric space, hence a 92 VIII. Normed Spaces topological space.) This metric is invariant under translations in E, that is, d(x, y) = d(x + a, Y + a) for all x, y, a E E. For every x E E, d(x, 0) = Ilxll. 8.1.7. Examples. Starting with the norms 8.1.2, one recovers the usual metrics on Rn or en (1.1.15). In 1 the distance between two sequences (Xi) and (Yi) is SUp(IXl - yd, IX2 - Y21 •... ). In 11, the distance between two sequences (x;) and (Yi) is IXl - Yll + IX2 - Y21 + .. '. If f and 9 are two bounded, real-valued or complex-valued functions on a set X, then the distance between them deduced from the norm of 8.1.3 is 00 , dU, g) = sup If(x) - g(x) 1; .xeX this is the metric of uniform convergence. 8.1.8. Theorem. Let E be a normed space. (i) The mapping (x, y) 1-+ X + yofE x E into E is continuous. (ii) The mapping (..1., x) 1-+ AX ofR x E (or e x E) into E is continuous. Let xo, Yo E E and 8> O. If x, Y E E are such that Ilx - xoll ~ 8 2 and Ily - Yoll ~ 8 2' then lI(x + y) - (Xo + Yo)11 = II(x - Xo) + (y - Yo)1I ~ Ilx - xoll + Ily - Yoll ~ 8. This proves (i). Let Xo E E, ..1.0 E R (for example) and 8 > O. Set '7 = inf(l. 1 + IAo~ + Ilxoll) > O. Let x E E, A E R be such that IA - Ao I ~ 11 and II x - Xo II ~ 1'/. Then IIAx - Aoxoll = 11(..1. - AoXx - xo) + Ao(x - xo) + (A - Ao)xoll ~ 1..1. - Aollix - xoll + l..1.o lllx - xoll + 1..1. - Aolllxoll ~ '7 2 + 1..1.0111 + Il xoll'7 ~ 11(1 + 1..1.01 + IlxolI) ~ E. This proves (ii). 8.1.9. IfE is a normed vector space, then every linear subspace ofE, equipped with the restriction of the norm of E, is automatically a normed vector space. For example, let X be a topological space and F the set of continuous, bounded real-valued functions on X. Then F, equipped with the norm of uniform convergence, is a normed linear subspace of the normed space defined in 8.1.3. 93 8.2. Continuous Linear Mappings 8.1.10. Product of Seminormed Vector Spaces. Let E 1, ... , En be seminormed vector spaces (all real or all complex). Let E = E1 X ••• x En, which is a real or complex vector space. For x = (x 1, ... , X n ) E E, set One verifies in the usual way that this defines a seminorm on E. HE b . . . , En are normed spaces, then this seminorm is a norm, and the metric space defined by the norm of E is the product of the metric spaces E l' ... , En in the sense of 3.2.3. Other useful norms can be defined on E; for example, Ilxll Ilxll = Ilxlll + ... + Ilxnll, = sup(llxlll,···, Ilxnll). On Rn or en, one recovers the norms of 8.1.2. 8.1.11. The Normed Space Associated with a Seminormed Space. Let E be a seminormed space. Let F be the set of x E E such that I x II = 0. If x, y E F then Ilx + yll ::::;; Ilxll + Ilyll = 0, therefore x + y E F. Obviously Ax E F for every scalar A.. Thus F is a linear subspace of E, and one can form the quotient vector space E' = ElF. Let x' E E'. Choose a representative x of x' in E. The number Ilxll depends only on x' and not on the choice of the representative x. For, every other representative of x' is of the form x + u with u E F; then Ilx + u II::::;; Ilxll + Ilull = Ilxll, and similarly Ilxll::::;; Ilx + ull + Ilull = Ilx + ull, thus Ilx + ull = Ilxll· We may therefore set Ilx'll = Ilxll. Since x r-+ Ilxll is a seminorm on E, one verifies easily that x'r-+ Ilx'll is a seminorm on E'. This seminorm is a norm: for, if x' E E' is such that I x' I = 0, and if x is a representative of x' in E, then Ilxll = 0, therefore x E F, therefore x' = 0. We say that E' is the normed space associated with the seminormed space E. The study of the properties of E is practically equivalent to the study of the properties of E'. 8.2. Continuous Linear Mappings ° 8.2.1. Let E, F be normed spaces. Let u be a linear mapping of E into F. Let B be the closed ball in E with center and radius 1. We define: (1) Ilull = sup Iluxll E [0, xeB + 00]. We have (2) Iluyll::::;; Ilu1111Y11 forallYEE 94 VIII. Normed Spaces (with the convention that O· + 00 = 0). For, (2) is clear if y = o. If y i= 0, let x = Ilyil-ly. Then y = Ilyllx, therefore uy = Ilyllux, and so IluYIl = Ilylllluxll:::;; lIyllliull becausexEB. 8.2.2. More precisely, II u II is the smallest of the numbers a E [0, that IluY11 (3) ~ allyll + 00] such for all y E E. For, if a satisfies (3) then, in particular, Iluxll Ilull :::;; a. ~ a for all x E B, therefore 8.2.3. Let S be the sphere in E with center 0 and radius 1. If E i= 0, then every element of B may be written Ax with 0 s: A. ~ 1 and XES, therefore (4) Ilull = sup Iluxll. xeS .. 8.2.4. Theorem. Let E, F be nomled spaces and u a linear mapping of E into F. The following conditions are equivalent: (i) (ii) (iii) (iv) u is continuous at 0; u is continuous; u is uniformly continuous; Ilull < + 00. (iii) =:> (ii) =:> (i). This is obvious. (i) =:> (iv). Suppose that u is continuous at O. There exists an tf > 0 such that y E E and Ilyll ~ tf imply IluY11 :::;; 1. Then, if x E E, we have Ilxll :::;; 1 =:> Iltfxll ~ tf therefore Ilull < + 00. (iv) =:> (iii). Suppose Ilull < Ilx - =:> lIu(tfx)1I ~ 1 = Iluxll :::;; tf- 1 , + 00. Let x, y E E and e > O. Then yll :::;; M =:> Ilux - uY11 = Ilu(x - y)11 :::;; E Ilullllx - yll ~ E, thus u is uniformly continuous. 8.2.5. Condition (iv) of 8.2.4 means, in the notations of 8.2.1, that u(B) is bounded. The expression 'bounded linear mapping' is used as a synonym for 'continuous linear mapping'. 8.2.6. Example. Let E = ~([O, 1], R), equipped with the norm of uniform convergence. The mapping fH f(O) of E into R is linear; it is continuous because If(O) I :::;; Ilfll for allfEE. Let F be the linear subspace of E formed by the differentiable functions, equipped with the norm induced by that of E. The mapping ff-+ 1'(0) of F into R is linear; it is not continuous, since, for every number A > 0, one can 95 8.2. Continuous Linear Mappings construct a function f E F such that If (x) I ::;; 1 for all x E [0, l] but 1'(0) (for example. f(x) = Axl(l + Ax». ~ A 8.2.7. Let E, F be normed spaces. We denote by .P(E, F) the set of all continuous linear mappings of E into F. If u, V E .P(E, F) and Ais a scalar, then u + v E .P(E, F) and AU E .P(E, F) by 8.1.8. Thus .P(E, F) is in a natural way a real or complex vector space. When E = F, one writes .P(E) = .P(E, E). The identity mapping idE, also denoted 1 or I, is an element of .P(E). If u E .P(E) and v E .P(E) then u v E .P(E), so that .PCE) is in a natural wayan algebra over R or C with unity element 1. 0 8.2.8. Theorem. Let E, F, G be normed spaces. Ci) The mapping u 1--+ II u II of .PCE, F) into [0, + (0) is a norm on .P(E, F). Cii) IfuE.PCE,F)andvE.P(F,G), then Ilvoull::;; Ilvllllull. (iii) IlidEIl = 1 if E # O. (i) Let u, v E .P(E, F). For every x E E, II(u + v)(x) II = Ilux + vxll ::;; Iluxll + Ilvxll ::;; Ilullllxll + Ilvllllxll = Cllull + Ilvll)llxll, therefore Ilu + vii::;; Ilull + Ilvll. Also, IIAU II = sup IICAU)(x)1I = sup IAIlIuxll IIxll,,; 1 Ilxll,,; 1 = IAI sup Iluxll = IAlllull· II xII ,,; 1 Finally, if Ilull = 0 then Iluxll = 0 for all x E E, therefore ux x E E, thus u = o. (ii) Let u E .PCE, F), v E .P(F, G). For every x E E, = 0 for all IICv u)(x) II ::;; Ilvlllluxll ::;; Ilvllllullllxll, 0 therefore Ilv 0 ull ::;; Ilvllllull. (iii) This is obvious. 8.2.9. Thus, .P(E, F) is in a natural way a normed vector space. The norm defines a metric and a topology on .PCE, F). This topology is called the norm topology on .PCE, F). By 8.1.8, the mappings (u. v) 1--+ U (A, u) 1--+ A.U + v of .PCE, F) x .P(E, F) into .PCE, F), of R x .P(E, F) (or C x .P(E, F» into .PCE, F), are continuous. It follows easily from 8.2.8(ii) that the mapping (u, v) 1--+ V is continuous. 0 u of .P(E, F) x .P(F, G) into .P(E, G) 96 VIII. Normed Spaces 8.2.10. Let E be a real or complex vector space. The elements of g(E, R) or geE, C) are the continuous linear forms on E. The space g(E, R) or geE, C) is a normed vector space, called the dual of E and often denoted E'. 8.3. Bicontinuous Linear Mappings 8.3.1. Theorem. Let E, F be normed spaces, u a linear mapping of E onto F. The following conditions are equivalent: (i) u is bijective and bicontinuous; (ii) there exist numbers a, A E (0, + 00) such that allxll :::;; Iluxll :::;; Allxll for all x E E; (iii) there exist numbers a, A E (0, for all x EE with + 00) such that a :::;; Iluxll $; A Ilxll = 1. Suppose u is bijective and bicontinuous. There exist Iluxll $; Allxll for all xEE and Ilu-Iyll $; Bllyll for all YEF (8.2.4). Let xEE. Set y = ux, so that x = u-1y. Then Ilxll :::;; Bllyll, that is, {l/B)llx I :::;; Iluxll, and liB> since B < + 00. (ii) ~ (i). Suppose condition (ii) is satisfied. Then u is continuous (8.2.4). Next, ux = implies allxll = 0, therefore x = (because a> 0); this proves that u (which is surjective by hypothesis) is bijective. Finally, let y E F. Set x = u-1y. Then y = ux, and the inequality allxll :::;; Iluxll may be written a Ilu - I YII :::;; IIYII, or Ilu- 1 YII $; (lla) Ilyli. Thus u- 1 is continuous (8.2.4). (ii) ~ (iii). This is obvious. (iii) ~ (ii). Suppose that condition (iii) is satisfied and let us prove that allxli $; Iluxll $; Allxll for all x EE. This is clear if x = 0. If x i= 0, let x' = x/llxll. Then Ilx'll = 1, therefore a:::;; Ilux'll $; A. Now, ux = u(llxllx') = Ilxllux', therefore allxll :::;; Iluxll $; Allxll· (i) ~ (ii). A, BE(O, +00) such that ° ° ° 8.3.2. Theorem. Let x t--> I X III and x t--> I X 112 be two norms on a vector space E. The following conditions are equivalent: (i) the topologies defined by the two norms on E are the same; (ii) the identity mapping ofE I into E2 (where E1, E2 denote the vector space E equipped with the norms x t--> Ilxlll' X t--> Ilxllz) is bicontinuous; (iii) there exist numbers a, A E (0, + 00) such that allxlll for all x E E. $; Ilx!12 $; Allxll l 97 8.3. Bicontinuous Linear Mappings The equivalence (i) ~ (ii) is obvious for any pair of topologies on a set (cf. 2.4.7). The equivalence (ii) ~ (iii) follows from 8.3.1. 8.3.3. Definition. If two norms on a vector space satisfy the conditions of 8.3.2, they are said to be equivalent. This is clearly an equivalence relation among norms. 8.3.4. For example, it is well known that, on R" or defined in 8.1.2 are equivalent. More generally: ~ Theorem. On R n ar e, the three norms cn, all narms are equivalent. Let us treat for example the case of R". Let x t--+ Ilxll be a norm on R". Let xt--+llxll' be the norm (xl, ... ,xn)t--+lxll + ... + IXnl. It suffices to prove that these two norms are equivalent. In the proof, we are going to use topological concepts in R"; the only topology on R n we shall use will be the usual topology, which is defined by the norm x t--+ Ilxll' (cf. 8.1.7,1.1.15,1.1.2). Let (ej, ... , en) be the canonical basis ofR". Set A = sup(llelll, ... , lien II)· If x = (Xl' ... , X") E R", then Ilxll = Ilxlel + ... + x"enll :::; Ix111lelil :::; A(lxll + ... + Ix"1) = Allxll'· It follows from this that the function x, Xo E R" and B > 0, then Ilx - xoll' :::; i => Let S be the set of X = t--+ Ilx lion R n is continuous; for, if Ilx - xoll :::; e (Xb ... , Ilxll' X = IXII + ... + Ixnlilenil => Illxll - Ilxolll :::; B. x n ) E R n such that + ... + Ixnl = 1. This is a closed set in R (1.1.12), clearly bounded, hence compact (4.2.18). The function X t--+ Ilxll is continuous on S by what was shown earlier, and it does not vanish on S, therefore there exists a number a > 0 such that Ilx II ~ a for all XES (4.2.14). Since, moreover, Ilxll :::; A for all XES, we see that the two norms are equivalent. n 8.3.5. Corollary. On a finite-dimensional real ar camplex vectar space, all norms are equivalent. For, such a vector space is isomorphic to R n or en for some n. 8.3.6. Thus, on a finite-dimensional real or complex vector space E, there exists a 'natural' topology :T: the topology defined by any norm on E. If u 98 VIII. Normed Spaces is any isomorphism of the vector space Rn or cn onto E. then :T is the transport by u of the usual topology of Rn or C". 8.3.7. Let E, F be normed vector spaces, u a linear mapping of E into F. If E is finite-dimensional. then u is automatically continuous. For, we can suppose by 8.3.6 that E = Rn (or C"). Let (e l , . . . , en) be the canonical basis of Rn (for example). Let ai = u(e i ) E F. For every x = (XI' ... , xn) ERn, we have u(x) = xla l + ... + xna n. The mapping Xf--+X I of R n into R is continuous (3.2.8). The mapping A. f--+ A.a I of R into F is continuous (8.1.8). Therefore the mapping X f--+ X I a t of R" into F is continuous. Similarly for the mappings x f--+ X2a2," ., X f--+ x"a n . Therefore the mapping x f--+ (xlal,"" x"an ) of R n into F" is continuous (3.2.7). Consequently, the mapping x f--+ Xtaj + .. , + X"a n of R" into F is continuous (8.1.8). However, if E is infinite-dimensional then u may be discontinuous, as we saw in 8.2.6. 8.3.8. If E and Fare normed spaces of finite dimensions m and n, then. by 8.3.7, ,P(E, F) is the vector space of all linear mappings of E into F. This vector space has dimension mn. so by 8.3.6 it possesses a natural topology :Y. If one chooses bases in E and F. there is a canonical linear bijection u f--+ M. of 'p(E. F) onto the vector space Mn. m of real or complex matrices with n rows and m columns. Under this bijection, the topology !Y corresponds to the natural topology of Mn. m; the latter topology is defined, for example, by the norm ((Xi)I'; i';n. I ,;j,;m f--+ Ii,.i 1Q(ij I· 8.4. Pre-Hilbert Spaces 8.4.1. Definition. A complex pre-Hilbert space is a complex vector space E equipped with a mapping (x, y) f--+ (x y) of E x E into C, called the scalar product (or 'inner product'), satisfying the following conditions: 1 (i) (x 1y) depends linearly on y, for fixed X; (ij) (xly) = (ylx) for x,YEE (so that (xly) depends 'conjugate-linearly' on x, for fixed y); (iii) (xix) ~ 0 for x E E. There is an obvious notion of isomorphism of pre-Hilbert spaces. 8.4.2. Example. For X=(XI •... ,Xn)EC" set and y=(yt' ...• yn)EC", 99 8.4. Pre-Hilbert Spaces One verifies immediately that (x, y) H (xIY) is a scalar product on en, called the canonical scalar product. 8.4.3. Example. Let E be the complex vector space whose elements are the sequences (Ab ,1,2, ... ) of complex numbers that are zero from some index onward. For set (this sum involves only a finite number of nonzero terms). One verifies immediately that (x, y) H (x Iy) is a scalar product on E. 8.4.4. Example. More generally, let I be a set. Let E be the complex vector space whose elements are the families (AJiEI of complex numbers such that Ai = 0 for almost all i E I (cf. 3.3.1). For x = (AJiEI E E and Y = (flJiEI E E, set (xly) = I iEI Iifli' This is a scalar product on E. If I = {1, 2, ... , n}, one recovers Example 8.4.2. If I = {l, 2, 3, ... }, one recovers Example 8.4.3. 8.4.5. Example. Let E be the set of continuous complex-valued functions on [0, 1]. For g, hE E, set (g Ih) = f g(t)h(t) dt. This is a scalar product on E. 8.4.6. Example. Let X be the vector space of all sequences (..1.1' A2, ... ) of complex numbers. We denote by l~, or simply [2, the set of sequences (AI' A2,"') E X such that I:~l 1..1..12 < + 00. Let us show that this is a linear subspace o/X. It is clear that if s E [2 and A E C, then AS E P. Let Recall that if et, pEe then let + PI 2 + let - PI 2 = (a + m(et + f3) + (a - /3)(et = 2aet + 2/3P = 21etl 2 + 21P1 2 ; P) 100 VIII. Normed spaces therefore 10( + tW ~ 210(1 2 + 21/W. This established, we have <Xl <Xl L liln + Ilnl 2 ~ n=L1(21il 12 + 211l.1 2 ) n= 1 n <Xl '= + 2L IIlol 2 < 10(1 2 + I/W - 211X11PI, therefore thus s + tE 12. Moreover, 0 ~ (Iexl - IPI)2 00 '= 2 L lilnllll.1 ~ .=1 so that the series L:'= 1 00 2 L lilnl2 + <t), <Xl L (lilnl2 + 11l.1 2) < +<Xl, .=1 l.lln is absolutely convergent. Set (sit) '= ll1l1 + I21l2 + .... We obtain in this way a scalar product on 12. 8.4.7. Let E be a pre-Hilbert space, E' a linear subspace of E. The scalar product ofE, restricted to E', is a scalar product on E'. Thus E' automatically becomes a pre-Hilbert space. For example, the pre-Hilbert space of 8.4.3 is a pre-Hilbert subspace of /2. 8.4.8. Let E be a pre-Hilbert space, and x, y E E. We say that x, yare orthogonal if (xly) '= O. This relation between x and y is symmetric. We say that subsets M, N of E are orthogonal if every element of M is orthogonal to every element of N. If M is orthogonal to N, then every linear combination of elements of M is orthogonal to every linear combination of elements of N. 8.4.9. Let M c E. The set of elements of E orthogonal to M is a linear subspace of E that is denoted M 1. and is called, through misusf: of language, the linear subspace of E orthogonal to M. 8.4.10. Theorem (Cauchy-Schwarz Inequality). Let E be a pre-Hilbert space. For all x, y E E, l(xlyW ~ (x Ix)(y Iy)· For all il E C, we have o ~ (x + ilylx + ily) (1) = lil(yly) + l(ylx) + il(xly) + (xix). Multiply through by (Yly); after calculation, we obtain (2) 0 ~ [I(YIY) + (x Iy)] [il(YIY) + (ylx)] + (xlx)(yly)'- Suppose first that (Yly) of; O. We can then set il '= -(ylx)/(yly) (xly)(ylx). 101 8.5. Separated Pre-Hilbert Spaces in (2). obtaining o ~ (xlx)(yly) - (x Iy)(y Ix), which is the inequality of the theorem. If (x Ix) #- 0, we need only interchange the roles of x and y in the preceding. Finally, if (xix) = (yly) = 0 then (1) reduces to ° ~ I(ylx) (3) + A(X Iy). Taking A = -(Ylx) in (3), we get ° ~ -(xly)(xly) - (xly)(xly) = -21(xly)1 2 , therefore (x Iy) = 0 and the inequality of the theorem is again verified. 8.5. Separated Pre-Hilbert Spaces 8.5.1. Theorem. Let E be a pre-Hilbert space, and x an element of E. The following conditions are equivalent: (i) x E E.L (in other words, (x Iy) = 0 for all y E E); (ii) (xix) = o. (i) => (ii). This is obvious. (ii)=> (i). Suppose (xix) =0. By 8.4.10, for every YEE we have I(x Iy) 12 = 0, therefore (x Iy) = O. 8.5.2. The linear subspace E.L of E is called the kernel of the scalar product. If E.L = 0, we say that the scalar product is non-degenerate, or that the preHilbert space is separated (cf. 8.5.8). By 8.5.1, this amounts to saying that (x Ix) > 0 for every nonzero vector x of E. 8.5.3. One verifies easily that the pre-Hilbert spaces defined in 8.4.2-8.4.6 are separated. However, if E is a nonzero complex vector space and if one sets (x Iy) = 0 for all x, y E E, then the pre-Hilbert space so obtained is not separated. 8.5.4. Let E be a pre-Hilbert space, E.L its kernel, E' the complex vector space E/E.L. We are going to define a scalar product on E'. Let x', y' EE'. Choose representatives x, y of x', y' in E. Then the number (x Iy) depends only on x', y' and not on the choice of representatives. For, any other representatives are of the form Xl = X + U, Yl = Y + v with u, v E E.L, whence (xII}'!) = (xly) + (xlv) + (uly) + (ulv) = (xly) + 0 + 0 + = (xly). ° 102 VIII. Normed Spaces We can therefore define (x'li) = (xly). The fact that (x,y)~(xly) is a scalar product on E implies easily that (x', i) I-> (x' Iy') is a scalar product on E'. Thus E' is a pre-Hilbert space. Let us show that E' is separated. Let x' be a nonzero element of E'. Let x be a representative of x' in E. Then x ¢ El. (otherwise, x' = 0), therefore (x Ix) > 0 and consequently (x' Ix') > O. We say that E' is the separated pre-Hilbert space associated with E. This construction reduces most problems about pre-Hilbert spaces to problems about separated pre-Hilbert spaces. 8.5.5. Theorem. Let E be a pre-Hilbert space. For every x E E, set Ilxll = JCxlx), Then x~ Ilxll is a seminorm on E. For this seminorm to be a norm, it is necessary and sufficient that the pre-Hilbert space be separated. If x, Y E E and), E C, then IIhl1 2 = (hlh) = I).(xlx) = 1).1 2 1IxI1 2 , Ilx + yf = (x + ylx + y) = (xix) + (yly) + 2 Re(xly) ~ Ilxlll + Ilylll + 21(xly)1 ~ Ilxlll + Ilylll + 211xlillyll by 8.4.10 = (11xll + Ilyll)2, therefore x I-> Ilxll is a semi norm. For this seminorm to be a norm, it is necessary and sufficient that x =f. 0 =:> Ilxll > 0, in other words that x =f. 0 =:> (xix) > O. 8.5.6. Theorem. Let E be a pre-Hilbert space and x, y E E. Then: (i) Ilx + yl12 + Ilx - ylll = 21!x112 + 211yl12 (parallelogram law), and (ii) 4(xly) = Ilx + yl12 - II-x + yliZ + illix + ylll - ill-ix + ylll (polarization identity). For Ilx + Ylll + Ilx - yf = = = (x + ylx + y) + (x - y x - y) (xix) + (xly) + (ylx) + (yly) + (xix) - (xly) - (ylx) + (yly) 2(xlx) + 2(yly), 103 8.5. Separated Pre-Hilbert Spaces and + ylx + y) = -( -x + Yl-x + y) = i(ix + ylix + y) = -i( -ix + Yl-ix + y) = (x + (xly) -(xix) + (xly) i(xlx) + (xly) -i(xlx) + (xly) (xix) + (ylx) + (yly), + (ylx) - (yly), - (ylx) + i(yly), - (ylx) - i(yly)· 8.5.7. Theorem (Pythagoras). Let E be a pre-Hilbert space, and pairwise orthogonal elements ofE. Then Xl' ... , xn For, 8.5.8. Let E be a separated pre-Hilbert space. By 8.5.5, E has a norm x ~ II x II. Therefore, by 8.1.6, E has a metric d(x, y) = Ilx - yll = (x - ylx - y)I/2, hence a topology. This topology is separated (1.6.2(a», which in some measure justifies the expression 'separated pre-Hilbert space'. 8.5.9. Examples. If en is equipped with the canonical scalar product, one recovers the norm and metric already considered. In (2, one has d«A.lo A2'" .), (;iI' J12," .» = (IA1 - J111 2 + IA2 - J121 2 + .. -)1/2. In the pre-Hilbert space ~([O, 1], C) of 8.4.5, one has dU, g) = (1 )1/2 ( J If(t) - g(t)12 dt ; o the corresponding topology is called the topology of convergence in mean square. If a sequence Un) tends to f for this topology, one says that Un) tends to f in mean square. 8.5.10. Theorem. Let E be a separated pre-Hilbert space. The mapping (x, y) ~ (x Iy) of E x E into C is continuous. The mapping x t-> Ilxll = d(x, 0) is continuous (5.1.1). Moreover, the mappings (x, y) ~ x + y, (A, x) t-> AX are continuous (8.1.8). The theorem then follows from the polarization identity (8.5.6). 104 VIII. Normed Spaces 8.6. Banach Spaces, Hilbert Spaces 8.6.1. Definition. A normed space that is complete (as a metric space) is called a Banach space. A complete, separated pre-Hilbert space is called a Hilbert space. 8.6.2. Example. Let E be a finite-dimensional normed space. Then E is a Banach space. For, one can suppose that E = R n or en. The given norm is equivalent to the usual norm of Rn or en (8.3.4). Every Cauchy sequence for the given norm is therefore a Cauchy sequence for the usual norm, hence has a limit (5.5.9). Thus E is complete. In particular, a finite-dimensional separated pre-Hilbert space (for example, en equipped with the canonical scalar product) is a Hilbert space. 8.6.3. Example. Let X be a set, E the vector space of bounded real-valued functions on X, equipped with the norm of uniform convergence. Then E is a Banach space. For, ff(X, R) is complete for the metric of uniform convergence (6.1.6). If a sequence (fn) of functions in E tends uniformly to a function f E ff(X, R), then SUPxeX Ifn(x) - f(x) I :::;; 1 for n sufficiently large. therefore f is bounded. Thus E is closed in ff(X, R) hence is complete. In particular, /00 is a Banach space. 8.6.4. Example. Let us show that 12 is complete (hence is a Hilbert space). For n = 1, 2, ... , let Xn E /2. Then Xn = (Anl, An2 , An3,".) with Ir;l IAnil2 < + 00. Suppose that Ilxp - xqll --+ 0 as p, q --+ 00, and let us show that (xn) tends to an element of [2. We have Ir; 1 IApi - Aq;/2 --+ 0 as p, q --+ 00. A fortiori, for every fixed positive integer i, Api - Aqi --+ 0 as p, q --+ 00, therefore (Ani) tends to a complex number Ai as n --+ 00. Let e > O. There exists a positive integer N such that 1 =I 00 p. q 2 N i= 1 IApi 1 - Aqi 12 :::;; e. Let A be a positive integer, provisionally fixed. We have A p, q 2 N = L IApi - Aq;]2 :::;; e. i=l Fix p 2 N and let q --+ 00. We obtain: A P2 N = L IApi i= 1 A;l2 :::;; e. 105 8.6. Banach Spaces, Hilbert Spaces This inequality being true for every positive integer A, we deduce that <Xl (1) P~ N => L IApi - i= 1 Ad 2 ~ e. This proves, first of all, that for p ~ N the sequence (Api - Ai) belongs to [2. Since Ai = Api - (Api - Ai), we deduce that (Ai) is an element x of [2. The relation (1) can now be written Thus xp --+ x as p --+ 00. 8.6.5. Example. An almost identical proof shows that [I is a Banach space. 8.6.6. Example. The separated pre-Hilbert space of 8.4.3 is not complete (cf. 8.7.1). 8.6.7. Theorem. (i) A closed linear subspace of a Banach space is a Banach space. (ii) A finite product of Banach spaces is a Banach space. Assertion (i) follows from 5.5.6, assertion (ii) from 5.5.8. 8.6.8. Theorem. Let E be a normed space, F a Banach space. Then 2(E, F) is a Banach space. Let Un) be a Cauchy sequence in 2(E, F). Let x Ilf",(x) - fn(x)1I ~ IIf", - E f..llllxll E. We have 0 --+ as m, n --+ 00. Since F is complete, there exists an element f(x) of F such that J.(x) --+ f(x) as n --+ 00. We have thus defined a mapping f of E into F. The equality fn(x + y) = J.(x) + fn(y), valid for all n, yields f(x + y) = f(x) + f(y) in the limit; similarly f(AX) = Af(x) for every scalar A. Thus f is linear. Let e > O. There exists an N such that m, n ~ N => IIfm - fnll ~ e, that is, IIfm(x) - fn(x)1I ~ e for all x E E such that IIxll ~ 1. Fixing x and m, and letting n tend to infinity, we obtain Ilfm(x) - f(x)II ~ e for IIxil ~ 1. From this, we deduce first of all that IIf(x)II ~ e + IIfm(x)II ~ e + IIf",1I for therefore f E 2(E, F). Moreover, we see that" fm m ~ N. Thus (I.) tends to fin 2(E, F). IIxll f II ~ ~ 1, e, and this for all 106 VIII. Normed Spaces 8.6.9. Corollary. Let E be a normed space. Its dual is a Banach space. For, the dual is either 2'(E, R) or 2'(E, C), and R, C are complete. *~ 8.6.10. Theorem (Banach-Steinhaus). Let E be a Banach space, F a normed space, (U;);EI a family of continuous linear mappings of E into F. The following conditions are equivalent: (i) SUP;EI IluJ < + 00; (ii) for each x E E, SUP;EI Ilu;xll < + w. = (i) (ii). This is immediate since Ilu;xll ::;; Ilu;llllxll. Suppose that condition (ii) is satisfied. The functions x 1-+ Ilu;xll on E are continuous, and their upper envelope is finite. By 7.4.15, there exist a closed ball B in E with center a and radius p > 0, and a constant M > 0, such that Ilu;xll::;; M for all x E Band i E I. Since Ilu;(x - a)11 ::;; Ilu;xll + Ilu;all, there exists a constant M' > such that Ilu;yll ::;; M' for Ilyll ::;; p and for all i E I. Then, for Ilyll ::;; 1, we have ° therefore * Iluill ::;; p-iM', and this for every i E I. 8.6.11. Corollary. Let E be a Banach space, F a normed space, and (Ui' U2,"') a sequence of continuous linear mappings of E into F such that, for every x E E, UnX has a limit ux in F. Then u is a continuous linear mapping of E into F. The fact that u is linear is immediate. By 8.6.10, there exists a finite constant M such that Ilunll ::;; M for all n. For every x E E, we have Ilunxll ::;; Mllxll for all n, whence Iluxll ::;; Mllxll on passage to the limit. Therefore u is continuous. 8.7. Linear Subspaces of a Normed Space 8.7.1. Let X be a normed space, E a linear subspace of X. If E is finitedimensional, then E is closed in X (8.6.2 and 5.5.7). In particular, in a finitedimensional normed space, all linear subspaces are closed. This is not so in general. For example, consider again the example E of 8.4.3, which is a linear subspace of [2. Let us show that E is dense in [2 (hence not closed, since it is distinct from [2). Let (An) E [2 and e > O. There exists a 107 8.7. Linear Subspaces of a Normed Space positive integer N such that Ln;>N 1An 12 ~ 8. Let (J.1n) be the element of E such that J.1n = An for n < N, J.1n = 0 for n 2. N. Then 00 II(An) - (J.1n)112 = I n=1 IAn - J.1n1 2 = I 1,1..1 2 ~ 8, n2:N which proves our assertion. In particular, E is not complete. 8.7.2. Theorem. Let E be a normed space, F a linear subspace of E. Then the closure F of F in E is a linear subspace ofE. Let x, Y E F. There exist sequences (x n), (Yn) in F such that Xn -> x, Yn -> y. Then Xn + Yn E F and Xn + Yn -> X + Y (8.1.8), therefore x + Y E F. If A is a scalar, then Ax n -> AX and AX n E F, therefore AX E F. 8.7.3. Theorem. Let E be a normed space, and AcE. Let B be the set of linear combinations of elements of A and let C = B. Then C is the smallest closed linear suhspace of E containing A. One knows that B is a linear subspace ofE, therefore C is a linear subspace of E (8.7.2), and it is clear that C ::;) A. If C' is a closed linear subspace of E containing A, then B c C' since C' is a linear subspace, therefore C c C' since C' is closed. 8.7.4. Definition. With the notations of 8.7.3, we say that C is the closed linear subspace of E generated by A. If C = E, we say that A is total in E. 8.7.5. Theorem. Let E, F be normed spaces, and u E 2(E, F). Then the kernel ofu is a closed linear subspace ofE. For, the kernel is u- 1 (O), and it suffices to apply 2.4.4. 8.7.6. However, the range of u is in general not closed in F. 8.7.7. Theorem. Let E be a normed space, E' a dense linear subspace of E, F a Banach space, and u' E 2(E', F). There exists one and only one u E 2(E, F) that extends u'. One has Ilull = Ilu'll. The uniqueness of u follows from 3.2.15. By 8.2.4, u' is uniformly continuous. By 5.5.13, there exists a continuous mapping u of E into F that extends u'. If x, Y E E, there exist sequences (x n), (Yn) in E' such that Xn -> x, Yn -> y. Then u'(x n + Yn) = u'x n + u'Yn, 108 VIII. Normed Spaces which gives u(x + y) = ux + uy in the limit. One sees similarly that u(b) = AuX for every scalar A.. Thus u is linear. Since u extends u', it is clear that Ilull ~ Ilu'li. On the other hand, the inequality Iluxll :S lIu'llllxll is true for every x E E', hence remains true for every x E E by passage to the limit, therefore Ilull :S Ilu'll. 8.7.8. Theorem. Let E be a separated pre-Hilbert space, and AcE. Then A1- is closed linear subspace of E. We already know that A1- is a linear subspace of E. On the other hand, for every x E E let F x be the set of y such that (x Iy) = O. Then F x is closed F x , therefore A1- is closed. by 2.4.4 and 8.5.10. Now, A1- = nxeA 8.8. Riesz's Theorem •• 8.8.1. Theorem (Riesz). Let E be a separated pre-Hilbert space, F a complete linear subspace ofE, x E E, and <5 the distance from x to F. (i) There exists one and only one y E F such that Ilx (ii) y is the only element of F such that x - y E F1-. (a) There exists a sequence (Yn) in F such that There exists a positive integer N such that n~N For m, n ~ => IIx - Ynl1 2 :s «5 2 - yll = <5. Ilx - Ynll --+ «5. Let e > O. + e. N, we then have 211x - Yml1 2 + 211x - Ynl1 2 :s 4<5 2 + 4e. Applying the parallelogram law (8.5.6) to the left side, we obtain 112x - Ym - Ynl1 2 + IIYm - Ynl1 2 :s 4<5 2 + 4e, or IIYm - Ynl1 2 :s 4<5 2 + 4e - 411x - Ym ; Y. Now, t(Ym + Yn) E F, therefore so that IIYm - Ynl1 2 :s 4e. r 109 8.8. Riesz's Theorem Since F is complete, the sequence (Yo) tends to an element y of F. Since Ilx - y.1I -+ IIx - yll, we have IIx - yll = ~. (b) Let z E F. For all A. E R, we have IIx - ylll ~ IIx - (y + .l.z)1I 2 = IIx - yl12 + .l.211z112 - 2 Re(x - yl.l.z) x whence o ~ .1. 211z112 - 2.1. Re(x - ylz). This requires that Re(x - ylz) = O. Replacing z by iz, we see that = O. In other words, x - y E Fl.. (c) Let y' be an element of F distinct from y. Then x - y is orthogonal to y - y', since y - y' E F. By the theorem of Pythagoras (8.5.7), we have (x - ylz) IIx - y'II 2 = IIx _ ylll + lIy _ y'1I 1 > IIx _ Yil l = ~l. This proves the uniqueness assertion in (i). Moreover, x - y' is not orthogonal to F, since (x - y'ly - y') = (x = - yly - y') II y - y'II 1 + (y - y'ly - y') > O. This proves the uniqueness assertion in (ii). * 8.8.2. With the notations of 8.8.1, we say that y is the orthogonal projection ofx on F. *. 8.8.3. Theorem. Let E be a Hilbert space. (i) IfF is a closed linear subspace olE, then (pl.)l. = P. (ii) Moregenera/ly, ifAcE then (Al.)l. is the closed linear subspace generated by A. (i) It is clear that F c (Fl.)l.. Let x be an element of E that does not belong to F. Let y be its orthogonal projection on F. Then x - y E Fl., therefore (ylx - y) = 0, consequently (xix - y) Thus x ~ (Fl.)l.. = (x - ylx - y) > 0 because x#- y. 110 VIII. Normed Spaces (ii) Let B be the set of linear combinations of elements of A and let C = B (cf. 8.7.3 and 8.7.4). It is clear that B.L = A.L. We have C.L = B.L on account of 8.5.10. Therefore (A.L).L = (C.L).L. But (C.L).L == C by (i), thus (A.L).L = C. *~ 8.8.4. Theorem. Let E be a Hilbert space, and AcE. The following conditions are equivalent: (i) A is total in E; (ii) 0 is the only element ofE orthogonal to A. Set B = A.L. To say that A is total means, by 8.8.3(ii), that (A.L).L = E, in other words that B.L = E. But the latter condition is equivalent to B = {O} since E is a separated pre-Hilbert space. *~ 8.8.5. Theorem. Let E be a Hilbert space. (i) For every x E E, the mapping y 1--+ (x Iy) of E into C is a continuous linear formfx on E. (ii) The mapping x 1--+ fx of E into E' is bijective and conjugate-linear. One has Ilfxll = Ilxil for all x E E. It is clear that fx is a linear form on E. We have Ifx(y) I = l(xly)1 s IlxlIIIYII. therefore fx is continuous and ilfxll s Ilxll. Let us show that Ilfxll = Ilxll for all x E E. This is obvious if x = O. Suppose x -# 0; then Ilxllz = fix) s Ilfxllllxll, thus Ilxll s IIIxl1 after cancelling Ilxll. Let XI' Xz E E and AI, Az E C. For all Y E E, fAIXI+A2X2(Y) = = = = (AIXI + A2 x 21y) II(Xlly) + IZ(Xlly) Idxl(Y) + I 1 Ix2(y) (XdXI + Izfx')(y), thus fAIXI +A2X2 = ).Ifxl + I 2f x2' In other words, the mapping x 1--+ fx of E into E' is conjugate-linear. In view of the equality Ilfxll = Ilxll, it is injective. Let us show that it is surjective. Let fEE' and let us prove that there exists an x E E such that F = fx. This is obvious if f = O. Assume f -# O. Then the kernel F of I is a closed linear subspace of E distinct from E. Therefore F.L -# 0 (8.8.3). Choose a nonzero element t of F.L. Then t ¢: F (otherwise (t It) = 0). thus f(t) -# O. Multiplying t by a suitable scalar, we can suppose thatf(t) = 1. Let x = Iltll-1t.ForallYEE, fey - I(y)t) = fey) - f(y)f(t) = 0, III S.S. Riesz's Theorem thus y - f(y)t E F. Consequently fx(y) = (xly) = (xlf(y)t) = f(y)(xlt) = f(Y)lltll- 2 (tlt) = f(y). 8.8.6. Let E be a separated pre-Hilbert space. One calls orthonormal family in E a family (e;)i~1 of elements of E such that lIeili = 1 for all i E I and (ede) = for i E I,j E J, i =F j. Such a family is linearly independent, because if ° for scalars Ait' .... Ain ' then the Pythagorean theorem implies that 0= Ilk e· + ... + k whence Ail = ... = Ai = 0. '1 ~1 In e·In 112 = I).. 12 + ... + I).. 12 11 In , n If E has finite dimension p, the cardinal number of I is therefore at most equal to p. However, in general there exist infinite orthonormal families. For example in 12 , let en = (0,0, ... ,0, 1,0,0, ... ), where the 1 appears in the n'th place. Then the sequence (e!. e2."') is orthonormal. 8.8.7. Let E be a separated pre-Hilbert space. One calls orthonormal basis of E a total orthonormal family in E, that is (8.7.4), an orthonormal family (ei)i~1 such that the linear combinations of the ei are dense in E. Suppose E is finite-dimensional. By 8.7.1, an orthonormal basis of E is an orthonormal family that is a basis in the algebraic sense. For example. in e" equipped with the canonical scalar product, the canonical basis is an orthonormal basis. In general, an orthonormal basis is not a basis in the algebraic sense (but there is almost never any risk of confusion). For example in P, the sequence (en) of 8.8.6 is an orthonormal basis by 8.7.l. .. 8.8.8. Theorem. Consider E = ~([o, 1], C) as a separated pre-Hilbert space with the scalar product (f, g) 1-+ g f(t)g(t) dt (8.4.5). For every nEZ, let en be the function t 1-+ e 2"in! on [0, 1], which is an element of E. Then the family (en)nez is an orthonormal basis ofE. g (a) We have (em Iell) = fb e- 2"im'e 2 "in, dt = e 2"i(n-m), dt. H n = m, the value of the integral is 1. If n =F m, then the function t 1-+ e 2 "i(n - m), admits the primitive e 2 "i(n-m)I/2ni(n - m), which takes on the same value at t = and t = I, therefore the integral is O. Thus, the family (en)nez is orthonormal. (b) Let E' be the set offE E such thatf(O) = f(I). HfE E' thenfmay be extended in a unique way to a function 9 of period 1 on R, and 9 is continuous. By 7.5.6, there exists a sequence (fp) of trigonometric polynomials that tends to 9 uniformly on R. Then fA If(t) - fitW dt --+ 0, thus (fp) tends to f in the pre-Hilbert space E. Now,fp is a finite linear combination of the en' Thus, if we denote by A the linear subspace of E generated by the en, then A=> E'. ° 112 VIII. Normed Spaces (c) Let hE E. For n = 1,2, 3, ... let h n be the function that coincides with h on [lin, 1J, such that hn(O) = hn(l), and which is linear on [0, lin]. The Ihn I are bounded above by a fixed constant M (for example, M = sUPre[O,lllh(t)I). Then d(h, hn )2 = = r Ih(t) - hn(t) 12 dt 1 Ih(t) - hn(tW dt :::; - (2M)", o n i l~ therefore hn tends to h in E. Now, hn E E'. Therefore F = E. Then = E, whence A = E, and the family (en) is an orthonormal basis. *. A ::J P 8.8.9. Theorem. Every Hilbert space has an orthonormal basis. Let E be a Hilbert space. Consider the orthonormal subsets of E. They form a set flJ ordered by inclusion. Let (P;) be a totally ordered family of elements of [lJ. Each PAis an orthonormal subset of E. Moreover, for any A and fl, either P;, ::J PI' or PI' ::J P A; it is then clear that th€: union of the p;. is an orthonormal set, containing all the p;.. By Zorn's theorem, there exists a maximal orthonormal subset P of E. If P is not total in E, then there exists a nonzero x in E orthogonal to P (8.8.4); replacing x by x/llxll, one can suppose Ilxll = 1. Then P u {x} is an orthonormal subset of E, which contradicts the maximality of P. Thus P is total in E, hence is an orthonormal basis. CHAPTER IX Infinite Sums The student already knows the definition of a convergent series Xl + + ... of real numbers, and the definition of the sum of such a series. In this chapter, we generalize in two different directions: (1) Instead of the Xi being real numbers, we take them to be vectors in a normed space. This is a fairly superficial generalization (though useful in certain contexts-see §5). (2) Instead of the Xi being indexed by the integers 1,2, ... , we assume that the set of indices is arbitrary. A lot of very concrete questions lead in fact to the case where the set of indices is N 2 ('double series'), or NP ('p-fold series'), or Z, etc., and the best thing is to study at one stroke the general situation. We shall then see why we spoke of 'limit along a filtering set' in 7.2.3. Although the case of series is a good point of reference. it is well to be prudent: for example, compare Theorem 9.4.6 with the well-known fact that a convergent series is not always absolutely convergent. X2 9.1. Summable Families 9.1.1. Definition. Let E be a normed space, (X;)iEI a family of elements of E. Let A be the set of finite subsets of I; it is an increasingly filtering ordered set, thus one can speak of limit along A (7.2.3). For every J e A, let Let seE. 114 IX. Infinite Sums The family (X;}'EI is said to be summable, with sum s, if the family (Sj)jEA tends to s along A. We then write s = I'E! x,. In other words, (X;}'E! is summable with sum s if, for every f, > 0, there exists a finite subset Jo ofI such that, for every finite subset J of! containing Jo , one has III'Ej x, - sll ~ c. 9.1.2. Theorem. Let (X,)iEI. (Y;)ieI be summable families in E, with sums s, t. Then the family (Xi + J'i)iE! is summable with sum s + t. For J E A, let Sj = Iie! Xi, tj = IiEj Yi' Then, along A, Sj tends to sand tends to t, therefore (Sj, t j ) E E x E tends to (s, t) E E x E, therefore Sj + tends to s + t. Now, Sj + t J = Iid (Xi + y;). tj tj 9.1.3. Similarly, if A is a scalar, then the family (AX;}iEI is summable with sum AS. We thus have the formulas 9.1.4. Theorem. Let (X;}iEI be a summable illfinite family ()f elements of E. Then Xi tends to 0 along the filter of complements offinite subsets of 1. Let s = IiE' Xi' Let c > O. There exists J J' E A, J' ::J E A such that J = lis)' - sll ~ e 2 Then . IE I - J = IlsJuli} - sll e s2 and 9.1.5. Example. If(x t • x z ,"') is a summable sequence of elements ofE, then the sequence (x") tends to 0 as n ..... 00. It is well known, from the example E = R, that the converse is not true. ~ 9.1.6. Theorem (Cauchy's Criterion). Let E be a Banach space, (X;)iEI a family of elements of E. The following conditions are equivalent: (i) the family (Xi)iE! is summable; (ii) for every e > 0, there exists a finite subset J o of I such that, for every finite subset K ofl disjointfi-om J o , one has IlsKi! s e. 115 9.2. Associativity, Commutativity Suppose (Xi)ieI is summable with sum s. Let e > O. There exists J o E A such that, if J E A and J ::::l J o , then IlsJ - sll ~ e/2. Let K E A with K ('\ J o = 0. Then IlsJo - sll e ~"2 and therefore I!sKi! = IlsJouK - sJol1 ~ e. Suppose that the condition (ii) is satisfied. Let e > O. There exists J o E A such that, for K E A satisfying K ('\ J o = 0, one has IlsKi! ~ e/2. If J, J' E A and J, l' ::::l J o , then therefore IlsJ - SJ' 1/ ~ e. Thus. the set of SJ, for J E A and J ~ e. Therefore (SJ)JeA has a limit along A (5.5.11). ::::l J o , has diameter 9.1.7. DefinitioD. Let E be a normed space, (Xi)ieI a family of elements of E. The family (Xi)ieI is said to be absolutely summable if the family (lixdl)ieI is summable in R. ~ 9.1.8. Theorem. Let E be a Banach space, (Xi)ieI an absolutely summable family of elements of E. Then (Xi)ieI is summable. Let e > O. There exists J o E A such that therefore (Xi)ieI is summable by 9.1.6. 9.2. Associativity, Commutativity 9.2.1. Theorem. Let E be a Banach space, (X;)ieI a summablefamily ofelements ofE. Let J c I. Then (X;)ieJ is summable. Let e > O. There exists a finite subset J o of I such that, if K is a finite subset ofI disjoint from J o , then IlsKI! ~ e. Then J ('\ J o is a finite subset of J, and if K' is a finite subset of J disjoint from J II J o then IISK,II ~ e. Cauchy's criterion applied to (X;)ieJ proves that this family is summable. 9.2.2. Theorem (Associativity). Let E be a Banach space, (Xi)ieI a summable family in E. (I/)/eL a partition ofl. For every IE L, set y/ = LieI, Xi, which is meaningful by 9.2.1. Then thefamily (Y/)leL is summable and Lie) Xi = LleL Yt~ 116 IX. Infinite Sums Set s = Liel Xi' Let e > 0. There exists a finite subset 10 ofI such that, if 1 is a finite subset of I containing 10 , then IlsJ - sll :::; e/2. Let Mo c L be the set of IE L such that II intersects 10 , This is a finite subset ofL. Let M be a finite subset ofLcontaining Mo. We are going to show that (1) which will establish the theorem. Let n be the number of elements of M. For each IE L, there exists a finite subset I; of II such that Ily, - sl;l1 :::; F./2n, and we can require that I; contain 10 n II by enlarging it if necessary. The union of the Ii, as I runs over M, is a finite subset 1 of I, and 1 ~ 10 , We have IlsJ - sll :::; e/2, that is, (2) Now, Ily, - sd :::; e/2n for every IE M, therefore (3) L YI Il leM - L SI;II :::; e12. leM The inequality (1) follows from (2) and (3). 9.2.3. Example. Let it is summable, then (Xmn)m,n= 1,2, ... be a double sequence or real numbers. If m'~IXmn = m~1 (~/mn) n~1 C~IXmn). 9.2.4. Let E be a Banach space, (X;)iel a family of elements of E, (I,)'eL a partition of!. Suppose that each su bfamily (Xi)i el, is summable with sum YI, and that the family (YI)leL is summable. This does not imply that the family (X')iel is summable. For example, take E = R and consider the sequence of real numbers (1, -1,2, - 2, ... , n, - n, ... ). Each subfamily (n, - n) is summable with sum 0, and the family (0.0,0, ... ) is summable, yet the original sequence is not summable (for example by 9.1.5). 9.2.5. However, one has the following result: Theorem. Let E be a normed vector space, (X;)iel a family of elements of E, (II)' e L a partition of I with L finite. Assume that each subfamily (X;)iel, is summable with sum YI' Then (X;}'el is summable with sum LleL YI' 117 9.3. Series Let e > 0. Let n be the number of elements ofL. For every 1E L, there exists a finite subset J1 of I, such that, if l' is a finite subset of I, containing J/ then IlsJ' - YIII :$; eln. Let J = UIEL J It which is a finite subset of I. If J' is a finite subset of I containing J, then l' is the disjoint union of the J' ( l I" and J' ( l I, is a finite subset of I, containing J/. Therefore IIsJ'nl, - YIII :$; eln, and this for every 1E L. Since SJ' = L/EL sJ'nh' we have IIsJ' y,II :$; e, whence the theorem. L/EL 9.2.6. Theorem (Commutativity). Let E be a normed space, (Xi)i,,1 a summable family of elements of E with sum s. Let (J be a bijection of I onto I. Then the family (X<7(i»i,,1 is summable with sum s. Let e > 0. Let J o be as in 9.1.1. Then (J-I(J O) is a finite subset ofl. Let J be a finite subset ofIcontaining(J-l(J o)' Then (J(J) => J o, therefore II LiE<7(J) Xi - sll :$; e, that is, whence the theorem. 9.3. Series 9.3.1. Definition. Let E be a normed space, and (Xl' X 2 , ... ) a sequence of elements of E. We say that the series with general term Xn is convergent with sum s (where sEE) if Sft = L7~ 1 Xi tends to s as n tends to infinity. We then . WrIte s = "co £..,n: 1 X •• In other words, the series is convergent with sum s if, for every e > 0, there exists an N such that n ~ N => lis. - sll :$; e. The series with general term Xn is said to be absolutely convergent if the series with general term IIx.II is convergent. 9.3.2. By means of proofs analogous to those of 9.1, one establishes the following results: (a) If the series with general terms Xn and Yn are convergent, then the series with general terms Xn + Yn and AX. (A a scalar) are convergent, and co co 00 n=1 .:1 L (xn + Y.) = L Xn + L Y., n=1 00 00 .:1 n~l LAX. = AL X n• (b) If the series with general term Xn is convergent, then tends to infinity. (The converse is not true.) Xn tends to 0 as n 118 IX. Infinite Sums (c) (Cauchy's Criterion.) Let E be a Banach space, (Xl' X2, of elements of E. The following conditions are equivalent: .•. ) (i) the series with general term Xn a sequence is convergent; (ii) for every e > 0, there exists an N such that (d) Let E be a Banach space, (XI' X2"") a sequence of elements of E. If the series with general term Xn is absolutely convergent, then the series is convergent. (The converse is not true.) 9.3.3. Theorem. Let E be a normed space, (Xl' X2, .•. ) a sequence ofelements of E. If the sequence is summable with sum s, then the series with general term xn is convergent with sum s. Let E > O. There exists a finite subset J o of {1, 2, ... } such that, if J is a finite subset of {1, 2, ... } containing Jo. then lis - LiEJ xiii s: E. Let N be the largest of the integers in Jo . If n ~ N then Jo c {1, 2, ... , n}, therefore lis I xiii::; e. This proves that I Xi -+ S as n -+ 00. D= L?= 9.3.4. The converse of 9.3.3 is false, as will follow from the example in 9.3.5. 9.3.5. Let E be a normed space, (Xl' X2, ... ) a sequence of elements of E, (J a permutation of {1, 2, 3, ... }. If the sequence (Xl> Xl,"') is summable, then L:'= I Xn = Lnoo= I x,,(n) by 9.3.3 and 9.2.6. However, consider the series l-!+t-i+t-· .. , which one knows to be convergent with sum log 2. By a suitable permutation of the order of the terms, one obtains the series This series is easily seen to be convergent, with the same sum as ! - i + i - i + /0 - /2 + ... ; this sum is therefore !log 2. The sum has changed after rearrangement of the terms. In particular, the sequence (1, -!, t, -i, ... )is not summable (cf. also 9.4.6). 119 9.4. Sum mabIe Families of Real or Complex Numbers 9.4. Summable Families of Real or Complex Numbers • 9.4.1. Theorem. Let (XJiEI be afamity of real numbers z O. Consider the finite partial sums SJ = IiEJ Xi' where J E A (the set of all finite subsets of I). Let s = SUPJEA sJ E [0, + 00]. (i) If s < (ii) If s = + 00, then the family (X;)iEI is summable with sum s. + 00, then SJ tends to + 00 along A. The mapping J f-+ SJ of A into R is increasing, since the Xi are Z O. The theorem therefore follows from 7.2.4. 9.4.2. Thus, for a family (X;}iEI of numbers z 0, the symbol IiEI Xi is always meaningful: it is a finite number if the family (X;)iEI is summable, + 00 otherwIse. 9.4.3. Theorem. Let (Xi)iEI, (Y;)iel be two families of numbers that Xi ::;; yJor all i E I. Then IiEI Xi ::;; IiE'Y;. In particular, (Y;)iEI is summable, then the family (Xi)i EI is summable. z 0; assume if the family Let A be the set of finite subsets of I. If J E A, then IiEJ Xi ::;; IiEJ YiPassing to the limit along A, we obtain IiEI Xi ::;; IiEI Yi. 9.4.4. Theorem (Associativity). Let (Xi),el be afamily of numbers a partition ofl. Then I Xi = I iEI leL z 0, (I/)/EL (Ixi). iel1 If the family (X;)iEI is summable, this follows from 9.2.2. Otherwise, IiE' Xi = + 00. For every finite subset J of I, we have I Xi = I iEJ tEL (I ie],nJ : ; I (I leL ieh Xi) Xi) by 9.4.3. This being true for all J, we conclude that thus the equality of the theorem is again true. 120 IX. Infinite Sums 9.4.5. Theorem. Let (Xi)ie" (Yj)jeJ befamilies of numbers ~ O. Then 2: (i.j)eIxJ XiYj = (We make the convention that O· (2:iel Xi) (LieJ Yj). + 00 = 0.) By 9.4.4. we have L (i.j)elxJ Now. LieJ XiYi = Xi LieJ Yi' XiYj = L (L XiYj)' ieI ieJ Setting t = LeJ Yi' we thus have L =L (i.i)eIxJ XiYj iel Xit. It t < + 00. this is equal to t(Liel Xi), whence the theorem. If t = + 00, we distinguish two cases. If all of the Xi are zero, then tXi = 0 for all i, therefore Liel tX i = 0 = t Liel Xi' If one of the Xi is > 0, then tXi = + 00 for such an i, therefore Liel tXi = + 00; on the other hand, Liel Xi> 0, therefore t Liel Xi = +00 . • 9.4.6. Theorem. Let (Xi)ieI be a family of real or complex numbers. The following conditions are equivalent: (i) the family (Xi)iel is summable; (ii) the family (Xi)iel is absolutely summable. (ii) => (i). This follows from 9.1.8. (i) => (ii). Suppose that the family (Xi)i e[ is summable. If the Xi are all real, let 11 (resp. 12 ) be the set of i E I such that Xi ~ 0 (resp. Xi < 0). The families (Xi)iel, and (X;)iel, are summable (9.2.1). Therefore the families (IX;!)ie" and (IX;!)ie!, are summable (9.1.3). Therefore the family (IXiILI is summable (9.2.5). If the Xi are complex, then the family (Xi) is clearly summable. Since Re Xi = 1(Xi + X;), the family (Re x;) is summable (9.1.2). Similarly, the family (1m x;) is summable. Then the families (I Re Xi I), (11m Xi I) are summable by the first part of the proof, therefore so is the family (I Re x;! + 11m x;!). Finally, since Ix;! ~ IRe xd + 11m x;!, the family (I Xi I) is summable (9.4.3). 9.4.7. Series with terms ~ O. If Xl' X2' • .• are numbers ~ 0, then the sequence (L;'= 1 X;) is increasing, therefore has a limit in R. equal to its supremum s, which is denoted Lr; 1 Xi' This number is equal to Lie(I.2 ....} Xi' For, ifthe sequence (x;) is summable, this follows from 9.3.3. Otherwise, there exist arbitrarily large finite subsums LieJ Xi (9.4.1), therefore arbitrarily large sums L1'= 1 Xi' therefore 00 LXi i=1 = + 00 = LXi' i.(1,2 .... ) 121 9.4. Summable Families of Real or Complex Numbers 9.4.8. To sum up, if (Xi)i e I is a family of elements of a normed space E, one has the following diagrams: I arbitrary, E complete: absolute summability => summability. I arbitrary, E = R or C: absolute summability -¢> summability. I = {I, 2, ... }. E complete: absolute summability => summability ~ JJ absolute convergence => convergence. I = P, 2, ... }, E = R or C: absolute summability -¢> summability ~ JJ absolute convergence => convergence. 9.4.9. Let ex E R One knows that the number 1 1 1 2~ 3~ 4~ 1+-+-+-+ .. ·= I XEN-{O) x~ is finite if ex > 1 and infinite if ex ::::;; 1. This amounts to saying that 1 -<+oo-¢>ex>L I Ixl~ XEZ-{O) Here is an important generalization: Theorem. Let x ~ Ilxll be a norm on RP and let ex E R Then 1 -II II~ < + I X XEZP-{O) 00 -¢> ex > p. Since all norms on RP are equivalent, it suffices (in view of 9.4.3) to carry out the proof for the norm (x), x 2 , .... x p ) ~ sup(lx!l, Ix 2 1. . . · , IXpl)· On ZP - {OJ, this norm takes on the values 1,2,3, .... For n = 1,2,3, ... , let An be the set of x E ZP such that Ilxll = n. Then An = where Bni - n ::::;; x j ::::;; Bn! U Bn2 U··· u Bop, is the set of (x)' x 2 , . . . , x p ) E ZP such that Xi = ±n and n for j #- i. It is clear that Card(BnJ = 2(2n + 1)P- 1, therefore nP-l ::::;; Card An::::;; 2p(2n + 1)P-l ::::;; 2p(3n)p-l = 2· 3P - 1 • pnP - 1• 122 IX. Infinite Sums Consequently, By 9.4.3 and 9.4.4, we deduce from this that L L -II 1-II' < -~-p+! -1- < . p- ! . P L -'-p+l' -_ - 2 3 XEZP.-{O) X nEN-{O} n nEN-{O) n from which I L -- < XEZP-{O) Ilxll' 1 +00 <=> L ._p+! nEN-{O) n < +cx:.: <=> ex - p + 1 > 1 <=> ex > p. 9.4.10. The Hilbert Space /2(1). Let I be a set. We denote by J~(I), or simply 12(1), the set of families x = (XJiEI of complex numbers such that LiEI Ix;!2 < + 00. Let us show that this is a linear subspace of g;(I, C). It is clear that if x E [2(1) and A E C, then AX E 12(1). Let x = (Xi);.I E [2(1) and Y = (Y;)i E I E [2(1). Then L IXi + y;l2 :$; L (21xi12 + 21Yi1 2) leI = 2 L IXil2 + 2 L ly;l2 ieI < thus x by 9.4.3 ieI by 9.1.2, 9.1.3 ieI + x, + Y E J2(1). Next, 2 L Ix;lIYil ::; L (lxi l2 + iel ieI IYiI 2) < +00, so that the family (XiY;)iEI is absolutely summable, therefore sum mabie (9.1.8). Set (xIY) = LiEIXiYi' One verifies easily that (x,Y)H(xIY) is a scalar product on [2(1). Imitating 8.6.4 step by step, one sees that [2(1) is a Hilbert space. (In the proof of 8.6.4, consideration of the sum L~ 1 must be replaced by that of a sum LiEJ' where J is a finite subset of I.) If I = {I, 2,3, ... }. then [2(1) = /2. If I = {I, 2, ... , n}, then [2(1) = en equipped with the canonical scalar product. Let io E I. Consider the family eio = (Xi)iEI such that Xi = 0 for i"# i o• Xi = 1 for i = io . Then eio E /2(1). The family (eJiEI in /2(1) is orthonormal. As in 8.7.1, one verifies that the linear combinations of the ei are dense in 12 (1), thus (eJiEI is an orthonormal basis of f(I), called the canonical orthonormal basis. 123 9.5. Certain Summable Families in Hilbert Spaces 9.5. Certain Summable Families in Hilbert Spaces • 9.5.1. Theorem. Let E be a separated pre-Hilbert space, (X;)iel afamily of pairwise orthogonal elements of E. (i) If the family (Xi)iel is summable with sum s, then I IIxdl 2 < +00 ieI (ii) If Iie! IIxil12 < summable. + 00 and IIsll 2 = Ilixil12. ie:1 and E is a Hilbert space, then the family (X;)iel is Let A be the set of finite subsets of I. For every J E A. let SJ = IieJ Xi' If (X;)iel is summable with sum s, then SJ tends to s along A. therefore IIsJI1 2 tends to IIsl12 along A. Now, IIsJI1 2 = IieJ IIxill2 (8.5.7). therefore IIsll2 = IiEI II xill 2 • Suppose IiEl IIxdl 2 < + 00. For every 6> 0. there exists J E A such that, if K E A and K n J = 0. then IieK IIxdl 2 ~ e2 (9.1.6), that is, IlsK11 ~ e. If, moreover, E is a Hilbert space, this implies that (X;)iEI is summable (9.1.6). • 9.5.2. Theorem. Let E be a separated pre-Hilbert space, (e;)iel an orthonormal basis of E. (i) Let x E E, and set Ai = (e;l x). The family (A;e;)iel is summable in E, and x = IiE' Aiei' _ (ii) IfYE E and J.li = (elly), then thefamily (Ai J.li)i E[ is summable and (xIY) = Iie, IiJ.li· In particular, Ilxf = Iie,I A/12. Let 6 > O. There exist a finite subset J o of I and a linear combination x' of the ei for i E J o, such that Ilx - x'il ~ e. Let J be a finite subset of I such that J ::::> J o • and set For j E J, we have (ejlx - z) = (ejlx) - (ejl i~ Ai e) = Aj - Aj = 0, therefore x' - z (which is a linear combination of the ej for j gonal to x - z. Consequently IIx - zl12 whence IIx - zll sumx. ~ + liz - x'1I 2 = IIx - x'1I 2 ~ E J) is ortho- 6 2, e. This proves that the family (A.ie;)ie! is summable with 124 IX. Infinite Sums Since LiEJ Aiei and LiEJ Iliei tend to x and y along A, the number (LiEJ AiedLiEJ Ili e;) = LiEJ Ii Ili tends to (x Iy) along A. 9.5.3. Under the conditions of 9.5.2, we say that the Ai are the coordinates of x with respect to the orthonormal basis (ei)iE!' If E is finite-dimensional, we recover the usual concept of coordinates with respect to the basis (ei)iE!, since x = Lie! Aiei' ~ 9.5.4. Corollary. Let f tC([O, 1], C). For every k E Z, set E Ak = f f(t)e-21tikt dt Lk= (' Fourier coefficients of f'). Then -n Ake21tikt tends in mean square to f. as n tends to infinity. If also 9 E tC([O, 1], C) and Ilk = JA g(t)e-21tikl dt, then i l o f(t)g(t) dt = L AkJik' nEZ This follows from 8.8.8 and 9.5.2. ~ 9.5.5. Theorem. Let E be a Hilbert space, (ei)iE! an orthonormal basis of E. For every x = (Ili)iel E [2(1), let f(x) be the element Liel Iliei of E (which is defined, by 9.5.1(ii». Then the coordinates of f(x) with respect to (ei)ie! are the Ili. andfis an isomorphism of the Hilbert space [2(1) onto the Hilbert space E that transforms the canonical orthonormal basis of 12(1) into (e;}iEI' (a) Let A be the set of finite subsets of I. Let j E I. If J ':; A and J ~ {j}, then (ejlLiEJ Ilie;) = Ilj; since LieJ Iliei tends to f(x) along A, we see that (ejlf(x» is equal to Ilj. (b) It is clear that f is a linear mapping of 12(1) into E. For every Y E E, let g(y) be the family of numbers «eiIY»iel> which belongs to [2(1) by 9.5.2. Then 9 is a mapping of E into P(I), and f(g(y» = y by 9.5.2. On the other hand. 9 (f(x» = x for all x E /2(1) by (a). (c) Thus, f and 9 are linear bijections inverse to one another. By 9.5.2, 9 preserves the scalar product. Thus f is an isomorphism of the Hilbert space [2(1) onto the Hilbert space E. (d) It is clear that f transforms the canonical orthonormal basis of P(I) into (ei)iEI' *~ 9.5.6. Corollary. Every Hilbert space is isomorphic to a space [2(1). This follows from 8.8.9 and 9.5.5. CHAPTER X Connected Spaces This chapter, which is very easy, could have come before Chapter III (but there were so many questions urgently requiring study!). The problem is to distinguish, by various methods, those spaces that are 'in one piece' (for example a disc, or the complement of a disc in a plane) and those which are not (for example, the complement of a circle in a plane). 10.1. Connected Spaces 10.1.1. Theorem. Let E be a topological space. The following conditions are equivalent : (i) there exists a subset of E, distinct from E and 0, that is both open and closed; (ii) there exist two complementary non empty subsets of E both of which are open; (iii) there exist two complementary nonempty subsets of E both of which are closed. This is clear, since if A is a subset of E, we have A open and closed ¢> A and E - A open ¢> A and E A closed. 10.1.2. Definition. If a topological space E satisfies the conditions of 10.1.1, it is said to be disconnected. In the contrary case, it is said to be connected. 126 X. Connected Spaces 10.1.3. Theorem. R is connected. Let A be an open and closed subset of R. Assuming A and R - A nonempty, we are going to arrive at a contradiction. Let x E R - A. One of the sets A n [x, + 00), An (- 00, x] is nonempty. Suppose, for example, that B = A n [x, + 00) oF 0. Then B is closed. Since B = A n (x, + 00), B is also open. Since B is closed, nonempty and bounded below, it has a smallest element b (1.5.9). Since B is open, it contains an interval (b - e, b + e) with e > O. Thus b cannot be the smallest element of B. 10.1.4. However. R - {O} is not connected. For, ( - 00,0) and (0, + 00) are complementary nonempty open sets in R - {O}. 10.1.5. Definition. Let E be a topological space and AcE. We say that A is a connected subset of E if the topological space A is connected. 10.1.6. Theorem. Let E be a topological space, (Aj)ieJ a family of connected subsets ofE, A = UiEJ Ai' If the Ai intersect pairwise, then A is connected. Suppose A is not connected. There exist, in the topological space A, subsets E I, V I n Ai and V 2 n Ai are open in Ai and complementary in Ai' Since Ai is connected, VI n Ai = 0 or V 2 n Ai = 0. Let II (resp. 12 ) be the set of i E 1 such that Ai c V I (resp. Ai c V 2)' Then V I (resp. V 2 ) is the union of the Ai for i E 11 (resp. i E 12 ), therefore there exist an Ai and an Aj that are disjoint. contrary to hypothesis. VI' V 2 that are complementary, nonempty and open. For every i 10.1.7. Theorem. Let E be a topological space, A a connected subset of E, B a subset ofE such that A c B c A. Then B is connected. In particular, A is connected. Suppose B is the union of subsets V 1, V 2 that are disjoint and open in B. We are to prove that one of them is empty. There exist open sets 0'1> V 2 in E such that V I = B n V'!> V 2 = B n V 2. The sets A n V I and A n V 2 are open in A, disjoint, with union A. Since A is connected, we have for example A n V I = 0, therefore A n 0'1 = (A n B) n 0'1 = A n (B n 0'1) = A n V I = 0, in other words AcE - U'I' Since E - U'I is closed, we infer that AcE - U'I, whence B n 0'1 = 0, that is, VI = 0 . .. 10.1.S. Theorem. Let X, Y be topological spaces, f a continuous mapping of X into Y. If X is connected. then f(X) is connected. 127 10.2. Arcwise Connected Spaces If f(X) is not connected, there exist in f(X) sets U b U 2 that are open, complementary and nonempty. Then f -leU I), f- I (U 2 ) are open, complementary and nonempty in X, which is absurd . .. 10.1.9. Theorem. Let A c R. Thefollowing conditions are equivalent: (i) A is connected; (ii) A is an interval. We can suppose that A is nonempty and not reduced to a point. Let A be an interval. If A is open in R then A is homeomorphic to R (2.5.5), hence is connected (10.1.3). If A is an arbitrary interval, let I be its interior in R. Then I is an open interval in R hence is connected, and I c AcT, so that A is connected (l0.1.7). Let A be a connected subset of R and let us show that A is an interval. By means of the increasing homeomorphism x f--> tan x of ( - n/2, n/2) onto R, we are reduced to the case that A c ( - n/2, n/2). Then A admits a supremum bE R and an infimum a E R. We have A c [a, b]. We are going to show that A ::J (a, b); it will then follow that A is one of the four intervals (a, b), (a, b], [a, b), [a, b], and the proof will be complete. Arguing by contradiction, let us suppose that there exists an Xo such that a < Xo < band Xo ¢ A. Then A is the union of the sets A n ( - 00, xo) and A n (xo, + 00), which are open in A. Since A is connected, one of these two sets is empty, say A n (xo, + 00). Then x < Xo for all x E A, which contradicts the fact that b is the least upper bound of A. 10.1.10. Theorem. Let X be a connected topological space, fa continuous rea/valued function on X, a and b points ofX. Then f takes on every value between f(a) and feb). For, f(X) is a connected subset of R (10.1.8), hence is an interval of R (10.1.9). This interval contains f(a) and feb), hence all numbers in between. 10.2. Arcwise Connected Spaces 10.2.1. Definition. Let X be a topological space and a, bE X. A continuous mapping f of [0, 1] into X such that f(O) = a, f(1) = b is called a continuous path in X with origin a and extremity b. If any two points of X are the origin and extremity of a continuous path, X is said to be arcwise connected. 10.2.2. Example. If E is a normed vector space, then E is arcwise connected. For, if a, bEE, the mapping t f--> f(t) = a + t(b - a) 128 X. Connected Spaces of [0, I] into E is continuous (8.1.8), and f(O) Rn is arcwise connected . = a, f(1) = b. For example, • 10.2.3. Theorem. Let X be an arcwise connected topological space. Then X is connected. Let Xo E X. For every x E X, let fx: [0, 1] ~ X be a continuous path with origin Xo and extremity x. Since [0, 1] is connected (10.1.9), the set Ax = fi[O, 1]) is a connected subset of X (10.1.8). Now, x E Ax, thus the union of the Ax is X. Since Xo belongs to all of the Ax, X is connected (10.1.6). 10.3. Connected Components 10.3.1. Theorem. Let X be a topological space, and x E X. Among the connected subsets of X containing x, there exists one that is larger than all the others. There exists at least one such set, namely {x}. The union ofall of the connected subsets of X containing x is connected (10.1.6) and is obviously the largest of the connected subsets of X containing x. 10.3.2. Definition. The subset of X defined in 10.3.1 is called the connected component of x in X. 10.3.3. Theorem. Let X be a topological space. (i) Every connected component of X is closed in X. (ii) Two distinct connected components are disjoint. In other words, the dif- ferent connected components of X form a partition ofX. (i) Let Ax be the connected component of x. Then Ax is connected (10.1.7). But Ax is the largest connected subset of X containing x, therefore Ax = Ax. (ii) Let Ax, Ay be connected components that are not disjoint. Then Ax u Ay is connected (10.1.6). Since x E Ax u Ay, we have Ax u Ay c Ax, whence Ay c Ax. Similarly Ax cAy, therefore Ax = A y • 10.3.4. Examples. A connected space has only one connected component. The space R - {O} has two connected components. namely (- 00, 0) and (0, + ro). Exercises Chapter I 1. Let .1. be the set of (x, y) E Rl such that Xl + yl :s: 1. Let S be the set of points of Rl of the form (x, 0) with 0 :s: x :s: 1. Let A = .1. - S. Find the interior, exterior, boundary and closure of A (relative to R2). 2. Let X be a topological space. If A is a subset ofX. we denote by Bd(A) the boundary of A. (a) Show that Bd(A) c Bd(A), Bd(A) c Bd(A). Show by means of an example (try X = R and A = Q n [0. 1J) that these three sets can be distinct. (b) Let A, B be subsets of X. Show that Bd(Au B) c Bd(A) u Bd(B); show by means of an example (try X = R, A = Q n [0. 1J and B = [0, IJ - A) that these two sets can be distinct. If A n B = 0, then Bd(A u B) = Bd(A) u Bd(B). 3. (a) Show that, on a set with two elements, there exist four topologies. (b) On a finite set, the only separated topology is the discrete topology. 4. An open subset of R is the union of a sequence of pairwise disjoint open intervals. 5. One ordinarily identifies R with the subset R x {O} of R2. Then [0, 1J has nonempty interior relative to R, but empty interior relative to RZ. Chapter II 1. Let X be a topological space and A a nonempty subset of X. A subset V of X is called a neighborhood of A if there exists an open subset U of X such that A cUe V. (a) The set of neighborhoods of A is a filter ff. (b) Give a necessary and sufficient condition for the identity mapping of X into X to have a limit along ff (assuming X is separated). (c) Let X = R, A = N. Show that there does not exist a sequence VI' V z , V 3 , · · · of elements of ff such that every element of ff contains one of the Vi' 130 Exercises 2. Define a mapping f of R into R by f(x) :; (1 + ~ sin x. Find the adherence values off: (a) as x --+ - 00; (b) as x --+ + 00; (c) as x --+ O. 3. Let X be the set R equipped with the discrete topology. Show that the identity mapping of X into R is continuous, but is neither open nor closed. 4. Let X, Y be topological spaces,! a mapping of X into Y. The following conditions are equivalent: (a)fis continuous and closed; (b)f(A) :; f(A) for every subset A of X. Chapter III 1. Let E be a topological space and I = [0. 1]. On the product space E x I. consider the equivalence relation R whose classes are: (i) the sets with one element ({x. t)}, where x E E, tEl, t -:ft 1; (ii) the set E x {l}. The topological space C = (E x I)/R is called the cone constructed over E. (a) For x E E, denote by f(x) the canonical image of (x, 0) in C. Show that f is a homeomorphism of E onto f(E). (b) Show that E is separated if and only if C is separated. 2. Let X be a topological space. L a subset of X and x E L. We say that L is locally closed at x if there exists a neighborhood V of x in X such that L n V is a closed set in the subspace V. We say that L is locally closed in X if it is locally closed at each of its points. (a) Show that the following conditions are equivalent: (i) L is locally closed; (ii) L is open in L; (iii) L is the intersection of an open set and a closed set in X. (b) The inverse image of a locally closed subset under a continuous mapping is locally closed. (c) If Ll and L2 are locally closed in X. then Ll n L2 is locally closed in X. (d) If Ll is locally closed in L 2, and L2 is locally closed in L J , then Ll is locally closed in L 3 • (e) Suppose that L is locally closed in X. Let 0/1 be the set of open subsets U of X such that LeU and L is closed in U. Let F be the boundary of L with respect to L. Then X - F is the largest element of 0/1. 3. Let X. Y be topological spaces. (a) Let x, Xl' X2,'" E X and y. Yl' h •... E Y. If the sequence «x., admits (x, y) as adherence value in X x Y, then the sequence (x.) (resp. admits x (resp. y) as adherence value in X (resp. V). (b) Show that there exists in R2 a sequence that admits no adherence value. even though each of the sequences (x.) and (Y.) has an adherence value in It. (Y.» Y.» «x., Y.» 4. The canonical projections of a product of topological spaces onto the factor spaces are open mappings. 5. Let X, Y be topological spaces, A c X and BeY. The following topologies on A x B coincide: (a) the topology induced by the product topology on X x Y; (b) the product of the induced topologies on A and B. 6. On R", define an equivalence relation 3i in the following way: (Je l , X2' •.•• x.) and (YI' Y2"'" Y.) are equivalent if Xj - Yj E Z for all i. Show that the quotient space R'/3i is homeomorphic to 1"". Exercises 131 7. Let P be the canonical mapping ofR onto T. Letfbe the mapping x ...... (p(x), p(x}2» ofR into T2. Show thatfis injective and continuous, but thatfis not a homeomorphism of R onto feR). 8. Let X, Y be separated topological spaces,f a continuous mapping of X into Y. The graph offis a closed subset of X x Y. 9. Let E be a topological space. F and G subsets of E such that G c F. For G to be closed in F, it is necessary and sufficient that G n F = G, where G denotes the closure ofG in E. Chapter IV 1. In R2 equipped with the usual metric, let D be an open disc with center Xo and radius IX > 0, and let A be a compact set contained in D. Show that there exists IX' E (0, IX) such that A is contained in the open disc with center Xo and radius IX'. 2. Let E be a separated space, (x lo X2, ••• ) a sequence of points of E that tends to a point x of E. Show that {x, x 1> X 2, •.. } is compact. 3. Let E be a separated space. Suppose that for every set X, every filter base rJI on X, and every mappingf of X into E,f admits an adherence value along rJI. Then E is compact. 4. The topological spaces (0, 1) and [0, 1] are not homeomorphic. 5. Let E I, E2 be nonempty topological spaces. If EI x E2 is compact, then EI and E2 are compact. 6. Let A = R n+ I - {O}. Define an equivalence relation f7i n on A in the following way: two points x and y of A are equivalent if there exists t E R - {O} such that y = tx. The quotient space A/f7i. is denoted P.(R) and is called the real projective space of dimension n. (a) Let It be the canonical mapping of A onto P.(R). Show that It is open. (b) Show that It is not closed. (c) Let r be the set of (x, y) E A x A such that x is equivalent to y. Show that r is closed. From this, deduce that P.(R) is separated. (d) Let 9'" be the equivalence relation on Sn obtained by restriction of f7i n • Show that the quotient space S.!9'" is compact. (e) Let cp be the restriction of It to Sn. Show that cp is continuous and defines a homeomorphism ofS.!9'n onto P.(R), so that Pn(R) is compact. (I) Let 9 be the mapping of SI into R2 such that g(x, y) = (x 2 - y2, 2xy) for x, y E R, x 2 + y2 = I. Show that g(SI) = SI and that 9 defines a homeomorphism o[SI/.9(' onto SI' so that P l (R) is homeomorphic with SI' Chapter V 1. Show that the conclusion of 5.5.12 may fail if X is not assumed to be complete. (Take X = Q. Let rl' '2, ... be the elements of Q arranged in a sequence. Take Un = Q - {r.}.) 2. Let X be a metric space and A, B, C subsets of X. Show that one does not necessarily have d(A. C) ~ dCA, B) + deB, C). (Take X = R, A = [0, 1], B = [1, 2], C = [2, 3].) 132 Exercises 3. In R·, equipped with the usual Euclidean metric, the diameter of an open or closed ball is twice its radius. ° 4. Let X be a set. For x, Y E X, set d(x, y) = 1 if x ::j: y, d(x, y) = if x = y. Show that d is a metric on X and that the corresponding topology is the discrete topology. The diameter of a ball of radius < 1 is 0. 5. Let X be a metric space, and x, XI, X2, X3, •.. points of X. Show that the following conditions are equivalent: (i) x. -+ x; (ii) every subsequence of (XI> X2"") has a subsequence tending to x. 6. In 5.5.10, there are four hypotheses: X complete, the F. closed, the F. decreasing, b. -+ 0. Show that if anyone of these hypotheses is omitted, it can happen that FI n F2 n··· is empty. 7. Show that in 5.5.13, f may fail to exist iff' is only assumed to be continuous (but not uniformly continuous). (Take X = Y = R, X' = R - {O}, f(x) = sin(1/x) for X EX'.) 8. Let X be a metric space, A a closed subset of X, B a compact subset of X. Assume that A n B = 0. Show that d(A, B) > 0. °° 9. Letfl,f2, ... be continuous functions;;:: on [0, 1], such that: (rx)j~ ;;:: f2 ;;:: f3;;:: ... ; ({3) the only continuous function;;:: on [0, 1] that is majorized by all of the J" is the function 0. (a) For every x E [0, 1], let l(x) = lim.~<Xl J,,(x). For every integf:r n ;;:: 1, let O. = {x E [0, I]ll(x) < I/n}. Show that 0. is a dense open set in [0, 1]. (b) Show that I vanishes on a dense subset of [0, 1]. Chapter VI 1. Let C be the set of continuous real-valued functions on [0, 1], (:quipped with the metric of uniform convergence. If n is an integer > 0, we denote by A. the set off E C for which there exists t (*) E [0, 1 - l/n] with the following property: If(t') - f(t) 1 :::; n(t' - t) t' forall E [t, t + l/n). (a) Letf E C be differentiable at at least one point of [0,1). Show thatf E A. for some n. (b) Let (fl' f2' ... ) be a sequence of elements of A. tending uniformly to an element f of C. Assume that there exists in [0, 1 - l/n] a sequence II. t 2 , .•• having a limit t, such that If,{t') - .Ht i) I :::; n(t' - t i) for all t' E [ti' ti + l/n], (rx) Show thatf,(t i ) -+ f(t). ({3) Show that I f(t') - f(t} I :::; n(t' - t} for t' E (t, t + l/n). (y) Show that t has the property (*) relative tof. (c) Show that A. is closed in C. (d) Let fEe be a continuously differentiable function and let B > 0. Let n be an integer >0. Set M = sUPosxsllf'(x)l. 133 Exercises Let 0 = IXO < IXI < IXZ < ... < IXp_ 1 < IXp = 1 be real numbers such that each interval [lXj' IXj+ I] has length :s; e/(M + 2n). Let 9 be the function that is linear on each of the intervals [lXj' IXj+ I]' and is such that g(lXo) = g(1X2) = g(1X4) = ... = 0, g(IXI) = g(1X3) = g(lXs) = ... = E. (IX) Show that Ig(t') - g(t) 1~ (M + 2n)lt' - tl if t and t' belong to the same interval [lXj' IXj+ I]. (f3) Show that d(j; f + g) :s; B. (y) Show that f + 9 ¢ An. (Argue by contradiction.) (e) Show that the interior of An (relative to C) is empty. (Argue by contradiction. One uses (d) and the fact that every continuous real-valued function on [0, 1] is the uniform limit of continuously differentiable functions; on this subject, see 7.5.5.) (f) Show that the intersection of the complements C - AI' C - A2, ... is nonempty. Show that Al U A z U ... # C. (g) Show that there exists anf E C that is not differentiable at any point of [0, 1]. 2. This exercise is a commentary on Dini's theorem. (a) Construct on [0, 1] an increasing sequence of continuous real-valued functionsfl.fz, ... that tends simply to a functionfthat is not continuous (so thatf. does not tend uniformly to f). (b) Construct on [0. IJ a sequence of continuous real-v,,; ....:d functions that tends simply to but does not tend uniformly to 0. (c) Construct on R an increasing sequence of continuous real-valued functions that tends simply to the function 1 but does not tend uniformly to 1. ° 3. Let F be a continuous real-valued function on [0, lY Let C = f6'([0, IJ, [0, IJ). Throughout the exercise, one utilizes the metric of uniform convergence. (a) Letf E C. Show that for every S E [0, IJ. the function t H F(s, t,f(t» on [0, 1] is continuous. Set g(s) = f F(s, t. f(t» dt. Show that 9 E f6'([0. 1]. R). (b) As f varies, we have thus defined a mapping f H 9 of C into 'C([O, 1J, R). Let ep denote this mapping. Show that ep is uniformly continuous. (c) Show that ep(C) has compact closure in f6'([0, 1], R). Chapter VII l. Let f: R --> [0, + co) be a lower semicontinuous function. Show that f is the upper envelope of a family of continuous functions ~ O. 2. Let X be a topological space, Xo E X, and f, 9 finite numerical functions on X that are lower semicontinuous at Xo. Suppose that f + 9 is continuous at Xo. Show that f and 9 are continuous at Xo· 134 3. Let (U I , U2, m, n ~ 1. Let Exercises U 3 , ••• ) be a sequence of real numbers such that u•• + n :s; um + Un for . Un / = I1m sup-, "~a;l n l' = lim inf~. "--fro n (a) Show that Ui,+i'+".+in $; Ui, + Ui, + ... + uin for all i l , ••• , in, and that mUn for all positive integers m, n. (b) Fix an integer p ~ 1. Let 0: = sup(O, Ul' Uz, .•.• Up-I)' (0:) For every integer n ~ 1, write n = k.p + rn with :s; 'n < P (Euclidean division of n by p), Show that U mn $; ° Un knP Up 0: -$;-.-+-. n n p n (Ii) Show that / $; Up/po What can be deduced about I' from this? (c) Show that uJn has a limit A as n -+ 00, and that uJn ~ A for aU n. 4. Consider the topological space x= [0, 1] x [0, 1] x [0, IJ x .... An element of X is an infinite sequence (Xl, X2, X3, ...) of numbers in [0, 1]. (1) Let n be an integer ~ 1. If / E'I([O, 1]·, R), one defines a fUllction g on X by setting g(Xl' X2' X3"") = /(Xl' Xl' ... , X.) for any (Xl' X2' X3' ...) E X. Show that g Erc(X, R). Let C n be the set of continuous functions on X obtained in this way as/runs over '1([0, l]n, R). (2) Show that C l C C 2 C C 3 C . . . . (3) Show that C l U C 2 U C 3 U ..• is dense in '1(X, R) for the topology of uniform convergence. S. For a, b, C E Rand a < 0, denote bY/.,b.c the function X H every continuous complex-valued function on R tending to limit on R of linear combinations of the functions f.. b, c' °e= 2 +bx+,' on R. Show that at infinity is the uniform Chapter VIII 1. Let (e l , e2"") be the canonical orthonormal basis of /2. Show that this sequence has no convergent subsequence. From this, deduce that [2 is not locally compact. 2. Show that there exists no scalar product on l~ for which the corresponding norm is the norm of 8.1.5. (Show that the parallelogram law fails.) Analogous question for Ie. 3. Let E be a separated pre-Hilbert space, C a complete convex subSl.,>! of E, X a point of E. Show that there exists one and only one Y E C such that d(x, y) ,= d(x, C). 4. For every X = (Xl' X2"") E I., define a linear form/.. on I~ by the formula /"«Yl' Yz, Y3' ...» = XlYI + X2YZ + X3Y3 + ... I. for all (Yl> Yz, Y3' ...) E l~ (note that the series on the right side is absolutely convergent). Show that / .. E (l~' and that the mapping X H /" of into (Ii)' is a linear bijection such that 11/,,11 = Ilxll for all X E I•. 135 Exercises 5. Let E be a normed space and u E Sf(E). Show that (Use Exercise 3 of Chapter VII.) Ilu'II!I' has a limit as n ---> 00. 6. Letfl,fZ' ... be continuous complex-valued functions on [0, 1]. Consider the following conditions: (a)f. -+ 0 uniformly; (b)f...... 0 in mean square; (c)f. tends to 0 simply. Then (a) ~ (b), (a) ~ (c). Show that the implications (b) ~ (a), (c) ~ (b), (b) ~ (c) are false. Chapter IX 1. If, in a normed space E, every absolutely convergent series is convergent, then E is complete. 2. Let E be a Banach space and u E Sf(E). Show that the series UZ u3 3! l+u+-+-+··· 2! is absolutely convergent in Sf(E). Its sum is denoted eO. Show that eU is bicontinuous, with inverse e-". Show that eU+ v = eUe Vif u, v E Sf(E) and uv = vu. Show that Ileull :;; e lluli . 3. Let a, b E [0, 1). Show that the family (amb')(m .• )ENx N is summable in R. 4. Let (ai)ie I be a summable family of numbers ~ O. Then the family (aDie I is summable. 5. Let (Xl' Xz, X3 • ... ) be a sequence of real numbers. Suppose that for every sequence (Yl' Yz, Y3' ... ) of real numbers tending to 0, the series XIYI + XzY2 + X3Y3 + '" is convergent. Show that IXII + IX21 + IX31 + ... < + 00. Chapter X 1. Let X, Y be topological spaces. Consider the following conditions: (a) X and Yare connected; (b) X x Y is connected. Then (a) ~ (b). If X and Yare nonempty, (b) ~ (a). 2. Let Q be an open subset of R". The following conditions are equivalent: (a) Q is connected; (b) Q is arcwise connected. 3. Let X be a topological space, and a, b E X. The relation: there exists a continuous path in X with origin a and extremity b is an equivalence relation in X. The equivalence classes for this relation are called the arcwise-connected components of X. 4. A topological space X is said to be locally connected if every point admits a fundamental system of connected neighborhoods. If this is the case, then all of the connected components of X are open in X. 5. The topological spaces X = (0, 1) and Y = [0, 1) are not homeomorphic (compare the complements of a point in X and in V). 136 Exercises 6. One wants to show that it is impossible to find a sequence of pairwise disjoint. nonempty closed subsets F I , F 2 , ••• of [0, I] whose union is [0.1]. One argues by contradiction, on supposing that such subsets have been constructed. (a) If Bd(Fn) denotes the boundary of Fn in [0, 1], show tnat F = Bd(F I) U Bd(F 2) U ... is closed in [0, 1]. (b) Show that the interior of Bd(Fn) in F is empty. Deduce a t:ontradiction from this by applying Baire's theorem. Index of Notations (The references are to paragraph numbers) N Z Q R C the set of nonnegative integers the ring of integers the field of rational numbers the field of real numbers the field of complex numbers A 1.4.1 A 1.5.1 "C(X, Y) 2.4.1 Sn 2.5.7 T 3.4.3 U 3.4.5 R 4.4.1 d(z, A), dCA, B) 5.1.4 g;;(X, Y) 6.1.1 lim sup 7.3.1 lim inf 7.3.1 sup(ft), inf(};) 7.4.5 sup(f, g), inf(f, g) 7.4.5 IR' Ie, I'" 8.1.4 Ii, I~, [I 8.1.5 Ilull (u a linear mapping) ~(E, F), 2'(E) 8.2.7 I~, 12 8.4.6 M-L 8.4.9 IW), [2(I) 9.4.10 8.2.1 Index of Terminology (The references are to paragraph numbers) absolute convergence 9.3.1 absolutely convergent series 9.3.1 absolutely summable 9.1.7 adherence value 2.6.1 adherent point 1.5.1 adjunction of a point at infinity 4.S.9 Alexandroff compactification 4.5.9 arcwise connected 10.2.1 Ascoli's theorem 6.3.1 Baire's theorem 5.S.12 Banach space 8.6.1 Banach-Steinhaus theorem 8,6.10 bicontinuous 2.S.2 Borel-Lebesgue theorem 4.1.4 bounded linear mapping 8.2.5 canonical orthonormal basis 9,4.10 canonical scalar product 8.4.2 Cauchy-Schwarz inequality 8,4.7 Cauchy sequence 5,S.1 closed 1.1.8, 1.2.5, 2.4.6 closed ball 1.1. 9 closure I.S.1 coarse space 1.2.3 coarser 1.2.3 coarsest topology 1.2,3 compact 4.1.2 complete 5.S.3 connected 10.1.2 connected component 10.3.2 continuous 2.4.1 continuous at a point 2.3.1 continuous path 10.2.1 convergence in mean square 8.S.9 convergent series 9.3,1 coordinates 9,S.3 dense 1.5.7 diagonal 3.2.13 diameter 5.1.2 Dini's thoorem 6.2.9 directed set 7.2.1 disc 1.1.4,1.1.9 discrete 1.2.3 distance function 1.1.1 dual 8.2.10 elementary open set 3.2.1, 3.3.1 envelope 7.4.S equicontinuous S.4.1 equicontinuous at a point 5.4.1 equivalent norms 8.3.3 extended real line 4.4.1,4.4.3 exterior 1.4.7 exterior point 1.4,7 extremity 10,2.1 140 Index of Terminology filter 2.1.1 filter base 2.1.1 finer 1.2.3 Fourier coefficient 9,5,4 fundamental system of neighborhoods generated closed linear subspace 8.7,4 Hausdorff space 1.6. I Hilbert space 8.6.1 homeomorphic 2.5,4 homeomorphism 2.5.2 increasingly filtering 7.2.1 induced topology 3.1.2 infimum 7.1.1 inner product 8.4.1 interchange of order of limits interior 1.4. I interior point 1,4. I isolated 3.1.8 kernel 6.1.14 1.3.6 parallelogram law 8.5.(1 point at infinity 4.5.9 pointwise convergence 6.2. I polarization identity 8.5.6 pre-Hilbert space 8.4.1 product of metric spaces 3.2.3 product of normed spaCf'S 8.1.1 0 product topological space 3.2.2, 3.3.1 product topology 3.2.2.3.3.1 Pythagoras, theorem of 8.5.7 quotient topological space quotient topology 3.4." Riesz's theorem 8.5.2 least upper bound 1.5.9, 7. Ll limit 2.2.1 limit along a filtering set 7.2.3 limit inferior 7.3.1 limit superior 73.1 locally compact 4.5.2 lower envelope 7,4.5 lower semi continuous 7,4.1,7,4.10 metric I. 1.1 metric of uniform convergence metric space I. I. I orthonormal basis 8.8.'/ orthonormal family g.~.6 oscillation 7. L3 6. 1.2 n-dimensional torus 4.2.16 neighborhood 1.3. I non-degenerate 8.5.2 norm 8. l.l normal space 7.6.2 normally convergent series 6.1.9 normed space 8.1.1 normed space associated with a semi norm g.1.11 norm topology 8.2.9 open 1.1.3,1.2.1,2,4.6 open ball 1.1.4 origin 10.2.1 orthogonal 8,4.9 orthogonal linear subspace 8.4.10 orthogonal projection 8.8.2 3.4.2 8.g.1 scalar product 8.4.1 semicontinuity 7.4.1, 7.4.10 semi norm 8.1.1 seminormed space 8.1.' separated 1.6.1, g.5.2 separated pre-Hilbert space 8.5.2, 8.5,4 simple convergence 6.2. I slab 1.1.15 sphere 1.1.12 stereo graphic projection 2.5.7 Stone-Weierstrass theorem 7.5.3 subspace, metric 1.1.1 subspace, topological 3.1.2 successive approximations 5.7.2 sum 9.1.1 summable 9. 1.I supremum 1.5.9,7.1.1 topological space 1.2. I topology 1.2.1 torus 4.2.16 total 8.7.4 trigonometric polynomial 7.5.6 ultrafilter 4.3.1 uniform convergence 6.1.2 uniform convergence on every compact subset 6.1.15 uniformly continuous S.3.1 uniformly convergent series 6.1.8 uniformly equicontinuous 5.4.1 upper envelope 7.4.5 upper semicontinuous 7.4.1,7.4.10 Undergraduate Texts In Mathematics continued from ii Martin: The Foundations of Geometry aDd the Non-Euclidean Plane. 1975. xvi, 509 paaes. 263 iIlus. Martin: Transformation Geometry: An Introduction to Symmetry. 1982. xii, 237 paaes. 209 iUus. MiUmanlParker: Geometry: A Metric Approach with Models. 1981. viii, 355 paaes. 259 iIlus. Owen: A First Course in the Mathematical Foundations of Thermodynamics 1984. xvii, 178 paaes. 52 iUus. PrenowitzlJantosciak: Join Geometries: A Theory of Convex Set and Linear Geometry. 1979. nii, 534 paaes. 404 lius. Priestly: Calculus: An Historical Approach. 1979, xvii, 448 paaes. 335 iUus. Protter/Morrey: A First Course in Real Analysis. 1977. xii, 500 paaes. 135 iUus. Ross: Elementary Analysis: The Theory of Calculus. 1980. viii, 264 paaes. 34 iUus. Siller: Alaebra. 1976. xii, 419 paaes. 27 illus. Simmonds: A Brief on Tensor Analysis. 1982. xi, 92 paaes. 28 iUus. Singer/Thorpe: Lecture Notes on Elementary Topoloay and Geometry. 1976. viii, 232 paaes. 109 iUus. Smith: Linear AIaebra. 1978. vii, 280 pages. 21 illus. Smith:. Primer of Modem Analysis 1983. xiii, 442 paaes. 45 illus. Thorpe: Elementary Topics in Differential GeometrY.· 1979. xvii, 253 paaes. 126 illus. Troutman: Variational Calculus with Elementary Convexity. 1983. xiv, 364 paaes. 73 illus. WhyburnlDuda: Dynamic Topoloay. 1979. xiv, 338 pages. 20 illus. Wilson: Much Ado About Calculus: A Modem Treatment with Applications Prepared for Use with the Computer. 1979. xvii, 788 PIIIes. 145 iUus.