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# 1984 Book GeneralTopology

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```Undergraduate Texts in Mathematics
Editors
F. w. Gehring
P. R. Halmos
C. DePrima
I. Herstein
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.
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
Jacques Dixmier
Univcrsite de Pari&quot; 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&quot;
Dixnner, Jacques.
Gf'nf'ral lOpolo~y
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&quot;l..:, @ Presses Uni\elsitaires de
France, Pari~, 191\ l.
&copy; 1984 Springer Science+Business Media New York
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&quot;&quot;, 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&quot;; if U is open in R&quot;, then V is
open in R&quot;.
These theorems cannot be obtained by the methods of this course, but, having
seen their statements, some readers will perhaps want to learn something
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 &sect;l certain concepts already known concerning
metric spaces, we introduce topological spaces in &sect;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
&sect;&sect;4 and 5. Separated topological spaces are introduced in &sect;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&quot; and
Y = (Yl' Y2, ... , Yn) E R&quot;, set
One knows that d is a metric on R&quot;, and in this way R&quot; 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&gt; 0 such that every point x of E satisfying d(xo, x) &lt; 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) &lt; p. Then A is open. For, let Xo EA. Then dCa, xo) &lt; p.
Set e = p - dCa, xo) &gt; O. If x E E is such that d(xo, x) &lt; e, then
dCa, x) ::; dCa, xo) + d(xo. x) &lt; dCa, xo) + e = p,
thus x E A.
The set A is called the open ball with center a and radius p. If p &gt; 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
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 &gt; 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 &gt; 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 &gt; 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) &gt; p. Set e = dCa, x o) ~ p &gt; 0.
If x E E is such that d(xo, x) &lt; e, then
dCa, x) 2': dCa, xo)
~
d(xo, x) &gt; 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'.
&deg;
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 &lt; 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&gt; reduces to a set with 2 points. In the ordinary
plane, one says 'circle' rather than 'sphere',
&deg;
&deg;
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' &gt; 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 &gt; 0
such that every point x of E satisfying d(xo, x) &lt; 6 belongs to A. If x E E
satisfies d'(x O ' x) &lt; CIl, then d(xo, x) &lt; 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&middot;, 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,&middot;&middot;&middot;, Ix. - Ynl)&middot;
d&quot;(x, y)
One knows that d, d', d&quot; satisfy the conditions of 1.1.14, hence define the same
open subsets ofRn. However, for d&quot; 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&quot;, 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&quot; 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&quot; 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 &gt;
in V.
&deg;that is contained
(ii) =&gt; (i). This is clear since the ball considered in (ii) is open and contains x.
(i) =&gt; (ii). If V is a neighborhood of x, there exists an open subset U ofEsuch
&deg;
that x E U c V. By 1.1.3, there exists e &gt; 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 &lt; x &lt; 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) =&gt; (ii). Suppose A is open. Let x E A. Then x E A c A, thus A is a
neighborhood of x.
(ii) =&gt; (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' =&gt; 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 &gt; 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
&deg;
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 &deg;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&raquo;, therefore A nBc (A n BY by 1.4.3.
&deg;
1.4.6. However, in general (A u B)O '&quot; 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) '&quot;
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&lt;=&gt; there exists a neighborhood of x disjoint from A
&lt;=&gt; there exists a neighborhood of x contained in E ~ A
&lt;=&gt; x is interior to E ~ A
&lt;=&gt; 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 &lt;=&gt; 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 =:&gt; 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 &gt; 0 such that (x - e, x + e) c U. By
the definition of supremum (least upper bound), there exists YEA such that
x - e &lt; 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' &gt; x, set e = x' - x &gt; O. Then (x' - e, x' + e) is a neighborhood of x', therefore intersects A. Let YEA n (x' - e, x' + e). Since
y &gt; 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) &gt; 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) &lt; e/2 and d(y, z) &lt;
e/2, therefore d(x, y) &lt; 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 &sect;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 &sect;&sect;3 and 4. Two topological spaces are said to be homeomorphic (&sect;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 &sect;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 &quot;f/' of neighborhoods of Xo is a filter on X (1.3.4). If jf&quot; is a fundamental system of neighborhoods of xo, then jf&quot; is a filter base on X.
2.1.3. Example. Let Xo E R. The set of intervals (xo - 6, Xo + 6), where 6 &gt; 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 &gt; 0;
where
e&gt; 0;
the set of (xo - e, x o],
where e&gt; 0;
the set of (xo - e, xo),
where
e &gt; 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 &gt; 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 =&gt; 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 &gt; 0, there exists N such that n ;:::: N
=&gt;
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&gt;Xo
lim
f(x)
X-Xo, x.sXo
lim
f(x)
x ..... xO,x&lt;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&quot; .... f(x) = I;
(ii) for every j E J, there exists i E I such that feW;) c Vj'
&quot;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&quot; .... f(x) = I;
(ii) for every B &gt; 0, there exists '1 &gt; 0 such that
&quot;Q
X E
X, d(x, X o) ::; '1
=&gt;
d(f(x), l) ::;
B.
For, the closed balls with center Xo (resp. /) and radius&gt; 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&quot; E f?J such that B&quot; c B ! l B/. Then f(B&quot;) c
V !l V'. Since B&quot; oF 0, one has f(B&quot;) 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 &gt; 0, there exists '1 &gt; 0 such that
x E X and d(x, x o) S '1 =&gt; d(f(x), f(x o&raquo; 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&middot;
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) =&gt; (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&middot;
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) =&gt; (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) =&gt; (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) =&gt; (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&gt; 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'&raquo; 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&middot;+ I joining a and x is
the set of points of the form
(Ax l ,
... ,
Ax., 1 + .Ie(Xn+l - 1&raquo;
with .Ie e R. This point is in R&middot; when 1 + .Ie(Xn+ 1 - 1) = 0, that is, when
.Ie = (1 - Xn+1)-1 (Xn+l&quot;# 1 because x&quot;# 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&gt; ... , 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; &gt; 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.
&deg;
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
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&quot; 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
&quot;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) =&gt; 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) =&gt; (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) =&gt; (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) =&gt; (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) =&gt; (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) =&gt; (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&quot; be sets such that
E ::J E' ::J E&quot;. Let :T be a topology on E, :Y' the topology induced by :T on E'.
:T&quot; the topology induced by :T' on E&quot;. Then .'T&quot; 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&quot;. There exists a subset U' ofE', open for
:T', such that U' n E&quot; = U&quot;. 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&quot; c E&quot; be an open set for .r~. There exists a subset V ofE, open for .r,
such that V n E&quot; = V&quot;. Set V' = V n E'. Then V' is open for .r', and V&quot; =
V' n E&quot;, thus V&quot; 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 &gt; 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 &gt; 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&quot;,
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&gt; ... , xn) E V,
there exist open neighborhoods V 1&gt; ••• , Vn of Xl' ... , Xn in E l , ... , En such
that VI X ... x Vn C U. There exists 6j &gt; 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) &lt; 6,
therefore d(x&quot; y;) &lt; 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 &gt; 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&gt; ... , 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&gt; 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&raquo;, 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~~::&quot;&quot;~-=-=-=-.-_-&quot;-1
-,-(X-)---'
Ez
/'
I
I
/, I
E
I
I
I
&quot; , ()()
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&raquo;, 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&raquo;, 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
&laquo;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
&laquo;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:
&laquo;x, y), z) H (x, y), &laquo;x, y), z) H x, &laquo;x, y), z) H y, &laquo;x, y), z) H z, &laquo;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 &lt; 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&raquo; 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---&gt; (};(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---&gt; (};(t&raquo;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\ &lt;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') &lt;=&gt; 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 (&sect;1) suggests no intuitive image, it is a
very fruitful definition (see the properties of compact spaces in &sect;&sect;2 and
3, and the applications in nearly all the rest of the course). In &sect;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 &sect;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 &gt; 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 &lt; 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 &lt; 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.) '&quot; 0. This contradiction proves
that A '&quot; 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
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&middot;
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&middot;
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
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&middot;&middot;&middot; 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&raquo;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 &gt;0. There exists rt. &gt; 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. &gt; 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.&middot;
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,,&middot;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&quot;, the compact sets are the closed bounded sets.
(A subset of R&quot; is said to be bounded if its n canonical projections onto R
are bounded.)
Let A be a compact subset ofR&quot;. Then A is closed in R&quot; (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&quot;. Since B is bounded. one has
Be [al. b l ] x [az. b2] x ... x [a., b&quot;] = 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&middot; is not compact.
The sphere Sn (2.5.7) is bounded in R&quot;+ I, and closed in R&quot;+ I (1.1.12),
hence is compact (4.2.18).
With the notations of 2.5.7, the space S&quot; - {a} is homeomorphic to R&quot;,
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 :::&gt; 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;;&quot; 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, &sect;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 &cent; 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 :::&gt; Fl n Y,G 1 :::&gt; F2 n Ywith
Flo Fl E:F, therefore G l n G 2 :::&gt; (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; :::&gt; .'F and that Y E 'If, therefore ~&sect; &quot;# .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
&lt;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 &lt;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 &lt;fI such that f(U) c V. Thus f tends to Xo along &lt;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; &quot;# 0. Now, the F J form a filter
base. Let &lt;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 &lt;fl. Fix
i E I. Let V be a neighborhood of xo' Let U E &lt;fI be such that U c V. Then
F; n U &quot;# 0 since F; E &lt;fI and U E OU. Therefore F; n V &quot;# 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&middot;
* 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&raquo;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 &lt; + 00 and - 00 &lt; X;
(3) one sets - 00 &lt; + 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
--&lt;x&lt;-
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 &lt; b &lt; 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&gt; t. Let a E R be such that I&gt; a &gt; 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) &gt; a,
X E
B'
= rex) &lt;
a.
B ('\ B', one sees that f(x) &gt; 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) =&gt; (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 &gt; 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&raquo; 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.&raquo;.
IEJ
Now, njEK C j is compact in X (4.2.9(ii&raquo; 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&raquo;.
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&quot;,! 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 &middot;
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&quot; 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 &sect;1, we introduce some concepts that are quite geometrical:
diameter of a set, distance between two sets.
In &sect;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 &sect;3, and the somewhat analogous
concept of equicontinuity in &sect;4.
The very important concept of complete metric space is studied
in &sect;&sect;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 &gt; 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
&quot;2 + d(xo, Yo) + &quot;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
+ &quot;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:
&gt;
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 &gt; 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,&quot;') 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.&quot; x. 2 '
to x.
••• ),
where nl &lt; n2 &lt; .. &quot; that tends
Suppose condition (ii) is satisfied. Then x is an adherence value of
(x.&quot; 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 &gt; n l such that d(x. 2 , x) :::;; Then there exists n3 &gt; n2
such that d(x.J' x) :::;; t, etc. The sequence (x.&quot; 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&raquo;
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&gt;
such that, for every '1 &gt; 0, there is ayE X with d(x, y) :-s; '1 yet
d(f(x), fey&raquo;~ ~ 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&raquo; ~ s. Then
Yn --&gt; X but f(Yn) does not tend to f(x) .
&deg;
.. 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 &gt; 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'&quot; .). 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&raquo;; etc. If the process stops, the theorem is
established. Otherwise, there exists a sequence (Xl&gt; X2,&quot;') 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' &gt; n such that X n ' E B(x, a/2). Then d(xn' x n') &lt; 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,&quot;') be a sequence of
points of A. Let Xi E A be such that d(Yi' Xi) :::;; 1/i. Some subsequence
(x. &quot; x. 2 ' ••• ) tends to a point x E X. Then x E A (5.2.1). On the other hand,
d(y.&quot; x) :::;; d(y.&quot; x.)
+ d(x.&quot;
1
x) :::;; ;.
+ d(x n&quot;
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: &gt; 0, there exists an '1 &gt; 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 &gt; 0 such that
Xl, Xl E
Now, let x, = 1/'1, X2
R and lx, - x 2 1 :::;; '1 =&gt; 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 &gt;
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 &gt; O. We are to construct an '1 &gt; 0 with the property of 5.3.1. For
every x E X, there exists an '1x &gt; 0 such that
x' E X, d(x', x) ~ '1x
=:&gt;
d(f(x'),f(x&raquo; ~
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&quot; t'1x2&quot;'&quot; t'1xJ &gt; O.
Now let x', XU E X be such that d(x',
x' E B x &quot; in other words,
XU) ~
'1. There exists an i such that
d(x;, x') &lt; t'1x,.
(1)
Then
thus
d(x;, x&quot;) &lt; '1x,.
(2)
The inequalities (1) and (2) imply
d(f(x'), f(x;) :::;; }'
whence d(f(x'), f(x&quot;&raquo; :::;;
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 &gt; 0, there exists an 17 &gt; 0 such that
d(x, x o) ~ '1 =:&gt; d(f(x), f(xo&raquo; ~ 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 &gt; 0, there exists an '1 &gt;
such that d(x, x') :::;; '1 =:&gt; d(f(x), f(x'&raquo; ~ e.
(a)
&deg;
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 &gt; 0 such that
for all a.
Xo
if, for every
E
&gt; 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 &gt; 0, there exists an I] &gt; 0 such that
d(x, x') :::;; '1 =&gt; d(h(X), h(x'&raquo; :::;; 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 &gt; 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 &gt; 0
such that
Ix - xol :::;; '1
=&gt;
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 &gt; 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 &gt; 0, there exists an N such that m, n
d(a m , an) :::;; e.
~
N =&gt;
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 &pound; &gt; O. There exists a positive integer N
such that n ~ N =&gt; d(a&quot;, a) :s; &pound;/2. Then
m, n
~
N
=&gt;
d(a m, a) :s;
&pound;
e
:2 and d(a&quot;, a) :s; :2
=&gt;
d(an&quot; an) :s;
&pound;.
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&quot; x n2 ' ... ) a subsequence. If the sequence (xn.) has a limit I, then the
sequence (xn) also tends to l.
n
Let &pound; &gt; O. There exists an N such that m, n
N. For ni ~ N, we have
~
N
=&gt;
d(xm' xn) ::;
&pound;.
Fix
~
(1)
As i --- x, we have X n , --- I, therefore d(x&quot;&quot; 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,&quot;') 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&quot;'&quot; 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&gt; Y.I) ~ d(Ym' Yn)
-+
0
as m, n -+ 00, therefore (Yll, Y21, Y3l&gt; ... ) is a Cauchy sequence in XI'
Since XI is complete, this sequence has a limit II in XI' Similarly,
(YI2, h2' Y32,&quot;') has a limit 12 in X2 , &middot; &middot; &middot; , (YIP' Y2p, Y3p' ... ) has a limit
Ip in Xp. Therefore (YI' Yz, ... ) tends in X to the point 1= (/1&quot;'&quot; 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&middot; is a complete space.
5.5.10. Theorem. Let X be a complete metric space. Let (F 1&gt; 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 &gt; O. There exists an N such that bN ~ e. If m, n ~ N then am, a&quot; 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 &gt; 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 &raquo; ~ 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&raquo; ~ t and B3 c B2 ,
etc. By 5.1.3, (j(f(Bi&raquo; :\$; 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 &gt; 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.&raquo;
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 (&quot;) V 2 &quot; •.. is dense in X.
Set V = V 1 &quot; V 2 &quot; .. '. Weare to prove that for every open ball B
with radius&gt; 0, V&quot; B is nonempty. Denote by B(x, p) (resp. B'(x, p&raquo; the
open ball (resp. closed ball) with center x and radius p.
The set V 1 &quot;B is open and non empty, therefore it contains a ball
Bl = B'(Xl' Pl) such that 0 &lt; Pl S 1. The set V 2 &quot; B(Xl' Pi) is open and
nonempty, therefore it contains a ball B2 = B'(X2' P2) such that 0 &lt; P2 S l
The set V 3 (&quot;) B(X2, P2) is open and nonempty, therefore it contains a ball
B3 = B'(X3' P3) such that 0 &lt; P3 S t, etc.
By construction, B :::l Bl :::l B2 :::l B3 :::l • &quot; . 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 &gt; O.
There exists an '1 &gt; 0 such that
Zl,
Z2 E
X' and d(zl&gt; Z2)
s
'1 =&gt; d(f'(Zl),r(Z2&raquo;
s
e.
Now, (xn) is a Cauchy sequence, therefore there exists an N such that
m, n
Then
m, n
~
N =&gt; d(x m , xn) S 1'/.
~ N =&gt;
d(f'(xm)' r(x n&raquo; S
Il.
Thus (f'(xn&raquo; 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
=&gt;
d(u n , vn) s'1
=&gt;
d(f'(u n), r(vn&raquo; S e.
Letting n tend to infinity, one obtains d(f(u), f(v&raquo; S e. Thus
u, v E X and d(u, v) S
~
=&gt; d(f(u), f(v&raquo; 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 &gt; 0, there exists a jinite covering of X by
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 &gt; 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
&plusmn;;
&plusmn;
yi. yS. YL&middot;&middot;&middot;
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 &gt; 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 &middot;
(i) A is compact;
(ii) for every c &gt; 0, one can cover A by a finite number of halls of X with
=
(i) (ii). This is obvious.
(ii) = (i). Suppose condition (ii) is satisfied. Let e &gt; O. There exist closed
balls B 1 , ..• , Bn in X with radius e that cover A. Then A c: Bl u&middot;&middot;&middot; 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&raquo;~
~
Then
thus (1 - A)d(a, b) ~ 0. Since 1 - A &gt; 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&middot;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&raquo; ~ U(X., X.+ 1)
~ AAnd(xo, Xl) = .1.&quot;+ 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.&quot; + .1.&quot;+1 + ... + A&middot;+ P - 1)d(xo, Xl)
d(x.,
~ .1.&quot;(1
X.+l)
+ A+
.1. 2
+ .. .)d(xo, Xl) = A. d~~ ~l).
Since 0 ~ A &lt; 1, A&quot; -+ 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&raquo; -+ 0, that is, d(xn+ h f(a&raquo;
-+ 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)! &lt; 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)&quot; ...
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. &lt; 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&raquo; 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&raquo;
therefore f(x)
= g(x); thus f
= g.
hE jO(X, Y) then, for every x E X,
d(f(x), hex&raquo;~ ~ d(f(x), g(x&raquo;
= 0,
It is clear that d(f, g) = d(g, f). Finally, if
+ d(g(x), hex&raquo;~
~
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&raquo; ~ 0, in other words: for every
e &gt; 0 there exists an N such that
n
~
N
=&gt;
d(f,,(x), f(x&raquo; ::; 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 &gt; 0, there exists B E 81
such that
A. E B =&gt; dCf;.(x), f(x&raquo; ::; 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 &gt; O. There exists
a E R such that A. ~ a=&gt; e-). ::; e. Then (provided a ~ 0):
A.
~
a
=&gt;
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&raquo; ::; dCfm' f,,)
~
0 as
m, n ~
00,
thus Cf.(x&raquo; 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 &gt; O. There exists an N such that
m, n
~
N
=&gt;
=&gt;
dCfm' f,,) ::; e
dCfm(x), f,,(x&raquo; ::; 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&raquo; ::; e.
This being true for all x E X, we have dCfm, f) ::; e. Thus,
m
~
N
=&gt;
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&quot;(X, C). Let s e ff&quot;(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 &gt; 0, there exists an N such that
n~ N
=&gt;
lu1(x)
+ uix) + '&quot; + 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 &gt; 0, there exists an N such that
6.1.9. Let Uh U2' ••• e ff&quot;(X, C). We say that the series with general term Un
converges normally if there exists a sequence aI' a2&quot;&quot; of numbers ~ Osuch
that
an &lt; + 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,'&quot;
eff&quot;(X, C). Letahaz, .. , be numbers ~ OsuchthatLf' an &lt;
an for all nand x. Let B &gt; O. There exists an N such that
+ 00 and Iu.(x) I ~
n
~
m
~
N =&gt;
IXm
+ IXm+ 1 + ... + IX.
~ B.
Then
n~ m~ N
=&gt;
Ium(x) + '&quot; + 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&quot;(X, Y) uniformly along fll. Thenfe ~(X, V).
Let Xo e X and e &gt; O. There exists A. e A such that
d(fix), f(x&raquo; ~
B
3
for all x e X. Next, there exists a neighborhood V of Xo in X such that
X E
V =&gt; 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 &gt; 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 =&gt; d(g(t, x'), g(to, x&raquo; :::;;
~.
As x runs over X, the W x form an open covering of X, therefore there exist
,XnEX such that X=Wx,v .. &middot;vWXn ' Set V=Vx,n .. &middot;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&raquo; :::;; e/2 by (1). Similarly, (1) implies that
d(g(to, z), g(to, x;) :::;; e/2. Then
Xl&gt; ...
d(g(t, z), g(to, z&raquo; :::;; e.
This being true for every Z E X, we have d(I&quot; I,o) :::;; e. Thus t E V=&gt; d(I&quot; 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 &sect;(T, V). We make the following assumptions:
each!s has a limit I. along
~;
!s tends uniformly along [JJ to an f
E
&sect;(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
&gt; O. There exists BE fJI such that s E B =&gt; d(/s, f) :s;: 8/3. Fix So E B.
8
:S;::3
d(!saCt), f(t&raquo;
for all t E T.
There exists e E ~ such that t, t' E C =&gt; d(!so(t), !soCt'&raquo; :s;: &lt;:/3. Then if t, t' E e,
we have
d(f(t), f(t'&raquo; :s;: d(fCt), j~o(t&raquo;
8
&lt;:
8
:S;::3 +:3 +:3
+ dCfs/t), .fso(t'&raquo; + 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&raquo;
E
Y
X
Y
tends to (is&quot;, 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 &gt; OandA &gt; O.Then
x E [-a, a] and n 2 A
+a
=&gt;
(x - n)2 2 A2 =&gt;
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&gt;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&gt; ... , 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,&quot;&quot; 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---&gt; 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---&gt; f(x o) of 3F(X, Y) equipped with ~ into Y is continuous. Now, this is
clear since
d(f(xo), g(xo&raquo; :::;; 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: &gt; 0. Let Xn be the set of x E X such that g.(x) ~ 1:. Then Xl =&gt; X 2 =&gt;
X3 =&gt; ... 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.
&deg;: :;
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 &gt; 0, A can be covered by a finite number of balls of
By 5.4.4, there exists an 'I &gt; 0 such that
x, x'
E
X, d(x, x') ::; 1'/, f
E
A
~
d(f(x), f(x'&raquo; ::;
8
4'
We can cover X by a finite number of open balls with centers Xl&gt; ... , 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&raquo;; 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) &lt; 1'/.
Therefore
d(f(x), f(x i&raquo;
e
::; 4'
d(g(x), g(x i &raquo; ::;
8
4'
Moreover,
Therefore d(f(x), g(x&raquo; ::; 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 &sect;&sect;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 &sect;3.
In &sect;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 &sect;5 we give a very general result that encompasses these earlier
results. It is applied in &sect;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 &lt; a, there exists an x
E
X such that f(x) &gt; 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&raquo;,
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) &lt; + 00,
bounded below if infxEx f(x) &gt; - 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&raquo;).
=
Set a = SUPXEX f(x), ai = SUPXEX.!(X). It is clear that ai :::;; a for all i E 1.
Let b &lt; a. There exists x E X such that f(x) &gt; b. Next, there exists i E I
such that x E Xi' Then b &lt; 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&raquo;)
(x. y)e X x Y
XEX
=
YEY
sup (sup f(x, y&raquo;).
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&raquo; ::; SUPxeX f(x)
(ii) 1/ f ~ 0 and g ~ 0, then
+ SUPxeX g(x).
SUp (f(x)g(x&raquo; ::; (suPf(X&raquo;)(SUpg(X&raquo;).
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&raquo;
+ 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&raquo; + k.
(ii) If k ~ 0, then SUPXEX (kf(x&raquo; = k(SUPXEX f(x&raquo;.
Set a = sUPxExf(x). By 7.1.6,
sup (f(x)
xeX
+ k)
::; a
+ k.
+ oc, equality clearly holds; suppose k &lt; + 00. Now let b &lt; a + k.
We have b = c + k with c &lt; 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&raquo;
(ii) If f ~ 0 and g ~ 0, then
=
SUPXEX f(x) + SUPYEY g(y).
sup (f(x)g(y&raquo; = (sup f(X&raquo;) . (SUpg(y&raquo;).
xeX.yeY
yeY
xeX
Set
a
=
c
=
supf(x),
XEX
b
=
supg(y),
YEY
sup (f(x) + g(y&raquo;.
xeX,yeY
73
7.2. Limit of an Increasing Numerical Function
By 7.1.5,
c
= sup [sup (f(x) + g(y&raquo;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&quot; E X such
that x&quot; ;::: x and x&quot; ;::: 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&quot; of x and x', and one has
Bx&quot; c Bx n Bx&quot;
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 &gt; - 00. Let V be a neighborhood of I in R; it contains a
neighborhood of the form [a, b], where a &lt; 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&raquo;.
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 &gt; lim sup f(x)
Then there exists B
E
n &lt; lim inf f(x).
and
!!J such that
xEB =&gt; n &lt;f(x) &lt; 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&quot;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 &lt; b. Let I/. E (a, b).
As in 7.3.3, there exists B E f!J such that x E B =&gt; f(x) &lt; 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&raquo;
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&raquo; ::; (limiJ/JSUP f(X&raquo;) . (limiJ/iSUP g(X&raquo;).
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&raquo;
::; 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&raquo;
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&raquo;;
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&raquo;.
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&raquo;
=
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. &lt; 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. &gt; 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 &cent; &gt; f is both lower and upper semicontinuous
at Xo'
7.4.4. Example. For x E R, set f(x) = x if x &quot;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 /;&quot; 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&middot; 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. &lt; f(xo) + g(xo). There exist {t, v E R such that {t + v = A.,
{t &lt; f(xo), v &lt; g(xo). Next, there exist neighborhoods V, W of Xo in X
such that
X E
V =&gt; f(x)
~ {t,
X E
W =&gt; g(x)
~
v.
Then
X E
V n W =&gt; 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
&lt;for example, continuous at xo), then f is lower semicontinuous at Xo&middot;
Let A.' &lt; f(xo). We have f(xo) = SUPie'/;(XO)' Therefore there exists an
I such that A. &lt; fi(xo}. Next, there exists a neighborhood V of Xo such that
x E V =&gt; /;(x) ~ A.. Then x E V =&gt; f(x) ~ A..
i
E
7.4.10. Definition. Let X be a topological space and f E \$&gt;(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) &gt; Ais open.
Since f-1([ -00, A]) and f- 1&laquo;A, +00]) are complementary 10 X,
conditions (ii) and (iii) are equivalent.
(i) =&gt; (iii). Suppose f is lower semicontinuous on X. Let
A
=
f- 1 &laquo;A, + 00]).
If Xo E A then f(x o) &gt; A; since f is lower semicontinuous at x o , there exists a
neighborhood V of Xo such that x E V =&gt; f(x) &gt; A. Therefore V c A. Thus,
A is a neighborhood of each of its points, consequently is open.
(iii) =&gt; (i). Suppose condition (iii) is satisfied. Let Xo E X. Let A &lt; f(xo).
The set A = f- \(..1., + cx;]) is open, and Xo E A, thus A is a neighborhood of
Xo' Since f(x) &gt; 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) &gt; A. If A &lt; 0, then X,1 = X. If
O:S; A &lt; l, then X&quot; = Y. If A ~ 1, then X&quot; = 0. The sets X and 0 are open
in X. By 7.4.11, we thus have
cp lower semicontinuous &cent;&gt; 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 &gt; 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&quot;' n ... n X,1n = X,,). Since X is compact,
X&quot; is nonempty. Let Xo
be a point of this intersection. For every A &gt; m, we have Xo E X&quot;' that is,
f(xo) :S; A. Therefore f(xo) :::;; m. Also f(xo) ~ m, thus f(xo) = m.
n,1&gt;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) &lt; + 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) &lt; + 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) &gt; 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) &gt; 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
&gt; 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) &gt; 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, , &quot; &quot; 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) &gt; 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) &lt; 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&quot; ... ,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) &lt; 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
&deg;~
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 - !&lt;Pn(t) + Jt&raquo;.
Jt)
Jt
&deg;
Now, Pn(t) +
~
therefore 1 - !&lt;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&raquo; is increasing and bounded above
by
therefore has a finite limit f(t) ~ that satisfies
&deg;
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 &lt;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 -&gt; U, Vn -&gt; v uniformly; then u&quot; + Vn -&gt;
U + v uniformly, whence u + v E Yf. Similarly, uv E Jf; and if A E R, then
AU E &pound;. Thus, every polynomial in u, that is, every function of the form
..1.0
+ A1U + A2u 2 + '&quot; + Anu&quot;,
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&raquo;.
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 &gt; 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 &lt;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 &quot;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 &pound;:
9
7.5.5. Corollary. Let X be a compact subset of Rn, and f
E &quot;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&quot; whose restrictions to X form a subset Jf of &quot;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 ~ &quot;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&raquo; 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&raquo; = e Z7tix for all x E R. Let h = 9
E
&quot;6'(U, C). For all x E R, f(x) = g(p(x&raquo; = h(e- 1 (p(x&raquo;) = h(e Z7tiX ).
Now, U is a compact subset of R2 = C. Let E &gt; O. There exists a polynomial
am.xPny' in x and y, with complex coefficients, such that
e
0
e
Lm,&quot;
Ihex + iy) - ~&quot; amnxPny' I ::5:
for every point x
E
+ iy ofU (7.5.5). Consequently, for every t E R, we have
I h(eZ&quot;i!)
-
~n am.(cos 2m)m(sin 2m)&quot; I ::5: E,
that is,
Since cos 2m
function
= t(e 27tit + e- 2 &quot;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&raquo;. 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&raquo;. Now let hE 16'0 and e&gt; 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) =&gt; (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
&deg;
U=g-l&laquo;-oo,t&raquo;,
V
=g-l&laquo;t, +00&raquo;.
Then U and V are disjoint open sets in X, and A c: U, B c: V.
(i) =&gt; (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) =&gt; (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 &lt; d' =&gt; 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&raquo;. Suppose the U(d) already chosen for d
2/2&quot;, ... ,2&quot;/2&quot; = 1, in such a way that
U(k/2&quot;) c U&laquo;k
= 0,
1/2&quot;,
+ 1)/2&quot;) for 0 ~ k &lt; 2&quot;.
Let us define U(k/2&quot; + 1) for k =0,1, ... ,2&quot;+1. For keven, U(k/2&quot; + 1) has
already been chosen. For k odd, thus of the form 2h + 1, we choose an open
set U &laquo;2h + 1)/2&quot; + 1) such that
U(2h/2&quot; + 1) c U&laquo;2h
+
1)/2&quot;+1) c U&laquo;2h
+
1)/2&quot;+1) c U&laquo;2h
+ 2)/2&quot;+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 &cent; 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 &gt; O.
If 0 &lt; a &lt; 1, then there exist d, d', d&quot; E D such that
a- e
~
d &lt; d' &lt; a &lt; d&quot;
~
a
+ e.
If one had x &cent; U(d&quot;), it would follow that f(x) ~ d&quot;, which is absurd; thus
x E U(d&quot;). On the other hand, f(x) &gt; d', therefore x &cent; U(d'), therefore
x &cent; U(d) by (1). Consequently, if one sets V = U(d&quot;) n (X - U(d&raquo;, V is an
open neighborhood of x. Let y E V. Then y E U(d&quot;), therefore fey) ~ d&quot;;
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 &lt; a &lt; 1.
If a = 1, there exist d, d' E D such that a - e ~ d &lt; d' &lt; 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&quot; E D such that 0 = a &lt; d&quot; ~ a + B. One sees as
above that V = U(d&quot;) is an open neighborhood of x and that, for every
y E V, one hasf(y) ~ d&quot;, 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,&middot;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 =
u(x)
=
@n+ 1(f(X) - 9n(X&raquo;
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)&quot;+ Iv(x)1
E
1
v(x)1 = (~-)&quot;+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 \$ &yen;~)&quot;+ 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) &lt; d(x, B) (resp. d(x, B) &lt; d(x, A&raquo; is
openinX.ItisclearthatU n V = 0.Ifx E A then d(x, A) = Oandd(x,B&raquo;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.
&deg;
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 .
&deg;
&deg;
&deg;
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&raquo;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&raquo;, 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,'&quot; ,J.. E CC(X, R), 01&quot;&quot; ,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.
&sect;&sect;1 to 5 are already familiar, excepting possibly Theorem 8.3.4 on
equivalent norms. In &sect;&sect;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---&gt; 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).
&deg;
It follows from (a) that x = 0 =&gt; Ilxll = 0. If, conversely, Ilxll = =&gt; 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.&middot;&middot;&middot;. xn)11
= IXII
II(x l , ... , x.)11
= sup(lxll,&middot;&middot;&middot; .lx.I)&middot;
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--+
&quot;I&quot; 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,.&middot;
.).
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 &lt; +00.
This is a linear subspace of 100. For, if s = (x t , x 2 , • •• ) E [t and t =
(Yt, Y2,&quot;') E [I. then
00
00
00
L Ix. + Ynl:::; n=l
L Ix.1 + I
.=1
(1)
n=1
IYnl &lt;
+00.
thus S + t E P. It is clear that AS E /t for every scalar A.
For s = (Xt, X2,&quot;') E [I, set
IIsll =
Then s t--+
IIsll
00
I
Ix.l.
is a norm on Jl (the triangle inequality results from
(1&raquo;.
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&gt; 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 &gt; O. Set
'7 = inf(l. 1 +
IAo~ + Ilxoll) &gt; 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,&middot;&middot;&middot;, 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&middot; 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
&deg;
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&middot; + 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 &lt; + 00.
(iii) =:&gt; (ii) =:&gt; (i). This is obvious.
(i) =:&gt; (iv). Suppose that u is continuous at O. There exists an tf &gt; 0 such
that y E E and Ilyll ~ tf imply IluY11 :::;; 1. Then, if x E E, we have
Ilxll :::;; 1 =:&gt; Iltfxll ~ tf
therefore Ilull &lt; + 00.
(iv) =:&gt; (iii). Suppose Ilull &lt;
Ilx -
=:&gt;
lIu(tfx)1I
~ 1 = Iluxll :::;; tf- 1 ,
+ 00. Let x, y E E and e &gt; O. Then
yll :::;; M =:&gt; 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 &gt; 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&raquo;.
~
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&middot;
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&raquo;
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&gt; since B &lt; + 00.
(ii) ~ (i). Suppose condition (ii) is satisfied. Then u is continuous (8.2.4).
Next, ux = implies allxll = 0, therefore x = (because a&gt; 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&middot;
(i)
~
(ii).
A, BE(O,
+00) such that
&deg;
&deg;
&deg;
8.3.2. Theorem. Let x t--&gt; I X III and x t--&gt; 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--&gt; Ilxlll' X t--&gt; 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&quot; 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&quot;. Let x t--+ Ilxll be a norm on R&quot;.
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&quot;; 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&quot;. Set
A = sup(llelll, ... , lien II)&middot;
If x = (Xl' ... , X&quot;) E R&quot;, then
Ilxll = Ilxlel + ... + x&quot;enll :::; Ix111lelil
:::; A(lxll + ... + Ix&quot;1) = Allxll'&middot;
It follows from this that the function
x, Xo E R&quot; and B &gt; 0, then
Ilx - xoll' :::;
i
=&gt;
Let S be the set of X =
t--+ Ilx lion R n is continuous; for, if
Ilx - xoll :::; e
(Xb ... ,
Ilxll'
X
=
IXII
+ ... + Ixnlilenil
=&gt;
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 &gt; 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&quot;.
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&quot;). 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&quot; into F is continuous. Similarly for
the mappings x f--+ X2a2,&quot; ., X f--+ x&quot;a n . Therefore the mapping x f--+
(xlal,&quot;&quot; x&quot;an ) of R n into F&quot; is continuous (3.2.7). Consequently, the mapping x f--+ Xtaj + .. , + X&quot;a n of R&quot; 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&middot;
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&quot;
set
and
y=(yt' ...• yn)EC&quot;,
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,&quot;') E X such that I:~l 1..1..12 &lt; + 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
&lt;Xl
&lt;Xl
L liln + Ilnl 2 ~ n=L1(21il 12 + 211l.1 2 )
n= 1
n
&lt;Xl
'=
+ 2L
IIlol 2 &lt;
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
+ &lt;t),
&lt;Xl
L (lilnl2 + 11l.1 2) &lt; +&lt;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)&middot;
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
&deg;
~ I(ylx)
(3)
+ A(X Iy).
Taking A = -(Ylx) in (3), we get
&deg;
~ -(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) =&gt; (ii). This is obvious.
(ii)=&gt; (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) &gt; 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).
&deg;
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-&gt; (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 &cent; El.
(otherwise, x' = 0), therefore (x Ix) &gt; 0 and consequently (x' Ix') &gt; O. We say
that E' is the separated pre-Hilbert space associated with E. This construction
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-&gt; Ilxll is a semi norm. For this seminorm to be a norm, it is
necessary and sufficient that
x =f.
0
=:&gt;
Ilxll &gt;
0,
in other words that
x =f.
0 =:&gt; (xix) &gt; 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)&middot;
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&raquo;, 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&laquo;A.lo A2'&quot; .), (;iI' J12,&quot;
.&raquo;
=
(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-&gt; Ilxll = d(x, 0) is continuous (5.1.1). Moreover, the
mappings (x, y) ~ x + y, (A, x) t-&gt; 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,&quot;.) with Ir;l IAnil2
&lt; + 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 &gt; 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
&lt;Xl
(1)
P~ N
=&gt;
L IApi -
i= 1
~ 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&quot;,(x) - fn(x)1I
~
IIf&quot;, -
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 &gt; O. There exists an N such that m, n ~ N =&gt; 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&quot;,1I
for
therefore f E 2(E, F). Moreover, we see that&quot; 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 &lt; + 00;
(ii) for each x E E, SUP;EI
Ilu;xll &lt; + 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 &gt; 0, and a constant M &gt; 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' &gt; such that Ilu;yll ::;; M' for
Ilyll ::;; p and for all i E I. Then, for Ilyll ::;; 1, we have
&deg;
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,&quot;') 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 &gt; O. There exists a
107
8.7. Linear Subspaces of a Normed Space
positive integer N such that Ln;&gt;N 1An 12 ~ 8. Let (J.1n) be the element of E
such that J.1n = An for n &lt; 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 -&gt; x, Yn -&gt; y.
Then Xn + Yn E F and Xn + Yn -&gt; X + Y (8.1.8), therefore x + Y E F. If A is a
scalar, then Ax n -&gt; 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 -&gt; x, Yn -&gt; 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 &lt;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
~
=&gt;
IIx - Ynl1 2 :s
&laquo;5 2
- yll = &lt;5.
Ilx -
Ynll
--+
&laquo;5. Let e &gt; O.
+ e.
N, we then have
211x - Yml1 2 + 211x - Ynl1 2 :s 4&lt;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&lt;5 2 + 4e,
or
IIYm - Ynl1 2 :s 4&lt;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 &gt; 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')
&gt; 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) &gt; 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 &cent;: 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
&deg;
for scalars Ait' .... Ain ' then the Pythagorean theorem implies that
0= Ilk e&middot; + ... + k
whence Ail = ... = Ai = 0.
'1
~1
In
e&middot;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.&quot;') 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&quot; 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&quot;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&quot;im'e 2 &quot;in, dt =
e 2&quot;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 &quot;i(n - m), admits the
primitive e 2 &quot;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=&gt;
E'.
&deg;
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)&quot;,
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 &sect;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, &gt; 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 &gt; 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 ,&quot;') is a summable sequence of elements ofE, then
the sequence (x&quot;) 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 &gt; 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 &gt; 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
~&quot;2
and
therefore I!sKi! = IlsJouK - sJol1 ~ e.
Suppose that the condition (ii) is satisfied. Let e &gt; 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 &gt; 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 &gt; 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 &gt; 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 &gt; 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&quot;
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&lt;7(i&raquo;i,,1 is summable with sum s.
Let e &gt; 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) =&gt; J o, therefore II LiE&lt;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 = &quot;co
&pound;..,n: 1 X ••
In other words, the series is convergent with sum s if, for every e &gt; 0, there
exists an N such that n ~ N =&gt; 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 &gt; 0, there exists an N such that
(d) Let E be a Banach space, (XI' X2&quot;&quot;) 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 &gt; 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&gt; Xl,&quot;') 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-&middot; .. ,
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 &lt;
(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&quot; (Yj)jeJ befamilies of numbers ~ O. Then
2:
(i.j)eIxJ
XiYj =
(We make the convention that O&middot;
(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 &lt; + 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 &gt; 0, then tXi = + 00 for such an i,
therefore Liel tXi = + 00; on the other hand, Liel Xi&gt; 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) =&gt; (i). This follows from 9.1.8.
(i) =&gt; (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 &lt; 0). The families
(Xi)iel, and (X;)iel, are summable (9.2.1). Therefore the families (IX;!)ie&quot; 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 =&gt; summability.
I arbitrary, E = R or C:
absolute summability -&cent;&gt; summability.
I = {I, 2, ... }. E complete:
absolute summability =&gt; summability
~
JJ
absolute convergence =&gt; convergence.
I
= P, 2, ... }, E = R or C:
absolute summability -&cent;&gt; summability
~
JJ
absolute convergence =&gt; convergence.
9.4.9. Let ex E R One knows that the number
1
1
1
2~
3~
4~
1+-+-+-+ .. &middot;=
I
XEN-{O)
x~
is finite if ex &gt; 1 and infinite if ex ::::;; 1. This amounts to saying that
1
-&lt;+oo-&cent;&gt;ex&gt;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~ &lt; +
I
X
XEZP-{O)
00
-&cent;&gt; ex &gt;
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. . . &middot; , IXpl)&middot;
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&middot;&middot;&middot;
u
Bop,
is the set of (x)' x 2 , . . . , x p ) E ZP such that Xi = &plusmn;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&middot; 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' &lt;
-~-p+!
-1- &lt;
. p- ! . P L -'-p+l'
-_
- 2 3
XEZP.-{O) X
nEN-{O} n
nEN-{O) n
from which
I
L -- &lt;
XEZP-{O) Ilxll'
1
+00 &lt;=&gt;
L ._p+!
nEN-{O) n
&lt; +cx:.:
&lt;=&gt;
ex - p + 1 &gt; 1
&lt;=&gt;
ex &gt; 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 &lt; + 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
&lt;
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) &lt; +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&quot;# 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 &lt; +00
ieI
(ii) If Iie! IIxil12 &lt;
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 &lt; + 00. For every 6&gt; 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&middot; In particular, Ilxf = Iie,I A/12.
Let 6 &gt; 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 ::::&gt; 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&raquo;. 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&raquo; 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 &laquo;eiIY&raquo;iel&gt; which belongs to [2(1) by 9.5.2.
Then 9 is a mapping of E into P(I), and f(g(y&raquo; = y by 9.5.2. On the other
hand. 9 (f(x&raquo; = 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 &cent;&gt; A and E - A open
&cent;&gt; 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 &gt; 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&gt; V 2 in
E such that V I = B n V'!&gt; 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--&gt; 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 &lt; Xo &lt; band Xo &cent; 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 &lt; 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--&gt;
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 , &middot; &middot; &middot;
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,'&quot; E X and y. Yl' h •... E Y. If the sequence &laquo;x.,
as adherence value in X x Y, then the sequence (x.) (resp.
adherence value in X (resp. V).
(b) Show that there exists in R2 a sequence
value. even though each of the sequences (x.) and (Y.) has an adherence value in It.
(Y.&raquo;
Y.&raquo;
&laquo;x., Y.&raquo;
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&quot;, define an equivalence relation 3i in the following way: (Je l , X2' •.•• x.) and
(YI' Y2&quot;'&quot; Y.) are equivalent if Xj - Yj E Z for all i. Show that the quotient space
R'/3i is homeomorphic to 1&quot;&quot;.
Exercises
131
7. Let P be the canonical mapping ofR onto T. Letfbe the mapping x ...... (p(x), p(x}2&raquo;
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 &gt; 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&gt; 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'&quot; be the equivalence relation on Sn obtained by restriction of f7i n • Show
that the quotient space S.!9'&quot; 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&middot;, equipped with the usual Euclidean metric, the diameter of an open or closed
&deg;
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 &lt; 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&gt; X2&quot;&quot;) 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&middot;&middot;&middot; 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) &gt; 0.
&deg;&deg;
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&quot; is the
function 0.
(a) For every x E [0, 1], let l(x) = lim.~&lt;Xl J,,(x). For every integf:r n ;;:: 1, let O. =
{x E [0, I]ll(x) &lt; 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 &gt; 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 &gt; 0. Let n be an
integer &gt;0. Set M = sUPosxsllf'(x)l.
133
Exercises
Let 0 = IXO &lt; IXI &lt; IXZ &lt; ... &lt; IXp_ 1 &lt; 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 &cent; 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.
&deg;
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&raquo; on [0, 1]
is continuous. Set
g(s) =
f
F(s, t. f(t&raquo; 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 --&gt; [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&middot;
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-,
&quot;~a;l
n
l' = lim inf~.
&quot;--fro
n
(a) Show that Ui,+i'+&quot;.+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 &lt; P (Euclidean
division of n by p), Show that
U mn \$;
&deg;
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]&middot;, R), one defines a fUllction g on X by
setting
g(Xl' X2' X3&quot;&quot;)
= /(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 &lt; 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'
&deg;e=
2
+bx+,' on R. Show that
at infinity is the uniform
Chapter VIII
1. Let (e l , e2&quot;&quot;) 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.,&gt;! 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&quot;&quot;) E I., define a linear form/.. on I~ by the formula
/&quot;&laquo;Yl' Yz, Y3' ...&raquo; =
XlYI
+ X2YZ + X3Y3 + ...
I.
for all (Yl&gt; 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 /&quot; 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 ---&gt;
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+-+-+&middot;&middot;&middot;
2!
is absolutely convergent in Sf(E). Its sum is denoted eO. Show that eU is bicontinuous,
with inverse e-&quot;. 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 + '&quot; is
convergent. Show that IXII + IX21 + IX31 + ... &lt; + 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&quot;. 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
&quot;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'&quot; 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
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.&quot;
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
continued from ii
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