North-Holland Mathematical Library Board of Advisory Editors:
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NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM * LONDON
Riesz Spaces
Volume I
W. A. J. LUXEMBURG
California Institute of Technology, Pasadena, California
and
A. C. ZAANEN
Leiden University, Leiden, The Netherlands
1971
NORTH-HOLLAND PUBLISHING COMPANY -AMSTERDAM
*
LONDON
@ NORTH-HOLLAND PUBLISHING COMPANY
-
1971
All rights reserved. No part of thispublication may be reproduced,stored in a retrieval system, or transmitted, in anyform or by any means, electronic, mechanical,photocopying, recording or otherwise, without the prior permission of the copyright owner
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Preface
In the last decades there has been published a large number of books on functional analysis, either textbooks or books on a more advanced level, some of these with the main emphasis on normed vector spaces and others devoted to more general topological vector spaces. In most of these books the notion of a (partially) ordered vector space receives little or no attention, although a fair proportion of the standard examples of vector spaces is partially ordered. We recall that a (partially) ordered real vector space L is a vector space over the real numbers which is at the same time a partially ordered set such that the vector space structure and the order structure in L are compatible. This means that, for any pair of elements f and g in L satisfyingf 5 g, it follows thatf+h 6 g + h holds for all h in L and af,< ag holds for all real numbers a 2 0. If, in addition, the order structure in L is a lattice structure (i.e., if any pair of elements in L has a least upper bound with respect to the order), then L is called a vector lattice or also a Riesz space. Familiar examples are the vector space of all real continuous functions on a given topological space, all the spaces L, (p > 0) in integration theory, and certain linear subspaces of the vector space of all Hermitian operators on a Hilbert space. The present book is devoted to the theory of Riesz spaces. The first volume contains what may be called the algebraic part of the theory, whereas the second volume will be more analytic in character. The theory of Riesz spaces was founded, independently, by F. Riesz, H. Freudenthal and L. V. Kantorovitch in the years around 1935, and it is interesting to observe now, more than thirty years later, the different methods of approach. F. Riesz was interested primarily in what is at present called the order dual space of a given ordered vector space, and he presented an extended version of his short 1928 Congress note ([l], International Mathematical Congress at Bologna) in a 1940 Annals of Mathematics paper [2], a translation of a 1937 Hungarian paper. H. Freudenthal, in 1936, proved a V
VI
PREFACE
“spectral theorem” for Riesz spaces [1 1, the significance of which is illustrated by the fact that the Radon-Nikodym theorem in integration theory as well as the spectral theorem for Hermitian operators in Hilbert space are corollaries, although it was not until early in the fifties that a direct method was indicated for deriving the spectral theorem for Hermitian operators from the abstract spectral theorem. Finally, around 1935, L. V. Kantorovitch ([I], [2], [3]) began an extensive investigation of the algebraic and convergence properties of Riesz spaces, with applications to linear operator theory. He was soon joined, in Leningrad, by several other mathematicians, of which we mention A. G. Pinsker, A. I. Judin and B. Z. Vulikh. A few year later, between 1940 and 1944, important contributions to the subject were published by H. Nakano ([l], [2], [3], [4], [ 5 ] ) , T. Ogasawara [l], K. Yosida ([l], [2], [3]) in Japan and S. Kakutani and H. F. Bohnenblust (with papers about concrete representations of abstract L-spaces and M-spaces) in the United States. After this first period it has still lasted a relatively long time before the results and terminology of the various centers of research (mainly in Japan, the Soviet Union and the United States) began to grow together. As an illustration we mention the important paper by I. Amemiya ([2], 1953), an extension of earlier work by H. Nakano. On account of the “Nakano terminology” in this paper it is not immediately visible that we have to do here with prime ideal theory in Riesz spaces, to some extent similar to M. H. Stone’s prime ideal theory in distributive lattices ([2], 1937). It is one of our objectives in the present book to collect the principal results, unify the terminology, and draw the reader’s attention to similaritiesas the one mentioned above. As is often the case with workof this kind, this has led to certain new results (not always explicitly indicated as new in the text). This is perhaps an appropriate point to say a few words about some other books devoted, in part or wholly, to ordered vector spaces, or, more specifically, to Riesz spaces. In the book by G. Birkhoff on lattice theory ([l], first edition in 1940, revised editions in 1948 and 1967), and also in the book by L. Fuchs on partially ordered algebraic structures ([l], 1966), the partially ordered vector spaces appear only as rather special examples. In the Birkhoff book, as indicated by the title, the principal interest is in general lattices without any other algebraic structure, and in the book by Fuchs a Riesz space is considered mainly as an example of the more general notion of a (not necessarily commutative) lattice group. The book by H. Nakano ([6], 1950) is pretty well restricted to the author’s own research. There is also a book by T. Ogasawara (1948), the contents of which are not so easily ac-
PREFACE
VII
cessible, because the book is written in Japanese, and no translation has been made. In the Soviet Union, in 1950, three mathematicians of the Leningrad school (L. V. Kantorovitch, B. Z. Vulikh and A. G. Pinsker) published a large monograph, called “Functional Analysis in Partially Ordered Spaces”, and in 1961 B. Z. Vulikh [2] published a smaller and more modern textbook on the same subject which is very readable, but still to a large extent devoted mainly to the research done in the Soviet Union. Vulikh‘s textbook has been translated into English (1967). Finally, we mention the more recent books by A. L. Peressini (Ordered Topological Vector Spaces, 1967) and G. Jameson (Ordered Linear Spaces, 1970); the emphasis in both of these books is on topological ordered vector spaces (Le., ordered vector spaces in which a topology is introduced such that the topology is compatible with the vector space structure as well as with the order structure). Spaces of this kind (in particular those in which the topology is generated by a Riesz norm) will be treated in the second volume of the present book. Readers who desire to restrict themselves at first to the topics of foremost importance (in particular the readers interested primarily in Freudenthal’s spectral theorem and its applications) are advised in the introductory Chapter 1 to pay attention only to sections 1-3 and the very beginning of section 9, and then to omit in Chapter 2 everything after Theorem 16.4 as well as in Chapter 4 the sections 26 (on atoms) and 32 (on the Dedekind completion). Chapter 5 on prime ideals may then also be omitted completely, and in Chapter 6 (on Freudenthal’s spectral theorem) it is enough to study only sections 38, 39, 40 until after Theorem 40.2, and then section 41. This will prepare the reader to understand all of Chapter 8 (on Hermitian and normal operators in Hilbert space), apart from a few isolated references to the spectral representation theory in Chapter 7. Finally, any reader who decides to take up also the study of prime ideals (Chapter 5) and spectral representations (Chapter 7) is advised to look first or simultaneously at sections 5-8 of Chapter 1. In 1967 we prepared a “preprint” for the present volume which was distributed on a small scale. The preprint was much more concise; for example, most of Chapter 1 and all of the Chapters 5 and 7 were not included, and also the number of bibliographical references was very small.
A major role in the second volume will be played by the properties of linear mappings from one Riesz space into another, by the order dual space of a given Riesz space, and by topological Riesz spaces (in particular by
VIIl
PREFACE
normed Riesz spaces). The interplay between topological continuity and order continuity will be an important feature. Some of this material can be found in a series of notes published by us in the Proceedings of the Netherlands Academy of Science, Amsterdam, Vols. 66-68 (1963-65). A further remark is in order here. The spaces in sections 9-10, the normed Kothe spaces and the Orlicz spaces, serve to illustrate some phenomena discussed in the abstract theory. This will be the case more frequently in the second volume than in the first. After some hesitation we have decided, however, to maintain these sections in the first volume. We express our thanks to the National Science Foundation of the U.S.A. and to the Netherlands Organization for the Advancement of Pure Research (Z.W.O.) for financial support at various stages during the preparation of this work. We also wish to thank Mrs. L. Decker and Mrs. A. Y. Hudson who, in Pasadena and Leiden respectively, gave us their valuable help in preparing the typed manuscripts of the present book as well as, back in 1967, of the preprint. Finally, we express our appreciation for the assistance and cooperation received from the staff at the North-Holland Publishing Company. W. A. J. Luxemburg A. C. Zaanen
CHAPTER 1
Distributive Lattices and Normed Function Spaces
We present a survey of several definitions and theorems (on distributive lattices, in particular Boolean algebras and Boolean rings, and on certain normed linear function spaces), which will be needed in what follows. Most of what is said in the sections 9 and 10 on function spaces is not immediately required.
1. Partial ordering Let X be a non-empty set; the elements x , y , . . .of X will be called the points of X . The set of all ordered pairs (x, y ) of points of X is called the Cartesian product of X by itself, and is denoted by X x X . By a relation in X we shall understand a non-empty subset of X x X ; the relation is sometimes denoted by R, and we shall write xRy whenever ( x , y ) is an element of the subset R of X x X which defines the relation. Well-known examples are the equivalence relations; R is called an equivalence relation whenever (i) it follows from xRy and yRz that XRZ (the relation is transitive), (ii) xRx holds for all x E X (the relation is re$exive), (iii) it follows from xRy that yRx (the relation is symmetric). If the equivalence relation R has the property that the subset of X x X which defines R consists only of all points ( x , x ) for x E X , then R is the relation of equality. The relation R in X is called a partial ordering of X whenever R is transitive, reflexive and anti-symmetric, i.e., whenever (i) it follows from xRy and yRz that x R t , (ii) xRx holds for all x E X , (iii) it follows from xRy and yRx that x = y . If R is a partial ordering in X , we will usually write x S y (or, equivalently, 1
2
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.
1,8 1
y 2 x) for xRy. Elements x, y of X for which either x S y or x 2 y holds are said to be comparable; if neither x S y nor x 2 y holds, then x and y are said to be incomparable. If every two elements of X are comparable, the partial ordering is called a linear ordering. The other extreme case is that every two different elements of X are incomparable, and so the partial ordering states now only that x 5 y holds if and only if x = y . If X is partially ordered and Y is a non-empty subset of X , then Y is partially ordered in a natural manner by the partial ordering which Y inherits from X . If the inherited partial ordering in Y is a linear ordering, then Y is said to be a chain in X . If X is partially ordered, Y a non-empty subset of X , and xo E X satisfies xo 2 y for ally E Y, then xo is called an upper bound of Y. If xo is an upper bound of Y such that xo 5 xb for any other upper bound xb of Y, then xo is called a least upper bound or supremum of Y. In this case xo is uniquely determined (in other words, any non-empty subset of X has at most one supremum). Indeed, if both xo and xb are suprema of Y, then xo S x&and x& 5 xo, and so xo = xb . If xo is the supremum of Y, this will be denoted by xo = sup Y or by xo = sup ( y :y E Y ) . The notions of lower bound and greatest lower bound or infimum are defined similarly. Notation: xo = inf ( y :y E Y ) if xo is the infimum of Y. The element xo of the partially ordered set Xis called a maximal element if it follows from x E X and xo S x that xo = x (observe that this is not the same as requiring that xo 2 x holds for all x EX). If there exists an element xo E X such that xo 2 x holds for all x E X , then xo is called the largest element of X , and in this case xo is also a maximal element. Actually, xo is now the only maximal element of X. In the converse direction, if xo is the only maximal element of the partially ordered set X , then xois not necessarily the largest element of X. Similar remarks hold for minimal elements and the possibly existing smallest element. We recall the following well-known and frequently used lemma.
Zorn’s lemma. If every chain in the partially ordered set X ha$ an upper bound, then X contains at least one maximal element. We proceed with some definitions.
Definition 1.1. Let X be a partially ordered set. (i) The set X i s called order complete if every non-empty subset of X has a supremum and an infimum.
CH.1,s 21
3
LATTICES
(ii) The set X i s called Dedekind complete i f every non-empty subset which is bounded from above has a supremum and every non-empty subset which is bounded from below has an injimum. (iii) The set X is called Dedekind a-complete i f every non-empty finite or countable subset which is bounded from above has a supremum and every non-empty jinite or countable sub$et which is bounded from below has an infimum. (iv) The set X is called a lattice i f every subset consisting of two elements has a supremum and an injimum. For Dedekind completenessa one-sided condition is sufficient,as follows.
Theorem 1.2. The partially ordered set X is Dedekind complete i f and only i f every non-empty subset which is bounded from above has a supremum. Proof. Let every non-empty subset of X which is bounded from above have a supremum, and assume that Y is a subset of X which is bounded from below. We have to prove that inf Y exists. To this end, observe first that the set L(Y) of all lower bounds of Y is non-empty and bounded from above, so lo = supL(Y) exists. Since y holds for all I E L ( Y )and all y E Y, any y E Y is an upper bound of L(Y), and so 1, 6 y. This shows that lo is a lower bound of Y, i.e., lo = sup L ( Y )is itself a member of L(Y). It is evident now that lo is the greatest lower bound of Y. In a partially ordered set with a smallest and a largest element the notions of order completeness and Dedekind completeness are evidently identical. Conversely, of course, any order complete partially ordered set has a smallest and a largest element.
Is
2. Lattices
Let X be a lattice. We shall denote the supremum of the set consisting of the elements x, y E X by sup (x, y ) , or by x v y if this is notationally more convenient. Similarly, the infimum of the set consisting of x and y will be denoted by inf (x, y ) or by X A Y . By induction it follows easily that in a lattice every finite subset has a supremum and an infimum. If the elements in the finite subset are x1,. ., x,,, its supremum is denoted by sup (xi, . .,x,,) or x1 v v x,, or v l=lxi, and its infimum by inf (xl , . . ., x,,) or x1 A . . .AX,, or A : = ~ x i .
. ..
.
.
Definition 2.1. The lattice X i s called distributive i f (Yl VY,) = holdsfor all x, y , ,y , E X . x A
h Y 1 ) V (XAY,)
4
[CH. 1,s 2
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
We can interchange suprema and infima in this definition, as shown by the following theorem. Theorem 2.2. The lattice X i s distributive if and only if X V ( Y l W 2 ) = ( X V Y d A (XVY,)
hoIds for all x , y 1 ,y,
E
X.
Proof. Assume that X i s a distributive lattice, and denote by 1 and r the left hand side and right hand side respectively of the formula to be proved. Also, write z = x v y , . Then r = Z A ( X V ~ , )=
(ZAX)V(ZAY,)
=~ ( X V Y l ) ~ ~ ~ V ~ ( X V Y l ) ~ Y Z ~ = (XA
4 v (Y, A X ) v ( X A Y , ) v (Yl A h )
= xv(vlAX)V(XhY,)v(y,hy,)
where we have used in the last line that y , proof in the converse direction is similar.
= XV(YlAY,) =
AX
6 x and
17
XAY,
s x. The
Theorem 2.3. In order that the lattice X is distributive, it is necessary and z and x A y , 5 z implies x A (y, v y,) 5 z. suficient that x A y , Proof. The necessity is evident from the definition of distributivity. Conversely, let X be a lattice such that x ~ y 6, z and x ~ y s, z implies X A (y, v y,) z. We have to prove that
s
X A 0 1VY,)
= ( X A Y d V (XAYZ)
holds for x , y , ,y , E X . Denoting the left hand side and right hand side of the formula to be proved by 1 and r respectively, we have x A y , 5 1 and x A y , 1, so r 1. On the other hand, since x A y , r and X A y z r, we have by hypothesis that X A (yl v y , ) 5 r, i.e., 1 5 r. It follows that 1 = r. If a lattice X has a smallest and (or) a largest element, these are sometimes called the null and (or) the unit of X ; we shall denote the null and the unit by 6 and e respectively. If X is a distributive lattice with null and unit, and if x , x’ E X satisfy X A x’ = 6 and x v x’ = e, then x’ is called a complement of x. Of course, x is now also a complement of x’. Note the possibility that X consists of only one element; now 6 = e.
s
s
s
s
Theorem 2.4. If the element x in the distributive lattice X with null and unit
CH. 1 , 8 3 1
5
BOOLEAN ALGEBRAS
has a complement then x' is uniquely determined (in other words, every element in X has at most one complement). XI,
Proof. Assume that x" is also a complement of x. Then x" = x " v 8 = X - V ( X A X ' ) =
(X-VX)A(X"VX')
Similarly x' = x' v x". Hence x" = x'.
= eA(x-vx') = x- vx'.
If X is a lattice with null 8, and x ~ =y 8 for the elements x, y of X , then x and y are called disjoint elements. If Y is a non-empty subset of X , the set of all x E X such that x is disjoint from all y E Y is called the disjoint comple-
ment of Y, and this set is denoted by Yd. If Xis a lattice with null 8, and Z is a subset of X with the property that zl, z2 E Z implies z1 v z2E 2 and z E 2 implies z' E Z for all z' satisfying z' 5 z, then Z is called an ideal in X . The condition that z E Z implies z' E Z for all z' S z is equivalent to z A x E Z for all z E Z, x E X. Note that the set (81, i.e., the set consisting of 8 only, is an ideal. Also, if Y is a non-empty subset of the distributive lattice X with null, then the disjoint complement of Y is evidently an ideal in X. 3. Boolean algebras
By definition, a Boolean algebra is a distributive lattice with null and unit such that every element in the lattice has a complement. By Theorem 2.4 the complement is uniquely determined. Note the possibility that the Boolean algebra consists of only one element.
x
Theorem 3.1. If x and y are elements in the Boolean algebra X such that
5 y , then
xx,y = (z: x 5 z
y),
with the partial ordering inheritedfrom X, is also a Boolean algebra with x as null and y as unit. Proof. We denote the null and the unit of X by 8 and e respectively. Evidently, Xx,yis a distributive lattice (with respect to the partial ordering inherited from X ) with x as null and y as unit. It remains to show that every z E X x , yhas a complement in X x , y .Let z' be the complement of z in X, and
6
[CH.I,§ 3
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
set z* and
= (z’ A y) v x.
Then
ZAZ* = ZA{(Z’A~)VX]
=
{ZA(Z‘A~))V(ZAX)
zvz* = Z V ( Z ’ A ~ ) V X = Z V ( Z ’ A ~ ) =
= evx =
(ZVZ‘)A(ZV~) =
which shows that z* is the complement of z in X x , y .
x
eAy = y,
Theorem 3.2. For any x in the Boolean algebra X , denote the complement of Then the following holds. (i) For any x, the element x’ is the largest element in X disjoint from x. (ii) I f x 5 y , then x’ 1 y‘. (iii) We have (x vy)’ = x’ A Y ’ for all x, y in X (and hence ( X A y)’ = x’ v y’). (iv) If {x,: T E {z}] is an indexed subset of X such that x = sup x, exists, then x‘ = inf xi. x by
XI.
Proof. (i) We have X A x’ = 8 and x v x’ = e. Assume that x“ is also disjoint from x, so X A x“ = 0. Then y = x’ v x“ satisfies and
xvy = xvx‘vx“ = e XAy = xA(x‘vx“) = (xAx’)v(xAx-) =
e.
Hence y = x’, i.e., x‘ v x“ = x’. It follows that x“ 5 x‘, which shows that x’ is the largest element disjoint from x. (ii) Let x 5 y. Then y’ A x 6 y‘ A y = 8, so y’ is disjoint from x. It follows now by means of (i) that y’ 5 x’. (iii) It follows from x v y 2 x that (x v y)‘ 5 x’. Similarly (x v y)’ 5 y‘, and so (x v y)’ 5 x’ A y’. We have also (X’Ay’)A(xvy) = (x’~y’Ax)v(x’Ay’Ay) = eve = so x‘ A y‘ is disjoint from x v y. This implies, by part (i), that x’ A Y ’ 5 (x v y)’.
e,
The final result is, therefore, that
(XVy)’ =
X’AY’.
(iv) Evidently, x’ is a lower bound of the set of all x:. Let y be another lower bound, i.e., y 5 xi for all T. Then y‘ 2 x:‘ = x, for all z, so y‘ 2 x. It follows that y” S x’, i.e., y 5 x‘. This shows that x’ = inf xi.
We list some examples. Given the non-empty point set X,the collection r of subsets of X is called a ring whenever it follows from A, B E r that A v B E r and A -B E r, where A -B denotes the set theoretic difference of A and B. It can easily be verified that in this case h i t e unions and finite
CH. 1,s 41
7
BOOLEAN RINGS
intersections of sets of r are again members of r. The ring r is partially ordered by set inclusion, andevidently r is thus a distributive lattice with the empty set as smallest element. The ring r of subsets of X is called afield (algebra) whenever the set X itself is a member of r. In this case r is a Boolean algebra with X as unit (i.e., X i s the largest element of r),and for any A E r the set X-A is the complement of A . Another example is obtained by considering a measure space ( X , A, p). This means that X is a non-empty point set, A is a a-field (a-algebra) of subsets of X (the field A is called a a-field whenever countable unions of sets of A are again members of A), and p is a non-negative countably additive measure on A. Let A, be the subcollection of A consisting of all sets of measure zero. The subcollection A, is an ideal in the Boolean algebra A, and it is easilyverified that if we say that the setsA andB in A are A,-equivalent (or, in other words, A and B are p-almost equal) whenever the symmetric difference ( A - B ) u (B-A) is an element of A,, then this defines indeed an equivalence relation in A. The set A/& of all equivalence classes is partially ordered in the natural manner, that is to say, if [A] and [B] are equivalence classes, and A, B are arbitrary elements from [ A ] and [B] respectively, then [ A ] 5 [ B ] by definition whenever A is included in B except for a set of measure zero. It follows easily that A/Ao is a Boolean algebra with respect to this partial ordering; A / A , is called the measure algebra corresponding to the measure space ( X , A, p). In practice, A, is often neglected, i.e., elements of A in the same equivalence class modulo A, are identified. The set A/A, is then again denoted by A, and an element of this new A is spoken of as if it were a subset of X , whereas in reality we mean the complete equivalence class of which the subset we consider is a member. 4. Boolean rings
Let X be a distributive lattice with smallest element 8. For any x E X , we shall denote the set ( y : y E ~e , y x)
s s
by Xo, and we shall say that Xe, is an initial segment of X. Suppose now that, for every x E X,the initial segment Xe,x is a Boolean algebra, i.e., for any pair of elements x , y in X satisfying y x there exists an element z satisfying z A y = 0 and z v y = x. The element z is appropriately called the complement of y with respect to x, and we shall use the notation z = x 0y. It is evident that x 0y = 8 if and only if x = y. We will prove that it is
s
8
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.1 , 8 4
possible to define in X an addition and a multiplication such that with respect to these operations X becomes a commutative ring (in the ordinary algebraic sense) with some extra properties. The definitions are simple; the sum x +y is defined as the complement of x A y with respect to x v y, and the product xy is defined as x ~ y We . divide the proof that Xis a ring with respect to these operations into several smaller parts. y
Lemma 4.1. We have ( x 0y)z = (xz) 0 (yz) for all x , y , z E X such that
s x.
Proof. For brevity, write x 0y = 1, and ( x 0y)z = 1, so 1 = 1,z. In order to prove that 1 is the complement of yz with respect to xz, we have to show that lyz = 8 and l v (yz) = xz. This follows from iyz = i,zyz = (zlv)z =
1 v (r.1
=
( I , z) v (YZ)
= (11 v (YZ>>{Z
ez = e,
” (YZ)l = (4 v
N
I
v
=
x(l, v z ) z = xz.
The definition for x + y , i.e., x + y = ( x v ) 0 (v),
implies that x+y = y + x for all x and y, x+z = x for all x if and only if z = 8, and x + y = 8 if and only if xy = x v y , i.e., if and only if x = y (note that both x and y are between xy and x v y). It follows now also that (x+y)z = xz+yz for all x , y, z. Indeed, the lemma above shows that (x+y)z = { ( x v y ) 0
(.Y)>Z
= { ( x v v ) z > 0 (xyz)
= {(xz)v (yz)} 0 (xzyz) = xz+yz.
It remains to prove that addition obeys the associative law.
Lemma 4.2. We have for all x, y , z E X .
(x+y)+z = x + ( y + z )
Proof. Set e = x v y v z. The set Xo, is a Boolean algebra; in this Boolean algebra we will denote the complement of any t E Xo,e by t’. Given 8 5 t - s 5 e, the complement of t with respect to s is now t‘s (cf. the proof of I Theorem 3.1). Hence x + y = ( x v y ) 0 ( x y ) = ( x y ) ’ ( x v y ) = ( x ’ v y ’ ) ( x v y ) = ( x ’ y ) v (xy’),
so
(x+y)’ = (x’y)’(xy’)’ = ( x v y’)(x’ v y ) = (xu) v (x‘y’).
CH. I , § 41
9
BOOLEAN RINGS
It follows that ( x + y ) + z = {(x +y)’z} v { ( x + y ) z ’ } = (xyz) v (x’y’z) v (x‘yz’) v (xy’z’).
Hence, by symmetry, (x+y)+z = x+(y+z).
It has been proved thus that Xis a commutative ring in the algebraic sense with respect to x + y and xy defined as the sum and the product of x and y. The smallest element 8 of X i s the null element of the ring (i.e., x + 8 = x for every x). In general, the ring has no unit element; if, however, X is a Boolean algebra, then the largest element e of X is the unit element of the ring, i.e., xe = x for every x E X . The ring X has the extra properties that x2 = x and x + x = 8 hold for every x E X. We observe that these extra properties are not independent. Indeed, if R is a ring (in the algebraic sense; null element 8) such that every x E R is idempotent (i.e., x2 = x for every x ) , then R is commutative and x + x = 8 for every x. The proof follows by observing that ( ~ + y = ) ~x + y implies xy+yx = 8; substitution of y = x yields x + x = 8, so x = - x , and this implies then that xy-yx = 8. Any ring R of this kind is called a Boolean ring. We have proved, therefore, that any distributive lattice X with smallest element 8 and such that every initial segment Xo, is a Boolean algebra can be made into a Boolean ring by defining that X+y
= (XVy) 0 ( X h y ) ,
Xy = X h y .
We shall say that the Boolean ring structure is derived from the given lattice structure. Some simple additional remarks follow. If x and y are disjoint, i.e., if x ~ =y 8, then x + y = x v y . For arbitrary x and y, the elements x + y and xy are complements with respect to x v y ; hence, x + y and xy are disjoint, so x+y+xy
= (x+y)v(xy) = xvy.
Finally, observe that x g y holds if and only if xy = x, and this is equivalent to the statement that x = yz for some z. Indeed, if the last condition is satisfied, then xy = y2z = yz = x. Naturally, the question arises now whether any Boolean ring R can be partially ordered in such a manner that R becomes a distributive lattice with the null element 8 of R as smallest element, such that every initial segment RB,xis a Boolean algebra and such that the Boolean ring structure derived from this lattice structure is again the given Boolean ring structure.
10
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH.
1,$4
The answer is affirmative. Given the Boolean ring R, we define x 5 y to hold whenever xy = x holds. It is easily verified that this defines a lattice structure with 8 as smallest element, xy as infimum and x + y + x y as supremum of x and y. By way of example, we show that z = x + y + x y is indeed the supremum of x and y. It follows from x z = x and yz = y that z is an upper bound of x and y. If z‘ is another upper bound (i.e., if xz’ = x and yz’ = y), then zz’ = xz’+yz‘+xyz’
= x+y+xy
= z,
so z 5 z’, which shows that z is the supremum of x and y. Itis now immediately verified by computation that the lattice is distributive and that, for 8 5 y 5 x, the element x+y is the complement of y with respect to x in the initial segment Re, , so Rd, is a Boolean algebra. Hence, R has the desired lattice structure. In the Boolean ring structure derived from this lattice structure multiplication is defined by xy = x A y , which is exactly the originally given multiplication, and the sum of x and y is defined as the complement of xy with respect to x v y = x + y + x y , so (by what was observed just above) the “new” sum of x and y is x + y + x y + x y = x + y , which is exactly the originally given sum. The new Boolean ring structure in R is, therefore, identical to the old one. We show, finally, that a subset I of the Boolean ring R is an ideal in the order sense if’and only if it is an ideal in the algebraic sense. The definitions for these notions agree in so far as it concerns the condition that x E I, y E R implies xy E I. The difference is that in one definition it is also required that x, y E I implies x + y E I, and in the other one that x, y E I implies
XVYEI.
Given that I is an algebraic ideal, it follows from x, y E I that x + y E I and xy E I, so x v y = x + y + x y E I. Given that I is an ideal in the order sense, it follows from x, y E I that x v y E I; hence x +y E I on account of x + y 5 x v y (note that x + y 6 x v y follows immediately from the definition of x + y as the complement of xy with respect to x v y ) . We recall that (for algebraic ideals as well as for order ideals) a maximal ideal is a proper ideal not properly included in any other proper ideal, and a prime ideal I is an ideal such that xy E I implies that one at least of x E I and y E I holds. Theorem 4.3. (i) Any proper ideal in the Boolean ring R is a prime ideal i f and only if it is a maximal ideal. (ii) Any ideal I in the Boolean algebra X is aproper prime ideal if and only if,for any x E X,the ideal I contains either x or the complement x’, but not both.
CH. 1,s 41
11
BOOLEAN RINGS
Proof. (i) First, let Z be a proper ideal which is prime, and assume that x and y are elements of R not contained in Z. Then xy is not contained in Z since Z is prime. This implies now that y -x E Z. Indeed, assume that y -x is not contained in Z. Then, since xy is also no member of I, we would obtain the result that x y ( y - x ) is no member of Z, i.e., xy-xy no member of I, which is obviously false. Hence, to summarize, if x and y are not in Z, then y = x + i for some i E Z. Assume now that Zis not maximal, i.e., Zis properly included in some proper ideal J. Choose elements x and y such that x is in J but not in Z and y is in R but not in J. Since x and y are not in Z, we have y = x + i for some i E Z. This implies that y E J. Contradiction. Hence, Z is maximal. For the converse, assume that Zis an ideal which is not prime, so there exist elements x and y, not in Z, such that xy E Z. Consider the ideal J generated by I and x; J consists of all elements i+ ax+n x for some i E I, some a E R and some non-negative integer n (we recall that n x denotes a sum of n terms, each term equal to x). We have Z c J c R with Z # J (on account of x being no member of I). Also J # R since y is no member of J. Indeed, if y were in J, then y would be of the form i + a x + n x with ax+n * x not in I, so
-
-
-
xy = xi+ax+n
x
-
not in I, because xi is in Z and ax+n x is not in Z. But this contradicts xy E I, which was established above. Hence Z # J # R, showing that if I is not prime, then I is not maximal. (ii) Let Z be a proper prime ideal in the Boolean algebra X with null element 8 and unit e. We have xx’ = 8 for every x E X, so x or x’ is a member of I, but not both, since otherwise e = x v x‘ E I, and then Z would not be proper. Conversely, let Z be an ideal in X with the stated property. Then e = 8’ is no member of I, so Zis proper. Also, if x and y are in the set theoretic difference X-I, then x’ and y’ are in I, so x’ v y‘ E I, i.e., xy E X-Z. Hence, given x and y in X such that xy E I, we must have x E Z or y E Z. This shows that Z is prime. Theorem 4.4. Given the ideal Z in the Boolean ring R and the point xo E R such that xo is no member of Z, there exists a maximal ideal J 3 Z such that xo is no member of J. Hence, every ideal Z in R satisfies
Z=
(J: J maximal ideal, J
I>
I).
12
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
Zn particular, since
{ e } is the smallest
[CH.1, 8 4
ideal in R, we have
n (J: J maximal ideal)
=
{e}.
Proof. By means of Zorn's lemma it is easily derived that there exists an ideal J 3 I such that xo is not contained in J, but xo is contained in every ideal in which J is properly included. We shall prove that J is maximal or, equivalently, J is prime. Assume that J is not prime, so there exist elements x and y, not in J such that x y E J. The ideal generated by J and x is properly larger than J, and so xo is contained in this ideal, i.e., xo = j , + a , x + n * x for some j , E J, some a, E R and some non-negative integer n. Actually, since a, x + n x = ( a , + n * x)x = a; x, we may assume that xo = j , + a , x for some j , E J and some a, E R. Similarly xo = j 2 + a 2y . It follows that
-
2 xo = xo = (jl+alx)(j2+a2y)EJ
on account of xy E J. Contradiction. Hence, J is maximal. We proceed with some definitions. The ideal Z in the lattice X with null 8 is called a band whenever, for every subset D of I possessing a supremum in X , this supremum is already a member of I (i.e., if D c Z and do = sup D exists, then do E I). It follows immediately from the definitions of ideal and band that any set theoretic intersection of ideals (or bands) is again an ideal (or band). Given the non-empty subset D of X , the intersection of all ideals including D is called the ideal generated by D . Similarly, the intersection of all bands including D is called the band generated by D. Note that {e} and X itself are bands in X . We recall that elements x, y in the lattice X with null 0 are called disjoint whenever they satisfy X A Y = 8 (cf. section 2). Sometimes this is denoted by x l y . Given the subsets D , and D2 of X , the notation D , ID2 means that d , Id2 holds for all d , E D , and all d2 E D , . Given the non-empty subset D of X , the set of all x E X satisfying x 1d for all d E D is called the disjoint complement of D , and this set is denoted by Dd. The set (0')' will be denoted by Ddd.Similarly for Dddd,and so on. Some simple results about disjointness are collected in Theorem 4.5, and a few more sophisticated results in Theorem 4.6.
Theorem 4.5. Let X be a distributive lattice with null 0. (i) For any non-empty subset D of X , the disjoint complement Dd is an ideal satisfying Dd = Dddd. (ii) The ideal Z in X satis'JiesZ = Ddfor some D c X ifand only i f Z = Zdd. (iii) The ideals I , , I, in X satisfy I , 1Z2 if and only if Z , n Z2 = { O } .
CH.I , § 41
BOOLEAN RINGS
13
Proof. (i) If x and y are members of Dd, then x A z = y A z = 8 for every z E D, so ( x v y ) A z = 8 by the distributive law, and hence x v y E Dd. It follows immediately that Dd is an ideal. Obviously D is included in Ddd, so (applying this to the set D d )we obtain Dd c Dddd.On the other hand, it follows generally from D, c Dz that D: 3 D;; hence D c Dddimplies Dd 3 oddd. The final result is now that Dd = Dddd. (ii) If Z = Dd for some D c X , then
Zdd = Dddd= Dd = 1. Conversely, if Z = Zdd, then Z = Dd for D = Id. (iii) If Z, , Z2 are ideals satisfying ZIII z , then any x E Zln Zz satisfies x Ix, i.e., x A x = 8, so x = 8. Conversely, if Zl and Zz are ideals having only 8 in common, then x A y = 8 for all x E Z, and all y E I,. Indeed, x A y is an element in I , n I, and hence must be equal to 8.
Theorem 4.6. Once more, let X be a distributive lattice with null 8. (i) For any ideal I in X , the set Zdd is the largest among all ideals A in X with the property that for every 8 # x E A there exists an element 8 # y E Z such that y S x. (ii) If Zl, . . ., In are ideals in X , then
(
n n
1
li)dd
=
n p. n
1
(iii) The ideal I , vZ, generated by the ideals Z, and I, in X consists of all elements x v y with x E I , and y E I , , and ( I , v = Zp n I;. In particular, if Z is an ideal in X , then (Zv Id)>"= {O}, and so (Zv Id)dd= X . Proof. (i) We prove first that Zddhas the stated property. Given 8 # x E Zdd, assume no y satisfies 8 # y E Z and y 6 x. Then x A z = 8 for every z E Z (if yo = x A zo # 8 for some zo E Z,then 8 # yo E I and yo 5 x). It follows that x E Id. But we have also x E Idd.Hence x E Zd n Zdd = {O}, contradicting x # 8. Now assume that A is an ideal with the stated property, and assume that A c Idddoes not hold. Then there exists an element x E A such that x is not disjoint from Id, so x A y # 8 for some y E Id. It follows that 8 # p = x A y E Zd n A . Now, by hypothesis, there exists 8 # q E Z such that q 5 p. But then 8 # q E Z n Id, which is impossible. (ii) It is trivial that Zi)" c I?. For the converse, it is sufficient to prove that Z = I , n I , satisfies Cdn Zid c Zdd. Given 8 # x E Ifd n Zid, there exists by part (i) an element 8 # y E I , such that y 5 x. Since 8 #
(n;
n;
14
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH.1, 5 4
y E Z, n Zid,there exists now also an element 8 # z E I , such that z 5 y. Hence 8 # z E Zln Z2 = Z and z 5 x. It follows, once more from part (i), that x E Zdd. (iii) The ideal ZlvZ, must contain all elements of the form x v y with x E Z, and y E 1,. It is easily verified that the set of all these elements is itself already an ideal (note that z 5 x v y implies z = zA ( x v y ) = (z A x ) v ( z A Y ) = x' v y' with x' E Zland y' E Z,), and hence this is the smallest ideal including Zland Z2 Evidently, (Ilv is included in If and in Zi,and so is included in Zfn Zi Conversely, if x E Zfn Zi ,then x is disjoint from Zland Z2,so (by the distributive law) x is disjoint from Zlv I,.
. .
Everything proved in Theorems 4.5 and 4.6 holds in particular in a Boolean ring. We present a few extra disjointness properties holding in a Boolean ring. Theorem 4.7. (i) Every disjoint complement (i.e., every set of the form Dd for some subset D ) in a Boolean ring R is a band. (ii) Any proper ideal I in a Boolean ring R cannot contain an element x E R and its disjoint complement { x } simultaneoudy. ~
Proof. (i) Since Dd = (I ( { x } :~x E D),it is sufficient to prove that { x } ~ is a band for every x E R . For this purpose, assume that { y , : o E {o}} is an indexed point set in R such that y , Ix for all o and such that yo = sup y , exists. We have to prove that y o Ix. Denote the complement of x with respect to z = x v yo by x', i.e., x and x' are complementary elements in the Boolean algebra Re,=. All y , are in Re,= and y , ~ x= 8 for all o. Hence, since x' is the largest element in Re, I disjoint from x , we have y , 5 x' for all o, so yo 5 x'. It follows then from x A x' = 8 and yo 5 x' that x AY, = 8, so yo Ix . (ii) Assume that Z is an ideal in R and x is a point of R such that x and { x } are ~ contained in Z.We have to prove that Z = R. Evidently, every y 5 x satisfies y E Z. If y > x and z is the complement of x with respect to y , then z A x = 8, so z E { x } c ~ Z.This irnpliesy = x v z E Z.Finally, ify is arbitrary, then w = x v y 2 x , so w E I, and hence y E Z on account of y iw. It follows that every y E R is a point of Z,so Z = R . For any ideal in the distributive lattice X with null 8, we shall denote by
{I} the band generated by I,i.e., {I} is the intersection of all bands which include Z. The ideal Zl is said to be order dense in the ideal Z2 whenever {I,} 3 Z, , and the ideal Z is simply called order dense whenever { I } = X . Since the band generated by {I} is equal to { I } itself, it is evident that {I,} 3 I , is equivalent to {I,} 3 {Z2},and so Z, is order dense in Z, if and
CH. 1,s 41
15
BOOLEAN RINGS
only if {I,} =I { I 2 } .It follows now also immediately that if Zlis order dense in Z2 and Z2 is order dense in Z3,then Zlis order dense in Z3.There is still another notion, called quasi order denseness, about which similar statements can be made, and even a little more. The ideal Zl is said to be quasi order dense in the ideal I2 whenever Ifd =I 12,and the ideal Z is simply called quasi order dense whenever Idd = X . The following holds. Theorem 4.8. By Z,I,, Zl, . . ., Z, we denote ideals in the distributive lattice X with null 0. (i) Thefollowing statements are equivalent: (a) Zlis quasi order dense in I,, i.e., fid 3 12, (b) Zfd 3 Zid, (c) Ifd is quasi order dense in ZF. (ii) IfZl c I,, then Il is quasi order dense in I , if and only if I? = IF. (iii) IfZl is quasi order dense in I, and I2 is quasi order dense in Z3,then Zlis quasi order dense in 13. (iv) IfIl, . . ., I,,are quasi order dense in I,, then Zi is quasi order dense in I,. (v) For any I, the ideal Z v I d is quasi order dense.
0;
Proof. The statements in (i), (ii), (iii) follow immediately from the definitions by observing that Idddd= Zdd. The statement in (iv) follows from Ii)" = Zy,and (v) was proved in Theorem 4.6 (iii).
(0:
0:
In a Boolean ring we can prove more. We will show that, for any ideal Z in a Boolean ring, the band { I }generated by Z is exactly equal to Zdd. It will follow then that the notions of order denseness and quasi order denseness in a Boolean ring are identical. The following theorem presents the details. Theorem 4.9. In a Boolean ring R the band {I} generated by an ideal I is equal to Zdd, and we have
{I} = I d d = { x : x = s u p y f o r a l I y ~ I , y ~ x } . Proof. We set
S = { x : x = sup y for all y
E Z,y
x}.
It follows immediately from the definition of a band that S c { I } . The set
Zdd is a band by Theorem 4.7, and Idd includes I; hence Zdd includes { I } . Combining these results, we obtain that S c { I } c Zdd. It remains to prove that every x, E Zddis a member of S. To this end, set V = ( y : y E Z,y 5 x,). If x, = sup V does not hold, there exists an upper bound xb of V such that x,, 5 xb does not hold. Then y o = x, A xb is an upper bound of V such that
16
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH.1,s 4
yo 5 x,, yo # x,. Let zo be the complement of yo with respect to x,, so zo ~ y =, 8, zo v y , = x, and zo # 8. It follows from x, E Iddthat z, E Idd,
.
so (by Theorem 4.6 (i)) there exists an element z1 # 8 in I satisfyingz1 6 zo The element zlis also a member of the set V, so z1 S y o . But then z1 6 zo A yo = 8, so z1 = 8, which contradicts z1 # 8. Hence x, = sup V holds, which shows that x, E S. This is the desired result.
Corollary 4.10. The ideal I in the Boolean ring R is order dense if and only if every x E R satisjes x = sup ( y : y E I, y 5 x). Order dense ideals and maximal ideals are not the same, even in Boolean algebras. An order dense ideal in a Boolean algebra is not necessarily maximal. Indeed, let S be an uncountable point set and X the Boolean algebra of all subsets of S, partially ordered by inclusion. The collection of all finite subsets of S is an order dense ideal in X , but I is not maximal since I is properly contained in the proper ideal Il consisting of all finite or countable subsets of S. Also, a maximal ideal in a Boolean algebra is not necessarily order dense. Indeed, let X be the Boolean algebra of all subsets of the point set S, where S consists of at least two points. Let x, be a fixed point of S, and I the ideal in X consisting of all subsets of S not containing the point x, . The ideal I is maximal but not order dense.
Exercise 4.11. Let R be the Boolean measure algebra corresponding to Lebesgue measure in the real line. Show that the ideal I of all bounded “subsets” is order dense, although I is no maximal ideal. Show the same for the ideal of all “subsets” of finite measure. Exercise 4.12. If X is a lattice with the property that, for every subset (y, : CT E { c T } ) for which supy, exists and for every x E X , the element sup ( x A y,) exists and satisfies SUP ( x A Y u )
=
x A (SUP Y,),
(1)
then we say that the infinite distributive law (1) holds in X . Similarly, if the lattice X has the property that, for every subset ( y , : CT E { c T } ) for which inf y , exists and for every x E X , the element inf ( x v y,) exists and satisfies inf ( x v y,)
=
x v (inf y,),
(2)
then the infinite distributive law (2) holds in X . Let X be the collection of all open subsets (with respect to the ordinary topology) of the open interval E = (- 1, 1) in the real line. Show that X , partially ordered by inclusion, is a distributive lattice with the empty set as
CH. 1, 8 41
17
BOOLEAN RINGS
smallest element and E as largest element. For any subset (0,:CT
E
{(T})of
X the set sup 0, is the union of the sets 0, and inf 0, is the interior of the intersection of all 0,.It follows easily that the infinite distributive law (1) holds. Show that (2) does not hold in X,even for sequences. Hint: Let 0 = (-1,O) and 0, = (-n-', 1) for n = 1 , 2 , . .
..
Exercise 4.13. Show that in a Boolean ring R the two infinite distributive laws hold. Hint: For the proof of sup ( x A ~ , )= X A (supy,), set yo = supy,. Evidently x A yo is an upper bound of the set of all x A y, . If x A yo is not the supremum, there exists an upper bound w o S x ~y~ ,w o # x A yo. The complement zo of w o with respect to x ~ y satisfies , now zo # 8 with zo x ~ y , and zo A ( x h y,) = 8 for all (T. It follows that zo A x ~ y = , 8, since the disjoint complement of zo A X is a band. But zo 6 x ~y~ implies then that z O ~ z= O 8, so zo = 8. Contradiction. The other distributive law follows by taking complements with respect to an appropriate upper bound. Exercise 4.14. Show that the ideal I in the Boolean algebra R with unit e is order dense if and only if e = sup(y:y~I). Hint: Given that e = sup ( y :y E I), use one of the infinite distributive laws to prove that x = sup (y : y E I, y 5 x ) for every x E R. Exercise 4.15. There is another method of proving that X A (sup yo) = sup ( x ~ y , )holds in a Boolean ring. Without loss of generality we may assume that any finite supremum of elements y , is itself one of the y,. Assuming this, let I, be the ideal of all z 5 x and I, the ideal of all z satisfying z S yo for at least one (T. Then I , n I, is the ideal of all z satisfying z S X A ~ ,for at least one (T. Now complete the proof by using that ( I , n I , ) ~=~ n lid. Exercise 4.16. The open subset 0 of the topological space S is called regularly open if the interior of the closure of 0 is again 0, i.e., if int 0 = 0. The closed subset F of S is called regularly closed if int F = F. Show that 0 is regularly open if and only if the set theoretic complement F = S - 0 is regularly closed. Note that S and the empty set are regularly open. For any subset A of S, denote the set int A by A*. Note that int A c A* holds for any A . Show that if A* c A (equivalently, A* c int A ) , then int A is regularly open. In particular, int F is regularly open for any closed F, and so 0 is regularly closed for any open 0. Derive from these results that the interior ~
18
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH. 1, $ 4
of an arbitrary intersection of regularly open sets is regularly open. In particular, any finite intersection of regularly open sets is regularly open. Note that a finite union of regularly open sets is not necessarily regularly open. Show that if 0 = 0 , , where all 0, are regularly open, then O* = int 0 is the smallest regularly open set including all 0,. Obviously the collection X of all open subsets of S, partially ordered by inclusion, is a distributive lattice. Show that the collection Y of all regularly open subsets of S, partially ordered by inclusion, is also a lattice, such that 0, A 0, = 0, n 0,and 0,v O2 = (0, u 02)*hold for all 0, and 0, in Y. Show now that the mapping 0 --f O* of X onto Y is a lattice homomorphism (precisely, show that the empty set and S are mapped onto themselves, and 0, n 0,and 0, u 0, are mapped onto 0;A 0;and 0: v 0; respectively). Derive from this result that Y is a distributive lattice. Finally, show that Y is an order complete Boolean algebra. Hint: If A* c A , then A* = int A c int A , so int A = int A. Hence it follows from c A that
u,
int (int) c int A = int A ,
so int A is regularly open. =
u 02)* For the proof that, for arbitrary open sets 0 , , O,, we have (0, 0:v O ; , note that
0:v O*2- - int (0; u 0;)= int (int 0,u int 0,) = int (int Olu int 0,)= int (0, u 0,)= int (0, u 0,)= (0, u 02)*, ~
where it has been used that O1 and 0,are regularly closed. n O,)* = 0: n O ; , note first that although For the proof that (0, 0, n 0, may be properly included in 0,n these sets have always the same interior. Indeed, if Vis a non-empty open subset of int (0,n then V c 0,, and so, if xo is any point of V, the set V is an open neighborhood of the point xo E 0,. It follows that V contains a point of O , , i.e., V , = V n 0, is not empty. But then it follows from V , c o2in the same way that V n 0, n 0, is not empty. In other words, given any point so o f int (0, n any neighborhood of so contains points of 0,n O , , so so E 0, n 0,. This shows that
a,,
a,),
--
so
int (0, n 0,)c 0,n O , , int (0,n 0,)c int (0, n 0,).
It follows that these sets are equal, i.e., 0: n 0;= (0, n 02)*.
a,),
CH. 1,s 51
PRIME IDEALS
19
Exercise 4.17. The element a # 8 in the Boolean ring R is called an atom if it follows from 8 # b S a that b = a. Show that the band I in R is a maximal ideal if and only if there exists an atom a such that I = {a}d. Hint: If a is an atom, then Z = {a}dis a band by Theorem 4.7,and Z is a proper subset of R . It remains to prove that Z is a maximal ideal. Take x E R, and set x1 = X A a. Then either x1 = 8 (so x E I ) or x1 = a (so x 2 a). Hence, if x is no member of Z, then x = a v y with y E I. Thus, if J is any ideal properly including I, then J contains a; hence, J includes the ideal generated by I and a, so J = R. Conversely, if the band I is a maximal ideal, then Id # { O } , since otherwise I = Zdd = R, against the definition of a maximal ideal. Hence, there exists an element a such that 8 # a E Z d . If there also exists an element b such that 8 # b S a, b # a, the ideal generated by Z and b is properly between I and R, so this is impossible. It follows that a is an atom, i.e., every element # 8 in I d is an atom. If both a and b are atoms in I d , then a A b = 8, and so the ideal generated by I and b is properly between I and R, which again is impossible. Hence, I d consists only of 8 and one atom a # 8. It follows that I = Zdd = {a}d. Exercise4.18. The mapping m of the Boolean ring R into the real numbers is called a state whenever (i) m(0) = 0, (ii) x y implies m(x) S m(y), (iii) x A y = 8 implies m(x v y ) = m ( x ) m(y),(iv) m(x) # 0 for some x E R. The set I,,, = ( x : m(x) = 0} is called the null set of the state m. Show that I,,, is a proper ideal in R. Show that the proper ideal Z in R is prime if and only if there exists a two-valued state such that Z is its null set. Show also that m ( x v y ) + m ( x ~ y )= m ( x ) + m ( y ) for all x , y holds for every state m.
+
Exercise 4.19. Let X be a distributive lattice with smallest element, and let x E X . Show that the ideal I, generated by x (i.e., the smallest ideal containing x ) is the set ( y :y <= x). Now, let X be the collection of all open subsets of the real line (ordinary topology), partially ordered by inclusion. Given the open subset E of the real line, the ideal ZE is, therefore, the set ( E l : El c E, El open). Show that, in general, Zid is properly larger than I E . Note that in a Boolean ring we have Ztd = I, for every x (by Theorem 4.9).
5. Prime ideals Let X be a distributive lattice with smallest element 8. We recall that the ideal P in X is called prime whenever it follows from x A y E P that one at
20
DISTRIBUTIVE LAlTICES A N D NORMED FUNCTION SPACES
[CH.1, 5 5
least of x E P or y E P holds. Note that X itself is a prime ideal. We present some simple facts about prime ideals. Theorem 5.1. (i) The ideal P is prime if and only if, for any ideals A , B satisfying A n B c P, one at least of A c P or B c P holds. (ii) Zf xo E X and P is an ideal in X, maximal with respect to the property of not containing xo (i.e., any ideal Q 3 P such that x,, is no member of Q satisfies Q = P), then P is prime. (iii) Every maximal ideal in X is prime. Proof. (i) Let P be prime, and let A , B be ideals such that A n B c P. If neither A c P nor B c P holds, there exist elements x E A and y E B such that x and y are no members of P. It follows that X A Y E A n B c P, so x E P or y E P since P is prime. Contradiction. Hence, one at least of A c P or B c P holds. Conversely, let the ideal P have the property that if A , B are ideals such that A n B c P, then A c P or B c P. We have to prove that X A Y E P implies x E P or y E P. To this end, let A and B be the ideals generated by x and y respectively, i.e., A = { z :z 5 x } and B = { z :z 5 y } . Then A n B = { z : z S x A y } , and so A n B c P on account of ~ A ~ E ItP . follows that A c P or B c P, so X E P or Y E P . (ii) Let xo E X , and P an ideal in X maximal with respect to the property of not containing xo. If P is not prime, there exist elements y , z, not in P, such that y A z E P. The element xo is in the ideal generated by P and y , so xo = p 1 v y , for some p1 E P and some yi 5 y . Similarly, xo is in the ideal generated by P and z, so xo = p2 v z l for some p 2 E P and some z1 5 z. Setting p 3 = p 1 v p 2 , we have xo 5 p 3 v y , and xo 5 p 3 v z , , so
xo 5 ( P 3 V Y l ) A ( P 3 V Z l ) = P3V(YlAZl)EP, which implies xo E P, thus contradicting our hypotheses. It follows that P is prime. Compare the present proof and the proof of Theorem 4.4. (iii) If P is a maximal ideal, and x,, is any element of X not in P, then P is maximal with respect to the property of not containing xo, so P is prime by part (ii). The following theorem is now an easy consequence. Theorem 5.2. Given the ideal Z and the element x,, not in Z, there exists an ideal P 3 I such that P is maximal with respect to the property of not containing xo. Hence, by the preceding theorem, P is prime. It follows now that, for any ideal Z in X (alsofor Z = X ) , we have I = ( P :P 3 Z, Pprime ideal).
0
CH. I , # 51
In particular
21
PRIME IDEALS
n ( P :Pprime ideal)
=
{O}.
Proof. The set of all ideals which include I and do not contain xo is partially ordered by inclusion. Each chain in this set has an upper bound (the union of all elements in the chain). Hence, the set has a maximal element, i.e., there exists an ideal P =I I such that P is maximal with respect to the property of not containing xo. By the preceding theorem P is prime. Any subset S of X not containing 8 such that x, y E S implies x A y E S is called a lower sublattice of X . Given xo # 8, the set of all y 2 xo is a lower sublattice. The empty set is also a lower sublattice. It is immediately evident that if P is a prime ideal, then its set theoretic complement S = X - P is a lower sublattice with the extra property that x E S, y 2 x implies y E S. In the converse direction, if S is a lower sublattice (even one with the above mentioned extra property) its complement S’ = X - S is not necessarily an ideal (x, y E S’ may fail to imply that x v y E 5”). However, given that P is an ideal and S = X-P is a lower sublattice, it follows immediately that the ideal P is prime. A lower sublattice is called maximal whenever it is not properly included in any other lower sublattice. Given any lower sublattice S, the set of all lower sublattices including S is partially ordered by inclusion. Any chain in this partially ordered set has an upper bound (the union of all elements in the chain). Hence, the set has a maximal element, i.e., the given lower sublattice S is included in a maximal lower sublattice. Note that if x,, is a member of the maximal lower sublattice S,, then any y 2 xo is a member of S, . Also, if zo is no member of S, ,then zo A x= 8 holds for at least one X E S,. Indeed, if z o ~ # x 8 for all X E S,, we add all these elements zo A X to S,,, (note that no zo A x is a member of S,, since otherwise zo would be a member of S,,,), and then the thus properly enlarged set is still a lower sublattice, contradicting the fact that S, is maximal. We proceed with another definition. Given the ideal I and the prime ideal M 3 I, we say that M is a minimalprime ideal with respect to I whenever any prime ideal M‘ for which I c M‘ c M holds satisfies M‘ = M. The prime ideal M is simply called a minimal prime idealif M is minimal with respect to the ideal (81. The following theorem follows now easily.
Theorem 5.3. (i) Given a chain (with respect to partial ordering by inclusion) Po is again a prime ideal. of prime ideals (Po: a E {a}), the intersection (ii) Given the ideal I and the element xo not in I, there exists a prime ideal
22
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.1,s 5
M 2 Zsuch that xo is no member of M and such that M is minimal with respect to 1. Hence, for any ideal Z in X (alsofor Z = X ) , we have
n ( M : A4 Z, Mprime ideal minimal with respect to I ) . In particular n (A4: M minimalprime ideal) ($1. Z=
2
=
Hence, given xo # 8, there exists a minimal prime ideal not containing x o . Zt follows also that X itself is not a minimal prime ideal, unless in the trivial case that X consists of 8 only.
u,
Proof. (i) The complements S, = X-P, are lower sublattices, so S, is also a lower sublattice. In other words, the complement of the ideal P, is a lower sublattice. This implies that P is prime. P= (ii) Let I be an ideal in X and xo an element of X such that xo is no member of Z. The set of all prime ideals which include Z and do not contain xo is non-empty (cf. the preceding theorem). This set is partially ordered by antiinclusion (i.e., P, 5 Pz whenever P, 2 Pz), and any chain in this partially ordered set has an upper bound (the intersection of the elements of the chain; this is a prime ideal by part (i)). Hence, the partially ordered set has a maximal element, i.e., there exists a prime ideal M which includes I, does not contain xo , and is minimal with respect to I. Some further results, connected with the existence of maximal lower sublattices, follow.
Theorem 5.4. ( i ) The subset S of X is a maximal lower sublattice ifand only if the complement X - S is a minimal prime ideal. (ii) Every prime ideal includes a minimal prime ideal. (iii) Any minimal prime ideal M cannot contain an element x E X and its disjoint complement { x } ~simultaneously, unless X consists of 8 only. (iv) Zf M is a minimal prime ideal and x is a member of M , then { x } is ~ ~ contained in M. Proof. (i) Assume first that S is a maximal lower sublattice. Write M = X-S. Since s E S, t 2 s, implies t E S, we have that x E M , y 5 x implies y E M. In order to prove that x , y E M implies x v y E M , observe that x E M implies the existence of an element S E S satisfying X A S = 8. Similarly, y E M implies the existence of an element t E S satisfying y A t = 0. Then s1 = S A is ~ an element of S and X A S ~= Y A S , = 0. It follows that ( x v y ) ~ s=, 8, showing that x v y is no member of S. In other words, x v y is a member of M. It has thus been proved that M is an ideal. The com-
CH.1 , 8 51
PRIME IDEALS
23
plement of M is the lower sublattice S, and so M is a prime ideal in view of one of the remarks made above. If M would properly include another prime ideal M * , then S would be properly included in the lower sublattice X- M*. This is not so since S is maximal, and so M is a minimal prime ideal. Conversely, assume now that Mis a mhimal prime ideal. Then S = X - M is a lower sublattice (simply because M is a prime ideal). If S is not maximal, there exists a maximal lower sublattice S* which properly includes S. Then, by what has just been proved, X- S* is a minimal prime ideal properly included in M . This is impossible, and so S is a maximal lower sublattice. (ii) Given the prime ideal P,the complement S = X-P is a lower sublattice. S is included in a maximal lower sublattice s*,and so the minimal prime ideal M = X- S* is included in P. (iii) Assume that M is a minimal prime ideal in X (hence M # X)and x is a point of X such that x and { x } ~ are contained in M . The set S = X - M is a maximal lower sublattice, so x E M implies that x ~ =y 6' for some y E S. It follows from x ~ =y 8 that y E { x } ~c M . Hence y E S and y E M hold simultaneously, which is impossible. But then x and { x } are ~ not simultaneously contained in M. (iv) Let M be a minimal prime ideal, and let x E M . We may assume that M # X. By part (iii) there is an element y E S = X- M such that y E { x } ~ . If { x } ~ were ~ not wholly contained in M , there would exist an element z E { x } such ~ ~ that z E S. Then both y and z are members of S, and y A z = 8. This is impossible since S is a lower sublattice. ~~ Note that the statement in (iv) is of interest only in the case that { x } is properly larger than the ideal generated by x (cf. Exercise 4.19). In the case that the distributive lattice with smallest element which we consider is a Boolean ring, the notions of proper prime ideal, proper minimal prime ideal and maximal ideal are the same (by Theorem 4.3). It follows that a subset S of the Boolean ring R is a maximal lower sublattice if and only if the complement R - S is a maximal ideal. Also, for example, Theorem 5.2 (where it was proved that I = n(P:P 3 I, P prime) for any ideal I, so (P: P prime) = {O}) and Theorem 5.3 (containing similar statements for minimal prime ideals) become identical now with Theorem 4.4 (where in a Boolean ring the same results are proved for maximal ideals). Finally, Theorem 5.4 (stating that a proper minimal prime ideal cannot contain x and { x } simultaneously) ~ becomes now the statement in Theorem 4.7 that in a Boolean ring any proper ideal cannot contain x and { x } ~simultaneously.
n
24
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH.I , S 6
6. Hulls and kernels; Stone's representation theorem As before, let X be a distributive lattice with smallest element 0. By B we denote the set of all proper prime ideals in X , b y 4 the set of all proper minimal prime ideals (if X contains at least one element x # 8, then every minilnal prime ideal is automatically proper), and by 9 the set of all ideals Q for which there exists an element x, not in Q and depending upon Q, such that Q is maximal with respect to the property of not containing x. As proved in Theorem 5.1, any ideal Q of this kind is prime, and in Theorem 5.2 it was actually proved that Z = ( Q : Q 3 Z,Q E 2 ) holds for every proper ideal I . In particular, we have (I (Q : Q E 9 ) = {O} unless X consists of 8 only, in which case the set 9 is empty. The sets .A%and 9 are subsets of 8, and if the lattice Xis a Boolean ring we have 9 = A = 22 = 2,where f denotes the set of all maximal ideals. Given the non-empty subset W of 9 and the element x E X , we let { R } x denote the set of all R E W such that x is no member of R. Note that {R}@ is empty and ( { R } x : x E X) = 9. In case X has a largest element e, we have { R } == W.
n
Definition 6.1. Let W be a non-empty subset of 8.For any non-empty subset W,of .9,the kernel k ( 9 , ) of W,is defined by
k(W,) =
n(R :
R E
w,),
and for W , empty we define k ( 9 , ) = X. Note that k ( W , ) is an ideal in X . For any non-empty subset D of X , the hull h(D) of D is defined by h(D) = ( R :R E W ,R
2
0).
Obviously,it depends not only upon the above definitions,but also upon the choice of W,what k ( W , ) and h(D) will turn out to be. The following simple facts will be proved first.
Theorem6.2. (i) W e have h ( k ( W l ) )3 9,for every W,c W,and k(h(D))3 IDfor every D c X,where ZD denotes the ideal generated by D. (ii) If 9 , = h(D)for some D c X,then h ( k ( 9 , ) ) = 9,. (iii) ZfZ = k(W,) for some 9, c 9,then k(h(Z)) = Z. (iv) If W = B or W = 2, then k(h(Z)) = Z holdsfor every ideal Z in X.
Proof. (i) Evident from the definitions. (ii) Assuming that g1= h(D), it is sufficient to prove that h ( k ( 9 , ) ) c W,.To this end, note that 9, = h(D) implies k ( W , ) = k(h(D))3 D, so
h(k(W,)) c h(D) = 9,.
CH. 1,§ 61
HULLS AND KERNELS; STONE'S REPRESENTATION THEOREM
25
(iii) Similarly. (iv) Let 9 = 8. Given the proper ideal Z in X , it follows from Z= n ( p : p 3 z , p E q that Z = k ( 8 , ) for Pl= (P: P 2 Z, P E P), so k(h(Z)) = Z by part (iii). A direct verification shows that k(h(Z)) = Z holds also for Z = X . The proof for 9 = 9 is similar. We will now assume that the subset 9 of B which we consider satisfies 0 ( R : R E 9 )= {O}. The subsets 9 = 9,9 =& and 9 = 9 satisfy this condition. Furthermore, given g1c 92, we denote the complement 9 - 9 1 by 9;.
Theorem 6.3. Let 9 c 8 satisfy the condition that the intersection of all R E W is {O}. Then given any ideal Z in X , we have Zd = k(h(Z)c).
Proof. If Z = X , then h(Z) is empty, and so h(Z)c = 92. It follows that k ( h ( z y ) = {el = z d . If I # X , let y E Zd and R E h(Z)". Then I is not included in R,and hence there exists an element x E Z such that x is no member of R. Since x E I and y E I d , we have x AY = 8. It follows that y E R (since R is prime and x is not in R).Hence, any y E I d is contained in any R E h(Zy. This shows that Id c k(h(Zy). Conversely, given any y E k(h(Z)"),we have to prove that X A Y = 8 holds for every x E I. It is sufficient for this purpose to prove that x A y E R for every R E 9 (since this implies that X A Y E ( R : R E 92) = {O}). If R E h(Z), then x E Z c R, SO X A Y E R. If R E h(Zy, then y E R, so x~ y E R. Hence x A y E R in either case.
n
Corollary 6.4. (i) Given again that the intersection of all R E 92 is {O}, we have { X I d = k({R}x) for every x E X . (ii) Zn the case that 9 =A,we have for every x E X .
h(k({M)x))= { M I X
Proof. (i) Given x E X , let Z be the ideal generated by x. Then Id = { x } ~ and h(I)' = {R}x.Hence, the formula to be proved is a special case of the formula in Theorem 6.3.
26
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.1 , s 6
(ii) We may assume that x # 8. It follows from { x } = ~ k ( { M } , ) that h({Xld)
= h ( k ( { M } x ) )=
{WX.
(1)
On the other hand, if the proper minimal prime ideal M satisfies M E h ( { ~ } ~ ) , i.e., if { x } c ~ M , then x E M is impossible by Theorem 5.4 (iii). It follows ) that x is no member of M , so M E {M},. This shows that h ( { ~ } c~ {M},, and so formula (1) becomes h ( { 4 d ) = h ( k ( { M } x ) )= { M I X .
Theorem 6.5. (i) We have {P},c {P},ifand only i f x 5 y . Hence {PI, = = y. (ii) We have { Q } , c { Q } , i f and onZy i f x 5 y. Hence { Q } , = { Q } , ifand onZy if x = y. (iii) We have { M } x c { M I , ifandonly if{^}^^ c {y}". Hence { M } , = { M } , ifand only i f { x } = ~ {y}" ~ or, equivalently, ifand only i f { x } = ~ {Y}~.
{P},ifand only i f x
Proof. (i) It is evident that x 5 y implies {P},c {P},.Conversely, assume that {P},c {P},holds, but x 5 y does not hold. Then x is not contained in the ideal Z, generated by y , so there exists a prime ideal P 3 Zysuch that x is no member of P. It follows that P E {PI, c {P},,so y is no member of P. This contradicts y E Z, c P. Hence x 5 y must hold. (ii) Similarly. (iii) { M } x c { M } , implies k ( { ~ } , 2) k ( { M } , ) , i.e., { x } = ~ { Y } in ~ view of Corollary 6.4 (i). It follows that { x } c~ {y}". ~ Conversely, { x } c ~ {y}dd ~ implies {x}ddd= {y}ddd,i.e., { x } ~ { Y } ~or, in other words, k ( { M } , ) = k ( { M } , ) . Then M I X ) )
=~
(ww,))Y
so { M } x c { M } , by Corollary 6.4 (ii).
Given the distributive lattices X and Y with smallest elements 8, and 8, respectively, the mapping n of X into Y is called a lattice homomorphism when n(8,) = O,, and n ( x l ) = y l , n ( x 2 ) = y z implies that n ( x l v x 2 ) = y1 v y , and n ( x l A x 2 ) = y1 ~y~ for all xl, x 2 E X . If the lattice homomorphism n is one-one and onto Y, i.e., if x1 # x2 implies n ( x l ) # n ( x 2 ) and every y E Y is of the form y = n ( x ) for some x E X , we say that n: is a lattice isomorphism of Xonto Y. It is a natural question to ask whether any distributive lattice Xwith smallest element 8 is lattice isomorphic to some lattice the elements of which are subsets of a point set, those subsets being partially ordered by inclusion and the empty set playing the role of smallest
CH.I , 0 61
HULLS AND KERNELS; STONE'S REPRESENTATIONTHEOREM
27
element. An affirmative answer is due to M. H. Stone ([l], 1936, for the case that X is a Boolean ring, and [2], 1937, for the more general case). Details follow. Once more, let W be a non-empty subset of 8, and for every x E X let {R},denote the set of all R E W such that x is no member of R. Note that is empty, and
" {R>y = {Rlxv,,
"
(2) for all x, y E X . It follows that the subsets {R},of W , for x running through X , form a distributive lattice Y (with respect to partial ordering by inclusion) with the empty set as smallest element. If X is a Boolean ring, and we have 0 5 y 5 x with z the complement of y with respect to x, then {R}=is the ordinary set theoretic complement of {R},with respect to {R},. If X is a Boolean algebra with largest element e, then {R}== 9 is the largest element in the image lattice Y , and Y is now also a Boolean algebra. W
X
{R>x
W
Y =
{RIXAY
Theorem 6.6 (Stone's representation theorem [2], 1937, for the case that
9 = 9').Given the distributive lattice X with smallest element 0 and the nonempty subset W of 9,the mapping x
{R>x
-i
is a lattice homomorphism of X onto the lattice Y of all Sets {R},.The mapping is a lattice isomorphism if W = 9' or W = 9 , but not necessarily so if a =A. If X is a Boolean ring, the mappings x -+ {P},,x -+ {Q}, and x -+ { M } , are all the game, and this common mapping is now a lattice isomorphism of X onto the Boolean ring of all subsets { P I x of 8.If X is a Boolean algebra, the sets {P},form also a Boolean algebra with B as its largest element.
Proof. It follows from the formulas (2) above that x -+ { R } xis a lattice homomorphism. It follows from Theorem 6.5 (i), (ii) that the mapping is a lattice isomorphism for W = 8 or W = 2. In the case that W = A , the lattice of all sets { M } , can be a Boolean ring without X being so. If for every pair of elements x , y in X satisfying B 5 y 5 x there exist y , , y , in X such that y , v y , = x, y , AY, = B and y , , y have the same disjoint complement, then { M } , = {M},, by Theorem 6.5 (iii), and so { M } Y 2is now the set theoretic complement of { M } , with respect to { M } x .This shows that the set of all { M } , is now a Boolean ring. We proceed with some properties of the sets { P } x ,{Q}, and { M } , which, in the next section, will give rise to topological compactness properties.
28
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH. 1, $ 6
Theorem 6.7. (i) Given the point xo E X and the set {x, : T E { T } } of points in X such that {P},, c {P},,,there exist indices z1, . ., T" in the index set { T } such that xo S sup (xTl, . ., xJ, and so { P } x , c ul=l{P}xz,. (ii) Given thepoint x o E Xand the set { x , : z E { T } } ofpoints in Xsuch that {Q}x, c { Q}x,, there exist indices z1, . . ., z, in the index set { T } such that xo 6 sup (x,, , . . .,x,,), and so {Q},,c u1= 1 {Q},,,.
u,
u,
.
.
u,
Proof. (i) Assume that { P } x , c {P},,,and for any z, let y, = xo A x,. Then 8 6 y, 6 xo holds for all T, and {P},,= {P},..We have to prove that xo is equal to the supremum of a finite number of the y,. If not, xo is no member of the ideal I generated by the set of all y,. Hence, in this case there exists a prime ideal P such that P =I I and xo is no member of P. It follows that P E {P},,,but for no T the ideal P is a member of {P},,. , simply because y, E I c P for all T. This contradicts {P},,= {P},,,.Hence, we must have xo I sup (x,, , . . .,x,") for appropriate z1, ., z,. (ii) The proof is similar, since the prime ideal P in part (i) can be chosen so as to be maximal with respect to the property of not containing xo
u,
u, ..
.
Theorem 6.8. Let % . ? be a non-empty subset of 9' such that&? c 9, and let { x , : z E {T}} be a set ofpoints in Xsuch that the collection of all {R},%has the
finite intersectionproperty (i.e., everyjinite intersection of sets in the collection is non-empty). Then the intersection of all {R},* is non-empty.
n;=l
Proof. Since any finite intersection {R},%,is of the form for y = inf (xT1 ,. . .,xTn),it follows from the finite intersection property of the collection of all {R},c that the collection of all points x,, together with their finite infima, is a lower sublattice in X . This lower sublattice is included in a maximal lower sublattice S and so M = X - S is a minimal prime ideal. we have Since no x, is a member of M and since M is a member of 9, ME {R}x*,i.e., the intersection of all { R } x sis non-empty. As observed above, it can happen that the lattice of all { M } xis a Boolean ring without X itself being so. In this case a theorem analogous to Theorem 6.7 holds, as follows.
0,
Theorem 6.9. r f the lattice of all { M } x is a Boolean ring, and i f the points E X (z E {z}) satisfv { M } , c { M } x s ,there exist indices z l , ., r, in the index set { T } such that {M},, c Ulil {M}xsd. xo E X , x,
u,
..
Proof. Replacing, if necesary, all x, by xo AX,, we may assume without loss of generality that 0 S x, 6 xo holds for all r, and so { M } x o= { M } x s .Since the set of all { M } xis a Boolean ring, the set theoretic com-
ur
CH. 1.8 71
THE HULL-KERNEL TOPOLOGY
29
or u;=l
plement of any {M},* with respect to {M},,, is of the form {M},,,; it follows now from {M},,, = {M},* that { M } , s is empty. But then, by the preceding theorem, there must be indices z1 , . . ., T,, such that nl=l{ M}yri is already empty, i.e., { M } x o= {M)xri.
ut
Exercise 6.10. Let X = (xo, x, ,. . ., x,,) with n 2 2, linearly ordered by defining that xi 5 xk f o r j 4 k. Show that Xis a distributive lattice with xoas smallest element and x,, as largest element, but X is no Boolean algebra. Show that every ideal in Xis of the form ( y :y 5 xk) for some k. Show that every ideal is a prime ideal, the ideal ( y :y 4 x,,-~) is the only maximal ideal, every ideal ( y :y xk) is maximal with respect to not containing x k + , and ( y :y = xo) is the only minimal prime ideal. Given x E X , determine explicitly what {P},,{Q}, and { M } , are. Check the statement that = {M},, holds if and only if {x}dd= {y}ddholds, and show that {x}" = {y}dd is not the same as x = y. Finally, determine explicitly the Stone representation of X .
,
7. The hull-kernel topology As in the preceding section, let 9 'be the set of all proper prime ideals in the distributive lattice X with smallest element 8, and let 9 be a non-empty subset of 9. Also, just as before, given x E X , let {R}xbe the set of all R E 92 such that x is no member of R. We recall that {R}#is empty and ( { R } , : x E X )= 9.
We introduce a topology in 92 by choosing all sets of the form {R}, as a {R}xiis of the subbase for the topology. Since any finite intersection form {R}xofor xo = inf (x, , . . ., x,,), the sets { R } , form actually a base for the topology. Theorem 7.1. Let 9be topologized as explained above. For any ideal Z in X, the hull h(Z)={R:RE92,R=Z} is a closed set in this topology. Conversely, every closed set is the hull of some ideal I. Furthermore, for any subset W,of 9, the closure of 9, is the set h(k(W,)), and for this reason the topology is usually called the hull-kernel topology in 9. The hull-kernel topology in 9 is the relative topology of the hull-kernel topology in 9.
30
x
DISTRIBUTIVE LAlTICES AND NORMED FUNCTION SPACES
[CH.I , $ 7
Proof. The closed sets are the intersections of sets of the form { R :R E W , the closed sets are the sets
E R}, i.e.,
h(D)={R:REW,DcR}, where D is an arbitrary non-empty subset of X . Observing now that h ( D ) = h(ID),where ID is the ideal generated by D, we obtain the result that a subset of 9is closed if and only if it is the hull of some ideal in X . For the proof of the statement about the closure of a subset of 9, let 9, be an arbitrary subset of 9. Evidently 9, is contained in the closed set h ( k ( 9 , ) ) ; in order to prove that h ( k ( 9 , ) ) is the closure of 9, we have to Any 9,of this show that h ( k ( 9 , ) ) is included in any closed set g23 9,. kind is of the form W , = h ( I ) for some ideal I in X , so h ( I ) 2 9,. Then
W W ) =) h(k(91))9 i.e., h ( I ) 3 h ( k ( 9 , ) ) by Theorem 6.2 (ii). But h(1) = W 2 , and so .g2ZJ h(k(W,)) which is the desired result. Finally, it follows immediately from the definitions that the hull-kernel topology in W is the relative topology of the hull-kernel topology in 9.
Corollary 7.2. Let the subset 9 of B satisfy the extra condition that the intersection of all R E W is {e}. Then, in order that the subset Y of X be the disjoint complement I d of some ideal I , it is necessary and suficient that Y be the kernel of some open subset of 9. Proof. Follows from the formula I d = k ( h ( I Y ) , proved in Theorem 6.3. We recall that a topological space is called a To-space (Kolmogorof space) whenever, given any two different points in the space, one at least of these points has a neighborhood not containing the other one.
Theorem 7.3. The hull-kernel topology in B is a To-topology, the base sets { P } xof which are open and compact. UP,and P, are points of B such that, considered as ideals in X , neither of them is included in the other one, then P, and P, can be T,-separated, i.e., each of P , and P2 has an open neighborhood not containing the other point. This holds in particular if P , and P, are minimal prime ideals or if P, and Pz are both maximal with respect to the property of not containing the same point x,, E X . Proof. If PI # P,,there exists a point x , E P, such that x , is no member of P, or there exists a point x2 E P2 such that x2 is no member of P,. In the first case { P } x ,is an open neighborhood of P, such that P, is not contained in this neighborhood. In the second case { P } x , is an open neighborhood of
CH.
1,
6 71
THE HULL-KERNEL TOPOLOGY
31
P, such that P, is not contained in it. It follows that the hull-kernel topology in 9 is a To-topology. Evidently, every set {P},is open by definition. By Theorem 6.7 (i) every set {P},is compact. If PIand P, are prime ideals such that neither of them includes the other, there exist points x1 E PI and x2 E P, such that x , is no member of P, and x2 no member of P,.Then { P } x ,is an open neighborhood of P, not containing P, and {P},,is an open neighborhood of P, not containing P, . Theorem 7.4. (i) The hull-kernel topology in 2 is a To-topology, the base sets { Q } , of which are open and compact. (ii) The hull-kernel topology in Jl is a Hausdorf topology, the base sets { M } , of which are open as well as closed. The topology is, therefore, completely regular. Proof. (i) Similarly as in the preceding theorem. The compactness of the sets {Q}, follows from Theorem 6.7 (ii). (ii) The hull-kernel topology i n d is a T,-topology because M , # M , implies that neither of M , and M 2 is included in the other one. The base sets { M } , are closed by Corollary 6.4 (ii), stating that h ( k ( { M l x ) )= { M I X holds for every x E X . This implies that M1 has a closed and open neighborhood not containing M , . But then the complement of this neighborhood is an open and closed neighborhood of M , . Hence, the topology is Hausdorff. In order to show that the topology is completely regular, observe that if the point Mo €4 and the closed subset f o f J l are given such that Mo is no member of $, then the open s e t A - f contains a neighborhood of M o of the form { M } , o for some xo E X . The functionf( M ) , satisfyingf(M) = 1 for all M E andf( M ) = 0 for all other M , is now continuous on 4 on account of { M } x obeing open as well as closed. We recall that a topological space is sometimes called totally disconnected if there exists a base consisting of sets which are open as well as closed. Note that in this case the collection of all sets which are open as well as closed is also a base. According to the last theorem the hull-kernel topology in A?is a totally disconnected Hausdorff topology.
Theorem 7.5. (i) In the particular case that the lattice of all { M } x is a Boolean ring, the hull-kernel topology in A is a Hausdorftopology, the base sets { M } = of which are open as well as closed and compact. The topology is, therefore, totally disconnected, locally compact and completely regular. Every
32
[CH.1,s 7
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
subset ofA? which is open as well as closed and included in one of the { M} , , say in { I W } ~is~itserfof , the form { I W } ~for , , some y o 5 x o . (ii) In the particular case that Xis a Boolean ring, the hull-kernel topology in B = 9 =A? has all the properties mentioned in (i). Proof. (i) The compactness of the sets { M } x follows from Theorem 6.9. For the proof of the last statement in (i), let be a subset of A,open and closed and such that A? c {M}x,,.Since a is a closed subset of the compact set { M } x o the , set &? is compact. Also, since B is open, 9 is the union of a collection of base sets {M},- with x, ixo for all z. By the compactness 93 . But then is already a finite union of base sets, say of { M } x z l ,.. ., a = { M } y ofor yo = sup (x,,, . . ., x,,). (ii) Note that it was observed in the final paragraph of section 5 that for the prime ideals in a Boolean ring we have B = 2 = A . Let the topological space S with points p , q, . be a To-space. For any p , q E S, set p S q whenever q E { p } , where (as usual) { p } denotes the closure of the set consisting of the point p only. It is easily verified that this defines a partial ordering in S (the hypothesis that S is a To-spaceis used in order to prove that p S q, q 5 p implies p = q). Obviously, the partial ordering is the relation of equality if and only if the space is a T,-space (i.e., if and only if every point is a closed set). We apply these remarks to the topological space 9 of all proper prime ideals in the distributive lattice X with smallest element 8, the topology being the hull-kernel topology.
..
_ .
Theorem 7.6. (i) Introducing a partial ordering in B by defining that P1 I P2whenever P2E {PI}, we have P , 5 P , ifand only i f P , c P,, and hence P, E {P,}ifand only $PI c P , .
(ii) For any x # 8 we have P E (p},i f and only i f there exists a minimal prime ideal M E { P } x such that M c P . It follows that
and so certainly
mx u (0: mx u =
=
M E d , ME {p>x),
((P>:P€95,PE{P},).
-
Proof. (i) Given that P, c P,, we have to prove that P, E {Pi}, i.e., we have to prove that every neighborhood of P, contains P,,and it is sufficient to do this for any neighborhood of P, of the form { P } x .Hence, assume that P2E {P},,i.e., x is no member of P,. Then x is no member of P,,so P, E ___ { P } x . This is the desired result. Conversely, if P, E {P,},then every {P},
CH. 1,g 71
33
THE HULL-KERNEL TOPOLOGY
satisfying P2E { P } x contains P,,so if x is no member of P2 then x is no member of P, In other words, x E PI implies x E P2,i.e., P, c P2. (ii) Let first x # 8, M c P and M E { P } x . Then, by part (i), we have __ P E { M } , and so it follows from M E { P } x that P E { M } c { P } x . Conversely, assume now that Po E { P } x . Then every neighborhood of Po has a non-empty intersection with { P } x ; in particular, every set {P}, for which y is no member of Po has a non-empty intersection with { P } x . It follows that the collection of all sets
.
~~
({P } , n { P } x :y no member of Po) has the finite intersection property, and so by Theorem 6.8 the intersection of all these sets is non-empty. This non-empty intersection contains at least one element (i.e., at least one prime ideal), and hence it contains at least one minimal prime ideal M (since every prime ideal contains a minimal prime ideal). Then M E { P } x and also M E { P } , for ally not in P o , i.e., y is not in M if y is not in Po.In other words, we have M E ( P } x and M c Po. The theorem shows that in the topology of 9 the closure of any set { P } x is simply the union of the closures of each of its points. It is even so that the closure of { P } xis already the union of the closures of the points in { P).r that are minimal prime ideals. Since 9 = { P } x ,it follows that
u
s=U(p}x=(J((M):MEA), where the closures are to be taken, of course, in the topology of ?P.It follows immediately that JZ (as a subset of 9) is dense in 9, i . e . 2 = 9.
Exercise 7.7. Show that A, in its hull-kernel topology, is a Baire space (i.e., any countable intersection of open and dense sets is dense). Hint: Given that the sets 0, ( n = 1, 2 , . . .) are open and dense, it is xo # 8, has a non-empty intersection sufficient to prove that every 0,. Every 0, is of the form 0, = {M}:). Since 0, is dense, with { M } x oand 0, have a non-empty intersection, so y , = inf (xo, xs:)) # 8 for some XI:). Since 0,is dense, { M } , , , and O2 have a non-empty intersec(2) ) # 8 for some We thus obtain a lower tion, so y 2 = inf (xo, x,,( 1 ) ,x,,
n
u .':5~
sublattice of which xo and all x5",'are members. The set theoretic complement (with respect to X ) contains a minimal prime ideal M such that M is in the intersection of {M}x,,and 0,. x,
Exercise 7.8. We use the notations of Exercise 6.10. Show that for x (k 2 1) we have { P } x = ( P : P = (x,,, . . ., x,) with m < k).
=
34
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.1,$8
and for x = xo the set {P},is empty. Show now that the open sets in the hull-kernel topology of 9 are exactly the sets {P},,and so the closed sets are the empty set and the sets of the form (P: x E P)for some given x E X . Derive from these facts that the hull-kernel topology in 8 is a To-topology, but no Tl-topology. Given Po = ( x o , . . ., x k ) in 8, show that
{ P o } = (P: P = ( x o , . . ., x,) with m 2 k ) ,
-
and so theonly minimal prime ideal M = ( x o ) satisfies { M } = 8. Also, show that {P},= 8 for x # xo in X . Compare these results with the statements in Theorem 7.6.
8. Compactness and separation properties of the hull-kernel topology We use the same notations as in the preceding section, and we recall it was proved that all sets { P } x are compact in the hull-kernel topology of 9, all sets { Q } , are compact in the hull-kernel topology of Q , but a set of the form { M } , need not be compact in the hull-kernel topology of&. We will investigate now under which conditions we have (i) compactness of 8, (ii) compactness of 9, (iii) compactness of {M},,, for a given xo E X , (iv) compactness of all { M } , , (v) compactness of&. We also recall that, in general, 8 is a To-spacebut not necessarily a Tl-space. It will be investigated under which conditions we have that (vi) 8 is a TI-space, (vii) 9 is a Hausdorff space, (viii) B is a compact Hausdorff space. Theorem 8.1. (i) 8 is compact in its hull-kernel topology i f and only if X has a largest element. (ii) 2 is compact in its hull-kernel topology if and only if X has a largest element.
uy=l
u,
Proof. (i) If 9is compact, it follows from 8 = { P } xthat B is already a finite union of sets {P},,say B = {P},,.But then 9 = {P},,holds for xo = sup (xl, . ., xn). Hence, since { P}, c B = { P},, holds for every x , we have x 5 xo for every x E X (cf. Theorem 6.5), which shows that xo is the largest element of X .
.
CH. l , § 8)
COMPACTNESS AND SEPARATION PROPERTIES
35
Conversely, if X has a largest element e, it follows from x 6 e that { P } x c {P},for all x, so 9 = { P } x = {P},.Every { P } x is compact by Theorem 7.3; hence, in particular, 9 = {P},is compact. (ii) The proof for 2 is similar. Before proceeding with our program, we introduce some further terminology and notations. The elements x, y in X will be calledd-equivalent whenever { M } x = { M } , , i.e., whenever { x } = ~ ~ { y } d d(cf. Theorem 6.5). We will write x 3 y(&) in this case. Note that x = O(&) implies x = 0, and x1 E Yl(&), x2 G y z ( d ) implies x1vx, E y , v y z ( A ) and x1 AX, E
u
Y1 A Y Z ( 4 .
The element xo E X is said to have the A-complementation property if, given any y E X satisfying 0 5 y 5 xo,there exist y l , y , in X such that y1 G y ( A ) , y 1 A y 2 = O and y 1 v y 2 = x o ( A ) . It is evident from the remark above that, if so desired, we may assume without loss of generality that y 1 = y . It follows immediately that xo has the A-complementation property if and only if, for any y such that { M } y c { M } x o ,there exists an element y , E X such that { M } x ois the disjoint union of { M } , and { M } y I .In other words, xo has the &-complementation property if and only if the initial segment ({MI, : = {MIXO)
{my
of the lattice of all { M } x is a Boolean algebra. Theorem 8.2. The set { M } x ois compact in the hull-kernel topology of i f xo has the &-complementation property. In this case, every subset of {M}x,, which is open as well as closed is of the form for some x E
if and only
x.
Proof. Assume first that xo has the &-complementation property, i.e., the initial segment ({MI, : { M I ,
=W
L O )
is a Boolean algebra. In order to prove that { M } x ois compact, it is sufficient to prove that if { M I x oc { M } x s ,then { M } x ois already covered by a finite union of the {M}.+. Replacing, if necessary, every x, by x0hxI, we may assume without loss of generality that O x, 5 xo holds for all z, and so = { M } x c .Since xo has the A-complementation property, the set theoretic complement of any { M } , - with respect to { M I x ois of the form { M } , - ;it follows that { M } y zis empty. By Theorem 6.8 there exist indices zl, . . .,z, such that {M},ri is already empty, i.e., { M } x o =
u,
u
U = l
n,
{~lx,*.
nyZl
36
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH. 1,s 8
Conversely, assume that { M } x ois compact. Then, given any { M } , satisfying { M } , c { M } x o ,the set theoretic difference D = { M } x o - { M } yis an open and closed subset of { M } x o .Since D is open, we have D = {M},* for suitable elements x,; since D is closed and { M } x ois compact, we have that D is compact. Hence, similarly as above, D is a finite union of sets { M } x , ,. . ., { M } x nand , so D = { M } y 2 for y 2 = sup ( x l , . . . , x n ) . This shows that xo has the A-complementation property. The last part of the proof shows also that any subset of { M } x owhich is open as well as closed is of the form { M } xfor some x E X.
u
Corollary 8.3. AN sets { M } xare compact in the hull-kernel topology of JZ
if and only if the set of all { M } x is a Boolean ring.
Proof. Follows immediately from the last theorem. Note that one half of the present corollary was proved already in Theorem 7.5. Theorem 8.4. The following conditions are equivalent. (i) A is compact in its hull-kernel topology. (ii) All sets are compact in the hull-kernel topology o f A and there exists an element xo E X such that { x ~=}X ~(an~ element xo of this kind is sometimes called a weak unit in X ) . (iii) The set of all { M } xis a Boolean algebra.
-
Proof. (i) (ii) Every set { M } x is a closed subset of the compact space .A, so is compact. Furthermore, it follows from the compactness of A and f r o m A = { M I x that& is already a finite union of sets { M } x , say .A = { M } x , . But then d Z = { M } x ofor xo = sup ( x i , . ., x,,). Hence, since { M } x c A = { M } x ofor every x E X , we have x E { x } c~ ~ { x , , } ~for ~ every x E X (cf. Theorem 6.5), and so { x ~=}X.~ ~ (ii) * (iii) All { I V }are ~ compact, so the set of all { M } x is a Boolean ~ x~E X , we ring by the preceding corollary. Since { x } c~ X~ = { x ~for} all have { M } x c { M } x ofor all x E X , so .A = U { M } x = { M } x owhich , shows that { M } x ois the largest element in the Boolean ring of all In other words, the Boolean ring is a Boolean algebra. (iii) * (i) Let { M } x obe the largest element in the Boolean algebra of all = { M } x = { M } x o .Since the set of all { M } xis certainly sets { M } x ,SO a Boolean ring, all sets { M } xare compact by the preceding corollary. Hence d = { M } x ois compact.
uy=l
u
u
Theorem 8.5. The following conditions are equivalent. (i) Every proper prime ideal in X is a maximal ideal. (ii) 9 = A.
.
CH.1 , § 8 )
COMPACTNESS AND SEPARATION PROPERTIES
37
(iii) 9 is a Hausdorff space in its hull-kernel topology. (iv) 9 is a T,-space in its hull-kernel topology.
Proof. (i) * (ii) Let every proper prime ideal in Xbe maximal, and assume that P1 and P2 are proper prime ideals such that P, c P,.Then P, = P2 because P2 is maximal. This implies that any proper prime ideal P1 cannot properly contain another prime ideal, i.e., P1 is a minimal prime ideal. (ii) => (iii) Follows from Theorem 7.4 (ii). (iii) =-(iv) Evident. (iv) * (i) Let B be a T,-space, and assume that P, and P, are different proper prime ideals. Then P, c P2 is impossible. Indeed, if P1 c P2,then P, E { PIx implies P, E { P } x,so every neighborhood of P, is also a neighborhood of P,, contradicting the assumption that P, has a neighborhood not containing P,. Similarly, P, c P, is impossible. Hence, neither of Pi and Pz is included in the other one. This implies now that every proper prime ideal is a maximal ideal, because if there exists a proper prime ideal P1which is no maximal ideal, then P, is properly included in a proper ideal Z. But, by Theorem 5.2, Z is included in a proper prime ideal P2,so P1 c P, with proper inclusion. This yields a contradiction.
Theorem 8 . 6 . 9 is a Hausdorfspace in its hull-kernel topology ifand only if X is a Boolean ring. Proof. Assume first that @ is a Hausdorff space. Then 9 = .,# holds by the preceding theorem. It follows that the lattice of all { P } xis the same as the lattice of all { M } x ,and the lattice is lattice isomorphic to X by Stone’s re= { P } xare compact in presentation theorem (Theorem 6.6). Now, all the hull-kernel topology of .A = 9 by Theorem 7.3, so the set of all { M } x is a Boolean ring by Corollary 8.3. Hence, the isomorphic lattice Xis also a Boolean ring. Conversely, if Xis a Boolean ring, we can apply Theorem 7.5 (ii), according to which 9 = 9 = .,# holds and all properties mentioned in Theorem 7.5 (i) hold; in particular, the hull-kernel topology in 9’= L2 = .,# is Hausdorff. Corollary 8.7. 9 is a compact Hausdorff space in its hull-kernel topology i f and only if X is a Boolean algebra.
Proof. According to Theorem 8.1 9 is compact if and only if X has a largest element. According to the last theorem B is Hausdorff if and only if Xis a Boolean ring. Hence, 9 is compact and Hausdorff if and only if X is a Boolean algebra.
38
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.
I,§ 8
Theorem 8.6 and Corollary 8.7 are due to M. H. Stone (Theorems 17 and 18 in [2], 1937). For most of the results in sections 4-8 there exist analogous results in Riesz spaces, as will be shown in the next chapters. This holds in particular for the results about prime ideals; several theorems about the sets and Jl and the hull-kernel topologies in these sets are quite similar to the corresponding theorems for Riesz spaces. A great part of these theorems for Riesz spaces is due to D. G. Johnson and J. E. Kist ([l], 1962), extending earlier results of K. Yosida, H. Nakano and I. Amemiya. For details we refer to sections 33 and 35-37.
Exercise 8.8. In this exercise, Xis a distributive lattice with null element 8. Notations are the same as before. Given the non-empty subset D of X , we consider the open subset { P } Dof 8,defined by =
u
({P>x
:
D)-
(i) Show that { P } D = { P}i,, where ID is the ideal generated by D . (ii) Show that for ideals 11,I2in X we have { P } I , = { P } l , if and only if Il = I,. Furthermore, show that every open subset 0 of 8 is of the form 0 = { P } l for some ideal I in X; more precisely,
O = { P } i for I = ( x : { P } x c 0). Finally, show that if I is the ideal generated by the element x E X (i.e., I = ( y : y j x)), then { P } I = { P } x .
(iii) The set of all ideals in X , partially ordered by inclusion, is a lattice with I1 A 1 2 = 11n 1 2 . 11 V 1 2 = ( X I V X2 : X1 E 1 1 , X2 E 1 2 ) , The set of all open subsets of 8, partially ordered by inclusion, is also a lattice with supremum and infimum of two open sets equal to union and intersection respectively. Show that the mapping Z --+ { P } I is a lattice isomorphism of the lattice of all ideals in X onto the lattice of all open subsets of 8. Note that these lattices are order complete. (iv) For any subset d of 8,we denote the set theoretic complement of d by dc, the interior of d by do, and the closure of d by d - .Show that, for any non-empty subset D of X , we have {P}Dd =
{P};,
{P}Ddd
=
{P};".
(v) We recall that a subset A of a topological space is called regularly open whenever A = A - " . Show that the ideal I in X satisfies I = Zdd if and only if { P } i is regularly open. We recall that the collection of all regularly
CH. 1 , § 81
COMPACTNESS AND SEPARATION PROPERTIES
39
open subsets of 9 is an order complete Boolean algebra (cf. Exercise 4.16). Show that the lattice of all ideals Zin XsatisfyingZ = Iddis lattice isomorphic to the Boolean algebra of all regularly open subsets of 9. Hence, the lattice of all ideals Z in X satisfying Z = Zdd is also an order complete Boolean algebra. (vi) The ideal Z in X is called a projection band if X = I v Z d , i.e., every x E Xis of the form x = x1 v x, for some x1 E Z and some x, E I d . Show that x, and x2 are uniquely determined by x. Also show that if x = x1 v x 2 , y = y , v y , , then x y holds if and only if x, S y1 and x, 6 yz hold. Show now that X = Zv I d implies indeed that Z is a band, i.e., if x = sup x, and all x, are in Z,then x is in I. Show that the ideal I is a projection band if and only if {P},is an open and closed subset of 9. Show that the following statements for the ideal I are equivalent. (a) I d is a projection band. (b) Zdd is a projection band. (c) { P};' is open and closed. (d) { P}; is open and closed. Hint: For the proof that { P } I , = {P},,implies I, = Z,, observe that if {P},,= {P},,then any { P } xwith x E I, is covered by the union of all {P},, with y E Z2 so it follows from the compactness of { P } x that there exists a finite subcovering {P},,,. Then x 6 yo for yo = sup ( y , , . . ., y,), and so XEI,. For (vi), assume that X = ZvZd and x = x1 vx, = x i vx; with x, , x i E Z and x,, x; € I d . Then xAxl = xi = xi A X , and X A X ; = x i = x1 A x i , so x1 = x i , and similarly x, = x i . Now assume x = x, v x,, y = y , v y , and x S y . Then x1 S y , so x1 A Y = xl. But then x1 ~ y =, xl, i.e., x1 S y , . Finally, for the proof that Z is a band, let x = sup x, with all x, in I. Set x = xI v x, with x1 E Z, x, E I d . Then x1 2 x, by what has just been proved, so x1 is an upper bound of the set of all x,. On the other hand we have x, 6 x = sup x,. It follows that x = x, so x E I.
uI=,
Exercise 8.9. In this exercise Xis a distributive lattice with null element 0. Notations are the same as before. We assume that the hull-kernel topology in 9 has the property that incomparable points P,, P, of 9 have disjoint neighborhoods. (i) Show that for every Po E 9 the set {Po}(i.e., the set of all P E 9 satisfying P =I Po)is linearly ordered. (ii) Show that for any given M E { L I ~there } ~ ~exists exactly one prime
40
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH. 1, $ 8
ideal Q 3 M such that Q is maximal with respect to the property of not containing x, Denote this ideal Q by QM,xo. Show that for any prime ideal Q, maximal with respect to the property of not containing x,, there exists a minimal prime ideal M E { M } , o such that Q = QM,xo. For x, fixed, we denote Q M , xo briefly by Q M . The mapping M -, QMis, therefore, from { M } x o onto {Q}xO, where { Q } , O denotes the set of all prime ideals maximal with respect to the property of not containing x,. Show that, given the prime ideal P not containing xo ,there exists a minimal prime ideal M in { M } , o such that M c P c QM.In fact, this holds for every minimal prime ideal M included in P. (iii) Show that the following statements are equivalent. (a) Every proper prime ideal includes a unique minimal prime ideal. (b) For all x , y E X, if { P } x n {P},is empty (i.e., if X A =~ O), then { P } xn { P}, is empty. (c) For every xo # 8, the mapping M -+ QM of { M } x oonto { Q } , O is a one-one mapping. (iv) For every x E X , let I, denote the ideal generated by x (i.e., I, = ( y :y S x ) ) . Under the additional assumption that I:' is a projection band for every x E X , show that (a), (b), (c) in part (iii) hold. Hint: For part (i) assume that P1 3 Po, P2 3 Po and P, ,P, are incomparable. Then, by hypothesis, there exist disjoint neighborhoods { P } x , of PIand { P } x , of P, . It follows that x , A x , = 8 and hence one at least of x , and x , , say x1, is an element of the prime ideal Po.But then x, E Po c P,, contradicting P, E {P},, . Hence P, 2 Po,P, Po implies that P, and P, are comparable, so {Po}is linearly ordered. For part (iii) we prove(a) *(b) => (c) * (a). For (a) => (b), assume { P } x n { P}, empty, but { P}, n { P } y contains an element P. By Theorem 7.6 (ii) there exist minimal prime ideals M , and M , , included in P,such that M I E { P } x and M , E { P } y , so M I # M , since {P},and {P},aredisjoint. On the other hand we have M 1 = M , by hypothesis (a). For (b) => (c) we have to prove that M , # M , in { M } x oimplies QM1# Q M 2It. follows from M , #M 2 that M , and M , have disjoint neighborhoods { P } x and { P}, in 8. Then {P},and {P},are disjoint by hypothesis, so { M , } and {M,} are disjoint, which implies that Q M ,# Q M , . For (c) (a) assume that P is a minimal prime ideals included in P. Let proper prime ideal and M , , M , are xo be a point of X not in P. Since { P} is linearly ordered there exists a unique Q, E { Q } , O such that Q, 3 P.In the mapping M -+ QM of {M},o onto {Q}xO both M , and M , have Q, as image. Hence, by (c), we have M , = M,.
.
= J
CH.1,s 91
41
NORMED KOTHE SPACES
For (iv), observe that if 2 is a projection band for every x, then { P I x is open and closed for every x (cf. the precedingexercise).It is sufficient to prove that if { P } x and {P},are disjoint, then { P } xand {P},are disjoint. Without any restrictions, - it follows from the disjointness of the open sets { PIx and {PI,, that } x and { P},are disjoint. Since { P } x is open, it follows - { Pthen that { P I x and {P},are disjoint. ~
9. Normed Kothe spaces Let ( X , A , p) be a measure space, as defined in section 3. We shall assume that the CarathCodory extension procedure has been applied already to p, so that A is, therefore, the a-field of all p-measurable subsets of X. It will also be assumed that sets differing only a set of measure zero are identified. In other words, A is identified with the Boolean algebra A / A o ,where A , is the ideal of the sets of measure zero (cf. section 3). The set of all realvalued (and finitevalued) p-measurable functions on X will be denoted by M ( r ) ( X , p ) ,or briefly M") whenever confusion is impossible. More precisely, functions which differ only on a set of measure zero are identified, and M") is the set of the thus obtained equivalence classes of real p-measurable functions. Incidentally, this identification of p-almost equal functions is a natural consequence of the identification of p-almost equal subsets of X . It can be allowed now that a representing function in some equivalence class of M") assumes the values co or - co on p-null sets. Evidently, M'" is a real linear space with respect to addition and multiplication by real numbers defined pointwise p-almost everywhere. Similar remarks hold for the set M = M ( X , p) of all complexvalued p-measurable functions. The set M is a complex linear space. Given the real number p satisfying 0 < p < CO, the set L, consists by definition of all p-measurable complex functionsf satisfying If(x)IPdp< 00, where the integration is over the whole set X . The set L, is a linear subspace of M, and for 1 S p < co the spaceL, is a Banach space with respect to the norm
+
Ilfll,
=
(J
Xl f l P q p .
The space L, consists by definition of all essentially bounded p-measurable functions (the functionf is called essentially bounded if there exists a finite number C > 0 such that If(x)l S C holds for all x E X , except at most on a p-null set). The space L, is a Banach space with respect to the norm
Ilfll,
=
ess SUP If(x)l.
42
DISTRIBUTIVE LATTICES A N D NORMED FUNCTION SPACES
[CH.1, 5 9
The spaces L, (1 5 p 5 03) are special examples of normed Kothe spaces. We present a brief survey of the main definitions and properties of these spaces. For a more detailed account with proofs we refer to Chapter 15 of the book on integration by A. C. Zaanen ([I], 1967). For the sake of simplicity, we will assume that the measure p is o-finite, i.e., Xis the union of a finite or countable number of sets of finite measure. By P we will denote the set of all non-negative p-measurable functions, where we will permit also that a function in P assumes the value 03 on a set of positive measure. We assume now that to each f E P there corresponds a number p(f) such that (a) 0 5 p(f) S co (the value +03 is, therefore, admitted), p(f) = 0 if and only iff(.) = 0 for p-almost every x , p(af) = a p ( f ) for every finite constant a 2 0, p ( f + g ) 5 p ( f ) + p ( g ) for allf, g E P, (b) i f f , g E P and f(x) 5 g ( x ) for p-almost every x , then p(f) 5 p(g). The function p( f),thus defined on P and mapping P into the non-negative numbers, is called afunction norm. The definition of p(f) is extended to complexvalued p-measurable f by defining that p ( f ) = p(lf1). We denote by L, the set of allfsatisfying p ( f ) < 00. It is easily shown that any f E L, is p-almost everywhere finitevalued, and this guarantees that L, is a linear subspace of the linear space M which was introduced above. The function p is a norm on Lp; the normed linear space L, is now called a normed Kothe space. The function norm p is said to have the Riesz-Fischer property if, for any sequencef, (n = 1 , 2 , . . .) in Lp such that EF p ( f , ) < 03, we have Zy If.1 E L , , i.e., p( EF)l.f1 < co. It was proved by I. Halperin and W. A. J. Luxemburg (1956) that the space Lp is norm complete (i.e., Lp is a Banach space) if and only if p has the Riesz-Fischer property (cf. Zaanen [I], Theorem 64.2). The function norm p is said to have the Fatou property if it follows from 0 S f i S fi S . .tf, with all f,E P, that p(fn) p(f). It can be proved easily that the Fatou property implies the Riesz-Fischer property, but not conversely (cf. Zaanen [l], Theorem 65.1). Hence, if p has the Fatou property, then L, is norm complete. Note that the L, norm, for 1 p 6 03, has the Fatou property. For any p-measurable subset E of X , we shall denote the characteristic function of E by x E . If E is a p-measurable subset of X such that p ( x F ) = co for every subset F of E satisfying p ( F ) > 0, then E is called a p-purely infinite subset. If there exist no p-purely infinite subsets, the norm p is said to be saturated. It follows easily that any set E of positive measure is p-purely
+
.
CH. 1,s 91
43
NORMED KOTHE SPACES
infinite if and only if any f E L, vanishes p-almost everywhere on E. Any p-purely infinite set is of no importance, therefore, for the study of the space L,; hence, p-purely infinite subsets of X can just as well be removed from X . It can be proved that there exists a maximal p-purely infinite subset of X . Removing this subset from X , we arrive at the situation that p is saturated. Without loss of generality we may assume, therefore, that p is saturated from the very beginning (cf. Zaanen [I], Theorem 67.2). For any f E P, the number p ’ ( f ) is defined by
(
p’(f) = SUP Ixfg & :9 E P Y p(g)
s 1)
*
Due to the hypothesis that p is saturated, it follows easily that p’ is a function norm with the Fatou property (cf. Zaanen [l], Theorems 68.1 and 68.4). The corresponding normed Kothe space L,, is called the associate space of L,. For functions f E L, and g E L,.,the Holder inequality
holds (cf. Zaanen [l], Theorem 68.2). If g E L,. is given and we define G on L, by r
for everyf E L,, then G is a bounded linear functional on L, the norm IlGll of which satisfies IlGll = p ‘ ( g ) . It follows that L,, can be embedded as a closed linear subspace into the Banach dual (L,)* of L, (cf. Zaanen [l], Theorem 69.2). The Banach dual (also called the adjoint space or conjugate space) of L, is the Banach space of all bounded linear functionals C on L, with norm IlGll = SUP (IG(f)l : P(f) 1).
s
The bounded linear functional G on L, is called an integral whenever it follows from 0 5 f , EL, (n = 1,2, . .), f , 10 pointwise on X , that G(f,) -+ 0. It is an important theorem that G is an integral if and only if G EL,,, i.e., if and only if there exists a function g E L,, such that G ( f ) = s f g d p holds for allfe L, . In this case, G and g determine each other uniquely; in fact, G and g are identified under the embedding of L,. into L: (cf. Zaanen [l], Theorem 69.3). The bounded linear functional G on L, is said to be positive if G ( f ) 2 0 holds for every f E L, satisfyingf(x) 2 0 for all x E X , and G is said to be
.
44
DISTRIBUTIVE LAITICES A N D NORMED FUNCTION SPACES
[CH.1, 9
real if G(f ) is real for every realvalued f E L,. Every real G is of the form G = G , -G2 with G , and G2 positive. Any arbitrary G E L , * admits a decomposition G = ( G , - G , ) + i(G3- G4) with all components G , (i = 1, 2, 3,4) positive. The set of all real G EL,*is partially ordered by defining that G , 5 G , whenever G z - G , is positive. The positive bounded linear functional G is called singular if it follows from 8 6 G , 5 G and G , EL,. that G , = 8, where 8 denotes the null functional. In other words, the positive bounded linear functional G is singular whenever the only positive integral majorized by G is the null functional. The arbitrary functional G EL,*is now said to be singular if there exists a decomposition for G with positive components (as explained above) such that all these components are singular. The set of all singular functionals is a closed linear subspace of L,*,and there is an important decomposition theorem, stating that every G E L , * has a unique decomposition G = G,+G, such that G, is an integral and G, is singular. In other words, Lz is the direct sum of the closed linear subspaces of all integrals and all singular functionals respectively (cf. Zaanen [l], Theorem 70.2). As observed above, it is not difficult to prove that the associate p' of the saturated function norm p is again a function norm. It is a much deeper fact that p' is also saturated (cf. Zaanen [l], Theorem 71.4), but once this has been established it follows immediately that all higher associates p'") (n = 2, 3, . .) are saturated function norms, where p'") is the associate of p ( " - l ) . It is easy again to prove that p("+') = p'") for n 2 1, but it may happen that p and p(') are not equal. The norms p and pCz)always satisfy p(') 5 p , and = p holds if and only if p has the Fatou property (cf. Zaanen [l], Theorem 71.1). This key result is due to W. A. J. Luxemburg and, independently, to G. G. Lorentz. The function f E L, is said to be of absolutely continuous norm if p( fxE,) 10 for every sequence {En: n = 1,2, . . .} of p-measurable subsets of X such that En descends to a set of measure zero. It can be proved that the set L; of all functions of absolutely continuous norm is a norm closed linear subspace of L, with the extra property that iff E Lz, g is measurable and Ig(x)l 4 If(x)l on X , then g E L ; (cf. Zaanen [l], Theorem 72.3). If L; = L,, we say that the norm p is absolutely continuous. All L, norms (1 5 p < co) are absolutely continuous, but the L , norm is not. If L, = L,(X, p) with X the real line and p Lebesgue measure, then L; consists only of the null function; if L, is the sequence space I , , then L; is the subspace of all null sequences. The subspace L; is related to the subspace (Lf)s of L,* consisting of all
.
CH. 1, § 101
45
ORLICZ SPACES
singular functionals. It turns out that Li is the inverse annihilator of (L,*),, i.e., f E Li holds if and only if G(f)= 0 for all G E (L,*)s(cf. Zaanen [I], Theorem 72.4). As a consequence, it follows that the norm p is absolutely continuous if and only if L,*consists only of integrals. Although Li is the inverse annihilator of (L;),, it is not necessarily true that (L,*)sis, conversely, the annihilator of Lz.This is true only in the case that the carrier of Li is the whole set X , i.e., if there is no subset of X of positive measure such that all f E Li vanish on this subset (cf. Zaanen [l], Theorem 72.6). It is also true that the carrier of Li is the whole set X if and only if the Banach dual (Li)*of Li can be identified with L,.in the sense that every G E (Li)* is of the form I-
for some g E Lp,and all f E Lp, and such that IlGll = p ‘ ( g ) (cf. Zaanen [l], Theorem 72.7). As a final theorem we mention that the space L, is reflexive if and only if the norms p and p’ are absolutely continuous and p has the Fatou property (cf. Zaanen [l], Theorem 73.2). 10. Orlicz spaces
For a more detailed account of the contents of the present section we refer to W. A. J. Luxemburg’s thesis on Banach function spaces ([l], 1955) and to the book by M. A. Krasnoselskii and Ya. B. Rutickii on convex functions and Orlicz spaces ([l], 1961). Let u = cp(u), u 2 0, be a non-decreasingfunction of u such that cp(0) = 0, cp does not vanish identically, and cp is left continuous at all points u > 0 (Example: cp(0) = 0, cp(u) = 1 for all u > 0). Furthermore, let u = $ ( u ) be the left continuous inverse of u = cp(u), such that $(O) = 0, $ ( u ) = uo for u1 < u 6 u2 if cp jumps at uo from u1 to u z , and if lim q ( u ) = I is finite as u + co,then $ ( u ) = co for u > 1 (in the above example we have $(u) = 0 for 0 5 u 6 1 and $ ( u ) = 00 for u > 1). Dehition10.1. If cp and $ satisfy the above conditions, then @(u) and Y(u), defined for u 2 0 and u 2 0 by
@(u)
ru
= J cp(t)dt, 0
Y(u)=
rv J O
$(t)dt,
are called complementary Young functions ( W. H . Young, 1912).
46
DISTRIBUTIVE LATTICES AND NORMED FUNCTION SPACES
[CH. I , § 10
In the example above, we have @(u) = u and Y(v) = 0 for 0 5 u 5 1, Y(u) = co for u > 1. If p is a real number satisfying 1 < p < co and q is defined by p - ' + q - ' = 1, and if q ( u ) = u p - ' , then +(u) = u 4 - l , so @(u) = p-'uP and Y(u) = q-'u4. It is not difficult to prove that the real function @(u),defined for u 2 0, is a
finitevalued Young function if and only if @(O) = 0, @(u) 2 0 for u @ is not identically zero, and @ is convex.
2 0 but
Theorem 10.2 (Young's inequality; Luxemburg [ 11, Theorem 11, 1 , 1; Krasnoselskii and Rutickii [I], section 2). I f @ ( u )and Y(u)are complementary Young functions, then uu
5
@(u)+ Y(u)
holdsfor all u 2 0, v 2 0, and equality occurs if and only if one at least of the equalities u = q(u), u = + ( v ) holds.
The Young function @ is said to have property 6 , if @(u) > 0 for all u > 0 and if there exist constants a > 0, m > 0 such that @(2u) 5 m@(u) for 0 S u 6 a, and @ is said to have property A , if there exist constants b > 0, M > 0 such that @(b) < co and @(2u) 5 M@(u)for all u 2 b (in this case, therefore, @(u) is finite for all u). If @ has both these properties,
i.e., if there exists a constant M > 0 such that @(2u) 5 M@(u)holds for all u 2 0, then @ is said to have the property (6,, A,) (cf. Luxemburg [l], Definition 11, 1, 3; Krasnoselskii and Rutickii [l], section 4). Assume now that p is a a-finite measure in the point set X , exactly as in the preceding section. In order to avoid a great number of different subcases, we shall assume in addition that p has no atoms. Furthermore, let @ and Y be complementary Young functions. By P, = P,(X, p) we will denote the set of all complex p-measurable functions f ( x ) on X such that
=I
wo-)
X
@(lf'l)& <
The set Py is defined similarly. It can be proved that for p ( X ) < co the set P, is a linear space if and only if @ has A,, and for p ( X ) = co the set P, is linear if and only if @ has (a, A,). Similarly for Py (cf. Luxemburg [I], Theorem 11, 1, 3; Krasnoselskii and Rutickii [l], Theorem 8.2). As in the preceding section, let P be the set of all non-negative p-measurable functions. To every f E P we assign the number p,(f)
=
inf (k-I : k 2 0, M,(kf) 6 1).
The number p y ( f ) is defined similarly. It follows easily that p , and p y are
CH.
1,s 101
ORLICZ SPACES
47
saturated function norms; the corresponding normed Kothe spaces are called Orlicz spaces, and denoted by L, and LP respectively. These spaces, for the particular case that q ( u ) and @ ( u ) are continuous for all u 2 0 and u 2 0, strictly increasing and tending to infinity as u, u -,00, were introduced by W. Orlicz ([l], 1932). The norms p , and pP have the Fatou property, and hence Lo and Ly are norm complete spaces (cf. Luxemburg [11, Theorem 11, 2, 1 ; Krasnoselskii and Rutickii [l], section 9). The norm pP is related to the associate norm p& of p , . These norms are equivalent; more precisely, we have
PP
s P& s a,.
The associate space of L, is, therefore, the space LP , but provided with the norm p b instead of the norm pP (cf. Luxemburg [l], Theorem 11, 2, 3; Krasnoselskii and Rutickii [l 1, formula 9.24). For the discussion of the subspace L: of all functions of absolutelycontinuous norm in L,, we shall restrict ourselves to the case that @(u)is finitevalued for all u. It can be proved now that L: consists of allfE Lo such that n
is finite for all constants k > 0, and also that L: is the norm closure of the linear subspace of allf that are bounded and vanishing outside a set of finite measure (this set depending uponf). The carrier of L: is the whole set X , and so it follows from the results in the preceding section that the Banach dual of L: can be identified with the associate space of Lo (cf. Luxemburg [l], Theorem 11, 3, 1 ; Krasnoselskii and Rutickii [l], Theorem 10.3). It follows easily that the norm p , is absolutely continuous (i.e., L: = L,) if and only if the Young class P,, defined above, is equal to the whole of L,, i.e., if and only if P, is linear (cf. Luxemburg [l], Theorem 11, 3, 3; Krasnoselskii and Rutickii [l], section 9, subsection 5). By way of example, consider the case that @(u) = e"-u- 1. Then Y(u) = (u+ 1) log (u+ 1)-u, so Q, has 6, but not A , and Y has (d2, A 2 ) . It follows that PP is linear but P, is not. Hence L$ = LIp but L: # L,. Finally, we mention that for p ( X ) < 00 the space ,?&is reflexive if and only if @ and Y both have A , , and for p ( X ) = 00 the space L, is reflexive if and only if Q, and Y both have (a,, A , ) (cf. Luxemburg [l], Theorem 11, 3, 5 ; Krasnoselskii and Rutickii [l], section 14).
CHAPTER 2
Elementary Properties of Ordered Vector Spaces and Riesz Spaces
The contents of the present chapter, except for the last section (on order convergence and relatively uniform convergence), are mostly of an elementary nature. Many proofs are so obvious that brief indications will be sufficient. The only notable exceptions are in Theorem 11.9 and in the discussion of Example 11.2(x). The notions introduced here appeared already in the earlier development of the subject in the years between 1935 and 1942, a development mainly due to F. Riesz, L. V. Kantorovitch, H. Freudenthal, G. Birkhoff, K. Yosida, H. Nakano and T. Ogasawara. 11. Ordered vector spaces and Riesz spaces
We begin immediately with a definition.
..
Definition 11.1. The real linear space L, with elementsf, g, ., is called an ordered vector space if L is partially ordered in such a manner that the partial ordering is compatible with the algebraic structure of L, i.e., (i) f 5 g implies f i - h 5 g+h for every h E L, (ii) f 2 0 implies af 2 0 for every real number a 1 0. The ordered vector space L is called a Riesz space iffor every pair f and g in L, the supremum sup (f,9 ) with respect to the partial ordering exists in L. The terminology in the above definition is taken from Bourbaki (espace de Riesz). A Riesz space is also often called a vector lattice or, in the Soviet literature, a K-lineal. In the terminology of H. Nakano and his school, a Riesz space is called a semi-ordered linear space. I t will be proved in Theorem 11.5(v) that a Riesz space is indeed a lattice. We list some examples of Riesz spaces.
Example 11.2. (i) Let R" (n 2 1) be the real linear space of all real ntuples f = ( f l , . . .,f.)with coordinatewise addition and multiplication by real numbers. If we define thatf 5 g means that& 5 gk holds for 1 S k 5 n, then R" is a Riesz space with respect to the thus introduced partial ordering. 48
CH.2, 5 111
ORDERED VECTOR SPACES AND RIESZ SPACES
49
(ii) The partial ordering as defined in (i) above is not the only manner to make R" (n 2 2 ) into a Riesz space. There exists also the lexicographical ordering, as follows. For R2,letf S g forf = (fl ,fz)and g = (gl,g2)if either fl < g1 or f l = g1,f2 5 g 2 . The ordering is now a linear ordering, and so RZ is a Riesz space with respect to the lexicographical ordering. Similarly for n > 2. (iii) If L is the real linear space of all real finitevalued functionsf(x) on the arbitrary non-empty point set X with pointwise addition and multiplication by real constants, then L is a Riesz space with respect to the partial ordering introduced by defining thatf S g means thatf(x) 5 g(x) holds for every x E X. The linear subspace of all real bounded functions on X , with the same partial ordering, is a Riesz space by itself. If Xconsists of a finite number of points, say n points, we obtain essentially the example in part (i), i.e., the space R" with coordinatewise ordering. If X consists of a countably infinite number of points, we obtain the sequence space (s) of all real sequences and the subspace I , of all bounded real sequences. (iv) If X is a non-empty topological space and C ( X ) is the real linear space of all real continuous functions on X , then C ( X ) is a Riesz space with respect to the partial ordering introduced by defining thatf 5 g means that f ( x ) 6 g(x) holds for every x E X . The linear subspace of all real bounded continuous functions on X , with the same partial ordering, is a Riesz space by itself. If Xis a locally compact space, then the linear subspace of all real continuous functions on X with a compact carrier (also called compact support) is a Riesz space by itself. (v) As in section 3 and section 9, let ( X , A, p) be a measure space, i.e., p is a countably additive non-negative measure on the a-field A of subsets of the non-empty point set X . We assume that sets of measure zero are neglected; more precisely, A is identified with the Boolean measure algebra A/&, where A , is the ideal of sets of measure zero. By M(') = M'"(X, p) we denote the set of all real p-almost everywhere finitevalued p-measurable functions on X , with identification of p-almost equal functions, and pointwise addition and multiplication by real numbers. Then M(') is a real linear space, partially ordered by defining that f 5 g means that f ( x ) 5 g ( x ) holds for p-almost every x E X . The space M'"(X, p) is a Riesz space with respect to the thus defined partial ordering. (vi) Let M") = M")(X, p) be the same as in the preceding part (v). Many linear subspaces of M") are Riesz spaces by themselves. By way of example,
50
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH. 2 , s 1 1
-= -=
let p be a real number satisfying 0 p 00, and let L, = L,(X, p) be the well-known linear space of all (rea1)fE M"' satisfying n
Then L, is a Riesz space with respect to the partial ordering inherited from M"'. The same holds for the space L , consisting of all essentially bounded f i n M"'. (vii) The spaces Lp (1 5 p 5 co) are normed linear spaces with respect to the norm
less sup If(x)l
for p = co.
These spaces are special examples of Orlicz spaces, and Orlicz spaces are special examples of normed Kothe spaces. All these spaces are Riesz spaces with respect to the partial ordering inherited from M"'(X, p). (viii) Let p be a finitely additive signed measure (also sometimes called a signed charge) on the field (also called algebra) r of subsets of the nonempty point set X such that the number
llpll = SUP (IP(4I :A E r )
is finite. Under the natural definitions of addition and multiplication by real numbers the set L of all p of this kind is a real linear space, partially ordered by defining that p1 p z means that pl(A) _S p z ( A ) holds for every A E r. The space L is a Riesz space. Indeed, the ordering is compatible with the algebraic structure and furthermore, given pl and p z in L, the element v = sup (pl ,p z ) exists in L and is given for any A E r by
v(A) = sup { p l ( B ) + p z ( A - B ) :A
=,
B E r}.
(1)
For the proof, note that it is not difficult to see that v, as given by (l), is finitely additive and llvll 5 llplll + llpzll, so v EL. Furthermore, it is evident that v ( A ) 2 p l ( A ) as well as v(A) 2 p z ( A ) for every A E I', so v is an upper bound of p1 and p z Finally, if z is also an upper bound of pl and p z and if A E r, then z ( A ) = z ( B ) + z ( A - B ) 2 pi(B)+pZ(A-B)
.
for every B E I' satisfying B c A, and hence Z(A)
2 sup { p l ( ~ ) + p 2 ( ~ :- A~ )
This shows that v = sup (pl,pz).
BE
r} = V ( A ) .
CH.2, 3 111
ORDERED VECTOR SPACES AND RIESZ SPACES
51
(ix) Let H be a Hilbert space (over the complex numbers) with elements . . and inner product ( x , y), and denote by% the real linear space of all bounded Hermitian (i.e., self-adjoint) operators A , B, . . . on H. The space&' is an ordered vector space with respect to the partial ordering in% defined by saying that A S B means that (Ax, x ) =< (Bx, x ) holds for all x E H. It can be proved that % is not a Riesz space unless in the trivial case that H is one-dimensional. There exist, however, many linear subspaces of 3? that are Riesz spaces with respect to the ordering inherited from%'. Specifically, let 9 be a non-empty subset of 2 such that the elements of 9 are mutually commuting (i.e., A B = BA for all A , B E g ) ,let Y'be the set of all elements of % that commute with 9 (i.e., each element of v'commutes with each element of 9),and let Y" be the set of all elements of % that commute with V'. It can be proved that Y" is a Riesz space such that 9 c Y" c Y'(for further details and proofs, cf. Chapter 8). (x) This example is somewhat technical in nature; the reader who is not interested in harmonic functions is advised to proceed immediately to Definition 11.3. Let G be a region in the plane and L the ordered vector space of all functions f ( x , y ) in G such that f = u1 - u2 with u l , u2 harmonic and nonnegative in G. The ordering is the pointwise ordering. The null element of L is the function identically zero, so the subset L+ of allfe L greater than or equal to the null element consists of all non-negative harmonic functions in G. Every non-negative constant function is a member of L + . Note that any u E L+ is either strictly positive or identically zero in G. In order to prove that L is a Riesz space, it is sufficient to show that among all harmonic upper bounds of two given functions u1 and u2 in L+ there is a smallest one. Indeed, suppose this has been proved and suppose that f = u1 -u2 and g = u3-u4 are arbitrary functions from L. Then f l = f + ( u 2 + u 4 ) and g1 = g (u2 u4) are in L+,and so the least upper bound hl offl and g1 in L+ exists. It follows that h, - (u2 + u4) is the least upper bound off and g in L. Hence, let uI and u2 in L+ be given. We will prove that the pointwise infimum s = s ( ~ y, ) of the set H of all continuous superharmonic functions h in G with h 2 u l , h 2 u2 is the desired harmonic least upper bound of u1 and u 2 . By definition, the continuous function h in G is superharmonic if, for any point A in G and any circle in G we have h(A) 2 h"(A), where h" is the continuous function harmonic in the interior of the circle and having the same values as h on the boundary of the circle and outside the circle. We will prove first that h E H implies h" E H . It is sufficient to show that if x, y ,
.
+ +
52
ORDERED VECTOR SPACES AND RlESZ SPACES
[CH. 2 , g 11
the first circle is called C, and C , is any circle in G intersecting C, and h# is constructed from h and C1 exactly as h" was from h and C , then (h")# 5 h". For any point A in the interior of C1 but not in the interior of C we have h"(A) = h(A), and hence it follows from h" 5 h that
(h")#(A)
5 h#(A) 5 h ( A ) = h"(A).
It remains to prove that (h")#(A) S h"(A) for all points A in C n C1. Now, d = h" - (h")# is harmonic in the interior of C n C1 and non-negative on the boundary of C n C , . Since the function d attains no extremum in the interior of C n C1, we must have d 2 0 in C n C1. This is the desired result. We return to the proof that the pointwise infimum s of all functions h E H is the harmonic least upper bound of the given functions u1 and u2 in L + . Given the point A in G and E > 0, there exists a function h E H such that h(A) < s(A) E. Then h(P) < s ( A ) E holds for all P in a certain neighborhood of A since h is continuous, and hence
+
+
s ( P ) 5 h(P) < S ( A ) + E
holds for all P in this neighborhood. In order to derive an inequality in the converse direction, let C be a circle in G with center A and radius r ; for any h E H let h" be the continuous function harmonic in the interior of C and having the same values as h on the boundary of C and outside C. For any point P in the interior of C we have, by the Poisson integral formula, that h"(P) = /"K(P, 0 Q ) h ( Q ) d q,~ where
-
1 r2-AP2 K(P,Q ) = 2n PQ2
*
Note that K(A, Q) = (2n)-', and that K(P, Q ) , as Q varies on the boundary of C, varies between
_1 . r - A P
2n r + A P
and
r+AP 2n r - A P '
---1
Hence, given E > 0, there is a neighborhood of A such that if P lies in this neighborhood, then K(Py Q) 2 (1 -E)K(A, Q )
CH.2,$11]
ORDERED VECTOR SPACES AND RIFSZ SPACES
53
holds for all Q on the boundary of C. It follows that
h(P) 3 h”(P) = S,”K(p, Q ) h ( Q ) d v ~
holds for every h E H. Hence, taking the infimum on the left, we obtain that s(P) 2 (1 -&)s(A) holds for all P in the neighborhood under consideration. Combining all results, it is evident now that s is continuous in G . Observing that s = i d (h : h E H ) , it is also evident that s is superharmonic. Hence, s is continuous and superharmonic, so s E H. This implies that s is the smallest continuous and superharmonic upper bound of u1 and u2 in the region G. In order to show that s is the smallest harmonic upper bound of u1 and uz in G , it will be sufficient to prove now that s is harmonic, i.e., to prove that s ( ~ )= (2n)-1JoZns(~)dqQ
holds for every point A in G and for Q varying on any circle C in G with center A . This is so because if
> (2n1-1 [ : S ( Q ) ~ ~ Q would hold for some circle with center A , the corresponding function s“ would satisfy s“ E H and s ” ( A ) < $(A), which is impossible. This finishes the proof. The present proof, wherein it is shown first that s is continuous and secondly that s is harmonic, is due to C . Visser. The example of the harmonic functions is due to F. Riesz ([2], 1937), and he refers to a proof by means of Poincare’s method of balayage. Having proved that L is a Riesz space, we will show that, in general, not every harmonic function in the region G is a member of L. For this purpose, assume that G is the open unit circle in the complex plane, u non-negative and harmonic in G, point P is in G, and C is a circle with the origin 0 as center and radius r satisfying OP < r < 1. For Q varying on the boundary of C, we have
54
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH.2 , s 11
and so, for r 7 1, we obtain that
#(P){1 - OP} s 2 4 0 ) .
Hence, givenf E L, the set of numbersf(P){1 - OP}is bounded for P varying in the unit circle. Now, let F ( z ) = (1 -z)-' for IzI < 1. The real part f of F is harmonic in G , but the above boundedness condition is not satisfied, and so f is no member of L. A somewhat different proof for the existence of a smallest harmonic upper bound of two given non-negative harmonic functions u1, u2 in a region G is as follows. Once more, let H be the set of all continuous superharmonic functions h in G satisfying h 2 u1 and h 2 u 2 , and let the function s be the pointwise infimum of all h E H in G . Furthermore, let S be a fixed closed circular disc in G, and for each h E H let h" be the continuous function harmonic in the interior of S and equal to h outside S and on the boundary of S. The set H" of all h" is a subset of H, and s = inf (h" :h" E H " ) . Given h ; , h i E H " , let the continuous function h i in G be defined by for every point ( x , y ) outside S and on the boundary of S, and inside S the function h i is harmonic. Then h i E H", and h i hl- as well as h i S h i (indeed, within the set H" the function h i is the greatest lower bound of hl- and h i ) . We will use the notation h i = hl- A h i . Since all numbers
s
.
are 2 0, the infimum a of this set of numbers exists. Let {A,"; n = 1, 2, . .} be a sequence in H" such that SSsh,dxdy + a
as n
-,m.
On account of the property proved just above we may assume that hl- 2 h; 2 * * -.It follows that there exists a function qs 2 0 in G such that h,- 1qps holds pointwise in G . By Harnack's well-known theorem, qs is harmonic in the interior of S. We will prove that qs(x,y ) = s(x, y ) holds for all points ( x , y ) in the interior of S; it will follow that s is harmonic in the interior of S, and SO (since S may be any circle in G ) the function s is harmonic in G . For the proof that qs = s holds in the interior of S, it is sufficient to prove that 'ps 5 h" holds in the interior of S for every h" E H " . If not, there exists a
CH.2,§11]
ORDERED VECTOR SPACES AND RIESZ SPACES
55
function h i E H" such that h,' < cps holds in a subregion of the interior of S. Let kn- = hn- ~ h , 'for n = 1,2, . . ..We have kn' E H" for all n = 1,2, ..., and kn- 1'pl in the interior of S with cpl 6 cps and cpl 6 h,' in the interior of S, so 'pl(x,y) < cps(x,y) for the points of the subregion referred to. But then
which is impossible. For further details about this example, cf. Example 23.3. (xi) Let X = {x : u < x < p} be a bounded open interval in the real axis, and L the ordered vector space of all real linear functions on X. Evidently, we may just as well assume that Xis the closed interval {x : ct 5 x 6 p). The space L is a Riesz space; givenf and g in L, the function h = sup (f,g ) is the linear function determined by h(a) = m a x {f ( u ) , g(ct)} and h(P) = max {f (p), g(p)}. The present example is the one-dimensional simplification of the preceding example. For further details, cf. Example 23.3. Definition 11.3. Given the ordered vector space L, the subset Lf = { f : f E L,f 2 O] is called the positive cone of L. Elements of L' are calledpositive elements. Theorem 11.4 (Cone properties). The positive cone L i of the ordered vector space L has the following properties. ( i ) f , g E L + impIiesf+gEL+. (ii) f E L' implies af E L'for every real number a 2 0. (iii) f, - f E L+ impliesf = 0. Conversely, if L' is a mbset of the real linear space L such that L' has the properties (i), (ii) and (iii), then L is made into an ordered vector space by defining that f 6 g holds i f and only if g -f E L', and L' is then exactly the positive cone of L with respect to this partial ordering. Proof. The routine proof is left to the reader.
Theorem 11.5. Let L be an ordered vector space with positive cone L'. The following holds now. (i) f 2 g i f a n d o n l y i f f - g E L f . (ii) f 2 g ifandonly i f f = sup (f,g ) , and also ifand only i f g = inf (f,9). (iii) f 2 g i f and only if af 2 ag for a > 0, and also if and also i f a f 6 ag for a < 0. (iv) If sup (f,g ) exists, then inf (-f, - g ) exists, and
inf(-f, - 9 )
=
-sup (f,9).
56
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH.
2, 5 11
(v) Iff, g E L ,then sup (f,g ) exists in L ifand only ifinf (f,g ) exists in L, and in that case we have SUP ( f h, 9 +h ) = SUP (f, s)+h, inf ( f + h , g+h) = inf (f, g)+h for any h E L. In particular, i f L is a Riesz space then sup (f,g ) and inf (f,g ) exist for any pair of elements f, g E L. I t follows also that i f L is an ordered vector space such that sup (f,0 ) existsfor every f E L, then L is a Riesz space. (vi) If sup (f,g ) exists, then for a 2 0, SUP (af, a s ) = a SUP (f, g) for a 5 0, sup (af,ag) = a inf (f,g ) inf (af,ag) = a inf (f,g ) for a 2 0, for a 5 0. inf (af, ag) = a sup (f,g) (vii) I f L is a Riesz space andf, g, h E L , then SUP {SUP (A 91,h} = SUP {SUP (f,h), SUP (9, h)} = SUP (f,9 7 h). Similarly for inf (f,g , h). Hence, any finite set of elements in a Riesz space has a supremum and an infimum in the space.
+
Proof. (i) and (ii) are evident. (iii) We will only prove that f 2 g and -f 5 -g are equivalent. This follows easily by observing that both inequalities are equivalent tof - g 2 0. (iv) It follows from part (iii) that if h is an upper bound offand g, then -h is a lower bound of -f and -g. Hence, if h = sup (f,g ) exists, then -h = inf (-A -g). (v) If sup (f,g ) exists, then sup (f,g ) +h is evidently an upper bound of f+ h and g +h. If k is also an upper bound off +h and g +h, the element k - h is an upper bound off and g , so k - h >= sup (f,g). It follows that k 2 sup (f,g)+ h. This shows that sup (f,g ) +h is the supremum off +h andg+h, i.e., SUP ( f + h , g+h) = SUP (f,g)+h. In particular, substituting h = -f-g,
we obtain
SUP (-A -9) = SUP (f,s)-f-s. But then, in view of the preceding part (iv), the element inf (f,g ) exists and satisfies inf (f,g) = -sup (-A -g). It follows now exactly as above that g)+h inf ( f + h , g+h) = inf (f, holds for every h EL. (vi) and (vii) are now evident.
CH.2,8 111
57
ORDERED VECTOR SPACES AND RIESZ SPACES
Let f be an element of the ordered vector space L with the property that sup (f,0) exists. Then, as shown in the last theorem, inf (f,0) exists, and so sup (-A 0) exists. Furthermore, the element sup (2f, 0) exists, so that by subtraction off we obtain the result that sup (f, -f) exists. Definition 11.6. I f L is an ordered vector space, and the eIement f of L has the property that sup (f,0 ) exists, we will write
Note already that inf (f,0 ) = -sup
(-I; 0) = -f -.
Theorem 11.7. If L is an ordered vector space, and the elementf of L has the property that sup (f,0 ) exists, then the following holds. (i) f +, f - EL+; f' = (-f)- and f - = (-f)+;
If1 = I-fl.
(ii) f = f'-f-
with inf(f+,f-) = 0 ; If1 = f + + f - , and so IflEL'. (iii) 0 6 If1 and 0 5 f - S Ifl. (iv) -f- 5 f 5 f + . (v) (af)' = af' and (af)- = af- for a 2 0 ; (af)' = -af- and (af)- = -af+ for a 5 0; lafl = la1 If1 for real a. (vi) Iff, g E Land both f + = sup (f,0) and g+ = sup (9,0) exist, then g+ and f - 5 g - hold, f 5 g holds if and only iff
f's
-
+
Proof. (i) Evident from the definitions. (ii) We have f -f = sup (f,0)-f = sup (0,-f) = f -, and so f = f '-f -. In addition, 0 = -f-+f- = inf(f,O)+f- = inf(f+-f-,o) +f-= inf(f+,f-). Furthermore, we have If1 = sup(f, -f) = sup(2f,O)-f = 2f+-(f+-f-) = f + + f - . (iii) It follows from f +, f - EL+ and If 1 = f + +f - that 0 S f S If 1 a n d 0 S f - 5 Ifl. (iv) It follows from f = f'-f- that -f- = f-f' S f as well as +
+
fSf+.
(v) Follows from Theorem 11.5 (vi).
58
[CH. 2, 5 11
ORDERED VECTOR SPACES AND RIESZ SPACES
(vi) Assuming that f 6 g holds, we have f + = SUP (f,0) 5 SUP ( 9 , O ) = 9+, f - =sup(-f,o)~sup(-g,o)=g-. Conversely, iff' S g + and f - 2 g-, then f = f + - f - S g + - g - = g. Theorem 11.8. If L is an ordered vector space, and the elementsf and g of L have the property that sup (f, g ) exists, then ( f - g ) ' , (g-f)' and If - g l exist, and 0) SUP (f,9)= ( f - s ) + + g = (9-f)+ +f, (4 i n f ( L 9 ) = f - ( f - g ) + = s - ( g - f ) + , (iii) SUP (f,g)+inf(f, 9) = f + g , (i.1 SUP (f, 9)- inf (f, 9) = If (v) 2 S U P ( f , d =f+s+If-sl, (vi) 2inf(f,g) =f+g-If-gl. 9 1 9
Proof. (i) We have SUP (f, 9) = SUP ( f - 9 , O)+s = ( f - d + +9. (ii) We have inf(f,g) = f+inf(O,g-f) = f - ( g - f ) - = f - V - g ) ' . (iii) Follows from (i) and (ii) by addition. (iv) Subtraction of the formulas in (i) and (ii) yields SUP (f,d - i n f ( f , 9) = ( f - d + +(s-f)+= (f-s)++(f-s)- = If-91. (v) and (vi) are obtained from (iii) and (iv) by addition and subtraction respectively. From the formulas proved in the last theorem we will derive some interesting identities, holding in any Riesz space (holding in particular, therefore, for real numbers). Theorem 11.9. If L is a Riesz space and f and g are elements of L, then
2 SUP (Ifl,191) = If+sl+lf-sl9 2inf(lfl, 191) = llf+9l-lf-9ll~ Proof. Repeated application offormula (v) in the preceding Theorem 11.8 shows that 2 SUP 191) = 2 SUP (f,-A -9) = SUP (2 SUP (f, -91,2 SUP (9,-f )> = SUP {f-s+If+sl, s-f+If+sll = If+9l+lf-sl* In order to obtain the second formula we will writef = p q and g = p -q, ( I f 1 9
9 3
+
CH.2, 5 111
ORDERED VECTOR SPACES AND RIESZ SPACES
59
so f + g = 2p and f - g = 24. Using the just obtained first formula twice, we thus obtain 2 inf (IfL Igl) = 2lfl+2Ig1-2 SUP (IfL Id) = 4 f l + 2lsl- lf+gl - If-sl = 21P +41+2lP - 41 -2lPl-2141 = 4 SUP (IPIY l4l)-21Pl-214l = 2(lPl+ 141+ llPl - 1411)-21PI -2141 = 211Pl-lqll = 112Pl-124ll =
IIf+sl-If-glI.
An alternative proof of the first formula, due to G. J. 0. Jameson, is as follows. It has to be proved that SUP (IP+4L IP-41) = lP1+141 holds for all p , 4 E L.We have SUP (IP+4L IP-41) = SUP (P+47 - P - 4 , P - 4 , -P+4)7 and by Theorem 11.5 (v) we have SUP ( P + 4 , - P + 4 ) = SUP (P7 - P ) + 4 = lPl+47 suP(-P-qyP-4) = SUP ( - P , P ) - 4 = IPI-4, so SUP (IP+41, IP-41) = SUP M + 4 Y IPI-4) = IPI+141Y once more by Theorem 11.5 (v). Observe that it follows immediately from the second formula that inf (Ifl, 191) = 0 holds if and only if If+gl = lf-91. It is not difficult to prove by more elementary means (i.e., without using the second formula) that inf (Ifl, Is]) = 0 implies If+gl = If-gl (cf. Theorem 14.4 (i)), but until rather recently the converse did not seem to be mentioned in the literature. In 1964 it was set as a problem by the present authors in “Wiskundige Opgaven”, a collection of problems edited by the Dutch Mathematical Society, to prove that If+gl = If-gl implies inf (Ifl, Igl) = 0 (Wiskundige Opgaven 21 (1964), problem 192). The original rather inelegant solution was much improved by basing the proof on the second formula in Theorem 11.9. This was done, independently, by A. de Leeuw van Weenen (Mrs. A. van Arkel), M. A. Kaashoek and N. G. de Bruyn. The elements of the positive cone L+ of the ordered vector space L will occasionally be denoted by u, v , w,. . .. The positive cone L+ is called generating if everyf E L can be written in the form f = u - v for appropriate u and v in L+.The positive cone is not always generating. By way of example, let L be real two-dimensional number space RZ with L+ = {f:f=(fi,fZ),fZ = OJ-1 2 01, i.e., Lf is the positive horizontal axis. The positive cone Lf of a Riesz space I, is always generating. Indeed, given the elementf of the Riesz space L,we
60
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH. 2, $ 11
have f = f '-f - with f +, f - EL'. In the converse direction, if the positive cone ' L of the ordered vector space L is generating, and if sup (f,g) exists for allf, g EL', then L is a Riesz space. Indeed, given arbitrary elements f and g in L,we have f = ui-u2 and g = u3-u4 with ui EL' (i = 1,2,3, 4), and so fl = f + ( u 2 + u 4 ) and g1 = g+(u2+u4) are in L + . Then h, = sup (fl,gl)exists by hypothesis. It follows that SUP (f,9) = h1-
(.2
+ U4)Y
so sup (f,g) exists for all f and g in L . This shows that L is a Riesz space. It will be proved in the next theorem that the decompositionf = f '-f has a certain minimal property. Theorem 11.10. Let L be an ordered vector space. (i) Ifu, u E L+ and sup (u, u ) exists, then 0
5 inf (u, u ) 5 sup (u, u ) 5
u+u.
(ii) Zff = u - u with u, UEL' and iff' = sup (LO)exists, then f ' S u and f - 5 v. Hence, the decomposition f = f -f- as a diference of two positive elements is a "minimal" decomposition. It was proved in Theorem 11.7 (ii) that inf (f+, f -) = 0. Conversely, iff = u-v with u, u E L + and if inf (u, u ) exists and satisfies inf (u, u ) = 0, then f = sup (f,0) exists and we have u = f + and u = f In other words, iff = u- u with inf (u, u ) = 0, thenf = u - u is the minimal decomposition.
'
-.
+
Proof. (i) Follows immediately by observing that inf (u, u ) and sup (u, u ) are in L+ and inf (u, u ) + sup (u, u ) = u + v. (ii) Let f = u - u with u, u E L + ,and let f = sup (f,0) exist. Then we have f = sup (f,0) 6 u because f 5 u as well as 0 S u. Hence also f '-f 5 u - f , i.e., f - 5 u. For the second part, assume that f = u-v with inf (u, u ) = 0. Then it is obvious that u and u are in L+,and from Theorem 11.8 (ii) it follows that +
0 = inf (u, u ) = u-(u-u)+
and so
f '
= u. But then u =
u-f = f'-f
=
u - f +,
=f - .
Exercise 11.11. Let fl,. . .,f, be elements of the Riesz space L.Prove that ISUP (fl
Y
Iinf(f19
- - .,f,)l * *
5 SUP (IfilY * * lf,l>Y .,f,)l 5 SUP (Ifll, * - KI). .Y
.Y
CH.2,511]
ORDERED VECTOR SPACES AND RIESZ SPACES
61
Exercise 11.12. Let f, g be elements of the Riesz space L. Prove that 2 SUP {inf(f, 9)s inf(-f, -911 = If+sl-lf-91, 2SUP {inf(f, -d,inf(--fgN = If-sl-lf+sl. Compare this result with the result in Theorem 11.9.
Exercise 11.13. Let L be an ordered vector space with positive cone L+. Prove that L is a Riesz space if and only if for every pairf, g E Lthere exists an element h E L such that
(f+L+) n (g+L+)= h+L+, where the addition signs denote algebraic sums (i.e., f+L+ is the set of all
f+ u for u varying in L'). Prove that L is not necessarily a Riesz space if the last condition is only satisfied for allf, g EL+,but L is a Riesz space if, in addition, ' L is generating.
Exercise 11.14. Letf, g, p , q be elements of the Riesz space L. Prove that iff +g = p q, then SUP (AP)+ inf (9,4 ) = f + 9 .
+
Exercise 11.15. Let L be the set of all real functions of finite variation on the bounded interval [a,b] in the real line with the extra property that f(a) = 0 holds for every f E L. The set L is a real linear space under pointwise addition and multiplication by real constants. Show that the set L+ of all f E L such that f is non-decreasing in [a, b ] has the cone properties. Hence, L is an ordered vector space with positive cone L + . Show that L is a Riesz space. More precisely, show that if
P = {u = t, < t , < ... < t"
= xo
5 b}
is a partition of [a, xO],and we divide the set of subintervalswhich form P into disjoint subsets P',P'' and make the sum
c'
2'
where and denote summation over the subintervals in P' and P" respectively, then
for P running through all partitions of [a, xo] and P',P" still variable for fixed P.
62
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH. 2,s 12
The example is due to L. V. Kantorovitch [l]. Note the analogy with Example 11.2 (viii). Hint: Assume first that g ( x ) = 0 for all x. It is well-known that in this particular case the number sup,.,,,, s(P', P") is exactly the positive variation P f ( x o )off at the point xo. It is also well-known that f ( x ) = P f ( x )- N,-(x), where N,- is the negative variation off, is a representation off as a difference of two functions ofL+, and also that if f ( x ) = p l ( x ) - p 2 ( x ) is another representation of this kind, then p 1 2 Pf and p 2 1 Nf. Hence f = Pr - N,is the minimal representation, so Pr = f v 0 and Nf = (-f)v 0 by Theorem 11.10 (ii). The general result about f v g follows by observing that
f v s = {(f-g)vOI+g. 12. Inequalities and distributive laws in a Riesz space
In the present section L denotes a Riesz space. Theorem 12.1 (Triangle inequality). For all f, g E L , we have
sf
(f+9)+
+
+9+,
( f + s ) -S f - + 9 - , Ilfl-lsll s If+sl s Ifl+lsl. 2 f+g and f + + g + Z 0, so f + +9+ 2 SUP (f+s,0) = ( f + 9 ) + .
Proof. We have f
+
+g+
It follows that and so
f - + g - = (-f>++(-s)+
1 (-f-9)+
= (f+9)-,
If+sl = ( f + s ) + + ( f + ss) f- f + 9 + + f - + = 9 Ifl+lsl. This implies that If1 - Igl 5 1f - g l and 191- 1 f I S I f - g l , so IIfl-lsll = suP(Ifl-191,191-lfl) S If-91This holds as well if - g is substituted for g, and so
IIfl-lsll 5 If+sl. Theorem 12.2 (Distributive laws). If D is a subfet of L such that f o = sup ( f :f E 0)exists in L , then
inf ( f o , 9) =
holds for every g in L.
SUP
{inf (f, s) : f E D>
CH.2,s 121
INEQUALITIES AND DISTRIBUTIVE LAWS IN A RIESZ SPACE
63
Similarly, i f f l = inf ( f :f E D ) exists in L, then
SUP ( f l , 9) = inf {SUP (f,9): f E Dl holdsfor every g in L.
Proof. Assume that f o = sup ( f :f E 0)exists and let g be an arbitrary element of L. Sincef o 2f holds for allf E D, we have inf (f o ,g ) 2 inf (f,g ) for all f E D, and so inf ( f o , g ) is an upper bound of the set of elements {inf (f,g) :f E D } . Let m be another upper bound of this set. Then m 2 inf (f,9) = f +9- SUP (f,91, so m-gfsup (f,g ) 2 f holds for all f E D. It follows that m-g+sup
(fo, 9) 2f
holds for all f E D, and so i.e., This shows that
m-g+sup ( f o , 9) B SUP ( f : f E 0) = f o ,
m 2 f o + g - s u P ( f o , 9 ) = inf(f0,g). inf (So,g ) = sup {inf (f,g ) :f E D } .
The second formula can be proved similarly or can be derived from the first formula by observing that f l = inf ( f :f E 0)is equivalent to -f l = sup (-f:f E 0). Corollary 12.3 (Finite distributive laws). For brevity, denote sup (f,g ) and inf (f,g ) in the present corollary by f v g and f ~g respectively. Then the formulas (fV9)Ah= (fAh)V(9Ah), hold for allf, g , h E L.
(fA9)Vh = ( f V h ) A ( V h )
It has been shown thus that every Riesz space L is a distributive lattice with respect to the partial ordering. There is in L no smallest and no largest element (unless in the trivial case that L consists only of the null element). The positive cone L+ by itself is a distributive lattice with the null element of L as smallest element; it follows that all results proved in Chapter 1 for a distributive lattice with smallest element hold in L'. In particular all results about prime ideals and the hull-kernel topology hold. We will say more about this in the next chapters.
64
[CH.2, 8 12
ORDERED VECTOR SPACES AND RIESZ SPACES
Theorem 12.4. (i) (Birkhofidentity.) For a l l f , g, h in L we have ISUP
(9, h)l
(f, +sup
+ linf (5h)-inf
(9, h)l = lf-91.
(ii) (Birkhofinequalities.) For allf, g, h in L we have ISUP (f, h)-suP (9, h)l 6 If-gl, Iinf(f, h)-inf(g, h)l 6
If-sl.
(iii) For aNf, g in L we have
If+-s+l6 lf-91
and r - g - 1 6 If-91.
Proof. (i) Applying the formula k - 4 1 = SUP @, q)-inf(p, 4 )
to the case that p = sup (f,h) and q = sup (9, h) as well as to the case that p = inf(f, h) and q = inf (9, h), we obtain
ISUP
h)l = h) A (9 V h)+ ( f A h ) V (g A h ) -fA h A g A h.
(f, h)-sup (9, h)l +linf(f, h)-inf(g, =f V h V g V h
- (fV
By means of the distributive laws, applied to the terms in the middle, this is seen to be equal to f V g V h- (fA
g) V h+ (fV g) A h - f A g A h = = { ( f v s)v h =
+ ( f v d A h ) - {(f.
{ ( f v s > + h )- {(ff4+h)
9 ) v h + (f. 9) A h )
= f v g - f A s = If-sl.
(ii) and (iii) are now evident. The theorem is due to G. Birkhoff cf. Theorem 7.8 on p. 109 of the first edition of his book on lattice theory [l](1940); the formula in (i) is not mentioned in the second edition (1948) and appears again in the third edition (1967);the formulas in (ii) and (iii) are mentioned in all the editions. Many of the formulas in this chapter, not involving scalar multiplication, hold also in ”lattice groups” even when these are not commutative, but the proofs are then sometimes more difficult. Until rather recently (1956, J. A. Kalman [l]) a proof for the Birkhoff identity in this more general case was unknown. We observe, incidentally, that the identities in Theorem 11.9 hold in a lattice group if and only if the group is commutative.
CH.2,s 121
65
INEQUALITIES AND DISTRIBUTIVE LAWS IN A RIESZ SPACE
Theorem 12.5. For all u, v, w B' L we have inf(u, w)+inf (u, w)-w 6 inf (u+v, w) 6 inf (u, w)+inf and
sup (u, u )
(0,
w)
- inf (u, v)+w _<
s sup (u, w) + sup
Ifinf ( v , w)
=
0, then
(0,
(1)
w) - w 6 sup (u + 0, w) 5 sup (24, w)+ sup ( 0 , w). (2)
inf (u+ v, w) = i d (u, w).
If inf (u, u ) = 0,then
inf (u+u, w) = inf (u, w)+inf (v, w). Proof. The last inequality in (1) is trivial. Indeed, setting 1 = inf (u, w)+ inf (u, w), we have trivially that 15 u + w and 1 U + W , so
s
1I inf(u+w, u+w) = inf (u, v)+w.
The first inequality in (1) follows then immediately. It remains only to prove that inf (u + u, w) 5 inf (u, w) + inf (u, w). (3) In virtue of one of the Birkhoff inequalities we have
o 6 inf(u+v, w)-inf
(u, w) = linf (u+v, w)-inf(u,
and it is obvious that Hence
O
w)I
6 I(u+u)-ul = 101 = v,
s inf (u+u,
w)-inf (u, w)
o 6 inf (u+v,
5 inf (u+v,
W)
5 W.
w)-inf (u, w) 6 inf (0, w),
which is the same as (3). Similarly, the only non-trivial part in the proof of (2) is the proof that sup (u, w) + sup ( 0 , w) -w I sup (u
+
0,
w).
(4)
Subtracting (3) from the identity u + u + w = (u+w)+(u+w)-w, we obtain (4). If inf (v, w) = 0, we obtain inf (u+Y, w) 6 inf (u, w) from (1). On the other hand, it follows from u + u 2 u that inf (u+u, w) 1 inf (u, w). Hence inf (u+u, w) = inf (u, w).
66
[CH.2,o 13
ORDERED VECTOR SPACES AND RIESZ SPACES
If inf (u, u ) = 0, it follows from the general formula u+v = sup (u, u)+ inf (u, u ) that now u+u = sup (u, u). Similarly u1 + u , = sup (ul, v l ) for u, = inf (u, w ) and u1 = inf (u, w ) . Hence inf(u+u,w)
= ( u + v ) ~ w= ( U V U ) A W = ( U A W ) V ( U A W )
= u1 v uI = u1
+ u1 = inf (u, w)+inf (u, w).
Exercise 12.6. Let u , , . . .,u,,, u E L+ and wi = inf ( u i , v ) for i = 1, Show that inf (ul . . . +u,, u ) = inf (wl + . . . +w,,, u).
. . .,n.
+
Exercise 12.7. Show that for allf, g , p , q E L we have inf(f, g)+inf(p, 4 ) = i n f ( f + p , f + q , 9+P, S + d , 9)+SUP (P, 4) = SUP ( f + P , f + % 9+P, 9 f 4 ) . SUP (f, 13. Suprema and infima of subsets of an ordered vector space
Let L be a n ordered vector space. Any subset of L can be considered as an indexed subset {f, : T E {T}}, where {T} is an appropriate index set.
Theorem 13.1. (i) I f f = supf, exists in L, then -f = inf (-A). Similarly, = inff, exists, then -f’ = sup (-A). (ii) Iff = supf , exists in L, then af = sup (af,) for a 2 0 and af = inf (ah)for a 5 0. Similarly, ifinff, exists. (iii) I f f = supf, exists in L, then f + h = sup ( f , + h ) for every h E L . Similarly, if inff, exists. (iv) Iff = supf, and g = sup gb exist in L, then
iff’
f+s
= sup(f,+g,). Ga
Similarly, if inf f , and inf go exist. (v) I f f = supf, exists in L, and if h E L has the property that sup (f,h ) and all sup ( f , , h ) exist in L, then SUP (f,h ) = SUP {SUP (f, h ) } . 9
T
’
In particular, iff = supf , holds and i f f and allf J exist, thenf = supf ?, . Similarly, iff’ = inff, exists in L, and if h E L has the property that inf (f’,h ) and all inf (A, h ) exist in L, then +
inf (f’,h) = inf {inf (f,,h)}. T
CH.2 , s 141
67
DISJOINTNESS
Proof. (i), (ii) and (iii) are evident. (iv) Let f = supf, and g = supg,. It is evident that f + g is an upper bound of the set of all elementsf,+g,,. If h is another upper bound of this set, we have h 2 f+g, for every Q, and so h 1f+g. This shows that
f + s = sup(f,+g,,). 7.
(v) Let f = supf,, and let h E L have the property that sup (f,h ) and all sup (A, h ) exist in L. It is evident that sup (f,h ) is an upper bound of the set of all elements sup (A, h). If k is another upper bound of this set, we have k 2 sup (f,,h ) 2 f, for all z, and so k 2 supf 7 = f. Since k 2 h is also satisfied, we have k 2 sup (f,h). This shows that SUP !f,h) = SUP {SUP (f, h ) } . 9
7
The next theorem goes in the converse direction.
Theorem 13.2. r f f = supf, and f + g tisfies g = sup g,,.
=
sup,, ,, (f7+g,,) exist, then g sa-
Proof. For every fixed Q, we have f7+g,, S f + g for every z, and so supf,+g, 5 f+g, i.e., f+g, S f + g , so g,, 2 g. This shows that g is an upper bound of the set of all go. If h is another upper bound of this set, then g,, S h for all 6,so f , +go 5 f +h for all z and Q, which impliesf +g 5 f h, and so g S h. This shows that g = sup g,,.
+
Theorem 13.3. Let L be a Riesz space. r f f = supf,, then f = supf,' and f - = inff,-. Similarly, i f g = inf g7, then g+ = inf g: and g- = SUP g7- * +
Proof. Letf = supf,. It was proved in Theorem 13.1 (v) that f and from one of the distributive laws we derive
+
= sups,',
inf(f, 0 ) = SUP (inf(S7, O)}, t
i.e., -f - = sup (-L-). It follows that f - = inff,-.
14. Disjointness In the present section we assume again that L is a Riesz space.
Definition 14.1. The elements f and g of L are called di$joint ifinf (If I, Igl) 0. This will be denoted by f Ig.
=
68
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH.2,8 14
Note that for elements of L' the present definition is in agreement with the definition of disjointness in section 2 for elements in a lattice with null element. Theorem 14.2. (i) I f f l g and lhl 5 Ifl, then h I g. (ii) Ig and a is a real number, then af I g. (iii) Iffi I g andfz 19,then (A +fz)I 9. (iv) We have flg if and only iff '1g and f - I g.
vf
Proof. (i) We have
0 6 inf(Ihl, Igl) 6 inf(Ifl, Igl) = 0. (ii) Note first that for elements u, u in L' and any number b > 0 we have inf (bu, bu) = b inf (u, u ) by Theorem 11.5 (vi), so u lv implies bu Ibv. Now, flg is equivalent to If1 I 191, and so
(la1+ 1)lfl I (la1 + 1)lSl holds for every real number a. Hence, since
- If1 5 (la1 + l)lfl, it follows from part (i) that a f l (la1 + 1)lgl. Similarly, since Igl 6 (la1 + 1)1g1, lafl = la1
it follows that a f l g . (iii) Letf, Ig andf2 Ig. An application of Theorem 12.5, formula (l), shows that
inf(If1+f2l, Isl) 6 inf(If1I+lf2l, 191) inf(If1L Isl)+inf(If21,191) = o+o = 0, so (f1 + f 2 1I 9. (iv) Iff I g, thenf' Ig on account of I f ' ! = f' 5 Ifl. Similarly we find that f - I g. Conversely, iff' Ig and f - Ig, then (f' +f-)lg, i.e., If1 Ig, and s o f l g.
s
The elementfof L is said to be disjoint from the non-empty subset D of L iff Ig holds for every g E D. Furthermore, the subset D of L is said to be solid if it follows fromfE D and lhl 5 If1 that h E D. Note now already for later purposes that in virtue of (i), (ii), (iii) in the last theorem the set of all elements disjoint from a fixed nonempty subset D of L is a solid linear subspace of L. This solid linear subspace is called the disjoint complement of D , and is denoted by Dd.This terminology is in agreement with that introduced in section 2.
CH. 2,@141
69
DISJOINTNES
Theorem 14.3. If D is a subset of L such that fo = sup (f :f E D) exists in L, and iff _L g holds for all f E D, then fo Ig. Proof. We have f 'J- g and f - J- g for all f E D by Theorem 14.2 (iv). Furthermore,wehavef,f = s u p ( f ' : f ~ D ) a n d f t = i n f ( f - : f ~ D ) b y Theorem 13.3, and so inf (f:,
Igl) = sup {inf (f', 191) :f E D} = 0
by one of the distributive laws, as well as 0
5 inf(f;, Id) 5 inf(f-, Igl) = 0
for all f E D. It follows that f: I g as well as f;
Ig, so f o Ig.
Note that it follows from this theorem that the disjoint complement Dd of the subset D of L has the property that if E is a subset of Dd and fo = sup (f :f E E) exists in L,then fo E Dd. Theorem 14.4. (i) Iff J- g, then (ii)
if+sl = If-gl Iff Ig, then
= Ifl+lgl =
Ilfl-lsll
= SUP (IfL
Isl).
(f+g)+ = f + + g + = sup (f +,g'), (f+g)- =f - +g- = sup (f-,9-).
Proof. (i) It follows from and
0 = 2inf(Ifl, Isl) =
lfl+l~l--llfl--l~ll
s If+sl 5 lfI+lgL IIfl-lsll 5 If-sl 5 Ifl+lsl Ilfl-lgll
that
If+gl = If-sl = lfl+Igl = Ilfl-lsll.
Finally, we have
If I + Is1 = SUP (Ifl, Isl)+inf (If I, Isl) = SUP (IfI, Isl). (ii) It follows from inf(If1, Igl) = 0 together with 0 6 f' 5 If1 and 0 5 g' 5 191 that inf (f ', g') = 0. Hence f 'Ig', and so f '+g' =
sup (f ', 9') by part (i) of the present theorem. Similarly we obtain that f- Ig- and f -+g- = sup (f-,g-).
In order to show that (f+g)' = f + + g ' and (f+g)- = f - + g - we recall that by Theorem 12.1 we have (f+g)' 5 f'+g' and (f+g)- 5
70
ORDERED VECTOR SPACES A N D RIESZ SPACES
[CH. 2, 6 15
f - + g a y holding for arbitrary f, g E L . It will be sufficient, therefore, to prove that f l g implies f f + g + 5 ( f + g ) + and f - + g - 5 ( f + g ) - . The first inequality follows from f++g++(f+$
5f++gf+f-+9-
= Ifl+lgl = If+sl = ( f + g ) + + ( f + s ) - ,
and similarly for the second inequality.
In the converse direction we have the following theorem.
Theorem 14.5. (i) Zflf + g l = I f - g l , then f lg. (ii) Zf If+gI = sup ( I f I, Igl), then f lg. Similarly, of course, if If-gl = SUP ( I f I, Igl), thenf 19. Proof. (i) Follows from the formula 2inf(IfL Isl) = IIf+sl-If-gIl.
(ii) If I f + g l = sup (If 1, Igl), it follows from
2sup ( I f l , Igl) =
If+sl+lf-sl
that I f + g l = I f - g l , and so f l g by part (i). It will be evident from the foregoing that the notion of disjointness has some properties similar to the properties of orthogonality in Hilbert space or Euclidean space. Here is another property of that kind.
Theorem 14.6. I f ( f j : i = 1 , . . ., n ) is a system of nonzero and mutually disjoint elements of L, then this is a linearly independent system. Proof. If the system fails to be linearly independent, one of the f i , sayf,, is a linear combination of the other elements in the system, say f l = a2f2 +anfn.By Theorem 14.2 we have
+ . ..
fi
I(a2f2 + .
* *
+anL),
i.e.,fl If l . This means that inf (If l l , If l l ) = 0, i.e., If i l = 0, and sofl = 0. But this contradicts the hypotheses.
15. Monotone sequences and directed sets We begin with a definition.
Definition 15.1. The sequence (f n : n = 1 , 2, . . .) in the ordered vector space L is called increasing i f f l 5 f 2 5 . . . and decreasing l f f l 1f i 2 . . ..
CH.2 , s 151
71
MONOTONE SEQUENCES A N D DIRECTED SETS
This will be denoted by f n t or fnJ respectively. If f n t andf = supf n exists in L, we will write f, f f, or f,f, f if necessary. I f f , J and f = inf f n exists in L, we will write f, J f. Theorem 15.2. (i) I f f , f f , then A, t f for every subsequence (S,: k = 1, 2, . . .) such that n , < n2 < . . .. This holds in particular for the subsequence (f,: n 2 no), where no is an arbitrary but jixed natural number. (ii) I f f n t f and a 2 0 is real, then af. t a$ (iii) I f f n t f andgn t 9, thenfn+gn t f + g (iv) If f n t f, grit andf, +gn t f +9, then gn t 9-
Proof. (i) Let f n tf, and let (f,,:k = 1,2, . . .) be a subsequence such that n, < n2 < . . .. Evidentlyf is an upper bound of the subsequence. If g is another upper bound of the subsequence, and f , is an arbitrary member of the sequence (f n : n = 1,2, . . .), there exists a member nk, 0 of the index set (n, :k = 1, 2, . . .) such that m S n k . 0 , and hence fm
5 f n k , 5 9' 0
The inequality f , S g holds thus for all m yso f = supf , 5 g . It follows that f = supf n k . (ii) Evident. (iii) Evidentlyf n + g n is increasing and f + g is an upper bound of the set of all elementsf,+g,. If h is another upper bound of this set, and if m and n are natural numbers, say m S n, then f m + g n4 f n + g n h, and hence (in virtue of Theorem 13.1 (iv)) we have f + g = suP(fm+gn) Ihm. n
This shows that f + g = sup ( f , + g n ) ,s o f , + g , t f + g . (iv) For any fixed n we have f , + g n f m + g mfor all m fm+gn ISUP ( f m + g m : m 2 n ) m
3 n, and so
=f+g
holds for all m 2 n. It follows that SUP (fm m
: m 2 n)+gn
S f+g,
i.e.,f + g , S f + g , so gn 5 g . This shows that g is an upper bound of the set of all g,,. If h is another upper bound of this set, thenf,+g, 5 f+h holds for all n, which implies f + g 5 f + h , and so g 5 h. This shows that g = SUP 9 . 3 i.e.9 gn t 9-
72
1CH. 2.8 15
ORDERED VECTOR SPACES AND RIESZ SPACES
Theorem 15.3. If L is a Riesz space, and iff,, t f and g,, t g, then
gn1.t SUP (f, g), inf ( f n 2 gn) t inf (f, g)*
SUP ( f n ,
In particular, f n
tf
implies
f
+
and f.- 1f - .
Proof. It is evident that sup (A,g,,) is increasing and sup (f,g ) is an upper bound of the sequence of all elements sup (f,,g,,). Let h be another upper g,,) 5 h holds for all n, so bound of the sequence. Then f, 5 sup (f,,, f S h. Similarly g 5 h, and hence sup (f,g ) 5 h. This shows that sup (f,g ) is the supremum of the sequence of all elements sup (A,g,,), i.e., sup (A,g,,) t SUP (f,s). Set p = sup (f,g ) and q = inf (f,g ) , and similarly set p,, = sup (&, g,,) and q,, = inf(A,g,,) for n = 1,2,. .. Observe that p + q = f + g and p,,+ q,, = f,+gn for every n. We have p,, p as just proved, q,, is increasing and Pn+qn =fn+gn t f + g = P + q
.
by part (iii) of the preceding theorem. It follows then from part (iv) of the preceding theorem that qn t q, i.e., inf (f,,g,,) t inf (f,g). Definition 15.4. I f L is an ordered vector space, ifu,, E L f for n = 1,2, and s, = u1 + . . . + u,, satisfies s,, s, we will write u,, = s.
1:
.. .
Theorem 15.5. Let L be a Riesz space. (i) Given that u, v , z E L', u+u = z and 0 5 z,, z, there exist sequences 0 S u,, t u and 0 S v,, t v such that u,,+v,, = z,, for all n. In particular, if u, v E L f and u+v = zi with zi EL' for all n, there exist elements u;, vi EL' such that u = ui, v = v; and u ~ + v=~zi for all n. (ii) This can be extended to a greater number of terms; we present the formulation for three terms. I f u, v , w, z E L f , u + v + w = z and 0 5 z,, t z, there exist sequences 0 u,, t u, 0 5 v,, T v and 0 5 w,, t w such that u,,+v, w,, = z, for all n.
1: 1;
17
+
Proof. (i) Setting u,, = inf (u, z,) for all n, we have
0 5 u,, t inf (u, z) and it is evident that v, = z,,-u,, satisfies v,, Since
0 4 u,,+,-u,,
= u,
2 0 and u,,+v,,
= inf (z,,+~,u)-inf (z,,, u )
=
Iz,,+l-z,,
z,, for all n.
CH. 2,s 151
73
MONOTONE SEQUENCES AND DIRECTED SETS
by Birkhoff’s inequality, we obtain that
5 Zn+1-un+1 = un+1, so 0 5 u,,T. It follows then from u,, t u, v,,t and u,,+u,, = z,, t z vn
vn
= Zn-un
t v-
= u+
u that
(ii) Set u+u = s, so s+ w = z and 0 5 z,, t z. By part (i) there exist sequences 0 5 s,,t s and 0 5 w,, t w such that s,,+w,, = z,, for all n. Once more by part (i), there exist sequences 0 5 u,, t u and 0 5 u,, u such that u,,+ v,, = s,, for all n. The desired result follows now.
Corollary 15.6. Every Riesz space L has the following properties. ( i ) (Dominated decomposition property.) Given that
..
0 5 u 5 zi+
... +zp
with z,, E Li for n = 1, .,p , there exist elements u l , . . ., upE L i such that u = u l + . . +up and u,, 5 z,,f o r n = 1,. .,p. (ii) (Riesz interpolation property.) If u, u, z l , z2 E L+ and u+ u = z l + z 2 , there exiAt u l , u 2 , v l , u2 E L+ such that
.
.
u , + u , = u, u , + u , = z1,
u,+u, = u, uz+u2 = z2.
Definition 15.7. The indexed subset {f, : T E {z}} of the ordered vector space L is called directed upwards if for any pair zl, t 2E { t }there exists r 3 E { T } such thatf,, 2 f,,andf,, 2 f,, hold simultaneously, and {f , : t E { t } is } called directed downwards i f f o r any pair t l ,t 2E { t }there exists t 3 E { t }such that c, 5 f,, and f,, Sf,,hold simultaneously. This is denoted by f,t or f,J respectively. I f f,t and f = supf, exists in L, we will write f , t S, or f , t, f if necessary. Iff,l and f = inff, exists in L, we will writef, Jf. The upwards directed sets {f,} and {g,},indexed by means of the same index set {z}, will be called equidirected iffor any pair z1 , t 2E {T} there exists t 3 E {z} such Ithat f,, 2 f,, andf,, 2 f,, as well as g,, 2 g,, and g,, 2 g,, hold simultaneously. The upwards directed sets {f,} and {g,} are equidirected in particular iff,, 5 f,, holds if and only if g,, S gr2holds (note, however, that this is not a necessary condition for being equidirected). A similar definition holds for downwards directed sets. Monotone sequences are special examples of directed sets. Directed sets have many properties similar to properties of monotone sequences, as shown in the next theorems. Theorem 15.8. (i)
IfA t f
and a 2 0 is real, then af, t af.
74
[CH. 2 , § 15
ORDERED VECTOR SPACES A N D RIESZ SPACES
(ii) Iff, t f and g U t Y g then ( f , + g U ) t T , U (f+g)' (iii) I f {f,} and {g,} are equidirected,f, t f and g, t g, then f, g, t f g. (iv) Iff, tf, g u t and (f,+ gu) t r , u (f+g), then g u t 9. (v) If {A} and {g,} are equidirected,f,tf andf,+g, t f + g , then g, t g.
+
+
Proof. (i) Evident. (ii) Let f,t f and g,, t g. Note first that the set of all elementsf,+g,,, indexed by the set of all pairs (z, a), is directed upwards. Indeed, given (zl,al)and (z2, 02),there exists (z3,03)such that f,, 2 f,, and f,, 2 Az as well as gu, 2 g,,, and g,,, 2 g,,, and hence f,,+g,,, 2 f r , + g u l and f,, +g,,, 2f,,+guz. It was proved already in Theorem 13.1 (iv) that f+ g = sup,,,, &+go). Hence, combining these facts, we obtain that (f, +g u )
tr, u
( f +g).
(iii) Let {f,} and {g,} be equidirected,f, t f and g, t g. The set of all elementsf,+g, is now directed upwards, and evidentlyf+g is an upper bound of the set. If h is another upper bound of the set and z1,z2 E {z} are given, there exists z3 E {z} such thatf,, 2 f,, and g,, 2 gr2,and hence f,,+gzz
Sfr,+gr, S he
By Theorem 13.1 (iv) we havef+g = (f,,+g,,), so on account of f,,+grz5 h holding for all z1 ,z2 we find now that f + g S h. This shows that
f+
=
"P
r
(fT
+
ST)'
(iv) In Theorem 13.2 it was proved that i f f = supf, and f + g = sup,,,, (L+g,,), then g = sup gu.Hence, iff, If, g,, is directed upwards and (L+gu) t r , u (f+g), then g u t 9. (v) Let {f,}and {g,} be equidirected, f, f and (f,+g,) t (f+g). We will prove first that f +g is an upper bound of the set of all elementsf,, +gTz. For this purpose, given z1and z2, let z3 be such thatf,, 2f,,and g,, 2 grZ. Then f,,+g,z S f T 3 + g T 3 r f + g , which is the desired result. Sincef + g is already the supremum of the set of all elements fr+g,, it follows that
f + g = SUP (fz, +grz). T I . 72
Hence, by part (iv) above, we have g,
t g.
CH. 2 , s 151
MONOTONE SEQUENCES AND DIRECTED SETS
75
Theorem 15.9. Let L be a Riesz space. ( 9 Iff, t f and 9, t 9,then
"P (f, SU) tT, U "P (f,g), inf (f,,S U ) t T , U inf (f,g). 9
f, t f implies that sup (f,, g) t sup (f,g) and inf (A,g) inf (f,g) for every g E L. As a still more particular case, we have that f, t f impliesf,' t f f a n d f ; 1f - . (ii) I f {A} and {g,} are equidirected,f, t f and gr t g, then
I n particular,
"P (f,9,)t "P (f,g)? inf (f,ST) 7 inf (f,9). 7
9
Proof. (i) The proof for sup (f,, g,) t,, ,sup (f,g) is very similar to the proof of the corresponding statement for sequences in Theorem 15.3. For the second formula, set p = sup (f,g) and q = inf (f,9);similarly, set p,,, = sup (A,g,) and q,, ,= inf (A,g,) for all z, CT.Observe that p+q = f + g and p,, +qr,,= h + g , for all r, CT.We have p,, ,t p as observed above; furthermore. q,, ,is directed upwards and
PT,U+qT,O = (f,+gU)
t ( f + g ) = P+q
by part (ii) of the preceding theorem. It follows then from part (iv) of the preceding theorem that g,, , q, i.e., inf (A,g,) t,, inf (f,9). (ii) Similarly. Theorem 15.10. Let L be a Riesz space. Given that u, v, z E L + ,u+u = z and 0 5 z, t z, there exist directed sets 0 5 u, T u and 0 S v, t u, equidirected with {z,} in the sense that z,, S z, implies uT15 u,, and vI1 S v,,, and such that u,+ v, = z, holds for all T. This can be extended to a greater number of terms. Proof. Set u, = inf (u, z,) for all r. Then we have 0 5 u, t inf (u, z ) = u, and z,, S zI2implies uI1 S u,,. Furthermore, it is evident that v, = z,-u, satisfies v, 2 0 and u,+v, = z, for all z. For zTl5 z,, we have
0
s u,,-u,,
=
inf (z,,, u)-inf (zTl, u ) S zT2-zTI
by Birkhoff's inequality, so =
5 zT,-uUIZ
=,,T'
and hence 0 5 v,t. It follows finally from u, t u, v,t and u,+v, = z, z = u+v that u, t v.
t
76
ORDERED VECTOR SPACES AND RIESZ SPACES
[CH. 2,515
Theorem 15.11. (i) Given the indexed subset {f,} of the Riesz space L, there exists an upwards directed subset {f,} of L such that {fu} c {A} and {f,} has the same upper bounds as {f,}. (ii) If {f,} is an upwards directed subset of the Riesz space L andAo is an arbitrary member of {A},then the set {f, :f,2 f,,} is still upwards directed and has the same upper bounds as {f,}. In particular, the set {A}has a supremum in L if and only i f the set {f,-Lo:f, 1f,,} has a supremum in L+
.
Proof. (i) For any finite subset z = (a1,. . ., 0,) of the index set {a}, let
f, = fu,,....
U"
= SUP (fu, 7
* *
.,fun).
Evidently {fu} is a subset of the thus obtained new set {A}, and {f,} and {A} have the same upper bounds. In addition, {f,} is directed upwards. Indeed, given we set
= (01 Z"
7
*
., a,),
= (a1,
Z'
= (0; ,
. . .) a;),
. . .)a, ,a; , . . .)0;),
and then we havef,., 2f, andf,.. 2 A,. (ii) It is evident that the subset {f, :f,2 A o }is still upwards directed. It remains only to prove that any upper bound h of {f, :f,Zf,,}is also an upper bound of {f,}. Given an arbitrary f,,E {A}, there exists an index z2 E (T} such thatf,, 2 f,, andf,, 2 f,,. Hence,f,? is a member of the subset {f,:f,5 f,,}, which implies thatf,, 5 h. It follows that f,, S f,, S h, and this shows that h is indeed an upper bound of {f,}. We recall that a partially ordered set is called Dedekind complete whenever every non-empty subset which is bounded from above has a supremum and every non-empty subset which is bounded from below has an infimum. It was proved in Theorem 1.2 that it is sufficient for Dedekind completeness that every non-empty subset which is bounded from above has a supremum. The first part of the last theorem shows that in a Riesz space L it is sufficient for Dedekind completeness that every non-empty upwards directed subset which is bounded from above has a supremum, and the second part of the same theorem shows that it is even sufficient that every non-empty upwards directed subset in L+ which is bounded from above has a supremum. Exercise 15.12. Show that an ordered vector space has the dominated decomposition property if and only if the space has the Riesz interpolation property (cf. Corollary 15.6 for the definitions of these properties). For any u 2 0 in the space, denote the subset (w : 0 5 w 5 u ) by [O, u ] .
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MONOTONE SEQUENCES AND DIRECTED SETS
Show that the dominated decomposition property is equivalent to the property that [O, u ] [O, u ] = [O, u u J
+
+
holds for all u, u 5 0, where the addition sign on the left denotes the algebraic sum (i.e., the set of all w 1+ w2 with w1 E [0, u] and w2 E [0, u ] ) . a
Exercise 15.13. Let f, g and h be real polynomials on the interval ( x :
5 x I; b) with a and b finite, such that h 2f and h 2 g hold pointwise
on the interval and such that h is not identically equal to eitherfor g. Show the existence of a real polynomial k such that k 2f, k 5 g and k 5 h hold pointwise on the interval, and k is not identically equal to h. It follows that the ordered vector space of all real polynomials on the interval, partially ordered pointwise, is not a Riesz space. Hint: Let k = h - c ( h - f ) ( h - g ) with the positive constant c so small that 1- c(h -f)and 1 - c(h -g) are positive on the whole interval.
Exercise 15.14 (F. Riesz, [2]). Let a and b be finite real numbers, and L the ordered vector space of all functions
on ( x :a 5 x 5 b), wherep and q are real polynomials such that q(x) > 0 for all x in the interval, and where the algebraic operations and the partial ordering are pointwise. Show that the positive cone L' is generating (i.e., everyf E L is of the formf = u-u with u, u EL') and L has the Riesz interpolation property, but L is not a Riesz space. Hint: Given u l , u2, ul , u2 E L + such that u1+ u2 = u1 u2 = w, let
+
xi
if
W(X)
xi
=0
for i,j = 1,2. Then w i j ( x ) = ui(x) and w i j ( x ) = uj(x). Note that wij(x) is indeed a function in L', i.e., if a 5 xo 5 b and xo is a root of multiplicity n of the equation W ( X ) = 0, then the polynomial ui(x)ui(x)contains at least the factor ( X - X ~ ) " + ~ Actually, . ui(x)uj(x)has at least the factor ( X - X ~ ) ~ " .It follows by means of the result in the preceding exercise that L is no Riesz space. Exercise 15.15. In Exercise 11.14 it was required to prove that iff+g p + q in the Riesz space L, then
=
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[CH. 2.5 16
SUP (f, PI + inf (999) = f +9. Show that this result can also easily be derived by means of the Riesz interpolation property. 16. Order convergence and relatively uniform convergence
The properties discussed in the preceding section are very much analogous to the convergence properties of monotone sequences of real numbers. An extension to not necessarily monotone sequences follows. Theorem 16.1. It is said that the sequence (f.: n = 1,2, . . .) in the Riesz space L is order convergent to the element f E L whenever there exists a sequence p,, 10 in L such that If -f,,l 6 p,, holdsfor all n. This will be denoted by f.+f. The following properties hold. (i) I f f . +f and& + g , then f = g. (ii) ? f f n l f o r f , tf, thenf. +f. (iii) If f,,? or fJ a n d f . +f, then f.? f or f.1f respectively. (iv) I f f,, +f, g,, + g, and a and b are real numbers, then af,+bg,, + af+ bg, sup (fn,g,,) + sup (f,g ) and inf (f,,, g,,) -, inf (f,9). In particular, iff, +f, thenf: + f + + , ff -i andIf,,l + Ifl. (v) Iff, -,f, thenf,, +f for every subsequence {f,,, : k = 1,2, . . .} such that nl < n, < . . ..
Proof. For (iii), observe that iff.? andf,, +f, then ( f -f,)- is monotonely 4 p,, 10. increasing, whereas on the other hand 0 S ( f - f . ) - S If-f.1 Hence ( f - f . ) - = 0, so f - f . = If-f.1. It follows that f-f,, 10. It is not true in every Riesz space L that iff. + f i n L and a,, -, a in real r,umber space, then a,,& + af. By way of example, let L be the lexicographically ordered plane, f. = f = (1, 1) for all n and a,, = n-'. The sequence {a,,&} is now decreasing, but it is not true that a,,& 10.There is an important class of Riesz spaces, however, for which f,, +f and a,, + a implies that a,f. + af. This is the class of Archimedean Riesz spaces. The Riesz space L is called Archimedean if it is true for every u E L+ that the decreasing sequence (n-lu :n = 1,2, . . .) satisfies n - ' ~10.It follows then immediately that E,,U 4 0 for every sequence (E,, : n = 1, 2, . . .) of positive real numbxs satisfying E,, 10. Properties of Archimedean Riesz spaces will be discussed in the next chapter; here we will restrict ourselves to the observation that the Riesz space L is Archimedean if and only if, given u, v E Lf such that nv 5 u holds for n = 1,2, . . ., we have v = 0. Indeed, if L is Archimedean
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and 0 S nu S u holds for n = 1,2, . . ., then
0
sv
n-'u 5.0,
so v = 0. Conversely, suppose that 0 5 nu 6 u (n = 1 , 2 , . . .) implies v = 0 and, for an arbitrary uo EL', consider the sequence (n-lu0 : n = 1,2, . . .). We have to prove that any lower bound w of the sequence satisfies w 5 0, since this will imply immediately that 0 is the infimum of the sequence. For this purpose, observe that v = sup (w, 0) is still a lower bound of the sequence, so we have 0 6 nu 5 uo for n = 1,2, . . .. This implies v = 0, i.e., sup (w, 0) = 0, and hence w 6 0. One final remark about terminology. The nonzero element f in L is called infinitely small with respect to the element g in L whenever nl f I 5 Igl holds for n = 1,2, . . .. The Riesz space L, therefore, is Archimedean if and only if L contains no infinitely small elements. Besides order convergence we introduce another kind of convergence for sequences in a Riesz space. Theorem 16.2. Given the element u 2 0 in the Riesz space L, it is said that the sequence (f,: n = 1,2, . . .) in L converges u-uniformly to the element f E L whenever, for every E > 0, there exists a natural number N, such that If - f , l 5 E M holds for all n 2 N,. It is said that the sequence (f,:n = 1,2, . . .) in L converges relatively uniformly tof wheneverf, converges u-uniformly tof for some u E Li (note that we may have different ufor direrent sequences). Relatively uniform convergence o f A to f will be denoted b y f , +f(ru). In the Soviet literature relatively uniform convergence is called convergence with respect to a regulator. The following properties hold. ( i ) I f L is Archimedean a n d f , converges relatively uniformly to f, t h e n h converges in order to$ Hence, if L is Archimedean andf, converges relatively uniformly to f as well as to g, then f = g. (ii) Iff, +f (ru), gn + g(ru), and a and b are real numbers, then a f , +bg, -,af + bg(ru), SUP (f,,gn) SUP (f, g)(ru), inf (f,, g,) inf (f,g)(ru). +
+
In particular, f, -, I
f I(4.
--f
f ( r u ) implies that f,' +f '(ru), f l -,f -(ru) and
1S.I
Proof. Routine. Relatively uniform convergence is stable, i.e., it has the property that for
80
[CH.2,O 16
ORDERED VECTOR SPACES AND RlESZ SPACES
any sequence f, + 0 (ru) there exists a sequence of real numbers (A, : n = 1,2, . . .) such that 0 6 A, 00 and 1,Z.f.+ 0 (ru). Indeed, given that f, + 0 (ru) there exists a sequence of positive real numbers (E, : n = 1,2, . . .) and an element u E L + such that 8, J. 0 and If,l 5 E,U for all n, and so A, = c,,-* satisfies the abovementioned conditions. Order convergence is not necessarily stable. By way of example, letf,(for n = 1,2, .) be the element of the space 1, with the first n coordinates zero and all other coordinates equal to 1. Thenf, LO, but for any sequence of real numbers 1, satisfying 0 5 A,, t 00 it is impossible that 1,f,, + 0, simply because 1,f,is not bounded from above. The following theorem is, therefore, of some interest.
..
Theorem 16.3. In an Archimedean Riesz space order convergence is stable if and only if order convergence and relatively uniform convergence are equivalent. Proof. In an Archimedean space relatively uniform convergence implies order convergence by the preceding theorem. Assuming stability of order convergence, it will be sufficient, therefore, to prove that order convergence implies relatively uniform convergence. To this end, letf, + 0. Since order convergence is stable by hypothesis, there exists a sequence 0 < A, t 00 such that 1,f, + 0. It follows that 1,IjJ + 0, and hence the sequence
(1,,If,l : n = 1,2,. ..) is bounded, i.e., there exists an element U E L +such that A,,lf,l 5 u holds for all n. Then If,! 5 A;' u holds for all n, and so f. + 0 (ru). Conversely, if order convergence and relatively uniform convergence are equivalent, then order convergence is stable because relatively uniform convergence is so. The subset S of the Riesz space L is called order closed if for every order convergent sequence in S the order limit of the sequence is also a member of S. It is immediately verified that the empty set and the space L itself are order closed and that arbitrary intersections and finite unions of order closed sets are order closed (note that if Si, . . .,S,, are order closed and the sequence (f,, :n = 1,2,. . .) is contained in Si, then at least one of the Si contains an infinite subsequence; then use Theorem 16.1 (v)). Hence, the order closed sets are exactly the closed sets of a certain topology in L, the order topology.
u;
Theorem 16.4. (i) The order topology in L satisfies the TI-separationaxiom, i.e., every subset of L consisting of one point is order closed. (ii) The subset V of L is open in the order topology ifand only i f , for every
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.
sequence (f,: n = 1,2,. .) in L converging in order to a point f havef, E Vfor all but afinite number of thef,.
E
V, we
Proof. (i) Evident. (ii) Assume fist that V c L is open,f, +fand f E V. If it is not true that f, E V holds for all but a finite number of thef. , there exists an infinite subsequence (&, :k = 1,2,. . .) in the complement S of V, and so it follows fromf,, +fand from the fact that Sis closed thatf E S. But this is impossible since we have f E V by hypothesis. Conversely, let V c L have the property that if any sequence ( f . :n = 1,2, ., .) converges in order to an elementf E V, thenf. E Vholds for all but a finite number of thef,. In order to show that the complement S of V is closed, assume that f,E S for n = 1,2, . . . and f , + J Iff E S does not hold, thenfE V and s o f , E V for all sufficiently large n by hypothesis. This is impossible sincef, E S holds for all n. Hence we must have f E S, and this shows that S is closed. The remaining part of this section is rather technical; the reader who is not immediately interested in further properties of the order topology and the corresponding relatively uniform topology is advised to proceed to the next chapter. It follows from the last theorem that if the sequence (fn :n = 1,2, . . .) converges in order tof, thenf, converges to f in the order topology (Le., any neighborhood o f f in the order topology contains all but a finite number of the f,). In view of the definition of the order topology in L, according to which a subset o f L is closed whenever it contains all its sequential limit points with respect to order convergence, we can easily prove now that a mapping cp of L (provided with the order topology) into an arbitrary topological space X is continuous if and only iff. +f always implies that cp(f,) converges to cp(f) in the topology of the space X . Indeed, if the mapping cp is continuous, i f f , +f and if 0 c X is an open neighborhood of cp(f), then the inverse image cp-'(O) is an open neighborhood off, sofn E ~ ~ ' holds ( 0 )for all but a finite number of thef,. But then cp(f,) E 0 holds for all but a finite number of the cp(f,), which shows that cp(f,) converges to cp(f) in the topology of X . Conversely, given thatf, +falways implies that cp(f,) converges to cp(f) in the topology of X, we have to prove that cp is continuous, i.e., we have to prove that the inverse image cp-'(O) of any open set 0 c Xis open in L. To this end, letfbe a point in cp-'(O) and let f , -.f. Then cp(f,) converges in X to cp(f), so all but a finite number of the cp(f,) are in 0. It follows that all but a finite number of thef, are in cp-'(O),
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so in view of the above Theorem 16.4 the set rp-'(O) is open. As a particular case we obtain the result that if rp is a mapping of L into itself such thatf, +f always implies rp(f,) + rp(f), i.e., order convergence o f f , to f implies order convergence of rp(f,) to rp(f), then cp is continuous with respect to the order topology. This remark can be used in the proof of the following theorem. Theorem 16.5. (i) The order topology in ihe Riesz space L is translation invariant, i.e., for any fo E L the mappingf +f +f o of L onto itself is a homeomorphism. (ii) For any real a # 0 the mapping f + af of L onto itself is a homeomorphism. (iii) For any$xed g E L the mappings f + sup (5 g ) and f + inf (f,g ) of L into itse,f are continuous. (iv) The mappingsf +f -+ ,f +f - andf -+ If 1 of L into itselfare continuous. Proof. For (iii) and (iv), observe that fn + f implies sup sup (f,g ) , and similarly for the infimum.
(A, g ) +
For any subset S of L , the pseudo order closure S' of S is the set of all f E L such that there exists a sequence (f,:n = 1,2,. . .) in S converging in order tof. It is permitted that f , =f,for m # n. The order closure of S, i.e., the closure of S in the order topology, will be denoted by cl ( S ) . Evidently we have s c S' c cl ( S ) . Taking closures, we obtain cl ( S ) c cl (S') c cl ( S ) , so cl ( S ' ) = cl ( S ) . Hence, replacing S by Sf in the first formula, we obtain also that S' c (S)'c cl ( S f ) = cl (S),
and by induction it follows that
s c S'
c (9)'c S" c
. . . c cl ( S ) .
(1)
Further information is contained in the following theorem. Theorem 16.6. (i) The subset S of L is order closed if and only i f S = S', i.e., S = S' already implies that S = cl (5'). (ii) The subset S of L satisfies S' = cl (5') ifand only i f S' = (S')'. Proof. (i) We have to show that S = S' implies that S is order closed. For this purpose, assume that S = S', f.E S for n = 1,2, . . . and f, +f. Then f E S' by the definition of S', and so f E S on account of S = S'. This shows that S is order closed.
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(ii) If it is given that S’ = (S’)’,then S’is order closed by part (i) of the present theorem, i.e., S‘ = cl (S’). But cl ( S ’ ) = cl ( S ) holds without extra assumptions, so under the present hypotheses we have S‘ = cl (S). Conversely, given that S‘ = cl ( S ) , it follows immediately from formula (1) above that S’ = (9)’. We will prove now that, for every subset S of L, the pseudo order closure and the order closure of S are equal if and only if the space L has another remarkable property, called the diagonal gap property. The Riesz space L is said to have the diagonalproperty whenever, given any double sequence (f,k: n, k = 1, 2, . . .) in L,any sequence (fn :n = 1 , 2 , . . .) in L and any fo E L such that fnk+ f, for all n (“as k + a”)and f, +f o , there exists for every n an appropriate k = k(n) such that fn,k(,) +f o . The space L is said to have the diagonal gap property if, under the same hypotheses that f& +f, for every n a n d f , +fo, there exists a sequence f,,,k(,,) with n , < < n2 < . ., i.e., an infinite sequence containing at most one member from each sequence (fnk :k = 1,2,. . .), such that
.
(”as a”)* The diagonal property, as well as the notion of relatively uniform convergence and of stability of order convergence, was introduced by L. V. Kantorovitch ([2], 1936). He proved that in a very special class of Archimedean Riesz spaces, called regular spaces, order convergence is stable (and so order convergence and relatively uniform convergence are equivalent) and the diagonal property holds. The diagonal property and diagonal gap property will receive further attention in later chapters; at the present we will only show that the diagonal gap property is related to the order topology as follows. fn,,k(ni) + f O
+
Theorem 16.7. We have cl ( S ) = S’for every subset S of L if and only if L has the diagonal gap property. The diagonal property is, therefore, a
sujicient condition in order that cl ( S ) = S’ shall hold for every subset S of L. Actually, it will be proved later (in Theorem 68.8) that the diagonal gap property and the diagonal property are equivalent.
Proof. Assuming first that L has the diagonal gap property, we have to prove that (S‘)’ = S‘ holds for every subset S of L. Givenf o E (S‘)’,there is a sequence (f, : n = 1,2, . . .) in S‘ such thatf, +fo,and for every n there is a sequence ( f & : k = 1, 2,. . .) in S such that fnk+f, (as k + m). Then, since L has the diagonal gap property, there is a sequence f,,,k(,,) +fo,and sofoE S’.Hence, anyf, E (S‘)‘satisfiesfoE S’. This shows that (S’)’ = S‘. Conversely, assuming that (S’)’ = S’ holds for every S c L, we have to
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prove that L has the diagonal gap property. Assume, therefore, that f, +f,, as k -P co for n = 1,2,. . and f, +fo. There are three cases to consider, (i) f, # fo for all n = 1,2, . . ., (ii) f,,= fo for only finitely many n, (iii) f,,= fo for infinitely many n. In the first case there are, for each fixed n, only a finite number off& equal tof,. These we may omit, i.e., we assume that f , k # fo holds for all n, k. Now, let S = (fnk :n, k = 1, 2, . . .). Then fo E (S)’= S, so there is a sequence in S converging in order to fo. For every n there are at most finitely many k such thatf, is a member of this sequence (otherwise there would exist a subsequence converging in order to fo as well as tof,, which is impossible on account o f f , # fo).For each n for which there exists at least onefnk in the sequence, let k(n) be the largest k such that fd is in the sequence. There are infinitely many such n, say n , , n,, . . .; we may assume that nl < n, < . ., and we have now that fn,,f(,,,) +fo. In the second case we may omit all f, and all ,Lfor the values of n satisfyingf, = fo . In the third case we restrict ourselves to the values of n satisfying f, = fo . We may assume, therefore, that fnk -P fo as k -P co for all n. Now take aa auxiliary sequence p,, 10 such that all p,, are strictly positive (if p , -1 0 is possible only whenever p,, = 0 for all sufficiently large n, our theorem is trivially true), and set g,,k = fnk +p,, for all n, k. In view of what has been proved for the first case, there exists a sequence gnf,k ( n f ) +fo. It follows easily that f,,, k(,,,) +fo . Combining all results, it has been proved thus that L has the diagonal gap property.
.
.
In is an interesting fact that tverything we have done now for order convergence can be done exactly analogously for relatively uniform convergence. The only additional phenomenon to keep in mind is that in a non-Archimedean space the limit of a relatively uniformly convergent sequence is not necessarily uniquely determined. It can happen, therefore, that f, +f ( r u ) and f, + g(ru) without f and g being equal. We briefly collect the main definitions and theorems. The subset S of the Riesz space L is called (relatively) uniformly closed whenever, for every rdatively uniformly convergent sequence in S, all relativdy uniform limits of the sequence are also members of S. The empty set and L itself are uniformly closed, and arbitrary intersections and finite unions of uniformly closed sets are uniformly closed. Hence, the uniformly closed sets are exactly the closed sets of a certain topology in L, the relatively uniform topology. If L is Archimedean, the relatively uniform topology satisfies the TI-separation axiom, i.e., every set consisting of one point is closed. Conversely, if every set consisting of one point is relatively uniformly closed,
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then L is Archimedean. Indeed, if not, there exist strictly positive elements u and v in L such that no u holds for n = 1,2,. . .. It follows that the sequence ( f . :n = 1,2, . . .), with f . = 0 for all n, satisfies If.-vl S n-lu for all n, s o f , + v(ru) as well as (trivially)& + O(ru). This contradicts the hypothesis that the set ( 0 ) is relatively uniformly closed. To summarize, we have that L is Archimedean if and only if the set ( 0 ) is relatively uniformly closed. Assuming again that L is arbitrary (i.e., not necessarily Archimedean), the subset Vof L is open in the relatively uniform topology if and only if, for every sequence (f,:n = 1,2, . . .) in L which converges relatively uniformly to a point f E V,we havef. E V for all but a finite number of thef,. It follows that iff, +f(ru), thenf, converges to f in the relatively uniform topology (i.e., any neighborhood off in the relatively uniform topology contains all but a finite number of thef.). Also, a mapping cp of L (provided with the relatively uniform topology) into a topological space Xis continuous if and only i f f , +f ( r u ) always implies that cp(f,) converges to cp ( f ) in the topology of the space X. In particular, if cp is a mapping of L into itself such that f , +f ( r u ) always implies that cp(f.) + cp(f)(ru), then cp is continuous. It follows that, for f o E L and the real number a # 0 fixed, the mappings f +f +f o and f + af are homeomorphisms of L onto itself, for g E L fixed the mappingsf + sup (f,g ) and f + inf (J, 9 ) of L into itself are continuous, and finally, the mappings f +f +,f +f - and f + If I o f L into itself are continuous. For any subset S of L, the pseudo uniform closure S:, of S is the set of all f E L such that there exists a sequence (f,: n = 1,2,. . .) in S converging relatively uniformly to$ The closure of S in the relatively uniform topology will be denoted by 3. Evidently, we have
s c s;, c (s;,);,
c
. . . c s.
The diagonal property and the diagonal gap property for relatively uniform convergence are defined analogously as for order convergence. The following theorem holds now. Theorem 16.8. (i) The subset S of L is relatively uniformly closed if and only if S = SA,i.e., S = S:, already implies that S = S. (ii) The subset S of L SatisJes S:, = 3 if and only if S:, = (S;,):,,. (iii) In an Archimedean Riesz space L we have = Si, for every subset S of L if and only if L has the diagonal gap property for relatively uniform convergence. The diagonal property for relutively uniform convergence in an Archimedean space L is, therefore, a suficient condition in order that S = S;, shall holdfor every subset S of L . Actually, it will be proved later (in Theorem
s
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ORDERED VECTOR SPACES AND RlESZ SPACES
[CH.2,s 16
72.2) that, for relatively uniform convergence, the diagonal gap property and the diagonal property are equivalent. Proof. In (iii) the hypothesis that L is Archimedean is used in the part of the proof where it is assumed that S:, = (S;,):,, for every S, and where f n k -tf,(ru) and fn +fo(ru) withf, #fo for all n. Just as in Theorem 16.7 we may assume that.f,, # fo holds for all n, k. We choose S = (fnk :n, k = 1, 2, . . .). Then foE (S:,):, = S:,, so there is a sequence in S converging relatively uniformly to f o . For any fixed n, there are at most finitely many k such thatf,, is a member of the sequence. Indeed, if this is not the case, the sequence has a subsequence converging relatively uniformly to f,. The subsequence also converges relatively uniformly tofo . But we havef, # fo, and on the other hand, since L is Archimedean, every uniformly converging sequence in L has a unique limit. Hence, the assumption that for a fixed n the sequence contains infinitely manyf,, leads to a contradiction. Just as before, for each n for which there exists at least in the sequence, let k(n) be the largest k such thatf,, is in the sequence. There are infinitely many such n, say n , , n 2 , . . .; we may assume that n, < n2 < . . ., and we have now that f,,, k ( n i ) +fo. The rest of the proof is exactly as in Theorem 16.7. Let L be Archimedean. Then relatively uniform convergence of a sequencef, tofimplies order convergence off, tof, which implies immediately that every order closed set in L is relatively uniformly closed. Hence, in an Archimedean Riesz space the relatively uniform topology is stronger than the order topology. Iff, + f ( r u ) and f,+fare equivalent, the pseudo uniform closure S:, and the pseudo order closure S' of any set S are identical, and the same holds then for the topological closures S and cl (5'). Indeed, S c S c cl (S) holds always (in an Archimedean space) and, on account of the two modes of convergence (for sequences) being equivalent, the uniformly closed set S is also order closed, so must be equal to cl (S). Thus, in an Archimedean Riesz space where order convergenceand relatively uniform convergence for sequences are equivalent, the order topology and the relatively uniform topology are identical. A necessary and sufficient condition in an Archimedean space for equivalence of the two modes of convergence is that order convergence is stable (cf. Theorem 16.3). We will prove now that the diagonal gap property for order convergence is a sufficient condition.
s
Theorem 16.9. If L is Archimedean and has the diagonal gap property for order convergence, then order convergence in L is stable, and so order convergence and relatively uniform convergence are equivalent, and the order topol-
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ogy and the relatively uniform topology are identical. Furthermore, this topology has the property that the closure of any set consists of the set itself plus all its sequential limit points. Proof. It is sufficient to prove that order convergence is stable under the given conditions. Hence, given a sequence (f n :n = 1,2, . . .) in L such that f,, + 0, we have to prove the existence of real numbers A,, such that 0 A,, t co and A,fn -,0. Sincef,,+ 0 implies the existence of a sequence (p,, : n = 1, 2, . . .) in L such that Ifnl 5 p,, 10, it will be sufficient to find 0 Ant co such that A,,p,, 0. To this end, set P,,k = npk for n, k = 1, 2 , . . .. For n fixed we havep,, J. 0 as k + co;hence, in view of L having the diagonal gap property, there exists a sequence
s
s
with n, < n, <
pnr,k(nr) -, 0
. . ..
Sincep,, is decreasing in k for n fixed, we may assume here that k(ni) increases strictly as n, increases. To summarize, we have +0
with n , < nz < . . ., k(nl) < k(n,) <
. . ..
Now, for every natural number n satisfying k(ni) 5 n < k(ni+,), let A,, t 03. Then
n i , so 0
s
IAnfnl
5
lAnpnl
A,, =
5 niPk(nr),
and so it follows easily that A,, f,,+ 0. This shows that order convergence is stable. Finally, it follows from Theorem 16.7 that the closure of any set consists of the set itself plus all its sequential limit points. In the following list several situations which may occur in an Archimedean Riesz space L are presented. The first column indicates whether order convergence and relatively uniform convergence are equivalent or not in L. In the second column is written “good” or “bad” according as pseudo order closure and order closure are the same for every subset of L or not. Similarly equiv. equiv. non-equiv. non-equiv. non-equiv. non-equiv.
good bad good good bad bad
good bad good bad good bad
Exercises 16.15 and 16.16 Exercise 16.17 impossible by Th. 16.9 impossible by Th. 16.9 Exercise 16.18 Exercise 16.19
88
[CH. 2 , @16
ORDERED VECTOR SPACES AND RlFSZ SPACES
in the third column for the pseudo uniform closure and the uniform closure. The final column shows where an example may be found or why the listed situation is impossible. Exercise 16.10. Show that iff. + f i n the Archimedean Riesz space L and a,, + a in real number space, then ad,, + af in L. Exercise 16.11. We use the notions and notations as introduced in this section. (i) Show that if L is Archimedean, and cl (S) = S' holds for every S c L, then order convergence and relatively uniform convergence in L are equivalent, and the order topology and relatively uniform topology are identical. (ii) Show that if in the Archimedean Riesz space L the order topology and the relatively uniform topology are identical and S:, holds for every S c L , then order convergence and relatively uniform convergence are equivalent. Hint: (i) follows by combining Theorems 16.7 and 16.9. For (ii), note that the hypotheses of (ii) imply those of (i). Indeed, under the hypotheses of (ii) we have S:, = 3 = cl ( S ) and S:, c S' c cl ( S ) for every S c L , so S' = cl (S).
s=
Exercise 16.12. With the same notations, show that if S' = cl ( S ) for every S c L,and the decreasing sequence (p,, :n = 1,2,. . .) in L+ converges to zero in the order topology (i.e., every neighborhood of zero contains all but a finite number of thep,,), thenp,, 10. Show similarly that if S:, = S holds for every S c L,and the decreasing sequence (p,, :n = 1,2,. .) in L' converges to zero in the relatively uniform topology, then p,, + 0 (ru).
.
Exercise 16.13. It was proved in this section that the Riesz spaceL is Archimedean if and only if the set (0) is uniformly closed. It was observed by T. Ito [1] that L is Archimedean if and only if the positive cone L+ is uniformly closed. Present a proof. Hint: Assume L+ uniformly closed. We have to prove that (0)is uniformly closed. If not, there exists an element f # 0 in {0}iu,i.e., there exists u E Lf and a sequence of numbers E,, 4 0 such that If1 S E,,u for all n. Sincef # 0, one at least off and f - is not zero, say f+ # 0. We have['fI 5 E,,u for all n, so f E (0) and also -f + E (0) c = Lf,which is impossible. Conversely, if L is Archimedean, u,, E Lf for n = 1,2, . . . and u,, +f ( r u ) , then u,, + A so there exists a sequencep,, 10 in Lf such that u,, -f =< p,, for all n. Then 0 j u,, j f+p, for all n, sof 2 -p,, 0. This shows thatfe L'. +
+
v)
CH.2,O 161
89
ORDER CONVERGENCE AND RELATIVELY UNIFORM CONVERGENCE
Exercise 16.14. Show that in an Archimedean Riesz space L the following proputies are equivalent. (i) L has the diagonal property for order convergence. (ii) Order convergence in L is stable and, for any sequence (u,, : n = 1,2, .) in L + , there exists a sequence (A,, : n = 1,2, . .) of strictly positive numbers such that (A,,u, :n = 1,2,. .) is bounded from above. As observed earlier in this section, L. V. Kantorovitch [2] has introduced a special class of Archimedean Riesz spaces, called regular spaces, and he proved that these spaces have the diagonal property. He proved also that regular spaces satisfy condition (ii) of the present exercise. Hint: If (i) holds, then order convergence is stable by Theorem 16.9. Given the sequence (u,, :n = 1, 2,. . .) in L + , set fnk = k-'u,, for n, k = 1,2,. .. Then fnk J. 0 as k -+ co for every n, and so fn,&(,,) + 0 for appropriate k(n). Hence, setting A,, = {k(rz)]-', we have Anu, 0, which implies that (A,,u, :n = 1,2,. .) is bounded. Conversely, assuming (ii) to hold, we have to prove that if f n k + f , as k -+ co and f, +fo , then f,,k(") -P fo for appropriate k(n). Since order convergence and relatively uniform convergence are equivalent in view of the stability of order convergence, there exists for each n = 1,2,. . an element u,, E Lf such that f& + f , holds u,,-uniformly. Then there exist numbers A,, > 0 such that (A,,#,,:7z = 1,2, .) is bounded, say AnunS u. It follows that fnk -P f.holds u-uniformly for every n, so there exists for every n a natural number k(n) such that
..
.
.
.
-+
.
.
..
Ifn,k(n)-fnl
Finally, it follows fromf,,,(,,)-f,
+0
5 n-'u. and f , - f o
+0
that f , , & ( , , ) - f O
-+
0.
Exercise 16.15. Let L be the Riesz space of all real functions on a finite point set X,the ordering is pointwise. Show that order convergence, relatively uniform convergence and uniform convergence are equivalent, and L has the diagonal property with respect to this mode of convergence. It will be proved later (cf. section 26) that every finite dimensional Archimedean Riesz space is "isomorphic" to a space of the kind considered in the present exercise; hence, the same statements about convergence hold in every finite dimensional Archimedean Riesz space. Exercise 16.16. Let (s) be the Riesz space of all real sequences (cf. Example 11.2 (iii)), the ordering is coordinatewise. Show that order convergence and relatively uniform convergence are equivalent, and (s) has the diagonal prop-
90
ORDERED VECTOR SPACES A N D RIESZ SPACES
[CH. 2 , s 16
erty with respect to this mode of convergence. Hence, pseudo closure and closure of a set in the corresponding topology are the same. Hint: In view of Theorem 16.9 it is sufficient to prove that (s) has the diagonal property for order convergence. Hence, by Exercise 16.14, it is sufficient to prove that order convergence is stable and that for any sequence (u,,:n = 1, 2, . . .)in (s)' there exists a sequence (A, :n = 1, 2, . . .) of strictly positive numbers such that (A,,u, :n = 1,2,. . .) is bounded. For the stability of order convergence it is sufficient to prove that, given f, 10, there exists u E (s)' such that f, 4 0 holds u-uniformly. The sequence u = {u(l), u(2), . . .}, defined by u(n) = nfl(n), is easily seen to satisfy this condition. Given the sequence (u, :n = 1,2,. . .) in (s)', determine A1 > 0 such that Alul(l) 6 1, then Az > 0 such that Azuz(l) S 1 and Azuz(2) -< 1, and so on. Hence, A, > 0 satisfies A,,u,(k) S 1 for k = 1,2,. . ., n. The desired majorant v of (A,,u, :n = 1,2, . . .) is found now by defining
.. u(1) = 1, v(2) = max (1, A1u1(2)}, u(3) = max (~y~lul(3),Azuz(3)}y..
Exercise 16.17. Let L be the Riesz space of all real sequencesf = {f(l), f(2), . . .} with only finitely many nonzero coordinates; the ordering is coordinatewise. Show that order convergence and relatively uniform convergence in L are equivalent, and show also that L fails to have the diagonal gap property for order convergence as well as for relatively uniform convergence. Hint: The proof that order convergence in L is stable is similar to the corresponding proof in Exercise 16.15. For the second statement it is sufficient to prove that L does not have the diagonal gap property for relatively uniform convergence. For n = 1,2, . . ., let u,, EL' have its first n coordinates equal to 1 and all other coordinates zero, and letf,, = k-lu,, for n, k = 1, 2, . . .. Then fnk 4 0 (ru) as k co for every n. Any diagonal gap sequence taken from (f,,: n, k = 1,2,. . .), however, is not bounded from above. Exercise 16.18. Let L be the Riesz space C([O,11) of all real continuous functions on the interval [0, 11, where [0, 11 is provided with its ordinary topology. Show that order convergence in L is not stable and L does not have the diagonal gap property for order convergence. In other words, order convergence and relatively uniform convergence are not equivalent,and there exists a subset of L for which the pseudo order closure is not the same as the order closure. Show that relatively uniform convergence of a sequence is the same as the familiar uniform convergence of the sequence, the relatively uniform closure of a set is the familiar uniform closure, and the relatively
CH. 2 , s 161
ORDER CONVERGENCE AND RELATIVELY UNIFORM CONVERGENCE
91
uniform topology is the norm topology with respect to the uniform norm. It follows immediately that the pseudo closure and the closure of any set with respect to this topology are the same. Derive this also by proving that L has the diagonal property with respect to uniform convergence. Show that the same facts hold in the real sequence space I , . Hint: For the proof that order convergence is not stable, consider for example the sequence (h:n = 2 , 3 , . . .), where f,,(x) = 1 on [0, n-'I, f n ( x ) = 0 on [2n-', 11 and linear on [n-', 2n-']. Thenjf, 4 0, but there exists no sequence 0 6 A,, t co such that A& -+ 0. For the proof that L does not have the diagonal gap property for order convergence, take the same sequence (fn : n = 1,2,. . .) and set fnk = nfk. Then f n k 10 as k -+ co for every n, but there is no sequencef,,,,k(nr)-,0.
Exercise 16.19. Let L b: the Riesz space of all real continuous functions with compact carrier on the real line, where the real line is provided with its ordinary topology. Show that order convergence and relatively uniform convergence are not equivalent, and show also that L fails to have the diagonal gap property for order convergence as well as for relatively uniform convergence. Hint: The proof that order convergence is not stable is similar to the proof in the preceding exercise. The proof that L does not have the diagonal gap property is (for order convergence as well as for relatively uniform convergence) similar to the proof in Exercise 16.17.
CHAPTER 3
Ideals and Bands; Archimedean Riesz Spaces
In the present chapter certain linear subspaces of a Riesz space L are in troduced. Any linear subspace A of L which is solid (i.e., f E A and Ig[S If1 impliesf E A ) is called an ideal (or order ideal). An ideal A is called a band if, whenever a subset of A has a supremum in L, that supremum is an element of A. The linear mapping n of the Riesz space L into the Riesz space M is called a Riesz homomorphism if the mapping preserves finite inlima (and hence preserves finite suprema). The kernel (null space) A, of the Riesz homomorphism n is an ideal in L. If the Riesz homomorphism is one-one onto M , then L and M a r e called Riesz isomorphic. Any Riesz isomorphism of L onto M is a one-one positive linear mapping (the mapping is called positive because it maps L+ into M + ), but not every one-one positive linear mapping of L onto M is a Riesz isomorphism. The mapping is a Riesz isomorphism if and only if the inverse mapping is also positive. Given the ideal A in the Riesz space L, the quotient space L/A is a Riesz space with respect to the natural partial ordering in L/A. Denoting by Lf] the member of LIA that containsf, the mappingf -P Lf] is a Riesz homomorphism of L onto LIA, and A is the kernel of the homomorphism. Hence, every quotient space of L (with respect to an ideal) is a Riesz homomorphic image of L. Conversely, if n is a Riesz homomorphism of L onto M with kernel A, then If]+ nf is a Riesz isomorphism of L / A onto M, and so every Riesz homomorphic image of L is Riesz isomorphic to a quotient space of L (with respect to an ideal). The Riesz homomorphism n of L into M is called a normal Riesz homomorphism if n preserves arbitrary suprema and infima. It will be proved that the Riesz homomorphism n of L onto M is normal if and only if the kernel of n is a band. If D is any subset of L, there is always a smallest ideal, and also a smallest band, which contains D. These are called the ideal and the band generated by D. The disjoint complement od of any non-empty subset D is always a band and Oddis a band including D, but odd is not always equal to the band 92
CH. 3 , s 171
RIESZ SUBSPACES, IDEALS AND BANDS
93
generated by D. Denote by { D } the band generated by D. We have { D } = Ddd for every non-empty subset D of L if and only if L is Archimedean. The set of all disjoint complements, partially ordered by inclusion, is always an order complete Boolean algebra, but the set of all bands, partially ordered by inclusion, is a Boolean algebra if and only if L is Archimedean.
17. Riesz subspaces, ideals and bands In the present section L denotes a Riesz space. Certain linear subspaces of L, possessing extra properties related to the lattice structure of L, will be distinguished by special names, as follows. Delinition 17.1. (i) The linear subspace K of L is called a Riesz subspace of L if,for every pair of elementsf, g in K, the elements sup (f,g) and inf (f,g) are also in K (where sup (f,g) and inf (f,g ) are the supremum and infimum o f f and g that exist in L by hypothesis). Note already that any Riesz subspace K, with the linear space structure and the order structure inheritedfrom L, is a Riesz space by itself. (ii) The linear subspace A of L is called an ideal in L (also called an order ideal in L if necessary to avoid confusion with the algebraic notion of an ideal in a ring) whenever A is solid, i.e.. whenever it follows from f E A , g E L and Igl 5 1f I that g E A. (iii) The ideal A in L is called a a-ideal whenever it follows from A E A (n = 1,2,. . .) and f = sup& in L that f E A. (iv) The ideal A in L is called a band in L if, whenever a subset of A has a supremum in L, that supremum is an element of A , i.e., if it follows from D c A and f = sup D thatf E A. Obviously, the notion of a a-ideal is intermediate between the notions of an ideal and a band. Unfortunately, there is little universal agreement about the terminology. In the Bourbaki terminology a Riesz subspace was originally called “un sous-espace propre”, an ideal was “un sous-espace e‘pais” and a band was “une bande”. More recently (N. Bourbaki [l], 1965) a Riesz subspace has become “un sous-espace corkticulc2” and an ideal has become “un sous-espace isolk”. In the Nakano terminology a Riesz subspace and an ideal are called a linear lattice manifold and a semi-normal subspace respectively, and the Nakano definition of a normal subspace coincides with the present definition of a band in the important special case that L is Dedekind complete. Finally, in the Soviet literature, a Riesz subspace and an ideal are called a sublineal and a normal sublineal respectively, and a band is called a component.
94
IDEALS AND BANDS; ARCHIMEDEAN RlESZ SPACES
[CH.3, S 17
There is another point which should be kept in mind. As observed earlier, the positive cone L+ is a distributive lattice with the zero-element as smallest element. The present definition of an ideal is not to be confused, therefore, with the earlier definition in Chapter 1 of an ideal in the lattice L'. If A is an ideal in the Riesz space L,then the positive part A + = ( f :f E A , f 2 0) of A is an ideal in the lattice sense in L+.Indeed, it is evident that u E A + and 0 S u 6 u implies u € A + , and also that u, u E A" implies sup (u, u ) E A + (on account of 0 S sup (u, v ) 5 u+ v E A'). Conversely, however, if D is an ideal in the lattice sense in L+,there does not necessarily exist an ideal A in L such that A + = D. By way of example, given u # 0 in L', the ideal D (in the lattice sense) generated by u consists of all elements u satisfying 0 v 6 u. If there were an ideal A in L such that A + = D, the positive part A + = D would contain not only all u satisfying 0 u 5 u, but also all positive multiples of u. In the sequel, by an ideal we will understand an ideal in the Riesz space sense. If it is necessary to consider an ideal in the lattice sense this will be explicitly stated.
Theorem 17.2. (i) Every band in L is a a-ideal, and every ideal in L is a Riesz subspace. (ii) The ideal A in L is a band if and only i f , whenever an upwards directed set of positive elements in A has a supremum in L, that supremum is in A (i.e., ifand only if it follows from 0 S u, t u and u, E A for all z that u E A). Similarly, the ideal A is a a-ideal if and only if, whenever an increasing sequence of positive elements in A has a supremum in L, that supremum is in A. (iii) Any Rie$z subspace of a Riesz subspace of L is a Riesz subspace of L. Any ideal in an ideal in L is an ideal in L. Any a-ideal in a a-ideal in L is a o-ideal in L. Any band in a band in L is a band in L. Proof. (i) We need only prove that any ideal A in L is a Riesz subspace. Since A is a linear subspace, it is sufficient to verify that sup (f,g ) E A holds for every pairf, g in A. We have 2 SUP (f, 9) =f+s+If-gl,
and so it is sufficient to prove that If I E A holds for everyf E A. This follows immediately from the definition of an ideal. (ii) We need only prove that if A is an ideal with the property that 0 u, 1 u with u, E A for all z implies u E A , then A is a band. To this end, let A be an ideal possessing this property. We have to prove that for any sub-
CH.3 , § 171
95
RIESZ SUBSPACES, IDEALS A N D BANDS
set of A possessing a supremum in L, that supremum is in A. Hence, let
(f,: a E { n } ) be a subset of A such that f = supf, exists. It has to be proved that f E A. Enlarging the set of allf, by all elements of the form SUP (fu,
7
-
* .,fO.),
we obtain an upwards directed set (f,: t E {t}) such thatf, 7 f (cf. Theorem 15.1 1 (i)), and it is obvious from part (i) of the present proof that the larger set (L: z E {t)) is still a subset of A. It follows now that f:,f; E A for all z, since 0 5 f 5 If,l and 0 5 f; 5 If,]and A is an ideal. Furthermore, f, Tfimplies 0 5 f: f +,sof' E A by our present hypothesis on A. Also, we havef; .If- 2 0, and hencef - E A simply because f;E A for all t and A is an ideal. The final result is, therefore, that f = f -f - E A, which was to be proved. (iii) Let K , be a Riesz subspace of the Riesz subspace K of L, and let f, g E Kl . We have to show that h = sup, (f,g) E K , , where sup, denotes that the supremum is taken in L. It follows fromf, g E K , c K that h E K, and obviously h is the supremum off and g not only in L but also in K, i.e., h = supK(f,g). But then we have h E K , on account of Kl being a Riesz subspace of K. The proofs of the remaining statements are trivial. +
We present some simple examples of Riesz subspaces, ideals and bands.
Example 17.3. (i) Let L be the Riesz space C([O,11) of all real continuous functions on the interval (x :0 5 x 5 1). The linear subspace of all polynomials is no Riesz subspace; the linear subspace of all constants is a Riesz subspace, but no ideal; the linear subspace of all f E L satisfyingf(0) = 0 is an ideal, but no band; the linear subspace of allfe L satisfyingf(x) = 0 for 0 6 x 5 3 is a band. (ii) The linear subspace of I,, consisting of all real multiples of a fixed element u,, in the positive cone of ,Z such that uo has at least two strictly positive coordinates, is a Riesz subspace but no ideal. The space C([O,11) is a Riesz subspace of, but not an ideal in, the space of all real functions which are Lebesgue summable over (x : 0 5 x l), under the convention that Lebesgue almost equal functions are identified. (iii) Let (s) be the Riesz space of all real sequences and (co) the Riesz space of all real null sequences; the ordering is coordinatewise. The space ZI is a proper ideal in (co), the space ( c o ) is a proper ideal in I , , and I , is a proper ideal in (s). None of these ideals is a a-ideal, and hence none of them is a band.
96
[CH.3 , b 17
IDEALS AND BANDS; ARCHIMEDEAN RIESZ SPACES
.
(iv) The subset of (s), consisting of all f = (f(l), f ( 2 ) , . .) satisfying f(1) = 0, is a band in (s). If L is the lexicographically ordered plane, then the “vertical axis” (i.e., the set of all f = (f’, fi)satisfyingfl = 0) is a band in L. Note that any other “line through the origin” is a Riesz subspace but no ideal. The following theorem is an immediate consequence of the definitions.
Theorem 17.4. (i) In any Riesz space L the linear subspace {0}, i.e., the set consisting only of the zeroelement, is a band in L. Similarly, L itseIf is a band in L. (ii) Any arbitrary non-empty set theoretic intersection of Riesz subspaces (or ideals, or a-ideals, or bands) is a Riesz subspace (or ideal, or a-ideal, or band). Given the non-empty subset D of L, the set of all Riesz subspaces which include D is non-empty since the space L is a member of that set. Hence, the intersection of all the members of the set is the smallest Riesz subspace including D. A similar argument holds for ideals, a-ideals and bands.
Definition 17.5. The smallest Riesz subspace (or ideal, or a-ideal, or band) including a given non-empty subset D of L is called the Riefz subspace (or ideal, or o-ideal, or band) generated by D. I f D consists of one elementf, the Riesz subspace (or ideal, or a-ideal, or band) generated by f is called a principal Riesz subspace (orprincipal ideal, or principal o-ideal, or principal band). The ideal AD generated by the non-empty subset D can easily be described explicitly. Indeed, AD consists of all g E Lsuch that 191 5 lalfiI+
+laJnl holds for appropriate fi, .. .,f,E D and appropriate real a l , . . .,a, (where n is also variable, of course). For the proof, note that any element *
lalfil+ ... +laJ,I;fi,...,f,ED, must be in A,, and hence also any g such that 191 is majorized by such a linear combination. On the other hand, it is easy to see that the set of all such g is already an ideal including D . In particular, the principal ideal generated by the element f E L consists of all g satisfying 1st lafl for an appropriate real a. Before presenting the final theorem in this section we recall that the algebraic sum A + B of the linear subspaces A and B of a linear space consists of allf+g such thatfE A and g E B. If A and B have only the zeroelement in common, A +B is said to be the direct sum of A and B, and in this case we write A @ B.
CH. 3 . 3 171
97
RlESZ SUBSPACES, IDEALS AND BANDS
Theorem 17.6. Let A and B be ideals in the Riesz space L. (i) The algebraic sum A +B is an ideal. Givenf 2 0 in A + B, there exists a decomposition f = fl +f2 such that f l 2 0,f2 2 0 and f l E A , f 2 E B. (ii) We have A IB (i.e., f l If 2for all f l E A , f2 E B ) ifand only i f A n B = { 0 ) , i.e., ifand only i f A + B is a direct Jum A Q B. In this case, therefore, the decompositionf = f l+f2 ,fl E A ,f 2 E B, of any f E A 0 B is unique, and f 2 0 implies that f l 2 0 and f2 2 0. (iii) If A IB, andf, g are elements in A @ B having decompositionsf = fl +f2 and g = g1+ g 2 , thenf S g implies that f l S g1 andf 2 6 g 2 . Proof. (i) A + B is a linear subspace; hence, in order to show that A + B is an ideal we have to prove that f E A + B and Igl 5 If I implies g E A +B. Assume, therefore, that f = fl f2 with fl E A and f z E B, and let 191 S If I. Then
+
9+ 5 191
5 If1 6
Ifll+lf21,
so by the dominated decomposition property there exists a decomposition
g + = g‘+g” such that 0 5 g’ 5 Ifll and 0 5 g” 5 If21. It follows now from If l l E A and If21 E B that g’ E A and g” E B. This shows that = g’+g”EA+B.
Similarly we have g - E A + B , and so g = g f - g - E A + B . This is the desired result. Assume now that f 2 0 and f = f’+f”withf’ E A andfl’ E B. Then 0
5 f S If’l +If%
so by the dominated &composition property there exists a decomposition f = fl+f2 such that 0 S f l 5 If’l and 0 5 f 2 5 If”l. Since If’l E A and If”l E B, it follows that f l E A and f2 E B, and so f = f i +fz is a decomposition of the desired kind. (ii) Assume first that A B. Given f E A n B, we have now t h a t f l f , so inf (If I, If I) = 0, and hence f = 0. This shows that A n B = (0). Conversely, assuming that A n B = { 0 } , it has to be proved that inf(If1L
If211 =
0
holds for all f l E A , f2 E B. This is evident because inf (If l l , and so must be the null element. (iii) We have g -f 2 0, and g - f has the decomposition
s-f
= (s1-f1)+(92-f2).
If21)
E A n B,
98
IDEALS AND BANDS; ARCHIMEDEAN RIESZ SPACES
[CH.3 , s 18
Hence g 1-fi 2 0 and g 2 -fi 2 0 in virtue of what has been proved already. Exercise 17.7. Let u, v be positive elements in the Riesz space L , let w = inf (u, u ) and A , , A , , A,,, the principal ideals generated by u, u, w respectively. Show that A,,,= A , n A , . Now, let z = sup (u, u ) and A, the principal ideal generated by z . Show that A , = A, + A , .
18. Riesz homomorphisms and quotient spaces It will be proved in the present section that the ideals in the Riesz space L are exactly the kernels (null spaces) of linear mappings of L into another Riesz space M that preserve finite suprema, i.e., for which the image of any finite supremum of elements in L is the supremum (in M) of the images. For mappings of L onto M of this kind the kernel is a band if and only if the mapping preserves suprema of arbitrary sets. Linear mappings of this kind are called Riesz homomorphisms. Given the ideal A in the Riesz space L, the quotient space LIA can be made into a Riesz space in a natural manner. and the canonical mapping of L onto L/A is a Riesz homomorphism having A as kernel. We present the necessary definitions and proofs. In the whole section L and M will denote Riesz spaces. Definition 18.1. The linear mapping n of L into M is called a Riesz homomorphism whenever inf (nf,ng) = 0 holds for every pair of elementsf, g E L satisfying inf (f,g ) = 0. It will be shown first that a linear mapping n of L into M is a Riesz homo-
morphism if and only if n preserves finite infima or, equivalently, if and only if n preserves finite suprema. Theorem 18.2. The linear mapping n of L into M is a Riesz homomorphism if and only if n{inf ( f , 9 ) ) = inf (nf, n s ) holds for aNf, g
E
L . Equivalently, n is a Riesz homomorphism if and only i f
4SUP (f,s)>= SUP (nf,4 holds for a l l f , g
E L.
Proof. For the first part it will be sufficient to prove that a Riesz homomorphism n satisfies n{inf ( f , g ) > = inf (nf,719)
for all f, g E L . To this end, let f, g E L be given. Setting h = inf (f,g ) ,
CH. 3 , s 181
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RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
we have
inf (f-h, g - h ) = 0,
so
inf (nf- nh, ng - nh)
by hypothesis, i.e.,
=
0
inf (nf,n g ) = nh = n{inf (f,9)). The proof for sup (f,g ) follows by observing that
g). SUP (f,9 ) = f +9 - inf (f,
In the next theorem we collect some properties of a Riesz homomorphism. Theorem 18.3. Let n be a Riesz homomorphism of L into M. (i) f 2 g in L implies nf 2 ng in M. In particular, n is a positive linear mapping, i.e., n maps L+ into M + . (ii) W e have
.(f+)
=
(.f)+,4 f - 1
= (xf)--, .(lfI) =
bfl.
Hence f Ig in L implies nf Ing in M. (iii) The kernel (null space) A, of n is an ideal in L.
Proof. (i) Let f 2 g , so f
=
sup (f,9). It follows that
nf = n(sup (f,9 ) ) = SUP (nf,w),
which implies that nf (ii) We have
.(f+)
=
2 ng.
n{sup (f,0)) = sup (nf,no) = sup (nf,0) = (nf)'.
Similarly for the other formulas. (iii) It is evident that the kernel A , of n is a linear subspace of L. It follows from n(lfl) = Infl thatfE A , holds if and only if If[ E A , holds. Hence, iff E A , and Igl S If [, then 0
so 19)E A,, which implies g
S n(lsI>I; .(If0 E A,.
= 0,
This shows that A , is an ideal.
Definition 18.4. The Riesz $paces L and M are called Riesz isomorphic if there exists a Riesz homomorphism n of L onto M such that n is a one-one mapping. Evidently, the inverse mapping n-l is then a one-one Riesz homomorphism of M onto L. The mapping A is now called a Riesz isomorphism of L onto M , and evidently n- is then a Riesz isomorphism of M onto L.
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IDEALS AND BANDS; ARCHIMEDEAN RlESZ SPACES
[CH.3 , s 18
As observed earlier, any Riesz homomorphism K of L into M is a positive mapping (i.e., f 1 0 in L implies nf 2 0 in M ) . Hence, any Riesz isomorphism of L onto M is a one-one positive linear mapping of L onto M. It is not true, however, that every one-one positive linear mapping of L onto M is already a Riesz isomorphism. The following theorem and example will clarify this.
Theorem 18.5. The one-one positive linear mapping IL of L onto M is a R i e u isomorphism if and only if.- is also positive. Proof. Assume that
71-l
is also positive. We have to show that
4 s u p (f,9)) = SUP (.A ng> holds for allf, g E L . Since K is positive, it follows from sup (I; g ) 1 f and SUP (5 g ) 2 g that
(I; 9)) 2 SUP (nf,ns>. is positive, it follows from sup (nf,ng) 5 nf and sup (nf,ng) +UP
Since a-1 2 ng that and so
71-
l{SUP
(Kf, 719))
5 SUP
SUP (76ns) 1 +UP
r f 7
s),
(A s>>.
Example 18.6. Let L be two-dimensional plane with the ordinary coordinatewise ordering (i.e., (xl, x2) ( y , ,y 2 ) whenever x1 5 y 1 and x2 S y 2 ) , and let M be the same plane with the lexicographic ordering (i.e., (xl , x2) 5 ( y l , y 2 ) whenever x1 c y 1 or x1 = y , , x2 5 y 2 ) . The identity mapping of L onto M is a one-one positive linear mapping, but the inverse mapping is not positive. The spaces L and M in this example are not Riesz isomorphic, i.e., not only that the identity mapping is not a Riesz isomorphism, but it is true that there exists no Riesz isomorphism of L onto M at all. Indeed, any one-one linear mapping of M onto L transforms the halfplane {(yl, y 2 ) :y1 > O] into a halfplane, so there must be positive elements of M that are transformed into non-positive elements of L. Another proof is by observing that L is Archimedean and M is not.
Given the ideal A in L, the (algebraic) quotient space of all equivalence classes modulo A will be denoted by LIA, and the element of L/A containing the elementf of L will be denoted by Lf]. Hence we have [fl] = Lf2]if and only iffl -fiE A. It will be proved now that LIA can be partially ordered
CH.3, $181
RlEsZ HOMOMORPHISMS AND QUOTIENT SPACES
101
in such a manner that L / A is a Riesz space with respect to the partial ordering.
Definition 18.7. Given I f ] and [ g ]in L / A , we shall write I f ] 5 [ g ] whenever there exist elements flE I f ] and g1 E [ g ]satisfying f, S 9,. The following lemma is now evident. Lemma 18.8. Thefollowing conditionsfor I f ] and [ g ]in L f A are equivalent.
(i) I f 1 5 Esl. (ii) For any fl E there exists an element g1 E [ g ]satisfying fl 5 g l . (iii) For eoery flE I f ] and every g1 E [ g ] there exists an eZement q E A such thatg, -f1 2 4.
v]
Theorem 18.9. Given the ideal A in the Riesz space L , the quotient space LIA is partially ordered with respect to the relation introduced in Definition 18.7,and LIA is a Riesz space with respect to the partial ordering. Proof. We prove first that L / A is partially ordered. It is evident that I f ]
2 I f ] holds for every I f ] . If I f ] 5 [ g ] and [ g ] 5 [h],there exist by the preceding lemma elements q l ,q2 E A satisfying g-f 2 q1 and h-g 5 q 2 ,
so h-f 2 q1 +q2 E A, which shows that I f ] 5 [h]. Finally, assume that I f ] 5 [ g ] and [ g ] 5 I f ] hold simultaneously. Then there exist elements ql,q2EAsatisfyingg-f2qlandf-g2q2,sof-g5 - 4 , andg-f 5 - q 2 . Writing 4 = SUP (-q1, --4A
we have q E A and I f-gl 5 q. Since A is an ideal, it follows that f - g E A , so I f 1 = k l . The order structure is compatible with the algebraic structure. Indeed, it is obvious that I f ] 2 [O] implies a w l = [ a f ] 2 [O] for every real a 2 0. Furthermore, if I f ] 5 [ g ] ,there existflE I f ] andg, E [ g ]such thatf, 5 g l . Hence, if [h]is given and h is an arbitrary element of [h],we have fl + h 5 g1+h, so Ifl + h ] 5 [gl +h] by Definition 18.7. It follows that IfI+[hl =
rflI+Pl= Vl+hI
s [s1+hI
= [g,I+[hl= [gl+[hl.
In order to show that L / A is a Riesz space we will prove that sup ( I f ] , [ g ] ) exists for every pair I f ] , [ g ] ;more precisely, we will prove that holds for all f , g. It is evident that we have [sup ( f , g ) ] 2 I f ] and [sup (f,g ) ] 2 [g]. Hence, it will be sufficientto prove that any upper bound
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IDEALS AND BANDS; ARCHIMEDEAN RIESZ SPACES
[CH. 3 , s 18
[h] of I f ] and [g] satisfies [h] 2 [sup (f,g ) ] . Given such an upper bound [h], there exist elements q l , q2 E A such that h - f 2 q1 and h-g 2 q 2 , so h - f 2 q and h - g 2 q for q = inf ( q l , q2)E A. But then h 2 SUP ( f + q , 9 + 4 ) = SUP (f, 9)+4, so 12-sup (f,g) lows.
2 q, from which the desired result [h] 2 [sup (f,g ) ] fol-
On account of f + g = sup (f,g)+inf (f,g) we have immediately the additional result that inf ( I f 1 9 b l ) = W(f, 911. The Riesz space LIA is called the quotient Riesz space of L with respect to the ideal A. The mapping f + I f ] is a Riesz homomorphism of L onto L/A,and A is the kernel (null space) of the homomorphism. This shows that every quotient space of L with respect to some ideal is a Riesz homomorphic image of L. Conversely, let n be a Riesz homomorphism of the Riesz space L into the Riesz space M. The image n(L)is a Riesz subspace of M , so n is a Riesz homomorphism of L onto the Riesz space n(L).Let A bz the kernel of the mapping n.Then A is an ideal in L by Theorem 18.3 (iii), and the mapping If]+ n ( f ) is now a Riesz isomorphism of LIA onto n(L).Hence LIA and n(L)are Riesz isomorphic. This shows that every Riesz homomorphic image of a Riesz space L is Riesz isomorphic to the quotient space of L with respect to the kernel of the homomorphism. The situation is, therefore, completely analogous to the situation in algebraic ring theory, where we have ring homomorphisms and quotient spaces with respect to ideals in the ring theoretic sense. We recall that the ideal A in L is called a a-ideal in L whenever it follows fromf, E A (n = 1,2, . . .) and f = supf, in L that f E A (cf. Definition 17.1 (iii)). By Theorem 17.2 (ii) an ideal A is a a-ideal if and only if, whenever an increasing sequence of positive elements in A has a supremum, that supremum is in A. The a-ideals in L act as kernels of what we will call Riesz a-homomorphisms. Definition 18.10. The Riesz homomorphism n of L into M is called a Riesz a-homomorphism if n preserves countable suprema, i.e., if it follows from f = supf, (n = 1,2, . . .) in L that nf = sup nf, holds in M .
A sufficient condition for the Riesz homomorphism n to be a Riesz a-homomorphism is that 0 sf,t f in L+ implies 0 5 nf, t nf in M + . Evidently, the kernel of a Riesz a-homomorphism is a a-ideal in L. For Riesz
CH.
3, 6 181
103
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
homomorphisms of L onto M the converse is also true, as shown by the following theorem.
Theorem 18.11. For a Riesz homomorphism n of L onto M the following conditions are equivalent. (i) x is a Riesz a-homomorphism. (ii) For any a-ideal N in the space M the inverse image n-'(N) is a a-ideal in the space L. (iii) The kernel K, of n is a a-ideal in L. Proof. (i) * (ii). Assume that n is a Riesz a-homomorphism and N a a-ideal in M . For the proof that n-'(N) is a a-ideal in L,assume that 0 5 u,, t u with u, E n-'(N) for n = 1 , 2 , . . .. Then nu,,E N for all n, and 0 I nu,, 7 xu since n is a a-homomorphism. It follows that nu E N since N is a a-ideal, and hence u E n - ' ( N ) . This shows that n-'(N) is a a-ideal. (ii) * (iii). This is evident because K, is the inverse image of the a-ideal (0)in M. (iii) => (i). Assume that K, is a a-ideal, and 0 u,, t u in L. We have to nu show that 0 5 nu,, xu in M. For this purpose, assume that nu,, 5 xf holds for somefc L and all n, where it is no loss of generality to assume that f 5 u (because otherwise we may replace f by inf (5 u)). For n = 1 , 2 , . . ., let zn = SUP (un,f)-J Then 0 S z n t SUP ( ~ , f ) - f= u-A
s
s
and z,, E K, holds for all n on account of
xz,, = sup (nu,, n f ) - n f
=
nf-nf
=
0.
Hence, since z, E K, for all n and K, is a a-ideal, it follows from z,, t u -f that u-f E K,, and so nf = nu. This shows that nu,,t nu.
Definition 18.12. The Riesz homomorphism n of L into M is called a normal Riesz homomorphism if n preserves arbitrary suprema, i.e., fi it follows from f = supf , in L (where a runs through an arbitrary index set) that nf = sup n f, holds in M. A sufficient condition for the Riesz homomorphism n to be normal is that 0 Sf,t f in L+ implies 0 nf, t nf in M f . Evidently, the kernel of a normal Riesz homomorphism is a band in L. For Riesz homomorphisms of L onto M the converse is also true, as shown by the following theorem.
s
104
IDEALS A N D BANDS; ARCHIMEDEAN RIESZ SPACES
[CH.
3 , g 18
Theorem 18.13. For a Riesz homomorphism n of L onto M the following conditions are equivalent. (i) n is a normal Riesz homomorphism. (ii) For any band N in the space M the inverse image n-’(N) is a band in the space L. (iii) The kernel K, of n is a band in L. Proof. Similar to the proof of Theorem 18.11.
Exercise 18.14. The last two theorems do not hold for Riesz homomorphisms of L into M (i.e., these theorems cease to hold if “onto” is replaced by “into”), as shown by the following example. (i) Let rl , r2, . . . be the sequence of numbers +, +, i,*, g,,; . . .,and let 0 < E < 1. Set E, = ~/2”+’for n = 1 , 2 , . . .. For n = 1 , 2 , . . ., let un(x) be the real continuous function on [ 0 , 1 ] , zero outside the interval [rn-E,,, r,,+E,,], satisfying un(rn)= 1 and being linear in [r,,-E,,, r,] and in [r,,, r,,+~,,].Furthermore, let
t,
Y,,
= sup ( u l , .
. .)u,)
..
for all n. Show that, for each x E [0, 11, the sequence (v,,(x) : n = 1,2, .) is increasing, and the pointwise limit w(x) satisfies w(r,,) = 1 for all n, but w ( x ) is not identically equal to 1 , since the set ( x : W ( X ) > 0 ) is of Lebesgue measure at most equal to E. Show that, in the Riesz space C( [0, 1 1 ) of all real continuous functions on [0, 11, we have 0 Iv,, t e, where e = e ( x ) = 1 for all X E [0, I ] . Hence, the supremum of the sequence in C([O,11) is not the pointwise supremum. (ii) Let L be the Riesz space C([O,I ] ) of all real continuous functions on [0, 11 and M the Riesz space of all real (finitevalued) functions on [0, I ] . Let n be the identity mapping of L into M . Show that n is a Riesz homomorphism with kernel K, = {0}, but x is no o-homomorphism and hence n is not normal.
Exercise 18.15. Show that if a is a Riesz homomorphism of L onto M , then the image n ( A ) of any ideal A in L is an ideal in M . Show also that for ideals A and B in L we have “ ( A n B ) = n ( A ) n n(B). Hint: In order to show that n ( A ) is an ideal, it is sufficient to prove that if 0 5 u E A and 0 5 m 6 nu in M , then m E n(A). Since 0 S m = nf for some f E L, it follows easily that m = nf +,so w = inf ( u , f ’ ) E A and nw = inf (nu, nf +) = inf (xu, m ) = m.
This shows that m E n(A).
CH. 3 , s 191
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105
For ideals A and B in L, we have to prove that n ( A ) n n(B) c n(A n B). Given f E n ( A ) n n(B), we have 1f I E n(A) n n(B), so let If1 = nu with 0 5 u E A and If l = nu with 0 5 v E B. Then w = inf (u, 0 ) satisfies 0 w E A n Band nw = If I. It follows that f E n(A n B). 19. Disjoint complements
It has been shown in section 17 that every non-empty subset D of the Riesz space L generates a band, the smallest band including D . There is still another band attached to D, the disjoint complement of D . We recall the definition. Definition 19.1. If D is a non-empty subset of the Riesz space L , the set of all f E L satiflying f D (i.e., f 1g for all g E D ) is called the disjoint complement of D , and denoted by Dd.The set (Dd)dwill be denoted by Ddd. Theorem 19.2. For any non-empty subset D of the Riesz space L, the following holds. (i) Dd is a band in L. (ii) D c Ddd,and heme the band generated by D is included in the band D". Furthermore, we have Dd = Ddddand Dd n Ddd = (01, so the ideal Dd+odd is a direct sum Dd @ Ddd.
Proof. (i) It was already observed in the remarks following Theorems
14.2 and 14.3 respectively that Dd is a band. (ii) It is evident that D c Oddholds. Hence, substituting Dd for D , we obtain that Dd c Dddd.On the other hand, since D , c D2 implies D f =I D f , we derive from D c Dddthat Dd XI Dd". It follows that Dd = Dddd.Finally, D d I Ddd implies by Theorem 17.6 (ii) that od n Ddd = (0).
It can happen that the band generated by a subset D of the Riesz space L is properly smaller than Ddd,as shown by the following example. Let L be the lexicographically ordered plane, and D the vertical axis, i.e., D = { f = ( f l , f i ) :f i = O}. Then D is a band such that Dd = (0). Hence, the band generated by D is D itself, but Dddis the whole spaceL . It will be proved in section 22 that in an Archimedean Riesz space the band generated by any non-empty subset D is always equal to Ddd. Theorem 19.3. Let L be a Riesz space. (i) For any non-empty subset D of L, the set Ddd is the largest subset of L having the same disjoint complement as D. (ii) If0 and E are non-empty subsets of L such that D IE, then D" I Edd.
106
IDEALS A N D BANDS; ARCHIMEDEAN RIESZ SPACES
[CH. 3,
9 19
In particular, D 1E implies that the bands generated by D and E are disjoint. (iii) For any ideal A in L, the set Add is the largest among all ideals B in L with the property that f o r every 0 # f E B there exists 0 # g E A such that 191 4
Ifl.
(iv) If A , ,
. . ., A,, are ideals in L, then
Proof. (i) We have Dd = Dddd,and so D and Ddd have the same disjoint complement. In order to show that Ddd is the largest subset of L having the same disjoint complement as D , assume that D , is a non-empty subset of L such that Df = od. Then D , c D!d = Ddd,which is the desired result. (ii) If D 1E, then D c Ed = (Edd)d,and so D l Edd. Repeating the argument, we obtain DddL Edd. (iii) It will be shown first that Addhas the stated property. Let 0 # f E Add, and assume there exists no g such that 0 # g E A and 191 S 1f I. This implies that inf (If 1, lhl) = 0 holds for every h E A (if go = inf (Ifl, Ihol) # 0 for some ho E A , then 0 # go E A and lgol S If I). It follows that f E Ad. But we have also f E Add.Hence f E Ad n Add = (0}, contradicting f # 0. Now assume that B is an ideal with the property that for every 0 # f E B there exists an element 0 # g E A such that 191 5 Ifl, and assume that B c Add does not hold. Then there exists an element f E B not disjoint from Ad, so k = inf ( I f l , lhl) # 0 for some h E Ad. It follows that 0 # k E B n Ad. Now, by hypothesis, there exists an element 0 # g E A such that 191 5 k, so 0 # g € A n Ad. This is impossible. (iv) It is evident that Ai)ddc A f d .For the converse, it is sufficient to prove that Ad;' n Aid c ( A , n A2)dd.
(n;
n;
The proof is based upon the property proved in part (iii). Given 0 # f E A:d n Aid, there exists by part (iii) an element 0 # h E A , such that Ihl =< If I. Since 0 # h E A , n A:d, there exists now also an element 0 # g E A 2 such that 191 5 lhl. Hence 0 # g E A , n A 2 and 191 5 If I. It has been proved thus that the ideal Afd n A? has the property that for every f # 0 in the ideal there exists an element 0 # g E A , n A 2 such that Igl 5 Ifl. But then A? n Aid is included in ( A , n A2)ddby part (iii). This finishes the proof.
Corollary 19.4. (i) I f A is an ideal in the Riesz space L, then Ad = (0},i.e. Add = L, ifand only iffor every 0 # f E L there exists an element 0 # g E A such that Igl S If I.
CH. 3 , 8 201
107
THE BAND GENERATED BY AN IDEAL
(ii) If A is an ideal in the Riesz space L, then ( A 0 Ad)d = { 0 } , and so ( A 0 A d y = L.
Proof. Part (i) is evident, and (ii) is proved by observing that ( A 0 AdJd is included in Ad and in Add,so in Ad n Add = (0). The results proved in this section are analogous to the results for ideals in distributive lattices proved in Theorems 4.5,4.6 and 4.7. Exercise 19.5. Show that if (D, : o E {o}) is an arbitrary set of non-empty subsets of the Riesz space L, then
In particular we have (n A,)dd c n Atd if all the A , are ideals. The last inclusion may be proper even for a countable intersection (for a finite intersection there is equality by Theorem 19.3). If every A,is of the form A , = Dt, then we have (nA,)dd = n Atd. Show this. Hint: In order to show that proper inclusion can occur for a countable number of ideals, let L = C([O, I]), let rl , r 2 , . . . be the rational numbers in [0, 11, and A , the ideal of allf E L satisfyingf(r,) = 0. Then A, = {0}, so A,)dd = (0). On the other hand A,d = {0} for every n, so Atd = L. If A,, = Dt for every o, then Atd = Dtdd = Dt = A , for every o, and so n Atd = n A , c (n A,)dd.
(n;
0;
n?
20. The band generated by an ideal We bsgin with a notational definition.
Definition 20.1. If D is a non-empty subset of the Riesz space L , the band generated by D will be denoted by {D}.
As shown in the preceding section, { D } is included in Dddbut not always equal to Ddd.A further investigation of the structure of { D } is necessary, therefore. It is obvious that the ideal A , generated by D satisfies {A,} = {D}, and hence (since the structure of A , is known) we may restrict ourselves mainly to the investigation of the band { A } generated by a given ideal A . The structure of { A } is what one might expect; the positive elements in { A } are exactly the elements in L+ that are suprema of subsets of A . The details follow in part (i) of the next theorem. Part (ii) of the same theorem shows that { A } and Add have an interesting property in common. Just as we have
108
(n;
IDEALS AND BANDS; ARCHIMEDEAN RIESZ SPACES
[CH.3, $ 2 0
0;
AJdd = Af' for a finite number of ideals (cf. Theorem 19.3 (iv)), so we have ((-);Ai}= ( ) ; ( A i } .
Theorem 20.2. (i) I f A is an ideal in the Riesz space L, then every 0 5 u E { A } satisfies u = sup(u:uEA,O 5 u 5 2.4). Conversely, by the definition of a band, any u EL' satisfying this condition is an element of { A } . (ii) I f A , , * * A , are ideals in L, then a,
{
1
Ail =
ij 1
{~i).
Proof. (i) Denote by B the set of all UEL' satisfying 0 6 u, T u for some upwards directed set (v, :T E (T}) in A' = A n L', and denote by A" the set of all u1 - u2 with u l , u2 E B. Then
A c A" c { A } ,
(1)
and we will prove first that B consists of all u EL" satisfying u = sup(u:uEA,O
6v5
u).
(2)
Indeed, if u E Lf satisfies this condition then u is a member of B since the set (v : v E A , 0 6 u u ) is an upwards directed subset of A'. Conversely, u E B implies that we have 0 u, ? u for some upwards directed set (u, : z E {z}) in A + , and so any upper bound of the set (v :v E A, 0 5 u - u ) is at least u. On the other hand u is an upper bound of (v : u E A , I 0 5 u 5 u), and hence (2) holds. Now note that u E B and 0 5 up 5 u implies that ur E B. Indeed, if 0 4 u, t u with all u, in A , then
s
0 6 inf (u,, u ' ) t inf (u, u ' ) = u' with all elements inf (uz, u') in A , and this shows that ur E B. It follows then also from u E B and 0 u' u that u-u' E B. This, however, implies that any positive element of A" is already in Byi.e., B = (A")'. The proof will b: complete, therefore, if we show that A" is a band, since this will imply by (1) that A" = { A } , so B = (A")' = (A}', and B = (A}' is exactly what we have to prove. In order to show that A" is a linear subspace, it is sufficient to prove that u l , u2 E B implies u1+ u2 E B. Ifp is an arbitrary upper bound of the set (v : v E A , 0 I; u
5
111
+ u2),
(3 1
CH.3 , s 211
109
ORDER DENSE IDEALS
+
then p 2 v1 v2 holds for every pair ul ,v2 E A satisfying 0 5 ul 5 u1 and 0 5 u2 5 u2, and so p 2 u1 +u2 follows by taking suprema. On the other hand ul + u2 is an upper bound of the set (3), and so u, + u2 is the supremum. This shows that u, +u2 E B. For the proof that A" is an ideal, assume that f E A", s o f = ul -u2 with 0 6 u l , u2 E B. On account of 0 6 f 5 u1 and 0 5 f - 5 u2 it is evident thatf+ andf- are in B, so I f l = f ++f- E B. Hence, iff E A" and Igl 5 Ifl, then g+ and g- are in B (on account of g+ 5 Igl 5 If 1 E By and similarly for g-), and so g = gf - 9 - E A". Finally, in order to show that A" is a band, assume that 0 5 u, u with u, E B for all z. Then +
u = sup u, = sup,(sup (v : u E A , 0 5 u
=sup (u : u E A , 0 5
5 u,
2 24,)) for some z).
Since the last set is directed upwards, we have u E B. This completes the proof. For the converse, it is sufficient (ii) It is evident that A,} c to prove that { A , ) { A J = {A, ~ ~ 1 .
{n;
n
n
n
To this end, assume that 0 5 u E {A,} {A2}. There exist directed sets (u, : z E {z}) in A l and (u, : o E {o}) in A 2 such that 0 5 u, f u and 0 5 u, t u ; hence, wT,u= inf (u?, u,) satisfies wT,uE A , A2 and 0 5 w,, , u by Theorem 15.9 (i). It follows that u E { A , A 2 } .
n
n
For the o-ideal A , generated by the ideal A we have a similar theorem, and the proof is also similar. Theorem 20.3. (i) If A,, is the o-ideal generated by the ideal A , then euery 0 5 u E A , sati@es u = sup u,, for some sequence (u,, :n = 1,2, . . .) in A such that 0 <= u, t u. Conversely, any u EL' satisfying this condition is an element of A,. (ii) ZfAo , . . ., A,, are ideals in L , then
(
n n
1
n (Adon
Ai)u
=
1
21. Order dense ideals, quasi order dense ideals and super order dense ideals
In this section L denotes a Riesz space. Definition 21.1 (i)
If A and B are ideals in L , then A
is said to be order
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IDEALS A N D BANDS; ARCHIMEDEAN RIESZ SPACES
[CH.3, $21
dense in B i f { A } 3 B, i.e., i f f o r every 0 5 u E B there exists a directed set 0 S u, 7 u such that u, E A holdsf o r all T. In particular, A is order dense in L (or simply order dense) if ( A } = L . (ii) If A and B are ideals in L , then A is said to be quasi order dense in B if Add =I B, i.e., i f for every 0 # f E B there exists 0 # g E A such that 191
5 Ifl.
In particular, A is quasi order dense in L (or simply quasi order dense) if Add = L . (iii) I f A and B are ideals in L, then A it said to be super order dense in B i f A , 3 B, i.e., if for every 0 5 u E B there exists a sequence 0 S u, t u such that u, E A holds f o r all n = 1,2, . . .. In particular, A is super order dense in L (or simply super order dense) if A, = L . It is evident that super order denseness implies order denseness, and order denseness implies quasi order denseness. The next theorem is three theorems in one. The term n-order dense must be read consistently as either order dense, quasi order dense or super order dense, and accordingly A , must be read as { A ) , Addor A , respectively.
Theorem 21.2. Let A , B, C, A , , . . ., A , be ideals in L . (i) Thefollowing conditions for A and B are equivalent: A is n-order dense in B, A is n-order dense in B,, A , is n-order dense in B, A , = B, A, 3 B,. (ii) I f A c B, then A is n-order dense in B ifand only $ A , = B,. (iii) I f A is n-order dense in B and B is n-order dense in C, then A is n-order dense in C. (iv) If A , , .. ., A , are n-order dense in B, then A , is n-order dense in B.
0;
Proof. (i) Evident upon observing that (A,), = A , . (ii) Evident; observe that A c B implies A , c B,. (iii) By hypothesis we have A , 2 B, and B, 3 C,; it follows that A , 2 C,, i.e., A is n-order dense in C. (iv) By hypothesis we have ( A i ) , 3 B for i = 1, 2,. . ., so (-);(Ai),=) B. But (-);(Ai),= ((-):Ai), by Theorems 19.3 (iv), 20.2 (ii) and 20.3 (ii) respectively, so A i ) , 3 B. This shows that A i is n-order dense in B.
(n;
0;
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111
Theorem 21.3. For any ideal A in L , the ideal A 0 Ad is quasi order dense.
Proof. This was essentially proved in Corollary 19.4 (ii); we repeat the proof. It is evident that ( A 0 Ad)d is included in Ad and in Add, so in Ad Add = (0). This shows that ( A 0 Ad)d= {0}, so ( A 0 Ad)dd= L . In other words, A 0 Ad is quasi order dense.
n
Note that the results in Theorems 21.2 and 21.3 are analogous to results proved in Theorem 4.8 for ideals in a distributive lattice.
Definition 21.4. (i) The element e in L' is called a strong unit in L if the principal ideal generated by e is the whole space L , i.e., if for every f E L there exists a positive number a, depending uponf, such that If I 5 ae. (ii) The element e in L' is called a weak unit in L if the principal band generated by e is the whole space L , i.e., if the principal ideal generated by e is order dense. (iii) The element e in L f is called a quasi unit in L if the set E, consisting of the single element e, satisJies Edd = L , i.e., if the principal ideal generated by e is quasi order dense. It will be shown in the next section that in an Archimedean Riesz space L the principal band generated by e E Li is always equal to Edd,so in an Archimedean Riesz space weak units and quasi units are the same. In the final theorem of the present section we present a necessary and sufficient condition for an element u E L +in order that for every ideal A such that u is in Addthe element u is already in { A } . This will prepare the way for the investigation of Archimedean Riesz spaces in the next section.
Theorem 21.5. Let u EL' be given, and consider the decreasing sequence ( n - l u :n = 1,2, . . .). We have n-'u 10 ifand only if, for every ideal A such that u is in Add,the element u is already in {A}.
Proof. Let n-'u 10, and assume that u E Addholds for some ideal A , but u E { A }does not hold, i.e., u is not the supremum of the set Mu = ( v : V E A , O5 v 5 u). This implies that the set Mu has an upper bound w' such that u j w' does not hold. Then w = inf (u, w') is also an upper bound of the set Mu, and 0 5 w < u holds, where w < u denotes that w 5 u and w # u. Since, therefore, 0 < u-w E Add and since A is quasi order dense in Add,there exists an element z E A such that 0 < z 5 u - w. For every u in the set Mu we have now that z + v ~ Aand 0 < z+v 5 z+w 5 u,
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+ u is also a member of the set Mu.Repeating the argument, we find that nz is a member of Mu for n = 1,2, . . ., so nz 6 u for all n, i.e., so z
0 < z j n-'u.10. This is impossible, and hence we must have u E { A } . For the proof of the converse, assume that u E { A } holds for every ideal A satisfying u E Add,but n - l u 10 does not hold. The last assumption implies that there exists an element u # 0 in L+ such that 0 < nu 5 u holds for n = 1,2,. . .. Let A , be the principal ideal generated by u, and set A = A , @ A t . Since A is quasi order dense by Theorem 21.3, we have Add = L, so u E Add.The desired contradiction will be obtained if we can prove that u is not in { A } . For this purpose, write
Mu= ( w : w E A , O j w
j
24).
Proving that u is not in { A } is the same as proving that u = sup Mu does not hold. Let w be an arbitrary member of Mu. Then w = w 1+w2 with w 1 E A , and w2 E A t . Observe that 0 5 wl j w and 0 S; w2 6 w by Theorem 17.6 (ii). Since the principal ideal A , consists of all f E L satisfying If1 5 nu for some natural number n, we have w1 j nu for some n, so w,+u
5 (n+l)u s
u
+
with w 1+ u E A,. We have also that w2 j w j u, and w2 I(wl u ) on account of w1 + u E A , and w2 E A t , so ( w , + u ) + w , = sup ( w , + v , wz) j u.
It follows that w 1+ w 2 j u-u, i.e., w S u-u. This holds for every w E Mu, so u - u is an upper bound of the set Mu. Since u > 0, this shows that u = sup Mu does not hold. It follows immediately from the last theorem that n-'u L O (as n -+ m) holds for every u EL' if and only if Add = { A } holds for every ideal A in L. Riesz spaces where this holds are called Archimedean, and will be discussed in the next section. 22. Archimedean Riesz spaces We recall the definition of an Archimedean Riesz space, as presented in section 16. Definition 22.1. The Riesz space L is called Archimedean i f it is true for
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ARCHIMEDEAN RIESZ SPACES
every u E L+ that the decreasing sequence (n-lu : n = 1,2, n-'u 10.
. . .)
satisfies
The lexicographically ordered plane is an example of a non-Archimedean Riesz space. We will present several necessary and sufficient conditions for a Riesz space to be Archimedean; some of these conditions are rather trivial, but some others are definitely not.
Theorem 22.2. (i) The Riesz space L is Archimedean if and only if it is true for every u E Li that E,U 5.0holds for every sequence of real numbers (E, : n = 1,2, . . .) satisfying E, 10. (ii) The Riesz space L is Archimedean if and only if, given u, v E L' mch that nu 5 u holds for n = 1,2,. . ., we have v = 0. (iii) The Riesz space L is Archimedean if and only if, given f, g E L such that ng S fholds for n = 1,2,. . ., we have g 5 0. (iv) Any Riesz subspace of an Archimedean Riesz space is Archimedean.
Proof. (i) Evident. (ii) Let L be Archimedean, and assume that 0 6 nu 5 u holds for n = 1, 2,. . .. Then 0 5 v 5 n-'u 10, so u = 0. Conversely, suppose that 0 6 nu 6 u (n = 1, 2, . . .) implies v = 0, and, for an arbitrary uo EL', consider the sequence (n-'uo :n = 1,2,. . .). We have to prove that any lower bound w of the sequence satisfies w 5 0, since this will imply immediately
that 0 is the infimum of the sequence. For this purpose, observe that v = sup (w, 0) is still a lower bound of the sequence, so we have 0 5 nu 5 uo for n = 1,2, . . .. This implies v = 0, i.e., sup (w, 0) = 0, and hence w 5 0. (iii) If L is Archimedean and ng 5 f holds for n = 1,2, . . .,then ng j f holds for all n, so g 5 n-lf' 10, which shows that g 6 0. Conversely, if ng 6 f for n = 1,2,. . . implies g 5 0, then it follows in particular from no 5 u(n= 1,2, ...)f o r u , v ~ L + t h a t =O,andsoLisArchimedeanbypart(ii). v (iv) Evident. +
Iff and g are elements in the Riesz space L such that f # 0 and nl f I 5 191 for n = 1 , 2, . . ., then f is called infinitely small with respect to g. The element f E L i s called an infinitely small element if there exists an element g E L such that f is infinitely small with respect to g. Evidently, L is Archimedean if and only if L contains no infinitely small elements. It is also evident that in an arbitrary Riesz space the set l o ( L )consisting of the zeroelement and all infinitely small elements is an ideal. The next theorem is the main theorem on Archimedean Riesz spaces.
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[CH. 3,B 22
Theorem 22.3. The following conditionsfor a Riesz space L are mutually equivalent. (i) L is Archimedean. (ii) Every non-empty subset D of L satisJies {D}= Ddd. (iii) Every band A in L satisjies A = Add. (iv) For every ideal A in L the ideal A @ Ad is order dense. (v) For every ideal A in L the ideal { A } 0 Ad is order dense. (vi) An ideal in L is order dense i f and only if it is quasi order dense. In other words, the ideal A in L is order dense i f and only if Ad = {0},i.e., if and only iffor every 0 # f E L there exists an element g E A such that 0 # 191 5
Ifl.
Proof. (i) c> (ii). It was observed in the last remark of the preceding section 21 that L is Archimedean if and only if { A } = Addholds for every ideal A in L. This is equivalent to { D } = Dddholding for every non-empty subset D of L. Indeed, denoting by AD the ideal generated by D, we have { D } = {AD} and Ddd = (AD)dd. (ii) * (iii) Evident. (iii) * (iv) Given the ideal A in L , set B = A 0 Ad. Then { B } d = Bd = ( A 0 Ad)d= {0},
so {B}dd= L. It follows now from (iii) that { B } = L, i.e., B = A 0 Ad is order dense. (iv) * (v) Evident. (v) * (vi) We need only prove that quasi order denseness of an ideal implies order denseness. Hence, assume that (v) holds and the ideal A is quasi order dense, so Ad = (0). Then { A }is order dense by (v), i.e., the band generated by { A }is equal to L . But { A } itself is already a band, so { A } = L . This shows that A is order dense. (vi) * (ii) Assuming that (vi) holds, it will be sufficient to show that { A ) = Addholds for every ideal A. Given the ideal A , the ideal B = A 0 Ad is quasi order dense by Theorem 21.3, so B is order dense by the present hypothesis (vi), i.e., { B } = L. This implies that every u EL' satisfies u = S U P ( V + W : O 6 V E A , O5 W E A ~ , V +SWu). This holds in particular if 0 5 u E Add.But in this case the conditions 0 S w E Ad, w S u imply that w = 0, so it follows for 0 5 u E Addthat u = SUP
(V
:0
S v E A , v 5 u).
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This shows that u E { A } . It has been proved thus that Add c { A } . Since { A } c Addis always satisfied, we obtain the final result that { A } = Add. It was proved in the last theorem that if { A } = Addholds for every ideal A in the Riesz space L, then L is Archimedean. It may be asked whether { A } = Add,holding for every principal ideal A in L , is already sufficient for L to be Archimedean. The answer is negative, as shown by the following example, due to L. J. M. Waayers ([l], Ch. 2, section 4, Example 4).
Example 22.4. Let L be the linear space of all real functionsf on the interval [0, 11 such that f (x) # 0 holds for only finitely many x in [0, 11. We introduce a partial ordering in L by defining that f g means that either f ( 0 ) < g ( 0 ) or f ( 0 ) = g(O), f ( x ) 5 g(x) for all x E [0, 11. Evidently, L is a non-Archimedean Riesz space with respect to this partial ordering. The positive cone L+ consists of allfe L satisfying eitherf(0) > 0 or f ( 0 ) = 0, f ( x ) 2 0 for all x E [0, 11. Iff€ L+ and f ( 0 ) 0, the ideal A, generated by f satisfiesA , = L , and so
s
=-
{A,} = A , = L
=
(A,)?
If 0 # f E L+ and f ( 0 ) = 0, then (A,)d consists of all g E L satisfying g(0) = 0 and g ( x ) = 0 at all x where f ( x ) > 0. Hence (A,)dd consists of all h E L satisfying h ( x ) = 0 at all x wheref(x) = 0. It follows that (A,)dd = A,, and so we must surely have that {A,} = (A,)dd.
If (f,: t E {t}) is an upwards directed set in the Riesz space L such that the set is bounded from above by the elementf o E L , we shall write In this case the set
f,t
Sfo.
G = ( g : g 2 f,
for all t)
is non-empty, and G is directed downwards because g l , g2 E G implies that inf ( g l , g 2 ) E G . It follows that the set
( 8 - f , :9 E G, T E {TI) is directed downwards. We will prove now that Archimedean Riesz spaces are characterized by the property that in the thus described situation there is no gap between the upwards directed set of all f, and the downwards directed set G .
Theorem 22.5. Thefollowing conditionsfor the Riesz space L are equivalent. (i) L is Archimedean. (ii) Given any directed set f,t fo in L , and writing G = ( g : g 2 A
s
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IDEALS AND BANDS; ARCHIMEDEAN RIESZ SPACES
[CH.
3,022
for all T), the downwards directed set b-f,
satisjies g -f, 3.0.
Proof. (i)
:9 E G, 7 E (4)
=- (ii) Let L be Archimedean,f, 5 fo and G = ( g :g Zf, for all z).
Assume now that w EL' satisfies 0 4 w g - f , for all g E G and all T. The proof will be complete if we show that w = 0. Sincef, 5 g - w holds for every z and every g E G, we have g- w E G for every g E G. It follows by induction that g - k w E G holds for k = 1,2, . . and every g E G, so in particular, choosing g = f o and choosing also a fixed r0 E {z}, we obtain f o - k w Zf,,i.e.,
.
kw S fo-fr0
for k = 1 , 2 , . . ..
Since L is Archimedean, this implies that w = 0. (ii) * (i) Assume that (ii) is satisfied, and let 0 5 nu 5 u,, for n We have to prove that this implies u = 0. Writing
V
= (v : t, 2 nu
=
1, 2, .. ..
for n = 1 , 2 , . . .),
we have v-nu J,,,O by hypothesis (where J,,, denotes that o runs through V and n through the natural numbers). Since
v-nu = v - ( n + l ) u + u 2 u for all v E V and all n, and since v -nu 1 0, we must have u = 0. In any Riesz space L the set d of all bands is partially ordered by inclusion, and the same holds for the subset 4? of all bands that are disjoint complements of some subset of L. We will prove that with respect to this partial ordering d is an order complete distributive lattice possessing a smallest and largest element, and a is an order complete Boolean algebra. Furthermore, &' = 37 holds if and only if L is Archimedean.
Theorem 22.6. The set d of all bands in a Riesz space L is an order complete distributive lattice possessing a smallest and a largest element. Proof. Given the bands A and B, the bands A n B and { A + B } evidently satisfy A n B = inf (A, B), { A + B } = sup (A, B) with respect to the partial ordering in d .This shows that
is a lattice. It is
CH. 3 , s 221
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ARCHIMEDEAN RlESZ SPACES
also evident that {0} is the smallest element and L is the largest element in a?. Given the set 9 = ( A , :cr E { c r } ) of elements in a?’, the band generated by all elements contained in at least one of the A, is the supremum of 9 and A , is the infimum of 9.This shows that a?’ is order complete. It remains to show that the lattice &’ is distributive. In view of Theorem 2.3 it will be sufficient to prove that if A , B , ,B , , C are bands such that A n B , c C and A n B2 c C , then A n { B , +B 2 } c C. For this purpose, it is even sufficient to prove that A n ( B , + B 2 ) c C. Indeed, once this has been proved, it follows that
0.
A n {B,+B2} = { A } n {B, +B,} = { A n (B,+B2)} c { C } = C.
In order to prove now that A n ( B , + B,) c C, assume that f E A n ( B l + B 2 ) . For the proof that f E C, we may assume without loss of generality that f 2 0. Then, on account of Theorem 17.6, there exists a decompositionf = fl +fz such that f, 2 0, f2 5 0 and fl E B, ,fiE B,. Sincef E A , it is evident now that f,and f2 are members of A , and hence we have f,E A n B, c C and f2E A n B, c C. Itfollows that f = f , + f , E C, which is the desired result. Now consider the subset of A? consisting of all bands that are disjoint complements. In other words, consists of all bands A such that A = Dd holds for some subset D of L. This implies that a consists of all bands A satisfying A = Add. Indeed, if A E 37, then A = Dd holds for some (nonempty) subset D of L, so that in view of the fact that od = Dddd holds by Theorem 19.2, we have A = Dd = pdd = (Dd)dd= Add. Conversely, if A = Addholds, then A is the disjoint complement of Ad, and so A is a member of a.
Theorem 22.7. The set 37 of all disjoint complements in a Riesz space L, partially ordered by inclusrion, is an order complete Boolean algebra.
Proof. We observe first that, for all ideals A and B in L, we have ( A + B)d
= Ad n Bd, and so
( A +B)dd = (Ad n Bd)d.
Another simple remark is that if A is an ideal included in the member C of a,then Addis also included in Cdd= C. Hence, given A and B in L8,any upper bound C of A and B in 98 includes A +B, and so includes ( A B)dd. But ( A +B)ddis itself a member of a, so we have
+
sup ( A , B ) = ( A +B)dd = ( A d n BdY.
118
[CH. 3,
IDEALS A N D BANDS; ARCHIMEDEAN RlESZ SPACES
8 22
a,
Still assuming that A and Bare elements of the band ( A n B)ddis included in Add = A as well as in Bdd = B, so ( A n B)dd c A n B. The converse inclusion is evident, so ( A n B)dd = A n B, and hence A n B E A?. It is then evident that inf ( A , B ) = A n B holds in A?. What has been shown now for two elements in dY can immediately be generalized to an arbitrary subset 9 = ( A , : o E (o}) of Denoting by A the band generated by all elements contained in at least one A , , the band Addis the supremum of 9 in and A , is the infimum of 9 in A?. It follows already that Li? is an order complete lattice with (0) as smallest and L as largest element. In order to prove the distributivity of the lattice we have to show that if A , B , , B,, C E 93 satisfy A n B , c C and A n B, c C, then A n (sup ( B , , B , ) ) c C holds, i.e., A n ( B , B,)dd holds. For this purpose, it is sufficient to prove that A n ( B , +B,) c C . Indeed, once this has been proved, it follows that
a
a.
nu
a,
+
A n ( B , +B,)dd = Addn ( B , + BZ)dd= [ A n ( B , +B2)Iddc Cdd= C .
The proof that A n ( B , + B 2 ) c C holds is exactly the same as in the proof of the preceding theorem. It remains to prove now that every element A in has a complementary element in 93,i.e., an element B E satisfying sup ( A , B ) = L and inf ( A , B ) = (0). The element B = Ad satisfies these conditions, as shown by sup ( A , Ad) = ( A 0 A d ) d d = L, inf ( A , A ~ =) A n Ad = (0). That there exists a connection between the set at' of all bands, the set A? of all disjoint complements and the Archimedean property is shown in the next theorem.
Theorem 22.8. Let a? and a be the same sets as abooe. Thefollowing conditions are now equivalent. (i) L is Archimedean. (ii) at' and Li? are identical, i.e., every band if a disjoint complement. (iii) The set at' of all bands, partialIy ordered by inclusion, is a BooIean algebra.
Proof. (i) (ii) Assuming that L is Archimedean, we have to prove that every band A in L satisfies A = Add.This was proved in Theorem 22.3. (ii) * (iii) Evident, since 9Y is a Boolean algebra by the preceding theorem.
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ARCHIMEDEAN RIESZ SPACES
(iii) =. (i) By hypothesis, d’ is a Boolean algebra. Hence, given the band A , there exists a band B such that A n B = {0} and { A + B } = L. It will be proved first that this implies B = Ad. Since A n B = {0}, we have A 1B, so B c Ad. Also, once more on account of A n B = {0}, we have A + B = A @ By and so { A @ B } = L. It follows that every u E Lf satisfies u = s u ~ ( u + w : O S U E A , OWS E B , U + W u). S In particular, if 0 S u E Ad, then any u E A such that 0 satisfy u = 0, and so
5 u S u holds must
u=su~(w:OSWEB,W~U) holds for 0 u E Ad. It follows from this formula that u is a member of B (because B is a band), and so we have Ad c B. Combined with B c Ad which was proved above, we obtain B = Ad. Since { A 0 B } = L, we have shown thus that { A @ Ad} = L holds for every band A in L, i.e., A 0 Ad is order dense in L for every band A . But then condition (v) in Theorem 22.3 is satisfied, so L is Archimedean.
Exercise 22.9. Let L be the lexicographically ordered plane. List all bands in L, and show directly that the set d’ of all bands, partially ordered by inclusion, is no Boolean algebra. Also, find an upwards directed set f,T fo in L with the set G of all upper bounds of the set (f,: T E {T}) such that the downwards directed set ( 9 - f , : 9 E G9 7 E
(4)
does not satisfy g - f , J. 0.
Exercise 22.10. Let X be a compact Hausdorff space and L the Riesz space C ( X ) of all real continuous functions on X . As a particular case, keep in mind the case that X is the bounded interval [a, b ] in the real line with its ordinary topology. Given the ideal A in L, let F be the closed subset of X , defined by F = (x : x E X,f(x) = 0 for allfE A ) . For the present purposes, the complement of F, i.e., the open set 0 = X - F, will be called the open carrier of the ideal A . Show (for the case that L = C( [a, b ] ) ) that different ideals may have the same open carrier. Given the ideal A with open carrier 0 and complement F = X - 0 , there does not necessarily exist a non-negative function u(x) in the ideal A such that U ( X ) = 0 for all x E F and U(X) > 0 for all x E 0, although of course for every xo E 0 there exists a function,fE A such thatf(x,) > 0.
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[CH.3 , s 22
Show that, given the open set 0 c X , there exists a largest ideal A having 0 as its open carrier, namely, A = ( f : f L~, f ( x ) = 0 for all x
E
F = X-0).
In the particular case that X i s a compact metric space and A is the largest ideal having the given open subset 0 of Xas its open carrier, show that there exists a function u(x) 2 0 in A such that u(x) > 0 for all x E 0. Returning to the general case, show that if 0,and O2 are open subsets of X with A , and A , as the largest ideals having 0,and O2 as open carriers, then A , # A 2 holds if and only if 0, # 0 2 .Show also that the largest ideal having 0 as its open carrier is not necessarily a band. Every band, however, is such a largest ideal. Next, show that for any ideal A the open carrier of the disjoint complement Ad is the interior of the set F = ( x : f ( x ) = 0 for all f E A ) . Finally, show that if A is the largest ideal having 0 as its open carrier, then A is a band if and only if 0 is regularly open (for the definition and some properties of regularly open sets, cf. Exercise 4.16). The set of all bands in L and the set of all regularly open subsets of Xare both order complete Boolean algebras (partial ordering by inclusion), and assigning to each band its open carrier, we thus obtain a lattice isomorphism between these Boolean algebras. Hint: On the interval [- 1, 11, let u , ( x ) = 0 for x 5 0 and ul(x) = x for x 2 0, and let uz(x) = 0 for x S 0 and u,(x) = x 2 for x 2 0. The ideals in C( [ - 1, 1I), generated by u, and u, ,are different and have the same open carrier. If A is a band, then A = (Ad)dholds because C ( X ) is Archimedean, and so the open carrier of A is the interior of a closed set. This implies that the open carrier of A is regularly open. Conversely, if A is the largest ideal having the regularly open set 0 as its open carrier, we introduce the largest ideal B having the interior of F = X - 0 as its open carrier. Then A and Bd have the same open carrier 0, and they both are the largest ideal having 0 as open carrier, so A = Bd. This shows that A is a band. Exercise 22.11. Let (X,A , p) be a measure space such that p ( X ) is finite, and let M") = M("(X, p) be the Riesz space of all real p-almost everywhere finitevalued p-measurable functions on X , with identification of p-almost equal functions (cf. Exercise 11.2 (v)). Given the ideal A in M"), let r be the set of all p-measurable subsets E of X such that every f E A vanishes p-almost everywhere on E, and let
ARCHIMEDEAN RIESZ SPACES
121
Show that there exists a set E, E r such that p(E,) = a, and hence no subset of X- Eo of positive measure is a member of r. Show that E, is uniquely determined (except for a set of measure zero) by the conditions that E, is a member of r and no subset of X- Eo of positive measure is a member of r. The set C = X - E , is called the carrier of the ideal A . Show that different ideals may have the same carrier, Giveo the ideal A, there does not necessarily exist a function u ( x ) in A such that u ( x ) is strictly positive on the carrier of A. Given the ideal A and the set E, as above, show that E, is the carrier of the disjoint complement Ad. Show that for any given p-measurable subset C of X there exists a largest ideal having C as carrier, namely, the band A having C as its carrier, i.e., A = (f:f E M('),f(x) = 0 for p-almost every x E X- C). Extend these results to the case that the measure ,u is o-finite, i.e., X is a countable union of sets of finite measure.
CHAPTER 4
Dedekind Completeness and Projection Properties
Some Riesz spaces are Dedekind complete (i.e., every subset which is bounded from above has a supremum) and others are Dedekind a-complete (i.e., every countable subset which is bounded from above has a supremum) without being Dedekind complete. Some Riesz spaces have the projection property (i.e., every band has the property that the direct sum of the band and its disjoint complement is the whole space) and some have only the principal projection property (i.e., the above direct sum property holds only for principal bands). One of the main theorems in elementary Riesz space theory is that the following implications hold (with self-evident abbreviations):
PDed*a-comp1. princ. proj. prop. Ded. compl. . % proj. prop. /
3
Archim.,
and no inverse implication holds. Furthermore, Dedekind o-completeness and the projection property are independent. Dedekind a-completeness and the projection property together imply Dedekind completeness. The proof of the last result is postponed until the chapter on spectral theory, since the proof is based upon the spectral theorem in Riesz spaces. It will be shown that any Archimedean space of finite dimension, say of dimension n, is Riesz isomorphic to n-dimensional real number space R" with the usual ordering; the proof that will be presented is essentially due to A. I. Judin (1939). The proper ideal J in the Riesz space L is called maximal whenever there is no ideal properly between J and L. It will be proved that the ideal J is maximal if and only if L/J is one-dimensional. If L has a strong unit, the intersection of all maximal ideals (usually called the radical of L)is the ideal consisting of all infinitely small elements. Hence, if L is Archimedean and has a strong unit, the intersection of all maximal ideals is the zeroelement. This result is due to K. Yosida and M. Fukamiya (1941). If L does not pos122
CH. 41
DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
123
sess a strong unit, it may occur that L does not have any maximal ideals at all, even in the case that L is Archimedean and has a weak unit. Thc Riesz space L is said to be order separable whenever every set in L having a supremum contains a finite or countable subset having the same supremum. It will be shown that an Archimedean Riesz space is order separable if and only if every set of mutually disjoint strictly positive elements which is bounded from above is at most countable. Dedekind complete Riesz spaces having the last mentioned property were investigated by A. G. Pinsker (1950), and in the Soviet terminology these spaces are called Kspaces of countable type. The Riesz space L is said to have a countable order basis if there exists an at most countable set of elements such that the band generated by this set is the space L itself. If L has a countable order basis and L has the projection property, then every band in L has also a countable order basis, but it is not necessarily so that every ideal in L has now a countable order basis. It is a pleasant and perhaps somewhat surprising result that if L is Archimedean and has a countable order basis then every ideal in L has a countable order basis if and only if L is order separable. Any band A in the Riesz space L such that A 0 Ad = L holds is called a projection band. In this case, every f E L has a unique decomposition f = fl +fi with fl E A and fiE Ad. Writing fl = P A L the mapping PA of L into L is a projection as well as a Riesz homomorphism. Conversely, if P is a projection of L into itself such that 0 5 Pu S u holds for every u E L + , there exists a projection band A such that P = P A . The set of all projection bands, partially ordered by inclusion, is always a Boolean algebra, and the set of all principal projection bands is an ideal in this Boolean algebra. Furthermore, for any given Boolean algebra 9 there exists a Riesz space L such that 9 is lattice isomorphic to the Boolean algebra of all projection bands in L. Given the Dedekind complete Riesz space K with the Riesz subspace L, the space K is called a Dedekind completion of L if every f E K satisfies
It will be proved that the Riesz space L possesses a Dedekind completion if and only if L is Archimedean. The Dedekind completion is uniquely determined except for isomorphism. Furthermore, the Archimedean Riesz space L is order separable if and only if its Dedekind completion is so.
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[CH.4,§ 23
23. Dedekind completeness We recall the definitions of Dedekind completeness and Dedekind a-completeness (cf. Definition 1.1), we apply these notions to Riesz spaces and we also introduce several related notions. Definition 23.1. (i) The Riesz space L is called Dedekind complete if every non-empty subset of L which is boundedfrom above has a supremum. Equivalently, L is Dedekind complete if every non-empty subset of L which is bounded from below has an injimum. (ii) The Riesz space L is called Dedekind a-complete if every non-empty at most countable subset of L which is boundedfrom above has a supremum. (iii) The Riesz space L is called order separable if every non-empty subset D possessing a supremum contains an at most countable subset possessing the same supremum as D. (iv) The Riesz space L is called super Dedekind complete i f L is Dedekind complete and order separable. Evidently, Dedekind completeness imp!ies Dedekind a-completeness. In the Soviet literature a Dedekind complete Riesz space is called a K-space, and a super Dedekind complete Riesz space is called a K-space of countable type. An order separable Riesz space is sometimes called a Riesz space possessing the countable sup property, because the supremum of a subset, if existing, is already the supremum of a finite or countable subset of the given subset. In the Bourbaki terminology a Dedekind complete space is called "un espace compl2tement re'ticulP and in the Nakano terminology it is called a universally continuous space. Theorem 23.2. (i) The following conditions for the Riesz space L are equivalent. (a) L is Dedekind complete. ( b ) Given the upwards directed set (u, :z E (t}) such that 0 S u,t 5 u holds in L, there exists an element uo in L such that u, uo. (c) Given the downwards directed set (u, : t E (71) such that 0 S u,J holds in L, there exists an element uo in L such that u, 1uo. (ii) The following conditions for the Riesz space L are equivalent. (a) L is Dedekind a-complete. ( b ) Given the increasing sequence (u,, :n = 1,2, . . .) such that 0 S u,t 5 v holds in L, there exists an element uo in L such that u, t uo . ( c ) Given the decreasing sequence (u, :n = 1, 2, . . .) such that 0 5 u,l holds in L, there exists an element uo in L such that u,, 1uo.
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DEDEKIND COMPLETENESS
(iii) The Riesz space L is order separable if and only i f , given any upwards directed set (u, : z E { r } ) and any element uo in L + such that 0 6 u, 1uo, there exists a sequence (u,, :n = 1, 2, . . .) in the set of all u, such that u,, uo
.
Proof. (i) It is evident that (a) implies (b), and it was already observed in the remark following Theorem 15.11 that (b) implies (a). For the proof that (b) implies (c), assume that (b) holds and that we have 0 5 u,J. in L. If uT0is one of the u,, the set (u, : u, 6 u,,) is directed downwards and has the same lower bounds as the set of all u,. It will be sufficient, therefore, to prove that the set (u, : u, 5 uT0)has an infimum. To this end, note that the set satisfies and so w = sup (u,,-u,)
exists by hypothesis (b). But then we have uTo-w = inf u,.
The proof that (c) implies (b) is trivial. The proofs for parts (ii) and (iii) are similar. Before presenting some examples, we observe already that Dedekind a-completeness of a Riesz space L implies that L is Archimedean. Indeed, assume that 0 6 nu 5 v holds for n = 1, 2, . . .. By the Dedekind u-completeness of L the element uo = sup (nu : n = 1,2, . ..) exists, and evidently we have 0 5 u 6 uo. On the other hand we have also uo = sup (2nu : n = 1,2,.
. .) = 2 sup (nu :n = 1,2,. ..) = 2u0,
so uo = 0, and hence u = 0. It has thus been shown that 0 6 nu n = 1, 2 , . . implies u = 0, i.e., L i s Archimedean.
.
5 v for
Example 23.3. (i) The n-dimensional real number space R", with the usual coordinatewise ordering, is a super Dedekind complete space. The plane R2 with lexicographical ordering is non-Archimedean, and hence fails to be Dedekind a-complete. (ii) The Riesz space C([O, 11) of all real continuous functions on the interval [0, I ] is not Dedekind a-complete. Indeed, it is easy to define an increasing sequence of non-negative continuous functions, converging pointwise to 1 for 0 S x < and to zero for s x 1. The sequence, therefore, has no supremum in the space C( [0, 11). (iii) The Riesz space L of all real functions on the uncountable point set
+
+
s
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4, !j 23
X, with the ordinary pointwise ordering, is Dedekind complete but not super Dedekind complete. The supremum of any subset bounded from above is the ordinary pointwise supremum. For any xo EX, define the function ux,(x) E L by uxo(xo)= 1 and uxo(x)= 0 for x # x,. The set (ux : x E X ) has the supremum e, where e ( x ) = 1 for all x E X , but no countable subset has e as its supremum. (iv) Let ( X , A, p ) be a measure space such that p ( X ) is finite, and let M(') = M(')(X, p ) be the Riesz space of all real p-almost everwyhere finitevalued p-measurable functions on X, with identification of p-almost equal functions (cf. Example 11.2 (v)). The Riesz space M"' is super Dedekind complete. In order to prove this, assume that 0 5 u,? 5 u in M(", where we assume first that u = u ( x ) is a bounded function (so Jxu dp is finite). Then the set of numbers
is bounded, and so P = sup, JxuTdp exists as a finite number. It follows that there exists an increasing sequence (uTn:n = 1, 2, . . .) in the set of all u, such that S,.,,,dp
P as n
+
co.
The pointwise supremum uo(x) = sup uTn(x)is then a p-measurable function such that Jx uodp = P, and it follows easily that every u, satisfies u,(x) 5 u0(x) for p-almost every x (the exceptional set depending upon z, of course). This shows that uo = sup u, holds in M(". If now 0 5 u,? 5 u holds in M'", where u is not necessarily a bounded function, we consider for n = 1, 2, . the functions u,,,, = inf (u,, n), and then u, = sup (u,,,, : z E {z]) exists in M"' for every fixed n. It is also evident that uo = sup (u,, : n = 1, 2, . . .) exists in Mc", and it follows easily that
..
uo = sup (u,,,, : z
E
{z], n = 1,2, . . .) = sup (24, : z
E {Z})'
This proves the Dedekind completeness of M@).Since the supremum of the set (ur : z E {z]) is obtained as the supremum of an appropriate countable subset, the proof shows at the same time that M(') is super Dedekind complete. Note that, for any number p satisfying 1 5 p 5 coy the space Lp(X,p ) is an ideal in M(')(X, p). It follows immediately that Lp(X,p ) is also a super Dedekind complete Riesz space. More generally, any normed Kothe space
CH.4,s 231
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DEDEKIND COMPLETENESS
LJX, p ) is an ideal in M(')(X, p), and so LJX, p ) is a super Dedekind complete Riesz space. The super Dedekind completeness proof for M"' can immediately be extended to the case that Xis the union of a countable number of sets of finite measure. If X fails to be such a countable union, the proof that M"' is Dedekind complete can still be carried through if the measure p is locally finite (i.e., every set of positive measure has a subset of finite positive measure) and localizable (i.e., the Boolean algebra of all p-measurable subsets of X is order complete), but M(') is now no longer super Dedekind complete. The spaces L J X , p ) for values of p satisfying 1 6 p < 03, however, are still super Dedekind complete under these more general conditions. (v) Let L be the Riesz space of all finitely additive signed measures p on the field r of subsets of the non-empty point set X such that llpll = SUP (IP(41 :A E 0
is finite (cf. Example 11.2 (viii)). The Riesz space L is Dedekind complete. For the proof, assume that 0 6 pL,f6 1 holds in L.It is not difficult to show now that if the number po (A) is defined for every A E r by p o ( A ) = sup ( p z ( A ):Z E {z}), then p,, is an element of L and po = sup holds in L. In general, the space L is not super Dedekind complete. For further details, cf. Example 25.3. (vi) Let G be a region in the plane and L the Riesz space of all functions f = u1- u2 with u1 ,u2 harmonic and non-negative in G (cf. Example 11.2 (x)). The Riesz space L is super Dedekind complete. For the proof, assume that 0 5 uzJ holds in L. It was shown essentially in Example 11.2 (x) that the pointwise infimum uo = inf (uT: z E {z}) is harmonic in G, and this implies already that L is Dedekind complete. The proof that uo is harmonic was based on the observation that if S is a circular disc in G, then there exists a sequence (urn :n = 1,2, . .) in the set (uT: t E {z}) such that uTn1uo holds pointwise in the interior of S. Let now (& : k = 1,2, . . .) be a sequence of open circular discs covering G, and for each k let (uki, uk2 , .) be a sequence from (uT: z E {t}) such that y,,1uo as n + 03 holds pointwise in the interior of S,. It follows that uo = inf (ukn: k, n = 1,2, . .) holds pointwise in G. Ordering now the double sequence into asingle sequence (0, : n = 1, 2,. .), there exists a sequence (w,,: n = 1 , 2 , . . .) in (u, : T E {z}) such that w,J and w,,6 u, for each n, hence w,,1uo This shows that L is super Dedekind complete. For further details, cf. Example 41.2. (vii) Let X = ( x : c( S x S p ) be a bounded interval in the real axis, and L the Riesz space of all real linear functions on X (cf. Example 11.2 (xi)).
.
..
.
.
.
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 23
It is easy to see that L is super Dedekind complete. For later purposes we list all bands in L. Let the element e E L+ be defined by e(x) = 1 for all x E X,the element el E L+ by el(.) = 1, el(/?) = 0, and the element e, E L + by .,(a) = 0, e,(p) = 1. It is evident that el + e , = e, and el Ie, .Let Eland E, be the bands generated by e , and e, respectively. It is easy to see that ({0}, E l , E 2 ,L ) is the set of all possible bands in L. For further details, cf. Example 41.3. The following condition on a Riesz space, denoted by property (*), is somewhat stronger than order separability.
Definition 23.4. The R i e u space L is said to have the property (*) whenever every non-empty subset D of L which is bounded from above contains an at most countable subset having the same upper bounds as D. The space L has property (*) if and only if every non-empty and upwards directed set (y : z E {T}) in L+ which IS bounded from above contains an at most countable subset having the same upper bounds as (u, :z E {z}). The proof is as in Theorem 23.2 (iii). We will prove now that the property (*) and order separability are equivalent if L is Archimedean. Theorem 23.5. I f L is Archimedean, then L has property (*) ifand only i f L is order separable. Proof. We need only prove that in an Archimedean space order separability implies property (*). To this end, assume that L is Archimedean and order separable, and let 0 5 u,? 5 v,, in L with V = (v :v 2 u, for all z).
We have to show the existence of a sequence (uTn:n = 1,2, . . .) in the set (u, : T E {T}) such that the sequence has V as the set of its upper bounds. Since L is Archimedean, Theorem 22.5 can be applied, so the set (v-u, :u E
v,z E {T})
satisfies w,,,,= v - u, J,, ,0, and hence in view of the order separability there exists a sequence (w, : n = 1,2, . . .) in the set of all w,,,, such that inf w, = 0. We may write w, = v,-utn, and without loss of generality it may be assumed that v, is decreasing. Indeed, if necessary, replace v2 by v; = inf (vl ,v,), replace v 3 by u j = inf ( v ; , v 3 ) , and so on. Since (y :z E {z}) is directed upwards, we may similarly assume that urnis increasing. We assert now that (uTn: n = 1,2, . . .) is the desired sequence having V as the set of
s
CH. 4, 231
129
DEDEKIND COMPLETENESS
its upper bounds. To prove this, it will be sufficient to show that any w E L satisfying w 2 urn for all n satisfies w 2 u, for all z. Hence, assume that w 2 urnholds for all n. Then 0
s sup (v,,,w ) - w =
sup (on, w)-sup (urn,w ) 5
U,-Urn
10,
so sup (v,,,w ) 1 w. Observing now that wS
SUP
(ur, w) 5
SUP
(on,
W)
holds for all z and all n, we obtain by taking the infimum over n that w = sup (u,, w) for all z, i.e., w 2 u, for all z. By definition, the Riesz space L is super Dedekind complete if L is Dedekind complete and order separable. It is one of the pleasant consequences of the last theorem that we can prove now easily that Dedekind a-completeness combined with order separability already implies super Dedekind completeness.
Theorem 23.6. If the Riesz space L is Dedekind a-complete and order separable, then L i.s super Dedekind complete.
Proof. It will be sufficient to show that L is Dedekind complete. To this end, assume that 0 6 u,t v holds in L. We have to show that sup u, exists. Since L is Archimedean on account of the Dedekind a-completeness and since L is order separable by hypothesis, the preceding theorem shows that L has the property (*). Hence there exists a sequence (urn :n = 1 , 2 , . . .) in the set (u, : z E {T}) such that (urn : n = 1,2, . . .) and (u, : z E (71) have the same upper bounds. The element u = sup urn exists since L is Dedekind a-complete, and so u is an upper bound of (u, : z E (z}) because u is an upper bound of (urn: n = 1 , 2 , . . .). But then u must be the supremum of the set (y : z E {z}), since u is already the supremum of the sequence of all urn.The existence of sup u, shows that L is Dedekind complete.
s
We insert another theorem about order separability.
Theorem 23.7. Consider the following conditionsfor the Riesz space L. (i) L is order Jeparable. (ii) Every a-ideal in L is a band. (iii) Every order dense ideal in L is super order dense. In any Riesz space L we have (i) (ii) * (iii). In an Archimedean Riesz space the three conditions are actually equivalent, but the proof of that fact is postponed until Theorem 29.3.
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[CH.4,823
Proof. (i) + (ii) Let A be a a-ideal in L and assume that 0 5 u E { A } , where ( A } denotes the band generated by A . It will be sufficient to prove that u E A . On account of 0 S u E { A } there exists an upwards directed set (u, : z E {z}) in A such that 0 5 u, t u, and hence, by the order separability of L , there exists a sequence (uzn:n = 1,2, . .) in the set (u, : z E (7)) such that 21 = sup u,,. This implies u E A on account of A being a a-ideal. (ii) (iii) Given the order dense ideal A , we have on account of (ii) that the a-ideal generated by A is already equal to { A } = L.It follows that every u EL' is the supremum of an appropriate sequence in A . This shows that A is super order dense.
.
In general, the direct sum of two disjoint bands in the Riesz space L is not again a band, but merely an ideal. By way of example, if A and B are the bands in C([O,11) consisting of all f E C([O, 11) vanishing on the closed intervals [0,3] and [+, 11 respectively, then A 0 B is the ideal of allf vanishing at the point +.Under certain conditions, however, the direct sum of two disjoint bands is again a band. This holds in particular if L is Dedekind complete. Theorem 23.8. I f L is a Dedekind complete Riesz space, the following holds. (i) If A , and A , are bands in L such that A , 1A , , then A , 0 A , is a band in L. (ii) Every band A in L satisjies A 0 Ad = L.
Proof. (i) Let 0 6 u, t u with u, E A , 0 A z for all z E {z}. We have to prove that u E A , 8 A , . For every z there is a decomposition u, = u: + u:' with u: E A , and u:' E A , . It follows immediately from Theorem 17.6 that u,, S u,, implies u:, S ui2 and u:: u::, and so (u: : z E and (ui' : z E {T}) are equidirected with (u, : z E {z}). This implies that 0 5 u: ts u and 0 5 u:'T S u hold. Since L is Dedekind complete, there exist elements u' and u" in L+ such that 0 6 u: 7 u' and 0 u:' t u", and so we have u' E A , and 11" E A , because A , and A , are bands. It follows now by Theorem 15.8 (iii) from the equidirectedness of (u: : z E {z}) and (u;' : z E {z}) that
{TI)
u, = u;+u:'
t u'+"'.
On the other hand we have u, t u by hypothesis, and so u = u'+u". This shows that U E A l 0 A , . (ii) As noted earlier, any Dedekind a-complete space, and hence any Dedekind complete space, is Archimedean. Hence, given the band A in the Dedekind complete space L, the quasi order dense ideal A @ Ad is order dense by Theorem 22.3, i.e., the band generated by A 0 Ad is equal to L.
CH.4 , § 241
PROJECTION PROPERTIES
131
But A @ Ad itself is a band by part (i) of the present theorem, so A 0 Ad = L. Generalizing the result in part (i), it will be proved in section 30 that in a Dedekind complete space the algebraic sum of two (not necessarily disjoint) bands is always a band.
24. Projection properties The final theorem in the preceding section shows that to every band A in a Dedekind complete Riesz space there corresponds another band B (namely, B = A d )such that A @ B = L. In view of this result it is a natural question to ask what can be inferred about ideals A and B in a n arbitrary Riesz space L if A 0 B = L.The answer is contained in the following theorem.
Theorem 24.1. I f A and B are ideals in the Riesz space L such that A 0 B
= L, then B = Ad and A = Bd, so A = Add and B = Bdd. Hence, A and B
are bands, each the disjoint complement of the other one.
Proof. It follows from A 0 B = L that A n B = (01, so A IB, and hence B c Ad. In order to prove now that Ad is included in B, assume that 0 5 U E Ad. By hypothesis u has a decomposition u = ul+uz with 0 2 u1 E A and 0 S uz E B. Since
05
u1
5 ueAd
holds, we have u1 E Ad. On the other hand we have u1 E A . It follows that u1 = 0, and so u = u2 E B. It has thus been proved that Ad c B. The final result is that B = Ad. There are several variants of the last theorem. In one of these A is only assumed to be a linear subspace, but in return B is now immediately assumed to be the disjoint complement of A .
Theorem 24.2. If A is a linear subspace of the Riesz space L such that A @ Ad = L, then A = Add. Proof. It will be sufficient to prove that Addis included in A . To this end, let f E Add be given. By hypothesis there exists a decompositionf = fl+fZ with fl E A c Add and fzE Ad. By the uniqueness of the decomposition for any element in the ideal Add@ Ad we have then that f = fl +fi must also be the decomposition off in the direct sum Add@ Ad, so f2 = 0 on account of f E Add. This shows that f = fl E A , and so Add is included in A . There is another more sophisticated version. For the formulation we need
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 24
the following definition. The ideal C in the Riesz space L is called the order direct sum of the linear subspaces A and B whenever C = A @ B and whenever, for every element u 2 0 in C, the (unique) decomposition of u into an element u E A and an element w E B is such that u 2 0 and w 2 0. If A and B are disjoint ideals and C = A @ B, then this is an order direct sum by Theorem 17.6. In the converse direction we have the following theorem. Theorem 24.3. If the ideal C in the Riesz space L is the order direct sum of the linear subspaces A and B, then A and B are ideals. Hence, i f L is the order direct sum of the linear subspaces A and B, then A and B are bands such that A = Bd and B = Ad. Proof. Let the ideal C be the order direct sum of the linear subspaces A and B. We prove first that u E A+ and 0 I; v 5 u implies u E A'. For this purpose, set w = u-u, so u = u + w with u 2 0 and w 2 0. Then, since u E A + c C' and C is an ideal, we have u, w E C+,and so there exist decompositions u = u1 +u, and w = w1 + w 2 with u l , w1 E A' and u 2 , w z E B'. It follows that u2+w2 = u - ( u , + w , ) ~A n B = {0},
so v z + w 2 = 0, and hence uz = wz = 0. This shows that u = u1 E A'. In the next step we prove that everyf E A has a decomposition of the form f = p 1 - q l with p l , q1 E A'. Indeed, given f E A, we have f E C, and so f ' and f - are members of C' since C is an ideal. Let f = p 1 +pz and f = q1 +q2 be the corresponding decompositions with p l , q1 E A' and p z , q2 E B'. Then +
so
f =f+-f-
= (Pl-ql)+(Pz-qz),
Pz-42 = f - ( P r 4 1 ) E A n B = (01. It follows that f = p 1 - q l with p l , q1 E A'. Finally, observe that any f E A satisfiesf = p 1 - q l with p l , q1 E A', and also f = f ' - f - with 0 5 f' 5 p 1 and 0 f- 6 q l . Hence' f andf- are members of A + by what has already been proved, so If] E A + . It is an immediate consequence now that A is an ideal. Similarly, B is an ideal. If the space L itself is the order direct sum of the linear subspacesA and B, we have now that A and B are ideals in L such that A 0 B = L.Hence, by Theorem 24.1, A and B are bands such that A = Bd and B = Ad.
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PROJECTION PROPERTIES
In view of the foregoing theorems the following definition is now appropriate. Definition 24.4. Any band A in the RieJz space L, having the property that A @ Ad = L, is called a projection band. In this case, iff = f l + f 2 is the decomposition of an arbitrary f E L as the sum of f i E A and f 2 E Ad, then f l and f 2 are called the components off in A and Ad respectively.
If A is a projection band, then Ad is a projection band. The converse is not necessarily true (if L is the lexicographically ordered plane and A is the vertical axis, then Ad = (0)is a projection band, but A is not because A = Add is not satisfied). If, however, L is Archimedean, then A is a projection band if and only if Ad is so. Indeed, if Ad is a projection band then Addis a projection band, and Add = A holds for every band A because L is Archimedean. In the next theorem we present more information about projection bands. Theorem 24.5. (i) The band A in the Riesz space L is a projection band if and only if,for any u E L', the element UI =
S U ~ ( U : U E A5, Ov
5 U)
exists, and in this case u1 is the component of u in A. Similarly, u2 = SUP
(W
: w E Ad, 0
5w5
U)
is then the component of u in Ad.In other words, ifu, and u2 are these suprema, we have u = ul + u2 with u1 E A and u2 E Ad. (ii) For any f E L, denote the component o f f in the projection band A by PAf. The mapping PA of L into itself has the following properties: (a) PA is linear and idempotent (i.e., Pi = PA),so PA is aprojection. (b) 0 5 P A u 5 u holdr for every u E L'. The mapping PA is called the order projection on the projection band A. (iii) In the converse direction, i f P is aprojection (mapping L into itsev) such that 0 6 Pu 6 u holdi for every u E L+,there exists aprojection band A such that P is the order projection on A. (iv) A s an immediate consequence of (i), the band A in the Riesz space L is a projection band if, for every principal band B, the intersection A n B is a projection band. Proof. (i) Assume first that A is a projection band, so L = A 0 Ad. and let u E L + have the decomposition u = u1 + u 2 . Set
v=
( u : v E A , O 5 v 5 u).
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[CH.4,8 24
For any v E V, the element u--0 2 0 has the decomposition
u-u = ( u l - u ) + u z , and so (since positive elements have positive components) we have u1 - u >= 0, i.e., v 6 u l . This shows that u1 is an upper bound of V . On the other hand u1 is a member of V, and so u1 = sup V. Assume now, conversely, that A is a band such that 241
= sup (u : u E A , 0 5 u
6
2.4)
exists for any given u E L + .Since u1 is then a member of A (on account of A being a band), it will be sufficient for the proof of A 0 Ad = L to show that u2 = u - u1 is a member of Ad. If not, we have p = inf (u2 ,z ) > 0 for some Z E A (we recall that p > 0 means that p 2 0 and p # 0). Then 0 < p E A , and also p 5 u 2 . It follows that u1 + p E A and u1 + p 5 u1+ u z = u, so u , + p E ( u : U E A , O 5 v 6 24). But then u , + p ~ s u p ( u : u E A , O ~ u ~ Uu)1 ,=
s o p 6 0, which contradicts p > 0. The final result is, therefore, that uz E Ad. (ii) Evident. (iii) This is essentially a reformulation of Theorem 24.3. Let A be the range of the given mapping P , i.e., A = (Pf :f E L). Similarly, let B be the range of the mapping I - P , where I denotes the identity mapping, so B = ( f - P f :f E L). Evidently, A and B are linear subspaces of L. Furthermore, we have A n B = (0). Indeed, any f E A n B satisfies f = P g for some g E L as well as f = h-Ph for some h E L , and so f = P g = P z g = P(h-Ph) = Ph-Pzh = 0.
+
The algebraic sum A B is therefore a direct sum A 0 B. As evident from f = Pf+ ( f - P f ) , holding for every f E L, we have A 0 B = L. Finally, since u 2 0 implies Pu 2 0 and u-Pu 2 0 by hypothesis, we see that L is the order direct sum of the linear subspaces A and B. It follows from Theorem 24.3 that A and B are projection bands, and obviously P is now the order projection on A . We recall that, by Definition 18.1, the linear mapping n of the Riesz space L into the Riesz space M is called a Riesz homomorphism whenever inf (nf,n g ) = 0 holds in M for every pair of elements f, g E L satisfying inf (f,g ) = 0. Necessary and sufficient for the linear mapping n of L into
CH. 4,§ 241
135
PROJECTION PROPERTIES
M to be a Riesz homomorphism is that 4 i n f (f,g)> = inf (nf,n g ) holds for allf, g E L ,or also that 7+P
(f, s)>= SUP (nf,v )
holds for allf, g E L (cf. Theorem 18.2). In particular, if n is a Riesz homomorphism, then nf' = (nf)', n f - = ( n f ) - and n(lf1) = Infl.
Theorem 24.6. The order projection PA on the projection band A in the Riesz space L is a Riesz homomorphism of L into itself. Proof. Let inf (f,g ) = 0 in L. Then we havef, g EL', so it follows now from 0 5 P A f S f and 0 S P,g 5 g that O
<= inf (PAX PAg) <= inf (f,g ) = 0,
so inf ( P A LP A g ) = 0. We observe that if PA is the order projection on the projection band A , then the disjoint complement Ad is the kernel (null space) of the mapping P A , so denoting by [f]the element of L/Ad which containsf, the mapping Lf] + PAf is now a Riesz isomorphism of L/Ad onto A . We recall (cf. Definition 17.5) that every ideal (or a-ideal or band) generated by one single element of L is called a principal ideal (or principal a-ideal or principal band). Sincef and If1 generate the same ideal (namely, the set of all g E L such that 191 5 nlfl holds for some natural number n, depending on g ) , we may assume that every principal ideal (and every principal a-ideal and principal band) is generated by an element of L f . Principal projection bands can be characterized in a special manner, as follows.
Theorem 24.7. (i) The principal band A , generated by the element v E L ' , is a projection band if and only if,for every u E L , the element +
u1 = sup (inf (u, nu) : n = 1 , 2 , .
. .)
exists, and in this case u1 is the component of u in A , i.e., u1 = PAu. (ii) If A is the principal band generated by v E L ' and u is a positive element in A , then u = sup (inf (u, nu) :n = 1,2,. . .). Hence, every principal a-ideal is already a band. In other words, the notions of principal a-ideal and principal band are the same.
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DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
[CH.4,s 24
Proof. (i) Given the principal band A generated by v E L + and given
u E Lf, we set
W = ( w : w E A , O5 w 5 u), W' = (inf (u, nv) : n = 1,2, . . .). It is evident that W' is included in W,and so any upper bound of W is an upper bound of W'. In order to prove the converse we note that, given w E W,there exists an upwards directed set (w,: z E {z}) in the ideal generated by v such that 0 5 w, t w. Each w, satisfies w, 5 n,v for some natural number n,; hence, it follows now from w, 5 w 5 u and w, 5 n,v that w, 5 inf (u, n,u). This shows that any upper bound of W'is an upper bound of all w,, and so of w,and hence of W. It has been proved thus that Wand W' have the same upper bounds. In particular, the supremum of W'exists if and only if the supremum of W exists, and these suprema are then the same. By Theorem 24.5 (i), A is a projection band if and only if u1 = sup W exists, i.e., if and only if u1 = sup W' = sup (inf (u, nu) : n = 1,2,
. . .)
exists, and in this case u1 is the component of u in A. (ii) If we consider A as a Riesz space by itself, then A is still the principal band generated by v , and A is now a projection band. Hence, by part (i), u = sup (inf (u, nu) : n = 1 , 2 , .
holds for every element u satisfying 0
5
. .)
u E A.
Definition 24.8. The Riesz space L is said to have the projection property if every band in L is a projection band, and L is said to have the principalprojection property if every principal band in L is a projection band. The final theorem in this section is now an immediate consequence of several theorems proved earlier.
Theorem 24.9. (i) Every Dedekind complete Riesz space has the projection property. (ii) The Riesz space L has the projection property fi and onIy if, for every u E L +and every band A in L, the element s u p ( v : v ~ A , O5 v
5 u)
exists, and this element is then the component of u in A.
CH. 4,s 251
THE MAIN INCLUSION THEOREM
137
(iii) The Riesz space L has the principal projection property $and only if, for every pair u, v E L+,the element sup (inf (u, nu) : n = 1,2, . . .) exists, and this element is then rhe component of u in the principal band generated by v. (iv) As an immediate consequence of (ii), the Riesz space L has theprojection property if and only if every band contained in a principal band is a projection band. Proof. The statement in part (i) was proved in Theorem 23.8 (ii). Exercise 24.10. (i) It has been shown that any order projection PA on a projection band A is a Riesz homomorphism. Show that if P is a projection (i.e., P is a linear idempotent mapping of L into L ) such that P is a Riesz homomorphism, then P is not necessarily the order projection on a projection band. (ii) It has been shown that any order projection PA on a projection band A satisfies 0 g PAu 5 u for every u EL'. Show that if T is a Riesz homomorphism of L into itself such that 0 5 Tu 5 u holds for every u EL', then T is not necessarily a projection. Hint: For (i), let L be two-dimensional real number space ordered in the usual coordinatewise manner, let A be any line through the origin having a half-line in common with the positive cone, but not the positive horizontal or vertical axis, and let P be the projection on A in the direction of the horizontal axis. For (ii), let Tf = t.f for all f E L. Exercise 24.11. Let PA be the order projection on the projection band A. Show that 0 5 u, u in L implies 0 5 PAu, t PAu in L. Hint: If v is an arbitrary upper bound of the set of all PAu,, then u+ (u-PAu) is an upper bound of the set of all u,. 25. The main inclusion theorem The properties concerned with Dedekind completeness and order projections on bands are connected, as follows. Theorem 25.1 (Main inclusion theorem). With obvious notational abbreviations, the following implications hold in any Riesz space L.
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH. 4,s 25
No implication in the conuerse direction holds, and Dedekind o-completeness and the projection property are independent properties. Finally, Dedekind a-completeness and the projection property together imply Dedekind completeness; the proof of this last statement is postponed until after the proof of the spectral theorem in Riesz space (cf. Theorem 42.8).
Proof. It is evident that super Dedekind completeness implies Dedekind completeness and Dedekind completeness implies Dedekind a-completeness. It was proved in Theorem 24.9 (i) that Dedekind completeness implies the projection property, and it is evident again that the projection property implies the principal projection property. Hence, we are left with the proof of the following implications: Ded. o-comp. =. princ. proj. prop.
=r
Arch.
For the proof of the first implication we assume that L is Dedekind o-complete. In view of Theorem 24.9 (iii) we have to show that, for every pair of elements u, u E L ', the element sup (inf (u, nu) : n
=
1,2, . . .)
exists. Writing WJ, = inf (u, nu) for n = 1 , 2 , . . ., we have 0 2 writ 2 u, and so sup (w, :n = 1,2, . . .) does indeed exist because L is Dedekind ocomplete, For the proof of the last implication, assume that L has the principal projection property, and let 0 5 nu 5 u hold for n = 1,2, .. We have to prove that u = 0. It follows from 0 5 nu 5 u that inf (u, nu) = nu for n = 1,2, . . ., and it follows from the principal projection property that the element uo = sup (inf (u, nu) : n = .)
..
exists. In other words, uo = sup (nu :n = 1,2,. 2u0 = sup 2nu
=
.
. .) exists. But then
sup nu = uo,
so uo = 0. It follows that u = 0. This concludes the proof that all implications indicated by the arrows hold. The proof that no implication in the converse direction holds consists, naturally, of a list of counterexamples.
CH. 4,g 251
THE MAIN INCLUSION THEOREM
139
(i) A Dedekind complete space that is not super Dedekind complete. Let X be an uncountable point set, and L the Riesz space of all real bounded functions on X (with pointwise ordering). If D is the subset of L consisting of all f such that f (x) = 0 at all x E X except one point x f where f ( x f ) = 1 (the point xf running through all points of X ) , then sup D is the function f o satisfyingf,(x) = 1 at all x E X , and there is no countable subset of D having the same supremum. (ii) A Dedekind a-complete space without the projection property, and heme without the property of being Dedekind complete. Let L be the Riesz space of all real bounded functions f on ( x :0 5 x 5 1) such thatf(x) # f ( 0 )holds for at most countably many x, with pointwise ordering. Given that 0
5
unf
5
0
holds in L , the function sup u,, exists in L,and is simply equal to the pointwise limit of the sequence (un :n = 1,2, .). Hence, L is Dedekind a-complete. In order to show that L does not have the projection property, consider the band A in L consisting of allfE L vanishing on (x : 0 5 x 5 +), so that Ad is then the band of all f E L vanishing on ( x : 3 < x 5 1). Any f E A satisfiesf(x) # 0 for at most countably many x; the same holds for any f E Ad, and hence also for any f in A 8 Ad. It follows that A 0 Ad # L , and so L does not have the projection property. It follows from the Dedekind a-completeness that L has the principal projection property, so this example shows also that the principal projection property does not always imply the projection property. (iii) A space with the projection property that is not Dedekind a-complete, and hence not Dedekind complete. Let L be the Riesz space of all real bounded functions f on X = (1,2,. . .), assuming only a finite number of different values, with pointwise ordering. Given the arbitrary band A in L, we set
..
X,
= (x
:f ( x ) # 0 for at least one f E A )
and X , = X - X , . Then Ad consists of all f E L vanishing on XI, and so A = Addconsists of allf E L vanishing on X,. Givenf E L, the decomposition
f
=f X X l
+fxx2
(with xxl and xxI the characteristic functions of XI and X , respectively) shows that f x x , E A and fxx, E Ad, so L = A 8 Ad holds. It follows that L has the projection property. In order to show that L is not Dedekind a-complete, consider the sequence (u,, : n = 1 , 2 , . .) in L , defined by u,,(x) = x-' for x = 1,2,. . ., n and
.
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 25
u,,(x) = 0 for x = n + 1, n + 2 , . . .. Then we have u,, E L for all n and 0 5 u,,t 5 uo with uo(x) = 1 for all x, but the sequence (u,, :n = 1,2, .)
..
has no supremum in L. Finally, we observe that obviously L has the principal projection property, so this example shows also that the principal projection property does not always imply Dedekind a-completeness. (iv) A space with the principal projection property, but without the projection property and without the property of being Dedekind a-complete. Let L be the Riesz space of all real bounded functionsf on the point set X = (1,2, . . ., co) such that f ( x ) # f ( 0 0 ) holds for at most finitely many x, with pointwise ordering. Let A be an arbitrary principal band in L, generated by u EL', and write XI = ( x : f ( x ) # 0 for at least onefe A ) = (x : u ( x ) > 0).
If u(co) = 0, then X, consists only of a finite number of points, and X , = X-XI contains the point co. If u(m) > 0, then X, conbists of all but a finite number of points, and XI contains the point co. Hence, givenfc L, we have in both cases that
f =f X X l + f x x , with fxx, E A and fxx2E Ad. It follows that L = A @ Ad, and so L has the principal projection property. The same sequence as in the preceding example shows that L is not Dedekind a-complete. In order to show that L does not have the projection property, consider the band A in L defined by Then so
A = ( f : f ~ L , f ( x ) = Ofor x = 1,3,5 ,...).
Ad = ( f : f ~ L , f ( x=) 0 for x
=
2 , 4 , 6 , . . .),
A @ A d = ( f EL,f ( m ) = 0 ) # L.
(v) An Archimedean space without the principal projection property. We shall, actually, present two different spaces of this kind, differing with respect to the number of principal projection bands in the space. The first space is C([O,11). It is evident that the space is Archimedean, and it will be proved that (0) and the space itself are the only projection bands in the space. This implies then immediately that the space does not have the principal projection property. In order to show that (0) and the space itself are the only projection bands, consider first an arbitrary band A # (0) in C([O,11). The set 0 = ( x : f ( x ) # 0 for at least one f E A)
CH.4.5 251
THE MAIN INCLUSION THEOREM
141
is an open non-empty subset of [0, 11 and, denoting the closure of 0 by X,, we have Ad = (f: f E C([O, l]),f(x) = 0 on 0) =
( f : f eC([O,lI),.f(x)
=
0 on Xl).
If X , = [0, 11, then Ad = {0}, and so A = Add = C([O,l]), which shows that in this case A is a projection band. Assume now, however, that A is a projection band different from (0)and from C( [0, 11). Then the corresponding set X , is non-empty as already observed, and X , is now also different from [0, 11. It follows that the complementary set X2 = [0, 1]-X, is nonempty, and for every point xo E X 2 there exists at least one f E Ad such that f ( x , ) # 0. This implies that A = Add = ( f : fC([O, ~ l]),f(x) = 0 on A',). Since A is a projection band, we have A 6 Ad = L, and hence the function e(x), satisfying e ( x ) = 1 for all X E [0, 11, has a decomposition e ( x ) = e,(x)+e2(x) with el E A and e2 E Ad. This implies that el(x) = 0 on X , and e2(x) = 0 on X , , so e,(x) = e(x)-e2(x) = 1 on X,. But then el(x) assumes only the values zero and one, both values on non-empty sets, and this contradicts the continuity of the function e,(x). Hence (0)and the space itself are the only projection bands in C( [0, 1I). For the second example, let X be an uncountable point set and L the Riesz space, with pointwise ordering, of all real functions f on X for which there exists a finite number f ( m ) such that, for any given E > 0, we have If(x)-f(co)l 2 E for at most finitely many x. In other words, if Xis topologized by the discrete topology, then L consists of all real continuous functions on X tending to a finite limit as x tends to infinity (this last expression to be understood in the sense that the one-point compactification of X is introduced). It is evident that everyfe L is bounded, and f ( x ) # f ( c o ) holds on an at most countable subset of X. It is also evident that L is Archimedean. We will show now that not every principal band is a projection band. To this end, let E = (x, ,x 2 , . . .) be a countably infinite subset of ,'A and define u(x) on X by u(xk)l= k-' for k = 1,2, . , . and u ( x ) = 0 outside E. Furthermore, let e(x) = 1 for all X E X .Evidently, the functions u ( x ) and e(x) are members of L'. Assuming that
w = sup (inf (e, nu) :n = 1,2, . . .)
exists in L, we have w ( x ) = 1 for all x E E, and so w(co) = 1. Since Xis uncountable and the set of all x satisfying w ( x ) # w(00) = 1 is at most count-
142
[CH.4,§ 25
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
able, there exist many points y outside E where w ( y ) = 1 holds; let y o be one of these points. We introduce the function w’E L satisfying w‘(yo) = 0 and w’(x) = w ( x ) for all x # y o . Then w’ < w (i.e., w‘ 6 w and w’ # w), and w’ is still an upper bound of the set of all elements inf (e, nu). Hence, w is not the supremum of the set of all elements inf (e, nu). This is a contradiction, and so the principal band generated by the element u is no projection band. There are, however, plenty of principal projection bands in L besides {0} and L itself. If u E L+ satisfies u(x) > 0 for only finitely many x, then the band generated by u is a projection band. In particular, if u(x) > 0 holds at one point only, then the band generated by u is a projection band. It is evident, therefore, that every nonzero band in L includes a nonzero principal projection band. We will prove now that all properties of the Riesz space L that occur in the main inclusion theorem are inherited by ideals in L. Theorem 25.2. If the Riesz space L has one of the properties in the main inclusion theorem (super Dedekind completeness, Dedekind completeness, Dedekind o-completeness, the projection properry, the principal projection property, the Archimedean property), and A is an ideal in L, then A has the same property. Proof. In order to prove that Dedekind completeness is inherited by the ideal A , we assume that L is Dedekind complete and (u, : T E {T}) is an upwards directed set such that 0 6 u,t u with u E A (and so u, E A for all T ) . The element uo = sup u, exists in L, and it follows from 0 5 uo u and u E A that uo E A. This shows that A is Dedekind complete. If L is super Dedekind complete, there exists a sequence (u,,, :n = 1,2, . .) in (u, : T E {T}) such that u,,, T uo. Since u,,, E A holds for all n, it is evident thus that A is super Dedekind complete. The proof for Dedekind o-completeness is similar. Assume now that L has the projection property, B is a band in the ideal A , and u is a positive element in A . In order to show that A has the projection property we have to prove that the set
s
s
.
V = (u:O 6 v
5 U,UEB)
has a supremum in A . It will be sufficient to prove that the set V has a supremum uo in L such rhat uo is a member of A , because then uo is obviously the supremum of V in A . Let { B } be the band in L generated by B. Since L has the projection property, the set
CH. 4,s 251
143
THE MAIN INCLUSION THEOREM
W = (w:O
s w s U, W E { B } )
has a supremum uo in L, and evidently uo is a member of A (because uo 2 11 and u E A ) . It will be sufficient, therefore, to prove that the sets V and Ware identical. If 0 5 w E { B } , there exists a directed set (ul : 1E {A}) in B such that 0 vA7 w. Hence, if we have in addition that 0 S w 6 u E A , then w and all uA are members of A , and so w = sup vA is a member of B because B is a band in A . This shows that the conditions 0 5 w E { B } and w u E A together imply that w E B, and so the sets Vand Ware indeed identical. The proof for the principal projection property is similar; note that if B is the principal band in A generated by the element u E A', then { B } is the principal band in L generated by u. The proof for the Archimedean property is trivial.
s
s
To conclude this section, we derive some further properties of the Riesz space the elements of which are the finitely additive signed measures on a given a-field of subsets of a point set X. Particular attention will be paid to the subspace of all countably additive signed measures.
Example 25.3. Let r be a o-field (a-algebra) of subsets of the non-empty point set X , and let L be the Dedekind complete Riesz space of all finitely additive signed measures p on r with the property that
11PIl = SUP (IP(A)I :A E r )
is finite (cf. Example 11.2 (viii) and Example 23.3 (v), where it was assumed only that r is a field). It is well-known that if p is a countably additive signed measure on r, then the condition that llpll is finite is automatically satisfied (due not only to the countable additivity of p, but also to the hypothesis that r is a o-field, and not merely a field). It follows that the set of all countably additive signed measures is a linear subspace L, of L. Given the countably additive signed measure p, it is well-known that the positive and negative variations of p are also countably additive. In other words, if p E L , , then 'p and p- are also members of L,, and hence 1p1 = p' + p - is a member of L,. Also, it is not difficult to see that if v E L ,p E L, and 0 5 v p, then v E L , . It follows already that L, is an ideal in L. We will prove that L, is a band in L. For this purpose, assume that we have 0 5 p r t po in L with pr E L, for all z E {T}. Furthermore, assume that (En :n = 1,2, .) is a sequence of sets in r satisfying En t Eo in r. We have to show that po(En)7 po(Eo). Given E > 0, determine zo E { r } such that
s
..
0 5 (Cro-PL,,)(Eo-EJ
< 8.
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
holds for all n
=
1,2,
[CH. 4, 5 25
. . ., and we have also that pzo(Eo-En)+ 0 as n -+
00.
It follows that 0 5 po(Eo-E,) < 2.5 for n sufficientlylarge. This is the desired result. Hence, L, is a band in L. It follows from the Dedekind completeness of L that L has the projection property, and so L = L, 8 (L,)”. The elements of (L,)d are called purely finitely additive measures on r. The band L, in L may be considered as a Riesz space on its own, Dedekind complete in view of the last theorem proved. The space L, has the following properties. (i) If p, v E L, , then we have p Iv if and onIy if there exiJt disjoint sets A , B E r such that A u B = X and IvI(A) = lpl(B) = 0. (ii) I f p , v E L,, then v is an element in the band generated in L, by p ifand only if Ivl is absolutely continuous with respect to lpl, i.e., if and only ifany E E r satisfying lpl(E) = 0 also satisfies Ivl(E) = 0. For the proof of (i) we observe first that, given any Q E L , , there exist disjoint sets A, B E r such that A u B = X and o - ( A ) = a + ( B ) = 0. This is the well-known Hahn decomposition of X with respect to Q . Given p, v E L, such that p Iv, we consider the element Q = 1p1- Ivl E L,. Since Q is thus the difference of two disjoint positive elements, the decomposition Q = IplIvI is minimal in the sense of Theorem 11.10 (ii), so c+ = 1p1 and Q- = IvI. Hence, according to the theorem about the Hahn decomposition, there exist disjoint sets A, B E r such that A u B = X and I v @ ) = I,ul(B) = 0. Conversely, if there exist A, B E r satisfying A u B = X and Ivl(A) = lpl(B) = 0, then Q = inf (Ipl,Ivl) satisfies a(E) = a(E n B ) + o ( E n A )
5 Ipl(B)+Ivl(A) = 0
for every E E r, so Q = 0, i.e., p J- v. For part (ii), it follows easily from the definitions that if v is an element in the band generated by p, then IpI(E) = 0 implies Ivl(E) = 0. Conversely, let Ipl(E) = 0 imply Ivl(E) = 0, and let I v l = v l + v 2 with v1 E B,, and v2 E (BJd, where B,, is the band generated by p. Since Ipl Iv 2 , there exist sets A, B E r such that A u B = Xand v 2 ( A ) = Ipl(B) = 0. Then we have Ivl(B) = 0 by hypothesis, so v2(B) = 0 on account of 0 5 v2 5 I vI. But
CH. 4,s 261
ATOMS AND FINITEDIMENSIONAL RIESZ SPACES
145
then v2(A u B) = 0, i.e., v 2 ( X ) = 0, so v 2 is the zero element of L,. It follows that Ivl = v1 E B,, and so v E B,,. Exercise 25.4. Let L be the Riesz space of all real functions f on the uncountable point set X such that f( x ) tends to a finite limitf ( 0 0 ) as x tends to infinity (cf. Theorem 25.1, part (v)), and let A be the ideal in L consisting of allf E L satisfyingf ( 0 0 ) = 0. Show that A is order dense in L (although not super order dense), and A by itself is super Dedekind complete (although L does not even possess the principal projection property). Exercise 25.5. Let L be the Riesz space of all real bounded functions f on [0, 1] such that f ( x ) # f (0)holds for at most countably many x (cf. Theorem 25.1, part (ii)), and let A be the ideal in L consisting of all f E L satisfying f ( 0 ) = 0. Show that A is order dense in L (although not super order dense), and A by itself is super Dedekind complete. Hence, the Dedekind a-complete (but not Dedekind complete) space L contains an order dense ideal that is super Dedekind complete. This is of interest since it will be proved later (cf. Theorem 29.5) that if a Dedekind a-complete space contains a super order dense ideal that is super Dedekind complete, then the space itself must be super Dedekind complete. 26. Atoms and finitedimensional Riesz spaces We start with the definitions of an atom and a discrete element. Definition 26.1. The element f # 0 in the Riesz space L is called an atom whenever it follows from 0 5 u 5 If 1, 0 5 v 5 If I and u 1v that u = 0 or v = 0. The element f # 0 in the Riesz space L is called a discrete element whenever every element in the principal ideal generated by f is a real multiple off. Before presenting examples we prove a simple lemma. Lemma 26.2. (i) I f f is an atom or a discrete element, rhen either f > 0 or f c 0. (ii) Every discrete element is an atom, but an atom is not necessarily a discrete element.
Proof. (i) Let f be an atom. Since 0 5 f ' 5 If/, 0 6 f- 5 If1 and f i.If-,we have eitherf = 0 0r.f- = 0. Hencef > 0 or f c 0. Let f be a discrete element. Since If I is a member of the ideal generated by f, we must have If I = af for some real a. Here a # 0 on account of If1 # 0. +
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,826
Hence f = u-'lfl, which shows t h a t f > 0 or f c 0 according as u > 0 or u < 0. (ii) In order to show that the discrete elementf is an atom, we may assume without loss of generality that f > 0. Let 0 5 u 5 f, 0 S u S f and u 1u. We have to show that u = 0 or u = 0. Since f is discrete, we have u = uf and u = pf, so inf (21, u ) = yf with y = min (a, p). But inf (u, u ) = 0, so y = 0, which implies immediately that u = 0 or u = 0. In order to show that an atom is not necessarily a discrete element, consider the lexicographically ordered real plane. Every point except the origin is an atom, but unless the point is on the vertical axis it is not discrete. I f f i s an atom (or a discrete element), then af is an atom (or a discrete element) for every real a # 0. Iff is an atom (or a discrete element) and 0 S 191 S Ifl, then either g = 0 or g is an atom (or a discrete element). In the space (s) of all real sequences, the elements
(LO, 0 , . . .), (0,1,0,. . .), . . .
are discrete elements, and hence they are atoms. In n-dimensional real number space R", ordered coordinatewise, the elements (1,
o,o,. . .), (0,
1 , 0 , . . .), .
. ., (0,. . .,o,
1)
are discrete elements, and hence they are atoms. The space C([O,11) does not have any atoms. If p is a non-atomic measure in the point set X and if p is a real number satisfying 0 c p < 00, the Riesz space L,(X, p) does not have any atoms. We will prove first that any Riesz space of strictly positive finite dimension has an atom.
Theorem 26.3. I f the Riesz space L is of jinite dimension n, where n 1 1, and if0 c u E L,there exists an atom e in L such that 0 < e 2 u. Furthermore, every set of mutually disjoint atoms in L has at most n elements. Proof. We assume that L is of dimension n with 1 S n < co, and u is an element of Lf satisfying u # 0. We will prove the existence of an atom e in L satisfying 0 < e 5 u. If u is itself an atom there is nothing more to prove; if not, there exist elements u l , u2 such that 0 c u1 S u, 0 c u2 S u and u1 Iu 2 . Note that ( u l , u 2 ) is a linearly independent set in L (since disjointness implies linear independence by Theorem 14.6). If u1 or u2 is an atom there is nothing more to prove; if not, there exist elements u l l , u12, uZ1,uz2 such that 0 < u l l 5 u l , 0 < u12 2 u1 with u l l Iu12 and 0 c uZ1 S u z , 0 < uZ2 S u with uzl Iu Z 2 . Note that ( u l l , u l z , u z l , u z z ) is a linearly
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ATOMS A N D FINITEDIMENSIONAL RIESZ SPACES
independent set in L since the elements in the set are nonzero and mutually disjoint. Proceeding in this manner, the procedure breaks off after a finite number of steps since the maximal number of linearly independent elements in L is n. Hence, there exists an atom e in L such that 0 < e S u. Any set of mutually disjoint nonzero elements in L is linearly independent, and hence contains at most n elements. In particular, therefore, any set of mutually disjoint atoms in L contains at most n elements. In an Archimedean Riesz space we can say more about atoms. In particular, atoms and discrete elements are now the same. Theorem 26.4. In an Archimedean Riesz space L the following holds. (i) I f f is an atom in L and 0 S u S If 1, then u = af,for some real a. (ii) I f f and g are atoms in L, then either f Ig or f = ag for some real a # 0. (iii) I f f is an atom in L, then the principal band B, generated by f consists of all real multiples o f f , and B, is a projection band. Note that B, is equal, therefore, to the principal ideal generated by f, andf is a discrete element.
Proof. (i) Without loss of generality we may assume that f > 0. Assume that 0 5 u 6 f.If u = 0, then u is a real multiple off, so we may assume now that 0 < u S f. The set of numbers (p : pu S f)is non-empty and bounded from above since L is Archimedean (indeed, if this set were unbounded, we would have 0 < nu 5 f for n = 1, 2, . . ., which is impossible in an Archimedean space). Let a = sup (p : pu 5 f ) . Then 1 5
c1
< co and
au
S A the last inequality holds since p f
pu 7 au in L in view of the fact that
a implies
L is Archimedean. We will prove now thatf = mu. Indeed, otherwise we have u = f - au > 0, and then, on account of ( u - E u ) ' ~ u for E 4 0, there exists a number E such that 0 < E < a and (u-EU)' > 0. It follows that 0 < (U-&u)+ = (f-(a+E)u)+
f+ =f,
and so, since a 2 1, we have surely that 0 < (f-(.+E)u)+
S 2aJ
(1)
We next note that 0 < (f-(a+E)u)-, since otherwise (a+&)u S J against the definition of a. Hence 0 < (f-(a+E)u)- = ((a+E)u-f)+
5
( ~ + E ) u 2au
5 2af.
(2)
148
[CH.4,s 26
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
But (1) and (2) contradict each other since the element 2uf is an atom. It follows that f = uu, where u satisfies 1 2 a < 00 as observed above. Hence u= which is the desired result. (ii) Assuming that f and g are atoms, we set If u = 0, then f 1g. If u > 0, then the result in part (i) shows that f = u1 u a n d g = u2uwitha, # O a n d ~ ~ # O , a n d s o f = u , u ~ ~ g . (iii) By part (i) every element in the ideal generated by the atom f is a real multiple off. Now assume (as we may) that the atom f is positive, and let v E L f be an element in the band B, generated byf. Since B, is the same as the a-ideal generated byf (cf. Theorem 24.7 (ii)), there exists an increasing sequence (u, : n = 1, 2,. .) of positive elements in the ideal generated by fsuch that u, t v. Every u, is of the form u, = a, f (a, real, IX, 2 0), and so a, f u, which implies that u, is increasing (i.e., non-decreasing) as n increases. It is impossible that u, 00, because this would imply 0 < nf 6 u for n = 1,2, . . ., which is impossible in an Archimedean space. It follows that u, a. for some finite uo, and so a, f t a, f (once again because L is Archimedean). Combining now that u, f t u and a, f a o f , we obtain u = uof , so every positive element in B, is a real multiple off. The same is true, therefore, of every element in B,. For the proof that Bf is a projection band, it is sufficient to show that
.
sup (inf (u, n f ) :n exists for every u EL'. For n = 1,2,
=
1,2,.
. .)
. . . we have inf (u, nf)
E
Bf, so
inf (u, n f ) = u, f for an appropriate increasing sequence (a, : n = 1,2, . . .) of real numbers. Since u, f 5 u holds for all n, it is impossible that u, t co. Hence u, t uo < CO, and so inf (u, nf) = anf t .of, where we have used again that L is Archimedean. This shows that the desired supremum exists. The band B, generated by an atom (or even by a discrete element) in a non-Archimedean space is not necessarily a projection band. By way of example, if L is the lexicographically ordered plane, then f = (0, 1) is a discrete element, but the corresponding band Bf is the vertical axis, and this is no projection band.
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149
There are some remarkable relations between the presence of atoms in a Riesz space and the presence of maximal bands. The Theorems 26.6 and 26.7 that follow are essentially due to S. Yamamuro ([l], 1967); in the original formulation of Theorem 26.7 the space L is assumed to be Dedekind complete. The present more general formulation is essentially due to F. van Schagen. We first introduce maximal ideals and maximal bands.
Definition 26.5. The ideal A in the Riesz space L is called a maximal ideal i f A # L, and if there is no ideal in L properly between A andL (i.e., any ideal Al such that A c Al c L holds satisjies either A , = A or A , = L ) . The band B in the RieJz space L is called a maximal band if B # L , and i f there is no band in L properly between B and L. The band B in the Riesz space L is called a hypermaximal band i f B is a maximal ideal besideJ being a band. It follows immediately from these definitions that any band B is hypermaximal if and only if B is simultaneously a maximal band and a maximal ideal.
Theorem 26.6. I f f is an element of the Riesz space L such that fhe disjoint complement ( B f ) dof the band Bf generated by f is a maximal band, theri f is an atom. Proof. If (B,y is a maximal band, we have immediately that f # 0. In order to prove that f is an atom, it is sufficient to show that If I is an atom. If If1 is no atom, there exist elements u and v in L" such that 0 < u S If I, 0 < u 6 If I and u 1u. Then B,, c Bf,so (BJ)dc (Bu)d,and the last inclusion is proper since v is a member of (B,,)dbut not of (Bf)d.We have also (Bu)d# L because u is no member of (B,)d. The band (B,,)dis, therefore, properly between (Bf)dand L, which contradicts the hypothesis that ( B f ) d is a maximal band. It follows thatfis an atom.
The converse of the statement in the last theorem is not generally true. In the lexicographically ordered plane every elementf # 0 is an atom, but the disjoint complement ( B f y satisfies ( B f y = {0}, so ( B f y is no maximal band. In an Archimedean space, however, the situation is much nicer, as shown by the following theorem.
Theorem 26.1. Let L be an Archimedean Riesz space. (i) The following conditionsfor a band A in L are equivalent. (a) A is a hypermaximal band. (b) A is a maximal band. (c) There exists an atom f in L such that A = (B,)d.
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[CH.4, !j 26
(ii) The intersection of all maximal bands is the zeroelement if and only i f the zeroelement is the only element in L disjoint from all atoms. (iii) L has no maximal bands i f and only i f L has no atoms.
Proof. (i) It is evident that (a) implies (b). We will prove now that (b) implies (c), so we assume that A is a maximal band. Then Ad # (0) since Ad = (0) would imply that A = Add = L. If Ad contains elements u > 0 and u > 0 such that u Iv , then the band B generated by A and u is properly between A and L, which is impossible. Hence, such elements u and v do not exist. It follows that every elementf # 0 in Ad is an atom, and by the same argument it follows that Ad cannot contain disjoint atoms. Hence, by Theorem 26.4 (ii), there exists an atom f such that Ad = B f , and so A = Add = ( B fId* For the proof that (c) implies (a) we assume that f is an atom, and we have to prove that A = (Bf)dis a hypermaximal band. For this purpose, assume that C is an ideal satisfying C 3 (Bf)' with proper inclusion. It will be sufficient to prove that C = L. Since the inclusion is proper, there exists an element g contained in C but not in ( B f ) d Hence, . since Bf is a projection band by Theorem 26.4, we have g = g1+g2 with 0 # g1 E Bf and g 2 E ( B f ) d . Then g1 = c1ffor some real number c1 # 0. It follows that
f = aM1gl = a-1(g-g2), and so f E C on account of g E C and g 2 E c C. But f E C implies Bf c C, and since we have also (by hypothesis) that (By)' c C, the final result is that Bf 0 ( B f ) dc C, i.e., C = L. This shows that ( B f ) dis a hypermaximal band. (ii) It follows from what has just been proved that the intersection of all maximal bands is the set of all elements that are disjoint from all atoms. Hence, the intersection of all maximal bands is the zeroelement if and only if the zeroelement is the only element in L disjoint from all atoms. (iii) Evident from part (i). It was already observed that, in a non-Archimedean space, even though f is an atom, the band A = ( B f ) dis not necessarily maximal. The converse may also fail to hold. Indeed, if L is again the lexicographically ordered plane, then the vertical axis is a maximal band A in L (even a hypermaximal band), but A is not of the form A = (Bf)"for any$ S. Yamamuro, in his paper referred to above, uses a definition of an atom different from ours. In his terminology an elementf # 0 is called an atom if it follows from f = fi + f 2 with fl 1fithat fi = 0 or f i = 0. We shall call
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ATOMS A N D FINITEDIMENSIONAL RIESZ SPACES
any f possessing this property an indecomposable element. Obviously, an atom is indecomposable but the converse does not hold. The space C([O,11) has no atoms, but it has plenty of indecomposable elements. We will prove now that, in any Riesz space with the principal projection property, atoms and indecomposable elements are the same.
Theorem 26.8. In any Riesz space L with the principal projection property atoms and indecomposable elements are the same. Proof. It is sufficient to prove that iff is no atom, then f is not indecomposable. We assume, therefore, that f is no atom. Then there exist elements u and u in L+ such that 0 c u S If 1, 0 < u 5 If I and u 1u. The band B, generated by u is a projection band by hypothesis; let f = f l+f2 withf, E B, and fi E (B,)d. Then f l # 0 because otherwise f = f2 E (BJd, so f iu, which contradicts 0 < u 5 Ifl. We have also f z # 0, because otherwise f = f, E B,, so If I E B, and hence u E B, on account of 0 c u S 1f 1, but this contradicts u IB, . It follows from f l # 0 andfz # 0 that f is not indecomposable. We now turn our attention to Archimedean Riesz spaces of finite dimension.
Theorem 26.9. If ( e l , . . ., en) is a set of mutually disjoint atoms in the Archimedean Riesz space L with the property that there exiJtJ no nonzero element in L disjointfrom ( e l , . . ., en),then L is n-dimensional and ( e l , . ..,en) is a baJis of L in the algebraic sense. The algebraic decomposition of any f E L as a sum of real multiples of the basis elements is exactly the decomposition of f as a hum of components off in the bands generated by the bahis elements. Proof. Since ( e l , . . ., en) is a linearly independent system, the dimension of L is at least n. The bands B , , . . ., B,, generated by e l , . . ., en are projection bands by Theorem 26.4; given u E L + ,let ccl e l , . ., u,en be the components of u in B, , . . ., B,,. Then
.
and so
0 5 uiei 2
zi
for i
=
1 , . . ., n,
. . . +a,,e,, = sup (cc,el, . . ., a,,e,,)
0 6 cc,e,+
u,
where we have used that the sum and the supremum of a finite number of disjoint positive elements are the same. Thus
w
and so 0
=
s w s u-cciei
u-(cc,e,+
for i = I , .
. . . +cc,,e,,) 2 0,
. ., n. But u-aiei
is the component of
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,8 26
the band BP; it follows that w E BP, i.e., w 1e, for i = 1, 2, . . ., n. Hence w = 0 in view of our hypotheses, so u = a, el + . . . + a, en. The desired results follow immediately. u in
Theorem 26.10. the Archimedean Riesz space L has the properly that an-v system of mutually disjoint nonzero elements is jinite, then L is of jinite dimension, say of dimension n, and there exists a basis ( e l , .,en)of mutually disjoint atoms.
..
Proof. Assuming that L does not consist exclusively of the zeroelement, we will prove first that L contains an atom. The proof is similar to the proof of Theorem 26.3. Indeed, L contains an element u > 0, and if u is an atom we are ready. If not, there exist ul, u2 E Li such that 0 < u1 u, 0 < u2 5 u and u1 Iu2. If one of u1 or u2 is an atom we are ready; if not, we proceed and obtain nonzero and mutually disjoint elements ull , u12, u21, u 2 2 . The procedure breaks off after a finite number of steps, since by hypothesis there exists no infinite disjoint system of nonzero elements. Hence, L contains an atom e , . Let B , be the corresponding principal projection band. If B: = {0}, the proof is complete. If B: # {0), it is proved similarly that B; contains an atom e 2 , and so on. This procedure again breaks off after a finite number of steps, and the desired result follows then from the preceding theorem.
It was proved in Theorem 18.5 that a one-one positive linear mapping 71 of the Riesz space L onto the Riesz space M is a Riesz isomorphism if and only if the inverse mapping n-' is also positive. This will be used to prove now that, up to Riesz isomorphism, there is only one n-dimensional Archimedean Riesz space, namely, n-dimensional real number space R" with the usual coordinatewise ordering. The following proof is a variation upon the proof due to A. I. Judin ([l], 1939). Theorem 26.11. Any Archimedean Riesz space L ofJinite dimension n, where n 2 1, is Riesz isomorphic to n-dimensional real number space R" with the usual coordinatewise ordering. Proof. L satisfies the hypotheses of the preceding theorem, so since it is given in advance that the dimension of L is n, there exists a basis (el . . .,en) of positive and mutually disjoint atoms. Everyf E L satisfiesf = PI el . . . +/?,en,where the terms on the right are the components off in the bands generated by e l , . . ., en respectively. We have f 3 0 if and only if all components o f f are positive, i.e., if and only if pi 2 0 for i = 1,. . ., n. It follows that the one-one linear mapping
+
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MAXIMAL IDEALS
x : P I el
+ - - +Pnen *
+
(PI
9
*
-
Pn)
of L onto R" is positive, and the inverse x-' is also positive. Hence, n is a Riesz isomorphism of L onto R". Exercise 26.12. Let L be an Archimedean Riesz space of finite dimension n, where n 2 1. (i) Show that if (el , ., en)and (e; ,. . ., ei) are sets of mutually disjoint atoms in L, then each e; is a nonzero real multiple of one of the e j , and conversely. In other words, except for appropriate real nonzero constants, the second set is a permutation of the first set. (ii) Show that every ideal B # (0) in L is a direct sum of one or several of the bands generated by e l , ., en respectively. In the converse direction, show that every direct sum of this kind is a projection band.
..
..
27. Maximal ideals In the preceding section maximal bands in a Riesz space were investigated, and it was shown that, at least in an Archimedean space, there exists at least one maximal band if and only if there is an atom in the space. Spaces without atoms, therefore, do not possess maximal bands. The present section will be devoted to a more detailed study of maximal ideals. Loosely speaking, there are of course more maximal ideals than maximal bands, but in spite of that there exist Riesz spaces (even Dedekind complete ones) having no maximal ideals. We recall (cf. Definition 26.5) that the ideal J i n the Riesz space L is called maximal if J is a proper ideal (i.e., J # L) and if there is no ideal properly between J and L. We will also say that the ideal A in L is a non-trivial ideal if (0) # A # L. First of all we characterize now the Riesz spaces having no non-trivial ideals. Theorem 27.1. Let L be a Riesz space, L # (01,such that L has no nontrivial ideals. Then L is Archimedean, every u > 0 in Li is an atom, and the ideal generated by this atom u is already the whole space L, i.e., every f E L is of the form f = au for some real a. Conversely, if there exists an element u > 0 such that every f E L is a real multiple of u, then L # (0)and L has no non-trivial ideals. Proof. If L were non-Archimedean, there would exist elements v and w in L+ such that 0 < nu 4 w holds for n = 1,2,. , and so the principal
.,
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DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
[CH. 4,8 27
ideal generated by v would be non-trivial. This is impossible by hypothesis, so L is Archimedean. Now take an arbitrary element u > 0 in L'. In order to show that u is an atom, assume that 0 5 v 5 u, 0 5 w 5 u and v Iw. We have to prove that v = 0 or w = 0. The ideals A, and A , generated by v and w respectively are disjoint and hence, since L has no non-trivial ideals, one at least of A , and A , must be equal to {0}, say A , = (0). But then v = 0. This is the de;ired result, showing that u is an atom. Furthermore, u > 0 implies A , # {0}, so A , = L. It follows that everyf E L is a real multiple of u (cf. Theorem 26.4 (iii)).
Theorem 21.2. Let A and B be ideals in the Riesz space L such that A c B c L. For any f E L the element of LIA containing f will be denoted by I f ] , exactly as in section 18, so [f ] = [ g ]i f and only i f f -g E A. The elements I f ] for which f E B holds form now an ideal [B]in LIA, and [B] is non-trivial if and only i f B is properly between A and L. Conversely, i f B* is an ideal in LIA, then the set of all f satisfying L f ] E B* is an ideal B in L such that A c B c L and [B] = B*. Proof. Assume that A and B are ideals in L such that A c B c L. First of all, for the above definition of [B],note that iffl, f2E I f ] and one of these is a member of B, sayfiE B, then also f2E B. Indeed, we havef i -f l E A c B, so f i E B impliesf 2 E B. We prove that [B]is an ideal in L/A. It is evident that [B] is a linear subspace of L/A, so it remains to prove that [f]E [B]and I[gIl 5 I[f]limplies [ g ]E [B].For the proof, observe that l I f ] l = [If I] and I[g]l = [Igl], so I[g]l 5 I[f]l i$ equivalent to [lgl] 6 [Ifl].It follows that 191 S If I + q for some q E A (cf. Lemma 18.8), and so we have 191 E B on account of If 1 + q E B. But then we also have g E B, and hence [ g ]E [B]. This is the desired result, showing that [B]is an ideal in L/A. All of the other statements are evident. Before presenting the next theorem, we recall that the positive element e in the Riesz space L is called a strong unit in L if the principal ideal generated by e is the whole space L , i.e., if every f E L is majorized in absolute value by an appropriate multiple of e (cf. Definition 21.4).
Theorem 21.3. (i) For any maximal ideal J in the Riesz space L the space L / J has nonzero elements, but LIJ has no non-trivial ideals. Hence if u E L + is not a member of J, then the element [u]in LIJsatisJies [ u ] > [O], andevery element of LIJ is a real multiple of [u]. (ii) I f L has a strong unit e > 0 and J is a maximal ideal in L , then LIJ
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MAXIMAL IDEALS
155
consists of all real multiples of [el. In other words, given f E L, there exists a real number 0: such that Lf] = a[e], i.e., f - ae E J. We will denote this number ci by "f(J). The mapping
:f
".I
Af(J)
of L onto the real numbers is a Riesz homomorphism such that the image "e(J) of the strong unit e satisfies "e(J) = 1. Conversely, if n is a Riesz homomorphism of L onto the real numbers such that n(e) = 1, then there exists a maximal ideal J in L such that n is exacfly the mapping KJ
:f + "f(J).
(iii) I f L has a strong unit e > 0 and J , , J z are diflerent maximal ideals in L, and if a and b are given real numbers, there exists an elementf E L such that " f ( J 1 ) = a and " f ( J z ) = b. Proof. (i) Follows immediately from the two preceding theorems. (ii) The first statement in part (ii) follows from the result in part (i) by observing that the strong unit e is no member of any maximal ideal J. We need only prove, therefore, the last statement. Let n be a Riesz homomorphism of L onto the real numbers such that n(e) = 1 , and let J be the kernel of 71. By Theorem 18.3 (iii) the kernel J is an ideal, and the quotient space L / J is now Riesz isomorphic to the space of real numbers. The space of real numbers has no non-trivial ideals, so the same holds for L / J . This shows, by Theorem 27.2 above, that J is maximal. Finally, writing n( f ) = a and "f ( J ) = p for an arbitrary f E L , we have f - a e E J a n d f - p e E J, so c( = p. This shows that the mappings A and nJ :f + " f ( J ) coincide. (iii) It follows from Ji # Jz that one of these, say J 1 , contains an element g such that g is no member of Jz.Then "g(J,) = 0 and " g ( J Z )# 0. We may assume without loss of generality that " g ( J z ) = 1. It follows thatf = ae +(b-a)g satisfies " f ( J 1 ) = a and " f ( J z ) = b.
So far, we have not yet proved that there really exist maximal ideals. We will do so now for a Riesz space having a strong unit.
Theorem 27.4. I f the Riesz space L has a strong unit e > 0, then every proper ideal in L is included in a maximal ideal. It follows that an element f E L is contained in no maximal ideal at all if and only if the ideal generated by f is the space L itseif(i.e., ifand only if1f I is a strong unit in L ) . Proof. Assume that A is a proper ideal in L, and let (A, : r E { r } ) be the set of all proper ideals A, satisfying A, =I A ; this set is partially ordered by
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH. 4 , # 27
inclusion. Given the chain ( A , :u E {c}) in the set (A, : z E {z}), the set A , = U , A , is an ideal in L . The ideal A , is proper since e is no member of any A,, and so e is no member of A , . It follows that A , is the supremum of the chain with respect to the partial ordering in the set of all A,. Hence, Zorn’s lemma can be applied, so there exists a maximal ideal J satisfying J I A. It is evident that anyf E L such that the ideal A , generated by f is equal to L is contained in no maximal ideal at all. If, however, A, is a proper ideal, then A, is included in a maximal ideal J, and sof E J. It is hardly necessary to draw attention to the similarity between the present results and those for maximal ideals (in the ring theoretic sense) in a commutative ring R with unit element. Every proper ideal in R is included in a maximal ideal, and the element f E R is contained in no maximal ideal at all if and only iff has an inverse element, i.e., if and only if the ideal generated by f is the ring R itself. The proof is similar to the proof of Theorem 27.4. In some cases, in particular if R is a complex commutative Banach algebra with unit element, the quotient space R/Jfor any maximal ideal J i s isomorphic to complex number space, and now the number “ f ( J ) may be defined exactly as in Theorem 27.3 (ii). The mapping nJ :f + “f(J) of R onto the complex numbers is now a multiplicative linear functional such that nJ(e) = 1. Conversely, if n is a nonzero multiplicative linear functional, then n(e) = 1, the kernel of n is a maximal ideal J, and n is exactly the functional nJ. Furthermore, the intersection of all maximal ideals (called the radical of R) consists of all elementsf€ R satisfying Ilf”ll”” + 0 as n + a.The radical, therefore, consists of all elements which, in a certain sense, are very small. Something similar happens in the Riesz space case, as we will prove now. The following theorem is due to K. Yosida and M. Fukamiya ([l], 1941). The present proof is somewhat different from the original proof. Theorem 21.5. If L is a Riesz space with a strong unit e > 0, if I, is the ideal consisting of the zeroelement and all infinitely small elements in L (cf. section 22), and (J, : u E {u]) is the set of all maximal ideals in L , then I, = J,. The intersection of all maximal ideals is sometimes called the radical. In a Riesz space with a strong unit, therefore, the radical is equal to the ideal of all infinitely small elements.
0,
Proof. Given the infinitely small element f E L , there exists an element g E L+ such that nlf 1 5 g holds for n = 1,2, . . .. Since g 5 noe for some natural number no, it follows now from nn,( f I S g 5 noe for n = 1 , 2 , . . . that nlfl 5 e holds for n = 1 , 2 , . . ..
CH.4.8 271
157
MAXIMAL IDEALS
n,
In order to show that I, c J, holds it is sufficient to prove that every infinitely small element f is a member of every maximal ideal J. Since nlfl 6 e holds for n = 1,2, . .., we have n"lfl(J) 6 "e(J) = 1
for n = 1,2,.
. .,
"Ifl(J) = 0. This implies that If I E J, so f E J. For the inverse inclusion J, c I. we have to prove that for any nonzero non-infinitely smallf E L there exists a maximal ideal J o f whichf is no mernber. Note first that, given the non-infinitely small element f, there exists a natural number no such that nolfl 6 e fails to hold. Under these circumstances, if nolf I 5 e holds, then 1f I (and hence f ) is contained in no proper ideal at all, and hencef is contained in no maximal ideal. It is also possible, however, that neither nolf I 6 e nor nolf I 2 e holds. In this case we have (noI f I -e)' > 0 and (noIf I -e)- > 0, and so the principal ideal A generated by p = (no If1 -e)- is now nontrivial. Let J be a maximal ideal satisfying J =I A. Then p E A c J, so "p(J) = 0. We have also that so
so
0,
P = (noIfl-e)- = (e-fioIfl)+
2 e-noIfl,
0 = "p(J) 2 "e(J)-no "Ifl(J) = 1-no
"If
I(J).
But then no "If I(J) 5 1, so "1 f l(J) > 0, which implies that If1 is no member of J. It follows that f is no member of J. This completes the proof. The next theorem is an immediate corollary,
Theorem 27.6. In an Archimedean Riesz space L with a strong unit the intersection of all maximal ideals consists of the zeroelement only. It follows that for every element f # 0 in L there is a maximal ideal J (depending upon f ) such that "f( J ) # 0. In other words, " f ( J ) = 0 holdsfor all maximal ideals J if and only i f f = 0. Once more, assume that L is a Riesz space with a strong unit e > 0. For each f E L and each maximal ideal J i n L, let the real number " f ( J ) be defined as indicated in Theorem 27.3 (ii). For f fixed and J variable, we thus obtain a real Ifunction "f on the set $T of all maximal ideals. Doing this for everyf, we obtain a set "L of real functions ^f on $, and the mapping f + " f of L onto "L is a Riesz homomorphism (since the mapping f -+ "f (J) for J fixed is a Riesz homomorphism of L onto the real numbers). It is understood here that the partial ordering in "L is the ordinary pointwise ordering. The strong unit e is mapped onto the function "e satisfying "e(J)
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DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
[CH.4,s 27
= 1 for all J. According to the last theorem the mapping f + "f is a Riesz isomorphism if and only if L is Archimedean. An Archimedean Riesz space with a strong unit, therefore, is Riesz isomorphic to a Riesz space the elements of which are real functions "f on a point set $. It will be proved later that $ has a natural topology (the hull-kernel topology) such that $ is Hausdorff and compact and every "f is continuous with respect to this topology. The analogy with the Gelfand representation theory for a commutative Banach algebra with unit element is thus seen to be almost complete.
Example 27.7. Let X be a compact Hausdorff space and C ( X ) the Riesz space of all real continuous functions on X . The space C ( X ) is Archimedean and the function e, satisfying e ( x ) = 1 for all x E X,is a strong unit. We will determine all maximal ideals J in C ( X ) and all functions "f on the set f of the maximal ideals. Given the point xo E X , the subset of all f E C ( X ) vanishing at xo is evidently a maximal ideal in C(X).We will prove now that every maximal ideal J in C ( X ) is of this kind. Indeed, suppose there is no point of X at which all functions of J vanish. Given x, E X , there exists then a function f,, E J such that fx,(xo) # 0. It follows that lfxol E J a n d Ifxo(xo)l> 0, so lfxo(x)l >0 holds for all x in an open neighborhood N,, of x,. A finite number of such neighborhoods, say N,, ,. . ., NXp,covers X , so = SUP ( l f x , l ,
-
-7
lL,l> E J
and u ( x ) > 0 for every x E X . It follows that the function u-l is continuous on X,so there exists a constant M > 0 such that
0 < (.(x)}-'
sM
holds for all x . But then we have u(x) 2 M-' for all x , which shows that the constant M - l is an element of the ideal J. This, however, implies that everyfin C ( X )is already in J , which is impossible. The final result is that there must exist a point xo E X such that f ( x o ) = 0 holds for all f E J, and so J consists of all f~ C(X)satisfying f ( x o ) = 0. Now, let Jobe the maximal ideal corresponding thus to the point xo E X , and letfe C(X).The number a = "f(Jo)is uniquely determined by the condition that f - ue E J,, i.e.,
f ( x o ) - ue(xo) = 0. This shows that u = f ( x o ) , i.e,. "f(Jo)= f ( x , ) . Hence, identifying evzry
CH. 4, 8 271
159
MAXIMAL IDEALS
point x,, E X with the corresponding maximal ideal J o , the Riesz isomorphic spaces C ( X ) and ^{C(X)} can be regarded as identical.
Example 27.8. We present an example of a Dedekind complete Riesz space possessing a weak unit, but possessing no maximal ideals. Let ( X , A , p ) be a measure space such that 0 < p ( X ) < co and p has no atoms, and let L, = L , ( X , p ) be the corresponding space of all real p-summable functions (with identification of p-almost equal functions). As explained in Example 11.2 (vi), the space L , is a Riesz space with respect to the partial ordering according to which f 5 g means by definition that f(x) I;g(x) holds for p-almost every x E X . The space L is super Dedekind complete by Example 23.3 (iv), and the function e satisfying e ( x ) = 1 for all x E X is a weak unit. We will prove now that L , has no maximal ideals. Indeed, assume that J is a maximal ideal in L For anyf E L , let lf] be the corresponding element o f L , / J ,and let n be the mapping n :f+ If]of L, onto LJJ. Furthermore, let u E L1 be such that u is no member of J. Then [u] > [O], and so according to Thecrem 27.3 (i) every element Lf] of L,/J is a real multiple of [u]. Denoting this multiple by cp(f), we have
,
,
+
.(f)
=
{cp(f)1[4
for all feL,. Obviously, cp(f) is a positive linear functional on L 1 , and since n is a Riesz homomorphism of L onto L , / J we have
,
,.
.
for allf, g E L , In particular, we have cp(lf1) = Iq(f)l for everyfE L First of all we will prove now that the functional cp is normbounded on L Indeed, if not, there exists for every n = 1,2,. a function 0 S & E L1 such that Ilf,l = 1 and cp(f,) 2 n3. In this case, letf(x) be the pointwise sum of the series C?n-’&(x) of non-negative functions on X , and denote the partial sums C:n-Y,(x) by s,(x). The sequence (s, :k = 1, 2, .) is evidently a Cauchy sequence in L , , so sk converges in norm to a function g E L .It follows that some subsequenceconvergespointwise palmost everywhere to g. On the other hand every subsequence converges pointwise to .f Hencef = g holds p-almost everywhere; this shows thatfis not only the pointwise limit but also the norm limit of the sequence (sk :k = 1,2,. .). Thenfis a member of L ,and so q ( f ) must be a finite non-negative number. Butf 2 n-’&, so cp(f) 2 n-’cp(f,,) 2 n for n = 1,2,. . .. This is impossible, and hence cp is normbounded.
,.
..
..
,
,
.
160
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,@28
Since cp is normbounded, there exists now a function g EL, (X, p) such that holds for allfe L 1 . In addition, since cp is positive, we must have g(x) 2 0 almost everywhere on X. The functional cp is not the zerofunctional,and so the set E = ( x :g ( x ) > 0 ) is of positive measure. Let El and E, be disjoint subsets of E such that El u E, = E and p ( E l ) = &?,) = +p(E); sets El and E, of this kind exist since p has no atoms. The characteristic functions x1 and x2 of El and E2 satisfy now cp(X1)
and
=/
cp{suP (xl
EI
’0,
x2)>
=dXE) =
gdP
9
d X 2 )
/
E
=
f ’ E2
SdP
0
g d p = dx1>+(P(X2)m
This contradicts formula (I), since the sum of two strictly positive numbers cannot be equal to the largest of these numbers. It follows that L possesses no maximal ideals.
Exercise27.9. Let L = L (X, p) be the same space as in the last example, and let R be the space of real numbers. The Cartesian product L 1 x R is a Riesz space if the algebraic operations and the ordering are taken coordinatewise. Show that L x R has only one maximal ideal, and deterrnine this ideal. 28. Order bases and disjoint systems The present section and the next section are devoted to several aspects of the finer structure of some Riesz spaces.
Definition 28.1. If A is an ideal in the Riesz space L and (h: Q E {o}) is a set of elemevts in A , then this set is called an order basis of A (or a quasi order basis of A ) whenever the ideal generated by (f,: (i E {.>) is order dense (or quasi order dense) in A .
If A is a principal ideal or principal band generated by the elementfE L, then the set consisting of the single elementfis an order basis of A. Any order basis of the ideal A is a quasi order basis of A, and the ideal A, considered as a Riesz space by itself, is Archimedean if and only if every quasi order basis of A is an order basis of A.
CH.4,5 281
161
ORDER BASES AND DISJOINT SYSTEMS
Any quasi order basis of A is also a quasi order basis of Add,and hence of { A } (where, as before, { A } denotes the band generated by A ) . Any order basis of A is also an order basis of { A } . The set (f,:a E {a}) is a (quasi) order basis of the ideal A if and only if the set (If,] :a E {a}) is so. In this case, the finite suprema of the elements If,l i.e., the elements sup (I f,I . ., l f J ) for n running through the natural numbers and al,. . ., a, through the index set {a), form an upwards directed set(u, :z E {z}) in L', and this upwards directed set is also a (quasi) order basis of A . Note that the ideal generated by the set (u, :z E {z)) consists of all f E L satisfying If I 5 cu, for an appropriate non-negative number c = cr and an appropriate z = zJ. If the ideal A has a countable order basis (f.: n = 1 , 2 , . . .), then the set ( I j J : n = 1,2, . .) is also an order basis of A , and so
.
.
(u, :u, = sup
..., lf.l),
(Ifl[,
.
n = L 2 , . .)
is an order basis of A satisfying 0 v,,t. Similarly for a quasi order basis. Iffl,. ..,f N is a finite order basis of the ideal A , then (vn
: un = SUP
(If119
* *
-9
If.I),
n = 192, *
- -
-
-9
N)
is an order basis of A satisfying 0 I; vl 5 * 5 0,. Setting now = trN for all natural numbers n > N , the set (v, :n = 1,2, .) is an order basis of A satisfying 0 5 unt. By this simple device it will be possible to treat the finite and countable cases simultaneously in some of the theorems to follow.
..
ti,
In Theorem 24.7 (ii) it was proved that if u and v are elements of Lf such that u is a member of the principal band generated by v, then u
=
sup (inf (u, nu) :n
=
1,2,. . .).
Furthermore, it was shown that every principal a-ideal is already a band. These results will now be extended to bands having a finite or countable order basis.
..
Theorem 28.2. If thejnite or countable set of elements (u, : n = 1, 2, .) in the Riesz space L satisfies 0 5 v,t, and if A is the band generated by this set (i.e., the set of all u, is an order basis of A ) , then every u 2 0 in A satisfies u = sup (inf (u, nu,) :myn
=
1,2, . . .) = sup (inf (u, nu,,) :n = 1,2,. . .).
Hence, the a-ideal generated by an at most countable set is already equal to the band generated by the set. In other words, every a-ideal with afinite or countable order basis is a band.
162
DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
[CH.4, !j 28
Proof. Denote by A,,, the ideal generated by the set (u,, : n = 1,2, . . .). Then A is the band generated by the ideal A,,, . Any u 2 0 in the band A is the supremum of the set of all w E A,,, satisfying 0 5 w 5 u. In other words, we have u = sup W, where W = ( W : W E A , , )0, 5 ~ 5 ~ ) . We set
W‘ = (inf (u, nu,) :m, n = 1,2, . . .), W“ = (inf (u, nu,,) : n = 1, 2 , . . .).
It is evident that W“ c W‘ c W, so in order to show that u is also the supremum of W’ and W”, it will be sufficient to show that any upper bound of W ” is also an upper bound of W . Given w E W, we have w E A,,,, so there exist a natural number k = k , and an index m = m, such that 0 5 w 5 ku,. Hence for n = max ( k , m ) , 0 5 w 5 nu, which implies that
0 5 w 5 inf (u, nu,,),
where we have used that w 6 nu,, as well as w 5 u. It follows that any upper bound of W ’ is an upper bound of every w E W, and so is an upper bound of W. It was proved in Theorem 25.1 (the main inclusion theorem) that if the Riesz space L is either Dedekind a-complete or L has the projedon property, then every principal band is a projection band. This result will now be extended to bands having a finite or countable order basis. Theorem 28.3. Let the Riesz space L be either Dedekind a-complete or let L have the projection property. Furthermore, let the set (v,, :n = 1,2, .) of elements in L satisfy 0 5; u,f, and let A be the band generated by thiJ set. Then A is a projection band, and for any u E L’ the component PAu in A is given by
..
PAu = sup (inf (u, nu,) :m, n =
. .) =
.
sup (inf (u, nu,,):n = 1,2,. .). Proof. Exactly as in the preceding proof, let A,,, be the ideal generated by the set (v,, : n = 1,2, . .) and, given u E L + , let W, W’ and W” be defined by
.
CH.4,s 281
163
ORDER BASES AND DISJOINT SYSTEMS
W W'
s
: w E A,,,, 0 w u), = (inf (u, nu,,) : m,n = 1, 2 , . = (W
W" = (inf (u, nu,) : n
= 1,
2 , . . .).
. .),
The proof that W, W' and W" have the same set of upper bounds is exactly as above. We also define the set W , by
w, = ( w : w E A , O 6 w
5 u).
Since W t W , ,any upper bound of W, is an upper bound of W. Conversely, since any element of W , is the supremum of a set of elements in W, any upper bound of W is an upper bound of W , . Hence, the sets W , W ' , W" and W, have the same upper bounds. Assume now that L has the projection property. Then the band A is a projection band, and so PAu = sup W, holds by Theorem 24.5 (i). It follows that PAM = sup w' = sup w", which is the desired result. Next, assume that L is Dedekind a-complete. Then sup W" exists, and so sup W, exists. But, once again by Theorem 24.5 (i), this implies that A is a projection band and PAu = sup W , Hence also
.
PAU
= sup
w' = sup w".
Definition 28.4. (i) Every set ( f , : a E {a}) of nonzero mutually disjoint elements of the Riesz space L is called a disjoint system in L. (ii) The disjoint Jystem ( f , : a E {a})in L is said to be maximal i f (f, : a E {o}) is a quasi order basis of L (i.e., f If,for all a E {a} impliesf = 0). (iii) If the disjoint system (f,: a E {a}) is at the same time an order basis of L , then (f,: a E {a}) is calleda disjoint order basis o f L . These notions are related, as follows.
Theorem 28.5. Every Riesz space L , containing at least one nonzero element, has a maximal disjoi1.t system. Every disjoint order basis is a maximal disjoint system, in an Archimedean Riesz space every maximal disjoint system is a disjoint order basis. Proof. Assume that L contains at least one nonzero element. The set of all disjoint systems in L is then non-empty and partially ordered by inclusion, and obviously every chain in this partially ordered set has a supremum. Hence, by Zorn's lemma, the set has a maximal element, i.e., L has a maximal disjoint system.
164
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 29
The other statements follow by noting that order denseness implies quasi order denseness, and that in an Archimedean space these notions are identical. Theorem 28.6. Let L be a Riesz space with the principalprojection property, and let (0, : n = 1,2, . .) be afinite or countable order basis of L , where we may assume without loss of generality that 0 5 v,,T holh. Then L has a finite or countable disjoint order basis (w, : n = 1,2, .) such that, for every n, there exist basis elements w l , . ., w, with m 5 n and generating the same band as u, .
.
.
..
Proof. For n = 2,3, . . ., let v; be the component of on in the band generated by 0,Set u1 = u1 and u,, = v,-vL for n = 2,3,. . .. The system (u, : n = 1,2,. . .) is an order basis, the elements of which are mutually disjoint, such that for every n the elements u l , .. ., u,, generate the same band as v,. Some of the elements u,,, however, may be thezeroelement. Omitting the zeroelements and renumbering, we obtain the desired disjoint order basis (w, :n = 1, 2,. . .).
Exercise 28.7. If L does not have the principal projection property, the conclusion in Theorem 28.6 does not always hold. To show this, let L be the space of all real functions f on the uncountable point set X such that f ( x ) tends to a finite limit f (00) as x tends to infinity (cf. part (v) in Theorem 25.1), and let E = (xl, x 2 , . . .) be a countably infinite subset of X . Define u l ( x ) by vl(xk)= k-’ f o r k = 1, 2, . . . and vl(x) = 0 for all x outside E, and define v z ( x ) = 1 for all x E X . Then ( v l , v 2 ) is an order basis of L such that 0 =< v1 5 v 2 . Show that there exists no disjoint order basis ( w l ,w 2 ) such that w1 generates the same band as vl. 29. Bands or ideals with a countable order basis and the relation with order separability If the Riesz space L has a countable order basis, it is a natural question to ask under which conditions any given band or ideal in L will also possess a countable order basis. It is not very surprising that the projection property is a sufficient condition in order that every band in L has now also a countable order basis. The condition is by no means necessary for this purpos=, however, and also the condition is not sufficient to imply that every ideal in L has a countable order basis. It is perhaps a little more surprising that for an Archimedean space L with a countable order basis there exists a simple
CH.4,8291
BANDS OR IDEALS WITH A COUNTABLE ORDER BASIS
165
necessary and sufficient condition in order that every ideal in L will also have a countable order basis. The condition is that L shall be order separable (cf. Definition 23.1 (iii)). Since order separability and the projection property are independent, it follows that each of these properties is sufficient, but not necessary, in order that every band in L shall have a countable order basis.
Theorem 29.1. (i) I f the Riesz space L has a finite or countable order basis, then every projection band in L has afinite or countable order basis. (ii) IfL has the projection property, then every band included in a band with afinite or countable order basis has also afinite or countable order basis. (iii) I f L has the projection property and L has afinite or countable order basis, then every band in L has a finite or countable order basis.
Proof. (i) Let (u,, : n = 1,2,. ..) be an order basis in L , where we may assume that 0 5 u,t, and let A be a projection band in L . For every n we set u, = v,,+ w,, with v,, E A and w,,E A", and we will prove that (u,, :n = 1, 2, . . .) is an order basis of A . Every z E L + satisfies
.
z = sup (inf (z, nu,,) : n = 1,2,. .) = sup (inf (2,nv,+nw,,) : n = 1,2,
. . .)
by Theorem 28.2. In the particular case that z is a member of A + = A n L ', we have z Iw,, for all n, and so (cf. Theorem 12.5) inf (z, nu,,+nw,,) = inf (z, nu,,) for all n, which implies z = sup (inf (z, nu,,) : n = 1,2,. . .). This shows that every z E A + is an element of the band generated by (v, : n = 1,2,. . .), so A is included in this band. Conversely, the band generated by (v, :n = 1,2,. .) is evidently included in A . It follows that A is equal to the band generated by (v,, :n = 1,2,. . .). In other words, (v,, :n = 1,2, .) is an order basis of A . The statements in (ii) and (iii) follow immediately.
..
.
As a preliminary result to the main theorem in this section, we have the following lemma.
Lemma 29.2. I f the Riesz space L has the property that every ideal included in a principal ideal has an at most countable quasi order basis, then L has also the property that for any directed set (ur :z E {z}) satisfying 0 S u, t u and for any number E satisfying 0 < E < 1 there exists an increasing sequence (urn : n = 1, 2, . . .) in (uI : z E {z}) such that (urn-E U ) - 10. It follows then that
166
[CH.4,5 29
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
0
and
s u-urn 6 (l-E)u+(uTn-Eu)inf (urn,E U ) t
with (urn-Eu)- 1 0 EU.
Proof. Before beginning with the proof, note that in an Archimedean space the hypothesis is that every ideal included in a principal ideal has an at most countable order basis, and that the conclusion of the lemma comes very near to order separability. For the proof, assume that we have 0 g U, u and 0 c E < 1. We set u, = (u,-Eu)+
for all z E {z},
and we denote by A the ideal generated by the set (u, : z E (7)). Since 0 5 2 u, 6 u holds for all z, we have A c A , , where A , denotes the principal ideal generated by u. We conclude from 0 5 u, t (1 - E)U that u is a member of the band generated by (0, :z E {z}), so the ideal A , is included in this band. In other words, A is order dense in A , . By hypothesis, the ideal A has an at most countable quasi order basis, i.e., there exist elements z,, ( n = 1,2, . .) in A such that the ideal generated by the set (z,, :n = 1,2,. .) is quasi order dense in A , and hence quasi order dense in A , . This means that if we have w E A , and w I z, for all n, then w = 0. Without loss of generality we may assume that 0 5 z,?. We have z,, E A for any given natural number n, and hence there exists a number c,, 2 0 and an index z, E { r } such that z,, 5 c,,uTn.We use here, of course, that the system (0, :z E is upwards directed. Keeping in mind now that urn = (u,,-Eu)+, it is evident that we may assume without loss of generality that the sequence (uTn:n = 1,2,. .), and hence the sequence (u,,, : n = 1,2, . .), is increasing. It follows that wTn= (u,--Eu)- is decreasing, and from we,,I urnand z, 5 c,,u,,, we infer that wen Iz,, holds for n = 1 , 2 , . .. It follows easily that wTn10. Indeed, assume that 0 6 w 5 wen holds for all n. Then w is a member of A , (since all w,, are members of A,), and w Iz,, holds for all n (since 0 S w 6 wTmand wTnIz,), so w = 0 by what was observed above. This shows that wTn10, i.e., u,
.
.
{TI)
.
.
.
(uTn-EU)-
lo,
and thereby concludes the first part of the proof. It follows immediately that 0
6
U-u,"
= (l-E)U+(EU-u,,)
5 (I-E)U+(EU-U~,,)+
= (1-e)u
+ (uTn-&u)-.
CH.4,8 291
BANDS OR IDEALS WITH A COUNTABLE ORDER BASIS
167
Furthermore, since (u,,-Eu)- = -inf ( U ~ , , - E U , 0) = -{inf it follows from ( u r n - ~ u ) -5.0 that
inf (urn,E U )
(U,,,EU)-EEU}
= Eu-inf (u~,,, EU),
t EU.
The last lemma would not be very interesting or important, were it not for the fact that it plays a crucial role in the proof of the following main theorem (cf. W. A. J. Luxemburg [2], 1968). For Dedekind complete spaces it was proved already by A. G . Pinsker (cf. the book by L. V. Kantorovitch, B. Z. Vulikh and A. G. Pinsker [l], 1950) that property (vi) of the theorem implies property (i), i.e., if in the Riesz space L every bounded disjoint system of positive elements is at most countable, then L is order separable. Dedekind complete Riesz spaces with the property that every bounded disjoint system of positive elements is at most countable, are called K-spaces of countable type in the Soviet literature. It follows from the main theorem below that a space of this kind has all the properties listed in the theorem. In our terminlogy, therefore, the space is a super Dedekind complete Riesz space.
Theorem 29.3. In an Archimedean Riesz space L the following conditions are equivalent. (i) L is order separable (i.e., any set in L having a supremum contains an at most countable subset having the same supremum). (ii) Any non-empty set D in L which is bounded from above contain., an at most countable subJet having the same upper bounds as D. (iii) Every o-ideal in L is a band. (iv) Every order dense ideal in L is super order dense. (v) For any diJjoint system ( f , : o E {o}) in L and any g E L we have inf (Igl, If,l) # 0 for at most countably many (r E (vi) Every disjoint system of positive elements which is boundedfrom above is at most countable. (vii) Every ideal included in an ideal with a countable order basis has an at most countable order basis. (viii) Every ideal included in a principal ideal has an at most countable order basis. In particular, i f L is Archimedean and has an at most countable order basis, a necessary andsuficient condition in order that every ideal in L shall have an at most countable order basis is that L is order separable.
{GI.
168
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 29
Proof. (i) e (ii) This was proved in Theorem 23.5. (i) 3 (iii) * (iv) This was proved in Theorem 23.7. (iv) * (v) Given the disjoint system (f,: cr E { u } ) in L and the element g EL,denote for every cr separately by A , the ideal generated by f,,and let A be the ideal generated by the system (f,: cr E {cr}). Note that every f E A is a finite sum f = C k f k with fk E Auk. Since L is Archimedean, the ideal A 0 Ad is order dense, and so Igl = SUP (U : u E A
6 Ad, 0 5 u
IS[).
In view of the hypothesis that every order dense ideal is super order dense, there exists a sequene (u,, : n = 1,2,. . .) in A 6 Ad such that 0 5 u,, 1Igl. Now, every element h EA 0 Ad is a finite sum h = hk + h’ with hk E A , and h’ E Ad, and hence we have h If,for all but a finite number of thef,(these exceptional cr depending upon h, of course). In particular, every u,, satisfies ti, If, for all but a finite number of the f,.But then each f,,except for an at most countable number of exceptions, satisfiesf, Iu,, for all n simultaneously, and so f, I191 on account of 0 6 u,, 1191. In other words, inf (Igl,)[,fI # 0 holds only for at most countably many cr E {cr}. (v) => (vi) Let (u, :cr E {c}) be a disjoint system of positive elements in L such that u E L+ is an upper bound of the system. Then u, = inf (u, u,) holds for every cr. By the hypothesis (v) we have inf (u, u,) # 0 for at most countably many cr. Hence u, # 0 holds for at most countably many cr. On the other hand, since (u, :cr E {o}) is a disjoint system, we have u, # 0 for every c E {cr}. It follows that the system (u, :cr E {o})is at most countable. (vi) (vii) Let B be an ideal with a countable order basis (f,: n = 1, 2, . . .) and A an ideal included in B. For the present purpose we may assume that A # (0). By Theorem 28.5 A has a maximal disjoint system (y :z ~ { z } ) ; we may assume that all u, are positive. For any fixed n, consider all u, satisfying u, = inf (Ifnl, u,) > 0. The system (0, : u, > 0) is a disjoint system of positive elements bounded from above by ,1I.f so the system is at most countable by hypothesis. This argument can be repeated for every n separately and so, except for an at most countable number of the u,, each u, is disjoint from allf, simultaneously. But u, If,for all n implies that u, I B; on the other hand we have u, E A c B, and so u, = 0 which is impossible since u, is a member of a disjoint system. It follows that the system (u, : z E {z}) is an at most countable system (u,, : n = 1, 2, . . .). Since A is Archimedean, every maximal disjoint system in A is an order basis of A . Hence, (un : n = 1, 2, . . .) is an at most countable order basis of A . (vii) 3 (viii) Evident.
xk
CH.4,s 291
BANDS OR IDEALS WITH A COUNTABLE ORDER BASIS
169
(viii) * (i) In order to show that L is order separable, it is sufficient to prove that if 0 5 u, t u holds in L , then there exists a sequence (uTn:IZ = 1,2,. . .) in (u, :z E (7)) such that urnf u. Assume, therefore, that 0 5 u, f u holds in L . In view of hypothesis (viii) we can apply the preceding Lemma 29.2. According to the lemma, there exist for every fixed natural number m sequences (T,,,,,:n = 1,2, . . .) c ( 5 ) and 0 5 v,, tsL+(n = 1,2,. . .) such that 0
5 u-u,,,
5 m-'u+v,,,, with
It is sufficient for our purpose to prove that
v,,
.
sup (uTmn :myn = 1,2,. .) = u.
1,O.
(1)
(2)
To this end, assume that
5 u,,, 5 v u holds for all m, n. In view of u-v 5 U - U , , ~ we infer then from formula (1) that u - v 5 m-'u holds for all m = 1,2,. . .. Since L is Archimedean, 0
this implies that u-v = 0. Hence u = u, so formula (2) holds. This completes the proof. We present some consequences of the last theorem. Theorem 29.4. If the Riesz space L is Archimedean, and A is a super order dense ideal in L which is order separable, then L itself is order separable.
Proof. In view of the preceding theorem it will be sufficient to prove that every order dense ideal in L is super order dense in L . For this purpose, let B be an order dense ideal in L . Then A n B is an order dense ideal in the given ideal A . By hypothesis A is order separable, so it follows from the preceding theorem that A n B is super order dense in A . But A is super order dense in L by hypothesis, so A n B is super order dense in L by Theorem 21.2 (iii). It is obvious that B is then also super order dense in L . This is the desired result. In Theorem 23.6 it was shown that any Dedekind o-complete space which is order separable is super Dedekind complete. The final theorem in the present section generalizes this result. Theorem 29.5. If the Dedekind a-complete RieAz space L ha3 a super order dense ideal that is order separable, then L is super Dedekind complete. Proof. In view of Theorem 23.6, referred to above, it is sufficient to show
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DEDEKINDCOMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 29
that L is order separable. This follows from the present hypotheses by means of the preceding Theorem 29.4. We recall (cf. Exercise 25.5) that a Dedekind a-complete space L may contain an order dense and order separablr. ideal without L being super Dedekind complete. It is interesting to compare this with the situation that K is a Riesz subspacc of L such that every f E L satisfies as well as
f = s u P ( g : g E K , g Sf)
(3)
zf).
(4)
f = inf(h:hEK,h
It will be proved in Theorem 32.9 that under these conditions order separability of K implies order separability of L.If K is an ideal in L the condition (3) states that K is order dense. The extra condition (4), however, holding for an ideal K implies that K = L,so the theorem stated above is trivial if Kis an ideal. For a Riesz subspace Kformula (4)makes sense also if K Z L , and (3) and (4) together guarantee now that order separability of K implies order separability of L . At the beginning of the present section it was observed that order separability and the projection property are independent, and that in an Archimedean Riesz space with a countable order basis each of these conditions is sufficient, but not necessary, in order that every band in L shall have a countable order basis. This is illustrated by the examples in the following exercises. Exercise 29.6. Show that every ideal in C([O,11) has an at most countable order basis. As a consequence, C([0, 11) is order separable. This is an example, therefore, of a space with a countable order basis and being order separable, but failing to have the projection property. Exercise 29.7. Let L be the Riesz space of all real bounded functionsf on the uncountable point set X , with pointwise ordering. Show that L is an example of a space with a countable order basis and having the projection property (because L is Dedekind complete), but failing to be order separable. It follows from the results proved in this section that there exists an ideal without a countable order basis. Show, explicitly, that the ideal A of allfe L that vanish except on a finite subset of X does not have a countable order basis. Show that the ideal A , considered as a Riesz space by itself, is an example of an order separable space without a countable order basis.
CH.4 , § 301
THE BOOLEAN ALGEBRA OF ALL PROJECTION BANDS
171
30. The Boolean algebra of all projection bands
If A is a projection band in the Riesz space L , the order projection on A will be denoted by PA (i.e., forf E L , iff = fl+f2 with fl E A and f 2 E Ad, then fi = PAf; the same notation was used in Theorem 24.5). The set of all projection bands in L will be denoted by B (L ).The set B ( L )is partially ordered by inclusion. The mapping A -+ PA of B ( L ) onto the set of all order projections on projection bands is one-one, and if a partial ordering in this set of order projections is introduced by defining that PA 2 PB holds if and only if A c B holds in B ( L ) , then the one-one mapping A c,PA is order preserving in both directions, and thus the set B ( L ) and the set of all order projections on projection bands are order isomorphic. Any projection band A satisfies A = Add (cf. Theorem 24.2), so A is a disjoint complement. The set B ( L ) is, therefore, a subset of the Boolean algebra 37 of all disjoint complements, with the ordering in P ( L ) inherited from the ordering in &’. Since (0) and L are projection bands, the smallest element (0) and the largest element L of &’ are at the same time the smallest and largest elements of B(L). If A is a projection band, then so is Ad. In an Archimedean space the converse holds as well, i.e., if A is a band such that Ad is a projection band, then A is a projection band (on account of A = Addholding now for every band). For any two elements A and B of &’ we have that A n B = inf ( A , B ) i n 37 and ( A + B)dd = sup ( A , B ) in (cf. the proof of Theorem 22.7). It will be shown first that if A and B are elements of B ( L ) , then these formulas continue to hold in B(L). Theorem 30.1. (i) If A and B are projection bands, then A n B is a projection band, and so A n B = inf ( A , B ) holds in P(L).The corresponding projections satisfy
and f o r any u E L’ we have PAnBu= inf (PAu,PBu). Furthermore, we have PA 2 PB(i.e., A c B ) if and only i f P A u for every u E L +.
PBu holds
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH. 4,6 30
(ii) If A and B are projection bands, then the algebraic sum A + B is a projection band, and so A + B = SUP ( A , B ) holds in B(L). The corresponding projections satisfy P A + B= P A + P B - P A ~ B , and for any u E Li we have PA + B u = SUP (PAU,PBu). Hence, P A + B= PA+ P B holds i f and only if A n B ={O), i.e., i f and only i f A l B. (iii) If A and B are projection bands such that A c B, then PB-PA is the projection on the projection band Ad n B. In other words, B is the direct sum of the projection bands A and Ad n B. Proof. (i) For any UEL' the element P B u is a member of B. Hence, since 0 5 PAPBu5 PBu, the element PAPBuis also a member of B. Since it is obvious that PAPBuis also a member of A , we have already that PAPBu E A n B. In order to show that A n B is a projection band satisfying PAnB = PAPBit is sufficient, therefore, to prove that ( U - P A P B U )(~A n B).
For this purpose it is sufficient to prove that if 0 5; v 5; u-PAPBu
and v~ A n B,
(1)
then v = 0. Given an element v satisfying (l), we have O so
v =
0
PAU
2v
5 PAU-P~PBU
=
PAU-PAPBU
= PBV 5; PBu-P;u = P,u-P,u
5 u-PBu, = 0,
which shows that indeed v = 0. It has been proved thus that A n B is a projection band and PAnB= PAPB.By symmetry we have also that PAnB = PEP,. Assume now that the projection bands A and B are such that PA 5 P B holds. This means that A is included in B, so A = A n B. It follows that PA = PAnB = PAPB, which implies that PAu = PAPBu5 PBu holds for every u E L '. Conversely, let PAu 5 PBu hold for every u E L', and consider in particular an element u satisfying 0 u E Bd. Then PBu = 0, SO
s
CH. 4.5 301
173
THE BOOLEAN ALGEBRA OF ALL PROJECTION BANDS
P A u = 0, i.e., u E Ad. This shows that Bd is included in Ad, and hence A = Add c Bdd = B, which is equivalent to PA5 PB. Assume now again that A and Bare arbitrary projection bands, let u E Li, and denote inf(PAu, P,u) by wo. Then P A n B u 6 w o by what was observed above. On the other hand, since wo E A n B and 0 <= wo (= u, and since PAnBu
= sup (w : 0
w
5
u, W E A n B),
we have wo 5 P A n B u . Hence P A n B u= wo. (ii) Observe first that, for projection bands D, E and F, we have F = inf(D, E) if and only if = sup(@, Ed). Assume now that A and B are projection bands. By part (i) the formula Ad n Bd = inf(Ad,Bd) holds in B ( L ) ,and the order projection on Ad n Bd is P A d n B d = (pL-pA)(pL-pB) = PL-(PA+PB-PAnB)* Hence, by the remark above, sup(A, B) exists in 9 ( L ) and is equal to (Ad n Bd)d. The corresponding order projection is P , + P B - P A , B . It remains only to show that (Ad n Bd)d= A + B. This is not difficult. The band C = (Ad n BdYcontains Add = A as well as Bdd= B, and so contains A + B. On the other hand, we have Pcf = PJ+PJ-P,,JEA+B
+
for every f E L (since each of the three terms is a member of A B), and so C = (Ad n BdY c A + B . For any u E Li we have P A +B u
= P A u + P B U- P A ,
Bu
= P A fJ + P B &!
- inf(PA
u y
P B U) =
sup(pA u y P B u).
(iii) If A and B are projection bands, then Ad n B is a projection band by part (i), and the order projection on Ad n B is pB(pL-pA)
= pB-pAn
B-
Hence, if A c B holds, the order projection on Ad n B is PB-PA. The following theorem is now an easy consequence.
Theorem 30.2. The set B ( L ) of all projection band.s in the Riesz space L , partially ordered by inclusion, is a Boolean algebra. Proof. It has already been shown that B ( L ) is a lattice with (0) as smallest and L as largest element. In order to prove the distributivity of the lattice
174
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4, g 30
B ( L ) , we have to show that if A , B , , B, and C are members of B ( L ) satisfying A n B, c C and A n B, c C, then A n (SUP (Bi, B2)) c C
holds (cf. Theorem 2.3). We have to show, therefore, that A n (B, +B2)c C holds. The proof for this is exactly as in Theorem 22.6 (where it was proved that the lattice of all bands is distributive). Finally, it remains to show that every element A of B ( L ) has a complementary element in B ( L ) , i.e., ’ an element B E B ( L ) such that sup(A, B ) = L and inf(A, B ) = (0). The element B = Ad satisfies these conditions, as shown by sup(A, Ad) = A + A d
=
A @ Ad = L,
inf(A, A d ) = A n Ad = (0). In any Riesz space L with the projection property the set a ’ of all bands, the set of all disjoint complementsand the set B ( L )of all projection bands are identical. In the space C([O,I]) the only projection bands are (0) and the space itself (cf. part (v) of the proof of Theorem 25.1), and so 9 ( L )is here a proper subset of the set sd = B’. For a few more examples we refer to the exercises at the end of this section. For the further discussion of the properties of B ( L )it will be convenient to introduce a property for the Riesz space L which is intermediate between the principal projection property and the Archimedean property.
a
Definition 30.3. The Riesz space L is said to have suficiently many projections i f every nonzero band contains a nonzero projection band. Note that L has already sufficiently many projections if every nonzero principal band contains a nonzero projection band. Indeed, assume this condition to be satisfied, and let A be an arbitrary nonzero band. Then A contains an element f # 0, so the principal band Bf generated by f satisfies (0) # B, c A . By hypothesis B, contains a nonzero projection band, which is then also contained in the given band A. Theorem 30.4. (i) If L has the principal projection property, then L has sufficiently many projections. (ii) I f L has suficiently many projections, then L is Archimedean.
Proof. (i) It was observed above that L has sufficiently many projections if every nonzero principal band contains a nonzero projection band. This condition is surely satisfied if every principal band is a projection band. (ii) Assume that L has sufficiently many projections, and let u, v be ele-
CH.4,0301
175
THE BOOLEAN ALGEBRA OF ALL PROJECTION BANDS
ments in L' satisfying 0 5 v 5 n-'u for n = 1 , 2 , . . .. We have to prove that v = 0. It follows from the hypotheses that every element f in the ideal generated by v satisfies If 1 5 n-'u for n = 1 , 2 , . ., and so every elemniit g in the band B, generated by u satisfies 191 5 n-'u for n = 1,2,. .. Assume now that u # 0. By hypothesis there is then a projection band B c B, such that B # {0}, and so the component u1 of u in B exists and satisfies
.
u1
= SUP(Z
:z E B, 0
.
5 z 5 u).
But every positive z in B satisfies 0 5 z 5 u (every positive z in B even satisfies 0 6 z 5 n-'u for n = 1 , 2 , . .), so u1 = sup(z : Z E B'). It follows that
.
u1 = SUP(Z : z E B') = S U P ( ~ Z:z E B') = 2u1,
so u1 = 0. But then, since u1 = sup ( z : z E B'), we have z = 0 for every z E B', i.e., B = (0).This contradicts the previous statement that B # (0).
In order to see that the property to have sufficiently many projections is properly intermediate between the principal projection property and the Archimedean property, observe that C( [0, 1 1) is Archimedean but does not have sufficiently many projections, whereas the space of all real functions f o n the uncountable point set X such that f ( x ) tends to a finite limit f ( m ) as x tends to infinity has sufficiently many projections but does not have the principal projection property (cf. part (v) of the proof of Theorem 25.1). In the next theorem we present some results about the existence of suprema of infinite sets of elements in 9 ( L ) . Before stating the theorem we note that if (A, :z E {z}) is an upwards directed set in B ( L ) and u E L + ,then (PA,u : T E {T}) is an upwards directed set in L by Theorem 30.1 (i).
Theorem 30.5. (i) Let (A, : z E {z]) be an upwards direcred subJet o f B ( L ) such that sup(PAIu : z E {T}) exists in Lf for every u E L'. Then A = {u A,} is a projecrion band, A,f A holds in B ( L ) , and PA,u f P A u holds in L for every u E L'. We recall that { u A,} denotes the band generated by the set v A , , i.e., the smallest band which includes all bands A,. (ii) r f L has suficiently many projections and if A , f A holds in B ( L ) , then A = {u A,} and PA,u f PAu holds in L for every u E Lf.
Proof. (i) For every f
EL
we define the element Pf by
176
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,5 30
so in particular PU = sup ( P ~ , u:z E {z}) for every u E L + .We will prove first that P(u+u) = Pu+Pu holds for all u, u E L'. Indeed, the sets (PA,u : z E {z}) and (PA,u : z E (2)) are equidirected, so P,,(u+u) = P A . U + P A ~ V tPu+Pu, i.e., P(u+u) = Pu+Pu. It follows then that P(u-u) = Pu-Puif 0 5 u 5 u. Hence, iff and g are now arbitrary elements of L,we have for that and so
0 5 w =f++g+-(f+g)+ =f-+g--(f+g)Pw = P ( f ++ g + ) - P ( f + g ) + = P ( f - + g - ) - P ( f + g ) - , Pf +Pg+ - P ( f + g ) + = P f - + p g - - P ( f + g ) - . +
It follows that (Pf - P f - ) + ( P g + - P g - ) = P ( f + g ) + - P ( f + g ) - - , +
i.e., Pf+Pg = P ( f + g ) . Since, evidently, P(af) = aPf holds for any f E L and any real number a, it has been shown thus that P is a linear mapping of L into L. We proceed to derive further properties o f P. Since 0 5 PA,u 5 u holds for every z and every u EL', we have 0 5 Pu 5 u for every u E L + .We will prove now that P 2 = P. Given u E L + ,it follows from 0 5 Pu 6 u that 0 5 P2u 6 Pu. On the other hand, we have PA,# = p & ( p A , U )
5 PA,(PU) 5 p 2 u
for every z, and so Pu 5 P2u. It follows that P2u = Pu holds for every u EL', so P2 = P. Combining all results it has been proved that P is a projection (of L into L ) such that 0 5 Pu 5 u holds for every U E L ' . But then, by Theorem 24.5 (iii), there exists a projection band A such that P = PA. Since PA,u 5 PA# holds for every T and every u EL', we infer from Theorem 30.1 (i) that A , c A holds for all z, and so {u A,} is included in A . On the other hand, it follows from PAU = SUp(PAl# :
E
{z})
E
(u At},
holding for every u E L+, that A is included in {u A z } . It follows that A = { u A,}, and so A, t A holds in B ( L ) . Finally, from the definition of P in the first line of the proof it is evident that PASut PAu holds in L for every u EL'.
CH. 4,5 301
177
THE BOOLEAN ALGEBRA OF ALL PROJECTION BANDS
(ii) Assume that L has sufficiently many projections, and let A, f A hold in B ( L ) . The complementary elements in the Boolean algebra B ( L ) satisfy A: 1Ad, so taking infima with respect to A we obtain A n Af.1 A n Ad = (0). Hence, writing B, = A n Af for every r, we have B, E S ( L ) and B, 1(0) holds in B ( L ) . But then n B, = {0}, because otherwise there exists by hypothesis a projection band C # (0) in n B,, and this would imply that (0)# C c B, holds for every r, contradicting B, 1(0). Hence n B, = {0}, so that using now that n A: = (u AJd by formula (1) in Exercise 19.5, we obtain
(0) = (IB, =
n( A n A:) = A n ( (IA:)
= A n(
u A,)',
which shows that (u A J d I A, i.e., (u AJd c Ad. It follows that A = Add c (u A,)dd. On the other hand, since every A, is included in A, we have (u A,)dd c Add = A. Hence A = (u A,)dd,and since L is Archimedean, this is equivalent to A = {u A,}. In order to prove that PA,u f PAu holds for every u E L + , we observe that A is the direct sum of A, and B,, so PA-PA,= PB,.We will prove, therefore, that P,*u 5.0for u E L + .If, for some u EL', there exists an element v such that 0 6 v 5 PBIuholds for all T, then v is contained in all B, on account of PBsuE B,. Hence v E n B, = (01,which shows that v = 0. It follow^ that PBZuJ- 0. We will apply the just obtained results in the proof of the next theorem, relating certain properties of the Riesz space L to certain properties of the Boolean algebra B ( L ) . Theorem 30.6. (i) I f L is Dedekind a-complete, then so is P ( L ) . (ii) L has the projection property i f and only i f L has suficiently many projections and S ( L ) is Dedekind complete. In particular, i f L is a Riesz space with suficiently many projections, then L has the projection property if and only if B ( L )is Dedekind complete.
Proof. (i) We assume that L is Dedekind a-complete and we have to prove that every at most countable subset of B ( L ) has a supremum. To this end, let A, E 9 ( L ) for n = 1,2, . . ., and set B, = SUP(A, ,
. . ., A,,) = A l + . . . +A,,
.
for every n. Then B,, E P ( L ) for every n, and the sequence (B,, : n = 1,2, . .) is increasing and has the same upper bounds as the sequence (A, :n = 1, 2, . . .). In the proof that sup A, exists in B ( L ) we may assume, therefore, that the sequence (A, : n = 1,2, . . .) is increasing. Then PA,u t 6 u holds
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH. 4,s 30
for every u E L', and so sup PA,u exists in L since L is Dedekind a-complete. But then, by part (i) of the preceding theorem, we have A,, t A in B ( L ) with A = (u A,,}. This shows that B ( L )is Dedekind a-complete. (ii) Assume first that L has sufficiently many projections and B ( L ) is Dedekind complete (i.e., order complete). We have to show that every band in L is a projection band. Given the band A in L , let ( A , : z E {T}) be the set of all projection bands included in A . This set is directedupwards in B ( L ) . Hence, since B ( L ) is Dedekind complete, we have A , t B for some projection band B. It follows now from part (ii) of the preceding theorem that B = { u AT},and so B c A . If the inclusion is proper, then A n Bd # (0) , because A n Bd = ( 0 ) implies A IBd, so A c Bdd = B. Hence, still assuming that B is properly included in A , the nonzero band A n Bd contains a nonzero projection band C. Then C c Bd, but on the other hand C is one of the A , , so C c B. This is impossible for C # (0). It follows that B = A , and so A is a projection band. Conversely, assume that L has the projection property. Every band is now a projection band, so we have B ( L ) = d = 9Y,where &' and A? are the set of all bands and the set of all disjoint complements respectively. The set 9, however, is an order complete Boolean algebra (cf. Theorem 22.7). It follows that B ( L ) is order complete (i.e., Dedekind complete).
It may be asked whether the Boolean algebra B ( L )of all projection bands in a Riesz space L has any special properties which an arbitrary Boolean algebra need not have. The answer is negative, as shown by the following theorem. Theorem 30.7. Given any non-degenerate Boolean algebra 9 (i.e., 9 consists of at least two elements), there exists a Riesz space L # (0)such that B and B ( L ) are lattice isomorphic.
Proof. Assume first that 9 is a field A of subsets of a non-empty point set X , partially ordered by inclusion. The characteristic function of any set E E A will be denoted by x E . Let L be the Riesz space of all real step functions with respect to A , i.e., L consists of all real functionsf on Xof the form
f
P
=
n= I
c,,xE,;
c,, real, En E A
for n = 1 , . . ., p ,
(1)
where p is also variable of course; the algebraic operations and the partial ordering are pointwise. Every f E L has many different representations of the form (1); for f # 0 there is one standard representation where all c, are nonzero and mutually different, and all En are non-empty and disjoint. If
CH. 4,s 301
THE BOOLEAN ALGEBRA OF ALL PROJECTION BANDS
f # 0 has the standard representationf = C : c j E , , then I f 1 satisfies If I
Cf IcnIxE,,.
179
=
Given E E A, the set of allf E L vanishing on the set theoretic complement E' = X - E is evidently a band BE in L . The band BE is a projection band because the decomposition f = fxE+fxE' shows that L is the direct sum of BE and B i . In this way there corresponds to every set E E A a projection band BE in B ( L ) . Different sets E give different bands BE; for E the empty subset of X we have BE = {0}, and for E = X we have BE = L . The mapping E + BE of A into B(L) is, therefore, a one-one mapping, and obviously the mapping preserves infima and suprema of pairs of elements. For the proof that A and S ( L ) are lattice isomorphic it remains only to show that the mapping is onto B (L ). For this purpose, assume that A is an arbitrary projection band in L , and let 1 = p l ( x ) + p , ( x ) be the decomposition of the function identically one into p 1 E A and p 2 E Ad. By the properties of components in projection bands we have 0 5 pi(.) S 1 ( i = 1,2) and inf ( p l ( x ) ,p z ( x ) )= 0 for every x E X . It follows that p 1 and p 2 can only assume the values zero and one. so there exists a set E E A such that p 1 = xE and p 2 = x E t .Let BE be the projection band corresponding to the set E. We will prove that the given projection band A satisfies A = BE. Since p1 = xE E A and since BEis easily seen to be the ideal generated byxE,we have immediately that BE c A . For the proof that A c BE holds, assume thatf = C ~ C J E A , where f is written in standard representation. Then If I = CfIcnlxE,E A , so each term is a member of A , which implies that xEnE A for n = 1, . . .,p , and SO xE, E A . Also, since the sets En are disjoint, we have l f x E , , ( x )S 1 for every x E X . Taking components in A , it follows that C:xEn p 1 = z E , and so
1:
This shows that If I E BE. It has been proved thus that every f E A satisfies If1 E BE, and hence satisfiesf E BE,so A c BE. The final result is that A = BE. Combining everything we proved it follows that A and 9 ( L )are lattice isomorphic. If 9is an arbitrary non-degenerate Boolean algebra, there exists by Stone's representation theorem (Theorem 6.6) a field A of subsets of a non-empty point set X such that 9 and A are lattice isomorphic. Once again, let L be the Riesz space of all real step functions with respect to A. Then B ( L ) and A are lattice isomorphic, so B ( L ) and 9 are lattice isomorphic. This completes the proof.
~,
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[CH.4,s 30
Exercise 30.8. In Theorem 30.6 (ii) it was proved that if L is a Riesz space with sufficiently many projections, then L has the projection property if and only if B ( L ) is Dedekind complete. This is no longer true if L fails to have sufficiently many projections. Show that if L = C( [0, 1I), then P ( L )is Dedekind complete but L does not have the projection property. Exercise 30.9. Any set
A = ( ( x , y ) : a < x 6 b, c < y 5 d ) in real two dimensional number space R2 is called a cell. The empty set is also considered to be a cell. Show that the set of all finite unions of cells included in a fixed cell A , , partially ordered by set inclusion, is a Boolean algebra 9 such that 9 is not Dedekind a-complete. Hint: Observe that every finite union of cells can be written as a finite union of disjoint cells. This is based on the fact that if A and Al ,. . .,A , are cells, then A - A , (the set theoretic difference) is a finite union of disjoint cells (perhaps empty), and so by induction A - (ulAi) is a finite union of disjoint cells (perhaps empty). Note also that any open (in the ordinary topology) subset of A. is the supremum (in the set theoretic sense) of a sequence of elements from the Boolean algebra 9. Hence, if the open subset is, for example, the interior of a circle lying in the interior of A , , then the approximating sequence of elements from 9 has no supremum in 9. Exercise 30.10. Let L be the Riesz space of all real functionsf on the uncountable point set X such thatf(x) tends to a finite limitf(co) as x tends to infinity (cf. part (v) of the proof of Theorem 25.1). Given any subset E of X , the set of all f E L vanishing on the complement X - E is a band BE. Show that every band in L is of the form BE for some subset E of X . Show also that BE is a projection band if and only if E is finite or the complement X - E is finite. Show that the Boolean algebra B ( L ) is not Dedekind complete. Exercise 30.11. Let L be the Dedekind o-complete Riesz space of all real bounded functions f on X = ( x : 0 6 x 5 1) such that f ( x ) # f ( 0 ) holds for at most countably many x (cf. part (ii) of the proof of Theorem25.1). Given any subset E of X , the set of allfe L vanishing on the complement X - E is a band BE. Show that any band in L is of the form BE for some subset E of X . Show also that BE is a projection band if and only if E is countable or the complement X - E is countable. Show that the Boolean algebra B ( L ) i s not Dedekind complete.
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31. Principal projection bands
The element f of the Riesz space L is called a projection element if the principal band generated by f is a projection band. This principal band will then be denoted by Bf = Blfl, and the set of all principal projection bands by P p ( L ) .The order projection on the principal projection band Bf = Blf1 will be denoted by Pf = P I , I .It is evident that P P ( L )is a subset of g ( L ) and that (0)is a member of g p ( L ) , but L is not necessarily a member of .%(L)It follows easily from Theorem 30.1 that if u, v E Lf satisfy Bu,Bu E P p ( L ) , then suP(Bu, B,) = Bu+Bu = Bu+u = Bsup(u,u) E @P(L)' inf(Bu, Bu)= Bu n Bu = B i n f ( u , v ) E g P ( L ) , and so P P ( L )ordered , by inclusion, is a sublattice of B ( L ) .Note that and
P i n f (u, u )
Psup(u,u) =
=
Pup, = Pu Pu
Pu+u = P u + P u - P i n f ( u , u ) ,
so Pu+u= Pu+P, holds if and only if u l v. I f f € L has the property thatf' andf- are projection elements, thenfis a projection element, and we have We recall that the subset 4 of the Boolean algebra 23 is called an ideal in 23 if sup(b, b') E 4 holds for all b, b' E f and if inf(b, b,) E Y holds for all b E 3,b, E 9 (cf. the last paragraph of section 2). The ideal 9 is said to be order dense if ( 4 )= 52, where ( 4 )is the band generated by 4 (cf. the paragraph in section 4 preceding Theorem 4.8). By Theorem 4.9 we have (4)= 4dd in a Boolean algebra, so 9 is order dense whenever 9dd = 9.According to Theorem 4.6, this is equivalent to saying that f is order dense if, for any bl # 8 in 9 (where 8 is the smallest element of g ) , there exists an element b2 # 8 in f such that b2 5 b , . Theorem 31.1. (i) q P ( L )is an ideal irr g ( L ) , and g P ( L )is order dense in @(L) i f and only i f every nonzero projection band contains a nonzero principal projection band. In particular, g p ( L )iJ order dense in @ ( L ) if L has the principal projection property. (ii) g p ( L )= B ( L ) holds if a i d only if L itself is a principal band (i.e., if and only if L has a weak unit).
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Proof. (i) It was observed already that if B,, and B, are members of Bp(L), then sup(B,,, B,) is also a member of P,(L). Now, let A be a projection band and u E L+ a projection element such that A c B,,. For the proof that .",(I,) is an ideal, we still have to show that A is then a principal band. We will prove that A is the principal band generated by u, = PAu. For this purpose, note that PA = PAPuon account of A c B,,, so for any UEL' we have P Av = PAP,u = PA[sup (inf (u, nu)) : n = 1,2, . . .] = sup ,,[PA(inf (u, nu))] = sup ,,(inf (PAu, nP, u)) = sup ,,(inf (P,u, n u l ) ) = sup ,,(inf (P,u+(P,-P,)u,
nu1))
= sup ,,(inf ( 0 , nu,)),
where we have used first that 0 5 W, t w implies PAw, f PAw, secondly that PA is a Riesz homomorphism, and thirdly that inf ( w l + w z ,w 3 )= inf ( w l , w3) holds for w 1, w 2 ,w j E L' if w 2 Iw 3 . The result obtained shows that, for any v EL', the element sup (inf (u, nu1) :n
=
.
1 , 2 , . .)
exists, so the band B,,, generated by u1 is a projection band by Theorem 24.7 (i). But then we have sup(inf(v,nu,):n= 1 , 2,...) = P , , , u , and so PAu = P,,, u holds for every u E L'. It follows that A = B,,, , which is the desired result. The other statements in (i) are now evident. (ii) If the space L itself is a principal band, then L is a member of P p ( L ) . Since.P,(L) is an ideal in P(L),it follows that every member of B ( L )is also a member of S,(L), i.e., P p ( L )= P(L).In other words, every projection band is now a principal projection band. Conversely, if P p ( L )= B ( L ) holds, then L is a member of B p ( L ) ,i.e., L is a principal band. The following facts are easy consequences.
Corollary 31.2. (i) If the elements u, u of L+ are projection element3 such that B, c B,, , then the component u1 of u in B, satisfies B,,, = B,. In particular, if L has the principal projection property and if 0 6 u 5 u, then the component ul of u in B, satisfies B,,, = B,.
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(ii) I f L has suficiently many projections and if vo and (v, : t E ( 5 ) ) are projection elements in L' such that B,, t B,, holds in B ( L ) , then the components uo and (u, : t E {z}) of any u E L + in B,, and in (Bur: t E {z}) satisfy 0 =< u, t ug. (iii) If u and (u, : t E {z}) are projection elements such that 0 S u, t u holds in L,then B,,= B,, holds in B ( L ) .
Proof. (i) This follows immediately from the fact, demonstrated in the proof of the last theorem, that if A is a projection band included in the principal projection band B,,, then the component u1 of u in A satisfies B,,, = A . (ii) In Theorem 30.5 (ii) it was shown that if L has sufficiently many projections, and if A, A holds in Y ( L ) ,then PA,u 7 PAu holds in L for every u E L+. In the present case, therefore, we have 0 S u, t u,, for every u E L'. (iii) We need only prove that B,, = sup Bur holds in B ( L ) . Evidently the band {u BUz}is included in B,,. Conversely, it follows from 0 5 u, t u that M E {u BUT},and so B. c {u Bur}. This shows that B,, = {u Bur}, and hence B,, = sup B,,=holds in S ( L ) . Note that B,, = sup B,,=holds even in the larger set d of all bands. As observed above, S J L ) is order dense in 9 ( L ) if and only if every nonzero projection band contains a nonzero principal projection band. For brevity, denote this property by Pr(0.d.). Evidently, Pr(0.d.) is weaker than the principal projection property, but unlike the property to have sufficientlymany projections the present property is not intermediate between the principal projection property and the Archimedean property. Actually, there exist non-Archimedean spaces having Pr(o.d.), as shown by the lexicographically ordered plane R 2 , where B ( L ) = Bp(L)consists only of (0) and L . Restricting ourselves to Archimedean spaces, we may ask whether Pr(0.d.) is really weaker than the principal projection property. The answer is positive, since for the space C([O,I]) we have again that B ( L ) = Y p ( L ) consists of (0) and L . It may be asked next whether there exist Archimedean spaces that do not have Pr(0.d.). The answer is again positive, as shown by the space of all real continuous functions on ( x : -00 < x < 0 0 ) having a compact carrier (i.e., for anyfin the space the closure of the set ( x : If ( x ) l > 0) is compact). In this space there are no principal projection bands except (0). It is an open question whether the property to have sufficiently many projections implies the Pr (0.d. ). In Theorem 30.6 several order completeness properties of B ( L ) were
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shown to be related to certain properties o f L . In the next theorem it is shown that, similarly, several order completeness and order denseness properties of g P ( L )are related to certain properties o f L. For simplicity we assume that L has the principal projection property, which implies, therefore, already that B p ( L )is order dense in B ( L ) .
Theorem 31.3. Let L be a Riesz space with the principal projection property. The following holds now. (i)P p (L ) is Dedekind complete if and only if every band included in a principal band is a projection band, i.e., by Theorem 24.9 (iv), if and only if L has the projection property. (ii) 2 p ( L )is Dedekind a-complete if and only if every band with a countable order basis and included in a principal band is a projection band, i.e., if and only if every band with a countable order busis is a projection band. By Theorem 28.3 Dedekind o-completeness of L is a suflcient condition for this. (iii) B p ( L )is super order dense in B ( L ) if and only i f L has a countable order basis. Super order denseness of P P ( L )in P ( L ) means that for every projection band A there exists a sequence Bun( n = 1, 2, . . .) of principal projection bands such that Bunf A holds in B ( L ) . (iv) As an immediate consequence of (i) and Theorem 30.6 (ii), we have that B ( L ) is Dedekind complete if and only i f g P ( L )is Dedekind complete. Proof. (i) Assume first that B p ( L )is Dedekind complete. Let u E L + ,denote by B,, the principal band generated by u, and let A be an arbitrary band included in B,,. We have to prove that A is a projection band. Since Bu 7
5 Bu
holds for v running through A + and since P P ( L )is Dedekind complete, the element sup(B, : u E A') exists in B p ( L ) ,say sup B, = B,. But then sup B, = B, holds as well in B ( L ) because B p ( L )is an ideal in B ( L ) . In view o f Theorem 30.5 (ii) it follows now from B, = sup B, that BW={uB,:u~Af},
i.e., B, is the band generated by the set of all v € A + . In other words, B, = A. This shows that A is a projection band. Conversely, assume that every band included in a principal band is a projection band, and let the principal projection bands (Bus: z E ( 2 ) ) and B,,
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satisfy the condition that Butc B,, holds for every z. We have to show that sup Busexists in P p ( L ) .For this purpose, consider the band A = (u Bus]. Since A is included in B,,,it follows from the present hypotheses that A is a projection band, so A = sup Busholds in B(L). Since the projection band A is included in B,, and since B p ( L )is an ideal in B (L), we have A E P P ( L ) , i.e., A = B,, for some v e L'. It follows that B,, = sup Bus holds in P p ( L ) , which is the desired result. (ii) Similarly as above for the fist statement. For the second statement, observe that if A is a band with order basis (v,,: n = 1,2, .. .) and B is any principal band, then A nB is a band included in the principal band B and A n B has the order basis (w,, :n = 1,2,. .), where w, is the component of v,, in the projection band B. (iii) Let L have a countable order basis (u,, :n = 1,2,. ..), where we may assume that 0 S v,, T holds, and let A E B ( L ) . We have to prove the existence of a sequence (u,, : n = 1,2, . . .) in L+ such that Bun A holds in P ( L ) . Setting u, = PAvnfor n = 1,2,. .., we have 0 S u,, t, and it was shown in the proof of Theorem 29.1 (i) that (u, : n = 1,2,. . .) is an order basis of A, i.e., we have A = {u B,,,). This implies evidently that Bunt A holds in B ( L ) . Conversely, assume that for every A E Y ( L ) there is a sequence (un : n = 1, 2,. . .) in Lf such that Bunt A. In particular, there is then such a sequence for A = L, so Bunt L. It follows now from Theorem 30.5 (ii) that L = {u Bun},which shows that (u,, : n = 1 , 2 , . .) is an order basis of L.
.
.
32. Dedekiod completion of an Archimedean Riesz space We begin with a definition. Definition 32.1. The Dedekind complete Riesz space K is called a Dedekind completion of the Riesz space L if the following conditions hold. (i) There exists a Riesz subspace L A of K such that L and LA are Riesz isomorphic. (ii) Every f E K satisfies
"
f
" = sup(g" : g" E LA,g" 4 f ") = inf(h" :hA
E LA,hA
2f").
Note first that any Riesz subspace of a Dedekind complete space is Archimedean, and so the Riesz space L can have a Dedekind completion only if L is Archimedean. The main result in the present section will be that any Archimedean Riesz space has indeed a Dedekind completion. Secondly,
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it is evident that if L has a Dedekind completion K, then K is uniquely determined except for an obvious Riesz isomorphism, i.e., if Kl is also a Dedekind completion of L , then K and Kl are Riesz isomorphic in such a manner that if g" E L Aand g: E L; correspond to the same element g E L , then g" and g r are corresponding elements in the isomorphism of K and Kl . Hence, if we identify L and its Riesz isomorphic image L Ain K, we can consider L as a Riesz subspace of its Dedekind completion K, i.e., L is embedded in K as a Riesz subspace. In order to obtain a somewhat more general background for what follows, we present some definitions. Given the non-empty partially ordered sets X and Y, we shall say that X and Yare order isomorphic if there exists a one-one mapping cp of X onto Y such that cp and its inverse cp-l are order preserving, i.e., such that cp(xl) 5 cp(x2) holds if and only if x1 S x2 holds. Furthermore, we shall say that the mapping I) of X into Y preserves suprema and injima if, whenever holds in X , then
xo = sup(x : x E A c X ) I)(xo) =
SUP(I)(X)
:x E A )
holds in Y, and similarly for infima. Consider now the following conditions for the one-one mapping cp of X into Y. a) The one-one mapping cp of X into Y has the property that X and its image q ( X ) are order isomorphic. (b) The one-one mapping cp of X into Y preserves suprema and infima. We shall prove that neither of these conditions implies the other. However, if X i s a lattice, then (b) implies (a), and if X and Yare lattices satisfying another special condition mentioned below, then (a) and (b) are equivalent.
Theorem 32.2. Let X and Y be non-empty partially ordered sets and let cp be a one-one mapping of X into Y. Then the following holds. (i) The condition (b).for cp does not necessarily imply the condition (a) for cp. However, i f X is a lattice, then (b) implies (a). (ii) The condition (a)for cp does rtot necessarily imply the condition (b) for cp. However, i f X and Yare lattices and ifevery y E Y satisfie& Y = SUP(cp(X) : 44x1 5 u) = inf(cp(x) : P(X) 2 Y ) , then (a) and (b) are equivalerrt. It follow,, in particular that if the Riesz space L possesses a Dedekind completion K, then suprema and injima in L are preserved in K.
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Proof. (i) Let Y be a partially ordered set in which the partial ordering is not the trivial relation of equality, i.e., there exists at least one pair y , ,y , in Y such that y1 # y, and y1 I y , . Let Xconsist of the same points as Y but with the partial ordering the trivial one, i.e., x, 5 x2 implies x1 = x, . For the mapping cp of X into Y we take the identity mapping. It is obvious that cp satisfies (b) but not (a). Now assume that X is a lattice and cp satisfies condhon (b). In order to show that cp satisfies condition (a), we need only prove that cp(xl) 5 cp(xz) implies x, 5 x,. Assume, therefore, that cp(xl) 5 cp(xz) holds. Since X is a lattice and cp preserves suprema, we have cp(x1 v 4 = cp(x1) v cp(x2) = cp(xd,
so x, v x, = x 2 on account of the mapping cp being one-one. It follows that
xt
5 x,.
(ii) Let X = C([O, l]), the Riesz space of all real continuous functions on [0, 11, and Y the Riesz space of all real functions on [0, 11. For cp we take the identity mapping of X into Y, so cp satisfies condition (a). That cp does not satisfy condition (b) is shown by the example presented in Exercise 18.14. Now assume that X and Y are lattices and assume also that cp satisfies condition (a) and the extra condition mentioned above. We need only prove that cp satisfies condition (b), i.e., we have to prove that xo = sup(x, : a E { a ] ) in X implies cp(xo) = sup cp(x,) in Y. If not, there exists y , E Y such that in Y we have yl < cp(xo) and y1 2 cp(x,) for all a. Then, in view of y , = inf(cp(x) : ~ ( x 2 ) yl), there exists an element x, E X such that cp(x,) 2 y1 2 cp(x,) holds for all a, but not cp(xl) 2 cp(xo). This is impossible, since cp(xl) 2 cp(x,) for all a implies that x1 2 x, holds for all a, so x1 2 sup x, = x,, which implies cp(xl) 2 cp(xo). This contradiction shows that cp(xo) = sup cp(x,) must hold. It will be proved now that for every non-emptj partially ordered set X there exists an order complete lattice Y and a one-one mapping cp of X into Y such that cp preserves suprema and infima. For convenience, we shall assume that X has no smallest and no largest element (if X has a smallest and (or) largest element, the same arguments hold with small modifications). Subsets of X will be denoted by A , B, . ., and points in A , B y . by a, b, . . . respectively, if it is convenient to do so. Given the subset A of X , the subset A" of X is defined by
.
A" = ( x :x
2 a for all a E A ) .
..
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We have A" = X if and only if A is empty (here it is used that X has no smallest element); the set A" is empty if and only if A is not bounded from above. Given the subset B of X , the subset B' of X is defined by
B'
=
( x :x
5 b for all b E B).
We have B' = X if and only if B is empty (here it is used that X has no largest element); the set B' is empty if and only if B is not bounded from below. For every A c X we denote (A")' by A"'. Evidently A c A"' holds for every A c X; the empty set 0 satisfies 0"' = 0 and X satisfies X"' = X . Any subset A of X satisfying A = A"' is called a cut in X . The set Y of all cuts is non-empty (because 0 and X are cuts), and Y is partially ordered by inclusion. Note that for X the set of all rational numbers with the usual ordering, the set Y is the set of all lower classes of Dedekind cuts in the rational numbers (with 0 and X extra added), i.e., apart from 0 and X the set Y is order isomorphic to the set of all real numbers. For any subset A of X the set A"' is a cut, and A"' is the smallest cut including A. In order to prove this, observe that A" c A"'" holds trivially. Also, it follows from A c A"' that A" 3 A"'". Hence A" = A"'", which implies that A"' = (A"')"'. This shows that A"' is a cut. If B is any cut including the set A, we have B 3 A, so B" c A", which implies that A"' c B"' = B. This shows that A"' is the smallest cut including A. We will prove now that the set Y of all cuts contains, besides 0 and X , elements of a special kind. First observe that since X has no smallest or largest element, X must contain at least two elements. Given xo E X , we denote the set consisting of the point xo only by (xo). Then (xo)" = ( x : x 2 xo), so (x0)"' = ( x :x 5 xo).The set (xo)u' is the smallest cut containing the point x o , and obviously (xo)"' is neither empty nor equal to X. If xo # x1, then (xo)"' # (xl)"'. Hence x + (x)"' is a one-one mapping of X into the set Y of all cuts, and (x)"' is for no x equal to either the smallest element 0 or the largest element X of the set Y. After these preliminary remarks it is now easy to prove the following theorem.
Theorem 32.3 (H. M . MacNeille [l], Secrion 11, 1937). Given the nonempty partially ordered set X containing no smallest or largest element, there exists an order complete lattice Y and a one-one mapping cp of X into Y such that cp preherves suprema and in$ma. More precisely, Y is the set of all cuts of X , partially ordered by inclusion, and the mapping cp is given by cp ( x ) = (xy'for every x E X. In addition, every cut A satisfies
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A = sup((.)”’
189
: (x)”’ c A) = inf((x)”’ : (x)”’ 3 A).
Proof. We prove first that the set Y of all cuts, partially ordered by inclusion, is order complete. For this purpose, assume that (A, : a E { a } ) is an arbitrary subset of Y; each A, is, therefore, a subset of X satisfyingA:’ = A,. The set A = (u,A,)”’ is a cut, an upper bound in Y of the set of all A,. If C is another upper bound, then A, c C holds for all a,so u , A , c C, which implies A = (UA,)”’ c C”’ = c. This shows that A = sup(A, : a E {a}). In order to show that inf(A, : a E {a}) exists in Y,we consider D = nA,. We have D’” c A,“’ = A, for every a,so D”’ c n A , = D. Since D c D’” is trivially satisfied, it follows that D”’ = D. This shows that D is a cut, and obviously D satisfies now D = inf (A, : a E {a}). It has thus been proved that every subset of Y has a supremum and an infimum, and so Y is order complete. In particular, Y is a lattice. Next, we show that the mapping cp of Xinto Y, defined by cp(x) = (x)”’, preserves suprema and infima. Let E be a subset of X such that xo = sup E exists. It was proved already that A = {u((x)y’ :x
E E)}”’
is the supremum in Y of the sets (x)”’ for x running through E. To show that cp preserves suprema, we must prove that A = (xo)”’. This follows by observing that E” = (u(x)”’ :x E E)” and xo is the smallest element in E”, so xo is the smallest element in (u(x)”’ :x E E)”. But then
A = (u(x)”’ : x E E)”’ = ( x :x 5 x o ) = (xo)u’. For the proof that cp preserves infima, assume that E is a subset of X such that x1 = inf E exists. It was proved already that B = n((x)”’ : x E E) is the infimum in Y of the sets (x)”’ for x running through E. We must prove that B = (xl)”’. Now, z E B is equivalent to z E (x)”’ for all x E E, so equivalent to z 5 x for all x E E. But this is equivalent to z 5 x1 and so to 2 E (XI)”’. Finally, we have to prove that every cut A satisfies A = sup((x)”’ : (x)”’ c A) = inf((x)”’ : (x)”’
13
A).
The first part of the formula holds trivially if A is the empty set of X, assume,
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therefore, that A is not the empty set and observe that, by the definition of a cut, xo E A implies x E A for all x S x,, In other words, we have x E A if and only if (x)”’ is included in A . It follows that
.
A
= U((X>”’:xEA)=
U((x)”’:(x)”’cA),
so A = sup((x)”’ : (x)”’ c A ) . The second part of the formula holds trivially if A = X , so we may assume that A # X. We observe that A c (x)”’ holds if and only if x is an upper bound of A . It follows that
A = A”’ =
n ((q’ : A“) = n ((xy: (x)”’
and so A = inf((x)”’ : (x)”’
=I A ) .
A),
Example 32.4. It is interesting to see what happens in the following extreme case, mentioned already in part (i) of the proof of Theorem 32.2. Let X consist of at least two points with the partial ordering in X the trivial one, i.e., x1 5 x2 implies x1 = x2. If A is the empty subset of X , then A”’ is also the empty subset; if A consists of one point, then A”’ consists of that same point, and if A consists of more than one point, then A”’ = X . Hence, the set Y of all cuts consists of all points of the set X with an extra smallest element and an extra largest element added to it, and this makes Y an order complete lattice with respect to partial ordering by inclusion (in X ) . The property that suprema and infima in Xare preserved in Y is a trivial property in this example, because there is nothing to preserve but the relation that x 5 x holds for every x. Even although under the mapping x + q ( x ) = (x)”’ of X into Y no element q ( x ) is equal to the largest element of Y, the above example shows that if we omit the largest element of Y the remaining set may no longer be a lattice. Note that in this example the remaining set is a Dedekind complete partially ordered set of a very trivial nature. A similar remark holds for the smallest element of Y. If, however, X i s a lattice and we delete the largest and smallest elements from Y, then the remaining set 2 is a Dedekind complete lattice. The Dedekind completeness is evident; it will be sufficient to prove that if A and B are cuts in X such that A # X and B # X , then sup(A, B ) # X . This follows by observing that there exists elements x1 and x2 in X such that A c (xl)”’ and B c (x2)”’, and so
sup(A, B ) c suP((xl)ulY(x2)Y’) = suP(cp(xl)Y cp(x2))
= q ( x , v x 2 ) = (XIv x2yz# X ,
where we use again that X has no largest element. By Theorem 32.2, the
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191
lattice Xcan be considered as embedded in the Dedekind complete lattice 2 such that suprema and infima are preserved. We apply this to the case that X is an Archimedean Riesz space in order to show that the corresponding set Z is now its Dedekind completion under appropriate definitions for addition and multiplication by real constants. The case that X is an Archimedean Commutative group was investigated by A. Clifford ([l], 1940), and the extension to Archimedean Riesz spaces is due to H. Nakano ([3], 1941) and A. I. Judin ([2], 1941).
Theorem 32.5. Every Archimedean Riesz space L has a Dedekind completion K. Proof. For L = (0) the theorem holds trivially; we may assume, therefore, that L # (0). Then L has no largest or smallest element, so we can apply the results proved earlier in this section. We let L play the role of X , and we denote by K the set of all cuts in L with the smallest cut 0 and the largest cut L deleted. According to the results established earlier, Kis a Dedekind complete lattice and L is embedded in K with preservation of suprema and infima, and such that every f # E K satisfies
f#
= sup(g : g E L, g
Sf")= inf(h :h E L, h zf").
It remains to prove that we can introduce addition and multiplication by real numbers in K such that these operations together with the already existing partial ordering make Kinto a Ricsz space. Of course the algebraic operations in K have to be defined so as to agree on the subspace L with the existing operations in L. The elements of K are cuts in L; for the purposes of the present proof we shall denote these cuts by A , B, ., and the elements in A , B, . . by a, b, . if convenient. Given cuts A , B E K, let
..
.
. .
P = (a+b:aEA,bEB). We prove first that P"' # L, i.e., we prove that the cut P"' is an element of K. Indeed, on account of A # L and B # L there exist elements fiE A" and f2E BUYso fi+f2E P". Sincef, +f2 is not the largest element of L (in other words, since there are plenty more elements in P"),this implies that P"' # L. Secondly, we consider the special case that cuts A and B correspond to elementsf and g of L, i.e., A = (f)'" and B = (g)UI. Then
P = (a+b:aEA,bEB)=(a+b:a5Jbbg)c(f+g)"'. Conversely, given c E (f+g)U', we have c
f+g, so c = f+ ( c - f ) with
192
DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4, $ 3 2
f E (f)"' and c-YE (g)"', which shows that c E P. It follows that P = (f+g)"'. Once these facts are established, we define the sum A B of the elements A,BEKtobethecut
+
A+B = ( a + b : a ~ A , b ~ B ) " .
It follows then already that A +B E K, and that
for allf, g EL, so the addition in K agrees on the subspace L with the addition in L. It is also evident that A + B = B+A holds for all A, B E K. We proceed to prove the associative law
(A+B)+C = A+(B+C) for all A, B, C E K. For this purpose, let V = (a+b+c: U E A , bEB, C
E
C).
It is evident that V is included in ( A + B ) + C , so V"' is also included in (A+B)+C. In order to prove the inverse inclusion, consider an element gE V". The inequality a+b+c 5 g holds for all a E A, b E B, C E C, so a + b 5 g-c. This shows that g - c E (A+B)", so s 5 g - c holds for all s e A + B and all C E C. But then s+c 5 g holds for all s e A + B and all c E C, which shows that g E {(A + B ) + C]".It has been proved thus that V" is included in {(A +B)+ C}", and so ( A + B ) + C = {(A+B)+C}"' c V"'.
+ +
+
It follows that (A B) C = V"'. By symmetry we have also that A (B+ C) = V"', and so the associative law holds. Our next purpose is to show that A, = (0)"' = (f:f5 0) is the zeroelement for the addition, i.e., A + A , = &+A
= A
holds for every A E K. Note that a E A and f E A . implies a+f equality only forf = 0. Hence,
5 a with
w = ( a + f : a E A , f E A,) satisfies W" = A", so W"' = A"' = A. But W"' = A + A o = A o + A
bij definition, and so A + A o = A o + A = A . We observe, incidentally, that the zeroelement for the addition is uniquely
CH. 4,s 321
DEDEKIND COMPLETION OF AN ARCHIMEDEAN RIESZ SPACE
193
determined, because if Ah is an element of Ksatisfying A +Ah = Ah + A = A for every A E K, then A . + A ; is equal to A . as well as to Ah , so Ah = A o . Before proving now that every A E K has an 'inverse', we show that if A is a n element of K, then B = (b : - b E A")is also an element of K. Indeed, we have c ~ B " o c b2 f o r a l l b E B o - c s - b forall -bEAU o - c 5 f for all f E A " e
--cEA"' =
A
(where o denotes equivalent statements). In order to prove now that B is an element of K, it will be sufficient to show that B"'is included in B. To this end, assume that d E B"', so d c for all c E B".But c E B" holds if and only if c = - a for some a E A, so d S - a holds for all a E A. It follows that - d E A", so d E B. This is the desired result. Now, given A E K, let B be the element of K just introduced, i.e., B = (b : -b E A"). We will prove that A +B = A . holds. By definition, we have A +B = R"',where
s
R = (a+b:aEA,bEB)=(u-c:uEA,cEA"), so every element of R satisfies a- c 5 0, and hence R c A o . In other words, the positive cone L' of L is included in R".We prove that L+ is exactly equal to R".To this end, let f be an arbitrary element of R".Sincef 2 r and 0 2 r for all r E R, we have inf (f,0) 2 r for all r E R,i.e., -f - 2 a-c. for all
s
a E A and all c E A". It follows that a + f c for all a E A and all c E A", so a + f - EA"' = A. But then a+nf- E A for n = 1 , 2 , . . by induction, s o i f w e b a E A a n d cEA'wehavenf- s c - a f o r n = 1,2, .... SinceL is Archimedean, this is possible only for f - = 0. Henc6, any f E R" satisfies f - = 0, i.e., R" c L'. This shows that R" = L', so R"' = ( f :f 5 0) = A . It has been proved thus that A B = A o , and the fact that L is Archimedean has been used in the proof. It is easy to see that A +B = A + B' = A . implies B = B', so the 'inverse' of A is uniquely determined. Note that if B, and Bz are the inverses of A , and A z respectively, then B1 +B2 is the inverse of Al + A 2 . The set K is, therefore, an abelian group with respect to the addition as defined above. We now define multiplication of any element A E K by an arbitrary real number a as follows.
.
+
.
[
(MU : a E A ) 0) MA = A . = (f the inverse of A la1 ( - - A )
:fs
-
for a > 0, for a = 0, for a = - 1; we denote this inverse now by -A, fora
194
DEDEKIND COMPLETENESS A N D PROJECTION PROPERTIES
[CH.4,s32
Note that aA is an element o f Kfor all A and all a, and a(f)ul = (af)"'holds for all f E L, so the multiplication by real numbers in K agrees on the subspace L with the multiplication in L. For the proof that a ( A + B ) = u A + a B holds for a > 0, note first that if we extend the definition of multiplication by a positive number a to any subset P of L by defining that UP = (ap :p E P ) , then the equality (aP)"'= CIP"'holds. Applying this to the case that
P = (a+b:aEA,bEB),
+
+
where A and B are elements o f K, we obtain aA aB = a(A B). For a = 0 the equality is trivially true and for a < 0 we have m(A+B) = Ial{-(A+B)}
=
laI{(-A)+(-B)} =,la]. ( - A ) + l a l . (4) = aA+aB.
Next, we have to verify that (a+P)A = aA +PA holds for all A E K and all real a, P. For a, p > 0 it is evident that (.+/?)A is included in aA+PA. For the converse, note that for a , , a2 E A we have
5 (a+P)(a, v a 2 )E (a+P)A. The cases that a, P < 0 or one at least of a, P equals zero follow easily. It ma, +Pa2
remains to investigate the case that a > 0, we may assume that a 1 IPI. Then
< 0. Without loss of generality
(x+P)A = (a--IPI)A+Ao = (a-IPI)A+lPlA-lPlA =
=
aA-IfiIA = aA+PA.
The proof for a(PA) = (aP)A is similar, and finally it is evident that aA = A for CI = 1 and all A . Hence, Kis a real linear vector space with respect to the operations introduced, and L (with the operations existing already in L)is a linear subspace. The final remark, which completes the proof, is that the algebraicand order structures in K are compatible, i.e., A c B implies A + C c B + C and A =) A , implies aA 3 olA, for any real a 2 0. By definition, the Dedekind complete Riesz space K is the Dedekind completion o f the Riesz space L if the following conditions hold. (i) L is Riesz isomorphic to a Rie3z subspace LA of K. (ii) Every f E K satisfies
'
'f
= sup(g":g"~L",g"
5 f") = inf(h":h"EL",h" If").
CH.4,8 321
195
DEDEKIND COMPLETION OF AN ARCHIMEDEAN RIESZ SPACE
It is sometimes useful to know that condition (ii) can be replaced by another condition which is easier to check. Theorem 32.6. The Dedekind complete space K i5 the Dedekind conipletion of the Riesz space L if and only if, beside5 condition (i), thefollowing condition is satisfied. (iii) For every 0 < f " E K there exist elements g", h" E L" such that 0 < g" 5 f " 5 h".
Proof. It is evident that (i), (ii) implies (i), (iii). Conversely, assume that (i), (iii) holds, and let 0 5 f E K. We will prove that
"
f " = sup(g":g"EL",g"
5 f").
(1)
Forf" = 0 this is evident, so letf" > 0. The element 24"
=
sup(gA:gAELA,gA Sf")
exists in K since K is Dedekind complete. It is obvious that u" 5 f # , and we have to show that u" = f". Iff -u" > 0, the condition (iii) implies the existence of v" E L" such that 0 < v" 5 f - u". But then
"
=
U"+Y"
"
su p (g A+v A: g A € L h , g hSf")
s
24
#
,
sinceg"EL",g" 5 f'implies g"+v"EL" andg"+u" S u"+u" 5 f". Hence u" +v" 5 u", so v" 5 0. This contradicts v A > 0. Hence formula (1) holds for every f 2 0. Iff E K is arbitrary, we set f = (f ")+- (f ")-. There exists an element h,^ EL" such that (f")- 5 -h,^, so h,^ -(f")Then f"-h: 2 0, so
"
"
"
s
f"-h:=~~p(g":g"€L",g~
sf".
s f"-h,^)
holds in view of what has already been proved. This implies
f
" = sup (9" :g"
E LA,gA
5 f").
But then we have also
" = sup (9" :g" LA,g" 5 -f"), f " = inf (h" :h" E LA,h" 2 f ").
-f SO
E
Theorem 32.1. The Dedekind complete space K is the Dedekind completion of the Riesz space L i f and only if, besides the condition that L is Riesz isometric to a Riesz subspace L A of K, the following conditions are satisfied.
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH.4,s 32
(iv) For every 0 < f # E K there exists an element g" E LA such that 0 < g" j f # . (v) The ideal in K generated by L A is ihe whole of K.
Proof. Note first that the ideal in K generated by the Riesz subspace L" is the set of all f # E K satisfying If #I 5 h A for some h" E L". Indeed, it is easy to see that this set is an ideal in K, and obviously it is the smallest ideal including LA. If K is the Dedekind completion of L, then condition (iv) holds, as proved in the preceding theorem. In order to show that (v) holds, assume .that f # is an arbitrary element of K. Then 1f '1 6 h A for some hA E LA by the definition of the Dedekind completion, sof # is a member of the ideal generated by L A . Hence, the ideal is the whole of K, i.e., condition (v) holds. Conversely, assuming that (iv) and (v) hold, we will prove that condition (iii) of the preceding theorem holds, and this will imply that K is the Dedekind completion of L. We have to show that for every 0 < f # E K there exist g", h" E L" satisfying 0 < g" 5 f # 6 h". The left hand inequality holds by (iv). For the right hand inequality, assume that 0 < f # E K. Then, by (v), f # is a member of the ideal generated by LA, so there exists an element h" E L A satisfyingf # j h". Corollary 32.0. If L is a Riesz subspace of the Dedekind complete space K such that for every f > 0 in the ideal D generated in K by L there exists an element g in L satisfying 0 < g 6 f, then D is the Dedekind completion of L.
Theorem 32.9. Let L be an Archimedean Riesz space with Dedekind completion K. Then K is order separable if and only i f L is so. It follows that if either K or L has any of the properties listed in Theorem 29.3, then the other one as well has all these properties. Proof. We need only prove that order separability of L implies order separability of K. Without loss of generality we may assume that L is a Riesz subspace of K. Elements in ' L and Kf will be denoted by u and v respectively. We have to prove that if 0 v, t v holds in K, there exists a sequence (urn :n = 1,2, .) in the set of all v, satisfying sup urn = v. T o this end, consider for each v, the set (u : u E L ,0 5 u j v J , and take the union of all these sets, so
..
= ( u : U E L 0, j u j v, for some T).
% is an upwards directed subset of L, bounded from above in L (there exists an element uo E Lsuch that uo 2 v; any uo of this kind is an upper bound of
CH. 4,s 32 ]
DEDEKIND COMPLETION OF AN ARCHIMEDEAN RIESZ SPACE
197
4 in L). The set 4,as a subset of K, satisfies sup Q = u. Since L is order separable, there exists an increasing sequence (uun: n = 1,2, . . .) in 4 such that the sequence has in L the same upper bounds as Q. It follows easily that in K we have 0 S uun u. For every n there exists an element urn in the set of all u,, not necessarily uniquely determined, such that uun 5 urn S v. It is evident now that sup urn = u. This completes the proof. Exercise 32.10. Let X be a Boolean algebra with smallest element 8 and largest element e. The complement of x E X will be denoted by x', and accordingly we write A' = (x' :x E A ) for any non-empty subset A of X. For any non-empty subset A of X we form the set A" of all upper bounds of A and the set A' of all lower bounds of A. Note that A" and A' are never empty. Prove that A" = (Ad)' and A' = (A')d, so A"' = Add.Derive from this result that the set of all cuts (i.e., the set of all A"' for non-empty A) is the same as the set of all disjoint complements. Finally, by combining results proved in Theorems 4.7 and 4.9, show that a non-empty subset A of Xis a disjoint complement if and only if A is a band.
Exercise 32.11. Let X be the same as in the preceding exercise. As proved in the present section, the set Y of all cuts in X i s an order complete lattice (partial ordering by inclusion) with 8 as smallest and X as largest element. Prove that Y is actually an order complete Boolean algebra. Hint:The proof is similar to the proof of Theorem 22.7. Exercise 32.12. Let X be a Boolean algebra, 9 'the set of all proper prime ideals in X , and { P } x the set of all P E B that do not contain the element x E X. More generally (cf. Exercise 8.8), for any non-empty subset D of X , let {PIDbe the set of all P E B such that P does not include the whole of D, i.e., (P>D = ({P>x :
u
Since {PI, = { P } I , , where ZD is the ideal in X generated by D, we may restrict ourselves to sets {P>I, where Z is an ideal in X. In the present exercise we will even restrict ourselves to sets { P } l for ideals Z satisfying Z = Zdd, i.e., to sets { P I I for l a band in X. Note that, by Exercise 8.8 (ii), the mapping I + {P>Iis one-one. Also, by Exercise 8.8 (v), the sets { P } l are exactly the regularly open subsets of 8 in the hull-kernel topology of 8.The collection of all bands in Xis partially ordered by inclusion, and the same holds for the collection of all regularly open subsets of 8.Once more by Exercise
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DEDEKIND COMPLETENESS AND PROJECTION PROPERTIES
[CH. 4,g 32
8.8 (v), both collections are order complete Boolean algebras under this partial ordering, and the mapping1 + {PII is a lattice isomorphism between these Boolean algebras. By means of the one-one mapping
x+Ix=(y:yEX,ySx) the Boolean algebra X may be considered as a subset of the set of all bands (this is exactly what we do also when embedding X in its completion by cuts; note that Z, is indeed a band, and x1 S x2 if and only if Ix, c I,,). The mapping x + Z, is a lattice isomorphism of X onto its image in the Boolean algebra of all bands. By Stone’s representation theorem, the mapping x + {P},is also a lattice isomorphism but now of X onto its image in the Boolean algebra of all regularly open subsets of 9. Hence, I, + { P I x is a lattice isomorphism of the Boolean algebra of all Z, onto the Boolean algebra of all {P},. By the preceding exercise, the Boolean algebra of all bands is the completion by cuts of the Boolean subalgebra of all bands of the form I,. Show that, accordingly, the Boolean algebra of all regularly open subsets of B is the completion by cuts of the Boolean algebra of all sets of the form {P},.In other words, the Boolean algebra of all regularly open subsets of 9’ can be regarded as the completion by cuts of X .
CHAPTER 5
Prime Ideals
The ideal P in the Riesz space L is called a prime ideal whenever it follows from inf (f,9)E P that at least one of Y E P or g E P holds. The ideal P is prime if and only if it follows from inf (f,g) = 0 that one at least o f f € P or g E P holds. Also, P is prime if and only if LIP is linearly ordered. Every maximal ideal is prime. Every ideal containing a prime ideal is prime itself, and the set of all prime ideals containing a fixed prime ideal Po is linearly ordered. If Po is an ideal maximal with respect to the property of not coctaining a given elementf E L,then Po is prime. By B we denote the set of all proper prime ideals, by the set of all ideals maximal with respect to the property of not containing the element YEL,and by 2 the union of all 2'. Then 2 is, therefore, a subset of 9. satisfying Given the ideal A andf, not in A , there exists an ideal Q E Q 2 A . Hence, A is the intersection of all Q E 2 satisfying Q =) A ; in particular, (0) is the intersection of all Q E 2. It follows that A is the intersection of all P E 9 'satisfying P 3 A ; in particular, (0) is the intersection of all P E 9.The prime ideal Mis called minimal if M = MI holds for every prime ideal MI c M . We prove that (0) is the intersection of all minimal prime ideals. Furthermore, every prime ideal contains a minimal prime ideal. If Xis a non-empty compact Hausdorff space and C ( X ) is the Riesz space of all real continuous functions on X , then every proper prime ideal Po in C ( X ) has one and only one point xo E X at which all f E Po vanish. We call xo the zeropoint of Po.The intersection of all prime ideals having xo as zeropoint is the ideal of all functions in C ( X ) that vanish in a neighborhood of xo. We will prove also that every prime ideal in C ( X ) is a maximal ideal if and only if X consists of a finite number of points, i.e., if and only if C ( X ) is Riesz isomorphic to R" (with the usual coordinatewise ordering) for some natural number n. Given f E L and the non-empty subset W of 9, we denote by { R ) / the set of all R €93' such that f is no member of R. The set of all {R}", u E L', partially ordered by inclusion, is a distributive lattice with the empty set as 199
200
PRIME IDEALS
[CH. 5 , g
33
smallest element. For any non-empty subset W,of W the kernel k(g,) of W 1 is the intersection of all R E Wl; for any non-empty subset D of L, the hull h(D) of D is the set of all R E 9 such that R 3 D. For the case that W = B or W = 9, we have {R},= {R},,if and only if A , = A , (where A , and A , denote the ideals generated by u and v respectively). For the case that W is the set& of all minimal prime ideals, we have { M } , = { M } , ,if and on14 it' (A,)dd = Also, the equality holds for every u EL'. Given the non-empty subszt 92 of 8,the sets {R},,u EL', form a base for a topology in 9,the hull-kernel topology in W.The closure of any subset Wl of 92 is the set h ( k ( g 1 ) ) .The hull-kernel topologies in B and A! are Totopologies, the base sets {P},and {Q}, of which are open and compact. The hull-kernel topology in A is a Hausdorff topology, the base sets of which are open and closed. The lattice of all base sets { M } , is a Boolean ring if and only if every { M } , is compact in the hull-kernel topology of A, and the lattice is a Boolean algebra if and only if A itself is compact. We finally mention the remarkable theorem that the hull-kernel topology in 9 is Hausdorff and compact if and only if there exists a field r of subsets of a point set X such that L is Riesz isomorphic to the Riesz space of all real step functions on X assuming each of its values on a set of r. Much of what is discussed in the present chapter is analogous to what was observed in Chapter 1 about prime ideals in distributive lattices. Many results in sections 33, 35 and 36 are due to D. G. Johnson and J. E. Kist ([l], 1962), extending and clarifying earlier results of I. Amemiya ([2], 1953). The particular subset of all maximal ideals in a Riesz space was inv stigated already by K. Yosida ([3], 1941) and by K. Yosida and M. Fukamiqa ([l], 1941); cf. section 27. Prime ideals in a Dedekind a-complete space (with a somewhat different terminology, however) were considered by H. Nakano ~ 3 1 1941). , 33. Prime ideals In many respects the contents of the present section are analogous to the contents of section 5, wherewe dealt with prime ideals in distributive lattices with a null element. Throughout this section L will denote a Riesz space.
Definition 33.1. The ideal P in L is called a prime ideal if it follows from inf (f,g ) E P that one at lea.st o f f E P or g E P holds.
CH. 5 , § 331
PRIME IDEALS
20 1
Note that the space L itself is a prime ideal. Theorem 33.2 (compare Theorem 5.1 (i)). For an ideal P in L, the following conditions are equivalent. (i) P is prime. (ii) Ifinf (f,g ) = 0, then one at least off E P or g E P holds. (iii) The quofient Riesz space LIP is linearly ordered. (iv) For any ideals A , B satisfying A n B c P one at least of A c P or B c P holds. Proof. (i) => (ii) Evident. (ii) (iii) It will be sufficient to prove that any If] E LIP satisfies one at least of the inequalities If] S [0] or If] 2 [O]. Given [f]E LIP, let f be an arbitrary member of If]. Since inf ( f +, f -) = 0, it follows from hypothesis (ii) that f E P or f - E P, i.e., If]' = If'] = [O] or [f]-= [ f - ] = [O]. In the first case we have If]S [O]; in the second case we have If] 2 [O]. (iii) =-(iv) We denote the mappingf + If] ofL onto LIP by n. The mapping 7t is a Riesz homomorphism; the images n(A) and n(B)of ideals A, B in L are ideals in LIP, and n(A n B ) = n(A) n n(B) (cf. Exercise 18.15). If it is given, therefore, that A n B is included in P, then n(A) n n(B) = { [O]} holds in LIP. In any linearly ordered Riesz space, however, the ideals are linearly ordered with respect to inclusion. According to hypothesis (iii) the space LIP is linearly ordered; it follows, therefore, from n(A) n n (B ) = { [0]} that n(A) = { [ O ] } or n(B) = { [O]}, and so A c P or B c P. (iv) ( i ) We note first that if we have u, u E L +and w = inf (u, v), and A , , A , , A,,, are the principal ideals generated by u, v, w respectively, then A, = A , n A , (cf. Exercise 17.7). Hence, under the present hypotheses, w = inf (u, u ) E P implies A, c P, i.e., A , n A , c P, and so A , c P or A, c P, which implies that u E P or v E P. Given the arbitrary elements f, g E L such that h = inf (f,g ) E P, we have inf (f-h, g - h ) = 0 E P, so f - h E P or g - h E P in view of what was proved already. Since h E P holds by hypothesis, it follows that f E P or g E P. In other words, P is prime. This concludes the proof. +
-
Theorem 33.3. (i) Every maximal ideal J in L is prime (compare Theorem 5.1 (iii)). (ii) I f P is a prime ideal and P I ,P2 are ideals including P, then P I and P2 are prime, and P I includes P2 or conversely. Hence, every ideal including a given prime ideal is prime, and the set of all ideals including a given prime ideal is linearly ordered (compare Exercise 8.9).
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PRIME IDEALS
[CH. 5 , s 33
(iii) The ideal (0) is prime if and only i f every ideal in L isprime. Equivalentl-v, ( 0 ) is a prime ideal if and only if L is linearly ordered.
Proof. (i) Let J be a maximal ideal in L.By Theorem 27.3 (i) the space LIJ is Riesz isomorphic to real number space, so L/J is linearly ordered. It follows, by the preceding theorem, that J is prime. (ii) Let P i , P , be ideals, both including the prime ideal P. We first prove that P, is prime. For this purpose, assume that inf (f,g ) = 0. Then, since P is prime, we have f E P c P , or g E P c P , , so P , is prime. Similarly, P , is prime. Now assume that neither of P , or P , is included in the other. Then there exist elements f , E P: and f 2 E P l such that f , is no member of P2 and fz is no member of P , Writing h = inf (fl ,f,), we have inf (fl-h, f , - h ) = 0, so f l - h E P or f , - h ~ P , say f , - h e P . Then f , - h ~ P , , so because h E P, we have also that f , E P , . This contradicts the assumption that f , is no member of P,. It follows that P , is included in P , or conversely. (iii) The first statement follows from part (ii). The second statement follows from Theorem 33.2 by observing that (0) is prime if and only if L/{O) = L is linearly ordered.
.
Theorem 33.4. If f o E L and P is an ideal in L, maximal with respect to the property of not containing f o (i.e., any ideal Q 3 P such that f o is no member of Q satisfies Q = P ) , then P isprime (compare Theorem 5.1 (ii)). Proof. If P is not prime, there exist elements g, h E L + ,not in P, such that inf (9, h) = 0. The element f o is in the ideal generated by P and g , so f o = p1+ g l for some p 1 E P and some g 1 in the principal ideal generated by g. Similarly,f o is in the ideal generated by P and h, so f o = p z + h , for some p , E P and some h, in the principal ideal generated by h. Note that g Ih implies g 1 Ih , . Setting p3 = SUP (hl, I P , ~ ) , we have lfol 6 p3+Igi1 and l f o l 6 P3+lhll, so
if01 2 inf(P3+bil,P3+Ihil)
=P3 EP, =P3+inf(bil, which impliesf o E P, thus contradicting our hypotheses. It follows that P is prime. The following theorem is now an easy consequence. It shows, among other things, that although it may easily happen that a Riesz space has no maximal ideals (cf. Example 27.8), every Riesz space has plenty of prime ideals.
Theorem 33.5. Given the ideal A and the element f o not in A , there exists an ideal P 2 A such that P is maximal with respect to the property of not con-
CH. 5 , s 331
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PRIME IDEALS
taining f o . Hence, by the preceding theorem, P is prime. It follows now that, for any ideal A in L (alsofor A = L), we have
In particular,
0( P :P A, Pprime ideal). 0 (P : Pprime ideal) = (0).
A =
2
In the speLial case that A is prime andf o is not in A , the prime ideaI P
A, maximal with respect to the property of not containing f o , is uniquely determined. Infact, P is now the union of the set of allprime ideals P, 2 A that do not contain So. This set is linearly ordered, i.e., the set is a chain, and P = P, is the largest member of the chain (compare Theorem 5.2). 2
uo
Proof. The set of all ideals which include A and do not contain fo is partially ordered by inclusion. Each chain in this set has an upper bound (the union of all elements in the chain). Hence, the set has a maximal element, i.e., there exists an ideal P 2 A such that P is maximal with respect to the property of not containing fo By the preceding theorem P is prime. All the other statements follow now easily.
.
Any subset S of L+ not containing the zeroelement and such that u, v E S implies inf (u, v ) E S is called a lower sublattice in L. Given uo > 0, the set of all v 2 uo is a lower sublattice. The empty set is also a lower sublattice. It is immediately evident that if P is a prime ideal with positive part P + = P n L+, then the set theoretic difference S = L+-P+ is a lower sublattice with the extra property that u E S, v 2 u implies v E S. In the converse direction, if S is a lower sublattice (even one with the abovementioned extra property), the set theoretic difference S’ = Lf-S is not necessarily the positive part of a n ideal (u, u E S’ may fail to imply that u+u E S‘). However, given that P is a n ideal and S = L+ -P+ is a lower sublattice, it follows immediately that the ideal P is prime. A lower sublattice is called maximal whenever it is not properly included in any other lower sublattice. Given any lower sublattice S, the set of all lower sublattices including S is partially ordered by inclusion. Any chain in this partially ordered set has an upper bound (the union of all elements i n the chain). Hence, the set has a maximal element, i.e., the given lower sublattice S i s included in a maximal lower sublattice. Note that if uo is a member of the maximal lower sublattice S,,,, then any v 2 uo is a member of S,,,. Also, if wo 1 0 is no member of S,,,, then inf (wo, u ) = 0 holds for at least one u E 3,.
We proceed with another definition. Given the ideal A and the prime ideal
204
[CH.5 , 5 33
PRIME IDEALS
M 13 A , we say that M is a minimal prime ideal with respect to A whenever any prime ideal M' for which A c M' c M holds satisfies M' = M. The prime ideal M is simply called a minimalprime ideal if M is minimal with respect to the ideal (0). The following theorem follows now easily. Theorem 33.6. (i) Given a chain (with respect to partial ordering by inclusion) of prime ideals (P, :a E (a)), the intersection P, is again aprime ideal. (ii) Given the ideal A and the element f o not in A , there exists aprime ideal M 2 A such thatf o iJ no member of M and such that M is minimal with reJpect to A . Hence, for any ideal A in L (alsofor A = L), we have
n,,
n ( M : M A, M prime ideal minimal with respect to A). In particular n ( M M minimalprime ideal) = (0). A =
3
1
Hence, given f o # 0, there exists a minimal prime ideal not containing fo. It follows also that L itself is not a minimal prime ideal, unless in the trivial case that L = ( 0 ) (compare Theorem 5.3). Proof. (i) The set theoretic differences S,, = L+-P: are lower subis also a lower sublattice. In other words, the complement lattices, so (with respect to L + )of the positive part of the ideal P = P, is a lower sublattice. This implies that P is prime. (ii) Let A be an ideal in L and fo an element of L such that f o is no member of A . The set of all prime ideals which include A and do not contain f o is non-empty (cf. the preceding theorem). This set is partially ordered by antiinclusion (i.e., PI P2 whenever PI =I P,), and any chain in this partially ordered set has an upper bound (the intersection of the elements of the chain; this is a prime ideal by part (i)). Hence, the partially ordered set has a maximal element, i.e., there exists a prime ideal M which includes A , does not contain f o , and is minimal with respect to A .
u,,s,,
n,,
s
Some further results, connected with the existence of maximal lower sublattices follow. Theorem 33.1. ( i ) The subset S of L+ is a maximal lower sublattice if and only if the set theoretic di'erence L+- S is the positive part of a minimal prime ideal. (ii) Every prime ideal includes a minimal prime ideal. (iii) Any minimal prime ideal M cannot contain an element f E L and its
CH. 5 , s 331
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205
disjoint complement {f } d simultaneously, unless L = {0} (compare Theorem 5.4).
Proof. (i) Assume first that S is a maximal lower sublattice. Write M + = L + - S . Since w E S, z 2 w implies Z E S, we have that M E M + , 0 5 v 5 u, implies u E M ' . In order to prove also that u, v E M + implies au+bu E M + for constants a, b 2 0, observe that u E M + implies the existence of an element w E S satisfying inf (u, w) = 0, i.e., u Iw. Similarly u E M + implies the existence of an element z E S satisfying v Iz. Then w 1 = inf (w,z ) is an element of S such that u Iw1 and v 1w 1 hold. It follows that, for all non-negative numbers a and b, we have au+bv E L+ and (au+bv) Iwl.The last formula shows that au+bu is no member of S. In other words, au+bu is a member of M + . It has thus been proved that M + is the positive part of an ideal M ; M is simply the set of all f = u - u with u, v E M + . The set theoretic difference L+- M + is the lower sublattice S, and so M is a prime ideal in view of one of the remarks made above. If M would properly include another prime ideal M * , then S would be prop\ rly included in the lower sublattice L+ - ( M * ) + .This is not so since S is maximal, and so M is a minimal prime ideal. Conversely, assume now that M is a minimal prime ideal. Then S = L+- M is a lower sublattice (simply because M is a prime ideal). If S is not maximal, there exists a maximal lower sublattice S* which properly includes S. Then, by what has just been proved, L+- S* is the positive part of a minimal prime ideal M* such that M* is properly included in M. This is impossible, and so S is a maximal lower sublattice. (ii) Given the prime ideal P,the set theoretic difference S = L + - P + is a lower sublattice. 5' is included in a maximal lower sublattice S*, and so the minimal prime ideal M , satisfying M + = L+- S*, is included in P. (iii) Assume that L # (0) and M is a minimal prime ideal in L (hence M # L). Furthermore, letfbe an element of L such thatfand { f } dare contained in M. The set S = L + - M + is a maximal lower sublattice, so If1 E M + implies that inf (Ifl, u ) = 0 holds for some u E S. It follows then that u E { f } d c M . Hence u E S and u E M hold simultaneously, which is impossible. But then f and { f } dcannot be simultaneously included in M. +
We introduce some notations that will be used repeatedly in the sections to follow. By B we denote the set of all proper prime ideals in L, by 4 the set of all proper minimal prime ideals (we recall that for L # {0} every minimal prime ideal is automatically proper), and by 3 the set of all ideals Q for which there exists an elementf o , not in Q and depending upon Q, such
206
PRIME IDEALS
[CH. 5 , # 33
that Q is maximal with respect to the property of not containingfo .As proved in Theorem 33.4, every ideal Q of this kind is a proper prime ideal, and in Theorem 33.5 it was actually proved that
A =
n( Q : Q
3
A,QE~?)
holds for every proper ideal A . In particular, we have
n (Q : Q
E
2 ) = {OI,
unless L = {0}, in which case the set 22 is empty. The sets A and 2 are subsets of 8,and the same holds for the set $of all maximal ideals (as we have seen, the set $ may be empty). Given the elementf # 0 in L, the set of all ideals, not containing f and maximal with respect to the property of not containingf, is denoted by &. Obviously, @ . is a subset of 9 ;actually, we have
2
=
u (3f:f#
0).
Given the non-empty subset W of B and the element f E L, we let { R } / denote the set of all R €9such that f is no member of R. Note that for f = 0 the set {R}ois empty. Furthermore, we have
u
( { R } f :f E L ) = 92.
Note that, according to these definitions, if e > 0 in L and the arbitrary element f in L are given, {Qe}f denotes the set of all prime ideals Q, not containing e andf, and such that Q is maximal with respect to the property of not containing e. For a fixed e > 0 in L, we consider the set {P},.Given Po in this set, all prime ideals P 3 Po in the set form a chain; the chain has a largest element Q , maximal with respect to the property of not containing e, so Q E 9,. We shall refer to this chain as the chain corresponding to Po. If M is a minimal prime ideal not containing e, the chain corresponding to M cannot be made longer within the set {P},;let us call a chain of this kind a complete chain in {P},. Every chain in {P},is part of a complete chain; this follows simply by observing that any prime ideal contains a minimal prime ideal. It can happen that different chains in {P},have the same largest element or, perhaps, have a part in common that consists of more than only the largest element. For an example we refer to the next section. If L has a strong unit e > 0 (i.e., the ideal generated by e is the whole of L), the set {P},is equal to the set B of all proper prime ideals and 9, is equal to the set f of all maximal ideals. The set B is now the union of all complete chains of prime ideals, each chain having a minimal prime ideal as smallest element and a maximal ideal as largest element.
CH. 5 , s 341
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COMPACT AND HAUSDORFF
207
Exercise 33.8. Given the ideal A in the Riesz space L, we denote the elements of L / A by [f], [ g ] , . ., and for any ideal B 3 A , we denote by [B]the ideal in L / A corresponding to By i.e., [ B ] is the set of all [f] such thatfis a member of B. Show that [ B ] is prime in L / A if and only if B is prime in L. Prove also that if A is a prime ideal (i.e., L / A linearly ordered), then LiA is Archimedean if and only if A is a maximal ideal.
.
Exercise 33.9. Prove that the prime ideal P is minimal if and only if for every u E P+ there exists an element v EL' -P+ such that u 1v. Exercise 33.10. Any ideal containing a prime ideal is a prime ideal. Similarly, show that if Vis a linear subspace of the Riesz space L such that Vcontains a prime ideal, then V is a Riesz subspace of L with the property that if f, g E L and inf (f,g ) E V, then one at least off and g is in V. 34. Prime ideals in C(X) for X compact and Hausdorff
In this section we assume that X is a non-empty compact and Hausdorff topological space, and C ( X ) the Riesz space of all real continuous functions on X . By e we denote the function in C ( X ) satisfying e(x) = 1 for all x E .'A Evidently, e is a strong unit in C ( X ) .It was proved in Example 27.7 that for any xo E X the set of all functions in C ( X )vanishing at xo is a maximal ideal J,, in C ( X ) ; conversely, given any maximal ideal J in C ( X ) , there exists a point xo E X such that J = J,,. For any xo EX, let .N(xo) be the system of all (not necessarily open) neighborhoods of xo, and let the ideal I,, be defined by
I,, = ( f :f E C ( X ) , (x :f ( x ) = 0 ) E M ( X O ) ) . Evidently, I,, is included in Jx,. It will be our purpose to prove that, in general, there are more prime ideals in C ( X )than maximal ideals; more explicitly, we shall prove that every prime ideal is maximal if and only if X consists of a finite number of points.
Theorem 34.1. Every prime ideal Po in C ( X ) has one and only one point
xo E X at which all functions in Po vanish. Call this point xo the zero point of P o . For every xo E X, the ideal I,, is the intersection of all minimal prime ideals having xo as zero point, and so I,, is also the intersection of all prime
ideals having xo as zero point. It follows that every prime ideal in C ( X ) is maximal if and only if I, = J, holds for every x E X . Proof. Given the prime ideal Po in C ( X ) , the set of all prime ideals
208
[CH. 5 , s 34
PRIME IDEALS
P 2 Po is a chain with a maximal ideal, say J,,, as largest element. It follows thatf(xo) = 0 for allf E Po.If there would exist a point x1 # xo such that f(x,) = 0 for all f E Po,this would imply that Po is included in Jx, . But then Jx, would be an element of the chain corresponding to Po,so J,, c Jx,. This is impossible. Hence, xo is the only point at which allfe Po vanish. The next thing we will prove is that I,, is included in every prime ideal Po having xo as zero point. Indeed, let u be a function in Ix’,. There exists an open neighborhood of xo at the points of which u vanishes; let v be a nonnegative function in C ( X ) , vanishing outside that neighborhood and satisfying v(xo) = 1 (note that a function v satisfying these conditions exists because X is a normal topological space). We have inf (u, u ) = 0, so u E Po or v E P o . But v is evidently no member of Po (since u(xO)= l), so u E Po holds. This shows that Zx, c P o . Hence, I,, is included in the intersection of all prime ideals having xo as zero point. In the converse direction we prove that if I,, is included in a prime ideal P I , then P, has xo as its zero point. Indeed, for any xi # xo there exists a neighborhood of xo not containing x1, and so there exists a function in Zx, (and hence in P,)that does not vanish at x , It follows that x, cannot be the zero point of P,,so xo must be the zero point of PI.Now, by Theorem 33.5, Zx, is the intersection of all prime ideals that include I,,, i.e., I,, is the intersection of a certain set of prime ideals all of which have xo as zero point. From the results found in the last two paragraphs we infer now that Ix, is the intersection of all prime ideals having xo as zero point. Among these prime ideals are all the minimal prime ideals having xo as zero point. Hence, Ix, is already the intersection of all minimal prime ideals having xo as zero point. This completes the proof.
.
Note that if Ix, is itself a prime ideal, then all minimal prime ideals with 9’corresponding to I,, (i.e., the chain of all proper prime ideals P 3 I,,) is the only chain having J,, as largest element. Conversely, if there is only one such chain, then the smallest element in the chain is necessarily equal to I,,. In other words, Zx, is no prime ideal if and only if there are at least two different minimal prime ideals M , and M 2with xo as zero point; the chains corresponding to these minimal prime ideals have, therefore, the same maximal ideal Jx, as largest element. By way of example, let X be the interval [0, 11 in the real line with its ordinary topology. For any xo E (0, l), the ideal Ix, is no prime ideal. Indeed, if
xo as zero point are necessarily equal to I,,, and so the chain in
CH. 5 , s 341
u(x) =
PRIME IDEALS IN
C ( X ) FOR X
for x 5 x o , for x 2 x o ,
209
COMPACT AND HAUSDORFF
v(x) =
(2-x
for x 6 x o , for x 2 x o ,
then inf (u, t.) = 0, but neither u nor v is a member of Ix,. Hence, there are are at least two different minimal prime ideals in C([O, 11) having xo as zero point, and so there are at least two different complete chains in B having J,, as largest element. Returning to the general case, we recall that a point xo in a topological space Xis sometimes called a P-point whenever any countable intersection of neighborhoods of xo is again a neighborhood of xo. This notion of a Ppoint plays a role in the next theorem. Theorem 34.2. Given again that X i.s Hausdorff and compact, and given xo E X , we have Zxo = Jxo i f and only if xo is a P-point.
Proof. Assume first that xo E X is a P-point. We have to prove that any f E Jx, satisfiesfe I,,. LetfE J,,, s o f ( x o ) = 0, and for n = 1,2, . define the neighborhood Onof xo by
..
On = { x : If(x)l
Then the set
< n-’1.
Z ( f )= { x : f ( x ) = O}
satisfies Z ( f ) = r). On,and so Z ( f ) is a neighborhood of xo since xo is a P-point. This shows that YE Ixo. For the converse, assume that Ix, = Jx, and let (On:n = 1,2, . . .) be a sequence of neighborhoods of x o . We have to prove that (0, : n = 1,2, . . .) is also a neighborhood of xo . Since the topology in Xis Hausdorff and compact (and hence completely regular), the topology is equal to the ”weak topology” generated by the functions of C ( X ) ; hence, given the neighborhood 0 of x o , there exist a number E > 0 and functionsfl , . . .,& in C ( X ) such that
0
n
(7 ( X
Then
i=1
dx)
: I.fi(x)-fi(xo)I < 8 ) c 0. =
max Ifi(x)-fi(xo)l
i = 1..
..,n
satisfies 0 5 g E C ( X ) , g ( x o ) = 0 and ( x :g ( x ) 5 -)e) c 0. It follows that h(x) = ( g ( x ) - + ) + satisfies 0 S h(x) E C ( X ) ,h(xo) = 0 and Z ( h ) = (X : h(x) = 0 ) =
(X
:g ( x )
5 38) c 0.
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[CH.5, 5 34
PRIME IDEALS
Setting now h , ( x ) = min (h(x), l), we have h, E C ( X ) , 0 6 h, 5 1 , h l ( x o ) = 0, and Z ( h , ) c 0. Applying this to every 0, in the given.sequence of neighborhoods of xo , we obtain functions h, E C ( X ) for n = 1 , 2 , . . . such that 0 5 h,, S 1 , h,,(xo) = 0 and Z(h,,) = ( x :h,(x) = 0 ) c 0,.Hence h = n-2h,, satisfies h E C ( X ) by uniform convergence and
xT
On the other hand, since h(x,) = 0, we have h E J,, = I,, , so Z ( h ) is a neighborhood of x o . It follows now from Z ( h ) c 0, that 0, is a neighborhood of xo. This is the desired result.
n
n
The main theorem in this section follows now. Theorem 34.3. Zf C ( X ) is the Riesz space of all real continuous functions on the non-empty compact Hausdorfspace X , then every prime ideal in C ( X ) is maximal if and only i f X consists of a$nite number of points, i.e., if and only if C ( X )is the space R" (with the ordinary pointwise ordering)for some natural number n. Proof. Assume first that X consists of a finite number of points. The topology in X is now the discrete topology, i.e., every point of X IS an open set, and so every point of X is a P-point. It follows that I, = J, holds for every x E X , so by Theorem 34.1 every prime ideal in C ( X ) is maximal. Conversely, if every prime ideal in C ( X ) is maximal, we have I, = J, for every x E X , so every point of X is a P-point. We have to prove that the number of points in Xis finite. If not, let (x, : n = 1,2, . .) be a sequence of mutually different points in X . Given any point P E X such that p is different from all x,, (if such a point p exists), there obviously exists for every natural number k a neighborhood ofp not containing x l , . ., x, . The intersection of these neighborhoods is a neighborhood ofp not containing any of the points x , . Similarly, for any n, there exists a neighborhood of x, not containing any x , for m # n. Since Xis compact, Xis equal to the union of a finite number of the thus introduced neighborhoods, and so X contains only a finite number of the points x , (n = 1 , 2 , . . .). This yields a contradiction. The final result is, therefore, that X consists of a finite number of points.
.
.
We may ask how in a space of type C ( X ) ,with Xcompact and Hausdorff, the notions of ideal and prime ideal in the Riesz space sense are related to the notions of ideal and prime ideal in the algebraic sense (i.e., in the ring theoretic sense). Any order ideal Zis also an algebraic ideal, i.e., f E Z and g E C ( X )
CH. 5 , s 341
PRIME IDEALS IN
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COMPACT AND HAUSDORFF
211
impliesfg E Z. The proof is derived by observing that there exists a constant LY such that Ig(x)l S a holds for all x . An algebraic ideal Z is not necessarily an order ideal. As an example, let X = [0, 1 ] and Z the algebraic ideal generated by x, so Z consists of all functions x f ( x ) withfcontinuous. Then Z does not contain the continuous function g(x), defined by g(x) = x sin x - l for x > 0 and g(0) = 0. Since Ig(x)l 5 x , this shows that Z is no order ideal. Furthermore, we observe that although any order prime ideal is an algebraic ideal, it is not necessarily an algebraic prime ideal. By way of example, let X = [0,1] and A the set of all f E C ( X ) such that If(x)l 5 a f x for all x and some constant a / , depending 0n.f. The set A is an order ideal as well as an algebraic ideal in C ( X ) , and evidentlyfo(x) = xi is no member of A . Hence, by Theorem 33.5, there exists an order prime ideal P 2 A such that fois no member of P.Observing now thatf; E A c P,we conclude that P is not prime in the algebraic sense. Finally, in the converse direction, although an algebraic ideal is not necessarily an order ideal, it is true that an algebraic prime ideal is an order prime ideal. For the proof, let Z be an algebraic prime ideal in C ( X ) , and assume thatfE Zand 191 S Ifl. Thenf’ E Z, so If1 E Z, and similarly If/* E Z. Also, the function glfl-*, if we define this function to be zero at all points where f is zero, is continuous on X , i.e., glfl-* E C ( X ) . Since lfl* E Z, it follows that 9 = glfl-* lfl* E 1-
-
It has been shown thus that forfE Z and Igl 5 If1 we have g E Z, i.e., Z is an order ideal. Now, let inf (f,g) = 0. Then f(x)g(x) = 0 for all x E X , so f~ Z or g E Z since Z is an algebraic prime ideal. It follows that Z is prime as an order ideal. This proof is due to D. H. Fremlin. In spite of these differences, the Theorems 34.1,34.2 and 34.3 hold also for algebraic prime ideals. It is a well-known fact (and the proof is very similar to the proof for order ideals) that the maximal algebraic ideals are the same as the maximal order ideals, i.e., the maximal algebraic ideals are the sets Jx,
=
(f:fE C(X),f(xo)
= 0).
For some further properties of algebraic ideals which will be mentioned, we refer to the book by L. Gillman and M. Jerison [ l ] for the proofs. Every maximal ideal is prime and every proper prime ideal is contained in a unique maximal ideal (cf. Theorem 2.11 in the book referred to). Hence, every proper prime ideal P has one and only one point xo at which allfE P vanish; we call xo the zero point of P.Also, the intersection of all prime ideals
212
[CH.5.8 35
PRIME IDEALS
containing a given ideal I is precisely the set of all elements of which some power belongs to I (cf. Theorem 0.18 in the book referr6.d to). Applying this to the ideal lXO
=
( f : f E C(X),( x : f (XI = 0)E Jqxo)),
and observing that a.power off belongs to Ix, if and only iff itself belongs to I,,, we obtain the result that Ixo is the intersection of all prime ideals having xo as zero point. It follows that every proper algebraic prime ideal in C ( X )is maximal if and only if I, = J, holds for every x E X . This takes care of Theorem 34.1. The Theorems 34.2 and 34.3 hold unchanged. 35. Hulls and kernels
We use the notations introduced in the final paragraphs of section 33. Hence, B and JY are the sets of all proper prime ideals andall proper minimal prime ideals in the Riesz space L,3fis the set of all prime ideals maximal with respect to the property of not containing the element f # 0, and 3 is the Finally, fl is the set of all maximal ideals in L . union of all Given the non-empty subset W of 9 and the element f E L,we denote by {R}/the set of all R €99such that f is no member of R. Note that {R}/= { R } I /for I everyfe L , and for f = 0 the set {R}/= {R}ois empty. Furthermore, we have
u ({R}/f L) :
E
= 3.
Also, for all u, v E L + ,we have the equalities { R l s u p (u, u )
={ N u
{%lf
=
(u, ")
{Nu
" {%
Y
n {RIu.
This shows that the set of all {R}u,u EL', partially ordered by inclusion, is a lattice with the empty set as smallest element. The lattice is dstributive, because the distributive laws hold in L+.If the Riesz space L has a strong unite, then {R}== W is evidently the largest element of the lattice (note that e is contained in no proper ideal at all).
Definition 35.1. Let W be a non-empty subset of 9'. For any non-empty subset W ,of 9, the kernel k ( 9 , ) of Blis defined by k(Wl) = and for
n ( R :R EB,),
Wlempty we define k(9Z1) = L. Note that k ( W l ) is an ideal in L.
CH. 5 , s 351
21 3
HULLS AND KERNELS
For any non-empty subset D of L, the hull h(D) of D is defined by h(D) = ( R :R E W ,R
3
0).
Obviously, it depends not only upon the above definition, but also upon the choice of 9, what k ( 9 , ) and h(D) will turn out to be. Compare the present definition with Definition 6.1 for hulls and kernels in a distributive lattice.
Theorem 35.2. (i) We have h(k(W,)) 3 W,for every B1c 9, and k(h(D)) 3 A, for every D c L , where A Ddenotes the ideal generated by D. (ii) ZfWl = h(D)for some D c L, then h(k(W,)) = 9,. (iii) ZfA = k(Wl)for some 3, c W,then k(h(A)) = A. (iv) ZfW = 8 or 9 = 9, then k(h(A)) = A hold$for every ideal A in L. Proof. (i) Evident from the definitions. (ii) Assuming that Wl= h(D), it is sufficient to prove that h ( k ( g 1 ) ) c 9,. T o this end, note that 9,= h(D) implies k ( W l ) = k(h(D)) XI D, so
h ( k ( 9 , ) ) c h(D) = 9,. (iii) Similarly. Given the proper ideal A in L, it follows from (iv) Let 9 = 9. A = ~ ) ( P : P ~ A , P E B ) that A = k ( 9 , ) for 8, = (P: P 2 A , P E 8),so k(h(A)) = A by part (iii). A direct verification shows that k(h(A)) = A holds also for A = L. The proof for W = 9 is similar. Note that the proof of Theorem 33.5 shows actually that A = (Q : Q 2 A, Q € 9 )
n
holds for every proper ideal A.
We will now assume that the subset W of B which we consider satisfies
0 (R: R €9) = (0). The subsets W = 9, 92 = 9 and W = 4 satisfy
this condition. Furthermore, given W , c 9, we denote the set theoretic difference 9 - W l by For the next theorem and its corollary, compare Theorem 6.3 and Corollary 6.4.
e.
Theorem 35.3. Let W c B satisfy the condition that the intersection of all R E W is (0). Then, given any ideal A in L, we have Ad = k(h(A)’).
Proof. If A = L, then h ( A ) is empty, and so h(A)’ k(h(A)’) = (0) = Ad.
=
W.It follows that
214
[CH.5 , s 35
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If A # L, let g E Ad and R E h(A)’. Then A is not included in R, and hence there exists an element f E A such that f is no member of R. Since f E A and g E Ad,we have inf (Ifl, 191) = 0. It follows that g E R (since R is prime and f is not in R). Hence any g E Ad is contained in any R E h(A)’. This shows that Ad c k(h(A)’). Conversely, given any g E k(h(A)”),we have to prove that inf (If I, Igl) = 0 holds for everyf E A. It is sufficient for this purpose to prove that the element inf (Ifl, 191) is a member of every R €9(since this implies that inf (If I, Igl) is contained in ( R :R €9) = (0)). If R E h(A), then f E A c R, so inf (If I, Igl) E R. If R E h(A)”, then g E R, so inf (If I, 191) E R. Hence, inf (If I, Igl) E R holds in either case.
Corollary 35.4. (i) Given again that the intersection of all R E W is {0}, we have (Auld = k({Rl,) for every u E L’, where A, denotes the ideal generated by u. (ii) In the case that 92 = A,we have for every U E L + .
h(k({M)u))= {MI,
Proof. (i) Note that h(AJ = {R],,, so the formula to be proved is a special case of the formula in Theorem 35.3. (ii) We may assume that u # 0. It follows from (AJd = k ( { M } , ) that h((A,Y) = h(k({Ml,))
= {MI,.
(1)
On the other hand, if the proper minimal prime ideal Msatisfies M Eh((A,Y), i.e., if (A,Jd c M , then u E M is impossible by Theorem 33.7 (iii). It follows that u is no member of M, so M E{ M } , . This shows that h((A,Jd)c {M},,, and so formula (1) becomes h((AIJd)= h ( ~ ( { M l , ) = ) {MI,. For the next theorem, compare Theorem 6.5.
Theorem 35.5. (i) Given u, v EL’, we have {P}, c {P},, if and only i f A, c A, , where A, and A , denote the ideals generated by u and v respectively. Hence {P}, = {P}, i f and only i f A, = A,,. (ii) Given u, v EL’, we have {Q}, c {Q}, i f and only if A, c A,,. Hence {Q}, = {Q},if and only i f A, = A,.
CH. 5 , s 351
HULLS A N D KERNELS
215
(iii) Given u, u E L', we have { M } , c { M } , i f and only i f Atd c A?. Hence { M } , = { M } , if and only i f Aid '= A:. If L is Archimedean, then { M } , c { M } , holds ifand only i f B , c B,, where B, and B, denote the bands generated by u and v respectively. Hence, if L is Archimedean, then { M } , = { M } , ifand only V B , = B,. Proof. (i) If A, c A,, the element u is a member of A,, so every ideal containing v contains u. Hence, given the prime ideal P such that u is no member of P , it follows that v is no member of P . In other words we have {PI, c {PIu. Conversely, assume that { P } , c { P } , holds, but A, c A, does not hold. Then u is not contained in A,, so there exists a prime ideal P 2 A, such that u is no member o f P . It follows that P E { P } , c {P},, so u is no member of P. This contradicts v E A, c P . Hence A,, c A, must hold. (ii) Similarly. I> (A,)d in (iii) {M}, c { M } , implies k ( { M } , ) =I k ( { M } , ) , i.e., Conversely, view of Corollary 35.4 (i). It follows that (A,Jddc (A,)ddc (A,)dd implies (AJddd=I (A,)ddd,i.e., (A$ =I (A,)d or, in other words, k ( { M } , ) =I k ( { M } , ) . Then
h(k({M}u)) = h ( w W u ) h i.e., { M } , c { M } , by Corollary 35.4 (ii). The remaining statements follow easily by observing that in an Archimedean Riesz space we have (AJdd = B, for every u € L f . Once more, let W be a non-empty subset of 8. As observed above, the set of all {R},,u E L', partially ordered by inclusion, is a distributive lattice Y with the empty set as smallest element, and the mapping +
{R),
is a lattice homomorphism of ' L onto Y. It may be asked whether the homomorphism can ever be an isomorphism, since in that case we would obtain a theorem analogous to Stone's representation theorem for lattices. The answer, however, is negative for L # (0). Indeed, for u > 0 we have 2u # u, but { R } 2 ,and {R},are always the same. Another question is under which conditions for L and (or) W the lattice Y of all {R},is a Boolean ring or even a Boolean algebra. A sufficient condition for the case that W =&? is presented in the following theorem.
Theorem 35.6. If L has the principal projection property, the lattice of all sets {M}", u EL', is a Boolean ring. For any fixed e > 0, the subset of all
216
[CH. 5 ,
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5 35
{hi},,0 5 u 5 e, i A now a Boolean algebra with the empty set as smallest element and { M } =as largest element. I f e is a strong unit in L , the lattice of all sets { M } , ,u E L + ,is a Boolean algebra. Proof. We have to prove that for { M } , c { M } , there exists an element w E L+ such that { M } , n { M } w is empty and { M } , u { M } w = { M } , . Without loss of generality we may assume that 0 6 v 5 u, because if necessary v may be replaced by v1 = inf (u, v ) . Indeed, we have { M } u l = { M } , n { M } , = {M},. Hence, we assume that 0 6 v S u holds. On account of the principal projection property the component v2 of u in the band B, generated by v exists, and we have B,, = B, (indeed, v2 E B, implies B,, c B, and v 5 u implies that their components v and v2 in B, satisfy u 5 v 2 , so B, c B,,). It follows that { M } u z= {M},,. Let w = u - v 2 . Then w EL+,u = v2+w, inf ( v 2 , w ) = 0,
sup ( v 2 , w )
= u,
"
so { W U n { M I w = {MIu2n { M I wis empty and { M } , { W W = {W,, u {MIw = {MI,. We proceed with some properties of the sets {P},,{Q}, and { M } , which, in the next section, will give rise to topological compactness properties. Before stating these properties, we make a few simple observations about the ideal generated by a given system (uT: z E {z}) of elements of L+. This ideal consists of all f satisfying n
for appropriate real ai(i = 1, . . ., n ) and appropriate uT,(i= 1, . . ., n). Since it is evident that, for v l , . . ., v,, E L + ,we have v1
+ . . . +v,
6 n sup ( v l ,
. . ., v,,),
the ideal under consideration may also be described so as to consist of all
f such that
If I
.
6 sup,iT'(
'Y
'Tn)
holds for appropriate uT,, . . ., uTnand for an appropriate real number c (depending upon f). For the following theorems, compare Theorems 6.7, 6.8 and 6.9.
Theorem 35.7. (i) Given e E L + and the set (ur : z E {z}) in L+ such rhat {P}=c u,{P},,, there exist indices zl,. . ., T,,in the index set {z} andapositive number c such that e
5 c SUP
(ur,,
--
-3
urn),
CH.
5 , s 351
HULLS A N D KERNELS
217
and SO {P},c Uy=1 {P},,,. (ii) Given e E L +and the set (u, : z E {z}) in L' such that { Q } , c U,{Q},, there exist indices zi, . ..,z, in the index set {z} and a positive number c such that e Ic SUP (%,, U,"),
--
-9
andso {Q>e c Uy=1 {Q}uzi* (iii) Given e E L +and the set (u, : z E {z}) in L+ such that 2 ' c U,{Q'},, there exist indices zl, .. z, in the index set {z] and a positive number c such that e c sup (u,,, . . ., u,,,).
.,
and so 9, c
uy={Q'},,,.
Proof. (i) Let {P},c u T { P } u eand , assume that e is not included in the ideal A generated by the system of all u,. Then there exists a prime ideal P =I A such that e is not contained in P. It follows that P E {P},,but for no z the ideal P is a member of {P},.,simply since u, E A c P. This contradicts {P},c u,{P}us.Hence, e must be in the ideal A generated by the system of all u,, i.e., there exist u,, ,. . ., u," and a positive number c such that e
c sup (u,,,
. . ., urn).
(ii) and (iii) The proof is similar since the prime ideal P in part (i) can be chosen so as to be maximal with respect to the property of not containing e.
Theorem 35.8. Let W be a non-empty subset of 9 such that A c 9, and let (u, :z E {z}) be a set in L+ such that the collection of all {R},,*has the finite intersection property (i.e., every finite intersection of sets in the collection is non-empty). Then the intersection of all {R},=is non-empty. Proof. Since any finite intersection n ~ = l { R } uisr lof the form {R}" for v = inf (u,,, . . ., uTn),it follows from the finite intersection property of the collection of all {R}u, that the collection of all elements u,, together with their finite infima, is a lower sublattice in L'. This lower sublattice is included in a maximal lower sublattice s, and so the set theoretic difference L+ - S is the positive part of a minimal prime ideal M. Since no u, is a member of M and since M is a member of 9, we have M E n,{R},,,, i.e., the intersection of all {R}u, is non-empty. As observed above (in Thcorem 35.6), it can happen that the lattice of all { M } , , ,u E L', is a Boolean ring. In this case a theorem analogous to Theorem 35.7 holds, as follows.
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Theorem 35.9. Ifthe lattice of all { I V }is~ a Boolean ring, and ifthe elements e E L + ,u, E L+(zE {z}) satisfy { M } e c u , { M } u r ,there exist indices z l , . ., z, in the index set {z} such that { M } e c {M},,<.
.
u:=
Proof. Replacing, if necessary, all u, by inf (e, u,), we may assume without loss of generality that 0 5 u, 6 e holds for all z, and so { M } e = {M},z. Since the set of all { M } , is a Boolean ring, the set theoretic complement of any with respect to ( M } e is of the form { M } u r ;it follows now from {M}e= {M},r that { M } , * is empty. But then, by the preceding theorem, there must be indices zl,. . ., z, such that {M}u,t is already empty, i.e., { M } e = ul=l{M}u,i.
u,
u,
n,
36. The hull-kernel topology As in the preceding section, let 8 be the set of all proper prime ideals in the Riesz space L, and let 9 be a non-empty subset of 8.Also, just as before, given u E L+,let { R } , be the set of all R E W such that u is no member of R. We recall that {R}ois empty and
u ({R}u:
# E L + )= 9.
n!=l
We introduce a topology i n 9 by choosing all sets of the form { R } , as a subbase for the topology. Since any finite intersection {R},, is of the form {R},o for uo = inf ( u l , . . .,u,,), the sets { R } , form actually a base for the topology. This method of introducing a topology i n 9 is, of course, analogous to the method used in section 7, where W is a set of prime ideals in a distributive lattice. For the following theorem and its corollary, compare Theorem 7.1 and Corollary 7.2.
Theorem 36.1. Let 9be topologized as explained above. For any ideal A in L, the hull h(A) = ( R : RE^, R 3 A )
is a closed set in this topology. Conversely, every closed set is the hull of some ideal A. Furthermore,for any subset g1of 9, the closure of 9,is the set h(k(Wl)),and for thi$ reason the topology is usually called the hull-kernel topology in 9. The hull-kernel topology in 9 is the relative topology of the hull-kernel topology in 8. Proof. The closed sets are the intersections of sets of the form (R :R E 9,
u E R), i.e., the closed sets are the sets
h(D) = (R : RE&?, D c R),
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THE HULL-KERNEL TOPOLOGY
219
where D is an arbitrary non-empty subset of L. Observing now that h ( D ) = h(AD),where AD is the ideal generated by D, we obtain the result that a subset of W is closed if and only if it is the hull of some ideal in L. For the proof of the statement about the closure of a subset of W ,let W , be an arbitrary subset of W . Evidently, 9, is contained in the closed set h(k(Wl));in order to prove that h(k(W,)) is the closure of W , we have to show that h ( k ( 9 , ) ) is included in any closed set W , = W ,. Any 9,of this kind is of the form W, = h(A) for some ideal A in L, so h(A) 3 W l . Then W ( h ( A ) ) )=
)>Y
i.e., h(A) =) h(k(W,)) by Theorem 35.2 (ii). But h ( A ) = W,, and so W , = h ( k ( 9 , ) ) which is the desired result. Finally, it follows immediately from the definitions that the hull-kernel topology in W is the relative topology of the hull-kernel topology in 8.
Corollary 36.2. Let the subset W of B satisfy the extra condition that the intersection of all R €93'is (0). Then, in order that the subset B of L be the disjoint complement Ad of some ideal A , it is necehsary and suflcient that B be the kernel of some open subset of 9. Proof. Follows from the formula Ad = k(h(A)E)proved in Theorem 35.3.
We recall that a topology in a point set Xis called a To-topology (Kolmogoroff topology) whenever for any two different points in X cne at least of these points has a neighborhood not containing the other one. The topology is called a T,-topology whenever for any two different points in X each of the points has a neighborhood not containing the other point, and a Hausdorff topology (as well-known) whenever each two different points have disjoint neighborhoods. The next theorem is analogous to Theorem 7.3. There is a significant difference now between the lattice situation in Theorem 7.3 and the present Riesz space situation. In the lattice situation, if P, and P2 are prime ideals, neither of them included in the other one, P, and P2 can be Tl-separated; in the present situation they can be Hausdorff separated.
Theorem 36.3. The hull-kernel topology in 9 is a T,-topoIogy, the base sets (P}. of which are open and compact. ZfP, andP2 arepoint3 of B such that, considered as ideals in L, neither of them is included in the other one, then P1 andP2 can be Hausdorf separated, i.e., there exist neighborhoods of P, andP, that are disjoint. This holds in particular is P1 and P, are minimal
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PRIME IDEALS
[CH.5 , s 36
prime ideals or if P, and P, are both maximal with respect to the property of not containing the same element uo E L + ,and (even more specially) ifP, and P2are maximal ideals.
Proof. If P1 # P2,there exists an element u1 E P, such that u, is no member of P, or there exists an element u2 E P, such that 1.4, is no member of P, . In the first case {P},, is an open neighborhood of P, such that P, is not contained in this neighborhood. In the second case {P},, is an open neighborhood of P , such that Pz is not contained in it. It follows that the hull-kernel topology in 9 'is a To-topology. Evidently, every set {P}, is open by definition. By Theorem 35.7 (i) every set {P}, is compact. In general we can expect no better result than To-separation, because for P1 included in P, every neighborhood of P2 is also a neighborhood of P , . If P, and Pz are prime ideals such that neither of them includes the other, there exist u, E P1 and u2 E P, in L+ such that u1 is no member of P, and u2 no member of P1. Note that inf (u, ,24,) E P, n P,. Now, let v1 = u,-inf ( u l , u,),
v, = u2-inf (u,, u,).
Then v , , v2 E L + ,v1 is in P1 but not in P,, and v, is in P2 but not in P,. Furthermore, inf (vl ,v,) = 0,so
{Plvl n { P I u z
= { P l i n r ( u 1 . u z ) = {PI0
i.e., {P},,, and {P}uzare disjoint. Hence, {P}u,and {P}v,are disjoint neighborhoods of P2 and P, respectively. The next theorems are analogous to Theorem 7.4 and 7.5. Theorem 36.4. (i) The hull-kernel topology in 9 is a To-topology, the base sets {Q}, of which are open and compact. (ii) For any e > 0 in L+,the space 2?e is a compact Hausdorffspace in its hull-kernel topology; the ba$e sets {2?e}, are open. The topology is, therefore, ' is eactly the set f of all maximal normal. If e is a strong unit in L, then 2 ideals. Hence, i f L has a strong unit, the set f of all maximal ideals is a compact Hausdorff space in its hull-kernel topology. (iii) The hull-kernel topology i n d is a Hausdorff topology, the base sets { M } , of which are open as well as closed. The topology i3, therefore, completely regular.
Proof. (i) Similarly as in the preceding theorem. The compactness of the sets {Q}, follows from Theorem 35.7 (ii). (ii) By definition, the base sets {p},, u E L + ,are open. The compactness
CH. 5 , s 361
221
THE HULL-KERNEL TOPOLOGY
Qr
of the space 9 ' follows from Theorem 35.7 (iii). Given and QZ in 9e, neither of these is properly included in the other one (otherwise the smaller one would not be maximal with respect to the property of not containing e), so Q; and Qi have disjoint neighborhoods (by the same argument as in the preceding theorem, or by simply observing that the hull-kernel topology in is the relative topology of the hull-kernel topology in 9).Hence, is a compact Hausdorff space in its hull-kernel topology. (iii) Given MI and M2 in A, neither of these is properly included in the other one, so MI and M 2 have disjoint neighborhoods. The hull-kernel topology in A is, therefore, Hausdorff. The base sets {M}, are closed by Corollary 35.4 (ii), stating that h(k({Mlu))= {MIu holds for every u E L+.In order to show that the topology is completely regular, observe that if the point Mo E Aand the closed subset 9 of A are given such that Mo is no member of 9,then the open set .A - 9 contains a neighborhood of Mo of the form {M},o for some uo EL'. The function f ( M ) , satisfying f ( M ) = 1 for all M E {Mju0and f ( M ) = 0 for all other M , is now continuous on A on account of {M},o being open as well as closed.
We recall that a topological space is sometimes called totally disconnected if there exists a base consisting of sets that are open as well as closed. Note that in this case the collection of all sets that are open as well as closed is also a base. According to the last theorem the hull-kernel topology in A is a totally disconnected Hausdorff topology. Theorem 36.5. In the particular case that the lattice of all { M } , is a Boolean ring, the base sets { M } , of the hull-kernel topology in d are not only open and closed, but also compact. The topology is, therefore, a totally disconnected and locally compact Hausdorfltopology. Every subset of d which is open as well as closed and included in one of the { M } ,, say { M } , o , is itself of the form { M } v ofor some vo EL+satisfying 0 5 vo uo.
Proof. The compactness of the sets { M I , follows from Theorem 35.9. For the proof of the last statement, let g be a subset of 4, open and closed and such that 9 c {M},,o. Since 93 is a closed subset of the compact subset { M } , o , the set 93 is compact. Also, since is open, 93 is the union of a collection of base sets { M ) , with 0 5 u, 5 uo for all T. By the compactness 93 is already a finite union of base sets, say { M ) , , , , . . ., { M } , , . But then 93 = { M } o ofor uo = sup (u,,, . . ., u,,).
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Let the topological space S with points p , q, . . . be a To-space. For any p , q E S ,setp 5 q whenever q E (p},where (as usual) (p) denotes the closure of the set consisting of the point p only. It is easily verified that this defines a partial ordering in S (the hypothesis that S is a To-space is used in order to prove that p =< q, q 5 p implies p = q). Obviously, the partial ordering is the relation of equality if and only if the space is a T,-space (i.e., if and only if every point is a closed set). We apply these remarks to the topological space 8 of all proper prime ideals in the Riesz space L, the topology being the hull-kernel topology. The next theorem is similar to Theorem 7.6.
Theorem 36.6. (i) Introducing a partial ordering in 8 by dejining that P , S P2 whenever P, E {Pi}, we haue P1 S P , if and only if P, c P 2 , and hence P2E (pl> if and only if P , c P 2 . (ii) For any u > 0 in L' we haue P E (p), if and only if there exists a minimal prime ideal M E {P}, such that M c P. It follows that
(p>, and so certainly
=
-
(J ((M) : M E & ,
{P},=
M E {P},),
u ((p}
:P E 8,P E {P},).
Proof. (i) Given that P, c P2,we have to prove that P2E {p1>,I.e., ' we have to prove that every neighborhood of P2 contains P,,and it is sufficient to do this for any neighborhood of P2of the form {P}, . Hence, assume that P, E {P},, i.e., u is no member of P,. Then u is no member of P,, so PI E {P},.This is the desired result. Conversely, if P, E (pl>, then every {P}, satisfying P2 E {P},contains P,,so if u is no member of P2 then u is no member of P, In other words, u E P,implies u E P,, i.e., P, c P,. (ii) Let first u > 0, M c P and M E {P},.Then, by part (i), we have P E (M},and so it follows from M E {P}, that P E (M) c {p>u. Conversely, assume now that Po E Then every neighborhood of Po has a non-empty intersection with {P},;in particular, every set {P}"for which u is no member of Po has a non-empty intersection with {P},.It follows that the collection of all sets
.
mu.
({P}"n {P},: u no member of Po) has the finite intersection property, and so by Theorem 35.8 the intersection of all these sets is non-empty. This non-empty intersection contains at least one element (i.e., at least one prime ideal), and hence it contains at least one minimal prime ideal A4 (since every prime ideal contains a minimal prime
CH. 5 , s 371
COMPACTNESS AND SEPARATION PROPERTIES
223
ideal). Then M E {P},and also M E { P } ufor all v not in P o , i.e., u is not in M if v is not in P o . In other words, we have M E {P},and M c P o . The theorem shows that in the hull-kernel topology of 8 the closure of any set {P},is simply the union of the closures of each of its points. It is even so that the closure of {P},is already the union of the closures of the {P},,it follows points in {P},that are minimal prime ideals. Since B = that B = U(p>" = ((M) : M E & ) ,
u
u
where the closures are to be taken, of course, in the topology of 8.It follows immediately that (as a subset of 8)is dense in 8,i.e., 2 = 8.
Exercise - 36.7. Show that, in the hull-kernel topology of 2, we have {Q,} if and only if Q , c Q , . Show also that, for any u > 0 in L', we have Q2 E
(e>.= U((~>:QE%WQ},).
Exercise 36.8. Show that A, in its hull-kernel topology, is a Baire space (i.e., any countable intersection of open and dense sets is dense). Hint:cf. Exercise 7.7. Exercise 36.9. It was proved in Theorem 36.4 (iii) that the base sets {M}, in the hull-kernel topology of A are open as well as closed. A partial converse is as follows. Let W be a subset of 9 such that the intersection of all R E W is equal to {0}, and all base sets {R}, of the hull-kernel topology i n 9 are open as well as closed. Show that W is a subset of A. In particular, if all base sets {P},of the hull-kernel topology in 8 are closed, then every proper prime ideal is minimal (compare parts (i) and (iv) of Theorem 37.6). Hint: Take a fixed R, €9. If R, = {0}, then R, is certainly minimal, so we may assume that R, # (0). Given u > 0 in R,, we have that R, is no member of the closed set {R},, so there exists a base set {R}" such that Ro E {R},, and { R } , , are disjoint (i.e., inf (u, v ) = 0). Hence, for every u E R:, there exists an element v EL' - R,f such that u Iv. Now use Exercise 33.9. 37. Compactness and separation properties of the hull-kernel topology We use the same notations as in the preceding section, and we recall that it was proved that all sets {P},are compact in the hull-kernel topology of 8, all sets {Q}, are compact in the hull-kernel topology of 2, but a set of the
224
[CH. 5,s 37
PRIME IDEALS
form { M } , need not be compact in the hull-kernel topology of A. We will investigate now under which conditions we have (i) (ii) (iii) (iv) (v)
compactness of 9, compactness of 9, compactness of {M},,, for a given uo E L + , compactness of all { M } , , compactness of A.
We also recall that, in general, 9 is a To-space but no T,-space. We will make some remarks about conditions under which 9is a T,-space, a Hausdorff space or a compact Hausdorff space respectively. The first theorem which follows is analogous to Theorem 8.1.
Theorem 37.1. (i) B is compact in its hull-kernel topology ifand only i f L has a strong unit. (ii) 9 is compact in its hull-kernel topology i f and only i f L has a strong unit.
u
Proof. (i) If B is compact, it follows from B = {P},that 9is already a finite union of sets {P},,, say B = {P},,, . But then B = { P } , holds for e = sup ( u l , . . uJ. Hence, since {P},c 9 = {P}=holds for every u EL', the element u is contained in the ideal generated by e (cf. Theorem 35.5 (i)), which shows that e is a strong unit in L. Conversely, if e is a strong unit in L, it follows (once more from Theorem 35.5 (i)) that {P},c { P } e holds for every u EL', so B = {P},= { P } e . Every {P},is compact; hence, in particular, B = {P}=is compact. (ii) The proof for 9 is similar.
.,
u;=
u
Before proceeding with our program, we introduce some furtber terminology and notation. The elements u, u in L+ will be called .&-equivalent whenever { M } , = { M } " , i.e., whenever (AJdd = (A,)dd, where A , and A , denote the ideals generated by u and v respectively (cf. Theorem 35.5 (iii)). We will write u 3 .(A)in this case. Note that u = O ( A ) implies u = 0, and u1 E u l ( A ) , u2 = v 2 ( A ) implies sup ( u l , u z ) = sup ( v l , v2)(.&)
and inf (ul, u 2 ) = inf ( v l , v z ) ( A ) .
In this case we have also that u1 +uz E ui + v Z ( A ) . Indeed, u1 +uz and sup ( u l , u z ) generate the same ideal, so u1 + uz = sup (ul ,uz)(.&). Similarly v1 v z = sup (ul , vz)(.&). Hence we have u1 uz = v1 ~z(A). The element u,, EL' is said to have the .&-complementation property if, given any u E L f satisfying 0 6 v 6 u o , there exist v l , vz in L t such that
+
+
+
CH. 5 , § 371
225
COMPACTNESS AND SEPARATION PROPERTIES
= u(&%), inf ( v l , v 2 ) = 0 and sup (ul ,v 2 ) = uo(&).
Instead of sup (v1,v2) we may just as well say that u l + v 2 = uO(&). It is evident from the remarks above that, if so desired, we may assume without loss of generality that vl = v. Hence, uo has the &%-complementationproperty whenever, given that 0 5 v 5 u,, , there exists an element v2 E L +such that u2 Iv and v + v2 E uO(&). The element v2 is now called a pseudo complement of v with respect to uo. It follows immediately that uo has the A-complementation property if and only if, for any v E L+ such that { M } , , c { M } , o ,there exists an element u2 E L + such that is the disjoint union of { M } uand { M } u 2 In . other words, uo has the &-complementation property if and only if the initial segment ({W": { W U = { W U O ) u1
= uo(&)
of the lattice of all {M}, is a Boolean algebra. For the next results, compare Theorem 8.2, Corollary 8.3 and Theorem
8.4.
Theorem 37.2. The set { M } u ois compact in the hull-kernel topology of JI if and only q u o has the &-complementation property, i.e., if and only if every v satisfying 0 5 v 5 uo has a pseudo complement with respect to uo. In this case, every subset of { M } u owhich is open as well a b closed is of theform { M } , for some u EL+. Proof. Assume first that uo has the &%-complementationproperty, i.e., the initial segment ( { W U :{ W U = { W U O ) is a Boolean algebra. In order to prove that is compact, it is sufficient to prove that if { M } , o c {M},*,then { M } u ois already covered by a finite union of the {M}"*.Replacing, if necessary, every u, by inf (uo, u,), we may assume without loss of generality that 0 5 u, 5 uo holds for all 7 : and so {M}, = {M}"*.Since uo has the &-complementation property, the set theoretic complement of any {M},* with respect to {M}, is of the form { M } u r ;it follows that { M } u sis empty. By Theorem 35.8 there exist indices zl,. . ., z, such that nl=l{M}uzlis already empty, i.e., {M},,o =
u,
uz
nt
uConversely, =: assume that 1{'}u71
'
{ M } , is compact. Then, given any { M } , ,satisfying { M } u c {M},o, the set theoretic difference D = { M } , o - { M } u is an open and closed subset of { M } u o .Since D is open, we have D = {M},for suitable elements u, E L+;since D is closed and { M } , o is compact, we
u,
226
[CH.5, $37
PRIME IDEALS
have that D is compact. Hence, similarly as above, D is a finite union of sets { M } , , , . ., {M},", and so D = { M } v 2for v2 = sup ( u l , ., u,,). This shows that u,, has the .4Z-complementation property. The last part of the proof shows also that any subset of {M},o which is open as well as closed is of the form { M } , for some u E L'.
.
..
Corollary 31.3. The following conditions are equivalent. (i) All sets { M > ,are compact in the hull-kernel topology of A. (ii) The lattice of all { M } , is a Boolean ring. (iii) For all u, v E L +satisfying 0 v 5 u, there exists a pseudo complement of v wirh respect to u. Proof. Follows immediately from the last theorem. Note that it was proved already in Theorem 36.5 that (ii) implies (i).
Theorem 31.4. The following conditions are equivalent. (i) Jf is compact in its hull-kernel topology. (ii) All sets { M } , are compact in the hull-kernel topology of A, and there exists an element e E L' such that (Aeyd = L (we recall that an element e of [his kind is called a quasi unit in L ; cf. Dejinition 21.4). (iii) The lattice of all { M } , is a Boolean algebra. (iv) For any u E L' there exists an element v E Li such that v 1u and z = u+v is a quasi unit in L. Proof. (i) * (ii) Every set { M } , is a closed subset of the compact space A, so { M } , is compact. Furthermore, it follows from the compactness of .A and from A = U { M } , that A!? is already a finite union of sets { M } , , say A = ul=l{ M } , t . But then A = { M } e for e = sup ( u l , . . ., u,). Hence, since { M } , c A = { M } e holds for every u EL', we have u E c (AJdd for every u E L+ (cf. Theorem 35.5 (iii)), and so (AeYd = L. (ii) (iii) All { M } , are compact, so the lattice of all { M } , is a Boolean ring by the preceding corollary. Since (A,)dd c L = (Ae)'' for all U E L ' , we have { M } , c { M } e for all U E L ' , so .A = { M } , = { M } e , which shows that { M } e is the largest element in the Boolean ring of all { M } , . In other words, the Boolean ring is a Boolean algebra. (iii) (iv) Let { M } ebe the largest element in the Boolean algebra of all { M } , , so { M } e = A!?. Given u E L', the set theoretic complement of { M } , with respect to A = { M } e is of the form { M } " for some v E L'. It follows that inf (u, v ) = 0 and w = sup (u, v ) = e ( A ) , i.e., v 1u and (Aw)dd= L. Observing now that w = sup (u, v ) and z = u+v generate the same ideal, we obtain = L.
-
u
CH.5, 8 371
CoMPACTNES$ AND SEPARATION PROPERTIES
227
(iv) =-(i) Note first that (iv) implies (iii). It remains to prove that (iii) implies (i). Let e be a quasi unit in L, i.e., let { M } e be the largest element in the Boolean algebra of all sets { M } , , so A? = {M}, = { M } e . Since the set of all {M}, is a Boolean algebra by hypothesis, and hence is a Boolean ring, the sets {M}, are compact by the preceding corollary. Hence, .A? = { M } eis compact.
u
as the set of all P E B such that For any f E L we defined {P}/= {P},,, f, and hence If 1, is no member of P. We now generalize this as follows. For any non-empty subset D of L,let {P}, be the open subset of 9, defined by {P>D =
u ({PI, :f
E 0).
It follows from the definition that {PID = {P}AD,where AD denotes the ideal generated by D. Furthermore, given the ideals A and B in L, we have { P } A = { P } Bif and only if A = B. Indeed, if { P } A = {P}B and iff E A, then the compact set {P},is covered by ( { P } # : g E B), so {PI,is already covered by a finite union {P}#,. But this finite union is equal to {P},, for h = sup (Igll, . ., 1g.l) E By so {P}, c {P},, with 0 S h E B. It follows that f is contained in the ideal Ah c By so f E B. Similarly, any g E B satisfies g € A . Thus, the mapping A -, { P } A is one-one. Our next remark is that every open subset 0 of B is of the form 0 = { P } A for some ideal A in L; more precisely, 0 = { P } A for A = (f:{P},c 0).
.
uy='=l
u
The mapping A --+ { P } A is, therefore, a one-one mapping of the set of all ideals in L onto the set of all open subsets of 9. Note that for A a principal ideal, say A = A,, we have { P } A = {P},. The set of all ideals in L and the set of all open subsets of B are lattices with respect to partial ordering by inclusion. In the set of all ideals in L we have
A l v A 2 = A 1 + A 2 = (f+g:fEAl,gEA2),
A l ~ A= 2A,nA2,
and in the lattice of all open subsets of B supremum and infimum of two open sets are their union and intersection respectively. Note now that the one-one mapping A + { P } Ais a lattice isomorphism, i.e., { P I A ~ V A= ~ {f'}AlhA2
{P}AI u { P I A Z y
= { P I A , n {P}AZ-
For any subset d of 9,denote the set theoretic complement of d by d", the interior of d by do, and the closure of d by d - .Then
228
[CH. 5, $37
PRIME IDEALS
holds for any non-empty subset D of L. For the proof of the first formula, note that ({P}D)eOis the largest open set disjoint from {P},, SO
u
u
({P}DYo= ( { P } :~g If for a1lj-E 0)= ( { P }:~ The second formula in (1) follows now immediately.
E
0")= {P}Dd*
Theorem 37.5. (i) The ideal A in L satisfies A = Addifand only i f { P } , is a regularly open subset of 8 (we recall that a subset S of a topological space X is called regularly open whenever S-" = S). (ii) The ideal A in L is a projection band (i.e., A @ Ad = L ) if and only i f {P }, is not only open but also closed in the hull-kernel topology of 8.In particular, the principal ideal A, is a projection band if and only i f { P } , is a closed subset of 9. Proof. (i) This follows immediately from the second formula in (1). (ii) Let A be a projection band in L,so A @ Ad = L, which (in the lattice structure of the set of all ideals in L) is the same as A v Ad = L.It follows that {PIAu {P},d = 8 with {P}, and {P},d disjoint. This shows that {P}, is the set theoretic complement of the open set {P},d, so {PIAis closed. Conversely, if {PI, is closed, it follows from the first formula in (1) that {P}, and {P},d are set theoretic complements in 8,so A v Ad = L, i.e., A @ Ad = L.
The results in the last theorem are analogous to results in Exercise 8.8. For the next theorem, compare Theorems 8.5 and 8.6. Theorem 37.6. The following conditions are equivalent. (i) 9 = A. (ii) Every proper prime ideal in L is a maximal ideal, i.e., 9 = $. (iii) The hull-kernel topology in 9 is Hausdorfl (iv) For every u E L', the set {PIuis not only open but also closed in the hullkernel topology of 9. ( v ) Every principal ideal in L is a projection band. (vi) The hull-kernel topology in 8 is a T,-topology. (vii) The quotient Riesz space LIA is Archimedean for every ideal A in L. Also, i f one (and hence each) of these conditions is satisfied, then 8 = 2 holds. Proof. (i) => (ii) Let 8 = A, and assume that P is a prime ideal and A an ideal such that P c A c L holds with A # L. We have to show that A = P. The ideal A is prime (since every ideal containing a prime ideal is itself a prime ideal); every prime ideal is minimal by hypothesis, so A is a minimal prime ideal. But then it follows from P c A that A = P.
CH.5 , s 371
229
COMPACTNESS AND SEPARATION PROPERTIES
(ii) =- (iii) If every proper prime ideal is maximal, and P,,P2are different proper prime ideals, then neither of them is included in the other, so P, and P, are Hausdorff separated by Theorem 36.3. (iii) * (iv) For every u EL', the set {P},, is a compact set in the Hausdorff space 8,so {P},, is closed. (iv) o (v) This equivalence was proved in part (ii) of the preceding theorem. (iv) =- (vi) It will be sufficient to prove that every point Po E 8 is a closed set in the hull-kernel topology of 9. Since
n ( { P } :~f not in Po)
=
(P: P prime, P
c
Po),
and since all {P}/are closed, the set of all prime ideals P c Po is closed. Call this set 9.It follows that the closure of the set (Po) consisting only of Po is contained in 9, i.e., W P O N c 233.
Observing now that h{k(Po)}consists of all prime ideals P =) Po,we conclude that (P: P prime, P 3 Po) c (P:P prime, P c Po). This shows that the only member of the set on the left is Po itself, i.e., h{k(Po)} F (Po).Hence, (Po)is closed. (vi) (i) Let 8 be a T,-space, and assume that P,, P, are different proper prime ideals. Then P, c P, is impossible. Indeed, if P1 c P2 holds, then P, e {P}/implies PIe {P}/,so every neighborhood of P, is also a neighborhood of P,, contradicting the assumption that P, has a neighborhood not containing P,.Similarly,P, c P, is impossible. Hence, neither of Piand P, is included in the other one. This implies that every proper prime ideal is minimal, because otherwise there exists a proper prime ideal P, which is not minimal, and then PI properly contains a minimal prime ideal P,. It remains to prove now that condition (vii) is equivalent to conditions (i)-(vi). We will prove that (ii) and (vii) are equivalent. For this purpose, we first observe that for any given ideal A in L there exists a one-one correspondence between the ideals B containing A and the ideals [ B ]in LIA, as follows. Given the ideal B 3 A , we let [ B ] consist of all L f ] E L / A such thatfE B holds; conversely, given the ideal [ B ] in L/A, we let B be the set of allfe L such that [ f ] e [ B ] holds (cf. Theorem 27.2). It follows easily that B is a prime ideal containing A if and only if [ B ]is a prime ideal in L/A. Also, B is a maximal ideal containing A if and only if [ B ]is a maximal ideal in LIA.
230
PRIME IDEALS
[CH.5,s 37
Hence, if B = 3 holds in L , then every prime ideal in L / A is a maximal idealin LIA, and so every principal ideal in L / A is a projection band. This shows that L / A has the principal projection property, so LIA is surely Archimedean. Conversely, let LIA be Archimedean for every ideal A , and assume that there exists a proper prime ideal that is not maximal. In other words, assume that there exist two different proper prime ideals P , and Pz satisfying P , c P z . Then LIP, contains the non-trivial ideal [ P z ] .Also, L/P,is linearly ordered (since P1 is prime). Any linearly ordered Riesz space containing a non-trivial ideal cannot be Archimedean (every nonzero element in the ideal is infinitely small with respect to every element not in the ideal), so LIP, is non-Archimedean, contradicting our hypothesis. Hence, every proper prime ideal in L is maximal. Finally, if all conditions are satisfied, it follows immediately from B = f that 9 = 9 holds. Of course, it follows now immediately that 9is a compact Hausdorff space in its hull-kernel topology if and only if all the conditions in the last theorem are satisfied and, in addition, L has a strong unit (because, by Theorem 37.1, the compactness of 9is equivalent to the existence of a strong unit). We can say more, however. In the first place, if all the conditions in the last theorem are satisfied and, in addition, L has a weak unit e (equivalently, a quasi unit e), then e must be a strong unit. Indeed, weak units and quasi units are the same since L is Archimedean, so assume that e is a weak unit, i.e., the band generated by e is the space L itself. According to our hypotheses, the ideal generated by e is a projection band, so the ideal generated by e is equal to the band generated bye, i.e., the ideal generated bye is L itself. This shows that e is a strong unit. In the following characterization of spaces L for which B is Hausdorff and compact we may assume without loss of generality that L # (0). Theorem 31.1. The hull-kernel topology in the set 9 of all proper prime ideals of the Riesz space L is Hausdorfland compacr if and only i f there exists a field (algebra) r of subsets of a non-empry point set X mch rhar L is Riesz isomorphic to the Riesz space M of all finite linear combinations (with real coeficients) of characteristicfunctions of sets belonging to r (partial ordering in M pointwise).
Proof. Assume first that M is a Riesz space of step functions as described in the statement of the theorem. The function e, satisfying e ( x ) = 1 for all
CH.5 , s 371
23 1
COMPACTNESS A N D SEPARATION PROPERTIES
x E X, is obviously a strong unit in M. Furthermore, given any f
C ( f ) be the set in r, defined by
E M,
let
C ( f ) = (x : x E X , f ( X ) # O), and let N ( f ) be the set theoretic complement of C ( f ) with respect to X. Given the function g E M, the functions gx,-(,) and gx&(,) are again in M; more precisely, they are members of the principal ideal A, and the disjoint complement A; respectively. It follows that A, is a projection band. Hence, M has all the properties listed in the preceding theorem; in addition, M has a strong unit. As a consequence, the space M (and then also any Riesz isomorphic space) has the property that the hull-kernel topology in the set of all proper prime ideals is Hausdorff and compact. Assume now, conversely, that L is a Riesz space such that the hull-kernel topology in the corresponding set 8 is Hausdorff and compact. In view of what was proved in the preceding theorems, L is then surely an Archimedean Riesz space with a strong unit, so the results proved in section 27 about maximal ideals are applicable. Since, in the present case, every proper prime ideal is maximal, this simply means that the results proved in section 27 hold for the set 9 of the proper prime ideals. According to these results (cf. in particular the paragraph immediately following upon Theorem 27.6) L is Riesz isomorphic to a Riesz space "L of real functions defined on the set B = $. Iff E L corresponds to "f E "L and if P E 8,then "f (P)is defined to be the (unique) real number a satisfyingf - ae E P. We shall prove now that, under the present hypotheses, each function "f assumes only a finite number of values. Indeed, for any given real number I, consider the set (P: " f ( P ) = I). This is the set theoretic complement (in 9) of the set
(P:"f(P) # A )
= {P},-ae,
so, since { P } f - a , is open as well as closed, the same holds for the set (P : "f(P) = A). Now, 9 is the union of all sets (P: " f ( P ) = I) for A varying over all real numbers, and S is compact. It follows that 9 is a finite union of these sets, say
B
n
=
(J (P : " j ( P ) = I i ) .
i= 1
.
This shows that "f assumes only the values I,, . ., I,. We proceed to collect further information about the function "f E "L, assuming the values I,, ., I,. Without loss of generality we may assume that I l > A2 > . . > I". For i = 1, . ., n, let gi be the set of all P E 9
.
..
.
232
[CH.5,s 37
PRIME IDEALS
such that "f(P)= 1,holds. We will prove that the characteristic functions of 8,,.. ., 8"are functions in "L.One possible proof for this fact is as follows. By addition to "f of a suitable multiple of the function "e (we recall that "e(P) = 1 for all P),we obtain a function "f,E "L with values 1; > 1; > . . . > 1; such that 1; > 1 and A: < 1 for i = 2 , . . ., n. Then
"f,= sup ("f,,"e)
E
"L,
and "f, assumes only the values 1; (on 8,)and 1 (elsewhere on 9). But then "f3 = "f,- "e is also in "L, and obviously the characteristic function x1 of 9,is a multiple of "f3.Hence x, E "L.Returning to the function "fi, we may assume without loss of generality that A; > 1 and 0 < 4 < 1, so "f,-A;-x, has its largest value equal to A;. Repeating the argument used before, we obtain xz E "L (where xz is the characteristic function of 8,), and so on. Every "fE "L thus gives rise to a finite number of subsets of 9 such that the characteristic functions of these subsets are elements of "L and "fis a finite linear combination (with real coefficients) of these characteristic functions. Let I' be the collection of all subsets of 8 thus obtained. It remains to prove that r is a field of subsets of 8,i.e., we have to prove that 8 E r and that 9,,8,E r implies 8, u 8, E r and 8,-9, E I'. This follows easily by observing that the characteristic function of B is exactly "e, the characteristic function of 8, u 8, is the supremum of the characteristic functions of 8, and 8, and the characteristic function of 8,- 8, is the infimum of the characteristic functions of 8, and 8, subtracted from the characteristic function of 8,. Since the characteristic functions of 8, and 8, are functions in "L,the same holds then for the characteristic functions of 8, u 8, and 9, - B,, so the desired result follows.
As an example where the hull-kernel topology in 9 is Hausdorff but not compact (i.e., an example where all conditions of Theorem 37.6 are satisfied but L has no strong unit) we mention the case that r is a ring, but not a field, of subsets of a non-empty point set X , and L is the Riesz space of all finite linear combinations (with real coefficients) of characteristic functions of sets belonging to r. Particular examples are obtained by taking for X any infinite point set and for r the ring of all finite subsets of X , or for X a point set equipped with a measure ,u such that ,u(X) = 00 and for r the ring of all subsets of finite measure (p-almost equal sets are identified). It was proved in Theorem 35.6 that if L has the principal projection property, then the collection of all sets {M}u, u EL', is a Boolean ring, and in Corollary 37.3 it was proved that the collection of all {MIuis a Boolean ring
CH. 5 , § 371
COMPACTNESS AND SEPARATION PROPERTIES
233
if and only if every {M}, is compact (in A).Hence, if L has the principal projection property, then every {M>yis compact in the hull-kernel topology of .A. Spaces with the principal projection property have another remarkable property. We will prove that in a space of this kind every proper prime ideal contains only one minimal prime ideal (in sharp contrast, for instance, to the space C([O,l]), where every maximal ideal contains infinitely many minimal prime ideals; cf. Exercise 37.13). In the converse direction, it is true for an Archimedean space L that if every { M } "is compact (in .A) and every proper prime ideal contains only one minimal prime ideal, then L has the principal projection property. We begin by proving some lemmas, the first of which is analogous to Exercise 8.8 (vi).
Lemma 37.8. Given the ideal A in the Riesz space L, rhefollowing conditions are equivalent. (i) Ad is a projection band. (ii) Addis a projection band. (iii) {P};' is open and closed. (iv) {P}; is open and closed. Proof. If Ad is a projection band, then so is its disjoint complement Add. If Addis a projection band, then so is Addd= Ad. In Theorem 37.5 (ii) it was proved that the ideal B is a projection band if and only if { P } Bis open and closed. Applying this to B = Ad, we thus find that Ad is a projection band if and only if {P},d = ({P},$O is open and closed. The complement of ({P}A)Co is {P};,so Ad is a projection band if and only if {P}; is open and closed. Applying the above theorem to B = Add,we find that Add is a projection band if and only if { P } , d d = {P};O is open and closed. We will say, by definition, that the Riesz space L has the quasiprincipal projection property if, for every principal ideal A,, the disjoint complement A; is a projection band, i.e., if L = A; 8 A: holds for every f E L. In virtue of the last lemma L has the quasi principal projection property if and only if every subset of B of the form {P}; is open and closed in the hull-kernel topology of 8.For any Archimedean space the quasi principal projection property is the same as the principal projection property (and so, for example, the space C( [0, 11) does not have the quasi principal projection property). It can happen that a non-Archimedean space has the quasi principal projection property. Indeed, it is easily verified that the lexicographically ordered plane has the quasi principal projection property. As observed above, the principal projection property implies that every
234
[CH. 5,537
PRIME IDEALS
set { M } , is compact in the hull-kernel topology of A. The same follows already from the quasi principal projection property. Lemma 37.9. Zf L has the quasi principalprojection property, then every $et { M } , is compact in the hull-kernel topology of A. Proof. We have to prove that every set { M } , is compact. According to Theorem 37.2, this is equivalent to proving that, for u, u EL' satisfying 0 S u u, the element u has a pseudo complement u2 with respect to u (i.e., u, Iu and { U + U ~ }=~ {u}"). ~ Hence, assume that 0 4 v 5 u and let u = u, + u, be the decomposition of u in the direct sum A f d @ A t , so u1 E Afd and u2 E A t . We will prove that u2 is a pseudo complement of v with respect to u. Since u2 Iu is evident, we need only prove that { u + u , } ~=~ {u}". The element u has the decomposition v = u + O in A: @ A t , so 0 u 6 u implies for the components in Atd that 0 S v u1 holds. Then v+u, 5 u, + u , = u, and so { u + u , } ~c~ {u}". It remains to prove that { u } c ~ ~ { v + u , } ~ ~which , is the same as ( U + U , } ~ c { u } ~Let . w be an arbitrary element of { V + U ~ } ~Then . w Iu, and w Iu, so w E A t , which implies w 1u l . It follows that w is disjoint from u,+u, = u, so w E { u } ~Hence, . {u+u,}~ c { u } holds. ~ The next lemma is analogous to Exercise 8.9, parts (iii) and (iv). Lemma 37.10. (i) If L has the quasi principal projection property and u, v EL' are such that {P},and { P } u are disjoint (i.e., inf (u, u ) = 0), then the closures {P}; and {P}; are also disjoint. (ii) Zn any Riesz space L the following conditions are equivalent. (a) For all u, v E L+ such that { P } , and {P}"are disjoint, the closures {P}; and {P}; are also disjoint. (b) Every proper prime ideal in L contains a unique minimal prime ideal.
Proof. (i) L has the quasi principal projection property, i.e., for anyfe L the band A: is a projection band. Hence, by Lemma 37.8, for every u EL' the set {P}; is open and closed in the hull-kernel topology of 9'.Assume now that inf (u, u ) = 0, i.e., {P},and {P}"are disjoint. Since { P } u is open, it follows immediately that {P}; and { P } uare disjoint. Since {P}; is open, it follows similarly that {P}; and {P}; are disjoint. (ii) We first prove (a) * (b). To this end, assume that (a) holds, but not (b). Then there exists a proper prime ideal Po containing two different minimal prime ideals M , and M , . It follows that there exist elements u1 E MI and u2 E M 2 such that u, is not in M , and u, is not in M , . Set u1 = u,-inf
( u , , u2),
v 2 = u,-inf
( u l , u,).
CH. 5 , s 371
COMPACTNESS AND SEPARATION PROPERTIES
235
Then inf ( v , , u 2 ) = 0, u , E M , and v2 E M2 , but v , is not in M 2 and v2 is not in M,. It follows that { P } y ,and {P}u, are disjoint, so {P}; and {P}; are disjoint by hypothesis. On the other hand it follows from Po I> M , that Po is in the closure of M , , and so (since M , E {P}u,)the prime ideal Po is surely contained in {P};. Similarly, Po is contained in {P};. Contradiction. Hence, (b) holds. For the proof that (b) implies (a), we assume that every proper prime ideal contains a unique minimal prime ideal. Now assume also that there exist elements u, u E L+ such that inf (u, v ) = 0, so {P},and { P } u disjoint, but {P}; n {P}; contains an element Po.It follows then from Po E {P}; that there exists a minimal prime ideal M , c Po such that M , E {P},, (cf. Theorem 36.6 (ii)). Similarly, there exists a minimal prime ideal M 2 c Po such and { P } uare disjoint, we have M , # M 2 . On the that M 2 E { P } u .Since {P},, other hand, MI and M2 are minimal prime ideals contained in the proper prime ideal Po,so M , = M2 since Po contains only one minimal prime ideal. Contradiction. Hence, {P}; and {P}; must be disjoint, i.e., (a) holds. Theorem 37.11. The following conditionsfor the Riesz space L are equivalent. (i) L has the quasi principalprojection property. (ii) All sets {M},,, u E L + ,are compact in the hull-kernel topology of A? and, in addition, every proper prime ideal contains a unique minimalprime ideal.
-
Proof. (i) * (ii) By Lemma 37.9 all sets { M } , are compact. By Lemma 37.10 every proper prime ideal contains a unique minimal prime ideal. (ii) (i) Assuming that (ii) holds, we have to prove that, for every u EL', the set {P}; is open in the hull-kernel topology of 8.To this end, let Po E {P}; be given, and denote by Mo the unique minimal prime ideal contained in Po.By Theorem 36.6 (ii) we have M o E {P},,so M o E { M } , holds in A?. Next, note that
at
=
n ( { P } :~v
EL+,
u not in Po)
is the set of all prime ideals P' satisfying P' c Po, so
at It follows that the set
=
(P': P' prime, M o c P' c Po).
n ( { M } :~u E L + ,v not in Po)
consists only of the one element M,, so on account of Mo E { M } , we have that the intersection
236
n( { M } :~u
[CH.5 , @37
PRIME IDEALS
E L+, u
not in P,) n { M } :
is empty, where { M } : denotes the set theoretic complement of {M}, with respect to A. Hence, denoting { M } , n { M } : by S,, the intersection of all S,, for u not in Po,is empty. Since all S, are closed and compact in A, a finite intersection of the S,, say S,,, must already be empty. Setting uo = inf ( u l , . . ., u,,), it follows now that uo is not in Po and {MIu,,is contained in { M I " . Hence, with all closures taken in 8, we have
Po E
P > U ,
= (MI, = (MI; = (PI;.
This shows that every point of (PI; is an interior point of {P};,so (P}; is open. As a corollary we will prove now that if L has the quasi principal projection property and e EL' is not the zeroelement, then the spaces ( M } e and 22e, equipped with their hull-kernel topologies, are homeomorphic. The proof is by no means trivial; we will have to combine quite a number of facts established earlier in an appropriate way. Some preliminary remarks follow. In Theorem 36.6 (ii) it was proved that the closure of the subset {P},of 9' satisfies {PI; = (J ((M) : ME&, M E {P},).
Now, M E { P } , , is the same as M E {M},, and
0= {MI;
= {PI;
holds for every M E {M}, , so it follows that (P};
=
{M};
=
u ((M) M E :
{M},,).
As a second remark we observe that (by definition) the sets { M } , n { M } e , u E L', form a base for the hull-kernel topology in { M } e .But
{MI, n { M } e = so the sets { M } , , ,0 5 u
{Mlinf(u,e)
9
5 e, form a base for the topology in { M } e .
Corollary 37.12. (i) Given e > 0 in the Riesz space L, there exist3 for every M E a unique prime ideal Q E 2?esuch that Q 2 M . Conversely, for every Q E 22e there exists at least one M E { M } e such that M c Q . We denote the Q thus corresponding to M by Q M . The mapping Z:M+QM
CH. 5 , s 371
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COMPACTNESS AND SEPARATION PROPERTIES
of { M } , onto 9, has the property that the image n({M), ) is closed for any subset { M } , o f { M } ' . (ii) If the Riesz space L has the quasi principal projection property, then the mapping a : M + Q M of {M}' onto 9' is a homeomorphism.
Proof. (i) Let M E { M } , be given. Since e is no member of M there exists a prime ideal Q, containing M and maximal with respect to the property of not containing e, so Q E 9'. This ideal Q is uniquely determined. Indeed, if both Q,and Q, contain M and are members of 9",then (since the set of all prime ideals containing M is linearly ordered) one of Q, and Q, must be contained in the other, which is possible for elements of 9 ' only if Q, = (I2. Conversely, given Q E 9,, every minimal prime ideal M c Q satisfies M E { M } , . The mapping n :M + Q M is, therefore, from {M}' onto 9',and the image a ( M ) of the element M E { M } , satisfies
-
a ( M ) = { M } n 9,,
-
where { M } denotes the closure in 9 '.Hence, for any subset of {M}' of the form { M } , , we have n({W,) =
u
( 4 M ) :M =
E
{MI,) =
(J((M) : M E { M } , ) n 9'
=
{MI;
n 9",
which shows that n({M},) is closed in the hull-kernel topology of 9'. (ii) We assume now that L has the quasi principal projection property, so by the last theorem all sets { M } " , u EL', are compact, and the mapping n :M + Q M of {M}" onto 9' in part (i) is now a one-one mapping. Since all { M } , are compact in A, the lattice of all sets { M ) , is a Boolean ring, and so (for any u satisfying 0 S u 5 e ) the set theoretic complement {M}: of { M } , with respect to {M}' is of the form { M } ufor some o EL', 0 5 o 5 e. Every closed subset of { M } " (in the hull-kernel topology of { M } , ) is an intersection of finite unions of sets { M } : ,i.e., every closed subset 9of { M } , is an intersection of sets { M > u Since . a is one-one, it follows that the image n ( 4 ) satisfies Every n({M},) is closed in 9" by what was proved in part (i), so n ( 9 ) is closed in 9'. The one-one mapping n maps, therefore, closed sets onto closed sets, so it follows from the continuity definition that the inverse mapping 7 c - l is continuous. But then n-' is a continuous one-one mapping from the compact space 9' (cf. Theorem 36.4 (ii)) onto the Hausdorff space
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PRIME IDEALS
{ M } = ,so it follows from a well-known theorem in topology that the inverse mapping of x - ' , i.e., x itself, is also continuous. This completes the proof that x is a homeomorphism.
As a final remark, note that every Dedekind a-complete space L has the principal projection property, and so L has all the properties derived from the quasi principal projection property in Lemma 37.10, Theorem 37.1 1 and Corollary 37.12. Exercise 37.13. Let (xk : k = 1,2, . . .) be a sequence of points in the closed interval [O, 13 such that xk J. 0 as k + 00, let I k = [xk+ xk] and let each I k be divided into n equal subintervals I k l , . ., I k n . For i = 1, . . .,n, letfi be a non-negative continuous function on [0, 11, strictly positive in the interior of Z l i , I,,, and zero at each point of all intervals & , j # i. (i) Let A . be the ideal in C([0, 11) consisting of allfe C( [0, 11) vanishing in a neighborhood of the point xo = 0. Show that the functionsf,, . .,fn are mutually disjoint, and none of them is a member of A . . (ii) Let Pi ( i = 1, .,n) be a prime ideal in C([0, 11) containing A . and not containingf,. Show that P i containsf;. f o r j # i, so all Piare mutually different. (iii) Prove similar facts for any arbitrary point xo in [0, 13, and derive from this that every maximal ideal in C( [O, 11) contains infinitely many different minimal prime ideals.
.
',
...
.
..
Exercise 37.14. (i) Let L be a Riesz space having the projection property and 9 'the set of all proper prime ideals in L. The set B is equipped with the hull-kernel topology. Prove that 9 'is an extremally disconnected space (i.e., the closure of every open subset of 9 'is open). (ii) Prove the same for the hull-kernel topology in 9,where W is any nonempty subset of 8. Hint:For (i), let d be an open subset of 9'.Then d = { P } Afor A = (f:{P}I c &), so { P } , d d = d-' by formula (1) in section 37. By hypothesis Addis a projection band, so by Theorem 37.5 (ii) the set &-' is closed. It follows then from d c &-' c d -that d - "= d - ,so &- is closed and open. For (ii), note that the closure of any subset & of 9 in the hull-kernel topology of W is the intersection of W and the closure of d in the hullkernel topology of 9'. Exercise 37.15. Show that 1, is the Dedekind completion of the Riesz space ( c ) of all real convergent sequences. Show that 1, is also the Dedekind
CH. 5.8 371
COMPACTN@SSAND SEPARATION PROPERTIES
239
completion of the Riesz space L of all real sequences f = (f(l),f(2) ,. . .) such that the range offis finite (i.e., L is the space of all real step functions on the set of all natural numbers). Exercise 37.16. Let L be an Archimedean Riesz space and K the Dedekind completion of L. For any ideal A in L, we set
A* = ( f *
:f*EK,lf*l
5 uforsomeu~A').
(i) Show that A* is an ideal in K such that A* n L = A. Hence, the mapping a : A + A* is a one-one mapping of the set of all ideals in L into the set of all ideals in K. (ii) Show that the mapping a is onto the set of all ideals in K if and only if for each u* E K + there exists an element u E Lf and a number tl > 0 such that tlu 5 u* S u. (iii) Let L be the Riesz space of all real sequences f = (f(l), f ( 2 ) , . . .) such that the range o f f is finite, so u* = (1, +,3, . . .) is an element in the Dedekind completion I , of L. Show that the condition in part (ii) is not satisfied, so in this case the mapping a is not onto the set of all ideals in 1,. (iv) In the Riesz space (c), let A be the ideal (co) of all sequences converging to zero. Show that A is a prime ideal in (c), but A* = a @ ) = A is no prime ideal in the Dedekind completion 1, of (c). Hint: For (ii), assume first that every ideal in K is of the form A* for some ideal A in L. Given u* E K, let A(u*) be the ideal in K generated by u*, and let A = A(u*) n L. By hypothesis A# = A(u*), so there exists an element u E A such that u* S u. It follows from u E A that u E A(u*), and so there is a positive number tl-l such that u tl-'u*. For the converse, assume that for each u* in K + we have u E L+ and tl > 0 such that au 5 u* 5 u. Let A" be an ideal in K. Set A = A" n L. Given u* 2 0 in A", we have u E L+ and tl > 0 such that tlu 5 u* 6 u. It follows from au 5 u* that u E A, and so it follows from u* 6 u that u* E A*. Hence A" c A*. The converse is evident, so A" = A*.
Exercise 37.17. Let L be an Archimedean Riesz space and K the Dedekind completion of L. For any ideal A in L, the ideal A* in K is defined as in the preceding exercise. Assume now that L has the projection property. Show that P* is a prime ideal in K if and only if P is a prime ideal in L. Show that P* is a minimal prime ideal in K if and only if P is a minimal prime ideal in L. Hint: It is easy to prove that if P* is prime in K,then P is prime in L. For the converse, assume that P is prime in L, and let inf (u", v " ) = 0 in K.
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[CH. 5 , s
PRIME IDEALS
37
Set D = (u : U E L, 0 5 u 6 u " ) and E = (0 : U E L0, 5 v 5 v"). Note that D IE. Since Dd n Ddd = {0} holds in L, we have Dd c P or Ddd c P. Assume that Dd c P holds. Prove that in this case v" E P" holds, as follows. Take w E L such that u" 5 w. On account of L possessing the projection property, we have w = w 1+ w 2 with w1 E Dd and w 2 E odd. Since E c od, we have w 2 Iu for all v E E, so w 2 Iu" on account of v" = sup E. But then 0"
= inf ( v " ,
w) = inf (u", w1+w,)
= inf (v",
wl),
5 w 1 E Dd c P,which shows that v" E P". If Ddd c P holds, prove similarly that u" E P". Assume now that M is a minimal prime ideal in L. By Exercise 33.9 this is equivalent to assuming that for every u E M' there exists an element S E S= L'-M' such that u l s . Given 0 5 u" E M " , there exists u E M' satisfying 0 5 u" 5 u, so there exists s E S such that u Is. It follows that u" 1s, and also S E S c S" = K + - ( M " ) ' . This shows that M # is a minimal prime ideal. The converse follows easily. so v #
Exercise 37.18. Under the same hypotheses as in the preceding exercisz, let A" be the set of all M" = a ( M ) , where M runs through the set A of all minimal prime ideals in L. The set .A" is, therefore, a subset of the set of all minimal prime ideals in K. We equip A and A" with their hull-kernel topologies. Show that the mapping a :M + M" from A onto A" is a homeomorphism. Hint: a is a one-one mapping, and the image of any open base set { M } , is the set { M " } , , so a maps open sets onto open sets. This implies that a-' is continuous. Thus, a-' is a continuous one-one mapping of the compact space ( M " } , onto the Hausdorff space { M } , , so a is also continuous on { M I , . This holds for every u EL'. It follows that a is continuous on A. Exercise 37.19. Show that in an Archimedean Riesz space every proper prime band P (i.e., P is simultaneouslya band and a proper prime ideal) is a maximal ideal (i.e., a maximal band; cf. Theorem 26.7). Show that this does not hold in the lexicographicallyordered plane. Hint:Assume L is an Archimedean Riesz space and P a proper prime band. It is easily seen that every nonzero element in Pd is an atom, and that there exist no disjoint atoms in Pd. Hence, if u > 0 is one of the atoms in Pd, then Pd is equal to the ideal A , generated by u. By Theorem 26.4, Pd = A , is a projection band, so L = P @ Pd. It follows that P is a maximal ideal.
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COMPACTNESS AND SEPARATION PROPERTIES
24 1
Exercise 37.20. Let L be an Archimedean Riesz space and W a non-empty subset of B such that the intersection of all R E W is (0). The set W is equipped with the hull-kernel topology. Show that the following conditions are equivalent. (i) W is a discrete topological space (i.e., every point ofW is an open set). (ii) Every R E W is a band in L. (iii) As soon as one point Ro is omitted from 9, the intersection of the remaining R €95' is no longer equal to (0). Hint: (i) * (ii) By Corollary 36.2 the kernel of every open subset of W is a band. Hence k ( { R , } )is a band for every point Roin W.But k({R,}) = R,, so Ro is a band. (ii) =. (iii) Omit Ro from W. On account of Ro n = (0}, every R , # Ro in 9%'satisfies Ro c R , or Rd, c R1. But Ro c R1 is impossible by the preceding exercise, so Rd,c R1. This shows that Ri is contained in the intersection of all R1 # R o y so this intersection is not equal to (0). By hypo(iii) (i) Let Ro E W and let W , be the set of all other R E 9. thesis, the kernel k(Wl) is not equal to (0). It follows that k(9,) is not contained in R o yi.e., Ro is no member of h ( k ( 9 , ) ) . Then h ( k ( 9 , ) ) = W , , so 9, is closed. It follows that the set consisting only of the point Ro is open.
fi
Exercise 37.21. Show that the lexicographically ordered plane satisfies condition (ii) in the preceding exercise, but not condition (i). Exercise 37.22. If the Archimedean Riesz space L has an order basis consisting of disjoint atoms, then L is sometimes called a discrete Riesz space. Examples are the space (s) of all real sequences and all its Riesz subspaces (such as all I, spaces, (c) and (c,)). (i) Let L be a discrete Riesz space. Show that the order basis of L consisting of disjoint atoms is uniquely determined. (ii) Let (u, : a E { a } ) be an order basis of disjoint atoms in the discrete is a proper Riesz space L. Show that, for every a E { a } , the band J, = prime band, so J, is a maximal ideal. Show that, conversely, every maximal ideal in L is of the form { u , } ~for some a E (a}. Hence, the set of all J, = ( u , } ~is exactly the set $ of all maximal ideals inL. Show that the intersection of all J, E f is {0}, so the conditions of Exercise 37.20 are satisfied for
W = $.
Exercise 37.23. Let L be an Archimedean Riesz space and let K be the Dedekind completion of L. (i) Show that if u E L+ is an atom in L,then u is an atom in K .
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[CH.5 , s 37
PRIME IDEALS
(ii) Show that if u# E K + is an atom in K, then u# is actually an element of L,and u# is an atom in L. (iii) Show that L IS a discrete Riesz space if and only if K is so. The results in Exercixes 37.15-37.23 are due to J. J. Masterson ([l], 1968).
Exercise 37.24. Let 9 be the set of all proper prime ideals in the Riesz space L, and W a non-empty subset of 9. We recall that the collection of all sets { R } / = ( R : R E W ,no~ member of R ) is a base for the hull-kernel topology in W.For any non-empty subset D of L, we define the open subset { R } , of W by =
u
( { R } / :fE O).
Show that { R } , = { R } A Dwhere , AD is the ideal generated by D. Show also that, for arbitrary ideals A and B in L,we have {R}A+B
=
"
{R}AnB =
n
"
In the remaining part of the exercise, assume that ( R : R €9) = {0}, which is equivalent to assuming that { R } / is empty only forf = 0. Show that, for any non-empty subset D of L,we have iR>,d
= (iR}D)E0,
{R}Ddd
=
Derive from these formulas that if the ideal A in L satisfies A = Add, then { R } A is regularly open in the hull-kernel topology of W.Derive also that if A is a projection band in L, then { R } Ais closed. Show (by considering the case that W = .A) that the converse of the last statement is not necessarily true. For 9 = 8,however, the converse holds, i.e., if A is an ideal such that { P } A is closed, then A is a projection band (cf. Theorem 37.5).
Exercise 37.25. We use the same notations as in the preceding exercise. Once again, let 9be a non-empty subset of 9,and D a non-empty subset of L. We will say that an element f of L is in the 9-extension of D whenever { R } / c { R } , holds. The 9-extension of D will be denoted by (9-ex)(D). Note that for D = (0)the set {R}(,l is empty and (W-ex)({O})
=
( R : R €9) =I ~ ;
the ideal Z9 is called the core of W.Prove the following facts. (i) (9-ex)(D) is an ideal containing D, and (9-ex)(9-ex)(D) = (9-ex)(D).
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COMPACTNESS AND SEPARATION PROPERTIES
(ii) {R}(a-cx)(D) = { R } D ,and (9-ex)(D) is the largest ideal A which { R } A = { R } , holds. if and only if (W-ex)(Dl) = (9-ex)(D,). (iii) {R}D,= (iv) For W = 9J and 9 = 2 we have (P-ex)(D) = A D ,
=)
D for
(%ex)(D) = AD,
where AD is the ideal generated by D. (v) If W consists of one prime ideal P o , then (%ex)(D) = Po if D c P o , and (9-ex)(D) = L otherwise. (vi) For W lc W 2 we have (W2-ex)(D) c (9,-ex)(D).
Hint:It is easy to prove that (9-ex)(D) is an ideal containing D. Then prove first that { R } D= {R}(a-ex)(D). We have {R}(a-ex)(D) =
u
( { R } f :f (W-ex)(D))
and the inverse inclusion is trivial. The remaining part of (i) follows now. Next prove (iii). It follows immediately from the definitions that {R}D,= {R}D2implies (W-ex)(D,) = (9-ex)(D,). For the converse, use that { R } , = {R}(a-ex)fD). For the still remaining part of (ii) observe now that if A is an ideal satisfying { R } A = { R } D ,then (9-ex)@) = (@-ex)(D), so A cannot be properly larger than the ideal (g-ex)(D).
Exercise 37.26. We use the same notations as in the preceding exercise. Let W be a non-empty subset of LP such that the intersection of all R E W is (0). Show that (9-ex)(D) c odd holds for every non-empty subset D of L. Hint:If, for some D, the formula fails to hold, there existsfe (9-ex)@) such thatfis no member of Ddd.Then there exists an element go E Dd such that p = inf (Ifl, Igol)> 0. It follows that p E (g-ex)(D) as well as p E D”, and { R } pnon-empty. Let Ro be an element of {R},,. Sincep is not in Ro and Ro is prime, Ro contains {P}~3 Ddd,so Ro contains D. It follows that Ro E {R}#holds for no g E D, so Ro is not contained in
u
({RIB : g E 0).
On the other hand it follows from P E (9-ex)(D) that Ro E {RIP c
u ({RIB
:9 E D ) .
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[CH. 5 , g 37
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Exercise 37.27. We recall that .A is the set of all minimal prime ideals. It was proved in the present chapter that (A-ex)(A,) = A? holds for every principal ideal A,. (i) Show that, in general, (A-ex)@) = Add does not hold for every ideal A, i.e., generally speaking, (.A-ex)(A) is properly included in Add. More precisely, (A-ex)(A) is between Addand the smaller ideal B generated by the collection of all sets Atd, for u running through A+. (ii) Assume that all sets {M},,, u E L+, are compact in the hull-kernel topology of A. Show that, for every ideal A, the extension (.A-ex)(A) is equal to the ideal generated by the collection of all sets Aid, for u running through A'. Hint: For (i), observe that (A-ex)(A) = Add,holding for the ideal A, is equivalent to the statement that if M is a minimal prime ideal such that A is included in M, then Addis also included in M. This statement is not generally true. Counterexample: L = C([O,l]), xo some point in [0, I], A the ideal Ix, of all f E L vanishing in some neighborhood of xo. Then Add = L,and there are many minimal prime ideals containing A, but not containing Add = L.
For (ii), it is sufficient to prove that for everyfin (A-ex)@) there exists an element u E A' such that f E Atd. Now, f E (.A-ex)(A) implies
u
({M>8
'
A),
so { M } , is covered by a finite union of the { L M } ~say , for gl,. . ., 9.. Then { M } / c {M},,for u = sup (Igll, . . ., Ignl), so u E A + and f E A?. Exercise 37.28. We present a brief account of I. Amemiya's method of treating prime ideals (although the notion of a prime ideal is not explicitly mentioned in his paper [2]). Let T be the subset of the extended real number system consisting of - co,0 and + co,provided with the usual ordering, addition and multiplication by finite real numbers, with (+ a)+ (T co) undefined. The set T is, therefore, a lattice, but T is no Riesz space. Now, let L be a Riesz space. According to I. Amemiya, the mapping cp from L onto T is called a hpectral function if (a) cp(af) = acp(f) for allfe L and all real numbers a. (b) cp(fv9) = max ( c p ( f
1, cp(9)) for all f, 9 EL.
It follows immediately that c p ( f ~ g = ) min (cp(f),cp(g)) for allf, g EL. In particular, cp is order preserving. Note that cp(2f) = q ( f ) for all f E L.
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COMPACTNESS A N D SEPARATION PROPERTIES
(i) Show that cp(f+g) = cp(f)+cp(g) if cp(f)+cp(g) makes sense. (ii) Show that the null space (f: q ( f ) = 0) of the spectral function cp is a prime ideal in L. (iii) Let P be a prime ideal in L. For anyf E L, we have exactly one of the cases If1 E P,f + not i n P orf- not in P. Set p(f) = 0 if If1 E P, c p ( f ) = co iff is not in P and cp( f) = - co iff - is not in P. Show that cp is a spectral function such that the null space of cp is exactly P. (iv) Show that different spectral functions have different null spaces, and different prime ideals have different spectral functions attached to them by the method described in (iii). Hence, there is a one-one correspondence between prime ideals and spectral functions. It follows that every theorem about prime ideals may be expressed in terms of spectral functions, and conversely. Hint: For (i), observe first that
+
+
min ( c p ( f ) , cp(g)) = c p { 2 ( f 4 ) 5 cp(f+g)
5 cpw-vg)) =
=
max (cp(f),cp(g))7
so cp(f+g) = cp(f)+cp(g) if q ( f ) = cp(g). Now assume, for example, that q ( f ) = +co and q ( g ) = 0. It follows from the inequalities above that cp(f+g) = -co is impossible. It is also impossible that cp(f+g) = 0, because then cp(f+g) = q ( g ) = 0 would imply that
+
Hence, cp(f+g) = co. For (iii), observe that since P is a proper ideal, we have ~ ( f # ) 0 for at least onef, and so cp is a mapping from L onto T. The formula cp(uf) = ucp(f) follows by observing that lufl = la1 * If[, (uf)' = uf for u > 0 and (uf)' = -uf - for u < 0. In order to show tbat q ( f )S cp(g) forf 5 g , consider separately the cases that cp(g) = 0 and cp(g) = -m. It follows then already that c p ( f v g ) 2 max (cp(f),cp(g)). For the converse inequality, it is sufficient to observe that if ~ ( fand ) q ( g ) are both less than +a,then ( f v g ) + = f v g + E P , so c p ( f v g ) < +co; if c p ( f ) = ~ ( g ) = - 0 0 , then ( f v 9 ) - = f - A g - is not in P, so c p ( f v g ) = - co. It follows that q ( f v g ) = max (cp(f), cp(g)). It is evident that P is the null space of cp. +
+
The following list shows the relations between several theorems in the present chapter and corresponding theorems in the papers of I. Amemiya [2] and D. G. Johnson-J. E. Kist [l].
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[CH. 5.8 37
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Amemiya Th 33.2 Th 33.5 Th 33.6 (i) Th 33.6 (ii) Th 33.7 Exerc 33.9
Th 1.1 Th 1.1 §6
Cor 35.4 Th 35.5 (i) Th 35.5 (iii) Th 36.1 Th 36.3 Th 36.4 (ii) Th 36.4 (iii) Th 36.6 Cor 37.3 Th 37.5 Th 37.6 Th 37.7 Lemma 37.10 (i) Th 37.11 Cor 37.12 (ii)
Johnson-Kist Th 3.1 Lemma 6.6 Lemma 6.3 Lemma 6.4 Lemma 5.2 Th 7.2
Ths 2.1, 2.2 Th 5.1 Th 6.1 06 $6 Th 3.2 $6 Th 6.3 §6
§3
Th 2.2
CHAPTER 6
Freudenthal’s Spectral Theorem
The main theorem in this chapter is H. Freudenthal’s spectral theorem (1936) in Riesz spaces possessing the principal projection property. In order toexplain the theorem, we first recall that, given the element e in the positive cone L+ of the Riesz space L, the sequence (f,: n = I , 2, . .) in L is said to converge e-uniformly to f E L whenever, for every E > 0, we have If-f.1 S ee for all n 2 N , . The spectral theorem is formulated first for the simple case that f is a given element in the ideal generated by e E L f . For any real a, the component of e in the principal projection band generated by (Me- f)’ is now denoted by p a , and the system (pa : - 0 0 < a < 0 0 ) is called the spectral system off with respect to e. We have p a t p s as a t 8, pa e as a t 00 and pa 1 0 as a J. - 00. Since f is in the ideal generated by e, there exist real numbers a, b and a positive number E > 0 such that ae S f S (b-e)e holds. Now, let n = n(a,, a l , . . ., u,), with a = uo < al < . . . < a, = b, be a partition of [a, b]. The sums
.
n
t(n;f>=
c n
k= 1
ak(pUk-PP,k-,)
are called the lower and upper sums off with respect to the partition n (and with respect to e). It is not difficult to see that
+ ; f ) S f S t(.;f). The spectral theorem states that, for the maximal interval length in x tending to zero, the sums s(n;f ) and f ( n f) ; tend monotonely and e-uniformly tof. This is sometimes written as b
f =J)dP.. 247
248
FREUDENTHAL'S SPECTRAL THEOREM
[CH.6
Iffis an element in the band generated by e, but not necessarily in the ideal generated by e, then a similar result holds with e-uniform convergence replaced by order convergence. Some further properties of the spectral system of an element f will be proved, one of which (going back to F. Riesz) is as follows. Writing (me- f)' = u, for any real m and o , , h = h - l ( ~ , - ~ , - h for ) any k > 0,we have that v,. h increases as h decreases and suP(%.h:hlo)=P,, i.e., the left derivative of u, exists and is equal top,. It will be shown that the Radon-Nikodym theorem in measure theory and the Poisson formula for harmonic functions in an open circle are particular cases of the spectral theorem. It will be proved in a later chapter that the spectral theorem for Hermitian operators (and for normal operators) in Hilbert space is also a particular case of the spectral theorem in a Riesz space. Given e E L', the sequence (f.:n = 1,2, . .) in L is an e-uniform Cauchy sequence if, for every E > 0, we have If,-fJ S ce for all m yn 2 N , . The Riesz space L is called uniformly complete whenever, for every e E L', every e-uniform Cauchy sequence has an e-uniform limit. It will be proved now that in a uniformly complete space Dedekind a-completeness and the principal projection property are equivalent and that, similarly, Dedekind completenessand the projection property are equivalent.The spectral theorem is the main tool in the proof. Since it is not difficult to verify that any Dedekind a-complete space is uniformly complete, it follows in particular that Dedekind a-completeness and the projection property together are equivalent to Dedekind completeness, a result already announced but not yet proved in Theorem 25.1 (the main inclusion theorem). It will also be proved that if the Boolean algebra of all projection bands is Dedekind complete, then the projection property and the principal projection property are equivalent and, similarly, Dedekind completeness and Dedekind a-completeness are equivalent. The above results will be applied to the spaces C ( X ) and c b ( X ) of all real continuous functions and all real bounded continuous functions on a topological space X . The spaces C ( X ) and c b ( X ) are uniformly complete, and C ( X ) is Dedekind complete if and only if c b ( X ) is so. Similarly for Dedekind o-completeness. A subset E of X is called a cozero set if there exists a functionfe c b ( X ) such that E = ( x : f ( x ) # 0) holds. A subset of Xis called closed-open if the set is closed as well as open. The subset E of X
.
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SPECZRAL SYSTEMS
249
is said to have a closed-open hull if among all closed-open sets containing E there is a smallest one, and this smallest one is then called the closed-open hull of E. It will be proved now that C,(X) is Dedekind a-complete if and only if every cozero set has a closed-open hull such that disjoint cozero sets have disjoint closed-open hulls. A sufficient condition for Dedekind o-completeness is that the closure of every cozero set is open. If X is completely regular, the condition is also necessary (L. Gillman and M. Jerison). Another sufficient condition is that the closure of every open F,-set is open. If X is Hausdorff and normal, the condition is also necessary (H. Nakano). A sufficient condition for Dedekind completeness is that the closure of every open set is open. If X is completely regular, the condition is also necessary (H. Nakano).
38. Spectral systems In this section we assume that the Riesz space L has the principal projection property; just as in earlier sections, the principal band generated by f E L will be denoted by B f , and the order projection on Bf will be denoted by Pf , so 9 = Pfg+(g--Pfg); PfSEBJ, g-PfgEB;, holds for every g EL. Furthermore, e will be a fixed nonzero element of L'.
Definition 38.1. The element p E L is caIIed a component of e whenever p and e - p are disjoint, i.e., whenever p I(e-p). The above definition is justified by the following lemma. Lemma 38.2. The eIement p E L is a component of e i f and onIy if there exists an element v EL' such that p = P,e. Hence, any component p of e satisfies 0 5 p e. Proof. Ifp is a component of e, we havep 1( e - p ) , and so it follows from e = p + ( e - p ) that p = Ppe. In particular, since e > 0, the components p and e - p are elements of the positive cone, so 0 5 p 5 e. Conversely, for any principal projection band (generated by f, say), the element P f e is a component of e. Besides the fixed nonzero element e in L' we will consider another element fin L such that f is a member of the band generated by e. The elementf will be kept fixed in the present section.
250
[CH.6,g 38
FREUDENTHAL'S SPECTRAL THEOREM
For any real a satisfying - 00 < a < co,we set u, = (ae-f)';
the band generated by u, will be denoted by B, for brevity, the order projection on B, by Pa, and the component of e in B, by p a , so pa = P,e. It follows immediately frGm Corollary 31.2 (i) that the band generated by pa is again B,. The following simple lemma is an immediate consequence of the definitions.
Lemma 38.3. W e have -a
<
5 /3 < 00
=> U,
S
US
* B,
c B,
* p a 5 PO.
In the next theorem we collect several properties of the system of projections Pa and the system of components p a .
Theorem 38.4. (i) We have Paf 5 ap, for all a; in particular, Paf for a 5 0. (ii) W e have a(e-pa) 5 f -Paf for all a ; in particular, Paf f for a (iii) For -00 < a 5 /? < 00, we have
s
s0
2 0.
5 (P,-Pa)(f) S P(Pp-Pa)(iv) W e have B, t B, and pa t p , as a t /3 (i.e., pa is left continuous). Furthermore, P,v 4 0 as a 1 - co holds for every v 2 0, and P,v t v as a t co Mbp-Pa)
holds for every v 2 0 in the band generated by e. In particular, pa 1 0 as u 4 - co and pa 7 e as a t co.
Proof. (i) From the general formula Pg+(g)= g' 2 0 it follows, by choosing g = ore-f, that P,(ae-f) 2 0, i.e., Paf 5 uP,e = up,. (ii) From the general formula
g-P,+(g)
=
50
g-g+ = -9-
it follows, by choosing g = ae-f, that
ae-f-P,(ae-f)
S 0,
i.e., a(e-pa) 5 f-P.5 (iii) It follows from - 0 0 < a S /3 < 00 that B, c B,,, and so Pap, = PpPa = Pa. Applying now the projection operator P,, to both sides of the inequality a(e-pa) If -P,f, we obtain
d ~ - P ap ) S
(Pp-Pa)(f
1.
CH.6,8 381
251
SPECTRAL SYSTEMS
The other inequality to be proved follows from = ( + - P a ) ( q 9 f ' )5 (P#-Pu)(BPp) = B(P#--PU).
(P,-P,)(f)
(iv) Since L is Archimedean, we have that (ae-f)'
t (Be-f)+
as
a t B,
and so B, is the band generated by the set of all bands B. satisfying a < p, .e.,
1
But then B, is the supremum of the set (B, : a < p), so B, t B, as a t p, and it follows from Theorem 30.5 (ii) that also P,u t Psu for any u E L'. In particular, pa t p,, as a t 8. In order to show that P,v 10 as a 1 00 holds for every D 2 0, it is sufficient to prove that B, = (0). For any element w satisfying 0 5 w E B,, we have
-
nu
nu
w
I( a e - f ) -
= (f-ae)'
= (f+Iale)+
for all a 5 0, and so w I(e+ 1aI-y)' for all a < 0. This implies ( a 1 - 0 0 ) that w J- e (note that L is Archimedean). On the other hand, w is a member of every B,, so in particular (take a = 0) w is a member of the band generated byf, and hence a member of the band generated by e. It follows that w = 0, and so B, = (0). Finally, in order to show that P,v t D holds for every u 2 0 in the band generated by e, it is sufficient to prove that 0, R, = {0}, where R, is the band generated by e-pa. If 0 5 W E R,,we have w I(ae-f)' for all a, so w I(e-a-y)' for all a > 0. This implies, similarly as above, that w = 0,so R, = (01.
n,
nu
nu
Corollary 38.5. I f f is a member of the ideal generated by e, i.e., if there exist real numbers a and b such that ae S f 5 be, then pa = 0 for a 5 a and pa = e for a > b. Proof. For a 5 a we have ae-f
5 (a-a)e 5 0, and so u,
= 0. This shows that B, = {0}, sop, = 0.
For a > b we have ae-f 2 (a-b)e, and so u, = (ie-f)'
= ae-f
2 (a-b)e.
This shows that B, is now the band generated by e, so pa = e.
= (ae-f)'
252
FREUDENTHAL'S SPECTRAL THEOREM
[CH.
6 , s 39
-=
The set (pa : - co 0: < co) of components of e is called the spectral system of the elementf with respect to the lixed element e.
39. Uniform Caucby sequences We assume that e is a fixed nonzero element in the positive cone ' L of the Riesz space L. First of all we present a definition, the first part of which was already given in Theorem 16.2.
Definition 39.1. The sequence (fn :n = 1,2, . . .) in L is said to converge e-uniformly to the element f E L whenever, for every E > 0, there exists a natural number no = no(&)such that I f-f.1 S Ee holds for all n 2 no. The sequence (fn : n = 1,2, . .) in L is called an e-uniform Cauchy sequence whenever,for every E > 0, there exists a natural number n, = n,(e) such that 1f,-fnl 6 Ee holds for all myn 2 n,
.
.
It is evident that every e-uniformly convergent sequence is an e-uniform Cauchy sequence. Given that (fn :n = 1,2,. . .) is an e-uniform Cauchy sequence, the set (f,: n = 1,2, . . .) is bounded from above and from below. Indeed, there exists a natural number n, such that lfn-fnll 5 e holds for all n 2 n , , and so lfnl 5 e+lfn,l holds for all n 2 n , . But then
-
IfnI 5 e+sup (Ifil,Ifil, * * ) Ifnil) holds for all n. If L is Archimedean, the e-uniformlimit of a sequence, if existing, is unique. Indeed, assuming thatfn converges e-uniformly to f as well as to g, we have If-gl 5 Ee for every E > 0, and so lf-91 = 0 since L is Archimedean, i.e., f = g. Furthermore, as observed already in Theorem 16.2 (i), any sequence in an Archimedean Riesz space converging e-uniformly to f is also order convergent to f.The following lemma, for monotone sequences, goes partially in the converse direction. Lemma 39.2. If (S, : n = 1,2, . . .) is an increasing e-uniform Cauchy sequence in the Archimedean Riesz space L, then fn is e-uniformly convergent to the elementf if and only if the sequence has a supremum, and this supremum is then equal to$ A similar statement holdsfor decreasing sequences. It forlows that in a Dedekind 0-complete space every monotone e-uniform Cauchy sequence is e-uniformly convergent.
Proof. Assume first that fn is increasing and fmis e-uniformly convergent to$ For n fixed and m > n, we have Ifn-inf(f,fn)I = Iinf(fm,fn)-inf(f,f,)I
S If,-fI
S ~e
CH. 6 , s 401
FREUDENTHAL'S SPECTRAL THEOREM 0
253
for any given E > 0, provided m is sufficiently large. Hence (since Ee 4 0 as 10 on account of L being Archimedean) we havef, = inf (Lf.). This shows thatf. 5 f holds for all n. It followsnow from0 4f-f. 5 ee for n sufficiently large, and from Ee 10 as E 10, that f - f . 10, so f = supf.. Conversely, iff. f f and (f.:n = 1, 2, . . .) is an e-uniform Cauchy sequence, it follows from E
5 f m - f . 5 Ee for m 2 n 2 no(&) and ( f , -f.) fm (f-f,) that 0 5 f -f. 5 Ee holds for n 2 no , and sof. con0
verges e-uniformly to f.
Definition 39.3. The Riesz space L is called e-uniformly complete whenever every e-uniform Cauchy sequence has an e-uniform limit. Theorem 39.4. I f L is Archimedean, then L is e-uniformly complete if and only ifevery monotone e-uniform Cauchy sequence has an e-uniform limit.
Proof. It is sufficient to prove that if L is Archimedean and every monotone e-uniform Cauchy sequence has an e-uniform limit, then any arbitrary euniform Cauchy sequence has an e-uniform limit. Hence, let L be Archimedean and let every monotone e-uniform Cauchy sequence have an euniform limit. Now, let ( f n :n = l ,2, . . .) be an arbitrary e-uniform Cauchy sequence in L. There exists a subsequencef., ,f.,,. . . such that n, < n2 < . . . and lfnk+l-fnkl 5 2-ke for k = 1, 2, . . .. The partial sums of the series
f2 + ( f n z - f n J +
+(fn3-fnJ+
f n y +( f n z
+(f",- f n z ) -
-fn1)-
+** +**
*, *
form increasing e-uniform Cauchy sequences, having therefore e-uniform limits g1 and g2 respectively. It follows that the partial sums of converge e-uniformly to g = g1-g2. In other words, f n k converges euniformly to g as k + m. It follows easily that fn converges e-uniformly to g a s n + co. 40. Freudenthal's spectral theorem
Exactly as in section 38, we assume that L is a Riesz space possessing the
254
PREUDENTHAL’S SPECTRAL THEOREM
[CH.6 , s 40
principal projection property, and e is a fixed nonzero element in L + . Furthermore, given the element f in the ideal generated by e, we choose the real numbers a and b such that, for an appropriate E > 0, we have ae S f
5 (b-E)e.
Hence, if (pa : - 00 < a < 0 0 ) is the spectral system off with respect to e, we have by Corollary 38.5 that pa = 0 for a I a and pa = e for a 2 b. By n = n ( a o , a l , . . ., a,,), where a
= a.
< a1 < . .. <
an =
by
we denote a partition of the closed interval [a, b ] , and In1 = max (Iak+l-ctkl : 0
s k 5 n-1)
denotes, as usual, the maximal subinterval length in the partition n. Given the partitions n l ,n2 of [a, b ] , we will write nl 5 n2 whenever n2 is finer than nl (Le., n2 contains all points of nl). The set of all partitions of [a, b ] is thus partially ordered. For any partition .(ao, a l , . . ., an) of [a, b ] , we introduce now the elements s(n;f ) and t(n;f ) of L, defined by
’(‘;f)
n
=
ak-l(pak-pEk-l),
t(.;f) =
c %(Pak-Pak-*),
k=l n
k= 1
called the lower sum and upper sum off with respect to n. In the following lemma we collect some preliminary results about lower and upper sums.
Lemma 40.1. (i) For all partitions x i , n2 we have
&;f) S f (ii) I f n l 5 n 2 , then
s t(n2;f).
s(n1;f) 5 s ( n 2 ; f ) 5 t ( n 2 ; f )
(iii) then
If
ae S fl
s t(n1;f).
f 2 5 (b--E)e for appropriate real a, b and some
s(n;f i )
5 s(n;f,) and t(n;f l )
hold for every partition n of [a, b ] .
s t ( n . f,)
E
> 0,
CH. 6,s 401
255
PREUDENTHAL’SSPECTRAL THE.OREM
(iv) We have
In particular, we have for all partitions nl, n2 of [a, b ] . Proof. We observe first that the statement in (iii), although intuitively plausible, is the only statement not trivial to prove. (i) It is sufficient to prove that s(n;f ) 5 f 5 t(n;f ) holds for every partition n = n(ao,. . ., an). From
c n
‘b
=
=
‘,-‘a
k=l
it follows that
(puk-puk-l)
n
f = ‘bf
=
1
k= 1
(‘Uk-‘Uk-I)f,
and so, by Theorem 38.4 (iii), we have n
n
..
(ii) We may assume that n , is the partition n(a,, ., an)and n2 has one additional point 8, say between ak and ak+1 , so ak < fl < ak+1. For the proof of s(nl; f ) S s(n2;f ) we have now only to show that ak- l ( p 4 k - P U k -
1)
5
‘k-
1 ( P p -p 4 k -
1)
+
p(p4k
-Pp),
and this is evidently true. The proof for t(n2;f ) 6 t(n,;f ) is similar. (iii) We assume first, as in the preceding parts, that we have one elementf satisfying ae S f (b-&)e, and we denote the spectral system o f f by (pa: - m < a < a).Given the partition n(ao, ., an) of [a, b ] and the p 1 5 . . 6 p,,, we consider the sum numbers 0 I
..
.
n
=
k=l
pk(PUk-PUk-I)*
We also introduce, for any real a, the element r, = e-p,, and we observe that PUk-pUk-1
=
rak-l-rak,
256
[CH. 6 , s 40
FREUDENTHAL'S SPECTRAL THEOREM
and r, = e for a 6 a and r, = 0 for a terms in
2 b. Observing now also that all
n
s ( n ; f ) - a e = C (akq1-a)(pUk-pa-J = SUP {(ak-l -a)ruk-l : 1 k= 1
5 k 6 n}.
Finally, assume now that we have
ae 6 fl 6 fz 5 (b-E)e for appropriate real a, b and some E > 0, and let (p: : - 00 < a < 0 0 ) and ( p r : - 00 < a < 00) be the spectral systems offl and fzrespectively. Since
(ae-fJ+
2 (ae-fz)+
holds for every a, the band generated by (ae-fJ' is included in the band generated by (ae-fl)', and sop: =< p i , i.e.,ri =< r: for all a. It follows that
s ( n ; f 1 ) - a e = sup {(ak-l-a)r:k-l :1 6 k 6 n } sup {(ak-l-a)r:L-l : 1
s
k 5 n } = s(n;f,)-ae,
and so s ( n ; f l )4 s(n;fz). The proof for t ( n ; f l )5 t(n;fi) is similar. (iv) We have
0Sf-s(n;f)
r t(n;f)-s(n;f)
c n
=
(ak-ak-l)(pUk-pUk-l)
k= 1
and similarly 0
5 t(n;f)-f
s Inle.
s bl c (pUk-pak-1) n
k=l
=
Inle,
CH.6, § 401
FREUDENTHAL’S SPECTRAL THEOREM
251
The following main theorem is now an immediate consequence of the last lemma and of the remarks on monotone e-uniformly convergent sequences in the preceding section.
Theorem 40.2 (H. Freudenthal’sspectral theorem, 1936). Let L be a Riesz space posseAsing the principal projeciion property, and let e be a fixed nonzero element in L + . For any f in the ideal generated by e andfor any partition 7c of the inrerval [a, b ] , where the real numbers a and b are such that ae S f S (b-8)e holdsfor some E > 0, let s(n;f ) andt(n;f ) be the lower and upper sums o f f with respect to n. Then sup s ( n ; f ) = f = inf t(n;f). z
X
More precisely, if (n, :n = 1,2, . . .) is a sequence of partitions of [a,b ] such that z,+~ is a rejinement of n, for each n and such rhat 1n.l 40,then hold e-uniformly.
s(nn;f) t
f
and t(nn;f)I
f
A suggestive notation, derived from the theory of integration, is to write
f
=Jadpa. b
Sincep, = 0 for all tx 5 a andp, = e for all a
f
=Im
2 bywe may just as well write
adp,.
-W
The notation is justified by the similarity of the present procedure and the procedure of H. Lebesgue for defining the integral of a bounded real measurable function. For the particular case that L is the Riesz space of all real functions on a point set X, and e is the function satisfying e(x) = 1 for all x E X, the spectral theorem expresses the simple fact that every bounded real function on Xcan be approximated uniformly (in the ordinary sense) by step functions, from below as well as from above. It will be proved in the next section that the Radon-Nikodym theorem in measure theory and the Poisson formula for bounded harmonic functions in an open circle are special cases of the spectral theorem. Furthermore, in section 56, we will prove that the spectral theorem for Hermitian operators in Hilbert space is also a special case of the present theorem. First, however, we extend the spectral theorem (at least partially) to the case that f is an element in the band generated by e, but not necessarily in the ideal generated by e. As before, let
258
FREUDENTHAL'S SPECTRAL THEOREM
[CH. 6 , s 40
( p a : - 00 < u < 00) be the spectral system off with respect to e. We may not expect now that f can be approximated e-uniformly by appropriate finite sums C tLk(pak-pak-,); we will prove that approximation in the sense of order convergence is still possible, as follows.
Theorem 40.3 (Spectral theoremfor the unbounded case).Let f be an element in the band generated by e with spectral system (pa : -00 < a < a),let ([a,,,b,,] :n = I , 2, . . .) be a sequence of closed intervals such that 0 2 a,, 1 - 00 and0 5 b,, t 03, and let n,(u,,,, . . ., unmn)be apartition of [a,,,b,,]such that, for every n, the restriction of z,,~ to [a,,, b,,] is a refinement of n,,,and j?nally, let lnnl 10 as n -P 03. For every n, set mn
sn
=
C an,k-l(Pa*-Pan,k-l)+ k=l
Then s, converges in order t o f , i.e., there exists a sequence (u,, : n = 1,2, . . .) in L+ such that If-snl 5 O n 1 0.
Furthermore, we have s,' f f '. Alternatively, if we let n(uo, . . ., u,) run through all partitions of afixed interval [a,b ] and if II'
then
=
'k-1(pUk-pUk-l),
t, (P*f--Paf 1. t f as0 5 b t a andPaf t 0 as 0 2 a 1 -a. SII
Furthermore, Pbf
Proof. We have mn
f =kC= l (pU.k-PUn,k-I)(f)+Panf+(f-Panf) and, by Theorem 38.4 (iii),
Hence, it follows that
If-%!
5 I%le+ IPanf I fI f - P b n f l 5 Ilrnle+Pa,f + + P a n f - + ( f + - P b n f
+)+(f--Pb,f-)-
Denoting the right hand side (i.e., the sum of the last five terms) by u,,, we
CH.6 , 5 401
259
FREUDENTHAL'S SPECTRAL THEOREM
have 0 S v,, 3.0 as n + 00 by Theorem 38.4 (iv). Thus If-snl
I v n 10,
i.e., s,, converges in order tof. In order to show that ;s t f +,we observe that the terms ins,, are mutually disjoint, and so (cf. Theorem 14.4 (ii), where it is proved that (f+g)' = f + g + holds for f 1g ) we have +
where the summation is over the values of k satisfying a,,,kthat ;s increases as n increases. Also,
If+-CI I If-snI
1 0. It follows
iV n 3 . 0 ,
which implies that s', S f holds for all n (indeed, if not, we have (sn',-I*+)+ = wo > 0 for some no ,and so (s,' -f +)+2 wo for all n 2 no ,which implies +
2
I 2 Is,' -f'l
2 (s; -f +)+2 wo for all n 2 no, and so contradicts o, J- 0). But then If -s: I = f v,,
IS,,-f
+
so we have
0 6 f +-s;
2 v,, 10.
+
-s,:
and
+.
This shows, in view of s', being increasing, that s,' t f As regards the alternative statement, it follows similarly as above that 0 =< Pbf-paf-sx
5 Inle,
and so s, t, (Pbf - P a f ) holds with e-uniform convergence. Finally, observe that for a = 0 the corresponding projection Pa = Po is the order projection on the band generated by (- f)' = f -, and so Pof = 0. This implies that Paf = 0 for all a 6 0, and so +
+
P a f = P a f + - P a f - = - P a f - 7 0 as 0
2 a 1 -a.
Similarly, since Pof -- = f -, we have Pb f - = f - for all b 2 0, and so Pbf = P b f + - P b f - = P b f + - f - t f ' - f -
=f
as 0
5 b f a.
On the basis of the result just proved we could write again
f
=Sm -m
adpa,
where the integral is to be interpreted as a kind of improper integral, i.e.,
260
FREUDENTHAL'S SPECTRAL THEOREM
j Z m rdp, is the limit (as a 1 - co and b 7 gence of the integral
Isba
dpa = pb
00)
[CH. 6,540
in the sense of order conver-
f - p a f.
We shall make no use of this notation. The spectral theorem is due to H. Freudenthal ([l], 1936). He assumed the Riesz space L to be Dedekind a-complete, and so the present theorem, in which L is only assumed to have the principal projection property, is somewhat more general. This is not a mere technicality, as will be shown in Theorem 42.5. In the same year 1936 a paper by S. W. P. Steen [1] was published, in which the spectral theorem was proved for elements in a Dedekind complete Riesz space L such that L is at the same time a commutative ring in the algebraic sense. The spectral theorem was also proved for Riesz spaces of a particular kind in a paper by F. Riesz, originally written in Hungarian (1937), and published in French (in the Annals of Mathematics) in 1940 121. I n this paper F. Riesz considers an ordered vector space L (cf. Definition 11.1) such that the positive cone L+ is generating (i.e., everyfE L is the difference of two elements in L") and such that the dominated decomposition property (cf. Corollary 15.6) holds. Every Riesz space has these properties, but as shown in Exercise 15.14, a space L having these properties is not necessarily a Riesz space. The real linear functional cp on L is called positive whenever cp(f) 2 0 holds for a l l f z 0 in L. The set L" of all linear functionals cp = cpl - cpz with cpl ,cp, positive is evidently a real vector space, and if we define cp 6 JI for elements cp, JI EL" to mean that JI - cp is positive, it is easily seen that L" is an ordered vector space. The space L" is called the order dual of L, and it can be shown that L" is actually a Dedekind complete Riesz space. It was for a Riesz space L" of this kind that F. Riesz proved the spectral theorem. Finally, we mention a paper by H. Nakano ([4], 1941), in which the spectral theorem was proved for a Dedekind a-complete space (as part of a general theory of integration in such spaces). In order to prove some other properties of the spectral system (pa :
- co < a < 00) of an elementfin the band generated by the nonzero ele-
ment e in L", we will introduce (just as before) the notation r, = e-pa for any real a, and the order projection on the band generated by r, will be denoted by R,. Evidently, P,g+R,g = g holds for every g in the band generated by e.
CH. 6, # 401
26 1
FREUDENTHAL'S SPECTRAL THEOREM
Theorem 40.4. ( i )
Iff
is an element in the band generated by e,
if (pa :
- a < u < a)is the spectral system off and r, = e-pa for every u, then f + = sup ur,. It is even true that f = sup unranfor anrunning through any countable and dense subset of the set of all real numbers. More generally, if1 +
is any real number, then
( f - h ) + = sup, (u-A)r,.
(ii) Conversely, fi uq 4 f holdsfor some real u > 0 andsome component q of e, then q 5 r,. More generally, given u and A. such that lo< u, if +
("-1o)q
4 (f-Aoe)+
holds for some component q of e, then q 6 r,. Proof. (i) By Theorem 38.4 (ii) we have ur, 4 R,f for every u. Furthermore, we have R,f = R , f ' - R , f 5 R,f+ 6f ' . It is evident, therefore, that f is an upper bound of the set of all elements (ur, : - a < u < a).On the other hand, as shown in the proof of the preceding theorem, we have +
f+ =
+
= "P
=
n Mn,k-180
=
n
fl
un,k-l(PadJ,pan,k-l)
un,k-l(ran,k-]-rMd)
ISUP%, k k
c
Un,k-IhO
k-
l-ra,,,)],
so f is already the supremum of a sequence of elements each of which is less than or equal to an appropriate ctr,. It follows that f = sup urn. If (u, :n = 1,2, . . .) is a countable and dense subset of the set of all real numbers, we can restrict the points in all partitions x, to this subset, and the same argument shows then that f + = sup unrun. In order to prove that +
+
(f-Ae)+ = sup, (u-A)r, holds for every real 1, we observe that for 1 fixed the elementf' = f - Ae has the spectral system (pi : -a < u < a )withpi = and so r: = e-pi = r,+Afor all u. It follows that (f')" = sup uri, i.e., (f-Ae)+ = sup, ur,+A = sup, (u-n)r,.
262
[CH. 6.5 40
FREUDENTHAL'S SPECTRAL THEOREM
(ii) Assume first that aq 5 f ' holds for some real a ponent q of e, and let 0 < A < a. Then
aq-le 5 f + - l e =f'-(Ae)'
=- 0 and some com-
5 (f-Ae)',
g -h' _I (g-h)' holds for all g, h EL. It follows where we have used that ' that (a-A)q-I(e-q) 5 (f-Ae)'.
Taking positive parts, and observing that q and e-q are positive and disjoint, we obtain ( a - I ) q 5 (f-Ae)', implying that q is in the band generated by (f-le)' = ( I e - f ) - . This shows that q is disjoint to the band generated by (Ae-f)', and so q IpA.This holds for all A satisfying 0 < A < a. Hence, since pA pa as A t a, we have q I p , . It follows that q is in the band generated by r, = e-pa. But then, since q and r, are components of e, we must have q 5 r,. Now, let A, < a and (a-Ao)q 5 (f-l,e)' for some component q of e. Writing a-A,, = a' andf-Aoe =f',we have, therefore, that a'q 5 (f')'. Hence, denoting the spectral system off' by ( p i : -a < p < a)and writing ri = e-pi, it follows already that q 5 r;, = But rl = rp+ao holds for all p, so q 5 r,. This concludes the proof.
r
It is appropriate at this point to make some remarks about a certain ambiguity in the method of obtaining a set of components of e having the properties of a spectral system. Given e and f as before and given the real number a, we have definedp, as the component of e in the band Ba generated by (ae-f)', and the complementary component r, is then defined by r, = e-pa. The thus obtained spectral system (pa : - 00 < a < a)is left continuous, i.e., a T /? implies pa t ps. Another possibility, working just as well in principle, is to define rl as the component of e in the band generated by (f- ae)' = (ae-f)-, and then to introduce the complementary component p: by p: = e-r& . It turns out that (p: : - co < a < co) has much the same properties as (pa : -a < a < a). More specifically, a 5 /? implies pi 5 p i , and p: 10 as a 1 - co and p: t e as a t 00. Instead of left continuous, however, p: is now right continuous, i.e., a 5- /? implies p: 1pi. Furthermore, in the case thatfis contained in the ideal generated by e, we have (a+E)e 5 f 5 be for appropriate real a, b and an appropriate E > 0. The lower sum s ' ( n ; f ) and upper sum t ' ( n ; f ) off with respect to the partition n of [a, b ] may now be defined in terms of (p: : - 00 < a < co) exactly as was done before in terms of (pa : - co < a < co). If (n, : n = 1,2, . . .) is a
CH.6, § 401
FREUDENTHAL'S SPECTRAL THEOREM
263
sequence of partitions of [a, b], each n n f l a refinement of z, and if lnnl 10, then again s'(n,; f ) t f and t'(n,; f ) 1f hold e-uniformly. Iff is contained in the band generated by e, but not necessarily in the ideal generated by e, then a statement in terms of (p: : -00 < a < a)holds, completely similar to the statement in terms of (pa : -00 < a < co) which was proved in Theorem 40.3. All this is not surprising, since the sytems (pa : - 00 < a < co) and (p: : - co < a < 0 0 ) are closely related, although not necessarily identical. The following lemma holds.
Lemma 40.5. We have pa S pi S pa for a < B. In other words, we have ra 2 ri 2 rs for a < B. Proof. It follows immediately from (ae-f)' 1(f-ae)' that pa 1rd, and so the bands generated by pa and rl are disjoint. But then the band generated by pa is included in the band generated by p: = e - ri ,sop, 5 p: . In order to show that pi 6 ps holds for a < B, we have to show that e-ri 5 p s , i.e., r: +ps 2 e. For this purpose it is sufficient to prove that sup ( r i , p s ) = e, i.e., it is sufficient to prove that the algebraic sum of the bands generated by ri and ps respectively contains the element e. In other words, it is sufficient to prove that e is contained in the algebraic sum of the bands generated by ( f - ae)' and (Be- f)' respectively. Now, in view of the 2 (g+h)', we have general formula g'+h' (f-ue)'
+(Be-f)+
>= ( P - E ) ~ ,
from which the desired result follows. Analogous to Theorem 40.4 we have the following theorem, the first part of which is a little better than the first part of Theorem 40.4.
Theorem 40.6. (i) We have f ' = sup ar&. More generally, if1 is any real number, then (f-Ae)' = sup(a-A)r:. a
(ii) Conversely, if aq If 'holdsfor some real a > 0 and some component q of e, then q 5 r; holdsfor every 1 < a. More generally, given a and 1, such that a > 1, , if ( a - I,)q 5 ( f -I . e)' holds for some component q of e, then q 4 r; holdsfor every 1 -c a.
Proof. (i) Since f ' = sup ctr, holds by Theorem 40.4 and since ri 6 ra holds for every a, it is evident that f ' is an upper bound of the set of all elements arh. It will be sufficient now to show that
264
[CH.
FREUDEIWHAL'S SPECTRAL THEOREM
6 , s 40
f + = sup(uri : u 2 0). For this purpose, choose a number E > 0. Given any u that a < B 4 a+&. Then ur;
2 0, choose p such
2 ur,., 2 (B-E)~,.,= Br,.,-Ers 2 Br,.,-Ee.
If a runs through the set of all non-negative numbers, then B runs through the set of all positive numbers, so that on account of f we obtain
+
= sup Brs = sup (Pr,.,: B
sup (Brs-Ee :
> 0)
=- 0) = f +-ee.
This shows that any upper bound of (ari : a 2 0) is at least f'-ee. But E > 0 is arbitrary, and inf (Ee : E > 0) = 0, so f + = sup uri. (ii) If uq I ;f for some u > 0 and some component q of e, then q 5 r, by Theorem 40.4, and so q I ;r, 5 r; for A < u. +
Given again the elementf in the band generated by e, the spectral system (pa : - a < u < a)off has the following properties: (i) p , is a component of e for every u. (ii) u 6 B implies that p , 4 p s . (iii) p , I p s as u t B,p, 10 as u 1-a and pa t e as a t a. (iv) If 0 2 a, 1 - a and 0 4 b,, t a,and n,(u,,, ., u,,,) is a partition of [a,, b,] such that, for every n, the restriction of n,+ to [a,, b,] is a refinement of n,, and if lnnl 10, then
..
c ma
s, =
an*k-l(Pu~-Pan,k-J k= 1
is order convergent to$ We will show that the system (p, : - a < a < a)is uniquely determined by these properties. Before doing so, we present a simple lemma. Iff, f f o orf, foand allf, are contained in some band B, then f o is in B (by the definition of a band). We shall prove now that iff. is order convergent tofo and allf, are in the band B, then f o is in B.
Lemma 40.7. I f B is a band in the Riesz space L, iff. E B for n = 1,2, . . . andf, is order convergent to f o , then f o E B.
Proof. There exists a sequence (u,, :n = 1,2, . ..) in L+ such that Ifo-f,,l
- u, 5.0.Hence I
CH.6 , s 401
0 Sf,'-inf(f,+,f,+) and so inf, {f
265
FREUDENTHAL'S SPECTRAL THEOREM
=
linf(f,+,f,+)-inf(f,+,f,+>l
s If,'
-fn+ I 5 If0 -fnI 6 un 10,
0'- inf ( f i,f :)>= 0, which implies that fo+ = SUP {inf(f,+,f,+)>. n
But inf ( f :,f.') is in B for every n on account o f f : f l E B. Similarly we derive that E B, and so f o E B.
fo
E B. It follows that
Theorem 40.8 (Uniqueness theorem).Let f be an element in the band generated by e, let (pa : - co < u < co) be the spectral system of S, and let (qa : 03 < u < 0 0 ) be a set of elements in L having the following properties: (i) qa is a component of e for every u. (ii) a 5 B implies rhat qa 6 qs. (iii) q a T q s a s a T B , q a J O a s u J-00 a n d q a f e a s a f c o . (iv) If0 2 a,, 1- co and 0 5 b, t 00, and 7c,(uno, u,,,,,) is apartition of [a,, b,] such that, for every n, the restriction of z,+~ to [a,, b,] is a refinement of n,, and i f ln,l J. 0, then
-
. ..,
is order convergent t o 5 Then (qa : - 00 < u < 00) is the spectral system of S, i.e., qa = pa for every a.
Proof. From property (iii) of the system of all qa it follows immediately that
as n + 00, where the horizontal arrow denotes order convergence. Hence, given any fixed real 12, it follows then from property (iv) that
On account of
it is evident now that m,
266
FREUDENTHAL'S SPECTRAL THEOREM
(CH.
6 , 8 40
is also order convergent to I e - f . Hence, taking positive parts, we obtain the result that (n-~nk)(qa,k-qa,*k-l) (Je-f)'.
c
+
anksA
For every n, all terms in the last sum on the left are contained in the band generated by g A (in view of the hypothesis that a 5 p implies ga I g p ) , and so it follows by means of the preceding lemma that (Ie-f)' is also contained in the band generated by 41. This immediately implies that pA qa . In order to obtain the inverse inequality, we introduce the component ra complementary to qa, i.e., ta = e - g , for every real a. Then qUnk
-q a n , k -
1
-
'an, k - 1
-
and so we infer from (1) that m.
which implies
k=l
ASUn.k-1
(a.,k-l-A)(tan,k-i-tank)
(an, k - 1 -n>(tan,
+ f - I e 9
k - 1 -ta,>
--t
(f-Ie)+-
For every n, all terms in the last sum on the left are contained in the band generated by t A (since t, decreases as a increases), and so it follows by means of the preceding lemma that (f-Ae)' is also contained in the band generated by rA. This implies that r; S t a , where (as before) r; is the component of e in the band generated by (f-Ae)'. It follows that p i = e-r; satisfiesp; 2 q A . It follows from Lemma 40.5 that p i t pr as I t p, and by hypothesis we have S p ; holding for all p , we obtain qa t q,, as A p. Hence, in view of qr S p,,. This completes the proof.
Exercise 40.9. The system (ga : - 00 < a < co) of elements in the Riesz space L is called convex (with respect to a ) if, for each pair of real numbers a1 a2 and for each a such that
-=
a = tal+(l-r)a,
<= t <= 1, we have g a S tga, + (1 - t)ga,
holds for some t satisfying 0
*
(i) Show that if (ga : - 00 < c1 c 0 0 ) is convex, then ( g z : - co < a < is convex. In particular, iff is an element in the band generated by the element e > 0 in L', and we write ua = ( a e - f ) + for every real a, then ( u a : -GO < a < a)is convex. 00)
CH. 6 , § 401
FREUDENTHAL'S SPECTRAL THEOREM
(ii) Show that (g, : - 00 < a <
00)
267
is convex if and only if
holds for all triples a1, a 2 , a3 satisfying a1 < a2 < a 3 . Show also that (9, : -co < a < 0 0 ) is convex if and only if (a2- al)-'(ga2-gUl)
5 (a3- a2)-l(gu3-ga2)
holds for all triples a l , a 2 , a3 satisfying a1 < a2 < a3. (iii) Assume that L is Archimedean, let (9, : - 03 < a < a)be bounded on every bounded subinterval [a, b ] of the real line, and let 9.)(.1+.2)
5 HS,, + 9 a 2 )
hold for all tll , a2. Show that g, is order continuous with respect to a (Le., for any a and for any sequence of real numbers h, + 0 the sequence ga+h, is order convergent to g,), and show that (g, : - co < a < co) is convex. (iv) Show that any convex system (g, : - 00 < a < co) is bounded on any bounded interval of the real line. Hence, if L is Archimedean, any convex system is order continuous. Hint:For (iii), let a be in the interior of [a, b ] , and let Ig#l j uo for all /3 E [a, b]. Given the natural number n, choose the number h (positive or negative) such that a-nh and a+nh are in [a, b]. Since 2ga 5 ga+h+g#-hy it follows that ga-ga_,, 5 g,+h-g,, so ga-ga-nh
n(ga+h-ga)
s ga+nh-gah.
This shows that 1g,+h-gU1 5 2n-'u0, and so (since L is Archimedean) the order continuity of g, follows. Given now that a1 a2 and tl = f a , + (1 - t)a2, it follows easily from the hypotheses that 9, tg,, + (1 - t)ga2
-=
s
-
holds for every rational number t = k 2-" (k = 0, 1,2, . . ., 2"). The set of these rational numbers is dense in [0, 11, so the order continuity of ga implies that the same inequality holds for every t in [0, 11 (for this last remark, note that f , 5 g,,, f, +fo and gn + go implies fo 5 go). For (iv), let [ a l , a 2 ]be a bounded interval, choose a3 such that a3 < al , and write = (a1-a3)-1(gUl-gU$)' Then h
( a - a l ) ~ l ( g , - g , l ) for all a such that a1 < a j a 2 , so g, 2
268
FREUDENTHAL'S SPECTRAL THEOREM
[CH.6,8 40
ga,+h(a- al). It follows that 5% B gal - (a2 - a1)lhl
for all a E [a1, a,], so g , is bounded from below. Since g , 4 sup (g,,, gat) by the convexity, g , is also bounded from above in [al, a,].
Exercise 40.10. Let ( g , : - 00 < a < co)be a convex system of elements in the Dedekind o-complete space L. Show that, for any fixed a, the quotient h - l ( g a + h - g a )is decreasing as h > 0 is decreasing, so the right deriuatiue
exists. Similarly, the Zeft derivative
s
exists. Show that D - g , 5 D'g, 4 0 - g s D + g s holds for a < B. Show that D - g , = D'g, at each point a where either D - g , is right continuous or D'g, is left continuous.
Exercise 40.11. Let the Riesz space L have the principal projection property, and let f be an element in the band generated by the nonzero element e in L'. For any real a, let u, = (xe-f)' and let pa be the component of e in the band B, generated by u,. Finally, let r, = e-pa and let C, be the band generated by r,. Similarly, let ri be the component of e in the band CL generated by u, = ( a e - f ) - , andp: = e-ri the complementary component We recall that pa is left continuous, p: is right continuous and pa 5 p: 5 pp for a < p. Show that the left and right derivatives D-u, and D'u, exist and satisfy D-u, = pa and D'u, = pi for every a. Hint: Note that, for every a and for h > 0, we have
+ '(0,-0,
h - (ua + h - ua) h -
+h)
= e.
Denoting the two terms on the left by tl and t , respectively, we have 0 4 tl 5 e and t , is in Ba+h,so 0 5 t , 5 p a + h ,whereas 0 5 t , 5 e and tz is in CL, so 0 5 t 2 5 r:, i.e., f , = e--t, 2 p:. It follows then from p: 5 tl 5 P : + h and the right continuity ofp; that D+ua = p:.
Exercise 40.12. Let L have the principal projection property, let e be a fixed nonzero element in L + , and assume that the band generated by e is
CH. 6,s 411
THE RADON
- NYKODYM
THEOREM AND THE POISSION FORMULA
269
order separable. Given the element f in the band generated by e, let ( p , : - 00 < a < 0 0 ) be the spectral system off with respect to e. Show that p , , as a function of a, has an at most countable number of discontinuities (i.e., pa # p: for at most countably many a). Hint: Write v(c1) = p:-pa for every a. If a. and a1 are different points of discontinuity for p a , then u(ao)and u ( a l ) are nonzero and disjoint. Hence, if pa has uncountably many discontinuities, there exists an uncountable disjoint system having e as an upper bound. This is not compatible with the order separability of the band generated by e. 41. The Radon-Nikodym theorem and the Poisson formula
In the present section it will be shown that the Radon-Nikodym theorem in measure theory and the Poisson formula for harmonic functions in an open circle are particular cases of the spectral theorem in a Riesz space. Example 41.1. In Example 23.3 (v) we introduced the Riesz space L of all finitely additive signed measures p on the field I' of subsets of the point set X , having the property that llclll = SUP (IP(4I :A E r ) is finite, and we proved that L is Dedekind complete. In Example 25.3 the extra assumption was made that r is a o-field (and not merely a field) and the set L, of all countably additive members of L was proved to be a band in L. Furthermore, the following two facts were proved. (i) If p, v E L , , then p J- v holds i f and only i f there exist disjoint sets A , B E r such that A u B = X and Ivl(A) = Ipl(B) = 0. (ii) v E L , , then v is in the barid generated by p i f and only i f IvI is absolutely continuous with respect to lpl, i.e., any E E r satisfying lpl(E) = 0 also satisfies IvI(E) = 0. We assume now that E~ is a fixed nonzero element in L: and we first prove the following result. (iii) The measure IL is a componenr of c0 (i.e., n satisfies 0 5 n 5 e0 and n I( E ~ n)) i f and only i f there exists a set A E r such that
up,
holdsfor every E E r, where xA denotes the characteristic function of A . For the proof, assume first that a is a component of E ~ By . property (i)
270
FREUDENTHAL'S SPECTRAL THEOREM
[CH.
6,s 41
there exist disjoint sets A, B E r such that A u B = X and (8, - n ) ( A ) = n(B) = 0. It follows that n(E) = eO(E)for any E E r such that E is a subset of A. Hence, if E E r is now arbitrary, we have K ( E ) = n(E n A ) = E,(E n A ) =
JE
xAd ~ , .
Conversely, if A, B E r are given such that A v B =,'A and we set n(E) = all E E r, then K is evidently a countably additive measure on r such that 0 5 n 5 E, and such that
jE XAdEOfor
c
P
holds for all E E r. It follows immediately, by means of property (i), that n I(go-n). The set A E r in (iii) is not uniquely determined, because if Eo E r satisfies eO(EO)= 0, then Eo may be added to A or deleted from A. It follows easily from the results obtained so far that the components n1 and nz of E, are disjoint if and only if the corresponding sets A, and A z can be chosen such as to be disjoint. Given a = uini, with n,, . . ., n, components of E, and u, , . . ., un real numbers, every nI corresponds by (iii) to a function xAI, and n
4 ~=)iC= l uixAI(x) satisfies
a(E) =
JE
S(X)d&,
for every E E r. It follows that B = I ; u i= n i is a member of L,' if and only if n
4 ~=)C
aixAI(x) I=1
20
holds almost everywhere. Hence a1 5 az holds if and only if s l ( x ) I s z ( x ) holds &,-almost everywhere. We prove finally that the Radon-Nikodym theorem is a special case of the spectral theorem. (iv) Let E, be the same nonzero measure as before, let 0 g p E L,, and let p be &,-absolutely continuous (i.e., in view of property (ii) p is a member of the positive part of the band generated by 8,). Then there exists a non-negative
CH.6 , s 411
271
THE RADON-NIKODYM THEOREM AND THE POISSONFORMULA
e,-summable function g ( x ) on X such that holds for every E E r. Conversely, i f g ( x ) is non-negative and 8,-summable on X,then the formula (1) dejnes an element p in L: such that p is &,-absolutely continuous. For the proof, assume that 0 p E L , and p is &,-absolutelycontinuous, so p is contained in the positive part of the band B generated by E , . By the spectral theorem (unbounded case) there exists a sequence (a, :n = 1,2, . . .) in B such that 0 a, p, each a, a finite linear combination of components of E , , so
s
s
an(E) =
1 E
sn(x)d&o
for an appropriate non-negative step function s,(x). Since 0 6 L,) implies that
Q,
tp
(in
0 5 an(E) t P(E)
holds for every E E r, it follows that
where g ( x ) is the pointwise limit (&,-almost everywhere) of the increasing sequence s,(x); n = 1 , 2 , . . .. Evidently we have g ( x ) 2 0 for &,-almost every x E X , and the c,-summability of g follows by observing that p ( X ) = j x g(x)d&,is finite. It follows from property (iv) that the band generated bye,, as a Riesz space by itself, is Riesz isomorphic to the space of all real 6,-summable functions, where the isomorphism can be determined such that E, itself corresponds to the function e,(x) satisfying eo(x) = 1 for all x E X. The presentation given here is related to the exposition in a paper by K. Yosida ([2], 1940).
Example 41.2. Let G be a region in the plane and L the Riesz space of all functionsf = u1 - u2 with u1 , u2 harmonic and non-negative in G. In Example 23.3 (vi) it was proved that L is super Dedekind complete. In the present example we restrict ourselves to the case that G is the open unit circle (i.e., the center of G is the origin 0 and the radius r satisfies r = 1). The boundary of G will be denoted by X (so X can be written as X = (q :0 6 cp < 2n)) and the collection of all Bore1 subsets of X will be denoted by r. The collec-
272
SPECTRAL THEOREM
FREUDENTHAL'S
[CH.
6 , I 41
tion r is, therefore, the smallest a-field of subsets of X containing all intervals. By M we denote the Riesz space of all finitevalued countably additive signed measures on I'. One of the members of Mis Lebesgue measure m on X . We have m ( X ) = 2n, so the normalized Lebesgue measure E, = (2n)-'m satisfies e O ( X )= 1. Furthermore, given any p E M , we have (uniquely) that p = Pl+P2,
where pl is &,-absolutely continuous (i.e., pl is a member of the band generated by Lebesgue measure) and p2 IE,. Corresponding to pl there exists a Lebesgue summable function g,(cp) on X such that r
holds for every E E r. For the points P and Q in or on the boundary of the circle G we denote by K(P, Q) the Poisson kernel
There exists a one-one correspondence between the harmonic functions in the Riesz space L and the measures in the Riesz space M , as follows. Given f E L, there exists a measure p E M such that f(P) =
1
QEX
K(P, Q>&
holds for every point P in G. Conversely, given p E M , the above formula defines a functionfe L . For these facts, cf. e.g. the book by K. Hoffman on Banach spaces of analytic functions ([l], Ch. 3). The correspondencebetween L and M is one-one, linear and order preserving in both directions. Hence, L and M are Riesz isomorphic. In particular, it follows that Mis super Dedekind complete. Given f E L, the corresponding p E M is of the form p = pl+p2 with pl absolutely continuous with respect to E, and p2 IE,, so f(P)=
1
QEX
K(P, Q)g,(Q)&
+
QEX
K(P, Q)&
-
Denoting the terms on the right byf,(P) and f2(P)respectively, the function f, is the component off in the band generated by the function e, E L satisfying eo(P) = 1 for all P , andfz is the component off in the disjoint complement.
CH. 6 , g 411
THE RADON-NIKODYM THEOREM AND THE POISSON FORMULA
273
We observe finally that there is also a theorem (cf. the book referred to above; Ch. 3) stating thatfe L has the radial boundary value g,(Q) for E,almost every point Q on X. In other words, denoting the radial boundary function off E L byf", we havef"(Q) = g,(Q) at &,-almost every point Q on X . We now restrict our attention to the band Beo generated in L by eo (we recall that eo is the function satisfying e,(P) = 1 for all points P in G). The band Beo consists, therefore, of all functions equal to the ordinary Poisson integral of their boundary value function, i.e., Beo consists of allfE L satisfying r
The band Be, is Riesz isomorphic to the band generated in M by E,, i.e., Beois Riesz isomorphic to the space of all (real) Lebesgue summable functions f * on X. It is evident, therefore, that if S is any subset of X (with characteristic function xs) such that S E r, then ~ s ( P= )
1
QeX
jBs.(.. Q)~&o
K(P, Q ) X S ( Q ) ~=~ O
is a component of e, . Conversely, every component of e, must correspond to a component of E , , and so must be of the above kind. It follows easily that the components e,, and e,, are disjoint if and only if S1and S, are disjoint subsets of X. Givenf in the band Beo and given the real number u, the component pa of e, in the band generated by (ae, -f)+ is now given by pa = e,, where S is the subset of X defined by S = (Q : Q E X ,f "(Q) < a).
For u < p, we have p @-pa = eT with
so
T = (Q : Q E Xa S f " ( Q ) < B),
holds for every point P in G. It follows now easily that the spectral theorem for f is expressed by the Poisson formula
f(P)=
1
QsX
K(P, Q ) ~ " ( Q ) ~ E ~ .
274
FREUDENTHAL'S SPECTRAL THEOREM
[CH.
6 , § 41
We will make a few additional remarks. In the first place we present an example of a nonzero functionfin the space L such thatfis disjoint frcm the band generated by eo . For this purpose, let Q, be a fixed point in X.We set f ( P ) = K(P, Q,) for every point P in G. Evidentlyfis a non-negative harmonic function in G ; the corresponding Bore1 measure p is defined for any E E r by p ( E ) = 1 if Q, is a point of E and p ( E ) = 0 otherwise (i.e., p corresponds to a unit point mass in the point Q,).Since p IE , , it follows thatf is disjoint from the band generated by e,. It is evident from the procedure of constructing the function f that there exists an uncountable disjoint system (fr : z E {z}) in L+,eachf, disjoint from e,. In order to show, finally, that the space L possesses no weak unit (and hence possesses no strong unit), we first prove the following simple result. (i) V L is an order separable Archimedean Riesz space with a weak unit e, then any disjoint system (u, : z E {t}) in L+ is at inost countable. The proof follows immediately from Theorem 29.3, according to which (since L is Archimedean and order separable) we have inf ( e , u,) > 0 for at most countably many t. On the other hand, since e is a weak unit, we have inf (e, u,) > 0 for all z. Hence, the index set {z} is at most countable. The space L in the present example is super Dedekind complete (and so L is surely Archimedean and order separable). Hence, if L would possess a weak unit, then any disjoint system would be at most countable. We have seen, however, that L contains an uncountable disjoint system. Hence, L has no weak unit. The band Beo has, of course, the element e, as a weak unit. We observe that Bpocontains no strong unit. Indeed, given any element u, 2 0 in Beo,the boundary function u; is Lebesgue summable over the boundary X of the circle G . Let u; be another non-negative Lebesgue summable function on X such that the quotient u;/ui is unbounded (note that there exists such a function u ; ) , and let u1 be the harmonic function in G, equal to the Poisson integral of u;. Since u, and u1 converge radially (Lebesgue almost everywhere) to u; and u; respectively, it follows that ul/uo is unbounded in G . This shows that u, cannot be a strong unit in Be,.
Example 41.3. Let X = (x : a x p ) be a bounded interval in the real axis, and let L be the Riesz space of all real linear functions on X . In Example 23.3 (vii) it was proved that L is super Dedekind complete. The element e E Lf is defined by e ( x ) = 1 for all x E X , the element el E L+ by e l ( a ) = 1, e l ( p ) = 0, and the element e, E L + by .,(a) = 0, e2(p) = 1. The bands generated by el and e, are denoted by El and E, respectively. All bands in
CH. 6,5411
THE RADON-NIKODYM THEOREM AND THE POISSON FORMULA
275
L are projection bands, and the Boolean algebra of all bands (in this case the
same as the Boolean algebra of all projection bands) consists of {0}, E l , E2 and L. Let , f L~; f(a) = a and f ( p ) = b with a c b, say. Furthermore, for any real 1,let p1 be the component of e in the band B1 generated by (1.-f)'. We have (0) for 1 a, BA = El for a < 1 S b, L for b < 1, and, accordingly, 0 for 1 =< a, e , for a < 1 5 b, e for b < 1.
[
Now, let [a,, b,] be an interval containing [a,b ] in its interior and let n(&, A,, .,1,) be a partition of [ a , , b,] such that a is contained in the interior of [&, Ak+,] and b in the interior of [ A j , L j + , ] . The corresponding lower sum s ( n ; f )satisfies
..
s ( n ; f ) = &el +Aj(e-e,)
=
Ake, +Aje2,
so s ( n ; f )is the linear function on the interval [a, p ] having the value 1k at a and the value at /I.The spectral theorem, for this example, is the statement that for1.1 tending to zero, the lower sum s ( n ; f )tends uniformly to ae, +be, = f. Similarly for the upper sum. Note that the Riesz space L is Riesz isomorphic to the Riesz space M consisting of all real functions "f on a point set S consisting of two points s1 and s2, the correspondence between f E L and "fE M being such that the values of "fin s, and s2 are equal to the values offin a and jl respectively. In other words, L is Riesz isomorphic to the plane R2 with ordinary coordinatewise ordering. Note that the elements el and e2 are atoms in L.
Exercise 41.4. Let L be the Riesz space of all real bounded functions on X = (1,2, . . .), assuming only a finite number of different values; the ordering is pointwise. This space has the principal projection property, but is not Dedekind a-complete. The function e, satisfying e(x) = 1 for all x E X, is a strong unit in L. Given anyf E L,state the spectral theorem forfwith respect to e. Exercise 41.5. In Exercise 11.15 we introduced the Riesz space of all real
276
FREUDENTHAL'S SPECTRAL THEOREM
[CH. 6,B 42
functionsf of finite variation on the interval [a, b] satisfying the extra condition thatf(a) = 0. The positive cone exists of all non-decreasingfin the space. Here we will consider the special case that the interval is (cp :0 S cp < 2n) and that the functions are continuous from the left besides being of finite variation and satisfyingf(0) = 0. Show that the thus obtained space K, with the positive cone defined as abcve, is a Riesz space. There is a oneone correspondence between the functions g E Kand the (countably additive) signed measures p on the Borel sets of [0,2n],as follows. Given g E K, we set p( [0, cp)) = g(cp) for any [0, q ) contained in [0, 2n), and p is then extended in the natural manner to every Borel set. Conversely, given the signed measure p, the function g E K is defined by g(cp) = p([O, cp)). Show that K and the Riesz space M of all (countably additive) signed measures on the Borel sets of [0, 2n) are Riesz isomorphic under this correspondence. Hence, if L is the Riesz space of all functionsf = u1- u2 in the unit circle with u l , u2 non-negative and harmonic, and M and K are the spaces mentioned above, then L, M and K are Riesz isomorphic. The function identically equal to one in the unit circle corresponds to normalized Lebesgue measure in M, and to g(cp) = (2n)-'cp in K. 42. Uniform completeness, Dedekind completeness and the spectral theorem
The first definition in the present section is based upon the notions of an e-uniformly convergent sequence and an e-uniform Cauchy sequence, as introduced in Definition 39.1. Definition 42.1. The Riesz space L is said to be uniformly complete whenever, for every e E L', any e-unifbrm Cauchy sequence has an e-uniform limit. Theorem 42.2. The Riesz space L isuniformly complete if and only f , for every e E L + , any monotone e-uniform Cauchy sequence has an e-unform limit.
Proof. Follows immediately from Theorem 39.4, according to which every e-uniform Cauchy sequence has an e-uniform limit if and only if this holds for monotone e-uniform Cauchy sequences. As observed in Lemma 39.2, every Dedekind o-complete space is uniformly complete, and it may be asked whether, at least for Archimedean spaces, the converse is also true. The answer is negative, as shown by the space C([O, 11) of all real continuous functions on the interval [0, 11. Hence, a Riesz space must possess still other properties besides uniform completeness
CH. 6 , s 421
DEDEKIND COMPLETENESS AND THE SPECTRAL THEOREM
277
in order to be Dedekind a-complete or Dedekind complete. Appropriate other properties are to be found in the list of properties in Theorem 25.1 (the main inclusion theorem), where the following implications were proved to hold: 47
Ded. conipl.
Ded. a-comp!.
Q proj.
%A.
pnnc. proj. prop. prop. -#,
It will be shown in the present section that in a uniformly complete space the principal projection property is equivalent to Dedekind a-completeness,and the projection property is equivalent to Dedekind completeness. The main tool in the proof is the spectral theorem. As a consequence, we will obtain the result (announced already in the inclusion theorem) that Dedekind a-completeness and the projection property together imply Dedekind completeness. It will also be proved that in a Riesz space L in which the Boolean algebra B ( L ) of all projection bands is Dedekind complete, the principal projection property is equivalent to the projection property and Dedekind a-completeness is equivalent to Dedekind completeness. In order to facilitate the proofs, we first prove two lemmas. We recall that the principal band generated by f E L is denoted by B,, and the set of all principal projection bands, partially ordered by inclusion, is denoted by YP(L>. Lemma 42.3. If the Riesz space L has the principalprojection property arid L is un$ormly complete, then g p ( L ) is Dedekind a-complete.
.
Proof. Let Bunc Be for n = 1,2, . .. For the proof that sup Bunexists in B p ( L )it may be assumed without loss of generality that all the u,, and e are elements in L f . Furthermore, we may assume that u, is increasing as n increases (if not, replace Bunby B,,, +. ..+,, for every n). Under these assumptions, let en be the component of e in Bunfor n = 1,2, . ., and observe that Bun= Be,. We have to prove, therefore, that sup Ben exists in P p ( L ) . Now, let E, be a convergent series with strictly positive real terms, and let
.
1:
Then (s,, : I Z = 1,2, . . .) is an increasing e-uniform Cauchy sequence, as follows from
o 5 s,,-s,
m
g ( C ek)e m+ 1
for n > m.
278
FREUDENTHAL'S SPECTRAL THEOREM
[CH. 6 , s 42
Hence, by the uniform completeness, s,, has an e-uniform limit u, and we have t u by Lemma 39.2. The band generated by u is a projection band on account of the principal projection property, and obviously this is the smallest band including all the bands generated by all the 5, (n = 1,2, . . .). Hence BSnt B, holds in g p ( L ) .But Bsn = Ben holds for all n, so Bent Bu holds in Bp(L).This is the desired result. s,,
Lemma 42.4. Once again, let the Riesz space L have the principal projection property and let L be uniformly complete. Furthermore, let 0 5 u,, t S e. Given thefixedpartition 'IT = z(ctO,al , . . ., a,) of [0,2], the lower sum of u,, (with respect to 'IT and e ) is denoted by s,, = IT; u,,). Then there exists an element v EL' such that 0 6 s,, t v. Proof. We denote by (p:) : - co < a < 00) the spectral system of u, with respect to e, by PF' the order projection on the band B$" generated bypz), and by R r ) the order projection on the band C$) generated by r p ) = e -pr). Then r S,,
=
r
ak-l(Pg)-P::)-l)(e) = 2 tCk-1(Rt)-,-R:')((e). k= 1
k= 1
Hence, by the same argument as was used for proving formula (1) in the proof of Lemma 40.1 (iii), we obtain
. . ., ar-
s,, = sup (aoRz)(e),
Rt)-l(e)).
Since p:) is the component of e in the band generated by (ue-u,,)', it is evident that for a fixed p:) decreases as n increases, so r:) increases as n increases. It follows that @) increases as n increases, and so the corresponding bands C$" in B,(L) satisfy C:) t 6 Be. In view of the preceding lemma there exists now a principal projection band, say Caysuch that Cp' 7 C, as n + co. Denoting the order projection on C, by R,, we have R f ' u 7 R,u for every u 2 0 by Theorem 30.5 (ii), and hence sn
t SUP (a0 R,(e),
* *
-9
a r - 1 Rap-i(e))*
Denoting the element on the right by v, we have 0 the proof:
5 s,, t u. This concludes
Theorem 42.5. The Riesz space L is Dedekind a-complete if and only i f L has the principal projection property and L is unvormly complete. Proof. It will be sufficient to prove that the principal projection property and uniform completeness together imply Dedekind a-completeness. As-
CH. 6 , s 421
DEDEKIND COMPLETENESS AND THE SPECTRAL THEOREM
279
sume, therefore, that L has the principal projection property and L is uniformly complete, and let 0 6 u, t 6 e hold in L. It will be sufficient to show that sup u, exists in L. To this end, let (x, : p = 1, 2, . . .) be a sequencz of partitions of the interval [O,21 such that, for everyp, x p + lis a refinement of x p , and such that lnpl 5 p - l . Every u, (n = 1, 2, . . .) has a spectral system ( p f ) : -co < a < co) with respect to e ; let $') = ~(n,;u,,) be the corresponding lower sum of u, with respect to xp (and with respect to e). The double sequence p , n = 1 , 2 , . . .,
s p = s(7rp; u,);
is increasing i n p by Lemma 40.1 (ii) and increasing in n by Lemma 40.1 (iii). Furthermore, we have
o _I u,-ssIp) 5
6 p-le
by Lemma 40.1 (iv), and so
-'?sI
s,(dI
5 (p-l+q-l)e
holds for all natural numbers p, q, n. As proved in the preceding lemma, there exists for each p an element up E Lf such that
o 5 sjp) t up
as n -+
00.
The sequence (up : p = 1,2, . . .) is increasing since sAp)is increasing in p . Given the natural numbers p and q, we have
Iup- uql 6 I up - s y l + Is?' - @)I+ I s2' - uql 6 I U ~ - S ! , P ) I + ( p - l +q-l)e+lu,-s?)I for every n, so that in view of 0 5 up-$') 10 and 0 Iuq -$)
1 0 we obtain
Iup-vql 6 (p-'+q-')e.
It follows that (up :p = 1 2, . . .) is an increasing e-uniform Cauchy sequence, so sup up = uo exists. In view of $2)5 up 5 u,,, holding for all n andp, it is evident that we have u, = sup s?' P
5 uo
y
and so uo is a n upper bound of the sequence of all u, . If u* is another upper bound, then &" 5 u* holds for all n andp, and so up = supslp' I: u* n
280
FREUDENTHAL’S SPECTRAL THEOREM
[CH.
6, S 42
holds for all p , which implies that u,, = sup up 5 u*. Hence we have u,, sup u,, and the proof is complete.
=
If, besides the conditions that L is uniformly complete and has the principal projection property, we impose any further condition upon L which will imply that Sp(L) is Dedekind complete (and not only Dedekind a-complete), then the same proof will work to show that L is Dedekind complete. The statement analogous to Lemma 42.4 will be now that if (u, :T E {T}) is a directed set in L such that 0 5 u, t 5 e holds, then the corresponding lower sums s, = s(n; u,) satisfy 0 j s, T u for some u EL’. The Dedekind completeness proof is then quite similar to the proof in Theorem 42.5. The next theorem illustrates the situation.
Theorem 42.6. The Riesz space L is Dedekind complete i f and only i f L is uniformly complete and L has the projection property. Proof. Follows immediately from Theorem 31.3 (i) according to which, in a space with the principal projection property, B,(L) is Dedekind complete if and only if L has the projection property. Theorem 42.7. The Riesz space L is Dedekind complete if and only i f L is Dedekind a-complete and B p ( L )is Dedekind complete. Proof. If L is Dedekind complete, then L has the projection property. Hence, by Theorem 31.3 (i), P p ( L )is Dedekind complete. The converse is evident in view of the remarks made immediately before stating Theorem 42.6. Theorem 42.5 and Theorem 42.6 were also proved by A. I. Veksler ([3], 1966), and Theorem 42.7 is contained (with a different proof) in H. Nakano’s book on Riesz spaces and modulared linear spaces ([6], 1950, Theorem 13.1). Another important consequence of Theorem 42.6 is the result, stated already in Theorem 25.1 (the main inclusion theorem) without proof, that Dedekind a-completeness and the project ion property together are equivalent to Dedekind completeness.
Theorem 42.8. The Riesz space L is Dedekind complete if and only if L is Dedekind a-complete and L has the projection property. Proof. Assume that L is Dedekind a-complete and has the projection property. As observed several times already, the Dedekind a-completeness implies uniform completeness. Hence, by Theorem 42.6, L is Dedekind complete.
CH.6, 421
281
DEDEKIND COMPLETENESS AND THE SPECTRAL THEOREM
Recapitulating some of the results proved, we have the following facts. (i) I f L is uniformly complete, then Ded. a-compl. %prim. proj. prop. %proj. prop.
Ded. coiilpl.&7
(ii) More precisely, we have
Ded. a-compl. o princ. proj. prop. and unif. compl. Ded. compl. e-proj. prop. and unif. compl.
The example of a space L which is Dedekind a-complete (and hence is uniformly complete), but which does not have the projection property shows that, even if L is uniformly ccjmplete, the two remaining arrows in diagram (i) do not go in the converse direction. In the next paragraph we will investigate some conditions under which these remaining arrows point both ways, whereas the arrows which now point both ways in diagram (i) will then again point only one way. It has been proved in Theorem 30.6 (ii) that if L has sufficiently many projections, then L has the projection property if and only if the Boolean algebra B ( L ) of all projection bands is Dedekind complete. In particular, if L is a space with the principal projection property, then L has the projection property if and only if B ( L )is Dedekind complete. Briefly, proj. prop. o princ. proj. prop. and B ( L ) Ded. compl. The picture will now be made complete by proving the following theorem, which (with a different proof) is also contained in the book by B. Z. Vulikh on partially ordered spaces ([2], 1961, Theorem V.4.3.) Theorem 42.9. The Riesz space L is Dedekind complete if and only i f L is Dedekind o-complete and B ( L ) is Dedekind complete.
Proof. We need only prove the non-trivial part of the theorem. Let L be Dedekind a-complete and let B ( L ) be Dedekind complete. Then L has the principal projection property, and this in combination with the Dedekind completeness of B ( L ) shows already that L has the projection property. The projection property and the Dedekind a-completeness together imply then that L is Dedekind complete. Corollary 42.10. The Riesz space L is Dedekind complete all the following conditions are satisfied.
if and only if
282
FREUDENTHAL’S SPECTRAL THEOREM
[CH. 6, J 43
(i) L has the principal projection property. (ii) L is uniformly complete. (iii) B ( L ) is Dedekind complete. Recapitulating, we have the following. If B ( L ) is Dedekind complete, then L P
Ded. compl.
Ded. a-compl.
% proj. prop.
- .
prrnc. proj. prop.
The example of a space L having the projection property (so that B ( L ) is then Dedekind complete), but which is not Dedekind a-complete shows that, even if B ( L ) is Dedekind complete, the two remaining arrows in the last diagram do not go in the converse direction. 43. The spaces C(X) and C,(X)
Let X be an arbitrary topological space. The Riesz space of all real continuous functions on X will be denoted by C ( X ) , and the ideal of all bounded functions in C ( X ) by cb(X).
Theorem 43.1. The Riesz spaces C ( X ) and cb(X) are uniformly complete. Proof. Let 0 fo E C ( X ) , and let (f,: n = 1,2, . . .) be an fo-uniform Cauchy sequence, i.e., for every E > 0 there exists a natural number no = no(&)such that
If(x)-f,(x)I
S &f0(x) for x E X ; m, n 2 no.
It follows that the pointwise limit f ( x ) = lim,,_.,&(x) exists for all x E X. In order to prove that f is continuous, fix xo E X and E > 0, and take a neighborhood V of x,-, such that Ifo(xo)-fo(y)I 5 E holds for all y E V and lf,o(xo)-f,,(y)I S E holds for all y E V and no = no(&).It follows now from If(x0
1-f(Y 1I s If,
(xo
1--LO ( Y )I +E l f 0 (xo ) +fo (Y 11 s E +4 2 fo (xo ) +4
9
holding for ally E V, thatfis continuous. This shows that C ( X )is uniformly complete. The proof for cb(X) is similar since it is evident that if allf.(n = 1,2, . . .) andf, are bounded, then the limit functionfis bounded. We first consider conditions for Dedekind a-completeness.
Theorem 43.2. Let C ( X )and C,(X) denote the same Riesz spaces as above. The following conditions are now mutually equivalent.
CH.6 , 5 431
(i) (ii) (iii) (iv)
THE SPACES
c(x)AND c,(x)
283
C ( X ) has the principal projection property. C ( X ) is Dedekind a-complete. c b ( X ) has the principal projection property. c b ( X ) is Dedekind a-complete.
Proof. C ( X ) and c b ( X ) are uniformly complete. Hence, by Theorem 42.5, the conditions (i) and (ii) are equivalent, and the same holds for (iii)
and (iv). Furthermore, it is obvious that Dedekind a-completeness of C ( X ) implies Dedekind a-completeness of c b ( X ) , and so it remains only to prove that if c b ( X ) is Dedekind a-complete, then so is C ( X ) . For this purpose, let us assume that 0 6 unt 5 u holds in C ( X ) . For any natural number k , set u,,k = inf (u,, , k), and denote by w k the function SUP,, u,,k, which exists by hypothesis. Note that w k is not necessarily the pointwise supremum of the increasing sequence of functions (u,,k : n = 1, 2, . . .). It is evident that 0 _I wk(x) 6 k holds for all x , and of course we have also that wk(x) _I v ( x ) and wk(x) 5 w k + l(x) holds for all k and all x. We prove now that w&O) = w k + l(xo) holds at every point xo where wk(xo)c k . Indeed it follows from u,,,k + 7 ,, wk+ 1, by taking the infimum with the function identically equal to k , that u,,k inf ( w k + 1, k ) . On the other hand we have u,,k t,, wk, and so w k = inf (wk+ , k ) . This implies that W k ( x 0 ) = wk+ l(xo) at every point xo where Wk(x0) < k . It follows from these facts that at each point x the values wk(x), for k = 1,2, . . ., are mutually equal from a certain k = k , onwards (one can take, for example, fork, the smallest k satisfying k > u(x)), and so the pointwise limit u ( x ) = limk+mwk(x) exists at every point x E X. Given xo and the natural number k > u(xo), we have k > u(x) in an appropriate neighborhood of xo, so u coincides with w k on that neighborhood. It follows that u is continuous at the point xo . This holds for every xo E X , so u E C ( X ) . It is now an immediate consequence that u = sup u,, holds in C ( X ) , which shows that C ( X ) is Dedekind a-complete. There is an analogous theorem for Dedekind completeness.
I,,
Theorem 43.3. Let C ( X ) and c b ( X ) be the same Riesz spaces as above. Thefollowing conditions are now mutually equivalent. (i) C ( X ) has the projection property. (ii) C ( X ) is Dedekind complete. (iii) cb(x)has the projection property. (iv) c b ( X ) is Dedekind complete. Proof. Similar to the proof of the preceding theorem, by making use of Theorem 42.6, according to which (i) and (ii) are equivalent, and also (iii) and (iv) are equivalent.
284
FREUDENTHAL'S SPECTRAL THEOREM
[CH. 6 , § 43
There is a connection between the projection bands in the Riesz space Cb(X)and the subsets of X that are simultaneously closed and open. Such subsets will be called closed-open sets. The set of all projection bands in cb(x),partially ordered by inclusion, is a Boolean algebra, and evidently the collection of all closed-open subsets of X , partially ordered by inclusion, is also a Boolean algebra. We will show that these Boolean algebras are lattice isomorphic. For this purpose, denote by e(x) the function in cb(x) satisfying e ( x ) = 1 for all x E X and note, incidentally, that e is a strong unit in cb(x),and so every projection band in cb(x)is a principal projection band. Given such a projection band B, let eB be the component of e in the band B. Then 0 5 eB 5 e and inf (eB,e - e B ) = 0, so the function eBassumes only the values zero and one, i.e., eB is the characteristic function xF of some subset F of X . Since eBis a continuous function, it follows immediately that F is a closed-open set. Since B is the band generated by x F , every function contained in B must vanish outside of F, so (x :f(x) # 0) is contained in F for every f E B, and for f = xF there is equality. The set F is called the closed-open carrier of the projection band B (compare Excercise 22.10). Conversely, given the closed-open subset F of X , the functions xF and x ~ are- continuous, ~ so these functions are complementary components of e in the Riesz space cb(x).The band B generated by xF is a projection band. Indeed, for any f E cb(x) we have the decomposition f = fxF+fxX-F as a sum of continuous functions, the first one in the band B and the second one disjoint from xF (and hence contained in Bd). Evidently, the closedopen carrier of B is exactly the given set F. Hence, we have proved the following theorem.
Theorem 43.4. The mapping which assigns to each projection band in C,(X) its closed-open carrier is a lattice isomorphism between the Boolean algebra of all projection bands in cb(x) and the Boolean algebra of all closed-open subsets of X . Consequently, i f one of these Boolean algebras is Dedekind complete, then so is the other one. It will be convenient to introduce some definitions. If E is a subset of X such that E = (x :f ( x ) = 0)for somef E cb(x), then E is called a zero set; any set theoretic complement of a zero set is called a cozero set. Every closed-open subset F of X is a cozero set because the complement X - F is the zero set of the continuous function x F . Given the subset E of X , consider the collection of all closed-open sets containing E. This collection is non-empty, because the set X is always a
CH. 6 , s 431
THE SPACES
c(x)AND cb(x)
285
member of the collection. If the collection, partially ordered by inclusion, has a smallest element, then this smallest element is called the closed-open hull of E, and is denoted by clp (E). Evidently every closed-open set is its own closed-open hull. We prove some simple lemmas about closed-open sets and cozero sets. Lemma 43.5. (i) If the closure of every open subset of X is closed-open (i.e., if the topological space X is extremally disconnected), then the Boolean algebra of all closed-open sets is Dedekind complete. (ii) If E and F are disjoint open subsets of X such that the closures of E and Fare closed-open, then these closures are disjoint. Proof. (i) Let (F, :T E {T}) be a collection of closed-open subsets of X; we have to prove that sup F, exists in the Boolean algebra of all closed-open sets. The set 0 = u,F, is open, so the closure 8 of 0 is closed-open by hypothesis, and hence 8 is an upper bound of the set of all F,. Let S be another upper - bound. Then S F, for all T, so S I> u,F, = 0.It follows that S = S 3 8. This shows that 8 is the required least upper bound. (ii) Denoting the set theoretic complement X - S of any subset S of X bij S', we have E c Fc with Fc closed, so E c F', which shows already that E and F are disjoint. But then we have F c (E)' with (E)' closed, so F c (Er,i.e., E and F are disjoint. Lemma 43.6. (i) Every cozero set is an open F,-set (we recall that an Fa-set is a countable union of closed sets). (ii) If the topological space X is normal, then conversely every open F,-set is a cozero set.
Proof. (i) Given the coZero set E, there exists a bounded continuous functionf such that E = ( x : f ( x ) # 0), so E=
(X
It follows that E is open and E =
: If(x)l > 0).
u:=l F,,, where
F,, = ( x : If (x)l 2 n - ' ) for n
=
1,2,
. . ..
Since each F,, is closed, this shows that E is an open F,-set. (ii) We assume that X is normal and E is an open set satisfying E = F,, with all Fnclosed. We have to prove that E is a cozero set. For each n, the sets F,, and EC are disjoint closed sets, so (since X is normal) there exists a continuous functionf,, on X satisfying 0 S fn(x) 5 2-" for all x , fn(x) = 2-" for all x E F,, and ,f,(x) = 0 for all x E E'. The function
286
FREUDENTHAL'S SPECTRAL THEOREM
[CH.6 , § 43
f ( x ) = C F f n ( x ) is continuous on X (by uniform convergence), and evi-
dently E is the cozero set 0f.f. In the next theorem we state necessary and sufficient conditions for Dedekind a-completeness of C ( X ) and c b ( X ) . Since C ( X ) is Dedekind a-complete if and only if c b ( X ) is so (by Theorem 43.2), we restrict ourselves to c b ( X ) .
Theorem 43.1. The space c b ( X ) is Dedekind a-complete if and only if every cozero set has a closed-open hull such that disjoint cozero sets have disjoint closed-open hulls. Proof. Assume first that c b ( X ) is Dedekind a-complete. By Theorem 43.2 this is equivalent to assuming that c b ( X ) has the principal projection property. In other words, every principal band is a projection band. Given f E c b ( X ) , let B, be the band generated byf. Since B, is a projection band, B, has a closed-open carrier F (by Theorem 43.4). It is evident that the cozero set (x : f ( x ) # 0 ) is contained in F, and ( x :f ( x ) # 0 ) is not contained in any closed-open set F' properly smaller than F, because otherwise the projection band B' corresponding to F' would contain f and yet be properly smaller than the band B, , which is impossible. Hence, F is the closed-open hull of the cozero set ( x : f ( x ) # 0). Given the cozero sets El = ( x : f ( x ) # 0) and E2 = ( x : g ( x ) # 0) such that El and E2 are disjoint, the functions fand g are disjoint elements in the Riesz space cb(X)y so the corresponding principal projection bands B, and Be are disjoint. Denoting the closed-open carriers of B, and Be by F and G respectively, we have xF E Bf and xC E Be, SO xF 1xG, i.e., F and G are disjoint point sets. Hence, disjoint cozero sets have disjoint closed-open hulls. Now, for the converse, assume that every cozero set has a closed-open hull such that disjoint cozero sets have disjoint closed-open hulls. We have to prove that c b ( X ) has the principal projection property. To this end, let f E c b ( X ) be given, let F be the closed-open hull of ( x :f ( x ) # O), and let B be the projection band having F a s its closed-open carrier. Evidently B containsS, so B contains the band B, generated byJ We want to show that B = B,, so it remains to prove that B is contained in B, = B Y , and for this it will be sufficient to prove that B is disjoint from any g in B;. For any g in B/"the sets ( x : f ( x ) # 0) and ( x : g ( x ) # 0 ) are disjoint, so by hypothesis the corresponding closed-open hulls are disjoint, so it follows that the corresponding projection bands B and C are disjoint. Since the given function g is included in C, it follows that B is disjoint from g. This concludes the proof.
CH.6 , s 431
THE SPACES
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287
There are various sufficient conditions for Dedekind a-completeness of
cb(x)which become necessary and sufficient if the topological space X has
certain extra properties. We present some of these conditions.
Theorem 43.8. A sufficient condition for Dedekind o-completeness of
cb(x) is that the closure of every cozero set is a closed-open set. If X
is
completely regular, then the condition is necessary and sufficient (cf. L. Gillman and M . Jerison [1], Problem 3 N). Proof. Assume first that the closure of every cozero set is closed-open. Evidently, for any cozero set E, its closure E is then the smallest closed-open set containing E, i.e., E is the closed-open hull of E. Furthermore, if E and F are disjoint cozero sets, then their closures are disjoint by Lemma 43.5, i.e., their closed-open hulls are disjoint. It follows, therefore, from the preceding theorem that cb(x)is Dedekind o-complete. Now, assume conversely that Xis completely regular and cb(x)is Dedekind a-complete. By the preceding theorem every cozero set has a closedopen hull, and disjoint cozero sets have disjoint closed-open hulls. We will prove that the closed-open hull of any cozero set E is the closure E of E, and this will show then that the closure of any cozero set is closed-open. Hence, letfe C , ( X ) be given with E = ( x :f ( x ) # 0) as the corresponding cozero set. Given any point xo of X such that xo is not in E , there exists (by the complete regularity) a function g E cb(X) such that g(xo) = 1 and g ( x ) = 0 for all x E E . Hence, since the cozero sets E and El = ( x : g ( x ) # 0) are now disjoint, their closed-opzn hulls must be disjoint. But xo is a point of El, so xo is surely no membcr of the closed-opx hull of E. This shows that any point not in E is also not in the closed-opm hull of E. Hence, the closed-open hull of E is the closure E . Theorem 43.9 ( H . Nakano [l], 1941). A sufficient conditionfor Dedekind a-completeness of cb(x)is that the closure of every open F,-set is a closedopen set. If X is a normal Tl-space (equivalently, i f X is Hausdorfand normal), then the condition is also necessary.
Proof. Assume first that the closure of every open F,-set is closed-open. Since every cozero set is an open F,-set by Lemma 43.6 (i), we have in particular that the closure of every cozero set is closed-open. Hence, by the preceding theorem, c b ( X ) is Dedekind a-complete. Now, assume conversely that X is a normal Tl-space and C b ( X )is Dzdekind a-complete. Then X is surely completely regular, so by the preceding theorem the closure of every cozero set is closed-open. But, in view of X
288
[CH.6 , 8 43
FREUDENTHAL'S SPECTRAL THEOREM
being normal, every open F,-set is a cozero set (cf. Lemma 43.6 (ii)), so the closure of every open Fa-set is closed-open. We proceed to derive conditions for Dedekind completeness of
cb(x).
Theorem 43.10. The space cb(x) is Dedekind complete i f and only i f Dedekind o-complete and, in addition, the Boolean algebra of all closed-open subsets of X is Dedekind complete. cb(x)is
Proof. As observed several times already, Dedekind completeness and Dedekind o-completeness of cb(x)are equivalent if and only if the Boolean algebra of all projection bands in cb(x)is Dedekind complete. By Theorem 43.4 this is equivalent to Dedekind completeness of the Boolean algebra of all closed-open subsets of X .
Theorem 43.11 (H. Nakano [l], 1941). A suficient conditionfor Dedekind completeness of cb(x)is that the closure of every open set is a closed-open set (equivalently, X is extremally disconnected). If X is completely regular, the condition is also necessary. Proof. Assume first that the closure of every open set is closed-open. In view of Theorem 43.9 the space C,,(X) is surely Dedekind a-complete, so in view of the preceding theorem we need only prove that the Boolean algebra of all closed-open sets is Dedekind complete. For this purpose, assume that (A, :z E {z}) is a collection of closed-open subsets of X . We have to prove that sup A, exists in the Boolean algebra referred to. The set 0 = UJ, is open, so the closed-open set B is an upper bound of the set of A, = 0, so B = B =I 0. all A,. Any other upper bound B satisfies B =I This shows that 0 = sup A,. For the converse, assume that X is completely regular and c b ( X ) is Dedekind complete. Given any open set E in X and a point xo in E, there exists (by the complete regularity of X ) a function fo E cb(x) such that f o ( x o ) = 1 andfo(x) = 0 for all x E E'. Varying xo in E (and hence varying the correspondingf,),it is evident from this that E is a union of cozero sets. Also, fixing xo again, the non-empty open set V = ( x : f o ( x )> 3) has its closure contained in the set (x :fo(x) 2 *), so xo E V c E. As observed, any open set is a union of cozero sets, so Vcontains a cozero set C such that xo E C c V, and hence we have xo E c B c E. The set is closed-open (because Xis completely regular and c b ( X ) is surely Dedekind ocomplete, so the closure of every cozero set is closed-open by Theorem 43.8). Since xo is an arbitrary point of the open set E, it follows therefore from xo E E c E with C closed-open that the open set E is a union of closed-open sets. In other
e
c
CH.6 , s 431
T H E SPACES
c ( x )AND cb(x)
289
words, the closed-open sets form a base for the topology in X (i.e., X is totally disconnected). In order to prove now that the closure E of the open set E is open, we write E as a union E = U , A , of closed-open sets A , . It follows from the Dedekind completeness of cb(x)that the Boolean algebra of all closed-open subsets of X is Dedekind complete, so the set sup A, exists in this Boolean algebra. From sup A, =I U , A , = E it follows that sup A, = sup A,
XE.
The set theoretic difference sup A , - E is open; assume this open set to be non-empty. Then, in view of what was observed above, there exists a nonempty closed-open set A . contained in sup A,-E. It follows that the set sup A , - A . is closed-open and contains all A,, so the set is an upper bound of the set of all A,, properly smaller than sup A,. This is impossible. Hence, sup A,-E is empty, i.e., E = sup A,. But then E is open, which is the desired result.
Exercise 43.12. As defined, a subset E of the topological space X is a cozero set whenever there exists a function f~ c b ( X ) such that E = ( x : f ( x ) # 0). Show that the subset E of X is a cozero set if and only if there exists a function f E c b ( X ) and a real number a such that E = ( x : f ( x ) > a). Exercise 43.13. Let X bc a completely regular topological space. Show that c b ( X ) is Dedekind o-complete if and only if the following conditions are satisfied. (i) Every countable union of closed-open sets has an open closure. (ii) Every cozero set is a countable union of closed-open sets. Hint:If c b ( X ) is Dedekind a-complete, the closure of every cozero set is open by Theorem 43.8. For (i), note that every closed-open set is a cozero set, and a countable union of cozero sets is a cozero set. For (ii), note that the cozero set ( x : f ( x ) > 0) is the union of the closures of the sets ( x : f ( x ) > CL,), where (an: n = 1,2, . . .) is any sequence of real numbers an 10.
Conversely, if (i) and (ii) are satisfied, then every cozero set has an open closure, so c b ( X ) is Dedekind a-complete.
Exercise 43.14. Let X be a completely regular topological space. Show that cb(x)is Dedekind complete if and only if the following conditions are satisfied. (i) Every union of closed-open sets has an open closure.
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FREUDENTHAL'S SPECTRAL THEOREM
[CH.6 , § 43
(ii) Every cozero set is a union of closed-open sets. The results in this exercise and the preceding exercise are due to M. H. Stone ([4], 1949). Hint: If cb(X) is Dedekind complete, then every open set has an open closure, so (i) is obviously satisfied. By the preceding exercise condition (ii) is satisfied in the even stronger form that every cozero set is a countable union of closed-open sets. Conversely, if (i) and (ii) are satisfied, note first that every open set is a union of cozero sets, and deduce from this by means of (ii) that every open set is a union of closed-open sets. Hence, according to (i), every open set has an open closure.
Exercise 43.15. The following example of a topological space X for which cb(X) is Dedekind a-complete, but not Dedekind complete, is due to H. Nakano [l]. Let X be an uncountable point set, one particular point of which is denoted by x, . (i) Consider the collection of all subsets of X that either do not contain x , or that contain x , and have an at most countable complement. Show that the sets in this collection form the open sets of a normal Hausdorff topology in X (so that, therefore, the topology is also completely regular). (ii) Show that the subset E of X is an F,-set if and only if E either contains x, or E is an at most countable set not containing x,. Show that E is an open F,-set if and only if E or the complement of E is an at most countable set not containing x, . Show that the closure of any open F,-set E is E itself. Derive from this fact that cb(X) is Dedekind a-complete. (iii) Let El and E2 be disjoint uncountable subsets of X such that E, E2 = X , and let x, E E l . Show that E2 is open, and the closure E 2 satisfies E 2 = E, {x,}, so E , is not open. Derive from this fact that cb(X) is not Dedekind complete.
u
u
Exercise 43.16. If every point in the topological space X is open (i.e., if we have the discrete topology), then every real function on X is continuous, so C ( X )and cb(X) are Dedekind complete. In order to present an example of a topological space X without any open points for which cb(X) is Dedekind complete, we combine the following facts from the theory of Boolean rings. (i) According to Excercise 4.17, the element a # 0 in the Boolean ring R is called an atom if it follows from 8 # b 5 a that b = a. It was proved in the exercise referred to that the band Zin R is a maximal ideal (equivalently, a prime ideal) if and only if there exists an atom a such that I = {a}d.
CH. 6 , s 431
THE SPACES
c(x)AND cb(x)
29 I
Derive from this fact that if P is a maximal ideal in R,and {P}denotes the band generated by P, then either {P}= R or P = {P}= { u } ~ for some atom a. Hence, if R contains no atoms, then {P} = R holds for every maximal ideal P in R,i.e., every maximal ideal is order dense. (ii) According to Excercise 4.16, the collection of all regularly open subsets of a topological space is an order complete Boolean algebra. For the particular case that the topological space is the real line with its ordinary topology, show that the Boolean algebra of all regularly open subsets has no atoms. (iii) Let X be the Boolean algebra from part (ii), so according to Theorem 7.5 the hull-kernel topology in the space 9 of all prime ideals in X is Hausdorff and compact, the base sets { P } x are closed-open, the space B itself is a base set, and any closed-open set is a base set. Note that X and the Boolean algebra of all base sets are lattice isomorphic by Stone’s representation theorem. Show that the closure of every open set in B is a base set, and derive from this fact that C b ( 9 )is Dedekind complete. (iv) Finally, show that no point of B is an open set. Hint: For (iv), assume that the point Po is an open subset of 9. Then Po = {P}xofor some point xo E X . Since X has no atoms, there exist disjoint elements x, , x2 E X such that 8 # x1 5 x,, and 0 # x, S xo. Then there also exist prime ideals PI,P, such that x, is in P, but x2 is not, and x, is in P2 but x1 is not. It follows that P, and P, are different elements of { P } x , ; this contradicts {P}xo= P o .
CHAPTER 7
Spectral Representation Theory
The present chapter is devoted to spectral representation theorems for Riesz spaces. In order to explain what is meant by a spectral representation of a Riesz space, we first define what an extended real-valued continuous function on a topological space Xis. By R" we understand the system of extended real numbers (i.e., the system of real numbers with + 03 and - 03 added to the system, and with the natural neighborhood systems for the finite real numbers as well as for a).Any continuous mapping f of the topological space Xinto R", with the extra property that the set on whichf is finite is dense in X , is called an extended real-valued continuous function on X. The set of all suchfis denoted by Cm(X).In general C"(X) is no linear space, because for givenf, g E Cm(X)the functionf + g need not be well defined within the class Cm(X). Certain subsets of Cm(X) can be Riesz spaces, however, with respect to addition, multiplication by real numbers and the lattice operations defined pointwise. The Riesz space "L,Riesz isomorphic to a given space L,is now said to be a spectral representation of L if "L consists of extended realvalued continuous functions on some topological space X. For Archimedean Riesz spaces there exist several variants of spectral representation theorems, depending upon the choice of the topological space X. One of the most general theorems of this kind is the Johnson-Kist variant (1962), as follows. Let P be a prime ideal in the Archimedean Riesz space L and e a nonzero element in L+ such that e is no member of P. For any f~ L the extended real number "f,(P) is now defined by " f ( P ) = sup ( M : (Me -f)+ E P).
Keeping P and e fixed, the mapping f -P " f ( P ) of L into R" is a Riesz homomorphism (except for a small complication due to the values +a). For the Johnson-Kist representation, let W be a set of proper prime ideals in L such that the intersection of all R in 9 is (O}, and let (en : cr E {cr}) be a disjoint order basis of strictly positive elements in L. We recall that, for any f E L,we denote by {R)/the set of all R E W such that f is no member of R. 292
CH. 71
SPECTRAL REPRESENTATION THEORY
293
In the subset g1= U,{R},u of 9 we introduce the hull-kernel topology. This topological space 9lis the topological space on which the functions of the spectral representation are defined, as follows. Given any R E 91, there is exactly one e, such that R is a point of {R},*, so this particular e, is no member of the prime ideal R. The extended real number " f ( R ) is now, by definition, the number "Lu(R),as defined above. For f fixed and R variable in a,, we thus obtain an extended realvalued function "fon 9,.It will be proved that "fis an extended realvalued continuous function on 9, ,and the set "L of all "fis a Riesz space, Riesz isomorphic to L. In general, the functions of the representation "L do not separate the points of 9,,i.e., given two different points R , and R2 of W,,there does not always exist an element EL such that " f ( R , ) # "f(R,). Hence, generally speaking, there are too many points in 9,. The situation is satisfactory in this respect if L is Archimedean with a strong unit e. The disjoint order basis consists now of the element e only, and for the system 9 of prime ideals we can take the system $ of all maximal ideals. This situation was investigated by K. Yosida (1941), who already observed the similarity with the Gelfand representation for Banach algebras with a unit element. The topological space f is Hausdorff and compact in the hull-kernel topology, and the representation "L of L is a Riesz subspace of the space C($) of all real continuous functions on f such that " L is uniformly dense in C($). In particular, the functions of "L separate the points of $. It follows also that "L is equal to C ( $ ) if and only if L is uniformly complete. If it is given in advance that L has the principal projection property, then " L = C ( $ ) if and only if L is Dedekind o-complete. In one of the examples L will be the space C,(X) of all real bounded functions on a locally compact Hausdorff space X , and it will be shown that $ can be identified with the Stone-Cech compactification of X , and for any f E C,,(X)the corresponding "fis now the unique extension off to a continuous function on f . We will also consider the case that L is Archimedean, but without a strong unit (K. Yosida, 1942). As in the general case, we have again a disjoint order basis (e, : o E {a}), and instead of the system fl we now have the system uuPu, where PUis the set of all prime ideals in L maximal with respect to the property of not containing e,. The Ogasawara-Maeda theorem (1942) is a representation theorem of a somewhat different kind. For L Archimedean, the functions f" of the representation L" are now defined on the space of all maximal ideals in the Boolean algebra W ( L )of all bands in L; the topology in D is the hullkernel topology. Since W ( L )is a Boolean algebra, 0 is compact and Haus-
294
SPECTRAL REPRESENTATION THEORY
[CH.
7,
5 44
dorff, and since g ( L ) is order complete, L2 is also extremally disconnected (i.e., every open subset has an open closure). This implies that Cm(Sa) is a Riesz space which is Dedekind complete as well as universally complete (i.e., every disjoint system of positive elements has a supremum). The Riesz isomorphic representation L A of L is a Riesz subspace of Ca(L2), the ideal generated in Cm(8)by L Ais order dense and it is also a Dedekind completion of L, and the space Cm(8) itself is what is called a universal completion of L. Even in the case that L has a weak or strong unit and L has the principal projection property, the functions of L Ado not necessarily separate the points of 0. If, however, L has a weak unit and L has the projection property, then LA separates the points of 8. A variant of the Ogasawara-Maeda theorem is obtained by considering, instead of the Boolean algebra of all bands in L,the sublattice of all principal bands. For the case that L has the principal projection property, the functions f of L A are now defined and continuous on an appropriate subset of the locally compact space Sap, the points of which are the maximal ideals in the sublattice g P ( L )of all principal bands. This sublattice is now a Boolean ring. For the case that L is Dedekind o-complete, this representation IS due to H. Nakano (1941). In a Riesz space L with the principal projection property it is true that a prime ideal P is minimal if and only i f f € P implies Bf c P,where B, is the band generated by f. As a consequence, there is a homeomorphism between aPand d (both sets equipped with the hull-kernel topology). Hence, Nakano's representation can also be considered as a representation by functions on the space of all minimal prime ideals. Finally, given any e > 0 in L,there exists for every M E { M } e a unique prime ideal Q M E 2esatisfying Q M 3 M. If L has the principal projection property, the mapping M + QM is a homeomorphism of { M } e onto and so Nakano's representation and Y osida's representation are essentially the same. In the final section of this chapter it will be proved that if Q is a proper prime ideal in the Riesz subspace K of the Riesz space L,then there exists a prime ideal P in L such that the intersection of P and K is exactly Q. There is an analogous theorem (the Cohen-Seidenberg theorem) for prime ideals in a commutative ring R with subring S, but in this case we have the extra hypothesis that R is integrally dependent upon S. A
44. The Johnson-Kist representation theorem
By Rm we will denote the two-point compactification of the set R of
CH.
7, 8 441
295
THE JOHNSON-KIST REPRESENTATION THEOREM
real numbers. This means that, as a point set, R" consists of all real numbers together with the points + 00 and - 00. The rules for addition and multiplication are extended to & co in the usual way, where it is understood that the multiplication obeys 0 (kco) = (fco) 0 = 0. Furthermore, as usual, the linear ordering of the real numbers is extended to R" by defining that - 00 < a < + 00 holds for every finite real a.The set R" is now topologized in such a way that the bounded open intervals
-
(LX:
-
- co c u c ~ < b a <),
together with the intervals
(a:-co S a < b c
00)
and (a:-co < a < a
5 co)
form a base for the topology. The set R" is sometimes called the extended real number system. Given the topological space X , any continuous mappingf of X into R" such that the set R(f) = ( x : If(x)l c co) is dense in Xis called an extended (realvalued) continuous function on X . The set of all extended (realvalued) continuous functions on X will be denoted by Cm(X). Givenf, g E Cm(X) and the finite real number a,the functions inf (f, g), sup (f,g) and af, all members of C"(X), are defined by
, {inf (f,g)}(x) = inf { f ( x ) g(xN, {SUP (f,g)l(x) = SUP {f(x), g ( x ) ) , (Kf)(x) =
.f(>
for all x E X , where it is understood (as defined above) that O.( 5 a) = 0. For any f E Cm(X),the set R(f) = ( x : If(x)l < co) is open and dense. If f, g, h E Cm(X) and h ( x ) =f(x)+g(x) holds for all x E R(f) n R(g), then (by definition) It is called the sum off and g (notation h = f+g). Since the set R(f)n R(g) is dense, the function h = f + g is uniquely defined if it exists. As an example thatf+g does not necessarilyexist for allf, g E C"(X), we mention the case that X is the set ( x : 1 5 x 5 co) with the usual topology (i.e., X is the one-point compactification of the interval [l, a)), f ( x ) = n on [2n- I , 2n], f(2n+3) = 2n, f linear in [2n, 2n+$] and in [ 2 n + 3 , 2 n + l ] for n = 1 , 2 , . . ., andf(+co) = +a. All this can be done such that f is extended continuous. Furthermore, let g(x) = - f ( x 1) for all x E X . It is impossible now to definef+g at the point x = + co in such a way thatf+g E Cm(X)holds. This example shows, therefore, that Cm(X) is not always a real linear space, simply because Cm(X) is not necessarily closed with respect to the operation of addition.
+
296
SPECTRAL REPRESENTATION THEORY
[CH.
7,s 44
A subset of Cm(X),closed under the operations of addition, multiplication by finite real constants and the taking of finite suprema and infima, is obviously a Riesz space with respect to pointwise ordering. Accordingly, any subset of C*(X) of this kind is called a Riesz space of extended real continuous functions on X (the condition that every function in the space is finite on a dense set is tacitly included in the definition, therefore).
Lemma 44.1. Let P be a prime ideal in the arbitrary Riesz space L, and let f be an arbitrary element of L. Then exactly one of the following conditions is satisfied. (i) f E P (so f + and f - are members of P). (ii) f + E P, but f - is no member of P. (iii) f - E P, but f is no member of P. +
Proof. I f (i) does not hold, then at least one off + and f - is no member of P. On the other hand, since inf (f', f -) = 0, one at least o f f and f - is a member of P. Hence, if (i) does not hold, then exactly one of (ii) and(iii) +
holds. Once more, let P be a prime ideal in the Riesz space L,let e be a nonzero element in the positive cone L+ such that e is no member of P and, finally, let f be an arbitrary element of L . By the last lemma the set of all finite real numbers can be decomposed into three disjoint sets A = ( a : (ae-f)' E P, but ( a e - f ) - not in P), B = ( a : ( a e - f ) E P) , C = ( a : (ae-f)- E P, but (ae-f)' not in P).
Observe that a E A and b 5 a implies j? E A . Also, a E C and p 2 a implies /3 E C. The set B consists of at most one number (since e is no member of P). I f B is non-empty and a. is the one number in B, then we have a E A for a < a. and a E C for a > cto. It can happen that every real number is contained in the set A , so B and C are then empty. Similarly, it can happen that every real number is in C.
Definition 44.2. Let the prime ideal P and the elements e E L +and f be the same as above. The extended real number " f ( P ) is now defined by " f ( P ) = sup ( a : (ae-f)+ E P ) = sup ( a : (ae-f)- not in P) = inf ( a : ( a e - f ) - E P) = inf (a : (ae-f)+ nor in P).
This has to be understood in the sense that ifthe set
( a : (ae-f)+ E P )
EL
CH. I, 5 441
THE JOHNSON-KIST REPRESENTATION THEOREM
29 7
is emp1.y or contains all real numbers, then " f ( P ) = - 00 or " f , ( P ) = + 00 respectively. Note that thefour numbers which defive " f ( P )are indeed the same. We will prove now that, except for a small complication due to the values & m, the mapping f + " f ( P ) of L into R" is a Riesz homomorphism.
Theorem 44.3. Keeping P and efixed, let " f ( P ) be defined for every f E L by Definition 44.2 above. Since no misunderstanding can arise as long as e is kept fixed, we will write f( P ) instead of " f ( P ) . Then the following holds. (i) "(bf)(P) = b"f(P)for every finite real /3 and eoery f E L, where (we repeat that) it is understood that O*(f 0 0 ) = 0. (ii) Iff, g E L are such that "f(P)+"g(P) makes sense (i.e., "f ( P ) and "g(P)are not both infinite and of opposite sign), thenf +g satisfies A
" ( f+g)(P) = "f (PI+ (iii) I f h = inf (f,g ) in L , then
"h(P) = inf { "f(P),"g(P)}. Similarly for sup (f,g). In particular, we have " ( f + ) ( P )= SUP {"f(P),01, "(f-)p) = -inf {"f (P), O}, "Ifl(P) = I"f(P)I.
(iv) We have "e(P) = 1, and f
Ie implies "f(P)= 0.
Proof. (i) Since "g(P) = 0 holds for g = 0, it is evident that the formula to be proved holds for b = 0 and arbitrary$ For 0 < b < m we have ( a : (ae-f)+ E P ) = ( a : (aBe-bf)+
Taking suprema, we obtain " f ( P ) = P - l
EP)= (y/b:(ye-bf)+ EP). "(Pf)(P), so
b"f(P) = "(bf)(P).
In order to complete the proof, it will be sufficient to consider the case that
b=
-1. We have
" ( - f ) ( P ) = sup ( a : (ae+f)+ E P ) = sup ( a : (-ae-f)- E P ) = = sup ( - y : (ye-f)- EP) = -inf ( y : (ye-f)- E P ) = -"f(P).
(ii) Given f, g E L such that "f (P)+"g(P) makes sense, we will prove that "f (PI + 6 "(f+g)(P) (1)
298
[CH. 7, 8 44
SPECTRAL REPRESENTATION THEORY
holds. For "f(P)+ " g ( P ) = - co the inequality is trivially true, so we may assume that "f(P)+" g ( P ) > - co, i.e., "f(P)> - co and "g(P) > - co. Let a be a finite real number satisfying a < "f(P)+"g(P). Then there exist finite real numbers ai and az such that a = a1 az with a1 < "f(P) and az < "g(P), and so it follows from ( ~ 1 ~ e - f ) E ' P, ( a z e - g ) + E P and
+
{ae-(f+g)}+
=
{(ale-f)+(aze-g)}+
5 (ale-f)++(aze-g)+
that { c r e - ( f + g ) } + EP, so a 5 " ( f + g ) ( P ) . This holds for every a < "f(P)+"g(P); it follows that the inequality (1) holds. Applying this result to -fand -9, we obtain the inequality in the converse direction, and so ^ ( f + g ) ( P ) = "f(P)+ "g(P).
I
(iii) Let h = inf (f,g ) in L . It is evident from the definitions that ^h(P) "f(P)and "h(P) 5 ^g(P), so "h(P) 5 inf { "f(P)," g ( P ) } .
If the infimum is - coy there is nothing left to prove. We may assume, therefore, that the infimum is greater than - 00. In order to prove now that "h(P) is greater than or equal to this infimum, we observe first that (ae-h)+
{ae-inf (f,g ) } + = {ae+sup (-A - g ) } + = {sup (Me-f, a e - g ) } + = sup {(me-f)', (me-g)+}.
=
For any m < inf {"f(P)," g ( P ) } the elements ( a e - , f ) + and (Re-g)' are bothmembers ofP, so the just established formula shows now that ( a e - h ) + is also a membsr of P, i.e., a 5 "h(P). This holds for all so which is the desired result. The proof for sup ( f , g ) is similar. (iv) It follows from the definitions that "e(P) = 1 . If fle , then inf (Ifl, e ) = 0, so inf ("Ifl(P), 1 ) = 0. It follows that " I f l ( P ) = 0, i.e., I"f(P)l = 0, and so " f ( P ) = 0. We will prove now that, given the Archimedean Riesz space L, there exists a topological space such that L is Riesz isomorphic to an appropriate Riesz space of extended real continuous functions on this topological space. This is a beautiful result, the usefulness of which is restricted, however, by a large arbitrariness in the choice of the topological space. I n general, as we will see, the topological space will contain 'far too many' points. By a careful choice, the occurrence of this unpleasant phenomenon can be prevented,
CH. 7,8 441
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299
but only in some very special cases (for example, the case that the space L has a strong unit) the arbitrariness can be almost completely avoided. Let L be an Archimedean Riesz space, W a set of proper prime ideals in L such that ( R : R E W )= {0},
n
and (e, : cr E {a}) a disjoint order basis of strictly positive elements in L (i.e., e, > 0 for all a, e,, 1e,, for al # a2, and f = 0 for any f E L satisfying f Ie, for all a; cf. Definition 28.4). Exactly as in earlier sections (cf. section 36) we denote, for any f E L , by { R ) / the set of all R E W such that f is no member of R. Note that all sets {R},,, a E {a}, are mutually disjoint subsets of 99.In the subset 9,= U,,{R},=of W we introduce the hullkernel topology (cf. section 36). The sets { R } " ,where u runs through the set of all u EL' for which there exists one of the e, satisfying 0 6 v 6 e,, form a base for the topology. Given any R E W land anyf E L,we define the extended real number "f ( R ) as follows. By the definition of g1there is exactly one e, in the set of all e, such that R is a member of {R},,, so this particular e, is no member of the prime ideal R. We now take the extended real number "fe,(R) as defined in Definition 44.2 above, so "fe,(R) = sup ( a : (cre,-f)+
E
R),
and this is, by definition, "f(R). For f fixed and R variable in W l, we thus obtain an extended real function "f on Wl.We will call "f the JohnsonKist representation ofJ The range of "f (i.e., the set of values assumed by the function "f) is called the spectrum off (with respect to the system (e, :a E {a}) and the set 9of prime ideals). By way of example, iff is one of the elements e,, say f = eao,and if (e, : a E {a}) consists only of eao,then " f ( R ) = 1 for all R E W,, whereas if there exist elements e, with a # ao,then " f ( R ) = 1 for all R E {R}eooand "f ( R ) = 0 for all other R E g1. In the first case, therefore, the spectrum off = e,, consists only of the number one, whereas in the second case the spectrum off = e,, consists of the numbers zero and one.
Theorem 44.4 (Johnson-Kistspectral representation theorem, 1962). Exactly as above, let (e, : a E {a}) be a disjoint order basis of strictly positive elements in the Archimedean Riesz space L , let W be a set of proper prime ideals satisfying n ( R : R E 9) = {0}, and let 9,= U,(R},, be topologized by means of the hull-kernel topology.
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Finally, for any f E L , let " f be the Johnson-Kist representation off. Then "f is, therefore, an extended real function on the topological space 9,. Every "f is continuous on W , in the sense explained in the beginning of the present section, and the set of all "f is a Riesz space "L of extended realvalued continuousfunctions on W ,. Finally, the mapping f + "f is a Riesz isomorphism of L onto "L.
Proof. We will show first that, for ahyf, the function "f is continuous on W lin the sense defined for extended real functions. Assume first that e is one of the e,, Ro E { R } eand "f (R,) = a, ,where a, is a finite real number. Given any E > 0, the elements {(aO+E)e-f}+ and {(a,-e)e-f}- are no members of R,. We set u = inf (e, {(aO+e)e-f}+,((a,-e)e-f}-).
Then u 2 0, { R } , c {R}eand u is no member of Royso {R}, is an open neighborhood of Ro . Given now any R E {R},, the element u is no member of R , so {(aO+c)e-f}+ and {(aO-e)e-f}- are no members of R either, which implies that ao-e S " f ( R ) S a,+&. This shows that "f is continuous at Ro. Assume again that Ro E {R}e,but now " f ( R o ) = u, the element (ae-f)- is no member of R,. We set u =
+ CO. Given any real
inf (e, (ae-f)-).
Then u 2 0, { R } , c {R}e and u is no member of R o y so { R } , is an open neighborhood of Ro . Given now any R E {R},, the element u is no member of R, so ( a e - f ) - is no member of R either, which implies that ^ f ( R ) 2 u. This shows that " f is continuous at R,. Finally, if "f(R,) = -a,then "(-f)(R,) = +coy so "(-f) is continuous at R o . But "( -f ) = - "f, and it follows that " f is continuous at Ro. It has to be shown next that, for every "f,the set (R: I"f(R)I < CO) is dense in 9,. Assume that for some " f this fails to hold, and observe that we have ( R : I"f(R)I < CO) = ( R : "1 f I(R) < CO) on account of I"f(R)I = "1 f I(R) holding for every prime ideal R We assume, therefore, that the set
N
=
( R : "If I(R) =
CO)
E
9,.
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THE JOHNSON-KIST REPRESENTATION THEOREM
has a complement (relative to 9,) not dense in 9'. In other words, the s e t N includes one of the base sets of the topology, i.e., there exists an e,, call it e for brevity, and an element 0 < u 5 e such that {A},c JV. Then "Ifl(R) = 00 for every R E {R},,so (ne- If])+ E R for n = 1,2, . . and all R E {R},.Keeping n fixed for the moment, it follows that (ne- ')[fI is an element of the kernel k({R},) of the set {R},;we recall that
.
n
~({RI,) = (R : R E PI,) by definition. But k ( { R } , ) = A: by Corollary 35.4 (i), where A, is the principal ideal generated by u and A: is the disjoint complement of A,. It is here that we use that r)(R : R E 9) = ( 0 ) ; note that {R},, is the set of all R E .9Z such that D is no member of R (equivalently, {R},is the set of all R E 9,such that u is no member of A). Combining the results, we obtain ~ n = 1,2,. . ., i.e., that (ne-lf[)' E A for
for n = 1 ~ 2 ,....
( e - n - ' l f [ ) + E A:
Since ( e - n - ' l f [ ) + e as n + 00 (because L is Archimedean), we obtain that e E A:, so u 1e. On the other hand we have 0 < u S e. This yields a contradiction. Hence, the set (R: I"f(R)I < 00) is dense in 29' for every f€L. It has been proved thus that every "fis an extended (real) continuous function on 9'. Taking now into account the results of the preceding theorem and recalling that "h is (by definition) the sum of "fand "g whenever "h(R) = "f(R)+"g(R)holds for all RE (R : I"f(R)I <
al)
n (R : I"g(R)I <
al),
it follows that the set "Lof all "fis a Riesz space, and the mappingf-, "fis a Riesz homomorphism. All that remains to be proved, therefore, is that the mappingf-, "fis one-one. It is sufficient to prove that "f(R)= 0 for all R E 9,impliesf = 0. Since the proof can be given forf andf - separately, we may assume that f 1 0. Fix e,. For any R E {R},*we have "f(R)= 0, i.e., inf ( a : (ae,-f)- E R) = 0. +
It follows that (n-le,,-f)- E R for n = 1, 2, . . . and all R E {R},-.This implies that (n-'e,,-f)- is a member of the kernel of {R),-.Once more by Corollary 35.4 (i) the kernel of {R},-is equal to the disjoint complement of the ideal Aeu generated by e,,. Hence, we have (n-'e,,-f)'EA~u, i.e., (n-'e,,-f)- Ie,, or, equivalently, (f-n-'e,,)+
Ie,.
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SPECTRAL REPRESENTATION THEORY
[CH. 7, 8 44
Since L is Archimedean, we have (f-n-le,)' t f as n -,00, so f 1e,. This holds for all 0 , s o f = 0 on account of the system (e, : 0 E (01) being a disjoint order basis in L.This concludes the proof. We present some remarks, mainly in order to show that the very general approach as indicated in the above theorem has some undesirable features. Adhering to the notations introduced in the theorem, let Ro be one of the prime ideals in the set 9, = U a { R } e e Then . Ro is a member of exactly one of the sets {RIe-, say for 0 = 6,. Observe that "e,,(Ro) = 1 and "e,(R,) = 0 for all (r # 6,. The ideal Ro plays two different roles; in the first place Ro is a prime ideal in L , and secondly R, is a point of the topological space 9,on which the functions "fof "L are defined. Since L and " L are Riesz isomorphic, we can consider the isomorphic image "R, = ("f :f E R,) of Ro in "L; evidently, "R, is a prime ideal in " L . It is a natural question to ask what relation there is between the point R, of the topological space 9, and the prime ideal "R,. More specifically, since "Roconsists of functions "fdefined on 92, ,we ask whether the functions of "R, have somethingspecial in their behavior at the point R,. The answer is affirmative; all "f E "R, vanish at the point R,, i.e.,
"Ro c ("f: "f(R,) = 0). The proof is simple. We have "f E "Ro if and only iff E R o , and so "f E "Ro implies that cre-f E R, holds for a = 0, which (by definition) implies that "f(R,) = 0. It is tempting to conjecture that "Ro is actually equal to the set ( " f : "f(R,) = 0 ) , but this conjecture is false in general. Indeed, looking again at "R,, we observe that "R, is a prime ideal in "L such that "R, does not contain the element "e,,, but in general "R, need not be maximal with respect to the property of not containing "e,,. According to Theorem 33.5 A Ro is included in a prime ideal "R,,, of "L such that "R,,, is maximal with respect to the property of not containing "e,,, and "R,,, is uniquely determined since the set of all prime ideals in " L that contain "R, is linearly ordered. Looking now at "R1 = ("f:"f(R,) = 0), it is obvious that this is a prime ideal in " L , maximal with respect to the property of not containing "e,, (indeed, as soon as we have a function "f0 not in " R , , say "fo(R,) = a # 0, then "e,,-cr-'"f0 is in " R , , so the ideal generated by "R, and "fo contains "e,,). Hence, since "R, is contained in "R, , as we proved above, R1 is exactly the prime ideal "R,,,, so A
"R,
t
h
R,,, = ("f : "f(H0) = 0).
CH.
7, 5 441
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303
The following theorem is, therefore, a consequence of these considerations.
Theorem 44.5. With the notations of the preceding theorem, if R, is one of the prime ideals in the set 9$,= U,{R}e, and if e,, is the one element in the set of all e, for which R, E {R},- holds, then "Ro = ("f:"f(R0) = 0 ) holds if and only i f R o is maximal with respect to the property of not containing em* The problem discussed above is related to another problem which we will consider now. First we present a definition. Given a set S of (real or complex) functions on a point set X , it is said that S separates the points of X whenever, for any two different points x , ,x2 in X there exists a functionf E S such that f (x,) # f (x2).Returning again to the notations used in the JohnsonKist representation theorem, it is natural to ask now whether "L separates If " L does not, i.e, if there exist different points R , , R2 the points of 9,. in 9, such that " f ( R , ) = " f ( R 2 )holds for all f E L , we may just as well remove either R 1 or R2 from the point set 92, without destroying the representation. Similarly, if there is a whole subset of points RI(zE {.}) such that "f (Rr,) = "f (RT2) holds for all f E L and all rl, z2 E {z}, we may delete all but one of these points from 92,. This can indeed be done, but of course it does not contribute toward keeping the representation procedure as simple as possible. Hence, the given set 9of prime ideals should preferably be chosen such that the representation space "L of functions separates the points of the topological space 9,.The following lemma indicates how to achieve this purpose.
Lemma 44.6. Let e be a nonzero positive element in the Archimedean Riesz space L , and let P , and P2 be prime ideals in L such that P, # P2 and e is no member of either P, or P2. Then " f ( P 1 ) = "f,(P2)holds for every f E L if and only if e is no member of the algebraic sum P1 +P2.In other words, "f,(Pl) = "f,(P2)holds for every f E L i f and only if there existJ a prime ideal P, mch that PI and P2 are contained in P3 and e is no member of P3. Proof. Without fear for misunderstandings we may write " f ( P ) instead of "f,(P)for any prime ideal P not containing e. Assume first that there exists a prime ideal P, such as indicated above. Letf E L be arbitrary. Then
^f(P,) = sup ( a : (ae-f)+ Similarly "f(P,) = inf ( a : (ae-f)-
E P,)
5 sup ( a : (Me-f)' E P , ) = "f(P3).
E P,)
2 inf ( a : (ae-f)-
EP3)=
"f(P3).
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[CH.7,544
SPECTRAL REPRESENTATION THEORY
Hence "f(P,)= "f(P3).Similarly we obtain "f(Pz)= "f(P3),and so
"f(P1)=
"f (Pz).
For the converse, we assume that "f(P,) = "f(PZ)holds for all f E L , and we have to show that e is no member of P, Pz.Assume this to be false, so assume that e E PI+Pz.Then there exists a decomposition e = u+u with 0 S u E P, and 0 S u E Pz. It follows then from 0 S u S e that 0 5 "u(P) 6 1 holds for every prime ideal P not containing e; in particular, c1 = "u(P,) = "u(Pz) satisfies 0 5 a 5 1. Similarly, /? = "u(P1) = "u(Pz) satisfies 0 j? 2 1. Furthermore it follows now from e = u+u that
+
1 = "e(P,) = "u(P,)+"u(P,)= a+B.
But ~ E P implies ,
CI
= "u(P,) = 0, and U E P , implies
/?
= "u(Pz) = 0.
This contradicts a +/? = 1. Hence, the element e is no member of P,+Pz. Returning to the original question whether "L separates the points of W, or not, it follows from the lemma that the answer can be affirmative only whenever every R E W,= U {Rleu U
is maximal in the sense that if euois the one e, for which R E {R},-holds, then R is maximal with respect to the property of not containing eu0. We recall that, given the set W of prime ideals in L and given the element e E L f , we denote by We the set of all R E W such that R is maximal with respect to the property of not containing e (cf. the last part of the text of section 33). The final conclusion from the discussion in the present section can be formulated, therefore, as follows. Given the disjoint order basis (e, : CT E {o}) in the Archimedean Riesz space L , if we desire the corresponding representation "L to separate the points of the topological space 9, on which the functions of "L are defined, the best choice is to take
W
=
W 1=
u 2?e'-, U
where 2?eu consists of all prime ideals in L maximal with respect to the property of not containing eu. Of course, we must prove that this choice is legitimate in the sense that the intersection of all prime ideals in this particular collection W is equal to (0). We will do this in the next sections. Note that although W is now uniquely determined by the disjoint basis (e, :o E {o}), the arbitrariness in the choice of the disjoint basis remains. The general Johnson-Kist representation theorem is due to D. G. Johnson and J. E. Kist ([I], 1962); the representation for the case that 9 = was introduced already earlier by K. Yosida ([3], 1942).
uu2?eu
CH. 7. $451
THE YOSIDA THEOREM IN A SPACE WITH STRONG UNIT
305
45. The Yosida spectral representation theorem in an Archimedean Riesz space with strong unit We assume in the present section that L is an Archimedean Riesz space with a strong unit e E L + .The subset consisting of e only is now an order basis of L, and an ideal Z in L is maximal with respect to the property of not containing e if and only if Z is a maximal ideal. Hence, the set 9' of all ideals maximal with respect to the property of not containing e is the set # of all maximal ideals. According to Theorem 27.6 the intersection of all maximal ideals in an Archimedean space with a strong unit is {0}, so the JohnsonKist theory in the preceding section is applicable to the present situation for the case that the set W of prime ideals to be considered is the set f . According to Theorem 27.3, given the maximal ideal J in L and the element f EL,'there exists exactly one finite real number u0 such that aoe- f E J, and this number is denoted by "f ( J ) in section 27. In the Johnson-Kist representation the same notation "f ( J ) is used. Fortunately, the notations agree because the number a. for which aoe-fE J holds is at the same time the supremum of all c( for which (cre-f)+ E J holds. The Johnson-Kist representation "L of L is, therefore, exactly the set of functions "f on f which was already introduced in section 27. Note that since "e(J) = 1 for every J, and any f E L is included between two finite real multiples of e, we have that the corresponding "f is between two finite multiples of "e, i.e., every "f E "L is a bounded function. In section 27 the hull-kernel topology in f was not yet available. In the present discussion we assume $ to be topologized by means of the hullkernel topology. According to Theorem 36.4, the point set f becomes a compact Hausdorff space in this topology. The functions "f of " L are, therefore, bounded continuous functions on this topological space f , and so "L is a Riesz subspace of the space C ( f ) of all real continuous functions on $. We will prove in Theorem 45.4 that "L is equal to the whole space C($) if and only if L is uniformly complete. According to what was observed in the preceding section, " L separates the points of f . Actually, this was proved already in Theorem 27.3 (iii). Precisely, it was proved there that if J I and J2 are different points of # and a, b are given real numbers, then there exists an element f E L satisfying " f ( J 1 )= a and "f(J2) = b. Note that, according to Theorem 44.5, if Jo is a maximal ideal in L , then its Riesz isomorphic image "Jo in "L satisfies ^JO
=
("f:" f ( J 0 ) = 0).
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SPECTRAL REPRESENTATION THEORY
[CH. 7 , g 45
We recall that the spectrum of a given element f E L is t,he range of the function "f, i.e., the spectrum off consists of all real numbers u such that a = " f ( J ) holds for some J E $. Since "f is a continuous function on the compact Hausdorff spacs $,the range of "f is a compact (i.e., bounded and closed) subset of the set of all finite real numbers. Hence, the complementary set, i.e., the set of all real numbers not in the spectrum off, is open.
Theorem 45.1. Thefinite real number uo is not in the spectrum o f f if and only if there exists a neighborhood of uo such that B,, the band generated by (ue- f ) ' , is constant in this neighborhood. Proof. Assume first that uo is not in the spectrum o f f , i.e, the function f does not assume the value uo. Since the complementary set of the spectrum off is open, there exists a neighborhood ( a O - & , u O + & ) of go such that no point in this neighborhood is in the spectrum off, i.e., "f assumes no values in (uo - E , uo + E ) . It follows that the values assumed by the function ( u O e - f ) are either 2 E or S - E . Hence, " { ( u o e - f ) + } assumes the value zero at all points where A(rxOe-f) is less than or equal to -6, and A { ( u o e - f ) + } is greater than or equal to E at all other points. Obviously, the band "B, generated by " { ( u , e - f ) + ) is then the same as the ideal generated by the same function, and ^B, satisfies A
A
"B,,
=
("9 : " g ( J ) = Oat all Jwhere " ( u , e - f ) ( J ) < 0).
(I)
Now, restricting u to the interval ( c ~ , - $ E , a,++&), the points J where " ( a e - f ) ( J ) < 0 holds are the same as where " ( u O e - f ) ( J ) < 0 holds, so B, = "B,,, by formula (1). For every u we have that "B, is the isomorphic image of B,, so it follows finally that B, = B,, for every u in the neighborhood (c(,-+E, go++) of go, i.e., B, is constant in this neighborhood. For the converse we have to show that if uo is in the spectrum off, then there is no neighborhood of uo in which B, is constant. Assume that a, is in the spectrum off and that at the same time B, is constant in some interval [a1,or2] having a, in its interior. Since uo is in the spectrum o f f , there exists a maximal ideal J, such that "(uoe- f ) ( J o ) = 0. Then A
" ( u l e - f ) ( J , ) = u l - u o < 0 and A ( u 2 e - f ) ( J 0 ) = u 2 - u 0 > 0,
so there exists an open neighborhood V of J,, such that " ( u l e - f ) ( J ) < 0 and " ( u 2 e - f ) ( J ) > 0 holds for all J E V. It follows that
CH. 7 , § 451
THE YOSIDA THEOREM IN A SPACE WITH STRONG UNIT
307
" { ( a l e - f ) - } ( J ) > 0 and " { ( a 2 e - f ) + ) ( J )> 0 holds for all J E V . Since (a2e - f ) + E B,, = B,, , (a1 e - f ) - E B:, = B : ~ ,
the functions " { ( a 2 e - f ) + } and " { ( a l e - f ) - } are disjoint elements of " L , which contradicts the just established fact that both functions are strictly positive on the neighborhood V of J o . Hence, if a. is in the spectrum off, there is no neighborhood of a. in which B, is constant. For a completely different proof of the fact that if a0 is in the spectrum off, then there is no neighborhood of a. in which B, is constant, cf. Exercise 45.9. Rather unexpectedly, minimal prime ideals will play an important role in the proof. It was observed above that " L is a Riesz subspace of the space C ( $ ) of all real continuous functions on the compact Hausdorff space $. We may ask how "large" the subspace " L is. The following well-known lemma will provide information concerning this point. The lemma plays a role in the proof of the Stone-Weierstrass theorem (cf. the book by L. H. Loomis [1], Lemma 4C) and is related to a classical result due to M. Krein and S. Krein
(PI,
1940).
Lemma 45.2. Let A be a set of real continuous functions on a compact topological space X such that A is a lattice under pointwise ordering (i.e., f, g E A implies max (f,g ) E A and min (f,g ) E A). Furthermore,given x, y E X ( x # y ) and the real numbers a and b, let there exist a function f in A such that f ( x ) = a a n d f ( y ) = b. Then the uniform closure of A is the set C ( X ) of all real continuousfunctions on X . Proof. Let f
E
C ( X ) be given. For any pair of points x, y E X (also for E A satisfyingfX,,(x) = f ( x ) and f,,,(y) =
x = y ) there exists a function f,, f ( y ) . Given E > 0, let ux,y
= ( z :fx,y(z)
Vx,y = ( z :fx,y(z> > f ( Z ) - E ) .
Fixing y and varying x, the open sets Ux,ycover X , and so a finite subfamily of them covers X . Taking the minimum of the corresponding fX,,, we obtain a function f y E A such that f y < f + e on X and& > f-8 on the open set V,, which is the intersection of the corresponding sets V x , y .Note
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SPECTRAL REPRESENTATION THEORY
[CH. 7,s 45
that y E V,. Now varying y , the set X is covered by a finite numbzr of the V,. Taking the maximum of the corresponding f,, we obtain a function f, E A such that f-E
cf,c f+E
holds on X . This shows that the arbitrary functionf E C ( X )is in the uniform closure of A . Theorem 45.3. With the same notations as in the beginning of the present section, the set "L is a Riesz subspace of the Riesz space C( &) of all real continuous functions on $ such that "L is uniformly dense in C ( f ) , i.e., the uniform closure of "L is C($).
Proof. "L satisfies all conditions that A was supposed to satisfy in the preceding lemma. We will investigate now under which conditions "L is equal to the whole of C($). We recall that a Riesz space L is said to be uniformly complete whenever, for every UEL', every u-uniform Cauchy sequence in L has a u-uniform limit in L. Theorem 45.4. Once more, let L be an Archimedean Riesz space with a strong unit e E L+. Thefollowing conditions are now mutually equivalent. (i) L is uniformly complete. (ii) Every e-uniform Cauchy sequence in L has an e-uniform limit in L. (iii) " L = C($). Hence, i f Lis a Dedekind o-complete space with a strong unit, then "L = C( $1.
Proof. It is evident that (i) implies (ii). Given (ii), it follows that every "e-uniform Cauchy sequence in "L has an "e-uniform limit. In other words, since "e(J) = 1 for all J E $, every uniform Cauchy sequence of functions in "L has a limit in "L.This implies that "L is equal to its own uniform closure C($), i.e., "L = C($), so (iii) holds. Finally, assume that (iii) holds and let ( J , : n = 1, 2, . . .) be a u-uniform Cauchy sequence in L for some u E L+.Since u is majorized by an appropriate multiple of e, the sequence (fn : n = 1,2,. . .) is an e-uniform Cauchy sequence, so ("J, : n = 1,2, . . .) is a uniform Cauchy sequence (in the ordinary classical sense) in "L = C($). The pointwise limit function "fof the sequence ("fn : n = 1,2, . . .) is at the same time the "e-uniform limit function, and so "fis a member of C($). Similarly, "fis also the "u-uniform limit of "J,, which implies that f is the u-uniform limit off,. Thus, (iii) implies (i).
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If L is Dedekind a-complete, then L is Archimedean and uniformly complete (cf. Lemma 39.2), so if L has also a strong unit, then "L = C($). The condition (ii) in the theorem can also be formulated by saying that L is a Banach space with respect to the norm llfll
=
inf ( a : -ae
s f s ae).
The fact that (ii), thus formulated, implies "L = C ( $ ) was proved by S. Kakutani ([I], 1940), M. Krein and S. Krein ([l], 1940), and by H. Nakano (PI, 1941).
Theorem 45.5. Let L be a Riesz space possessing the principal projection property and having a strong unit. Then L is Dedekind a-complete if and only if "L = C ( $ ) holds. Proof. It is sufficient to prove that under the above hypotheses "L = C ( $ ) implies that L is Dedekind a-complete. It follows from "L = C ( $ ) that L is uniformly complete, and this jointly with the principal projection property implies the Dedekind a-completeness of L (cf. Theorem 42.5). Assuming that "L = C ( 9 ) holds, we observe that the set "L = C ( $ ) is not only a real vector space, but also a commutative ring in the algebraic sense with respect to ordinary pointwise multiplication of functions as ring multiplication. Ring multiplication and multiplication by real numbers are related by (a"f)(b"g) = (ab)("f"g). Hence, "L is a commutative algebra with "e as unit element. But then, since L and "L are isomorphic as linear spaces, we may introduce a ring multiplication for the elements of L , simply by defining that the product off and g is the (uniquely determined) element h E L satisfying "h = "fAg. The thus introduced multiplication makes L into a commutative algebra with e as unit element (i.e., ef = f e = ffor eveiyfe L ) . Multiplication and order are related by the following properties. (i) S, g E L + implies f g E L + . (4 lfgl = If1 191 for a1l.L 9 EL. (iii) f g = 0 if and only i f f 1 g. Given that L is Archimedean and has a strong unit e EL', we compare "Land ""L. The elements of ""L = " ( A L are ) bounded realvalued functions on the set " fof all maximal ideals "Jin " L, where "Jdenotes the maximal
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ideal in "L corresponding with the maximal ideal J in L under the isomorphism between L and "L. The function "(" f)is denoted briefly by ""J
Theorem 45.6. We have ^"f("J) = "f (J)for every f and every J. Proof. In view of the isomorphism we have ""f("J) = sup ( a : (a"e-"f)'
E
" J ) = sup ( a : (ae-f)'
EJ) =
"f(J).
The elements of "L and ""L are functions on the point sets $ and "f respectively, so comparing "L and ""L will include comparing the hull-kernel topologies in $ and "$ respectively. The hull-kernel topology in f is determined by the base sets {J),
=
(J:f not in J) = (J : " f ( J ) # 0),
so the hull-kernel topology in the isomorphic image "$ is determined by the base sets ("J: ""f("J) # 0), for f running through L. Identifying the point sets $ and "#' by identifying the point J in f with the point ^J in "2, the functions "f and ""fare then automatically identified in view of the above theorem, according to which we have ""f("J) = " f ( J ) for everyf and every J. It follows that the base sets for the topologies in $ and "$ (and, therefore, the topologies themselves) are identical. The spaces " L and ""L are then identical not only as Riesz spaces, but also as spaces of continuous functions on $ and "$ respectively. We will make some additional remarks concerning the case that "L = C ( $ ) holds. The space "L = C ( $ ) is now a commutative Banach algebra (with respect to the uniform norm); the ring multiplication is the ordinary pointwise multiplication of functions. The reader who is familiar with the theory of Banach algebras will remember that the Gelfand representation L* of "L can be identified with "L itself. We denote the elements in the Gelfand representation by f *, g*, . . .; these elements are realvalued functions on the set $* of all maximal ideals (in the algebraic sense) J* in "L. It is well-known that, given such a maximal ideal J*,there exists a unique point J E $ such that J* consists of all "fE "L vanishing at the point J. Conversely, the set of all "fthat vanish at a given point J E f is a maximal ideal in "L in the algebraic sense. Note, therefore, that maximal ideals in "L are the same in the algebraic sense and in the Riesz space sense. Thus, there is a one-one correspondence between the points J of $ and the points J* of $*. For any "f E "L,the corresponding function f * in the Gelfand representation turns out to satisfy f * ( J * ) = " f ( J ) for every point J*, and ,f* is topologized by introducing in $* the weakest topology that
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makes all f * continuous. Identifying J* and J for every J, the functions f * and "fare then automatically identified, and it turns out that the new topology in f * = j is the same as the original topology in f . To summarize, the spaces "L,""L and L* are identical not only as Riesz spaces or as Banach algebras with respect to the uniform norm, but also as spaces of continuous functions on the compact Hausdorff spaces f , " fand #* respectively. The representation of an Archimedean Riesz space with strong unit, discussed in the present section, is due to K. Yosida ([l], 1941). We finally mention that M. H. Stone, in a series of two notes ([3], 1940-41), briefly presents a spectral representation theory, applicable to Archimedean Riesz spaces with a strong unit and to lattice groups, certain aspects of which are similar to the Yosida theory in the present section. We proceed with some examples. Example 45.7. Let X be a compact Hausdorff space and C ( X ) the Riesz space of all real continuous functions on X . The function e, satisfying e ( x ) = 1 for all x E X , is a strong unit in C(X). We will determine the Yosida representation ^ C ( X ) of C(X). As shown earlier, there corresponds to every maximal ideal Jo in C ( X ) a point x, E Xsuch that Jo is exactly the set of all f E C ( X ) vanishing at x , , and conversely every subset of the form ( f : f ( x o ) = 0) is a maximal ideal in C(X). Hence, there is thus a one-one correspondence between the points x , E X and the maximal ideals Jo in C(X). Given f E C ( X )and the maximal ideal J , , the number u = ^ f ( J o ) is determined by the condition that ue-f should be a member of J,, i.e., ( u e - f ) ( x , ) = 0 for x , the point corresponding to J,. It follows that " f ( J , ) = f ( x o ) . Hence, if we identify x, and J o , the function f in C ( X ) is then automatically identified with its image "fin "C(X). Thus, algebraically, C ( X ) and " C ( X ) are identical. We will prove now that the hull-kernel topology in the set $ of all maximal ideals (i.e., in the set X after the identification) is the same as the original topology in X . The hull-kernel topology is determined by the base sets { J } , = (J : " f ( J ) # 0) = (X : f ( x ) # 0),
for f varying in C(X), and evidently each set of this kind is open in the original topology. Hence, the hull-kernel topology is weaker than the original topology. On the other hand, the original topology is a compact topology and the hull-kernel topology is a Hausdorff topology (actually, both topologies are Hausdorff and compact), and it is well-known that a compact topology is never properly stronger than a Hausdorff topology. Hence, the
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topologies are identical, and so C ( X )and " C ( X ) are also identical as spaces of continuous functions on a compact Hausdorff space. As a further illustration, we look at the case that X is the interval [0, 11 in the real line with the ordinary topology. Besides the Yosida rcpresentation by means of the maximal ideals, we also consider the representation by means of the minimal prime ideals. The intersection of all minimal prime ideals is {0}, so the Johnson-Kist theory in the preceding section is applicable to the set A of all minimal prime ideals. The subset A1= { M } =is A itself, so A is topologized by means of the hull-kernel topology, and the representation is by means of continuous functions "f on A. Each minimal prime ideal M , is contained in a unique maximal ideal J o y so if J, = (f:f (x,) = 0), then x, is the unique point in [0, 11 at which all functions in M , vanish (cf. section 34). Every maximal ideal contains an infinity of different minimal prime ideals (cf. Exercise 37.13), so to any given x, E [0, 13 there correspond infinitely many minimal prime ideals Mo, , (t E {t}) such that every f E L satisfies "f(M,, ,) = f ( x o ) ,where "f is the isomorphic image off. This clearly shows how badly the isomorphic image "L of L fails to separate the points of A.Note also that the inclusionin M , c (f: " f ( M , ) = 0) is proper for each M , ,because the right hand side is exactly the maximal ideal J, in which Mo is included.
Example 45.8. Let X be a Hausdorff and locally compact space such that Xis non-compact with respect to this topology. Hence, every point of X has a compact neighborhood not equal to X itself. Note already that if N ( x o ) is a compact neighborhood of the point x, , then there exists a real continuous functionf such that 0 S f ( x ) 5 1 for all x E X , f ( x o ) = 1 andf(x) = 0 for all x outside N(xo). The Riesz space c b ( X ) of all real bounded continuous functions on X has the function e, satisfying e ( x ) = 1 for all x E X , as a strong unit. We will determine the Yosida representation of cb(X). Given xo E X , the set Jxo
=(fI
f
cb(x),f
(x0)
=
O)
is a maximal ideal in cb(X). Besides the maximal ideals of the form Jx, there are other maximal ideals in cb(X). In order to prove this, observe first that the set of allfe c b ( X ) such that the set ( x : f ( x ) # 0) is contained in a compact set is a proper ideal A , in cb(X), and so there exist maximal ideals J 3 A , . Since at no point xo E Xall functions in A , vanish (cf. the remark above), it is evident that no J =I A , is of the form Jx, for some x, E X . (i) Every maximal ideal J in cb(x)i b either of theform Jxofor some x, E X or eIse we have J =I A , .
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For the proof, assume that J is not of the form J,,. It is sufficient to prove now that iffis a function in c b ( X ) such that ( x : f ( x ) # 0) is included in a compact subset C of X , thenfis a member of J. For this purpose we may assume that 0 5 f ( x ) 5 1 holds for all x E X. Given any x, E C, there exists a function uxoin J + satisfying u,,(xo) > 1, so u,,(x) > f ( x ) holds for all x in a neighborhood of x, .Since Cis covered by a finite number of such neighborhoods, it is evident now that f is majorized by a function in J, so f E J holds. This completes the proof. The maximal ideals of the form J, are sometimes called fixed maximal ideals in C,(X), and the maximal ideals J 2 A , are then calledfree maximal ideals. Given f E c b ( X ) and the fixed maximal ideal J,,, the number ct = "f (J,,) is determined by the condition that cte-f should be a member of J,,, i.e., (cw-f)(x,) = 0. It follows that "f(J,,) = f ( x o ) . Hence, if we identify x, and J,,, the function f i n c b ( X ) is then automatically identified with the restriction of " f to the set of all points J,,, i.e., if we consider X a s a subset of f , then "fis an extension of$ The hull-kernel topology in f is determined by the base sets { J } , = ( J : "f(J) # 0)
forfvarying in cb(X), so the relative topology in Xof the hull-kernel topology is determined by the base sets {JI,
n x = ( x :m# 01,
and evidently each set of this kind is open in the original topology of X . Hence, the relative topology in X of the hull-kernel topology in 2 is weaker than the original topology in X . In order to show that the topologiesare identical, it is sufficient now to show that, given the open set 0 in the original topology and given the point x, E 0, there exists a real continuous function f on X such that x, E ( x : f ( x ) # 0) = 0, since this will imply that 0 is a union of sets that are open in the relative hull-kernel topology, and so 0 will be open in that topology. Now, it is a well-known theorem in a locally compact Hausdorff space X that if C and 0 are subsets of X that are compact and open respectively, and such that C c 0, then there exists a real continuous function f on X such that 0 5 1 for all X E X f, ( x ) = 1 for all X E Candf(x) = 0 for all x outside 0 (cf. the book by L. H. Loomis [l], Theorem 3E). In particular, since any set consisting of one point is compact, we have that if x, E 0,
sf(x)
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SPECTRAL REPRESENTATION THEORY
[CH. 7,
8 45
then there exists a real continuous function f such that 0 S f ( x ) =< 1 for all x E X,f (x,) = 1 and f ( x ) = 0 for all x outside 0. Hence x,
E
( x :f ( x ) # 0) c 0.
This is the desired result. It has thus been shown that the relative topology in X of the hull-kernel topology in $ is exactly the original topology in X. To summarize,if we consider Xas a subset o f f , then the original (Hausdorff and locally compact) topology in X is the relative topology of the (Hausdorff and compact) topology in $, and every f E cb(X) is the restriction to X of the corresponding continuous function "f on $. In other words, every f E Cb(X)can be extended to a continuous function "f on the compact Hausdorff space f . The extension is unique in the sense that if the real continuous function g on $ is an extension of a function f E cb(X), then g must be equal to "f.In order to prove this, we first prove the following results. (ii) I f f E c b ( X ) vanishes outside a compact subset C of X , then " f ( J ) = 0 holds for aN free maximal ideals J. The proof is simple. By hypothesis we have f E A , (where, exactly as above, A , is the ideal of all f E cb(x)vanishing outside a compact set), so f E J f o r every free J. It follows that "f(J) = 0 for every free J. (iii) If Jo is a free maximal ideal and No is an arbitrary neighborhood (in $) of J o , and C is an arbitrary compact subset of X (with complement C' = X - C ) , then No contains at least one point of C'. For the proof we may assume that No is a base set for the topology, so No = (J : "f ( J ) # 0) for some f E cb(X). In particular we have then that "f (J,) # 0, and so, in view of (ii) above, the function f does not vanish for every x outside C. In other words, there is at least one x1 E C' such that f ( x l ) # 0, i.e., " f ( J x l )# 0. This shows that Jxl E N o , i.e., x1 E No if we identify x1 and J x , . We return to the proof that if the continuous function g on $ is an extension off E cb(X), then g = "f.This is equivalent to proving that if the continuous function g on f satisfies g ( x ) = 0 for all x E X,then g is identically zero on $. It follows from (iii) that in any neighborhood of any free maximal ideal Jo there are points J where g ( J ) = 0 holds. Hence, by the continuity of g , we have g ( J o ) = 0. This holds for every free J o , and so g is identically zero. The locally compact Hausdorff space X is embedded, therefore, in the compact Hausdorff space f such that Xis dense in f (follows immediately from (iii)) and every bounded continuous function on X can be extended
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uniquely to a continuous function on $.Given any other compact Hausdorff space $' 3 X such that X is dense in f' and every bounded continuous function on X can be extended uniquely to a continuous function on f ' , there exists an obvious homeomorphism of f onto f' leaving the points of X invariant. In other words, f is uniquely determined except for a homeomorphism leaving X invariant. The space $ is called the Stone-tech compactification of the locally compact Hausdorff space X. Any real continuous function f on X having the property that for every real a > 0 the set ( x : I f ( x ) l > a ) is compact is usually said to vanish at infinity (the name is explained by the fact that f can be extended continuously onto the one-point compactification of X , namely by setting f(x,) = 0 at the point of infinity x,). The name is also appropriate in the framework of the Stone-tech compactification, because iff is continuous and vanishes at infinity, then "f ( J ) = Oat every free maximal ideal J. This follows immediately from (iii), according to which every neighborhood of J contains points x E X at whichf ( x ) is arbitrarily small in absolute value. It has been proved now that under the Riesz isomorphism of cb(x)onto "(Cb(X)}the image "f of anyf E cb(x)is the continuous extension off from the space X onto the larger space 3.Differentf yield different "f, since the isomorphism is one-one. The space ^{cb(x)} is actually equal to the space C($) of all real continuous functions on f . Indeed, any g E C($) is bounded (since $ is compact), so the restrictionf of g to Xis a member of c b ( x ) y and this implies (as proved above) that g = "f. Hence Because of the Riesz isomorphism between Cb(x) and C ( f ) the maximal ideals in cb(x)and C($) are in one-one correspondence. If J is a fixed maximal ideal in cb(x),say J = J,,, then the image "J consists of all "f E C($) vanishing at J,,, i.e., all "$vanishing at xo (since xo and J,, are identified). If J is a free maximal ideal in cb(x),then "J is a maximal ideal in C ( / ) , but not of the form "J,, for any xo E X. On the other hand, since $ is compact and Hausdorff, any maximal ideal in C( f ) consists of all ^$vanishing at some point of $. Hence, if J is free in c b ( x ) y then the corresponding "J consists of all "f vanishing at some point in f -X . (iv) Let (x,, : n = 1,2, . . .) be a sequence of points in X tending to inj k i t y in the sense that each compact subset of X contains onlyjhitely many of the points x,, . Furthermore, let f o E cb(x)satisfy fo(xn)= 0 for all n. Then there exists at least one free maximal ideal Jo such that f o E Jo ,and so "fo(Jo) = 0.
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[CH.
7,s 45
For the proof, consider the ideal AoLfo]generated by the ideal A . (the ideal of all continuous functions vanishing outside a compact set) and the functionfo. Every function in this ideal vanishes at all but a finite number of the points x,,, and so the function e, satisfying e(x) = 1 for all x E X , is no member of AoLfo].It follows that A o u o ] is a proper ideal in cb(X), and so A o l f o ]is contained in a maximal ideal Jo . This implies that "fo(J0) = 0. There may be several free maximal ideals Jo such that fo E Jo holds. In particular this holds for all Jo in the closure (in 9) of the set (x,, :n = 1,2, . . .). The set of all such Jo is non-empty (note that the set of all x,, has at least one point of accumulation in the compact set j ,and no point of X is a point of accumulation since each compact set in X contains only a finite number of the points x,,). (v) If(x,, :n = 1,2,. . .)isthesamesetofpointsasin(iv)andiff, E cb(x) satisjies fl(x,,)+ 0 as n + 00, then "f,(Jo) = 0 for every free T. o in the closure ofthe set (x,, : n = 1,2, . . .). The proof follows immediately by observing that "fl is continuous. In many examples of locally compact Hausdorff spaces X (e.g., all real number spaces X = R" for n = 1,2, . . .) the space X is a countable union X = C,,, where each C,, is compact and contained in the interior of C,,, 1. Then each compact subset S of X is contained in a finite union of the C,,. Indeed, denoting the interior of any set A by A", we have X = C,O, so S c C,O. By the compactness S is already covered by a finite union of the C,O,and so S is covered by a finite union of the C,,.For a space X of this kind we prove a converse of (v), as follows. (vi) Assume that the locally compact Hausdorf space X has the special property mentioned above. Let fo E cb(x)and let Jo be afree maximal ideal such that "fo(Jo) = 0. Then there exists a sequence (x,, :n = 1,2, . . .) of mutually diflerent points in X such that the sequence tends to infinity and fo(x,,)-,0 as n -, co. For the proof, let S be a compact subset of X , and let E > 0. Observe that Jo is in the closure of X - S (by (iii)), so every neighborhood of Jo has a non-empty intersection with X - S . This holds in particular for the neighborhood
ur=l
u."=
u.".
of Jo. Hence, there exists a point x, E X - S such that Ifo(x,)l < E. For E = 1 and S empty this yields a point x1 E X such that Ifo(xl)l < 1. Let n, be the smallest natural number n such that x1 E C, = S,, holds.
u;=,
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Then S,,, is compact, so there exists a point x2EX- S,,,satisfying Ifo(xz)l<3. Let n2 be the smallest natural number n > n, such that x2 E S,, holds. Then S,,, is compact, so there exists a point x j E X - S,,, satisfying Ifo(xs)l <$. Continuing this procedure, we obtain a sequence (x,, :n = 1 , 2 , . . .) of mutually different points such that Ifo(x,,)l n-l holds. Each compact set is contained in a set of the form S,,, and so each compact set contains only a finite number of the points x,,.
-=
All results obtained are applicable to the case that Xis the set of natural numbers with the discrete topology (each point of X has itself as a compact neighborhood). The space Cb(X) in this case is the sequence space I,. Let Jo be a free maximal ideal in I, , and to each f~I , assign the value of "fat the point J,. Let us write cpdf) = "f(Jo). Then cp is obviously a (real) bounded linear functional on the normed linear space I , such that (i) if f = (f(l), f(2), . . .) is a convergent sequence, then cpdf) = limf(n) as n -,coy (ii) cp(f) does not change its value if we change a finite number of the coordinates off, (iii) lim inff(n) 5 cp(f) lim supf(n); in particular c p ( f ) 2 0 iff(n)zO for all n, so cp is a positive linear functional. The functional cp is sometimes called a Bunuch-Muzur limit for the space 1, *
s
Exercise 45.9. (i) Show that if M is a minimal prime ideal in the Riesz space L and f is a member of M , then the band generated byf is contained in M. (ii) Let L be an Archimedean Riesz space with a strong unit e. It was proved in Theorem 45.1 that, givenfe L, if the number a, is in the spectrum off, then there is no neighborhood of a, in which the band B, (i.e., the band generated by ( a e - f ) + ) is constant. We indicate a different proof. Assume that a, is in the spectrum off and that at the same time B, is constant in some interval [a1, a2] having a, in its interior. Since a, is in the spectrum off, there exists a maximal ideal Jo such that "f(Jo) = a,. Let Mo be a minimal prime ideal in Jo .Then, by Lemma 44.6,we have "fe(Mo)= " ~ ( J o ) = a,, i.e., sup ( a : (ae-f)+ EM,) = a,. Show now that B, c Mo holds for every a < g o , so according to assumption Be c M , holds for every a in [ a 1 , a * ] . Show that this yields a contradiction.
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Hint: For (i), note that by Exercise 33.9 there exists an element u in the set theoretic difference L+- M + such that u l l f I. Then the disjoint complement { o } ~satisfies {old c M since M is prime, and so the band generated by f is contained in M.
Exercise 45.10. Let L be an Archimedean Riesz space with a strong unit e. In Theorem 45.4 it was proved that the following conditions are equivalent. (i) L is uniformly complete. (ii) Every e-uniform Cauchy sequence in L has an e-uniform limit in L. (iii) "L = C($). The proof presented in Theorem 45.4 used that (i) =- (ii) => (iii) =- (i). We indicate how to prove that (ii) implies (i) without using (iii). Assume that (ii) holds, let u EL', and let (f,:n = 1,2, . . .) be a u-uniform Cauchy sequence. Show that (f,:n = 1,2,. .) is also an e-uniform Cauchy sequence, and so has an e-uniform limit f by hypothesis. In the particular case that fn is monotonely increasing, we have f = supf, by Lemma 39.2, and the same lemma shows now that f is also the u-uniform limit off,. For the general case, use the device employed in the proof of Theorem 39.4.
.
Exercise 45.11. Let L be a uniformly complete Archimedean Riesz space with a strong unit e. Then, according to Theorem 45.4, L is Riesz isomorphic to the function algebra "L = C($), so there is a commutative ring multiplication defined in L with e as unit element. As usual, the elementf-' is called the inverse off whenever f f = e. Show that f -'exists if and only if If1 2 ce for some strictly positive constant c, and in this case we have
-'
If-?
= Ifl-'.
Exercise 45.12. Let L be the Riesz space of all real bounded functions holds for at most finitely many x, with pointwise ordering (cf. Theorem 25.1, part (iv)). The space is Archimedean (actually, L has the principal projection property), and the function e, satisfying e(x) = 1 for all x E X , is a strong unit. Determine the structure of the space $ of maximal ideals and the structure of "L. In particular, show that $ may be identified with X topologized such that every point x # co is open and every subset containing the point co and having a finite complement is open. Identifying $ and X , the spaces " L and L are then identical as Riesz spaces. Indicate explicitly a function in C($) that is no member of "L.
f on X = (1,2,. . ., co) such that f ( x ) # : ( a )
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Exercise 45.13. Let L be the Riesz space of all real bounded functions f on X = ( x :0 6 x 6 1) such thatf(x) # f ( 0 ) holds for at most countably many x , with pointwise ordering (cf. Theorem 25.1, part (ii)). The space is Archimedean (actually, L is Dedekind 0-complete), and the function e, satisfying e ( x ) = 1 for all x E X,is a strong unit. Determine the structure of the space $ of maximal ideals and the structure of "L.In particular, show that $ may be identified with X topologized such that every point x # 0 is open and every subset containing the point 0 and having a finite or countable complement is open. Identifying y and X, the spaces "L and L are then identical as Riesz spaces. Check in a direct manner that every function in C($) is also in "L. Exercise 45.14. Let L be the Riesz space of all real bounded functions on X = (1,2, . . .), assuming only a finite number of different values, with pointwise ordering (cf. Theorem 25.1, part (iii)). The space is Archimedean (actually, L has the projection property), and the function e, satisfying e(x) = 1 for all x E X, is a strong unit in L. Obviously, L is a Riesz subspace of I , . Given the ideal I i n L,show that the set of all g E I , satisfying Igl 6 If I for some f~ I, is an ideal I' in I, such that I' L = I. Conversely, given the ideal Z' in I , , the intersection I' L is an ideal I in L such that any g EZ' is majorized in absolute value by some ~ E Z Show . that the thus defined correspondence between I and I' is one-one. It follows that the maximal ideals J in L correspond one-one to the maximal ideals J' in I , . Show that everyf E I, can be approximated e-uniformly, from above as well as from below, by functions in L. Hence, givenfe I , and the real number a, the set (J : " f ( J ) > a) is a union of sets of the form (J : ^f.(J) > a) for an appropriate sequence (f.: n = 1, 2, . . .) of functions in L. Show, that, therefore, the set of all maximal ideals in L,provided with the hullkernel topology, may be identified with the set of all maximal ideals in I,, , also provided with the hull-kernel topology. Hence, the topological space 2 of all maximal ideals in L is the Stone-Cech compactification of X , where X is provided with the discrete topology. Exactly as was the case for the space I , , every "fE ^L is a continuous extension off E L onto the larger (compact and Hausdorff) space $. The space "L is uniformly dense in C($), but not equal to C($). Indicate explicitly a function in C($) that is no member of "L.
n
n
46. The Yosida spectral representation theorem in an Archimedean Riesz space Let (e, : u E {c}) be a disjoint order basis in the Archimedean Riesz space
320
SPECTRAL REPRESENTATION THEORY
L, and let
[CH.7,8 46
W=(JP, U
where denotes the set of all prime ideals in L maximal with respect to the property of not containing e,. We will show that the Johnson-Kist representation theorem is applicable to 9, and (as already observed in section 44) for this purpose it is necessary (and sufficient) to show that the intersection of all R E W is equal to (0). Theorem 46.1. If L is Archimedean, if (e, : (r E { c T } ) is a disjoint order basis in L consisting of strictly positive elements, and if W = u,2?eu, then
n (R :
R E
W )= (01.
Proof. Let e, be one of the e,, and let I, be the ideal generated by e, . Assume there exists an elementf # 0 in I , such that f E R holds for all R e P o . Then u, = I f I satisfies 0 < u, E R n I, for every R E LPo.We will prove now that u, is a member of every maximal ideal in the Riesz space I,. For this purpose, let J' be a maximal ideal in I, .Then J' is an ideal in L not containing e,, so by Theorem 33.5 there exists an ideal Ro E 9" such that J' is contained in R, .The intersection R, n I , is an ideal in I, containing the maximal ideal J' but not containing e,, so R, n I, must be equal to J'. As proved above, u, E R n I, holds for every R E Po,so we have in particular that u, E Ro n I , = J'. This holds for every maximal ideal J' in the Riesz space I , , which implies that u,
E
n (J' :J' maximal ideal in I,).
The space I,, however, is Archimedean and has e, as a strong unit, so the intersection of all maximal ideals in I, is (0). This contradicts u, > 0. Hence, the intersection of all R E W contains no nonzero element of I , . Assume now that the intersection of all R E W contains an element g # 0. Then inf (Igl,e,) # 0 for at least one c, say for r~ = go. Write e, for eu0. Hence, we have f o = inf (191,e,) # 0 and f o is in the intersection of all R E 9. Also, f o is contained in the ideal I , generated by e, .This contradicts the result already established. It follows, finally, that
n (R:
RE
W )= (01.
For any f E L , let "f be the Johnson-Kist representation o f f with respect to the system (e, :c E (.}) and the system of prime ideals 92 = U , P U ,where W is provided with the hull-kernel topology. Every ^f is
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THE YOSIDA THEOREM I N AN ARCHIMEDEAN RIESZ SPACE
321
an extended realvalued continuous function on 9, and the mapping f + ^f is a Riesz isomorphism of L onto the Riesz space "L of all ^f. The sets 2!euare mutually disjoint subsets of 9, and by Theorem 36.4 every 2!eu is compact and Hausdorff. It follows that W is locally compact and Hausdorff. Note that because PUis exactly the set of all R E W not containing the element e,, it is evident from the definition of the hull-kernel topology that is open. Hence, for any a. E {a}, the set
u
(Li!e- :Q # ao) is open, and so 2!euo is closed. This shows that every 2!eu is open as well as
closed and compact. Any f in the ideal generated by e,, has the property that f vanishes on every 2!eufor a # ao, and "f is a bounded continuous function on 2!euo. Furthermore, every prime ideal R E 9 = uu2?eu satisfies R = ( f : " f ( R ) = 0), as follows from Theorem 44.5. As proved in section 44,the set "L of functions on W separates the points of 9.More precisely, if R 1 and R2 are points of W lying in different sets JeU,say for a, and az respectively, then ^euI(Rl)= 1 and "eu,(Rz)= 0. If R , and R2 are different points of W but in the same say for a = ao, then there exists an element f E L such that ^f ( R , ) # "f(R2).We will show that without loss of generality it may be assumed that this elementf is a member of the ideal Zuogenerated by eao.There are several cases to consider. (i) If " f ( R , ) = + 00 and "f(R,) = - 00, let g = inf (f,euo). Then "g(R,) = 1 and "g(R,) = - 0 0 , so h = g f = sup ( g , O ) satisfies ~ E Z , , with "h(R,) = 1 and "h(R,) = 0. (ii) If " f ( R , ) = +a and "f(R2) is finite, say ^f(R2)= a, let g = f-me,,. Then " g ( R , ) = +00 and "g(R,) = 0, so h = inf (Igl, e,,) satisfies h E Zmowith "h(R,) = 1 and "h(R,) = 0. Similarly if "f(R,)is finite and "f(R2) = - CO. (iii) If " f ( R , ) = a and ^f(R,) = p are finite, we may assume that a > B. Then g =f-Be,, satisfies "g(R,) = a-B > 0 and "g(R2) = 0. It follows that h = inf (Igl, eao)satisfies h E Zuowith "h(R,) > 0 and " ~ ( R z ) = 0. It follows that for every a the functions in the ideal Z, already separate the points of .PU. Hence, by Lemma 45.2, the image "I, of Z, is uniformly dense in the space C ( P U )of all real continuous functions on It follows from Theorem 45.4 that "I, = C(2'") if and only if Z, is uniformly oomplete. In particular, if Lis uniformly complete, then "I, = C(P') holds for every a.
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SPECTRAL REPRESENTATION THEORY
47. The space Cm(X)for X extremally disconnected As defined in section 44,if X is a topological space, then any continuous mappingf of X into RO'such that the set ( x : - co < f ( x ) < a)is dense in Xis called an extended real continuous function on X . The set of all suchfunctions is denoted by CO'(X);it was proved in section 44 that CO'(X)is not necessarily a linear space. In the present section we assume that the topological space X is extremally disconnected (i.e., the closure of every open set is open), and we will prove that in this case P ( X ) is a linear space (and even a Riesz space) under the appropriate definitions.
Theorem 41.1. Let f be afinitevalued real continuous function defined on the open subset 0 of the extremally disconnected space X . Thenf can be uniquely extended to an extended real conrinuous.functionf on the closure 0 of 0. Defining now,for example,f(x)= 0 on X - 0, thefunctionf is then an extended real continuousfunction on the whole of X . Proof. For every finite real t, define the subset Fr of 0 by
4
=
( x :X
E
0,f ( x ) < t ) .
The set 4 is open, and so the closure F, is closed and open. For t , S t2 we have F,, c F,,, so we have
0 c U F , c 0. 1
For any x E U , F , we set f ( x ) = inf ( t : x E FJ, and for any x E 0- U,F, we set f(x) = +a.We will prove that f is the required extension off. First we show thatf(x) = f (x) holds for every x E 0. To this end, observe that i f f ( x ) < t, then x E F,, and so x E F,, which implies that f ( x ) 5 t. This holds for all t such that t > f ( x ) , so it follows thatf(x) 5 f (x). For the converse inequality, note that for any real t the intersection of ( x :f ( x ) > t ) and F, is empty. Now, by a routine argument, if A and B are open sets such that 2 and B are open, and if A n B is empty, then A n B is empty. Hence, in the present case, the intersection (x
:f(x) >
o nF,
is empty. It follows that any point x E 0 at which f ( x ) > t holds is not a point of F,, so we have f ( x ) 2 t at that point. This shows that f ( x ) 2 f ( x ) holds for every point X E 0.The final conclusion in this part of the proof
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P ( X ) FOR ' A
EXTREMALLY DISCONNECTED
323
is, therefore, thatf(x) = f ( x ) holds for every x E 0, and s o l i s an extension off. In order to prove that3 is continuous on 0,we have to show that for every t satisfying - 00 c t 6 co the set (x :f(x) c t ) is open and for every t satisfying -a S t c co the set (x :3(x) S t ) is closed. This follows by observing that ( x : f ( x ) c t ) = u (F, : s c t), (x :3(x) t ) = n (Fs:s > t). In particular the theorem shows that any finitevalued real continuous functionf defined on an open dense subset of X can be uniquely extended to a functionf~C"(X). This will be used to prove the following theorem.
Theorem 47.2. If X is an extremally disconnected topological space, then C"(X) is a Riesz space.
Proof. Given f l and f2 in Cm(X), the sets O1 and O2 on which f l and f 2 are finite are open and dense. Then O1 n O2 is open and dense, and the function equal tof,(x)+f2(x) at every x E O1 n O2 is finitevalued and continuous on O1 n 0 2 .Hence, this function can uniquely be extended to an extended continuous function on X, and by definition (cf. section 44) this extended function is fi +f2.Similarly, iff E Cm(X)and the finite real number a are given, then af is the unique member of C"(X), equal to af(x) for all x in the set where f is finite. Note that for a = 0 the function afis identically zero. The argument for sup (f,9 )and inf (f,g ) , forf, g E Cm(X), is similar. Definition 47.3. The Riesz space L is called universally complete whenever every system (uz :T E { T } ) of mutually disjoint elements in L+ has a supremum. Dedekind completeness and universal completeness are independent. If L is the Riesz space of all real functions f on the uncountable point set X such that (x : f ( x ) # 0) is at most countable, then L is Dedekind complete but not universally complete. If L is the Riesz space of all real continuous functions on the real line, then L is universally complete but not Dedekind complete. Theorem 41.4. If X is an extremally disconnected topological space, then the Riesz space C" ( X ) is Dedekind complete and universally complete.
Proof. We show first that C"(X) is Dedekind complete. For this purpose, assume that (y :z E { r } ) is a set of functions in C"(X) such that 0 5 u,
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SPECTRAL REPRESENTATION THEORY
[CH.
7,s 47
5 u, holds for some u, E C m ( X )and for all z. Let 0, be the subset of X o n
which u, is finite. Then 0,is open and dense and every u, is finite and continuous on Oo, so uo and all u, are members of the space C(0,)of all real, finitevalued, continuous functions on 0,.Since 0, is extremally disconnected, the space C ( 0 , ) is Dedekind complete (indeed, the space C,(O,) is Dedekind complete by Theorem 43.11, and so C(0,)is Dedekind complete by Theorem 43.3). Let u = sup u, in C ( 0 , ) . Then u is a finitevalued continuous function on Oo, and so u has an extended realvalued continuous extension ii, defined on the whole set X. Since u, 5 u holds on 0, , the continuity implies that u, 5 ii holds on X,so ii is an upper bound (in C m ( X ) )of the set of all u,. If v is another upper bound, then u, 5 u holds on X, so u,(x) 5 u(x) holds in particular for all x E 0,.This implies that i i ( x ) 5 v ( x ) holds for all x E Oo, and so ii 5 v holds on X . It follows that ii = sup u, holds in Cm(X). Hence, C m ( X )is Dedekind complete. For the proof that C m ( X ) is universally complete, let (u, : T E {t}) be a set of mutually disjoint elements in { C m ( X ) } + For . each T, let 0, be the subset of Xon which u, is finitevaluedand non-zero. The sets 0, are mutually disjoint, and obviously the set
as follows. For each T, is open and dense. We define the function u on 0we set u ( x ) = u+(x) for x E Or, and for x E X - U,O, we set u(x) = 0. Then u is finitevalued and continuous on 0, and uz(x)5 u(x) holds for every x E 0. The unique extension ii of u to X is now a member of Cm(X),and it is easily seen that ii = sup u, holds in C a ( X ) . This shows that C m ( X )is universally complete. We will now present an example in which an extremally disconnected topological space comes up in a natural manner. For this purpose we recall some facts about Boolean algebras proved in Chapter 1. Let X be a Boolean algebra with null element 8 and unit element e, and let 9’ be the set of all proper prime ideals in X.For any X E X,the set {PIxis the set of all P E 9such that x is no member of P. The sets of the form { P } x are the base sets for the hull-kernel topology in 9’and, as proved in Theorem 7.3, the base sets { P } x are open and compact. Hence, in particular,B = {P},is compact. In Corollary 8.7 it was proved that 9 is also Hausdorff. According to Theorem 8.5, if the hull-kernel topology is Hausdorff, then every proper prime ideal is a minimal prime ideal. Hence, this applies in the present case we consider here.
CH.
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THE SPACE
Cm(X)FOR
x EXTREMALLY DISCONNECTED
325
Finally, by Theorem 7.5 any open and closed subset of some { M } x = { P } x is of the form { M } y = { P l y for some y S x. Using these facts, we can prove now the following theorem.
Theorem 47.5. The Boolean algebra X is Dedekind complete (or, equivalently, order complete) if and only if8 is extremally disconnected. Proof. Assume first that X is Dedekind complete, and let 0 be an open subset of 8.Then 0 is of the form 0 = u , { P > x s Since . Xis Dedekind complete, the element xo = sup x, exists in X. The set { P } x ois closed and open; it will be sufficient for our purpose, therefore, to show that {P}xo= 0. In any case, x, 5 xo implies { P } x , c {P}xofor all z, so 0 c {P}x,,and hence
0 = (p>x, = { P } x , . Assume now that 0 is properly included in {P>xo.Then the set theoretic difference { P } x , - O is open and non-empty, and so contains a set {P],, for some y # 8. The set theoretic difference {P}xo- { P } yis of the form {P}=, where z is the relative complement of y with respect to x,, so {P}xois the disjoint union of the sets { P } y and {P}=with z 5 x o , z # x,. Since in the point set 9 the subsets 0 and { P } y are disjoint, we have 0 c { P } z ,and so { P } x , c {P}, for every z. By Theorem 6.5 this implies that x, 5 z holds for every z, so xo = sup x, z.
s
This contradicts z 5 x o , z # x,. Hence 0 = {P}x,, which is the desired result. It has been proved thus that 9is extremally disconnected. Assume now, conversely, that 9’is extremally disconnected. For the proof that X is Dedekind complete, let (x, :z E {z}) be a subset of X . We have to prove that sup x, exists. For this purpose, consider the open set 0 = u , { P } x c .The set 0 is closed and open, so 0 = {Plx0for some xo E X . Obviously, we have {P}x, c {P}xofor all z, so (by Theorem 6.5 again) we have x, 5 xo , which shows that xo is an upper bound of the set of all x,. If y is another upper bound, then {PIy 3 0, so { P } y=
my 0 3
= {PjXO.
It follows that y 2 xo . This shows that xo = sup x,. Hence, X is Dedekind complete. This completes the proof.
The theorems on C m ( X ) for an extremally disconnected topological space Xare due to H. Nakano ( [ 5 ] , 1942) and T. Ogasawara ([I], 1942).
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SPECTRAL REPRESENTATION THEORY
[CH.7,s 48
Exercise 47.6. We indicate a different proof of Theorem 47.5. (i) Assume that X is a Dedekind complete Boolean algebra. For x E X , the complement of x (in the Boolean algebra sense) is denoted by x', and the principal ideal generated by xo E Xis denoted by Ix,. For any ideal I in X, show that I d = IX,. and Idd = I,,, where xo = sup (x :x E I). Show that I d is a projection band in X , i.e., X = I d v I d d(cf. Exercise 8.8, part (vi)), so {P};is open and closed. Since every open subset 0 of B is of the form {P}, for some ideal I, it follows that 0 is open and closed, so B is extremally disconnected. (ii) Assume that Xis a Boolean algebra such that the corresponding prime ideal space 9is extremally disconnected. Show that {P}; is open and closed for every ideal I i n X, so Id is a projection band for every I, i.e., Zddv I d = X. It follows that the unit'element e can be written as e = x1 v xz with x1 E Idd and xz E I d . Show that x , and xz are complements, so xz = x i . Show that any x E Zdd satisfies x 5 x,; in other words, show that Idd is the principal ideal generated by x1, i.e., Idd = Ix,. Since I is contained in Idd,it follows that x1 is an upper bound of I. Show now that x1 is the least upper bound of I (use Theorem 4.9). Finally, show that Xis Dedekind complete.
48. Subsets of the set of all bands An important role will be played in what follows by certain sets of bands in a Riesz space; in the present section we will derive some results about such sets of bands, besides those proved already in section 22. Let Jal be the set of all bands in the Riesz space L; the set Jal is partially ordered by inclusion. By Theorem 22.6 Jal is an order complete distributive lattice with smallest element (0)and largest element L. By B we denote the subset of Jal consisting of all disjoint complements. The band B satisfies B E 9?if and only if B = Bdd holds. The set 9? is also partially ordered by inclusion. By Theorem 22.7 B is an order complete Boolean algebra with smallest element (0)and largest element L. More precisely, if B, and Bz are members of a, then SUP
( B , , B,) = ( B , +Bz)dd = ( B f n B:)d,
inf (B, ,B,) = B , n B z , and the complement of B E 9? is the band Bd. According to Theorem 22.8 we . ? = d if and only if L is Archimedean. have % For anyfe L, let A, be the ideal generated by$ Let apbe the subset of &?
CH.7,s 481
SUBSETS OF THE SET OF ALL BANDS
321
consisting of all bands of the form A?. For brevity, we will denote A? by lf].Note that [f] = [If I] for everyf. Also, note that [O]consists only of the null element, and [f] = [O] if and only iff = 0. Finally, g E If]implies [gl = [fl, so 9 E If],h E [gl implies h E rfl.
Theorem 48.1. The set A?, is a distributive sublattice of B’. More precisely, we have sup (bl, [UI) = [u+vl = [SUP (u, 41, inf ([u], [ u ] ) = [inf (u, u)] for all u, u E Lf.Note that inf ( [ u ] ,[ u ] ) = [O] i f and only i f u 1u. Denoting by 4, the ideal generated by A?, in A?, the ideal 3, is order dense in A?, i.e., 3: = A?. Proof. For all u, u EL’ we have immediately that
4n 4 =
so It follows that
=
(Asup(u,uJd9
[uId n [uId = [u+uId = [sup (u, U ) ] ~ .
sup ( [ u ] , [ u ] ) = ([uId n [old)” = [u+uIdd = [u+u], and similarly sup ( [ u ] , [ u ] ) = [sup (u, u ) ] . Also, by means of Theorem 19.3 (iv), we find that inf ( [ u ] , [ u ] ) = [u] n [u] = ~i~n ~t~
= (A, n A , ) = ~ ~ ( A ~ , , ~ ( ~=, [inf(u, ~ , , ) ~u)]. ~
This shows that 93,is a sublattice of 9. Evidently, 3?, is distributive since B’ is so. The ideal Y Pin B’ generated by 93,consists of all elements of B’ majorized by an element of a,,and hence the disjoint complement 3; consists of all B E A? disjoint from A?,, i.e.,
4: = (B :B E A?, inf (B, [ u ] ) = [O] for all u EL’). In order to show that 9 : = A?, it is su5cient to show that 3; = [O]. If 3: contains a band B # [O],there exists an element uo > 0 in L such that uo E B. Since B E 9:, we must have inf(B, [Uol) = [OI, and so B n [uo] = [O].This implies [uo]n [uO] = [0], so [uo] = [O], which contradicts uo > 0. Hence 9; = a, i.e., 4, is order dense in A?.
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SPECTRAL REPRESENTATION THEORY
[CH.7,8 48
We recall that L is said to have the quasi projection property whenever Bd 0 Bdd = L holds for every band B in L; and that L has the quasi principal projection property, whenever [u] 0 [uId = L holds for every u E L+. Theorem 48.2. If L has the quasi principal projection property, then the set g pis a Boolean ring. Proof. Assume that L has the quasi principal projection property, and let Bpwith [u] c [u], where we may assume without loss of generality that 0 S u 5 u (because u, o E L + and [u] c [u] implies [u] = [u] n [ v ] = [inf (u, u)]). We have to show the existence of an element w EL' such that inf ( [ u ] , [w])= [O] and sup ([u], [w])= [u]. Now, since L has the quasi principal projection property, we have u = u1 + w with u1 E [ v ] and w I0, so w Iv already implies that inf ( [ v ] , [w])= [O]. On account of u1 1w we have sup ( u l , w) = u, so [u], [ v ] E
SUP ( [ U l l ,
[wl)
=
[SUP (u1,w)I = [ul.
Also, ul E [u] implies [ul] c [ v ] , so
On the otber hand, since [u] c [u] and [w]c [u], we have the converse inclusion, and so sup ([u], [ w ] ) = [u]. This is the desired result. The converse of the last theorem does not hold. Indeed, the space C(( - 00, 0 0 ) ) of all real continuous functions on (- 00, a)does not have the quasi principal projection property, but the corresponding set Bpis a Boolean ring. Theorem 48.3. I f L has the projection property, then Bpis an order dense ideal in 99.
Proof. Note first that under the present hypotheses L is Archimedean, so 93 = d,i.e., every band is an element of a.Also, for every UEL', the
band [u] = Atd is now the band generated by u. We need only prove that if B is a band contained in a band of the form [u] for some u E L ', then B = [ v ] for some v E L + .Since L has the projection property, the component of u in the band B exists. Let v be this component. Then it is obvious that [ v ] is included in B. Conversely, any element of B which at the same time is majorized (in absolute value) by a finite multiple of u is also majorized (in absolute value) by a finite multiple of v, and so is a member of [v]. Any
CH. 7, $491
THE OGASAWARA-MAEDA REPRESENTATION THEOREM
329
arbitrary element w of B + is a supremum of elements as just described, and so w E [u]. It follows that B = [u], which is the desired result. 49. The Ogasawara-Maeda representation tbeorem Let L be an arbitrary Riesz space, and let B ( L ) be the Boolean algebra of all bands in L that are disjoint complements. In the present section we will prove still another representation theorem for L, due to F. Maeda and T. Ogasawara ([l], 1942). More precisely, we will prove that if L is Archimedean, then L is Riesz isomorphic to a Riesz space of extended realvalued continuous functions on a topological space, the points of which are no longer prime ideals in L as in the preceding sections, but now are prime ideals in the Boolean algebra B(L). We proceed with the details. The Riesz space L is not immediately assumed to be Archimedean. As in the preceding section we write If]for the band A?, and we denote the set of all I f ] by g p ( L ) . Prime ideals in g ( L ) will be denoted by w, and the Stone representation of g ( L ) will be denoted by Q(L)or, briefly, by s;) if no confusion can occur. The elements of Sa, therefore, are the prime ideals w , and the topology in Sa is the hull-kernel topology, determined by the base sets
U, = (w : B no member of w ) , where B is an element of B ( L ) . As proved in Corollary 8.7, Sa is a compact Hausdorff space. The Boolean algebra B ( L ) is Dedekind complete, so (as proved in Theorem 47.5) the space Sa is extremally disconnected. Every base set U, is open, closed and compact, so the set U,, as a topological space by itself with the relative hull-kernel topology, is compact, Hausdorff and extremally disconnected. It follows now from Theorem 47.4 that the Riesz space C"(Sa) is Dedekind complete and universally complete. The same holds for C m ( U Bfor ) any base set U,. Definition 49.1. Let e be ajixed nonzero element in L+. For any f E L and any w E Sa, we dejine the extended real number f "(a) = f "(w; e ) by f A ( w ) = f " ( w ; e) = sup ( u : [ ( u e - f ) + ] E a),
with the convention that f "(a) = - 03 or f "(0) = + 03 if [(ole-f)'] E w holdsfor noJinite real a or [(ae -f)' ] E w holdsfor allJinite real u respectively. Regarding this definition, note that if [(ue-f)'] E w holds for some a , then [(pe-f)'] E w holds for all j3 < ct, so the definition as stated above makes good sense. Another point to observe is that, generally, I f ] , [ g ] E o
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SPECTRAL REPRESENTATION THEORY
[CH.7.5 49
implies [ f + g ] E o.Indeed, since w is an ideal, it follows from [ f ] ,[ g ]E w that sup ( [ l f l ] ,[Igl])E o,so [lf1+1gI] E o,and hence certainly r f + g ] = [lf+gl] E o.Also, [ f 3 = [If I ] E o implies that lf'], l f - ] E o.Hence, we have [ae-f] E w if and only if [(ae-f)'] E o and [ ( a e - f ) - ] E w. Since o is a prime ideal and since
[(ae-f)+I
I[(ae-J)-l
holds in g ( L ) , one at least of [(ae-f)+] and [ ( a e - f ) - ] is a member of w. [f [ae- f ] E w holds for more than one value of a,then by addition we obtain the result that [el E o.Hence, if it is given that [el is no member of w,then [ae-f] E w holds for at most one value of a. It follows from all this that if [el is no member of o,there exists one real a. (ao = +a not excluded) such that a. = sup ( a : [ ( a e - f ) + ]E o)= inf ( a : [(ae-f)-] E 0).
We have [ ( a e - f ) + ]E w for a < a. but not for a > ao, and we have [ ( a e - f ) - ] E o for a > a. but not for a < ao.
Theorem 49.2. Let e be a fixed nonzero element in L', and let w E D be such that [el is no member of o.For any f E L, we writef"(o) brieflyfor f "(0; e). Thefollowing holds. ( i ) f "(w) = ofor all f E L satisfying l f ] E o. (ii) e"(w) = 1. (iii) (pf)"(o) = p f ^ ( o ) f o r all f E L and allfinite real p . (iv) ( f + g ) ^ ( w ) = f ^ ( w ) + g " ( o ) f o r a l l f , g E L for whichf^(w)+g"(w) is well defined (i.e., f ^(w)and g"(w) are not both infinite and of opposite sign). (v) I f h = inf (f,g ) in L, then h ^ ( o ) = inf ( f ^ ( o ) ,g^(w)). Similarly for sup (f,g). In particular, we have
( f +)%)
= SUP
(f"(40)Y
(f-)^(w) = - i d (f^(w),O), I f I"(w> = If"(w)l.
Proof. (i) Let I f ] E w. For a < 0 we have [(ae-f)+I
= [(-f)+I
= rf-I
E 0,
so f"(o)2 0. For a > 0, it follows from ae 5 f+ (me -f )+ that we should
CH. 7.8 491
THE OGASAWARA-hlAEDA REPRESENTATION THEOREM
331
have [ae]E o if [(ae-f)'] E o.Since [ae]E o does not hold, the formula [(ae-f)+] E o does not hold either. Hencef"(o) = 0. (ii) Evident. (iii) Since g"(w) = 0 holds for g = 0, it is evident that the formula to be proved holds for B = 0 and arbitrary$ For 0 < B < co we have
( a : [(ae-f)+] E o)= ( a : [(aBe-Bf)+] E w )= (y/B : [(ye-Bf)+] E 0). Taking suprema, we obtain f " ( w ) = B-'(Bf)"(o), so
Pf"(0) = (Pf)"(o>. In order to complete the proof, it will be sufficient to consider the case that j? = -1. We have
(-f)"(w) = sup ( a : [(o!e+f)+]E o)= sup ( a i [ ( - a e - f ) - ] E o) = sup ( - y : [(ye-f)-] E o)= -inf (y : [(ye-f)-] E o)= -f"(o). (iv) Givenf, g E L such thatf"(o)+g"(o) is well defined, we will prove that f A ( 4 + g A ( w ) I( f + g ) W (1) holds. Forf"(w)+g"(o) = -a the inequality is true, so we may assume thatf"(w) > -a and g"(w) > -a.Let a be a finite real number satisfying a
[{me- (f+9)) + 1 =
[{(a1 e
-f)+ (a2 e -s)>1 = [(a1e -f)+ + (a2 e -g)+1
= SUP ([(a,e-f)+I,
+
[(a2e-g>+1)E 0 ,
and this implies a 6 (f+g)"(w). This holds for any o! < f " ( o ) + g " ( o ) ; it follows that the inequality (1) holds. Applying this result to -fand -9, we obtain the inequality in the converse direction, and so (f+g)"(o) = f W + s W . (v) Let h = inf (f,g) in L. It is evident from the definitions that h " ( o ) 5 f"(o)and h"(w) 5 g"(w), so
h"(w) 6 inf (f"(4, g"(0)). If the infimum is - 00, there is nothing left to prove. We may assume, therefore, that the infimum is greater than -a.In order to prove now that h"(w) is greater than or equal to this infimum, we observe first that
332
[CH.7,s 49
SPECTRAL REPRESENTATION THEORY
(ae-h)+
=
{ae-inf (f,g)}+ = {ae+sup (-f, -g)}+
=
{sup (ae-f, ae-g)}+
=
sup {(ae-f)+, (ae-g)+}.
For any a < inf ( f " ( w ) , gA(w)) the bands [(ae-f)'] and [(ae-g)'] are both members of w, so the just established formula shows now that [(ae-h)+] is also a member of w, i.e., a 5 h"(w). It follows now that inf ( f "(w), g"(w)) h"(w). This, together with the converse inequality, yields the desired result. The proof for sup (f,g) is similar. For e E L +fixed, f E L and w
E
Ute, = (w : [el no member of a),
we consider the number f ^(w) = f ^(w; e) as defined above in Definition 49.1. For f fixed and w variable, we thus obtain an extended real function f" on Ucel.We shall prove now that f is continuous on Urel,wheref A is to A
be considered as a mapping from the topological space Urel into R". It should be noted, however, that we will not prove that the set (0 : -a < f ^(a)< co) is dense, i.e., we do not prove that f ^ is an extended real continuous function in the sense as defined in section 44.
Theorem 49.3. We use the same notations as before. For every f corresponding function f " is continuous on Ute,.
E L,
the
Proof. For every finite real a, we have (w : f ^ ( w ) < a) =
/?
(w : [(Pe-f)']
no member of a),
where on both sides we restrict ourselves to elements w E Ucel.Each of the sets occurring on the right is the intersection of Uceland a base set, so this intersection is open. This shows that the set (w :f " ( o ) < a) is open. The same holds for a = + co on account of (0: f " ( w ) <
u a
+co) =
n=l
< n).
(0 : f A ( 0 )
Applying this to the function -f ", we obtain the result that (w :f ^(w) > a) is open for every finite real a and for a = - 00. It follows that f " is continuous on Urel. We will investigate now in what respect the Archimedean and non-Archimedean cases differ from each other. By way of example, we consider the case that L is the lexicographically ordered plane. The set g ( L ) of all disjoint complements consists of {0} and L, so the only prime ideal in g ( L ) is
CH.
7. 5 491
333
THE OGASAWARA-MAEDA REPRESENTATION THEOREM
the prime ideal wo consisting of (0)only. The Stone representation space Q(L)consists, therefore, of the one element wo. It depends now on the choice of the nonzero element e E L + what f "(ao)will be for any f E L. Choosing e = (1, e,), where e, is an arbitrary real number, we have f "(coo) = fl for anyf = (fi,f , ) E L, sofA(wo) depends in this case only on the first coordi= 0 for all points f = ( O , f i ) on nate off, and in particular we have f "(aO) the vertical axis, i.e., f "(coo) = 0 for allfthat are infinitely small with respect to e. Choosing e = (0,l), we havef"(w,) = +a for all f = ( f l yf , ) with f l > 0 and f "(wo) = - co for all f = (f l ,f , ) with f l < 0, whereas f "(coo) = f 2 f o r f = (0,fZ). In this example it was seen that elementsf that are infinitely small with respect to e satisfyf h(w; e ) = 0. This holds generally, as shown in the following theorem. For the formulation of the theorem, note that iff E L and 0 < e E L+ satisfy If1 S ae for all real a > 0, this simply means that either f = 0 or f is infinitely small with respect to e.
Theorem 49.4. I f f E [el, then If I 5 ae holds for all real a > 0 if and only i f f "(0) = f "(a; e ) = 0 holds for aZ1 w E Ucel.
Proof. If If I 5 ae for some a > 0, then - ae 5 f 5 re, and so it follows from Theorem 49.2 that - a 5 f"(w) 5 a holds for all w E Ucel. Hence, if If I 5 ae holds for all a > 0, then f "(0) = 0 holds for all w E Ucel. Conversely, assume that f "(0) = 0 holds for all w E Urel. This implies that for all a > 0 and all w E UCe,,the band [(ae-f)'] is no member of w . Now assume also that ( a o e - f ) - # Ofor some a. > 0. Then [ ( a , e - f ) - ] is properly larger than [O], so by Theorem 4.4 there exists an element o E D such that [ f a o e - f ) - ] is no member of w. On account of (aOe-f)- = (f-aoe)+ s f + , it follows that L f ' ] is no member of w. Furthermore,f E [el impliesf E [el, so I f ' ] c [el, and hence [el is no member of w. We thus obtain the result that there exists an element O E Urel such that neither [ ( a , e - f ) + ] nor [(aOe-f ) - ] is a member of w. This is impossible since w is prime. It follows, therefore, that ( a e - f ) - = 0 for all a > 0. In other words, we have ae-f 2 0 for all a > 0, i.e., f 5 ae for all a > 0. Applying this also to -J we find that I f I 6 ae holds for all a > 0. +
The following theorem is an immediate consequence.
Theorem 49.5. L is Archimedean if and only if it is true for every nonzero e ) = 0 for some f E [el andfor all w E Ure3implies element e in L that f "(a; f=O. +
334
SPECTRAL REPRESENTATION THEORY
[CH.
7,s 49
An inspection of the proof of Theorem 49.4 shows that the following generalization holds. a
Lemma 49.6. If 0 < u S e and i f f E [u], then If I 4 ae holds for all real > 0 ifand only i f f " ( w ) = f " ( o ; e ) = 0 holds for all o E ULul.
We will show now that if L is Archimedean and e is any nonzero element in L', then (for anyf E L ) the functionf "(a;e ) is finite on a dense subset of Ucel. This will show, therefore, that now f " is a member of the space Crn(Ucel>-
Theorem 49.1. I f L is Archimedean, if e is a nonzero element in' L andf is an e ) is finite on a dense subset arbitrary element of L, then the function f "(o; of Urel. Proof. We may assume, for the present purpose, that f is a member of L'. Assuming that the theorem is false forf, there exists an element u EL' such that 0 < u 5 e andf"(o; e ) = + 03 for all w E Ura. It is evident that Lf] E o holds for no o E Ucul(because Lf] E w for an o satisfying o E Vculimplies f "(o; e ) = 0 by Theorem 49.2 (i)). Hence we have Without loss of generality we may assume now that u
5 f, because
and so replacing u by inf (u,f ) does not change Urul.For any w E Uculwe have f "(o; e ) = +coy so [(ae-f)'] E o holds for all a > 0. By an earlier remark it follows then that for any a > 0 the band [(ae-fl-l = [(f--ae)+l
is no member of o,so [ ( a - t f - e ) ' ] is no member of w. On the other hand, for any a < 0 the band [(a-tf-e)'] = [O] is a member of o. Hence e"(w;f ) = 0 for every o E UCa.But then, on account of 0 < u 6 e, we have u " ( w ; f )= 0
for every o E Ucul.
An application of the last lemma shows now that u af holds for all a > 0, i.e., u is infinitely small with respect to f (since u is nonzero). This is impossible. Hence, the assumption that the theorem is false yields a contradiction. As an immediate consequence, we obtain the following theorem.
CH.7 , § 491
THE OGASAWARA-MAEDA REPRESENTATION THEOREM
335
Theorem 49.8. I f L is Archimedean and e is a nonzero element in L', then the image L A of L under the mapping f +f "(a;e ) is a Riesz subspace of the Riesz space C" (Urel). Before proving some further properties of L", we state a lemma.
Lemma 49.9. Let e be a$xed nonzero element in the positive cone Li of the e) Archimedean space L, and letf be an element of the band B in L. Thenf "(o; = 0 holdr for all UBd
u[e]
=
Uinf ( B d , [el)
=
UBdn[e] *
Proof. If w E Ugd, then Bd is no member of the prime ideal w, and so B is a member of w. On account of If]c B it follows then that If]E w , and so, if in addition w E Urel,we have f "(o; e ) = 0 by Theorem 49.2 (i). Theorem 49.10. Once again, let e be afixed nonzero element in the positive cone Li of the Archimedean Riesz space L. Furthermore, let 0 c u E L ' , and let the function g E Cm(Urel)have the property that g ( w ) 2 0for all o E Urel and 0 c g(w) u"(w; e )
s
holdsfor all w in a non-empty open subset 0 of Urel.Then there exists a nonzero element v in L f such that holdsfor all o E Urel.
~ " ( we;)
s g(w)
Proof. For every natural number n, consider the subset of 0 on which n-' < g ( o )
u"(w; e ) < n
holds. This is an open subset 0, of 0, and for n sufficiently large Oiis nonempty. Assuming this, 0, contains a base set U, with B # (0). Let p be an element of L satisfying 0 < p E B. Then Urplis non-empty and contained in O,,so n-' c g(w) 6 u " ( o ; e ) c n (1) holds for all w E Ucpl.On account of Urplc 0 c Urelwe may assume without loss of generality that p satisfies 0 < p 5 e (if necessary, replace p by i d ( p , e), which does not change Urpl).We will prove now that p and u are not disjoint. Indeed, p 1u would imply inf (u"(w; e),p"(o; e)) = 0
for all o E ULel,so p " ( o ; e ) = 0 for all w E Ucpl.By Lemma 49.6 this,
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SPECTRAL REPRESENTATION THEORY
[CH. 7.8 49
however, would imply p = 0 (note that L is Archimedean), contradicting p > 0. Hence, p and u are not disjoint, i.e., u is no member of the band [pld. Since [PI 0 [PId is order dense in L,there exists an upwards directed system (w, : z E {z}) in [ p ] 8 [pld such that 0 6 w, f n-'u (where n is the same natural number as in (I)). Hence, if w, = w:+w:' with w: E [PI and w:' E [pld,it is impossible that w: = 0 holds for all z, because this would imply w, E [PId for all T, and so u E [pld,which is not true. In other words, there exists an element u E [ p ] satisfying 0 < u 5 n-%. It follows that u"(w; e )
5 n-'u"(o; e ) for all o E Ucel.
Combining this with the inequality (l), we obtain
s
~"(o; e ) n-l < g ( w ) for all o E Ucpl, and by the preceding lemma we have u"(o; e ) = 0 for all o E Ucpldn Urel. Hence z,"(o; e ) 5 g ( o ) for all o E Urel. Finally, it follows from 0 < u E [ p ] c [el that u"(w; e ) is not identically zero on Urel(again by Lemma 49.6 or, alternatively, by Theorem 49.4). As a consequence, we have the following theorem, showing that at least for elements in the band [el the mapping f +f "(o; e ) does not only preserve finite suprema and infima, but arbitrary suprema and infima as well. Theorem 49.11. Again, let e be a $xed nonzero element in the positive cone (u, : z E {z}) be a subset of Lc such that u, E [elfor all z and u = sup u, exists. Then
'L of the Archimedean Riesz space L. Furthermore, let
holds in Cm(Urel).
u"(o; e ) = sup u:(o; e )
Proof. Evidently, U" is an upper bound in Cm(Urel)of the set of all u,". Furthermore, since the space Cm(Ure,)is Dedekind complete by Theorem 47.4, the function h(w) = sup u,"(o; e ) exists in Cm(Urel).We have g ( o ) = u"(o; e)-h(o) 2 0
for all o E Ucel,and we must prove that g ( o ) is identically zero. If g ( o ) is not identically zero, there exists, by the preceding theorem, a nonzero element u in Lf such that u^(w; e ) 5 g(o) holds for all o E Ucel.Actually, as the proof of the theorem shows, we have u E [el. Now, since z, > 0 and u = sup u,, we do not have u- (u,+u) 2 0 for all z, so there exists an index z, such that (u-(u,,+u))> 0. Yet, the image
CH. 7,8501
THE OGASAWARA-MAEDA REPRESENTATION THEOREM
337
{u-(uTo+u>}" of u-(uro+u) is a non-negative function on Urel because UG+U"s
h + g = uA,
and so the image of (u-(uro+u))- is identically zero on Ucel.Observing now that u - (uro u ) E [el, and so (u - (uro+ u))- E [el, it follows from Theorem 49.4 that (u- (uT,+u))- = 0 on account of its image being identically zero on Ucel.This contradicts (u- (uzo+u))- > 0. Hence, the final conclusion is that g ( w ) is identically zero.
+
50. The Ogasawara-Maeda representation theorem for an Archimedean space
with a weak unit or with a disjoint order basis
We assume first in this section that L is an Archimedean Riesz space with a weak unit e E L + (i.e., e is a nonzero element in L+ such that the band generated by e is L itself). The notations are the same as in the preceding section. In particular, g ( L ) is the Boolean algebra of all bands, prime ideals in g ( L )are denoted by w,the Stone representation of g ( L )is denoted by 0, and the band generated by f E L is denoted by Lf]. The hull-kernel topology in Q is determined by the base sets
U, = (w : B no member of 0). Since e is a weak unit, we have [el = L , and so Ucelis equal to the whole space Q. Note also that since f E [el is now automatically satisfied for any f E L , it follows from Theorem 49.4 thatf"(w;e) = 0 holds for all w E Q if and only iff is the zeroelement of L. Hence, we have the following theorem.
Theorem 50.1. If e is a weak unit in the Archimedean space L, then the mapping f -P f e ) is a Riesz isomorphism of L onto a Riesz subspace L A of Cm(Q).The isomorphism preserues arbitrary suprema and in$ma. Theorem 50.2. Under the same hypotheses as in the preceding theorem, the ideal D, generated in Cw(Q)by L A is a Dedekind completion of L, and DL is order dense in Cm(Q). Proof. The ideal DLconsists of all g E Cw(0)for which there exists an element u E L+ such that Ig(w)l u"(w; e ) holds for all w E 0. The space Cw(Q)is Dedekind complete, so the ideal DL,as a Riesz space by itself, is also Dedekind complete. Since L is Riesz isomorphic to the Riesz subspace L A of DL,it is sufficientfor the proof that DL is a Dedekind completion of L to show that for every 0 < g E DL there exist nonzero elements u, Y E L+ such that 0 u"(w;e ) g ( w ) 5 ~ " ( 0 e );
338
SPECTRAL REPRESENTATION THEORY
[CH.7,s 50
holds for all o E Sa (cf. Theorem 32.6). The existence of such an element u follows from the definition of DL, and the existence of such an element v follows from Theorem 49.10. DL is order dense in Cm(Sa),i.e., the band generated by DL is Cm(Sa),since the band generated by e is already equal to Cm(Sa). Corollary 50.3. I f L is Dedekind complete and e is a weak unit in L, then L A is an order dense ideal in Cm(0). Proof. L A is now Dedekind complete, and so L A is equal to its Dedekind completion DL. We now introduce the notion of the universal completion of a Riesz space, a notion very much similar to the notion of the Dedekind completion. Definition 50.4. The Dedekind complete and universally complete Riesz space L' is called a universal completion of the Riesz space L if the following conditions hold. (i) L is Riesz isomorphic to a Riesz subspace L A of L'. (ii) Everyf' E (L')+satisfies
f' = sup ( g A :g A E L A ,g A sf').
Any Riesz subspace of a Dedekind complete space is Archimedean, and so the Riesz space L can have a universal completion only if L is Archimedean. In this case L has a Dedekind completion, and evidently the Dedekind completion of L is then' included in the universal completion of L. We use here, tacitly, that if L has a universal completion L', then L' is uniquely determined except for an obvious Riesz isomorphism, i.e., if L' and L; are universal completions of L, then L' and L; are Riesz isomorphic in such a manner that iff E L A and f E L,^ correspond to the same element f E L, then f and f ,^ are corresponding elements in the isomorphism between L' and L; . Hence, if we identify L and its Riesz isomorphic image L A in L', we can consider L as a Riesz subspace of its universal completion L', i.e., L is :mbedded in L' as a Riesz subspace. Finally, note that in view of condition [ii) in the definition of the universal completion the embedding is such that the ideal DL generated by L is order dense in L'. Actually, DL is the Dedekind completion of L. Applying all this to the present situation, where L is Archimedean and has a weak unit, we have the following theorem. A
A
Theorem 50.5. If L is Archimedean with a weak unit, then the space Cm(Sa) is a universal completion of L.
CH. 7, $501
339
THE OGASAWARA-MAEDA REPRESENTATION THEOREM
Proof. The space L" is a Riesz subspace of the Dedekind complete and universally complete space Cm(B),and the ideal DLgenerated by L" is order dense in Cm(B). We will prove now that if, in addition, L has the projection property, then L A separates the points of Q. Theorem 50.6. I f L has the projection property and e is a weak unit in L, then L Aseparates the points of a. Moreprecisely, $al, w2 are any two diferent points of B and a, b are finite real numbers, there exists an element f E L such that f "(wl;e ) = a andf A(w2; e ) = b.
Proof. The topological space 0 is compact and Hausdorff (and hence normal), so that if wl,u2 are two given points of B, different from each other, there exists a real function g E C(Q) such that 0 5 g(w) 5 1 holds for all w E B,g ( o l ) = 1 and g(w2) = 0. The open set (w :g(w) 3) contains a base set U , such that w1E U,. Obviously w2 is a point of the complementary set uBd.By the projection property of L we have e = p1+p2 with p1 E B and p 2 E Bd. It follows then from Lemma 49.9 that p;(w; e ) = 0 for all w E u,d and p:(o; e ) = 0 for all w E U,. Since e"(w; e ) = 1 for all w , and e" = p;+p:, we have p:(w; e ) = 1 for all w E U, and p;(w; e ) = 0 for all 0 E u,d. In particular, p;(wl; e ) = 1 and p r ( w 2 ;e ) = 0.Given the real numbers a and b, the elementf = be+(a-b)pl satisfiesf " ( w l ; e ) = a and f " ( 0 2e;) = b. If L is Archimedean and has a strong unit e, then L Ais a Riesz subspace of the space C ( a ) of all real continuous functions on the compact Hausdorff space B. Assume now that L has also the projection property. Regarding C(B) as a normed linear space with respect to the familiar uniform norm, the space L Ais a subset of C(B) such that LAis a lattice under pointwise ordering and such that, for given wl,w2 E B and given real a, b, there exists a function f " in L Asuch that f "(a1)= a andf "(w2) = b. Hence, by Lemma 45.2, the uniform closure of L" is the space C(B), i.e., L Ais uniformly dense in C(Q). We have, therefore, the following theorem.
=-
Theorem 50.7. I f L has the projection property and e is a strong unit in L, then L A is a uniformly dense Riesz subspace of C(B). If L is Archimedean, but has no weak unit, we consider a disjoint order ' . For each e,, basis (e, : u E { u } ) of L,consisting of nonzero elements in L let UCeml be the corresponding open and closed base set in B. The union
0=
u
(u[ee]
:t~ E {.I)
340
[CH.7,g 50
SPECTRAL REPRESENTATION THEORY
is an open and dense subset of 8. Indeed, if 0 were not dense, the set 8- 0 would contain a non-empty subset of the form Uculfor some u EL'. Then Uculn UCeulis empty for every a, so [inf (u, e,)l =
PI
for every a. It would follows that u Ie, holds for all a, so u = 0, which contradicts the fact that Urulis non-empty. Hence, 0 is dense. Assume now that, for every a, we have U[e,] = A , u B u ,
where A , is an open and dense subset of Ute,,,and B, is the complementary subset. We prove that the union A = A , is dense in 0. If A is not dense in 0, there exists a point ooin the complementary set B = B, with a relatively open neighborhood V of w,, such that V contains no points of A . Note that V is also open in the topology of Sa. The point wo is a point of exactly one B,, say ooE B,,,. The intersection V, of V and U[e,ol is now an open subset of U[e,,l such that V , contains no points of Aa0. This is impossible since A,, is dense in Uceool. Hence, A = A , is dense in 0. Since 0 is dense in Sa, we obtain also the result that A is dense in 8. This is applied now to the following situation. Given f E L, we consider for each a E {a} the function f "(a;e,), defined on Ucemland belonging to Cm(Uce,,).The function f is then defined on 0 = Ucea1in such a manner that, for every a,f "(o;e,) is the restriction o f f ,,to Uceol.This can be done uniquely, since the sets Uceulare mutually disjoint. Denoting, for every a, by A , the subset of Ute,, on whichf ,,(a; e,) is finite, the set A , is open and dense in UreUl.Hence, the set A = A , on which f " is finitevalued is dense in 0, and so A is dense in a. Also, f is evidently continuous on A . The function f can be extended uniquely, therefore, to an extended realvalued continuous function on 8,which we will denote again by f ,. The set LA of all functionsf thus obtained is now a Riesz subspace of Cm(8),and the mappingf f of L onto L A is evidently a Riesz homomorphism. We will show that the mapping is a Riesz isomorphism.
0,
u,
u,
u,
A
u,
A
A
Theorem 50.8. Given the Archimedean Riesz space L wirh disjoint order basis (e, : a E {a]), let L A be the Riesz subspace of Cm(8)as defined above. The following holds. (i) The mapping f +f of L onto L A is a Riesz isomorphism. (ii) The ideal DL generated in C*(8) by L A is a Dedekind completion of L, and Cm(Sa) itserf is a universal completion of L. This shows once more that A
CH.
7 , s 511
THE N A K A N O REPRESENTATION THEOREM
341
every Archimedean space has a Dedekind completion, and it shows also that every Archimedean space has a universal completion. (iii) If L has the projection property, then L A separates the points of Uu U[eol. Proof. (i) We have to show that f ^(a)= 0 for all o impliesf = 0. We may assume that f 2 0. Iff > 0, then u, = inf (f,e,) > 0 for at least one Q, say for Q = ao.On account of 0 < uuo f we have
s
0
s u,h,(w; euo)s f " ( o ;euo)= 0
for all w E Ureuo3, so ubhg(o; euo) = 0 for all w E UCe,,]. It follows from Theorem 49.4that u,, = 0, which contradicts uu0 > 0. Hence, we must have f=O. (ii) Similar to the proofs of Theorem 50.2 and Theorem 50.5. (iii) If w1 , w2 are points of different Uceol,say for (rl and o2, then e: separates the points. If ol,w 2 are points of the same Uceel,the proof is similar to the proof of Theorem 50.6. Exercise 50.9. Show that if L A is the Ogasawara-Maeda representation of the space L = C([O,l]), then L A does not separate the points of Sa. Hint: Observe that the function e in L, satisfying e ( x ) = 1 for all x E [0, I], is a strong unit in L, so e"(w; e ) = 1 holds for all w E Sa. It follows that the uniform norm in L is transferred without change to the uniform norm in LA; hence, LA is norm complete because L is so. Observe also that if L A would separate the points of 8, then L A would be uniformly dense in C(Sa).Combining these facts, it would follow that L A = C(Sa). Finally, note that C(Sa)is Dedekind complete, but L is not. Exercise 50.10. In the preceding example the space L did not even possess the principal projtction property, and so it is not very surprising that L A does not separate the points of Sa. Even if L does have the principal projection property, it can happen that L A does not separate the points of Sa. More precisely, let X be a topological space such that the Riesz space L = Cb(X) of all real bounded continuous functions on Xis Dedekind a-complete, but not Dedekind complete (cf. Exercise 43.15). Show that the corresponding space LAdoes not separate the points of 0. 51. The Nakano representation theorem For Dedekind a-complete Riesz spaces there exists another representation theorem, due to H.Hakano ([3], 1941). This is a theorem similar to the
342
SPECTRAL REPRESENTATION THEORY
[CH. 7,
5 51
Ogasawara-Maeda theorem, and actually the main part of the theorem holds already in spaces possessing only the principal projection property. To begin with, we will even assume that L is an Archimedean Riesz space without any additional properties. As before, we denote by B ( L ) the Boolean algebra of all bands in L, and by B p ( L )the distributive sublattice of all principal bands. The principal band generated by f E L is denoted again by If].Prime ideals in g p ( L )will be denoted by w p ,and the set of all proper prime ideals, equipped with the hull-kernel topology, is denoted by Q,. Let e be a fixed nonzero element in L', and let upbe a prime ideal in S?,(L) such that [el is no member of 0,. In other words, we assume that w pE Urel,where Ucelis the open and compact base set of Sap consisting of all prime ideals of which [el is no member. For any f E L, the extended real number f "(0,;e ) is now defined by
f " ( w p ;e ) = sup ( a : [ ( a e - f ) + ] E up), with the usual conventions for the cases that [(ae-f)'] E w p holds for no finite real u or for all finite real a. Exactly as for the Ogasawara-Maeda situation in section 49, it follows from wp E Ucelthat there can be at most one real t~ satisfying [ ( a e - f ) ] E w,, and so the numberf "(up; e ) satisfies also f "(w,; e )
=
inf ( a : [ ( a e - f ) - ] E 0,).
It can be proved exactly as in Theorem 49.2 that the following holds. (i) f " ( o , ; e ) = 0 for allfe L satisfying L f ] E w,. (ii) e"(w,; e ) = 1. (iii) (pf)"(o,; e ) = pf"(w,; e ) for allfE L and all finite real p. (IV) (f+g)"(w,; e ) = f " ( o , ; e)+g"(w,; e ) for allf, g E L for which the right hand side is well defined. (v) If h = inf (f,g) in L, then hA(wp;e ) = inf ( f "(w,; e), g"(w,; e ) ) . Similarly for sup (f,g). In particular, we have
(f+)"(w,; e ) = sup (f"(w,; e), O), ( f -)^(up; e ) = -inf ( f " ( w p ;e), O), IflA(wp; e ) = If%,; 4 . For e E L' fixed, f E L and upE Ucel,we consider the number f "(up) = f ' (up; e). For f fixed and opvariable, we thus obtain an extended realvalued functionf" on Ute,. Exactly as in Theorem 49.3 it can be proved thatf" is
CH. 7,5 511
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THE NAKANO REPRESENTATION THEOREM
continuous on Ucel.The next thing to observe is that the equivalent of Lemma 49.6 holds, i.e., if 0 < u 5 e and f E [u], then f = 0 holds if and only iff"(w,) = f ^(wp;e ) = 0 holds for all w pE Ucul.We repeat the proof that f "(w,) = 0 for f~ [u] and all 0, E Uculimpliesf = 0. It follows from the definition off " that for all ct > 0 and all w pE Ucu,the band [(cte-f)'] is no member of w,. Assume that (ctOe-f)- # 0 for some cto > 0. Then [ ( a 0 e - f ) - ] is properly larger than [O], so by Theorem 5.2 there exists an element wpE 0,such that [ ( a 0 e - f ) - ] is no member of 0,. On account of
(ctoe-f)- = (f-ctoe)+ 5 f
+
it follows that If'] is no member of w,. Furthermore,fE [u] impliesf' E [u], so [ f ' ] c [u],and hence [u] is no member of w,. We thus obtain the result that there exists an element w p E Uculsuch that neither [ ( c t O e - f ) + ] nor [(ctoe-f)-] is a member of 0,. This is impossible since w p is a prime ideal in a,,. It follows, therefore, that (cte-f)- = 0 for all ct > 0, i.e., cte - f 2 0 for all a > 0. But then we have f S ae for all ct > 0, so f 5 0 (since L is Archimedean). Applying the same argument to -f, we find - f 5 0, and so f = 0. As a particular case (u = e ) we find that iff E [el, then f = 0 holds if and only iff "(a,; e ) = 0 holds for all w pE Ucel. It follows by means of the so obtained result that for anyf E L the function f A ( w p ;e ) is finitevalued on a dense subset of Ucel(exactly as in Theorem 49.7), and so the function f ", defined on Urel,is a member of Cm(Ucel). Summarizing, we have the following theorem.
Theorem 51.1. I f L is Archimedean and e is a nonzero element in L+,the mapping f +f"(w,; e ) is a Riesz homomorphism of L onto a subclass of Crn(UceJ Again, let L be Archimedean, and let (e, : n E {n}) be a disjoint order basis in L, consisting of nonzero elements in L'. For each e,, let Uceulbe the corresponding base set in QP. The union
u
X = (u,eu]: 0 E (01) is now an open and dense subset of 0,(proof as in section 50). Also, if for every t s we have a decomposition Uceul= A, v B,, where A, is open and A, is open and dense in Uceu1and B, is the complementary set, then A = dense in X , and so A is also open and dense in (proof as in section 50). This is applied to the following situation. Given f E L, we consider for each o E {o}the functionf"(o,; e,), defined on UceUl and belonging to Cm(UCeul). The function f ^ is then defined on X = Ucee1in such a manner that, for
u,
u,
344
SPECTRAL REPRESENTATION THEORY
[CH. 7 , s 51
every a,f "(up; e,) is the restriction off " to Uceul.This can be done uniquely since the sets UreUlare mutually disjoint. For each a, the set A , on which f "(w,; e,) is finite is an open and dense subset of Ureul.It follows, therefore, from the remarks above that the set A = A , on whichf " is finite is open and dense in X , and so f " is a member of C""(X).The mapping f +f " is then a Riesz homomorphism of L onto the set L" of all functions f", and LA is a subset of Cm(X).Actually, the mapping is a Riesz isomorphism, as shown by the following theorem.
u,
Theorem 51.2. Given the Archimedean Riesz space L wirh disjoint order basis (e, : c E {a}), let L A be the set of all functions f " on X = Uceul,as defined above. Then every f " is a member of C w ( X ) ,and the mapping f +f" of L onto L A is a Riesz isomorphism.
u,
Proof. We have to show that f " ( w , ) = 0 for all w, E Ximplies f = 0. We may assume that f 2 0. Iff > 0, then u, = inf (f,e,) > 0 for at least one c, say for a = a,. On account of 0 < uu0 5 f we have 0
5 u.^o(wp;eu0)5 f "(a,; e,,)
=0
for all w p E Ureao,, so ub^,(w,; eao)= 0 for all cop E UccUol. On account of uu0E [e,,] it follows that u,, = 0. This contradicts u,, > 0. Hence, we must have f = 0. We will assume now that L has the principal projection property. Then the set B J L ) of all principal bands in L is a Boolean ring (cf. Theorem 48.2), and so (by Theorem 7.5) the hull-kernel topology in 0,is now Hausdorff, and the base sets Uculare open, closed and compact. Furthermore, the collection of all base sets Urulis a Boolean ring with respect to the partial ordering in this collection defined by ordinary set inclusion. The set X = Ureulon which the functions of L A are defined is, therefore, a disjoint union of compact Hausdorff spaces. We will show that L" separates the points of X.
u,
Theorem 51.3. I f L has the principalprojectionproperty, and X = 0, (/reel is the topological space on which the functions of L A are defined, then L A separates the points of X . Proof. If w1 and w2 are points of different sets Uceol,say for c1 and a2, then .:(al) = 1 and e:(w2) = 0, so the points are separated. Assume now that wl and w2 are different points of the same Uceu3, say for a = go. For brevity, we write e, instead of eu0. Hence, [eo] is neither a member of w1 nor of w 2 . It follows from o1# w2 that one of these, say w l , contains a band [u'] for some u' EL' such that [u'] is no member of w 2 . Then u =
CH.
7,g 511
THE NAKANO REPRESENTATION THEOREM
345
inf (u', e,) satisfies [u]E wl, but [u] is no member of o2(since o2is prime). g P ( L )is a Boolean ring, so the complement of [u] with respect to [e,] exists in g P ( L ) let ; [u] be this complement. Since [u] is no member of w 2 , it follows from inf ( [ u ] ,[ u ] ) = [O] that [o] E w2. Since [u] E o, and [e,] = sup ( [ u ] ,[ o ] ) is no member of ol, it follows that [u] is no member of ol. Note that [O] # [u] # [e,] and [O] # [u] # [e,]. Hence, denoting the component of e, in [u] byp, we have 0 # p # e,. It follows now from the facts established that
and so
[(ore,-p)+] = [O]for a < 0, [(ue,-p)+] = [u]forO < a < 1, [(cre,-p)+] = [e,] for a > 1, p A ( o , ) = sup ( a : [(creo-p)+]E ol)= 0, p " ( 0 2 ) = sup ( a : [(ere,-p)+] E w 2 ) = 1.
This shows that o1and w2 are separated. We will prove now that in a space with the principal projection property there exists a one-one correspondence between the points of 52, and the minimal prime ideals. The theorem we prove first is even a little more general. We recall that, in an arbitrary Riesz space, I f ] = A?.
Theorem 51.4. (i) If M is a minimal prime ideal in the Riesz space L and i f f is an element of M , then [ f ] is included in M . (ii) I f the Riesz space L has the quasi principal projection property and P is a proper prime ideal in L, then P is a minimal prime ideal if and only if f E P implies I f ] c P. Proof. (i) Let f E M , where M is a minimal prime ideal in the Riesz space L. Then we have If I E M', and so there exists a nonzero element u in the set theoretic complement L+- M + such that v IIf I holds. The disjoint complement { u } ~is included in M since M is prime. The band [ f ] is included in { u } ~ and , so [ f ] is included in M (compare Exercise 45.9). (ii) Assume that L has the quasi principal projection property and P is a proper prime ideal inL. IfP is minimal, thenfE P implies If]c P by part (i). Conversely, assume that f E P implies I f ] c P. Given f E P, it follows then from I f ] @ I f l d = L that there exists an element v 2 0 in I f l d such that u is no member of P (and so u # 0). Hence, for each f E P there exists an element u E L+-P+ satisfying u 1If I. It follows immediately from this that the lower sublattice L+ -Pf is maximal, and so P is a minimal prime ideal.
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[CH. 7,
8 51
Theorem 51.5. Let L have theprincipalprojectionproperty. To each opE a, assign the subset M,, of L, defined by
Mu, = ( f : I f 1 E 0,). Then M,, is a minimal prime ideal in L, and the mapping op+ Mu, ib a homeomorphism of 8, onto the set A of all minimal prime ideals, when 8, and A are equipped with their hull-kernel topologies. Proof. It is immediately evident that, for every w pE a,, the set Mu,is a prime ideal in L . Given f E M,, and g E [ f ] , we have [ g ] c I f ] , and so g E M o p . It follows, therefore, from f~ M,, that [ f ] c M,,, and so (in view of the preceding theorem) Ma, is a minimal prime ideal. Different op give different Mu,. Indeed, if o1and w2 are different points of a,, then one of these, say wl, contains an element [fO] not contained in 02. Then fo is an element of M,, but not of M,,, and so M,, and M,, are different. Finally, every minimal prime ideal M is of the form Mu,for some w pE 8,. Indeed, given the minimal prime ideal My consider the set o of all If] corresponding to elementsf E M. It is easily seen that o is a prime ideal in a, and the corresponding M , consists of all g EL such that [ g ] = I f ] holds for somefE M. Again by the preceding theorem, [ g ] = [f]for some f E M implies g E M , so M , = M . We have shown thus that the mapping w p -,M,, is one-one from 8, onto A. The subset IY[,,~ of a, consists of all opnot containing [u],so VrUlcorressponds under the mapping with the set { M } , of all M E A not containing u. Hence, o, -+ M,, is a homeomorphism of 8, onto A when both sets are equipped with their hull-kernel topologies. Still under the hypothesis that L has the principal projection property, let (e, : r~ E ( 0 ) ) be a disjoint order basis of L,consisting of nonzero elements in Lf.Under the homeomorphism defined above, the sets X = umUce,l and Y = U , ( M } , u , equipped with their hull-kernel topologies, are homeomorphic. We have the Nakano representation L A of L on X , and on Y we have the Johnson-Kist type representation Z . Given f E L and opE UCeol, the value f "(o,)is given by
f A ( w , ) = sup ( a : [ ( a e , - f ) + ] E a,). If M,, is the point of Y corresponding with w p under the homeomorphism, then M,, E { M } e u ,and the value "f(M,,) is given by
" f ( ~ , , )= sup ( a : (cte.,-f)+
E
Map).
CH.7,s 511
347
THE NAKANO REPRESENTATION THEOREM
Since (ae,-f)+ E Mop holds if and only if [ ( a e , - f ) + ] E O holds, ~ it is evident that we have f"(w,) = "f(Moc), and so the representations L A and "L are not only Riesz isomorphic, but may be considered isomorphic as spaces of extended realvalued continuous functions on homeomorphic topological spaces. There is still another homeomorphism involved. According to Corollary 37.12, if L has the principal projection property and u 0 is given in L, there exists to every M E { M } ua unique prime ideal QME 2"such that QM includes M , and the mapping M + Q , from { M } , onto 9"is a homeomorphism when the sets are equipped with their hull-kernel topologies. Hence, if (e, : o E {o}) is the same disjoint basis as before, the sets Y = U , { M } , a and 2 = Uu2ea, equipped with their hull-kernel topologies, are homeomorphic. We have (as observed above) the representation "L on Y, whereas on 2 we have the Yosida representation, which we will call # L for the present purposes. Given f E L and M E { M } , - , we have
=-
''f(M)
=
sup ( x : ( x e , - f ) + E M ) ,
and if QMis the point of Zcorresponding with M under the homeomorphism, then "f(Q,) = SUP (a : (aeb-f)+ E Q M ) . It follows now from Lemma 44.6 that " f ( M ) = #f(Q,). Hence, taking the composition of the two homeomorphi3ms involved, we obtain the following theorem. Theorem 51.6. If the Riesz space L has the principal projection property, and (e, : c E {c})is a disjoint order basis in L consisting of nonzero elements in L + , then the Nakano representation L" of functions f " on the subset X = U,,U~,,,of Qp and the Yosida representation "L of functions on the of the set 9 of all prime ideals can be identiJied as spaces subset Z = uo9ea of extended realvalued continuous functions on homeomorphic topological spaces. 6 " f
It may be of some interest to quote from K. Yosida's paper ([3], 1942) in which he introduced the representation on 2 = U,Pa, as follows. Recently, by different approaches, H. Nakano and F. Maeda-T. Ogasawara treated a more general case where the existence of an Archimedean unit is not assumed. The purpose of the present note is to show that our method is also applicable to this case as a short-cut to the representation theory. Their representation space is totally disconnected and so their results will not be a direct extension of our preceding note. T. Nakaya'6
348
SPECTRAL REPRESENTATION THEORY
[CH. 7,
8 51
ma, who stressed the applicability of the Lorenzen-Clifford procedure to the representation of the vector lattice, kindly read the manuscript and discussed with me the difference of their method and that of ours. The conclusion may, in short, bz stated as follows. The point of their representation space is perhaps, so to spzak, a minimal prime ideal, while our point is a maximal prime ideal." No further explanation about the last sentence is presented in Yosida's paper. H. Nakano, in his representation thsory, uses a somzwhat different terminology. If L has the principal projection property, the set g,,(L) of all principal bands in L is a Boolean ring, as observed above. For each prime ideal o, in g P ( L ) ,consider the set theoretic complement y p = 99,(L)-wP. According to Theorem 5.4, each set y p is a maximal lower sublattice in 91p(L) and, conversely, every maximal lower sublattice in 99,(L) is one of the y,. Corresponding to the hull-kernel topology in the set 0,of all opthere is a topology in the set rpof all y,; the base sets are the sets of the form (y, : [u] E y,), corresponding to the base sets (0,
: [u] no member of
0,)
in 8,.The spaces a, and r,, equipped with these topologies, are then homeomorphic. Given the fixed nonzero element e in Lf and the maximal lower sublattice y p satisfying [el E y,, Nakano defines the number
to be equal to the numbsr f"(op; e ) defined above for cop = g,(L)- y,. For 7, variable in the set Vrel of all y p satisfying [el E y,, we obtain the functionf" on Vcel. The extension to a union of the form UUVCeml for a disjoint order basis (eu : o E {o})is evident. In Theorem 51.6 it was proved how, for a Riesz spacs with the principal projection property, the Nakano representation and the Yosida representation of the space may bs regarded as identical, i.e., the representations are isomorphic as spaczs of continuous extended realvalued functions on homeomorphic topological spaces. We shall now also indicate in which sense the Nakano representation (in Theorem 51.2) is related to the Ogasawara-Maeda representation (in Theorem 50.8). For a proper understanding of what we have in mind we shall first discuss the problem of determining the Stone representation of a sublattice X of a distributive lattice X (with null element 0) in terms of the Stone representation of X .
CH. 7,s 511
THE NAKANO REPRESENTATION THEOREM
349
Lemma 51.7. Let X be a distributive lattice with null element 8, and X' a sublattice containing 8. Furthermore, let w' be a proper prime ideal in X' and w an ideal in X having the property that w n X'
=
a',
such that w is maximal with respect to this property (i.e., any ideal in X which properly contains w has an intersection with X' properly larger than w'). Then w is a prime ideal in X . Proof. If w is not prime, there exist elements x and y of X such that x and y are no members of w, but x A y is a member of w. The ideal generated in X by w and x is properly larger than w, so this ideal contains an element x' E X ' - w'. The element x' can be written, therefore, in the form x' = p , v x1 with p 1 E w and x1 5 x. Similarly, the ideal generated by w and y contains an element y' E X - w ' , so y' = p 2 v y , with p2 E w and y , S y. Note that XlAY, 5 XAyEO. We set p 3 = p 1 v p , and z' = x' A y'. Then z' E X' - w'(z' E X' holds because x' and y' are in X ' and X ' is a lattice, and z' is no member of w' because w' is a prime ideal in A?). Also, we have z'
so z'
S
x'
6 (P3
S p 3 v x1 and z' 6 y' 5 p 3 v y,, xl) ( P 3
v Y l ) = P3
AYl)
(xl
It follows that z' E w, so that (in view of z' E X ' ) we have z' E w n X ' = 0'. On the other hand we have z' E X' - 0'. Contradiction. Hence, w is a prime ideal in X. Note that the proof is similar to the proof of Theorem 5.1 (ii), where it was shown that an ideal in X , maximal with respect to the property of not containing a given element of X , is a prime ideal. The lemma will be used now to prove the following prime ideal extension theorem.
Theorem 51.8. ( i ) Let X be a distributive lattice with null element 8, and X ' a sublattice containing 0. Furthermore, let w' be a proper prime ideal in X . Then there exists a prime ideal w in X such that
w n X' = w'. The prime ideal w can be taken such that w is maximal with respect to the property that its intersection with X' is w'.
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SPECTRAL REPRESENTATION THEORY
[CH. 7 , s 51
(ii) With the same hypotheses as above, if X ’ is an ideal in X , then w is uniquely determined by 0’. Explicitly, we have now that w = ( p :p E X , p
A
x’ E w’for all x’ E X’).
Proof. (i) Consider the set S of all ideals Z in X such that Z n X’ = a‘. The set S is non-empty because the ideal I,,, generated in X by w‘, is a member of S (indeed, I,. consists of all x E X satisfying x p‘ for some p’ E w’, and so if x E 1,. n X’, then x E a’).Also, S is partially ordered by inclusion. If (Ib: ~7 E {o]) is a chain in S, then the ideal uoZois also a member of S, the least upper bound of the chain. Hence, by Zorn’s lemma, there exists an ideal w in X having the property that w n X ’ = w‘ and such that w is maximal with respect to this property. It follows from the preceding lemma that w is prime. (ii) Now, let X‘ be an ideal in X and w‘ a proper prime ideal in X ’ . We write S = ( p : p E X , p ~ xE’ w‘ for all x‘ EX’). It is easily verified that S is a prime ideal in X such that S n X’ = w‘. Let w be an arbitrary prime ideal in X having the property that w n X’ = w’. For the uniqueness we have to prove that w = S. In order to show first that S is included in w, choose a fixed element xb E XI- w’. Note that x; is, therefore, no member of w . Every p E S satisfies PAX;
Em’C
0,
s o p E w since w is prime and xb is no member of w . This shows that S c w . For the converse, take any p E w . If p is no member of S, there exists an element yb E X‘ such that q = p A yb is no member of 0’. But we have q E w on account ofp E w , and we have q E X’ on account of y ; E X’, so q E w n X ’ =w‘. Contradiction. Hence, p E w implies p E S, i.e., w c S. It follows that w = S. This completes the proof. Note that the present proof of part (ii) does not depend on the result in part (i), and so does not depend on Zorn’s lemma. Once more, let X be a distributive lattice with null element 8, and let X’ be a sublattice containing 8. Proper prime ideals in X and X ’ respectively will be denoted by w and w’ respectively; the set of all proper prime ideals in X will be denoted by D and the set of all proper prime ideals in X’ by a’.The sets D and D’are equipped with their hull-kernel topologies. The base sets of D will be denoted by U, for x E X , where
U, = (w : x no member of w).
CH. 7,8 511
Similarly, the base sets of Q‘ are denoted by Ul for x The subset
351
THE NAKANO REPRESENTATION THEOREM
U:
E X’,
where
= (0’ : x no member of a’). QP
=
(J (U,: X E X ‘ )
of Q is precisely the set of all proper prime ideals in X that intersect X’ in proper prime ideals. Conversely, by the extension theorem, for every proper prime ideal w’ in X’ there exists a proper prime ideal w in X such that w n X ‘ = w’. Hence, if we write
cp(w) = w n X ’ f o r w E W , then cp is a mapping from Q p onto W . In order to investigate the properties of the mapping cp, we first observe that, besides the topology T induced in Q p by the hull-kernel topology, there exists in Q p also the topology z p generated by the base sets of the form U, for sume x E X’. Since Sap is the union of all sets U x ( x E X’ ) and since the sets U,(y E X) form a base for the hull-kernel topology in Q, it is evident that the sets y EX) U , n U, = U X h Y ( x E x’,
form a base for the induced topology z. Hence, every base set for the topology z p is also a base set for z, and so the topology z p is weaker than 7. In general, z p will be properly weaker than z, because in general a set ( I x A y , for x E X’ andy E X , will not be a union of sets Uxowith all xu E X’. However, if X’ is an ideal in A’, then x A y E X‘ for all x E X’ and all y E X , and so T and z p are identical in this case. Actually, this is the only case that 7 and r p are identical. Indeed, assume that z and z p are identical. Then, for any x E X’ and any y E X , the set U,,, is a union of sets Uxbwith all xu E X’. By the compactness of U,,, (i.e., by Theorem 6.7 (i)) the set U,,, is already a finite union of sets U,-,and so U,,, = U, for some z E X’. But then we have x ~ =y z by Theorem 6.5 (i), so X A y E X’ for all x E X’ and all y E X . This shows that X’ is an ideal in X . Hence, we have the following result.
Lemma 51.9. The topologies z and z p in Q p are identical if and only if X’ is an ideal in X. We will prove some properties of the mapping cp. As defined above, cp maps Sap onto W , and cp is one-one if X’ is an ideal in X (because then, by Theorem 51.8 (ii) above, the w in cp(w) = w’ is uniquely determined by
w‘).
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SPECTRAL REPRESENTATION THEORY
Lemma 51.10. The mapping cp of
Qp
[CH.7 , s 51
onto Q' is z,-continuous and r,-open.
Proof. In order to prove that cp is 7,-continuous, we have to show that for any x E X' the set cp-'(U:) is a 7,-open subset of Sap. From the definition of cp it follows that o E cp-'(Ui) if and only if cp(o)E U:. Since x E X', the statement that cp(o)E Ui is equivalent to saying that x is no member of o. It follows that cp-*(~:) = U, for all x E x', and U, is an open base set of the 7,-topology, again on account of x E X'. In order to complete the proof we have to show that if A is a 7,-open subset of Sap, then cp(A) is an open subset of Q'. For this purpose, observe that A = U(U,:xES)forsomeScX'. For every x E X' we have cp( U,) = U: , and so cp(A)= U ( c p ( u , ) : x E S ) = U ( U : : x E S ) ,
which is indeed an open subset of
a.
It is of importance to consider again the particular case that X'is an ideal in X. In this case (as observed above) cp is a one-one mapping, and in addition the 7,-topology in Q p is now identical with the induced hull-kernel topology z. Hence, we immediately derive the following theorem from the last lemma.
Theorem 51.11. If X' is an ideal in X , then the mapping cp of QP onto Q', defined by cp(o)= o n X' for every w E Q p , establishes a homeomorphiAm between Q p and Or,both sets equipped with their hull-kernel topologies. We return to the general case that the sublattice X' is not necessarily an ideal in X . Then the mapping cp is not necessarily one-one, and we introduce an equivalence relation in Q p by saying that the members o1 and o2 of Sap are cp-equivalent whenever cp(o,) = cp(wz). Evidently this is indeed an equivalence relation. The set of cp-equivalence classes will be denoted by Qp/cp, and the mapping of Q p onto Sap/cp, assigning to each element w of Q p the equivalence class to which o belongs, will be denoted by 6. The finest topology in SaP/qfor which 6 is still a r,-continuous mapping is usually referred to as the 7,-quotient topology of B P / qrelative to the cp-equivalence relation. Thus a subset A of O P / qis open in the quotient topology if and only if 6 - ' ( A ) is a z,-open subset of Q p . Concerning the quotient topology
CH. 7.5 511
THE NAKANO REPRESENTATION THEOREM
353
we have the following result, in a certain sense a generalization of the last theorem. Lemma 51.12. The space Qp/qequipped with the quotient topology is
homeomorphic to the space
a’ equipped with the hull-kernel topology.
Proof. We denote the elements of Qp/cp by S(w), o E Q p . Thus S(w = 8(w2) if and only if cp(w,) = q ( w z ) .The one-one mapping $ of QP/qonto Q’ is now defined by $(S(w)) = cp(o) for all o E Q p .
We show first that $ is continuous. To this end, choose x E X’ and consider the base set U: c Q’. We have to show that $-‘(U;) is open. This is not difficult; we have
S-l($-yu;)) = u,,
which is an open set in the r,-topology, so $-‘(U;) is open by the definition of the quotient topology. To complete the proof we have to show that is an open mapping. To this end, let A be an open subset of Qp/cp. We have $(A) = c p ( W - 4 ) ) .
Since by the definition of the quotient topology the set S - ’ ( A ) is r,-open and since the mapping cp is T,-open, the set $ ( A ) is open in the topology of Q’. The proof is thus complete. We shall now compare the Nakano representation and the OgasawaraMaeda representation of a Riesz space. For a space with the projection property we obtain a beautiful and simple result; if the space is only assumed to be Archimedean, there arise some complications. We assume that L is a Riesz space possessing the projection property. The Boolean algebra of all bands in L is denoted by a(L); the subset of g ( L )consisting of all principal bands in L is denoted by Pd,(L>. According to Theorem 48.3 the set a P ( L )is an ideal in g ( L ) . The Stone representation of a(L)is the set Q of all proper prime ideals w in B’(L), equipped with the hull-kernel topology. Similarly, the Stone representation of g,,(L) is the set Q’ of all proper prime ideals o’in a,(L), equipped with the hull-kernel topology. As before, if B is a band in L, the base set U, for the topology in Q is defined by
U s = (o: B no member of o).
354
[CH. 7 , s 51
SPECTRAL REPRESENTATION THEORY
Hence, i f f € L and If]is the band generated byS, then
U,,, = (w : [f] no member of 0). Similarly, the base set Ui,, for the topology in s;)' is defined by Uif1 = (0': [f] no member of
0').
Finally, the subset Sap of Sa is defined by
BP = (j (U,,]: f E L ) . In Sap we have two topologies, the topology T induced by the hull-kernel topology in s;) and the topology z p generated by the base sets U,,,( f e L ) . It follows from Lemma 51.9 that the topologies z and z p are identical (because a P ( L )is an ideal in g(L)). The mapping cp from QP onto s;)' is defined by cp(w) = w n g p ( L )for o E Sap.
According to Theorem 51.1 1 the mapping cp establishes a homeomorphism between BPand a', both sets equipped with their hull-kernel topologies. In order to introduce the Ogasawara-Maeda representation and the Nakano representation of L, let (e, : CT E { c T } )be a disjoint order basis of L, consisting of nonzero elements of L + . Then
Y
=
u
(U[eu]: IJ E
(GI)
is an open and dense subset of !2, and so Y is surely an open and dense subset of Sap. Also, X = (Uiee] : 0 E ( 0 ) )
u
is an open and dense subset of 8'. Furthermore, cp maps each Uceulonto the corresponding Uieul, and so Y and X are homeomorphic under the mapping cp. As explained in sections 49 and 50, the Ogasawara-Maeda representation LGM of L is a Riesz subspace of Cm(s;)).IffA is the function in L& corresponding to any given member f of L, we may just as well consider the restrictionf; off" to Y , sincef'' on Q is completely determined by its restrictionf;, We denote the set of allf; again by LgM.It follows that the mapping f -+ fy^ is a Riesz isomorphism between L and the Riesz subspace L G M of Cm(Y). GivenfE L and the point coo E Y, the point wo belongs to exactly one of the Ute,] , say for IJ = ; the valuef"(wo) is now given by
fA(%)
= SUP (. : [(.ee,-f)+
I E00).
CH. 7 , 5 511
355
THE NAKANO REPRESENTATION THEOREM
The Nakano representation L c of L is a Riesz subspace of Cm(X). More precisely, iff; is the function in LG corresponding to a given member f of L, then the mapping f -S f : is a Riesz isomorphism between L and the Riesz subspace L i of Cm(X). Given f E L and the point wb EX, the point ob belongs to exactly one of the U ; , , , say for CT = a,; the value f $(cob) is now given by
f$(ob)= sup (ci : [(ae,,-f)+] E cob). It follows immediately that if ooand wb are points of Y and X respectively, corresponding under the homeomorphism between Y and X (i.e., oh = oon g p ( L ) ) then , f"(o0) =
fiv^(ob).
Hence, the Ogasawara-Maeda representation and the Nakano representation of L may be regarded as identical, i.e., the representations are isomorphic as spaces of continuous extended realvalued functions on homeomorphic topological spaces. In view of Theorem 51.6 the same is true for the Nakano representation and the Yosida representation (here it is already sufficient that L possesses the principal projection property). Hence, we obtain the following final result.
Theorem 51.13. If the Riesz space L has the projection property, and (e, : CT E {cT})is a disjoint order basis in L consisting of nonzero elements of L', then the Nakano representation L c of functions f ; on the subset X = u,U;e,l of Sa', the Yosida representation "L of functions "f on the subset Z = uaL2ea of the set B of all proper prime ideals in L, and the OgasawaraMaeda representation L& of functi0ns.f" on the subset Y = Ur,,, of Q can be identifed as spaces of continuous extended realvalued functions on homeomorphic topological spaces.
ub
If we assume only that L is Archimedean, it is still possible to compare the Nakano representation and the Ogasawara-Maeda representation. We introduce W(L),g , ( L ) , Sa, Sa' and the subset Sap of Sa exactly as above. The set a , ( L ) is a sublattice of g ( L ) , but not necessarily an ideal in B(L). Hence, the topologies 7 and 7, in Bp are now not necessarily the same; 7, may be properly weaker than 7. Also, the mapping cp from Q p onto Sa' is not necessarily one-one. As explained in Lemma 51.12, the space QP/qequipped with the 7,-quotient topology relative to cp is now homeomorphic to Q' equipped with the ordinary hull-kernel topology. Thus, for Y and X defined as above, the spaces Y/cp and X are homeomorphic. It is of importance here to note that Y is saturated
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SPECTRAL REPRESENTATION THEORY
[CH.7 , s 51
for the cp-equivalence relation, i.e., if o1E Y and w2 is a point of Q p such that ~(0,) = cp(o,), then w2 E Y. Indeed, w1 E Y implies w1 E UCe,, for exactly one a E (a}, and so q ( w l )E U;,,,for that (r. But then q ( w 2 ) eU i e u j , so 0 2 E qe,, c Y. The Ogasawara-Maeda representation consists of functions f y^ defined on Y, and each f y ^ has the property that f p ( w l ) = f p ( w 2 ) holds for cpequivalent points o1and w 2 . Indeed, if o1(and hence also w 2 ) belongs to Ure,, for a certain 6,then which is equivalent to
holds for all f E L if cp(ol)= cp(oz). But It follows that fp(wl)= fp(o2) then we can define for any f E L a function f ^ / c p on Y/cp by means of the formula (f^/q)(S(w))=fc(o) for all w E Y, where 6 denotes the quotient mapping from I’ onto Y/cp. The functions f^/cp are continuous extended realvalued functions on Y / q (this includes
the statement that each f ” / q is finite on an open and dense subset of Y / q ) . The set of all f ^ / q is thus a Riesz subspace of C m ( Y / q ) ;we shall say that this Riesz subspace is the modified Ogasawara-Maeda representation of L. It follows now easily that if 6(0) and w’ are points of Y/cp and Xrespectively, corresponding under the homeomorphism between Y/cp and X , then holds for every f E L , and so the Nakano representation and the modified Ogasawara-Maeda representation of L can be regarded as identical, i.e., they can be identified as spaces of continuous extended realvalued functions on homeomorphic topological spaces.
Exercise 51.14. Let X be a distributive lattice with null element 8, and X ’ a sublattice containing 8. Let Q and SZ‘ be the Stone representations of X and X respectively, and let Sap be the subset of SZ, defined by sZp = u ( U x : x E X ’ ) . The set Sap is equippecl with the topology T~ generated by the base sets U, for x E X ’ . As explained in the paragraphs which precede Lemma 5 1.9, the mapping cp:w+onX’
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THE NAKANO REPRESENTATION THEOREM
357
maps f-2, onto Q'. For o1and w2 points of Sap, prove that cp(ol)c cp(02) if and only if w2 is in the r,-closure of the set consisting of the point ol, i.e., if and only if o2E 7,-closure { w l } . In particular, o1and o2 are cpequivalent if and only if
o1E 2,-closure { w 2 } and o2E 7,-closure {ol}, i.e., if and only if {ol} and {a2}have the same 2,-closure. Hint: Write cp(o,) = w; for i = 1, 2. Then 0;c o; if and only if for all x E X' such that w; E Ui we have w ; E U i , i.e., if and only if for all x E X' such that o2E U, we have o1E U,. This is equivalent to the statement that every 7,-neighborhood of o2contains ol,i.e., o2is in the 5,closure of ol.
Exercise 51.15. Let X be a Boolean algebra with null element 8 and unit element e, and let X' be a Boolean subalgebra of X with the same 8 and e. The Stone representations of X and X', equipped with their hull-kernel topologies T and TI, are denoted by Q and Q' respectively. The base sets U, (x E X ) of T are open, closed and compact (Theorem 7.5); similarly for the base sets U ~ ( X E X of ' ) 7'. Note that U, = Q and U : = Q'. Note also that the subset QP= U(Ux:xEX) of Sa satisfies Q p = U, = Q. As in Lemma 51.10 we introduce the mapping cp from Q = Sap onto Q', defined by q(o)= o n X'forwEQ = QP, and in Q = Op we introduce the topology T,, generated by the base sets U, for x E X'. By Lemma 51.10 the mapping cp is 7,-continuous and z,-open. Show that cp is 7,-closed. Hint: In the topology 7, the base sets U, (x E X ' ) are open, closed and compact. In particular, Q = U, is 7,-compact. Hence, any r,-closed subset F of 52 is 7,-compact, and so the image cp(F) is a r'-compact subset of a' (since cp is a continuous mapping). But the topology 7' is Hausdorff, so cp(F) is closed.
Exercise 51.16. Let X be the same as in the preceding exercise, and let X' be the Boolean subalgebra consisting of 0 and e. Show that a' consists of only one point, and the topology z, in Q has only the empty set and Q itself as open sets. Hence, T, is no Hausdorff topology as soon as X has at least one element different from 8 and e.
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SPECTRAL REPRESENTATION THEORY
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Hint: The topology z p fails to be Hausdorff as soon as Sa has at least two different points, i.e., as soon as X has at least two different proper prime ideals. If X has an element x such that 8 # x # e, then { 6 } is no prime ideal. The intersection of all proper prime ideals is { 6 } , so there must be at least two different proper prime ideals.
Exercise 51.17. Let X be a Dedekind complete Boolean algebra and let
Sa be the Stone representation of X , equipped with the hull-kernel topology. Note that SZ is an extremally disconnected compact Hausdorff space by
Theorem 47.5, and so Cm(D) is a Dedekind complete and universally complete Riesz space by Theorem 47.4. The point w E B is called bounded whenever f(w) is finite for every f E Cm(B). Show that w E Sa is bounded if and only if w is a P-point. We recall that w is said to be a P-point whenever every countable intersection of neighborhoods of w is a neighborhood of w (cf. Theorem 34.2). Hint: If wo is a P-point such that f(wo) = co for some f E Cm(Sa), then f ( w ) = + 00 for all w satisfying
+
w
E
D =
n
m
n= 1
(W
: f ( ~>) n).
But D is a neighborhood of wo, so f is infinite on a set containing a nonempty open set. This is impossible on account off E Cm(SZ). Conversely, let wo be bounded and assume wo is no P-point, so there exists a sequence (D, : n = 1,2, . .) of neighborhoods of oo such that n n D nhas no non-empty open subset. Each D, contains a base set 0, such that wo E 0, c D,, and so F = is a closed set containing wo such that F has no non-empty open subset. We may assume that 0, is decreasing as n increases. For each n, the characteristic function f, of 0, is continuous (since 0, is open and closed). Setf(w) = fn(o)for all w E D.At each point in the open and dense complement F C of F the function f is finite and continuous. Hence by Theorem 47.1, the restriction off on Fc can be ex tended uniquely to a function SE Cm(Sa), and it follows easily from the construction of 1 that, actually, 1 = f on the whole of D. Hence, we have f E Cm(Q)andf(wo) = co. This is impossible because wo is bounded.
.
0.0,
I,"=
+
52. Prime ideal extension in a Riesz space In Theorem 51.8 we proved an extension theorem for proper prime ideals in a distributive lattice with null element. There exists a similar extension
CH.7, 521
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PRIME IDEAL EXTENSION I N A RIESZ SPACE
theorem for proper prime ideals in a Riesz space. We first prove a lemma similar to Lemma 51.7.
Lemma 52.1. Let K be a Riesz subspace of the Riesz space L, and let Q be a proper prime ideal in K. Furthermore, let P be an ideal in L having the property that PnK=Q,
and such that P is maximal with respect to this property (i.e., any ideal in L which properly contains P has an intersection with Kproperly larger than Q). Then P is a prime ideal in L. Proof. If P is not prime in L, there exist elements g and h in Lf such that g and h are no members of P, but inf (9,h ) is a member of P. The ideal generated by P and g is properly larger than P, so this ideal contains an element foE K - Q. The element fo can be written, therefore, in the form fo = p 1 +gl with p 1 E P and g1 E I, (where Z, is the ideal generated in L by g). Similarly,the ideal generated by P and h contains an elementfb E K- Q, so fd = p 2 +hl with p z E P and hl E I,,. Note that inf (9,h ) E P implies inf (Isl!, lhll) E P. We set p3 = SUP (IPII, lp21i and
f6' = inf(Ifol, IfJ).
Since I fol and lfdl are members of K- Q, and since Q is a prime ideal in K, we have fg E K-Q. Also, we have
0 Sf6'
so
5 If01 = lP1+911 5 P3+191L 0 Sf6' 5 IfJ = IP2+hll 5 P3+Ihll, 0 gfd' 5 inf(P,+lgil, P3+Ihil) = P3+inf(Igil,
lhil)Ep-
It follows that ft E P, so that (in view off: E K) we have f EP n K = Q. On the other hand, as shown above, we have ft E K-Q. Contradiction. Hence, P is a prime ideal in L. The lemma will be used to prove the following extension theorem.
Theorem 52.2. (i) Let K be a Riesz subspace of the Riesz space L, and let Q be a proper prime ideal in K. Then there exists a prime ideal P in L such that PnK=Q.
The prime ideal P can be taken such that P is maximal with respect to the property thar its intersection with K is Q .
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[CH.7,
52
(ii) With the same hypotheses as above, if K is an ideal in L, then P is uniGwely determined by Q . Explicitly,
P+
=
(u:O S uEL,inf(u,u)EQ+foraNvEK').
Hence, in the special case that (0) is aproper prime ideal in the ideal K, the unique prime ideal P in L satisfying P n K = ( 0 ) is P = K d .
Proof. (i) Consider the set S of all ideals Z in L such that I n K = Q . The set S is non-empty because the ideal Zp,generated in L by Q , is a member of S (indeed, Ze consists of all f E L satisfying If I S q for some q E Q f , and so i f f € Zp n K , then f~ Q ) . Also, S is partially ordered by inclusion. If (Zc: a E { c T ) ) is a chain in S, then the ideal udZd is also a member of S, the least upper bound of the chain. Hence, by Zorn's lemma, there exists an ideal P in L having the property that P n K = Q , and such that P is maximal with respect to this property. It follows from the preceding lemma that P is prime in L. (ii) Now, let K be an ideal in L and Q a proper prime ideal in K. We write S
= (u
: 0 5 u E L, inf (u, v ) E Q + for all v E K + ) .
It is easily verified that S is the positive part of a prime ideal Po in L such that Po n K = Q . Let P be an arbitrary prime ideal in L having the propxty that P n K = Q . For the uniqueness we have to prove that P = P o , i.e., we have to prove that Pf = S. In order to show first that S is included in P', choose a fixed element vo E K + - Q + . Note that oo is, therefore, no member of P. Every u E S satisfies inf (u, v o ) E Q + c P, so u E P since P is prime and vo is no member of P.This shows that S c P '. For the converse, take any u E Pf.If U is no member of S, there exists an element vo E K + such that w = inf (u, v o ) is no member of Q'. But we have w E P+ on account of 0 S w u E P', and we have w E K + on acvo E K + , so W E P f n K f = Q + . Contradiction. count of 0 5 w Hence, U E P f implies U E S, i.e., P + c S. It follows that P f = S. This completes the proof. Note that the present proof of part (ii) does not depend on the result in part (i), and so does not depend on Zorn's lemma.
Independently of the theorem already proved, we will show now that a minimal prime ideal in the Riesz subspace K can b: extended to a minimal prime ideal in L.
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Theorem 52.3. Let K be a Riesz subspace of the Riesz space L, and let Q be a minimal prime ideal in K. Then there exists a minimal prime ideal P iri L such that P n K = Q . Proof. The theorem is trivially true for K = { 0 } , so we may assume that K # {0}, and then the given Q is a proper ideal in K . The non-empty set K + - Q + is now a lower sublattice in K + as well as in L + , and since Q is a minimal prime ideal in K , the lower sublattice K + - Q + is maximal in K', i.e., for any u E Q + there exists an element u E K + - Q + such that inf (u, u ) = 0. Any lower sublattice in L+ is contained in a maximal lower sublattice; let A be a maximal lower sublattice in L + containing K + - Q + . Note that the intersection A n K + is equal to K + - Q + , and not larger. Hence, P+ = L+ - A is the positive part of a minimal prime ideal P in L , and evidently we have P n K = Q . It is possible to derive the general prime ideal extension theorem (Theorem 52.2) from the extension theorem for minimal prime ideals as follows. Let, once more, Q be a proper prime ideal in the Riesz subspace K of the Riesz space L . We have to prove the existence of a prime ideal P in L such that P n K = Q . To this end, let ZQ b: the ideal generated in L by Q , and let L' = L/IQ. As on earlier occasions, the element of the quotient space L' that containsfE L is denoted by [f]. Thz set ([f]: fK~) is a Riesz subspace K' of L', and { [ O ] } is now a prime ideal in K'. Obviously, { [ O ] ) is then a minimal prime ideal in K'. It follows that there exists a minimal prime ideal P' in L' such that P' n K' = [O]. Writing
P
=
(f: f E L , Lf] E P'),
it is easily verified that P is a prime ideal in L , and the intersection P n X satisfies P n K = Q . The prime ideal extension is monotone in the sense explained in the following theorem.
Theorem 52.4. Let K be a Riesz subspace of the Riesz space L,and let Q , and Q , be proper prime ideals in K such that Q c Q , . Then there exist prime ideals P , and P, in L such that P , andP, intersect K i n Q , and Q,respectiuely and such that P , c P, . Hence, P , and P, are proper prime ideals in L,and i f Q , is properly included in Q , , then P , is also properly included in P2.
,
Proof. Let Pibe a prime ideal in L satisfying P, n K = Q , , and let now L' = LIP,. As on earlier occasions, the element of the quotient space L' which contains f E L is denoted by Lf]. The set (Lf] :f E K ) is a Riesz sub-
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[CH.7,g 52
SPECTRAL REPRESENTATION THEORY
space K' of L',and { [O]} is now a prime ideal in K' (as well as in L'). Also, Q; = ([f] :f E Q,) is a prime ideal in K', so thereexists a prime ideal Pi in L' such that Pi n K' = Q ; . Writing now
P,
= (f : f € L, [f]E P i ) ,
it is easily verified that Pzis a prime ideal in L such that P, =
Qz-
3
P, and P, n K
Corollary 52.5. (i) Let K be a Riesz subspace of the Riesz space L , and let P be a maximal ideal in L such that K is not completely contained in P. Then the intersection Q = P n K is a maximal ideal in K. (ii) Once more, let K be a Riesz subspace of the Riesz space L , and let P be a minimal prime ideal in L such thar K is not completely contained in P. Then the intersection Q = P n K is a minimal prime ideal in K. Proof. (i) The intersection Q = P n K is a proper prime ideal in K. Assume that there exists a proper ideal Q , in K which properly includes Q. Then Q, is a proper prime ideal in K, hence, by the above theorem, there exists a proper prime ideal P, in L which properly includes P.This is impossible since P is a maximal ideal in L . Thus there exists no such Q, ,and so Q is a maximal ideal in K. (ii) Similarly. In the converse direction, we may ask whether a maximal ideal in Kcan always be extended to a maximal ideal in L. This is not so, as shown by the following example. Example 52.6. Let (X, A, p) be a measure space such that 0 < p ( X ) < 00 and p has no atoms, and let L = L , ( X , p) be the corresponding space of all real p-summable functions (with identification of palmost equal functions). As explained in Example 11.2 (vi), the space L , is a Riesz space with respect to the partial ordering according to which f 5 g means by definition that f ( x ) 5 g ( x ) holds for palmost every x E X. It was proved in Example 27.8 that the Riesz space L has no maximal ideals. The subset K of all (p-essentially) bounded functions in L isanideal in L , , and the function e, satisfying e(x) = 1 for all x E X, is a strong unit in K. Heace, by Theorem 27.4, K has maximal ideals, and each maximal ideal Q in K has a unique extension to a proper prime ideal P in L , . The prime ideal P is no maximal ideal in L , however, because L possesses no maximal ideals. For minimal prime ideals the situation ismore pleasant; accordingtoTheo-
,
,
,
,
,
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PRIME IDEAL EXTENSION IN A RIESZ SPACE
363
rem 52.3 any minimal prime ideal in K can be extended tc a minimal prime ideal in L. The prime ideal extension theorem (is., part (i) of Theorem 52.2) was proved by I. Amemiya (Theorem 1.2 in [2], 1953), although the theorem is not immediately recognizable in Amemiya’s paper because of the terminology in terms of spectral functions instead of prime ideals (cf. Exercise 37.28 for the relation between spectral functions and prime ideals). It may be asked whether there exists a similar extension theorem for algebraic prime ideals in commutative rings. The answer is affirmative. There is a theorem, due to I. S. Cohen and A. Seidenberg ([l], 19461, as follows. Let S be a subring of the commutative ring R such that R is integrally dependent upon s,i.e., for every x E R there exists a natural number n and elements a,, . . ., a,,-l, in S such that xn+un-lxfl-l+
.. . + a ,
=
0.
Then, for any proper prime ideal Q in the subring S, there exists a prime ideal P in R such thaL P n S = Q. For Ln indication of the proof we refer to Exercise 52.14. If the subring S is an ideal in R,then the extension P is uniquely determined by Q, but this is not mentioned in the Cohen-Seidenberg paper because in this paper the extra hypothesis is made that the ring R has a unit element contained also in the subring S, and so if S is an idbal in R, then S and R coincide and all extension theorems are reduced to trivialities. That some cxtra hypothesis on the subring S (such as the integral dependence of R upon S) is necessary to make the extension work, has probably to do with the fact that if S is an (algebraic) ideal in the commutative ring R, and S1 is an ideal in S, then S , is not necessarily an ideal in R. For order ideals in a Riesz space it is immediately evident from the definitions that any ideal Zlin an ideal Z in the Riesz space L is an ideal in L. Once again, let K be a Riesz subspace of the Riesz space L.The set of all proper prime ideals in L, equipped with the hull-kernel topology, will be denoted by P ( L ) , and the notation B ( K ) is introduced similarly. The base sets for B ( L ) are the sets
{P},= (P: P E P ( L ) , fnot in P) forf an element of L+;similarly, the base sets for B ( K ) are the sets
{Q}, = (Q : Q ~ B ( K ) , f n o tin Q) for f an element in K’. We introduce the open subset
{P>K=
u ({PI/
:fE K + )
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SPECTRAL. REPRESENTATION THEORY
[CH.7,s 52
ofB(L). The set { P } Khas two topologies, the topology T induced by the hullkernel topology in B ( L ) and the topology z p generated by the base sets of the form {P},for f E K'. Since { P } Kis the union of all sets {P}, for f E K + and since the sets {P}e ( g EL') form a base for the hull-kernel topology in B(L), it is evident that the sets
{PI, n {Pie = { P I i n f (1,e)
(.fE
K + ,9 EL+)
form a base for the induced topology 7. Hence, every base set for the topology rP is also a base set for z, and so the topology z p is weaker than the topology z. In general, z p will be properly weaker than z, because in general a for f E K + and g EL', will not be a union of sets set of the form {P}inf(f,H), {P},, with all f,E K + . However, if K is an ideal in L, then inf (f,g ) E K' holds for all f E K' and all g E L + ,and so z and z p are identical in this case. Actually, it is easy to verify that z and z p are identical if and only if for any f E K' and any g EL' there exists an element h E K' such that inf (f,g ) and h generate the same ideal in L. Indeed, let z and z p be identical. Then, for any f E K' and any g E L + ,the set {P}i,f,,( is a union of sets {P],, with all f , E K'. By the compactness of {P}inf(f,e) (i.e., by Theorem 35.7 (i)) the (f, is already a finite union of sets {P},,, and so set {P},,, {Plinf(,,g)
= {PIh
for some h E K'. But then inf (f,g ) and h generate the same ideal in L by Theorem 35.5 (i). We state the result as a separate lemma.
Lemma 52.7. The topologies z and z p in { P } Kare identical ifand only iffor any f E K' and any g E L+ there exists an element h E K' such that inf (f,g ) and h generate the same ideal in L. The subset {PIK of B ( L ) is precisely the set of all proper prime ideals in L that intersect Kin a proper prime ideal. Conversely, for every proper prime ideal Q in K there exists a proper prime ideal P in L such that P n K = Q. Hence, if we write p(P) = PnKforPE{P}K, then q isamappingfrom{P}, onto B ( K ) . Completelysimilar to Lemma 51.10 and Theorem 51.11 we have the following theorem.
Theorem 52.8. The mapping cp of { P } Konto B ( K ) is z,-continuous and zpopen. If K is an ideal in L , then cp is one-one, and cp establishes now a homeomorphism between { P } KandB(K), both sets equipped with their hull-kernel topologies.
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PRIME IDEAL EXTENSION I N A RIESZ SPACE
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We introduce an equivalence relation in {P}Kby saying that the members P, and P, of {P}K are cp-equivalent whenever p(P1) = p(P2). The set of cp-equivalence classes is denoted by {P}K/rp,and 6 denotes the mapping from {P}Konto {P},/p assigning to each P E {P}Kits p-equivalence class. The set {P},/v is equipped with the 7,-quotient topology relative to the pequivalencerelation. Also, {P},/cp is mapped in a one-one manner onto 9 ( K ) by the mapping $, where $ is defined by $(d(p))
=
q(p) f o r P E {P}K*
Similarly as in Lemma 51.12 of the preceding section, we have the following result. Lemma 52.9. The space {P}K/pequipped with fhe T,-quotient topology is homeomorphic to the space 9 ( K ) equipped with the hull-kernel topology. It was proved already earlier in the present section that if K is an ideal in L,then the topologies z and 7, in {P}Kare identical, and also the mapping rp from {P}Konto B ( K ) ,defined by p(P) = P n K for all PE{P}K,IS ' one-one. We shall prove now as a final result that, generally, as soon as 7 and 7 , are the same, then cp is one-one. For this purpose we first introduce a topological formulation for the cp-equivalencerelation, similar to the formulation in Exercise 51.14. Lemma 52.10. For any twopoinrs PI andP, of {P}K we have p(Pl) c cp(P2) if and only ifP, is contained in the z,-closure of the set consisting of the point P, ,i.e., ifandonly ifP, E 7,-closure {P,}. In particular, P, andP, are cp-equivalent if and only if P, E z,-closure {P,} and P, E 7,-closure {P,},
i.e., ifand only if{P,} and {P,} have the same 7,-closure.
Proof. Write cp(P,) = Qi for i = 1,2. Then Q, c Q, if and only if for all u E K + such that Q2E { Q } . we have Q, E {Q}.,i.e., if and only if for all u E K + such that P, E {P}. we have P, E {P}". This is equivalent to the statement that every 7,-neighborhood of P, contains PI , i.e., P, is in the ~,-closure of P, . Theorem 52.11. If K is a Riesz subspace of the Riesz space L such that the 7,-topology in {P}K is identical with the topology z induced in {P}x by the hull-kernel topology of B ( L ) , then there exists for every proper prime ideal Q in K a unique prime ideal P in L such that Q = P n K, i.e., the mapping cp is one-one.
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SPECTRAL REPRESENTATION THEORY
[CH. 7,@52
Proof. We shall use that the 2-closure of a subset D of { P } Kis equal to the intersection of {P}K and the closure of D in the hull-kernel topology ofg(L). Hence, by the preceding lemma, assuming now that 2 and z p are identical, we have for the points P, and P, of { P } K that q ( P 1 ) c q ( P z ) holds if and only if P, is contained in the intersection of { P } Kand the hull-kernel closure of {P,}.The condition that P, is contained in { P } Kis always satisfied. By Theorem 36.6, the condition that P, is contained in the hull-kernel closure of {P,}is satisfied if and only if P, c P,. Hence, we have cp(Pi) c cp(P,) if and only if P, c P, holds. This shows that P, and P, are pequivalent if and only if P1 = P,,that is, if and only if cp is one-one. Exercise 52.12. (i) Let K be a Riesz subspace of the Riesz space L , and let Q , and Q be ideals in K such that Q, c Q.Furthermore, let P, be an ideal in L such that P, n K = Q , ,and let Zp be the ideal generated in L by Q. Finally, let Po be the ideal generated in L by P, and Zp , so Po is the algebraic sum of P, and Zp. Show that Po n K = Q . (ii) We indicate a proof of Theorem 52.4 without using quotient spaces. Hence, let Q , and Q be proper prime ideals in the Riesz subspace K of L such that Q , c Q. We have to show the existence of prime ideal extensions P, and P satisfyingP, c P. Let P, be any prime ideal in L such that P, n K = Q , . Now, let S be the set of all ideals I in L such that Z 3 P, and Q , c Zn K c Q .
Note that P, is a member of S. By Zorn’s lemma the set S has a maximal element P. Show by means of part (i) that P n K = Q . It follows that P is a prime ideal in L possessing the required properties. Hint: For (i), denote the intersection Po n K by Q , . It is evident that Q, 2 Q , so we have to show that Q , c Q . Any foE Q,’ is of the form fo = f,+f with 0 f,E P, and 0 5 f~ Zp (cf. Theorem 17.6 (i)), but it is conceivable thatf, andfare no members of K. However,fis majorized by an element f’ E Q, and we may assume that f 5 f’ 5 fo (because iff’ 5 fo is not satisfied, WL replacef’ by inf (f’,f,)).Writing now f; = fo we havef, = f;+f’with 0 6 f’ E Q and 0 g f; 5 f,E P,,sof; E P, .Note also that fo andf’ are in K, so f; is in K. Hence
s
-r,
.f; E P1n K = Q1 c Q .
It follows thatf, = f;+f’ E Q , so Q , c Q.
Exercise 52.13. Let Z be an ideal in the Riesz space L . Show that the set g(L/Z) of all proper prime ideals in L/Z, equipped with the hull-kernel to-
CH.7 , s 521
PRIME IDEAL EXTENSION IN A RIESZ SPACE
367
pology, is homeomorphic to the set theoretic difference D = B ( L ) - { P } , , where D is equipped with the topology induced by the hull-kernel topology in P(L)* Hint: Note that there is a one-one correspondence between the proper prime ideals in LIZ and the proper prime ideals in L which contain Z. Also note that { P } , c { P } Iif and only iff E I, so the induced topology in D isdetermined by the base sets { P } , n D for f not in Z. Also, note that { P } , n D = { P } gn D if f-g E Z .
Exercise 52.14. We indicate a proof of the Cohen-Seidenberg extension theorem for algebraic prime ideals. Let S be a subring of the commutative ring R such that R is integrally dependent upon S, i.e., for every x E R there exists a natural number n and elements a o , . . ., a,,- in S such that X”+a,-,x”-’+
... +a,
= 0.
(i) Let Q be a proper prime ideal in S, and let P be an ideal in R with the property that P n S c Q and such that P is maximal with respect to this property (that is, any ideal in R which properly contains P has an intersection with S containing an element of S - Q ) . Show that P n S = Q and P is a prime ideal in R. Show that the prime ideal extension theorem follows immediately. (ii) Let Q , and Q , be proper prime ideals in S such that Q , c Q , . Then there exist prime ideals P , and P, in R such that P , c P , , and P1 and P2 intersect S i n Q , and Q , respectively. Show this. Hint: For (i), assume P n S # Q , so P n S properly contained in Q . Then there exists an element x E Q - P . The ideal generated by P and x is properly larger than P , so its intersection with S contains an element y E S- Q . The element y is of the form y =p+rx+m
Thus we have
( p E P, r E R , m an integer). rx
There exist a, , . . ., a,-
so
-x
= y-mx(P).
,in S such that
r “ + a , - , r”-’ +
. . . +ao
(rx)”+a,-,(rx)”-’x+
=
0,
. . . +sox" = 0.
Hence, in view of (I), (y-mx)”+a,_,(y-mx)”-’x+
. . . +a,x” = O(P).
368
SPECTRAL REPRESENTATION THEORY
[CH. 7, 52
The left hand side, therefore, is an element of P.However, the left hand side is also an element of S (since x and y and all coefficients are in S). Hence, the left hand side is in P n S c Q. Also, since x E Q and Q is an ideal in S, we have u,,-,(~-~x)"-'x+
... + u ~ x " E Q ,
so (y-mx)" E Q. But Q is prime in S, so y - m x E Q. Contradiction, because x E Q, but y is no member of Q. It follows that P n S = Q. Now, assume P not prime in R. Then there exist x, y E R such that x and y are no members of P,but xy E P. The ideal generated by P and x contains
an element zoE S - P = S - Q , so zo = P1 +x,
i.e., x1 = r , x + n x (rl z& E S - Q such that
E
(P1 E p , x1 E ZJ9
R,n an integer). Similarly there exists an element
z& = PZ+YZ
(Pz E PY Y z E l y ) ,
i.e., y z = rzy+my (r2 E R, m an integer). Note that so
We have also zoz& E S, so z0zA E P n S = Q. Since Q is prime in S, this implies that one at least of zoand z; is in Q . Contradiction. For (ii), there is a proof similar to the proof for part (ii) in Exercise 52.12.
Exercise 52.15. Show that if S is an ideal in the commutative ring R, and Q is a proper prime ideal in S, tben there is a unique prime ideal P in R such that P n S = Q. Explicitly, show that
P
=
(x : x E R, xy
EQ
for all y
E S).
Note that integral dependence of R upon S is not necessary.
CHAPTER 8
Hermitian Operators in Hilbert Space
Given the Hilbert space H (over the complex numbers, finite dimensional or infinite dimensional; the elements are denoted by x, y, . . and the inner product of x and y is denoted by ( x , y)), the set 2 of all bounded Hermitian operators in H i s a real vector space, partially ordered by defining that A 5 B holds for A and B in 2 whenever (Ax, x ) S (Bx, x ) holds for all x E H . The set A? is not a Riesz space, except in the trivial case that the Hilbert space H i s one-dimensional. We will prove that A? has a property analogous to Dedekind completeness, namely, if (A, : z E {T}) is an upwards directed subset of A? which is bounded from above, then A = sup A, exists in A?. We denote the positive cone of 2 by 2'. It is well-known that every A E 2' has a unique square root in 2 +i.e., , there exists a unique B E A?' satisfying B 2 = A . Denoting, for any A E 2,the square root of A' by IAI, we set
.
A+
=
3(A+IA/),
and we will prove that, among all B E 2 commuting with A and satisfying B L A and B 2 0 (where 0 is the null element of @), the element A + is the smallest. We believe the proof is a little simpler than usually presented. Given the subset 9of 2 such that any two members of 9commute, the commutant W ( 9 ) of 9 is by definition the set of all B E 2 commuting with every A E 9, and the second commutant W'(9) is the commutant of %'(9).We will prove that V'(9 )is a Dedekind complete Riesz space, super Dedekind complete if the Hilbert space H i s separable. The classical spectral theorem for a bounded Hermitian operator A is now a fairly immediate consequence of Freudenthal's spectral theorem in the Riesz space %?"({A}), and also the spectrum of A in the classical sense is exactly the spectrum of A considered as an element of the Riesz space. The spectral theorem for a bounded normal operator follows easily. In order to further clarify the structure of the second commutant, it will be shown that if W is a subset of 2 such that W is a (real) linear subspace of 369
370
HERMITIAN OPERATORS I N HILBERT SPACE
[CH.8 , 8 53
8 as well as a ring (in the algebraic sense), and if 92 is closed under the operation of taking strong limits, then W is a Riesz space with respect to the partial ordering inherited from 2.In addition, if H is separable and W contains the identity operator, then U"(92) = W . Furthermore, it will be shown that any second commutant W ' ( 9 )is an abstract L, space, i.e., the operator norm satisfies IlSUP (1-41, 1Bl)Il
=
max (IIAIL 11B11)
for all A and B in W'( 9). The ordered vector space L is called an anti-lattice whenever, for any two elementsf, g E L ,the element inf (f,g ) exists in L if and only iff and g are comparable (i.e.,fz g or f 5 9 ) . We will prove R. V. Kadison's theorem (1951) that 8 is an anti-lattice. 53. The ordered vector space of Hermitian operators in a Hilbert space
Let H be a Hilbert space (over the complex numbers); the elements of H will be denoted by x, y , z, . and the inner product of x and y by ( x , y ) . As well-known, the bounded linear operator A in H (i.e., from H into H ) is called Hermitian (or self-adjoint) if
..
(Ax, Y ) = (x, AY)
holds for all x and y in H. The set of all bounded Hermitian operators in H will be denoted by H,arbitrary elements of H by A , B y . .., the identity operator and null operator by E and 8 respectively. The following lemma is well-known. Lemma 53.1. (i) If A and B are elements of 8, then the product A B (in the usual operator sense) is an element of Af i f and only if A and B commute (i.e., ifand only i f A B = BA). (ii) If A , E 8for n = 1 , 2 , . . . and A,, converges weakly to the bounded linear operator A (i.e., (A,x, y ) + (Ax, y ) as n -+ 00 for all x , y E H ) , then A E 8. Hence, if A,, E &for n = 1,2, . . and A , converges strongly to the bounded linear operator A (i.e., IIAx-A,xll + 0 as n -+ co for all x E H ) , then A E H.
.
Proof. (i) We have A B E 2 if and only if (ABx, y ) = ( x , ABy) holds for all x , y E H . Also, on account of A E Af and B E 8, we have (x, ABy) = (Ax, BY) = @AX, Y )
CH. 8, 531
for all x , y
HERMITIAN OPERATORS IN HILBERT SPACE
E H.
371
It follows that A B E 2 f if and only if (ABX,Y) = ( B A x , y )
holds for all x , y E H , and this is equivalent to A B = BA. (ii) Let A , E 2 for n = 1,2, . . ., and let A be a bounded linear operator such that ( A , x , y ) --f ( A x , y ) as n + 03 for all x , y E H . Then soA~2f.
(Ax,y )
=
lim ( A , x , y )
=
lim ( x , A n y ) = ( x , A y ) ,
The set 2 f is a real vector space under the usual algebraic operations. Introducing in 2 f a partial ordering by defining that A 2 B holds whenever ( A x , x ) 2 ( B x , x ) holds for all x E H, the set 2 f becomes an ordered vector space (i.e., A 2 B implies A + C 2 B + C for all C E2P and aA 2 aB for all real a 2 0). Note that it is not completely trivial to prove that A 2 B and B 2 A together imply that A = B . We briefly recall the proof. If A 2 B and B 2 A hold simultaneously, then ( A x , x ) = (Bx, x ) holds for all x E H . Replacing x by x +y in this inequality, we find that ( A x , r)+ (AY, x ) = (Bx, v)+ (BY, x ) holds for all x , y E H . Observing now that the number (Ay, x ) = ( y , A x ) is the complex conjugate of ( A x , y), we derive from the last formula that the real parts of ( A x , y ) and (Bx, y ) are equal for all x, y E H . Replacing y by iy, it follows that the imaginary parts of ( A x , y ) and ( B x , ~ are ) also equal. Hence, we have ( A x , y ) = ( B x , ~for ) all x , y E H. This implies A = B. The positive cone 2f' of 2 f consists of all A E 2 satisfying ( A x , x ) 2 0 for all x E H,the elements of 2' are usually called positive (semi-) definite Hermitian operators. If the Hilbert space H is one-dimensional, every element of 2 is a real multiple of the identity operator, and so every two elements of %' commute. Hence, 2 f is now a real commutative algebra (i.e., simultaneously a ring and a real linear space), isomorphic to the algebra of all real numbers and such that the isomorphism is order preserving in both directions. As soon as the Hilbert space H i s at least two-dimensional, 3? is no longer an algebra, as shown by the following simple lemma.
Lemma 53.2. (i) If A E Z , and A commutes with every member of 2fy then A is a real multiple of the identity operator E. (ii) rfthe Hilbert space H i s at least two-dimensional, then 2 f is no algebra. Proof. (i) If H is one-dimensional, the statement is true. Hence, assume
372
HERMITIAN OPERATORS I N HILBERT SPACE
[CH. 8 , 5 53
that H i s at least two-dimensional, and let A commute with evzry element of Z . Then A commutes in particular with every orthogonal projection P in the Hilbert space H, and so every x in the range of P satisfies A x = APx = PAX, which shows that A x is also in the range of P. Every closed linear subspace of H is the range of an appropriate orthogonal projection, so it follows that A leaves every closed linear subspace of H invariant, in particular A leaves every one-dimensional subspace invariant. This implies that for every X E H there exists a real 1, such that A x = I,x. It remains to prove that Ax = AY for x and y linearly independent. To this end, given linearly independent x and y , let (z, t } be an orthonormal basis of the subspace spanned by x and y , and let B be the Hermitian operator satisfying Bz = t and Bt = z on this subspace, while B is identicallyzero on the orthogonal subspace. Then
1 , t = At = ABz
=
BAz
=
1,Bz = 1,t,
and so 1, = 1,. It follows easily that 1, = 5. (ii) If H is at least two-dimensional, the set Z contains an element A that is no real multiple of the identity E, and so by part (i) there exists an element B E 2 such that A and B do not commute. But then, by part (i) of the preceding lemma, A B is no element of Z .This shows that Z is no algebra. We recall the generalized Schwarz's inequality, according to which
](AX,Y)12 5 (Ax7 x)(Ay,V ) holds for all A E 2' and all x , y E H. The proof is based upon the inequality
0
5 ( A ( x - ~ Y )x, - 1 ~ )= ( A X ,~ ) - 1 ( A y x)-X(AX, , y)+AX(Ay, y ) ,
holding for all complex 1.If (Ay, y ) # 0, we choose
I = (x7 AY)/(AY7 Y)7 and the inequality to be proved follows immediately. If (Ay, y ) = 0, we have to prove that (Ax, y ) = 0. Assuming that (Ay, y ) = 0, but (Ay, x ) = reiq with r > 0 and cp real, we choose 1 = Rediq with R positive and so large that Rr > (Ax, x ) . This immediately yields a contradiction in the above inequality, and so (Ax, y ) must be zero. We also observe that A E 2' implies A" E #' for n = 2,3, . Indeed, for n even, say n = 2k, we have
..
( A Z k x , x )= llAkxl122 0
CH. 8,s 531
HERMITIAN OPERATORS IN HILBERT SPACE
373
for every x E H, and for n odd, say n = 2k+ 1, we have
(P+ 'x, x)
= (AA'x, Akx)
20
for every x E H. It follows that any polynomial in A with non-negative coefficients is an element of S+, and since the product of any two polynomials '.Furtherof this kind is again of the same kind, the product is again in S more, if A E &?+ is given and the number y A is defined by YA
= sup ((Ax,
x)/11x112
:x #
o},
then y A = 11A11. Indeed, it follows from (Ax, x)
that that
I IIAxll * llxll 5 IlAll
*
llX1l2
5 11A11. Conversely, we have by the generalized Schwarz's inequality llAX1l2 = (Ax, Ax)
5 ('4x9 X)+(AAX, Ax)+ 5 Y ~ l l ~ l l Y ~ l l ~ ~ l l ~
and so llAxll 5 ~ ~ l l xiflAx l # 0. The same inequality holds if A x = 0, and hence we have llAll 5 y A . Combining the results, we obtain y A = 11A11. In particular it follows from 6 5 A 5 E that j l A 5 1, so IlAll 6 1. But then we have 11A"11 5 1 for n = 1,2,. ., and so 6 5 A" 6 E in view of the just established results.
.
Lemma 53.3 (J. P. Vigier [l], 1946). such that 6 I A -B 5 E, then IIAx-Bx114
5
If A andB are Hermitian operator3
{(Ax, x)-(Bx,
x))11x112
holdsfor every x E H.
Proof. Writing A - B = C, we have C
2 6 (i.e., C E &+), and so
I Icx114= (cx,cx)2 I (cx,x ) ( ~ 2 x , cX) by the generalized Schwarz's inequality. In view of the remarks above, it follows from 6 5 C E that llCll 5 1, and so we have
llCX1l4 6 (CX, x)11x112, which is the desired result. We will prove finally in this section that 3" has a property analogous to Dedekind completeness of a Riesz space.
374
HERMITIAN OPERATORS I N HILBERT SPACE
[CH. 8,§ 53
Theorem 53.4. (i) If (A, : t E {z}) is an upwards direcfedsubset of 8 such that the subset is boundedfrom above, then A = sup A, exists in 2,and for in ( t }such that every E > 0 and every x E H there exists an index IIAx-A,xll < E holdsfor all A, 1 (ii) If (A, : z E { 7 } ) is the same as above and the Hilbert space H is separable, there exists an increasing sequence(A, :n = 1, 2, . . .) in (A, : z E {z}) such that A,,x + A x holds for all x E H simultaneously (i.e., the sequence of operators ATnconverges strongly f o the operator A). Proof. (i) We choose z0 E {z}. The sets (A, : z E {z}) and (A, : z E {z}, A, 2 A,,) have the same upper bounds, so we may just as well assume that A, >= A,, holds for all z E {z}. But then, subtracting A,, and multiplying by an appropriate positive constant, we may restrict ourselves to the case that 8 5 A, 5 E holds for all T. Under this assumption, if A,. 1 A,, we have by Vigier's lemma that
II&x-A,xI14
I { ( A z , ~. ,) - ( A x , x)}IIxII'
holds for every X E H . Since, fixing x, the upwards directed set of real numbers (A,x, x ) has a finite supremum, it follows that for every e > 0 there exists an index t8,,in {z} such that IIA,.x-A,xlj
In particular, far n = 1,2,
<E
for all A,, A,.
2
. . ., there exists an index zX," in {z]
IIA,x- A,, ,,XII < n-l
for all A,
such that
2 A,*,-,
and we may assume that ATx,n+l
holds for all n. Then
2
*rx,n
I I A , x . , ~ - A , x , n ~ I
1 n, and so the sequence
.
(A,,,, x : n = 1 , 2 , . .)
in H converges to an element in H , which we denote by A x . Note that A x is uniquely determined in the sense that if (A,.*,,, : n = 1,2,
. . .)
is another increasing sequence such that
IIA~x-&,,,,xII
< a-1
for all A,
1 A,;,",
CH.
8,s 531
HERMITIAN OPERATORS IN HILBERT SPACE
375
then A,.x,,,x converges also to Ax. The proof follows by observing that there exists an operator A, satisfying A, 2 A,=, and A , 2 A,,x,,, simultaneously, and so I I A T X . n ~ - A r * x , n ~< I I 2n-I. It is easily verified now that the thus defined operator A in H is linear. Indeed, for the proof that A is additive, let x and y in H be given, set z = x + y , and observe that
1
llAx-A,xll < 2n-' llAy-ATyll < 2n-1 IIAz-A,zll < 2n-'
for all A, 2 A,*, n , for all A, 5 ATY,, for all A, 2 A,=, .
(1)
,,, ATY, and A,=.,, simultaneHence, for all A, greater than or equal to ATX, ously, the three inequalities in (1) hold simultaneously, and so in view of A,z = A,x+A,y we have IIAz-(Ax+Ay)lI < 6n-I.
..
This holds for n = 1, 2, .; it follows that Az = A x + A y . It is evident that A is a bounded operator. Indeed, for llxll = 1 we have
.
l l ~ x l sl IIA,,,,xII+~~-'
s 1+2n-'
for n = 1,2,. ., so A is bounded with IlAll 5 1. Furthermore, we have A E X , i.e., A is Hermitian. Indeed, given x and y in H, there exist increasing sequences Arx,,,and ATY, ,, such that Arx,,,x and ATY,y converge to Ax and Ay respectively, and hence (cf. the inequalities in (1)) there exists an increasing sequence (ATn:n = 1,2, .) such that
..
A,x + A x and ATny+ A y hold simultaneously. It follows that (A,,x, y ) = (x, A , J ) converges to ( A x , y ) as well as to (x, Ay), and so (Ax, y ) = (x, Ay), i.e., A is Hermitian. Finally, it follows from
AX, x ) - ( A , x ,
x) IS IIAx-A,xlI
*
llxll
5 2n-'IIxll
for all A, 2 A,*,, that (Ax, x) differs an arbitrarily small amount from the number sup, (A,x, x), and hence
). =
x,
holds for all x E H. In other words, A = sup A, holds in X .
376
[CH.8 , s 53
HERMITIAN OPERATORS I N HILBERT SPACE
(ii) We assume now that His separable, so there exists a countable dense set (x, : n = 1,2, . . .) in H. For each x, there exists a sequence (Ank : k = 1, 2 , . . .)
C
( A , : T E {TI)
such that Ankincreases as k increases and Ankx, -P Ax, as k + 00. More precisely, Ankcan be chosen such that
IIAx,-A,x,l( < k-‘ holds for all A, 2 Ank.We will prove the existence of an increasing sequence (& :k = 1,2, . . .) in (A, : z E {T)) such that Bkxn-+ Ax, as k + 00 holds for all x, simultaneously. Since it may be assumed by part (i) that 0 5 A - A , - E holds for all z, it is sufficient by Vigier’s lemma to show that (Ax,, xn)I (Bkxn,x,) tends to zero as k -+ 00 for every x, . The proof is, naturally, a variant upon the diagonal procedure. We choose B , = A,,, and then B2 E (A, : z E {t}) such that B, 2 A 1 2 , B2 2 A 2 2and B2 2 B,; generally, if B,, . . .,Bk- have been chosen already, we choose BkE (A, : z E {T}) such that Bk is greater than or equal to A l k , A Z k , . . ., A k k , B k - 1 . NCW, if n iS given, let k 2 n. Then Bk 2 Ank, and SO
,
0
5
( A x , , xn)-(Bkxn,
xn)
5 (Ax,, X n ) - ( A n k X n , xn) 5 IIAxn-Ankxnll
’
Ilxnll
5 k-lIIxnll*
As observed above, this shows that IIAX,-&x,,ll tends to zero a s k + 00 for every x,. Note also that on account of 0 5 Bk 5 E we have llBkll 6 1 for all k. Given now the arbitrary element x in H a n d the number > 0, choose x, from the dense subset such that ] ~ x - x , ,< ~ ~+E and then determine ko such that IIBkx,-Ax,ll < 3& holds for all k 2 ko.It follows then from
!IAx-
Bkxl
I6
IIAx-Axnll
+I IAxn-
I f II B k x n - Bkxll 6 IIx-xnll + I I A x n - B k x n l l
BkxnI
+
l~xn-xll
that IIAx-Bkxll c & for all k 2 k o . In other words, Bkx converges to Ax as k + co for every x E H, i.e., Bk converges strongly to A. Exercise 53.5. Let H be the Hilbert space consisting of all complexvalued functions x ( t ) on the real line such that x ( t ) # 0 for at most countably many t, and with Ix(t)l’ < 00. The inner product in H is
ct
t
where j j denotes the complex conjugate of y.
CH. 8,0541
THE SQUARE ROOT
377
OF A POSITIVE HERMITIAN OPERATOR
Show the existence in 2'P of an upwards directed set (A, :z E {z}) with supremum A such that for no increasing sequence (ATn: n = 1,2, . . .) in (A, : z E (z}) we have A,x + A x for all x E H simultaneously.
54. The square root of a positive Hermitian operator It is a well-known theorem that every A E 2' has a unique square root A* in &+,i.e., there exists a unique element A%in 2'P' such that (A*)2 = A . For the proof we may obviously restrict ourselves to the case that 8 5 A E holds, and we present C. Visser's construction of the square root ([l], 1937). We set A , = E and
s
for n = 0 , 1 , 2 , . . .,
A,,+1 = A,,+$(A-Ai)
and we will prove the existence of an operator A , in S ' satisfying 4,J A , and A t = A. For the proof, let B = E - A and B,, = E- A,, for n = 0, 1,2, . . . Then 8 5 B E as well as B, = 8 and B,,,, = +(B+B:)
for all n,
as follows from a simple calculation. We show first that, for every n, the operators B,, and B,,, -B,, are polynomials in B with non-negative coefficients. For n = 0 this holds in view of B, = 8 and B, = +B. Assume now that, for some value of n, we know already that B,,-, and B,,-B,,-l are polynomials in B with non-negative coefficients. Then B,,, and hence B,, +B,,- 1, are also such polynomials, and so
,
Bn + 1 -B, = +(B,Z -B.'-
1)
= HBn +Bn - 1)(Bn-Bn
,
- 1)
is also such a polynomial (note that we use here that B,,- and B,, commute). The induction proof is now complete; it follows that we have
e = B, s B, 5 B~ 6 ...t.
Since 0 = B, 6 E, and since 8 5 B,, 5 E implies 8 B,,,, = $(B+B:) 5 E, we have then that
e = B,
I;B,
6 B,
s ...t
5 B."
E, and so
E,
whence on account of Theorem 53.4 there exists an operator B, E A?' such that B,, 1B, and B,,x -P B,x for every x E H. Returning to the operators A,, = E-B,,, we conclude, therefore, that A , = E-B, satisfies
E = A , Z A , Z A ~ B. . . ~ ~ , z e
378
HERMITIAN OPERATORS IN HILBERT SPACE
[CH.
8,854
and A,x + A , x for every x E H . This implies A i x + A k x for every x E H. Indeed, using that IIA,I 1 5 1 holds for all n and writing y = A , x, we have
IIA,ZX-A:X~I
5
IIA,~X-A,A,X~~+IIA,A,X-A:X~~
= IIAn(Anx-Amx)Il+ ll(An-Am)AmxII
5
l l ~ n ~ - ~ m ~ l l + l l A n Y - ~ m Y l0 l +
as n + co. Taking limits now (n + a)in the formula A , + ~ x= A,x++(Ax-A,ZX),
we find that A: x = A x holds for every x
E H,
so A: = A .
As a first consequence we note that if soae Hermitian C commutes with A , then C commutes with every An,and henct. C commutes with A , . Secondly, it follows then that if A and C are commuting elements of &+,then A C = C A is again in 8 ' (note that A C = C A E & follows already from Lemma 53.1 (i)). For the proof, let A , be the just obtained square root of A which, therefore, commutes with C, and observe that (CAX,X) = ( C A , A,x,
X) =
( A , CA,x,
X) =
(CA,x, A , x ) 2 0
holds for every x E If. Uniqueness of the square root is not ntccssary for this proof. More generally, we have now that if A 2 B holds in 2,and C is an element of 2"' commuting with A and By then C A 2 CB. Indeed, since C and A - B are commuting elements in %+,we have C ( A - B ) E 2"+ by the result just established, so CA 2 CB. Note, as a particular case, that if A 2 B 2 8 and A and B commute, then A 2 2 A B 3 B 2 . In the above construction of the square root of A , where 8 have used A . = E and
..
An+1
5 A 5 E, we
= An++(A-A,Z)
as n = 0, 1,2, . for the approximating sequence. We introduce another sequence (A; : n = 0, 1,2, .) by defining A; = 8 and
..
= A;++{A-(A:)2)
for n = 0 , 1 , 2 , .
.. .
Similarly as above, writing again B = E - A and BL = E - A : , we have BA = E and % + l = 3(B+(B;)'),
CH. 8, $541
THE SQUARE ROOT OF A POSITIVE HERMITIAN OPERATOR
319
and so it follows by induction (and by means of the just established result that the product of commuting elements in A?’ is again in 2’)from
B;+, -B; = +(B;+B,’-,)(B;-B;-,)
that (B; : n = 0,1,2, . . .) is a decreasing sequence in A?’,possming therefore an infimum B&. In other words, writing A‘, = E- Bk , we have
8 = A; 5 A ; 5 A; 5 ...I A& 5 E and AAx + A’, x for every x E H. Exactly as above it follows that (A’,)’ = A , i.e., A‘, is also a square root of A . Collecting the thus obtained results, we have
. .. t A ; 2 ...1 A ,
8 = A; 5 A ; 5 A; 5
with (A&)’ = A,
E = A. 2 A , 2 A’
with A: = A .
Writing again B,, = E- A,, and B; = E- A: for all n, we hake Bo I;B ; , and so it follows by induction from B,,+l = +(B+Bz) and B:+l = +{B+(B:)’}
that B, 5 BA holds for all n, i.e., A: 5 A, holds for all n. But then (since A: is increasing and A, is decreasing as n increases) we have A; 5 A, for all m, n = 0,1,2, . . ., and so A’, 5 A , . It follows that A,
=
A’,+(A,-A’,)
with A , - & 2 8. Furthermore, since all A: are polynomials in A, and since A , commutes with any polynomial in A , it follows that A , and A’, commute; hence A: = (A:)’+2A&(Aw -A’,)+(A, -A’,)’.
But A: = (A’,)’ = A , and the other t e r n are elements of A?+.It follows that (A, -A’,)’ = 8, and so A , = A’, . This shows already that the square roots A , and A’, of A are equal. For the uniqueness proof, assume now that C E 2“ is another positive square root of A, so C 2 8 and Cz = A . Then C and A commute in view of C A = A C = C 3 ,and this implies that any polynomial in C and A commutes with all A , and all A:. Also, it follows from C’ = A 2 E that llCX11’ =
(CX,
for all x E H, so llCll
Cx) = (CZx, x) = (Ax, x)
5
1, and hence 8
llx11’
5 C 5 E. The sequence
3 80
(A: : n = 0,1,2,
HERMITIAN OPERATORS IN HILBERT SPACE
. . .) is now defined by A:
=
= A Y + + { P ~ - ( A : ' ) ~ ) for
C and
n = 0,1,2,.
[CH.
8,s 54
.. .
Introducing B:' = E- A:' for all n, we have
B ; , and so it is easily derived by induction that B,, 5 BA' and Bo 5 B t 5 BA holds for all n, i.e., A; 5 A: 5 A,, holds for all n. In the proof we use that BA', as a polynomial in A and C, commutes with all B,, and BA, so B,, 5 Bi' 5 Bi implies B.' S (B;')' 5 (B;)', and hence B,,+l 5 Bikl 5 Bi+ 1. Observing now that A:' = C holds for all n (in view of A: = C and C 2 = A), we find that A; S C 5 A, holds for all n, and so A& 5 C 5 A,. But A& = A,, so A& = C = A,, i.e., the positive square root of A is unique. One last remark for later purposes. All A; are polynomials in A without constant term (i.e., without a term which is a nonzero multiple of E), so the positive square root of A is the strong limit of a sequence of polynomials in A without constant term. There exist other proofs, even shorter ones, for the uniqueness of the square root. The present proof, however, shows also that the sequence (A, : n = 0,1,2, . . .) with An+1
= A~++(A-A,Z)
always converges to the square root of A no matter how A . is chosen, provided only that 8 5 A . 5 E holds and A , commutes with A (cf. Exercise 54.3 at the end of the present section). We consider now an arbitrary Hermitian operator A. The positive square root of A' will be denoted by IAI, and we set
+
A" = *(A IAI). This notation might cause confusion; we do not wish to suggest tbat A + = sup (A, 8) holds in A?. In the next section it will be proved, however, that this formula holds in an appropriate subspace of 2. We prove here that A+ is always an upper bound of the set consisting of A and 8.
Theorem 54.1. Given A E 2, we have A + ly, IA,1 2 A iutd IAI 2 - A .
2 8 and A + 2 A or, equivalent-
CH. 8 , s 541
THE SQUARE ROOT OF A POSITIVE HERMITIAN OPERATOR
381
Proof. It is evident that the statements are equivalent. Hence, since A’ and (-A)’ have the same square root [ A ! ,it is sufficient to prove that IAI 2 A holds. Multiplying, if necessary, by a positive constant, we may assume that -E A S E, and so O 5 A’ S E. We set B = E - A 2 and we define (B,, : n = 0 , 1 , 2 , . . .) by Bo = Oand
B , + ~= +(B+B,Z) As shown above, we have O = Bo
for n = 0,1,2,.
.. .
5 BI B B2 5...r B , 5 E,
and E-B, is the square root IAI. For the proof that IAI 2 A holds we have to show, therefore, that B, = E - ] A ] S E - A , and on account of B, T B, it will be sufficient to show that B,, 5 A - A holds for all n. For n = 0 this is evident. Assume that B,, 5 E - A has been proved for some value of n, and observe that B, and A commute. Then we have B: (E- A)’, which can be rewritten as
B,Z 5 E - 2 A + A Z = 2E-2A-(E-A’) and so
2B,,+, = B+B,Z
= 2E-2A-B,
5 2(E-A),
which is the desired result for n+ 1. This completes the proof. We will prove next that, among all Hermitian B, commuting with a given Hermitian A and satisfying B 2 A as well as B 5 8, the operator A’ is the smallest. Theorem 54.2. If A and B are commuting Hermitian operators such that B 2 A as well as B 2 8, then B 2 A’. Equivalently, if A and C are commuting Hermitian operators such that C 2 A as well as C 2 - A , then C 2 IAI. Proof. Let A and B be commuting Hermitian operators such that B 2 A as well as B 2 8. Setting C = 2B-A, we have C 2 A as well as C 2 - A , and C commutes with A . Hence, if we prove that C 2 IAI, it will follow that 2 B - A 2 IAI, i.e., B 2 A’. Assume, therefore, that C 2 A as well as C 2 - A , and C commutes with A . Then C- A and C + A are in Z?’ and they commute, so ( C + A ) ( C - A ) 2 8, i.e., C 2 2 A’. It follows then from the construction of the square root that (C”* 2 (A2)* = IAI.
Since C E Z?’ (because C obtain C 2 IAI.
2 A and C 5 - A implies 2C 2 A - A
= O), we
382
HERMITIAN OPERATORS IN HILBERT SPACE
[CH.8,s 55
Exercise 54.3. Let A and C be commuting Hermitian operators between 8 and E, and set
for n = 0,1,2,. . . . .A: = C, A,*,, = A,*++{A-(A:)'} Show that A,* converges strongly, although not necessarily monotonely, to the positive square root of A.
55. The second commutant of a commuting set of Hermitian operators Let 9 be a subset of i%? such that all elements of 9 commute mutually. The set @')( 9)of all B in 2 commuting with every element of 9 is called the commutant of B, and for n = 2,3, . . . the set Y ( " ) ( 9of ) all B in X commuting with every element of Y("-')(B) is called the n-th commutant of 9.Evidently, every V(")(9) is a real linear subspace of 2,and every real polynomial in the elements of 9, with a constant term consisting of a real multiple of E included, is contained in every V(")(9). Also, if the se9 )and Ak converges weakly quence (Ak :k = 1,2, . . .) is contained in e(")( to the bounded linear operator A (cf. Lemma 53.1 (ii)), then A E V(")(g). The sets %(')(a), V(')(9) and V(3)(9)will be denoted briefly by V', V" and V'".
Lemma 55.1. Wehave%?" c V ' a n d V ( " + 2 ) ( 9= ) V("l(L2)for everyn, i.e.,
. . ., 6= ) . .. .
fg' = C p ' S -- Y ( 9 = g t f
=~
( 4= ) ~ (
Furthermore, V" is a commutative ring, i.e., Y" is a ring in the algebraic sense the elements of which commute mutually.
Proof. Note first that, generally, g1c 5Bz implies u'(L21)13 V'(9'). Hence, 9 c W implies V' 13 W'. Similarly, 9 c W' implies F ' 3 V'". On the other hand, every element of %' commutes with every element of Y,', and so v' c v'".It follows that v'" = V', and this implies that V("") = V(") holds for every n = 1, 2, . In order to show that the elements of V" commute mutually, assume that A and B are in v". Then A E Y,' and B E V" c V', so A and B commute since every element of V" commutes with every element of v'. It follows, in view of this commutativity property, that the product of two operators from Y,' is Hermitian in the first place, and once this has been established it follows then also that the product of operators from Y" is again an element of v",so Y" is a ring.
. ..
CH.8 , s 551
THE SECOND COMMUTANT
383
By way of example, let 9 consist only of the identity operator E. Then the first commutant U' satisfies v' = JZ' and the second cornmutant Y" consists of all real multiples of E (cf. Lemma 53.2). This shows that the first commutant " ( 9 )is not necessarily a ring. Note, furthermore, that B1 c implies Y'(91) 3 W ( 9 2 ) ,and so v''(gl)c %?"(92). Adding the does not change identity operator E to 9,if E is not already contained in 9, v' and W'. There is enough material available now to show that, for any subset 9of mutually commuting elements of S, the second commutant of 9 is a Riesz space. It was already observed that W ' ( 9 ) is a real linear subspace of X , and with respect to the ordering inherited from S the space v''(9)is evidently an ordered vector space. Theorem 55.2. For any subset 9of mutually commuting elements of A?, the second commutant 9 )is a Dedekind complete Riesz Apace. If the Hilbert space H is heparable, the Riesz space U''(9 )is super Dedekind complete. %"I(
Proof. In order to show that U " ( 9 ) is a Riesz space, it is sufficient to show that sup (A, 8) exists in W'( 9)for every A E W'( 9).We will prove that the element A + = +(A+IAI) satisfies this condition. I n the first place, since [A1 is the strong limit of a sequence of polynomials in A', and since all these polynomials are elements of v"(9),we have IAl E W ' ( 9 ) , and so A + = -)(A+IAI) ~ v " ( 9 ) . Secondly, it is evident that v''(9)c v ' ( { A } ) , and so, if B E U ' ' ( 9 ) is an upper bound of the set consisting of A and 8, then B 2 A + by Theorem 54.2. This shows that A + = sup (A, 0) holds in B).As observed above, it follows that v''(9)is a Riesz space. If (A, :z E {z}) is an upwards directed subset of ""(9) such that the set is bounded from above, then A = sup A, exists in JZ' by Theorem 53.4. In order to show that %"(29) is Dedekind complete, it is sufficient to show that A EW'(B),i.e., it is sufficient to show that A commutes with every B E 'X'( 9). Tc this end, let x E H and B E U'(9)be given, and set y = Bx. As shown in the proof of Theorem 53.4, there exists an increasing sequence (ATn: n = 1,2, . . .) in (A, : z E {z}) such that A,,x + Ax and Army+ A y hold simultaneously. This implies that %'I(
ABx = Ay
=
lim Arny = lim A,,Bx = lim BA,,x = BAx.
Thus ABx = BAx holds for every x
E
H, which is the desired result.
384
HERMITIAN OPERATORS IN HILBERT SPACE
[CH.8,§ 55
If the Hilbert space H is separable, there exists an increasing sequence (Az, : n = 1,2, . . .) in (A, : z E {z}) such that A,x + Ax holds for all x E H simultaneously (cf. Theorem 53.4 (ii)). By the Dedekind completeness of V'(53) the element B = sup A,, exists in %"(9 )and A,, converges strongly to B (once again by Theorem 53.4). It follows that B = A, so A = sup A,,. Hence, the given set (A, : z E {z}) has a countable subset with the same supremum. This shows that W ' ( 9 )is super Dedekind complete. This theorem has been known for a long time, but it seems that until around 1952 no simple proof was available (i.e., a proof without making use of the spectral theorem for Hermitian operators). Then, for the case that 53 consists of one Hermitian operator, the theorem was proved in an elementary manner by V. I. Sobolev ([l], 1952); the proof was extended to the general case by V. D. Lyubovin ([l], [2], 1954). For any Hemitian operator A , let N ( A ) and R(A) denote the null space of A and the closure of the range of A respectively. As well-known, N ( A ) and R ( A )are mutually orthogonal closed linear subspaces of the Hilbert space H, and H i s the direct sum of N ( A ) and R(A). In the following series of lemmas we collect some simple properties of the Riesz space W ' ( 9 ) , where 9 is again a set of mutually commuting elements of %.
Lemma 55.3. (i) For A and B in al'(9)we have A IB if and only if AB = 8, and also if and only if A' 1B 2 . Also, A IB holds if and only if the closed linear subspaces R(A) and R(B) are orthogonal in H. (ii) For A , B and C in W ' ( 9 ) it follows from A IB that AC IBC. Proof. (i) For the proof that A IB holds if and only if AB = 8, we assume first that A 2 8 and B 2 8, and we write C = inf (A, B). Note that A, B and C commute mutually, since all elements of W ' ( 9 ) commute mutually. Assuming first that AB = 8, we have C 2 5 AB = 8, so C 2 = 8, which implies C = 8, i.e., A IB. For the proof of the converse it may be assumed without loss of generality that A IB and 8 5 A 5 E as well as 8 5 B S E. Then 8 5 AB 6 A and 8 AB 5 B, so
8 5 AB 5 inf (A, B ) = 8, i.e., AB = 8. Now assume that A and B are arbitrary elements of W'(52) satisfying A IB. This means that IAI 1IBI, and so IAI * IBI = 8 in view of what was proved already. Since IAI 1B1 and lABl are both equal to the positive square
CH. 8, S 551
3 85
THE SECOND COMMUTANT
root of A2B2,we have, therefore, that I A B= ~ I A ~I .B I= 8,
which implies that AB = 8. Conversely, if A and B are elements of %"( 9) satisfying A B = 8, then lABl = 8, so IAI IBI = 8, which shows that IAl 1 .IBI, i.e., A 1B. On account of these results we have now that
-
AIBoAB=eo(AB)Z=eoAzBZ= eoAzIBZ. Finally, lei again A and B be elements of U"(L3) satisfying A _L B, so AB = 8. Furthermore, let x and y be elements of the Hilbert space H such that xis in the range of A and y is in the range of Byi.e., x = Ax' and y = By' for appropriate x', y' E H. Then (x, y ) = (Ax', By') = (x', ABy')
= 0,
so the ranges of A and B are orthogonal in H. Hence, the closures R ( A ) and R(B) of these ranges are also orthogonal. Conversely,if A and Bare elements of W ' ( 9 )such that R(A) IR(B),then (Ax, B y ) = 0 for all x, y E H, i.e., ( x , ABy) = 0 for all x , y E H. This implies that AB = 8, so A J- B. (ii) If A, B and C are elements of U"(9) and if A IBy then AB = 8, so ABCZ = 8, i.e., (AC)(BC) = 8, which implies that AC IBC.
Lemma 55.4. (i) Given A E %'"(9), we have A + J- A - in the Riesz space sense, which implies that the closed linear subspaces R ( A + ) and R ( A - ) are orthogonal. It follows that
ll~x1I2= IIIAlxllZ = ll~+x112+11~-x11z holdsfor every x E H. Hence llAll = 11 IAI 11, and we have Ax = 0 ifandonly i f
[ A ~= x 0. (ii) FotAandBinV"(L3)wehave IAl 5 1B1 ifandonlyifIIAxJI 4 IlBxll holds for every x E H. (iii) I f A, ( n = 1,2, .) and A are elements of Y"(L3) such that A,xO + Ax, holds for some x, E H (not necessarily A,x + Ax for all x E H ) , then A:x, --r A+X0.
..
Proof. (i) By the preceding lemma R( 4 + ) and R ( A - ) are orthogonal subspaces of H , and so A + x IA-x holds for every x E H. Since A = A+-A-
and IAI = A + + A - ,
386
HERMITIAN OPERATORS I N HILBERT SPACE
it follows that
llAx1I2 = IIA+xl12 +llA-x1(2=
[CH.8,§ 55
11 IAIx112.
(ii) If A and Bare elements of U ” ( 9 )satisfying IAl 6 lBl, then A 2 5 B2, and so l l A ~ 1 1=~ (A’x, X ) 5 (B’x, X) = (IBx(I2 holds for every x E H. Conversely, if A and B are elements of U “ ( 9 ) such that IIAxll 5 IIBxll holds for every x E H, then A’ 5 B2, and so IAl 5 IBI. (iii) Since IA+-ATI 6 IA-A,I, we have by (ii) that
As observed earlier, the identity operator E is an element of %”( B), and it is obvious that the ideal in W ’ ( 9 ) generated by E is the whole space 9).
%’I(
Lemma 55.5. (i) I f P is a component of E in 9),then P is the orthogonal projection on some closed linear subspace of the Hilbert space H. (ii) Given A in %”( 9),let P be the component of E in the principal band generated in C“(9)by A . Then P is the orthogonal projection on R(A). In other words, we have N ( P ) = N ( A ) and R(P) = R(A). It follows that i f A and B are elements of %”( 9 )generating the same principal band in W’( 9), then N(A) = N(B) and R ( A ) = R(B). Conversely, i f A and B are elements of %”( 9 )such that R ( A ) = R(B) or, equiualently, N ( A ) = N(B), then A and B have the same disjoint complement by Lemma 55.3, and hence A and B generate the same band in W’( 9). (iii) I f P , and P, are components of E such that P , and P, are elemenrs of %”(9), then P , IP , holds in %“(9) if and only if R ( P , ) and R(P2) are orthogonal subspaces of H. %If(
Proof. (i) If P is a component of E in W ‘ ( 9 ) , then we have P I( E - P ) by definition, and so P ( E - P ) = 8 by Lemma 55.3. It follows that P 2 = P. Hence, P is a bounded Hermitian operator satisfyingP 2 = P. As well-known, this implies that P is the orthogonal projection on some closed linear subspace of the Hilbert space H. (ii) Under the given hypotheses, we need only prove that N ( P ) = N(A). Since A is disjoint from E - P , we have A ( E - P ) = 0, so A = AP. This shows that x E N(P) implies x E N(A). For the converse, observe that since P is an element of the band generated by A there exists an upwards directed set (B, : z E {z}) in the ideal generated by A such that 8 5 B, P. Hence, if
CH. 8 , § 551
387
THE SECOND COMMUTANT
x E N ( A ) , then A x = 0 and lAlx = 0, so (B, x , x ) = 0 for all t E {t}, and it
follows that
(Px, X) = SUP (B,x,
X)
= 0.
This implies Px = 0, so x E N(P). (iii) Follows from Lemma 55.3. In several parts of the lemmas in this section it was assumed that A and B are members of W ' ( 9 ) for some set 9 of mutually commuting Hermitian operators, and then several properties of A and B were proved depending upon various additional conditions imposed upon A and B. It may be asked whether these results depend upon the choice of 9. Fortunately this is not the case. In order to see this, note first that the assumption that A and B are members of W ' ( 9 )for some 9 of the kind referred to is equivalent to the assumption that A and Bare commuting Hermitian operators. Tndeed, given the commuting Hermitian operators A and B, we can take for 9the set conand 92 are subsets of sisting of A and B. Secondly, as observed earlier, if 9, 22' of mutually commuting operators such that g1c B2, then W'(Bl) c %"( 9,).Both of these are Dedekind complete Riesz spaces, so if Y is any subset of W'(91) which is bounded from above, then Y has a supremum in V'(9,)as well as in W'(9,). These suprema are the same. Indeed, if Y consists of two elements, A and B say, then sup ( A , B) is equal to A + (B-A)', which does not depend on any elements in U"(9,) but not in W ' ( 9 , ) . Hence, if Y is a finite set, the supremum depends only on the members of %"'(g1). Now, let Y = ( A , : T E {t}) be an infinite subset of %"'(g1), let B, be its supremum in W'( g1) and B, its supremum in%"'( 9,). In view of the preceding remarks we may assume that Y is directed upwards. Then, as shown in the proof of Theorem 53.4, we have (B1x, X) = SUP (A,x,
X)
= (B,x,
X)
for every x E H . Hence B, = B 2 . To mention one more example. in Lemma 55.4 (i), where we have one member A of W ' ( 9 ) , we can just as well state that we are given one Hermitian operator A , and the formula A + 1A - can now be understood to hold in the Riesz space V ' ( { A } ) , i.e., in the second commutant of the set consisting of A alone. As observed, A + 1A - holds just as well, however, in %"'(g) for any set 9 of mutually commuting Hermitian operators having A as one of its members. A similar remark holds for Lemma55.5 (ii). It has been proved in Lemma 55.1 of the present section that W'(9) is a commutative ring with respect to the ordinary multiplication of operators
388
HERMITIAN OPERATORS I N HILBERT SPACE
[CH. 8 , s 56
on the Hilbert space H . The identity operator E is the unit element with respect to this multiplication. It has been proved also (in Theorem 55.2) that W ’ ( 9 )is a Dedekind complete Riesz space and hence, according to the results in section 45, W ’ ( 9 ) is Riesz isomorphic to the Riesz space “{V’(9)} of all real continuous functions on a certain compact HausdorlT space, such that E corresponds in this isomorphism to the function identically equal to one. In “{W’(9)} we have the ordinary pointwise multiplication of functions. Transferring this back to W’(93), we obtain another commutative multiplication in W’(9), also with E as unit element. It is a reasonable conjecture that these two multiplications in V ‘ ( 9 )are the same. This is indeed the case, but it seems easier to postpone the proof until after the spectral theorem in the next section.
Exercise 55.6. If it is given that A and B are members of 2’ such that AB = 6, it follows easily that BA = 6, and so we derive from Lemma 55.3 (i) that A IB holds in the Riesz space $?’({A, B } ) , i.e., A + B = [ A-BI. It is interesting to observe that, without assuming in advance that A and B commute, it follows for Hermitian A and B from A + B = IA-BI that AB = BA = 6. In order to prove this, show first that A and B are members of &+; next, show that AB+BA = 6, which will imply then that AB = BA. Hint:Observe first that A+B-IA-BI
6 A+B-(A-B)
= 28,
and similarly A + B - I A - B [ 6 24. Hence it follows from A + B = [A-BI that A and B are members of 2’. Next, note that (A+B)’ = IA-BI’ = (A-B)’,
so AB+BA = 6. It follows readily from AB = -BA that A’B’ = B2A2, so B2 commutes with A’. But then B’ commutes with the positive square root of A’, i.e., B’ commutes with A, so AB’ = B’A. This implies, similarly, that AB = BA. It follows that AB = BA = 6. This proof, as presented in a paper by D. M. Topping ([l], 1965), is due to H. Leptin. 56. The spectral theorem for Hermitian operators and normal operators As in the preceding section, let 9 be a subset of 2 such that all members of 59 commute mutually, and let W’(9) be the second commutant of 9. We recall that 8 and E are the notations for the null operator and the identity operator respectively. It will be investigated first what the notion of E-uni-
CH. 8,6561
THE SPECTRAL THEOREM
389
form convergence of a sequence in %"'(9 )amounts to in terms of the usual Hilbert space terminology. We recall that the sequence (A, :n = 1, 2, . . .) in V ' ( 9 )is said to converge E-uniformly to A E V'(9 )whenever there exists for every E > 0 a natural number n, such that IA -A,I 5 EEholds for all n 5 n,. Lemma 56.1. If A , (n = 1,2, .. .) and A are members o f Q " ( 9 ) , then A, converges E-uniformly to A if and only if I IA -A,[ I -+ 0 as n -+ 00.
Proof. In the space V ' ( 9 )the sequence A, (n = I , 2, . . .) converges Euniformly to A if and only if, for every E > 0, there exists a natural number n, such that IA-A,I 5 ~ E f o all r n 2 n,, i.e., (IA -A&,
X)
S ~ ( xx) , for all x E H and all n 2 n,.
This is equivalent to
5 E for all n 2 n,, i.e., equivalent to IIIA-A,III 5 E for all n 2 n,. Since IA-A,I and A-A, sup {(IA-A,lx, x ) : llxll
5
1)
have the same operator norm (cf. Lemma 55.4 (i)), the last inequality is equivalent to (IA-.4,11 5 E for all n 2 n,. This shows that A, converges E-uniformly to A if and only if A, converges to A in the operator norm. Given the operator A in W ' ( 9 ) ,we will consider now the spectral system (Pa : - co < x < 03) of A with respect to E. We recall that Pa is the component of E in the band B, generated by (uE-A)'; the order projection on the band B, (in the Dedekind complete Rieszspace W'(9)) is denoted by 9,. Hence, we have Pa = 9, E for every a. We will write A, = 9, A for every u. It follows from Lemma 55.5 that Pa is an orthogonal projection in the Hilbert space Hsuch that Pa and (uE- A)' have the same null space. The range of Pa,which is the orthogonal complement of the null space of Pa,is then the closure of the range of (uE-A)+. Since Pa increases as u increases (i.e., c1 < fl implies Pa Ps in V'(9),so (Pax,x ) 5 (Psx,x ) for all x E H), it follows immediately that the null space N(P,) decreases and the range R(P,) increases as CL increases. Furthermore, there exist real numbers a and b such that aE 6 A 5 bE holds, and we have Pa = 8 for u 5 a and Pa = E for u > b, i.e., R(P,) consists only of the null element of H for u 6 a, and R(P,) is the whole space H for u > b. Before proceeding with the investigation of the projections Pa in the spectral system of A, we recall the general theorem that if Pl and P2are commuting orthogonal projections in the Hilbert space H, then P = P1P2is also
390
HERMITIAN OPERATORS IN HILBERT SPACE
[CH. 8,s 56
an orthogonal projection with range equal to the intersection of the ranges of PI and P2. We include a brief proof. It is trivial to see that P is Hermitian and P 2 = P, so P is an orthogonal projection. It is also evident that R(P,) n R(P,) is included in R(P). For the converse, assume that x E R(P), so x = P , P z x . Then P, x = P: P 2 x = P , P z x = x , so x E R(P,). Similarly, we have x E R(P2), so x E R(P,) n R(P2). Returning to the spectral system of A , we observe that, for - 00 < ci S p < 00, the element P,-P, is also a component of E (more precisely, P,-Pa is the component of E in the band B, n B:). It is evident from Pa S P, that Pa is an element of the band B,, so Pa I(E-P,), i.e., P,(E-P,)
=
(E-P,)P,
=
e,
and so Pa = PUP, = P,P,. It follows that PB-P, = P,(E-P,)
=
(E-P,)P,,
which implies (in view of what was observed above about commuting orthogonal projections) that the range of the orthogonal projection P, -P, in the Hilbert space H is the intersection of the ranges of P, and E-Pa, i.e., R(P,-P,)
= R(P,) n N(P,),
which is the same as the orthogonal complement of R(P,) with respect to R(P,). Finally, if -00 < a, S p, S ci2 5 p2 < 00, we have so the ranges of PB,-Pa, and PB2-Pa, are mutually orthogonal subspaces of H.
Lemma 56.2. (i) The operator A, A,
=
=
B,A satisfies
P,A = AP,,
and SO we have A,x = A x for all x E R(P,) and A,x = 0 for all x E N(P,). Furthermore, A leaves R(P,) and N(P,) invariant, i.e., x E R(P,) implies A x E R(P,) and x E N(P,) implies A x E N(P,). (ii) We have (Ax, x ) S a(x, x ) for all x E R(P,) and ( A x , x ) 2 a(x,x ) for all x E N(P,). Hence, for - 00 < a S p < a,we have a(x, x ) S ( A x , x) S B(x, x )
for all x E R(P,) n N(P,) = R(P, -Pa). Furthermore, A leaves R(P,-P,) invariant.
CH. 8,s 561
Proof. (i) Since A ,
I(E-Pa), we have A,(E-P,)
so A ,
=
A,P,
391
THE SPECTRAL THEOREM
=
=
(E-P,)A, =
e,
IPa, we have = P,(A - A , ) = e,
P,A,. Since ( A - A,)
( A -A,)P,
so AP, = A,Pa and P,A = P,A,. It follows that A , = P,A
=
AP,.
In order to show that A leaves R(P,) invariant, assume that x E R(P,). Then A x = AP,x = P,Ax, so A x E R(P,). In order to show that A leaves N(P,) invariant, assume that x E N(P,), i.e., Pax = 0. Then P,Ax = A P g = 0, so A x E N(P,). Note that since N(P,) and R(P,) are orthogonal complements in H , we can also derive the invariance of one from the invariance of the other one by means of the theorem that if a Hermitian operator leaves a linear subspace invariant, then the operator also leaves the orthogonal complement invariant. (ii) By Theorem 38.4 (i) we have 9 , A S aP, for all a, i.e., A , S aP,. It follows that (A,x, x ) 5 a(P,x, x ) for all x E H, i.e., (AP,x, x ) S a(P,x, x ) for all x E H.
In particular, for x E R(P,), it follows that (Ax, x ) By Theorem 38.4 (ii) we have A-Y,A
5 a(x, x).
2 a(E-Pa) for all a,
i.e., A - A , 2 a(E-Pa). It follows that ( A x , x ) - (AP,x, x ) 2 a(x, x ) - .(Pax, x ) for all x E H. In particular, for x E N(P,), it follows that (Ax, x ) 2 a(x, x). Since A leaves R(PB)and N(P,) invariant, A also leaves the intersection R ( P B )n N(P,) = R(PB-P,) invariant. Given now the real numbers a and b and the positive number E such that U E 5 A 5 (b-&)E,
we assign to any partition n = n(ao, a t , . . ., a,) of [a, b ] the lower sum s(n; A ) and the upper sum t(n;A), given by
n
392
[a. 8,B 56
HERMITIAN OPERATORS I N HILBERT SPACE
The sums s(n; A) and t(n;A) are Hermitian operators of the form B k Q k , where the operators Qk (k = 1,. . ., n) are orthogonal projections in the Hilbert space H with mutually orthogonal ranges such that the direct sum of these ranges is the whole Hilbert space H , and the B k (k = 1, .. ., n ) are real numbers. Hence, the numbers Pk are the eigenvalues of & B k Q k , and the ranges R(Qk) are the corresponding eigenspaces. Furthermore, we have s(n; A) 5 A 5 r(n; A), and e 6 t ( n ;A ) - + ; A ) 5 I ~ I E , where In1 is the maximal interval length in the partition
e 6 A - s ( ~ ;A ) 5 e 5 t(n; A ) - A 5
R.
Hence
lnl~, I~IE,
which implies that the operator norms IIA-s(n; A)ll and Ilt(n;A)- All are less than or at most equal to 1x1. The spectral theorem for the Hermitian operator A follows now from Freudenthal's spectral theorem in Riesz spaces (Theorem 40.2), as follows.
Theorem 56.3 (Spectral theorem for a Hermitian operator). If A is a Hermitian operator in the Hilbert space H with spectral system (Pa ; - co < u < co), and the real numbers a, b are such that aE
5 A =< (b-E)E
for some E > 0, and if (n,,:n = 1,2, . . .) is a sequence of partitions of [a, b], each of which is a rejinement of its predecessor and such that In1 . 10, then s(n,,;A ) 4 A 6 t(n,,;A ) holdsfor every n, and s(n,,;A ) as well as t(n,,;A ) converge to A in the operator norm, i.e., IIA-s(nn; A)II + 0, llA-t(nn; A)ll
+
0
as n 4 ax Furthermore, in the Riesz space %'"({A)) or, for that matter, even in the ordered vector space 2,the increasing sequence ~(n,,; A ) has A as its supremum, and the decreasing sequence t(x,,; A ) has A as its infimum. Proof. Follows by applying Freudenthal's spectral theorem in the Riesz
CH. 8,s 561
393
THE SPECTRAL THEOREM
space %“({A}) and by observing that, on account of Lemma 56.1, E-uniform convergence is the same as convergence in the operator norm (usually called uniform convergence of operators). As already observed in the discussion of Freudenthal’s spectral theorem, the spectral theorem is often expressed by writing A
J-mu dP, . a?
=
We shall make no use of this notation. Note, however, that in Exercises 56.11, 56.12 and 56.13 it is indicated how the spectral system (Pg: - 00 < u < 0 0 ) generates a projection valued “measure” in the real line; the integral udP, can be interpreted as a Stieltjes-Lebesguetype integral with respect to the thus generated measure. The uniqueness result in Theorem 40.8, when applied to the Hermitian operator A in the last theorem, yields the following result.
Theorem 56.4. Let (Q, : - co < u < co) be a set of orthogonalprojections in the Hilbert space H , all Q, members of the second commutant W‘({A}),
and such that (i) Q, is increasing as u increases, (ii) Q,TQ,asuTP, Q , = e f o r a S a a n d Q , = E f o r u > b , (iii) for every sequence of partitions K , , ( U , , ~ , a n l , . . ., a,,,,,,) of [a, b ], each n,+ a refinement of K,, and such that ln,,l 1 0, the sequence m,
=
C1
k=
un,k-l(Q a* -Q am,k-a )
satisfies s; 7 A, i.e., (stx, x ) (Ax, x ) for every x E H (since, in any case, si increases as n increases, this condition is satisfied in particular whenever si converges to A in the operator norm). Then (Q,: - 00 < u < co) is the spectral system of A. Generally, for bounded linear operators, if IIA-A,Il 0 and IIB-BnII + 0 as n + co,then (IAB-A,,B,II 4 0 as n 4 00. Applying this to the case that A is the same Hermitian operator as before and B is a power of A, it follows easily by induction that -+
llAk-?(nn; A)ll
-+
0 as n
+ 00
for any natural number k, where (s(K,; A) :n = 1,2, . . .) is the same sequence of lower sums as before. The k-th power of s(n,; A) is easily determin-
394
HERMITIAN OPERATORS IN HILBERT SPACE
[CH. 8,s 56
ed. Indeed, since
with mutually disjoint terms, we have
Similarly, where
-,o
I I ~ ~ - t ~ ( i qA, ,) ;I ~
as n
--f
00,
There are several further remarks. In the first place, given the set 9 of mutually commuting elements of X , we consider the Yosida representation " { W ' ( 9 ) } of the Dedekind complete Riesz space W ' ( 9 ) . The elements of "{%'"( are 9) the } real continuous functions on the compact topological space $, the points of which are the maximal ideals in W ' ( 9 ) . The vector space structure and the lattice structure in "{W'( g ) }are the usual ones, so " { V f ( 9 )is}a Riesz space. The operator A E W ' ( 9 ) is mapped onto the function "u E " { W f ( 9 ) } where , the value "u(J) of "u at the point J E $ is the (unique) real number L satisfying A -AE E J. This mapping is a Riesz isomorphism of %"( 9 )onto "{V'(9)}. We return to the problem (mentioned in the preceding section) whether the operator multiplication in W ' ( 9 ) is the same as the multiplication inherited from the pointwise multiplication of the functions in the Yosida representation " { W ' ( 9 ) } .First we compare operator multiplication of components P, and P2 of E (these components are members of %"(9), of course) and the pointwise multiplication of the corresponding functions "p, and "p2. The components PIand P2 are commuting orthogonal projections in the Hilbert space H,let R(P,)and R(P2) be their ranges. Then P = P,P2 = P2P, is the orthogonal projection with range R ( P l ) n R(P2),and obviously P is a lower bound of P, and P2. Any other member A of W f ( 9 ) ,satisfying 8 5 A 5 PI and 8 6 A 5 P2,must vanish on each of the null spaces N ( P l ) and N ( P 2 ) ,so the range of A is included in R(P) = R(P,)n R(P2),which implies that A vanishes on the orthogonal complement of R(P). From the hypotheses it follows that 0 5 (Ax, x) 5 (x, x) holds for all x E R(P), so in view of (Px, x) = (x, x) holding for all these x, it follows now immediately that A 5 P.This shows
CH.8,§ 561
THE SPECTRAL THEOREM
395
that P is the greatest lower bound of P, and P,, i.e.,
P,P, = P2Pl
=
inf (P,,P,),
where inf (P,,Pz)denoles the infimum of P, and P2 in the Riesz space W'(9). The corresponding functions "pl and "p2 can assume only the values zero and one (since "pl and "p, are components of the function identically equal to one), so A
PI "P2
= ^P2 "P1 = inf ("PI 9
"PZ), where inf ("p,, "p2)denotes the infimum of "p, and "p2 in the Riesz space "{W'( 9)}Since . inf (P,,Pz)and inf ("pl, "p2)are corresponding elements in the Riesz isomorphism, it follows that P,P2 and "pi "p2are corresponding. But then A B and " d b correspond for A and B finite linear combinations of components of E in W"(9). Observe, finally, that for arbitrary members A and B of ""(9) there exist sequences A,, and B,,(n = 1, 2, . . .) in W"(9), each A,, and each B,, a finite linear combination of components of E, such that llA- A,,I I and IIB- B,,I I tend to zero as n + 00. It follows that IIAB- A , B.11 tends to zero, and the functions "a,,"b,,, corresponding to A,,B,,, converge uniformly in "{W'(L2)} to " d b . This shows that A B and "a"b are corresponding elements under the Riesz isomorphism. Hence, operator multiplication in W ' ( 9 ) is the same as the multiplication inherited from the pointwise multiplication of functions in "{W'( 9)). Secondly, it will be proved that the definition of the spectrum of an element in an Archimedean Riesz space with a strong unit, as presented in section 44 (cf. also Theorem 45.1), is equivalent to the familiar definition of the spectrum of a Hermitian operator A when we apply the general definition to the Riesz space %"({A}) or, more generally, to any Riesz space U " ( 9 ) of which the given operator A is a member. We recall the general definition. Given the Archimedean Riesz space L with strong unit e, we consider the Yosida representation "L. The spectrum o f f € L (with respect to the unit e ) is, by definition, the range of the function "E' "L corresponding tof, i.e., the real number c1 is in the spectrum off whenever there exists a maximal ideal J i n L such that u e - f E J. Assuming now that the Riesz space L is Dedekind complete (or even only Dedekind a-complete), the representation " L consists of all real continuous functions on the compact topological space f of all maximal ideals, and so it follows easily that for any given g E L the number zero is not in the spectrum of g if and only if there exists a function "h in "L such that "g"h = "h"g = "e, i.e., "g(J)"h(J) = 1 for all points J E$. The element h in L , corresponding to "h, is then the inverse of g with respect to the multiplication in L inherited from the point-
396
HERMITIAN OPERATORS IN HILBERT SPACE
[CH.
8,s 56
wise multiplication of the functions in " L , i.e., h = g-' in the usual notation. Hence, givenf E L and the real number ao, we have that a. is not in the spectrum off if and only if aoe-fhas an inverse in "L with respect to the multiplication in L inherited from "L. We consider now the Dedekind complete Riesz space W'(9), where 9 is again a collection of mutually commuting Hermitian operators in the Hilbert space H , and we observe first that the identity operator E is a strong unit in C'(9).Applying the remarks made above to the present situation, we obtain the result that, for any A in C"(9) and any real number a. ,the number a. is not in the spectrum of A if and only if a. E-A has an inverse in W'(9) with respect to the multiplication in W'(9) inherited from the pointwise multiplication in A{W(9)}. This multiplication, however, is the ordinary operator multiplication, as proved in the preceding remark. Hence, a. fails to be in the spectrum of A if and only if (aoE-A)-' exists as a member of W'( 9).This is almost the familiar definition. Indeed, according to the usual definition in Hilbert space theory we say that a. fails to be in the spectrum of A whenever ( a o E - A ) - l exists as a bounded linear operator. In order to show that the two definitions are equivalent, it remains to show that if A is a member of C"( 9 )and A - ' exists as a bounded linear operator, then A-' is also a member of W ' ( 9 ) . It is trivial to prove that A-' is Hermitian, and so all we have to show is that A-' commutes with every member of W ( 9 ) . For this purpose, let B in C ' ( 9 )and x and y in H be given, and set z = A-'y, so y = Az. Then ( A - ' B x , y ) = (A-'Bx, Az) = (Bx, Z ) = (x, Bz) = (A-'x, ABz) = (A-Ix, BAz) = ( B A - l x , v). This holds for all x , y E H , so A - ' B = BA-'. The proof is thus complete. We will now extendthe spectral theorem to bounded normaloperators in the Hilbert space H. For this purpose we recall some elementary facts. For any bounded linear operator T (from H into H ) there exists a unique bounded linear operator T* (the adjoint of T) such that (Tx,y) = (x, T*y) holds for all x and y in H. We have T* = T if and only if T is Hermitian (Le., if and only if T is self-adjoint), and T is called normal whenever T and T* commute, i.e., whenever TT*= T*T. For any bounded linear operator T, setting
A = +(T+T*)and B = -$(T-T*), the operators A and B are Hermitian, and T = A + i B and T* = A-iB.
CH.8,o 561
397
THE SPECTRAL THEOREM
This is the decomposition of an arbitrary T into Hermitian components. The decomposition is unique (Le., if A +iB = A'+iB' with A , B, A', B' Hermitian, then A = A' and B = B'). Furthermore, if we have T = A + i B with A and B Hermitian, then T is normal if and only if A and B commute. Assume now that N i s a normal bounded linear operator, and let N = A + iB be the decomposition into Hermitian components. We consider the Riesz space %"({A, B ) ) or, more generally, any Riesz space W'( 9),where 9 is a set of commuting Hermitian operators of which A and B are members. Let (P,: - 00 < a < a)and (Pi : - 00 < B < m) be the spectral systems of A and B respectively. All members of these spectral systems are members of %'"(9), and so they commute mutually. Now,for brevity, we shall denote by [a1 , a,; B1, B z ) the half open rectangle in the a/?-plane, given by ( ( a y B ) :a1
5
a
< a t , Bi
sB<
B2).
The corresponding operator
Q c ~uz: .
= (p=~-pui)(Pi~-ph1)
~ 1 , ~ 2 )
is an orthogonal projection in H with range equal to the intersection of the ranges of the projections Pal -Pu,and Pil -Pil. It follows immediately that Q ~ u i . u z ; ~ i~, z )
and
Q[a;a4:B3.P4)
have mutually orthogonal ranges (i.e., their product equals 0) as soon as the rectangles [a1, a,; pl, /I2)and [a3, a4; B 3 , B4) in the aB-plane are disjoint. Choose the real numbers a, b and the positive number E such that aE 5 A I; (b-E)E and aE 6 B 5 (b-e)E hold, let n(aO, al,. ., an) and n'(po, PI , ...,B,) be partitions of [a, b], and let
.
m
be the corresponding lower sums. According to the spectral theorem for Hermitian operators, s(n; A ) and s(n'; B ) converge in the operator norm to A and B respectively as 1111 0 and ln'l '+0 ; hence,
+
s(n, n'; N ) = s(n; A ) i s(n'; B )
converges in the operator norm to N = A + i B as
1% n'l
= max (bl, In'l)
398
HERMITIAN OPERATORS
[CH. 8,g 56
IN HILBERT SPACE
tends to zero. In order to investigate the sum s(x, A'; N ) more closely, we set Qkj
= ~ p a k - P a k - ] j ( P i j - ~ ~ , - ,fjo r k = 1 , .
. ., n
and j = 1 , . . ., u i .
Observing that m
we find
s(n, n'; N ) = s ( x ; A ) + i s ( n ' ; B ) n
n
=
m
m
C C (ak-l k=l j=1
+iBj-I)Qkj.
The sum s(z, n'; N ) might appropriately be called, therefore, a kind of lower sum corresponding to the partition (n,z') of the square [a, b ; a, b ] .The partition (z, d)is determined by the partitions a and d along the axes. As observed above, we have IIN-s(n, d ;N)II
+0
as I(n, A')I
-,0.
Similarly, we have where
(IN-r(n, n'; N)II
4
n
t(n, X ' ; N ) = C k=l
0 as
((71,
.)'I
+ 0,
m
C (ffk+iSj)Qkj. j=1
More generally, the following holds. Theorem 56.5 (Spectral theorem for a normal operator). With the notations introduced above, we have n
m
where C k j is any complex number in the rectangle times this is written as N = fJabC
dQr.
[ f f k - 1, f f k ;
Pi-
1,
Pi]. Some-
CH. 8 , § 561
399
THE SPECTRAL THEOREM
Once more, let 93 bs a set of mutually commuting (bounded) Hermitian operators in the Hilbert space H , and let U " ( 9 ) be the second commutant of 9.Any operator N = A + iB, with A and B members of V'(Q), is normal. The set of all normal operators of this kind will be denoted by M . Evidently, N is a complex linear subspace of the complex linear space of all bounded linear operators on H . The members of N commute mutually, and N is a ring in the algebraic sense (i.e., if N , and N , are members of N ,then N1N2 = N2N , is a member of N). For A , B, . . . members of C"( 9),let a, "b, . . . be the corresponding members of the Yosida representation "{V'(9)}. For N = A + i B e M , we will say that the complex function n = "a+i"b represents N . Evidently, "n is a continuous complex function on the compact Hausdorff space $ of all maximal ideals J in u"(9). Conversely, given any continuous complex function on $, the real and imaginary parts "a and "b of "n correspond to members A and B of %''(.9), and so "n corresponds to the member N = A + iB of N.It follows immediately that JV is ring isomorphic to the ring of all continuous complex functions on $. For any N E N ,we define the spectrum of N to consist of all complex numbers in the range of the function "n which corresponds to N . It is evident from the definition that the number zero is not in the spectrum of N if and only if "n has an inverse, i.e., if and only if there exists a continuous complex function "m on $ such that A
A
"n(J)"m(J) = "m(J)"n(J) = 1
holds for all J E f . This is equivalent, therefore, to saying that zero is not in the spectrum of N if and only if there exists a member M of N such that M N = NM = E holds. If this is the case, then M is the inverse operator N-' of N . Conversely, if N is a member of N and the inverse N-' exists as a bounded linear operator, then N-' is also a member o f N , and so zero is not in the spectrum of N.The proof that N-' is a member of N is not completely trivial. The details follow. First of all, we infer from the wellknown formula "-')* = (N*)-' that
N-'(N-')* = N-'(N*)-'
=
(N*N)-' =
("*)-I
- (N*)-"-'
=
(N-')*N-',
so N-' is normal. Let N-' = P + i Q be the decomposition of N-' into Hermitian components, and note that ( N - ' ) * = P - i Q . Since Nis a member of N ,any Hermitian A in the first commutant W ( 9 )commutes with N , i.e.,
400
[CH.8,5 56
HERMITIAN OPERATORS IN HILBERT SPACE
A N = NA. Multiplying on the left and the right by N - ' , we obtain N - ' A = A N - ' , so A commutes with N-' = P + iQ. Similarly,anyA in%?'( G) commutes with ( N - ' ) * = P - i Q . Hence A commutes with P and Q separately. It follows that P and Q are members of W ' ( 9 ) and so N-' is a member of Jlr. Returning to the spectrum of N , we have proved thus that the number zero is not in the spectrum of N if and only if N-' exists as a bounded linear operator. It is an easy consequence that any given complex number c( is not in the spectrum of N if and only if (aE-N)-' exists as a bounded linear operator. Hence, our definition of the spectrum of a normal operator is equivalent to the familiar definition. It follows also that if N = A iB is the decomposition into Hermitian components of a normal operator Nand G is any set of commuting Hermitian operators of which A and B are members, then the spectrum of N is independent of the particular choice of 9. For any N = A + iB E N the , spectrum of N is the range of the continuous function "n = "a+i"b on the compact Hausdorff space $, and so the spectrum of N is a compact subset of the complex plane. The adjoint operator N* = A - i B corresponds to the complex conjugate function "n = "a-i"b, so the spectrum of N* is obtained from the spectrum of N by reflection with respect to the real axis. Since N is a ring, the operator NN* = N*N corresponds to the function " n 5 = l"n12. Hence, the positive Hermitian operator NN* has its spectrum equal to the range of the function I %I2. Now, and this holds for any Hermitian operator A , there is an obvious relation between the operator norm IlAll and the values of the function "a corresponding to A , as follows. Writing IlAll = m for brevity, we have -mE 5 A 5 mE, so - m 5 ^a(J) 5 m for all J E $, it., I^a(J)I 5 m for all J E $. Applying this to the Hermitian operator NN* = N*N, and observing that
+
we obtain the result that l"n12(J) IIN1I2 holds for all J, i.e., I"nI(J) 5 IlNIl for all J. This shows that the range of "n is contained in the subset (z : lzl 6 IlNll) of the complex plane. In other words, the spectrum of N is a compact subset of the set (z : Izl 5 IlNll). It was observed above that the spectrum of NN* = N*Nis the range of the function IAnlz. I n particular, if Nis a unitary operator, i.e., if NN* = N*N = E, then I "n1'(J) = 1 for every J E $, so the spectrum of N is a subset of the boundary of the unit circle in the complex plane.
CH. 8,s 561
THE SPECTRAL THEOREM
40 1
Exercise 56.6. (i) Show that if A E W'(9) corresponds to the function "a, then the uniform norm of "a is exactly equal to IlAll. Hence, one at least of the numbers llAll and -1IAll is in the spectrum of A. (ii) Show that if the normal operator N E JV corresponds to the function "n, then the uniform norm of "n is exactly equal to 11N11. Hence, the spectrum of N has at least one number of absolute value exactly equal to I lNlI. Hint: We may assume that A is positive. For brevity, write IlAll = m. If "a(J) < m for all J, then the function m- "a has an inverse, so mE- A has an inverse. It follows that, for some positive constantp, we have Il(mE-A)xll B pllxll for all x , so ((mE-A)x, x ) 2 p ( x , x ) for all x. This implies m- (Ax, x ) 5 p for all x with llxll = 1. Contradiction. Exercise 56.7. Let Pa(- co < a < co) be the spectral system of the Hermitian operator A. Show that a, is not in the spectrum of A if and only if there is a neighborhood of a, such that Pa is constant in this neighborhood, i.e., there exist real numbers a1 and a, such that a1 < a, < a, and PoIz-Palis the null operator. Hint: Combine Theorem 45.1 and Lemma 55.5 (ii). Exercise 56.8. Let N = A + i B be the decomposition into Hermitian components of the normal bounded linear operator N , and let (Pa : - co < a < a)and (Pi : - co < jl < co) be the spectral systems of A and B respectively. Also, let [xl,x,; y l , y z ]be a rectangle in the complex plane, completely outside the circle (z : IzI 5 IlNlI), and let Qo = (pxz-pxl)(p~2-p~,)
be the corresponding orthogonal projection. Show that Q , is the null operator. Hint: With the notations of Theorem 56.5 we have
so holds for every x E H. Choose E > 0 such that the given rectangle is contained in (z : IzI > IINII + E ) . Assume now that Q , is not the null operator. Then there exists x,, E H such that Ilxol I = 1 and Q , x, = xo , so (Q,x, ,x,) = 1. With the notations of Theorem 56.5, choose partitions 'II and n' such that the given rectangle is inside the square [a, b; a, b] and such that xl, xz and y , ,y , are among the endpoints of the subintervals of n and n' respectively.
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HERMITIAN OPERATORS I N HILBERT SPACE
[CH.
8,s 56
For ( x , x ' ) we have c = C C lckj12(Qkjxo, =
XO)
=
(CC )'1ckj12(Qkjxo,xo),
C C ICkjI'(QkjQox0,
XO)
where the last summation is exclusively over the combinations (k,j ) for which the corresponding rectangle lies inside [ x , , x,; yl ,y 2 ] .Hence
On the other hand, we have
Inequalities (2) and (3) together contradict (1). Exercise 56.9. Let N = A + iB be the same as in the preceding exercise, and also let Paand Pi be the same. For any rectangle p = [xl ,x,; yl ,y z ] in the complex plane, we shall denote the corresponding orthogonal projection
briefly by Q ( p ) . Assume now that the origin of the complex plane is not in the spectrum of N. Show that there exists a rectangle p around the origin such that Q ( p ) is the null operator. Hint: There exists a number C > 0 such that IINxlI 2 Cllxll for all x E H , i.e., (N*Nx, x ) 2 C211x112for all x E H. Take a rectangle p around the origin and contained in (z : IzI 5 *C), and show (similarly as in the preceding exercise) that Q ( p ) is the null operator. Exercise 56.10. For the same operator N as in the prueding exercise, assume that the complex number zo = xo+ iyo is not in the spectrum of N . Show that there exists a rectangle p around zo such that Q @ ) is the null operator. Hint: Observe that the origin is not in the spectrum of N-zoE, and apply the result of the preceding exercise. Exercise 56.11. Let Pa(- m < a < m) be the spectral system of the Hermitian operator A in the Hilbert space H . For any half open interval D = [a1 , a,) and any x E H , write P ( D ) = Pa,-Pa,and p x ( D ) = (P(D)x,x ) .
CH. 8, 6 561
THE SPECTRAL THEOREM
403
Since the real function g,(a) = (Pax,x ) is monotonely increasing in a and left continuous at every point ao(on account of P, t Pa,,as a 1cr,), it follows along familiar lines that on the semi-ring of all intervals D = [al , crz) the set function p, is a non-negative and countably additive measure. The measure p, is extended by means of the Carathbodory extension procedure. Every Borel subset of the real line is then px-measurable. This is done for every x E H , and hence every p x is a Stieltjes-Lebesgue measure on the a-field (a-algebra) of all Borel sets. Since p,(D) = (P(D)x,x ) 5 (x, x ) holds for every D = [a1, a2), it follows easily that p J E ) 5 Ilxll’ holds for every Borel set E, in particular for E equal to the real line itself. Now, keep the Borel set E fixed, and show that the square root of px(E),as a function of x, is a seminorm in the Hilbert space H , obeying the parallelogram law and majorized by the norm llxll in H, i.e., pc,,(E) = I c 1 2 p x ( ~for ) a11 complex c, {PX+Y(E)l3 P x +Y@)
PJE)
s {Px(E)13+ {PY(E)I*Y
+P.x - y(E) = 2P,(E) +2PY(E)Y
(4)
5 IlXllZ.
Hence, by a well-known theorem of J. von Neumann and P. Jordan (for a proof we refer to Wilansky [I], section 8.3), the functional qE(x, Y ) = i { ~ x + y ( E )-A - y ( E l + i ~ xd+E ) - i ~ x- iy(E)>
is a semi-inner product (i.e., it has all the properties of an inner product except that q E ( x ,x ) m a y be zero without x being the null element). Also, rp,(x, x ) = p,(E) for all x , and so I ~ E ( X U)l2 Y
5 p ~ ( E ) p y ( ~ )llxl12 ’ IlYl12
holds for all x , y in H. Hint: The formulas (4) hold for E = D = [al,a2), and by familiar measure theoretic arguments can then be shown to hold for every Borel set E.
Exercise 56.12. This is a continuation of the preceding exercise. For E and y fixed, q E ( x , y )is a bounded linear functional on H, so there exists y1 E H such that q E ( x ,y ) = ( x , yl) holds for all x. Evidently, y1 depends on y and E. Writingy, = T(E)y,it follows immediately that T ( E ) is a bounded linear operator satisfying IIT(E)II 5 1. Hence, q E ( x , y )= ( x , T (E )y )holds for all x , y in H, and IIT(E)I I 5 1. Also, since qE(x,y ) and q E ( y ,x ) are complex conjugate numbers, it follows that T ( E ) is Hermitian, so q E ( x , y )=
404
[CH. 8,§ 56
HERMITIAN OPERATORS IN HILBERT SPACE
(T(E)x,y ) holds for all x , y in H. For all x
E Hand
(T(D)x,x ) = CpD(X, x) = A ( D ) =
all D = [ a , , r 2 )we have
( W ) X ,
x),
so (since T ( D ) and P ( D ) are Hermitian) we may conclude that T ( D ) coincides with the orthogonal projectionP(D) = Pa,-Pa,. Show that if E is afinite union of half open intervals (open on the right), then T ( E ) is also an orthogonal projection, and note that the collection of all such finite unions is a ring. Finally, by observing that the smallest monotone family containing a given ring is the same as the smallest a-ring containing the same ring (cf. the measure theory book by P. Halmos [l], section 6), show that T ( E )is an orthogonal projection for every Borel set E. Hint: Observe that a finite union of half open intervals can be written as a disjoint finite union of such intervals. For the last statement, fix x E H and prove that the collection of all Borel sets F satisfying
(T(F)x, T(F)x) = (T(F)x,x ) is a monotone family, which, therefore, must coincide with the collection of all Borel sets.
Exercise 56.13. This is a continuation of the preceding exercise. For any Borel set E, write P(E) instead of T ( E) . Thus the set function P,defined for the Borel sets, is now an extension of the original set function P, defined
only for the half open intervals. The set function P is a projection valued measure on the a-field of the Borel sets (i.e., P(+) is the null operator, P ( E ) is an orthogonal projection for every Borel set E, and P is a-additive, which means that P ( E ) = P(E,)+P(E,) for E = El u E2 with El and E2 disjoint and P(En)t P( E) for En t E). Furthermore, if the real numbers a, b have the property that the given Hermitian operator A is between a times and b times the identity operator, then P(E) is equal to the identity operator for any Econtaining [a, b] in its interior. Show that, for any Borel sets El and E 2 , we have
P(El u &)+f‘(E1 n E2) = P(E1)+P(E2)7
P(E1 n E2)
= P(El)P&),
(5)
so in particular P(E,)P(E,) is the null operator for El and E2 disjoint. Given a partition n = (. . ., LL,, a,,, a , , . . .) of the real line, A is approximated in norm by
CH.8 , § 571
so
405
THE STRUCTURE OF THE SECOND COMMUTANT
( A x , x ) = lim
1a,(P(D,)x, x) = lim C a,$,(&)
Inl-10
Inl-tO
The sum on the right is exactly the Stielljes-Lebesgue integral (with respect to the measure p x ) of the step function equal to on Dk for each k. Hence, we obtain
s
( A x , x ) = adp,. Since p, is generated by the monotone function g,(cx) be written as
=
(Pax,x), this can
( A x , x) = / a d g x ( a ) = J a d ( P a x , x). Show, similarly, that for x and y fixed the set function p* ( E ) = (P(E)x,y ) is a complex measure on the a-field of the Bore1 sets, and show that
( A x , y) = s a d p - = s a d ( P a x ,y). Hint: In order to prove the first formula in (5), add P(El n E,) to both sides of P(E, u E,) = P(E1)+P(E, - € 1 ) .
From P(El n E,)
5 P(E,) S P(E, u
E,) it follows that
P(E,)P(E, n E,) = P(E, n E,) and P(El)P(E1u E,) = P(El). Hence, multiplying the first formula in ( 5 ) by P(E,), we obtain P(El)+P(El n Ed = P(El)+P(E,)P(E,)Y which proves the second formula in (5).
57. The structure of the second commutant In this section we will prove some additional theorems about the second commutant %”( g ) ,where 9 is once more a set of mutually commuting Hermitian operators in the Hilbert space H . For each A E 9 the null space N ( A ) of A is a closed linear subspace of H, and so H’ = n ( N ( A ) : A
E
9)
is also a closed linear subspace of H. Let H” be the orthogonal complement of
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HERMITIAN OPERATORS IN HILBERT SPACE
[CH.8,s 57
H', and let P' and P" = E-P' be the orthogonal projections in H having as their ranges the subspaces H' and H" respectively. It is a reasonable conjecture that the behavior of an element of %"( 9 )on the subspace H cannot be of great importance; we will prove that this is indeed true. We first recall that if the Hermitian operator B leaves a closed linear subspace H 1 invariant, then B leaves the orthogonal complement of H , invariant. Indeed, given that Bx E H , holds for all x E H I , any y IH , satisfies (By, x ) = ( y , B x ) = 0 for all x E H , ,so By IH I . Furthermore, if A and B are commuting Hermitian operators, then B leaves the null space N ( A ) and its orthogonal complement R ( A ) invariant (we recall that, by our definitions, R ( A ) is the closure of the range of A). Indeed, x E N ( A ) implies ABx = BAx = 0, so Bx E N(A). This shows that B leaves N ( A ) invariant, so that, by the preceding remark, B leaves R ( A ) invariant as well.
Theorem 57.1. As above, let 9 be a set of mutually commuting Hermitian operators in the Hilbert space H , let H' = n ( N ( A ) : A E 9 )and let H" be the orthogonal complement of H . Finally, let P' and P" be the orthogonalprojections on H and H' respectively. Then the following holds. (i) Each B E V(B) leaves H and H' invariant, and among the elements of W( 9 )are all Hermitian B which vanish on H" (i.e., Bx = 0for all x E H I ) . These are exactly the Hermitian B satisfying A B = BA = 8for all A E 9. (ii) Each B E "'(23) leaves H' and H" invariant, atid the restriction of B on H' is a real multiple of the identity operator on H'. In other words, there exists a real number a such that B - aP' is the null operator on H' (and, of course' B - UP'is equal to B on H ' ) . In the converse direction, any real multiple of P' is an element of W ' ( 9 ) . (iii) The principal band d generated by P' in W ' ( 9 ) consists of all real multiples of P'; the band L3 generated in V ' ( 9 )by the elements of 9 is the disjoint complement of d ,and we have
at @ 9l
=
W'(9).
Proof. (i) Since H and H" are orthogonal complements, it is sufficient to show that each B E V ( 9 )leaves H invariant. Given B E %'(9 we ), have A B = BA for every A E 9, and so B leaves N ( A ) invariant for every A E 9. But then B leaves H' = n ( N ( A ) :A E 9 ) invariant. Now, let B be Hermitian such that B x = 0 for all x E H'.This is equivalent to saying that H' is included in N(B), i.e., R(B) c H . Observing that
CH.8,s 571
THE STRUCTURE OF THE SECOND COMMUTANT
407
H c N(A) or, equivalently, R(A) c H'holds for all A E 9,we derive from R(B) c H and R(A) c H' for all A E 9 that R(B) is orthogonal to all R(A), and so AB = BA = 8 for all A E 9. It follows in particular that B commutes with every A E 9, and so B E W ( 9 ) . Conversely, if B is Hermitian such that AB = BA = 8 for all A E 9, then R ( B ) c N(A) for all A E 9, so R(B) c n (N(A) :A
E
9)= H'.
This implies that HI' c N(B), i.e., B vanishes on H'. (ii) The operator P' is Hermitian, and P'x = 0 holds for every x E H". It follows from part (i) of the present proof that P' is a member of W( 9). Each B E W ' ( 9 ) commutes therefore with P', and so B leaves the null space H'of P'invariant. But then B leaves the orthogonal complement H' invariant as well. The operator B in C"( 9)commutes with every C in W( 9), and so the restriction of B on H commutes with the restriction of C on H . But on account of what was proved in part (i), if we let C run through V ( 9 ) ,the restriction of C o n H runs through the set of all Hermitian operators in the Hilbert space H'. Hence, by Lemma 53.2 (i), the restriction of B on H must be a real multiple of the identity operator on H . In other words, there exists a real number a such that B-UP' is equal to the null operator on H'. Of course, B-UP' is equal to B on H ' . In the converse direction, every real multiple of P' is evidently an element of W ' ( 9 ) . (iii) If B is a member of the ideal generated in W ' ( 9 ) by P', then B is between two multiples of P', and so B vanishes on H ' . Furthermore, by part (ii), B is on H a real multiple of the identity operator. These facts together show that B is a real multiple of P' on the whole space H. It follows now that if B is a member of the band d generated in u"(9)by P', then B is also a real multiple of P'. Since AP' = P'A = 8 holds for every A E 9, we have P' IA for every A E 9, and so P' IB holds for every B in the band 3? generated in %"(9 ) by the elements of 9. This shows that the bands CQZ and 3? are disjoint. Since "'(9)is Dedekind complete, the direct sum d'0 3? is a band in %"'(9 and ), not merely an ideal. Let CEC ' ( 9 )be disjoint to d 0 3?, so C I P' and C I A for all A E 9. Then CA = AC = 8 for all A E 9, i.e., R(C) is contained in N(A) for every A E 9.It follows that R ( C ) is contained in the intersection of all N(A), i.e., R(C)c H . But C IP' implies that CP'= P'C = 8, so R(C)is contained in the null space H ' of P'. But then R ( C ) is = g"(9). contained in H' n H", so C = 8. It follows that d 0
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HERMITIAN OPERATORS IN HILBERT SPACE
[CH.8 , s 57
As observed earlier, every polynomial in the elements of 9 and with real coefficients, a constant term consisting of a real multiple of E included, is a member of %"(9), and it may be asked in a rather vague sense how dense the set of these polynomials is lying in Evidently, the band genersimply because the ideal ated by these polynomials is the whole of %'"(9), generated by E alone is already the whole of W ' ( 9 ) . The band generated by the polynomials without constant term is the band &? introduced in part (iii) of the last theorem (note that this makes a difference only if E is no member of 9).The following result, for the case that the Hilbert space H is separable, is more precise.
%''(a).
Theorem 57.2. If H is separable, then for any member A of %"( 9 )there exists a sequence (A,, : n = 1,2, .) in the Riesz subspace generated by die polynomials in the elements of 9 such that A,, converges strongly to A , i.e., IIAx-A,,xll --+ 0 for every X E H .
..
Proof. Polynomials in the elements of 9, with real coefficients, will be denoted by q( 9), q 1 ( 9 ) ,q2(B),. . .. The proof is divided into several parts. (i) Given x E H, let H [ x ]be the closed linear subspace of H spanned by all elements of the form q ( 9 ) x . Then we have H [ y ]IH [ x ] if y IH [ x ] . For the proof it is sufficient to show that (zl , z 2 ) = 0 for z1 = q 1 ( 9 ) y and z2 = q2(9 ) x . This, however, is evident on account of
(z1,z 2 ) = (41(=wY> 4 2 ( 9 ) x ) = ( A 9 1 P ) q 2 ( 9 ) x ) = (YY q 3 ( 9 ) x )
= 0.
(ii) The space H i s the direct sum of at most countably many mutually orthogonal subspaces H[x,,](n= 1,2, .). Indeed, since H is separable, it is evident in the first place that the number of mutually orthogonal H [ x ] , with each x # 0, is at most countable. That H is indeed the direct sum of such subspaces follows then from Zorn's lemma. The use of Zorn's lemma can be avoided by starting with a countable dense set (y,, : n = 1,2, .) in H, and defining x1 = y l , then x2 the component of y 2 orthogonal to H [ x , ] , then x3 the component of y 3 orthogonal to H [ x , ] 0 H [ x 2 ] ,and so on. In order to show that H i s then the direct sum of the H[x,,](for those values of n for which x,, # 0), it is sufficient to show that z l H[x,,]for all n implies that z = 0. This is easy; if z 1H[x,,]for all n, then z Iy . for all n, and so z = 0. (iii) Any A E W ' ( 9 ) leaves every H [ x ] invariant, i.e., y E H [ x ] implies that A y E H [ x ] . For the proof, let x be given, and denote the orthogonal projection on H [ x ] by P. Every B E 52 leaves H [ x ] invariant (in view of the definition of H [ x ] ) ,so PBP = BP, which shows that BPis Hermitian (simply
..
..
CH.8, $571
409
THE STRUCTURJ? OF THE SECOND COMMUTANT
becausePBPis Hermitian). But then Pand Bcommute by Lemma 53.1 (i), i.e.,
P E C ( 9 ) .It follows, on account of A E W'(g), that A and P commute.
Hence, y E H [ x ] implies that Ay = APy = PAY, so Ay E H [ x ] . (iv) Now, let A E W ( 9 )be given. According to part (ii), H is the direct sum of the mutually orthogonal subspaces H[x,,](n= 1,2, . . .). We may assume that Ilx,,ll = 1 for all n. We set x, = n-lx,,, which implies Ax, = n-'Ax,,. Observe that, in view of what was proved in part (iii), the components of Ax, in the mutually orthogonal subspacesH[x,,]are exactly the elements n-'Ax,. Once more by part (iii) we have Ax, E H [ x , ] ; hence, by the definition of H [ x , ] , there exists a sequence of polynomials ( q k ( 9 ) : k = 1,2, . . .) such that q k ( 9 ) x o + Ax, as k --t a.
x:
zy
The component of qk(9)xo in H[x,,]is n-'q,(g)x,, and, as observed above, the component of Ax, is n-'Ax,,. Hence i.e.,
n-'q,(Q)x,,
+ n-'Ax,,
q k ( 9 ) x , , + Ax,, as k
--t
as k
co for n
+ 00,
= 1,2,.
..
But then, for n fixed, q k ( 9 ) x + Ax holds for every x of the form x = q( 9)x,,, i.e., the formula q J 9 ) x + Ax holds for a set of x dense in H[x,,]. Since this is true for n = 1,2, . . ., the formula holds for a set of x dense in H. Let K be this dense set. Corresponding to the given operator A E W ( 9 ) ,there exists a number m > 0 such that -mE 6 A 5 mE. The operators R, = sup ( q k ( 9 ) , -mE) are members of the Riesz subspace of W"( 9)generated by the polynomials q@), and RkX = (sup ( q k , -mE))x
+ (SUP
( A , -mE))x = AX
holds for all x E K by Lemma 55.4 (iii). Similarly, the operators Ak = inf (R,, M E )are members of the same Riesz subspace, and A,x + Ax holds for all x E K.Inaddition, thenorms llAkll satisfy llAkll 5 m for k = 1, 2, . . .. Now, since K is dense in H, since A,x + Ax holds for all x E K and llAkll 5 m for all k,it follows easily that Akx + Ax holds for all x E H. This completes the proof. The above theorem and the proof of it are related to a theorem of J. von Neumann, F. Riesz, Y. Mimura and B. Sz. -Nagy (cf. the book by F. Riesz and B. Sz.-Nagy [l], section 129). In the terminology of the theorem referred
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HERMITIAN OPERATORS I N HILBERT SPACE
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to, we have proved that every A in W'(B) is a "function" of the elements of B. Given the set 9 of mutually commuting elements of 2,it was shown in Lemma 55.1 that the second commutant W'( 9), besides being a linear subspace of 2,is also a commutative ring. In addition, it is easy to see that if the sequence ( A , :n = 1,2, .) is containec in W'(9 )and A, converges strongly (or even only weakly) to the boundedlinear operator A , then A is in V'(9). These conditions are, therefore, necessary in order that a subset of 2 be the second commutant of some other subset of mutually commuting elements of 3".It may be asked whether these conditions are also sufficient. We will show now that, in a separable Hilbert space, this is indeed the case.
..
Theorem 57.3. Let W be a subset of 8 satisfying the following conditions. (i) W is a (real) linear subspace of X . (ii) W is a ring in the algebraic sense, i.e., A , B E W impIies that A B E 9 (and hence all members of W are mutuaIIy commuting). (iii) If A , E W f o r n = 1,2, . . . and the sequence (A, : n = 1,2, . . .) converges strongly to the bounded linear operator A , then A E 9. Then W is a Riesz space with respect to the ordering inherited from 2. In addition, if the Hilbert space H is separable and if the identity operator E is a member of 9, then %"'(W) = 9.
Proof. For the proof that W is a Riesz space with respect to the ordering inherited from X , it is sufficient to prove that for every A E 92 we have A + E W and A + = sup ( A , 8 ) with respect to the ordering referred to. As before, A + is here defined by A" = + ( A IAI). For the purpose of this part of the proof, let A E W be given. Then A' E 92,and so any polynomial in A' without constant term is also a member of W . Since IAl is the strong limit of a sequence of polynomials of this kind, it follows that IAl E 9, and so A" = + ( A+ IAI) E W.Hence, since A + 2 A and A" 2 8 hold in 2, the This shows already that A + is an upper bound same inequalities hold in 9. of A and 8 in W . In addition, we have by Theorem 54.2 that A + is the smallest among all Hermitian operators that commute with A and are upper bounds of A and 8. Since any member of W commutes with A , it follows that A" is the smallest among all members of W that are upper bounds of A and 0. In other words, A + = sup ( A , 0) holds in 9. This completes the proof that W is a Riesz space. Note that W is a linear subspace of W'(W),and that for every A E W
+
CH.8,s 571
THE STRUCTURE OF THE SECOND COMMUTANT
41 1
the equality A + = sup ( A , 8) does hold not only in W but also in W'(W). This shows that 9is a Riesz subspace of W'(W). Assume now that the Hilbert space H i s separable, and E E 9. Then every polynomial in the elements of W is again in W and so, in view of W being a Riesz subspace of W'(W), it is evident that the Riesz subspace of %"(W) generated by the polynomials in the elements of W is the space 9 itself. By the preceding theorem every member of %'"(a) is the strong limit of a sequence of elements in this Riesz subspace, i.e., every member of W ' ( W ) is the stronglimit of a sequence of members of B.By hypothesis the strong limit of a sequence of members of W is a member of W ; it follows that c 9. It is obvious that W c W'(W), and hence = W. For the first part of this theorem, cf. also B.Z. Vulikh ([l], 1957).
%"(a)
%''(a)
Any Riesz space L which is at the same time a normed linear space is called a normed Riesz space whenever norm and order in L are compatible, i.e., whenever llfll 5 llgll holds for every pairf, g E L satisfying If1 5 Igl. Note that this is equivalent to requiring that 11 If1 11 = l l f l l holds for all f~ L and IIf1 I 6 1 Igl I holds for every pair f, g E L satisfying 0 6 f 5 g. Normed Riesz spaces will be investigated in later chapters. For the present we restrict ourselves to observing that, for any subset 9 of mutually commuting elements of 2,the Riesz space W ' ( 9 ) is a normed Riesz space with respect to the ordinary operator norm. Indeed, it was proved in Lemma 55.4 (i) that II IAl II = llAll holds for every A E W ' ( 9 ) and it follows from Lemma 55.4 (ii) that 8 5 A 5 B implies IlAll 5 11B11. Indeed, it was proved in Lemma 55.4 (ii) that, for A , B E %"( 9),we have IAl 5 IBI if and only if 11A.xIl 5 IlBxll holds for every x E H. The normed Riesz space L is said to be an abstract L, space whenever
holds for every pair f, g E L ' . It is evident that the name is derived from the fact that the usual norm in any L , space satisfies this condition. It will be shown now that W'(9) is an abstract L, space. The proof could be based on the Yosida representation "{W'( g ) }by showing that the operator norm in %"( 9 )corresponds with the uniform norm in the space "{W'( 9))of continuous functions (cf. Exercise 56.1). We prefer to present an elementary proof, based on the spectral theorem.
Lemma 57.4. Let PI,. . ., P,,be orthogonalprojections in the Hilbert space
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HERMITIAN OPERATORS I N HILBERT SPACE
[CH.
8,g 57
H such that P i p j = 8 for i # j , and let S = x a i P i and i=l
..
T = ZziPi, I= 1
.
where a1, ., a, and z1 , . .,7, are non-negative numbers. It i s evident that S and T are Hermitian and commuting, and so SUP
holds in W'({Pl ,
( S , T ) = *(S+T+IS-TI)
...,P,}).The norms satisfy
IlSUP (S7 Till = max (IlSll, IITll).
Proof. It follows from
(S- T), =
that IS-TI =
Hence, observing that m a
we obtain
(Ci, T i )
c (ai-zt)2Pi n
t= 1
x loi-rJPt. n
i= 1
= t.(ot+zi+lai-ziI),
sup ( S , T ) =
c max n
i= 1
(ai,
zI)Pt.
Without loss of generality we may assume now that P i # 6 for i = 1, . . .,n. It is evident then from the definition of S that llSll = max (ai : 1 6 i
5 n),
and similar formulas hold for IIT1 I and for I lsup (S, T)I I. Hence IlSUP
(S T)lI
= max {max (Qi, 1Sisn
Ti)}
= max (IlSll, IlTll).
.
Lemma 57.5. If A , S1, S , ,. . are Hermitian and mutually commuting, and i f llA-S,,Il + 0, then 11 lAl-IS,,l 11 + 0. Proof. The second commutant W ' (9)of 9 = {A, S,, S2, . . .} is a Riesz space containing A and lAl, all S,, and all IS,,l. Hence
IlAl-lSnIl 6 IA-Snl,
CH.8,§ 571
THE STRUCTURE OF THE SECOND COMMUTANT
413
and so, keeping in mind that 11 IBI 11 = llBl1 holds for any Hermitian B, we obtain II IAl-ISnl I I 5 IIA-SnII 0. +
.
Corollary 57.6. If A, B, S, , S, , . . and TI, T 2 ,. . .are Hermitian and mutually commuting, and if I IA - S,,l I + 0 and I I B - T,I I + 0, then
IISUP
(A, B)-~uP ( S n y G)II
+
0.
-
Proof. The sequence S,,- T,,converges in norm to A B. Hence, by the above lemma, IS,,-T,,I converges in norm to IA -BI. It follows immediately that sup (S,,, T,,) converges in norm to sup (A, B). Theorem 57.7. If 9 is a set of mutually commuting Hermitian operators, and if A and B are members of W ( 9 )such that A 2 8 and B 2 8, then lbUP (A, B)II = max (IIAIL IlBll). Hence, W’( 9 )is an abstract L, space.
Proof. We may assume without loss of generality that 8 5 A 0 5 B 5 E. The spectral systems of A and B will be denoted by
5 E and
(P,(A) : -a < 1 < a)and (PJB) : -a < ,u < a) respectively. Any PA@)and any PJB) is a member of W ‘ ( 9 ) , and so they commute, i.e., PA(A)P,(B) = ~/I(B)~A(A). Consider now two partitions of the interval [0,2], defined by 0 = A0 < A1 <
... < 1, = 2,
0 = / l o < p1 < ... < & = 2,
and let
.
Qij
= { ~ A , + , ( A ) - P A , ( A ) } { i(B)-pp,(B)I P,+
.
for i = 0, 1, . .,n- 1 and j = 0, 1, . .,m - 1. Every Q i j is an orthogonal projection in H , and Q,Q,, = 8 for ( i , j ) # (k,1). Hence, if aij and 7 i j are non-negative numbers, and if ij
we have by Lemma 57.4 that
ij
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HERMITIAN OPERATORS I N HILBERT SPACE
This holds in particular if a i j = l i and z i j = P j for all i and j , so
i.e., S and Tare the lower sums of A and B respectively with respect to the partitions introduced above. Consider now a sequence of such partitions, each one a refinement of its predecessor, and let S,, T, (n = 1,2, . . .) be the corresponding lower sums of A and B. By the spectral theorem S, and T, converge in norm to A and B respectively and, as elements of V’(9),these operators all commute among each other. Hence, it follows from the last corollary that sup (S,,, T,) converges in norm to sup ( A , B). But then ~ ~ S llTnll , , ~ ~and , llsup (S,, T,)II converge to 11A11, IlBll and llsup ( A , B)II respectively, sc from
IISUP
(Sn, Tn)II = max (IISnIIy IITnII),
which holds for n = 1,2,
. . ., we derive finally that
llsup (A, B)II = max (IIAIL IIBII).
Exercise 57.8. It was shown in Theorem 57.2 that if the Hilbert space H is separable and 9 is a set of mutually commuting members of 2,then any member of V ’ ( 9is ) the strong limit of some sequence in the Riesz subspace generated in V ’ ( 9 by ) the polynomials in the elements of 9. In order to show that this does not necessarily hold if H is non-separable, let H be the same Hilbert space as in Exercise 53.5, i.e., H consists of all complex functions x ( t ) on the real line satisfying x ( t ) # 0 for at most countably many t, and such that xtlx(t)lz < co. The inner product is ( x , y ) = C t x ( f ) J ( t ) .Given the real number t o ythe one-dimensional linear subspace H(to) of H is defined by H(t,,) = ( x :x
E H,
~ ( t= ) 0 for all t #
to),
and P(to) is the orthogonal projection on H(to).Let 9 = ( P ( t ) : --co < t < a).
The set 9 is a subset of 2 (i.e., all P ( t ) are Hermitian) and P(s)P(t) = P(t)P(s) = 8 for s # t. Show that any A E V ( 9 )is of the form ( W t )= 4 M t )
CH.8 , 5 581
KADISON'S ANTI-LATTICE THEOREM
415
for some real bounded function a ( t ) and that, conversely, any a(t) of this kind thus determines a member of W ( 9 ) . At the next step, show that U"(9) = W ( 9 ) . Show also that if A, B E W ' ( 9 ) correspond thus to a(t) and P(t) respectively, then sup ( A , B ) corresponds to the function max (a(t),p(t)). Finally, show that if An(. = 1, 2, . . .) and A are members of C"(9),and A , converges strongly to A , then a,(f) converges pointwise to u(t). Show now that the Riesz subspace of W ' ( 9 ) generated by 9 consists of all members of U"(9) corresponding to real a ( t ) such that act) # 0 for at most finitely many t. Hence, the Riesz subspace generated by the polynomials in the elements of 9 (i.e., the Riesz subspace generated by 9 and the identity operator E ) consists of all members of W ' ( 9 ) corresponding to real a(t) of the form a ( t ) = ae(t)+a,(t), where a is a real number, e(t) = 1 for all t, and a,(t) # 0 for at most finitely many t. It follows that if the member A of W'(9 )is the strong limit of some sequence in this Riesz subspace, then the function a(t) corresponding to A is of the form a ( t ) = ae(t)+a,(t), with a2(t) # 0 for at most countably many t. This shows that not every member of W ' ( 9 ) is such a strong limit. 58. Kadison's anti-lattice theorem Once more, let &' be the ordered vector space of all bounded Hermitian operators in the Hilbert space H . It may be asked whether i@ is a Riesz space with respect to the ordering. Following R. V. Kadison ([I], 1951) we will prove that i@ is no Riesz space unless the Hilbert space H is onedimensional. We will prove, in fact, that &' is what is called an anti-lattice. The definition follows. Definition 58.1. The ordered vector space L is called an anti-lattice whenever, for every pairf, g E L, the element inf (f,g ) exists i f and only i f f and g are comparable elements (i.e.,f 2 g or f 4 g ) . Note that if L is an ordered vector space with a strong unit, then L is an anti-lattice whenever, for every pair u, v in the positive cone L ', the element inf (u, v) exists if and only if u and u are comparable. We turn to the proof that 2 is an anti-lattice.
416
HERMITIAN OPERATORS IN HILBERT SPACE
[CH. 8,s 58
Lemma 58.2. Let P I and P , be orthogonal projections in the Hilbert space H such that inf ( P , , P,) = 8 holds in 2. Then P , and P2 have mutually orthogonal ranges, i.e., P , P2 = P2P I = 8. Proof. Let the ranges of P , and P , be R ( P , ) and R(P,) respectively, and denote the orthogonal projection on the closure of the algebraic sum R(P,) R(P2) by P , . It is evident that P , 5 P 3 and P , 5 P 3 . Write D = P 3 - P 2 . It follows from D 2 8 that P , - D 6 P , , and it follows from P , 5 P 3 that
+
Hence
so
Pi-D
-
PI D PI
5 P3-D
= P2.
5 inf (Pi ,P 2 ) = 8, 5D
= P3-PZ.
This shows that the range of P I is included in the range of the orthogonal projection P 3 - P 2 . Since the ranges of P,-P, and P p are mutually orthogonal subspaces, it follows that the ranges of P I and P2 are mutually orthogonal. In the next theorem we present a much stronger result. Theorem 58.3. If P , and P , are orthogonal projections in the Hilbert dpace H such that inf ( P , , P 2 ) = 8 holds in 2,then P , = 8 or P2 = 6 .
Proof. Assume that P , # 8 and Pz # 8. Then there exist elements x # 0 in R(P,) and y # 0 in R(P,). We have x Iy by the preceding lemma, and we may assume that llxll = llyll = 1. It follows that P , x = x , P l y = 0 and P 2 x = 0, P 2 y = y. Let M be the subspace of H spanned by x and y , and let P be the orthogonal projection on M . The projections P , and P , leave M invariant, and the restrictions of P 1 and P2 on M have the matrices
with respect to { x , y } as an orthonormal basis. Now, note that the matrix
with real coefficients and with respect to the same basis, corresponds to a
CH. 8,s 581
KADISON’S ANTI-LATTICE THEOREM
417
positive definite operator on M if and only if a 2 0, c 2 0 and b2 5 ac. It follows that the operator A on M , corresponding to the matrix
is not negative definite on M . In other words, the member AP = PAP of 2 does not satisfy AP 6 8. On the other hand, since P , - A corresponds on M with the matrix
(-$
-f5),
it follows that P,- A is positive definite on M , and this implies that AP S P I . Indeed, given z E H , we set z = x + y with x E M and y 1M , and so
(APz, Z) = (AX,x + y )
=
( A X ,X )
5 (PI x , x ) 5 (P,X , x ) + (Ply, u) = (PI z, z), where it is used in the last equality that P I leaves M and the orthogonal complement of M invariant. It follows now that AP 5 P i , and similarly it is proved that AP S P 2 . But then we have AP
5 inf ( P , , P 2 ) = 8,
contradicting the earlier result that AP 5 8 does not hold. Thus we arrive at a contradiction, and so P , = 8 or P2 = 8 must hold.
Theorem 58.4. 8 is an anti-lattice. Proof. It is evident that if A and B are members of 8 such that A 5 B or A 2 B holds, then inf ( A , B ) exists in 2. Conversely, let A‘, B‘ E 8 be given and assume that inf (A’, B’) = C exists in H.Writing A’- C = A and B’- C = B, we have A , B E 8 and inf ( A , B ) = 8. The proof will be complete if we show that one at least of A and B is equal to 8. To this end, assume that A # 8 and B # 8. Then. by the spectral theorem, there exist orthogonal projections P,,P2 and positive numbers a,, a2 such that 8 < 6,P,
5 A and 8 < B2P2 6 B
(for 6, P , we can take one of the nonzero terms in an appropriate lower sum for A , and similarly for b2P2).It follows from inf ( A , B ) = 8 that inf ( 6 , P , , g 2 P 2 ) = 0, and hence inf ( P , , P2)= 8. But then, in view of the
418
HERMITIAN OPERATORS I N HILBERT SPACE
[CH. 8 , s 58
preceding theorem, we have PI = 8 or P, = 8. Contradiction. Hence, one at least of A and B must be equal to 8. As visible from the proof of Theorem 58.3, the proof that 2 is an antilattice in tbe general case is reduced to the case that the Hilbert space H is two-dimensional. It is reasonable to conjecture that if the Hilbert space H is real and two-dimensional (i.e., H is the ordinary real plane), then the fact that 3 is an anti-lattice has a simple geometrical meaning. This is indeed so. For any A # 8 in 2, consider the set of all x E H satisfying (Ax, x ) 5 1, and call this the indicatrix of A . For A 8 the indicatrix 'I of A is an elliptic disk with the origin as center, where the area between two different parallel lines is also to be regarded as an elliptic disk. If 8 < B 5 A , the indicatrix I A of A is contained in the indicatrix IB of B, i.e., if a member of 2+becomes larger, then its indicatrix becomes smaller. The anti-lattice theorem states that if A and B are incomparable in 2, then inf (A, B ) does not exist or, equivalently, sup (A, B ) does not exist. Since we can always add a sufficiently large multiple of the identity operator E to A and By this is the same as stating that if A 2 E, B 2 E and A , B are incomparable, then sup (A, B ) does not exist. The indicatrix of E is the unit circle in the plane, so A 2 E implies that the indicatrix IA of A is an elliptic disk with the origin as center and contained in the unit circle. Similarly for the indicatrix IB of B. Since A and B are incomparable, neither of IA and I B is contained in the other, i.e., the boundary of IA n Is is no ellipse. Assume now that C = sup (A, B ) exists. This means that C 2 A and C 2 By and any other upper bound C' of A and B satisfies C' 2 C. Hence, the indicatrix Ic of C is an elliptic disk (with the origin as center) contained in IA n I B , and any other elliptic disk with the origin as center and contained in' I n IB is also contained in Ic. This, however, is evidently false. Hence, sup (A, B ) does not exist in 8.
=-
Exercise 58.5. Find the geometrical meaning (in a real two-dimensional Hilbert space) of Theorem 58.3. Hint: The indicatrix of the orthogonal projection on a one-dimensional subspace is the area between two different parallel lines (equidistant from the origin). Also, if A and 8 are incomparable, the boundary of the indicatrix of A is a hyperbola.
CHAPTER 9
Riesz Homomorphisms and Quotient Spaces
We recall that the linear mapping n of the Riesz space L into the Riesz space M is called a Riesz homomorphism if the mapping preserves finite suprema (and hence preserves finite infima). The kernel (null space) of the Riesz homomorphism 'II is an ideal in L. Images and inverse images of Riesz subspaces are Riesz subspaces; in particular, the image n(L)of the space L itself is a Riesz subspace of M. Images and inverse images of ideals are ideals in n(L)and L respectively. The image of the principal ideal generated by an element u in L+ is the principal ideal generated in n(L)by the image AU of u. Finally, the image of a projection band in L is a projection band in
n(L).
We recall that the Riesz space L is called uniformly complete whenever, for every e in L', every e-uniform Cauchy sequence has an e-uniform limit. It will be proved that any Riesz homomorphic image of a uniformly complete space is again unifcrrnly complete. In other words, any quotient space of a uniformly complete space is uniformly complete. We recall that the subset S of L is uniformly closed (i.e., closed in the relatively uniform topology) if, for every e in L+ and for every e-uniformly convergent sequence the members of which are members of S, all e-uniform limits of the sequence are again members of S. It will be shown for any ideal A in L that the quotient space L/A is Arcbimedean if and only if A is uniformly closed (A. I. Veksler (1966); W. A. J. Luxemburg and L. C. Moore Jr. (1967)). In particular, if L is Archimedean and A is a o-ideal in L, then L/A is Archimedean (A. I. Veksler (1958); L. Brown and H. Nakano (1966)). These conditions refer to one particular ideal A and the corresponding quotient space LIA, but it may be asked also under which conditions every quotient space of L is Archimedean. The condition that every quotient space of L is Archimedean is related to some conditions to be satisfied by the hull-kernel topology in the set B of all proper prime ideals in L. It was proved already in Chapter 5 (cf. Theorem 37.6) that the hullkernel topology in B is Hausdorff if and only if every quotient space of L 419
420
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.
9
is Archimedean, and also if and only if every principal ideal in L is a projection band. In the present chapter we give a simpler proof (i.e., without reference to the hull-kernel topology in 9) of the fact that every quotient space of L is Archimedean if and only if every principal ideal in L is a projection band. We also prove that L is uniformly complete and has the property that all its quotient spaces are Archimedean if and only if L is Riesz isomorphic to the space of all real functions f on some point set X such that f vanishes outside some finite subset of X (the subset varying with f). The set lo(L), consisting of all infinitely small elements together with the null element, is an ideal in L, and Io(L) is included in the smallest uniformly closed ideal Zl(L) which exists in L. If L is Archimedean, then Zo(L) and I l ( L ) consist only of the null element, but there exist nonArchimedean spaces where I. (L) is a proper subset of Il(L). We will prove that if L has a strong unit, then Zo(L/A) = I,(L/A) holds for every ideal A . The seminorm p in L is called a Riesz seminorm if p(I f I) = p ( f ) holds for everyfe L and if 0 5 u j v implies p ( u ) S p(u). It will be proved that if p is a Riesz seminorm in L and A is an ideal in L, then the quotient seminorm in L/A, defined in the usual manner, is also a Riesz seminorm. If n is a Riesz homomorphism from L onto M, where L and M are equipped with their relatively uniform topologies, then n is a continuous mapping. Also, for any subset S of M, the set S is uniformly closed in M if and only if 71:-'(S) is uniformlyclosed in L. Finally, if the pseudo uniform closure of every set in L is uniformly closed, then the same holds in M, i.e., the pseudo uniform closure of every set in M is uniformly closed. We recall that the Riesz homomorphism 71: from L onto M is called a Riesz a-homomorphism if 71: preserves countable suprema (and infima). It was proved (in section 18) that 71: is a o-homomorphism if and only if the kernel of 71: is a a-ideal. We will prove now that a Riesz o-homomorphism 71: preserves the principal projection property and preserves Dedekind a-completeness. Also, 71: is a continuous mapping if L and M are equipped with their order topologies. If, in addition, L is Dedekind a-complete, then the subset S of M is order closed in M if and only if n -' (S ) is order closed in L. Finally, under the same hypotheses, if the pseudo order closure of every set in L is order closed, then the same holds in M. We recall that the Riesz homomorphism 71: from L onto M is called a normal Riesz homomorphism if 71: preserves arbitrary suprema (and infima). It was proved (in section 18) that 71: is normal if and only if the kernel of
CH.9, § 591
ADDITIONAL RESULTS ON RIESZ HOMOMORPHISMS
42 1
n is a band. We will prove now that a normal Riesz homomorphism preserves the projection property and preserves Dedekind completeness. 59. Additional results on Riesz homomorphisms
We recall the definition of a Riesz homomorphism as presented in section 18. Given the Riesz spaces L and M , the linear mapping a of L into M is called a Riesz homomorphism whenever inf (nf,ng) = 0 holds for every pair of elements f, g E L satisfying inf (f, g) = 0. According to Theorem 18.2 the linear mapping a of L into M is a Riesz homomorphism if and only if n preserves finite infima or, equivalently, preserves finite suprema. In other words, n is a Riesz homomorphism if and only if a is linear and n{inf (f,g ) } = inf (nf,w),
or, equivalently,
(f,s>> = SUP (nf,4 holds for allf, g E L . In the now following theorems we list several properties of a Riesz homomorphism.
Theorem 59.1. Let II be a Riesz homomorphism of L into M. (i) f 2 g in L implies nf 2 ng in M . In particular, a is a positive linear mapping, i.e., n maps L+ into M + . (ii) We haoe
.(f
‘1
=
( n f ) + ,n ( f -1 = ( n f ) - , .(If
I1 =f . 1 I.
Hence, f Ig in L impIies that nf Ing holds in M . (iii) The kernel (null space) A, of 7c is an ideal in L. (iv) We have nf 2 0 if and only if there exists an element u 2 0 in the kernel A, such that f + u 2 0. It follows that nf 2 ag holds if and only if there exists an element f l in L such that f l 2f,f l 2 g and nfl = nf. (v) We have Inf I 5 lngl if and only if there exists an element f’ in the kernel A, such that If’l 5 If I and If-f’I 5 (91. Proof. The parts (i), (ii), (iii) were proved in Theorem 18.3. (iv) Tf nf 2 0, then nf = (nf)’ = n ( f + ) , so a(f +-f) = 0, i.e., n( f -) = 0. This shows that the element u = f - satisfies 0 S u E A , and f+u=f+f-=f+Zo.
Conversely, if 0 S u E A, and f + u 2 0, then af = n ( f + u ) 2 0.
422
[CH.
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
9,8 59
It follows that nf 2 ng holds if and only if there exists an element 0 6 U E A , such that f - g + u 2 0. In other words, setting f + u = f l y we have nf 2 ng if and only if there exists an element fl in L such that fl 2 L f l 2 9 and nfl = nf. (v) If Inf I S Ingl, then n(lf I) S n(lgl), and so there exists an element 0 6 U E A , such that If1 5 Igl+u. But then it follows from u 2 Ifl-lgl and If I 2 If I - Is1 that i.e.,
inf (If I,
If 1 6 191+inf(lf I, u )
2 If 1 -191,
with inf (If
I,
u ) E A,.
This shows that in the inequality If I 5 Igl + u we may assume that u E A , and 0 6 u 5 Ifl. It follcws from the last inequality by means of the dominated decomposition property (cf. Corollary 15.6) that there exist elements u1 and u2 in L+ such that u = u1 +u2 and u1 5 f +,u2 6 f -. Then
0 6 Ifl-u = (f
+
-ud+(f
--.2>
= I(f+-u1)+(f--uz)l
= I(f+-u1)-(f--
U2)IY
where the last equality follows from Ip+q1 = Ip-ql, holding for p Iq. Hence, setting u1-u2 =f’,we can rewrite the formula 0 5 If l - u 6 191 as 1f-f’l 5 Igl withf’ E A , on account of u1 ,u2 E A,. Furthermore, we have
If’l
=
lU1-U21
= u,+u, = u
6 Ifl.
Conversely, if there exists an elementf’ E A , satisfying If-f’l it follows first from
6
191, then
IIf-TI-lfll 5 If’l
that I f-f’lIf I is a member of A,, and so n(1f I) = n(I f-f’l). Hence, on account of I f-f’l 6 Igl, we have now that n(1f I) 5 n(Igl), i.e., Inf 1 5 Ingl. This concludes the proof. For any mapping x of L into M , the image of any subset D of L will be denoted by n(D),and the inverse image of any subset E of M will be denoted by x-’(E). Theorem 59.2. Let x be a Riesz homomorphism of the Riesz space L into the Riesz space M . (i) For any RieJz subspace K of L , the image n(K)is a Riesz subspace of M . In particular, the image x(L) of the space L itselfis a Riesz subspace of M . (ii) For avy Riesz subspace N of M y the inverse image n-l(N)is a Riesz subspace of L.
CH.
9, 591
ADDITIONAL RESULTS ON RIESZ HOMOMORPHISMS
423
(iii) For any ideal B in L, the image n(B) is an ideal in n(L).Furthermore, i f B l and B2are ideals in L, then
n(B, n B,)
=
n(B,)n n(B2).
(iv) For any ideal C in n(L),the inverse image n-'(C) is an ideal in L. It follows that, for any ideal C in M , the inverse image n- ' ( C )is an ideal in L. (v) If B,, is the principal ideal in L generated by an element u E L+,then n(B,,) is the principal ideal generated in n(L)by the element nu. (vi) For any wbset D of L, we have
n(0") = {n(D)Id(vii) The image of a projection band in L is a projection band in n(L).
Proof. (i) Evident. (ii) Let N be a Riesz subspace of M. It is evident that n-'(N) is a linear subspace of L. Furthermore, given f and g in n-'(N), the images nf and ng are members of N , so sup (nf,ng) E N , i.e., n(sup (f,g ) } E N,which implies that sup (f,g ) E n-'(N). This shows that n-'(N)is a Riesz subspace of L. (iii) Let B be an ideal in L. In order to show that n(B)is an ideal in n(L), it is sufficient to prove that if 0 5 u E B and 0 5 m S nu with m E n(L), then m E n(B). On account of m E n(L)we have m = nffor somefe L; in view of m 2 0 we have then that
m = m+ = (nf)'
= n(f+).
But then w = inf (u,f ') is a member of B and
nw = inf (nu, nf +) = inf (nu, m ) = m. This shows that m E n(B),which is the desired result. For ideals B, and B2 in L, we have to prove that N B , ) n n(B2) =
n B2).
GivenfE n(B,)n n(B2), we have If I E n(B,)n n(B2),so let If I = nu with 0 5 u E B, and If I = nu with 0 6 v E B 2 . Then w = inf (u, u ) satisfies 0 5 w E B, n B2 and nw = Ifl. It follows that I E n(B, n B2),and so f E n(B1 n B2). For the results in this part, cf. also Exercise 18.15. (iv) Let C be an ideal in n(L).We have to show that n-'(C) is an ideal in L. It is sufficient to prove that if we have 0 5 u 5 u with u E a- '(C),
If
424
RIESZ HOMOMORPHISMS A N D QUOTIENT SPACES
[CH.
9,g 59
then u E n-'(C). This is immediate; it follows from 0 6 nu 6 nu E C that nu E C, and so u E n-'(C). If C is an ideal in Mythen C n n(L) is an ideal in n(L), and so n-'(C) = n-'(C n n(L)) is an ideal in L. (v) Let B, be the principal ideal in L generated by the element u E L'. According to part (iii) the image n(B,) is an ideal in z(L). It is evident that nu is a member of n(B,),and so the ideal generated in n(L) by nu is included in n(B,,). For the proof of the converse inclusion, assume that 0 5 m E n(B,,), so m = nu for some u satisfying 0 S u E B,,. Then u S ku holds for some real number k > 0, and so m = nu 5 knu, i.e., m is a member of the ideal generated in n(L)by nu. This shows that n(B,)is included in the ideal generated by nu. (vi) If m E n(Dd),then m = nf for somefE Dd. It follows fromf Ig for all g E D that nf Ing holds for all g E D , i.e., m Ing for all g E D . This shows that m E { R ( D ) } ~ . (vii) Let B be a projection band in'L, and C the disjoint complement of B, so B 0 C = L. By part (iii) the images n(B)and n(C)are ideals in n(L),and we have n(B) In(C). Given the arbitrary element m in n(L), we have m = nf for somef E L ; let f = fl+fz with fl E B and fzE C. It follows that
m = nf = n f l + n f z , which shows that n(B)@ n(C) = n(L). In other words, the Riesz space n(L) is the direct sum of the ideals n ( B )and n(C ) .But then, by Theorem 24.1, n(B)and n(C)are bands in n(L),and in view of the definition of a projection band (cf. Definition 24.4), z(B)and n(C)are projection bands in n(L). There is no result in the converse direction for (vii). Indeed, any ideal in L can act as null space for a Riesz homomorphism, i.e., any ideal in L can be the inverse image of the projection band (0) in n(L). We recall briefly that if the Riesz homomorphism n from L into M is one-one and onto Mythen n is called a Riesz isomorphism, and the spaces L and M a r e now said to be Riesz isomorphic. Under these circumstances the inverse mapping n-' is a Riesz isomorphism from M onto L. We also recall that if n is a Riesz homomorphism from L into M with kernel A,, then L/A, and n(L) are Riesz isomorphic under the mapping If]c+ nf, where Lf] denotes the equivalence class off. Given e E L+ and the sequence (fn :n = 1,2, . . .) in L, the sequence is said to converge e-uniformly tofE L whenever for every E > 0 there exists
CH. 9,s 591
425
ADDITIONAL RESULTS ON RlESZ HOMOMORPHISMS
a natural number no(&)such that If-LI 5 Ee holds for all n 2 no(&).The definition for an e-uniform Cauchy sequence is similar. These definitions were introduced in Theorem 16.2 and Definition 39.1. Furthermore, the sequence (f,: n = 1,2, . . .) in L is said to converge relatively uniformly to f E L whenever f, converges e-uniformly to f for some e EL'. Similarly, (f,: n = 1,2,. . .) is called a relatively uniform Cauchy sequence if the sequence is an e-uniform Cauchy sequence for some e E L'. Finally, by Definition 42.1, the Riesz space Z is said to be uniformly complete whenever, for every e EL', any e-uniform Cauchy sequence has an e-uniform limit. The following results, taken from a paper by W. A. J. Luxemburg and L. C. Moore Jr. ([I], 1967), hold for relatively uniform Cauchy sequences and their images under a Riesz homomorphism. For convenience we assume the homomorphism to be from L onto M instead of merely into M . Theorem 59.3. Let n be a Riesz homomorphismfrom L onto M. (i) The image of a relatively uniform Cauchy sequence in L is a relatively uniform Cauchy sequence in M. (ii) If (nf, :n = 1,2, . .) is a relatively uniform Cauchy sequence in M , then there exists a subsequence (nf, ; k = 1,2, . . .) and a corresponding sequence (fi : k = 1,2, . .) in L such that nh' = nh., for all k and : k = 1,2, . .) is a relatively uniform Cauchy sequence in L. (iii) I f L is uniformb complete, then so is M. Zn other words, any Riesz homomorphic image of a uniformIy complete space is uniformly complete.
.
.
.
Proof. (i) If If,-f,l
(,&I
6 ce for all m, n 2 no(&),then
Inf,-nf,l
5 me for all m, n 2 no(&).
(ii) Let (nL: n = 1,2, . . .) be a relatively uniform Cauchy sequencs in M. This implies that, for some e in L', the sequence (nf,: n = 1,2, . . .) is a ne-uniform Cauchy sequence. Let (nf,., : k = 1,2, . .) be a subsequence such that n, < n, < . . . and
.
for k = 1,2, .
Inf.,,, -nfnkl I2-kne
...
.
By Theorem 59.1 (v) there exist elements (gk : k = 1,2,. .) in L such that ngk = n(fnk+,-fnk) and lgkl g 2-ke It follows that
fk' =fnl+gI+
-
*
+gk-l
for all k.
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RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.9,s 60
satisfies nfk' = nf, for all k, and the sequence (fk' :k = 1,2, . . .) is an euniform Cauchy sequence in L. (iii) We assume that L is uniformly complete and that (exactly as in the preceding part (ii)) the sequence (nf. :n = 1,2, . . .) is a ne-uniform Cauchy sequence in M. By part (ii) there is a subsequence (nf., : k = 1,2, . . .) and a corresponding sequence (h':k = 1,2, . . .) in L such that nfk' = nf,, for all k and (fk' :k = 1,2, . . .) is an e-uniform Cauchy sequence. Since L is uniformly complete by hypothesis, f i converges e-uniformly to some f E L, and so nfi converges ne-uniformly to nf. In other words, nf., converges neuniformly to nf as k + co. It follows easily that nf, converges ne-uniformly to nfas n -+ co. This shows that M is uniformly complete. Corollary 59.4. If the Riesz space L is uniformIy complete, then any quotient space L/A is uniformly complete.
Exercise 59.5. Show that the Riesz space L is uniformly complete if and only if every principal ideal in L is uniformly complete.
60. Archimedean quotient spaces One of the problems arising in the theory of Riesz homomorphisms is to determine conditions under which the Riesz homomorphic image of an Archimedean Riesz space L is Archimedean. In other words, the problem is to determine which properties an ideal A in L should have in order that L/A be Archimedean. In general, L/A is not Archimedean, even in the case that L is super Dedekind complete, as shown by the following example. Example 60.1. (i) Let L be the Riesz space of all real sequences f = (f(l), f(2), . . .) and A the ideal of all bounded sequences. The element of L/A which contains f will be denoted by If]. Definef, E Lby f , ( n ) = n and f 2 E Lby f 2 ( n ) = n2 for n = 1,2, . . . Given the natural number k, we have f i ( n ) 6 k-y2(n) for all n 2 k, and so Lfi] S k-'If2] with If,] # [O]. This holds for all k = 1,2, . . ., so If,] is infinitely small with respect to v2]. It follows that L/A is not Archimedean. As a second example, let L be the real sequence space I , and A the principal ideal generated by the element u = (u(l), u(2), . . .) with u(n) = n-' for n = 1,2,. ... Let e = (a, 1 , . . .) and v = (u(l), u(2), . . .)with v(n) = n for n = 1,2, . . .. Given the natural number k, we have v(n) S k-'e(n) for all n 2 k, and so [u] 5 k-'[e] with [v] # [O]. Tbis holds for all k = 1, 2 , . . ., and so LIA is not Archimedean.
CH.
9,5 601
ARCHIMEDEAN QUOTIENT SPACES
427
(ii) In the above examples L is super Dedekind complete and LIA is not even Archimedean. The converse is also possible; a non-Archimedean Riesz space may possess a super Dedekind complete quotient space. By way of example, let L be the lexicographicallyorderedplane (i.e., (x, ,xz) j ( y , ,y 2 ) if either x , < y, or x , = y , , x2 5 y z ) and let A be the x2-axis. Then L is not Archimedean, A is a band in L, and L/A is Riesz isomorphic with the Riesz space of all real numbers, so L / A is super Dedekind complete. In order to present necessary and sufficient conditions for a quotient space to be Archimedean, we recall one more notion. According to the definition given in section 16, the subset S of the Riesz space L is called (relatively) uniformly closed whenever, for every relatively uniformly convergent sequence in S, all relatively uniform limits of the sequence are also members of S. The main result in the next theorem is now that if A is an ideal in the Riesz space L, then LIA is Archimedean if and only if A is uniformly closed. The details follow.
Theorem 60.2. Given the ideal A in the Riesz space L, the following conditions are mutually equivalent. (i) L / A is Archimedean. (ii) A is uniformly closed. (iii) If 0 6 u,, E A for n = 1,2, . . . and the sequence (u,, : n = 1,2, . . .) is increasing and relatively uniformly convergent to u, then u E A. (iv) r f u , v E L' and (nu-v)' E A for n = 1,2,. . ., then u E A. In particular, it foilows from the equivalence of (ii) and (iii) that the ideal A is already uniformly closed i f , for every monotone relatively uniformly convergent sequence in A, all relatively uniform limits of the sequence are members of A. Proof. (i) =s (ii). By hypothesis, the quotient space LIA is Archimedean, and we have to prove that A is uniformly closed. To this end, let e EL' be given and let (f, :n = 1,2,. . .) be e-uniformly convergent to f, with f.E A for all n. Given E 0, it follows from
=-
I f I 6 If-f,l+
1f.l 5 &e+1f.l
for n 2 no(&) that n(lf 1) 5 me, where n is the Riesz homomorphism from L onto L/A. The inequality n(lf1) j m e holds for every E > 0, and so n(lf1) = 0 since L / A is Archimedean. This shows thatfE A , and hence A is uniformly closed. (ii) =. (iii). Evident. (iii) * (iv). Let u, v EL' and (nu-v)' E A for n = 1,2, . . ..We assume
428
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH. 9,5 60
that A satisfies the condition in (iii), and we have to prove that u is a member of A. In view of
0
u-(u-n-'u)+
6 lu-(u-n-'u)l
= Iu+-(u-n-'u)+I
it is evident that the increasing sequence
w,,= (u-n-lu)';
n =
=
n-'u
. .,
converges u-uniformly to u. Hence, since w,,E A holds for all n, we have by hypothesis that u E A. This is the desired result. (iv) =- (i). We assume that the condition in (iv) is satisfied. In order to prove that L/A is Archimedean, we assume also that u, v E L + and
n m 5 nu for n where again
7c
= 1,2,.
..,
is the Riesz homomorphism from L onto LIA. Then
0 = (nnu-m)+ = n{(nu-u)+> holds for all n, and so (nu-u)' E A holds for all n. It follows by hypothesis (iv) that u is a member of A, i.e., nu = 0. This shows that L/A is Archimedean. As an immediate corollary we have the following theorem.
Theorem 60.3. If L is Archimedean and A is a a-ideal in L, then LIA is Archimedean. Proof. Let A be a a-ideal in the Archimedean Riesz space L. Assume that 0 5 u,, E A holds for n = 1,2, . . . and that the sequence (u,, :n = 1,2, .) is increasing and relatively uniformly convergent to u. Then we can apply Lemma 39.2, according to which the relatively uniform limit of an increasing sequence, if existing, must be equal to the supremum of the sequence. Hence, in the present case, we have u = sup u,,. But then u is a member of A, since A is a a-ideal. This shows that condition (iii) in the preceding theorem is satisfied, and so LIA is Archimedean.
..
Before stating another corollary, we recall that in an Archimedean Riesz space L, if e E L+ is given, every e-uniformly convergent sequence has a unique e-uniform limit. Conversely, if it is true for every e E L +that every e-uniformly convergent sequence has a unique e-uniform limit, then L is Archimedean (because, if not, there exist u, 0 EL' such that 0 < nu 6 v for n = 1,2,. ., and then the sequence (w,,:n = 1 , 2 , . .), defined by setting w,,= u for all n, converges v-uniformly to the null element as well as, trivially, to u itself). Hence, a necessary and sufficient condition in order that,
.
.
CH.9 , s 611
EVERY QUOTIENT SPACE IS ARCHIMEDEAN
429
for every e E L + ,every e-uniform Cauchy sequence has a unique e-uniform limit is that L be Archimedean and uniformly complete. Given that L satisfies this condition, it may be asked for which ideals A the quotient space LIA satisfies the same condition. Theorem 60.4. If L is Archimedean and uniformly complete, and if A is an ideal in L, then L/A is Archimedean and uniformly complete fi and only if A is uniformly closed.
Proof. The space L is uniformly complete. Hence, it follows from Corollary 59.4 that LIA is uniformly complete, without any extra condition on the ideal A. The only condition to be taken care of is, therefore, that L/A be Archimedean, and this is so if and only if A is uniformly closed. The theorems in this section were proved by W. A. J. Luxemburg and L. C. Moore Jr. ([l], 1967). The theorem that L/A is Archimedean if and only if A is uniformly closed was also proved by A. I. Veksler ([4], 1966). The result that if A is a a-ideal in the Archimedean space L then L/A is Archimedean is already contained in a paper by A. I. Veksler ([2], 1958) and in a paper by L. Brown and H. Nakano ([l], 1966). Finally, Theorem 60.4 is also an immediate consequence from Theorem 60.2 combined with Theorem 5 in a paper by A. I. Veksler ([3], 1966). 61. Riesz spaces with the property that every quotient space is Archimedean
From the results in the preceding section it follows that an Archimedean Riesz space can have many quotient spaces that are non-Archimedean as well as many that are Archimedean. It is natural, therefore, to ask which Riesz spaces have the property that every quotient space is Archimedean. An answer is contained in Theorem 37.6, where it was proved that the property referred to is equivalent to several other properties bearing upon the hullkernel topology in the set of all proper prime ideals, and also equivalent to the property that every principal ideal is a projection band. Let us denote the Riesz space we are discussing by L, and consider now in particular the following properties. ( a ) L/A is Archimedean for every ideal A in L. (B) Every principal ideal in L is a projection band. The proof in Theorem 37.6 that ( a ) and (p) are equivalent is not very simple, because (in both directions) the proof goes by means of a number of other properties. In one direction we have the following list.
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RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.9.8 61
LIA is Archimedeanfor every ideal A Every proper prime ideal is a maximal ideal 3 The hull-kernel topology in the set 8 of allproper prime ideals is Hausdor=> For every u EL' the set {PIu = ( P :P E 8,u not in P ) is closed => Every principal ideal is a projection band. In the converse direction the proof is even longer, as follows. Every principal ideal is a projection band => For every u EL' the set {P}. is closed => The hull-kernel topology in 9 is a TI-topology => Every proper prime ideal is a minimal prime ideal => Every proper prime ideal is a maximal ideal => LIA is Archimedean for every ideal A. We will indicate a more direct proof that (ct) and (p) are equivalent, without even mentioning the notion of a prime ideal. It will be used that, according to Theorem 60.2, L / A is Archimedean if and only if the ideal A is uniformly closed. =>
Theorem 61.1. Given the Riesz space L , the following conditions are equivalent. (i) For every ideal A in L the quotient space L / A is Archimedean, or, equivalently, every ideal in L is uniformly closed. (ii) Given u and v in L', there exists a natural number m (depending upon u and v) such that inf (u, nu) = inf (u, mv) for n = m+ 1,
. . ..
Proof. (i) => (ii). Let us assume that L / A is Archimedean for every ideal A and that, nevertheless, there exist elements u and v in L f such that the set (inf (u, no) :n = 1,2,. . .)
contains infinitely many distinct members. Let A be the ideal generated by the elements w,, = nu-inf (u, nu); n = 1 , 2 , . ..
..
.
The sequence (w,, : n = 1,2, .) is increasing, and sof E A holds if and only if there exist natural numbers n and k satisfying If1 _I kw,. We will show first now that v is no member of A. Indeed, if
u _I kw,, = k(nv-inf (u, nu)) would hold for appropriate natural numbers n and k , then
(kn- 1)v 2 inf (ku,knv) 2 inf (u, knv)
CH.9,s 611
43 1
EVBRY QUOTIENT SPACE IS ARCHIMEDEAN
would imply that and so
knv 2 inf {u+u, (kn+ l)u} 2 inf (u, (kn+l)v}, ju
2 inf {u, (j+l)u)
would hold for all natural numbersj 2 kn. It would follow immediately that inf (u,ju) = inf {u, ( j + 1)u) holds for a l l j 2 kn, contradicting the assumption that the set (inf (u, nu) :n = 1,2,
. . .)
contains infinitely many distinct elements. Hence, v is no member of the ideal A . Let n be the mapping f + [f]from L onto L/A, where Lf] denotes the equivalence class (modulo A ) which containsf. It follows from
nu
2 u+ {nu-inf
(u, nu)},
..
holding for n = 1,2, . . ., that nnu 5 nu holds for n = 1,2, . . Since u is no member of A , we have nu # [ O ] . Thus we have nnu 6 nu for n = 1,2, . . with nu # [O]. This contradicts the hypothesis that L/A is Archimedean. Hence, finally, we obtain the result that the set (inf (u, nu) : n = 1,2,
.
. . .)
can contain only a finite number of elements. (ii) => (i). We assume that the condition in (ii) is satisfied. Let A be an ideal in L, and let u, v E L+ satisfy (nu- u)' E A for n = 1,2, . . . In order to show that L/A is Archimedean, it will be sufficient by Theorem 60.2 to show that u is a member of A . Note that
.
(nu-u)+ = sup (nu-u, 0) = sup (nu,u ) - u = nu-inf (nu,u).
(1)
By hypothesis, there exists a natural number m such that inf (nu,u ) = inf (mu, u ) holds for all n 2 m.Hence, for n 2 m,we infer now from (1) that
{(n+l)U-vU)+-((nu-U)+ = u, and this implies that u E A , since the two terms on the left are members of A . The proof is thus complete. In the next theorem we compare condition (ii) of the last theorem with the condition that every principal ideal is a projection band.
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RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.
9,s 61
Theorem 61.2. Giuen the Riesz space L, the following conditions are equivalent. (ii) Giuen u and u in L+,there exists a natural number m (depending upon u and u ) such that inf (u, nu)
=
inf (u, mu) for n = mym + 1,
. . ..
(iii) Every principal idea2 in L is a projection band.
Proof. (ii) => (iii). We assume that the condition in (ii) is satisfied. Let
u be an arbitrary element of L + ,and denote by A, the principal ideal gener-
ated by u. By Theorem 24.7 (ii), every positive u in the principal band generated by u satisfies u = sup (inf (u, nu) :n = 1,2, . . .). In view of our present hypothesis, there exists a natural number m such that inf (u, nu) = inf (u, mu) for all n 5 m. It follows, therefore, that u = inf (u, mu), so u 5 mu, which shows that u is already a member of A,. It has thus been shown that the ideal A, is equal to the band generated by v. In order to show now that A, is a projection band, it is sufficient by Theorem 24.7 (i) to show that sup (inf (u, nu) :n = 1,2,
. . .)
exists for every u E L+. Given u EL', there exists (in view of our hypothesis) a natural number m such that inf (u, nu) = inf (u, mu) holds for all n 2 my and hence sup {inf (u, nu)} = inf ( u , mu). n
This is the desired result. (iii) =s (ii). We assume that every principal ideal is a projection band. This implies that for every u E Lf the ideal A, generated by u is equal to the band generated by u, and the band A,, is now a projection band. Let u EL' be given, with components u1 and u2 in A, and (A,)' respectively. Then inf ( u z , nu) = 0 for n = 1,2,. . ., and so inf (u, nu) = inf (ul ,nu) holds for n = 1,2, . . .. Furthermore, since u1 is a member of A,, we have u1 6 mu for some natural number m, and so inf (ul ,nu) = u1 = inf (ul ,mu)
CH. 9,s 61 ]
EVERY QUOTIENT SPACE IS ARCHIMEDEAN
433
holds for all n 2 m. It follows that inf (u, nu) = inf (ui,nu) = inf ( u l , mu) = inf (u, mu) holds for all n 2 m. This is the desired result. Combining the last two theorems, we get back the earlier result that L/A is Archimedean for every ideal A if and only if every principal ideal in L is a projection band. Let L be a Riesz space, not consisting exclusively of the null element, and B the set of all proper prime ideals in L. In Theorem 37.7 it was proved that if the hull-kernel topology in B is Hausdorff and compact (equivalently, all quotient spaces of L are Archimedean and L has a strong unit), then there exists a field (algebra) r of subsets of a non-empty point set X such that L is Riesz isomorphic to the Riesz space of all finite linear combinations (with real coefficients) of characteristic functions of sets belonging to r. In other words, under these conditions L is Riesz isomorphic to the space of real step functions corresponding to the field r. It is our purpose to prove now that we have a similar characterization if all quotient spaces of L are Archimedean (equivalently,the hull-kernel topology in B is Hausdorff) and L is uniformly complete. We shall see that, also in this case, L is Riesz isomorphic to a Riesz space of step functions, namely, the step functions corresponding to the ring of all finite subsets of a certain non-empty point set X. In other words, L is Riesz isomorphic to the Riesz space of all real functionsf on some nonempty point set X such that f vanishes outside an appropriate finite subset of X (the subset depending upon f).Any space of functions of this kind is, therefore, a standard example of a Riesz space having all of its quotient spaces Archimedean and being uniformly complete in addition. Evidently, all finite dimensional Archimedean Riesz spaces are very special cases. For a Riesz space having all of its quotient spaces Archimedean without being uniformly complete, we refer to Exercise 61.5 at the end of this section. First we prove a simple lemma. Lemma 61.3. Let L be a uniformly complete Riesz space such that every quotient space is Archimedean (i.e., such that every ideal in L is uniformly closed). Then, for any principal ideal in L , the number of mutually disjoint nonzero members of the ideal is finite. Proof. Assume that, for some v EL', the principal ideal A, generated by v contains an infinite sequence (u,, : n = 1,2, . . .) of mutually disjoint nonzero elements. We may assume that 0 < u,, 5 2-"v;
n = 1,2,.
.. .
434
[CH. 9.8 61
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
Let A be the ideal generated by the sequence (u, :n quence n
(s, =
k=l
uk : n = 1, 2,
=
1,2, . . .). The se-
. . .)
is an increasing v-uniform Cauchy sequence, so s = sup J', exists by the uniform completeness, and s is equal to the uniform limit of the sequence. Since the ideal A is uniformly closed, we have s E A , and so s 5 ms, holds for appropriate natural numbers m and n. But this implies that u , , ~6 ms,, which contradicts the fact that u , , ~and s, are disjoint.
Theorem 61.4. Let L be a Riesz space, not consisting exclusively of the null element. The following conditions are now equivalent. (i) L is uniformly complete and every quotient space of L is Archimedean. (ii) Every principal ideal in L is a finite dimensional Archimedean Riesz space (and so, by Theorem 26.1 1, given the principal ideal A # (0}, there exists a natural number n such that A is Riesz isomorphic to n-dimensional real number space R" with the usual coordinatewise ordering). (iii) L is Riesz isomorphic to the Riesz space of all real functions f on some non-empty point set X such that f vanishes outside an appropriate3nite subset of X , the subset depending on$ (iv) L is super Dedekind complete and every quotient space of L is Archimedean. Proof. (i) = (ii). We assume that L is uniformly complete and every quotient space of L is Archimedean. Let A be a principal ideal in L such that A # (0). Since L is Archimedean by hypothesis, it is obvious that A is Archimedean. Furthermore, in view of the preceding lemma, A has the property that any system of mutually disjoint nonzero elements is finite. It follows that Theorem 26.10 may be applied, according to which any Archimedean space possessing this property is of finite dimension. Hence, A is of finite dimension. (ii) = (iii). By Theorem 26.11, it follows from the hypothesis in (ii) that, given the principal ideal A in L, there exist disjoint atoms v l , .,v, in A (n varying with A ) such that every member of A is a real linear combination of these atoms. Let X be a maximal set of mutually disjoint atoms in L.The existence of a maximal set of this kind is immediately derived by means of Zorn's lemma. Every atom in L is a nonzero real multiple of one of the members of X , and so every f E L is a finite real linear combination of members of X. Hence, denoting the elements of X by x, and assigning to f E L the
..
CH. 9,s 621
THE IDEAL OF THE INFINITELY SMALL ELEMENTS
435
real functionf*(x) on X having as its value at each x E X the coefficient of x in the representation off, we have a Riesz isomorphism between L and the Riesz space of the real functions on X which vanish outside finite subsets of
X.
(iii) * (iv). We may assume that L is the Riesz space of all real functions
f on the point set X such thatfvanishes outside a finite subset. Obviously,
L is super Dedekind complete. The fact that every quotient space of L is Archimedean follows by observing that, for any pair u, v EL', there exists a finite subset of X such that both u and u vanish outside this subset, and so there obviously exists a natural number m such that inf (u, nu) = inf (u, mu) holds for all n 2 m. (iv) (i). We assume that L is super Dedekind complete and every quotient space of L is Archimedean. We have to show that L is uniformly complete. For this purpose it is sufficient to prove that, for any e EL', every increasing e-uniform Cauchy sequence has an e-uniform limit. By Lemma 39.2 this follows already from the Dedekind o-completeness of L.
-
The theorems in this section are taken from the paper by W. A. J. Luxemburg and L. C. Moore Jr. ([l], 1967) referred to earlier. Exercise 61.5. Let L be the Riesz space (coordinatewise ordering) of all real sequencesf = (f(l),f(2), . . .) such that for some no (varying with f) we havef(n) = f ( n o )for all n 2 no. Show that every quotient space of L is Archimedean, but L is not uniformly complete (compare with part (iv) in the list of counterexamples in the proof of Theorem 25.1). 62. The ideal of the infinitely small elements and Riesz seminorms with their quotient seminorms, in particular in spaces with a strong unit
As defined earlier, any element f # 0 in the Riesz space L for which there exists an element g E L such that nlfl 5 191 holds for n = 1,2, . . ., is called an infinitely small element. It is immediately evident that the set Io(L)of all infinitely small elements, together with the null element, is an ideal in L. If L is Archimedean, then Io(L)consists only of the null element, whereas if L is non-Archimedean, then Io(L)has also nonzero members. If A is any uniformly closed ideal in L, then Io(L) is included in A. Indeed, given f E Io(L),there exists an element g E Lsuch that nlfl 5 Igl for n = 1,2, . . ., and so (nlfl-lgl)+ = O E A f o r n = l,2,
... .
436
RIESZ HOMOMORPHISMS A N D QUOTIENT SPACES
[CH.9,g 62
By Theorem 60.2 it follows now from the uniform closedness of A that If 1 E A, and sofE A. This shows that Io(L)is included in A. Denoting the intersection of all uniformly closed ideals in L by I l (L ),the set Zl(L) is the smallest uniformly closed ideal in L, and in view of the preceding remarks we have Zo(L) c Zl(L). In any Archimedean space L we have evidently that ZO(L) = I d L ) = (01 holds, but (as will be shown presently) there exist non-Archimedean spaces in which Zo(L) is a proper subspace of Il(L). First we will show, however, that in spaces with a strong unit the equality l o ( L ) = I l ( L ) holds.
Theorem 62.1. If L possesses a strong unit, then Io(L) = Zl(L). More generally, we have Zo(L/A)= Zl(L/A)for every ideal A. Proof. Let e E L + be a strong unit in L. Given f E Zo(L), there exists an element g E L such that 1fI. 5 Igl holds for n = 1 , 2 , . . ., and g satisfies 191 S ce for some real constant c > 0. Hence nc-'l fI 5 e holds for n = 1,2, . . ., which implies that nlfI 5 e holds for n = 1,2, . . . . In particular, we have If I 5 e for every f E Zo(L). In order to prove now that Zo(L) = Zl(L),we have to prove that Zo(L)is uniformly closed, i.e., we have to show that if u, v EL" and (nu- v ) + E Zo(L) for n = 1,2, . . ., then u E Zo(L). By the remarks just made, (nu-v)+ E Io(L) for n = 1,2, . . . implies that (nu-v)" 5 e holds for n = 1,2, . . ., and so
nu 5 v + e
for n = 1,2,. . .
.
This shows that u is an infinitely small element, i.e., u E Io(L),which is the desired result. Given the ideal A in L, let n be the corresponding Riesz homomorphism from L onto L/A. Since e is a strong unit in L, it is evident that ne is a strong unit in L/A. Hence, we have ZO(L/A) = W / A )
by the result already proved. Example 62.2. This example is due to T. Nakayama, and it was published in a paper by K. Yosida and M. Fukamiya ([l], 1941). Let E be the vector space of all sequences of pairs of real numbers
f=
Y l )Z,. (
Y
YZ), *
*
4,
with the algebraic operations performed coordinatewise. The space E is a
CH.
9, $ 6 2J
THE IDEAL OF THE INFINITELY SMALL ELEMENTS
437
Riesz space with respect to the partial ordering defined by saying thatf 2 0 whenever for each k (k = 1,2, . . .) either x , > 0 or xk = 0, yk 2 0 . Let L be the Riesz subspace consisting of allfsuch that xk # 0 holds for only finitely many k. We introduce the subsets Soand S1 of L, defined by
So = (f: xk = 0 for all k , y, # 0 for only finitely many k ) ,
S, -- (f:x, = 0 for all k), and we will show that Zo(L)= So and Z,(L) = S , . This will prove then that Zo(L)is properly included in Z,(L). In the first place it is evident that So is included in Zo(L).In order to show that S , is included in Z,(L), let
f = ((03 Yl), be given. For n = 1, 2, . . ., let
(02Y 2 ) ,
*
.>E Sl
f, = {(O,Y?’), (0, Y‘;’), .> = 1, . . ., n and yr’ = 0 for k > n. Furthermore, * *
satisfy y p ) = y, for k
let
f ’ = ((0, y ; ) , (0, A), . .>
satisfy y; = kly,] for all k. Then f,E Zo(L)c Z,(L) for all n and If-f.1 6 n-’f’, s o f , converges relatively uniformly to$ It follows thatfe I l ( L ) . In order to prove that, conversely, Z,(L) is included in S1, we fix k temporarily, and set M = ( f : f E L ,xk = 0). Then M is an ideal in L such that M # L. Iff, E M for n = 1,2, . . ., and f.converges relatively uniformly to somefe L , thenfis a member of M . This shows that Z,(L) c M , i.e.,
Z,(L) c ( f : f € L ,x, = 0). The last inclusion holds for every k, and so I,(L) c S, . But then Zo(L)is also included in S , , and it follows easily that Io(L)is then also included in SO. Assume now that L is a Riesz space in which there is defined a seminorm. The seminorm will be denoted by p , i.e., the value of the seminorm at the element f will be denoted by p(f). Hence, to recapitulate, we have 0 6 p(f) c 00 for all f~ L , p(f) = 0 for f = 0, p(f+g) 5 p ( f ) + p ( g ) for allfand g in L , and p(af) = lalp(f) for allfE L and all real a. The seminorm p in L is called a Riesz seminorm whenever p ( f ) 6 p ( g ) holds for allf, g in L
438
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.9, $ 6 2
satisfying If1 6 191(i.e., whenever p(u) 5 p(v) for 0 5 u 5 v and p(f) = p(lf1) for allf). Of course, p is called a Riesz norm if, in addition, p(f) = 0 holds only forf = 0. We shall say that L is a normed Riesz space whenever
L is a Riesz space equipped with a Riesz norm. This agrees with what was
observed in section 57. Normed Riesz spaces and Riesz spaces with Riesz seminorms will be investigated more extensively in later chapters; for the moment we restrict ourselves to some remarks on Riesz seminorms (or norms) and their quotient seminorms (or norms). Theorem 62.3. Let L be a Riesz space with Riesz seminorm p, and let A be an ideal in L . Then the quotient seminorm A, de$ned in the usual manner in LIA by wl) = i.f(P(f) : f E V1)Y
is a Riesz seminorm in LIA. If A is p-closed, i.e., ifJ;, E A for n = 1, 2, . . . and p( f-f,) + 0 implies that f E A , then A is a Riesz norm in LIA.
Proof. It will be proved first that [O] 5 [u] 5 [u] implies A([u]) 6 A([.]). Given that [O] 6 [v] 6 [u] and given 8 > 0, there exists an elementf, E [u] such that P(f1)
For u1 = lfll we have
< @41)+&.
and Next, choose 0
5 v E [u], and let v1 01 E
and 0 5 u1
= inf (0,ul). Then
inf (bl, [UI) =
[Ol
5 u l , so P(Vl>
5 P ( U A < A([.])+&.
A(rv1)
5
It follows that P@l)
< A(hI)+E,
and so A( [ u ] ) 6 A( [u]). holds for every f We have to prove, secondly, that A ( [ l f l ] ) = Denoting, temporarily, elements of the ideal A by a, a', . . ., we have
A(u])
A([lfl I)
A(y.1)
=
inf {P(lfl - a ) : a E A } ,
= inf {p(lf-al)
:a E A } .
E
L.
CH. 9, S 621
On account of
THE IDEAL OF THE INFINITELY SMALL ELEMENTS
439
llf-4-If11 I la1 E A
we have If-al - If1 E A , which implies that any element If-al is of the form If1 -a’ for some a‘ E A . The set (If-a1 : a E A ) is included, therefore, in the set (If1 - a : a E A ) , and so For the converse, note that if .n is the Riesz homomorphismf + If]from L onto L/A, and a is a given member of A , then If1 and If1 - a have the same image under n, and so If1 and IIf1 - a1 have also the same image, i.e.,
[If11= [llfl -all. It follows by Theorem 59.1 (v) that there exists an element a’ E A such that so Taking infima, we find that Finally, we have to prove that if A is p-closed, then A(If]) = 0 implies that If]= [O]. The proof is familiar. In view of A( V]) = 0 there exists a sequence (f.:n = 1 , 2 , . . .) in [f]such that p(f.) + 0, i.e.,
P{f,-
-L>> 0.
(fl
+
All elementsf, -f.are in A , and so f, is in A on account of A being p-closed. Sincef, is arbitrary in Lf], this shows that [f]and A are identical, i.e., If]= 101. The next theorem will show that in an Archimedean Riesz space with a strong unit e we can define in a natural manner a Riesz norm pe, the euniform Riesz norm. Theorem 62.4. Let L be an Archimedean Riesz space possessing a strong unit e. Then p,(f) = inf (k :k 2 0, If1 I ke)
is a Riesz norm in L ,andfor any sequence in L the notions of norm convergence, e-uniform convergence and relatively uniform convergence are equivalent.
440
RIESZ HOMOMORPHISMS A N D QUOTIENT SPACES
[CH.9,s 62
Proof. It is easily verified that pe is a Riesz seminorm in L. In order to show that pe is a norm, let us assume that p e ( f ) = 0. Then 1f I 2 ke holds for all k > 0, and this implies that f = 0 because L is Archimedean. If p , ( f - f , ) + 0, then p , ( f - f , ) < E for all n 2 no(&),and so If-f,l 5 Ee for all n 2 no(&),which shows thatf, converges e-uniformly tof. Furthermore, e-uniform convergence obviously implies relatively uniform convergence. Conversely, if (f,:n = 1,2, . . .) is relatively uniformly convergent to f, then there exists an element u EL' such thatf, is u-uniformly convergent to f . Since u 5 ke holds for an appropriate constant k > 0, it follows that f, is e-uniformly convergent tof, and this evidently implies that p , ( f - f , ) + 0.
In the next theorem it will be proved that, under the conditions of the last theorem, if .4 is a uniformly closed ideal in L , then the quotient norm I in L/A of the e-uniform norm in L is exactly the [el-uniform norm in LIA. Theorem 62.5. Once again, let L be an Archimedean Riesz space with strong unit e, and let pe be the e-uniform norm in L introduced in the preceding theorem. Furthermore, let A be a uniformly closed ideal iri L . Then A is p,-closed, so (by Theorem 62.3) the quotient norm 1 of pe exists in L/A. Also, since L/A is Archimedeanand [el is a strong unit in LIA, the [el-uniform norm pCelexists in LIA. We have 1 = pCel.
Proof. The ideal A is uniformly closed, so A is closed under the operation of taking e-uniform limits, which implies that A is p,-closed. Hence, by Theorem 62.3, the quotient norm 1 of pc exists in L/A. Observe also that L/A is Archimedean because A is uniformly closed. It remains to prove that I = pCel.From the definition of pCclit follows that I[flI 5 (~ce~([fI)lCel holds for every [f]
E L/A,
and so
But, by the definition of the quotient norm, we have A([e]) 6 p,(e) we arrive already at the result that
=
1, so
For the converse inequality, observe that If I 6 ke implies [ I f I] S k[e], i.e., I lf]l 5 k[e]. Taking infima with respect to k, it follows that pCel([f])
CH.9,B 631
441
RIESZ HOMOMORPHISMS AND THE RELATIVELY UNIFORM TOPOLOGY
5 pe(f). Hence
~ceJ(Cf1)S
inf(pe(f)
: f Lf1) ~
=
4Cfl).
The final result is, therefore, that I = pcclholds.
63. Riesz homomorphisms and the relatively uniform topology WE repeat once more that, by the definition given in section 16, the subset S of the Riesz space L is called (relatively) uniformly closed whenever, for every relatively uniformly convergent sequence in S, all relatively uniform limits of the sequence are also members of S. The empty set and the space L itself are uniformly closed, and arbitrary intersections and finite unions of uniformly closed sets are uniformly closed. Hence, the uniformly closed sets are exactly the closed sets of a certain topology in L,the relatively uniform topology. It was proved in section 16 that, given the subset V of L , the set V is open in the relatively uniform topology if and only if, for every sequence (f,:n = 1,2, .) in L which converges relatively uniformly to a pointfe V, we havef, E Vfor all but a finite number of thef,. It follows that iff, -+ f ( r u ) , i.e., iff, converges relatively uniformly tof, then f, converges to f in the relatively uniform topology (i.e., any neighborhood offin the relatively uniform topology contains all but a finite number of the f,). Alsc, a mapping cp of L (equipped with the relatively uniform topology) into a topological space Xis continuous if and only iff, - + f ( r u ) always implies that cp(fn)converges to ~ ( fin)the topology of the space X. In particular, if cp is a mapping of L into itself such thatf, -+f(ru) always implies that cp(f,) -+ cp(f)(ru), then cp is continuous. It follows that, forfo E Land the real number u # 0 fixed, the mappings f + f + f o and f -P uf are homeomorphisms from L onto itself, for g E L fixed the mappings f -+ sup (f,g) andf-+ inf g) from L into itself are continuous and, finally, the mappings f -+ f +,f -+ f - andf-+ If1 from L into itself are continuous. We recall that, for any subset S of L,the pseudo uniform closure S;,, of S is the set of allfE L such that there exists a sequence (f,:n = 1,2, . .) in S converging relatively uniformly to$ The closure of S in the relatively uniform topology will be denoted by 3. Evidently, we have
..
v,
.
s c s;, c (s;,);,c .
c
s,
and it was proved in Theorem 16.8 that S is relatively uniformly closed if and only if S = S:, , i.e., S = S;, already implies that S = 3. Furthermore, the subset S of L satisfies S;, = 3 if and only if S;, = (S;,,);,,.In other words,
442
[CH. 9,s 63
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
pseudo closure and closure of S are the same if and only if the pseudo closure of S is equal to its own pseudo closure. The operation of forming pseudo closure or closure preserves the property to be a linear subspace, a Riesz subspace or an ideal, as shown in the following theorem. Theorem 63.1. (i) If S is a linear subspace of L, then so are S:, and 3. (ii) If S is a Riesz subspace of L, then so are S;,, and 3. (iii) If S is an ideal in L, then so are S:,, and 3.
Proof. The proofs for S;, are easy; we present the proofs for 3. (i) Let S be a linear subspace, let f, E 3, and let V be an open neighborhood off+G. Then V - s is a neighborhood off, so there exists an element f~ S such that f E V - a . It follows that f+S E V, which shows that Vis an open neighborhood off+g. Then V-fis aneighborhoodofg, so there exists an element g E S such that g E V-J We have then that f + g E V, so in view off+g E S i t has thus been shown that any neighborhood o f f + s contains a point of S. It follows that f + g E 3. Similarly, afE 3 for any real a. Hence, 3 is a linear subspace. (ii) Let S be a Riesz subspace, and let f, E 3'. We have to show that sup (3, Q) E 3. Given the open neighborhood V of sup (f,g ) , there exists an open neighborhood V , off such that sup (f, E V holds for allfE Vl (we use here that the mappingf-, sup (f, is continuous for S fixed). Take a fixed f E V , such that f E S is also satisfied. Then V is an open neighborhood of sup (5 Hence, there exists an open neighborhood V , of such that sup (f,g) E Vholds for all g E Vz.Take a fixed g E V, such that g E S is also satisfied. Then sup (f,g) E S, and V is an open neighborhood of the element sup (f,g). It has thus been shown that any neighborhood of sup (3,s) contains a point of S, and so sup (f, E 3. (iii) Let S be an ideal in L . It will be sufficient to prove that if 0 5 S 5 f a n d f ~3, then E 3'. For this purpose, let V be an open neighborhood of S , and observe that the continuous mapping f -,inf (f', 9 ) maps f into S. Hence, there exists an open neighborhood V , off such that inf (f', 8) E V holds for allfE V , . Take a fixed f E V , such thatfE S is also satisfied. Then inf (f+,Q) E V, and it follcws fromfE S that f' E S, so
s)
s s)
s).
s
a)
s
a ) E S.
0 5 inf (f+, Thus, any neighborhood of
contains a point of S, and so
E
s.
In Theorem 16.2 (i) it was observed that in an Archimedean Riesz space
CH. 9,s 631
RIESZ HOMOMORPHISMS AND THE RELATIVELY UNIFORM TOPOLOOY
443
relatively uniformly convergent sequences have unique limits. We generalize this result, as follows.
Theorem 63.2. The following conditions in the Riehz space L are mutually equivalent. (i) L is Archimedean. (ii) f.-,f ( r u ) a n d f . -,g(ru) implies f = g. (iii) The relatively uniform topology in L is a TI-topology, i.e., every set consisting of exactly one point is closed. (iv) The relatively uniform topology in L is a To-topology, i.e., for any two diferent points of L there is a neighborhood of one which does not contain the other point. Proof. It was observed already that (i) implies (ii). Conversely, if (ii) holds and L is not Archimedean, there exist elements u, v E L+ such that 0 < u S n-'v for n = 1,2,. . . . Setting w, = u for n = 1,2,. . ., the sequence (w, :n = 1,2, .) converges relatively uniformly to the nullelement as well as to u. This contradicts (ii). Hence, if (ii) is satisfied, then L is Archimedean. Thus, (i) and (ii) are equivalent. We have f E {O}:,, i f and only if for some e in L+ the inequality If I I ;n-'e holds for n = 1,2, (in other words, {O}:,, is equal to the ideal of the infinitely small elements). Hence, {O};" = (0)if and only if L is Archimedean. In other words, in view of the fact that a subset S of L is uniformly closed if and only if S:, = S, we may conclude that (0)is uniformly closed if and only if L is Archimedean. This shows that (i) and (iii) are equivalent. Since (iii) always implies (iv), we have still to show that (iv) implies (iii). For this purpose, assume that (iv) is satisfied, and let 0 # f E L . Then there exists a neighborhood V of the null element such that either f is not in V or the null element is not in f V. Replacing V, if necessary, by V n (- V ) , we may assume that V is symmetric. Hence, iff is not in V, then -f is not in V, and so the null element is not in f + V. It follows that in either case there exists a neighborhood V of the null element such that the null element is not in f V. In other words, given 0 # f E L , there exists a closed set F containing the null element but not containing f . Thus (0)is closed, and so (iii) holds.
..
. ..
+
+
We investigate the relation between Riesz homomorphisms and the relatively uniform topology.
Theorem 63.3. Let L and M be Riesz spaces, equipped with their relatively uniform topologies, and let n be a Riesz homomorphihmfrom L into M. Then .n is a continuous mapping.
444
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH. 9,5 63
Proof. Let S be a uniformly closed subset of M , let f,E n-'(S) for n = 1, 2, . . ., and fn -+ f(ru). Then there exists an element e E L+ such that f, converges e-uniformly to f, and henc: nA converges ne-uniformly to nf, which implies in particular that nfn -+ nf (ru). All nA are membxs of S and S is uniformly closed, so nfE S, i.e., f e n-'S. It follows that n - ' ( S ) is unifcrmly closed. Since this holds for any uniformly closed subset S of M , the mapping TC is continuous.
Corollary 63.4. If. is a Riesz homomorphism from L onto M and S is a subset of M , then S is uniformly closedif andonly if TC- ( S )is uniformly dosed.
'
Proof. It follows from the theorem above that n-' ( S )is closed if S is closed. The proof in the converse direction is a variant of the proof of Theorem 59.3 (ii). Assume that for some subset S of M the set x - ' ( S ) is uniformly closed, and let gn+ g(ru) hold in M with all gn members cf S. We have to prove that g is a member of S. There exist elements f,E n - ' ( S ) for n = 1, 2, . . . and f E L such that gn = TC& for n = 1,2, . and g = R . Hence, for some e E L', the sequence ( T C : n~= 1,2, . . .) converges ne-uniformly to nf; and so there exists a subsequence n, < n, c . . . such that
..
f o r k = 1 ,2 ,. . .
Inf-nf,,! S 2-k7re
By Theorem 59.1 (v) there exist elements (hk : k = 1,2, . . .) in L such that nhk = Zf-TCf,, and lhkl 5 2-ke for all k. It follows that the elements (Ik = f-k, 1,2, . . .) satisfy I f-lkl 5 2-ke for k = 1,2, . . .,
so l k converges e-uniformly to f as k = nf-nhk
+
:k =
co. Observing that
= nfnk E S for k = 1, 2,
. . .,
we have 1, E TC-'(S)for all k , and so it follows from +f(ru) and from the hypothesis that z-'(S) is closed, that f E n - ' ( S ) , and hence g = nf E S. This is the desired result. Concerning the pseudo closure we have the following result.
Theorem 63.5. If n is a Riesz homomorphismfrom L onto M and S is a subset of M , then 7c-'(s:u) = {7c-1(s)};u.
Hence, the pAeudo closure of S is uniformly closed if and only fi the pseudo closure of n-'(S) is uniformly closed.
CH. 9 , § 631
RIESZ HOMOMORPHISMS AND THE RELATIVELY UNIFORM TOPOLOGY
445
Proof. First, let f E {n-'(S)};,,, so there exists a sequence (f,: rl = 1, 2 , . . .) in n-'(S) such t h a t f , +f(ru) holds. Then we have nf, nf(ru), which shows that nf E Si,,. It follows that f E n-'(Si,,). Thus
{n-ys)};,,c n-'(SA).
For the proof of the converse, assume that f E L satisfies f E ~-'(5':,,), so nf E Siu. Then there exists a sequence (f,: n = 1 , 2 , . . .) in n-'(S) such that nf, -P xf (ru) holds. In other words, there exists an element e E L+ such that nf, converges ne-uniformly to nJ Once again, there exists a subsequence n, c n, c . . . such that 2-'ne for k = 1 , 2 , . . .,
Inf-nf,,I
so by Theorem 59.1 (v) there exist elements space of the mapping n such that
If-fn,-fil
(fk' : k = 1,2, . . .) in the null
for k = 1 , 2 , . .
S 2-'e
..
Hence, we have so f~ {n-'(S)}k. It follows that
.-'(s:,,)
c
{n-ys)};,,.
Combining (1) and (2), we obtain that
n-'(S;,,) =
{n-ys)};,,.
From the preceding Corollary 63.4 it follows that S:, is uniformly closed if and only if n-'(Si,,)is uniformly closed, i.e., in view of the just obtained result, S:,, is uniformly closed if and only if {n-'(S)};,, is uniformly closed.
Corollary 63.6. Let n be a Riesz homomorphism from L onto M . If the pseudo uniform closure of every set in L is uniformly closed, then the same holds in M , i.e., the pseudo uniform closure of every set in M is uniformly closed. Proof. Assume that the pseudo uniform closure of every set in L is uniformly closed. Then it is true in particular that, for every subset S of M , the pseudo uniform closure {n-'(S)}k is uniformly closed. Hence, by the theorem above, S;,, is uniformly closed. This is the desired result. Exercise 63.7. Show that, in any Riesz space L, the pseudo uniform closure of the ideal {0}, consisting only of the null element, is the ideal
446
RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.9, p 64
I,(L) of all infinitely small elements, and the closure of (0) in the relatively uniform topology is the ideal Il ( L), the intersection of all uniformly closed ideals. Show by means of the example in Example 62.2 that the pseudoclosure {O}; can be properly included in the closure of (0). 64. Images of principal bands under a Riesz homomorphism
As before, let L and M be Riesz spaces, and let n be a Riesz homomorphism from L onto M . The kernel (null space) of n will be denoted by K,. The principal ideal generated by an element u E L + is denoted by A,, and the band generated by an ideal A is denoted by { A ) . For any U E L +we have A , c {A,,], and so
4%) = n({Au))*
(1)
Note that A , and {A,} are ideals in L , which implies by Theorem 59.2 that n(A,) and n ( { A u } ) are ideals in M. Furthermore, n(A,) is the principal ideal generated by nu (according to Theorem 59.2 (v)), so n(A,,) = An,,. It follows from (1) that {n(Au>> = { 4 { A u } N -
(2)
We will investigate now under which conditions the inverse inclusion { ~ ( { A U l N = {n(Au)>
bolds, either for one fixed u in L+ or for all u in L+ simultaneously. This inverse inclusion is equivalent to 74AUl)
= {x(A,>>= {Azu},
i.e., equivalent to the condition that if u 2 0 is any member of the principal band generated by u, then nu is a member of the band generated by nu. In other words, this means that 0 u E { A , } implies nu E {A,,], i.e., if then
05 u
= sup (inf (u, nu) : n = 1, 2,
. . .),
nu = sup {inf (nu, nnu)} = sup n{inf (u, nu)]. n
n
Theorem 64.1. For any u EL’, the following conditions are equiualent. (i) We have ~ ( { A U l ) = b(AU>> = {AZJY i.e., the image of theprincipal band {A,,}is included in theprincipal band {A,,,).
CH. 9,s 64)
IMAGES OF PRINCIPAL BANDS UNDER A RIESZ HOMOMORPHISM
447
(ii) For any 0 5 v E {A,,}, if w E Lf and
(w+inf (v, nu)-v)+ E K,
for n = 1,2,.
. ., then W E & .
Proof. We write down several conditions each of which is equivalent to the preceding and (or) next one. (a) The first condition is the condition in part (i) of the theorem. (b) As observed above, condition (a) is equivalent to the condition that zv = sup n{inf (v, nu)} n
holds for every v satisfying 0 5 v E {A,,}. (c) For any 0 5 v E {A,,}, if
.
0 5 m 5 m-z{inf (v, nu)}
for n = 1,2,. .,then rn = 0. (d) For any 0 5 v E {A,,},if w E L+ and
0 5 zw 5 nu-n{inf (v, nu)} forn = l , 2 , ...,t h e n w ~ K , . (e) For any 0 5 v E {A,,},if w E L+ and
z(w +inf (v, nu) - v ) 5 0
for n = 1,2,. . .,then W E & . (f) For any 0 5 v E {A,,}, if w E L+ and
z{(w+inf ( v , nu)-v)+} = 0
for n = 1,2,. . .)then w E K,. (g) This is the condition in part (ii) of the theorem.
Theorem 64.2. If L has the principal projection property, the following conditions are equivalent. (i) We have z({Au}) =
{4%J>= {AZJ
for every u E Li, i.e., for every u in L + the image of the principal band generated by u is the principal bandgenerated by nu. (ii) For any u E L +and any 0 5 v E {A"}, i f w E L +and (w+inf (v, nu)-v)+ f o r n = 1 , 2,..., t h e n w ~ K , .
E
&
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RIESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH.9, S 64
In addition, if these conditions are satisfied, then M has the principal projection property.
Proof. It follows from the preceding theorem that (i) implies (ii), also if L does not have the principal projection property. Conversely, if (ii) holds; then it follows from the preceding theorem that n({A,,}) c { n ( A u ) }holds for every u E L+. Hence, combining this inclusion and the inclusion in formula (2) we obtain the result that {n({AuIN = b(AU)l
holds for every u E L '. Observe now that {A,,} is a projection band in L since L has the principal projection property. It follows by Theorem 59.2 (vii) that n({A,,}) is a projection band in M , and so n ( { A u } )= { n ( { A u } ) } .The final result is that, for every u E L + ,we have n({Aul)
=
{n(4Jl-
It remains to show that if L has the principal projection property and (i) holds, then M has the principal projection property. Every principal band in M is of the form {Aff,,}for some u E L + and ,
{A,,,} = {n(Au)l
=
nn({Aul)
by (i), which shows that {A,,,} is the image of a projection band in L , and so {Anu}is a projection band in M . This is the desired result.
Exercise 64.3. We recall that if 0 u,,t holds in L , and Cis the ideal generated by the system (u,, : n = 1,2, . . .), then any u 2 0 in the band {C}satisfies u = sup,, {inf (u, nu,,)} by Theorem 28.2. Furthermore, i t is evident that n(C)is exactly the ideal generated by the system (nu,,: IZ = 1,2, . . .). Show that the following conditions are equivalent. (i) n({C>> = {ml. (ii) For any 0 v E { C } , if w E L +and (w+inf (u, nu,,)-u>+ E Kff
f o r n = 1,2, ..., t h e n w ~ K , . Exercise 64.4. Show that if either L is Dedekind a-complete or L has the projection property, then the following conditions are equivalent. (i) n({C})= {n(C)}for every ideal C generated by a finite or countable system.
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RIESZ U-HOMOMORPHISMS AND THE ORDER TOPOLOGY
449
(ii) If 0 5 unt in L , if u 2 0 is a member of the band generated by (un : n = 1,2, . . .), if w E L + and (w+inf
(0,~J-TI)'
E
K,
f o r n = 1 , 2,...,t h e n w ~ K , . 65. Riesz a-homomorphisms and the order topology Let L and M be Riesz spaces. We recall that, according to Definition 18.10, the Riesz homomorphism n fromL onto Miscalled a Riesz a-homomorphism if n preserves countable suprema, i.e., if it follows from f = sup& (n = 1,2, . . .) in L that nf = sup 7th holds in M . A sufficient condition for the Riesz homomorphism n to be a Riesz a-homomorphism is that 0 5 ATf in L+ implies O 5 n f f n f in M + (or, equivalently, unJO in L + implies nuJO in M'). It was proved in Theorem 18.11 that the Riesz homomorphism n is a Riesz a-homomorphism if and only if the kernel Kn of n is a a-ideal in L . Equivalently, n is a Riesz a-homomorphism if and only if for any a-ideal N in the space M the inverse image n-'(N) is a a-ideal in the space L. Note that if n is a Riesz a-homomorphism from L onto M and u 2 0 is a member of the band { A , } generated by a given element ti in L', i.e., then Hence
u = sup (inf (u, nu) : n = 1 , 2 , . . .),
nu = sup n{inf (u, nu)} = sup {inf (nu,mu)). n
n
~({Aul)c
{Anul
holds for every ti E L + . In other words, the conditions (i) and (ii) of Theorem 64.1 are satisfied for every u E L'. The following theorem is, therefore, an immediate corollary of Theorem 64.2. Theorem 65.1. If L has the principal projection property and n is a Riesz a-homomorphism from L onto M , then holdsfor every u E L + and, in addition, the space M has also the principal projection property.
It will be proved now that a Riesz a-homomorphism preserves not only the principal projection property, but preserves also Dedekind a-completenes. We will prove even more. In general, if n is a Riesz a-homomorphism from
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RIESZ HOMOMORPHISMS A N D QUOTIENT SPACES
.
[CH.
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= 1,2, . .) and u are elements in L+ such that nu,,t nu holds in M’, there need not exist elements (u: :n = 1,2, . .) and u’ in L+ such that nu,,’= nu,,for n = 1,2, . . ., nu’ = nu and u;tu’ in L . If L is Dedekind a-complete, however, such elements ui and u‘ do exist.
L onto M , and (u,, :n
.
Theorem 65.2. Let L be Dedekind a-complete, and let n: be a Riesz a-homomorphism from L onto M. Then, given (u,, :n = 1,2, . . .) and u in L+ such that nu,,fnu holds in M + , there exist (u: :n = 1,2, . .) and u‘ in L f such that nu; = nu,,for n = 1,2, ., nu‘ = nu and u;tu‘ in L. Furthermore, M is now Dedekind a-complete.
.
..
Proof. Let u and u,, (n = 1,2, holds in M + . We set
. . .) be elements of L+ such that nu,,fnu
u; = inf {sup ( u l ,
. .., u,,), u }
for n = 1,2, . . . . Then nu; = inf (nu,,,nu) = nu,,holds for all n, and it is evident that we have
0 6 u ; t 5 u.
Since L is Dedekind a-complete, there exists an element u‘ E L + satisfying 0 5 u;tu’, so nuifnu’ because n is a a-homomorphism. On the other hand we have nu; = nu,,fnu. Hence nu’ = nu. In order to show now that M is Dedekind a-complete, assume that
0 5 m,
tsm
holds in M . There exists an element u EL’ satisfying m = nu, and for each n = 1,2, . . . there exists an element u,, E Lf satisfying m,, = nu,,. Hence 0 S nu,,tS nu in M . Introducing the same elements u; as above, we have nu: = nu,, = m,
for all n, and 0 5 u;
7 6 u.
Consequently, there exists an element u‘ in Lf such that 0 and so 0 5 nu; t nu’, i.e., m, t nu’.
5 uAtu’ holds,
This shows that M is Dedekind a-complete. The last theorem is due to A. I. Veksler ([l], 1958). We recall that, by the definition given in section 16, the subset S of the Riesz space L is called order closed if for every order convergent sequence in S the order limit of the sequence is also a member of S. The empty set and the space L itself are order closed, and arbitrary intersections and finite unions
CH.9,s 651
RIESZ U-HOMOMORPHISMS AND THE ORDER TOPOLOGY
45 1
of order closed sets are order closed. Hence, the order closed sets are exactly the closed sets of a certain topology in L , the order topology. It was proved in section 16 that the order topology in L is a T,-topology (i.e., every set consisting of exactly one point is order closed). Furthermore, the subset V of L is open in the order topology if and only if, for every sequence (f,: n = 1, 2, . . .) in L converging in order to a point Y E V, we havef, E V for all but a finite number of thef,. It follows that iff. +f, i.e., iff, converges in order tof, thenf, converges tof in the order topology (i.e., any neighborhood off in the order topology contains all but a finite number of thef,). Also, a mapping cp of L (equipped with the order topology) into a topological space X is continuous if and only i f f , +f always implies that cp(f,) converges to p(f) in the topology of the space X . In particular, if p is a mapping from L into itself such that f, --+ f always implies cp(f,) + c p ( f ) , then cp is continuous. It follows that, forfo E Land the real number a # 0 fixed, the mappingsf-tf+fo and f-,afare homeomorphism from L onto itself, for g E L fixed the mappings f -+ sup (f,g) and f + inf (f,g ) from L into f -,f- and f + 1f I itself are continuous and, finally, the mappings f -,f +, from L into itself are continuous. We recall that, for any subset S of L,the pseudo order closure S’ of S is the set of all f E L such that there exists a sequence (f,: n = 1,2, . . .) in S converging in order to$ The order closure of S, i.e., the closure of S in the order topology, will be denoted by cl(S). Evidently, we have
s c S‘ c (S’)’ c .. . c cl ( S ) ,
and it was proved in Theorem 16.6 that S is order closed if and only if S = S’, i.e., S = S‘ already implies that S = cl (S). Furthermore, the subset S of L satisfies S’ = cl(S) if and only if S‘ = (S‘)’. In other words, pseudo closure and closure of S are the same if and only if the pseudo closure of S is equal to its own pseudo closure. The operation of forming pseudo closure or closure preserves the property to be a linear subspace, a Riesz subspace or an ideal, as shown in the following theorem. Theorem 65.3. If S is a linear subspace of L (or a Riesz subspace of L , or an ideal in L ) , then so are S’ and cl(5‘).
Proof. The proof is very similar to the proof of Theorem 63.1, where we dealt with the analogous properties for the relatively uniform topology. We will now investigate the relation between Riesz a-homomorphisms and the order topology. To this end, observe first that if n is a Riesz a-homo-
452
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RlESZ HOMOMORPHISMS AND QUOTIENT SPACES
morphism from the Riesz space L into the Riesz space M , andf, +f holds in L , then n , + nf holds in M . Indeed,& +f implies the existence of a sequence p,JO in L such that I f - j J 5 p,10 holds in L . Since n is a Riesz a-homorphism, we have np,JO in M , and so
5 nPJ0
Inf-nfl holds in M , i.e., nf.
--t
nf holds in M .
Theorem 65.4. Let L and M be Riesz spaces equipped with their order topologies, and let n be a Riesz a-homomorphism from L into M. Then R is a continuous mapping.
Proof. Let S be an order closed subset of M , letf. E n-'(S) for n = 1 , 2, . . ., and let f. + f. Then R . + nf and all nf, are members of S, so nf is a member of S since S is order closed. It follows thatfE n-'(S). This shows that n-'(S) is order closed. Since this holds for any order closed subset S of M , the mapping n is continuous. Before discussing any corollaries, we prove a simple lemma. Lemma 65.5. I f n is a Riesz homomorphism (not necessarily a a-homomorphism) from the Dedekind a-complete Riesz space L onto the Riesz space M and if (m, :n = 1,2, . . .) is a sequence in M + satisfying m,JO, then there exists a sequence (u, : fi = 1,2, . . .) in Lf such that u,JO and m, = nu,,for n = 1,2,. . ..
Proof. For every m, there exists an element u, in L+ satisfying rn, = nu,. For n = 1,2,. . ., let w, = inf (ul, . . .,on). Then nw, = m, for n = 1 , 2 , . . ., and 0 S w,J in L f . Since L is Dedekind a-complete, the element w = inf w, exists in L', and we have
0 5 nw 5 xw,,
=
m,
for all n.
Hence, on account of m,J0, it is evident that nw = 0. If, therefore, we set u, = w,- w for n = 1,2, . . ., then u,JO and
nu, = nw,-nw
= xw, = m,
for n = 1,2,.
. ..
This completes the proof. Theorem 65.6. Let n be a Riesz homomorphismfrom the Dedekind a-complete Riesz space L onto the Riesz space M , and let S be a subset of M such that n- ( S )is order closed. Then S is order closed.
'
CH. 9 , s 651
453
RIESZ 0-HOMOMORPHISMS A N D THE ORDER TOPOLOGY
Proof. Assume that g, + g holds in M with all g, members of S. We have to show that g is a member of S. It follows from g, + g that there exists a sequence (m, :n = 1 , 2 , . . .) in M + such that Ig-gnl
5 mn l o
holds in M . By the above lemma there exists a sequence (u, : n = 1,2, . . .) in L+satisfying u, 10 and nu, = m, for n = 1,2, . . .. Furthermore, since the mapping n is onto M , there exists an elementf E L such that g = nf and there exist elementsf, (n = 1 , 2 , . . .) in n - ' ( S ) such that g, = nf. Hence nu,
[nf-nfl
0 with u,J 0.
.
By Theorem 59.1 (v) there exist elementsf,'(n = 1, 2,. .) in the null space of the mapping n such that
lf-fn-fil
S
un
holds for every n. Hence
lf-(f,+fi)l S
u, 40 with f,+f,' E n - ' ( S ) for all n.
Since n - ' ( S ) is order closed, it follows that f E n - ' ( S ) , and so g This is the desired result.
=
nf E S.
Corollary 65.1. If n is a Riesz a-homomorphism front the Dedekind ocomplete Riesz space L onto the Riesz space M and i f S is a subset of M , then S is order closed if and only if n- ( S ) is order closed.
'
Proof. Follows by combining Theorem 65.4 and Theorem 65.6. We make several remarks. In the first place, it follows from Theorem 65.2 that under the conditions of the last corollary the space M is Dedekind a-complete just as well as L . Furthermore, it follows from Corollary 63.4 that, for any subset S of M , the set S is uniformly closed if and only if n-'(S) is uniformly closed. Now, in an Archimedean space and hence surely in a Dedekind o-complete space, the relatively uniform topology is stronger than the order topology, i.e., every order closed set is uniformly closed. In general, therefore, the collection of uniformly closed subsetsof M will beproperly larger than the collection of order closed subsets. It follows that in this case the collection of subsets of L that are simultaneously uniformly closed and inverse images of subsets of M is also properly larger than the collection of subsets of L that are simultaneously order closed and inverse images of subsets of M . Expressed somewhat differently, we have the following theorem.
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RlESZ HOMOMORPHISMS AND QUOTIENT SPACES
[CH. 9.5 65
Theorem 65.0. Let n be a Riesz a-homomorphism from the Dedekind acomplete Riesz space L onto the Riesz space M , and let L have thefurtherproperty that the order topology and the relatively uniform topology in L are the same. Then M is also Dedekind a-complete, and the order topology and relatively uniform topology in M are the same. Proof. It is sufficient to prove that every uniformly closed subset S of M is order closed. Given the uniformly closed subset S of M , it follows from Corollary 63.4 that z-'(S) is uniformly closed. Hence, in view of the hypothesis about the topologies in L , the set n-'(S) is order closed. But then, by Corollary 65.7, A is order closed. Concerning the pseudo order closure we have the following result. Theorem 65.9 (i) If n is a Riesz a-homomorphism from the Riesz space L into the Riesz space M , then
holds for every subset S of M . (ii) If n is a Riesz homomorphism from the Dedekind a-complete Riesz space L onto the Riesz space M , then
n-'(s')c (n-'(s))' holdsfor every subset S of M. (iii) I f n is a Riesz a-homomorphism from the Dedekind a-complete Riesz spaLe L onto the Riesz space M , then
n-'(s')= (n-'(s))' holds for every subset S of M . Hence, the pseudo order closure of S is order closed ifand only ifthe pseudo order closure of n:-'(S) is order closed. Proof. (i) If f E (n-'(S))', there exists a sequence (f,: n = 1,2, . . .) in n-'(S) satisfyingf, +f. Since n: is a a-homomorphism, it follows that nh + nf, so nf E S'. This shows that fE n-'(S'). Hence
,'(s))'c n- I@'). (ii) Let f E n-'(S'), so nf E S'.Then there exists a sequence (f,: n = 1, 2,. . .) in n-'(S) satisfying nf, --+ n - This implies the existence of a sequence (m, : n = 1,2, . .) in Mf such that
.
CH.9,s 651
RIESZ U-HOMOMORPHISMS AND THE ORDER TOPOLOGY
455
holds in M. By Lemma 65.5 there exists a sequence (u, : n = 1,2, . . .) in L i such that u,, 10 and m, = nu,, for n = 1,2, . .. Hence, we have
.
Inf-n$,l
5 R U , , ~0 with u,, 10.
By Theorem 59.1 (v) there exist elementsf,'(n = 1, 2, .. .) in the null space of the mapping n such that
If-fn-fil
5 un
holds for every n. Hence
If-(f,,+fJl
5 u, 5.0
withf,,+fi E n-'(S) for all n.
It follows thatfE (a-'(S))'. Thus, we have
n-'(S') c
(n-ys))'.
(iii) By combining the results in parts (i) and (ii) it is immediately seen that n-'(S') = ( R - y s ) ) ' holds for every subset S of M. From Corollary 65.7 it follows that S' is order closed if and only if n-'(S') is order closed, i.e., in view of the just established result, S' is order closed if and only if (n-'(S))' is order closed.
Corollary 65.10. Let R be a Riesz a-homomorphism from the Dedekind ccomplete Riesz space L onto the Riesz space M . If the pseudo order closure of every set in L is order closed, then the same holds in M y i.e., the pseudo order closure of every set in M is order closed. Proof. Assume that the pseudo order closure of every set in L is order closed. Then it is true in particular that, for every subset S of M, the pseudo order closure (n-'(S))' is order closed. Hence, by the theorem above, S' is order closed. This is the desired result. Exercise 65.11. Show that if n is a Riesz a-homomorphism from the Riesz space L onto the Riesz space M, then conditions (i) and (ii) of Exercise 64.3 are satisfied for every sequence 0 5 u,,f in L+.Show that if, in addition, L is either Dedekind a-complete or has the projection property, then n({C}) = {R(C)}holds for every ideal Cgenerated by a finite or countable system in L.
Exercise 65.12. Let L be Dedekind c-complete and let M be order sepa-
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RIESZ HOMOMORPHISMS A N D QUOTIENT SPACES
[CH.
9,s 66
rable. It is now easy to show that if n is a Riesz a-homomorphism from L onto M , then M must be super Dedekind complete. It is somewhat more difficult, however, to show that if n: is only given to be a Riesz homomorphism from L onto M with kernel K,, then the disjoint complement K," is super Dedekind complete. Hint: For the second part, observe that K," and its image n:(K,d)are Riesz isomorphic, so n ( K f ) is Dedekind a-complete. Also, n ( K f ) is order separable. Hence, n:(K:) is super Dedekind complete, so the same holds for K,.
66. Normal Riesz homomorphisms Let L and M be Riesz spaces. We recall that, according to Definition 18.12, the Riesz homomorphism n from L onto M is called a normal Riesz hcmomorphism if n preserves arbitrary suprema, i.e., if it follows from f = supf,(a E {a}) in L that nf = sup nf, holds in M. A necessary and sufficient condition for the Riesz homomorphism n: to be a normal Riesz homomorphism is that 0 5 fr ff in L+ implies 0 S n:fr ?.fin M + (or, equivalently, u,JO in L+ implies n:u,JO in M + ) . It was proved in Theorem 18.13 that thc Riesz homomorphism n is a normal Riesz homomorphism if and only if the kernel K, of n: is a band in L . Equivalently, n is normal if and only if for every bandN in the space M the inverse image n-I(N)is a band in the space L . We add another condition to the list.
Theorem 66.1. The Riesz Itomomorphisin n from L onto M is normal if only if n ( W H = {W)} holdsfor every ideal C iii L .
atid
Proof. It is evident that if n: is normal, then n ( { C ) ) c {n(C)}holds for every ideal C in L . It will be sufficient, therefore, to show that if n({C))c {n(C)}holds for every ideal C in L , then the kernel K, is a band. The kernel K, is an ideal, and so it follow from the present hypothesis that This shows that the band {K,} is included in the kernel K,, and so {K,} K,, i.e., K, is a band.
=
Since a normal Riesz homomorphism is a Riesz a-homomorphism, every normal Riesz homomorphism preserves the principal projection property and preserves Dedekind a-completeness. It will be shown now that a normal
CH.9 , s 661
NORMAL RlESZ HOMOMORPHISMS
457
Riesz homomorphism also preserves the projection property and Dedekind completeness.
Theorem 66.2. If L has the projection property and K is a normal Riesz homomorphism from L onto M , then M has the projection property and n({C>>= {.(C)}
holds for every ideal C in L . Proof. Given the ideal C in L , the band { C } is a projection band, and so n ( { C } )is a projection band in M by Theorem 59.2 (vii). It follows that On the other hand we have n({C}>c { K ( C ) }since n is a normal homomorphism. Hence, n(W>= {~(C>}.
For the proof that M has the projection property, we have to show that any arbitrary band Nin M is a projection band. The inverse image n - ' ( N ) is a band in L . Denoting n - ' ( N ) by C, we have, therefore, that C = {C}holds. Hence N = K ( { C } )is the image of a projection band, and so N itself is a projection band. Theorem 66.3. If L is Dedekind complete and n is a normal Riesz homomorphism from L onto M with kernel K,, then M and the disjoint complement (K,)d are Riesz isomorphic, and so M i.s Dedekind complete. Proof. This is evident in view of L = K, 0 (K,)d.
CHAPTER 10
The Egoroff Property and the Diagonal Property
The element f of the Riesz space L is said to have the Egoroff property if, given any double sequence ( U n k :n, k = 1 , 2, . . .) in L such that I f l z U n k J k O
forn=1,2,...,
there exists a sequence v , J O in L with the property that for every pair (m,n) of natural numbers there exists a natural number k(m, n) such that u, 2 #,,,k(,,,,) holds. The space L is said to have the Egoroff property if every member of L has the Egoroff property. It will be proved in the present chapter that in any Riesz space L the set of all elements possessing the Egoroff property is a a-ideal in L.The Egoroff property is independent of the Archimedean property; the lexicographically ordered plane has the Egoroff property without being Archimedean, whereas in the Archimedean space C([O,11) of all real continuous functions on the interval [O, 11 the zerofunction is the only member of the space having the Egoroff property. The Riesz space L is said to have the strong Egoroff property if, given any double sequence ( U n k :n, k = 1 , 2 , . . .) in L such that UnkJkO
fern= 1 , 2 , . . . ,
there exists a sequence v, 4 0 in L with the property that for every pair (m,n) of natural numbers there exists a natural number k(m, n) such that v, 2 un,k(,, ,) holds. Evidently, the stronq Egoroff property implies the Egoroff property. The sequence space I , is an example of a space possessing the Egoroff property, but not the strong Egoroff property. We now recall that, according to the definitions given in section 16, the space L has the diagonal property if, given that fnk + k f n
there exists a “diagonal sequence”
+x
-
: = 1, 2, * .) converging in order tof, and L is said to have the diagonal gap property if (fn,k(n)
458
CH.
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THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
459
it follows from the same hypotheses that there exists a "diagonal gap sequence" (f,,r,k(no : i = 1,2, . . .; n , < n, < . . .) converging in order to$ We shall prove that the strong Egoroff property, the diagonal property and the diagonal gap property are equivalent. According to an earlier result (Theorem 16.7), these properties are then also equivalent to the property that for every subset S of L the order closure and the pseudo order closure of S are the same. It is an interesting result, due to T. Chow (1969), that another equivalent condition is that for every convex subset S of L the order closure and the pseudo order closure of S coincide. For any non-empty point set X , we shall denote by L ( X ) the Riesz space of all real (and finitevalued) functions on X , the ordering is pointwise. It will be proved that, for X finite or countable, the space L(X) has the strong Egoroff property. Also, assuming the continuum hypothesis to hold, we shall prove that, for X uncountable, the space L ( X ) does not have the Egoroff property. From this we derive the far more general result that, still under the assumption of the continuum hypothesis, any Riesz space which is Dedekind complete and which has the Egoroff property must be super Dedekind complete. The last result is essentially due to A. G. Pinsker (1949) and I. Amemiya (1952). Another theorem we shall prove is that if .n is a Riesz a-homomorphism from the Dedekind o-complete space L onto another Riesz space M , and L has the Egoroff property or the diagonal property, then M has the same property. Given the non-negative and countably additive measure p in the point set X,we denote as before by M(') = M(')(X, p) the Riesz space of all real, p-almost everywhere finitevalued and p-measurable functions on X ; the ordering is the p-almost everywhere pointwise ordering. It will be proved that if p is (totally) a-finite, then M(') has the diagonal property, and so M(') has then certainly the Egoroff property. The classical Egoroff's theorem for sequences in M"' is related to this fact, as will be explained. We shall also present a more general theorem about M(')(X, p), including the case that p is not a-finite. Given any measure p as above (not necessarily o-finite), p is called locally finite if every p-measurable set of infinite measure contains a p-measurable subset of finite positive measure. Furthermore, p is called localizable if the measure algebra of ( X , p) is Dedekind complete. It will be proved now, under the assumption of the continuum hypothesis, that if p is locally finite and localizable, then M(')(X, p) has the Egoroff property if and cnly if p is a-finite.
460
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. lo,§ 67
Any Archimedean Riesz space with the diagonal property will be called a regular Riesz space. In the original definition of a regular Riesz space L, due to L. V. Kantorcvitch (1936), it was assumed in addition that L is super Dedekind complete. In order to discuss regular Riesz spaces, we recall some earlier results. Order convergence in the Riesz space L is said to be stable if for any sequence f.+ 0 in L there exists a sequence of real numbers (A,, : n = 1,2, . . .) such that 0 S A,, t co and + 0. It was proved earlier (Theorem 16.3) that in an Archimedean Riesz space L order convergence is stable if and only if order convergenceand relatively uniform convergence of sequences in L are equivalent. For our present purpose it will also be convenient to say that the Riesz space L has the a-property whenever, for any sequence (u,, :n = 1,2, . .) in L f , there exists a sequence (A,, : n = 1,2, . .) of positive numbers such that the sequence (Ann.,, :n = 1,2,. . .) is bounded from above. We shall prove that the Archimedean Riesz space L is regular if and only if L has the a-property and order convergence in L is stable. The a-property on its own (i.e., without stability of the order convergence) is also of some importance. In an Archimedean Riesz space L the a-property is equivalent to the diagonal property for relatively uniform convergence and also equivalent to the property that, for every subset S of L, the uniform closure and the pseudo uniform closure of S are the same. It is again a result of T. Chow (1969) that another equivalent condition is that for every convex subset S of L the uniform closure and pseudo uniform closure of S coincide.
.
.
67. The Egoroff property It is assumed throughout the present section that L is a Riesz space.
Definition 67.1. I f ( f , k :n, k = 1, 2, . . .) i$ a double sequence of elements of L and if the element g E L has the property thatf o r every n there exists a natural number k(n) such that g 5 f., ,+), then we will write g << (&).
Note already that if ( f . , ) and (Xk)are double sequences that are increasing as k increases, and we have g << ( f . k ) and g' << ( f i k ) , then
Definition 67.2. The element f E L is said to have the Egoroflproperty if, given any double sequence (unk: n, k = 1,2, . . .) in L such that 0 S
U,kfkIfl
forn
=
1,2,....
CH. 10,s 671
46 1
THE EGOROFF PROPERTY
there exists a sequence 0 S u,,, t If I such that u,,, << (unk) holds for every m (i.e.,for every m and n there exists a natural number k(m, n) such that
S
Un,k(rn,n)
holds). The space L is said to have the Egoroff property if every element of L has the Egoroffproperty.
The Egoroff property, although in a somewhat different form, was introduced by H. Nakano (cf. his book [ 6 ] , 9 14). He assumes that L has the principal projection property, and he calls L a totaIly continuous space whenever, given any double sequence (Bnk:n, k = 1,2, . . .) of principal bands such that Bnk B as k + co holds for n = 1,2, . . ., where B is also a principal band, there exists a sequence (Cm:rn = 1, 2, . . .) of principal bands such that C, 1B and C,,,<< (Bnk)for every m. It is evident that the Egoroff property is related to Nakano's notion of total continuity. The Egoroff property is also related to L. V. Kantorovitch's diagonal property, as will be explained in the next section. Theorem 67.3. The set A of all elements of L having the Egoroffproperty is a a-ideal in L.
Proof. We prove first that if 0 5 w 5 u, and u has the Egoroff property, then so has w. For this purpose, assume that 0
Then 0
5
wnk f k W
s (W,k+(u-w))
fern fk
u
=
..
1,2,.
for n
=
1 , 2 , . . .,
and so there exists a sequence 0 S urnt u satisfying (wnk
+ (-.
w))
for every m. But then we have 0
5
(Vm-(u-W))+
for every m, and also
<< ( w n k )
t (24-(u-w))+
= w.
This shows that w has the Egoroff property. Next, if the elements u and u of Lf both have the Egoroff property, and if 0
5
w,kfku+v
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[CH. 10,s
THE EGOROPP PROPERTY AND THE DIAGONAL PROPERTY
67
holds for n = 1,2, . . ., then (by the dominated decomposition property) there exist double sequences
0 5 unk f k u and 0 5
unk f k 0
such that unk+unk = wnk holds for all rz and k. Hence, there exist sequences 0 5 u, f u and 0 5 u, t u such that u, << ( u n k ) and u, << ( u n k ) for every m, and so w, = u,+u, (for m = 1,2,. . .) satisfies
0 5 w,
t u+u
and w, << (wnk) for every m.
This shows that u+u has the Egoroff property. Now, assume that f and g in L have the Egoroff property. Then If1 and 191 have the property, and so, in view of lf+gl 5 If1 191, the elementf+g has the Egoroff property. Also, it is evident that iff has the property, then so has uffor any real a. It follows already that the set A of all elements of L having the Egoroff property is an ideal in L. Finally, we have to prove that the ideal A is a o-ideal, i.e., we have to prove that every element u 2 0 in the a-ideal A , generated by A has the Egoroff property. To this end, let 0 5 u E A , , and
+
05
unk f k
for n = 1, 2,
u
.
a.
There exists a sequence 0 5 w, f u such that w, E A holds for p = 1,2, Then inf (w,, u n k ) f k inf (w,,u ) = w,, and so there exists, for every p , a sequence (u?) : m Set
o 5 u?)
=
. . ..
1,2, . . .) such that
f, w, and u!$ << (inf (w,,u n k ) ) .
. . ., ug))
u, = sup (u:),
for m = I, 2 , .
. ..
Then 0 6 u, t 5 u holds, as evident from 0 5 u, S w, 5 u. On the other hand, for each p , we have u, 2 YE)for m 2 p , and so every upper bound of the set of all u, is greater than or equal to SLIP,,, ):0 = w,. But sup wP = u, so every upper bound of the set of all u, is greater than or equal to u. It follows that 0 5 u, t u. Fixing m and n for the moment, note that for each p there exists a natural number k ( p ) such that u:)
It follows that
5
inf(wp
SUP (un. k ( l ) Y
*
*7
7
un, k(p))
u n , k(m))
5
un, k(p)
*
= %, max ( k ( l ) , . . .,k(m))
*
CH. 10,s 671
463
THE EGOROFF PROPERTY
This shows that v, << (unk),and so u has the Egoroff property. The proof is thus complete. Corollary 67.4. I f A is a super order dense ideal in L such that every element of A has the Egorofproperty, then the space L has the Egorofproperty.
Proof. Follows immediately from the above theorem since L is the a-ideal generated by A . It will be proved now that the Egoroff property is equivalent to several other properties.
Theorem 67.5. (i) The element u E Li has the Egorofproperty ifand only if, given any double sequence (u,,k : n, k = 1,2, . . .) in L such that 05
U,,k f k
u for n = 1,2,.
. .,
there exists a sequence 0 5 v,, f u such that, for every n, we have v,, 5 u,,,k ( n ) for an appropriate k = k(n). (ii) The element u EL' has the Egoroffproperty i f and only if, given any doubZe sequence (uflk: n, k = 1, 2, . . .) and sequence (u,, : n = 1,2, . . .) ia L such that 0 5 %k f k un t u, there exists a sequence 0 5 v, for an appropriate k = k(n).
u such that, for every n, we have v,,
s un,k(,,)
Proof. (i) If u has the Egoroff property, then the condition in the theorem is obviously satisfied. Conversely, assume that the condition is satisfied, and let 0 S u,,k u for n = 1, 2, . . .. Setting
unk = inf ( u l k ,. . ., unk); n, k = 1 , 2 , . . .,
we have 0 5 v,,k f k u for every n, and so (by hypothesis) there exists a sequence 0 u,, t u such that, for every n, we have v,, 6 v,,,k(,,) for some k(n). For m 2 n we have
s
un
and for m 5 n we have
s
s vm 5 un
5
k(m)
Vn,k(n)
5
5
k(m)
7
%.k(n)*
This shows that v,, urn,k(m, ,) holds with k ( m , n ) = k ( m ) for m 2 n and k(m, n ) = k ( n ) for m 5 n. Hence, u has the Egoroff property. (ii) If the condition in part (ii) is satisfied, then so is the condition in part (i), and so u has the Egoroff property. Conversely, assume that u has the
464
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
Egoroff property, and let
5
We set unk
lrnk f k un
= u,kfu-u,;
[CH. 10,s 67
7 u.
n, k = 1, 2 , . . ..
Then 0 5 D,k f k u holds for every n, and so (by part (i)) there exists a sequence 0 5 w, t u such that, for every n, we have w, 5 on,+) for an appropriate k = k(n). Then u, = (wn+un-u)+
satisfies 0 5
D, un
t u, and =
(wn+un-u)f
5
(un,k(n)fun-u)+
=
%,k(n).
This shows that the condition in (ii) is satisfied. Before presenting still other properties equivalent to the Egoroff property, we list some simple examples. Example 67.6. (i) Let L be the Riesz space of all real numbers, and assume that 0 < u E L . Note that any sequence (w, :n = 1, 2, . . .) in L satisfying w, 10 has the property that w,, converges u-uniformly to zero. Indeed, given E > 0, we have w,, 5 E u for all n 2 N(&).Given now that 0 5 u,,, f k u holds for n = 1,2, . . ., the sequence (D,, :n = 1,2, . . .) defined by u, = (l-n-1)u
satisfies 0 5 u, f u, and for every n we have u,, 5 u,,,(,,) for an appropriate k(n) simply because u-u,k 5 n-'u = u-0, holds for k sufficiently large. This shows that L has the Egoroff property. (ii) The lexicographically ordered plane is an example of a non-Archimedean Riesz space possessing the Egoroff property. Indeed, if u = (xo,yo) is a nonzero positive element of the space and if 0 5 u,,k tr u holds for n = 1,2, . . ., we may assume without loss of generality that for all rt and k the first coordinate of unk is x,, , and a similar argument as above shows then the existence of a sequence 0 5 u,, t u such that all u, have the first coordinate equal to xo and such that for every n we have u, 5 u,,,k(,,) for an appropriate
W).
(iii) The Riesz space C([O, 11) of all real continuous functions on the Interval [0, 11 is Archimedean, but no function f in C([O,l]), unless f is identically zero, has the Egoroff property. Indeed, let f not be identically
CH. 10, $ 6 7 J
zero, let (rn : n for each n, let
465
THE EGOROFF PROPERTY
=
1.2, . . .) be the set of all rational numbers in [O, 11 and,
5
where u,,(x) = 0 for x [O. 11. It is evident that
=
unk t k
in C([O,l]),
rl , . . ., r, and u,,,(x) If(x)l for all other x in 05
unt t k
If1
holds in the space C([O,11). On the other hand, if v is an element of C([O, I ] ) satisfying 0 5 v << (u,,), then v(r,) = 0 for all n, and so u(x) = 0 for all x since v is continuous. Hence, there exists no sequence 0 u,,, t If1 in C([O,11) such that v, << (u,,k) holds for every m. This shows thatfdoes not have the Egoroff property. (iv) We will prove now that the space L ( X ) of all real (finitevalued) functions on a finite or countable point set X , with pointwise ordering, has the Egoroff property. It will follow that the subspace B ( X ) of all bounded functions in L ( X ) has also the Egoroff property. In particular, the sequence space ,Z has the Egoroff property. Furthermore, for any natural number n, the space R" (equipped with coordinatewise ordering) has the Egoroff property (indeed, R" is the space L ( X ) for X consisting of n points). Hence, any Archimedean Riesz space L of finite dimension has the Egoroff property (because L is Riesz isomorphic to R" for somt n). For the proof that L ( X ) has the Egoroff property, denote the points of X by x 1 ,x2, . .,let u E L'(X) and 0 5 u,,, f k u for n = 1,2, . . ., i.e., for every n we have
.
0 5 u,,k(x) f k u(x) for all x E X .
.
For n = 1,2, . . .we define u, E L ( X ) by u,(x) = u(x)-n-' for x = x l , . ., x, and u,(x) = 0 for all other x . Then 0 5 v, u holds in L ( X ) , and for
every n we have u, 5 u,,, k ( n ) for an appropriate k u(x)-unk(x) 5
=
k(n). This follows from
fl-'?
holding at the points x = xl, . . ., x , simultaneously for k sufficiently large. Hence, u has the EgoroR property. (v) It will be proved in section 75 that, for X uncountable, the space L ( X ) of all real functions on X does not have the Egoroff property. The proof uses the continuum hypothesis. It is of some interest, therefore, to have an explicit example of an uncountable X where it can be proved by a direct method that L ( X ) does not have the Egoroff property. The following example is due to J. -4. R. Holbrook ( [ l ] ,1968). Let X be the set of all sequences x = ( x ( l ) , x(2), . .) with positive integer coordinates, and L ( X )
.
466
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10, g 67
the Riesz space of all real functions on X. It is evident that L ( X ) is Dedekind complete. We will prove that the function e(x), satisfying e(x) = 1 for all XEX,does not have the Egoroff property. For this purpose, consider the subsets Xnk (n,k = 1,2, . . .) of X defined by
k). Identifying subsets of X with their characteristic functions, we have obviously that Xnk fk X holds in L(X)for every n. Nevertheless, the double sequence has the property that for no choice of k(n) do we have Xnk =
n
( x : x(n)
Xn,k(n)
=
x-
..
Indeed, given k ( n ) for each n, we set xo = (xo(l),x0(2), .) with xo(n) = k(n) 1, and then xo is no member of any of the Xn,k ( , , ) . It follows from this that X (i.e., the function e ( x ) ) does not have the Egoroff property. Indeed, assume that we have a sequence (urn :m = 1,2, . . .) in L ( X ) such that 0 2 v, << (Xnk)for every m. Then there exists a natural number k(m), for every m, such that v, g X,, k(,). Hence
+
sup
sup Xrn,k(m) f
x~
in view of the remark just made. This shows that X does not have the Egoroff property. (vi) It will be shown in section 71 that, for a (totally) o-finite measure p in the point set X , the Riesz space M("(X, p ) of all real, p-almost everywhere finitevalued, p-measurable functions has the Egoroff property. The relation with the classical Egoroff's theorem will then also become clear (cf. Example 74.4). The example in part (v) shows that there exist Dedekind complete Riesz spaces not possessing the Egoroff property. We note already that in section 75 it will be shown that any Dedekind complete space with the Egoroff property must be super Dedekind complete (at least if the continuum hypothesis holds). In any vector space V over the real numbers, the convex hull ( S ) of any (non-empty) subset S of V is the set of all f E V of the form P
f = C1A f n , n=
c:=,
with A1, . . ., 1, real and non-negative and satisfying An = 1, and with fi,. . .,fp members of S. The number p is finite but not fixed (i.e., p may be
CH. 10,s 671
THE EGOROFF PROPERTY
467
different for different f E ( S ) ) . The notion of a convex hull will play a role in the following characterization of the Egoroff property.
Theorem 67.7. Let u be an element of L+.Then each of the following corlditions is equivalent to the condition that u has the Egoroflproperty. (i) If u 2 u,,k J k 0 in L, then there is a sequence v, J 0 in L with iheproperty thatfor every pair (m,n ) of natural numbers there exists a natural number k(m, n ) such that v, 2 u,,,k(,,,,) holds. (ii) Ifu 2 u,,k l k 0 in L,then there is a sequence v, 10 in L with theproperty that for every n there exists a natural number k ( n ) such that v,, 2 u,,,k ( n ) holds. (iii) If u 2 U,,k J k u,, 4 0 in L,then there is a sequence v,, 10 in L with the property that for every n there exists a natural number k(n) such that v,, 2 %, k(n)
(iv) I f t i 2 U,,k J k u,, 4 0 in L, then there is a sequence u, 4 0 in L such that, for every m, we have v, >= w, for some w, in ((u,,,)), where ((u,,k)) denotes the convex hull of the set of all unk.
Proof. It is evident that (i) is a reformulation of the Egoroff property for u, and (ii) and (iii) are reformulations of the conditions in parts (i) and (ii) of Theorem 67.5. Hence, the conditions in parts (i), (ii) and (iii) of the present theorem are all equivalent to the condition that u has the Egoroff property. Also, it is evident that (iii) implies (iv), so if u has the Egoroff property, then (iv) holds. It remains to prove that (iv) implies the Egoroff property for u. This is the difficult part of the proof. Assume, therefore, that (iv) holds. We want to prove that (i) holds. To this end, let u 2 u,,k J k 0 in L. For every fixed k we may assume that u,,k is increasing as n increases (if necessary, replace u,,k by sup (ulk, . . ., u,,,)). Furthermore, we may assume that u # 0 and that there exists a sequence (un: n = 1,2, .) in L+ such that u 2 u, 4 0 and u,, # 0 for all n (because otherwise condition (i) is trivially satistied). Hence, we have
..
>= sup (unk, un) J k u, 1O, so by (iv) there is a sequence u: 10 such that, for every m, we have v: 2 w, for some w, ill the convex hull of all elements sup (u,,k, u,,). Keeping m fixed for the moment, let wm
where 1;k (n, k = 1,2,
=
n, k
A?k
(unk
9
un),
. . .) are non-negative real numbers, zero except for
468
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.
10, $ 6 7
finitely many n and k,and such that En, k A:k = 1. We first consider the summation with respect to k. For this purpose, set
and let k(m,n ) be the largest k for which Ark is nonzero (for m and n fixed). :A >= 0 with :A = 0 except for finitely many n, and EnA; ;= 1. Furthermore, we have
Then
uz 2
wm
2
c A? SUP
( u n , k(m, n)
n
7
un>*
It will be shown now that the sequence (vm = 20: : m = 1,2, . . .) satisfies the condition in (i). Evidently, we have v, 10 in L, S O it remains to show that, for any given pair ( M , N ) of natural numbers, there exists a natural number h(M, N )such that uM
Given M and N , let y
We prove that y we have
= 1.
=
3
sup {
uN, h(M, N )
nhN
n:
*
: nl 2 M } .
It is clear that 0 6 y 6 1. Furthermore, for m 2 M ,
Since u: 3.0 and uN # 0, it follows that y = 1. Hence, there exists a natural number p 2 M such that Af: 2 fi, and so 2v:
2
Now, let
h(M, N ) Then 2v;
=
max ( k ( p , n) :Af: # 0, n 3 N ) .
2
where we have used that unk is increasing in n for fixed k. We thus find that vM
which completes the proof.
= 2v:
2 uN,h ( M , N )
3
CH.10,s 671
THE EGOROFT PROPERTY
469
The above theorem is due to T. Chow (Mrs. T. Dodds; [l], 1969). It was proved in Theorem 65.2 that Dedekind o-completeness of a Riesz space is preserved under a Riesz o-homomorphism. If, in addition, the space has the Egoroff property, then this property is also preserved, as shown by the following theorem.
Theorem 67.8. Let x be a Riesz o-homomorphism from the Dedekind ocomplete Riesz space L onto the Riesz space M . Under these hypotheses, i f f E L has the Egorofproperty, then JO has xf E M. In particular, i f L is Dedekind o-complete and has the Egorofproperty, then M as well has these two properties. Proof. Since Ixf I = .(If I) holds for everyf E L, it will be sufficient to prove that if u E L +has the Egoroff property, then so has nu. Assume, therefore, that u E L + has the Egoroff property, and let
0 S snk f, xu in M for n = 1 , 2 , . . ..
In virtue of Theorem 65.2 there exists, for every n, a sequence (u,, :k = 1, 2 , . . .) in L and an element ui in L such that with nu,, = s, for all (n,k ) and nu; = xu for all n. Hence, taking if necessary unk+(u- u;) instead of u,,, we may just as well assume that 0 S u,, f k u holds for all n. But then, since u has the Egoroff property, there exists a sequence 0 S v, t u such that, for every n, we have v, 6 u,,~(,,)for an appropriate k = k(n). It follows that 0 5 xu, t xu and nun
5 xUn,R(n) = Sn,k(n)
for every n*
This shows that xu has the Egoroff property. We recall that in Theorem 21.2 (iv) it was proved that any finite intersection of super order dense ideals in an arbitrary Riesz space L is super order dense in L. We shall prove now, as a final theorem in the present section, that in a space with the Egoroff property any countable intersection of super order dense ideals is super order dense.
..
Theorem 67.9. If (A, :n = 1,2, .) is a sequence of super order dense ideals in the Riesz space L possessing the Egorof property, then the ideal A , is super order deme.
0:
Proof. Given u E L + ,there exists for every natural number n a sequence
470 (Unk
THE EGOROFP PROPERTY AND THB DIAGONAL PROPERTY
: k = 1 , 2,
. . .) in L such that
0
=< Unk f k U
and
U,k E A,
for k = 1 , 2,
[CH.10,s 68
. ...
Hence, since u has the Egoroff property, there exists a sequence 0 5 v,,, t u such that urn<< (Unk) for every m. It follows that every v,,, is a member of every A,, so urnE A, for every m. Since 0 5 v,,, t u and u EL' was arbitrary, this shows that 07 A, is super order dense.
0;
68. The diagonal property
We begin by recalling the definition of the Egoroff property. The element f in the Riesz space L has the Egoroff property if, given the double sequence (unk :n, k = 1,2, . . .) satisfying 0 5 unk f k If I for n = 1,2, . . ., there exists a sequence 0 5 v, t If I such that, for any pair (m, n), we have urn 5 u,, k(m, ), for an appropriate k(m, n). We might now generalize the definition by dropping the condition that all the U,k and all the V, are between the null element and If I. In other words, we might say that f has a kind of strong Egoroff property if, given that fnk f k f holds for n = 1,2, . . ., there exists a sequence gm fsuch that, for any pair (m, n), we have g,,, 5 f,,k(m, ), for an appropriate k = k(m, n). Note that it is f itself, and not If 1, that occurs in the definition. It is immediately evident that if one particular elementf of L has this strong Egoroff property, then so has any other element of L. Hence, the strong Egoroff property is a property of the space L as a whole. Rewording the definition a little, we obtain the following.
Definition 68.1. The Riesz space L is said to have the strong Egorofproperty if, given any double sequence (U,k : n, k = 1,2, . . .) in L+ such that u,k l k 0 for n = 1,2, . . ., there exists in'L a sequence v, 0 with the property that, for every pair (m, n ) of natural numbers, we have v,,, 2 u,, k(m, ), for an appropriate k = k(m, n). Analogous to Theorem 67.7 for the Egoroff property, the following theorem holds for the strong Egoroff property.
Theorem 68.2. The following conditions are equivalent. (i) The Riesz space L has the strong Egoroflproperty. (ii) I f U , k J k 0 in L, then there is a sequence v, 10 in L with the property that for every n there exists a natural number k(n) such that v, 2 u,, k(n) holds. (iii) If unk u, 4 0 in L, then there is a sequence v, J 0 in L with the property thatfor every n there exists a natural number k(n) such that v, 2 u,, k(n) holdr. (iv) If U,k 3.k u, J 0 in L, then there is a sequence v, J. 0 in L such that, for
CH.
10,s681
47 1
THE DIAGONAL PROPERTY
every m , we have v, 2 w, for some w, in ((u,)), convex hull of the set of all Unk.
where
((Unk))
denotes the
Proof. Similar to the proof of Theorem 67.7. Evidently, any space possessing the strong Egoroffproperty has the Egoroff property. The converse is not true, as will be shown in Example 68.6, where we will see that the sequence space I, does not have the strong Egoroff property, although 1, has the Egoroff property. However, if the Riesz space L has the Egoroff property as well as the property that every double sequence, decreasing to zero, is eventually dominated by some element of L+,then L has the strong Egoroff property. To be precise, we have the following definition. Delinition 68.3. The Riesz space L is said to have the d-property whenever, given the double sequence (unk :n, k = 1,2, . . .) in L+ such that Unk J k 0 holdsfor n = 1 , 2, . . ., there is an element w E L +with the property that,for every n, we have w 2 u,,,k(,,) for an appropriate k = k(n).
The d-property is the same as the notion of compIete regularity introduced by H . Nakano (cf. [6],§ 17). Theorem 68.4. The Riesz space L has the strong Egorof property i f and onIy i f L has the Egorofproperty and the d-property.
Proof. Let L have the Egoroff property and the d-property, and let 0 for n = 1,2, . . .. By the d-property of L there is an element w E L+ such that, for every n, we have w 2 u , , ~ ( , ,for ) an appropriate k = k(n). We set wni = u , , ~ ( , , ) + ~ for n, i = 1 , 2 , . . .. unk
Then w 2 wni Ji 0 for n = 1,2,. . ., and so, by the Egoroff property of L, there is a sequence v, J- 0 in L such that, for every pair of natural numbers (m, n), we have u, 2 w,,, ), for an appropriate i = i(m, n). Then
2 w n , i(m, n )
= %, k(n)+i(m, n )
for all m and n, which shows that L has the strong Egoroff property. Conversely, it is evident that if L has the strong Egoroff property, then L has the Egoroff property as well as the d-property. In an Archimedean Riesz space the d-property alone is already equivalent to the strong Egoroff property. Theorem 40.5. The Archimedean Riesz space L has the strong Egorof property i f and onIy i f L has the d-property.
472
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.10,s 68
Proof. Assume that L is Archimedean and has the d-property, and let u,,k i k 0 in L for n = 1,2, ..Then nu,, i k 0 for n = 1,2, . ., and so by the d-property of L there is an element w EL' such that, for every n, we have w 2 nu,,,k(,) for an appropriate k = k(n). Let q, = n-'w for n = 1,2, .. Then v,, 10 since L is Archimedean, and v,, >= u,,,k(n) for n = 1,2, This shows that L has the strong Egoroff property.
..
.
. ...
..
We will have a second look now at the spaces in Example 67.6. Example 68.6. (i) The Riesz space of all real numbers obviously has the strong Egoroff property. (ii) The same holds for the lexicographically ordered plane. (iii) There is no function, except the function identically zero, in the space C([O,11) of all real continuous functions on [0, 13 that has the Egoroff property, and so surely C([O,11) does not possess the strong Egoroff property. (iv) It was proved in Example 67.6 (iv) that the space L ( X ) of all real functions on a finite or countable point set X , with pointwise ordering, has the Egoroff property. We prove now that L ( X ) has the strong Egoroff property. For this purpose, let (xi,x2 ,. . .) be the points of X , and let U,,k 0 in L ( X ) for n = 1,2, .... Since we wish to prove that L ( X ) has the d-property, it is no restriction of the generality to assume that, for fixed k, the elements U,,k are increasing as n increases (as on earlier occasions, we may replace u,,k by sup ( U l k , . . .,u,,k),if necessary). Now define the function w on X by for n = 1 , 2 , . . ..
w(xJ = u,,,(x,)+l
Then w E L+(X), and w dominates the function u,,, perhaps not at all points E X , but at least at the points x p f o r p 2 n. Indeed, forp 2 n we have
x
unn(xp)
6
6 upl(xp) < w(xp>*
Upn(xp)
Hence, if we determine the natural number k(n) such that k(n) > n and u,,,k(,,)(x)< 1 at the points x = xl, . ., x,,-,, we will have
.
Un,k(,,)(X)
< W(X)
for all x
E
X.
I
This shows that L ( X ) has the d-property. Since L ( X ) is Archimedean, it follows that L ( X ) has the strong Egoroff property. It follows in particular that L ( X ) has the strong Egoroff property if Xconsists of a finite number of points, i.e., if L ( X ) is the space R", with coordinatewise ordering, for some natural number n. In other words, any Archimedean Riesz space of finite dimension has the strong Egoroff property.
CH. 10,s 681
473
THE DIAGONAL PROPERTY
From the definition of the Egoroff property it is immediately evident that if the Riesz space L has the Egoroff property, then so bas any ideal in L. The same is not true, however, for the strong Egoroff property. In order to illustrate this, we prove now that the sequence space 1, does not have the strong Egoroff property. For n, k = 1,2, . . ., let U,,k be the sequence in 1, having its first k coordinates equal to zero and all other coordinates equal to n. Then u,,k l k 0 for n = 1,2, . ., but there is no element w in the positive cone of 1, such that, for all n, we have w 2 u,,,k(,,) for appropriate k = k(n),simply because the sequence (u,,, k(,,) :n = 1,2, . . .) fails to be bounded no matter how k(n) is chosen. It will be proved now that the strong Egoroff property is related to certain properties of order convergence. We recall that by the notationf, +f we mean that the sequence (f.: n = 1,2, . . .) in L converges in order tof, i.e., there exists a sequence p,, 10 in L such that If-f,l 6 p,, holds for all n. Theorem 68.7. The following conditions are equivalent. (i) The Riesz space L has the strong Egoroflproperty. (ii) 1 f f . k +f as k + 00 holds in L for n = 1,2, . . ., then there exists a natural number k ( n )for euery n such thatf., k(,,) +f holds. (iii) 1 f f . k + k f n + f i n L,then there is a natural number k(n)for euery n such thatf,, k(,,) +f holds. (iv) 1 f f . k * k f , + f i n L, then there is a sequence (gm: m = 1,2, . . .) in the convex hull such that gm +f holh. Proof. (i) => (ii). Letf., --+ f as k + co for n = 1,2, . . .. By the definition of order convergence there exists for every n a sequence u,,k l k 0 in L such that 1 f . k - f I 5 u,,k holds for all n and k. Since L has the strong Egoroff property, there is a sequence v,, 4 0 such that, for every n, we have u,,,k ( n ) 5 u,, for some appropriate k(n). But then
.
Ifn,k(n)-fI
so f.,k(n) *f.
5
%,k(n)
5 un 1O,
(ii) => (iii). Let&, -tkfn + f i n L.By the definition of order convergence there exists for every n a sequence u,,k 3-k 0 in L such that 1f.k-f.l 5 Unk holds for all n and k. Furthermore, there exists a sequence u,, 10 in L such that I f . - f l 5 u,, holds for all n. For n fixed, it is then certainly true that U~ + 0 as k * co, so by hypothesis (ii) we have u,,k(,,) + 0 for appropriate k(n). Let the sequence (u,, :n = 1 , 2, . . .) in L+ satisfy U,,,k(n) 5 u,, 3.0.Then w,, = u,,+u,, satisfies w, 10,and we have k(n)
* f.
,fI
k(n)
-fl
6
u n , k(n)
+un 5 On +
un
=
wn
1
474
-
THE EGOROFF PROPERTY A N D THE DIAGONAL PROPERTY
[ C H . 10,s 68
(iii) => (iv). Evident. (iv) (i). Given that (iv) holds, we have to show that L has the strong Egoroff property. For this purpose, by Theorem 68.2, it is sufficient to show that if u,,k l k u,, 10 holds in L, then there is a sequence u, 10 in L such that, for every m y we have v, 2 w, for some w, in the convex hull ((u,,k)). Hence, assume that U,k l k u,, 10 holds in L. Then u,,k + k u, + 0 holds in L and so, by hypothesis, there is a sequence (w, :m = 1,2, . . .) in ( ( u , , k ) ) such that w, + 0. But then there is a sequence v, 10 such that v, 2 w, holds for all m. This is the desired result. We recall that, according to the definition in section 16, the Riesz space L is said to have the diagonal property (for order convergence) whenever, given that fnk +kfn
+f
holds in L, there is a natural number k(n) for every n such thatf,,,(,,, +f holds in L. Also, L is said to have the diagonal gap property (for order convergence) if, under the same hypotheses that fnk + k f . for every n and & +f, there exists a sequence&,, k ( n i ) witb n, < n2 < . . . such that fnr,k(nr)
+f
(as i
+
~0).
Observe now, firstly, that condition (iii) in the last theorem is exactly the statement that L has the diagonal property and, secondly, that in a space with the diagonal gap property the condition (iv) in the last theorem is evidently satisfied (and hence, since (iii) and (iv) are equivalent, the space has the diagonal property). This shows that the strong Egoroff property, the diagonal property and the diagonal gap property are mutually equivalent properties. We state this formally as a theorem. Theorem 68.8. Thefollowing conditionsfor the RieJz space L are equivalent. (i) L has the strong Egorofproperty. (ii) L has the diagonalproperty for order convergence. (iii) L has the diagonal gap property for order convergence. Adhering to the notations of section 16, the order closure of any subset S of L (i.e., the closure of S in the order topology) will be denoted by cl (S), and the pseudo order closure of S by S’. The set S’, therefore, consists of all f E L such that there exists a sequence (f.:n = 1,2, .) in S satisfying f. -.f. It was proved in Theorem 16.6 that cl (S) = S’ holds for the subset S of L if and only if S’ = (S’)’, and in Theorem 16.7 that cl (5‘) = S’ holds for every subset S of L if and only if L has the diagonal gap property. We now generalize this as follows.
..
CH.10, 8691
THE EGOROFF PROPERTY IN ORDER BOUNDED SETS
475
Theorem 68.9. Thefollowing conditionsfor the Riesz space L are equivalent. (i) L has the diagonalproperty. (ii) cl ( S ) = S' holds for every subset S of L. (iii) cl ( S ) = S' holds for every convex subset S of L . Proof. It will be sufficient to show that (iii) implies (i). Hence, let (iii) be satisfied, i.e., let (S')' = S' hold for every convex subset S cf L . Since the diagonal property is equivalent to the strong Egoroff property, it is sufficient to show that, if fnk + k f . +f holds in L , then there is a sequence (gm: m = 1,2, . . ) in ((f.,)) such that gm +$ Given that f., +,f. + J we write S = ((&)). Then E S' ', so f E S' by hypothesis. It follows that there is a sequence (gm:m = 1,2, . . .) in ((f",))such that g,,, -$ This is the desired result.
f 0
The results so far in this section are due to T. Chow (Mrs. T. Dodds; [l], 1969). We conclude the section by the following theorem, analogous to Theorem 67.8. Theorem 68.10. Let n be a Riesz o-homomorphismfrom the Dedekind ocomplete Riesz space L onto the Riesz space M. Under this hypothesis, i f L has the diagonal property, then so has M .
Proof. One possible way of proving the theorem is as follows. The space L has the diagonal property, so by the preceding theorem the pseudo order closure of every set in L is order closed. By Corollary 65.10 the same holds then in M , i.e., the pseudo order closure of every set in M is order closed. Once again by the preceding theorem, it follows that A4 has the diagonal property. 69. The Egoroff property and the diagonal property in order bounded sets Given the elementsf l and fiin the Riesz space L such that f l shall say that the set
5 f 2 holds, we
(f: f 1 S f S f , )
is an order interva2 in L. The interval is sometimes denoted by L f i , f i ] ; compare Exercise 15.12. For any u EL' the interval [-u, u] is called a symmetric order interval in L. The interval V;,f 2 ] is contained in the symmetric interval [-u, u ] for u = sup (I f i l , If,!). The subset S of L is called order bounded whenever S is contained in an order interval. In other words, S is order bounded whenever there exists an element u E L+ such that If I u holds for all f E S.
476
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. lo,§ 69
We shall prove now that, in a certain sense, the Egoroff property is the strong Egoroff property restricted to order bounded subsets, i.e., the Egoroff property is the diagonal property restricted to order bounded subsets. Theorem 69.1. The following conditions are equivalent. (i) The Riesz space L has the Egorofproperty. (ii) Iff,,’ +f as k + co holds in L for n = 1,2, . . . and the set S = (fd : n, k = 1,2, . . .) is order bounded, then there exists a natural number k ( n )for every n such thatf., k(,,) +f holds.. (iii) Iff.k + k f . + f i n L and the set S = (hc:n, k = 1,2, . . .) is order bounded, then there exists a natural number k ( n )for every n such thatf,, k ( n ) +f v holds. (iv) I f h k *kf. + f i n L and the set S = (fnk :n, k = I , 2, . . .) is order bounded, then there is a sequence (g,,, : m = 1,2, . . .) in the convex hull (S) = ((fnk)) such that g,,, *f holds.
...,
Proof. (i) =. (ii). Letf., *f as k + co for n = 1,2, and let the set S = (fnk :n, k = 1,2, .) be order bounded, i.e., there exists an element wo E L+ such that If.kl 6 wo holds for allf,, By the definition of order convergence there exists for every n a sequence u,,k i k 0 in L such that 1 f . k - f I - unkholds for all n and k. Then S
..
.
f.k-%k 2.f
for all n and k , so
sfnk+Unk
5.f 6 w O + u n k , which implies (n fixed, k + co) that I f I 6 wo. Hence If.k-fI -wO-unk
5 2w0 for all n and k, and this shows that without loss of generality we may assume that 0 5 u, 5 2w, holds for all n and k. But then, since 2w0 2 u,,k 0 for all n and since L has the Egoroff property, there is a sequence vn 10 such that, for every n, we have Un,k(,,) 5 v,, for some appropriate k(n). It follows that
16..
k(n)-fI
6
%, k(n)
5 vn 1 OY
so f.,k(,,) +f. Observe that the proof is similar to the proof for the corresponding statement about the strong Egoroff property in Theorem 68.7. (ii) * (iii) =. (iv) (i). Similarly as in Theorem 68.7.
As a consequence, we have now the following theorem. Theorem 69.2. Thefollowing conditionsfor the Riesz space L are equivalent. (i) L has the Egoroffproperty. (ii) L has the diagonal property (for order convergence) in order bounded sets.
CH.
10,s 701
477
REGULAR RIESZ SPACES
(iii) L has the diagonal gap property (for order convergence) in order bounded sets. (iv) cl ( S ) = S' holdsfor every order bounded subset S of L. (v) cl (S)= S' holdsfor every order bounded convex subset S of L. Proof. Similarly as in Theorems 68.8 and 68.9. 70. Stability of order convergence and regular Riesz spaces
We recall the definition of relatively uniform convergence as presented in section 16. Given u E L + ,we say thatf, converges u-uniformly to f whenever for every E > 0 there exists a natural number n, such that If-f.1 5 EU holds for all n 2 n,, and it is said thatf, converges relatively uniformly to f if f,, converges u-uniformly to f for some u EL'. This is denoted byf, +f (ru). In a non-Archimedean space the limit of a relatively uniformly convergent sequence need not be uniquely determined, but in an Archimedean space relatively uniform convergence off, to f implies order convergence off, tof, and hence the limit f of the sequencef, is uniquely determined (cf. Theorem 16.2). Relatively uniform convergence is stable, i.e., it has the property that for any sequence f n + O(ru) there exists a sequence of real numbers (A,, : n = 1,2,. .) such that 0 A,, T co and &f, + O(ru). Indeed, given that f.-+ O(ru), there exists a sequence of positive real numbers (E,, :n = 1,2, . . .) and an element u E Lf such that E,, 10 and If,l 6 E,,U for all n, and so A,, = E;* satisfies the above mentioned conditions. Order convergence is not necessarily stable. By way of example, letf, (for n = 1,2, .) be the element of the space I , with the first n coordinates zero and all other coordinates equal to 1. Thenf, J. 0, but for any sequence of real numbers A,, satisfying 0 I;A,, f co it is impossible that A,,& + 0, simply because A n f , is not bounded from above. It is, therefore, of interest to have a condition under which order convergence is stable. For an Archimedean Riesz space such a condition is given in Theorem 16.3, where it is stated that in an Archimedean Riesz space order convergence is stable if and only if order convergence and relatively uniform convergence are equivalent. In an Archimedean Riesz space stability of the order convergence is also related to the diagonal property or, equivalently, to the strong Egoroff property. It was shown in Theorem 16.9 that the diagonal property implies stability of the order convergence; the converse is not true, as shown in Exercise 16.17 (where we considered the space of all real sequences with only finitely many nonzero coordinates, and where it was shown that in this space order convergence and relatively uniform convergence are equivalent, but
.
..
478
THE EGOROFP PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s 70
the space does not have the diagonal property). In order to explain the exact relation between stability of order convergence and the diagonal property we introduce the following property, called the a-property. Definition 70.1. The Riesz space L is said to have the a-property whenever, given the sequence (u,, : n = 1,2, .) in L+,there exiJts a sequence (A,, : n = 1,2, . .) of strictly positive numbers such that the sequence (A,,u,,: n = 1,2, . .) is boundedfrom above. The main theorem follows now (compare Exercise 16.14).
..
. .
Theorem 70.2. In an Archimedean Riesz space L the following properties are equivalent. (i) L has the diagonal property for order convergence. (ii) L has the d-property, as defined in DeJnition 68.3. (iii) L has the a-property and order convergence in L is stable. Proof. The equivalence of (i) and (ii) was proved in Theorem 68.5. For the proof that (i) and (iii) are equivalent, assume first that (i) holds, i.e., L has the diagonal property. Then order convergence in L is stable by Theorem 16.9, so it remains to prove that L has the a-property. For this purpose, given the sequence (u,, : n = l , 2, . . .) in L+,we set fnk = k-'u,, for n, k = 1,2, . .. Then fnk J 0 as k -00 for every n (note that L is Archimedean), and sof,, k(,,) + 0 for appropriate k(n) in view of the diagonal propcrty. Hence, setting A,, = { k ( n ) } - l , we have R,,u,, + 0, which implies that the sequence (A,,u,, :n = 1 , 2 , . . .) is bounded. This shows that L has the a-property. Conversely, assuming that L has the a-property and order convergence in L is stable, we have to prove that iff,, + 0 as k + 00 for n = 1,2, ., thenf,, k(,,) + 0 for appropriate k(n). Since order convergence and relatively uniform convergence are equivalent by the stability of order convergence, there exists for each n = 1,2, . an element u,, E L + such that fd + 0 as k + 00 holds u,,-uniformly. By the a-property there exist numbers A,, > 0 such that (A,,u,, :n = 1,2, .) is bounded, say A,,u,, 5 u for all n. It follows that f n k + 0 as k + 00 holds u-uniformly for every n, so there exists for every n a natural number k(n) such that
.
..
..
..
This shows that f.,
k(n) + 0,
I.fn,k(n)I 5 n-luand completes the proof of the theorem.
An Archimedean Riesz space possessing the diagonal property (or, equivalently, possessing the strong Egoroff property) is sometimes called a
CH.lo,§ 701
REGULAR RIESZ SPACES
479
regular Riesz space. Historically, regular Riesz spaces were introduced by L. V. Kantorovitch ([2], 1936) as Dedekind complete and order separable spaces possessing the a-property and for which order convergence is stable. Similar definitions are to be found in a paper by M. Orihara ([l], 1942) and a paper by T. Ogasawara ([2], 1949). It was proved in Example 68.6 (iv) that the Riesz space of all real sequences (with the usual coordinatewise ordening) is regular, but the linear subspace I , of all bounded sequences fails to be regular. From the viewpoint of our last theorem this is due to the fact that, although ,Z is Archimedean and has the a-property, the order convergencein I , is not stable. Taking the subspace of I , consisting of all sequences with only a finite number of nonzero coordinates, it is easy to see that this subspace is Archimedean and order convergence in the subspace is stable, but the space fails to have the a-property. Some important examples of regular Riesz spaces will be discussed in the next section. In an Archimedean Riesz space with a strong unit stability of the order convergence is a severe restriction upon the space. Precisely, the following theorem holds.
Theorem 70.3. Let the Archimedean Riesz space L have a strong unit, and let L be either Dedekind a-complete or have the projection property. Then order convergence in L is stable if and only i f L is offinite dimension. Proof. If L is Archimedean and of finite dimension, then L is Riesz isomorphic to R" for some natural number n, and so it is evident that L has a strong unit and order convergence is stable. Conversely, assume that the Archimedean space L has a strong unit e, and is either Dedekind a-complete or has the projection property. Assume also that order convergence in L is stable and that in L there exists an infinite system of mutually disjoint nonzero elements. Let (f.:n = 1,2, . . .) be a countably infinite subsystem, and denote by B, the band generated by&. It follows from our hypotheses that L has the principal projection property; for n = 1,2, . . .,letp, be the component of the strong unit e in the principal band B,,, and let s, = p 1+ . . . + p n . Evidently the sequence (s, :n = 1,2, . . .) is increasing, and we will prove now that p = sups, exists. If L is Dedekind a-complete, this is evident. If L has the projection property, and if B is the band generated by ( p l ,p 2 , . . .), then the component p of e in B satisfiesp = sup s,,, i.e., p -s, 5.0.Indeed, assume that 0 6 v 6 p - s , holds for all n. Then, sincep-s, has a zero component in the bands B,, . . .,B,, the same holds for v. It follows that v is disjoint from all bands B,, and so v
480
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s 70
is disjoint from B. On the other hand we have v E B on account of 0 5 v 5 p . Hence v = 0, i.e., p - s , 4 0. In any case, therefore, s, converges in order top. Observing now that order convergence and relatively uniform convergence are equivalent on account of the stability of order convergence, we obtain the result that s, converges relatively uniformly to p , which implies immediately that s,, converges e-uniformly to p (since e is a strong unit, so p -s, S EU for all n 2 no and some u EL' implies p - s , 5 Ece for some constant c > 0 and all n 2 no). Hence, given E such that 0 < E < 1, there exists a natural number N such that p - s , S Ee. Taking components in B, for any n > N , we find pn 6 EP,,, which is impossible on account of p , # 0. We have derived, therefore, a contradiction. It follows that every system of mutually disjoint nonzero elements in L is finite. Hence, by Theorem 26.10, the space L is of finite dimension. We will weaken now the hypothesis that L has a strong unit, and assume only that every principal band in L has a strong unit; we will prove that regularity of L implies now that L is of finite dimension. Theorem 70.4. Let the Riesz space L be either Dedekind c-complete or have the projection property, and let every principal band in L have a strong unit. Then L is regular ifand only i f L is ofjinite dimension. Proof. We need only prove that regularity implies finite dimensionalityObserve first that, by the preceding theorem, every principal band in L is of finite dimension. Furthermore, by Theorem 26.10, it will be sufficient to prove that any system of mutually disjoint nonzero elements is finite. Assume, therefore, that there exists a countably infinite system ( f , :n = 1,2, . . .) of mutually disjoint nonzero elements. By the regularity of L there exists a corresponding sequence (A,, : n = 1,2, . . .) of positive numbers such that (A,,If,l : n = 1,2, . . .) is bounded; say Anlf,l S u EL' for all n. It follows that all elementsf, are included in the principal band of finite dimension generated by u. On the other hand, since allf, are mutually disjoint, the system (f,: n = 1,2,. . .) is a linearly independent system. This yields a contradiction. Hence, L must be of finite dimension. The proofs of Theorems 70.3 and 70.4 are taken from a paper by A. C. Zaanen ([2], 1968). Theorem 70.3, under the extra hypothesis that L is Dedekind complete, was known earlier (cf., for example, Theorem VI.4.2 in the book by B. Z. Vulikh [2]). The proof was based on the Yosida representation of L as the Riesz space of all real continuous functions on a compact topological space. The present simple proof avoids this representation theorem.
CH. 10,8 711
48 1
FUNCTION SPACES POSSESS:NG THE DIAGONAL PROPERTY
71. Function spaces possessing the diagonal property In Example 11.2 (v) we have introduced a measure space ( X , A , p) and the corresponding Riesz space M(')(X, p) of all real p-almost everywhere finitevalued p-measurable functions on the point set X. Functions that are palmost equal are identified. In Example 23.3 (iv) it was proved that for p a (totally) o-finite measure in X , the space M("(X, p) is super Dedekind complete. In the present section the same space will be investigated further, and we will prove that, once again for a o-finite measure p,the space M(')(X,p) is regular. In other words, for a o-finite measure p the space M(')(X, p) has the diagonal property, and hence M(')(X,p) has surely the Egoroff property. In particular, for Xcountable and each point of X of unit measure we find again the result that the space of all real sequences has the diagonal property or, equivalently, this space has the strong Egoroff property (cf. Example 68.6 (iv)). In the following, let p be a o-finite measure in X. All spaces L,,(X, p), for 0 < p 5 00, are ideals in M(')(X,p), and so all these spaces have the Egoroff property. It is not true, however, that all these spaces are regular, because it is easy to see that, unless the measure space ( X , A , p ) is of a trivial nature, the space L,(X, p) fails to be regular because order convergence in L,(X, p) is not stable. We shall prove, however, that for 1 5 p < KI the space L J X , p) is regular. More generally, let Lp(X,p) be a normed Kothe space as introduced in section 9, having the following properties: (i) Lp is norm complete, i.e., the function norm p has the Riesz-Fischer property, stating that for any sequence (fn :n = 1,2, . . .) in Lp such that CPP P ( f n ) < we have P(C? 1f.l) < a; (ii) the function norm p has the property that for any sequence (f,: n = 1,2,. . .) in Lp such that fn(x) 4 0 holds for p-almost every x E X we have P ( L ) 5.0. We shall prove that any normed Kothe space LJX, p) with these properties is regular; the regularity of L,(X, p) for 1 5 p < 00 will follow as a particular case. In general, L, does not satisfy condition (ii), in accordance with the fact that, in general, L , is not regular. Let ( X , A , p) be a measure space and, unless otherwise mentioned, let the measure p be (totally) o-finite. We first prove two simple lemmas.
.
Lemma 71.1. I f f . E M(')(X,p) for n = 1, 2, . . and f,,(x) 0 hold&for p-almost every x E X , then p ( x ) = sup,, I f,,(x)I is a p-almost everywhere finitevalued p-measurable function, i.e., p E M("(X, p). --f
482
THE EGOROFP PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s 71
Proof. It is evident that the function p is p-measurable, so we need only prove that p is p-almost everywhere finitevalued. By hypothesis we have f,(x) -+ 0 for p-almost every x E X; let xo be one of these points. Then there exists a natural number no = no(xo) such that If,(xo)l 5 1 for all n > no. It follows that ~(xo= ) SUP n
Ifn(x0)I
6 SUP (1, Ift(xo)I,
* *
-9
I.fno(xO)I),
sop(xo)is finite. This shows thatp(x) is finitevalued for p-almost every x E X.
. ..
Lemma 71.2. Let Kl c K2 c t X with p(K,) < co for all n, and f c r each n let A, c K, with p(Kn-A,) < 2-". Then the set lim inf A, is p-almost equal to X, i.e., p(X-lim inf A,) = 0. Proof. Keep the natural number m fixed temporarily, and let D, = K,-A, for n = 1,2,. . .. Then p(Dn)< 2-" for all n 2 m, and so
n m
p(lim sup 0,) = p(
m
UD n )
k = l n=k
m
S P ( U 0,) < 2 - ( k - l ) , n=k
This holds for all k, so p(lim sup D,) = 0. It follows that the intersection
K,,, n (lim inf A,) is p-almost equal to K,. But this holds for all m, so lim inf A, is p-almost equal to X. We apply Lemma 71.1 in the proof of the following theorem.
.
Theorem 71.3. The sequence (f.:n = 1,2, . .) in M") ( X , p) converges in order to f E M(')(X, p) if and only i f f , ( x ) tends to f ( x ) for p-almost every x E X. Proof. For brevity, denote the Riesz space M(')(X, p) by L.We assume first thatf, converges in order tof, so
If-f.1 6 un 4 0 for appropriate u, E L + .We have u l ( x ) 2 uz(x) 2 x E X, and so it follows from u, 5.0in L+ that inf u,(x) = lim u,(x) = 0 holds for p-almost every x E X. It follows that If(x)-f,(x)I
6 un(x) 4 0
. . . for p-almost every
CH. 10,s 711
483
FUNCTION SPACES POSSESSINGTHE DIAGONAL PROPERTY
holds for p-almost every x E X, so L ( x ) tends to f ( x ) for p-almost every x E X. For the converse, we assume that f E L andf, E L for n = 1,2, and L ( x ) tends tof ( x ) for p-almost every x E X , i.e.,L -fconverges to zero pointwise p-almost everywhere on X . For n = 1,2, . ., we set
. ..,
.
Pn(x) = sup ( I f k ( x ) - f ( x ) l
:
2
By Lemma 71.1 we have p,, EL, and it is evident that p,,(x) 10 holds for p-almost every x E X . But then
If-Ll 5 P n l o holds in L, i.e.,f. tends to f in order. Theorem 71.4. Order convergence in M("(X, p) is stable. Proof. We may assume that we have a sequence u,, 10 in M("(X, p), and we have to prove the existence of a sequence (A,, : n = 1,2,. . .) of positive numbers satisfying A,, f oc) such that A,,u,, converges in order to zero. For this purpose, take sets K , c Kz c . . . f X such that all Kk (k = 1,2,. . .) are of finite measure. For each k , let A , be a subset of Kk such that p(Kk-A,) < 2-, and such that (u,, : n = 1,2, . . .) converges uniformly to zero on A,. This is possible by the classical Egoroff's theorem. Let the natural number n, satisfy u,,(x) k-' for all x € A k and all n 2 n,. w e may assume that n, < nz < . . .. Now define the positive natural numbers A,, (n = 1,2, . . .) by A,, = 1 for n < n,, and A,, = k for n k 5 n < n k + l ( k = 1,2,. . .). Then 0 < A,, t co, and for nk S n < nk+, and x € A k we have A,,u,,(x) 5 ku,,,(x) 5 k-'.
For any point x ~ l i m i n A,, f there exists a natural number k ( x ) such that x E A, holds for all k 2 k ( x ) . Hence, if k 2 k(x) and n 2 n k , then A,u,,(x) S k-' holds for this particular point x. This shows that the sequence (A,, u,,(x) : n = 1,2, . . .) converges pointwise to zero for every x E lim inf A,,, i.e., A,,u,,(x)converges pointwise to zero p-almost everywhere on X (note that lim inf A , is p-almost equal to X by Lemma 71.2). It follows by means of the preceding theorem that A,u, converges in order to zero. Theorem 71.5. The space M(')(X, p) has the a-property. Proof. We denote M(')(X, p) by L for brevity. Let u,, E L +for n = 1,2, .. We have to prove the existence of positive numbers A,, (n = 1,2, . .) such that the system (A,, u,, : n = 1,2, .) is bounded inL+. For this purpose,
..
..
.
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THE EGOROFP PROPERTY A N D THE DIAGONAL PROPERTY
[CH.10,s 71
let Kl c K, c . . . T' X with all sets K, of finite measure. For every n, there exists a set A , c K, and a number A, > 0 such that p(K,-A,) < 2-" and such that A,,u,(x) S n-' holds for all x E A,. For any x E lim inf A , there exists a natural number n(x) such that x E A , holds for all n 2 n(x), so A,u,(x) 6 n-' holds for all n 2 iz(x). This shows that A,u,(x) converges to zero for every x E lim inf A,, i.e., for p-almost every point of X . It follows by Lemma 71.1 that P(X) = SUP n
IAnUn(X>I
is p-almost everywhere on X finitevalued, and so p is an upper bound in L of the system (A,u, : n = 1,2, . . .). This is the desired result. As a direct consequence of the theorem that an Archimedean Riesz space L has the diagonal property if and only if order convergence in L is stable and L has the o-property, we obtain the following theorem. Theorem 71.6. The space M(')(X, p) has the diagonalproperty (or, equivalently, M(')(X, p) has the strong Egorofproperty). In other words, if the double sequence of functionsf,, in M(')(X, p ) is such that for every n the sequence ( f n k : k = 1,2, .) converges pointwise p-almost everywhere on X to a function f,E M(')(X, p) and if the sequence (f,: n = 1,2, . . .) converges pointwise p-almost everywhere on X to a function foE M(')(X, p ) , then there exists a "diagonal sequence"&, k ( n ) converging pointwise p-almost everywhere on X to fo.
..
Any ideal A in M(')(X, p) is a Riesz space in its own right. It follows immediately that, for f E A andf, E A for n = 1,2, . . .,we havef, +fin order if and only if p ( x ) = SUP, If,(x)I is an element of A andf, converges pointwise p-almost everywhere on X to$ This holds in particular if A is a normed Kothe space L,(X, p), where p is a function norm as introduced in section 9. It was stated in section 9 that Lpis norm complete (Le., L, is a Banach space) p ( f , ) < co if and only if the norm p has the Riesz-Fischer property (i.e., implies P(C? ILl) < a). Theorem 71.7. If L, = Lp(X,p ) is norm complete, then L, has the oproperty. Proof. Given the sequence (u, : n = 1,2, . . .) in Lp', we determine the sequence (A,,: n = 1,2, . .) of positive numbers such that p(l,u,) < 2-" for n = 1,2, . . .. Then p(A,,u,) < 00, so p ( C 7 A,u,) < co by the RieszFischer property, i.e., the function uo = A,u, is a member of L,,. This
.
CH.
10,s 71 ]
FUNCTION SPACES POSSESSING THE DIAGONAL PROPERTY
485
shows that uo is an upper bound of the sequence (A,,u,, : n = 1,2, . . .), and so Lp has the a-property.
Theorem 71.8. If L,, = L,,(X, p) is norm complete, and p has the property that p ( f , ) J. 0 for any sequence (A : n = 1,2, . .) in Lp satisfying f , ( x ) J. 0 for p-almost every x E X , then Lp has the diagonal property.
.
Proof. It is sufficient to prove that order convergence in Lp is stable. For this purpose we may assume that we have a sequence u,, 10 in L p , and we have to prove the existence of a sequence (A,, : n = 1,2, . . .) of positive numbers such that A,, co and A,,#,,converges in order to zero. Since u,, 10 in L,, implies u,,(x) J. 0 for p-almost every x E X,we have p(u,,) 10.Let (vk = unk: k = 1 , 2 , . .) be a subsequence such that n , < n, < . . . and p(vk) < k-’ 2-, for all k. Then p(kvk) < 2-,, so w = kv, is a member of L,, by the Riesz-Fischer property. It follows that
-
.
m
W(x>
=
k= 1
kvk(x)
is finite for p-almost every x E X, so kvk(x) converges pointwise to zero p-almost everywhere on X . Evidently, the sequence (kv, :k = 1,2, . . .) is majorized in L,, by w. Now, for n = 1 , 2 , . . ., let
A,, = 1 for ti < n , and A,, = k for n k 5 n < n k + , (k = 1,2,. . .). kv,. It follows Then 0 < A,, co, and for n k 5 n < n k + l we have A,,u,, that (A,,#,,: n = 1,2, . . .) converges pointwise to zero p-almost everywhere
on X , and the sequence is majorized in Lpby w. This shows that A,, u,, converges to zero in order in L p , and so order convergence in Lp is stable.
The result in the last theorem covers the case that L, = L, for 1 Ip < co. As observed earlier, the space L , is not included, because L, does not have the diagonal property. The spaces L, for 0 < p < 1 are not covered by the theorem either, since
is a norm for 1 5 p < co, but not for 0 < p < 1. Nevertheless, the spaces L, (0 < p < 1) have the diagonal property; a proof is indicated in the following exercise.
Exercise 71.9. Under the same assumption about X and p, and for 0 < p c 1, show that Lp(X,p) is a metric space with respect to the distance function
486
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.
10, 5 71
Show also that if we define the function p on the set P of all non-negative p-measurable functions by I-
then p has all the properties of a function norm except the homogeneity. Precisely, for any constant u > 0 we have p(uf) = uPp(f) instead of p(uf) = up(f). Furthermore, show that the thus introduced p has the Fatou property, i.e. it follows from 0 5 f, tf, with all f,E P, that p(f,) t p(f). This implies that for any sequence (f,: n = 1,2, . .) in P such that p(f,) < 00 we have p ( c T f , ) < m. Show now that the space Lp (0 < p < 1) has the diagonal property.
.
ZT
Exercise 71.10. Let Lp(X,p) be a normed Kothe space. It may be asked whether the Riesz-Fischer property for L, is perhaps a consequence from the property that any sequence (f,: n = 1, 2, . . .) in Lp satisfying f.(x) 10 0. By way of example, let X p-almost everywhere on X also satisfies p(f,) be the interval [0, 1 1 and p Lebesgue measure, and let the function norm p be defined on the set P of all non-negative p-measurable functions by p(f) = JA fdp iff is essentially bounded and p(f) = co otherwise. In other words, L, is the space L, equipped with the L,norm. Show that Lp does not have the Riesz-Fischer property, although Lp has the other property referred to above. Exercise 71.11. We investigate another function space. Under the same assumptions about X , p and the set P of all p-measurable non-negative functions, let
for allfe P,and p(.f) = p ( l f 1 ) for every (real) p-measurable functionf. (i) Show that the set Lp = (f:p(f) < m) is a real vector space. (ii) Show that L, is a complete metric space with respect to the distance function d ( f ,g) = p(f-g). (iii) Show that L, is a Riesz space with respect to the partial ordering inherited from M(')(X, p). Show that Lp is an ideal in M("(X, p), and so Lp is super Dedekind complete.
CH. 10,s 721
RELATIVELY UNIFORM CONVERGENCE
487
(iv) Show that ifS, E Lp for n = 1,2, . . . and f n ( x )1 0 holds for p-almost every x E X,then p(f,) 3.0. (v) Show that Lp has the diagonal property. 72. The diagonal property for relatively uniform convergence
It was proved in section 70 that in an Archimedean Riesz space the diagonal property for order convergence is equivalent to the a-property together with stability of order convergence. In the present section we begin by proving that in an Archimedean Riesz space the diagonal property for relatively uniform convergence is equivalent to the a-property alone. Theorem 72.1. Let L be an Archimedean Riesz space. The following conditions are now equivalent. (i) L has the a-property, i.e., given any sequence (u,, : n = 1, 2, . . .) in L f , there is a sequence (A,, : n = 1,2, . . .) of positive numbers and an element u E L ' such that u,, 2 Anu holds for all n. (ii) I f fnk -+ f ( r u ) as k + 00 holds in L for n = 1, 2, . . ., then there exists a natural number k(n) for every ii such that f,,k(,,) + f (ru) holds. (iii) I f f n k +f,(ru) a s k + 00 hoids in L for n = 1,2, . . . a n d i f f , +f ( r u ) , +f ( r u ) then there exists a natural number k ( n ) for every n such that jk,k(n) holds. (iv) +fn(ru)a s k + co holds in L for n = 1,2, . . . and iff,, --f f ( r u ) , such then there is a sequence ( g m: m = 1,2, . . .) in the convex hull ((Ak)) that gm+f ( r u ) holds. ,
(ii). Let f n k +f ( r u ) as k -+ 00 for n = 1,2, . . .. By the Proof. (i) definition of relatively uniform convergence there exists a sequence (u,, : n = 1 , 2, . .) in Lf and a double sequence (&,,k : n, k = 1, 2, . . .) of nonnegative numbers such that &,,k J. 0 as k + co for every n and
.
Ifnk-f I 5 &,,kU,, for all n and k.
Since L has the a-property, there exists a sequence (A,, : n = 1,2,. . .) of positive numbers and an element u EL' such that u,, 5 A,,u holds for all n. Hence, writing 6,,, = &&A,,, we have 6,,k J. 0 as k + co for every n and
If,k-fl
5 8,,ku for all TI and k.
Now, since the space of real numbers has the strong Egoroff property, there is a sequence y,, J. 0 of nonnegative numbers such that, for every n, we have y,, 2 8,,,k(") for an appropriate k(n). Then
488
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
IG , k ( n ) - f l 5 a n , k ( n ) U 5
[CH.
10,s 72
Ynu
with Yn 10, and S O f , , k ( , , ) + f ( r u ) . (ii) + (iii). Let f n k +f,(ru) as k + 00 for n = 1, 2,. . a n d f , + f ( r u ) . Then f,&-f, + O(ru) holds for n = 1, 2, . . ., so by hypothesis there is a natural number k(n) for every n such thatf,, k ( n ) -f, + O(ru).Alsof, +f ( r u ) by hypothesis. Hence, by addition, we obtain that f,,&)(, + f ( r u ) . (iii) * (iv). Evident. (iv) =. (i). Given (iv), we have to show that for any sequence (u,, : n = 1 , 2 , . . .) in L+ there exists a non-negative sequence (A,, :n = 1,2, . . .) and an element w E L+ such that u,, 5 Anw holds for all n. Without loss of generality we may assume that 0 u,, 7. For the purposes of the proof, let u be a nonzero element of 1 5' (if u = 0 for all u E Lf,then L consists only of the zeroelement, and the theorem to be proved is trivially true). WL set
.
wnk= k-'u,,+n-'u w, = n-'u for n
for n, k = 1,2,. . ., = 1,2,. . ..
Then w n k + wn(ru)as k + 00 for n = 1,2, . . ., and w,, + O(ru). Hence, by hypothesis (iv), there is a sequence (urn : m = 1, 2,. . .) in the convex hull ((wnk)) such that u,,, + O(ru). For each fixed nz, since u, is a member of ( ( w , , k ) ) , u,,, is a finite linear combination of elements from (wnk: n, k = 1,2, . . .) with non-negative coefficients the sum of which equals one. Hence, denoting by A: the sum of the coefficientsbelonging to the same n, we have 2: 2 0 for all n, :A = 0 except for finitely many n, In :A = 1, and for an appropriate k(m, n). It follows from urn+ O(ru) that there is an element w E L+ and a sequence of non-negative numbers E, 10 such that urn I E, w holds for all ni. Then
1A:wn,
k(m, n )
n
Em
holds for all m. We will show that w is the element in L + the existence of which we have to prove. Evidently, the set s? of all natural numbers n such that A: # 0 for some rn is non-empty. We prove that s? has infinitely many elements. Indeed, assume d to be finite, and let no be the largest number in d.Then, for every m,we have E,W 2 C A : W , , ~ ( , , ~ ) 2 C A m , n - 1 u 2 C A Y n i ' u = nilu; n
n
n
CH. 10, $' 721
RELATIVELY UNIFORM CONVERGENCE
489
on the other hand, E,W 3.0 as m + co since L is Archimedean. Hence n l ' u S 0, which contradicts u > 0. It remains to show that for any given natural number N there is a positive number AN such that U, 5 I,w. For this purpose, let N ba given. By the above argument that sd is an infinite set, there is a natural number no = no(N) in sd such that no 2 N , so A:o # 0 for some m = m ( N ) . For this particular m we have, therefore, that
so, writing 1,
=
~,,,(A:~)-lk(rn, no), we have ANw L
uno
L
uN,
where it is used in the last inequality that the sequence (un : n = 1 , 2 , . . .) is increasing. This completes the proof. The above proof is due to T. Chow (Mrs. T. Dodds; [I], 1969). Observe that condition (iii) in the last theorem is exactly the condition that L has the diagonal property for relatively uniform convergence. Furthermore, if L has the diagonal gap property for relatively uniform convergence, then condition (iv) in the last theorem is evidently satisfied and hence, since (iii) and (iv) are equivalent, the space has the diagonal property for relatively uniform convergence. This shows that in an Archimedean Riesz space the a-property, the diagonal property for relatively uniform convergence and the diagonal gap property for relatively uniform convergence are equivalent. We state this formally as a theorem (similar to Theorem 68.8 for order convergence).
Theorem 72.2. The following conditions for the Archimedean Riesz space L are equivalent. (i) L has the a-property. (ii) L has the diagonal property f o r relatively uniform convergence. (iii) L has the diagonal gap property f o r relatively uniform convergence. Adhering to the notations of section 16, the relatively uniform closure of any subset S of L (i.e., the closure of S in the relatively uniform topology) will be denoted by 3, and the pseudo uniform closure cf S by S;,,. The set S;,, therefcre, consists of all f E L such that there exists a sequence (f,: I b = 1, 2, . . .) in S satisfyingf, -+ f ( r u ) . It was proved in Theorem 16.8 (ii) that = Si,, holds for the subset S of L if and only if S;, = (S;,,);, and in Theorem 16.8 (iii) it was shown that in an Archimedean space = S;,,
s
490
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.
10, 5 72
holds for every subset S of L if and only if L has the diagonal gap property for relatively uniform convergence. We now generalize this as follows. Theorem 12.3. The following conditions f o r the Archimedean Riesz space L are equivalent. (i) L has the diagonal property f o r relatively uniform convergence or, equivalently, L has the a-property. (ii) = S;,, holds for every subset S of L . (iii) = S;, holdsfor every convex subset S of L . Proof. It is sufficient to show that (iii) implies (i). Hence, let (iii) be satisfied, i.e., let Si,, = (S;,,);,,hold for every convex subset S of L . We have to show that if f n k+f , ( r u ) as k -+ co holds for n = 1,2, . . . and iff, -+ f (ru), then there is a sequence (9,": rn = 1 , 2 , . ..) in ((A,)) such that g,,, +f (ru). Given thatf,, + f , ( r u ) for n = 1,2, . . . andf, -+ f ( r u ) , we write S = ((Ak)). Then f E (S;,,);,,,sof E Si,, by hypothesis. It follows that there is a sequence ( g m : rn = 1, 2, . . .) in ( ( f n k ) )such that g , +f ( r u ) . This is the desired result. We conclude the section by the following theorem, analogous to Theorems 67.8 and 68.10. Theorem 72.4. Let n be a Riesz homomorphism from the Archimedean Riesz space L onto the Archimedean Riesz space M . Under this hypothesis, i f L has the diagonal property f o r relatively uniform convergence, then so has M . Proof. One possible way of proving the theorem is as follows. The space L has the diagonal property for relatively uniform convergence, so by the preceding theorem the pseudo uniform closure of every set in L is uniformly closed. By Corollary 63.6 the same holds then in M , i.e., the pseudo uniform closure of every set in M is uniformly closed. Once again by the preceding theorem, it follGws that M has the diagonal property for relatively uniform convergence. Another proof is obtained by observing that, under the given hypothesis, the o-property is preserved. equiv equiv non-equiv non-equiv non-equiv non-equiv
good bad good good bad bad
good bad good bad good bad
Exercises 16.15 and 16.16 Exercise 16.17 impossible by Th. 16.9 impossible by Th. 16.9 Exercise 16.18 Exercise 16.19
CH. 10,s 731
THE EGOROFF PROPERTY IN BOOLEAN RINGS
49 1
Finally, we show once more the list in section 16, illustrating situations which may or may not occur in an Archimedean Riesz space L . The first column indicates whether order convergence and relatively uniform convergence are equivalent or not in L . In the second column is written ,,good” or ,,bad” according as pseudo order closure and order closure are the same for every subset of L or not. Similarly in the third column for the pseudo uniform closure and the uniform closure. The final column shows where an example may be found or why the listed situation is impossible. The same list may be presented with somewhat different entries. For this st. st. not not not not
st. st. st. st.
reg. not reg. reg. reg. not reg. not reg.
a not a a not a G
not a
Ex. 16.15, 16.16; Th. 71.6, 71.8; Ex. 71.9, 71.11 Ex. 16.17 impossible impossible Ex. 16.18 Ex. 16.19
purpose, observe that order convergence and relatively uniform convergence in an Archimedean space are equivalent if and only if order convergence is stable (indicated as ,,st.” in the first column). Also, in an Archimedean space L pseudo order closure and order closure are the same for every subset of L if and only if L is regular. Similarly, pseudo upiform closure and uniform closure are the same for every subset of L if and only if L has the a-property. Furthermore, observe that regularity is the same as stability of order convergence together with the a-property. In formula, reg. = st.
u
a.
This makes it immediately evident that the situations in the third and fourth lines of the list are impossible, and that in all other situations the entry in the second column is determined by the entries in the first and third columns.
73. The Egoroff property in Boolean rings The theorems about the Egoroff property and the diagonal property in a Riesz space have their analogs in a Boolean ring. We restrict ourselves to only a few theorems of this kind. Throughout the present section, let R denote a Boolean ring. Definition 73.1. If (bnk: n, k = 1,2,
. . .) is a double sequence of elements
492
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s 13
of R and if the element c E R has the property that for every n there exists a natural number k(n) such that c 5 b,. &(,) then we will write c << (bnk). The element b E R is said to have the Egoroff pfoperty if, given any double sequence (bnk: n, k = 1, 2, . . .) in R such that
b,,
fk
b for n = 1, 2 , . . .,
there exists a sequence b,,, f b such that b,,, << (bnk)holds for every m (i.e., for every m and n there exists a natural number k(m, n ) such that bm
bn,k(m,n)
holds). The Boolean ring R is said to have the Egoroffproperty ifevery element of R has the Egoroff property.
The ideal I in R is called a a-ideal if it follows from b,, E I for n = 1,2, . . . and b, f b that b E I. Furthermore the ideal I is said to be super order dense in R if the a-ideal generated by I is equal to the whole of R. We shall now prove several results, analogous to results obtained for Riesz spaces in Theorem 67.3, Corollary 67.4 and Theorem 67.5.
Theorem 73.2. (i) The set of all elements of R having the Egoroff property is a a-ideal in R. Hence, i f I is a super order dense ideal in R such that every element of I has the Egoroffproperty, then R has the Egoroffproperty. (ii) The element b E R has the Egoroff property if and only if, given any double Jequence (b,,, : n, k = 1,2, . . .) in R such that
b,k
fk
b
for
It =
1, 2 , . . .,
there exists a sequence b, f b mch that, for every n, we have b,, 5 b,. appropriate k(n).
k(n) for
an
Proof. (i) We prove first that if a 5 b and b has the Egoroff property, then so has a. For this purpose, assume that a,,&f k a for n = 1, 2, . . .. Denoting by a’ the relative complement of a with respect to b (i.e., a ~ a =’ 8 and a v a ’ = b ) and writing bnk = ankva’, we have b n k = a n k v a ’ T a v a ’ = b for n = 1,2,
...,
and so there exists a sequence b, f b such that b,,, << (bnk)for all m. But then the sequence (a, = b,,, A a : m = 1 , 2 , . .) satisfies a, f a and
.
(bnk A
a) =
CH. 10,s 731
493
THE EGOROFF PROPERTY I N BOOLEAN RINGS
on account of bnkAa= (ankva')Aa= (ankAa)v(a'Aa)= ankvO= ank. This shows that a has the Egoroff property. Next, assume that a and b have the Egoroff property, and let cnkf k ( U
V b)
for n = 1, 2, .
. ..
Then ank= cnkA a satisfies ankfka,and b,, = Cnk A b satisfies bnkfkb.Hence, there exist sequences a,,,f a and b,,,fb such that a,,,<< (ank)and b,,, << (bnk)for every m. But then the sequence (c,,, = a,,,v b,,, : m = 1 , 2 , . . .) satisfies c,,, f (a v b ) and cm
(ank
bnk) =
for every m. This shows that a v b has the Egoroff property. It follows already that the set of all elements of R having the Egoroff property is an ideal in R.The proof that the ideal is a a-ideal is similar to the corresponding proof for Riesz spaces in Theorem 67.3. (ii) Similarly as in Theorem 67.5 (i). We present an example of a Boolean algebra possessing the Egoroff property. Theorem 73.3. I f p is a cr-finite measure in thepoint set X (i.e., X i s the union of an at most countable number of sets offinite measure), and i f R is the measure algebra of ( X , p), i.e., R is the Boolean algebra of all p-measurable sers modulo sets of measure zero (cf. section 3 ) , then R has the Egoroffproperty.
Proof. It will be sufficient to show that any set of finite measure has the Egoroff property, since the sets of finite measure form a super order dense ideal in R. Hence, let p ( E ) < coy and assume that Enkf k E for n = 1,2, . . ., where all sets Enkare measurable. Then p(E-Enk) 0 as k
--f
00
for n
= 1,2,.
. .,
and so, given the natural numbers m and n, there exists a natural number j(m, n ) such that p ( E - E n , j(,,,,nJ 5 2-"m-'. We may assume that j ( m , n ) increases as m increases and n remains fixed.
494
THE EGOROFF PROPERTY A N D THE DIAGONAL PROPERTY
Now, let
[CH. 10,s
74
m
E m = inf En, j ( m , n ) = n
n=l
En. j ( m , n )
for every m. Then Em increases as m increases, and
u m
E-Em =
n=l
(E-En,
j(m.n,),
so p ( E - Em)g m-I. It follows that E,tE and Em c En, ,,) for all m and n. This shows that E has the Egoroffproperty, which is the desired result.
If p is a measure in the point set X , not necessarily a a-finite measure, then the collection o f all subsets of X of a-finite measure modulo sets of measure zero is a Boolean ring, and exactly as above it follows that this Boolean ring has the Egoroff property. For an arbitrary non-o-finite measure p in X we cannot always say whether the measure algebra of ( X , p) has the Egoroff property. However, if the non-a-finite measure p satisfies some reasonable extra conditions and if the continuum hypothesis is assumed to hold, then it can be proved that the measure algebra of ( X , p) does not possess the Egoroff property (cf. Theorem 75.6). 74. Egoroff's theorem
In the present section it will be shown that the Egoroff property lies at the basis of Egoroff's well-known theorem for sequences of measurable functions. More specifically, we will prove a general theorem for abstract Riesz spaces, and the familiar Egoroff's theorem will then be a special case. Before doing so, we recall (cf. sections 30 and 31) that for an arbitrary Riesz space L we denote by B ( L ) the Boolean algebra of all projection bands in L and by P p ( L )theideal in B(L)consisting of all principal projection bands. The order projection on the projection band A is denoted by P A , the principal , in the band generated by f E L is denoted by B,. (note that B,. = B I f l ) and case that Bf is a projection band the order projection on Bf is denoted by P,. (hence P,. = PI,.,). We also recall a special case of Corollary 31.2, as follows. Lemma 74.1. Let the Riesz Apace L have the principal projection property. (i) If the elements u, v of Lf are such that B,, c B,,, then the component
in B, satisJies B,,, = B,. (ii) If U E L +and vn EL' for n = 1,2,. . ., and ifB,,,fB,,in 9 ( L ) , then the components u,, of u in BUn(n= 1,2, . . .) satisfy 0 5 unTu. u1 of u
CH. 10,s 741
495
EGOROFF'S THEOREM
(iii) I f 0 S untu in L , then B,,fB, in g ( L ) . One of the results in this section will be that for a Dedekind a-complete Riesz space L the space L has the Egoroff property if and only if every element of .Yp(L)has the Egoroff property in B(L). In one direction it is already sufficient that L has the principal projection property. Theorem 74.2. Let the Riesz space L have the principal projeciion property. If L has the Egoroffproperty, then every eIement of g J L ) in the Boolean algebra P ( L ) has the Egoroff property. Proof. Take U E L'. In order to show that the band B , has the Egoroff property in P ( L ) , assume that the projection bands Ank(n,k = 1,2, . . .) satisfy A,,,?, B, for n = 1 , 2,. . .. Since P,,(L) is an ideal in P ( L ) , all Ankare principal bands, i.e., there exist elements Link E L + such that Ank = B,,,, and so
BUnkfkB, for rt = 1, 2,.
. ..
By Lemma 74.1 (ii) we may assume without loss of generality that 0 5 unk t u as k -, co for n = 1,2, . . .. Hence, since L has the Egoroff property, there exists a sequence 0 5 u,Tu such that urn<< ( U n k ) for every m. It follows then from Lemma 74.1 (iii) that BJB,, and evidently we have Bum<< (Bun,) for every m. Since Bunk= An, for all n and k , this is the desired result. Before proving that in Dedekind a-complete spaces the converse holds as well, we first prove an abstract Egoroff type theorem. The theorem holds in Riesz spaces L having the principal projection property and having the property that every member of P p ( L ) has the Egoroff property. Furthermore, there is a given fixed element e E L + the ; role of uniform convergence in the classical Egoroff's theorem will be taken over by e-uniform convergence. More specifically, uniform convergence on each of an ascending sequence of point sets in the classical theorem is replaced by e-uniform convergence in each of an ascending sequence of principal bands. Theorem 74.3 (Abstract Egorof type theorem). Let L have the principal projection property, and let every member of P J L )have the Egoroffproperty. Furthermore, let e be a fixed element in L'. Given now the sequence ukJO in L , there exists a sequence 0 5 e,te such that, for any m = 1,2, . . ., the sequence (Pe,(uk) : k = 1,2, . . .) converges e-uniformly to the zeroelement a s k - , co.
496
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s 74
Proof. Setting e,,k = (n-'e-uk)+ for n , k
=
1 , 2 , . . .,
we have e,,kfkn-'e for every n. Hence, denoting the band generated by e,,k briefly by Bnk,we have Bnkf k Be by Lemma 74.1 (iii). Since the band Be has theEgoroff property, there existsasequence(B, : m = 1,2, . . .)inqp(L) such that B, Be and B,,, << (Bnk)for every m,i.e., B, c B,,, j ( m , n ) for an appropriate natural number j ( m , n). Furthermore, by Lemma 74.1 (ii), there exists a sequence 0 S e, t e such that B, = Bernfor every m. For brevity, we denote the order projection on B, by P,. On account of B, c B,,,j(,,n) it follows now from the definition of Bnkthat Pm{( n - ' e - uj ( , , ,,))-> = 0,
and so
Prn(n-'e-Uj(rn,
n,)
= Pm{(n- ' e - u j ( m ,
This shows that
pm(uj(,,,,))=< n-'e
n,)'}
-Pm{(n-'e-Uj(m,
2 0.
for my n = 1 , 2 , . . ..
Hence, keeping m fixed, we have
Pe,(Uk) 5 n-'e
for all k 2 j ( m , n),
i.e., Pe,(uk) converges e-uniformly to the zeroelement as k
-,00.
Example 74.4 (The classical Egoroff's theorem). We will show that the classical Egoroff's theorem for sequences of measurable functions is a particular case of the general Egoroff type theorem proved above. To this end, let p be a (non-negative and countably additive) measure in the point set X such that p is a-finite, and let M(') = M'"(X, p) be the Riesz space of all real, p-almost everywhere finitevalued and pmeasurable functions on X, the ordering is the p-almost everywhere pointwise ordering (cf. Example 11.2 (v)). The Boolean algebra B ( M ( " ) of all projection bands in M(') is isomorphic to the measure algebra of ( X , p), i.e., to the Boolean algebra of all p-measurable subsets of X modulo sets of measure zero, and evidently B ( M ( ' ) )is equal to @,,(A#@)) since the characteristic function xx of X is a weak unit in M"), which implies that the space M(') itself, and hence also every projection band, is a principal projection band. It is evident also that M(') has the principal projection property (actually, M'"
CH. l o , § 741
497
EGOROFF'S THEOREM
is even super Dedekind complete by Example 23.3 (iv)), and by Theorem 73.3 the Boolean algebra B ( M ( ' ) ) = g P ( M ( ' ) )has the Egoroff property. This shows already that M(') satisfies the hypotheses made upon L in the preceding abstract Egoroff theorem. Now, let (S, :n = 1,2, . . .) be a sequence in M(') such that f. converges pointwise on X tofE M @ ) .Without loss of generality we may assume that all theyn and alsofnowhere assume the values + 00 or - 00. For every x E X we set u k ( x ) = SUP (I f ( X ) - f n ( X ) l : n = k , k + 1, . .).
.
.
Then u k E M " ) for all k = 1, 2, . . by Lemma 71.1, and evidently we have u k @ in M'". Now, let e E (M('))+be the characteristic function of X. Applying the general Egoroff type theorem, we may conclude that there exists a sequence 0 5 emfe such that, for every m = 1,2, . . ., the sequence (Pe,(uk) : k = 1, 2,
. . .)
converges e-uniformly to the zeroelement as k + co. In other words, there exists a sequence of measurable sets X , 1Xsuch that, for every m = 1,2,. . ., the sequence ( u k : k = 1,2, . .) converges uniformly to zero on X,. This implies that, for every m = 1,2, ., the sequence (fn : n = 1,2, . .) converges uniformly t o f o n X,. In the particular case that p ( X ) is finite, note that it follows from X,fX that, given E > 0, there exists X,, such that p(X-X,,,,) < E. Thus, since p ( X - X , , ) < E a n d f , converges uniformly to f on X,, , we have obtained the classical Egoroffs theorem.
.
..
.
It was proved in Theorem 74.2 that if L is a Riesz space with the principal projection property and with the Egoroff property, then every member of g P ( L )has the Egoroff property. It will be proved now that in a Dedekind acomplete space the converse holds as well. In particular this will imply that for a a-finite measure p in the point set X the Riesz space M " ) ( X , p) of all real, p-almost everywhere finitevalued and p-measurable functions has the Egoroff property. Theorem 74.5. A Dedekind a-coniplete Riesz space L has the Egorofproperty if and only if every element of g p ( L )has the Egorofproperty. Proof. We need only prove that if every element of g P ( L )has the Egoroff property, then every element of L has the Egoroff property. Assume, therefore, that every elemtnt of g P ( L )has the Egoroff property, and let 0 6 unk f k u hold in L for n = 1,2, . . ..We must prove that u has the Egoroff property, i.e., we must prove the existence of a sequence 0 5 u, ? u satisfying
498
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10,s
75
urn<< (unk)for every m. In view of the general Egoroff type theorem there exists, for n fixed, a sequence 0 5 unpT, u such that punp(u-unk)
lk
holds u-uniformly for fixed n andp. Since Bun)f p B, and B , has the Egoroff property, there exists a sequence 0 5 urn1u such that B,,,, << (Bun,)for every m (cf. Lemma 74.1), and this implies that Pu,(u - u n k )
lk
0
holds u-uniformly for fixed n and m. Hence, given n and my there exists a natural number j ( m , n) such that
0 6 Pu,(u-un,
j(rn,n))
S m
-1
u,
and we may assume without loss of generality that j ( m , n) increases as m increases and n remains fixed. Now, for m = 1, 2, . . ., let urn= inf (Pu,(un,j ( r n , n , ) : n = 1,2,
. . .).
Note that urnexists sinceL is Dedekindo-complete. Also note that urnis a member of the band generated by urn, so Pu,(urn)= urn.It is evident that 0 5 urn1 and that urn6 urn, j(rn,n) for all ni, n. (1) In addition, we have Pu,(u)-Pu,(urn)
so and hence
0
= Pu,(u)-~rn = SUP Pu,(u-un, n
j(rn,nJ,
s Pu,(u)-Pu,(urn) 5 m-'u,
0 5 u -urn = {U - p u , ( ~ ) }+ {Pu,(u)-Pu,(urn)} + {Pu,(um) - urn} 5 {u-Pu,(u)}+m ' u = ( ~ ~ - ~ , , ) ( u ) + m - 'J.uo as m + 00. This shows that 0 5 urnTu.Since also urn< : (unk)by formula (I), it follows that the sequence (urn: m = 1,2, .) hasall thedesiredproperties, and so u has the Egoroff property.
..
75. The Egoroff property and Dedehind completeness We begin by proving two simple theorems, partly recapitulations of earlier results.
CH. 10,s 751
THE EGOROFF PROPERTY AND DEDEKINDCOMPLETENESS
499
Theorem 75.1. Let X be a non-empty point set, L ( X ) the Riesz space of all real jinitevalued functions on X (with pointwise ordering), B ( X ) the Riesz space of all real bounded functions on X , a n d p ( X ) the Boolean algebra of all subsets of X (with the ordering by inclusion). Thefollowing conditions are now equivalent . (i) L ( X ) has the Egorofproperty. (ii) B ( X ) has the Egoroff property. (iii) p ( X ) has the Egoroff property. I f x i s ajinite or countable point set, then L ( X ) , B ( X )andp(X)have the Egoroff property. Proof. Note first that L ( X ) is Dedekind complete. Hence, by Theorem 74.5, L ( X ) has the Egoroff property if and only if p ( X ) has the Egoroff property, i.e., (i) and (iii) are equivalent. It is evident that (i) implies (ii). Conversely, (ii) implies(i) by Corollary 67.4, since B ( X ) is a super order dense ideal in L ( X ) . If Xis finite or countable, then L ( X ) has the Egoroff property, as shown in Example 67.6 (iv). It follows, therefore, that B ( X ) and p ( X ) have the same property. Theorem 75.2. Let ( X , A , p) be a measure space, i.e., X is a non-empty point set and p is a countably additive non-negative measure on the a-field (aalgebra) A of subsets of X . Let M(')(X, p) be the Riesz space of all real functions on X that are p-measurable and p-almost everywhere jinitevalued (with identijicaiion of p-almost equal functions and with p-almost everywhere pointwise ordering), B ( X , p) the subspace of all p-essentially bounded functions, arid p ( X , p) the measure algebra of ( X , p), i.e., the Boolean algebra of all p-measurable subsets of X modulo sets of p-measure zero. Thefollowirrg conditions are now equivalent. (i) M ( ' ) ( X ,p) has the Egoroflproperty. (ii) B(X, p ) has the Egoroflproperty. (iii) p ( X , p) has the Egorofproperty. If p is o-jinite, then M ( ' ) ( X , p ) , B(X, p) and p ( X , p) have the Egorofproperty. Proof. Note first that M(')(X,p) is Dedekind a-complete, so that Theorem 74.5 is applicable. Note also that there is a one-one correspondence between the members of p ( X , p) and the principal ideals in M ( " ( X , p), as follows. Given E E ~ ( Xp), , the corresponding principal ideal is the ideal generated by the characteristic function z E .In the converse direction, given the ideal generated by u E ( M ( ' ) ( X ,p))', the corresponding subset of X is the set
500
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.
10,s75
(x : u ( x ) > 0). It follows that p(X, p) and @'JM(')) may be identified and so, by Theorem 74.5, M("(X, p ) has the Egoroff property if and only if p ( X , p) has the same property, i.e., (i) and (iii) are equivalent. The equivalence of (i) and (ii) is again a consequence of the fact that B(X, p ) is a super order dense ideal in M(')(X,p). If p is a a-finite measure, then p ( X , p ) has the Egoroff property by Theorem 73.3, and so M(')(X, p) and B(X, p) have the same property. Note that, according to Theorem 71.6, the space M(')(X, p ) has even the strong Egoroff property.
Note that Theorem 75.1 is a special case of Theorem 75.2 (in the second theorem take for p the measure assigning to each individual point the unit value; this measure is sometimescalled the discrete measure, or also the counting measure). It was shown by S. Banach and C. Kuratowski ([l], 1929) that if the continuum hypothesis holds and Xis a point set of the power of the continuum (i.e., there is a one-one mapping from X onto the real numbers), then there exists a double sequence
(Xnk:n, k = 1,2,. . .) of subsets of X such that the following conditions hold. (i) Xnk f k X for n = 1, 2, . .. (ii) For every choice of the natural numbers k(n), n = 1 , 2 , . ., the intersection is at most countable. This is also listed as proposition Cll in W. Sierpinski's book on the connuum hypothesis ([l], 1934). It follows that, for every X l j the intersection
.
.
m
is at most countable, and so 00
is at most countable. By induction, the set m
0 Xn,k(n) n =m is at most countable for every m.
00
C H . 10.5 751
THE EOOROFF PROPERTY A N D DEDEKIND COMPLETENESS
501
We infer from this that the Boolean algebra p ( X ) of all subsets of X does not have the Egoroff property. Indeed, if there would exist a sequence X m t X such that X, << ( X n k ) holds for every m, then X , c Xm,k(,) for appropriate k(m),and so W
X m
=
0
n=m
03
X n
c
n=m
Xn,k(n)
uz,l
is at most countable, which contradicts X = X,. We have, therefore, the following complement to Theorem 75.1.
Theorem 75.3. Assume that the continuum hypothesis holds. Then, given the non-empty point set X , the Riesz space L ( X ) of all realjinirevaluedfunctions on X has the Egoroflproperty if and only i f X isjinite or countable. Proof. If X is uncountable, then X has a subset XI of the power of the continuum, and so p ( X l ) does not have the Egoroff property. But then, by Theorem 75.1, L ( X , ) does not have the Egoroff property, and so neither does LV). The next theorem and its corollaries are now consequences of the just obtained result.
Theorem 75.4. Assume that the continuum hypothesis holds, and let L be a Riesz space with the Egoroflproperty and the projection property. Then L is order separable. Proof. Before presenting the proof, we recall several facts. Firstly, L has the projection property by our hypotheses, so LY',(L) is Dedekind complete by Theorem 31.3 (i). Secondly, we recall that by Theorem 29.3 the space L (being an Archimedean space) is order separable if (and only if) for any disjoint system (u, : a E {a}) of nonzero elements of L+ and for any u EL' we have inf (u, u,) # 0 for at most countably many a E {a}. We will prove that, under the present hypotheses, L satisfies this last condition. Hence, let u E Lf and a disjoint system (u, : a E {a}) of nonzero elements in L+ be given. For brevity we will denote the index set {a} by S. For the purposes of the present proof we may assume that u, = inf (u, u,) # 0 holds for all r~ E S, and we have to prove that the set S is at most countable. Note that (v, : r~ E 5') is also a disjoint system of nonzero members of L'. For every a E S , let B, be the band generated by v,, and for every subset T c S, let B, = SUP (B, : a E T ) .
502
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. 10, 75
This supremum exists and is an element of 9 J L ) since 9',,(L) is Dedekind complete (as observed above) and since the band B, generated by u is an upper bound of the set of all B,. Note that TI # T2 implies BT, # BT2 on account of every B, being different from (0) and B,, IB,, for crl # cr2. Obviously, the largest band of this kind is obtained by taking T = S, so
Bs = SUP (B, : cr E S). For T the empty subset of S we set BT = (0). It is evident now that the set of bands (BT : T c S), ordered by inclusion, is a lattice with smallest element (0) and largest element Bs ,and by the one-one order preserving correspondencebetween the members of (BT :T c 5') and the members of the Boollean algebra p ( S ) of all subsets T of S the set (BT : T c 5') is a Boolean algebra isomorphic to p(5'). We will prove now that the Boolean algebra (BT : T c S) has the Egoroff property. By the isomorphism ~ ( 5 ' will ) then have the Egoroff property, and so it will follow from the preceding theorem that S must be finite or countable. We first make some further remarks about the embedding of the Boolean algebra (BT : T c in g p ( L ) .If BTzf,i.e., if (BT. : t E (t}) is an upwards directed system of members of (BT : T c S), then BT, = sup BTr exists in the Boolean algLbra (BT : T c S). Indeed, To consists of ali indices cr occurring in at least one of the T,. It follows now that BTo = sup BTz holds as well in B (L). Indeed, any upper bound in g ( L ) of all BTr must include B, for every cr occurring in at least one of the T,, and hence must include BTo. Similarly (by complementation with respect to Bs), if BTr J , then B,, = inf BTz exists in the Boolean algebra (BT : T c S), and BT, = inf BTr holds as well in B ( L ) . AS observed above, the proof will be complete if we can show that the Boolean algebra (BT : T c S ) has the Egoroff property. Since L has the Egoroff property as well as the principal projection property, it follows from Theorem 74.2 that in the Boolean algebra B ( L )every member of 9 J L ) has the Egoroff property. In particular, every member of the set ( B , : T c S ) has the Egoroff property with respect to the Boolean algebra S ( L ) . We have to show, however, that the Boolean algebra (BT : T c S ) has the Egoroff property with respect to itself. To this end, let BTo be given, and assume that
s>
BT,, f k BTo in (BT : T c S ) for n = 1, 2, . . .. Then, as observed above, BT,, f k BT, holds as well inB(L), and so there exists a sequence (B, : in = 1,2, . . .) in B ( L )such that B, f BT, and B, << (BTnk)
CH. lo,§ 751
THE EGOROFF PROPERTY AND DEDEKIND COMPLETENESS
503
for every m. In general, B,,, will not be a member of ( B , : T c S), and for that reason we will try to replace the sequence (B, :m = 1 ,2 ,. . .) by a sequence of members of the Boolean algebra (BT : T c S). In order to do this, observe that for m and n given we have for an appropriate natural numberj(rn, n), where we may assume thatj(m, n ) is increasing in m for n fixed. Now let BTm= infB,n,J(m,n) n
for
111 =
1,2,. ..
in g ( L ) . As observed above, the infimum exists and is a member of (BT : T c S ) . In addition, B , is increasing as m increases, and so it follows from B, c BTm c BT, and B, t BTo that BTm t BTo. This, combined with BTm
BTn.
j(m. n )
holding for all rn and n, shows that the arbitrary element BT, i n ( B , : T c S ) has the Egoroff property. Thus the Boolean algebra ( B , : T c S ) has the Egoroff property. As observed above, this implies that the Boolean algebra p ( S ) has the Egoroff property, and so S is finite or countable. The last theorem has important consequences. Theorem 75.5. (i) ( A . G . Pinsker [l], 1949; I. Amemiya [l], 1952.) Assume that the continuum hypothesis holds. Then every Dedekind complete Riesz space possessing the Egoroff property is super Dedekind complete. (ii) Assume that the continuum hypothesis holds, and let L be a Dedekind a-complete Riesz space possessing the Egoroffproperty. Then L is super Dedekind complete if and only i f g , ( L ) is Dedekind complete.
Proof. (i) If L is a Dedekind complete Riesz space possessing the Egoroff property, then L satisfies all conditions of the preceding theorem (indeed, every band is now a projection band b-cause Dedekind completeness implies the projection property). Hence, by the preceding theorem, L is order separable. But L is Dedekind complete by hypothesis, and so L is super Dedekind complete. (ii) Let L be a Dedekind a-complete Riesz space with the Egoroff property, and assume first that Y p ( L )is Dedekind complete. Since L has the principal projection property, Theorem 3 1.3 is applicable, and hence the Dedekind completeness of Y p ( L )is equivalent to the projection property for L . Thus L satisfies all conditions of the preceding theorem, and so L is
504
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH. lo,§ 75
order separable. But L is Dedekind a-complete by hypothesis, and so L is super Dedekind complete by Theorem 23.6. Conversely, if L is super Dedekind complete, then L has the projection property, and so B,(L) is Dedekind complete by Theorem 31.3. The first part of the last theorem is essentially due to A. G . Pinsker ([l], 1949) and I. Amemiya ([I], 1952). More precisely, in Pinsker's work it is assumcd that L is Dedekind complete and L has the diagonal property, and it is proved that any disjoint system of nonzero elements in L+ majorized by an element of L+ is finite or countable (as proved in Theorem 29.3, this is equivalent to order separability in Archimedean spaces). In Amemiya's work it is assumed that L is Dedekind complete and totally continuous in the sense of H. Nakano (cf. section 67 for the definition of total continuity). The statement in part (i) of the theorem cannot be made stronger by substituting Dedekind a-completenessfor Dedekind completeness in the hypotheses. In other words, Dedekind a-completeness together with the Egoroff property does not necessarily imply Dedekind completeness. As an example, let L be the Riesz space of all real bounded functionsf on ( x : 0 6 x I 1) such that f ( x ) # f ( 0 ) holds for at most countably many x , with pointwise ordering. This space L is the example (ii) in the proof of Theorem 25.1 (the main inclusion theorem), and it is shown in this proof that L is Dedekind a-complete, but not Dedekind complete. It is not difficult to verify that L has the Egoroff property. Hence, L is Dedekind a-complete and has the Egoroff property, but L is not Dedekind complete. According to part (ii) of the theorem g P ( L )is not Dedekind complete for this particular space L ;it is not difficult to verify this directly.
As in Theorem 75.2, let (A',A , p) be a measure space, let M"'(X, p) be the Riesz space of all real, p-measurable and p-almost everywhere finitevalued functions on X (with identification of p-almost everywhere equal functions and with p-almost everywhere pointwise ordering), and letp(X, p) be the measure algebra of ( X , p), i.e., the Boolean algebra of all p-measurable subsets of X modulo sets of p-measure zero. For any p-measurable subset E of X , we shall denote by E* the equivalence class of which E is a member (i.e., we are now precise, and we distinguish between the set E and the member E* of p ( X , p)). The measure p is called locally finite (or sometimes p is said to have the finite subset property) if every p-measurable set E of infinite measure has a subset F such that 0 < p ( F ) < CO. It is not difficultto verify that p is locally finite if and only if, for every p-measurable
CH. lo,§ 751
THE EGOROFF PROPERTY AND DEDEKIND COMPLETENESS
505
set E, we have
E* = SUP (F* :I; c E, p ( F ) < a).
(1)
We briefly indicate the proof that local finiteness implies (1). It is evident that E* is an upper bound of the set
(F* :F c E, p ( F ) < a).
(2)
Assuming that E: is another upper bound, we consider the set E - El.It will be sufficient to show that p ( E - E , ) = 0. If not, the set E - E l has a subset F of finite positive measure. Then F and El are disjoint, but on the other hand F is p-almost included in El since E: is an upper bound of (2). This is impossible in view of p(F) > 0. Hence p ( E - E l ) = 0. Even for a locally finite measure p it is not necessarily true that for every of p-measurable sets of finite measure the supremum collection (F, : E sup F,* exists inp(X, p). Any measure for which this supremum always exists is called a IocaIizabIe measure, a notion introduced by I. E. Segal ([l], 1951). It is evident now that if p is locally finite as well as localizable, then sup EF exists inp(X, p) for any collection (E, :1E of p-measurable sets (whether of finite measure or not). Sincep(X, p) andg,(M(')) may be identified (cf. the proof of Theorem 75.2), this means that g,(M(')) is now a Dedekind complete Boolean algebra. Note that every a-finite measure p is of course locally finite as well as localizable, and according to Theorem 75.2 the space M("(X, p) has the Egoroff property. As the final theorem in this section we will prove now, under the assumption that the continuum hypothesis holds, that among the locally finite and localizable measures the a-finite measures are the only ones for which M(')(X,p) has the Egoroff property.
{a})
{a})
Theorem 75.6. Assume that the continuum hypothesis holds, and let p be a locally finite and Iocalizable measure in the point set X. Then M'"(X, p ) has the Egorofproperty if and only if p is a-finite. Proof. It will be sufficient to prove that if M(')(X, p ) has the Egoroff property, then p is a-finite. Assume, therefore, that M("(X, p) has the Egoroff property. As observed above, Y , ( M ( ' ) ) is Dedekind complete which, according to Theorem 31.3, is equivalent to M(')(X,p) having tht projection property. Hence, by Theorem 75.4, the space M(')(X, p) is order separable. Now, let ( X , : a E { a } ) be a maximal disjoint system in p ( X , p) such that 0 < p ( X , ) < co for all a E { a } (disjointness means that X,, n Xu, is a set of p-measure zero for m1 # m2). Such a maximal system exists on account
506
THE EGOROFF PROPERTY AND THE DIAGONAL PROPERTY
[CH.
l o , § 75
of Zorn's lemma. It is evident that X * = sup X,* ,so by the order separability there exists a sequence
(x:" : n = 1 , 2 , . . .) c (x,*: u E { u } )
such that X * = sup X z . But then X and the p-measurable set are in the same equivalence class, and so p is a-finite.
=:u
Xan
Just as Theorem 75.1 is a special case of Theorem 75.2, so Theorem 75.3 is a special case of Theorem 75.6.
Exercise 75.7. We recall that if L is an Archimedean Riesz space with Dedekind completion K, then K is order separable if and only if L is so. This was proved in Theorem 32.9. In the proof it was also shown that if 0 j w,, t wo holds in K, then there exists a sequence 0 v,, 7 wo with all v,, in 1. and v,, 5 w,,for all n. Assume now that the continuum hypothesis holds. Show that the Dedekind completion K of the Archimedean Riesz space L has the Egoroff property if and only if L has the Egoroff property and L is order separable. Show that the space of all real bounded functions f on (x :0 5 x j 1) with f ( x ) # f ( 0 ) for at most countably many x is a Dedekind o-complete space with the Egoroff property, but its Dedekind completion does not have the Egoroff property. Hint: Suppose first that K has the Egoroff property. Then Kis super Dedekind complete by Theorem 75.5, and so K and L are order separable. Now, in order to show that L has the Egoroff property, assume that 0 _I u,,k t u as k + 00 holds in L for n = 1, 2, .. It is not difficult to verify that 0 5 u,,k f u as k 00 holds as well in K, so there exists a sequence 0 6 w,, t u in Ksuch that w, << (U,,k) for every m. According to what was observed above, there existsasequence 0 5 u, t u in L such that urn j w, for all m, so v, << (u,J for every m. Conversely, suppose that L has the Egoroff property and L is order separable. For the proof that K has the Egoroff property, let 0 5 w,,k t w as k + 00 hold in K for n = 1,2, . . Then, by our earlier remarks, there exist #,,k E L such that 0 5 u,,k f w as k -,co for n = 1,2, . .. From this point on the proof is very similar to the last part of the proof of Theorem 67.3. For the example where L is the space of all real bounded f on X = ( x : 0 - j x j 1) withf(x) # f ( 0 ) for at most countably many x, note that the Dedekind completion K of L is the space B ( X ) of all real bounded functions on X , and so (by Theorems 75.1 and 75.3) K does not have the Egoroff property.
s
..
..
.
CH. 10,s 751
T H E EGOROFF PROPERTY AND DEDEKIND COMPLETENESS
507
Exercise 75.8. Assume that the continuum hypothesis holds. Show that the Dedekind completion K of the Archimedean Riesz space L has the diagonal property if and only if L has the diagonal property and L is order separable. Hint: In an Archimedean space the diagonal property is equivalent to the d-property (cf. Theorem 68.5).
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Index
super D. complete 124; D. completion 185 diagonal, d. property 83; d. gap property 83 direct, order d. sum 132 directed, d. set 73 discrete, d. element 145; d. Riesz space 241 disjoint, d. complement 5, 68; d. elements 5, 67; d. order basis 163; d. system 163; maximal d. system 163 distributive, d. lattice 3; d. law 3, 16, 62,63 dominated, d. decomposition property 73
absolutely continuous, a.c. norm 44 abstract, a. Egoroff type theorem 495; a. L , space 411 algebra, a. of sets 7; Boolean a. 5 ; measure algebra 7; a-a. of sets 7 anti-lattice,415; Kadison’s a.4. theorem417 anti-symmetric, a.-s. relation 1 Archimedean, A. Riesz space 78 associate, a. norm 44;a. space 43 atom, 19, 145 Banach-Mazur, B.-M. limit 317 band, 12, 93; b. generated by a subset 12, 96; principal b. 96; projection b. 39, 133 basis, order b. 160; disjoint order b. 163; quasi order b. 160 Birkhoff, B. identity 64; B. inequalities 64 Boolean, B. algebra 5 ; B. ring 9 bounded, order b. set 475 carrier, 119, 120; closed-open c. 284 Cartesian, C. product 1 chain, 2 closed, uniformly c. set 427 closed-open, c.-0. carrier 284; c.-0. hull 285 commutant, c. of a system of operators 382 comparable, c. elements 2 complement, c. in a lattice 4; c. of an element with respect to another element 7; disjoint c. 5, 68 complete, Dedekind c. 3; Dedekind a-c. 3; super Dedekind c. 124; order c. 2; uniformly c. 276; universally c. 323 completion, Dedekind c. 185 component, 249 cone, generating c. 59; positive c. 55 convex, 266 cozero, c. set 284 Dedekind, D. complete 3; D. u-complete 3;
Egoroff, E. property 460; abstract E. type theorem 495 equidirected, e. sets 73 equivalence, e. relation 1 extended, e. continuous function 295 extremally, e. disconnected topological space 322 Fatou, F. property 42 field, f. of sets 7; a-f. of sets 7 fixed, f. maximal ideal 313 free, f. maximal ideal 313 Freudenthal, F.’s spectral theorem 257,258 function, extended continuous f. 295; f. norm 42; spectral f. 244; superharmonic f. 51; Young f. 45 generating, g. cone 59 greatest, g. lower bound 2 Holder, H. inequality 43 homomorphism, Riesz h. 98; Riesz u-h. 102; normal Riesz h. 103 hull, 24, 213; closed-open h. 285 hull-kernel, h.-k. topology 29, 218 ideal, 5,93; u-i. 93; i. generated by a subset 12,96; maximal i. 10, 149; fixed maximal
512
INDEX
i. 313; free maximal i. 313; order densei. 14, 110; quasi order dense i. 15, 110; super order dense i. 110; prime i. 10,200; minimal prime i. 21, 204; principal i. 96; non-trivial i. 153 idempotent, i. element 9 inclusion, main i. theorem 137 incomparable, i. elements 2 indecomposable, i. element 151 infimum, i. of a subset 2 infinitely, i. small element 79, 113 integral, 43; i. dependence 363 interval, order i. 475 isomorphism, order i. 186; Riesz i. 99
Johnson-Kist, J.-K. spectral representation theorem 299 K, K-lineal 48; K.-space 124; K-space of countable type 124 Kadison, K.’s anti-lattice theorem 417 kernel, 24, 212 Kothe, normed K. space 42 lattice, 3; distributive 1. 3; vector 1. 48 least, 1. upper bound 2 linear, 1. ordering 2 localizable, 1. measure 504 locally finite, 1.f. measure 504 lower, 1. bound 2; greatest 1. bound 2; 1. sublattice 21, 203; 1. sum 254 A, &-complementation property 35, 224
main, m. inclusion theorem 137 maximal, m. element 2; m. disjoint system 163; m. ideal 10, 149; fixed m. ideal 313; free m. ideal 313 measure, m. algebra 7 minimal, m. element 2; m. prime ideal 21, 204 monotone, m. sequence 70 Nakano, N. spectral representation theorem 341 norm, absolutely continuous n. 44; associate n. 44;function n. 42; saturated n. 42 normal, n. Riesz homomorphism 103 normed, n. Kothe space 42 null, n. element of a lattice 4 Ogasawara-Maeda, 0.-M. spectral representation theorem, 329, 337
513
order, 0 . basis 160; quasi 0. basis 160; disjoint 0 . basis 163; 0. bounded 475; 0 . complete 2; 0. convergence 78; 0. dense ideal 14, 110; quasi 0.dense ideal 15, 110; super 0. dense ideal 110; 0. direct sum 132; 0. interval 475; 0. isomorphic 186; 0. projection 133; 0. separable 124; 0. topology 80 ordered, 0. vector space 48 Orlicz, 0. space 47 partial, p. ordering 1 Poisson, P. formula 273 positive, p. cone 5 5 ; p. element 55; p. linear functional 43 principal, p. band 96; p. ideal 96; p. projection property 136; p. Riesz subspace 96 projection, p. band 39, 133; p. element 181; order p. 133; p. property 136; principal p. property 136; quasi principal p. property 233 property, p. d2 and A 2 46; diagonal p. 83; diagonal gap p. 83; dominted decomposition p. 73; Egoroff p. 460; Fatou p. 42; &-complementation p. 35, 224; projection p. 136; principal projection p. 136; quasi principal projection p. 233; Riesz-Fischer p. 42; Riesz interpolation p. 73 pseudo, p. order closure 82; p. uniform closure 85 quasi, q. order basis 160; q. order dense ideal 15, 110; q. principal projection property 233; q. unit 111 quotient, q. Riesz space 102 radical, 156 Radon-Nikodym, R.-N. theorem 270 reflexive, r. relation 1 regular, r. Riesz space 479 regularly, r. open 17; r. closed 17 relation, anti-symmetric r. 1; equivalence r. 1; reflexive r. 1 ; symmetric r. 1 ; transitive r. 1 relatively, r. uniform convergence 79; r. uniform topology 84 representation, Stone’s r. theorem 27 Riesz, R. homomorphism 98; R. a-homomorphism 102; normal R. homomorphism 103; R. isomorphism 99; R. interpolation property 73; R. seminorm
514
INDEX
437; R. space 48; Archimedean R. space 78; discrete R. space 241; quotient R. space 102; regular R. space 479; R. subspace 93; principal R. subspace 96 Riesz-Fischer, R.-F. property 42 ring, r. of sets 6; Boolean r. 9 a , o-algebra 7; a-field 7; Riesz a-homomorphism 102 saturated, s. norm 42 seminorm, Riesz s. 437 separable, order s. 124 singular, s. linear functional 44 solid, s. subset 68 space, associate s. 43; uniformly complete s. 276; universally complete s. 323; Kspace 124; K-space of countable type 124; normed KBthe s. 42; abstract L , space 411; Orlicz s. 47; Riesz s. 48; Archimedean Riesz s. 78; discrete Riesz s. 241; quotient Riesz s. 102; regular Riesz s. 479; ordered vector s. 48 spectral, s. function 244; s. system 252; s. theorem in Hilbert space 392, 398; Freudenthal’s s. theorem 257,258; Johnson-Kist s. representation thorem 299; Nakano s. representation theorem 341 ; Ogasawara- Maeda s. representation theorem 329, 337; Yosida s. representation theorem 305, 320 stable, s. convergence 79 state, s. on a Boolean ring 19 Stone, S.’s representation theorem 27 Stone-tech, S.-c.compactification 31 5 strong, s. unit 11 1
sublattice, lower s. 21, 203 subspace, Riesz s. 93 sum, order direct s. 132 super, s. Dedekind complete 124; s. order dense ideal 110 superharmonic, s. function 51 supremum, s. of a subset 2 symmetric, s. relation 1 system, disjoint s. 163; maximal disjoint s. 163; spectral s. 252 totally, t. disconnected topological space 3 1 transitive, t. relation 1 triangle, t. inequality 62 uniform, relatively u. convergence 79; relatively uniform topology 84 uniformly, u. closed set 427; u. complete space 276 unit, u. in a lattice 4; quasi u. 1I I; strong u. 111; weak u. 111 universally, u. complete 323 upper, u. bound 2; least u. bound 2; u. sum 254 vector, v. lattice 48 weak, w. unit 111 Yosida, Y. spectral representation theorem 305, 320 Young, Y. function 45; Y.’s inequality 46 zero point 207 Zorn, Z.’s lemma 2
