Chapter 7 Laterally Complete Riesz Spaces

CHAPTER

7

Laterally Complete Riesz Spaces

The final chapter of this book deals with laterally and a-laterally complete Riesz spaces, that is, with Riesz spaces having the property that the supremum of every disjoint subset or disjoint sequence of positive elements exists. The first part discusses the remarkable lattice properties of these spaces. For instance, it will be shown that Archimedean a-laterally complete Riesz spaces satisfy the principal projection property, that Archimedean laterally complete Riesz spaces satisfy the projection property, and that in Archimedean Riesz spaces lateral and Dedekind completeness are independent properties. In the second part we shall deal with locally solid topologies on laterally or a-laterally complete Riesz spaces. It will be seen that on a a-laterally complete Riesz space every Hausdorff locally solid topology must satisfy the preLebesgue and a-Lebesgue properties ; that a a-laterally complete Riesz space can carry at most one Hausdorff Fatou topology (necessarily Lebesgue) and that if it can carry this topology then every other Hausdorff locally solid topology is finer and agrees with this Fatou topology on an order dense a-ideal. 23. THE LATTICE STRUCTURE OF LATERALLY COMPLETE RlESZ SPACES

We start this section with the definition of laterally and a-laterally complete Riesz spaces. 165

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[Chap. 7

Definition 23.1. A Riesz space L is said to be

(i) a-laterally complete, if the supremum of every disjoint sequence of L+ exists; (ii) laterally complete, fi the .supremum of every disjoint subset of L+ exists. Note. In the terminology of Luxemburg and Zaanen a laterally complete Riesz space is referred to as a universally complete Riesz space [46, Definition 47.3, p. 3231; here a universally complete Riesz space will be defined to be a laterally and Dedekind complete Riesz space (see Definition 23.17).The term “laterally complete” is taken from the lattice group theory; see [12] and [19]. Observe that the lexicographic plane provides an example of a laterally complete Riesz space that is not Archimedean. Theorem 23.2. For a laterally complete Riesz space L the following statements hold;

(i) Every band of L is a laterally complete Riesz space in its own right. (ii) I f L is Archimedean, then L contains weak order units. (iii) I f L is Archimedean, then every band of L is a principal band. Proof. (i) Let B be a band of L and let S be a disjoint subset of positive elements of B. Then u = s u p s exists in L and u belongs to B, since B is a band of L. But then u = sup S holds in B, and this shows that B is laterally complete. (ii) Let {e,} be a complete disjoint system of positive elements of L+. Since L is laterally complete, e = sup(e& exists in L and since L is Archimedean, e must be a weak order unit of L. (iii) Let B be a band of L. By (i) B is laterally complete and so by (ii) B contains a weak order unit e. But then, since L is Archimedean, B = Be (the band generated by e in L ) holds, and the proof is finished. H The following result can be proven as in (i) above. Theorem 23.3. Every a-ideal of a a-laterally complete Riesz space is a a-laterally complete Riesz space in its own right. One of the most remarkable properties of the Archimedean laterally complete Riesz spaces is included in the next theorem. Theorem 23.4. For an Archimedean Riesz space L the following statements hold:

(i) I f L is a-laterally complete, then L satisjes the principal projection property.

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THE LATTICE STRUCTURE

167

(ii) If L is laterally complete, then L satisfies the projection property. Proof. (i) Let u p E L+. Set w,

=((n

+ 2)u - v)+ nu -

( n = O,1,2,. . .)

0)-

and observe that w, A w , + ~= I9 holds for all n and all p 2 2. Observe also that since I9 I w, I ( n + 2)u holds for all n, we have {w,}c B,. Now the relations [(n

+ l)((n + 2)u + (n + 2 ) ~ , + ~ + I)(( n + 2)u - u)+ + ( n + 2)w,+ v U)+]AU

2 [(n

A

+ I)(( n + 2)u - u)+ + ( n + 2){(( n + 3)u - u)+ A ( ( n + 1)u - u ) - } ] A u 2 [(n + 2)(( n + 3)u r u ) + ] A [ ( n + 1)(( n + 2)u - u) + ( n + 2)(u - ( n + I)u)] A u

= [(n =

[(n + 2)(( n + 3)u - u ) + ]

A

u

imply that n

C ( r + I)wr 2 [(n+ l)((n + 2)u

r=O

-

u)']

But since L is a-laterally complete and w, suprema

A

A

u 2 (nu - u)'

A

u.

w, = 6' if Irn - n( 2 2, the

+ 2 ) ~ * , + ~2: n0) both exist in L and obviously f and g belong to B,, and thus u (f + g) E B,. f = sup((2n + l)w,,:n 2 0)

and

g = sup((2n

A

We shall show next that u - u A (f+ 9 ) E Bud.To this end observe that for every nonnegative integer k we have /

k

< UA(U -u ~ ( ku u)+) I U A ( U- u r \ ( k u - u ) ) = UA(2U-2UAku)=(U+2u)A(U+ku)A20-2UAku

= 2UA(k -k 1)U - 2UAkU

and thus n [ u ~ ( i i - u ~ ( ( f + g ) )I] ~ ~ ~ ~ [ 2 l )uu -~2 u(n kku ]+ s 2u holds for n = 1,2, . . . . Now, since L is Archimedean, u A ( u - u A (f+ 9 ) ) = 8 and so u - u A (f+ g) E B,d. Thus L = B, 0 Budand this shows that L has the principal projection property. (ii) The claim follows immediately from Theorem 23.2(iii) and part (i) above. The next four results are applications of the preceding theorem. The first theorem characterizes the Riesz spaces that are Dedekind and laterally complete.

168

[Chap. 7

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Theorem 23.5 are equivalent:

For an Archimedean Riesz space L thefollowing statements

(i) L is Dedekind complete and laterally complete. (ii) L is a-Dedekind complete and laterally complete. (iii) L is uniformly complete and laterally complete.

-

Proof. (i) = (ii) and (ii) = (iii) are obvious. To see that (iii) (i) note first that by Theorem 23.4(ii) L has the projection property, and then that the uniform completeness and the projection property are equivalent to the Dedekind completeness, by Theorem 2.1 1. W Similarly, it can be shown that a Riesz space is a-Dedekind and a-laterally complete if and only if it is uniformly and a-laterally complete. We continue with a useful lemma. Lemma 23.6. Let L be an Archimedean a-laterally complete Riesz space and let u, 1 8 in L. Thenfor every E > 0 there exists u E L' (depending upon E ) such that 8 I u, I 2-"u + eul holds for all n. Proof. Let u, -1 8, E > 0, and let P, denote the projection on the band generated by (u, - E U ~ ) ' in L (the projection P, exists by Theorem 23.4). Put u, = P,(ul), w, = u1 - u, for all n, and observe that w, A (u, - E U ~ ) += 8. Since (u, - wl)+ = P,(u, - E U ~ = ) P,(u,) - EU,, we get 8 I E U , 5 u , ~and hence v, 8. Note next that {2"+1(v,- u , + ~ ) }is a disjoint sequence of L', and hence 8 I 2"' ' ( v , - vn+1) I u must hold in L for all n and some u. Also, 8 I (u, A w, - &u1)+I (u, - m1)+A w, = 8 implies u, A w, I &ul for all n, and thus 8 <

U, = U , A U 1 = U, A ( U ,

+ W,)

u,

EM1

+

= (0, - v k ) f

+

holds in L for all k > n. Since u, for n = 1,2,. . . . H

U, A U,

+ U, A W ,

uk

1 8, we obtain easily that 8 I

u, I 2 - k

+

E U ~ ,

Theorem 23.1. Let T :L -+ M he a positive linear mapping between two Archimedean Riesz spaces. If L is a-laterally complete, then T is a-order continuous. Proof. Let u, 1 8 and let E > 0. According to the previous lemma, there exists u E L + such that 8 I u, I 2-"u + m1holds in L for all n. Now if 8 I w I T(u,) holds in M for all n, then it follows from the positivity of T that 8 5 w I T(u,) I 2 - " T ( u ) + ET(ul) for all n. Now because

Sec. 231

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THE LATTICE STRUCTURE

M is Archimedean, 0 I w I &T(ul) holds for all E > 0 and thus by using the Archimedeanness of M once more we get w = 8. Hence T(u,) 1 8 holds in M , so that T is a-order continuous. Note that a a-order continuous positive operator between two Archimedean Riesz spaces need not be order continuous even if its domain is laterally complete. To see this let L = RX for some uncountable set X , and let A be the a-ideal of L : A = {f

E

RX:{x E X :f(x) # 0} is at most countable).

So, LIA is Archimedean. However, since A is not a band of L, the canonical projection of L onto LIA is not order continuous. Theorem 23.8. For an Archimedean Riesz space L thefollowing statements hold: (i) l f L is a-laterally complete and A is a o-ideal of L, then L / A is an Archimedean a-laterally complete Riesz space. (ii) I f L is laterallj~complete and A is a band of L, then LIA is an Arckimedean laterallj~complete Riesz space.

Proof. (i) Let {ti,) be a disjoint sequence of positive elements of L / A . We can assume {u,,) E L+. Set u1 = u l , and inductively if u l , . . . p, have been defined, put w, = u l + ' . . + u,, and define

u,+

1

= Un+

1

- suP{41+1 A mw,:m

2 1>= u,+

1

- Pw,(un+1).

Observe that {u,) is a disjoint sequence of L+. Indeed, for m < n we have and hence urn I w,- 1, so P,_(u,) IPwn~ ~ ~ , A ~ , ~ = ~ ~ A ( ~ , - P ~ , , _P,,,,(~~,))EB,,,, , ( ~ , ) ) ~nBt,,,= ~ , A ( {O], ~ , -

that is, urnA u, = 8. Now put t i = sup(u,} in L, and note that 6, = ti, for all n and that the canonical projection of L onto L/A is a a-Riesz homomorphism (since A is a a-ideal). Hence D = sup (ti,,) holds in LIA, so that LIA is a-laterally complete. The Archimedeanness of L / A follows from the fact that A is a a-ideal. (ii) By Theorem 23.4(ii) A is a projection band of L and hence L / A is Riesz isomorphic to Ad. The result should now be immediate. Note. Theorem 23.4 was proven first by Veksler and Geiler [69, Theorem 8, p. 321 via the representation theorems of the Archimedean Riesz spaces; the elegant proof presented here is due to Bernau [13, Theorem 1, p. 3201. Theorem 23.7 is a generalization of a result of Fremlin [26, Corollary 1.13, p. 781. Theorem 23.8(i) as well as the example preceding this theorem can be found in [26, pp. 78-79].

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[Chap. 7

An important tool for the study of laterally complete Riesz spaces is the notion of the dominable set, which is introduced next.

Definition 23.9. Let L be a Riesz space, and let A be a nonempty subset of L'. Then A is said to be a dominable set i f for every 8 < u E L there exists 8 < v E L and a positive integer k satisfying (ku - a)' 2 u for all a E A. Note. The notion of the dominable set is due to Fremlin [26, Definition l.l(c), p. 721.

We shall see later (Theorem 23.22) that the dominable sets of an Archimedean Riesz space are precisely the subsets of L' that are order bounded in its universal completion. We continue with a simple but very useful lemma. Lemma 23.10. Let L be an order dense Riesz subspace of a Riesz space M , and let A be a nonempty subset of L'. Then A is a dominable subset of L ifand only i f A is a dominable subset of M . Proof. Assume first that A is dominable in M . Let 8 < u E L. Then there exist 0 < u* E M and a positive integer k such that (ku - a)' 2 u* for all a E A . Now pick 6 < v E L with 8 < u I u*, and note that (ku - a)' 2 u for all a E A , that is, A is a dominable subset of L. On the other hand if A is a dominable subset of L and 8 < u* E M , pick u E L with 8 < u I u*. Now choose 0 < v E L and a positive integer k satisfying (ku* - a)' 2 (ku - a)' 2 v for all a E' A. Hence A is a dominable subset of M and we are done. H

It should also be clear that subsets of dominable sets are also dominable. More important properties of the dominable sets are included in the next results. Theorem23.11. Let L be a Riesz space with the principal projection property, and let A be a dominable subset of L'. Then for each 8 < LI there exists v, 0 < v I u, and a positive integer n satisfying P,(a) nv for all a E A . Proof. Let 8 c u. Since A is a dominable subset, there exists a positive integer n and 8 < w satisfying

(*I

a A nu I nu - w

for all a E A.

Observe now that 8 < w I nu implies w A u > 8 and so t' = Pw(u) satisfies 8 < v. It now follows from (*) that f,(a) A no I nu - w for all a E A . Now for every a E A the element f = (P,(a) - nu)' satisfies f E B, and f E Bwd [since f I(P,(a) - nu)- 2 w > 81, and so f = 8. Hence P,(a) I nu

Sec. 231

171

THE LATTICE STRUCTURE

for all a E A. Now since u E B, implies P,(a) E B, for all a E A , we get P,(a) = P,(P,(a)) = P,(P,(a)) 5 nP,(u) = nu and we are done.

for all a E A,

I

Theorem23.12. Let L be a Riesz space with the principal projection property and let A be a dominable subset of L +. Then there exists a complete disjoint system {e,:i E Z} of strictly positive elements of L such that for each i E Z there exists a positive integer ni (depending upon i ) satisfying P,,(a) 5 niei for all a E A . Proof. Consider the collection of sets of the form {ej:i E Z} where ei > 8, ei A ej = 8 if i # j , and such that for each i E Z there exists a positive integer n, (depending upon i ) satisfying Pei(a)5 niei for all a E A. By Theorem 23.1 1 this collection is nonempty and by Zorn's lemma there exists a maximal disjoint set with respect to the above property, say {ei:i E I } . We claim now that {e,:i E I } is a complete disjoint system of strictly positive elements, i.e., {e,:i E = L. Indeed, if 8 < u E L satisfies u A e, = 8 for all i E I, then by Theorem 23.1 1 there exists 8 < u I u and a positive integer n satisfying P,(a) Inu for all a E A . But then by adding v to the collection {e,:i E I } we contradict the maximality property of the collection. This completes the proof.

We are now in the position to characterize the dominable sets of the Archimedean laterally complete Riesz spaces. Theorem 23.13. Let L be an Archimedean laterally complete Riesz space, and let A be a nonempty subset of L'. Then the following statements are equivalent: (i) A is order bounded from above in L. (ii) A is a dominable set of L. Proof. (i) * (ii) Assume that 8 I a I u holds for all a E A. Let w > 8. Since L is Archimedean, there exists a positive integer k such that v = (kw - u)' > 8. But then (kw - a)+ 2 (kw - u)+ = u holds for all a E A , and hence A is dominable. (Note that the lateral completeness is not needed here.) (ii) (i) According to Theorems 23.4 and 23.12, there exists a complete disjoint system {e,:iE I) of strictly positive elements with the property that for each i E I there exists a positive integer n, satisfying P,,(a) I niei for all a E A. Now since L is laterally complete, u = sup{n,e,:i E Z} exists and so by

[Chap. 7

LATERALLY COMPLETE SPACES

172

Theorem 2.15 we have a = sup{P,,(a):iE I } I sup{n,e,:iE I } = u

for all a E A.

Thus A is order bounded, and the proof is finished. Among the most remarkable properties of the laterally complete Riesz spaces are the ones included in the next theorem. Theorem 23.14. Let L be an order dense Riesz subspuce of an Archimedean Riesz space M . Then we have the following:

(i) If L is laterally complete, then L is full in M . (ii) If L is laterally and Dedekind complete, then L = M . Proof. (i) Let u E M . Since L is order dense in M and M is Archimedean, there exists a net {u,} c L+ with u, t Iu( in M (Theorem 1.14). But then {u,} is a dominable subset of M, and hence by Lemma 23.10 a dominable subset of L. Thus according to Theorem 23.13 there exists w E L satisfying u, I w for all CI. It now follows that IuI I w holds, so that L is full in M. (ii) Use part (i) and Theorem 2.2. Lemma 23.15. Let L be an order dense Riesz subspace of an Archimedeun Riesz space M . I f L is Dedekind complete, then every element of M + is the supremum of a disjoint system of elements of L + . Proof. Note that L is an ideal of M (Theorem 2.2). Now let 6 < u E M . Since M is Archimedean and L is order dense in M , there exists a net {u,} of L with 0 I u, t u in M. It follows that {u,} is dominable in L + and so by Theorem 23.12 there exists a complete disjoint system {ei:iE I } of L + such that for each i E I we have P,,(u,) I niei for all CI and some fixed positive integer n,. Now for each i E I define wi = supa{Pei(ua)}I niei E L + ,and note that {wi:i E I } is a disjoint set of L'. To complete the proof we shall show that u = sup{wi: i E I } holds in M. Observe first that Pei(u,) I u, I u for all s( and i implies wi I u for all i E I. Now assume that wi I u E M holds for all i E I . Then Pei(ua)I u holds u for all i and CI.But then by Theorem 2.15 we get u, = sup(P,,(u,):i E I} I for all CI,and hence u Iu. The proof is now complete. 0 We continue with an important extension theorem. Theorem 23.16. Let cp: L -, K he a normal Riesz homomorphism from a Dedekind complete Riesz space L into an Archimedean laterally complete Riesz space K . I f L is an order dense Riesz subspace of an Archimedean Riesz space M , then cp has a unique normal Riesz homomorphic extension from M into K .

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THE LATTICE STRUCTURE

In particular, it follows that ifcp is a normal Riesz isomorphism (into), then so is its extension. Proof. Let 0 < u E M . By Lemma 23.15 there exists a disjoint system {u,} of L+ with u = sup{u,} in M . Then {cp(u,)} is a disjoint system of K + (since cp is a Riesz homomorphism), and hence since K is laterally complete, u* = sup{cp(u,)) exists in K . Now, if (u,} G L+ satisfies sup{u,} = u in M , then for each fixed A we have t', = sup, {v, A u,) in L and hence from the order continuity of cp we get cp(v,) = sup,{cp(zi, A u,)} I u*. On the other hand if cp(u,) 5 v* holds in K for all 1,then from u, = sup,{u, A v,} and the order continuity of cp it follows that cp(u,) I u* holds for all a and hence u* = sup{cp(v,)} holds in K . Thus the element u* is independent of the system chosen from L+. So, we extend cp to M + by q ( u ) = u*. Observe that this extension of cp from M + into K + satisfies cp(u v) = cp(u) cp(v) for all u,u E M + . Thus by Lemma 3.1 cp can be extended linearly

+

+

to all of M . The rest of the proof is left as an exercise for the reader.

We continue with the definitions of universally and a-universally complete Riesz spaces. Definition 23.17. A Riesz space that is both laterally and Dedekind complete is called a universally complete Riesz space. Similarly, a a-Dedekind and a-laterally complete Riesz space will be called a a-universally complete Riesz space.

Note. Again we remind the reader that in the terminology of Luxemburg and Zaanen a laterally complete Riesz space is referred to as a universally complete Riesz space [46, Definition 47.3, p. 3231. Fremlin, inspired by Theorem 23.14(ii), calls a universally complete Riesz space an inextensible Riesz space [26, Definition l.l(a), p. 721, while Vulikh calls the universally complete Riesz spaces extended Riesz spaces [70, pp. 140-1441, As an application of Theorem 23.16 we have the following result. Theorem 23.18. Assume that two universally complete Riesz spaces L and M contain two order dense Riesz subspaces L , and M respectively, that are Riesz isomorphic, via n say. Then n extends uniquely to a Riesz isomorphism between L and M .

,,

Proof. Note first that since M , is order dense in M , n:L1 -, M is a normal Riesz isomorphism, and then that the ideal generated by L , in L, say A, is the Dedekind completion of L,. Next extend 71 to A + by n(f) = sup{n(u):u E L + ; 8 5 u ~ f if }f~ A' and then to all of A by n ( f )= n(f +) - n( f - ) for f E A ; note that this extension is a normal Riesz isomorphism (into). Now by Theorem 23.16, 7~ can be extended to a normal Riesz

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LATERALLY COMPLETE SPACES

[Chap. 7

isomorphism from L into M. To complete the proof it suffices to show that the above extension of n is onto. To this end let 0 < u* E M. Then there exists a net {u,} of M I + with 8 I u, t u* in M. Pick a net { u Z ) of L: with u, = n(u,) for each a. Now observe that (u,} is a dominable subset of M and hence of M I . Thus {u,} is a dominable subset of L , and consequently of L. Theorem 23.13 implies now that t ) I u, f u* holds in L for some u* E L, and therefore u, = n(u,) t n(u*)= u*, that is, 71 is onto. Next we continue with the definitions of the universal and a-universal completions. Definition 23.19. A universal (resp. a-universal) completion of a Riesz space L is a universally (resp. o-universally) complete Riesz space K having an order dense (resp. super order dense) Riesz subspace M that is Riesz isomorphic to L. (We shall think of L and M as identical.) It should be clear that only Archimedean Riesz spaces can have universal completions and by Theorem 23.18 any two universal completions of a Riesz space must be Riesz isomorphic. Next we discuss the relation between universally complete Riesz spaces and extremally disconnected topological spaces. It is well known that for every topological space X the set Cm(X)consisting of all extended realvalued continuous functions f on X for which the open set {x E X : If(x)l < cc}is dense in X is in general not a Riesz space under the ordinary operations [46, p. 2951; [70, p. 1171. If, however, the topological space X is extremally disconnected (that is, if the closure of every open set of X is also open), then C m ( X )forms a Riesz space under the ordinary operations, which is universally complete [46, Theorem 47.4, p. 3231; see also [50, Chapter 11, p. 291 and [70, Theorem V.2.2, p. 1261. Thus Cm(X)forms a universally complete Riesz space for every extremally disconnected topological space X . (It should be noted that the property that makes Cm(X)a Riesz space is the is a continuous extended real-valued function following: I f f : 0 + [ - CO,OO] defined on an open subset 0 of an extremally disconnected topological space X , then j has a unique continuous extension to the closure 6 of 0 ;see [46, Theorem 47.1, p. 3221.) The Hausdorff compact extremally disconnected topological spaces appear naturally in connection with Dedekind complete Boolean algebras. We shall recall briefly this relationship and we shall refer the reader for details to [46, Chapter 11; see also Exercise 1 at the end of this chapter. An ideal of a Boolean algebra 93 is a nonempty subset I of 93 such that u v u E I , if u,u E I and if it follows from w I u with u E I, that w E I. An ideal I is called a prime ideal if u A u E I implies u E I or u E 1. Let R denote the

Sec. 231

THE LATTICE STRUCTURE

175

collection of all proper prime ideals of LB. Then the collection of subsets of R of the form R, = {aE Q : u $ w ) ( u E 98)forms a basis for a topology on R, called the hull-kernel topology. Now if in addition 3 is a Dedekind complete Boolean algebra, that is, if every nonempty subset of LB has a supremum, then R with its hull-kernel topology is a compact, Hausdorff, and extremally disconnected topological space, called the Stone space of the Dedekind complete Boolean algebra 249. The following result holds: For a given HausdorfS, compact, and extremally disconnected topological space X , there exists a unique (up to a Boolean isomorphism) Dedekind complete Boolean algebra LB whose Stone space is precisely X . Now if L is an Archimedean Riesz space then the collection of all bands (ordered by inclusion) forms a Dedekind complete Boolean algebra 91L. The following basic theorem is due to Maeda and Ogasawara [47] and deals with the existence of universal completions. A proof can be found in [46, Theorem 50.8, p. 3401. Theorem 23.20. Let L be an Archimedean Riesz space and let R be the Stone space of .BL.Then K = Ca(R) is a universal completion of L. Since by Theorem 23.18 any two universal completions of a Riesz space are Riesz isomorphic, it makes sense to talk about “the universal completion.” The universal completion of an Archimedean Riesz space L will be denoted by L”. Note also that if L is an order dense Riesz subspace of an Archimedean Riesz space M , then the universal completion M” of M serves equally well as the universal completion of L, since obviously L “seats” order densely in M”. Thus the universal completion L“ is Riesz isomorphic to MUby Theorem 23.18. So, if L is order dense in M , then the universal completion L”“contains” M as an order dense Riesz subspace. More precisely, we have the following theorem. Theorem 23.21. I f an Archimedean Riesz space L is Riesz isomorphic to some order dense Riesz subspace of un Archimedean Riesz space K , then K is Riesz isomorphic to an order dense Riesz subspace of L” (and hence L and K have the “same” universal completion). Proof. Assume n is a Riesz isomorphism between L and an order dense Riesz subspace, say K of some Archimedean Riesz space K . Note that K” is the universal completion of K , , and then use Theorem 23.18 to extend n to a Riesz isomorphism between L” and K ” . The result now follows easily.

In terms of the universal completior, the following characterization of the dominable sets can be given.

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[Chap. 7

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Theorem 23.22. Let L be an Archimedean Riesz space, and let A be a nonempty subset of L+. Then the following statements are equivalent: (i) A is a dominable subset of L+. (ii) A is order bounded in L" (and hence sup A exists in L"). Proof. Apply Lemma 23.10 and Theorem 23.13. H

A sequential analog of the preceding theorem follows. Theorem 23.23. For a o-Dedekind complete Riesz space L the following statements are equivalent : (i) L is o-laterally complete (and hence o-universally complete). (ii) Every countable dominable subset of L+ is order bounded in L (and hence it has a supremum).

Proof. (i) (ii) Assume first that L contains a weak order unit e. Let {u,} be a countable dominable subset of L+. Then by Theorem 23.22 u* = sup{u,} exists in L". Observe that for each u E L + , w = sup{u, A v} exists in L (since L is o-Dedekind complete) and so since L is order dense in L", U * A V = W A U E L. Now if w, = ( ( n 2)e - u*)+~ ( n e u*)- ( n = 0,1,. . .), then by what was just noted {w,} c L+. Now proceeding as in the proof of Theorem 23.4(i) (and working in L"), we see that the projection of u* on e (which equals u*) can be written as u* A (f g ) where

+

f

= sup((2n

+ l)w,,:n

+

2 0}

and

g

= sup

+

{(2n 2)w,,,+ :n 2 0 } ,

which by the o-lateral completeness of L both belong to L. Thus u* = u* A ( f + g ) I f + g E L , so that 0 I u, I f + g holds in L for all n, and the proof in this case is finished. We shall show next that {u,} is included in a principal band Be of L. We can then apply the above arguments to the o-universally complete Riesz space Be to show that {u,} is order bounded in Be, and hence in L. To this end, define v1 = u1 and then inductively if v l , . . . ,v, have been defined, put w, = v1 + + v,, and define u,+ = u,+ - Pw,(un+l). As in the first part of the proof of Theorem 23.8, it can be shown that {v,} is a disjoint sequence of L'. Put e = sup{ v,} in L, and note that since 0 I v, I w, = sup{v,, . . . ,v,} I e, we have Pw,(un+1), v, E Be for all n and thus u,+ = PW,(un+1 ) + v,+ E Be for all n. Hence { u,} E Be, and we are done. (ii) 3 (i) Let {u,} be a disjoint sequence of L'. Then {u,} is order bounded in L", and therefore {u,} is a dominable subset of L". It follows that {u,} is a dominable subset of L and thus order bounded by hypothesis. Therefore, sup { u,} exists in L and the proof of the theorem is complete. H The previous theorem has a number of important consequences, some of which are included in the next results.

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177

Theorem23.24. For a Dedekind complete Riesz space L the following statements are equivalent:

(i) L is universally complete, that is, L = L". (ii) L is a-universally complete and has a weak order unit. Proof. (i) 3 (ii) Obvious. (ii) =-(i) Since L is Dedekind complete, L is an ideal of L" (Theorem 2.2). Now let 6' < e E L be a weak order unit of L and let 6' < u* E L". Since e is also a weak order unit of L", we have 6' I u* A ne t u*. Now { u* A ne} c L and so {u* A ne} is a dominable sequence of L. By Theorem 23.23 {u* A ne} is order bounded in L and thus u* E L. Hence L = L" and the proof is finished. H The next example shows that the Dedekind completeness cannot be dropped from Theorem 23.24. Example 23.25. Consider the Riesz space L consisting of all real-valued Lebesgue measurable functions defined on [OJ] (with the pointwise ordering). Then L is a-universally complete, has a weak order unit ( e ( x )= 1 for all x E [0,1] is a weak order unit of L),and it is neither laterally complete nor Dedekind complete. H Another example also shows that the existence of the weak order unit in statement (ii) of Theorem 23.24 cannot be dropped. Example 23.26. Consider an uncountable set X and let L = f f E

RX:( x E X :f ( x ) # 0 } is at most countable}, with the pointwise ordering. It can

be seen easily that L is Dedekind complete and a-universally complete without weak order units. Note, however, that L is not universally complete. H

We are now in the position to characterize the Riesz spaces that have a a-universal completion. Theorem23.27. For a Riesz space L the following statements are equivalent: (i) L is almost a-Dedekind complete. (ii) L has a a-universal completion (determined uniquely up to a Riesz isomorphism). Proof. (i) +- (ii) Assume first that L is a-Dedekind complete. Put U ( L + )= {u* E L":3{u,} E L+ with 6' s u, t u* in L"}, and define M = {u* - v*:u*,v* E U ( L + ) } .We have clearly the following Riesz subspace inclusions L E M E L". We shall show next that M + = U(L+).To this end let 6' I u* - v* with u*,v* E U(L+).Pick two sequences {u,} and {v,} of L+ with 6' I u, t u* and 6' I v, 7 v* in L". Note that since L is a-Dedekind

178

[Chap. 7

LATERALLY COMPLETE SPACES

complete u*

A

u is in L for each u E L + . Now 6

and u, - u* defined by

A

U,

- U*

A U,,

= U* A U, - U*

A U,

5

U* - U*

u, .+ u* - u* in L" imply that the sequence {w,} of L+ (0)

w, = sup{ui - u * A u i : i = 1,. . . ,n}

satisfies 0 Iw, t u* - u* in L". Thus u" - u" E U ( L + )and so M + = U ( L + ) . It can be proven now easily that M is a o-universally complete Riesz space that contains L as a super order dense Riesz subspace. Now if L is almost o-Dedekind complete, embed L super order densely in a 0-Dedekind complete Riesz space K and then embed K as above super order densely to a o-universally complete Riesz space. Thus L has a 0universal completion. The uniqueness of the o-universal completion can be derived easily by applying Theorem 23.23; the details are left to the reader as an exercise. (ii) (i) It follows immediately from the definition of the almost o-Dedekind completeness. Note. The preceding three theorems are due to Fremlin [26, pp. 73-76].

By a stepfunction s on a topological space X, we shall mean any function of Cm(X)for which there exists a collection { &: i E I ) of mutually disjoint closed-open subsets of X whose union is dense in X and such that s is constant on each F. In other words, s is a step function on X if it has a cixv, with each closed and open, representation of the form s = & n F = @ i f i # j , a n d w i t h t h e o p e n s e t L o = U ( K : i E I } denseinX. If the step function s has an at most countable representation, that is, if s = c,xv,, then it will be called a o-step fincrion. We shall denote by S"(X) (resp. by Sum(X))the collection of all step functions (resp. the collection of all o-step functions) on a topological space X.

CieI

c."=

Theorem 23.28. Let X be an extremally disconnected topological space. Then the following statements hold: (i) Sm(X)is an Archimedean laterally complete Riesz space. (ii) Sum(X)is an Archimedean o-laterally complete Riesz space. (iii) The following Riesz subspace inclusions hold: Sum(X)E S"(X) E Cm(X),with Sum(X)order dense in Cm(X). Proof. (i) Note first that Sm(X)is a vector subspace of Cm(X).Indeed, ifs and tare in Sm(X)with representations s = cixv,and t = bjxu,, then As = ilcixv, E Sm(X) for each ilE R and

Xiel

ciEI

cjeJ

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179

since U { v n U j : i E I , j E J }= ( l J { v : i ~ n ~(}U){ U j : j E J } ) is dense in X . For the Riesz space structure of Sm(X)note that sv t =C

2 max{ci,bj}Xvinu,E Sm(X)

id j d

and sA t =

C 1min{ci,bj}XvinujE Sm(X). is1 j c J

Since clearly Sm(X)is Archimedean, it remains to be shown that Sm(X)is a laterally complete Riesz space. To this end let {s,} be a disjoint system of positive elements of S m ( X ) .For each LX pick a representation for s,. Then for each LX collect all the closed-open sets I/ appearing in the representation of s, for which s,(x) > 0 for x E I/. Now the collection of all those closed-open sets forms a mutually disjoint family that we shall denote by { E: i E I } . Note that for each V;. there exists ci > 0 and exactly one c( such that cixb appears in the representation of s,. Put 0 = U{V;.:iE I ) and U = X - 0(closedcixvi Oxv. on the open dense set 0 u U . Note that s open) and let s = can be extended to a function of Cm(X)[46, Theorem 47.1, p. 3221 and hence s E Sm(X).Now it should be clear that s = sup{s,} holds in S m ( X )and this completes the proof of part (i). (ii) Proceed as in (i). (iii) The proof is straightforward and is left to the reader.

xitl

+

The next theorem shows the relation between Archimedean laterally complete Riesz spaces and the spaces of the form Sm(X),with X extremally disconnected. Theorem 23.29. Let L be an Archimedean laterally complete Riesz space. Then there exists a compact, HausdorJY; extremally disconnected topological space R satisfying thefollowing Riesz subspace inclusions: Sm(R)G L L Cm(R). Proof. Let K = Cm(R)be the universal completion of L, where R is Hausdorff, compact, and extremally disconnected. Since L has a weak order unit e, the embedding of L into Cm(R)can be taken to carry e to the constant function 1 on R. Now since L has the projection property [Theorem 23.4(ii)], xv E L for each closed-open subset I/ of R (Theorem 2.10). It now follows easily from the lateral completeness of L that Sm(R)G L E Cm(R). Examples of Archimedean laterally and a-laterally complete Riesz spaces will be presented next. Example 23.30. ( A n Archimedean laterally complete Riesz space that is not Dedekind complete.) By a theorem of Gleason [28, Theorem 3.2, p. 4861

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[Chap. 7

there exists a Hausdorff, compact, and extremally disconnected topological space X and an onto continuous function cp:X + [OJ] that does not carry any proper closed subset of X onto [O,l]. Note that cp E C(X) c Cm(X),and we shall show next that cp 4 Sm(X).Indeed, if cp E Sm(X),then cp must be constant on some nonempty proper closed-open subset V of X. But then rp maps the closed (actually compact) subset X-V onto [0,1], in contradiction with the properties of rp. Now if Sm(X)is Dedekind complete then Sm(X)= Cm(X)by Theorem 23.14(ii), in contradiction to cp 4 Sm(X).Thus Sm(X)is the desired example. (Observe also that Cm(X)is the universal completion of C[O,l]; the mapping T :C[O,l] + Cm(X),defined by T ( f )= f 0 cp for f E C[O,l], embeds C[O,l] order densely into Cm(X).) Example 23.31. (An Archimedean a-laterally complete Riesz that is neither a-Dedekind complete nor laterally complete.) If K is the Riesz space of Example 23.25 and Sm(X)is as determined in Example 23.30, then the Riesz space L = K x Sm(X)has the stated properties. Note. The proof that cp 4 Sm(X)in Example 23.30 is due to J. Quinn (personal communication). We devote the rest of this section characterizing the laterally complete Riesz spaces of the form RX.Note that two such Riesz spaces RX and RY are Riesz isomorphic if and only if X and Y are of the same cardinality (i.e., if there exists a one-to-one mapping from X onto Y).The characterizations of the finite dimensional RX spaces are given first. Theorem 23.32. For an Archimedean a-laterally complete Riesz space L the following statements are equivalent: (i) L is Riesz isonzorphic to some R". (ii) L admits an order unit. (iii) L admits a Riesz norm. Proof. (i) * (ii) Obvious. (ii) =. (iii) If e is an order unit of L , then p(u) = inf{A > 0:lul I Ae} for u E L is a Riesz norm on L. (iii) * (i) Let p be a Riesz norm on L. If {u,,} is a disjoint sequence of positive elements of norm one, then {nu,,} is also a disjoint sequence of L+ and so u = sup{nu,,} exists in L . But then we have the contradiction n I p(u) < 00 for all n. Thus every set of disjoint elements of L is finite and hence L is Riesz isomorphic to some R" (see Exercise 10 of Chapter 1). Theorem 23.33. For an Archimedean laterally complete Riesz space L the following statements are equivalent: (i) L is Riesz isomorphic to some RX. (ii) L admits a HausdorfS locally concex-solid Lebesgue topology.

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TOPOLOGIES ON LATERALLY COMPLETE SPACES

(iii) (L:)' = { u E L : 1p,1(lul) = 0 for all p E (iv) L is a discrete Riesz space.

=

181

{O).

Proof. (i) (ii) The topology of pointwise convergence on RX is a Hausdorff locally convex -solid Lebesgue topology. (ii) => (iii) Let z be a Hausdorff Lebesgue topology on L that is also locally convex. By Theorem 9.1 the topological dual L' of (L,r) is an ideal of L; and the result follows from the Hahn-Banach theorem. (iii) => (iv) Let 0 < u E L. Pick 0 < cp E L; with cp(u) > 0, and note that N , = { v E L :q(Ivl) = 0} is a band of L and so by Theorem 23.4(ii)L = N , 0 C,; where C , = Nqd.But C , admits a Riesz norm [namely p(v) = cp(lvl)] and so by the previous theorem C , is Riesz isomorphic to some R". Now let v be the projection of u on C,. Then there exists a discrete element w of C,, and hence of L, such that 6 < w I ZI I u. (iv) * (i) By Theorem 2.17 L is Riesz isomorphic to an order dense Riesz subspace of some RX.But then since L is laterally complete, the Riesz isomorphism must be onto and the proof of the theorem is complete. Note. The equivalence (ii) o (iv) of Theorem 23.33 was established in [60, Theorem 3.1, p. 2391. 24.

LOCALLY SOLID TOPOLOGIES ON LATERALLY COMPLETE RIESZ SPACES

In this section we shall study the locally solid topologies that the alaterally complete or laterally complete Riesz spaces can carry. We start with the following theorem. Theorem24.1. For a a-laterally complete Riesz space L the following statements hold: (i) Every disjoint sequence of L' converges to zero with respect to any locally solid topology on L. In particular, every locally solid topology on L satisjes the pre-Lebesgue property. (ii) If in addition L is Archimedean, then every locally solid topology on L satisjes the a-Lebesgue property. Proof. (i) For the first part note that if {u,,} is a disjoint sequence of L+, then so is {nu,,} and hence u = sup{nu,,} exists in L. The result now follows immediately from the relation 0 Iu, I n-'u for all n. For the last part apply Theorem 10.1. (ii) Let z be a locally solid topology on L and let u, 4 6' in L. If I/ is a solid r-neighborhood of zero, choose another r-neighborhhod U of zero with U + U 5 V and then E > 0 such that E U . Now, by Lemma 23.6

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[Chap. 7

there exists u E L+ such that 0 I u, I 2 - " u + w1holds for all n. Pick a u, I positive integer k with 2-"u E U for all n 2 k. But then we have 0 I 2-94 E U ~E U + U E V for all n 2 k, so that u, 5 0, and the proof is finished. H

+

We continue with more properties of the locally solid topologies on o-laterally complete Riesz spaces. Theorem 24.2. Let (L,T)be a Hausdorff locally solid Riesz space with the Fatou property. Then the following statements hold: (i) Every increasing z-bounded net of L+ is dominable. (ii) If L is o-laterally complete, then z is a Lebesgue topology. (iii) If L is universally complete, then (L,z) is z-complete.

u, t is z-bounded and that u > 0. Pick a Fatou Proof. (i) Assume 0 I z-neighborhood V of zero with u # r/: and then choose a positive integer n such that n-'u, E V for all a. Since V is solid and order closed, there exists 0 < w E L such that n- '11, A u I u - w for all ct (otherwise, if 0 I n- ' u , A u t u then u E V , a contradiction). But then (nu - u,)' = n(u - n-'u, A u ) 2 nw > 0 for all a, so that (u,) is dominable. (ii) Note that by Theorem 24.l(i) z is a pre-Lebesgue topology. The result now follows immediately from Theorem 11.6. (iii) By Theorem 13.4L is order dense in its topological completion i and by Theorem 23.14(ii)we get L = t,i.e., that (L,T)is z-complete. H It is natural to ask when a Hausdorff Fatou topology on a Riesz space can be extended to a Hausdorff Fatou topology on its universal completion. Clearly such a topology must be necessarily Lebesgue. The following theorem tells us what Hausdorff Lebesgue topologies on a Riesz space have such an extension; this theorem will be the key result for the discussion in this section. Theorem 24.3. For a H a u s d o r - locally solid Riesz space (L,T)with the Lebesgue property the following statements are equivalent:

(i) z extends to a Lebesgue topology on L". (ii) T extends to a locally solid topology on L". (iii) z is coarser than any Hausdor-a-Fatou topology on L. (iv) Every dominable set of L is z-bounded. (v) Every disjoint sequence of L+ is z-convergent to zero. (vi) Every disjoint sequence of L + is z-bounded. (vii) T h e topological completion t of (L,T)is Riesz isomorphic to L", that is, t is the universal completion of L.

TOPOLOGIES ON LATERALLY COMPLETE SPACES

Sec. 241

103

Proof. (i) =. (ii) Obvious. (ii) => (iii) Let z* be a locally solid extension of z to L" and let z' be a Hausdorff 0-Fatou topology on L. By Theorem 12.14 it is enough to show that z c z' holds on every component Nd of the carrier C , . So, let N = V,, where [ V , } is a normal sequence of solid z-neighborhoods of zero. Note that every disjoint subset of strictly positive elements of Nd is at most countable. Indeed, if A is a disjoint subset of strictly positive elements of Nd, then A = A,, where A , = ( u E A : u # V,!, and so if some A , is infinite it must contain a sequence {u,} that must be z*-convergent to zero in L" [Theorem 24.l(i)] and hence t-convergent to zero in L, contradicting u, $ V, for all n ; thus each A , is finite and hence A is at most countable. By Zorn's lemma there exists an at most countable complete disjoint system. Now since Nd admits a metrizable Lebesgue topology (its basis is { V , n Nd))it has the countable sup property and hence the restriction of z' to Nd, say zl, has the Fatou property. By Theorem 12.4 there exists a metrizable Fatou topology, say ,z, on Nd that is coarser than zl. Next consider (Ad,?,), and note that since Nd is order dense in fid(use Theorem 13.4 or 17.4),and the universal completion of Nd is the band generated by Nd in L", we have the Riesz subspace inclusions Nd E Ad E L" (see Theorem 23.21). Now since F,,, is a Frechet topology, it follows from Theorem 16.7 that z* restricted to Ad is coarser than F,. Hence z is coarser than z' on Nd and we are done. (iii) * (iv) First use Theorem 11.10 to extend z to the Hausdorff Lebesgue topology zb on La. Let A be a dominable subset of L. Then A is a dominable subset of Lb. By Theorem 23.12 there exists a complete disjoint system (ei:i E I } of strictly positive elements of Lb such that for each i E I , P,,(a) Iniei holds for all a E A and some positive integer ni. Let Bi be the band generated by ei in Lb and then use Theorem 3.2 to embed L order densely in nBi. But if zi denotes the Hausdorff Lebesgue topology of the restriction of z6 to Bi, then nziis a Hausdorff Lebesgue topology on nBiand hence its restriction to L is also a Hausdorff Lebesgue topology. The result now follows by observing that A (as embedded in n B i ) is nIt,-bounded and that n z i restricted to L is finer than z by our hypothesis. (iv) + (v) If {u,] is a disjoint sequence of L+, then so is {nu,).. Now since (nu,). is order bounded in L" it is dominable and hence by hypothesis z-bounded. Now let I/ be a z-neighborhood of zero that is also solid. Pick a positive integer k such that nu, E kV for all n. But then u, E I/ for n 2 k and thus u, 1,0. (v) + (vi) Obvious. (vi) * (vii) Let (L,F) be the topological completion of (L,z). Then L is order dense in L (Theorem 13.4),(t,?) satisfies the Lebesgue property, and L is Dedekind complete (Theorem 10.6).So, we have only to show that is

o:=

Ul=

e

[Chap. 7

LATERALLY COMPLETE SPACES

184

laterally complete. Let {ii,} be a disjoint sequence of ,'f and let V and W be solid z-neighborhoods of zero such that W + W G V. Since L is order dense in f, and ? is a Lebesgue topology, for each n there exists u, E L+ with f3 Iu, 5 ii, and 6, - u, E I T'. Observe that {u,} is a disjoint sequence of L+ and hence z-bounded. Pick a positive integer k with {u,} E kW and note that we have

ii,

= (6, - u,)

+ u, E kW + kW G kP

for all n.

Thus {fin} is ?-bounded and so every disjoint sequence of ' i is ?-bounded. Now assume that {ii,} is a disjoint system of f,'. Let A be the set of all finite subsets of the indexed set {a). For each A E A put C, = i,. Then 0 I5, 7 holds in f,, and we claim that { C,} is in fact a ?-Cauchy net. Indeed, if {C,) is not a ?-Cauchy net, then there is a solid z-neighborhood I/ of zero such that for any A E A there exist &,A2 E A with &,A, 2 A and with 6, CL2 $ P. Thus there exists a sequence {A,} E A with A,-E An+ and G, = CLn+ ,- CLn $ P for all n. But {G,} is a disjoint sequence of Lf and hence so is {nG,}. Thus by what was shown above {nGn} is ?-bounded and so there exists k with {nG,,} E k P . In particular, it follows that G, E V for sufficiently large n, a contradiction. Hence (6,) is a 5-Cauchy net and so 5, f ii holds in f, for some ii. But then 6, ii holds also in i.It now follows that 6 = sup{iiu} holds in t,so that f, is laterally complete. (vii) * (i) Observe that z^ is a Lebesgue extension of z to f, (Theorem 10.6).

xaE,

Remark. If a Hausdorff Lebesgue topology z on a Riesz space L extends to a Lebesgue topology z" on its universal completion L", then any basis { V } of zero for z consisting of Fatou z-neighborhoods of zero extends to a basis { V " }of zero for z" consisting of Fatou z"-neighborhoods of zero, where V" = {u*

E Lu:3{uu} E

I/ with f3 Iu,

t (u*l in L"}.

Note, in particular, that z" is determined uniquely. We are now in the position to state the main topological properties of laterally complete Riesz spaces.

Theorem 24.4. If a a-laterally complete Riesz space L admits a HausdorfS Fatou topology z, then any other Hausdorff locally solid topology on' L is finer than z. Proof. Assume that z is a Hausdorff Fatou topology and z* a Hausdorff locally solid topology on a a-laterally complete Riesz space L. Then by Theorem 24.l(i) z satisfies the pre-Lebesgue property and hence the Lebesgue property (Theorem 11.6). Likewise, by Theorem 24.l(ii) z* satisfies

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TOPOLOGIES ON LATERALLY COMPLETE SPACES

185

the a-Lebesgue property and hence the a-Fatou property. But by Theorem 24.1 z satisfies condition (v) of Theorem 24.3 and hence condition (iii) of the same theorem holds. Thus z E z* holds and we are done. w As an obvious consequence of the last result we have that the Hausdorff Fatou topologies on o-laterally complete Riesz spaces are uniquely determined. Theorem 24.5. A o-laterally complete Riesz space admits at most one Hausdorff Fatou topology, which must be necessarily Lebesgue.

The next theorem follows immediately from the preceding result. Theorem24.6. An Archimedean Riesz space can admit at most one Hausdorff Lebesgue topology that extends to its universal completion as a locally solid topology.

The following two theorems deal with metrizable locally solid topologies on a-laterally complete Riesz spaces. Theorem 24.7. If a a-laterally complete Riesz space L admits a metrizable locally solid topology z, then z is the only Hausdorff locally solid topology L can carry, and in addition z is a Lebesgue topology. Proof. Let z be a metrizable locally solid topology on a a-laterally complete Riesz space L. Then by combining Theorems 24.1 and 17.8, we infer that z is a Lebesgue topology. In particular, it follows that L must satisfy the countable sup property. Thus every other Hausdorff locally solid topology on L must be a Lebesgue topology, and the result follows from Theorem 24.5. Theorem 24.8. For a laterally complete HausdorfS locally solid Riesz space (L,z) the following statements are equivalent:

(i) z is metrizable (and hence uniquely determined). (ii) L has the countable sup property. Proof. (i) 3 (ii) By Theorem 24.7 z is a Lebesgue topology and so L must have the countable sup property. (ii) (i) Note first that z is a Fatou topology. Hence by Theorem 12.4 (since L has a weak order unit), there exists a metrizable locally solid topology z' on L, and by Theorem 24.7 we have z = z', that is, z is metrizable, and we are done.

The next result gives some conditions for the normal Riesz homomorphisms to be continuous.

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[Chap. 7

Theorem 24.9. Assume that z is a HausdorfSo-Fatou topology on a Riesz space L, T‘ a Lebesgue topology on a o-laterally complete Riesz space M , and that T : ( L , z )+ (M,z‘)is a normal Riesz homomorphism. Then T is continuous. Proof. If { V } is a basis for zero for T‘ consisting of Fatou neighborhoods, note that since T is a normal Riesz homomorphism, { T - ’( V ) }is a basis of zero for a Lebesgue topology, say z l , on L. Clearly T : ( L , z , )+ (M,z’) is continuous. Observe also that z1 (which is not necessarily Hausdorff) satisfies condition (v) of Theorem 24.3. Indeed, if ( u n j is a disjoint sequence of L’, then since T is a Riesz homomorphism { T(u,)} is a disjoint sequence of M + and hence by Theorem 24.l(i) T(u,) 1;6, so that u, 2 6. Now put N = n { T - ’ ( V ) :V E { V }}, and note that N is a band of L and that z1 induces a Hausdorff Lebesgue topology on Nd. Thus by Theorem 24.3 z1 must be coarser than z if both are restricted to Nd,that is, T restricted to Nd is continuous. Obviously T restricted to N is continuous and hence T restricted to the order dense ideal N 0 Nd is continuous. Now let {u,} c L+ with u, 6 and V,U E { V ) such that V + V E U . Since N 0 Nd is order dense in L, for each o! there exists u, E N 0 Nd with 6 5 u, 5 u, and T(u,) - T(u,) E V [we use here that z1 is Lebesgue and that T:(L,z,) --, (M,z’) is continuous]. But then u, It B holds in L and since {u,} c N 0 Nd, it follows T(u,) 2 0 and thus T(u,) E V for all o! 2 u l . Hence T(u,) = [T(u,) - T(u,)] + T(u,) E V

+VGU

for u 2 a l ,

so that T:(L,z)+ (M,z’) is continuous, and the proof is finished. The next result describes a basic property of the unique Hausdorff Fatou topology on a a-laterally complete Riesz space. Theorem 24.10. l f a a-laterally complete Riesz space L admits a HausdorfS Fatou topology z (necessarily unique and Lebesgue) then T is coarser than any HausdorfS locally solid topology on L and finer than any Lebesgue topology on L. Proof. The “coarser” part is just Theorem 24.4. To see the other part let z* be a Lebesgue topology on L , and consider the identity mapping l:(L,z)3 (L,z*). Now apply Theorem 24.9 to get T* E z. We now state a theorem giving some conditions under which an order continuous operator is continuous. Theorem 24.11. Let T be a positive, order continuous linear operator from a o-laterally complete Hausdor- locally solid Riesz space (L,T)into a locally solid Riesz space (M,z‘)with the Fatou property. Then T:(L,z)+ (M,Ti) is continuous.

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187

Proof. Let { V } be a basis of zero for z’ consisting of Fatou t’-neighborhoods. For each I/ E { V ]define V* = {u E L : T(lu1)E V } ,and note that { V * } defines a Fatou topology z* on L (and hence t* is Lebesgue) such that T : ( L , z * )-+ (M,z’) is continuous. Now since z is a-Lebesgue, it follows from Theorem 24.9 that the identity mapping Z:(L,z)-P (L,z*) is continuous and hence T:(L,z) (M,z’) must be continuous. -+

We can ask at this point whether there exist Archimedean o-laterally complete Riesz spaces admitting no Hausdorff locally solid topologies. The following example gives an affirmative answer to this question. Example24.12. Let L be the universal completion of C[O,l]. Assume that L admits a Hausdorff locally solid topology z. Then by Theorem 24.l(ii) the topology z is a o-Lebesgue topology and hence z restricted to C[O,l] is also a Hausdorff o-Lebesgue topology, contradicting the fact that C[O,l] does not admit Hausdorff o-Lebesgue topologies; see Example 8.2. This contradiction establishes that L does not admit any Hausdorff locally solid topology. 4

Regarding the quotient Riesz spaces of o-laterally complete Riesz spaces we have the following result. Theorem 24.13. Let L be a a-laterally complete Riesz space carrying a (necessarily unique) Hausdorff Lebesgue topology z, and let A be a band of L. Then z l A is the unique Hausdorff Lebesgue topologji on the o-laterally complete Riesz space LIA. Proof. Note that L / A is a-laterally complete (Theorem 23.8), z / A is Hausdorff (since every band is z-closed) and z / A is a Lebesgue topology by Theorem 8.11, and hence unique by Theorem 24.5.

For the a-laterally complete Riesz spaces admitting the unique Hausdorff Fatou topology the following interesting result holds. Theorem 24.14. If a o-laterally complete Riesz space L admits a Hausdorf Fatou topology z (necessarilyunique and Lebesgue), then there exists an order dense a-ideal of L such that any other Hausdorff locally solid topology on L induces the same topology on this a-ideal as z. Proof. By Theorem 11.6, z is a Lebesgue topology, and by Theorem 12.3, the carrier C , of z is an order dense a-ideal of L with the countable sup property. In particular, note that C , is a a-laterally complete Riesz space in its own right. Now if z* is any other Hausdorff locally solid topology on L , then z* satisfies the o-Lebesgue property (Theorem 24.1) and so, since C , has the

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countable sup property, T* restricted to C, must be a Lebesgue topology. The result now follows immediately from Theorem 24.5. 1 As an illustration of the preceding theorem we give the following example.

Example 24.15. Consider the universally complete Riesz space L = RX, for some uncountable set X . Let T be the Hausdorff Lebesgue topology of pointwise convergence on L. An easy verification shows that the carrier C, of z consists of all functions of L whose support is at most countable, that is, C, = {f E RX:{x E X : f ( x )# 0 ) is at most countable}.

According to Theorem 24.14, every Hausdorff locally solid topology on L induces the topology of pointwise convergence on C,. 1 Note. Theorems 24.1, 24.2, 24.5, 24.8, 24.9, and 24.11 were proven by Fremlin for universally complete Riesz spaces in [26]. The generalizations and the approach of the results in this section are due to the authors [7]. EXERCISES

1. Let 93 be a Boolean algebra. (i) Show that a proper ideal I of B is prime if and only if I is a maximal ideal. (ii) If an ideal I of B does not contain an element u, show that there exists a prime ideal J of 93 such that I c J and u # J . (iii) Let 51 denote the collection of all proper prime ideals of 93. For each ~ € 9 put 3 R , = { w ~ R : u q i c o } Show . that { 5 1 , : u ~ 9 3 }forms a basis for a Hausdorff, compact topology T on 51, called the hull-kernel topology. (iv) Show that A G R is closed-open with respect to z if and only if there exists u E B with A = 51,. (v) Show that (0,~) is extremally disconnected if and only if 93 is a Dedekind complete Boolean algebra. (vi) Show that if X is a compact, Hausdorff, and extremally disconnected topological space, then there exists a unique (up to a Boolean isomorphism) ) homeoDedekind complete Boolean algebra B whose Stone space ( 5 1 , ~ is morphic to X . 2. Show that if L is an Archimedean Riesz space, then the collection BLof all bands of L ordered by inclusion forms a Dedekind complete Boolean algebra whose Boolean operations are given by A A B = A n B, A v B = ( A + B)dd,and A' = Ad for all A,B E B L . 3. Let L be the Riesz space consisting of all real-valued, bounded everywhere, Lebesgue measurable functions on [0,1] with the pointwise

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ordering (everywhere). Determine the a-universal and universal completions of L. 4. Let L be an Archimedean a-laterally complete Riesz space. Show that if cp E L", then C , is a projection band of L. [Compare this with Exercise 14(ii) of Chapter I.] 5. Let L be an Archimedean a-laterally complete Riesz space. Show that L has no discrete elements if and only if L; = lo). 6. Complete the details of the proof of Theorem 23.27. 7. (Bernau [13]). Let L be an Archimedean Riesz space. The lateral completion LA of L is defined as the intersection of all laterally complete Riesz subspaces between L and L". Show that (LA)'= (L')A= L". 8. Let L = RX for a nonempty set X. Show that 15: = L-. (Hint: Use Exercise 17 of Chapter 1 or use Theorem 23.7.) 9. A set X is said to have a nonmeasurable cardinal if there is no measure p defined on all subsets of X that vanishes on the finite subsets of X and with p(X) = 1. Show that if a set X has a nonmeasurable cardinal, then for every cp E (R')" there exist xl, . . . ,x, E X and real numbers A,, . . . ,A, (all depending upon cp) such that q ( f ) = l , f ( x i )for all f E RX and derive that in this case ( R x ) r = (R')'. [For more about this, see W. A. J. Luxemburg, Is every integral normal? Bull. Amer. Math. Soc. 73 (1967), 685-688.1 10. Let L = RX,where X is a nonempty set and let T be the topology of pointwise convergence on L. For a sequence {u,} of L show the following: (i) u, 5 6 if and only if u, 3 6 in L. (ii) If z1 and z2 are two Hausdorff locally solid topologies on L, then u, 2 0 if and only if u, 3 8. 11. Let (L,t)be a a-laterally complete locally solid Riesz space, { V,} a normal sequence of solid z-neighbwhoods of zero and N = (?{ V,:n = 1,2,. . .}. Show that Nd is an Archimedean laterally complete Riesz space with the countable sup property. 12. Give an alternate proof of Theorem 24.9 by using Theorem 12.14. 13. Verify that the carrier of Example 24.15 is given by C, = {f E R X :{x E X:f(x) # 0} is at most countable}. 14. Let ( X , C , p ) be a finite measure space whose finite subsets have measure zero and let L be the Riesz space of all real-valued p-measurable functions defined on X (with the pointwise ordering everywhere). (i) Show that L is a a-universally complete Riesz space whose unique Hausdorff Fatou topology z is the topology of pointwise convergence. (ii) Let A = { f L ~ : f ( x )= 0 a.e.1. Show that A is a a-ideal of L and that L / A is a universally complete Riesz space.

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(iii) For each n put

Show that { V,} is a basis of solid order closed neighborhoods of zero for the unique Hausdorff Fatou topology on LIA. (iv) What is zlA? Compare your last answer with Theorems 24.13 and 14.3. 15. If z is a locally solid topology on a o-laterally complete Riesz space, show that C, is a o-laterally complete Riesz space with the countable sup property and that z restricted to C, is a Hausdorff Lebesgue topology. 16. Show that a Hausdorff locally solid topology z on a o-laterally complete Riesz space is Lebesgue if and only if C, is z-dense. 17. Let (L ,t) be a Hausdorff 0-laterally complete locally solid Riesz space. Show that L admits a Hausdorff Fatou topology (necessarily unique) if and only if C, is order dense in L. 18. Give an example of a o-universally complete Riesz space with a Hausdorff locally solid topology that is not Lebesgue. 19. Let X be a nonempty set. If every Hausdorff locally solid topology on RX is Lebesgue, show that X has a nonmeasurable cardinal. (Hint: If p is a finite measure on all subsets of X and vanishing on the finite sets, then the Hausdorff locally solid topology on RX generated by the Riesz pseudonorms

is not a Lebesgue topology.) OPEN PROBLEMS

1. Is every Hausdorff locally solid topology on a universally complete Riesz space, necessarily Lebesgue? (See [26, Section 3, pp. 84-88], where this question is studied in connection with the nonmeasurable cardinal problem.) 2. If a universally complete Riesz space admits a Hausdorff locally solid topology does it follow that it must admit a Hausdorff Lebesgue topology also?