Chapter 1 Distributive Lattices and Normed Function Spaces

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

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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).