Title: Natural compactifications of lattices
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Title: Natural compactifications of lattices
Physical Description: vi, 119 leaves : illus. ; 28 cm.
Language: English
Creator: Smith, Linda Joe Wahl, 1944-
Publication Date: 1969
Copyright Date: 1969
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Subject: Lattice theory   ( lcsh )
Topology   ( lcsh )
Mathematics thesis Ph. D
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Thesis: Thesis - University of Florida.
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NATURAL COMPACTIFICATIONS OF LATTICES















By
LINDA JOE WAHL SMITH














A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY















UNIVERSITY OF FLORIDA
1969






















































UNIVERSITY OF FLORIDA


3 1262 08552 2935


































To Nevins













ACKNOWLEDGMENTS


The author wishes to extend her sincere and grateful appreciation

to Dr. G. E. Strecker and Dr. G. A. Jensen for their helpful suggestions

in the preparation of this paper, and to acknowledge Dr. J. de Groot,

Dr. W. E. Clark, and Dr. Gale Nevill for serving on her Supervisory

Committee.

To her husband, Nevins, and her parents, Mr. and Mrs. J. K. Wahl

and Mrs. Nevins Smith, she extends many thanks for all the love and

moral support they have given over the years and especially during her

work on this paper. For typing assistance, she is very much indebted

to Mrs. Margaret Parramore.














TABLE OF CONTENTS


Page

. iii


ACKNOWLEDGMENTS . . . .


KEY TO SYMBOLS . . . . . . .


v


1. INTRODUCTION. . . . .... . . . . . . . 1

2. PRELIMINARIES . . . .. . . . . . . .. 5

3. PRODUCTS OF SUPEREXTENSIONS . . . ... . . . .. 15

4. LATTICE SUPEREXTENSIONS . . . ... . . . . 20

5. THE COMPLETION BY CUTS. . . . .... . . . .. 29

6. TOPOLOGICAL LATTICE COMPLETIONS . . . . . . .. 48

7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS. . . .. .68

8. EXAMPLES. . . . . . .... . . . . . . 86


APPENDIX I KAUFMAN'S ORDERABLE COMPACTIFICATIONS. ... . .111

APPENDIX II LATTICE CLOSURES. . . . ..... . . 113


BIBLIOGRAPHY. . . . . .. . . . . . . . . 117


BIOGRAPHICAL SKETCH . . . . . . . . .


* 119













KEY OF SYMBOLS


X ............ a topological lattice (2.8 (iii))

XD(a) = {X'E X: x : a} and X (a) = {x E X: a s x}

D ........... (2.3 (ii)) the collection of all closed ideals of X

FI ............ (2.3 (ii)) the collection of all closed dual ideals of X

r = D U rI

TD ........... (2.3 (iii)) the collection of all principal ideals of X
together with X

TI ........(2.3 (iii)) the collection of all principal dual ideals
of X together with X

T = TD U T .. (4.1 (iii)) is the usual subbase for X

Z .......... an arbitrary admissible subbase for X (4.2 (i))

AX .......... (2.15) the superextension of X with respect to E

S (or sometimes S* to avoid confusion) ... (2.14) the set of all
maximal linked systems of E which contain S

+ (or sometimes E*) = {S+: S E Z}. (2.14)

e ........... (2.16 (v)) the natural embedding of X into X X

B X .......... (2.17) the topological closure of e [X] in X X
(called the de Groot compactification of X with respect
to Z)

n ........... the collection of arbitrary intersections of Y

2 ........... the power set of X





D(A) = n{X (a): A c X (a)} where A E 2X (5.3)

X............ (5.2) the completion of X by cuts

d: X X ..... (5.4) the map defined by a -+ D({a})

= {X (d(a)): a E X} U{X (d(a)): a E X} U{X} (5.7)

wX ........... the Alexandroff one-point compactification of X

z = Zla Zb (6.1)

Z2 = 2a r Z2b (6.1)

T = {YD(i(a)): a E X} V{Y (i(a)): a e X} U{Y} (where Y = i[X]I) (6.6)

E ) ......... (6.15)














1. INTRODUCTION


Armed only with the definition of compactification, and asked

to compactify the open disc in the plane, a novice topologist would

most likely simply add its boundary. However, among the "standard"

methods which have been developed for obtaining compactifications of

spaces (Stone-Cech, Alexandroff, Freudenthal, Wallman, etc.) not one
1 v
yields this seemingly "natural" compactification. The Stone-Cech

compactification (although it has nice mapping properties) adds so many

points that the disc's "intrinsic character" is not preserved, and it

is in some sense "lost" in the new space. At the other extreme,

the Alexandroff compactification does not add enough. What we seek,

then, is a method for producing compact extensions which yields

"natural" compactifications in the sense that the closed interval [a, b]

"naturally" compactifies the open interval (a, b), the extended real

line "naturally" compactifies the rationals, the closed disc "naturally"

compactifies the open disc,and the Hilbert cube "naturally" compactifies

Hilbert space. In this dissertation we define and investigate such a

method.


Frink [8] has developed a Wallman-type method for obtaining
many compactifications for a given space. It has been shown by
Alo and Shapiro [2] that the compactification of the open disc by the
closed disc is of Wallman-type.







The method which we use is due to de Groot. For a given space,

this method produces compact extensions (called superextensions) and

compactifications (called de Groot compactifications) which are de-

pendent upon both the space and a chosen subbase for its closed sets.

Hence,in general, for each space there may be many superextensions

and de Groot compactifications. (See Chapter 2 for details.) In

particular, for locally compact Hausdorff spaces, it is possible (by

astutely choosing the subbases) to obtain both the Stone-Cech and the

Alexandroff compactifications as de Groot compactifications. Another

aspect of the versatility of the de Groot compactifications is that

they (unlike any of the "standard" compactifications mentioned above)

are productive. Specifically, we will show in Chapter 3 that the

topological product of any collection of de Groot compactifications

:(respectively, superextensions) of a family of spaces is homeomorphic

with a natural de Groot compactification (respectively, superextension)

of the topological product of the family.

In order to obtain the desired "natural" compactifications, we

consider only topological products of topological lattices (i.e., lattices

with the interval topology) and restrict our attention, when dealing

with lattices, to those superextensions which arise with respect to

certain "admissible" subbases. This restriction is not overly severe

since, as can be seen in Chapter 8, for a given topological lattice

there are, in general, many "lattice superextensions" (i.e., superex-

tensions with respect to admissible subbases). In Chapter 4 we will








show how the order of the original lattice can be extended to a

partial ordering of any lattice superextension in such a way that

the superextension becomes a complete lattice (in the interval

topology). Thus each lattice superextension of a topological

lattice turns out to be not only a topological compactification, but

also a lattice completion--hence a very "natural" type of compacti-

fication.

In Chapter 5 we consider the algebraic structure of a

topological lattice X and show that its completion by cuts can be

realized as a particular lattice superextension. In Chapter 6 we

;attempt to algebraically characterize other lattice superextensions of

X. This is done using the notion of topological lattice completions

of X, these being certain complete lattices containing X as a sublattice.

We show that all lattice superextensions of X are in fact topological

lattice completions of X and that those topological lattice completions

of X satisfying a particular "density" condition are lattice super-

extensions of X.

In [12] Kaufman considers (totally) orderable compactifications

for totally ordered spaces X; i.e., compactifications Y of X for which

the given topology on Y is the same as the order (interval) topology

on Y. Among other things,he establishes the existence of a "largest"

and a "smallest" orderable compactification for each totally ordered

space. In Chapter 7 we consider the notion of "domination" for lattice

superextensions of topological lattices and generalize Kaufman's results

to these compactifications of lattices. Hence each topological lattice








has a lattice compactification which is maximal relative to mapping

properties (analogous to the Stone-tech compactification for more

general spaces) and a lattice compactification which is minimal rela-

tive to mapping properties (analogous to the Alexandroff compactifi-

cation for locally compact Hausdorff spaces).

Chapter 8 consists of examples of lattice superextensions of

various kinds of lattices. Also, examples are given which show the

necessity of certain restrictions which are made in the text.













2. PRELIMINARIES


The purpose of this chapter is to state most of the basic

definitions, propositions, and theorems which will be needed in the

following chapters. Algebraic terms which are not specifically defined

here can be found in Birkhoff [ 5] and topological terms in Kelley [13].



DEFINITION 2.1. (i) A set X is called partially ordered if

there is a binary relation 5 on X which is reflexive, antisymmetric,

and transitive.

(ii) If X is a partially ordered set with more than one element

in which each pair of elements a and b have a supremum (denoted by

a v b) and have an infimum (denoted by a A b), then X is called a

lattice.

(iii) If X is a lattice and each subset C of X has a supremum

in X and an infimum in X, then X is said to be complete.2

(iv) If X is a lattice and a, b E X, then a and b are said to

be comparable if either a < b or b < a.

(v) If X is a lattice and a, b E X, then a < b will mean that

a 5 b and a / b.



1We consider only non-trivial lattices in order to avoid the
possibility that vX = AX.

Throughout we will use the standard conventions that for a
lattice X, A0 = vX and v0 = AX, and for a topological space X, f0 = X.

5







(vi) X is said to be totally ordered if for every pair of

elements x and y of X, x and y are comparable; i.e., x y or y x.



To avoid confusion when more than one lattice is considered,

the symbols "v, A and 5" may be subscripted such as "vX, AX, and :X"

to denote supremum in X, infimum in X, and order in X, respectively.



DEFINITION 2.2. Suppose that X and Y are lattices and

i: X Y is a function.

(i) The map i is called a lattice homomorphism whenever i

satisfies the following: For each non-void subset C of X, if v C

exists in X, then v i[C] exists in Y and i(vXC) = v i[C]. Dually, if

A C exists in X, then A i[C] exists in Y and i(A C) = A i[C].

(ii) The map i is called a lattice inclusion and X is called

a sublattice of Y if i is an injective lattice homomorphism.

(iii) The map i is called a lattice isomorphism and X is said

to be lattice isomorphic with Y if i is a surjective lattice inclusion.


DEFINITION 2.3. Let X be a lattice and I a non-void subset of

X.

(i) I is-increasing if a E I whenever a E X and i a for some

i E I. I is decreasing if a E I whenever a E X and a S i for some i I.

(ii) I is an ideal of X if I is decreasing and a v b E I for

each pair a, b E I. I is a dual ideal if I is increasing and a A b E I
3
for each pair a, b E I.



We do not consider 0 to be an ideal or dual ideal in order to
simplify the statements of proofs.







(iii) I is a principal ideal of X if there is some fixed a E X

such that I = {x E X: x 5 al. I is a principal dual ideal of X if

there is some fixed b E X such that I = {x E X: b 5 x}.



PROPOSITION 2.4. Let X be a lattice.

(i) Arbitrary intersections of ideals of X are either void or

ideals of X, and arbitrary intersections of dual ideals of X are either

void or dual ideals of X.

(ii) Finite intersections of ideals of X are ideals of X, and

finite intersections of dual ideals of X are dual ideals of X.

(iii) If 1 is a finite collection of ideals of X such that

X i f, then Uyl X, and dually if is a finite collection of dual

ideals of X such that X then U 9 X.


PROOF. (i) Suppose C is a collection of ideals of X such that

n X 0. Let I = n If b E I and a e X with a < b, then since b E E

for each E E ; and each E is decreasing, a E E for each E E Hence

a-E I; i.e., I is decreasing. Finally, if a, b e I, then since a, b e E

for each E E C and each E is an ideal, it follows that a v b E E for

each E, so that a v b E I. By Definition 2.3 (ii), I is an ideal of X.

A dual argument shows that the intersection of an arbitrary collection

of dual ideals of X is either void or a dual ideal of X.


(ii) Suppose II, 12, 1 -, I are ideals of X. By definition,

I. I 0 for each j = 1,2,-.-,n so there exists x. E I. for each j. It

follows that A{x.: j = 1,2,-' ,n} e I = {I.: j = 1,2,---,n} since
] J








each I. is decreasing, and hence I / 0. By (i) above, I is an ideal

of X. A dual argument shows that the intersection of a finite collection

of dual ideals of X is a dual ideal of X.


(iii) Since X g for

x = v{xF: FE % }. If x E

F is decreasing and xF < x,

x E X U and so U X.


each F E there exists xF E X F. Let

U-', then x e F for some F E,, and since

we see that xF E F--a contradiction. Thus

A dual argument proves the other statement.


PROPOSITION 2.5. If X and Y are lattices and i: X Y is a lattice

homomorphism, then i preserves the order of X.


PROOF. Let i: X -+ Y be a lattice homomorphism and suppose

x, y E X such that x
lattice homomorphism i(y) = i(x vX y) = i(x) v i(y); therefore

i(x) < i(y) and so i preserves the order of X.


DEFINITION 2.6.

topological closure of A

confusion seems likely.


Let X be a topological space and A c X. The
in X will be denoted by A or when no
in X will be denoted by A or A when no


DEFINITION 2.7. Suppose X and Y are topological spaces and


f: X -+ Y is a function.

(i) f is called

continuous, and a closed

(ii) f is called

embedding and f- X = Y.


an embedding of X into Y if f is injective,

map onto its image.

a dense embedding of X into Y if f is an







(iii) f is called a homeomorphism and X is said to be

homeomorphic with Y whenever f is a surjective embedding.

(iv) Y is called a compactification of X if Y is compact and f

is a dense embedding.



DEFINITION 2.8. Suppose that X is a lattice.

(i) A closed interval of X is a subset A of X such that either

A is a principal ideal of X, A is a principal dual ideal of X, or A is

the non-void intersection of a principal ideal of X and a principal

dual ideal of X. (In the latter case, there exist a, b E X such that

A = {x E X: a x 5 b}.)

(ii) The interval topology on X is the topology on X which has

all closed intervals of X as a subbase for the closed subsets of X.

(iii) A lattice X will be called a topological lattice when it

is also considered as a topological space having the interval topology.

(iv) E c 2 will be called a subbase for X when E is a subbase

for the closed sets of the topological lattice X and X e E.



The following proposition is due to Frink [9].


PROPOSITION 2.9. If X is a topological lattice, then X is compact

if and only if X is complete.



DEFINITION 2.10. Suppose that X and Y are topological lattices

and f: X Y is a function. f is called an iseomorphism and X is said

to be iseomorphic with Y if f is both a lattice isomorphism and a

homeomorphism.







The remainder of this chapter contains definitions and some

fundamental results concerning superextensions of T1-spaces. These

results are due to de Groot, Jensen, and Verbeek [ 7 ]. It is within

the framework of superextensions that we will study lattices in the

following chapters.


DEFINITION 2.11. Suppose that X is a T1-space

subbase for the closed subsets of X.

(i) If A and B are subsets of X, then A and B

screened by a subcollection Y of Z and is called a sc

B if UV = X and no member of Y meets both A and B. Ti

and y are screened by T if {x} and {y} are screened by

x E X and a subset A c X are screened by V if A and {xE

by '.


and Z is a



are said to be

:reen for A and

Jo elements x

', and an element

}are screened


(ii) E is called a T -subbase for X if for each x e X,

r{S e Z: x E S} = {x}, and whenever x E X and S E Z with x J S, there

is some T E with x E T and T n S = 0.

(iii) E is said to be weakly-normal if each pair of disjoint mem-

bers of E can be screened by a finite collection from E.

(iv) A non-void subcollection X of E is said to be linked if

each pair of elements of / has a non-void intersection; X is called a

maximal linked system of E if X is maximal in E with respect to the

property of being linked.


The next two propositions are found in [7 ]. Their proofs are

included only to acquaint the reader with the terminology.








PROPOSITION 2.12. A snace X is Tl if and only if there is a

T -subbase for the closed sets of X.


PROOF. Suppose that X is a T -space and let E be the collection

of all closed sets of X. Clearly E is a subbase for the closed subsets

of X. Let x e X. Since X is T1, {x} E E, and it follows that

{x} c n{S E Z: x e S} c {x}; i.e., n{S E E: x E S} = {x}. If x E X

and S E E with x S, then clearly {x} E E, x E {x}, and {x} S = 0.

Hence E is a T -subbase for X (2.11 (ii)).

Conversely, suppose that E is a T -subbase for X. Let x E X.

To see that {x} is closed, let y E X such that y d {x}. Hence

y (1{S E E: x E S} since E is a T -subbase for X. Thus there is some

closed set S E E with {x} c S and y V S. It follows that {x} is a closed

subset of X.



PROPOSITION 2.13. If E is a subbase for the closed subsets of

a T1-space X, then E is a T -subbase for X if and only if for each x E X,

{S E E: x E S} is a maximal linked system of E.


PROOF. Suppose that E is a T -subbase for X and x E X. Let

S= {S E: x E S}. Clearly /f is a linked system of E. If T c E and

T i {, then x / T. Thus since E is a T -subbase for X there is some

S E such that x E S and T ( S = 0. It follows that S E : and so

U {T} is not linked; therefore fH is a maximal linked system of E.

Conversely, suppose now that E is a subbase for the closed

subsets of a T1-space X such that {S E: x E S} is a maximal linked

system of E for each x E X. Let x be a fixed but arbitrary point of X.







Clearly {x} c n{S E Z: x e S}. If y E X and y / x, then it follows

from the fact that X is a T1-space that there is some T E E such that

x E T and y V T. Therefore y V f{S E E: x e S} and so

({S E Z: x e S} = {x}. Finally, let x E X and T E Z with x V T. Thus

T i {S E E: x e S} and since {S E : x e S} is a maximal.linked

system of E, there is some S E E such that x E S and S n T = 0. Hence

by Definition 2.11 (ii), E is a T -subbase for X.



NOTATION 2.14. Let E be a T -subbase for a space X.

X X will denote { c : K is a maximal linked system of E}.

For each S E S+ will denote {Y E X X: S eZ}, and E+ will

denote {S : S E Z1.



DEFINITION 2.15. Suppose that E is a T -subbase for a space X.

The set X X equipped with the topology which has ZE as a subbase for

the closed sets is called the superextension of X with respect to E.



The following proposition states some results found in [7 1

which follow readily from the definitions. The proof is thus omitted.


PROPOSITION 2.16. Suppose that Z is a T -subbase for the space X.

(i) If S, T E E with S f T = 0, then S+ nf T = 0.

(ii) If E E X and S E E such that / U {S} is linked, then S .

(iii) If c is a linked system of E, then o is contained in some

maximal linked system of E.

(iv) If X E X X and CZ / 0, then f x = {x} for some x e X.








(v) If e : X X X is defined by e (x) = {S E : x e S},

then e is an embedding of X into X X, and e [S] = S+ r e [X] for

each S 5 E.

(vi) The space X X is TI.

(vii) The space X X is compact.

(viii) If S and T are in E and S U T = X, then S UT = X X.



DEFINITION 2.17. Let E be a T -subbase for a space X.

(i) Let 8 X denote the topological closure of e EX] in X X;

this will be called the de Groot compactification of X with respect to

E. Note that {B X n1 S : S E Z} is a subbase for the closed subsets

of B X.

(ii) If;f c E, then o is said to he prime if and only if

whenever S1, S2, **-, Sn E such that U{S.: i = 1,2,---,n} = X, there

exists some j E {l,2,***,n} such that S. e .

(iii) A subset e of E is said to be centered if each non-void

finite subcollection from Z has a non-void intersection; is called

a maximal centered system of E in case 9 is maximal in E with respect

to the property of being centered.



It should be clear from Definition 2.7 (iv) that 3 X is indeed

a compactification of X. However, & X is not necessarily Hausdorff

even if X is Hausdorff.


PROPOSITION 2.18. Suppose that E is a T -subbase for a space X.








(i) If q is a finite subcollection of Z, then Uf = X if

and only if U{B1XT +: T E } X.

(ii) If Z is weakly-normal, then B X is Hausdorff.

(iii) If e E X X, then d~ 8 X if and only if Z is prime.

(iv) If ; is any maximal centered system of E, then is prime.

*(v) If E is weakly-normal and 'e A XX, then / e B X if and

only if Z contains a maximal centered system of E.



The proof of Proposition 2.18 can be found in [ 7] Chapter 1,

Proposition 1.5 through Proposition 1.9.

Propositions 2.16 and 2.18 together with the fact that a T1-space

is completely regular if and only if it can be embedded in a compact

Hausdorff space outline a proof for the following statement from [ 6]:

A T1-space X is completely regular if and only if there exists

a weakly-normal T -subbase for the closed subsets of X.



EXAMPLE 2.19. Suppose that X is a completely regular T1-space

and E is the collection of all non-void zero sets of X. Then E is a
V
weakly-normal T -subbase for X, and E X is the Stone-Cech compactification

of X. (This follows from Proposition 2.16 (i) and Theorem 6.5 of [10].)














3. PRODUCTS OF SUPEREXTENSIONS


In this chapter we investigate the productivity of superexten-

sions of T1-spaces and show that a natural superextension of the product

of an arbitrary family of T1-spaces is the product of the superextensions

of the family. We also show that the corresponding de Groot compacti-

fication of the product is the product of the de Groot compactifications.

Throughout this chapter we let {Ya }aA be a family of T1-spaces, Ea

be a T -subbase for the closed sets of Ya, Y be the topological product

of the family {Y } ,E and for each a E A, a : Y Y be the projection
a aEA a a
function. Also without loss of generality we may assume that for each

a E A, the set Y is itself a member of the subbase .
a a


-1
a a 1
for the closed sets of Y.


PROOF. It is clear that E is a subbase for the closed sets

of the product space Y. Suppose that y, z E Y and z E nr{7i [T]: w [T] E E
a a
-1
and y E na [T]}. Thus for each a E A, a (z) E rl{T: T E E and a (y) E T}.
a a a a
Since Ea is a T -subbase for Y a, ({T E Za: 7a (y) E T} = { a(y)};

therefore T (z) = T (y) for each a E A and hence z = y. Now let y E Y
a a
an -i
and iT [S] E with y i -1 [S]. If follows that a (y) 1 S, and
a a a







since E is a T -subbase for Y there is some T E E
a 1 a a
-1
S(y) 'E T and S ( T 0. Therefore, T [T] e E, y E
a a
a [S] n Ta [T] = 0. Hence E is a T -subbase for Y.
a a 1
2.11 (ii)).


such that

n I[T] and
a
(cf. Definition


In view of the above proposition we may consider the super-

extension of Y with respect to E.


PROPOSITION 3.2. Let P be the topological product of

family {X Ya For each (Va ) E P, the set A = {Q E E:
a aEA a aEA
a
a E A, na [Q] E } is a maximal linked system of E.
a a --_ _ _ _ _


the

for each


PROOF. Suppose that Q, R E V. For each a E A, a [Q], T [R] E ~a

so that IT [Q] n iT [R] 0; therefore, Q n R $ 0 so that I is linked.
a a
-1 -1
Suppose now that Eb [S] E and 0 U{7b [S]} is linked. If a E A and
-1 -1
a / b, then T [ S]] = Y If a = b, then T [ib [S]] = S. Thus for
ab a a b
-1 -1
each a E A it follows that na [i [S]] E ja and so 7i [S] C f; i.e., ff
a b a b
is a maximal linked system of E.



THEOREM 3.3. There is a homeomorphism f from the product P of

the superextensions {XA Y } aA to the superextension X Y of the product
a
Y. Furthermore, if g is the unique continuous map which makes each of

the diagrams

g
Y --> P


A
a a
)f I


a E a
e a


commute,








then the diagram


P




e




commutes.


PROOF. By Proposition 3.2 there is a function f: P X\ Y

defined by f((a ) a) = {Q E: for each a e A, E [Q] E X }.
a aEA a a
We will show that f o g = e For each y E Y, f(g(y)) = {Q E E: for

each a E A, a [Q] E T (g(y))} = {Q e E: for each a E A, Ta [Q] e (Ta ())}
a
= {Q e E: for each a E A, a (y) E a [Q]} = {Q E : y E Q} = e (y).

To see that f is injective, let 8 and X be members of P such
A A
that f / Thus there must be some b E A such that rb() ^ hb(A/), and
A A
since Tb(Y) and nb( () are maximal linked systems of Eb, it follows that
A A
there exist T T b(0) and S E Tb () such that S n T = 0. Hence

Tb [S] n T] = 0 and so = [S] and [T] cannot be contained in the
b b b b
same maximal linked system of E. Recall that if a E A and a / b, then

Tn [ [T]] = Y and if a = b, then na [nb [T]] = T. It follows that for
a b a a b
-1 A
each a E A, a[ b [T]] E Aa(). Hence by the definition of f,
a b a
-1 -1
[T [T] E f(9). By a similar argument we see that wb [S] E f(A). Thus

f(T) / f(A).

To see that f is surjective, let E e X Y. For each a E A, let

Xa = {r [Q]: Q E /}. For each a E A, a is clearly linked, and if
-1
S E E such that / U{S) is linked, then it. follows that U{Tr [S]}1
a a







-1 -1
is linked and hence a1 [S] E C. Therefore S = a [Wa [S]] E i so that
a a a
ta is a maximal linked system of Ea. Thus ( a)aEA E P and by the defi-

nition of f, f((a ) ) = {Q E E: for each a E A, w [Q] E X } "
a aEA a a
= {Q e E: Q E } = 4.

The continuity of f is an immediate consequence of the fact that
-1 -1 +
for each b E A and each TE E f- [(T [T]) ] = {(a ) a P:
b b a aeA
f(( ) ) E ( [T])+}= {( P: [T] f(( ) )}
a )aEA b E } afaEA [P: Tb a aEA
-1
= { )aA E P: for each a E A, a E[v [T]] E a} = {()aeA E P:
-1 A-1 A-1i +
T(= b b [T]) E b = A b {E x Y : T E }] = b[T ] which is a

closed subset of P.

Finally we show that f is a closed map. Using the fact that f

is injective and applying f to the above equality we know that for each

a E A and T E E f[ 1[T+]] = f[f- [( T]) ]] = (T- [T]) It follows
a a a a
A-1 + + +
from the fact that {- E[T ]: a E A, T+ E Z+} is a subbase for the closed
a a
sets of P and that f is bijective,that f is a closed map.

Consequently, XEY is homeomorphic with P.




Let Q be the topological product of the family {7 Y : a E A}.
L a
a
Thus Q is a subspace of the product space P.



COROLLARY 3.4. There is a homeomorphism h from the product Q

of the compactifications {8 Y } a to the de Groot compactification
------ LtE a aeA ---
a
B Y of the product Y.








PROOF. Let h be the restriction of-f in Theorem 3.3 to Q.

Since f is a homeomorphism we need only show that h[Q] = B Y.

To see that B Y c h[Q] let e B6 Y. Recall that in the proof

-of Theorem 3.3 we showed that for each a E A, a = {n [S]: S E-} is
a a
a maximal linked system of E and that f((.a)) )= To see that
a a aEA

(a)aEA E Q we will show that for each a E A, a is prime and Hence

a E Ya (cf., Proposition 2.18 (iii)). Let a E A and suppose that
a
{T T **', T } c E such that U{T.: i = 1, 2, ***, n} = Y It
1 n a 1 a
-1 -1 -1
follows that {T [T ], Tr [T ], .., [ T ]} c E and
a 1 a 2 a n -
-1
U{T- [T.1: i = 1, 2, ***, n} = Y. Since e is in B Y and hence is
a 1
-1
prime, there is some j E {1, 2, **', n} such that nT [T.] E Thus
a ]
-1
7T [n [T.]] = T. E e and so Z is prime.
a a ] a a

To see that h[Q] c Y, let (X ) E Q. Thus for each a E A,
E a aEA

SE 8 Y and so a is prime. Suppose now that Y is a finite subset
a E a a
a
of E such that UV = Y. Recall that each member of Y is of the form
-1
aT [T] for some a E A and T E Let B = {a E A: there is some T E E
a a a
-1
with n- [T] e '}. Suppose that for each b E B there is some
a

Yb E Yb U{T e b: nb [T] e T}. Then we can pick y E Y such that
-1
b(y) = yb for each b E B. It follows that for each 7a [T] E T,
-1
y i Ta [T] and so y E Y UT--a contradiction. Hence there must be
a
-1
some b e B such that U{T Eb : T b 1[T] e T = Y Since is prime,
-l
there is some S E E such that ib [S] E and S E b. It follows from
b b b
-1
the definition of f that l[I S] f((a ) ). Thus f((a ) ) is prime
b a aEA a aEA
and so h((a ) ) = f((a ) ) e B X. Consequently, h is a homeomorphism
a aEA a aEA C
from Q onto B Y.














4. LATTICE SUPEREXTENSIONS


For the remainder of the paper we restrict our attention to

superextensions of topological lattices with respect to T1-subbases

which consist of ideals and dual ideals. For these "admissible"

subbases we show that the superextensions are actually de Groot

compactifications (Theorem 4.6). Moreover, (with respect to a

"naturally induced" ordering), they are shown to be complete lattice

extensions of the original lattice (Theorem 4.11).



NOTATION 4.1. (i) For each a E X, XD(a) = {x e X: x : a) and

X (a) = {x E X: a x}. (D denotes decreasing and I denotes increasing.)

(ii) r = {C c X: C is a closed ideal of X}, r = {C C X: C is

a closed dual ideal of X}, and r = FD U I.

(iii) T = {X} U {XD(a): a E X} U {X (a): a E X}. (We refer

to T as the usual subbase for X.)

(iv) If E is a subbase for the closed sets of X and E c r, then

E = E F and ED D'



DEFINITION 4.2. (i) If E is a T -subbase for the closed subsets

of X and E c r,then E is called an admissible subbase for X.

(ii) A lattice superextension of X is any superextension of

the form X X where E is an admissible subbase for X.

20








PROPOSITION 4.3. If T c E c F, then E is an admissible subbase

for X.


PROOF. It is clear from the definition of the interval topology

(2.8 (ii)) that T is a subbase for the closed subsets of X. It is also

clear that any collection of closed subsets of X containing T is a

subbase for the closed sets of X. Hence E is a subbase for X. To

see that Z is a T1-subbase for X, let a E X. Thus since T c Z,

{a} c f{SE E : a E S} c {U E T: a E U} X (a) X (a) = {a};

i.e., n{S E E: a E S} = {a}. Finally, suppose a E X and S E Z such

that a / S. If S E ZD, then X (a) E ZI, a E X (a) and X (a) r S = 0.

Dually, if S E Z_, then XD(a) E ZD, a E X (a) and XD(a) n S = 0. Thus

E is a T -subbase for X (cf., Definition 2.11 (ii)) and hence is an

admissible subbase for X (4.2 (i)).



COROLLARY 4.4. T and F are admissible subbases for X.'




By the above corollary X X and X X are lattice superextensions

of X. Examples of these can be found in Chapter 8. Next we establish

that all lattice superextensions of X are in fact compactifications of

X which can be considered as complete lattices having X as a sublattice.



PROPOSITION 4.5. Let Z be an admissible subbase for X.

(i) ED and E are centered systems of E.

(ii) Every linked system of E is centered.







PROOF. (i) is immediate from Proposition 2.4 (ii).


(ii) Suppose that 9 is a linked system of E and

FI, F2, ***, Fn e If all of the F. belong to ED, then by (i) above

n{F.: i = 1, 2, ***, n} / 0. Similarly, if all of the F. belong to
1 1
EI, then n{F.: i = 1, 2, .**, n} / 0. In the remaining case we may

assume, by at most rearranging the subscripts, that there is some integer

m such that 1 5 m < n 1, F. E D for i = 1, .**, m, and F. EI for

i = m + 1, *.*, n. For each i e {1, ..., m} and j e {m + ***, n} we

can pick f. F. F F. because Z is linked. Since the members of D

are decreasing and those of I are dual ideals, it follows that for each

j e {m + 1, ., n}, gj = A{f. .: i = 1, *.., m} e F. r
J,] ]
(n{F.: i = 1, ***, m}). In the same way, since the members of E

are increasing and n{F.: i = 1, ***, m} is an ideal (2.4 (ii)), it
1
follows that v{g.: j = m + 1, .**, n} e (rn{F.: i = 1, .**, m}) n

(f{F.: j = m + 1, ..*, n}); i.e., is centered.




THEOREM 4.6. Each superextension of X with respect to an admis-

sible subbase Z is a de Groot compactification of X; i.e., X X = E X.


PROOF. Suppose that E is an admissible subbase for X. It is

clear from the definitions in Chapter 2 that EX XA X; therefore, we

need only show that X Xc_ C X. If X is a maximal linked system of E, then

by Proposition 4.5 (ii), is centered; i.e., e is a maximal centered

system of E. Thus Y is prime (2.18 (iv)), and so e E X (2.18 (iii)).








In view of Theorem 4.6 every lattice superextension of X is

actually a topological compactification of X. To see that it is also

a complete lattice with the interval topology and that it has X as a

sublattice we continue as follows.



DEFINITION 4.7. Suppose that E is an admissible subbase for X and

X, /m x X. We say that 5 X whenever 7i E c D E or, dually,

whenever n EI c n n E.



Since containment is a partial ordering, it follows readily that

4.7 defines a partial ordering on the elements of A X. Throughout the

remainder of this paper, X X will be understood to be partially ordered

according to Definition 4.7 whenever E is an admissible subbase for X.

The following theorem shows that X X is a complete lattice whenever

E is admissible.




THEOREM 4.8. Suppose that E is an admissible subbase for X.

Every non-void subset of X X has a supremum in X X and an infimum in

X X. Specifically, if 0 / C c X X and if


CD = D{ r D: C} and

CI = n{r n zx : E e C},







then


v C = CD U{S E ZI: CDU{S} is linked} and

A C = CI U{S e D: C U{S} is linked}.


PROOF. It is easily seen that ) = CD U{S E : CD U{S} is

linked} is a linked system of E. Suppose that T E ZD and U{T} is

linked. Since n ZI c n ZI for each eE C, it follows that

X U{T} is linked for each Z e C. Thus T E for each ;e C and so

T e CD; hence T E ). If T E I and B U{T} is linked, then CD U{T}

is linked and so by definition, T E Therefore ) is a maximal linked

system of Z. A dual argument shows that- = CI U{S e ZD: CI U{S} is

linked} is a maximal linked system of Z.

We continue by showing that P = v C. Since e n Zi c_ ~ Zi

for each E C, then by Definition 4.7, f for each f C. If

me A X and 7W< ', then there is some T in (T ZD) ( D r D) Since

T i n ZD = CD, there is some e C such that T k n D. It follows

from this that Le 4 M; therefore, = v C. A dual argument shows that

,=: A C.




PROPOSITION 4.9. Let E be an admissible subbase for X. Then

the natural embedding e : X -+ X X defined by x -- {S E : x e S} is a

lattice inclusion; thus X is a sublattice of X X.


PROOF. That e is injective follows from the fact that it is

a topological embedding (Proposition 2.16 (v)). To see that it







preserves supreme, suppose that C c X such that vX C exists in X. If

S E e (VX C) n ED, then vX C E S, and since S is decreasing it follows

that C c S. Therefore S E e (c) n ED for each c E C and by Proposition 4.8,

S E v e [C]. Hence by Definition 4.7, v e [C] 5 e (vX C). Conversely,

if SE v e[C] E ,D then S E e (c) f E-D and c E S for each c E C.

Thus C c S and,since S is a closed ideal of X, it follows from Frink

[ 9; Theorem 12] that vX C E S. Hence S E e(vX C), and by Definition 4.7,

e (vX C) < v e [C]; i.e., e (vX C) = v e [C]. A dual argument shows

that if C c X and AC exists in X, then e (X C) = A e [C]. By
X E X
Definition 2.2 (ii), eE is a lattice inclusion.



PROPOSITION 4.10. For any admissible subbase E of X, the

following hold: (Recall that for each S E, S = { E X X: S EZ}.)

(i) If S E E and T E E, then S+ = {aE X: v S } and
D I
T = {t E X X: A T+ < a}.

(ii) For each ~ZE X X, { E X X: t 5 } r = n{S+: S E m ) E }

and {E E AX: m < ;} = f){S+: S E m n Er }.
-1
(iii) For each flE A X, e- [{1 E X X: x r (L}] = (t rL E )
-1
and e.[{C E XX: m < }] = n(r) n E).

(iv) For each T EE D, v T = v e [T], and for each T E I,

A T = A e [T].

(v) If T E D and S E then S r T / 0 if and only if

A S+ v T.

(vi) If X E A X, T E E, and S E then T E if and only if

t 5 v T+, and S E t if and only if A S+ < .

(vii) For each E A X, v {A S+: S n E } =

= A {v S+: S E e n }.







PROOF. (i) Note first that by Proposition 4.8, v S+ X X

and S E v S. Clearly, if e E S, then s v S. Conversely, if

S v S, thenS SEcVS+ ED C C D and so ES The second

equality is proved dually.


(ii) Suppose that MV X X. It follows from Definition 4.7 that

{ x e X X: Z} c {S +: S E f n E }. Conversely, if E X X and

A /, then again by Definition 4.7 there is some T E (V r ED) (V \ D).

Thus / T+ and so {S +:S E 7M N The second equality is proved

dually.


(iii) Recall from Proposition 2.16 (v) that for each T E E,

T + e [X] = e [T]. Thus since eE is injective, it follows that for

each T E T = e- [e [T]] = e [T ] e X]] = el [T+ Let 1 X X.
-1
By the above remark and (ii) we see that e E[{ E X X: ~'}]
-1 + -1 +
= e [ n{S : S c: n n zD}] = n{e [S : S E n ED

= n{S: S e NI = n( in %). The other equality is proved similarly.


(iv) Let T E D. Again using Proposition 2.16 (v) we see that
+ +
e [T] cT+ and so v e [T] < v T. If S E v e [T] n Z then it follows

from Proposition 4.8 that T c S and so S E v T+; therefore by

Definition 4.7, v T < v e [T] and hence v T = v e [T]. A dual argument

shows that if T E Zi, then A T = A e [T].


(v) If there is some x E T r S, then by (iv) above

A S e e (x) v T Conversely, suppose that X = A S < v T By

(i) above Z E S and E T Thus S n T 0 and hence S n T i 0

(2.16 (i)).







(vi) Suppose that e X X and T E ED. By (i) above X < v T

if and only if E T and by definition, fc T if and only if T E .

Thus Te Zi if and only if < v T. Dually, if S E E then A S + <

if and only if S E .


(vii) Suppose that E X X. Let = v {A S+: S n },

and X = A {V T : T E X E D}. By (vi) above, for each S E X) I

and each T E E ZD, A S V
R n ZD. Thus by (vi), < v R+. Since for each S E ) (h ZI

A S 5 <, it follows from (v) that S f R f 0 for each S E Z C E;

i.e., (n~ E Z ) U{R} is linked. Therefore U {R} is linked, and so

R ce Hence by (v) and the definition of R R E J. By Definition 4.7,

Z <_ and hence P = X = 4.


THEOREM 4.11. Any lattice superextension of X is a complete

topological lattice having X as a sublattice.


PROOF. By Theorem 4.8 and Proposition 4.9 any lattice super-

extension of X is a complete lattice having X as a sublattice. By

Proposition 4.10 (i), the given subbase for a lattice superextension

X X is a collection of principal ideals and principal dual ideals of

X X, and by Proposition 4.10 (ii) each principal ideal and principal

dual ideal of X X is closed in the given topology on X X. Thus the

topology on X X generated by + is the interval topology on X X;

i.e., X X is a topological lattice.







We can conclude this chapter by showing that in the case where

X is totally ordered, all of the lattice superextensions of X are also

totally ordered.



PROPOSITION 4.12. If X is totally ordered and E is an admissible

subbase for X, then E is weakly-normal and X X is totally ordered.


PROOF. To see that E is weakly-normal, let S, T E E such that

S n T = 0. Either S u T = X, in which case {S,T} is a screen for S

and T, or there is some x E X -(S U T). In the latter case, since

E is a T -subbase for X we may pick S', T' E such that x e S' r T',

S r S' = 0, and T n T' = 0. It now follows from the facts that X is

totally ordered and that S and T are not both in ED or EI that

X = X (x) U X (x) c S' U T'. Thus {S',T'} is a screen for S and T.

Hence by Definition 2.11 (iii), E is weakly-normal. To show that X X

is totally ordered, suppose that ) and kare members of X X with X .

Then there, exist T E E n ED and S E X h E such that S T = 0. Since

X is totally ordered it must be the case that t s for each t E T and

s E S. If R E X ED, then it follows from R n S 0 that T c R,

R E E D, and thus .
















5. THE COMPLETION BY CUTS


In this chapter we characterize the completion by cuts of

a topological lattice as a particular lattice superextension

(Theorem 5.13). In doing so we prove a theorem (5.11) which is

used extensively in proving the main results in this and the

succeeding chapter.

Throughout this chapter we let X denote a topological lattice.

We first characterize the completion of X by cuts as a particular com-

plete sublaitice of the power set, 2X, of X which is ordered by

inclusion. Then using the theorem mentioned above we show that the

completion of X by cuts is iseomorphic with the superextension of X with

respect to the usual subbase T. We conclude the chapter by considering

the case where X is totally ordered and investigating the conditions

on X under which the completion by cuts is homeomorphic with the

Alexandroff one-point compactification.



DEFINITION 5.1. A completion-operator on a lattice L is a map
L L
2: 2 2 which satisfies the following conditions:

Cl. A c O(A), for each A E 2

C2. O(A) = O(Q(A)), for each A E 2.

C3. If A and B are in 2L and A c B, then O(A) c O(B).







REMARK 5.2. It follows from Ward [15] that if D is a completion-

operator on L, then A-.= {A E 2 : A = $(A)} is a complete lattice,ordered

by inclusion. Moreover, for each T c A,

(i) A Y = n and

(ii) v y = (U\ ),

and the mapping from L to A defined by x 4({x}) is an injective,

order-preserving map.



Next, we consider a specific completion-operator on X.



PROPOSITION 5.3. The map D: 2X 2 defined by

A n{X (a): A c X (a)} is a completion-operator on X.


PROOF. Cl: Clearly, for each A 2X, A c r{X (a): A c X (a)}.

C2: For each A E 2 by Cl, D(A) c D(D(A)). Conversely, if

A c X (a) for some a E X, then by definition, D(A) c X (a); therefore,

D(D(A)) c D(A).

C3: Suppose that A c B c X. If D(B) = X, then clearly D(A) c D(B).

If, however, B c X (a) for some a e X, then,since A c B c X (a), it follows

that D(A) c D(B).




It should be noted that for each a E X, D({a}) = XD(a) and that

D(A) is the set of all predecessors of all successors of all members of

A; i.e., D(A) = {x E X: x S a for all a E X with the property that b a

for all b E A}.








It follows from Proposition 5.3 and Remark 5.2 that

X = {A c X: D(A) = A) (when partially ordered by inclusion) is a

complete lattice and that the map d: X X defined by a D({a}) is

injective and order-preserving. In view of the equivalence for D(A)

noted in the preceding paragraph and Birkhoff [ 5; Chapter V, 9],

X is the completion of X by cuts.



PROPOSITION 5.4. The injection d: X + X defined by x ~ D({x})

is a lattice inclusion.


PROOF. Since d is injective, we need only show that d is a

lattice homomorphism (2.2 (i)). Suppose that C is a non-void subset

of X such that v C exists in X. It follows from Remark 5.2 that

v d[C] = D(U{d(x): x E C}) = D(C). Let c = v C. Thus C c X (c)
X X D
and hence D(C) D({c}); i.e., v d[C] 5 d(v C). Finally, if a E X
X x
such that C c XD(a), then a is an upper bound of C; hence c = v C : a

so that c e X (a). It follows that d(c) = d(v C) 5 D(C) = v d[C].
D X X
Thus d(v C) = v d[C]. A dual argument shows that if 0 / C c X such
XX
that AXC exists, then d(A C) = A d[C]. Thus d is a lattice
X
homomorphism (2.2 (i)).




Next, we demonstrate that X (when it is equipped with its

interval topology) is also a compactification of X.








PROPOSITION 5.5. For each A E X

(i) A {d(a).: A : d(a)} = A = v {d(a): a E A),

(ii) D(A) = n{ D(d(a)): A r d(a)},

(iii) X (A) = n{X (d(a)): a E A}.


PROOF. Let A E X.

(i) Since a E A if and only if d(a) 5 A, it is clear that

v {d(a): a E A} A A {d(a)? A d(a)}.

To see that v {d(a): a E A} = A, let B E X such that B < A.

Since X is ordered by inclusion there is some a E A such that a / B;

therefore X (a) B so that d(a) i B. Thus v {d(a): a E A} = A.

To see that A {d(a): A d(a)} = A,suppose that B E X such that

A < B. Since D(A) / D(B), it follows from the definition of D that

there is some a e X such that A c X (a) and B X X (a). Thus A 5 d(a)

and B $ d(a). It follows that A = A {d(a): A < d(a)}.


(ii) Since A d(a) implies that XD(A) c X (d(a)), it is clear

that

XD(A) n{X (d(a)): A d(a)}.

To see the other inclusion,suppose that B E X such that B E n{XD(d(a)):

A d(a)}. Thus B : d(a) for each a E X such that A : d(a). By (i),

A = A {d(a): A 5 d(a)} and it follows that B A; i.e., B E X (A).


(iii) Since a E A if and only if d(a) A, (iii) can be proved

by dualizing the argument in (ii).








PROPOSITION 5.6. For each a E X, d[X (a)] = X (d(a)) and

d[XI(a)] = X (d(a)).


PROOF. Let a E X. Since d preserves the order of X, it is

clear that d[X (a)] c X (d(a)), and since X (d(a)) is closed,

d[X (a)] c X (d(a)). To see the other inclusion, let A E X (d(a))

and let W be a basic open set containing A.

If W is of the form X U{X (Bi): i = 1, -**, n} for some

finite set {BI, ***, B} c X, then since A E W, it must be the case

that for each i, A X < B.. Since, by Proposition 5.5 (i),
1
A X = A {d(x): A N : d(x)}, it follows that for each i E {l, -**, n}

there is some b. X such that AX < d(b.) and B. S d(b.). Let
1 1 1
c = (A {b.: i = 1, "*', n}) A a. Then c E X, c < a, and since d is

a lattice inclusion, d(c) : d(a). Thus d(c) e d[XD(a)]. Finally,

note that d(c) e W, for if not, then d(c) e X (B.) for some

i E {1, **', n}, and so B. d(c) : d(b.) -a contradiction. Hence
1 1
W n d[XD(a)] i 0.

Consider now the case that W has the form X U{D(C): j = *,

for some finite set {C, ***, C } c X. Since for each j = 1, **', m,

A e C., and since A < d(a), it follows that for each j, d(a) % C.. Thus

d(-a) e W d[X (a)]. Hence W n d[XD(a)] / 0.

Finally, we consider the case that W = V n U where

V = X U{X (Cj): j = 1, ***, m} and U = X U{X(.): i = 1, n}
Dj 3
for some finite set {B, .**, B, C *, C }c :. Since A c V, it must
1 n 1 m -
be the case that A / A X and so A i 0. Since A i C. and A = v {d(x): x e A}

(5.5 (i)), it follows that for each j {1, **., m} there is some







x. E A such that d(x.) % C.. Let x = v {x.: j = 1, **., m}. Since

d is a lattice inclusion, it follows that for each j, d(x.) d(x).

Also d(x) A, and since A < d(a), d(x) E d[X (a)]. Note that

d(x) e V, for if not, then there is some j E {1, **., m} with

d(x) E X (Cj) and thus d(x.) d(x) 5 C. -- a contradiction. Note

also that d(x) E U, for if not, then there is some i E {1, **', n} with

d(x) E X (B.); thus B. 1 d(x) A so that A i U -- again a contradic-

tion. Consequently, d(x) E W n d[X (a)]. Hence, in each of the

possible cases, W 0 d[XD(a)] X 0, so that, since W was arbitrarily chosen,

A E d[XD(a)]. A dual argument shows that d[X (a)] = X (d(a)).



LEMMA 5.7. The collection T = {(D(d(a)): a E X}
-1
U {X (d(a)): a X} U {X} is a subbase for X, {d [A]: A E T = T ,
-l
and {d-[A]: A T T = TI.


PROOF. Clearly T is a collection of closed subsets of X and

T is contained in the usual subbase for X. Thus to show that T generates

the interval topology on X we need only verify that all principal ideals

of X and principal dual ideals of X,which are not in T, are closed in

the topology generated by T. This follows immediately from Proposition 5.5

parts (ii) and (iii).

To see that {d-l[A]: A T} = TD and {d-l[A]: A T } = T ,
-1 -
note that d-[X] = X and since d is a lattice inclusion-,

d- [X(d(a))] = XD(a) and d-l[X (d(a))] = X (a) for each a E X.







PROPOSITION 5.8. The completion of X by cuts is a compacti-

fication of X; specifically, the mapping d: X X is a topologically

dense embedding.


PROOF. Since X is complete, we know that it is compact (2.9).

That d is continuous follows immediately from the facts that T is a sub-

base for the closed subsets of X and {d-1[A]: A E T} = T (5.7) is a

collection of closed subsets of X.

That d is injective follows from the fact that d is a lattice

inclusion (2.2 (ii)).

To see that d is an embedding, we need only show that d is

closed onto its image. Recall that d is a lattice inclusion and is

(consequently) injective. Thus for each a E X, d[XD(a)]

= D(d(a)) n d[X] and d[X (a)] = X (d(a)) n d[X]; therefore, d[XD(a)]

and d[X (a)] are closed in d[X]. Hence d is an embedding since T is

a subbase for the closed sets of X.

Finally, to see that d[X] is topologically dense in X, recall

that we require that X 1 0. Let A E X. If A X < A, then A / 0 and so

there exists some a E A. Since d(a) 5 A, it follows from Proposition 5.6

that A E d[X (a)] c dX]. If A = A, then let a E X. Since A = A X < d(a),

it follows again from Proposition 5.6 that A e d[X (a c d[X]. Hence

X c d[X]; i.e., d[X] is topologically dense in X.


We return now to the discussion of lattice superextensions.







PROPOSITION 5.9. T is an admissible subbase for X.


PROOF. By Lemma 5.7, T is a subbase for X. To see that

T is a T -subbase for X let A E X. By the definition of T,

n{U e T: A E U} = [n{XD(d(a)): A d(a)}] n [n{X (d(a)): d(a) A}] n X.

Since nr{XD(d(a)): A d(a)} = XD(A) (5.5 (ii)) and r {X (d(a)): d(a) s A}

= {{X (d(a)): a E A} = X (A) (5.5 (iii)), it follows that


l{U e T: A E U} = XD(A) n X (A) = {A}.


Finally, suppose.that A E X and U E T such that A / U. We

consider first the case that U = XD(d(a)) for some a E X. Since A i U,

it follows that A % d(a). Since A = v {d(x): x e A} (5.5 (i)) there is

some x E A such that d(x) % d(a). Thus X (d(x)) e T and it follows

that A E X (d(a)) and X (d(a)) r X (d(a)) = 0. Since x E A if and

only if d(x) A, we may dualize the above argument in the case that

U e T I

Hehce T is a T -subbase for X (2.11 (ii)). Since T consists

of ideals and dual ideals of X, it is an admissible subbase for X (4.2 (i)).



PROPOSITION 5.10. If a, b E X, then X (d(a)) n X (d(b)) = 0

if and only if X (a) n X (b) = 0.


PROOF. Since d[X (a)] c X (d(a)) and d[X (b)] c X (d(b)), it

is easily seen that X (d(a)) n X (d(b)) = 0 implies that

XD(a) n X (b) = 0. Conversely, if XD(a) n X (b) = 0, then since

b % a and d is a lattice inclusion, it follows that d(b) % d(a). Hence

XD(d(a)) n X (d(b))= 0.







The following theorem will have several applications in this

cn-ater and Ch .r7ers 6 and 8.



T 5.11. Suppose that L and M are topological lattices

wi.ch satisfy the following conditions:

(i) f: L M is a continuous lattice inclusion.

(ii) 0 and Y are admissible subbases for L and M, respectively,

such that D = {f-lV]: V E YD} and 0 = {f-[V]: V E }.
D I
(iii) If V1, V2 E such that V1 r V2 i 0, then

f[v ] n f-[v] i 0
f- 1 1 2 '
Then X L is iseomorphic with A M via an iseomorphism f" which

is an extension of f; i.e., such that the following diagram commutes.




e
L > OL


I
f f*


M > AM
e





PROOF. Definition of f*:: Suppose that e X L. Clearly,
.-l
l' = {V E Y: f-l[V] E t} is a linked system of Y. If V a V and

{V} U f' is linked, then by conditions (ii) and (iii), Z U{f-[ V]} is

linked. Since X is a maximal linked system of 0, f- [] e and so
-l
V E '. 7 u-. {V E : f-[EV] c } is a maximal linked system of Y.

We define f*': X L A M by f"*() = {V E Y: f-l[V] e } for each c XA L.







f* is an extension of f; i.e., e o f = f' o e : In view

of condition (ii) we see that for each x E X, f* o e (x)

= f'({W E 0: x E W}) = f*({f-l[V]: x E f-1[V], V E })

= {V e T: f(x) E V} = e (f(x)). Thus the diagram commutes.


f is surjective: For each Mte X M it follows from (ii) and

(iii) that {f- [V]: V E %} is a maximal linked system of 0. By the

definition of f*, f*({f-l[V]: V E 17}) = 1; therefore, f* is surjective.


f* is a lattice inclusion: To see that f* is injective suppose

j,J E X L such that O Thus it follows from (ii) that there are

V, V' E with f-1] E, f-1[V'] E Xf, and f-EV] n f-l[V] = 0. By

condition (iii), V n V' = 0 and since, by the definition of f*,

V E f*(;) and V' E f*(.), it follows that f*(P) / f*(~).

To see that f* is a lattice homomorphism, suppose that C is

a non-void subset of X L, and let ZC = v C. Let V E YD. By the

definition of f*, V E f*(C ) if and only if f-[V] E C. Since

~C = v C, f- EV E C if and only if f-[EV] et for each te C (4.8).

For each t E C, it follows from the definition of f* that f-[EV] c Z

if and only if V E f*(C). Finally, by Proposition 4.8, V E f*(Z) for

each Ce C if and only if V E v f*[C]. It follows from the above

equivalences that V E f*(oC) if and only if V E v f*[C]. Thus

f*() ) N 'D = v f*[C] n gD and consequently, f*({) = v f*[C] (4.7).

A dual argument shows that f* preserves infima of X L. Hence

f* is an injective lattice homomorphism; i.e., f* is a lattice inclusion

(2.2 (ii)).







f" is continuous: It follows from the definition of f": that

for euch f-[V] c 0 and each AeX L, f-1[V] c if and only if

V e f (2). Thus for each V c Y,

f'-l[V+] = { E L: f*(e) V +

E {e X L: V E f"'(g)} (2.14)

= {E X L: f-EV] EX} = (f-lEV)T

which is a member of the subbase for the closed sets of X L. Since

{V : V E ?} is a subbase for the closed subsets of X M, it follows

that fr is continuous.


f" is a closed map: For each V e y, it follows from the above

equality and the injectivity of f* that

f*-[(f-1E[])] = fC[f '*-l[V]] = V+

By condition (ii), {f-l[V]: V E T} = 0 and so {(f-l[V1)+: V E Y} = 0+

is a subbase for the closed subsets of X L. It now follows that the

image under f' of each subbasic closed set in A L is a subbasic closed

set in X M and since f* is injective, f": is a closed map.

Thus f: satisfies all of the conditions of Definition 2.10, and

so f'* is an iseomorphism of X L onto X M.




PROPOSITION 5.12. If L, M, f, 0, and Y satisfy the hypotheses

of Theorem 5.11 and M is complete, then M is iseomorphic with X L via

an iseomorphism which makes the following d a:re'. commutative:




LL

L I
I

NK^ ^







PROOF. In view of Theorem 5.11, we need only show that

e,: M - X M is an iseomcrphism; i.e., that e. is surjective. If

e XYM, then by Proposition 4.5 (ii), 4 is centered. Since M is compact

and X consists of closed subsets of M, it follows that nf / 0. By

Proposition 2.16 (iv), nh is a singleton; i.e., xh. = {a) for some

a c M. It follows that e = e (a).

Thus if f* is the iseomorphism defined in Theorem 5.11, then
-1
S= e o f*': X L M is an iseomorphism with the property that

Soe, e 0 f" o ee = e oe o f = f.




THEOREM 5.13. If X is a topological lattice, then the completion

of X by cuts is iseomorphic with the superextension of X with respect to

the usual subbase T; i.e., X is iseomorphic with XTX.


PROOF. Clearly X and X are topological lattices and d is a

continuous lattice inclusion (5.4 and 5.8). T and T are admissible
-l
subbases for X and X, respectively (4.4 and 5.9),and {d- [A]: A e T} = TD

and {d [A]: A T} = T (5.7). Suppose that VI, V2 T such that

VI 2 V2 0. If V1 and V2 are both in TD or both in T then it
-1 -1
follows from the definition of T(5.7) that d- [V ] (1 d- [V] 2 0.

In the remaining case we may assume that VI eD and V2 TI. It now

follows from the definition of T (5.7) and Proposition 5.10 that

d-1Vl ] 1 d- V] 2 0.








Thus X, X, d, T, and T satisfy the hypotheses of Theorem 5.11,

and since X is complete, it follows'from Proposition 5.12 that X is

iseomorphic with XTX and the following diagram commutes.



eT

X ------ TX






ST





If we restrict our attention to the case where X is totally ordered,

then T becomes a weakly-normal subbase (4.12), and the completion by cuts

(X X) is Hausdorff (2.18 (ii)). The question arises as to how we can

characterize the locally compact, totally ordered spaces for which the

Alexandroff one-point compactification (wX) is the same as the completion

of X by cuts. Before answering this question we need a few preliminary

results.


c
DEFINITION 5.14. For each A c X, D(A)c = {x e X: D(A) c X (x)}.



PROPOSITION 5.15. For any subset A c X, D(A) and D(A) are

closed subsets of X. Furthermore, D(A) duallyy, D(A)c) is either void

or decreasing duallyy, increasing).








PROOF. Let A c X. That D(A) is closed follows from the fact

that D(A) = n({X (a): A c X (a)} (5.3). By Proposition 2.4 (i), D(A)

is either void or an ideal of X; i.e., D(A) is either void or decreasing.

To see that D(A)c is a closed subset of X, we will show that

D(A)c = r{X (a): a E A}. Since A c D(A), it follows from the definition

of D(A)c that for each a E A and each x E D(A)c, a < x so that x E X (a).

Thus D(A) c c {X (a): a E A). To see the other inclusion, suppose
c
that x E X such that x i D(A) Thus A X (x) so there is some

a E A with a0 i x. Hence x i X (a ) and so x i l{X (a): a E A};
c
therefore, D(A) = l{X (a): a E A} is a closed subset of X. It now

follows from Proposition 2.4 (i) that D(A)c is either void or increasing.



LEMMA 5.16. Let X be a totally ordered topological lattice.

(i) For each A c X, D(A) r D(A)c / 0 if and only if v A exists

in X; moreover, if v A exists in X, then D(A) r D(A)c = {v A}.

(ii) For each x e X, v D({x}) = x.

(iii) For each A c X, if vX D(A) exists, then AX D(A)c exists

and vX D(A) = AX D(A)C

(iv) For each A c X, if any of V A, v D(A), or A D(A)c exists

in X, then they all exist and are equal.

(v) For each A c X, if A d[A] = d(x) for some x e X, then

AA = x.


PROOF. (i) Let A c X. Suppose that b = v A exists in X.

Thus A c X (b) so that D(A) c X (b) and hence b E D(A)c. If a E X such

that A c X (a), then b = v A e X (a). Thus b E A{X (a): A c X (a)}

= D(A).









c c
Hence b E D(A) r D(A)c. If x E X such that x E D(A) n D(A)c, then

since x E D(A)c and b E D(A), it follows from the definition of D(A)c

that b E D(A) c X (x) and so b x. In the same way since b E D(A)c

and x e D(A), it follows that x E XD(b); therefore, x < b. Thus x = b

and D(A) n D(A)c = {v A}.

Conversely, suppose now that D(A) h D(A)c 0 and let

b E D(A) n D(A)c. Thus A c D(A) c X (b) so that a < b for each a E A.

If x E X such that x < b, then since b / XD(x) and b E D(A), it follows

from the definition of D(A) (5.3) that A X (x). Thus there is some

a E A such that a X XD(x); i.e., a 4 x. It now follows that b = v A.

Thus D(A) n D(A)c = {v A}.


(ii) Since for each x E X, D({x)) = X (x), it is clear that

x = v D({x}).


(iii) Let A c X such that v D(A) exists in X. Let b = v D(A).

Thus D(A) c X (b) and by the definition of D(A)c, b E D(A)c. Since

D(A)c is increasing (5.15), it follows that X (b) c D(A)c. Recall that

D(A) is closed in the interval topology (5.15) and hence in the order

topology [ 9; Theorem 12]; hence, b = v D(A) E D(A). It follows that

for each x e D(A)c, b < x and so x e X (b); therefore, D(A) c_ X (b).

Thus D(A)c = X (b) and so A D(A)c = b = v D(A).


(iv) Let A c X and suppose that v A exists in X. Let a = v A.

By (i) a E D(A)c so that D(A) c X (a). Also by (i), a E D(A) and since

D(A) is decreasing (5.15), it follows that X (a) c D(A). Thus D(A) = XD(a)

= D({a}). By (ii), v D(A) exists in X and v D(A) = a. Now using (iii)

we see that A D(A)c exists in X and A D(A)c = a.







Suppose now that v D(A) exists in X. By (iii), A D(A)c

exists and A D(A)c = v D(A). Thus we need only show that v A exists

and v A = v D(A). Let a = v D(A). Since D(A) is closed in the

interval topology (5.15), it is also closed in the order topology

[ 9; Theorem 12] and so a = v D(A) e D(A). Since D(A)c is also

closed in the interval topology (5.15) and hence in the order topology,

it follows that a = A D(A)c E D(A)c. Thus a E D(A) n D(A)c and by

(i) v A exists in X and a = v A.

Finally suppose that A D(A)c exists in X and let a = A D(A).
c c c
Since D(A)c is closed, it follows as before that a = A D(A) E D(A)C

Thus D(A) c X (a). Now if x E X such that A c X (x), then by the

definition of D(A), D(A) c X (x) so that x e D(A)c. Thus a : x and

so a E X (x). Hence a E O{X (x): Ac X (x)} = D(A). Thus

D(A) r D(A)c X 0. By (i) v A exists in X and a = v A. By a previous

case v D(A) exists and v D(A) = a = v A = A D(A)c


(v) Suppose A c X such that A d[A] = d(x) for some x e X.

For each a E A, d(x) 5 d(a) and so x a since d is a lattice inclusion.

If c E X such that x < c, then since d is injective, d(x) < d(c) and

there is some d(a) e d[A] such that d(x) 5 d(a) and d(c) A d(a). Since

X is totally ordered (4.12 and 5.13) it must be the case that d(a) < d(c).

Thus since d is injective, x 5 a < c. It follows that x = A A.




The following lemma is well known and so its proof will be


omitted.







LEMMA 5.17. If H is any Hausdorff locally compact space, and

if Y and Z are two Hausdorff compactifications of H with

IY HI = IZ HI = 1, then Z and Y are homeomorphic.



The following theorem characterizes those situations in which

the completion by cuts is the Alexandroff one-point compactification.




THEOREM 5.18. Suppose that X is a locally compact, totally

ordered topological lattice. Then in order for the completion by cuts

of X (X) to be homeomorphic with the Alexandroff one-point compactifi-

cation of X (wX) it is necessary and sufficient that there exists

exactly one subset A c X such that v A and A (X A) do not exist in X.


PROOF. Necessity: If there does not exist B c X such that v B

does not exist in X, then for each B c X, v B = v D(B) = b for some

b E X (5.16 (iv)). Thus D(B) / 0, and since D(B) is a closed ideal

of X (2.4 (i)), it follows that D(B) = d(b). Therefore X = X is compact,

and since X / wX, X i wX -- a contradiction. Thus there is some B c X

such that v B does not exist in X. Let A = D(B). Since A is decreasing

(5.15) and X is totally ordered, it follows-that X A = D(A)c. Since

D(A) = D(D(B)) = D(B), it follows that D(A)c = D(B)c. Thus by Lemma 5.16

(iv), v A and A (X A) do not exist in X and so at least one such subset

of X exists. Suppose now that there is another set C c X such that v C

and A (X C) do not exist in X. Recall that D(C) and A are members of

X and that X is homeomorphic with wX. If D(C) E d[X], then there is







some x E X such that D(C) = d(x). Since d(x) = D({x}), it follows

that v D(C) = x (5.16-(ii)). Hence v C = x (5.16 (iv)), but this

contradicts the fact that v C does not exist in X. Thus D(C) i d[X].

By a similar argument, using the fact that v A does not exist in X,

we see that A e d[X]. Now since IX d[X]I = owX XI = 1 and

D(C), A E X d[X], it follows that D(C) = A. If C / D(C), then since

C c D(C), there is some x E D(C) C, and since v D(C) does not exist

in X, d(x) < D(C) in X. Let G = d[X C]. Then d(x) E G so that G

is a non-void subset of X. Since X is complete, A G exists in X and

A G 5 d(x). Thus A G / D(C), so it must be the case that A G E d[X]

and there is some z E X such that A G = d(z). Hence z = A (X C)

(5.16 (v)); i.e., A (X C) exists in X--a contradiction. Therefore,

C = D(C) = A, and A is the unique subset of X for which v A and A (X A)

do not exist in X.


Sufficiency: Let A c X be the unique set such that v A and

A (X A) do not exist in X. Thus v D(A) does not exist in X (5.16 (iv)),

and since X D(A) = D(A)C, A (X D(A)) does not exist in X (5.16 (iv)).

By the uniqueness of A, A = D(A) and so D(A) / D({x}) for all x E X

(5.16 (ii)); i.e., D(A) = A E X d[X]. If B E X such that B j d[X],

then v B does not exist in X, for suppose that there is some b E X such

that v B = b. Then since B is a closed decreasing subset of X (5.15),

it follows that b E B and XD(b) = B; therefore, B = D({b}) = d(b) E d[X]

--a contradiction. Finally, since v B does not exist in X and B = D(B),

it follows that A (X B) = A D(B)c does not exist in X (5.16 (iv)).

Hence by the uniqueness of A, A = B.








Thus X d[X] = {A} and since X is a Hausdorff compactification

of the locally compact Hausdorff space X formed by adjoining one point

to.X, X is homeomorphic with wX (5.17).





In view of the fact that the open n-cell in Euclidean n-space

can be characterized as the product of n-copies of the open interval

(0, 1), it follows from Theorem 5.13 and the Product Theorem (3.4), that

the closed n-cell (i.e., the n-cell with boundary attached) is a de Groot

compactification of the open disc. Likewise, since Hilbert space can

be characterized as the countably infinite product of the open interval

(0, 1) [3], it follows that the Hilbert cube is a de Groot compactifi-

cation of Hilbert space. (See Example 8.3 for details.)













6. TOPOLOGICAL LATTICE COMPLETIONS



Recall that Theorem 5.13 provided a superextension characteri-

zation of the completion by cuts. The purpose of this chapter is to

obtain algebraic characterizations of other lattice superextensions.

This is done via the notion of topological lattice completions.

Throughout this chapter X will denote a topological lattice.

The topological lattice completions of X are complete topological

lattices having X as a sublattice and a subspace. We show that all

lattice superextensions of X are actually topological lattice comple-

tions of X (Theorem 6.22). Conversely, all topological lattice

completions of X which satisfy a certain density condition are shown

to be lattice superextensions of X (Theorem 6.20).

Throughout this chapter Y will denote a complete topological

lattice (i.e., a complete lattice equipped with the interval topology)

and all subbases considered are assumed to be subbases for closed sets.



-NOTATION 6.1. Let Z be a non-void subset of Y.

(i) Z = Z.

(ii) Y {z E Z: z y y), if {z E Z: z : y) 1 0, or
Zla = {y E Y: y = Z, if {z E Z: z 5 y} = 0


^ {z E Z: y z}, if {z E Z: y 5 z} i 0, or
Zb
= {y E Y: y = Z, if {z E Z: y : z} 0


Z1 = Zla Zlb.







(iii) Z2a = (Zla)lb

Z2b =(Zlb)la

2 2a 2b
z z n z

(iv) 00 la = b 2a 2b



It should be noted that Definition 6.1 can be easily extended

to obtain Za for each ordinal a, and if we define the "lattice closure

of Z in Y" to be Z where a is the least ordinal such that Z = Zl,

then the map which takes Z onto its lattice closure in Y can be shown

to be a Moore closure operator, or completion-operator on Y as in

Definition 5.1. See Appendix II for the details.



PROPOSITION 6.2. Suppose that A C B c Y. Then

(i) Ala cBla

(ii) Ab c Bb

(iii) A1 c B1
1 la lb 2
(iv) A c A c A U A c A2


PROOF. (i) If y E Ala, then {a e A: a < y} c {b E B: b y}

and y = v {a E A: a } y} v {b e B: b y} 5 y; i.e., y e Bla


(ii) Dualize the argument for (i).


(iii) This is immediate from (i) and (ii) above and the defini-

tions of A1 and B1 (6.1 (ii)).


(iv) The proof is immediate if A = 0 (6.1 (iv)); so assume

that A / 0.








That A c A follows from the facts that for each w E A,

w = v a E A: a w} e Al and w = {a E A: w a a} e Al. Clearly,
1 la Ib
A c A la Ab (6.1 (ii)).
la 2 la
To see that Al c A, let w Aa. Thus
la latlb 2a lbTh
w =A {z E A : w z} e (Al ) = A Finally, since A c A it
la (bla 2b
follows from (ii) above that w E A c (Al) = A ; therefore,
2a 2b 2 lIb 2
w E A A = A That A c A follows dually. Thus
1 la lb 2
A c A-c Aa Ab c A2



PROPOSITION 6.3. Suppose that A c Y.
la
(i) If y E A and A n YD(y) X 0, then Y (y) = n({Y(a): a E A

and a y}.

(ii) If y e A and A r Y (y) X 0, then YD(y) = ({YD(a): a E A

and y a a}.


PROOF. Both parts (i) and (ii) follow immediately from the

definitions.



PROPOSITION 6.4. Suppose that A c Y.

(i) If A contains the supreme of all non-void finite subsets

of A and y Ala such that A n Y (y) i 0, then y E (A n Y (y)).

(ii) If A contains the infima of all non-void finite subsets

of A and y E Alb such that A n Y (y) / 0, then y E (A n Y (y)).


PROOF. (i) Let A satisfy the hypotheses of (i) and suppose

that y E Ala such that A n YD(y) / 0. To see that y E (A A YD(y))

let W be any basic open set of Y containing y.







If W = Y U{Y (zi): i = 1, ***, n} for some finite set

{z1 ,*, z ) Y, then since for each i, z. y, it follows that

YD(y) n (U{YI(z.): i = 1, "**, n)) = 0 and so
(A f YD(y)) n (U{Yi(z.): i = 1, **, n}) = 0. Hence

A. n YD(y) C W and W n (A n YD(y)) / 0.

If W = Y U{YD(wj): j = 1, ***, m} for some finite set

{w ***, w } c Y, then for each j = 1, **', m, y 4 w.. Since
I- m]
y = v {a E A: a 5 y}, it is possible to pick,. for each j E {1, **', m},

some x. E A with the property that x. y and x. 4 w.. Let

a = v {x.: j = 1, ***, m}. By the hypotheses on A and since each

x. e A, it follows that a E A and a 5 y; i.e., a E A F YD(y). If

a i W, then there is some j E {l, ***, m} with a E YD (W); therefore,

x. a 5 w. and x. w.--a contradiction. Thus a E W and so

W n (A n YD(y)) $ 0.

In the remaining case W = U C V where

U = Y U{YD(w ): j = 1, ***, m} and V = Y U{Y (2z): i = 1, ***, n}

for some finite set {wl, '*, w, zl, *, zn }c Y. By an argument

analogous to the one above, there is some a E U n (A n YD(y)) and

since A n YD(y) c V, it follows that a E W n (A n Y (y)).

Since W was arbitrarily chosen, it follows that y E (A n YD(y)).


(ii) Dualize the argument for (i).



PROPOSITION 6.5. If i: X Y is a lattice inclusion and if

i[X]1 = Y, then v i[X] = v Y and A i[X] = A Y.








PROOF. Clearly v Y e Y = i[X]1 c i[X]la (6.2 (iv)). Thus

v Y = v {i(x) E i[X]: i(x) : v Y} = v {i(x) E i[X]} = v i[X]. The

other equality is proved dually.



NOTATION 6.6. If i: X Y is a lattice inclusion such that

Y = i[X]1, then we let

= {Y} u{YD(i(x)): x E X} U{Y (i(x)): x E X).



PROPOSITION 6.7. If i: X Y is a lattice inclusion such that
1 A
i[X] = Y, then T is an admissible subbase for Y.


PROOF. Clearly T consists of ideals and dual ideals of Y. Since

Y = i[X] it follows from Proposition 6.3 that each member of the usual
A A
subbase for Y is an intersection of members of T; therefore T is a

subbase for the closed subsets of Y.
A
To see that T is a T -subbase for Y, suppose that z E Y. Then

since z E i[XI X] la ni[X]lb, it follows that
A
){T e T: z E T} = (n{YD(i(x)): x E X and z i(x)}) )

(n{Y (i(x)): x E X and i(x) z}) n Y

= YD(z) n Y (z) (6.3) = {z}.
A
Finally suppose that z E Y and T E T such that z i T. Clearly

T / Y. If T E D, then there is some t E X such that T = YD(i(t)).

Since z i i(t) and z E i[X]1 c i[X]la (6.1 (ii)), there is some x E X
A
such that i(x) z and i(x) 4 i(t). Thus z e Y (i(x)), Y (i(x)) E T,

and YD(i(t)) ) Y (i(x)) = 0. If T E then a dual argument shows that







A
there is some YD(i(x)) E TD with z E YD(i(x)) and T fi YD(i(x)) = 0.
A A
By Definition 2.11 (ii), T is a T -subbase for Y. Hence T is an

admissible subbase for Y (4.2 (i)).



PROPOSITION 6.8. If i: X Y is a lattice inclusion such

that Y = i[X]1, then {il[T]: T E TD TD and {i-l T]: Te E } = TI

where T is the usual subbase for X.


-1 A -1 A
PROOF. Clearly TD c {i [T]: T E T } and T c {i [T]: T E T .

To see the other containments, recall that i is a lattice inclusion;

therefore, for each a E X, i-[Y (i(a))] = {x E X: i(x) 5 i(a)}

-l
= {x E X: x : a} = XD(a) e TD and dually, i-[Y (i(a))] = X (a) e T



PROPOSITION 6.9. Suppose i: X Y is a lattice inclusion such
1 A -1 -1
that i[X] = Y. If S, T E T, then i [S] n i [T] / 0 whenever S n T X 0.


A
PROOF. Suppose S, T E T such that S r T / 0. If S and T are
A A
both in TD or both in TI, then the conclusion is immediate from the defi-
A
nition of T (6.6) and the fact that i is a lattice inclusion. For the

remaining case we may assume that there exist t and s in X such that

T = Y (i(t)) and S = YD(i(s)). Since S f T / 0, we see that

i(t) 5 i(s) and since i is a lattice inclusion, t S s. Thus

i-l[S] n i-l[T] = XD(s) n Xi(t) t 0.



DEFINITION 6.10. Suppose that X and Y are topological lattices,

Y is complete, and i: X Y is a function. Y will be called a topological







lattice completion of X with respect to i whenever the following

two conditions are satisfied:-
1 2
(i) Y = i[X] or Y = i[X]

(ii) i is a continuous lattice inclusion.



PROPOSITION 6.11. If i: X Y is a lattice inclusion and

i[X] = Y, then Y is a topological lattice completion of X with respect

:to i; i.e., the map i is continuous.

^-1 A
PROOF. Since T is a subbase for Y (6.7), and {i-l[T]: T E T}

is the usual subbase for X (6.8), it follows immediately that i is

continuous.



PROPOSITION 6.12. If Y is a topological lattice completion of

X with respect to i: X Y, then i is an iseomorphism onto its image;

i.e., i is an embedding.


PROOF. In view of the definitions of topological lattice

completions (6.10) and lattice inclusions (2.2 (ii)), we need only show

that the restriction i: X i[X] is a closed map. For each a E X,

i[X (a)] = {i(x): x E X and x a}.

= {i(x): x E X and i(x) i(a)} (since i is a lattice inclusion).

= i[X] fYD(i(a)).

Dually, for each a E X, i[X (a)] = i[X] r)Y (i(a)). Thus the image

of every member of Tis a closed set in i[X], and since i is injective,

it follows that the image of each closed set in X is closed in i[X].








THEOREM 6.13. If i: X Y is a lattice inclusion and

i[X]1 = Y, then Y is iseomorphic with the completion of X by cuts.


PROOF. By Proposition 6.11, the function i: X Y is a
A
continuous lattice inclusion. T and T are admissible subbases for

X and Y, respectively, such that (i-1 [T: T E } = T and

{i-l[T]: T E } = T (6.7, 6.8, and 4.4). If T, S E T such that

T r S / 0, then i-lT] n i-l[S] / 0 (6.9). Thus X, Y, i, T, and

satisfy the hypotheses of Theorem 5.11, and since Y is compact it

follows from Proposition 5.12 that Y is iseomorphic with XTX, the

completion of X by cuts (5.13).




REMARK 6.14. It follows from Theorem 6.13 that for any topo-

logical lattice X, there is only one completion Y of X for which

Y = X. This occurs just when X is a complete lattice. In the case
2 1
where X c Y and X2, but not X is a complete sublattice of Y, the pos-

sibilities are not so limited. In fact, in this case a lattice inclusion

need not be continuous (see Example 8.6, Part I); however, for the

remainder of this chapter we will consider only topological lattice

completions of X.

In order to use Theorem 5.11 (as was done in the proof of

Theorem 6.13) to characterize one of these "larger" topological lattice

completions as a lattice superextension, it is necessary to obtain

a subbase which satisfies certain conditions. Namely, if Y is a







topological lattice completion of X with respect to i, then we must

choose some subbase E for Y such that

(i) E is an admissible subbase for Y,
-1
(ii) {i-[S]: S E E) is an admissible subbase for X, and

(iii) if S, T E E with S 0 T / 0, then i- SI] r i-l[T] / 0.

Keeping in mind conditions (i) and (ii) we make the following

definition.



NOTATION 6.15. If Y is a topological lattice completion of X

with respect to i, then

S() = {Y} U{YD(y): y E (iX]la i[X]I) U i[X]}

u{ (y): y e (i[x lb i[X]l) U i[X]}.



PROPOSITION 6.16. If Y is a topological lattice completion of

X with respect to i and Y = i[X1, then (i) T.


PROOF. This follows immediately from the fact that when

Y = i[X]1, then i[X]1 = i[X]la = i[X]b and so i[X]la i[X]1

i[X]Ib i[X]1 = 0.



PROPOSITION 6.17. If Y is a topological lattice completion

of X with respect to i, then

(i) (i) is an admissible subbase for Y, and

(ii) {i-l[S]: S E Z ()} is an admissible subbase for X.






PROOF. (i) Clearly Z consists of principal ideals and prin-

cipal dual ideals of Y. In order to see that E ) is a subbase for Y,

we need only show that all principal ideals of Y and principal dual
(i)
ideals of Y which are not in E are closed in the topology on Y

which has (i) as a subbase. If YD(y) is a principal ideal of Y and

Y (y) Z(i), then it follows from the definition of (i) that either

(a) y E i[X] or

(b) y E i[X]2 (i[X]la U i[X]lb).

Dually, if z E Y such that Y (z) I ) then either

(c) z e i[X] or

(d) z e i[X]2 (i[X]la U i[X]b).


(a) If y E i[X] then y E i[X] and so by Proposition 6.3 (ii),

YD(y) = n{YD(i(x)): x E X and y 5 i(x)}.


(c) If z E i[X] then z i[X]la, and by Proposition 6.3 (i),

Y (z) = n{Y (i(x)): x E X and i(x) < z}.


Thus for each w e i[X] Y (w) and YD(w) are closed in the topology

generated by Z( Hence in view of the definition of ) (6.15), for
la lb
each w E i[X] U i[X] Y (w) and Y (w) are closed in the topology
(iD
generated by E

2 2a 2a ab
(b) If y E i[X]2, then y E i[X]2a, and since i[X]2 = (i[X]la)b,

YD(y) = M{YD(w): w E i[X]la and y w} (6.3 (ii)).


(d) If z E i[X]2, then z E i[X]2b, so that since i[X]2b = (i[X]lb) l

Y (z) = n{Y (w): w E i[X]lb and w 5 z} (6.3 (i)).







It now follows from parts (b) and (d) that if q e i[X]2 (i[X]la U i[X]b),

then YD(q) and Y (q) are intersections of closed sets in the topology on
M (i)
Y generated by E and hence are closed in that topology. Thus E

is a subbase for the interval topology on Y.

(i)
To see that E is a T -subbase for Y, let z e Y and

= {SE E ): z E S}. Clearly R is linked. To see that P is a

maximal linked system of ), suppose that T () such that Ti {.
(i)
Then z i T. We will assume that T E D and note that a dual argu-

ment should be used when Te Thus there is some y E (i[X]la i[X)

U i[X] such that T = YD(y) and z 4 y. "We consider the possibilities

for z.

(e) z e (i[X]lb i[X]1) U i[X].

(f) z e i[X la

(g) z e i[X]2 (iCX] a i[X]b).

lb 1 (i)
(e) If z e (i[X] i[XI) U i[X], then S = Y (z) E E z e S, and

S n Y(y)= Y(z) n Y (y) = 0.


(f) If z e i[X]a, then since z 4 y, it is possible to pick some

w E i[X] such that w < z and w 4 y. Thus S = Y (w) E z e S, and

S ) YD(y) = YD(w) n YD(y) 0.


(g) If z e i[X]2 (i[X]la U i[X]lb), then since z e i[X]2b and

z 4 y, there is some q e i[X]l such that q.s z and q i y. If

q / i[X]l, then S = YI(q) E E(), z e S, and S = YDI(q) 0 YD() .

If q e i[X]l i[X]a, then since q 4 y, there is some w E i[X] such that
(i)
w q and w f y. Hence S = Y (w) E z E S since w < q < z, and

S n Y() Y = (w) n Y (y) = 0.








It follows from the arguments in (e), (f), and (g) that in all
(i)
cases there is some S E( such that z E S and S f T = 0. Clearly

S E H so that f U{T} is not linked. Thus f is a maximal linked system

of E By Proposition 2.13, E is a T -subbase for Y; therefore,

E() is an admissible subbase for Y (4.2 (i)).


(ii) Let E = {i-1 T]: T E (}. Since i is continuous, it

is clear that E is a collection of closed subsets of X. Also, since
-1 (i)
i is a lattice inclusion, it follows that {i [T]: T E E } is a

collection of ideals of X and {i- [T: T E E i)} is a collection of

dual ideals of X. Thus E c F. It follows from the definition of

E ) (6.15) that E contains all the principal ideals and principal

dual ideals of X. Hence by Proposition 4.3, E is an admissible subbase

for X.





In view of Proposition 6.17, the subbase () which we have

chosen for Y (where Y is a topological lattice completion of X with

respect to i) satisfies the first two conditions in Remark 6.14. It

is not true, however, that Zi) necessarily satisfies the third condi-
(i)
tion; i.e., it is not always true that if S, T E ) such that

S r T / 0, then i- [S] f i-l[T] X 0. An example of this can be seen

in 8.8. Thus we are lead to define the notion of "lattice density."







DEFINITION 6.18. If Y is a topological lattice completion of

X with respect to i: X Y, then we say X is lattice dense in Y when-

ever the following condition is satisfied:

If y e i[X]lb i[X] and z e i[X]la i[X]1 with the property

that y < z, then there is some x E X such that y < i(x) < z.



It should be noted that if Y = i[X] then X is lattice dense

in Y since i[X]la i[X]1 = i[X]b i[X1 = 0.



PROPOSITION 6.19. Suppose that Y is a topological lattice

completion of X with respect to i: X -+ Y and X is lattice dense in Y.

If S, T E (i) such that S n T / 0, then i-l[S] n i-1T ] 0.

-l
PROOF. If either T = Y or S = Y, then since i- [Y] = X, the
(i) (i)
proof is trivial. If T and S are both in Z or both in E then

since {i- [R]: R E } i)} is a collection of closed ideals of X and

dually {i [R]: R E } is a collection of closed dual ideals of X,

the proof again is trivial (4.5 (i)). In the remaining case we may

assume without loss of generality that there exist t E (i[X] i[X] )

U i[X] and s (i[X]lb i[X]1) U i[X] such that YD(t) = T and

Y (s) = S. Since S T / 0, 'it must be the case that s t. If

s e i[X], then s = i(x) for some x E X and x e i-l[Y (i(x))] 0 i- YD(t)]

= i-1[S] n i-l[T]. Dually, if t E i[X], then t = i(x) for some x E X

and x e i [Y (t)] ) i [Y (s)] = i [CS] i-l[T]. Thus suppose that

s, t i[X]. Since neither s nor t is in i[X] and s s t, it follows







that s < t. Now since X is lattice dense in Y, there is some x E X
-1 -1
with s < i(x) < t; therefore, x e i [Y (t)] n i [Y (s)]
-1 -1
Si-l[ES] i-lT].




THEOREM 6.20. Given that Y is a topological lattice completion

of X with respect to the inclusion map i: X Y, X is lattice dense in

Y, and E = {i -[TI: T E ) }, then Y is iseomorphic with X X via an

iseomorphism f: X X Y with the property that f o e = i.


PROOF. By the definition of topological lattice completions

(6.10), i is a continuous lattice inclusion. E and ) are admissible

subbases for X and Y, respectively, such that {i [S]: SE E( } =
-1 (i) (i)
and {i [CS: S E E(} = E (6.17). If S, T E E such that

S r T / 0, then i- [S] r i-l[T] / 0 (6.19). Thus X, Y, i, E, and

E) satisfy the hypotheses of Theorem 5.11. Since Y is complete,

there is an iseomorphism f: X X Y with the property that f o e = i

(5.12).





We now consider the converse question; i.e., which lattice

superextensions are topological lattice completions of X in which X is

lattice dense?



PROPOSITION 6.21. If E is an admissible subbase for X, then
for eac+ e E[x la + lb
for each T E % duallyy, T E C ), VT e [XI duallyy, A T E e [XI ).
I







PROOF. Suppose that T E D. Thus v T = v e [T] (4.10 (iv)).

It follows that v T = v {e (x): e (x) e e [X] and e (x) v T };

therefore, v T E e [Xla. A dual argument shows that if T e E, then
AT [X]lb
A T E e e.EXl




THEOREM 6.22. If E is an admissible subbase for X, then

X X is a topological lattice completion of X with respect to e : X X X.


PROOF. Let E be an admissible subbase for X. Recall that e

is a continuous lattice inclusion (2.16 (v) and 4.9) and that for each

e X X,

v {A S: S e : } = n = A {V S S e: ED} (4.10 (vii)).

Thus since for each S E Er ,i A S+ < X (4.10 (vi)) and

A S E e EX]b (6.21), it follows that
b2b
S= v { E XX: e[X]lb and E '} E (e [X]lb)la = e [X]2b

Dually, since for each S E f n D, v S+ E e [X]la (6.21) and < v S+

(4.10 (vi)), then

S= A { : XEX: p ee [X]la and < .} e (eJ[X] la ) = e [X]2a
2a 2b X2
Thus for each e E XX, E e X] 2 e EXI = e [XE ; i.e., EX = e X]

and so X X is a topological lattice completion of X with respect to

e X t X.




We conclude this chapter by exhibiting those lattice super-

extensions of X in which X is lattice dense.







DEFINITION 6.23. Suppose that E is an admissible subbase

for X. Then for each e E X, we let

X(D,f) = {x e X: e (x) : Z} and

X(I,f) = {x E X: X 5 e (x)}.



REMARK 6.24. It is easily seen from Definition 6.23 that for

each I e X,

v e [X(D,X)] 3r/ A e [X(I,e)].

Hence exactly one of the following statements is valid:

(i) v e [X(D,X)] = A = A e [X(I,.)]

(ii) v e [X(D,i)] = Z < A e [X(I,4)]

(iii) v e [X(D,.)] < z = A e [X(I,.)]

(iv) v e [X(D,Z)] < Z < A e [X(I,Z)].

Clearly for each E X EX, X(D,;) is a closed ideal of X which

may or may not be in E and X(I,X) is a closed dual ideal of X which may

or may not be in E.



DEFINITION 6.25. If E is an admissible subbase for X, then E

will be called thick if and only if for each pair , ) E X X such that

/ satisfies condition (ii) in Remark 6.24 and V satisfies condition (iii)

in Remark 6.24, it follows that 0 = E U{X(D,X), X(I,w)} is an admissible

subbase for X and there exists an iseomorphism f: X X X X such that

f o e = e .







THEOREM 6.26. If E is a thick, admissible subbase for X,

then X is lattice dense in A X.


PROOF. To see that X is lattice dense in A X, let ', t~l A X

such that e e [X]la e [X]1, 1 e X]lb e[X]l, and < <. Thus

o = v {e (x): x e X and e (x) 5f} = v e [X(D,Y)] (6.23).

Since Z I e [X] it must be the case that

< A {e (x): x E X and j e (x)} A e [X(I,f)] (6.23).

Hence X satisfies condition (ii) of Remark 6.24. By a dual argument

we see that A satisfies condition (iii) of Remark 6.24. Thus since E

is thick, X X is iseomorphic with X X where 0 = Z U{X(D,Z), X(I,N)} (6.25).

Let f: X X A X denote the iseomorphism. Since f is a lattice

isomorphism, f o e = e0 (6.25), and X(D,.) E 0D, it follows that

f(L) = f(v e [X(D,V)]) = v f[e [X(D,Y)]] (2.2)

= v e [X(D,)] = v X(D,f)+ (4.10 (iv)).

Dually, since X(I,M) E 0I,

fO() = A X(I,A)+.

Thus X(D,Z) e f(X) and X(I,n) e f(y) (4.10 (vi)). Since f preserves

the order of A X (2.5) and Yt < V, it follows that f(m) f(Z) and so

X(I,m) e f(m) n 01 c f(x) n OI (4.7);

therefore X(D,C), X(I,z) e f() and so X(D,;) 0 X(I,') / 0.

Let x E X(D, ;) X(I,n). Then

f(4) S e (x) 5 f( ).

Since f o e e we have f(M) f(e (x)) < f(Y), and since f is a lattice

isomorphism, yZs e (x) X. Finally, since i~ e [X] and X1 e [X], it

follows that M< e (x) < Z. Consequently, X is lattice dense in A X.
ZJ









PROPOSITION 6.27. If Y is a topological lattice completion

of X with respect to i: X Y and X is lattice dense in Y, then

E = {i-l[S]: S E (i)} is a thick, admissible subbase for X.


PROOF. Recall that E is an admissible subbase for X (6.17 (ii)).

By Theorem 6.22, Y is iseomorphic with A X. To see that E is a thick

subbase for X, suppose that f, ln E X X such that Z satisfies condition (ii)

,of Remark 6.24 and )4 satisfies condition (iii) of Remark 6.24. Since

v {e (x): x E X and e (x) 5-}

= < A {e (x): x E X and 5 .e (x)} (6.23, 6.24),
la 1 (i)
it follows that X E e X -e X. Thus {e X: } E ZD (6.15),

and so X(D,X) = i-l[E{ XEX: Z}] e ZD. Dually, tE e [X]lb e [X]1,
(i) -1
SE XE X: A :5 } e and so X(I,M) = i 1[{ EX X: Y W] e E .

Consequently, 0 = E U{X(D,); X(I,%)} = E is an admissible subbase for

X, and there is an iseomorphism f: A X X @X, namely, the identity map,

such that f o e = e Hence E is thick.



PROPOSITION 6.28. The usual subbase T and the subbase F are

both thick.


PROOF. To see that T is thick, we recall that X X is the

completion by cuts of X so that X X = e [X] Thus it follows that

every member of A X satisfies condition (i) in Remark 6.24. Hence T

vacuously satisfies the definition of a thick subbase.








To see that F is thick, suppose that , mY A X such that

Satisfies condition (ii) of Remark 6.24 and M satisfies condition (iii)

of Remark 6.24. Since X(D,/.) is a closed ideal of X and X(I,Y, is a

closed dual of X, then X(D,4) E FD and X(I,M E e (4.1 (ii)). Thus

0 = r U{X(D,;), X(I,,)} = Fis an admissible subbase for X (4.4) and

the identity map f: AX X A X is an iseomorphism such that f o e = e .



COROLLARY 6.29. X is lattice dense in A X and A X.


PROOF. Since both T and r are thick (6.28), admissible sub-

bases for X (4.4), the proof is immediate from Theorem 6.26.




All of the admissible subbases which are considered in the

examples (Chapter 8) are thick subbases for X. Whether or not every

admissible subbase for X is thick remains an open question.




THEOREM 6.30. Suppose that X is a topological lattice and

0 is an admissible subbase for X. Then X is lattice dense in A X

if and only if there exists a thick admissible subbase E for X such

that AnX is iseomorphic with AX via an iseomorphism f: A X AXX

with the property that f o e = e .


PROOF. Let 0 be an admissible subbase for X. Thus A X

is a topological lattice completion of X with respect to

eg: X A X (6.22).




67

-1 6
If. X is lattice dense in X X, then E = {e [S]: S e 0}

is a thick admissible subbase for X (6.27) and X X is iseomorphic

with X X via an iseomorphism f: X X A X with the property that

f o e = e (6.20).

Conversely, if there is some thick admissible subbase E for

X and an iseomorphism f: XX X X such that f o e = eg, then since

X is lattice dense in X X (6.26), it follows that X is lattice dense

in X X.















7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS


In this chapter we will consider the possible existence of

certain maps between different lattice superextensions of the same

topological lattice. The results will then be used to establish the

existence of maximal and minimal lattice superextensions for all

topological lattices and in certain cases, largest and smallest lattice

superextenions (Theorem 7.15). Thus in the realm of lattice super-

extensions we have analogues to the Stone-Cech and Alexandroff compacti-

fications. Kaufman [12] obtained similar analogues for the totally

ordered case. Theorem 7.15 is a generalization of Kaufman's results.

In this chapter, as in the previous chapters, X will denote a

topological lattice. Also, to avoid confusion, we will sometimes use

the symbol "*" in place of "+" to denote the subbasic closed sets of

a lattice superextension; e.g., if 0 is an admissible subbase for X,

then for each T e 0,

T* = {f E X X: T e } and 0* = {T*: T e O}.



DEFINITION 7.1. Let 0 and E be admissible subbases for X.








(i) We say that X X algebraically dominates X X if and only

if there is a surjective, order-preserving function f: X X -+ X X such

that the diagram


X X






e I

Xx


commutes (i.e., f o e = e ) and such that f preserves the supreme

of ideals in X and the infima of dual ideals in X; i.e., if I is an

ideal of X and D is a dual ideal of X, then f(v e [I]) = v (f o e )[I]

and f(A e [D]) = A (f o e )[D]. Such a function is called a dominating

function of X X over X X.

(ii) We say that the superextension X dominates the super-

extension X X if and only if X X algebraically dominates X X via a

continuous dominating function.



It might be noted that although X X and X X are both lattices,

a dominating function f is not required to preserve all supreme and

infima of X X; i.e., f is not required to be a lattice homomorphism.

The reason for this is apparent from Example 8.9 in which a topological

lattice X is exhibited for which the lattice X X algebraically dominates

the lattice X X (and also for which the superextension X X dominates







the superextension X X), but for which there are X, E X FX such

that f( ) A f(nz) / f( A nt). (See 8.9 for details.) Example 8.7

shows that algebraic domination does not necessarily imply superexten-

sion domination, for in this example there exists a unique algebraic

dominating function f: A X X X which is not continuous.



PROPOSITION 7.2. Suppose that 0 and E are admissible subbases

for X such that X X algebraically dominates X X via a dominating

function f: XX X 0X. Then

(i) for each R E D duallyy, R E i), f(v R+) = v e[R]

duallyy, f(A R+) = A e [R]),

(ii) for each R EED OD duallyy, R E Z 0 I), f(vR+) = v R;

duallyy, f(A R+) = A R*), and

(iii) for each X E XEX, (t nO) c f(.).


PROOF. () Suppose that R E D. Then by the hypotheses on

f and since R is an ideal of X, it follows that

f(v R+) = f(v e [R]) (4.10 (iv))

= v (f o e )[R] = v e [R].

If R E I, then a dual argument shows that f(A R ) = A e [R].


(ii) Suppose that R E D n D. Then since f(v R+) = v e [R]

((i) above) and v e [R] = v R* (4.10 (iv)), we see that f(v R+) = v R*.

Dually, if-R E EI r 0 then f(A R ) = A R*.

(iii) Suppose that X E X X. Let R E (; n 0 ). Thus < v R+

(4.10 (vi)), and since f is order-preserving, f(G) < f(v R ) = v R*

((ii) above);i.e., R E f(Z) (4.10 (vi)). Hence (t fl 0D) < f(4), and by

a dual argument, (V n 0 ) c f()). Therefore, ( n 0) c f(A).







LEMMA 7.3. If E is an admissible subbase for X, then for

each a E X, v e [XD(a)] = e (a) and A e [X (a)] = e (a).


PROOF. Let a E X. Then for each x E XD(a), x : a so that

e (x) 5 e (a) since e is a lattice inclusion (4.9); therefore,

v e [XD(a)] < e (a). Since a E XD(a), it follows that e (a) = v e [XD(a)].

A dual argument shows that e (a) = A e [X (a)].



PROPOSITION 7.4. If 0 and E are admissible subbases for X such

that 0 c_ then A X algebraically dominates A X.


PROOF. We define f: A X A X by

f(i) = v {A e [S]: S (4 fE )} for each E C X.


foe e: Let x E X. For each S ee (x) n E x E S so
--L --
that A e [S] < e (x); therefore, f(e (x)) = v {A e [S]:

S E e (x) ) } < e (x). Now suppose that T E e (x) n 8E. Thus

x E T and since T e 0 I c it follows that T E e (x). Hence by

Proposition 4.10 (iv) and the definition of f, A T* = A e [T] < f(e (x));

i.e., T E f(e (x)) f 01 (4.10 (vi)) and so e (x) 5 f(e (x)) (4.7).

It follows that f o e (x) = e (x) for each x e X.


f preserves the order of A X: Suppose that Y, 4 E A X such that

/ r7. Thus (. E ) c () r ) (4.7) and so f(f) = v {A e [S]:

S E n Z } < v {A e [S]: S E 7 ,t E } = f(Y>).







For each E X X, ( t( 0) c f(f): Let e E X'X. If

T E t 01, then by the definition of f, A e [T] f(/). Since

A e [T] = A T* (4.10 (iv)), it follows that T E f(C) (4.10 (vi)),

and so (n A 0 ) c f(C). If T E ( r 0 D), then for each S E (.e 0r ),

T f S ? 0. Thus there is some x E S n T and so A e [S] 5 e (x) 5 v e [TI.

It follows that

f(R) = v {A e [S]: S E (X n zI)} V e [T] = v T* (4.10 (iv));

therefore, T E f(f) (4.10 (vi)) and so ( n 0 ) c f(4).


f is surjective: If PZE @0X, then Yis a linked system of E

and hence is contained in some maximal linked system ) of E. Since

/1 c_ (Y n 0) c f(e), it follows from the maximality of M that f(/) = r.


f preserves supreme of ideals and infima of dual ideals of X:

Suppose that I is an ideal of X and Z = v e [I]. Since f preserves the

order of X, we see that for each x e I, f(e (x)) = e (x) 5 f(;) and

so v e [I] 5 f(). If T E (v e [I] n 0 ), then for each x E I,

T e e (x) (4.8) and so x e T; consequently, I c T. Since T E ED, it

follows that T E v e [I] = Z (4.8). Thus T E (P n 0) c f(4) and so

v e [I] 0D c_ f(1). By the definition of the order on X X (4.7),

f(.) v e [I]; therefore f(W) = f(v e [I]) = V e [I] = v (f o e )[I].

A dual argument shows that f preserves the infima of dual ideals of X.


In view of the above arguments, X X algebraically dominates X X.







PROPOSITION 7.5. Suppose that 0 and E are admissible subbases

for X such that 0 c E.

(i) If for each f EA X, (f 0 0 ) is contained in a unique

maximal linked system of 0, then there is a unique dominating function

f: XX X+ X. Moreover, f may be defined by f(e) = {T 0:

(r n o) U{T} is linked} for each X E X X.

(ii) If for each t E XEX,(t n 0) is a maximal linked system

of 0, then there is a unique and continuous dominating function

f: XAX A X.


PROOF. (i) Since 0 c E, there is a dominating function

f: A X + X (7.4). Suppose that g: A X A X is also a dominating

function. Thus for each E Xe X, (f n 0) c g(t) and (' n e) c f(.C)

(7.2 (iii)). Since (9 n 0) is contained in a unique maximal linked

system of 0, it follows that g(V) = f(X) for each Z E X; i.e., f = g

and f is unique.

To-see that f can be defined as stated, let X E X X and

S= {T e 0: ( i n 0) u{T} is linked}. To see that f is linked, let

S, T H. Then (.t n ) U{S) is linked and ( 0n ) U{T) is linked.

Since ( A 0) is contained in a unique maximal linked system ) of 0,

it. follows that ( n 0) u{S} and ( n 0) u{T} are both subsets of -.

Thus S n T / 0 and ffis linked. To see that f is a maximal linked

system of 0, suppose that T E 0 such that 0 u{T) is linked. Thus

(0 0) U{T) is linked, and so by the definition of f, T E Sf.

Since both ff and f(Z) are maximal linked systems of which

contain ( An 0) and (X 0 0) is contained in a unique maximal linked

system of 0, it follows that f(d) = f.








(ii) If for each e XA X, (4 ( 0) is a maximal linked

system of 0, then (. 0 O) is contained in a unique maximal linked

system of 0. By (i) above, there is a unique dominating function

f: A X X X defined by f(t) = {T E 0: (x n O) U{T} is linked} for

each E A X. Since (i n 0) is a maximal linked system of 0, it follows

that f(1) = ( nl 0). Thus for each T e 0,
-l
f-l[T* = {X e X: f(4) e T*} = { e XX: T E f()}

= E X X: T E } = T+ E

Since 0* is a subbase for the closed sets of X X, it follows that f

is continuous.



DEFINITION 7.6. If 0 and Z are admissible subbases for X,

then we say that 0 is equivalent to E whenever X X is iseomorphic

with XX via an iseomorphism f: X X X such that f o e = e .



PROPOSITION 7.7. If 0 and E are admissible subbases for X such

that each of AOX and X X algebraically dominates the other, then 0 and

E are equivalent. Specifically, we will show that if f: XAX XAX and

g: X X X X are dominating functions, then

(i) g o f is the identity on X X and f o g is the identity on X X

(ii) f and g are injective

(iii) f and g are closed maps

(iv) f and g are homeomorphisms

(v) f and g are lattice isomorphisms

(vi) X X is iseomorphic with X X.
i, ---- --- --- 9







PROOF. The hypotheses describe the following diagram:


EX




g


where g o e = e and f o e = e .


(i) Let R E D. Thus R is an ideal of X. By the properties

of dominating functions (7.1 (i)) and Proposition 4.10 (iv),

g 6 f(v R ) = g o f(v e ER]) = g(v f o e[R]) =

g( v e [R]) = v go e [R] = V e [R] = v R+.

Dually, for each Re c g o f(A R+) = A R+. Now let c e X X. If

R E x E ZD, then Z v R (4.10 (vi)), and so f(Z) < f(v R ) since f

is order-preserving. Thus g o f(Y) : g o f(v R+) = v R since g is

order-preserving. Hence R e g o f(L) (4.10 (vi)); i.e.,Z n ED c g o f(g).

Dually, f E c g o f(0) so that = g o f(Z) and g o f is the identity
I-
on A X.

A similar argument shows that f o g is the identity on X X.


(ii) If t and YIare elements of X X such that f(Y) = f(7), then

by (i) above t = g o f(Z) = g o f(lz = Y. Thus f is injective.

A similar argument shows that g is injective.







(iii) Let R E D. If Z E R+, then R E and so X v Rt (4.10 (vi)).

Since f is order-preserving, f(;) f(v R+) = v e [R] (7.2 (i)). Thus

f[Rt] c Q E X0X: < V e [R]}.

Conversely, if a E X X such that I v e [R], then since f is surjective,

there is some o E X X with f( ) = .. Now since g is order-preserving

and g o f is the identity on X X, it follows that / = g o f(4) =

g(Y) 5 g(v e [R]). Thus by the properties of g (7.1 (i)),

S5 g(V e [R]) = v (g o eQ)[R] = v e [R] = v R+ (4.10 (iv)).

Hence R E Z (4.10 (vi)) and so $ ER : therefore

) E XX: g v e0[R]} = fE[R] = g-l[R+].

A dual argument shows that if Re E then

g-[1R+] = fE[R] = { e gX: A eO[R] :}.

Since E is a subbase for the closed sets of X X, it follows from the

above qualities that f-1 = g is continuous. Hence f is a closed map.

A similar argument shows that g is a closed map.


(iv) Since each of f and g is closed, continuous, and bijective,

each must be a homeomorphism.


(v) Suppose that C is a non-void subset of X X and = v C.

Since f preserves the order of X X, it is clear that /Z= v {f( ):

E C} I f(). To see that f(4) 179 let T X n E D. Thus M7Z< v T*

(4.10 (vi)) so that for each ) E C, f( ) v T*. Since g is order-

preserving and g o f is the identity on X X, it follows that for each

9E C,


) = g o f(O) < g(v T*) = v e [T] (7.2 (i)).







Thus Z = v {: 9 e C)} v e [T]. Finally, since f is a dominating

function (7.1 (i)), it follows that

f(X) 5 f(v e [T]) = v f o e [T] =v e T] = v T* (4.10 (iv)).

Hence, T E f(4) (4.10 (vi)). Therefore, N OD c f('-) so that

M1= f()); i.e., f(v C) = v f[C]. A dual argument shows that

f(A C) = A f[C] and so since f is bijective, f is a lattice isomorphism.

A similar argument shows that g is a lattice isomorphism.


(vi) That f and g are iseomorphisms follows immediately from

(iv) and (v) above.



COROLLARY 7.8. If 0 and Eare admissible subbases for X such

that 0 c E and X X algebraically dominates XAX, then 0 is equivalent

to E.


PROOF. Since 0 c E, it follows that X X algebraically dominates

SX (7.4). By the hypotheses, XoX algebraically dominates X X. Thus

0 is equivalent with E(7.7).



PROPOSITION 7.9. Suppose that 0 is an admissible subbase for X

and E = 0 U T. Then

(i) E is an admissible subbase for X, and for each f- X X,

(Zf 0) is a maximal linked system of 0, and

(ii) for each Z E XX and each a E X, if ( fl 0) e e (a),

then X (a) e 1, and dually, if e (a) (t 0 0), then X (a) e X.
--- D -- --- --0 --- I







PROOF. (i) Since T c Z= 0 U T c F, E is an admissible sub-

base for X (4.3). Let C EX X. Clearly ( r 0) is linked. To see

that (X r 0) is a maximal linked system of 0, let T E 0D such that

(c. n 0) U{T} is linked. If T i X then there is some U EE I r

such that T n U = 0. Since I = 0 U T and {T} U ( n 0I ) is

linked, it must be the case that U E T ; i.e., U = X (u) for some

u E X. Since u / T and 0 is a T -subbase for X, there is some S E 0

such that u E S and S ( T = 0. It follows that X (u) c S, and

S E (. n 0); therefore, (. 0) U{T} is not linked -a contradiction.

Thus T E X. Dually, if T E 0 such that ( ?n ) U{T} is linked, then

T e X, and hence (X F 0) is a maximal linked system of 0.


(ii) By (i) above and Proposition 7.5 (i) there is a unique

dominating function f: X X X aX which may be defined by f(X) = (X n 0)

for each E E X X. Suppose that X e E X and a E X such that

(X n 0) = f() < e0(a). For each S E (X r E ), A S+ .(4.10 (vi))

so that f(A S ) + f(P) since f is order-preserving. If S E 0 then

f(A S+) = A S* (7.2 (ii)) and since f(X) < e (a), it follows that

A S* < e (a); therefore a E S so that S X XD(a) / 0. If S Oi ,

then there is some b E X such that S = X (b). In this case

f(A S+) = A e [S] (7.2 (i)) = A e [X (b)] = e (b) (7.3). Thus

e (b) f(Z) < e (a); therefore b a since e0 is a lattice inclusion

(4.9) and S n XD(a) = X (b) A XD(a) / 0. It follows that t U{XD(a)}

is-linked and so XD(a) E 1. A dual argument shows the other statement.








PROPOSITION 7.10. If 0 is an admissible subbase for X, then

0 is equivalent to 0 VU T.


PROOF. Let E = 0 U T. Then E is an admissible subbase for

X (7.9 (i)). For each e X X, (X fC 0) is a maximal linked system

of 0 (7.9 (i)) so that X X algebraically dominates X X via a unique,

continuous function f: X X X X which may be defined by

f(X) = (K n 0) for each 4,e X X (7.5). To see that f is an iseomorphism,

we need only show that f is a closed, injective lattice homomorphism (2.10).


f is injective: Suppose that -, Y) X X such that / Y,7 Then

we may assume, without loss of generality, that there exist T E (f n Z) ,

and S E (M rn E D) such that T n S = 0. If both T and S are in 0, then

clearly f(g) = (V n 0) X (t'n 0) = f(). Suppose that T 1 0 ;

i.e., there is some b E X such that T = X (b). Note that e (b) % f(N,

for if e (b) 5 f(;), then T = X (b) E P(7.9 (ii))--a contradiction.

Thus e (b) i f(/) and there is some T' e 0 such that T' E e (b) and

T' I f(W) = (>? n 0) (4.7). It follows that X (b) c T',

T' E (i n 0) = f(Z), and hence f(X) / f(/y). In the case that T E 0I

and S 1 0D, a dual argument shows that there is some S' E 0D such that

S c S', S' E f(M)), S' i (f n 0) = f(Y-), and hence f(X) / f0f).


f is a lattice homomorphism: Suppose C is a non-void subset

of X X. Since f is order-preserving, it follows that for each e C,

f(Y) 5 f(v C); therefore, v f[C] : f(v C). If T E v f[C] n D, then








for each E C, T E f(t) = (t 0) (4.8); hence T E/ for each E C

so that T E V C (4.8)'. Thus T E (v C n 0) = f(v C); i.e.,

(v f[C] r D ) C (f(v C) n 0 ). It follows that f(v C) 5 v f[C],

and so f(v C) = v f[C]. A dual argument shows that f(A C) = A f[C].


f is a closed map: Let R E E. If R E 0, then

f[R+] = {f(f): R E } = {f(Y): R e (Y ) 0)}

= {f(): R E f = {f(C): f(4) E R*}

= R* since f is surjective.

If R i 0, then R E E Oc T. Suppose that R E TD; i.e., there is some

a E X such that R = XD(a). Thus

f[R]+ = {f(c): t : e (a)} (4.10 (vi), (iv) and 7.3)

= {f(t): (X n 0) = f(c ) 5 f o e (a) = e (a)} (7.9 (ii) and

since f is order-preserving)

= {) EA X: e (a)} (since f is surjective).

Dually, if R = X (a) for some a E X, then

f[R+] = { E XX: e (a) ~}.

Thus the image under f of each member of + is a closed subset of XOX,

and since f is injective, it follows that f is a closed map.


It follows from the preceding arguments that there is an

iseomorphism f: X X X X such that f o e = eg. Thus 0 is equivalent

with 9 U T = E (cf., 7.6).




In view of Proposition 7.10 we may now assume without loss of

generality that,for our purposes, all admissible subbases contain T.

The following theorem now follows immediately from Propositions 7.4 and 7.10.







THEOREM 7.11. If E is any admissible subbase for X, then

X X algebraically dominates XTX and is algebraically dominated by X.





We consider now some situations in which one lattice super-

extension of X dominates another lattice superextension of X not only

algebraically, but also as a superextension.



PROPOSITION 7.12. If 0 and E are admissible subbases for X such

that 0 c E and 0 is weakly-normal, then

(i) for each o E X X, (G 0 0) is contained in a unique maximal

linked system of 0, and

(ii) the superextension X X dominates the superextension X X.


PROOF. (i) Suppose that E X X and ( n 0) is not contained

in a unique maximal linked system of 0. Then there exist IW, >7 e X X

such that M / A (f ) 0) c ; and (O r 0) c Y. Since Y/ Y?, there

exist T E g and S E Ysuch that S n T = 0 (2.16 (ii)). By the weak-

normality of 0, there is a finite screen {T1, T2, **, Tn } c of S and

T (2.11 (iii)). Since B X = X X, it follows that X is prime (2.18 (iii))

and hence there is some j {1, 2, -**, n} such that T. E I. Therefore,

T r T. / 0 and S n T. 0--a contradiction to the fact that

{T1, ', Tn} is a screen for S and T. Thus (X n 0) is contained in

a unique maximal linked system of 0.







(ii) By (i) above and Proposition 7.5 (i), there is a unique

dominating function f: X XX which may be defined by

f(4) = {T E 0: (o t 0) U{T} is linked} for each E e X X.

To see that f is continuous, let T E 0 and E X X such that

2 f-1[T^]. Thus T i f( ) so that {T} u (C n 0) is not linked.

Hence there is some S E (f r 0) such that T r) S = 0. By the weak-

normality of 0, there is a finite screen c 0 of S and T. Let

C = U{W : W E T and W r S = 0}. Clearly C is a closed subset of

X X. Since S E c and S n W = 0 implies that S+ FW+ = 0 (2.16 (i)),

it follows that W+ for each W E T with W n S = 0; i.e., d C.

If y is any member of f-[T*], then since E X X = U{W+: W c }, there

is some We such that e V; i.e., e. Since W e 0, we see that

S(9 0) c f(9) (7.2 (iii)), and since T E f(g), it follows that

T n W f 0; therefore, S W = 0. Thus a e U {W+: We T and

W S = 0} = C. It follows that f-i [T* cC and X / C. Thus since

Swas an arbitrary point not in f- [T*], it follows that f- [T*] is

a closed subset of X and so f is continuous.





If T is a non-void collection of subsets of X, then Tn denotes

the collection of arbitrary intersections of members of Y.



PROPOSITION 7.13. If 0 is an admissible subbase for X and
S= then
r = T UT then
D I'







(i) for each Xe E XX, (V n T) is a maximal linked system of

T, and

(ii) the superextension X X dominates the superextension X X.


PROOF. (i) Let E X X. Clearly ( r T) is a linked system

of T. If U E TD such that (V n T) U{U} is linked, then either

U = X E (~ T) or there is some a E X such that U = XD(a). To see

that XD(a) E e, let T E n oe By the hypotheses, T = M{V E TI:

T c V}. For each V E TI such that T c V, it follows that V E (X n T),

V n XD(a) X 0, and a E V; therefore a E T so that T n XD(a) f 0.

Since T was an arbitrary member of (.t 0 i), it follows that

(V n O I) U{XD(a)} is linked and hence U{XD(a)} is linked.

Consequently, XD(a) E ( n T). Dually, if U E TI and (X \ T) U{U}

is linked, then U e (X n T). Thus (.f A T) is a maximal linked

system of T.


(ii) That the superextension X X dominates the superextension

X X follows immediately from (i) above and Proposition 7.5 (ii).




We say that the lattice superextension X X is "larger" than

the lattice superextension X X (or, equivalently, X X is "smaller"

than X X) if and only if the superextension X X dominates the super-

extension X X.

In the following theorem we sum up the preceding propositions

obtaining two situations in which X X is actually the smallest lattice

superextension of X. The proof is immediate from Propositions 7.12

and 7.13.







n n
THEOREM 7.14. If T is weakly-normal or if F = TD U TI, then

XTX is the smallest lattice superextension of X; specifically, if 0 is

any admissible subbase, then the superextension X X dominates the super-

extension X X.




In view of the relationship between T, F and any admissible

subbase E, namely, T c E c F,it is natural to ask if X X is always

the smallest or if X X is always the largest lattice superextension of

X. In general the answer is negative to both questions. As noted

previously there is a topological lattice X for which the unique dominating

function f: X X -- X X is not continuous; i.e., X X is not smaller than

X X and X X is not larger than X X (8.7). Even under the restriction
n n
that r = T U T X X is not necessarily the largest. In Example (8.6,

Part II) there is a topological lattice which satisfies this restriction,

but for which there is an admissible subbase Z such that there is no

continuous dominating function f: X -X X; i.e., X X is not the largest

superextension of X.

The theorem below follows immediately from Proposition 7.12 (ii)

and Theorem 7.14. It is stated here for completeness.




THEOREM 7.15. If all admissible subbases for X are weakly-normal,

then X X is the largest lattice superextension of X and X X is the

smallest lattice superextension of X.








DEFINITION 7.16. (Kaufman [12]) Y is called an orderable

compactification of a totally ordered topological lattice X if Y is

a topological compactification of X and the order on X can be extended

to Y so as to yield the given topology on Y as the order (interval)

topology.



COROLLARY 7.17. (Kaufman) If X is totally, then X has a

largest and a smallest orderable compactification.


PROOF. Suppose that X is totally ordered. Thus each admissible

subbase for X is weakly-normal (4.12). By Theorem 7.15, FX is the

largest and X TX is the smallest lattice superextension of X. It is

shown in Appendix I that Kaufman's orderable compactifications of X are

precisely the lattice superextensions of X; therefore, X X is the largest

orderable compactification of X and X X is the smallest orderable compacti-

fication of X.




In view of Kaufman's method for obtaining orderable compactifi-

cations, we may characterize the lattice superextensions of a totally

ordered space X as the topological closures of hE[X] in I where I is

the closed interval [0,1], E is any subset of {f: f is a continuous

increasing map of X into I such that A f[X] = 0 and v f[X] = 1} which
E
separates the points of X, and h : X -* I is defined by nf(hE(x)) = f(x)

for each f E E and each x e X.















8. EXAMPLES


1 1
EXAMPLE 8.1. Let X = (0, ) U (-,1) with the usual ordering.

Then the completion of X by cuts, [0,1] with its usual ordering, is

iseomorphic with XTX (5.13), and the largest superextension X X is
1 3
iseomorphic with [0,-] U [ ,1] with its usual ordering. (The maximal
1 1
linked system of r defined by {(0,-)} U {(0,a]: < a < 1} U
'2 2
1 1
{[a,l): 0 < a < -} corresponds to the point in A X, and the one defined
2 4 F
by {(-,1)} U {(0,a]: < a < 1} U {[a,l): 0 < a < -} corresponds to
2 2 2
in X X.)
4 F





EXAMPLE 8.2. Let X = (0,1) with its usual ordering. Then

T = r and consequently every lattice superextension of X is iseomorphic

with [0,1] with its usual ordering.





EXAMPLE 8.3. Let n be a positive integer. For each

j e (1, **, n}, let I. = (0,1) with its usual ordering. Thus each

'I. is a topological lattice.

Let Y be the topological product of {I.: j = 1, ***, n}, and

for each j = 1, ***, n, let E. = T. (Note that for each I., T = F.)







-1
Thus E = {W. CT]: T E. j E {1, ', n}} is a T -subbase for Y (3.1),
] ] 1
and the superextension (de Groot compactification) of the product Y

with respect to E, X Y (B Y), is homeomorphic with the product of the

superextensions {X I : j = 1, .**, n} (3.3) (de Groot compactifications


{8 I.: j = 1, ***, n} (3.4)).

Since each I. = (0,1), it follows from Example 8.2 that X Y

is homeomorphic with a closed n-cell in Euclidean n-space; therefore,

a de Groot compactification of an open n-cell in Euclidean n-space is

the n-cell with boundary attached. Note also that since the results in

Chapter 3 (Products of Superextensions) were given for arbitrary products

of T1-spaces, we can generalize this example to the case that Y is the

product of an arbitrary collection of copies of (0,1). In particular,

in view of Anderson's result [3], a de Groot compactification of Hilbert

space is the Hilbert cube.


EXAMPLE 8.4. Let I = [0,1] with its usual

X = {x e I: x is rational}. Then under the usual c

logical lattice, and since I is the completion of X

iseomorphic with I (5.13).

For each irrational y E I, let y. and yu b

in X which correspond to y. Let

Y = X UI{y: y e I and y is irrational} U

{yu: y I and y is irrational}.


Ordering, and let

ordering X is a topo-

by cuts, X X is



e distinct points not








We extend the ordering on X to one on Y in the following way:


(a) If y c I and y is irrational, then for every x e X

and x' e X such that x < y < x', we let x < yp < y < x'.

(b) If y, z e I are both irrational with y < z, then we

let y <' z .


We prove the following statements:



(i) Y is a complete (totally ordered) lattice, and X ip a

sublattice of Y.


PROOF. It is easily seen from the definition of the ordering

on Y that X is a sublattice of Y, and since I is totally ordered, it

follows that Y is also totally ordered. To see that Y is complete,

suppose that C is a non-void subset of Y. Let C' = (C n X) U{y e I:

y is irrational and either yg E C or y e C}. Thus C' is a non-void

subset of the complete lattice I so that there is some c E I such

that c = v C'. If c E X, it follows from the definitions that c = v C
I Y
and. c E Y. If c is irrational and cu C, then again from the definitions

we see that c = v C Y. Finally, if c is irrational and c i C,
u Y u
then v C = c CY Y. A dual argument shows that A C exists in Y.



(ii) iAX is iseomorphic with Y.


PROOF. Define i: X Y by i(x) = x for each x e X; i.e., i is

the injection map of X into Y. Since X is a sublattice of Y, i is a







lattice inclusion. The qualities below follow readily from the defi-

nitions and some properties of the real numbers:


(a) If y E I is irrational, then

i[YD(y)] = i- YD(Y)] = n{X (x): x E X and y I x} and

i-lCY (y)] =i-1[Y (y ) = n{X I(x): x E X and x I y).
-[ -l
(b) If x E X, then i EY(i(x))] = XD(x) and i [Y (i(x))] = X (x).


It follows from (a) and (b) that i is continuous; i.e., i is a continuous

lattice inclusion.


(c) i[X] = i[X], i[X]la = {y : y I and y is irrational} UiEX],
lb
and i[X] = {y : y E I and y is irrational} Ui[XI.


It follows from (c) that Y = i[X]la Ui[X]b c i[X]2 c Y, i.e., Y = i[X]2

Thus Y is a topological lattice completion of X with respect to i (6.10).

To see that X is lattice dense in Y, suppose that v, w e Y with

v iX]la i[X], w i[X]b i[X]1, and w
(c) that there exist y, z e I such that y and z are irrational with

w = yu and v = z Since yu < z it must be the case that y < z

and so there is some rational number x E X such that y < x < z;

i.e., w = y
It now follows that Y is iseomorphic with X X where E = {i- [S]:
E (i 20). To see that Y is iseomorphic with X, we need only
S e 1 } (6.20). To see that Y is iseomorphic with A X, we need only








show that E = F. Recall that

Y) = {Y} U{Y D(i(x)): x e X} U {YI(i(x)): x e X}

{YD(yD): y e I and y is irrational} U

{Y (yu): y e I and y is irrational} (6.15).
-l
If T is a closed ideal of X, then either T = XD(x) = i- Y (i(x))] for

some x E X or T = {x e X: x < y} = i- [YD ()] for some irrational

y e I. In either case it is clear that T E D. A dual argument shows

that T. E E whenever T is a closed dual ideal of X. Thus E = r and Y

is iseomorphic with X X.





EXAMPLE 8.5. Let A, B, C be subsets of the plane defined as

follows:

A = [0,1) x {0}
2 2
B = {(x,y): (x-1) + y = 1, 0 < x < 2, 0 5 y < 1}

C = (3,4] x {0}.

Since A, B, and C are copies of half-open intervals we may assign to

each of them a natural (total) ordering from left to right. Also since

A and B are both copies of [0,1), there is a natural lattice isomorphism

p: A B. Let X = A U B u C. X becomes a lattice by defining a

supremum and an infimum for each pair of elements of X. Let z, w E X.

We let

Sp(w) v z, if w E A and z e B

z V w = w z = w V z if w, z e A, w, z e B, or w, z C

z if w E A U B and z e C







-1
Sw A p (z), if w E A and z e B

z A W = W A w A z, if w, z E A, w, z c B, or w, z E C

w, if w c A U B and z E C.

(i) The lattice described above can be visualized as follows:


(0,0) A (1,0)


(2,0)


0
(3,--) (4,0)
(3,0) C (4,0)


(ii) A X can be visualized as follows:
T


point in A X which is

not in eT[X].


eT[A] eT[C]








(iii) If 0 = T U T, then 0 = T U{AUB, C} and X X can be
visliD Ias
visualized as follows:


points in X X which are

not in e [X]


e [B



e -----A]
e@[A]


e EC]


(iv) F = T U{A, AUB, C} and X X can be visualized as follows:



e [B] points in A X which are

S/ not in eF[X]





er[A] e [C]




From (ii), (iii), and (iv) we see that three quite natural sub-

bases yield different lattice superextensions.





EXAMPLE 8.6. Suppose that L and M are lattices. We define the

lattice direct product, X, of L and M to be the set L x M with the




93


ordering defined by (a,b)
It is easily seen that the lattice direct product is indeed a lattice.




Part I


Let X be the lattice direct product (0,1) x (0,1) where (0,1)

is the open interval with the usual ordering. Let Y1 be the lattice

direct product [0,1] x [0,1],.where.[0,1] is the closed interval with

the usual ordering. X may be visualized as follows:


(0,1) (1,1)










(0,0) (1,0)

X


(i) X is a sublattice of Y, and Yi = = X2a X2b but the

inclusion map i: X Y is not continuous. Thus Y is not a topological

lattice completion of X with respect to i.


PROOF. It is easily seen from the definition of lattice direct


~la lb 2 2 2a 2b
product that X is a sublattice of YI, and it follows readily that

Y C XlaU Xlb C X2 i Y so that Y, = X2 (= X2arX2b).
I -- 1




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