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 onepoint 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 (StoneCech, Alexandroff, Freudenthal, Wallman, etc.) not one
1 v
yields this seemingly "natural" compactification. The StoneCech
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 Wallmantype 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 Wallmantype.
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 StoneCech 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 completionhence 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 Stonetech 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 nontrivial 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 nonvoid 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 nonvoid subset of
X.
(i) I isincreasing 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
aE 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 Fa 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 nonvoid 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 T1spaces. 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 T1space
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 weaklynormal if each pair of disjoint mem
bers of E can be screened by a finite collection from E.
(iv) A nonvoid subcollection X of E is said to be linked if
each pair of elements of / has a nonvoid 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 T1space 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 T1space 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 T1space 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 nonvoid
finite subcollection from Z has a nonvoid 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 weaklynormal, 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 weaklynormal 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 T1space
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 T1space X is completely regular if and only if there exists
a weaklynormal T subbase for the closed subsets of X.
EXAMPLE 2.19. Suppose that X is a completely regular T1space
and E is the collection of all nonvoid zero sets of X. Then E is a
V
weaklynormal T subbase for X, and E X is the StoneCech 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 T1spaces and show that a natural superextension of the product
of an arbitrary family of T1spaces 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 T1spaces, 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 A1 A1i +
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
A1 + + +
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 off 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 UTa 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 T1subbases
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 T1subbase 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 nonvoid 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 ED 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 weaklynormal and X X is totally ordered.
PROOF. To see that E is weaklynormal, 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 weaklynormal. 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 onepoint compactification.
DEFINITION 5.1. A completionoperator 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,
orderpreserving map.
Next, we consider a specific completionoperator on X.
PROPOSITION 5.3. The map D: 2X 2 defined by
A n{X (a): A c X (a)} is a completionoperator 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 orderpreserving. 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 nonvoid 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 {dl[A]: A T} = TD and {dl[A]: A T } = T ,
1 
note that d[X] = X and since d is a lattice inclusion,
d [X(d(a))] = XD(a) and dl[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 {d1[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
cnater 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 = {flV]: 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: fl[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: fl[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*({fl[V]: x E f1[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*({fl[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 f1] E, f1[V'] E Xf, and fEV] n fl[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 nonvoid 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, f1[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: fEV] EX} = (flEV)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*[(f1E[])] = fC[f '*l[V]] = V+
By condition (ii), {fl[V]: V E T} = 0 and so {(fl[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
d1Vl ] 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 weaklynormal 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 onepoint 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 onepoint 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 onepoint 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 followsthat 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 nonvoid 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 Xa 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 ncell in Euclidean nspace
can be characterized as the product of ncopies of the open interval
(0, 1), it follows from Theorem 5.13 and the Product Theorem (3.4), that
the closed ncell (i.e., the ncell 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 nonvoid 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 completionoperator 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 Ac 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 nonvoid 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 nonvoid 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 {il 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
il[S] n il[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 {il[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 (i1 [T: T E } = T and
{il[T]: T E } = T (6.7, 6.8, and 4.4). If T, S E T such that
T r S / 0, then ilT] n il[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 il[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) {il[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 = {i1 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 il[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 il[S] n i1T ] 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 il[Y (i(x))] 0 i YD(t)]
= i1[S] n il[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] il[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
Sil[ES] ilT].
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 il[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 = {il[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) = il[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 StoneCech 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, orderpreserving 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, ifR 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 orderpreserving, 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.
Tosee 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
fl[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 orderpreserving. Thus g o f(Y) : g o f(v R+) = v R since g is
orderpreserving. 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 orderpreserving, 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 orderpreserving
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] = gl[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 f1 = 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 nonvoid 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 orderpreserving. 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)}
islinked 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 nonvoid subset
of X X. Since f is orderpreserving, 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 orderpreserving)
= {) 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 weaklynormal, 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. 0a 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 f1[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 fi [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 nonvoid 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 weaklynormal 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 weaklynormal,
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 weaklynormal (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 ncell in Euclidean nspace; therefore,
a de Groot compactification of an open ncell in Euclidean nspace is
the ncell with boundary attached. Note also that since the results in
Chapter 3 (Products of Superextensions) were given for arbitrary products
of T1spaces, 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 nonvoid 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 nonvoid
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
ilCY (y)] =i1[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): (x1) + y = 1, 0 < x < 2, 0 5 y < 1}
C = (3,4] x {0}.
Since A, B, and C are copies of halfopen 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
