Lattice Topologies

Western Michigan University Western Michigan University

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Master's Theses Graduate College

6-1967

Lattice Topologies Lattice Topologies

Alan A. Bishop Western Michigan University

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LATTICE TOPOLOGIES

by . .Alan A'. Bishop

A Thesis Submitted to the

Faculty of the School of Graduate Studies in partial fulfillment

of theEducational Specialist Degree

Western Michigan University Kalamazoo, Michigan

June 1967

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.

INTRODUCTION

Several different topologies have been defined on lattices. In this expository paper we have collected and organized the known results which up to now have been scattered throughout the literature. With one exception, namely the metric topology discussed in the appendix, we have classified the topologies as being either an intrinsic or non-intrinsic topology. In the course of organizing related results we have changed and simplified many of the original proofs. For example the original proofs of Corollaries 1.1.7, 1.1.8, and 1.1.9 were independent of Theorem 1.1.6. By connecting them in this order and giving new proofs for Corollaries1.1.8 and 1.1.9 we were able to unify these results. Other modifications are present throughout the paper, primarily in Chapter 1.

We assume the reader has a basic knowledge of topology and lattice theory. All basic definitions and theorems from these areas not formally stated here can be found in [11] or [16], and in [6].

In writing this thesis, the author has benefited from the encouragement and patient advice of his advisor Professor Erik A. Schreiner. He also acknowledges the

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suggestions of Professors Gary Chartrand and Joseph McCully. The author wishes to thank them, as well as Western Michigan University which provided him the support via a Fellowship to continue his graduate work. Finally, the author wishes to thank his wife, Jeannie, for her support and encouragement.

Alan Arnold Bishop


MASTER'S THESIS M-1264

BISHOP, Alan Arnold LATTICE TOPOLOGIES.Western Michigan University, Ed.S., 1967 Mathematics

University Microfilms, Inc., Ann Arbor, Michigan


TABLE OP CONTENTS

PAGEINTRODUCTION ................................. iCHAPTER

I INTRINSIC TOPOLOGIES .................. 11. The Interval Topology .......... 12. The Order Topology........ 123. Order Compatible Topologies . . . 19

II NONINTRINSIC TOPOLOGIES .............. 261. The Complete Topology.... 262. The Prime Ideal Topologies . . . . 32

APPENDIX........................................ 43BIBLIOGRAPHY .................................... 50

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CHAPTER I

INTRINSIC TOPOLOGIES

1. The Interval Topologyt

By an intrinsic topology, we mean a topology defined in some natural way using the order relation of the lattice. In this chapter we shall investigate several such topologies. We are interested in those conditions on the lattice which insure certain desired propertiesof the topology such as Hausdorff, compact, and normal.

/The converse question is also of concern. For completeness sake, we begin with the following definition.

1.1.1. Definition. A Lattice is a partially ordered set L in which every two elements have a supremum and infimum. If every subset of L has a supremum and infimum, then L is called a complete lattice. If the partial ordering is total, then L is called a chain. Throughout this paper, L will always be used to denote a lattice. If L has a smallest element and a largest element, they will be denoted by 0 and 1, respectively. (Notice that if L is complete, 0 and 1 are in L). In L,if a < b and a <_ x <_ b implies x = a or x = b, we say bcovers a. In case b covers 0, b is called an atom. Lis said to be atomic if for each x ^ 0, there exists an

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atom b <_x.The interval topology was first introduced by Frink

[10] to generalize the usual topology of the real line in which each interval [a,b] is closed. We define the interval topology as follows and note that on the lattice of the real line, it coincides with the usual topology.

1.1.2. Definition. The interval topology of a lattice L is that topology obtained by taking as a subbase for the closed sets, all intervals of L of the form [a,») = {x:x >_ a) and (-»,a] = {x:x _< a}.

We observe that any closed interval [a,b] ={x:a x £ b} of the lattice is closed in the interval topology.

1.1.3. Lemma. Any lattice L is a T-^-space with respect to the interval topology.

Proof: Each {x} = [x,x] is closed.If L is a chain, we are able to say much more.1.1.4. Lemma (Birkhoff [6]). Any chain L is a

normal Hausdorff space in the interval topology.Proof: Since a normal T1-space is Hausdorff, it is

sufficient to show L is normal. Let S and T be disjoint closed sets in L with S' and T' their respective set theoretic complements. Observe that the open intervals form a basis for the open sets. Thus S* fl T' is a union of open intervals W (a e A). Using the Axiom of


Choice, choose x e W for each a e A. For each p e Sa anot in the interior of S, p is separated from T on either side by some W . Adjoin to S the open interval (p,xa) or (xo,p) as the case may be. After repeating the construction for T, we have the required disjoint open sets containing S and T, respectively.

1.1.5. Example. Let L be the lattice of all subspaces of a two dimensional vector space. This lattice is not Hausdorff in the interval topology since any basic open set containing 0 or 1 must contain all but a finite number of points.

In view of the above example, it is reasonable to ask under what conditions will the interval topology on a lattice be Hausdorff. The following necessary and sufficient condition was first proven by Baer [4].

1.1.6. Theorem. The interval topology of a lattice L is Hausdorff if and only if for all a x}) (J ((JyeB(z:z < y}).

Proof: Let L be Hausdorff and a < b. Then there- exist disjoint basic open sets, U and V, with a e U,

b e V. Since their complements are basic closed sets,


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u’ = ( u yeAl{z:z i y » ~ u (u xeA2tz:z i x}) and

V' = (UyeB {Z;Z - y}) U (U XeB2{z:Z - x})> where

Als A2S Bl» and B2 ai*e finite sutise'ts of L*

Let A = {x:x is minimal in A2 y andB = {y:y is maximal in A-j y B1>. Since L = U* |J V 1,A and B satisfy (ii).If x e A2, then a < x or a is incomparable to x

since a I U’. If x e B2 and x <_ a, then x <_ b. This implies b e V*, which is a contradiction. Thus, a £ x forall x e A. By similar arguments, x / b for all x e A, and a / y, y £ b for each y e B. Hence, (i) is satisfied.

To prove the converse, first let a < b. By assumption, there exists finite subsets A and B satisfying (i)

and (ii)~ Let V = U xeA*z:z - x* and u = U yeB*z:z - y**By condition (i), a £ x for all x e A. Hence, a / V. Moreover, since A is finite, V is closed. Thus V* is an open set containing a. Similarly, b e U’ which is open. Finally, by condition (ii), U* 0 Vf = 0.

Now let p and q be incomparable. Let a = p A 01 and b = p V Q* Furthermore, let A and B be the finite subsets satisfying (i) and (ii) with respect to a and b.Let B* = B y {p} and A* = A y {q}. Observe that A* and B* satisfy (i) and (ii) with respect to a and b.Define A^ = (x:x e A*, x £ p} and B^ = {y:y e B*, p £ y}.


Similarly, define A and B . Now, letq q

U = (^xeA {z:z - x}) U (UyeB {z:z - y}) andV = ( UxeA {Z:Z - X}) U ( UyeB {z:z 1 y})‘ Then<1 qp e U’ and a e V’ and these two sets are open.

Let x e A*—A . Then x < o. If x < a, thenp — - — *

x f. P A q = a. But this contradicts a £ x for allx e A*. Hence, x e A and it follows that A 1) A = A*.q p u qSimilarly, B |J B = B*. Thus by (ii), U \J V = L.P ^This implies that U* 0 V* = 0 so that L is Hausdorff.

Northam [23] and Wolk [25] independently found conditions for the interval topology, to be Hausdorff. By modifying the original proofs, we have their results as corollaries.

1.1.7. Corollary (Northam). The interval topology of a Boolean algebra L is Hausdorff if and only if L is atomic.

Proof: Throughout this proof x* denotes the uniquecomplement of x, the distinction between the complement of an element x or a set x, both now denoted x*, being clear from the content.

First assume L is atomic and x ^ y. Since x \ y* ^ or y A x* ^ there is an atom a contained in exactly one of x and y. Let a <_ x and a A y = 0* The closed intervals [a,l] and [0,a'] are disjoint since a £ z £ 1 and 0 £ z £ a* imply a £ a', which contradicts a ^ 0.


Moreover, if x £ [a,l], then x A a = 0, and thus,a* = 0 V a' = (x A a) V a’ = x V a'. This impliesx <_ a'. Hence, L = [0,a*] y [a,l] and [0,a’]', [a,l]’are disjoint open sets containing x and y respectively.

Conversely, let L be a Hausdorff Boolean algebrawhich is not atomic. Then there is an interval [0,b]which is a Boolean algebra with no atoms. Since L isHausdorff, [0,b] is also Hausdorff. Thus, there existfinite subsets A and B of [0,b] satisfying (i) and (ii)of Theorem 1.1.6. Let X = (a:a e A y B o r a ’ e A y B>and let {c1} c2,..., cfc} be the set of minimal nonzeromeets of subsets of X. Observe that c. A c = 0 fori ji t4 j . Since [0,b] has no atoms, for each ^ choose anonzero d. where d. < c.. We now show d = VT ..d. is1 I X Vl=l 1incomparable to each a e A y B so that [0,b] ^

(yxeA{z:z i x}) y ( y yeB{z:2 1 y>), Which contradicts (ii) of Theorem 1.1.6.

Since for each a' such that a e A y B there exists d. < c. < a1, we have d A a1 > d. > 0. Thus d /■ a fori i — — i —each a e A y B.

We now show that d A a < a for each a e A y B. By reindexing, let {c.:l£i<_n}be the set of all c. <_ a. If n = 1, then a A d = a A (v^--,^) = (a A dx) - -

yA Vj =2(a ^ = since a A d = 0 for all d / a.


n 1Suppose n > 1. It follows that < a for if

Vj=lCJ = a ’ thSn ° = V3=l(°n A cj> = 'n A =

c A a = c >0. If v! -.d, = a, we would have a sub- n n ,3 = 1 ,3

lattice of the form shown in figure 1.

.n-1cJ

0

n

Fig. 1

This lattice is not distributive, which is a contra

diction. Hence, < a* Using distributivity,

d A a = A a) < a since d A a = 0 if d i a.

But this implies a d. Since d £ a, we have the desired

contradiction. Hence, L is atomic.

1.1.8. Corollary (Wolk). If a lattice L fails to

contain an infinite set of pairwise incomparable elements,

then it is Hausdorff in the interval topology.


Proof: Let a and b be two elements such that a < b.If b covers a, then let A1 be a maximal set of pairwise incomparable elements containing b and B1 a similar set containing a. Define A = (A1 y B1)-{a} and B = (A1 y B ^ —{b}. If a < c < b, let C be a maximal set of pairwise disjoint elements containing c, and let A = B = C. In either case, A and B satisfy the conditions of Theorem 1.1.6 and thus, L is-Hausdorff.

By a proof similar to the preceding, the following result, due to Matsushima [20], can be obtained.

1.1.9. Corollary. The lattice L is Hausdorff in the interval topology if each a e L has a finite a-separating set (i.e., a subset S of elements incomparable to a with the property that any element, which is incomparable to a, is comparable to an element of S).

Recall that a subset of the real line is compact if and only if it is both closed and bounded. Such sets are complete lattices. We shall need the following topological result due to Alexander [1], proofs of which may be found in [1], [5], and [10].

1.1.10. Lemma. A T-^-space X is compact if there exists a subbase B for the closed sets which satisfies the condition: B has the finite intersection property implies fl{A:A e B} ^ 0.


1.1.11. Theorem (Frink). A lattice L is compact in the interval topology if and only if it is complete.

Proof: Let L be complete and C be any set ofclosed intervals [aa,bc(], o e A, having the finite

. intersection property. Since Ca-a,ba] fl [ag,bg] ^ 0for each a, 6 e A, [aQ V ag»t,a A T6 0* Thus,a < b , for each a, & e A and hence, 0 ^ [V a , A.bJ a — p a ct p pC 0{[a .bl : a e A}. In particular, if C is the sub- — a a

base of closed intervals, then C satisfies the condition of Lemma 1.1.10, and hence, L is compact.

Now suppose that L is compact. Let X be a nonempty subset of L. The family of intervals of the form (y:y £ x> has the finite intersection property and since L is compact, 0 Y{y:y < x} ^ 0. Thus A = { a : o e I } ,

X C a Cl

the set of lower bounds for X, is not empty. Moreover, for each {ai :1 <_i £ n> £ A and {x : 1 <_ j <_ m} £ X, n njm[ai,xj] = CViai, A^Xj] ? 0. Hence {[aQ,x]: a e I, x e X} has the finite intersection property, and since L is compact, there exists b e fl „T v[a ,x]. But then a < b < x, for all

C( C 1 j X C X Cl 01

a e I, x e X, so that b = Ax. By duality, VX exists.1.1.12. Example. Let X be any infinite set. The

set P(X) of all subsets of X, partially ordered by inclusion, is a complete atomic Boolean algebra and, hence, is a compact Hausdorff space in its interval topology.


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Now, let La (a e I) be a family of lattices with the interval topology. In general, the topological product of the Lq is not the interval topology of the direct product, even if I is finite. However, under certain lattice conditions the topologies coincide.These results are due to Alo and Frink [2].

1.1.13. Lemma. The interval topology of the direct product of an arbitrary number of lattices is contained in the topological product of the interval topologies of the lattices.

Proof: Let L = naLQ (a e I) with projection mapsD :L -*■ L (a e I). Then for each a e L, the subbasic - a a ’closed set [a,®) = Pi [a ,®)P-1, where a = ap . s a a* a * aSince pQ is continuous, [a,®) is closed in the product topology. Similarly, each (-®,a] is closed.

1.1.14. Theorem. The interval topology of the direct product of an arbitrary number of bounded lattices is equivalent to the topological product of the interval topologies of the lattices.

Proof: If 0q and 1q denote the smallest and largestelements of L , then 0 = (0 ) _ and 1 = (1 ) T are the a a acl a aeismallest and largest elements of L = n L . Observe that° a aa subbasic closed set of the product topology is of theform n [a ,b ], where a = 0 , b = 1 except for a a a a a* a afinitely many o. But then na[aa>ba = Ca,b], where


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a = (a ) and b = (b ), and thus is closed in the intervala atopology of L. Hence, the product topology is contained in the interval topology of L. Since the reverse inclusion holds by Lemma 1.1.13, the proof is complete.

We observe that if L1 has no largest element and Lg has no smallest element, then [a,°°)x(-«3b] is not an interval and the proof fails. In fact, it is not closed and hence, the topologies do not coincide.

We may weaken the boundedness conditions somewhat by imposing certain other restrictions on the lattices.

1.1.15. Theorem. Let C be a chain and L be a bounded lattice. Then the interval topology of CxL is equivalent to the topological product of the interval topologies of C and L.

Proof: By Lemma 1.1.13, it is sufficient to showthat any subbasic closed set A of the product topology is closed in the interval topology. First, let A = [a,®)x[c,d] c CxL. For each x > a in C, define A(x) = [a,x]x[c,d] y [x,°°)x[c,»). Observe that A(x) is closed in the interval topology and A c A(x), for every x ^ a. If (e,f) e D xA(x), then for x > e, (e,f) £ [x,°°)x[c,»). Thus, (e,f) e [a,x]x[c,d] c A and A =O xA(x) is closed in the interval topology. Similarly,(-*,a]x[c,d] is closed.


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1.1.16. Theorem. The interval topology of a finite number of chains, each with a smallest or a largest element, is equivalent to the topological product of the interval topologies of the chains.

Proof: We shall verify the theorem for sincethe proof of the general case is entirely analogous. Let

°1 E C1 and °2 e C2* ot,serve C°1ja1]xC02,a2 and[ais<»)x[a2,®) are subbasic closed sets of C^xCg. Furthermore, [01,a1]x[a2,®> and [a1,®)x[02,a2] are closed bythe proof of Theorem 1.1.15* Thus, the product topologyis contained in the interval topology of C^xCg.

The case where each chain has a largest element follows dually.

2. The Order Topology

It is well known that Moore-Smith convergence is intimately related to the topology of a set (i.e., a subset X is closed if and only if every convergent net in X converges to a point in X). Hence, given a type of convergence, we may define a topology on the set by the condition: X is closed if and only if it containsall points to which its nets converge. Such a topology was defined by Birkhoff [6], using order convergence.


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1.2.1. Definition. A net {x } is uo-directed if a —*---------a < 8 inrolies x < x„, and {x } is dovm-directed if— * a — p o —--------------a < 8 implies x > x0. We write x +x, if x is up-— ^ a — 8 a adirected and x = V x . Dually, x +x means x is down-a a 17 ’ a adirected and x = A x .a a

A net in a lattice L order converges to a e Lif there exist nets t +a and u +a, such thata a 3

t < x -a and x -*-b imply — — a aa = b.

Proof: There exist nets s +a, t +a, u +b, and v +b----- a a ■ 3 a a

such that s < x < t , u < x < v for all a. Hence,a — a — a3 a — a — afor each a,s < s_ < xD < v„ for all 8 > a. Thus,a — p — p — p —s < Adv„ = b, s o a = v s < b. Similarly, b < a.a — 8 8 Ya a — —

1.2.3. Lemma. Let'{x } be a net in the completelattice L. Then x-*a if and only if V A (xQ: 8 > a} =a p —a = A V {xQ: 6 > a}, a p —

Proof: If x -*-a, then t < t. < x0 < uQ a. Thus, t < A(x0: 8 > a} < V{x„: 8 > a} a } <a a — a a — —A„ V {x : 6 > a} < A u = a. a p — — a a

Conversely, if V A {x„: 8 > a } = a = A V {xQ: 8 > a},c t p — o plet t =A{x„: 8 > a} and u = V(xe: 8 > a}. Then t +a, o p — a p — oand u +a. a


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1.2.4. Definition. A subset X of a lattice L is O-closed if for every net {xq} C X, x0â implies a e X. The order topology of L is the topology in which a set is closed if and only if it is O-closed.

It is not immediately obvious that the order topology is, in fact, a topology. The crucial questions are resolved in the following lemma.

1.2.5. Lemma. Arbitrary intersections and finiteunions of O-closed sets are O-closed.

Proof: Let C (a e I) be O-closed and {x } a net a ain D C . If x -*a, then a e C for each o. Hence,a a a aa e 0 C . a a

If the Ci(l <_ i <_ n) are O-closed, and {x^: a e A} is a net in y {C. :1 <_ i £ n}, then there is a cofinalsubset B C"A, such that {x : B e B} is a subnet in one of— Pthe C., If x (a e A)-»-a, then there exist nets t +a and 1 a au +a with t < x < u . Moreover, since B is cofinal,a a — a — afor each a e A there exists $ e B, such thatt < t„ < u„ < u , so a = V t < V0t. < A-U. < A u .o - B — 6 - a a a — B B — 6 8 - a aThis implies t.+a and uQ+a. Since C. is closed, p p 1a e C. c I) {C.:l<i<n}.x — KJ 2_ — —

Observe that we have shown that if a net x->-a, thenevery subnet of {xq} also converges to a.

In the lattice of Example 1.1.5 of subspaces of atwo dimensional vector space, the order topology is


discrete, whereas the interval topology is not evenHausdorff. Hence, the two topologies are distinct;however, they are related as follows:

1.2.6. Lemma. The interval topology is containedin the order topology, implying the order topology is T1«

Proof: It is sufficient to show each interval [a,b]is O-closed. Let {x } be a net in [a,b], such that x -*x.a aThere exists nets t +x, u +x, such that t < x < u .a ’ a 3 a — a — aBut then, t < x < b and a < x < u for each a. Hence,a — a — — a — aa < A u = x = Vt <b.— a a a —

Since order convergence is unique, one might expect the order topology to be Hausdorff. However, Floyd [9] showed that the complete Boolean algebra of all regular open subsets (i.e., those equal to the interior of their closure) of the unit interval is not Hausdorff with respect to the order topology. This occurs since order convergence and topological convergence, although related, are not equivalent. For convenience, let us useO-convergence to mean topological convergence with respect to the order topology.

1.2.7- Lemma. If a net in a lattice L order converges, then it O-converges to the same point with respect to the order topology.

Proof: Let x^+a and suppose x does not O-convergeto a. Then there is an O-open set U containing a such


that {x } is not eventually in U. This implies {x } is a afrequently in U', the complement of U. Hence, there isa subnet {y„} in U'. By the proof of Lemma 1.2.5, y„â

p p ■ •and since U* is O-closed a e U’ contrary to assumption.Thus{x } O-converges to a. a

1.2.8. Theorem (DeMarr [8]). Let L be a complete lattice. If O-convergence implies order convergence, then L is a regular Hausdorff space with respect to the order topology.

Proof: By Lemma 1.2.2, L is Hausdorff. For eachnonempty open subset U, let S(U) = {x: Au <_ x ^VU}. Observe U C S(U) and S(U) is closed by Lemma 1.2.6. Let N(w) be the family of open neighborhoods of w e L, and D = {(U,y): U e N(w), y e U}. Define £ on D by

Ul,yi - U2,y2 lf and °nly lf U2 - Ul* Then D is adirected set. For each a = (U,y) e D, let x = y, formingathe net {x } which O-converges to w. By hyoothesis, ax ->w so that V{AU: U e N(w)> = w = A{VU: U e N(w)}, asince L is complete by Lemma 1.2.3.

Now suppose L is not regular. Then there exists an element w z L and an open neighborhood V of w such that for every open neighborhood U of w, S(U) fl V* ^ 0. For each U e N(w), select h(U) z S(U) fl V*, forming the net {h(U): U e N(w)}. Since h(U) e S(U), AU ■< h(U) <_ V(U). Thus, h(U)->w. By Lemma 1.2.7, (h(U)} is eventually in V, contrary to h(U) e V* for all U z N(w). Hence, L is


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regular.The remainder of this section is devoted to finding

sufficient conditions for a lattice to be a compact Hausdorff space with respect to the order topology. The results were apparently first proven by Ward [24] using filters, but the proofs shown here using nets are due to Atsumi [3]. Using the convention established with0-convergence, we shall say a net I-converges if and only if it converges with respect to the interval topology.

1.2.9. Lemma. Let L be a complete lattice. Auniversal net {xq} in L I-converges to x if and only if

V A (x0: 6 > c t } < x < A V {xe: B > a}, a B — — — a B —Proof: Let {xq: a e A} I-converge to x, and suppose

V A{x„: B > a } / x o r x / A V {xQ: B > a}, a B — — — a B —If V A {x„: B > a} / x, there exists a e A such that a B — —t = A{x„: B > a} / x. But then {x } is not eventuallya p — — ain the complement of [t ,1], which is an I-open neighborhood of x, contradicting the hypothesis that x1-converges to x. Thus, VQ A {x : B ^ a} <_ x. Byduality, x < A V(xe: B > ct}.— a p —

Conversely, let V A {x„ : B > a < x <a B — _ _A V { x : B > a } . If x does not I-converge to x, thereCt p ct

exists a basic open neighborhood U of x, such that (xa> is not eventually in U. Since {x^} is universal, it is eventualxy in U’ which is a basic I-closed set.


Now U’ = (Ui=i U (U “=1 C0,y3]) so that {xq}

is eventually in some [y^l] or [0,y^]. If {xq} Iseventually in [y.,l], then for some a e A,y. < A { x „ : 3 > a} < v A { x Q : 3 > a} < x. Thus,1 — 3 — — a c — —x e [yj.,1] c U', which is a contradiction. Since wearrive at a similar contradiction if {x } is eventuallyain [0,yj], {xq} must I-converge to x.

As an immediate corollary, we have that if L is also Hausdorff in the interval topology, then, by Lemma1.2.3, a universal net {xq} I-converges if and only if it order converges.

1.2.10. Theorem. A lattice L is compact and Hausdorff in the interval topology if and only if every universal net in L is order convergent.

Proof; L is compact and Hausdorff in the interval topology if and only if every universal net I-converges to one point and L is a complete lattice (see Theorem1.1.11). By the preceding remark, this is true if andonly if every universal net order converges.

1.2.11. Corollary. If a complete lattice L is Hausdorff in the interval topology, then

(i) L is a compact Hausdorff space with respect tothe order topology, and

(ii) the interval topology and the order topology coincide.


Proof: (i) By Lemma 1.2.6, the order topologycontains the interval topology so that L is Hausdorff with respect to the order topology. By Theorem 1.2.10, it is compact.

(ii) The identity mapping from L with the order topology (which is compact) to L with the interval topology (which is Hausdorff) is a homeomorphism.

We observe in the above corollary L is compact (with respect to the order topology) in a very strong sense since every universal net is not only O-convergent but also order convergent.

We conclude this section by observing that if L is either a complete chain or the lattice P(X) of subsets of a set X, then L is a compact Hausdorff space in the order topology (see Lemma 1.1.4 and Example 1.1.12).

3. Order Compatible Topologies

We now vary our approach, in that we define and examine a class of topologies, rather than a single topology. This class, the order-compatible topologies, was first introduced by Wolk [25], whose results were later improved upon by Naito [22].

1.3.1. Definition. A subset S of a lattice L is called up-directed if for all x, y e S there exists z e S, such that x <_ z and y <_ z. Dually, S is called


20

down-directed if x, y e S imply the existence of z e S with z <_ x and z _< y.

A subset K of L is called Dedekind-closed or simply D-closed if for every up-directed S C K with VS e L and every down-directed T c K with AT e L, VS e K and AT e K.

We remark that if t +a, then the set of all t isa aan up-directed set. Dually, if Ug+a, then the set of all ua is a down-directed set. Conversely, if S c L is up-directed, then the identity map i:S->-L is an up- directed net. If S is down-directed, then by replacing the order relation of S by its dual, i:S-»-L is a down- directed net. Recalling the proof of Lemma 1.2.5, we note that finite unions and arbitrary intersections of D-closed sets are again D-closed. Hence, we may define a topology using these special sets.

1.3.2. Definition. The D-topology of a lattice Lis the topology in which K c L is closed if and only ifK is D-closed.

Now, since all up-directed nets t +a and down- 3 adirected nets u +a are order convergent, we observe each aO-closed subset is D-closed.

1.3.3. Lemma. The D-topology contains both the order topology and the interval topology.

1.3.4. Definition. An order-compatible topology


21

is one containing the interval topology and contained in the D-topology.

Our goal is to find sufficient conditions for a lattice to have a unique order-compatible topology.

1.3.5. Definition. Let {x } be a net in a lattice ---------- aL. For S C L, denote by U(S) the set of upper bounds ofS and by L(S) the set of lower bounds of S. IfP = M-L (x : o > 6} and 0. = H„U {x : a > 8}, then y is '-’B a — v g a —medial for if y e U(P) 0 L(Q).

1.3.6. Lemma (Ward [24]). If {x } is a net in a — — o

lattice L which I-converges to y, then y is medial for{x }. a

Proof: Let x1 e L such that y £ x1« ThenV = {x:x >_ x1) is an open neighborhood of y and txa> iseventually in V. For some 8, xq / X-j for all a >_ 8.Thus, for all 8, x1 £ L {xq: a >_ 8>, which impliesx1 I P. Hence, if x1 e P, then y >_ x1# implying thaty e U(P). Dually y e L(Q).

1.3*7. Lemma. Let {x : a e A} be a net in L which ------ aI-converges to y. If xq is incomparable to y for all a e A, then there exists an infinite set of pairwise incomparable elements contained in {x q : a e A}.

Proof: Since every net has a universal subnet andevery subnet of a convergent net converges to the samelimit, we may assume {x : a e A} is universal. If {x }3 J a a


22

I-converges to y, then y e U(P) H L(Q). Moreover,since each x is incomoarable to y, x £ P and x £ Q a a a

for each a e A. Now choose a1 e A. The sets{x:x > x > and {x:x < x } contain no set of the form

°1 °1{x : a > 3} for any 8. Since it is universal, {x } is a — aeventually in {x:x / x } and {x:x / x }. Thus, there- ° 1 - a x

exists B1 > o1 and y1 > a1 such that{x : a > 8,} C {x:x £ x } and {x : a > y.} c. {x:x £ x }.a — 1 — — c*! a — '1 — — cx

Now choose ag with ag >_ Bx and a2 — Yl* I-t is clear thatx is incomparable to x . In a similar manner, we may a2 °1choose an infinite sequence aÔg,^*... of elements of Asuch that x is incomparable to x for i j4 j.

°i1.3.8. Lemma. Let y be medial for a net {xa: a e A}

In L. If x < y for all a e A, then y = v x •a J 3 J *ct aProof: Let z > x for each a. Then z e Q and since — a

y e L(Q), y <_ z.1.3.9. Lemma. Let L be a lattice containing no

infinite set of pairwise incomparable elements. If{x : o e A} is a net in L and y = V{x.:BeC} for each a 8subnet {x„: 6 e C} of the net, then there exists an up-Pdirected set M c {xq: o e A} such that y = VM.

Proof: Let S = {x : a e A} and assume y ^ VM for ----------- aany up-directed set M1 c s. If M1 contains a subnet {x^} of {xa: a e A}, then y = VgXg £ VM1 <_ VS = y. Hence, M1


23

contains no subnet of {x^: « e A}. Therefore, thereexists an e A such that x e S-3VL for all a > an . Next 1 a ± — jlchoose a maximal up-directed subset M„ of {x : a > >.d a — ±

Since y ^ VMg, there exist an element e A and a maximal up-directed set M^ c {x^: a >_ o2> c (S-M1)-M2. Continuing this process, we obtain an infinite sequence Mi,M2,••• of pairwise disjoint maximal up-directed sets which satisfy the condition: x e M . , y e M , i < j implyx £ y.

Since M. is a maximal up-directed set, tj.M. is noti ^ 1 1up-directed. Thus, there exist a. e M, and bn e M suchr 1 k l nthat there exists no z e II.M. with z > a, and z > bn,i i — 1 — 1

which implies there exist infinitely many Mi containing no upper bound of ai or b ^ Without loss of generality, assume this is the case for a ^ Let Ci = ai and from those Mi containing no upper bound of ax, choose such that j >_max(k,n). Then Ui>j {M^Mi contains no upper bound of ai} is not an up-directed set, so we may continue the process constructing an infinite sequence c^Cg,..., which satisfies i < j impliesc. £ c . Since we have al-ready shown i < j implies c. £ c , the set {c.:i = 1,2,...}is an infinite set of pairwise incomparable elements contradicting the hypothesis, completing the proof.

1.3.10. Theorem. If a lattice L fails to contain an infinite set of pairwise incomparable elements, then L possesses a unique order-compatible topology.


24

Proof: It is sufficient to show any D-closed subset K is closed in the interval topology. Let the net~{x : o e A} in K I-converge to y. By Lemma 1.3.7, there aexists no subnet (xe: $ e B} with each x. incomparable

p pto y. Hence, there exists a subnet {x^: B £B} with each jc < y or with each xQ > y. If each x < y, thenP P Py = V{xQ: B e B} (Lemma 1.3.8). By Lemma 1.3*9, therePexists an up-directed set M c {x^: B e B} c k with y = VM. Since K is D-closed, y e K. The case where x^ > y follows similarly using the dual statements of Lemmas1.3.8 and 1.3*9, which also hold.

1.3.11. Theorem. A complete lattice L possesses a unique order-compatible topology if and only if it fails to contain an infinite set of pairwise incomparable elements.

Proof: The complete lattice L is compact in theinterval topology, by Theorem 1.1.11. If L possesses aunique order-compatible topology, then it is compact inthe D-topology. Suppose (a^ i = 1,2,...} is an infiniteset of pairwise incomparable elements. ThenFn = {ai: i >_ n} is closed in the D-topology and thefamily of all Fq has the finite intersection property,but O f ¥= 0 contradicting the comoactness of L. n n w

The converse is Theorem 1.3.10.Recall in Example 1.1.12 that P(X), the set of


25

subsets of an infinite set X, has interval and order topologies which coincide. The atoms of P(X) form an infinite set of pairwise incomparable elements. Therefore, the D-topology is distinct from the interval and order topologies.


CHAPTER II

NONINTRINSIC TOPOLOGIES

1. The Complete Topologyi

The nonintrinsic topologies of a lattice are those which, in general, are not related to the order- compatible topologies; however, under certain specified conditions, they may be order-compatible. In this section, we consider the complete topology as defined by Insel [12]. The results given here are from [12] and [13].

2.1.1. Definition. Let L be a lattice. A nonempty subset X c L is called complete if and only if for every nonempty subset A c X, V A e X and AA e X.The complete topology or simply the K-topology is the topology of L having the family of complete sets as a subbase for the closed sets.

2.1.2. Lemma. Let F be a nest of complete subsetsof a lattice L. Then 0 F y6 0.

Proof: Let S = (VN:N e F) and consider a fixedNq e F. Furthermore, let a = VNQ, and defineSQ = {s:s e S,s <_ a}. For each s e SQ c S, there existsN e F such that s = VN < a. Since F is a nest ofs s —complete sets, s e Ng c NQ. Thus, SQ C NQ.

26


Now let s, t e S. There exist N and N in P suchS vthat s = VN and t = VN . Since either N c N orS S “ uN, c N , we either have s < t or t <_ s. Hence, for each x e S, x <_ a or x > a. If x > a, then ASQ a < x since SQ c NQ. If x < a, then x e SQ and hence,ASq <_ x. Thus, ASq is a lower bound for S. If y < s for all s e S, then y <_ ASQ since SQ C S. Therefore,AS = ASq e Nq for each NQ e F and so AS e H P.

2.1.3. Lemma. Let A be any collection of complete subsets of a lattice L such that A has the finite intersection property. Then f1 A ^ 0.

Proof: The lemma is proved by transfinite induction on the cardinality of A. If A is finite, the lemma is true. Suppose it holds if A has cardinality less than a. Let r be the set of ordinal numbers which are less than the least ordinal number of cardinality a. Assuming the cardinality of r is a, we index A by r, letting B„ e A correspond to the ordinal B e r . DefinePCQ = D {B : ye r, y < 6} and note C ^ 0 by the inducts Y ~ ption hypothesis. Since {C.: B e r} is a nest of com-Pplete subsets of L, by Lemma 2.1.2 0 A = 0 o _C0 f1 0.p £ 1 p

2.1.4. Theorem. Any lattice L is compact in its K-topology.

Proof: The complete sets are a subbase for theclosed sets satisfying the condition of Lemma 1.1.10.


28Hence, L is compact.

2.1.5. Lemma. In any lattice L, the K-topology is contained in the order topology. Moreover, the interval topology is contained in the K-topology if and only if L is complete.

Proof: Let K be any complete set in L and suppose{xa> is a net in K order converging to x. There existnets t +x and u +x such that t < x a} < V{x„: 3 > a} a} < A„ V (x„: S > 0} < Au = x. a — a p — — a p — — aSince K is a comnlete set x = V A {x0: B > a} e K anda B —hence, K is O-closed.

If L is complete, the subbasic closed sets (with respect to the interval topology) of the form [a,l] and [0,a] are complete sets. Thus, the interval topology is contained in the K-topology. Conversely, if the interval topology is contained in the K-topology, it is compact and L is complete.

Observe that if L is not complete, then the K-topology does not contain the interval topology and hence, the K-topology is nonintrinsic.

2.1.6. Definition. Let L be a complete lattice.If A is a nonempty subset of L, then Aq is the smallest complete subset containing A.

We wish to find a sufficient condition for a


compact topology of a complete lattice to be contained in the order topology. Using the convention established earlier, we shall say a net K-converges to x if and only if it converges to x in the K-topology.

2.1.7. Lemma. If a net {xq} in a complete latticeL K-converges to x, then

(i) x e D {x_: 8 > a}*1, andP —(ii) V A{xe: B > a} < x < A V{x_: 8 > a}.CL p •“* ” CL p —Proof: (i) If txa> K-converges to x, then each

subnet of the form {x„: 6 > a} K-converges to x. How-P —ever, this subnet is a net in {x : 6 >_ a}q which is closed, so x e (xQ: 8 > a}*1. Since this is true forP —each a, x e fl {xD: 8 > a}*1.3 ' a 8 —

(ii) For each a, A{xe: 6 > a} = A{x„: 8 > a}11 < x.P — P —Hence, V A(x_: 8 > a} < x.

3 a 8 — —Dually, x < A V {x.: 6 > a}.— a p

2.1.8. Theorem. Let L be a complete lattice and suppose T is any compact topology of L. Then if the K-topology is contained in T, T is contained in the order topology.

Proof: Let C c L such that C is not O-closed.Then there exists a net {x } in C with x -»-x and x £ C.a aSince L is complete, Vq A(x^: 8 > a} = x =A V{x„: 8 > a} by Lemma 1.2.3. Since T is comoact, a B —there exists a subnet {ya) of {xq} which T-converges and


30

hence, K-converges to a point y e L. By Lemma 2.1.7,x = V A (x *. 8 > a} < V A(yQ: 8 > a} < y < a p — — a p — — —A V : 3 > a} < A V { x e: 8 > a} = x. Hence, y i Ca p — — a p —and C is not closed in the T-topology.

We now establish conditions for the interval and complete topologies to coincide.

2.1.9. Lemma. Let {x } be a universal net in a ------ acomplete lattice L. Then {xq} K-converges to x if andonly if x e H ^ {x^: 8 ^ a}q.

Proof: Suppose {xq> does not K-converge to x.There exists a basic open set U such that {xq} is noteventually in U. Since {x } is universal, {x } isa aeventually in L-U and hence, in a complete set C c L-U. There exists o such that {x„: 8 > a} c C which impliesP — —{x_: 8 > a}*1 c C. Since x I C, x t fl (x0: 8 > a}.p — — a p —

The -converse is Lemma 2.1.7(i).2.1.10. Lemma. Let {x } be a universal net in a — --- a

complete lattice and let A = H^Xg: 8 >_ a}q. Then(i) A V {x0: 8 > a} = VA, ando p —(ii) V A {x : 8 > a} = Aa .a p —Proof: (i) For each a, A c {x„: 8 > a}-. Let— — — P —

s = V{x_: 8 > a}. Since VA < s for each a,a 8 — — aVA <_ âsQ. Now let y be arbitrary. For each a, there exists 6 such that 8 >_ a and 6 >_ y. Thus,{xg: 8 >. 6} c {xg: 8 >_ a} and {xg: 8 >. 6} c {xg: 8 >_ y}q.


31

Hence, s, < s and s. e {xe: g > y}*1 for all 9 6 — a 5 g —6 >_ a, S >_ y. Since this is true for each a,Ms.: 6 > a, 6 > y ) < A s . The reverse inequality0 — — — CL OL

is obviously true; hence, s = Ats : 6 >_ a, 5 >_ y} e {x0: g > y>q- for all y. Thus, s e D {x0: g > y}q = AP — Y p —and A V {xc: g > a } = A s (v< VA. a p — a a —

By duality, (ii) follows.2.1.11. Lemma. A lattice L is complete if it is

Hausdorff with respect to its complete topology.Proof; If x, y e L, there exist basic open sets

U and V such that x e U, y e V and U D V = 0. There exists a finite number of subbasic closed (I.e., complete) sets such that L = . Now for eachnonempty S c L, S = y “=1(S 0 Ci). Since each is complete, V(S D C^) and A(S D C^) exist. Therefore,VS = V.(V(S fl C )) and AS = A±(A(S 0 C^)) exist.

2.1.12. Theorem. For any lattice L, the following conditions are equivalent;

(i) L is Hausdorff in its complete topology.(ii) Every universal net in L is order convergent.(iii) L is a compact Hausdorff space with respect

to its interval topology.Proof: (i)=>(ii). L is complete by Lemma 2.1.11.

Let be a universal net. Since L is compact and


Hausdorff, K-converges to exactly one point x. ByLemma 2.1.9, {x} = D (x_: g > a}^ Thus,a p —A v (xe: g > a} = V{x} = A{x} = V k i ^ 0’ 6 > a} by

CL p “ a l x p —Lemma 2.1.10. Applying Lemma 1.2.3, we have {x^} order converges to x.

(ii)=v(iii). This is Theorem 1.2.11.(iii)=>(i). Since the interval topology is compact,

it is contained in the K-topology by Lemma 2.1.5. If the interval topology is also Hausdorff, so is the K-topology.

We observe that by Corollary 1.2.11, any of the above equivalent conditions are sufficient to make the interval, order, and complete topologies coincide.

2. The Prime Ideal Topologies

As we observed in the remark following Theoremo1.1.14, the interval topology of R (where R is the set •

of real numbers) does not coincide with the usual topology of the real plane. In fact, none of the topologies thus far defined have this property. The primary purpose of this section is to define a lattice topology which, on the lattice Rn, coincides with the usual topology of Rn. The results in this section are due to Naito [21].


33

2.2.1. Definition. A subset I of a lattice L is an ideal if the following conditions hold:

(i) x e l , y e L , y<_x imply y e I.(ii) x e I, y e I imply x y y z I.If, in addition, x A y e I implies x e I or y e I,

then I is called a prime ideal.A prime ideal I is a complete prime ideal or

CP-ideal if 0 j* A c I and VA exists imply VA s I.For any a e L, there exists, by Zorn's Lemma, a

minimal prime ideal containing a. We shall say such anideal is an MP-ideal.

Dually, we define dual prime ideals, dual CP-ideals, and dual MP-ideals.

We note that the set theoretic complement of a prime ideal is a dual prime ideal, and conversely.

The P-ideal topology of a lattice L is that topology for which the family of prime ideals and dual prime ideals is a subbase for the closed sets. Similarly, the family of CP- and dual CP- ideals and the family of MP- and dual MP- ideals are subbases for the CP-topology and MP-topology, respectively. Observe that the P-ideal topology contains each of the other two topologies.

We remark that in any chain, the CP-topology and the MP-topology coincide with the interval topology, while the P-ideal topology is discrete. The following example


illustrates that these topologies are distinct and, in general, not compatible with the order.

2.2.2. Example. Let L = {(x,y): 0 < x < 1,0 < y < 1} (J {(0,0),(1,1)}, (a subset of R2). The prime ideals are subsets of L of the form ({(x,y): 0 _< x < a}, {(x,y): 0 £ x <■ a},{(x,y): 0 <_ y < a}, or {(x,y): 0 < y < a}. Since (1,1) is the join of each of these prime ideals, L is the only CP-ideal. Dually, L is the only dual CP-ideal. Hence, the CP-topology is indiscrete and properly contained in the interval topology.

The MP-ideals are sets of the form {(x,y): O ^ x ^ a and {(x,y): 0 <_ y _< a}. Consider the sequence (1-j, i) + (1,1) in the MP-ideal I = {(x,y): y <_ i}.Since (1,1) / I, I is not D-closed and hence, the MP-topology is not contained in the D-topology. In both cases, the topologies are not order compatible.

2.2.3. Theorem. In any lattice L, the operationsof meet and join are continuous with respect to the P-ideal topology, MP-topology, and CP-topology.

Proof: We shall prove the theorem for the P-idealtopology only, the other two cases being similar.

Let U(a V b) be any basic open set containing a V b. Then U(a V b) = D {I£: 1 <_ k <_ n}, where Ifc is a primeideal or a dual prime ideal. If Ik is a prime ideal,


35

then a V b e 1^ implies a e or b e 1^. If a e I£, let Ufc(a) = and Uk(b) = L. On the other hand, if a i I£, let Uk(a) = L and Ufc(b) = I£. If Ifc is dual prime ideal, then a V b el/ implies a e I’ and b e I’.

a . 1C &Hence, let U (a) = Ufc(b) = I£. Letting U(a) = flk_1Uk(a) and U(b) = rik=1Uk(b), we have open neighborhoods of a and b, respectively, such that x e U(a) and y e U(b) imply x V y e U ( a V b ) .

Dually, A is continuous.Recall that L is a 1^-space with respect to each of

the interval, order and D topologies. In the prime ideal topologies, this is not always the case. Consider the five element nonmodular lattice illustrated in Figure 2.

1

a

b

0

Pig. 2


36

The prime ideals and dual prime ideals are L, 0, {0,c}, {0,a,b}, {l,c}, {l,a,b}. The closure of {a} is equal to the closure of {b} so that L is not a TQ-space and therefore not a T1~space.

In view of this example it is of interest to find conditions under which the prime ideal topologies are a T1-space.

2.2.4. Lemma. A lattice L is a TQ-space with respect to its P-ideal topology if and only if for any a,b e L with a < b, there exists a dual prime ideal containing b, but not a.

Proof: Let p, a e L, p ^ a and assume p y q z 1 andp i I, then q e I, p e I* and I’ is an open set separating p and q. Thus L is a TQ-space.

If L is a TQ-space and a < b, there exists a subbasic closed set I containing exactly one of a and b.If I is a dual prime ideal, b e I and a I I. If I is a prime ideal, a e I and b £ I. But then I1 is a dual prime ideal containing b, but not a.

By a proof similar to that of the above lemma, one can verify the next result.

2.2.5. Lemma. A lattice L is a T -space in the P-ideal topology If and only if for any a, b e L witha < b, there exist a prime ideal containing a, but not b


and a dual prime ideal containing b, but not a.Observe that the above lemma holds for the MP- and

CP-topologies if prime ideals are replaced by MP-ideals and CP-ideals, respectively, and dual prime ideals are replaced similarly.

2.2.6. Theorem. In any lattice L, the following conditions are equivalent:

(i) L is a T1-space in the MP-topology.(ii) L is a TQ-space in the MP-topology.(iii) L is a TQ-space in the P-ideal topology.(iv) L is a distributive lattice.Proof: The implications (i) ■=> (ii) and

(ii) => (iii) are obvious.(iii) => (iv). For each x e L, let

E(x) = {I: x e I, I is a dual prime ideal}. By Lemma2.2.4, x = y if and only if E(x) = E(y). Observe that

I e E(x \f y) <==> x v y e I <==> x e I ory e I <=> I e E(x) \J E(y). Hence, E(x V y) =E(x) (J E(y). Similarly E(x A y) = E(x) 0 E(y). The mapping x -*■ E(x) is an isomorphism of L onto a sublattice of the Boolean algebra 2E where E is the set ofall dual prime ideals. Hence, L is distributive.

(iv) => (i). Let a < b. By Zorn's Lemma, there exists a maximal ideal I in [a,b] which does not contain b. Suppose x, y e [a,b], x A y e I and x I I. If


x v z for all z e I, then there exists a maximal ideal of [a,b] containing I and x, contrary to the maximality of I. Thus, there exists z e I such that x v z = b. Since z f ^ y < _ z e I 9 Y = b h y =

(x V z) A y = (x A y) V (z A y) e I. Hence, I is aprime ideal in [a,b] which does not contain b. LetJ = {x e L: x A b <_ w, for some w e I}. Clearly J is anideal of L. If x A y e J, then there exists w e I suchthat (x a y) A b ^ w. We have w = ((x A y) A b) V w = ((xA b) V w) A ((y Ab) Vw). Since I is prime, ( x A b ) V w e I o r ( y A b ) v w e l . Hence, x e J or y e J so J is a prime ideal containing a, but not b. Dually, there exists a dual prime ideal containing b, but not a. By Zorn's Lemma, there exists an MP-ideal containing a, but not b, and a dual MP-ideal containing b, but not a. Thus, L is a T1~space in the MP-topology.

We note that if L is a TQ-space with respect to the CP-topology, then L is distributive. The converse statement does not hold since the lattice L ={(x,y): 0 < x < l , 0 < y < l } (J {(0,0), (1,1)} c R2 of Example 2.2.2. is a distributive lattice but is indiscrete in the CP-topology.

2.2.7. Theorem. If a lattice L is a T^space in its CP-topology, then it is infinitely distributive.

Proof: Suppose V x exists. Let b = a A V x so------ a o a a


39

b > a A x for each a. If b ^ V (a A x ), there exists — o a a 9 —c e L such that a A xq <_ c < b for each a. By Lemma2.2.5, there exists a CP-ideal I containing c, but not b.Thus, a A xq e I for each a. Since I is prime, a e I orx e l for each a. If a e I, then since b < a, b e I. o 9 — 9

If x e l for each a, then y x e l and hence, b e I. a ’ a a ’In either case, we have a contradiction. Thus,a A V x = b = V (a A x ). Dually, a v A x = A_(a V x ).a a a a a a a a

2.2.8. Theorem. If a lattice L is Hausdorff with respect to its CP-topology, then L is regular.

Proof: It is sufficient to prove that for any sub-basic closed set I and any a £ I, there exist disjoint open sets U and V containing a and I, respectively. Suppose I is a CP-ideal. Since a is an upper bound of {x:x e I,x <_ a} and a t I, there exists an upper bound b of {x:x e I,x <_ a} such that b < a. Since L is Hausdorff, there exists a finite cover of L of subbasic closed sets{I : 1 <_j <_ n} such that each I contains at most one of

J J

a and b. Observe that a e I, implies I is a dualJ 0

CP-ideal. Let E be the union of all I which do not con-J

tain a, and let P denote the union of all I which con-tain a. Then E' and F' are disjoint open sets such thata e E'. We need only show I = F 1 . If x e P 0 I, thenx A a e I and x A a e {x:x e I,x <_ a}. Hence, x A a < b.Since x e P, there exists a dual CP-ideal I containing x.£


MOWe have b _> x A a e 1^, which is a contradiction, so I c p'. The case where I is a dual CP-ideal is proved similarly.

Recall that the interval topology of the cartesian product of the lattice Lq is not in general the topological product of the interval topologies of the Lq.

2.2.9. Lemma. A subset I of the cartesian productL = n L is a CP-ideal if and only if I = n I , where I a a a a ' ais a CP-ideal of L and I = L , except for at most one a.a a a 3

Proof: Let I be a CP-ideal of L and I the oro--------------------------------------- ajection of I in L . We first show I is a CP-ideal of L ." a a aLet a e I and b < a . There exists a e I such that a a a — aap = a where d :L-*L is the projection map. Let b be a * a athe element of L with bp = b and bpe = ap. for eacha ^8 88 ^ a. Then b < a e I and b e l .— 0 0

Now suppose M £ I and VM exists in Lq. Let a be afixed element of I. For each x e M, there exists x e Io 3such that xp = x . Denote by b(x ) the element of La a afor which b(x )p = x and b(x )p. = ao. for 6 ^ a. a a a r 8 ‘ 8Since b(x) < a v x e l , b ( x ) e l . Let d be the elementa — ’ ’ aof L which has dp = VM and dp. = ap. for 8 ^ a. Thena p 8d = V{b(x ):x e M}. Since each b(x ) e I, d e I and a a 01hence, VM e I .3 a

Now let a A b e l . There exists c e l such thata a acp = a A b . Let a and b be elements of L such that a a


a A b = c e I. either a e I or b e I and hence, a e I 5 3 a aor b e l . The above shows that I is a CP-ideal of a a aL .a

We now show I = n I 9 where I is projection of Ia a ain L . Let a e n I and for each a, let b(a) be an a a aelement of I such that b(a)p = ap . For each a,a aa A b(a) <_ a. Let a A b(a) £ x. For each a,a p = (a a b(a))p < xp so a < x. Thus,a a a — a —a = V(a A t> C a)) - Since a A b e l for each o9 a e I. a aThe reverse inclusion is obvious, so I = n I .a a

SuDpose L I and Le =£ I . Then there exist a a p Belements a e L -I and bc e L —Ie. Let d be anya a a p p p

element of I. Let a be the element of L such thatap = a ,ap, = dp. for 6 ^ o, and b the element sucha a o othat bp. = b ,bp = d for 6 ^ 6. Since a A L < d e I,p p ~ 6 0a e I or b e I which contradicts a £ I and b„ £ I..a a p pThus, L = I for all but at most one a. a a

2.2.10. Theorem. The topological product of any collection of lattices L^, each with its CP-topology,is homeomorphic to the cartesian product I^L^ with itsCP-topology.

Proof: The two topologies have the same subbasicclosed sets.


2.2.11. Corollary. The Euclidean space Rn is homeomorphic to the lattice Rn with its CP-topology.

Proof: In the lattice R, the CP-ideals are precisely the closed sets of the form (-®,a] or [b,~).


APPENDIX

In the previous chapters, we have collected, organized and presented several results on lattice topologies which have appeared in papers in the literature. There are still other topologies which have been defined. For completeness, we now discuss these topologies.

The oldest and one of the most important lattice topologies is the metric topology. Since Birkhoff £6] and Maeda [17] have given complete accounts of metric lattices, we have delayed our discussion of the metric topology until now. We present here only the main results.

A.I. Definition. A modular function or valuation on a lattice L is a real-valued function $ on L which satisfies the condition:

x<j> + y<j> = (x v y)<J> + (x A y)<f> for all x, y e L.If x > y implies x<j> > y<j>, then <j> is said to be

positive.A metric lattice is a lattice L on which there can

be defined a positive modular function. In this case, we define the distance from x to y by

d(x,y) = (x v y)<f> - (x A y H for each x, y e L.

43


44

A.2. Theorem. Let be a positive modular function defined on a metric lattice L. Then

(i) L is a modular lattice,(ii) (L,d) is a metric space,(iii) |a<j> - b<J>| <_ d(a,b) (i.e., <t> is continuous),(iv) d(a V x, b v x) + d(a A x, b A x) <_ d(a,b),

and(v) the lattice operations are uniformly

continuous.Proof: (i) If a < c, (a V b) A c >_ a V (b A c).

Using the definition of a modular function twice, we have ((a V b) A c)<j> - (ay(b A c)H. Since 4> is positive,(a V b) A c = a v(b Ac).

(ii) It follows immediately from the definitionthat d(a,b) = d(b,a), d(a,a) = 0 for all a e L, and a bimplies d(a,b) > 0 since <t> is positive. Using the identity d(a,b) = a<|> + b<j> - 2(a A b)$, we haved(a,b) + d(b,c) - d(a,c) > 2((b v(a A c) )<(> -((a A b) V (b A c))<j>) _> 0 since <j» is positive andb v (a A c) b (a A b) V (b A c). Thus, d satisfiesthe distance axioms.

(iii) Observe that d(a,b) - (a<{> - b(j>) =(a V b)4> - a4> + b<j> - (a A b)<|> >_ 0, and similarly d(a,b) - (b<|) - a$) >_ 0 .

(iv) Since (a v x) A (b V x) > (a A b)V x and


45

(a A x) V (a A b) 1 (a V b) A x, d(a V x, b V x) + d(a A x 9 b A x) = (a V b v x)<j> - ((a v x) A (b V x)) «J> +((a A x) V (b A x))<j) - (a A b A x)<j> <_ (a V b v x)(f> -((a A b) V x)<j> + ((a v b) A x)<f> - (a A b A x)<f> =(a V b)<|> + x<t> - (a A b)<f> - x<j> = d(a,b).

(v) From (ii) and (iv), we have d(a v cs b V d) <_1 d(a V c, b v c) + d(b V c, b v d) <_ d(a,b) + d(c,d). Similarly, d(a A c, b A d) <_ d(a,b) + d(c,d).

A.3. Definition. A metric lattice (L,d) is metrically complete if every Cauchy sequence in L converges to a point in L (i.e., if (L,d) is a complete metric space).

A.4. Theorem. Any metric lattice (L,d) can be embedded isometrically in a metrically complete lattice (L*,d*) such that L is dense in L*. Moreover, the positive modular function $ on L which determines the distance function d, can be extended to the positive modular function <f>* on L* which determines the distance function d*.

Proof: We use the usual procedure of completing ametric space. Since L is a metric space, { x ^ ^ y } -if and only if l^m d(xQ,yn) = 0 is an equivalence relation

on the set of Cauchy sequences in L. Let L* be the set of equivalence classes {x^}*. For any metric space, the mapping x-»-{x,x,... }* is an isometric embedding of L onto


46

a dense subset of L*, and L* is a complete metric space.By Theorem A.4 (v), if'{x } and {yn> are Cauchy sequencesof L, then {x y y } and {x A y } are Cauchy. Thus, we n n n M nmay define (xnl* V tyn>* = txn V yn** and {xn}* A tyn** ={x Ay}** Routine comoutation shows L* is a lattice, n A nDefine <}>* on L* by {x }*<)>* = lim x <|>. By Theorem A.2* n n(iii), <|>* is well defined. Moreover, it can be shown that is a positive modular function.

A.5. Definition. A lattice L is conditionally complete if every bounded subset of L has a supremum and infimum in L. If every sequence of a lattice L has a supremum and infimum, L is said to be g-complete. Should each bounded sequence have a supremum and infimum, L is conditionally g-complete.

In a conditionally o-complete lattice L, a positive modular function $ is o-continuous if for each sequence {an>, aQ+a Implies an<|>+a$ and aQ+a implies an<}>+a<j>.

A.6. Theorem. Any metrically complete lattice L is conditionally complete and <j> is o-continuous. Conversely, if L is a a-complete metric lattice in which $ is o-continuous, then L is metrically complete.

Since our purpose in the appendix is to record the major results, and since the proof is long and involved, we omit it.

Matsushima [18,19] defined the Between topology of a


47

lattice in an attempt to generalize the interval topology. Since it was necessary to impose extremely severe conditions on the lattice to obtain results, those obtained were either trivial or of little importance. Birkhoff [7] also generalized the interval topology by defining the new interval topology. We state without proof two significant results which have been obtained.

A.7. Theorem. A closed subset of a conditionally complete lattice is compact in the new interval topology if and only if it is bounded.

A.8. Theorem [2], The new interval topology of the cartesian product of a finite number of chaims is equivalent to the topological product of the new interval topologies of the C .

v

There are several questions in lattice topologies which remain open. One of the most important and least considered questions concerns the continuity of mappings.To this date, there are no results in the literature in this direction. We have obtained the following theorem which illustrates the type of result which may be expected.

A.9. Theorem. Let L and be lattices, each with its P-ideal topology. Then every lattice homomorphism and dual homomorphism of L into L1 is continuous. Hence, every isomorphism and dual isomorphism is a homeomorphism.


48

Proof: Let be a homomorphism. We shall showthat the inverse image of each prime ideal I1 of L1 is a prime ideal of L. If x e I = and y < x, theny<j> <_ x$ e 1^, so y e I. Now let x, y e I. Thenx<|>, y<j> e I1 implying (x V y)<f> = x<j> v e and hence> x y y e I, Finally, if x y e I, then x<f> A 7 (x a y)«t> e I1. Since I1 is prime, x<|» e I1 or y<j> e 1^.But this implies x e l o r y e l . By a similar proof, the inverse image of a (dual prime ideal of L1 is a dual prime ideal of L. Since the prime ideals and dual prime ideals are the subbasic closed sets, $ is continuous.

To complete the proof, we observe that if | is adual homomorphism, then the inverse image of a prime ideal is a dual prime ideal and the inverse image of a dual prime ideal is a prime ideal.

We note that the above theorem does not hold for the interval topology. Let $ be the homomorphism of the unit interval of the real line into itself defined by 1<}> = 1 and x$ = 0 for x ^ 1. Then the inverse image of the closed set {0} is [0,1) which is open. Thus, <[> is not continuous.

We close with a list of open questions.1. Find necessary and sufficient conditions for a

lattice to be normal (or regular) with respect to one of the topologies which have been discussed.


2. Find necessary and sufficient conditions for order convergence and convergence in the order topology to be equivalent.

3. Do there exist lattices with unique order- compatible topologies which also contain infinite sets of pairwise incomparable elements?

4. Under what conditions is a lattice dense in its completion by cuts?

5. Under what conditions is the embedding of a lattice into its completion by cuts continuous?

6. Find conditions for the complete topology to be contained in the interval topology.

7. Find conditions for certain classes of mappings to be continuous. In particular, when are the residuated mappings and strongly ranged closed mappings (for definitions, see [14,15]) continuous?


BIBLIOGRAPHY

1. J. W. Alexander, Ordered sets, complexes, and the problem of bicompactification. Proc. Nat. Acad. Sci. U.S.A. 25 (1939), 296-298.

2. R. A. Alo and 0. Frink, Topologies on lattice products. Canad. J. Math. 18 (1966), 1004-1014.

3. K. Atsumi, On complete lattices having the Hausdorff interval topology. Proc. Amer. Math. Soc. 17 (1966), 197-199.

4. R. M. Baer, A characterization theorem for lattices with Hausdorff interval topology. J. Math. Soc.Japan 7 (1955), 177-181.

5. G. Birkhoff, Lattice theory (Revised Edition). Amer. Math. Soc. Colloq. Publ. XXV, New York, 1948.

6. G. Birkhoff, Lattice theory. Harvard University mimeographed notes, 1963.

7. G. Birkhoff, A new interval topology for dually directed sets. Univ. Nac. Tucumin. Rev. Ser. A. 14(1962), 325-331.

8. R. E. DeMarr, Order convergence and topological convergence. Proc. Amer. Math. Soc. 16 (1965), 588-590.

9. E. E. Floyd, Boolean algebras with pathological order topologies. Pacific J. Math. 5 (1955),687-689.

10. 0. Frink, Topology in lattices. Trans. Amer. Math.Soc. 51 (1942), 568-582.

11. S. Hu, Elements of general topology. San Francisco, 1964.

12. A. J. Insel, A compact topology for a lattice.Proc. Amer. Math. Soc. 14 (1963), 382-385.

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A. J. Insel, A relationship between the complete tocology and the order topology of a lattice.Proc. Amer. Math. Soc. 15 (1964), 847-850.M. F. Janowitz, Baer Semigroups. Duke Math. J. 32(1963), 85-96.M. F. Janowitz, A semigroup approach to lattices.Canad. J. Math. 18 (1966), 1212-1223.J. L. Kelley, General topology. New York, 1955.F. Maeda, Kontinuierliche geometrien, Berlin, 1958.Y. Matshushima, Between topology on a distributive lattice. Proc. Japan Acad. 35 (1959), 221-225.Y. Matshushima, Between topology for lattices. J. Math. Soc. Japan 16 (1964), 335-341.Y. Matshushima, Hausdorff interval topology on a partially ordered set. Proc. Amer. Math. Soc. 11 (I960), 233-235.T. Naito, Lattices with P-ideal topologies. Tohoka Math. J. (2) 12 (I960), 235-262.T. Naito, On a problem of Wolk in interval topologies. Proc. Amer. Math. Soc. 11 (I960),156-158.E. S. Northam, The interval tocology of a lattice. Proc. Amer. Math. Soc. 4 (1953), 824-827.A. J. Ward, On relations between certain intrinsic topologies on a partially ordered set. Proc.Cambridge Philos. Soc. 51 (1955), 254-261.E. S. Wolk, Order-compatible topologies on a partially ordered set. Proc. Amer. Math. Soc. 9 (1958), 524-529.

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