Aspects of Priestley Duality

Aspects of Priestley Duality

Inauguraldissertationder Philosophisch-naturwissenschaftlichen Fakultat

der Universitat Bern

vorgelegt von

Dominic van der Zypen

von Meikirch

Leiter der Arbeit: Prof. Dr. J. Schmid,

Mathematisches Institut der Universitat Bern

Aspects of Priestley Duality

Inauguraldissertationder Philosophisch-naturwissenschaftlichen Fakultat

der Universitat Bern

vorgelegt von

Dominic van der Zypen

von Meikirch

Leiter der Arbeit: Prof. Dr. J. Schmid,Mathematisches Institut der Universitat Bern

Von der Philosophisch-naturwissenschaftlichen Fakultat angenommen.

Der Dekan:

Bern, den 1. April 2004 Prof. Dr. G. Jager

When I consider thy heavens, the work of thy fingers, the moon and the stars, which thouhast ordained: What is man, that thou art mindful of him?

– Psalm 8:3,4

I wish to thank my supervisor Prof. Jurg Schmid for his advice and his helpful remarks. Heshowed me what to focus on when I tried to do too many things at once. I am deeply gratefulfor the patience and encouragement offered by my parents Eugen and Francoise van der Zypenand my sister Veronique and for the love of my girlfriend Marie-Christine Chappuis.

Contents

I Basic notions 6

1 Partially ordered sets and lattices 71.1 Partially ordered sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Subsets of posets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Upper and lower bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Morphisms between lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Boolean lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.7 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.8 Prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.9 Congruence relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Topology 122.1 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Bases and subbases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Priestley Duality 143.1 Ordered topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 The functors D : D → P and E : P → D . . . . . . . . . . . . . . . . . . . . . 15

II Three kinds of completeness 16

4 Affine complete lattices 174.1 Affine complete lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Affine complete Priestley spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Products of affine complete lattices . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Free products of affine complete lattices . . . . . . . . . . . . . . . . . . . . . 194.5 Embedding lattices in affine complete lattices . . . . . . . . . . . . . . . . . . 204.6 Q01 as initial object in the category of affine complete lattices . . . . . . . . . 214.7 Open questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5 Fractionally complete lattices 245.1 Fractionally complete lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.1.1 Lattices of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4

5.1.2 The maximal lattice of fractions . . . . . . . . . . . . . . . . . . . . . 245.1.3 Fractionally complete lattices . . . . . . . . . . . . . . . . . . . . . . . 255.1.4 Another way to make a lattice fractionally complete . . . . . . . . . . 26

5.2 Priestley Duality of ideals and open down-sets . . . . . . . . . . . . . . . . . . 275.3 Fractionally complete Priestley spaces . . . . . . . . . . . . . . . . . . . . . . 285.4 Products of fractionally complete lattices . . . . . . . . . . . . . . . . . . . . 325.5 Free products of fractionally complete lattices . . . . . . . . . . . . . . . . . . 33

6 Order completeness 356.1 Dualising completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356.2 Products and coproducts of complete lattices . . . . . . . . . . . . . . . . . . 35

7 Comparing the three kinds of completeness 377.1 Affine complete vs fractionally complete . . . . . . . . . . . . . . . . . . . . . 377.2 Order complete vs affine complete . . . . . . . . . . . . . . . . . . . . . . . . 387.3 Order complete vs fractionally complete . . . . . . . . . . . . . . . . . . . . . 39

III Representability and order components 42

8 Joint work with M. E. Adams 438.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438.2 Proof of theorem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448.3 Proof of theorem 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468.4 Open questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

IV Maximal compactness 53

9 Joint work with H. P. Kunzi 549.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549.2 Main problem and related questions . . . . . . . . . . . . . . . . . . . . . . . 559.3 Some further results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599.4 Sobriety and maximal compactness . . . . . . . . . . . . . . . . . . . . . . . . 61

V Outlook: Between topology and order 64

5

Part I

Basic notions

6

Chapter 1

Partially ordered sets and lattices

The aim of this chapter is to recall the basic definitions connected to the subject of order.They will constantly be referred to throughout this thesis.

1.1 Partially ordered sets

DEFINITION 1.1. A pair (X, R) consisting of a set X and a relation R ⊆ X×X is calleda partially ordered set if, for all x, y, z ∈ X,

1. (x, x) ∈ R (i.e., R is reflexive)

2. if (x, y), (y, z) ∈ R then (x, z) ∈ R (i.e., R is transitive)

3. if (x, y), (y, x) ∈ R then x = y (i.e., R is antisymmetric).

We will mostly use the term poset as a shorthand for ”partially ordered set”. R is referredto as the ordering relation of the poset (X, R). Instead of (x, y) ∈ R we usually write x ≤R yor even drop the index R if it is clear which relation we refer to. Often, we denote a posetby (X,≤R) instead of (X, R) and sometimes the index R is dropped. To make the notationeven lighter, we say ”let X be a poset” with the tacit assumption that we denote the orderingrelation of the poset with ≤.There are two special kinds of posets that deserve a name of their own. If for a poset (X,≤)we have x ≤ y or y ≤ x for all x, y ∈ X then we say that all elements are comparable andcall (X,≤) a chain. An antichain is a poset X whose order is ∆X = {(x, x);x ∈ X}; so in anantichain, no two distinct elements are comparable.If P,Q are posets, then a map ϕ : P → Q is called order-preserving if for all x, y ∈ P withx ≤ y we have ϕ(x) ≤ ϕ(y). The map ϕ is an order-embedding if for all x, y ∈ P we havex ≤ y ⇔ ϕ(x) ≤ ϕ(y). Note that any order-embedding is injective. If an order-embedding issurjective, then it is called an order-isomorphism.

1.2 Subsets of posets

DEFINITION 1.2. A subset D of a poset X is called a down-set if for all d ∈ D and x ∈ Xwith x ≤ d we have x ∈ D (”D is closed under going down”). A subset that is closed under

7

going up is called an up-set.

A subset A of a poset is called (up)-directed if for all a1, a2 ∈ A there is r ∈ A such thata1 ≤ r and a2 ≤ r. A nonempty directed down-set of a poset is said to be an ideal.Moreover, let Y be any subset of a poset X. Then, let (Y ] = {x ∈ X | x ≤ y for some y ∈ Y }and [Y ) = {x ∈ X | x ≥ y for some y ∈ Y }. Should Y = {y} for some y ∈ X, then, forsimplicity, we will denote (Y ] and [Y ) by (y] and [y), respectively. It is easy to see that (Y ]is a down-set of X and [Y ) is an up-set. Finally, let [x, y] = [x) ∩ (y]. The set [x, y] is calledthe interval generated by x and y. It is called proper if it contains more than one element.If (X,≤) is a poset and X0 ⊆ X then it is easy to verify that (X0,≤|X0) is a partially orderedset, where

≤|X0 = ≤ ∩ (X0 ×X0).

The poset (X0,≤|X0) is called subposet of X.

1.3 Upper and lower bounds

DEFINITION 1.3. If P is a poset and A ⊆ P a subset, then b ∈ P is called an upperbound of A if

a ≤ b for all a ∈ A.

A lower bound is defined dually. The set of all upper bounds of A is denoted by Au, and theset of all lower bounds by Al. (The sets Au and Al can be empty).

If A is a subset of a poset P then an element a0 ∈ A is called the least element of A if a0 ≤ afor all a ∈ A. Note that every subset of P possesses at most one least element; it is uniquebecause the ordering relation of each poset is antisymmetric. The notion of a greatest elementis dual.

DEFINITION 1.4. Let A be a subset of a poset P . If the set Au possesses a least elementthen this is called infimum or meet and denoted by inf(A). In a dual way we define thesupremum or join: If Al has a greatest element, then it is denoted by sup(A).

1.4 Lattices

DEFINITION 1.5. A lattice is a poset L such that for all subsets of the form {x, y} themeet x ∧ y := inf({x, y}) and the join x ∨ y := sup({x, y}) exist.

A complete lattice is a lattice having arbitrary meets and joins. Note that for lattice tobe complete, it suffices that all meets (or all joins) exist. The least element of a lattice (ifit exists) is denoted by 0, and the greatest element is denoted by 1. Moreover, a latticepossessing 0 and 1 with 0 6= 1 is called bounded or (0, 1)-lattice.

DEFINITION 1.6. A lattice L is said to be distributive if for all a, b, c ∈ L

[D] a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).

8

Note that [D] is equivalent to the statement: for all a, b, c ∈ L we have

[D′] a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c).

EXAMPLE 1.7. Let X be any set and let (P(X),⊆) denote the set of subsets of X orderedby set inclusion. We usually call P(X) the power set of X. The infimum and the supremumof any subset A ⊆ P(X) is given by

⋂A and

⋃A respectively. So, P(X) is a complete lattice.

Its distributivity is also easy to verify.

In this thesis we will be dealing with distributive lattices only.

1.5 Morphisms between lattices

DEFINITION 1.8. Let L,L′ be lattices. A map f : L → L′ is a lattice homomorphism iffor all a, b ∈ L we have

f(a ∧ b) = f(a) ∧ f(b) and f(a ∨ b) = f(a) ∨ f(b).

Every lattice homomorphism is clearly order-preserving. A lattice isomorphism is a bijectivelattice homomorphism. Lattices and lattice homomorphisms form a category L.Lattice homomorphisms between bounded lattices preserving 0 and 1 are referred to as (0,1)-lattice homomorphisms (recall that 0 and 1 are required to be distinct by definition in boundedlattices).In this thesis will be dealing with a non-full subcategory of L:

DEFINITION 1.9. Let D be the category of bounded distributive lattices with (0,1)-homo-morphisms.

With this terminology we will usually be a bit sloppy and say ’lattice homomorphism’ butmean ’(0,1)-lattice homomorphism’.

1.6 Boolean lattices

DEFINITION 1.10. Let L be a bounded lattice and let b ∈ L. We say that b′ ∈ L is acomplement of b if b ∧ b′ = 0 and b ∨ b′ = 1.

Note that complements are unique for distributive lattices.

DEFINITION 1.11. Let L be a bounded distributive lattice. Then L is called Boolean ifevery element of L has a (unique) complement.

We mention two examples of Boolean lattices:

• Let X be any set and consider (P(X),⊆). As seen in example 1.7 it is a distributivelattice; moreover for A ∈ P(X) the complement is given by A′ = X \ A. Note thatP(X) is a complete Boolean lattice.

9

• Let X be any infinite set. Let CF (X) = {A ⊆ X;A is finite or X \ A is finite} beordered by set inclusion ⊆. For A,B ∈ CF (X), note that A ∧ B and A ∨ B are givenby A ∩ B and A ∪ B respectively. The lattice CF (X) is Boolean (the complement isagain the set complement) but non-complete for the following reason. There is a setY ∈ P(X) \ CF (X). Then the subset A = {{y}; y ∈ Y } of CF (X) has no supremumin CF (X).

1.7 Ideals

DEFINITION 1.12. An ideal of a lattice L is an ideal of the poset L as defined in section1.2. This amounts to saying that an ideal is a down-set that is closed under finite joins. Anideal is called proper if it does not coincide with L.

The dual notion of an ideal is that of a filter.

EXAMPLE 1.13. Let X be any set and consider the lattice P(X). Then the set J = {A ∈P(X);A is finite} is an ideal.

For any a ∈ L the set(a] = {l ∈ L | l ≤ a}

is easily seen to be an ideal; ideals of this form are called principal. If I is an ideal of L thenwe say that

PI = {x ∈ L; (x] ∩ I is principal}

is the principal extension of I. It is a fair exercise to show that PI is an ideal (this uses thedistributivity of L). Moreover, if J1, J2 are ideals of L then we set

J1 → J2 = {x ∈ L;x ∧ j1 ∈ J2 for all j1 ∈ J1}.

Again, it is straightforward to check that J1 → J2 is an ideal.Moreover, if P is a poset, a ≤ b ∈ P then we define the interval of a, b to be

[a, b] = {x ∈ P ; a ≤ x ≤ b}.

In addition we need a notion of ”bigness”, especially for ideals:

DEFINITION 1.14. Let L be a distributive lattice and S ⊆ L be any nonempty subset. Iffor all x ∈ L we have

sup(S ∩ (x]) = x

then S is called join-dense.

1.8 Prime ideals

DEFINITION 1.15. Let L be a bounded distributive lattice. A proper ideal P ⊆ L is saidto be prime if for all a, b ∈ L we have

a ∧ b ∈ P implies a ∈ P or b ∈ P.

10

An ideal is prime if and only if its complement is a (prime) filter.

EXAMPLE 1.16. Let ω denote the first infinite ordinal (also known as the set of naturalnumbers). Consider the lattice P(ω). Then the set Q = {A ∈ P(ω); 0 /∈ A} is a prime idealof P(ω). On the other hand, J = {A ∈ P(ω);A is finite} is not prime.

Any distributive bounded lattice contains at least one prime ideal (this is proved using Zorn’sLemma). Prime ideals will play a key role in representing lattices as topological spaces, whichwe will see in chapter 3. We denote the set of all prime ideals of L by Ip(L) .

1.9 Congruence relations

Let (L,≤) be a bounded distributive lattice and suppose that θ is an equivalence relation onL. Then θ is called a congruence relation if

(C) for all (a, b), (a′, b′) ∈ θ we have (a ∧ a′, b ∧ b′) ∈ θ and (a ∨ a′, b ∨ b′) ∈ θ.

Then, (L/θ,≤θ) is again a lattice where [a]θ ≤θ [b]θ if and only if a ≤ b. Note that (C) makessure that ≤θ is well defined. Moreover, (L/θ,≤θ) is even a bounded distributive lattice, andsuprema and infima are given by

[a]θ ∧ [b]θ = [a ∧ b]θ and [a]θ ∨ [b]θ = [a ∨ b]θ.

Again, (C) makes sure that this is well-defined.

11

Chapter 2

Topology

2.1 Topological spaces

DEFINITION 2.1. A topological space is a pair (X, τ) such that X is a set and τ is acollection of subsets of X satisfying the following requirements:

1. ∅, X ∈ τ

2. U ⊆ τ =⇒⋃U ∈ τ

3. U1, U2 ∈ τ =⇒ U1 ∩ U2 ∈ τ.

The members of τ are called the open subsets of X. A subset of X that is a complementof an open set is called closed. Note that arbitrary intersections and finite unions of closedsets are closed. A subset of X that is both open and closed is called clopen. If S ⊆ X andthere is an open set U with x ∈ U ⊆ S, then S is called a neighborhood of x. The elementx ∈ S is then called inner point of S. The set of all inner points of S is called the interiorof S and denoted by intX(S) or simply int(S). It is the greatest open set (with respect toinclusion) contained in S, since it can be written as the union of all open sets contained in S.By DeMorgan’s Laws, the set X\int(X\S) is the intersection of all closed sets containing Sand therefore the smallest closed set containing S. It is called the closure of S and denotedby clX(S) or simply cl(S). A set S with cl(S) = X is called dense.

DEFINITION 2.2. If (X, τ) is a topological space and A ⊆ X, then A inherits a topologyτA from X:

τA := {U ∩A;U ∈ τ}.τA is the subspace topology of A and (A, τA) is a subspace of (X, τ).

2.2 Bases and subbases

DEFINITION 2.3. If (X, τ) is a topological space, then B ⊆ τ is a base of τ if for all opensubsets U of X and all x ∈ U there is a B ∈ B such that x ∈ B ⊆ U . For any subset S ⊆ τwe define

B(S) := {S1 ∩ ... ∩ Sr; r ∈ N and Sj ∈ S for all j ≤ r}.S is called a subbase of τ if B(S) is a base of τ .

12

Finally, given a set X and some collection S of subsets of X, we may consider the set T oftopologies that contain S. (T is nonempty since it contains P(X) as an element). It is easyto see that τ :=

⋂T is the smallest topology containing S and the only topology having S

as a subbase. We say τ is generated by S.We can represent τ , the topology generated by S in another way. Similar to definition 2.3 let

B(S) := {S1 ∩ ... ∩ Sr; r ∈ N and Sj ∈ S for all j ≤ r}.

Then the topology generated by S is the set

{U ⊆ X;U is a union of members of B(S)} ∪ {∅, X}.

2.3 Continuous functions

DEFINITION 2.4. Let X, Y be topological spaces. A function f : X → Y is continuous iffor all open subsets V ⊆ Y the preimage f−1(V ) is an open subset of X.

Note that if f : X → Y is continuous and bijective, the inverse map need not be continuous.Thus, a bijection is called a homeomorphism if both itself and its inverse are continuous.Let X, Y be topological spaces and assume that S is a subbase of Y . In order to verify thata given map f : X → Y is continuous, it suffices to check that f−1(S) is open in X for everyS ∈ S. Compositions of continuous functions are continuous.

2.4 Compact spaces

DEFINITION 2.5. A topological space (X, τ) is said to be compact if the following holds:

Whenever U is a subset of τ such that⋃U = X then there is a finite subset

U0 ⊆ U with⋃U0 = X.

Compact spaces will play a key role in this thesis. The following are the most important factsabout compact spaces:

1. A closed subset of a compact space is compact.

2. If X is compact and f : X → Y is continuous, then Y is compact.

We will also be making frequent use of Alexander’s Subbase Theorem. We need the followingdefinition: Let X be a space with subbase S. Then X is called subbase compact with respectto S if the following holds: Whenever V is a subset of S such that

⋃V = X then there is a

finite subset V0 ⊆ V with⋃V0 = X.

Alexander’s Subbase Theorem. Let X be a topological space with subbase S. If X issubbase compact with respect to S then X is compact.This theorem thus says that in order to check compactness of some space, we need onlyconsider subbasic coverings.

13

Chapter 3

Priestley Duality

3.1 Ordered topological spaces

DEFINITION 3.1. An ordered topological space is a 3-tuple (X, τ,≤) such that (X, τ) isa topological space and (X,≤) is a partially ordered set.

The next definition presents one of the most usual ways to assign a topology to any partiallyordered set.

DEFINITION 3.2. Let X be a poset. Set S− = {X \ (x] | x ∈ X}, and S+ = {X \ [x) |x ∈ X}. Then S = S− ∪ S+ is an subbase for the so called interval topology τi on X(sometimes, in the interest of clarity, τi will be denoted τi(X) when we wish to emphasize theposet concerned).

The interval topology is a topology that exists on any given poset. This is different with othertopologies that may or may not exist on posets, namely Priestley topologies:

DEFINITION 3.3. An ordered topological space (X, τ,≤) is called a Priestley space if

1. (X, τ) is compact.

2. (X, τ,≤) has the Priestley separation property, ie.

If x 6≤ y there exists a clopen down-set C ⊆ X that contains y but not x.

Note that if (X, τ,≤) is a Priestley space, then the interval topology τi(X) is contained in τ .So, for example, if τi(X) is not compact, then there is no topology on X making it a Priestleyspace since that topology would need to be compact.

EXAMPLE 3.4. Let ω be the first infinite ordinal, ordered by its usual well-ordering. Thenτi(ω) = P(ω) which is not compact. So there is no topology making ω Priestley.

We call a poset representable if there is a topology making it Priestley. The reader will findan equivalent definition and indeed a whole chapter devoted to representability in chapter 8.In order to make Priestley spaces into a category, we have to say what the morphisms are.A (Priestley) morphism between Priestley spaces is a continuous and order-preserving map.With P, we denote the category of Priestley spaces and Priestley morphisms.

14

In any Priestley space X, the collection of all clopen up-sets and all clopen down-sets is asubbase for X. This directly implies that X has a basis consisting of clopen sets (such spacesare also called zero-dimensional). In fact, already

{D ∩ (X \D′);D,D′ clopen down-sets of X}

forms a base for X.

3.2 The functors D : D → P and E : P → D

H. Priestley proved that the category D of bounded distributive lattices with (0,1)-preservinglattice homomorphisms and the category P of compact totally order-disconnected spaces(Priestley spaces) with order-preserving continuous maps are dually equivalent (for instance,see [25]).The functor D : D → P assigns a Priestley space (D(L), τP (L),⊆) to each object of D, where

• D(L) is the set of all prime ideals of L, that is, Ip(L)

• τP (L) is the topology generated by the sets

{Xa; a ∈ L} and {Ip(L) \Xa; a ∈ L},

where for a ∈ L we set Xa := {P ∈ Ip(L); a /∈ P}.

• ⊆ is set inclusion (between prime ideals).

The functor E : P → D assigns the lattice (E(X),∪,∩) to each Priestley space X, where E(X)is the set of all clopen down-sets of X with set union and set intersection as lattice operations.Moreover, if f is a (0,1)-lattice homomorphsim, then D(f) is defined by P 7→ f−1(P ) for anyprime ideal P , and for any morphism ϕ, its dual E(ϕ) is defined by C 7→ ϕ−1(C) for any clopendown-set C. It was proven by H. A. Priestley that these functors form a dual equivalencebetween D, the category of distributive (0,1)-spaces to P and the category of Priestley spaces.

15

Part II

Three kinds of completeness

16

Chapter 4

Affine complete lattices

4.1 Affine complete lattices

A k-ary function f on a bounded distributive lattice L is called compatible if for any congru-ence θ on L and (ai, bi) ∈ θ, (i = 1, ..., k) we always have (f(a1, ..., ak), f(b1, ....bk)) ∈ θ. It iseasy to see that the projections pri : Lk → L are compatible. With induction on polynomialcomplexity one shows that every polynomial function is compatible (see [23]). A lattice L iscalled affine complete, if conversely every compatible function on L is a polynomial.

G. Gratzer [12] gave an intrinsic characterization of bounded distributive lattices that areaffine complete:

THEOREM 4.1. ([12]) A bounded distributive lattice is affine complete if and only if it doesnot contain a proper interval that is a Boolean lattice in the induced order.

Note that in particular, no finite bounded distributive lattice L is affine complete: Let x ∈ Lbe an element distinct from 1. Then x has an upper neighbor, ie, there exists y ∈ L such that[x, y] = {x, y} which is isomorphic to the 2-element Boolean lattice.

EXAMPLE 4.2. The bounded distributive lattices [0, 1] and [0, 1]× [0, 1] are affine complete.

Proof. First, take any x < y in [0, 1]. Then the element a = x+y2 ∈ [x, y] has no complement a′

in [x, y]: Otherwise we would have a∧a′ = x which would imply a′ = x, but then a∨a′ = a 6= y.So [x, y] is not Boolean, whence [0, 1] has no proper Boolean interval.Secondly, let (x1, x2) < (y1, y2) ∈ [0, 1]×[0, 1]. With a similar argument as before, the element

(x1 + y1

2,x2 + y2

2) ∈ [(x1, x2), (y1, y2)]

does not have a complement in [(x1, x2), (y1, y2)]. Thus, [0, 1]× [0, 1] has no proper Booleaninterval and is therefore affine complete.

4.2 Affine complete Priestley spaces

The aim of this section is to characterize the Priestley spaces corresponding to affine completedistributive (0,1)-lattices. Such spaces will be called affine complete Priestley spaces. In otherwords, a Priestley space X is affine complete iff E(X) is affine complete.

17

The following theorem provides a rather straightforward translation of the algebraic conceptof affine completeness in order-topological terms.

THEOREM 4.3. Let X be a Priestley space. Then the following statements are equivalent:

1. E(X) is affine complete.

2. If U ⊆ V are clopen down-sets and U 6= V , then the subposet V \ U of X is not anantichain, i.e. V \ U contains a pair of distinct comparable elements.

Proof. (1) =⇒ (2). Suppose V \U is an antichain. Let C ∈ [U, V ] ⊆ E(X). Take C ′ =U ∪ (V \C).Claim: C ′ is a clopen down-set of X.It is clear that C ′ is a clopen subset of X since V \ C = V ∩ (X \ C). Now, let c ∈ C ′ andassume x < c. Then if c ∈ U , we are done, since U is a down-set. Assume c ∈ V \U . Since Vis a down-set, we get x ∈ V , and the fact that V \U is an antichain tells us that x cannot be amember of V \U . Therefore x ∈ U ⊆ C ′ which proves that C ′ is indeed a (clopen) down-set.Moreover, C ′ is the complement of C in [U, V ], i.e. C ∩ C ′ = U and C ∪ C ′ = V . BecauseC was arbitrary, we see that [U, V ] is a proper Boolean interval of E(X), whence E(X) is notaffine complete.

(2) =⇒ (1). Suppose U ⊆ V are distinct clopen down-sets. By assumption, there areelements x, y ∈ V \U such that x < y. There is a clopen down-set A with x ∈ A and y /∈ A.Consider B = (A∩V )∪U . So B ∈ [U, V ] and y /∈ B. Now we show that B has no complementin [U, V ]: Take any C ∈ [U, V ] with C ∪ B = V . Then y ∈ C, but since C is a down-set, wehave x ∈ C, thus x ∈ (B∩C)\U and B∩C 6= U . So whatever C we pick, C is no complementfor B, i.e. B is not complemented, and consequently [U, V ] is not Boolean. It follows that noproper interval of E(X) is Boolean.

We can formulate the above result in a more concise way:

COROLLARY 4.4. A Priestley space X is affine complete if and only if each nonemptyopen set contains two distinct comparable points.

Proof. It follows directly from theorem 4.3 that if each nonempty open set contains twodistinct points that are comparable, then X is affine complete.

Conversely, suppose that U is a nonempty open set which is an antichain, then there existopen down-sets C1, C2 such that ∅ 6= C1∩(X\C2) ⊆ U . Then [C1∩C2, C1] is a proper intervalsuch that C1\(C1 ∩C2) = C1 ∩ (X\C2) is an antichain (as a subset of the antichain U). Thustheorem 4.3 implies that X is not affine complete.

Note that the proof works exactly the same way if each occurrence of ”open” is replaced by”clopen” (basically because each Priestley space is zero-dimensional). So we can state as well:

A Priestley space X is affine complete if and only if each nonempty clopen set contains twodistinct comparable points.

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4.3 Products of affine complete lattices

The aim of this section is to prove that arbitrary products of affine complete lattices areaffine complete. We don’t need Priestley duality to do this. Priestley duals of affine completelattices, i.e. affine complete Priestley spaces, will come into play when we consider coproductsof affine complete lattices.

THEOREM 4.5. If (Li)i∈I is a family of (bounded) affine complete lattices, then Πi∈ILi isaffine complete.

Proof. We prove the contrapositive of the theorem. Suppose that Πi∈ILi is not affine com-plete. Then it contains a proper interval [ξ, η] that is Boolean. There exists some k ∈ K suchthat ξ(k) < η(k). We claim that

[ξ(k), η(k)] ⊆ Lk

is a Boolean interval. Set x = ξ(k), y = η(k). Suppose l ∈ [x, y] and define λ ∈ Πi∈ILi by

λ(i) =

{l if i = k

ξ(i) if i 6= k

Because [ξ, η] is Boolean, there exists λ′ ∈ Πi∈ILi such that λ ∧ λ′ = ξ and λ ∨ λ′ = η. Thusit is easy to see that l′ := λ′(k) is the complement of l ∈ [x, y]. Therefore, [x, y] is a properBoolean interval of Lk and whence Lk is not affine complete.

EXAMPLE 4.6. Theorem 4.5 implies that [0, 1]N is affine complete.

4.4 Free products of affine complete lattices

In this section, our goal is to prove that any free product of affine complete lattices is affinecomplete. A convenient way to obtain this result is to dualise the problem into the categoryof Priestley spaces. Free products (that is, coproducts) in D correspond to products in Pand vice versa; this is stated in the following proposition in a more general way.

PROPOSITION 4.7. [22] Let A and B be categories, and assume that F : A → B andG : B → A are contravariant functors that form a dual equivalence. Then:

1. If A is a product of a family of objects (Ai)i∈I of A, then F(A) is a coproduct of(F(Ai))i∈I .

2. If A is a coproduct of a family of objects (Ai)i∈I of A, then F(A) is a product of(F(Ai))i∈I .

Moreover we have shown that affine complete lattices correspond to affine complete spacesunder the Priestley duality.

THEOREM 4.8. If (Xi)i∈I is a family of affine complete Priestley spaces, then Πi∈IXi isaffine complete.

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Proof. Suppose that Xi is affine complete for every i ∈ I. It suffices to show that everynonempty subset V of Πi∈IXi of the form

V = π−1i1

(U1) ∪ ... ∪ π−1ir

(Ur)

contains two distinct comparable elements (where Uk ⊆ Xik open, nonempty). Take U1. Itcontains elements a < b, because Xi1 is affine complete. Now pick ξ ∈ V . Define ξ1, ξ2 ∈ Vby

ξ1(i) =

{ξ(i) if i 6= i1

a if i = i1

and

ξ2(i) =

{ξ(i) if i 6= i1

b if i = i1

Clearly, ξ1, ξ2 are distinct comparable elements of V .

Applying the Priestley duality now yields:

COROLLARY 4.9. The class of (bounded) affine complete lattices is closed under freeproducts.

4.5 Embedding lattices in affine complete lattices

First we will stay away from affine completeness in the worst possible way: we will embedeach L into a powerset of some set, which, being Boolean, is as affine incomplete as it gets.

LEMMA 4.10. Let L be a distributive lattice (L need not be bounded). There is a set Xand a lattice embedding

j : L ↪→ P(X)

where P(X) is the powerset of the set X.

Proof. First, endow L with a smallest element and a greatest element. Call this new boundeddistributive lattice L01. By Priestley duality, there is a Priestley space (X, τ,≤) such that thelattice E(X) of clopen down-sets is isomorphic to L01. Since E(X) is a sublattice of P(X),we are done.

Next, we will embed that powerset in an affine complete lattice.

LEMMA 4.11. Let X be a set and let Q = {q ∈ Q; 0 ≤ q ≤ 1}. Then there is a latticeembedding

j : P(X) ↪→ QX .

Moreover, Q is affine complete.

Proof. Set j : S 7→ χS ∈ QX for every S ⊆ X, where χS is defined by

χS(x) =

{1 if x ∈ S

0 if x /∈ S

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It is easy to see that j is a lattice embedding. Next, we claim that Q is affine complete. Takeany x < y in Q. Then the element a = x+y

2 ∈ [x, y] has no complement a′ in [x, y]: Otherwisewe would have a ∧ a′ = x which would imply a′ = x, but then a ∨ a′ = a 6= y. So [x, y] is notBoolean, whence Q has no proper Boolean interval. Therefore, Q is affine complete.Moreover, by 4.5, QX is affine complete which concludes the proof.

Lemmas 4.10 and 4.11 now imply:

COROLLARY 4.12. Every distributive lattice (not necessarily bounded) can be embeddedin a bounded affine complete lattice.

Admittedly, the construction provided by 4.10 and 4.11 is highly non-unique and has nominimality properties.

4.6 Q01 as initial object in the category of affine complete lat-tices

The aim of this section is to show that the lattice Q01 = Q∩ [0, 1] can be embedded into eachaffine complete lattice, which amounts to saying that Q01 is an initial object of the categoryof affine complete lattices (with (0,1)-homomorphisms, i.e. a full subcategory of the categorybounded distributive lattices). The key will be the notion of a dense chain.

DEFINITION 4.13. A chain (X,≤) is called dense if for all x < y ∈ X there is z ∈ Xwith x < z < y.

The first tool we need here is a well known result of model theory. It states that the theoryof dense linear orders is complete and has (Q,≤) as prime model. We will state this result ina more primitive way and prove it.

PROPOSITION 4.14. If (X,≤) is a bounded dense chain, there is a (0, 1)-embedding

ϕ : Q01 ↪→ X.

Proof. Let a : ω → Q01\{0, 1} be a bijection. We will write ak instead of a(k) to simplifynotation and will inductively build a subset

f ⊆ (Q01\{0, 1})× (X\{0X , 1X})

that’s an injective function from Q01\{0, 1} to X\{0X , 1X} which is even order-preserving.n = 0: Choose b0 ∈ X\{0X , 1X} and set f0 := {(a0, b0)}.n → n + 1: Assume that fn has been defined in a way that for all k, l ∈ {0, ..., n} the relationak ≤ al implies fn(ak) ≤ fn(al) and that fn is an injective function from {a0, ..., an} toX\{0X , 1X}. Now consider the element an+1 ∈ Q01\{0, 1}.Case 1: an+1 ≥ ai for all i ∈ {0, ..., n}. Then, since X is dense, there is bn+1 ∈ X such that1X > bn+1 ≥ fn(ai) for all i ∈ {0, ..., n}. So,

fn+1 := fn ∪ {(an+1, bn+1)}

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is an injective order-preserving function that continues fn.Case 2: an+1 ≤ ai for all i ∈ {0, ..., n}. Proceed similarly as in Case 1.Case 3: There are k, l ∈ {0, ..., n} such that ak < an+1 < al. We may assume that thereis no k′ ∈ {0, ..., n} with ak < ak′ < an+1 and likewise that there is no l′ ∈ {0, ..., n} withan+1 < al′ < al. Consider bk = fn(ak) and bl = fn(al). Since fn is order-preserving andinjective by assumption, we get bk < bl. Because X is dense, there is an element bn+1 suchthat bk < bn+1 < bl. Then

fn+1 := fn ∪ {(an+1, bn+1)}

is easily seen to be an injective order-preserving map that continues fn.Now, it is easy to see that

f :=⋃n∈ω

fn

is an injective order-preserving function from Q01\{0, 1} to (X\{0X , 1X} which is even order-preserving. So

ϕ := f ∪ {(0, 0X), (1, 1X)}

is an order embedding from Q01 to X.

PROPOSITION 4.15. Let L be a bounded affine complete distributive lattice. Then

a) There is a maximal chain C ⊆ L, i.e., a chain that is not properly contained in anotherchain in L.

b) If C is a maximal chain of L then C is dense.

Proof. a) is a standard application of Zorn’s Lemma: If K is a set of chains of L such thatfor any C1, C2 ∈ K we either have C1 ⊆ C2 or C1 ⊇ C2, then

⋃K is easily checked to be a

chain in L: Let x, y ∈⋃K, then there are members C,D containing x, y respectively; now

since K is a chain with respect to ⊆, at least one of the statements x, y ∈ C or x, y ∈ D holds.Since C,D are chains in L, either statement leads us to x ≤L y or x ≥L y. So K is boundedin the poset of all chains of L, thus Zorn’s Lemma implies that there is a maximal chain.As for b), assume that C is a maximal chain such that x < y ∈ C but there is no z ∈ Cwith x < z < y. Now if there were no z in the whole lattice L such that x < z < y, then[x, y] = {x, y} is a proper Boolean interval of L which implies that L is not affine complete,leading to a contradiction. Thus there is such a z, whence C∪{z} is a chain of L that properlycontains C, contradicting the maximality of C.

Now the propositions 4.14 and 4.15 directly imply the following.

THEOREM 4.16. If L is an affine complete lattice, then there exists a (0,1)-embeddingϕ : Q01 ↪→ L.

Proof. Pick any maximal chain C in L. Note that by maximality of C we have 0, 1 ∈ Csince C ∪ {0, 1} is a chain. So the inclusion map ι : C ↪→ L is a (0, 1)-embedding as well asthe embedding from Q01 to C provided by proposition 4.15. Composing these two, we get a(0,1)-embedding from Q01 to L.

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4.7 Open questions

In chapter 4.5 we showed that ever bounded distributive lattice can be extended to an affinecomplete lattice. This was achieved by making use of Q01 which happens to be embeddablein any affine complete lattice, ie, the ”smallest” affine complete lattice. Now the question is:Is the construction carried out in chapter 4.5 in some way canonical? For an arbitrary latticeL, does its ’affine hull’ have any interesting universal properties?

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Chapter 5

Fractionally complete lattices

5.1 Fractionally complete lattices

In this chapter, we will not restrict ourselves to bounded distributive lattices but ratherconsider the whole class of distributive lattices.

5.1.1 Lattices of fractions

The concept of fractionally complete lattices derives from the notion of a lattice of fractions.For both concepts, motivation and an excellent introduction is given in [29]. We want to givea brief overview over the main definitions and ideas used here.

Let L,L1 be distributive lattices.

DEFINITION 5.1. L1 is said to be a lattice of fractions of L if L is a sublattice of L1 andfor all x1, x2, y ∈ L1, there exists a ∈ L such that

a ∧ x1 6= a ∧ x2 and a ∧ y ∈ L.

Instead of saying ”L1 is a lattice of fractions of L”, we will also write L < L1.

5.1.2 The maximal lattice of fractions

We have seen when dealing with affine complete lattices that every bounded distributivelattice L may be embedded in a larger affine complete lattice, but this embedding is by nomeans canonical in any way. This section shows that things are much nicer when we embedbounded distributive lattices in larger fractionally complete lattices.In [29], the following fundamental fact about lattices of fractions was proved. For everydistributive lattice L there exists a distributive lattice Q(L) that is a maximal lattice offractions of L in the following sense:

L < L1 if and only if there exists L′1 such that L′

1∼= L1 and L ⊆ L′

1 ⊆ Q(L).

This maximal lattice of fractions may be realized as:

Q(L) = ({J ∈ I(L) : PJ is join-dense and J = PJ → J},⊆)

24

with the embedding L → Q(L) given by x 7→ (x].

EXAMPLE 5.2. Let F ′(N) be the set of all finite subsets of the natural numbers, and letL = F (N) = F ′(N) ∪ {N}. Then

Q(F (N)) ∼= P(N),

where P(N) is the powerset lattice of N.

Proof. Let J be any ideal of F (N), let l ∈ F (N) be arbitrary. So l ⊆ N is finite, thus (l] is afinite subset of F (N). Therefore (l]∩ J is a finite ideal and thus has a maximal element, so itis principal. This implies that PJ = L which is certainly join-dense. But on the other hand,it is easy to see that L → J = J . So I ∈ Q(L) for all I ∈ I(L) which in turn implies that

Q(L) = I(L). (?)

Moreover, we claim thatΦ : I(L) → P(N); J 7→

⋃J

is an isomorphism: For A ∈ P(N) consider J = {A0 ∈ F (N);A0 ⊆ A} which is clearly anideal. Clearly Φ(J) = A. This shows that Φ is surjective.On the other hand, suppose that J1 6= J2 are ideals of L = F (N) such that Φ(J1) = Φ(J2).We may assume that l = {n1, ...nr} ∈ J1\J2. But since

⋃J1 =

⋃J2, we get ni ∈

⋃J2 for all

i = 1...r. Now since J2 is an ideal, this implies {ni} ∈ J2 for all i = 1...r, which in turn tellsus that

{n1} ∪ ... ∪ {nr} = {n1, ..., nr} = l ∈ J2.

This is a contradiction.

Finally, (?) and the fact that Φ : I(L) → P(N) is an isomorphism imply that Q(F (N)) ∼=P(N).

5.1.3 Fractionally complete lattices

It is now easy to figure out what we will mean by the term ”fractionally complete lattices”.

DEFINITION 5.3. L is called fractionally complete if L < L1 implies L ∼= L1.

With the results of [29] we can characterize the class of fractionally complete lattices in amore concrete way.

THEOREM 5.4. ([29]) A distributive lattice L is fractionally complete if and only if thefollowing condition holds:

(C) If J ⊆ L is an ideal satisfying

• PJ is join-dense

• PJ → J = J

then J is principal.

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Proof. If L satisfies (C) then the embedding x 7→ (x] from L to Q(L) is an isomorphism.Moreover [29], Theorem 14 tells us that Q(L) is always fractionally complete, whence L isfractionally complete.

Conversely, suppose that L is fractionally complete. By definition we get L ∼= Q(L). [29],Theorem 14(i) says that

ϕ : Q(L) → Q(Q(L)); x 7→ (x]

is an isomorphism, whence Q(L) satisfies C. Because L ∼= Q(L), this implies that L satisfies(C).

EXAMPLE 5.5. 1. Every finite distributive lattice is fractionally complete.

2. Every bounded chain is fractionally complete.

3. If X is an infinite set, then the lattice consisting of all finite subsets plus X is notfractionally complete.

4. For each set X, the powerset P(X) is fractionally complete.

Proof. (1). This is clear because every ideal of a finite lattice is principal.(2). Let C be a chain and suppose that J ⊆ C is a non-principal ideal. Let x ∈ C\J . Then(x]∩J = J is not principal. Whence PJ = J , thus PJ → J = J → J = C 6= J (C is principalsince we are dealing with a bounded chain, but J is not principal). These considerations showthat any ideal I satisfying PI join-dense and PI → I = I must be principal.(3). This follows from an easy generalisation of example 5.2.(4). If F (X) is the lattice of all finite subsets of X plus X itself a similar proof to 5.2 showsthat Q(F (X)) ∼= P(X). Now Theorem 14 of [29] tells us that every lattice of the form Q(L)is fractionally complete, we are done.

5.1.4 Another way to make a lattice fractionally complete

There is way which is different from the construction carried out in section 5.1.2 to obtain afractionally complete lattice from an arbitrary lattice. The key lies in the following lemma.

LEMMA 5.6. Let L be a bounded distributive lattice such that sup(L\{1}) < 1. Then L isfractionally complete.

Proof. It is clear that such a lattice does not possess a proper join-dense ideal. Thus if anideal J has the property that PJ is join-dense, then we have PJ = L which implies that J isprincipal.

This gives us an easy recipe to make an arbitrary bounded distributive lattice fractionallycomplete: add a new top to it! This is the message of the following corollary.

COROLLARY 5.7. Let L be a bounded distributive lattice. Then define

Ltop = L ∪ {L}

and define≤top=≤ ∪{(x, L);x ∈ L} ∪ {(L,L)}.

26

Then (Ltop,≤top) is a fractionally complete distributive lattice.

Proof. It is very easy to verify that Ltop is a bounded distributive lattice. Moreover since Lwas a bounded lattice, we get sup(Ltop \ {L}) = 1L < L. Therefore, lemma 5.6 implies thatLtop is fractionally complete.

5.2 Priestley Duality of ideals and open down-sets

The next aim is to translate the concept of fractionally complete lattices into the category P ofPriestley spaces. An important tool will be the duality between ideals of bounded distributivelattices and open down-sets of Priestley spaces.

Let X be a Priestley space throughout this chapter; let E(X) be the lattice of all clopendown-sets of X. With I(E(X)) we denote the set of ideals of E(X). Moreover let L be thelattice of open down-sets of X. As noted in [7], we can set up the following connection betweenI(E(X)) and L.

LEMMA 5.8. ([7]) The map Φ : I(E(X)) → L defined by J 7→⋃

J is a lattice isomorphism.

Proof. Define Ψ : L → I(E(X)) by W 7→ {U ∈ E(X);U ⊆ W}. We want to argue that Φ andΨ are inverses of each other.

Claim 1. Ψ(Φ(J)) = J for every J ∈ I(E(X)). - Let U ∈ J . So clearly U is an open down-setwith U ⊆

⋃J = Φ(J). Thus U ∈ Ψ(Φ(J)). Conversely, let U ∈ Ψ(Φ(J)). So U is clopen

and U ⊆⋃

J . So U is covered by members of J . Since U is compact, it is covered by finitelymany members V1, ..., Vr of J . So we get

U ⊆ V1 ∪ ... ∪ Vr ∈ J,

and, as J is an ideal, we have U ∈ J .

Claim 2. Φ(Ψ(W )) = W for every W ∈ L. - It’s clear that Φ(Ψ(W )) =⋃{U ∈ E(X);U ⊆

W} ⊆ W . In order to prove the other inclusion, it suffices to show that for every w ∈ Wthere is a clopen down-set Uw such that w ∈ Uw ⊆ W . If W = X, we are done. SupposeW 6= X; since W is a down-set, we have y 6≤ w for every y ∈ X\W . So for every y ∈ X\Wthere is a clopen up-set Ay that contains y but not w. Now, X\W is covered by the Ay wherey ranges over X\W . Since X\W is compact there exist y1, ..., yr such that

(X\W ) ⊆ Ay1 ∪ ... ∪Ayr .

Then Uw = X\(Ay1 ∪ ... ∪Ayr) is a clopen down-set with w ∈ Uw ⊆ W .

Finally, it is easy to see, that Φ and Ψ are order-preserving; thus Ψ is an order-isomorphismand whence a lattice isomorphism.

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5.3 Fractionally complete Priestley spaces

The aim of this chapter is to characterize the Priestley spaces arising from fractionallycomplete distributive (0,1)-lattices. We will call these spaces fractionally complete Priest-ley spaces. The other way round, a Priestley space X is fractionally complete iff E(X) isfractionally complete.

To achieve this goal, we will make heavy use of the duality between ideals and open down-sets(henceforth called ”ideal duality”). First, we will dualise some properties an ideal can haveinto the world of open down-sets and start off with some basic facts.

PROPOSITION 5.9. ([7]) Let X be a Priestley space.

1. An ideal J ⊆ E(X) is principal if and only if Φ(J) is clopen.

2. An open down-set W is clopen if and only if Ψ(W ) is principal.

Proof. (1). Assume that J is principal; say J = (U ] for some clopen down-set U ∈ E(X).Then Φ(J) =

⋃U = U is clopen. Conversely, if Φ(J) is clopen, then by the ideal duality we

get J = Ψ(Φ(J)) = {V ∈ E(X);V ⊆ Φ(J)} = (Φ(J)]. Whence J is principal.

(2). This follows directly from (1) [use the fact that Ψ and Φ are inverses].

The following definition tells us, when an ideal is considered ”big”:

DEFINITION 5.10. An ideal I of a lattice L is said to be join-dense if for all z ∈ L wehave

z = sup((z] ∩ I).

Equivalently, I is join-dense if for all z ∈ L the following holds:

If c ∈ L, c ≥ x for all x ∈ L with x ∈ ((z] ∩ I), then c ≥ z.

It is helpful to introduce another notion of ”bigness” of open down-sets in Priestley spaces.Let us call an open down-set W of a Priestley space topologically join-dense if for everyC ∈ E(X) the following holds:

If C1 ∈ E(X), C1 ⊇ A for all A ∈ E(X) with A ⊆ C ∩W , then C1 ⊇ C.

Next, we want to prove that the notions of ”denseness” and ”topological join-denseness”coincide. In order to do this, we first need the following tool:

FACT 5.11. Let Z be a Priestley space and U ⊆ Z a dense open down-set. Then there existsno clopen down-set D 6= X such that D ⊇ A for all clopen down-sets A that are contained inW .

Proof. Suppose there is such a clopen down-set D. The equation Φ(Ψ(W )) = W tells us thatfor each w ∈ W there is a clopen down-set A that contains w and is a subset of W . ThusD ⊇ W ; since D is closed we get cl(W ) ⊆ D 6= W . So D is not dense.

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LEMMA 5.12. An open down-set W of a Priestley space X is dense if and only if it istopologically join-dense.

Proof. First, suppose that W is not dense. So there is an open set V 6= ∅ such that V ∩W = ∅.Since the collection of open down-sets and their complements form a basis of the Priestleytopology, there are open down-sets C1, C2 such that ∅ 6= C1 ∩ (X\C2) ⊆ V . Now takeC ′ = C1 ∩ C2. Since ∅ 6= C1 ∩ (X\C2), we get C ′ 6= C1. Moreover, take any clopen down-setA ⊆ W∩C1. If there were an a ∈ A∩(X\C2) we would get a ∈ W∩C1(∩(X\C2)) ⊆ W∩V = ∅,clearly a contradiction. Thus A ⊆ C ′ = C1 ∩ C2. So we just proved the following: There isa clopen down-set C ′ properly contained in C1 such that C ′ ⊇ A for all clopen down-setsA ⊆ C1 ∩W . Thus, W is not topologically join-dense.

Conversely assume that W is dense. Let C be any clopen down-set of X.Claim. C ∩W is dense in C.Proof of Claim. Assume not. Then there is a nonempty set V that is open in C such that(C∩W )∩V = ∅. But then V is open in X (since C is open in X). So W ∩V = W ∩(V ∩C) =(C ∩W ) ∩ V = ∅, so W is not dense in X, a contradiction.

Note that C is a Priestley space in its own right (with induced topology and order). So, fact5.11 implies that there is no clopen down-set C ′ strictly contained in C having the property(P) that C ′ ⊇ A for all clopen down-sets A ⊆ C ∩W . Now if there were a clopen down-setC1 6⊇ C with the property that C1 ⊇ A for all clopen down-sets A ⊆ C∩W , then C ′ = C∩C1

would have property (P) which it can not. So, If C ′ ∈ E(X), C ′ ⊇ A for all A ∈ E(X) withA ⊆ C ∩W , then C ′ ⊇ C. Since C was arbitrary, W is topologically join-dense.

PROPOSITION 5.13. Let X be a Priestley space.

1. Let J ⊆ E(X) be an ideal. Then J is join-dense if and only if Φ(J) is dense.

2. Let W ⊆ X be an open down-set. Then W is dense if and only if Ψ(W ) is dense.

Proof. Assume that J is join-dense. Set W = Φ(J). Lemma 5.12 tells us that it sufficesto show that W is topologically join-dense. Assume that W is not topologically join-dense.Then there are clopen down-sets C ′ 6⊇ C such that

C ′ ⊇ A for all A ∈ E(X) with A ⊆ C ∩W (1).

Now let A ∈ J with A ⊆ C. Thus A ⊆ W =⋃

J . So, by (1), A ⊆ C ′. Whence C ′ ∈ ((C]∩J)u

with C ′ 6⊇ C, which implies that C 6= sup((C] ∩ J). Thus J is not join-dense.

Conversely, assume that W = Φ(J) is a dense subset of X. Suppose that J ⊆ E(X) is notjoin-dense. The definition of join-denseness implies that there are clopen down-sets C 6= C ′

such that C ⊇ C ′ and C ′ ∈ ((C] ∩ J)u. Now, U = C\C ′ 6= ∅ is open. Assume thatx ∈ Φ(J) ∩ U . So there is an element D ∈ J with x ∈ D (because Φ(J) =

⋃J). Since J is

an ideal, C ∩D ∈ ((C]∩ J), but x ∈ (C ∩D)\C ′, so C ∩D 6⊆ C ′ which is a contradiction. SoΦ(J) ∩ U = ∅, therefore W = Φ(J) is not dense.

(2). This follows directly from (1) [use the fact that Ψ and Φ are inverses].

Next, we want to dualise the concept of principal extension.

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DEFINITION 5.14. Let W be an open down-set of a Priestley space X. Then we definethe topological principal extension of W to be

PW =⋃{C ∈ E(X);C ∩W is clopen}.

The next proposition shows that the topological principal extension is closely connected tothe principal extension.

PROPOSITION 5.15. Let X be a Priestley space.

1. If J ⊆ E(X) is an ideal, then Φ(PJ) = P (Φ(J)).

2. If W ⊆ X is an open down-set, then Ψ(PW ) = P (Ψ(W )).

Proof. (1). If x ∈ Φ(PJ), then there exists C ∈ E(X) such that x ∈ C and (C]∩ J = (D] forsome D ∈ J . Since Φ is a lattice isomorphism, Φ((C]) ∩ Φ(J) = Φ((D]). Since Φ((C]) = Cand Φ((D]) = D, we get x ∈ P (Φ(J)).Conversely, pick x ∈ P (Φ(J)). There is C ∈ E(X) such that x ∈ C and C ∩ Φ(J) is a clopendown-set, say D. Since Ψ is a lattice isomorphism, Ψ(C) ∩ Ψ(Φ(J)) = Ψ(D) which can besimplified to (C] ∩ J = (D]. Thus C ∈ PJ , so x ∈ Φ(PJ).

(2). Surprisingly, this part uses compactness of X (whereas part 1 did not). Take C ∈Ψ(PW )). So C ⊆ PW . For all x ∈ C there is Dx ∈ Ψ(PW ) such that x ∈ Dx. So Dx ∩W isclopen for all x ∈ C. Since C is compact and C =

⋃x∈C Dx, there exist x1, ...xr such that

C ⊆ Dx1 ∪ ... ∪Dxr =: D (?).

Since D∩W = (Dx1 ∩W )∪ ...∪ (Dxr ∩W ) =: D1 which is clopen, we can apply Ψ in order toget (D1] = Ψ(W ∩D) = Ψ(W ) ∩Ψ(D) = Ψ(W ) ∩ (D]. Thus D ∈ P (Ψ(W )). Since P (Ψ(W ))is an ideal, (?) implies that C ∈ P (Ψ(W )).Conversely, pick C ∈ P (Ψ(W )). So (C] ∩ Ψ(W ) = (D] for some D ∈ Ψ(W ). So, Φing thisequation gives Φ((C]) ∩ Φ(Ψ(W )) = Φ((D]) which in turn yields C ∩ W = D. ThereforeC ∈ {A ∈ E(X);A ∩ W clopen}, ie, C ⊆

⋃{A ∈ E(X);A ∩ W clopen} = PW , thus C ∈

Ψ(PW ).

A word on this proof may be in order. The reader will have noted that in the first twopropositions, the proof of statement (2) was taken care of by arguing: ”This follows directlyfrom (1) [using the fact that Ψ and Φ are inverses].” Why didn’t this work above? The(informal) reason is this: the first two propositions only involved properties concerningideals and open down-sets respectively (”clopen”, ”principal”, ”join-dense” and ”dense”).Whereas the statement of proposition 5.15 involves operations on ideals resp. open down-sets, namely the principal extension and the topological principal extension .

As a final ingredient, we describe the equivalent of J1 → J2 of ideals within the open down-setsof a Priestley space.

DEFINITION 5.16. Let W1,W2 be open down-sets of a Priestley space X. Then we set

W1 W2 =⋃{C ∈ E(X);C ∩D ⊆ W2 (∀D ∈ E(X) with D ⊆ W1)}.

30

Now we prove that this definition does exactly what we want it to do.

PROPOSITION 5.17. Let X be a Priestley space.

1. If J1, J2 are ideals of E(X), then Φ(J1 → J2) = Φ(J1) Φ(J2).

2. If W1,W2 are open down-sets of X, then Ψ(W1 W2) = Ψ(W1) → Ψ(W2).

Proof. (1). Let x ∈ Φ(J1 → J2). Then there is C ∈ J1 → J2 such that x ∈ C. Now takeany clopen down-set D ⊆ Φ(J1). So Ψ(D) ⊆ Ψ(Φ(J1)) = J1 which shows that D ∈ J1. SoC ∩D ∈ J2 as C ∈ J1 → J2. Therefore C ∩D ⊆ Φ(J2) =

⋃J2. Since D was arbitrary, this

implies that x ∈ C ⊆ Φ(J1) Φ(J2).Conversely, let x ∈ Φ(J1) Φ(J2). There is C ∈ E(X) with x ∈ C and C ∩D ⊆ W2 for allD ∈ E(X) contained in W1). Choose D ∈ J1. It is clear that D ⊆ Φ(J1). Therefore, by ourassumption, C ∩D ⊆ Φ(J2); applying Ψ we obtain:

Ψ(C) ∩Ψ(D) ⊆ Ψ(Φ(J2)) = J2.

Now C∩D ∈ Ψ(C)∩Ψ(D), so we get C∩D ∈ J2. Since D was arbitrary, we have C ∈ J1 → J2,thus x ∈ C ⊆ Φ(J1 → J2).

(2). Again, this part will use compactness of X. Pick C ∈ Ψ(W1 W2). So, C ⊆ W1 W2

and, for all x ∈ C there is Dx ∈ E(X) such that x ∈ Dx and Dx ∩ A ⊆ W2 for all A ∈ E(X)with A ⊆ W1 (?). Obviously, C ⊆

⋃x∈C Dx; since C is compact, there are x1, ..., xr such that

C ⊆ Dx1 ∪ ... ∪Dxr =: D .

We claim that D ∈ Ψ(W1) → Ψ(W2). Let D′ ∈ Ψ(W1). So D∩D′ = (Dx1∩D′)∪...∪(Dxr∩D′)which belongs to Ψ(W2) by (?) and because Ψ(W2) is an ideal. Since D′ was arbitrary, weconclude that D ∈ Ψ(W1) → Ψ(W2). As C ⊆ D and Ψ(W1) → Ψ(W2) is an ideal, we getC ∈ Ψ(W1) → Ψ(W2).

Conversely, suppose that C ∈ Ψ(W1) → Ψ(W2). This implies that

C ∈ {C ′ ∈ E(X);C ′ ∩D ⊆ W2 for all D ∈ E(X) with D ⊆ W1}︸ ︷︷ ︸U

.

But indeed, W1 W2 =⋃U , whence C ∈ Ψ(W1 → W2).

Propositions 5.15 and 5.17 say that we translated the concepts of principal extension andJ1 → J2, the relative pseudocomplement for ideals, adequately into the dual category. Thisenables us to characterize fractionally complete Priestley spaces in terms of ”open down-sets”.Recall that a distributive (0,1)-lattice L is fractionally complete if the following conditionholds:

If J ⊆ L is an ideal satisfying

• PJ is join-dense

• PJ → J = J

then J is a principal ideal.

31

From propositions 5.15 and 5.17 we obtain:

THEOREM 5.18. A Priestley space X is fractionally complete if the following conditionholds:

If W ⊆ X is an open down-set satisfying

• PW is dense

• PW W = W

then W is clopen.

5.4 Products of fractionally complete lattices

We prove that the class of fractionally complete lattices is closed under arbitrary products.

THEOREM 5.19. If (Li)i∈I is a family of fractionally complete lattices, then Πi∈ILi isfractionally complete.

Proof. Assume that all Li are fractionally complete. Let J ⊆ Πi∈ILi be an ideal with

• PJ is join-dense

• PJ → J = J .

We want to show that J is a principal ideal.Let i ∈ I be arbitrary, and let K = πi(J). It is easy to see that K is an ideal of Li.

Claim 1. πi(PJ) = PK.Proof of Claim 1. (Note that the first P means the principal extension in Πi∈ILi whereas thesecond P denotes the principal extension in Li.) Let x ∈ πi(PJ). So there exists ζ ∈ Πi∈ILi

such that ζ ∈ PJ and πi(ζ) = ζ(i) = x. Thus (ζ] ∩ J = (ξ] for some ξ ∈ J . It is easy to seethat this implies (x] ∩K = (ζ(i)] ∩ πi(J) = (ξ(i)] in L, whence x ∈ PK.On the other hand, let x ∈ PK. So (x]∩K is principal, say (x]∩K = (z]. Consider µ ∈ Πi∈ILi

defined by

µ(j) =

{x if j = i

0 otherwise

It is then easy to check that (µ] ∩ J = (ν], where ν(j) =

{z if j = i

0 otherwise. So µ ∈ PJ . Since

πi(µ) = x, we have x ∈ πi(PJ).

Claim 2. PK is join-dense.Proof of Claim 2. Suppose not. So there are x0 > x ∈ L such that x ∈ ((x0] ∩ PK)u. UsingClaim 1, we see that ξ0, ξ ∈ Πi∈ILi defined by

ξ0(j) =

{x0 if j = i

0 otherwiseand ξ(j) =

{x if j = i

0 otherwise

32

satisfy ξ < ξ and ξ ∈ ((ξ0] ∩ PJ)u. Thus PJ is not join-dense contradicting our assumption.

Claim 3. PK → K = K.Proof of Claim 3. We always have PK → K ⊇ K. Suppose that there exists y ∈ (PK →

K)\K. Consider µ ∈ Πi∈ILi defined by µ(j) =

{y if j = i

0 otherwise. With Claim 1, it is easy to

see that µ ∈ PJ → J which contradicts our assumption.

Note that Claims 2 and 3 together imply that K is principal (since Li is fractionally complete).So we have proved that for each i ∈ I there exists mi ∈ Li such that πi(J) = (mi]. We defineξ ∈ Πi∈ILi by

ξ(i) = mi for all i ∈ I.

We want to show that J = (ξ].Claim 4. ξ ∈ PJ → J .Proof of Claim 4. This amounts to showing that, given ρ ∈ PJ we get ρ ∧ ξ ∈ J . Nowρ ∈ PJ implies that (ρ] ∩ J = (ρ0] for some ρ0 ∈ J . Pick i ∈ I, define ξ′ ∈ Πi∈ILi by

ξ′(j) =

{ξ(i) if j = i

0 otherwise. Now we see that ξ′ ∈ J . Moreover ρ∧ ξ′ ∈ (ρ]∩ J , so ρ∧ ξ′ ≤ ρ0.

So we get (ρ ∧ ξ)(i) = (ρ ∧ ξ′)(i) ≤ ρ0(i). Since i ∈ I was arbitrary, this means

(ρ ∧ ξ)(i) ≤ ρ0(i) for all i ∈ I.

Thus ρ ∧ ξ ≤ ρ0, whence ρ ∧ ξ ∈ J , proving Claim 4.

Finally, Claim 4 and the fact that PJ → J = J imply that ξ ∈ J . Moreover, ξ is easily seento be the greatest element of J by definition. Therefore J = (ξ] is principal.

5.5 Free products of fractionally complete lattices

In contrast to the class of affine complete lattice, the class of fractionally complete latticesturns out not to be closed under arbitrary free products. Using our translation of the conceptof fractional completeness to the category of Priestley spaces, we will give an example of afractionally complete Priestley space X such that X × X is not fractionally complete anymore.

Consider the topological space βN, that is the Stone-Cech compactification of (N,P(N)). βNis clearly compact, moreover, βN is extremally disconnected, i.e., Hausdorff and the closureof every open set is open (see [8], 6.2.27). This implies that if a, b are distinct points of βN,there exists a clopen subset C ⊆ βN containing a but not b. Let ∆ = {(x, x);x ∈ βN} bethe antichain order on βN. It is easy to see that X = (βN,∆) is a Priestley space, indeed,the Priestley space corresponding to the Boolean power set lattice P(N). Note that E(X) isthe set of clopen subsets, since every subset is a downset (with ∆ being the antichain order).Thus, B := E(X) is Boolean. Moreover, X is extremally disconnected which implies that Bis a complete Boolean lattice ([7], Ex. 10.12). By [29] p. 687, B is fractionally complete andX is thus a fractionally complete Priestley space.

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Consider X × X. Now, the space βN × βN is not extremally disconnected (see [8], 6.3.21);whence by [7], Ex. 10.12, B′ = E(X ×X) is not a complete Boolean algebra. By [29], B′ isnot fractionally complete, whence X ×X is not a fractionally complete Priestley space.

The bottom line is that we have constructed a fractionally complete Priestley space X whoseproduct with itself is not fractionally complete. This implies that E(X) is a fractionallycomplete lattice such that the free product of E(X) with itself is not fractionally complete.

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Chapter 6

Order completeness

6.1 Dualising completeness

Recall that a (distributive) lattice L is called (order-)complete if for every subset A ⊆ L itssupremum sup(A) exists in L. Note that this implies that arbitrary infima exist.In order to dualise this concept, we need a topological notion.

DEFINITION 6.1. A topological space X is extremally disconnected if for any open setU ⊆ X the closure clX(U) is open (and thus clopen).

It is well-known which class of Priestley spaces corresponds to the class of complete distribu-tive lattices . We will establish this correspondence in the next theorem.

THEOREM 6.2. ([7]) Let L be a bounded distributive lattice. Then L is complete if andonly if its Priestley space D(L) is extremally disconnected.

Proof. Suppose L is complete; assume that U ⊆ X := D(L) is open. Now since X has abasis of clopen sets, U is the union of the clopen sets contained in U . Consider A := {C ⊆U ;C clopen in U}. Since L ∼= E(X), we know that E(X) is complete and therefore A ⊆ E(X)has a supremum, B. It is easy to see that B is the closure of U in X, so the closure of U isopen.Conversely, suppose that X = D(L) is extremally disconnected. Let A ⊆ L be arbitrary.Consider I, the ideal generated by A. It suffices to show that I has a supremum. ConsiderΦ(I), an open subset of X. The closure C = cl(Φ(I)) is open by assumption and thereforean element of E(X). Under the canonical isomorphism from E(X) to L, C corresponds to thesupremum of I and therefore A. So L is complete.

6.2 Products and coproducts of complete lattices

The following theorem is folklore:

THEOREM 6.3. ([7]) Arbitrary products of complete lattices are complete.

On the other hand, things are different for coproducts (i.e. free products of lattices). Herewe will use a topological argument and the fact that we know exactly how Priestley spaces

35

arising from complete lattices look like. Consider the Priestley space βN, as defined in 5.5.Consider the Priestley dual of (βN), i.e. L = P(N). The statements that the coproduct of Lwith itself is complete is equivalent to saying that the product of βN with itself is extremallydisconnected – which it is not, by [8], 6.2.28. Therefore coproducts of even finitely manycomplete distributive lattices need not be complete.

36

Chapter 7

Comparing the three kinds ofcompleteness

7.1 Affine complete vs fractionally complete

When comparing affine complete lattices and fractionally complete lattices, it is easier to startoff with lattices that are fractionally complete but not affine complete, for there is a plethoraof examples!

EXAMPLE 7.1. Each finite distributive (0,1)-lattice is fractionally complete and not affinecomplete.

Proof. Let L be finite. Then every ideal in L is principal, so the requirement for fractionallydistributive that each ideal J satisfying

1. PJ is join-dense

2. PJ → J = J

be principal is trivially satisfied. Moreover, if L were affine complete, 4.16 would imply thatQ01 can be embedded in L, which is impossible since L is finite.

EXAMPLE 7.2. Let X be a nonempty set. Then P(X) is fractionally complete but notaffine complete.

Proof. P(X) is not affine complete because it is Boolean. Moreover, example 5.5.4 says thatP(X) is fractionally complete.

On the other hand, it is rather harder to find a lattice that is affine complete but not frac-tionally complete. The following example has this property and is even countable.

Let Q = {q ∈ Q; 0 ≤ q ≤ 1}. Furthermore consider

L′ := {ξ ∈ QN; ξ(n) = 0 for all but finitely many n ∈ N}.

Now L′ lacks a top element. Let 1 : N → Q be the constant 1-map and set L = L′ ∪ {1}.

37

LEMMA 7.3. L as defined above is affine complete.

Proof. We show that L contains no proper Boolean interval. Let ξ < η ∈ L. Then thereexists m ∈ N with ξ(m) < η(m). Now there is an element q ∈ Q such that ξ(m) < q < η(m).Define ζ : N → Q by

ζ(n) =

{q n = m

ξ(n) n 6= m

Now it is easy to see that ζ has no complement in [ξ, η]: Suppose ζ ∧ ζ ′ = ξ. Then ζ ′(m) =ξ(m), but this implies ζ(m)∨ζ ′(m) = q 6= η(m), whence ζ∨ζ ′ 6= η. So [ξ, η] is not Boolean.

LEMMA 7.4. The above defined L is not fractionally complete.

Proof. First it is easy to see that L′ = L\{1} is a join-dense ideal of L; the reason is that 1

is the supremum of all the functions of the form χk for k ∈ N where χk(n) =

{1 n = k

0 n 6= k.

Define the idealJ = {ξ ∈ L′; ξ(k) = 0 for all odd k ∈ N}.

It is easy to see that J is non-principal.

Claim 1. PJ = L′.Proof of Claim 1. Of course, PJ is contained in L′ (otherwise J would be principal). Nowpick α ∈ L′. So α(k) 6= 0 for only finitely many k ∈ N. Consider α0 ∈ J defined by

α0(k) =

{α(n) if n is even0 otherwise

Clearly, (α] ∩ J = (α0]. We infer α ∈ PJ .

Claim 2. PJ → J = J .Proof of Claim 2. Trivially, J ⊆ PJ → J . Take any x ∈ L\I. If x ∈ PJ = L′, thenx ∧ x = x /∈ J , so x /∈ PJ by definition of PJ . So the only possibility remaining is x = 1.But this is ruled out immediately, because χ3 ∈ PJ and 1 ∧ χ3 = χ3 /∈ J . So PJ → J = J .

Summarizing, we have a non-principal ideal J such that PJ is join-dense and satisfyingPJ → J = J . Thus L is not fractionally complete.

7.2 Order complete vs affine complete

Any complete Boolean lattice is complete but not affine complete. Examples abound ofdistributive lattices that are complete but not affine complete, abound. We want to exhibitsome of them. Remember that a pair (x, y) of elements of a lattice L are called a coveringpair iff [x, y] = {x, y}.

EXAMPLE 7.5. Any complete distributive lattice containing a covering pair is complete butnot affine complete.

38

Proof. This is clear, since a covering pair is a non-trivial Boolean interval contained in thelattice considered.

EXAMPLE 7.6. Every finite distributive (0, 1)-lattice is complete but not affine complete.

Proof. If L is finite, then it clearly is complete, and since it is a (0,1)-lattice, it contains atleast two distinct elements. So, since it is finite, it contains a covering pair which makes itnon-affine-complete as we have seen in the last example.

EXAMPLE 7.7. Every complete distributive lattice can be made non-affine-complete byadding a new top element.

There are also plenty of examples of distibutive lattices that are not complete but affinecomplete. Note that every affine complete contains an order dense chain (as shown in theorem4.16) and whence it is not surprising that non-complete order dense chains lie at the core ofconstructing any lattice that is affine complete but not complete.The easiest example of a lattice that is affine complete but not complete is this:

EXAMPLE 7.8. Any power of Q := Q ∩ [0, 1] is affine complete but not complete. Moregenerally, any product of bounded order dense chains, at least one of them non-complete, isaffine complete but not complete. Such a lattice is for example [0, 1]× ([0, 1]\Q).

However, we are able to generalize this example.

EXAMPLE 7.9. Let L be any affine complete lattice, and let C be any order-dense boundedchain. Then L× C is affine complete but not complete.

7.3 Order complete vs fractionally complete

An important notion in this context is the following:

DEFINITION 7.10. A bounded distributive lattice L is called conditionally upper continousif a ∧

∨j xj =

∨j(a ∧ xj) for any a ∈ L and any set {xj : j ∈ J} whose join exists in L.

We will need the following theorem:

THEOREM 7.11. ([30]) If L is complete and conditionally upper continous, then L isfractionally complete. In particular, every algebraic lattice is fractionally complete.

This theorem enables us to construct an example of a bounded complete distributive latticethat is not fractionally complete.

EXAMPLE 7.12. Consider the poset (N,�), where m � n if and only if m divides n. Notethat 1 is the smallest and 0 is the greatest element. Then we have

1. (N,�) is a distributive lattice.

2. The lattice (N,�) is complete.

3. (N,�) is not fractionally complete.

39

Proof. First, note that given two elements a, b ∈ N then gcd(a, b) is the infimum and lcm(a, b)is the supremum of a, b. Moreover, it is easily verified that distributivity holds, see for instance[7].As for completeness it suffices to prove that arbitrary suprema exist. Take an infinite set A ofnatural numbers not containing 0. It is clear that no positive number divides every numbercontained in A. So 0 is the only natural number deviding every member of A, whence

∨A = 0.

In order to show that (N,�) is not fractionally complete it suffices to find a non-principalideal J ⊆ N such that

• PJ is join-dense in (N,�).

• PJ → J = J .

Take J := {2n;n ∈ N}. This is clearly an ideal, for it is directed since J is a chain, and it is adown-set because any divisor of a power of 2 is also a power of 2 (or 1). Now take any naturalnumber n ∈ N\{0}. There is a greatest natural number s such that 2s divides n. Whence

↓ n ∩ J = {1, 2, ...2s} =↓ 2s is principal.

This implies n ∈ PJ and proves that

PJ = N\{0}.

It remains to show that PJ → J = J . Let n ∈ N\J be arbitrary. If n ∈ PJ then n∧n = n /∈ J .If n = 0 then of course for instance 0 ∧ 3 = 3 /∈ J . So PJ → J = J . This shows that (N,�)is not fractionally complete.

However, any non-complete bounded chain C is a lattice that is fractionally complete but notcomplete. Moreover, one could conjecture that chains are essentially the only examples oflattices that are fractionally complete but not complete. But the conjecture is easily seen tobe false, since for each cardinal κ ≥ 2 the product lattice Cκ is not a chain and fractionallycomplete but not complete.

EXAMPLE 7.13. The distributive lattice (R\Q)×Q with a top and bottom element addedis fractinally complete but not complete.

The following lemma provides a partial the converse for theorems 5.19 and 6.3. It shows howto destroy one or the other of the properties of lattices considered here.

LEMMA 7.14. Let (Li)i∈I be a family of bounded distributive lattices. Then

1. If some Li is not complete, then∏

i∈I Li is not complete.

2. If some Li is not fractinally complete, then∏

i∈I Li is not fractionally complete.

Proof. Let A ⊆ Li be a subset of Li which has no supremum. Then

A := {f ∈∏i∈I

Li; f(i) ∈ A}

is easy to see to have no supremum in∏

i∈I Li. This proves the first statement.

40

Let J ⊆ Li be a nonprincipal ideal of Li such that PJ is join-dense and PJ → J = J . Thenconsider the ideal

J := {f ∈∏i∈I

Li; f(i) ∈ A}

of∏

i∈I Li. Define ρ ∈∏

i∈I Li by ρ(i) = p and ρ(j) = 1 for all j 6= i. Since p ∈ PJ there isp0 ∈ J such that (p] ∩ J = (p0]. Whence, setting ρ0 to be p0 in i and 1 everywhere else, weget (ρ ∩ J = ρ0. This shows that

P J ⊇ {f ∈∏i∈I

Li; f(i) ∈ PJ}

which directly implies that P J is join-dense in∏

i∈I Li. Another easy calculation yieldsP J → J = J .

This shows how to construct a large class of bounded distributive lattices that are fractionallycomplete but not complete and vice versa.

COROLLARY 7.15. Let L be any fractionally complete distributive lattice. Then L×Q01

is fractionally complete but not complete.

COROLLARY 7.16. Let L be any complete distributive lattice and let N be the lattice of thenatural numbers ordered by divisibility. Then L×N is complete but not fractionally complete.

41

Part III

Representability and ordercomponents

42

Chapter 8

Joint work with M. E. Adams

8.1 Introduction

One of the most important questions concerning Priestley spaces is the so called representabil-ity question:

What posets are order isomorphic to the poset of prime ideals of some boundeddistributive lattice?

A poset that is order isomorphic to the poset of prime ideals of some bounded distributivelattice is called representable. By Priestley Duality, a poset is representable if and only if itcan be endowed with a topology such that the resulting ordered topological space is Priestley.The question of which posets are representable essentially dates back to Balbes [2] (see also,Balbes and Dwinger [3]) and has been considered by a number of authors since (see, forexample, the expository article Priestley [26].)I was wondering whether it suffices to consider connected posets for the representabilityquestion. This would be the case if the following hypothesis were satisfied:

(H) A disjoint union of a family of posets (Pi)i∈I is representable if and only ifeach Pi is representable.

The following is a presentation of the main results obtained in [1].Let (X;≤) be a poset. Consider its associated undirected graph (X, V≤) where V≤ :={{x, y};x ≤ y or y ≤ x}. Let

R = {(x, y); {x, y} ∈ V≤} ∪ {(x, x);x ∈ X}.

Moreover, define R′ to be the transitive closure of R. Then R′ is an equivalence relation. Anorder component of X is an equivalence class [x]R′ of the relation R′ for some x ∈ X. A posetist called connected if it has only one order component.Our principal result are theorems 8.1 and 8.2.

THEOREM 8.1 (van der Zypen). If the order components of a poset (X;≤) are repre-sentable, then so is X.

43

As for the converse, I was able to make a positive statement about the order components ofa representable posets (lemma 8.4), but failed to establish that they must be representablethemselves in general. After discussing the problem with M.E. Adams, he came up with acounterexample which is the key ingredient of theorem 8.2:

THEOREM 8.2 (Adams). There exists a representable poset with an order componentwhich is not representable.

The proof of theorem 8.1 will be given in paragraph 8.2, where we begin by lemma 8.3, show-ing that a poset is compact under its interval topology iff each order component is compactunder its respective interval topology. It follows readily from lemma 8.3 that each order com-ponent of a representable poset is compact with respect to its interval topology. We thenestablish, in lemma 8.5, that if every order component of a poset is representable, then sotoo is the poset. In paragraph 8.3, we define a countably infinite poset which we show to beorder-isomorphic to an order component of a representable poset in lemma 8.7, but which, aslemma 8.8 shows, is not itself representable.

For any undefined terms or additional background, we refer the reader to Gratzer [11] andKelley [16], with each of which our notation is consistent.

8.2 Proof of theorem 8.1

LEMMA 8.3. Let (Xk;≤k)k∈K be a family of pairwise disjoint nonempty posets. Then for(X;≤) where X =

⋃k∈K Xk and ≤=

⋃k∈K ≤k, the following are equivalent:

(i) for each k ∈ K, the space (Xk; τi(Xk)) is compact;

(ii) (X; τi(X)) is compact.

Proof. Assume that (i) holds and let U be an open cover of X =⋃

k∈K Xk. By Alexander’ssubbase lemma, we may assume that

U = {X\(a] | a ∈ A} ∪ {X\[b) | b ∈ B}

for some subsets A,B ⊆ X. We distinguish two cases:

First, there is some k ∈ K such that A ∪ B ⊆ Xk. In which case, consider UXk= {Xk\(a] |

a ∈ A} ∪ {Xk\[b) | b ∈ B}. Since (Xk; τi(Xk)) is compact by assumption, UXkhas a finite

subcover{Xk\(a1], ..., Xk\(ar]} ∪ {Xk\[b1), ..., Xk\[bs)},

so {X\(a1], ..., X\(ar]} ∪ {X\[b1), ..., X\[bs)} is a finite subcover of U . Secondly, there is nok ∈ K such that A ∪ B ⊆ Xk. Hence, there are w1, w2 ∈ A ∪ B such that w1 ∈ Xk andw2 ∈ Xk′ for some k 6= k′ ∈ K. If w1, w2 ∈ A, then {X\(w1], X\(w2]} is a finite subcoverof U . If w1 ∈ A,w2 ∈ B, then {X\(w1], X\[w2)} is a finite subcover of U (similarly for

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w1 ∈ B,w2 ∈ A). Finally if w1, w2 ∈ B, then {X\[w1), X\[w2)} is a finite subcover of U .

Thus, in any case, (X; τi(X)) is compact.

Assume that (ii) holds and let k ∈ K. Assume that U is an open cover of Xk. By Alexander’ssubbase lemma we may assume that

U = {Xk\(a] | a ∈ A} ∪ {Xk\[b) | b ∈ B}

for some subsets A,B ⊆ Xk. Consider the following open cover of X =⋃

l∈K Xl

U∗ = {X\(a] | a ∈ A} ∪ {X\[b) | b ∈ B}.

Then U∗ has a finite subcover {X\(a1], ..., X\(ar]} ∪ {X\[b1), ..., X\[bs)} since X is compactwith its interval topology. Thus {Xk\(a1], ..., Xk\(ar]} ∪ {Xk\[b1), ..., Xk\[bs)} is a finitesubcover of Xk.

If (X;≤) is representable, then, for some topology τ , (X; τ,≤) is a Priestley space. Inparticular, (X; τ) is a compact space and, as τi ⊆ τ , so too is (X; τi). Thus, the following isan immediate consequence of 8.3.

LEMMA 8.4. Each order component of a representable poset is compact with respect to itsinterval topology.

We now go on to show that if the order components of a poset are representable, then so isthe poset.

LEMMA 8.5. Let (Xk,≤k)k∈K be a family of pairwise disjoint nonempty representableposets. Then (X;≤) is representable, where X =

⋃k∈K Xk and ≤=

⋃k∈K ≤k.

Proof. If K is empty or a singleton, the statement is trivial. So we may assume that K hasmore than one element. For any k ∈ K, let τk be a topology making (Xk; τk,≤k) a Priestleyspace. Fix k∗ ∈ K and x ∈ Xk∗ . We now build a subbase for a topology on X in three steps.We set:

S1 =⋃

l∈K\{k∗} τl;

S2 = {U ∈ τk∗ | x /∈ U};

S3 = {U ⊆ X | x ∈ U and U ∩ Xk∗ ∈ τk∗ and, for some k′ ∈ K\{k∗}, U = [U ∩ Xk∗ ] ∪[⋃

l∈K\{k∗,k′} Xl]}.

Then let τ be the topology having S = S1 ∪ S2 ∪ S3 as a subbase. Using Alexander’ssubbase lemma we check easily that (X; τ) is compact using the fact that any subbase membercontaining x is, in some sense, large by virtue of the definition of S3 ⊆ S. Moreover, aninspection by cases shows easily that (X; τ,≤) is totally order-disconnected.

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8.3 Proof of theorem 8.2

In order to prove 8.2 we give an example of a poset (P ;≤) which is order isomorphic to anorder component of a representable poset, but is not representable itself.

On the set

P = {p} ∪ {pi0,...,in | 0 ≤ n < ω and 0 ≤ ij < ω for 0 ≤ j ≤ n},

inductively define an order relation ≤ as follows.

For 0 ≤ j < i < ω,p < pi < pj .

For 0 ≤ i0 < ω, 0 ≤ k ≤ i0, and 0 ≤ i < j < ω,

pi0,i < pi0,j < pk.

For 0 ≤ i0, i1 < ω, 0 ≤ k ≤ i1, and 0 ≤ j < i < ω,

pi0,k < pi0,i1,i < pi0,i1,j .

In general, let 0 < r < ω.

For 0 ≤ i0, i1, . . . , i2r < ω, 0 ≤ k ≤ i2r, and 0 ≤ i < j < ω,

pi0,i1,...,i2r−1,i2r,i < pi0,i1,...,i2r−1,i2r,j < pi0,i1,...,i2r−1,k.

For 0 ≤ i0, i1, . . . , i2r+1 < ω, 0 ≤ k ≤ i2r+1, and 0 ≤ j < i < ω,

pi0,i1,...,i2r,k < pi0,i1,...,i2r,i2r+1,i < pi0,i1,...,i2r,i2r+1,j .

To see that (P ;≤) is a poset, for 0 ≤ n < ω, let

P (n) = {p} ∪ {pi0,...,im | 0 ≤ m ≤ n and, for 0 ≤ j ≤ m, 0 ≤ ij < ω}.

Thus, P (0) and, for each 0 ≤ n < ω, P (n + 1) \ P (n) are clearly antisymmetric and transi-tive. Further, x ∈ P (n) is comparable with y ∈ P \ P (n) only if x ∈ P (n) \ P (n − 1) andy ∈ P (n + 1) \ P (n), where it is the case that x > y and x < y depending on whether nis even or odd, respectively. In particular, ≤ is antisymmetric. Moreover, if n is even, sayn = 2r, then x = pi0,...,i2r−1,k and y = pi0,...,i2r,i providing 0 ≤ k ≤ i2r and 0 ≤ i < ω, and ifn is odd, say n = 2r + 1, then x = pi0,...,i2r,k and y = pi0,...,i2r+1,i providing 0 ≤ k ≤ i2r+1 and0 ≤ i < ω. In particular, ≤ is transitive and, as claimed, (P ;≤) is seen to be a countableconnected poset. We also note in passing that, for 0 ≤ i0, . . . , in < ω, [pi0,...,in) and (pi0,...,in ]are finite chains depending on whether n is even or odd, respectively, a fact that we will referback to later.

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In order to show that (P ;≤) is order-isomorphic to an order component of a representableposet, we will define a suitable order � on a compact totally disconnected space (C; τ) whichitself is homeomorphic to the Stone space of a countable atomless Boolean algebra. To do so,we will need an explicit description of (C; τ), which we now give.

Let Q = (Q;≤) denote the rational interval (0, 1). Then (A,B) is a Dedekind cut of Q provid-ing that A and B are disjoint non-empty sets such that Q = A∪B and, for a ∈ A and b ∈ B,a < b. For a Dedekind cut (A,B) of Q, A is a gap providing A does not have a greatestelement and B does not have a smallest element and, otherwise, it is a jump. Let (C;≤)denote the set of all decreasing subsets of the rational interval (0, 1) ordered by inclusion.Thus, for I ∈ C, if I 6= ∅ or Q, then I is a jump precisely when I = (0, r) or (0, r] for somer ∈ Q. Intuitively, (C;≤) may be thought of as the real interval [0, 1] where every rationalelement 0 < r < 1 is replaced by a covering pair. The interval topology τi, denoted hence-forth simply by τ , on (C;≤) has as a base the open intervals C, [∅, I) = {J ∈ C : J ⊂ I},(I, Q] = {J ∈ C : I ⊂ J}, and (I, J) = {K ∈ C : I ⊂ K ⊂ J}. It is well-known that (C; τ)is a compact totally disconnected space, whose clopen subsets are precisely the sets ∅, C, andfinite unions of sets of the form [I, J ] = {K ∈ C : I ⊆ K ⊆ J} where I = (0, r] and J = (0, s)for r, s ∈ Q with r < s.

Let Q = (si : 0 ≤ i < ω) be some enumeration of Q. We inductively define a new partialorder on C as follows:

In C, choose gaps x and, for 0 ≤ i < ω, xi such that

x < xi < xj for 0 ≤ j < i < ω,

where x is a member of the closure of {xi | 0 ≤ i < ω}, denoted cl({xi | 0 ≤ i < ω}), and set

x ≺ xi ≺ xj .

Choose clopen intervals (Xi : 0 ≤ i < ω) such that xi ∈ Xi, Xi ∩Xj = ∅ whenever i 6= j, thelength of Xi, denoted ln(Xi), is ≤ 1

2 in the pseudometric obtained from the metric imposedon C by the real metric on (0, 1), and (0, s0), (0, s0] 6∈ Xi for any 0 ≤ i < ω.

For 0 ≤ i0 < ω, 0 ≤ k ≤ i0, and 0 ≤ i < ω, choose gaps xi0,i ∈ Xi0 such that

xi0,i < xi0,j < xk for 0 ≤ i < j < ω,

where xi0 ∈ cl({xi0,i | 0 ≤ i < ω}), and set

xi0,i ≺ xi0,j ≺ xk.

Choose clopen intervals (Xi0,i : 0 ≤ i < ω) such that xi0,i ∈ Xi0,i, Xi0,i ∩Xi0,j = ∅ for i 6= j,Xi0,i ⊆ Xi0 , ln(Xi0,i) ≤ 1

22 , and (0, s1), (0, s1] 6∈ Xi0,i for 0 ≤ i < ω.

For 0 ≤ i0, i1 < ω, 0 ≤ k ≤ i1, and 0 ≤ i < ω, choose gaps xi0,i1,i ∈ Xi0,i1 such that

xi0,k < xi0,i1,i < xi0,i1,j for 0 ≤ j < i < ω,

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where xi0,i1 ∈ cl({xi0,i1,i | 0 ≤ i < ω}), and set

xi0,k ≺ xi0,i1,i ≺ xi0,i1,j .

Choose clopen intervals (Xi0,i1,i : 0 ≤ i < ω) such that xi0,i1,i ∈ Xi0,i1,i, Xi0,i1,i ∩Xi0,i1,j = ∅for i 6= j, Xi0,i1,i ⊆ Xi0,i1 , ln(Xi0,i1,i) ≤ 1

23 , and (0, s2), (0, s2] 6∈ Xi0,i1,i for 0 ≤ i < ω.

In general, let 0 < r < ω.

For 0 ≤ i0, i1, . . . , i2r < ω, 0 ≤ k ≤ i2r, and 0 ≤ i < ω, choose gaps xi0,i1,...,i2r,i ∈ Xi0,i1,...,i2r

such that

xi0,i1,...i2r−1,i2r,i < xi0,i1,...i2r−1,i2r,j < xi0,i1,...,i2r−1,k for 0 ≤ j < i < ω,

where xi0,...,i2r ∈ cl({xi0,...,i2r,i | 0 ≤ i < ω}), and set

xi0,i1,...i2r−1,i2r,i ≺ xi0,i1,...i2r−1,i2r,j ≺ xi0,i1,...,i2r−1,k.

Choose clopen intervals (Xi0,i1,...,i2r,i : 0 ≤ i < ω) such that xi0,i1,...,i2r,i ∈ Xi0,i1,...,i2r,i,Xi0,i1,...,i2r,i ∩ Xi0,i1,...,i2r,j = ∅ for i 6= j, Xi0,i1,...,i2r,i ⊆ Xi0,i1,...,i2r , ln(Xi0,i1,...,i2r,i) ≤ 1

22r+1 ,and (0, s2r+1), (0, s2r+1] 6∈ Xi0,i1,...,i2r,i for 0 ≤ i < ω.

For 0 ≤ i0, i1, . . . , i2r+1 < ω, 0 ≤ k ≤ i2r+1, and 0 ≤ i < ω, choose gaps xi0,i1,...,i2r+1,i ∈Xi0,i1,...,i2r+1 such that

xi0,i1,...,i2r,k < xi0,i1,...i2r,i2r+1,i < xi0,i1,...i2r,i2r+1,j for 0 ≤ i < j < ω,

where xi0,...,i2r+1 ∈ cl({xi0,...,i2r+1,i | 0 ≤ i < ω}), and set

xi0,i1,...,i2r,k ≺ xi0,i1,...i2r,i2r+1,i ≺ xi0,i1,...i2r,i2r+1,j .

Choose clopen intervals (Xi0,i1,...,i2r+1,i : 0 ≤ i < ω) such that xi0,i1,...,i2r+1,i ∈ Xi0,i1,...,i2r+1,i,Xi0,i1,...,i2r+1,i∩Xi0,i1,...,i2r+1,j = ∅ for i 6= j, Xi0,i1,...,i2r+1,i ⊆ Xi0,i1,...,i2r+1 , ln(Xi0,i1,...,i2r+1,i) ≤

122(r+1) , and (0, s2r+2), (0, s2r+2] 6∈ Xi0,i1,...,i2r+1,i for 0 ≤ i < ω.

Elsewhere on C, let � be trivial. Thus, since (X;�) is order-isomorphic to (P ;≤), (C;�) isa poset whose order components consist precisely of X = {x} ∪ {xi0,...,in | 0 ≤ n < ω and 0 ≤ij < ω for 0 ≤ j ≤ n} and 2ω singletons.

LEMMA 8.6. (C; τ,�) is a Priestley space.

Proof. As (C;�) is a poset and (C; τ) is a compact totally disconnected space, it remainsto show that, for u, v ∈ C, whenever u 6� v there exists a clopen decreasing set U such thatu ∈ U and v 6∈ U .

Since (X;�) is order-isomorphic to (P ;≤), we set, for 0 ≤ n < ω,

X(n) = {x} ∪ {xi0,...,im | 0 ≤ m ≤ n and, for 0 ≤ j ≤ n, 0 ≤ ij < ω}

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and observe that, as [xi0,...,in) or (xi0,...,in ] is a finite chain depending on whether n is even orodd, respectively, it follows from the choice of elements in X\X(n) that, for 0 ≤ i0, . . . , in < ω,⋃

(Xi0,...,in−1,k : 0 ≤ k ≤ in) is clopen increasing or decreasing, accordingly.

Consider u, v ∈ C with u 6� v. In each case we will exhibit a clopen decreasing set U suchthat u ∈ U and v 6∈ U .

If u < v, then u ≤ (0, s] < v for some s ∈ Q. Since � is compatible with ≤, set U = (∅, (0, s]].Henceforth, we assume that u > v and, in particular, u and v are incomparable under �.

Suppose there is an infinite sequence (ik : 0 ≤ k < ω) such that u ∈ Xi0,...,ik for any 0 ≤ k < ω.Then, by choice, u is a gap and, since ln(Xi0,...,in) ≤ 1

2n , v 6∈ Xi0,...,in for some 0 ≤ n < ω.Without loss of generality, we may assume that n is even. Set U =

⋃(Xi0,...,in,l : 0 ≤ l ≤ in+1).

By the above observation, U is clopen decreasing, u ∈ U , and, since U ⊆ Xi0,...,in , v 6∈ U .

Likewise, if there is an infinite sequence (jk : 0 ≤ k < ω) such that v ∈ Xj0,...,jkfor

any 0 ≤ k < ω, then v is a gap and, since ln(Xjo,...,jm) ≤ 12m , u 6∈ Xj0,...,jm for some

0 ≤ m < ω. We may assume, again with no loss in generality, that m is odd. SetU = C \

⋃(Xj0,...,jm,l : 0 ≤ l ≤ jm+1). Then, U is clopen decreasing, v 6∈ U , and, since

U ⊆ C \Xj0,...,jm , u ∈ U .

Suppose, for some finite sequence (ik : 0 ≤ k ≤ n), u ∈ Xi0,...,in , but u 6∈ Xi0,...,in,l for any0 ≤ l < ω. Then, providing u 6= xi0,...,in , it is not hard to see that there exists a clopen set Usuch that u ∈ U , v 6∈ U , and each element of U is incomparable under � to any other elementof (C;�), whereby U is decreasing. Were it the case that u 6∈ Xl for any 0 ≤ l < ω, then asimilar set may be defined unless u = x.

Likewise, suppose it is the case that, for some finite sequence (jk : 0 ≤ k ≤ m), v ∈ Xj0,...,jm ,but that v 6∈ Xj0,...,jm,l for any 0 ≤ l < ω. Then, providing v 6= xj0,...,jm , there exists a clopenset V such that v ∈ V , u 6∈ V , and each element of V is incomparable under � to any otherelement of (C;�). In this case, set U = C \ V . Likewise, unless v = x, a similar set may bedefined whenever v 6∈ Xl for any 0 ≤ l < ω.

Thus, it now remains to consider the eventuality that u = x or xi0,...,in for some (ik : 0 ≤ k ≤n) and v = x or xj0,...,jm for some (jk : 0 ≤ k ≤ m). Observe that, by hypothesis, since v < u,u = x is impossible and, hence, we need only consider u = xi0,...,in for some (ik : 0 ≤ k ≤ n).Further, if v = x, then, by hypothesis, u = xi0,...,in for some n > 0. Since v 6∈ Xi0 and u 6= xi0 ,u ∈ U =

⋃(Xi0,k : 0 ≤ k ≤ i1) ⊆ Xi0 , which, as observed above, is clopen decreasing. Thus,

in addition, we may assume that v = xj0,...,jm for some (jk : 0 ≤ k ≤ m).

A number of possibilities still remain to be considered.

Suppose first that n ≤ m.

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Assume ik = jk for all 0 ≤ k ≤ n. Then, by hypothesis, m ≥ n + 2 and, since u > v, n iseven. Thus, V =

⋃(Xi0,...,in,jn+1,l : 0 ≤ l ≤ jn+2) is clopen increasing v ∈ V , and u 6∈ V . Set

U = C \ V .

Suppose ik = jk for all 0 ≤ k < n, but in 6= jn. Then, by hypothesis, m ≥ n + 1. Sup-pose n is even. Were it the case that in > jn, then it would follow that u < v, contraryto hypothesis. Thus, we may assume that in < jn. But then it follows that m ≥ n + 2.Thus, v ∈ V =

⋃(Xi0,...,in−1,jn,jn+1,l : 0 ≤ l ≤ jn+2) which is clopen increasing and, since

V ⊆ Xi0,...,in−1,jn , u 6∈ V . Suppose n is odd. Thus, v ∈ V =⋃

(Xi0,...,in−1,jn,l : 0 ≤ l ≤ jn+1),which is clopen increasing, and again, since V ⊆ Xi0,...,in−1,jn , u 6∈ V . In either case, setU = C \ V .

Consider, for some 0 ≤ k ≤ n− 1, il = jl for all 0 ≤ l < k, but ik 6= jk. If k is even, then u ∈U =

⋃(Xi0,...,ik−1,ik,l : 0 ≤ l ≤ ik+1) which is clopen decreasing and, since U ⊆ Xi0,...,ik−1,ik

and v ∈ Xi0,...,ik−1,jk, v 6∈ U . If k is odd, then v ∈ V =

⋃(Xi0,...,ik−1,jk,l : 0 ≤ l ≤ jk+1) which

is clopen increasing and, since V ⊆ Xi0,...,ik−1,jkand u 6∈ Xi0,...,ik−1,jk

, u 6∈ V . In this case, setU = C \ V .

It remains to consider n > m.

Suppose ik = jk for all 0 ≤ k ≤ m. Then, by hypothesis, n ≥ m + 2 and, since u > v, m isodd. Hence, u ∈ U =

⋃(Xj0,...,jm,im+1,l : 0 ≤ l ≤ im+2) which is clopen decreasing, whilst

v 6∈ U .

Consider ik = jk for all 0 ≤ k < m, but im 6= jm. By hypothesis, n ≥ m + 1. Suppose m iseven. Then, u ∈ U =

⋃(Xj0,...,jm−1,im,l : 0 ≤ l ≤ im+1) which is clopen decreasing, and, since

U ⊆ Xj0,...,jm−1,im , v 6∈ U . Suppose m is odd. Were im < jm, then it would follow that u < v,contrary to hypothesis. Thus, we may assume that im > jm and, so, n ≥ m + 2. Hence,u ∈ U =

⋃(Xj0,...,jm−1,im,im+1,l : 0 ≤ l ≤ im+2) which is clopen decreasing, and, since it is

also the case that U ⊆ Xj0,...,jm−1,im , v 6∈ U .

Finally, it remains to consider the case that, for some 0 ≤ k ≤ m− 1, il = jl for all 0 ≤ l < k,but ik 6= jk. However, the same argument holds, word for word, as given in the analogouscase when n ≤ m.

Since the order components of (C; τ,�) consist of precisely X = {x} ∪ {xi0,...,in | 0 ≤ n <ω and 0 ≤ ij < ω for 0 ≤ j ≤ n} and 2ω singletons and, by choice, (X;�) is order-isomorphicto (P ;≤), the following is an immediate consequence of 8.6.

LEMMA 8.7. (P ;≤) is order-isomorphic to an order component of a representable poset.

The proof of theorem 8.2 will be complete once we have established the following.

LEMMA 8.8. (P ;≤) is not representable.

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Proof. Suppose, contrary to hypothesis, that (P ;≤) is representable and let (P ; τ,≤) be aPriestley space for some topology τ .

We claim that, for x ∈ P , there is a sequence (xi : 0 ≤ i < ω) such that either, for0 ≤ j < i < ω, xi < xj and x is the greatest lower bound of {xi | 0 ≤ i < ω} or, for0 ≤ i < j < ω, xi < xj and x is the least upper bound of {xi | 0 ≤ i < ω}.

To justify the claim, we consider the various possibilities. If x = p, then setting xi = pi yields,for 0 ≤ j < i < ω, p < pi < pj . Moreover, for y ∈ P \ P (0), [y) ∩ P (0) is finite. In particular,p is the greatest lower bound of {pi | 0 ≤ i < ω}. Similarly, for x = pi0,...,in , let xi = pi0,...,in,i

for 0 ≤ i < ω. If n is even, then, for 0 ≤ i < j < ω,

pi0,...,in,i < pi0,...,in,j < pi0,...,in .

Since pi0,...,in is the greatest lower bound of [pi0,...,in,i) and, for y ∈ P \P (n+1), (y]∩P (n+1)is finite, it follows that pi0,...,in is the least upper bound of {pi0,...,in,i | 0 ≤ i < ω}. Likewise,if n is odd, then, for 0 ≤ j < i < ω,

pi0,...,in < pi0,...,in,i < pi0,...,in,j .

Since pi0,...,in is the least upper bound of (pi0,...,in ] and, for every y ∈ P \P (n+1), [y)∩P (n+1)is finite, it follows that pi0,...,in is the greatest lower bound of {pi0,...,in,i | 0 ≤ i < ω}.

Using the above claim, we now show that every x ∈ P is an accumulation point.

To see this, say x is the greatest lower bound of {xi | 0 ≤ i < ω} where, for 0 ≤ j < i < ω,xi < xj . For 0 ≤ i < ω, there exists a clopen increasing set Vi such that xi ∈ Vi and xi+1 6∈ Vi.Clearly, {Vi | 0 ≤ i < ω} is an open cover of S = {xi | 0 ≤ i < ω} with no finite subcover. Inparticular, S is not closed. Choose y ∈ cl(S) \ S. If y 6≥ x, then there is a clopen decreasingset U with y ∈ U and x 6∈ U , from which it follows that U ∩S = ∅, contradicting y ∈ cl(S). Ify > x, then y is not a lower bound of S, as x is the greatest. In particular, for some 0 ≤ n < ω,xn 6≥ y. It follows that there is a clopen decreasing set U with xn ∈ U and y 6∈ U . Thus,S ⊆ {x0, . . . , xn} ∪ U , which is a closed set. On the other hand, y ∈ P \ ({x0, . . . , xn} ∪ U),contradicting the fact that y ∈ cl(S). We conclude that y = x and, in particular, that, asclaimed, x is an accumulation point. As similar argument holds in the case that x is the leastupper bound of {xi | 0 ≤ i < ω} where, for 0 ≤ i < j < ω, xi < xj .

Suppose then that L is a bounded distributive lattice such that (D(L); τ(L),⊆) (recall thenotation introduced in §??) is homeomorphic and order-isomorphic to (P ; τ,≤). For a, b ∈ L,there correspond clopen decreasing sets A, B, respectively. Suppose a < b. Then A ⊂ Band it is possible to choose x ∈ B \ A. Since x is an accumulation point, there exists adistinct element y ∈ B \A. Say, without loss of generality, x 6≥ y. Then there exists a clopendecreasing set U with x ∈ U and y 6∈ U . Set C = A∪ (B ∩U). Then C is a clopen decreasingset such that A ⊂ C ⊂ B. In particular, C corresponds to an element c ∈ L such thata < c < b. We conclude that (Q;≤) the rational interval (0, 1) is embeddable in L, that is,(Q+;≤) the rational interval [0, 1] is a (0, 1)-sublattice of L. If one such embedding is denotedby f+ : Q+ −→ L, then f corresponds to continuous order-preserving map D(f) : D(L) −→

51

D(Q+) which is also onto. That is, there is a mapping from P onto D(Q+). Since D(Q+) isuncountable and P is countable, this is impossible and, as required, we conclude that (P ;≤)is not representable.

8.4 Open questions

In lemma 8.5, given Priestley topologies for each order component of a given poset P , wewere able to topologise P (i.e., the union of these components) by doing what can be called’one-component-compactification’. Of course, this construction does not yield a unique topol-ogy on P ; for instance we freely chose the component intended to perform one-component-compactification. With infinitely many components, taking different components for theone-component-compactification yields different topologies on P such that in general we willget non-isomorphic Priestley spaces. So the question remains whether this construction hasany interesting universal properties.As we have just seen, posets with infinitely many order components, all of them being rep-resentable, can be endowed with different Priestley topologies. This means that they are notuniquely representable. Note that there are two notions of unique representability:

1. A poset (P,≤) is called uniquely representable in the first sense (ur1) if there is exactlyone topology making it Priestley.

2. A poset (P,≤) is called uniquely representable in the second sense (ur2) if for any topolo-gies τ, τ ′ making P Priestley, the ordered spaces (P, τ,≤) and (P, τ ′,≤) are isomorphic.

It is clear that a ur1 space is ur2, but it is not clear whether there are Priestley spaces thatare ur2 but not ur1.Moreover, note that if the interval topology of a representable poset is Hausdorff, then itmust be the unique topology on the poset making it Priestley (first, the interval topology isalways contained in the Priestley topology, and second, any topology refining a given compactHausdorff topology is not compact any more). It is natural to ask the converse as has beendone in [10]: If P is a ur1 (or ur2!) poset, is the interval topology of P Hausdorff?Other questions may arise from considering ur1 (or ur2) posets in the context of categoricalconstructions like the product. Note that the coproduct (i.e. disjoint union) case has beendealt with: Any infinite disjoint union will be representable by the construction of one-point-compactification. Is it possible that an infinite disjoint union of ur1 (or ur2) posets is stillur2?

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Part IV

Maximal compactness

53

Chapter 9

Joint work with H. P. Kunzi

9.1 Introduction

Priestley spaces are compact Hausdorff spaces. Every compact Hausdorff space (X, τ) is bothmaximally compact and minimally Hausdorff, that is, any topology strictly coarser than τ isnot Hausdorff and any topology strictly finer that τ is not compact. (This property can beeasily proved using the fact that in Hausdorff spaces, compact subspaces are closed).It is natural to ask whether the converse holds: is every maximally compact space Haus-dorff? The answer was found quickly to be No: H. P. Kunzi remarked that the one-point-compactification of the rationals is maximally compact but not Hausdorff.Now an even more natural question came up: If (X, τ) is a compact space is there a maximallycompact topology on X containing τ? This question turned out to be open. Observe thatZorn’s Lemma is of no direct use here, because of the following example:

EXAMPLE 9.1. Let N = ω be the set of natural numbers, for any n ∈ N set

τn = P({0, ..., n}) ∪ {N}.

Then {τn : n ∈ N} is a chain of compact topologies that has no upper bound in the poset ofcompact topologies on X since ⋃

{τn : n ∈ N} = P(N).

A topological space is called a KC-space (compare also [9]) provided that each compact set isclosed. A topological space is called a US-space provided that each convergent sequence hasa unique limit. It is known [36] that each Hausdorff space (= T2-space) is a KC-space, eachKC-space is a US-space and each US-space is a T1-space (that is, singletons are closed); andno converse implication holds, but each first-countable US-space is a Hausdorff space.A compact topology on a set X is called maximal compact provided that it is not strictlycontained in a compact topology on X. It is known that a topological space is maximalcompact if and only if it is a KC-space that is also compact [27]. These spaces will be calledcompact KC-spaces in the following.Let us note that while there are many maximal compact topologies, minimal noncompacttopologies do not exist: Any noncompact space X possesses a strictly increasing open cover

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{Cα : α < δ} of X where δ is a limit ordinal and C0 can be assumed to be nonempty. Clearlythen {∅, X}∪ {Cα : 0 < α < δ} yields a base of a strictly coarser noncompact topology on X.

Maximal compact topologies need not be Hausdorff topologies [34] (see also [4, 31]). Astandard example of a maximal compact topology that is not a Hausdorff topology is givenby the one-point-compactification of the set of rationals equipped with its usual topology.Indeed maximal compact spaces can be anti-Hausdorff (= irreducible), as we shall next observeby citing an example due to van Douwen (see [35]).In order to discuss that example we first recall some pertinent definitions. A nonemptysubspace S of a topological space is called irreducible (see e.g. [13]) if each pair of nonemptyopen sets of S intersects. Furthermore a topological space X called a Frechet space (see [8,p. 53]) provided that for every A ⊆ X and every x ∈ A there exists a sequence of pointsof A converging to x. For the convenience of the reader we include a proof of the followingobservation (compare e.g. Math. Reviews 53#1519 of [28]).

LEMMA 9.2. Each Frechet US-space X is a KC-space.

Proof. Suppose that x ∈ K where K is a compact subspace of X. Because X is a Frechetspace, there is a sequence (kn)n∈N of points of K converging to x. Since K is compact, thatsequence has a cluster point c in K. Because X is a Frechet space, there is a subsequenceof (kn)n∈N converging to c (compare [8, Exercise 1.6D]). Hence x = c ∈ K, because X is aUS-space. We have shown that K is closed and conclude that X is a KC-space.

EXAMPLE 9.3. (van Douwen [35]) There exists a countably infinite compact Frechet US-space that is anti-Hausdorff. By the preceding lemma that space is a KC-space and hencemaximal compact. Thus there exists an infinite maximal compact space that is irreducible.

On the other hand, by the result cited above each first-countable maximal compact topologysatisfies the Hausdorff condition (compare [33, Theorem 8]).

9.2 Main problem and related questions

While it is known that each compact topology is contained in a compact T1-topology (justtake the supremum of the given topology with the cofinite topology) [33, Theorem 10], thequestion whether each compact topology is contained in a compact KC-topology (that is, iscontained in a maximal compact topology) seems still to be open. Apparently that questionwas first asked by Cameron [6, p. 56, Question 5-1], but remained unanswered.Of course, a simple application of Zorn’s lemma cannot help us here, since a chain of compacttopologies need not have a compact supremum: Consider the sequence (τn)n∈N of topologiesτn = {∅,N} ∪ {[1, k] : k ∈ N, k ≤ n} (n ∈ N) on the set N of positive integers.On the hand, for instance each infinite topological space X with a point x possessing onlycofinite neighborhoods is clearly contained in a maximal compact topology: Just considerthe one-point-compactification Xx of X \ {x} where X \ {x} is equipped with the discretetopology and x acts as the point at infinity.The problem formulated above seems to be undecided even under additional strong conditions.Recall that a topological space is called locally compact provided that each of its points hasa neighborhood base consisting of compact sets. Note that a locally compact KC-space is aregular Hausdorff space.

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PROBLEM 9.4. Is each locally compact (resp. second-countable) compact topology con-tained in a maximal compact topology?

The authors also do not know the answer to the following generalization of their main problem.

PROBLEM 9.5. Is each compact topology the continuous image of a maximal compacttopology?

In [33, Example 11] it is shown that a compact space need not be the continuous image ofa compact T2-space. In fact, a careful analysis of the argument reveals the following generalfact (also stated in [14, 3.6]).

PROPOSITION 9.6. A KC-space Y that is the continuous image of a compact T2-spaceX is a T2-space.

Proof. Let f : X → Y be a continuous map from a compact T2-space onto a KC-space.Clearly f is a closed map, since f is continuous, X is compact and Y is a KC-space. Theconclusion follows, since obviously a closed continuous image of a compact T2-space, is aT2-space.

In this context also the following observation is of interest.

PROPOSITION 9.7. Let f : X → Y be a continuous map from a maximal compact spaceonto a topological space Y. Then Y is maximal compact if and only if the map f is closed.

Proof. Suppose that f : X → Y is closed. Since f−1{y} is compact whenever y ∈ Y , wesee that f−1K is compact whenever K is compact in Y (compare e.g. with the proof of [8,Theorem 3.7.2]). Since f−1K is closed, we conclude that K = f(f−1K) is closed and henceY is a compact KC-space. For the converse, suppose that the map f : X → Y is not closed.Consequently there is a closed set F in X such that fF is not closed. Clearly the compactset fF witnesses the fact that Y is not a KC-space.

In connection with the preceding result we note (compare [5, Example 3.2]) that T1-quotientsof maximal compact spaces are not necessarily maximal compact.

PROBLEM 9.8. Are T1-quotient topologies of maximal compact topologies contained inmaximal compact topologies?

Next we want to show that a weak version of our main problem has a positive answer.

PROPOSITION 9.9. Let (X, τ) be a compact T1-space. Then there is a compact topologyτ ′ finer than τ such that (X, τ ′) is a US-space.

Proof. As usual two subsets A and B of X will be called almost disjoint provided that theirintersection is finite. Let M = {Ai : i ∈ I} be a maximal (with respect to inclusion) familyof pairwise almost disjoint injective sequences in X with a distinct τ -limit (that is, eachAi ∈ M is identified with {xn : n ∈ N} ∪ {x} where (xn)n∈N is an injective sequence in(X, τ) that converges to some point x different from each xn). For each i ∈ I and m ∈ N, letAm

i = {xn : n ∈ N, n ≥ m} ∪ {x}. Let τ ′ be the topology on X which is generated by thesubbase τ ∪ {X \Am

i : i ∈ I, m ∈ N}.

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We first show that τ ‘ is compact. Let C be a subcollection of Aτ ∪ {Ani : i ∈ I, n ∈ N} with

empty intersection. (Here, as in the following, Aτ denotes the set of τ -closed sets.) Denotethe intersection of C with Aτ by F . We want to show that there is a finite subcollection ofC with an empty intersection. Of course, it will be sufficient to find a finite subcollectionof C with finite intersection. If C = F , then such a finite subcollection of C must exist bycompactness of (X, τ). So in this case we are finished. If we have in our collection C \ F twosets An

i and Amj with i 6= j, then their intersection will be finite. So in that case we are also

done.Therefore we can assume that the set C \ F is nonempty and its elements are all of the formAm

i0= {xn : n ∈ N, n ≥ m} ∪ {a} for some fixed i0 ∈ I and n ∈ M where M is a nonempty

subset of N and a is the chosen τ -limit of the sequence (xn)n∈N.If a ∈ ∩F , then clearly a ∈ ∩C —a contradiction to ∩C = ∅. So there is F0 ∈ F such thata 6∈ F0. Since F0 is τ -closed and the injective sequence (xn)n∈N τ -converges to a, we concludethat F0 ∩ {xn : n ∈ N} is finite, since otherwise a ∈ F0. Hence for any m ∈ M we have thatF0 ∩Am

i0is finite and we are finished again.

We deduce from Alexander’s subbase theorem that the topology τ ′ is compact.Next we want to show that (X, τ ′) is a US-space. In order to reach a contradiction, supposethat there is some sequence (xn)n∈N that τ ′-converges to x and y where x and y are distinctpoints in X. Replacing (xn)n∈N if necessary by a subsequence, we can and do assume thatthe sequence (xn)n∈N under consideration is injective and that xn does not belong to {x, y}whenever n ∈ N. The claim just made is an immediate consequence of the fact that theoriginal sequence (xn)n∈N attains each value at most finitely many often, since (X, τ) andthus (X, τ ′) is a T1-space and (xn)n∈N has two distinct limits in (X, τ ′).Then (xn)n∈N is an injective τ -convergent sequence having a τ -limit distinct from each xn

and by maximality of the collection M there is some Ai = {zn : n ∈ N}∪{z} where z denotesthe chosen τ -limit of the sequence (zn)n∈N) belonging to M such that Ai ∩ {xn : n ∈ N} hasinfinitely many elements. Suppose that there is some p ∈ N such that x or y does not belongto Ap

i . Then X \Api is a τ ′-open neighborhood of x or y, respectively, which does not contain

infinitely many terms of the sequence (xn)n∈N which is impossible, because x and y are bothτ ′-limits of (xn)n∈N. So there is no such p ∈ N and it necessarily follows that x = z = y —acontradiction. We conclude that (X, τ ′) is a US-space.

COROLLARY 9.10. Each compact topology is contained in a compact US-topology.

REMARK 9.11. It is possible to strengthen the latter result further to the statement thateach compact topology is contained in a compact topology with respect to which each compactcountable set is closed.

In order to see this we need the following two auxiliary results. We recall that a topologicalspace is called sequentially compact provided that each of its sequences has a convergentsubsequence.

LEMMA 9.12. Let X be a US-space and let {Kn : n ∈ N} be a countable family ofsequentially compact sets in X having the finite intersection property. Then ∩n∈NKn isnonempty.

Proof. For each n ∈ N find xn ∈ ∩ni=1Ki. Then the sequence (xn)n∈N has a subsequence

(yn)n∈N converging to k ∈ K1, because K1 is sequentially compact. Suppose that there is

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m ∈ N such that k 6∈ Km. Since there is a tail of (yn)n∈N belonging to Km and Km issequentially compact, there exists a subsequence of (yn)n∈N converging to some p ∈ Km.Since X is a US-space, it follows that k = p ∈ Km —a contradiction. We conclude thatk ∈ ∩n∈NKn.

LEMMA 9.13. Each compact US-topology is contained in a compact topology with respectto which each compact countable set is closed.

Proof. Let (X, τ) be a compact US-space and let τ ′ be the topology generated by the subbaseτ ∪ {X \K : K ⊆ X is countable and compact} on X.

We are going to show that τ ′ is compact. In order to reach a contradiction, assume that C isa subcollection of Aτ ∪ {K ⊆ X : K is countable and compact} having the finite intersectionproperty, but ∩C = ∅. Since τ is compact, we deduce that some compact countable setK belongs to C. Hence by countability of K there must exist a countable subcollection Dof C such that ∩D = ∅. Replace in D each member F of D ∩ Aτ by its trace F ∩ K onK to get a countable collection D′ of compact countable sets having the finite intersectionproperty. By a result of [21], each compact countable space is sequentially compact and henceD′ is a countable collection of sequentially compact sets in a US-space. Since ∩D′ = ∅, wehave reached a contradiction to the preceding lemma. We conclude that τ ′ is compact byAlexander’s subbase theorem. Evidently each compact countable set in (X, τ ′) is τ -compactand thus τ ′-closed.

PROBLEM 9.14. Given some fixed cardinal κ > ℵ0. Is each compact topology contained ina compact topology with respect to which each compact set of cardinality κ is closed?

A modification of some of the arguments presented above allows us to answer positivelythe variant of the main problem (see [6, Question 8-1, p. 56]) formulated for sequentialcompactness instead of compactness.

THEOREM 9.15. Each sequentially compact topology τ on a set X is contained in a se-quentially compact topology τ ′′ that is maximal among the sequential compact topologies onX.

Proof. Since (X, τ) is sequentially compact and any convergent (sub)sequence has a constantor an injective subsequence, it is obvious that any sequence in (X, τ) has a subsequence thatconverges with respect to the supremum τ ∨ τc where τc denotes the cofinite topology on X.Therefore by replacing τ by τ∨τc if necessary, in the following we assume that the sequentiallycompact topology τ on X is a T1-topology.Define now a topology τ ′ on X in exactly the same way as above. We next show that(X, τ ′) is sequentially compact provided that (X, τ) is sequentially compact. Let (yn)n∈N beany sequence in X. It has a subsequence (sn)n∈N that converges to some point a in (X, τ),because (X, τ) is sequentially compact. If (sn)n∈N has a constant subsequence, then (yn)n∈N

clearly has a convergent subsequence in (X, τ ′). So by choosing an appropriate subsequenceof (sn)n∈N if necessary, it suffices to consider the case that (sn)n∈N is injective and thatsn 6= a whenever n ∈ N. By maximality of M there is Ai = {zn : n ∈ N} ∪ {z} belonging toM such that {sn : n ∈ N} ∩ Ai is infinite. Hence there is a common injective subsequenceof the injective sequences (sn)n∈N and (zn)n∈N in this intersection. By definition of τ ′ thatsubsequence converges to z, because any basic τ ′-neighborhood G∩

⋂nj=1(X \A

kj

j ) of z where

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G is τ -open, Aj ∈ M and kj ∈ N (j = 1, . . . , n) contains a tail of that subsequence, since(zn)n∈N τ -converges to z and Aj ∩ Ai is finite whenever j = 1, . . . , n. We conclude that(yn)n∈N has a τ ′-convergent subsequence and that (X, τ ′) is sequentially compact. As in thepreceding proof, one argues that (X, τ ′) is a US-space.We now define a new topology τ ′′ on X by declaring A ⊆ X to be τ ′′-closed if and only ifxn ∈ A whenever n ∈ N and (xn)n∈N converges to x in (X, τ ′) imply that x ∈ A. It is well-known and readily checked that τ ′′ is a topology finer than τ ′ on X with the property that anysequence (xn)n∈N that converges to x in (X, τ ′) also converges to x in (X, τ ′′). In particular, itfollows that the space (X, τ ′′) is sequentially compact, because (X, τ ′) is sequentially compact.Let K be a sequentially compact subset in (X, τ ′′). Suppose that xn ∈ K whenever n ∈ N andthat the sequence (xn)n∈N converges to x in (X, τ ′). Then there is a subsequence (yk)k∈N of(xn)n∈N that converges to r ∈ K in (X, τ ′), since K is sequentially compact in (X, τ ′′) and τ ′ ⊆τ ′′. Thus x = r, since (X, τ ′) is a US-space and hence x ∈ K. By the definition of the topologyτ ′′ we conclude that K is closed in (X, τ ′′). Therefore each sequentially compact subset of(X, τ ′′) is τ ′′-closed. By [5, Theorem 2.4] we conclude that τ ′′ is a maximal sequentiallycompact topology on X, which is clearly finer than τ.

Let us finally mention another possibly even more challenging version of our main problem.

PROBLEM 9.16. Which (compact) T1-topologies are the infimum of a family of maximalcompact topologies?

Evidently the cofinite topology on an infinite set X is the infimum of the family of maximalcompact Hausdorff topologies of the one-point-compactifications Xx (where x ∈ X) that wehave defined above. In Proposition 6 below we shall deal with a special answer to Problem 5.

9.3 Some further results

Let (X, τ) be a compact topological space. Denote by Aτ (resp. Cτ ) the set of all closed (resp.compact) sets of (X, τ).Note that if τ and τ ′ are two compact topologies on a set X such that τ ⊆ τ ′, then Aτ ⊆Aτ ′ ⊆ Cτ ′ ⊆ Cτ . Of course, a topology τ is a compact KC-topology if and only if Aτ = Cτ .

As usual, a collection of subsets of X that is closed under finite intersections and finite unionswill be called a ring of sets on X. We consider the set Mτ of all rings G of sets ordered byset-theoretic inclusion on the topological space (X, τ) such that Aτ ⊆ G ⊆ Cτ . Since Aτ issuch a ring, Mτ is nonempty. If K is a nonempty chain in Mτ , then

⋃K belongs to Mτ . By

Zorn’s lemma we conclude that Mτ has maximal elements.We shall call a collection C of subsets of a set X compact∗ provided that each subcollection ofC having the finite intersection property has nonempty intersection. We use this nonstandardconvention in order to avoid any confusion with the concept of a compact topology.

LEMMA 9.17. Let (X, τ) be a compact topological space. If G is a maximal element in Mτ

that is a compact∗ collection, then G = Aτ ′ where τ ′ is a maximal compact topology finer thanτ.

Proof. Suppose that G is a maximal element in Mτ that is compact∗. Then {X \K : K ∈ G}yields the subbase of a topology τ ′ on X. Observe that Aτ ⊆ G ⊆ Aτ ′ . Since G is compact∗,

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τ ′ will be compact, by Alexander’s subbase theorem. Because τ ′ is compact, τ ⊆ τ ′ impliesthat Aτ ′ ⊆ Cτ . Hence Aτ ′ belongs to Mτ . We conclude that G = Aτ ′ by the maximality of G.

It remains to be seen that τ ′ is maximal compact. If τ ′′ is a finer topology than τ ′ andcompact, then Aτ ′ ⊆ Aτ ′′ ⊆ Cτ . Hence by maximality of G, Aτ ′′ = G = Aτ ′ and so τ ′ = τ ′′.We have shown that τ ′ is maximal compact.

PROPOSITION 9.18. Let (X, τ) be a compact topological space such that each filterbaseconsisting of compact subsets has a nonempty intersection. Then τ is contained in a maximalcompact topology τ ′.

Proof. Let G be any maximal element in Mτ as defined above. Recall that G is closed underfinite intersections. Hence any nonempty subcollection G′ of G having the finite intersectionproperty generates a filterbase consisting of compact sets on X. It follows from our hypothesisthat G is a compact∗ collection. Furthermore by lemma 4 we conclude that G is equal to theset of closed subsets of a maximal compact topology τ ′ that is finer than τ.

It is known and easy to see (compare [20, Theorem 6]) that if X is a compact KC-space,then the product X2 is a KC-space if and only if X is a Hausdorff space. As an applicationof Proposition 4 we want to show however that the seemingly reasonable conjecture that theproduct topology of a large family of maximal compact topologies is no longer contained ina maximal compact topology is unfounded. In order to see this we next prove the followingresult.

LEMMA 9.19. Let (Xi)i∈I be a nonempty family of T1-spaces such that each Xi has theproperty that every filterbase of compact sets has a nonempty intersection. Then the productΠi∈IXi also has that property.

Proof. We can (and do) assume that I is equal to some finite ordinal or an infinite limitordinal ε. Let F be a filterbase of compact subsets on the product Πγ<εXγ .

For each α < ε we shall inductively find xα ∈ Xα such that the set Aα = {(yγ)γ<ε ∈ Πγ<εXγ :yγ = xγ whenever γ ≤ α} satisfies Aα ∩K 6= ∅ whenever K ∈ F .

Suppose now that for some δ < ε and all α < δ, xα ∈ Xα have been chosen such thatAα ∩K 6= ∅ whenever K ∈ F . Let us first establish the following claim.Caim: (∩α<δAα) ∩K 6= ∅ whenever K ∈ F .

If δ is a successor ordinal, then by our induction hypothesis Aδ−1 ∩K 6= ∅ whenever K ∈ F .Therefore the claim is verified, since the family {Aα : α < δ} is monotonically decreasing. Solet δ be a limit ordinal (possibly equal to 0) and fix K ∈ F . Since for each α < δ, Aα is closedbecause every space Xα is a T1-space, and since Aα ∩K 6= ∅ the claim holds by compactnessof K and the monotonicity of the sequence {Aα : α < δ}. (For the case that δ = 0 as usualwe use the convention that ∩∅ = Πγ<εXγ .)

Continuing now with the proof we next consider the filterbase {[prXδ(∩α<δAα∩K)] : K ∈ F}

of compact sets on Xδ. By our assumption on Xδ there exists some

xδ ∈ ∩K∈F [prXδ(∩α<δAα ∩K)].

It remains to show that for each K ∈ F , {(yγ)γ<ε ∈ Πγ<εXγ : yγ = xγ , γ ≤ δ} ∩K 6= ∅; butthis is an immediate consequence of xδ ∈ prXδ

(∩α<δAα ∩ K). Finally note that ∩α<εAα =

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{(xα)α<ε} and that — for ε exactly as in the case of the ordinal δ above — ∩α<εAα ∩K 6= ∅whenever K ∈ F . Hence the assertion of the lemma holds.

PROPOSITION 9.20. The product topology of a nonempty family of compact KC-topologiesis contained in a maximal compact topology.

Proof. Note first that in a compact KC-topology each filterbase of compact sets has anonempty intersection. We conclude by the preceding lemma and Proposition 4 that thecompact product topology of an arbitrary nonempty family of maximal compact topologiesis contained in a maximal compact topology.

COROLLARY 9.21. Let (Xi)i∈I be a nonempty family of spaces each of which is containedin a maximal compact topology. Then also their product topology is contained in a maximalcompact topology.

9.4 Sobriety and maximal compactness

Note that the closure of each irreducible subspace of a topological space is irreducible. Recallalso that a topological space is called sober (see e.g. [13]) provided that every irreducibleclosed set is the closure of some unique singleton. Clearly each Hausdorff space is sober.Furthermore a subset of a topological space is called saturated provided that it is equal to theintersection of its open supersets.A short proof of the following result is given in [15].Let {Ki : i ∈ I} be a filterbase of (nonempty) compact saturated subsets of a sober space X.Then

⋂i∈I Ki is nonempty, compact, and saturated, too; and an open set U contains

⋂i∈I Ki

iff U contains Ki for some i ∈ I.

COROLLARY 9.22. Let (X, τ) be a compact sober T1-space. Then τ is contained in somemaximal compact topology τ ′.

Proof. Since all (compact) sets in a T1-space are saturated, the condition stated in Proposition4 is satisfied by the result just cited. The statement then follows from Proposition 4.

PROBLEM 9.23. Characterize those sober compact topologies that are contained in a max-imal compact topology.

REMARK 9.24. Let us observe that the maximal compact topology τ ′ obtained in corol-laryollary 3 will be sober, since the only irreducible sets with respect to the coarser topologyτ are the singletons. Van Douwen’s example [35] mentioned earlier shows that a maximalcompact topology need not be (contained in) a compact sober topology.

EXAMPLE 9.25. Note that the closed irreducible subsets of the one-point-compactificationX (of the Hausdorff space) of the rationals are the singletons: Any finite subset of a T1-spacewith at least two points is discrete and hence not irreducible. Moreover any infinite subsetof X contains two distinct rationals and thus cannot be irreducible. We conclude that anarbitrary power of X is a compact, sober T1-space, because products of sober spaces are sober(see e.g. [13, Theorem 1.4]).

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In the light of the proof of Proposition 4 one wonders which compact sober T1-topologies canbe represented as the infimum of a family of maximal compact topologies (compare Problem5). Our next result provides a partial answer to this question. An interesting space satisfyingthe hypothesis of Proposition 6 is a T1-space constructed in [18]: It has infinitely many isolatedpoints although each open set is the intersection of two compact open sets. (It was noted inthe discussion [18, p. 212] that that space is locally compact and sober.)

Recall that a topological space X is called sequential (see [8, p. 53]) provided that a setA ⊆ X is closed if and only if together with any sequence it contains all its limits in X.

PROPOSITION 9.26. Each compact sober T1-space (X, τ) which is locally compact orsequential is the infimum of a family of maximal compact topologies.

Proof. Note that if K belongs to the closed sets of a maximal compact topology σ finer thanτ, then K is compact with respect to σ and thus with respect to τ . In order to verify thestatement, it therefore suffices to construct for any compact set C that is not closed in (X, τ)a maximal compact topology σ finer than τ in which C is not closed.So let C be a compact set that is not closed in (X, τ). In (X, τ) we shall next find a compactset K0 such that K0 ∩ C is not compact.Suppose first that X is locally compact.Then there is x ∈ X such that x ∈ clτC \C. Let F = {K : K is a compact neighborhood at xin (X, τ)}. Of course, ∩F = {x}, since X is a locally compact T1-space. Suppose that K ∩Cis compact in (X, τ) whenever K ∈ F .

Then {K ∩C : K ∈ F} is a filterbase of compact saturated sets in X. According to the resultcited above from [15], we have ∩F ∩ C =

⋂{K ∩ C : K ∈ F} 6= ∅. Since x 6∈ C, we have

reached a contradiction. Thus there is a compact neighborhood K0 of x such that K0 ∩ C isnot compact in (X, τ).Suppose next that X is sequential. Since C is not closed, there is a sequence (xn)n∈N in Cconverging to some point x ∈ X such that x does not belong to C. Assume that {({x}∪{xn :n ∈ N, n ≥ m}) ∩ C : m ∈ N} is a filterbase of compact sets. Clearly its intersection isempty, because τ is a T1-topology and (xn)n∈N converges to x—a contradiction. Hence thereis m ∈ N such that ({x}∪ {xn : n ∈ N, n ≥ m})∩C is not compact. Denote the compact set{x} ∪ {xn : n ∈ N, n ≥ m} by K0.

So our claim holds in either case.Note now that τ ∪ {X \K0} is a subbase for a compact topology τ ′ on X that is also soberand T1. By corollaryollary 3 there is a maximal compact topology τ ′′ finer than τ ′. Observethat X \ C 6∈ τ ′′ : Otherwise C ∈ Aτ ′′ and, since K0 ∈ Aτ ′′ , also K0 ∩ C ∈ Aτ ′′ . ThereforeK0 ∩C ∈ Cτ ′′ and K0 ∩C ∈ Cτ —a contradiction. Thus indeed X \C 6∈ τ ′′. We conclude thatτ is the infimum of a family of maximal compact topologies.

Observe that the argument above also yields the following results.

COROLLARY 9.27. Each locally compact sober T1-space in which the intersection of anytwo compact sets is compact is a KC-space (and therefore is a regular Hausdorff space).

COROLLARY 9.28. Each sequential sober T1-space in which the intersection of any twocompact sets is compact is a KC-space.

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We next give an example of a compact sober T1-topology that is not the infimum of a familyof maximal compact topologies.

EXAMPLE 9.29. Let Y be an uncountable set and let −∞ and∞ be two distinct points notin Y . Set X = Y ∪{−∞,∞}. Each point of Y is supposed to be isolated. The neighborhoodsof ∞ are the cofinite sets containing ∞ and the neighborhoods of −∞ are the cocountablesets containing −∞. Clearly X is a compact sober T1-space.Next we show that with respect to the defined topology τ a subset A of X is compact andnot closed if and only if A is uncountable, ∞ ∈ A and −∞ 6∈ A : Indeed, if ∞ ∈ A, then A isclearly compact and if A is uncountable and −∞ 6∈ A, then A cannot be closed. In order toprove the converse suppose that A is compact and not closed in (X, τ). Then A is certainlyinfinite. It therefore follows from compactness of A that ∞ ∈ A. Since A is not closed, weconclude that −∞ ∈ A \A and hence A is uncountable.Of course, if τ ′ is a maximal compact topology such that τ ⊆ τ ′, then Aτ ⊆ Aτ ′ ⊆ Cτ ′ ⊆ Cτ .Observe that the topology τ ′′ generated by the subbase {{−∞}}∪ τ clearly yields a compactT2-topology finer than τ. Obviously, Cτ \ Aτ ⊆ Aτ ′′ by the description found above of thenonclosed compact sets in (X, τ). Thus Aτ ′′ = Cτ . We conclude that τ ′′ is finer than anymaximal compact topology containing τ. Hence τ ′′ is the only maximal compact topology(strictly) finer than τ.

Let us recall that a topological space is called strongly sober provided that the set of limitsof each ultrafilter is equal to the closure of some unique singleton. Of course, each compactHausdorff space satisfies this condition.We finally observe that each locally compact strongly sober topological space (X, τ) possessesa finer compact Hausdorff topology; just take the supremum of τ and its dual topology (seee.g. [17, Theorem 4.11]). By definition, the latter topology is generated by the subbase{X \K : K is compact and saturated in X} on X.

No characterization seems to be known of those topologies that possess a finer compactHausdorff topology.

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Part V

Outlook: Between topology andorder

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In the category of posets there are different ”measures” of posets; since they seem to havemuch in common with the cardinal functions used in topology, we will call them posetcardinal functions. The following are examples of poset cardinal functions. Let (P,≤) bea partially ordered set. Recall that ≤ is called a linear order if for all x, y ∈ P we have x ≤ yor y ≤ x.

1. The width of a poset: Define

width(P,≤) = sup{card(A);A ⊆ P is an antichain}.

2. The height of a poset: Define

height(P ) = sup{card(C);C ⊆ P is a chain}.

3. The dimension of a poset. Let C(P ) be the set of linear orders on P . It is well-known(see [32]) that if (P,≤) is any partially ordered set, then

≤ =⋂{K ∈ C(P ); ≤ ⊆ K}.

So the following is well defined:

dim(P,≤) = min{card(K);K ⊆ C(P ) and ≤ =⋂K}.

A given poset (P,≤) can be endowed with many different topologies arising from the orderingrelation ≤. For instance consider u(P ), the so called upper topology which is generatedby the set {P\(x];x ∈ P}. Or dually, the lower topology (denoted byl(P )), generated by{P\[x);x ∈ P}. Note that the supremum of the upper and the lower topology is the intervaltopology τi(P ). Moreover there is the Alexandrov topology, consisting of all upper sets of P(recall that U ⊆ P is an upper set if u ∈ U and p ≥ u imply p ∈ U). All these topologies areexamples of induced topologies on a poset.Recall that also in the category of topological spaces there are certain measures which arecalled (topological) cardinal functions. Since posets and topological spaces can be linked bymeans of induced topologies, the following question arises:

What relation do the poset cardinal functions on a poset have with the (topologial)cardinal functions of the induced topological spaces of that poset?

The obvious genericity of this question suggests many problems. I want to conclude with twoinstances:

1. Is there a relation between the order dimension of the poset and any of the topologi-cal dimensions (for instance, covering dimension, inductive dimension...) of the uppertopology (or indeed, any of the induced topologies)?

2. If the ordering relation on a poset P is extended, in what way does this influence thecardinal functions of (P, τi(P ))?

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Bibliography

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[2] R.Balbes, On the partially ordered set of prime ideals of a distributive lattice, Canad. J. Math. 23 (1971),866–874.

[3] R.Balbes and P.Dwinger, Distributive Lattices, University of Missouri Press, Colombia, Mo., 1974.

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Curriculum Vitae of Dominic van der Zypen

1976 born in Berne, Switzerland on June 14. Citizen of Meikirch BE,Switzerland.

1983-1987 Primary school Meikirch, Switzerland.1987-1989 Secondary school Meikirch.1989-1996 High school Gymnasium Bern-Neufeld, Type A (Latin, Ancient Greek).1996 High school graduation (Maturitat).July-Oct 1996 Military service (boot camp) in the Swiss army.1996-2001 Mathematics studies at the University of Berne. Minors: Computer

Science, Physics, Philosophy and Biblical Hebrew.Feb 2000 Diploma in Biblical Hebrew.Jan 2001 Masters degree in Mathematics; thesis title ”A comparison of partial

orders on the set of isomorphsim classes of representation-finitealgebras” (supervisor Prof. Christine Riedtmann).

2001-2004 Doctoral thesis ”Aspects of Priestley Duality” (supervisor Prof. JurgSchmid).

Talks:

- ”Representable posets and their order components”, 2003 Summer Conference on Topologyand Its Applications, Howard University, Washington, DC, USA.- ”Aspekte der Priestley-Dualitat”, Kolloquiumsvortrage, University of Berne.

Language skills:

German (mother tongue), French (fluent), English (fluent), Italian (fair), Modern Hebrew(basic).

Favourite pastimes:

Playing cello (two orchestras, chamber music), singing (choir), studying languages, program-ming, reading.

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