Aspects of Priestley Duality
Inauguraldissertation der Philosophisch-naturwissenschaftlichen Fakult¨at der Universit¨atBern
vorgelegt von
Dominic van der Zypen
von Meikirch
Leiter der Arbeit: Prof. Dr. J. Schmid, Mathematisches Institut der Universit¨atBern Aspects of Priestley Duality
Inauguraldissertation der Philosophisch-naturwissenschaftlichen Fakult¨at der Universit¨atBern
vorgelegt von
Dominic van der Zypen
von Meikirch
Leiter der Arbeit: Prof. Dr. J. Schmid, Mathematisches Institut der Universit¨atBern
Von der Philosophisch-naturwissenschaftlichen Fakult¨atangenommen.
Der Dekan: Bern, den 1. April 2004 Prof. Dr. G. J¨ager When I consider thy heavens, the work of thy fingers, the moon and the stars, which thou hast ordained: What is man, that thou art mindful of him? – Psalm 8:3,4
I wish to thank my supervisor Prof. J¨urg Schmid for his advice and his helpful remarks. He showed me what to focus on when I tried to do too many things at once. I am deeply grateful for the patience and encouragement offered by my parents Eugen and Fran¸coisevan der Zypen and my sister V´eronique and for the love of my girlfriend Marie-Christine Chappuis. Contents
I Basic notions 6
1 Partially ordered sets and lattices 7 1.1 Partially ordered sets ...... 7 1.2 Subsets of posets ...... 7 1.3 Upper and lower bounds ...... 8 1.4 Lattices ...... 8 1.5 Morphisms between lattices ...... 9 1.6 Boolean lattices ...... 9 1.7 Ideals ...... 10 1.8 Prime ideals ...... 10 1.9 Congruence relations ...... 11
2 Topology 12 2.1 Topological spaces ...... 12 2.2 Bases and subbases ...... 12 2.3 Continuous functions ...... 13 2.4 Compact spaces ...... 13
3 Priestley Duality 14 3.1 Ordered topological spaces ...... 14 3.2 The functors D : D → P and E : P → D ...... 15
II Three kinds of completeness 16
4 Affine complete lattices 17 4.1 Affine complete lattices ...... 17 4.2 Affine complete Priestley spaces ...... 17 4.3 Products of affine complete lattices ...... 19 4.4 Free products of affine complete lattices ...... 19 4.5 Embedding lattices in affine complete lattices ...... 20 4.6 Q01 as initial object in the category of affine complete lattices ...... 21 4.7 Open questions ...... 23
5 Fractionally complete lattices 24 5.1 Fractionally complete lattices ...... 24 5.1.1 Lattices of fractions ...... 24
4 5.1.2 The maximal lattice of fractions ...... 24 5.1.3 Fractionally complete lattices ...... 25 5.1.4 Another way to make a lattice fractionally complete ...... 26 5.2 Priestley Duality of ideals and open down-sets ...... 27 5.3 Fractionally complete Priestley spaces ...... 28 5.4 Products of fractionally complete lattices ...... 32 5.5 Free products of fractionally complete lattices ...... 33
6 Order completeness 35 6.1 Dualising completeness ...... 35 6.2 Products and coproducts of complete lattices ...... 35
7 Comparing the three kinds of completeness 37 7.1 Affine complete vs fractionally complete ...... 37 7.2 Order complete vs affine complete ...... 38 7.3 Order complete vs fractionally complete ...... 39
III Representability and order components 42
8 Joint work with M. E. Adams 43 8.1 Introduction ...... 43 8.2 Proof of theorem 8.1 ...... 44 8.3 Proof of theorem 8.2 ...... 46 8.4 Open questions ...... 52
IV Maximal compactness 53
9 Joint work with H. P. K¨unzi 54 9.1 Introduction ...... 54 9.2 Main problem and related questions ...... 55 9.3 Some further results ...... 59 9.4 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 called a 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 referred to as the ordering relation of the poset (X,R). Instead of (x, y) ∈ R we usually write x ≤R y or even drop the index R if it is clear which relation we refer to. Often, we denote a poset by (X, ≤R) instead of (X,R) and sometimes the index R is dropped. To make the notation even lighter, we say ”let X be a poset” with the tacit assumption that we denote the ordering relation 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 and call (X, ≤) a chain. An antichain is a poset X whose order is ∆X = {(x, x); x ∈ X}; so in an antichain, 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 with x ≤ y we have ϕ(x) ≤ ϕ(y). The map ϕ is an order-embedding if for all x, y ∈ P we have x ≤ y ⇔ ϕ(x) ≤ ϕ(y). Note that any order-embedding is injective. If an order-embedding is surjective, 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 ∈ X with 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 that a1 ≤ 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, for simplicity, 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 called the 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 ordered set, 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 upper bound 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 the set 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 ≤ a for all a ∈ A. Note that every subset of P possesses at most one least element; it is unique because the ordering relation of each poset is antisymmetric. The notion of a greatest element is dual.
DEFINITION 1.4. Let A be a subset of a poset P . If the set Au possesses a least element then this is called infimum or meet and denoted by inf(A). In a dual way we define the supremum 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} the meet 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 to be complete, it suffices that all meets (or all joins) exist. The least element of a lattice (if it exists) is denoted by 0, and the greatest element is denoted by 1. Moreover, a lattice possessing 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
[D0] 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 ordered by set inclusion. We usually call P(X) the power set of X. The infimum and the supremum of any subset A ⊆ P(X) is given by T A and S 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, L0 be lattices. A map f : L → L0 is a lattice homomorphism if for 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 bijective lattice 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 bounded lattices). 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’ but mean ’(0,1)-lattice homomorphism’.
1.6 Boolean lattices
DEFINITION 1.10. Let L be a bounded lattice and let b ∈ L. We say that b0 ∈ L is a complement of b if b ∧ b0 = 0 and b ∨ b0 = 1.
Note that complements are unique for distributive lattices.
DEFINITION 1.11. Let L be a bounded distributive lattice. Then L is called Boolean if every 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 distributive lattice; moreover for A ∈ P(X) the complement is given by A0 = X \ A. Note that P(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} be ordered by set inclusion ⊆. For A, B ∈ CF (X), note that A ∧ B and A ∨ B are given by A ∩ B and A ∪ B respectively. The lattice CF (X) is Boolean (the complement is again the set complement) but non-complete for the following reason. There is a set Y ∈ P(X) \ CF (X). Then the subset A = {{y}; y ∈ Y } of CF (X) has no supremum in CF (X).
1.7 Ideals
DEFINITION 1.12. An ideal of a lattice L is an ideal of the poset L as defined in section 1.2. This amounts to saying that an ideal is a down-set that is closed under finite joins. An ideal 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 then we 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 the distributivity 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. If for 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 said to 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 natural numbers). Consider the lattice P(ω). Then the set Q = {A ∈ P(ω); 0 ∈/ A} is a prime ideal of 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’s Lemma). Prime ideals will play a key role in representing lattices as topological spaces, which we 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 on L. Then θ is called a congruence relation if
(C) for all (a, b), (a0, b0) ∈ θ we have (a ∧ a0, b ∧ b0) ∈ θ and (a ∨ a0, b ∨ b0) ∈ θ.
Then, (L/θ, ≤θ) is again a lattice where [a]θ ≤θ [b]θ if and only if a ≤ b. Note that (C) makes sure that ≤θ is well defined. Moreover, (L/θ, ≤θ) is even a bounded distributive lattice, and suprema 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 a collection of subsets of X satisfying the following requirements: 1. ∅,X ∈ τ 2. U ⊆ τ =⇒ S 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 complement of an open set is called closed. Note that arbitrary intersections and finite unions of closed sets are closed. A subset of X that is both open and closed is called clopen. If S ⊆ X and there is an open set U with x ∈ U ⊆ S, then S is called a neighborhood of x. The element x ∈ S is then called inner point of S. The set of all inner points of S is called the interior of S and denoted by intX (S) or simply int(S). It is the greatest open set (with respect to inclusion) 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 S and therefore the smallest closed set containing S. It is called the closure of S and denoted by 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 topology τA 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 open subsets U of X and all x ∈ U there is a B ∈ B such that x ∈ B ⊆ U. For any subset S ⊆ τ we 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 of topologies that contain S.(T is nonempty since it contains P(X) as an element). It is easy to see that τ := T 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 if for 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 that a given map f : X → Y is continuous, it suffices to check that f −1(S) is open in X for every S ∈ 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 S U = X then there is a finite subset S U0 ⊆ U with U0 = X.
Compact spaces will play a key role in this thesis. The following are the most important facts about 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 following definition: Let X be a space with subbase S. Then X is called subbase compact with respect to S if the following holds: Whenever V is a subset of S such that S V = X then there is a S finite subset V0 ⊆ V with V0 = X. Alexander’s Subbase Theorem. Let X be a topological space with subbase S. If X is subbase compact with respect to S then X is compact. This theorem thus says that in order to check compactness of some space, we need only consider 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) is a 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 partially ordered 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 the poset concerned).
The interval topology is a topology that exists on any given poset. This is different with other topologies 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 Priestley space since that topology would need to be compact. EXAMPLE 3.4. Let ω be the first infinite ordinal, ordered by its usual well-ordering. Then τi(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 find an 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 a subbase for X. This directly implies that X has a basis consisting of clopen sets (such spaces are also called zero-dimensional). In fact, already
{D ∩ (X \ D0); D,D0 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)-preserving lattice 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 any prime ideal P , and for any morphism ϕ, its dual E(ϕ) is defined by C 7→ ϕ−1(C) for any clopen down-set C. It was proven by H. A. Priestley that these functors form a dual equivalence between 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 is k easy to see that the projections pri : L → L are compatible. With induction on polynomial complexity one shows that every polynomial function is compatible (see [23]). A lattice L is called affine complete, if conversely every compatible function on L is a polynomial. G. Gr¨atzer [12] gave an intrinsic characterization of bounded distributive lattices that are affine complete: THEOREM 4.1. ([12]) A bounded distributive lattice is affine complete if and only if it does not 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 ∈ L be 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.
x+y 0 Proof. First, take any x < y in [0, 1]. Then the element a = 2 ∈ [x, y] has no complement a in [x, y]: Otherwise we would have a∧a0 = x which would imply a0 = x, but then a∨a0 = 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 x + y x + y ( 1 1 , 2 2 ) ∈ [(x , x ), (y , y )] 2 2 1 2 1 2
does not have a complement in [(x1, x2), (y1, y2)]. Thus, [0, 1] × [0, 1] has no proper Boolean interval 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 complete distributive (0,1)-lattices. Such spaces will be called affine complete Priestley spaces. In other words, 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 concept of 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 an antichain, 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 C0 = U ∪ (V \C). Claim: C0 is a clopen down-set of X. It is clear that C0 is a clopen subset of X since V \ C = V ∩ (X \ C). Now, let c ∈ C0 and assume x < c. Then if c ∈ U, we are done, since U is a down-set. Assume c ∈ V \ U. Since V is a down-set, we get x ∈ V , and the fact that V \U is an antichain tells us that x cannot be a member of V \ U. Therefore x ∈ U ⊆ C0 which proves that C0 is indeed a (clopen) down-set. Moreover, C0 is the complement of C in [U, V ], i.e. C ∩ C0 = U and C ∪ C0 = V . Because C was arbitrary, we see that [U, V ] is a proper Boolean interval of E(X), whence E(X) is not affine complete.
(2) =⇒ (1). Suppose U ⊆ V are distinct clopen down-sets. By assumption, there are elements 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 complement in [U, V ]: Take any C ∈ [U, V ] with C ∪ B = V . Then y ∈ C, but since C is a down-set, we have x ∈ C, thus x ∈ (B ∩C)\U and B ∩C 6= U. So whatever C we pick, C is no complement for B, i.e. B is not complemented, and consequently [U, V ] is not Boolean. It follows that no proper 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 nonempty open set contains two distinct comparable points.
Proof. It follows directly from theorem 4.3 that if each nonempty open set contains two distinct points that are comparable, then X is affine complete.
Conversely, suppose that U is a nonempty open set which is an antichain, then there exist open down-sets C1,C2 such that ∅= 6 C1 ∩(X\C2) ⊆ U. Then [C1 ∩C2,C1] is a proper interval such that C1\(C1 ∩ C2) = C1 ∩ (X\C2) is an antichain (as a subset of the antichain U). Thus theorem 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 two distinct comparable points.
18 4.3 Products of affine complete lattices
The aim of this section is to prove that arbitrary products of affine complete lattices are affine complete. We don’t need Priestley duality to do this. Priestley duals of affine complete lattices, i.e. affine complete Priestley spaces, will come into play when we consider coproducts of affine complete lattices.
THEOREM 4.5. If (Li)i∈I is a family of (bounded) affine complete lattices, then Πi∈I Li is affine complete.
Proof. We prove the contrapositive of the theorem. Suppose that Πi∈I Li is not affine com- plete. Then it contains a proper interval [ξ, η] that is Boolean. There exists some k ∈ K such that 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∈I Li by ( l if i = k λ(i) = ξ(i) if i 6= k
0 0 0 Because [ξ, η] is Boolean, there exists λ ∈ Πi∈I Li such that λ ∧ λ = ξ and λ ∨ λ = η. Thus it is easy to see that l0 := λ0(k) is the complement of l ∈ [x, y]. Therefore, [x, y] is a proper Boolean 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 affine complete. A convenient way to obtain this result is to dualise the problem into the category of Priestley spaces. Free products (that is, coproducts) in D correspond to products in P and 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 and G : 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 spaces under the Priestley duality.
THEOREM 4.8. If (Xi)i∈I is a family of affine complete Priestley spaces, then Πi∈I Xi is affine complete.
19 Proof. Suppose that Xi is affine complete for every i ∈ I. It suffices to show that every nonempty subset V of Πi∈I Xi of the form
V = π−1(U ) ∪ ... ∪ π−1(U ) i1 1 ir r contains two distinct comparable elements (where Uk ⊆ Xik open, nonempty). Take U1. It contains elements a < b, because Xi1 is affine complete. Now pick ξ ∈ V . Define ξ1, ξ2 ∈ V by ( ξ(i) if i 6= i1 ξ1(i) = a if i = i1 and ( ξ(i) if i 6= i1 ξ2(i) = 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 free products.
4.5 Embedding lattices in affine complete lattices
First we will stay away from affine completeness in the worst possible way: we will embed each 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 X and 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 bounded distributive lattice L01. By Priestley duality, there is a Priestley space (X, τ, ≤) such that the lattice 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 lattice embedding j : P(X) ,→ QX . Moreover, Q is affine complete.
X Proof. Set j : S 7→ χS ∈ Q for every S ⊆ X, where χS is defined by ( 1 if x ∈ S χS(x) = 0 if x∈ / S
20 It is easy to see that j is a lattice embedding. Next, we claim that Q is affine complete. Take x+y 0 any x < y in Q. Then the element a = 2 ∈ [x, y] has no complement a in [x, y]: Otherwise we would have a ∧ a0 = x which would imply a0 = x, but then a ∨ a0 = a 6= y. So [x, y] is not Boolean, 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 embedded in a bounded affine complete lattice.
Admittedly, the construction provided by 4.10 and 4.11 is highly non-unique and has no minimality 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 each affine complete lattice, which amounts to saying that Q01 is an initial object of the category of affine complete lattices (with (0,1)-homomorphisms, i.e. a full subcategory of the category bounded 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 ∈ X with x < z < y.
The first tool we need here is a well known result of model theory. It states that the theory of dense linear orders is complete and has (Q, ≤) as prime model. We will state this result in a 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 simplify notation 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 relation ak ≤ al implies fn(ak) ≤ fn(al) and that fn is an injective function from {a0, ..., an} to X\{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 that 1X > bn+1 ≥ fn(ai) for all i ∈ {0, ..., n}. So,
fn+1 := fn ∪ {(an+1, bn+1)}
21 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 there 0 0 is no k ∈ {0, ..., n} with ak < ak0 < an+1 and likewise that there is no l ∈ {0, ..., n} with an+1 < al0 < al. Consider bk = fn(ak) and bl = fn(al). Since fn is order-preserving and injective by assumption, we get bk < bl. Because X is dense, there is an element bn+1 such that 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 := fn n∈ω
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 another chain 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 that S for 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 ∈ S 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 bounded in 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 ∈ C with 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 properly contains 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)-embedding ϕ : Q01 ,→ L.
Proof. Pick any maximal chain C in L. Note that by maximality of C we have 0, 1 ∈ C since C ∪ {0, 1} is a chain. So the inclusion map ι : C,→ L is a (0, 1)-embedding as well as the embedding from Q01 to C provided by proposition 4.15. Composing these two, we get a (0,1)-embedding from Q01 to L.
22 4.7 Open questions
In chapter 4.5 we showed that ever bounded distributive lattice can be extended to an affine complete lattice. This was achieved by making use of Q01 which happens to be embeddable in 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 lattice L, does its ’affine hull’ have any interesting universal properties?
23 Chapter 5
Fractionally complete lattices
5.1 Fractionally complete lattices
In this chapter, we will not restrict ourselves to bounded distributive lattices but rather consider 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 give a 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 and for 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 distributive lattice L may be embedded in a larger affine complete lattice, but this embedding is by no means canonical in any way. This section shows that things are much nicer when we embed bounded distributive lattices in larger fractionally complete lattices. In [29], the following fundamental fact about lattices of fractions was proved. For every distributive lattice L there exists a distributive lattice Q(L) that is a maximal lattice of fractions of L in the following sense:
0 0 ∼ 0 L < L1 if and only if there exists L1 such that L1 = L1 and L ⊆ L1 ⊆ 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].
0 EXAMPLE 5.2. Let F (N) be the set of all finite subsets of the natural numbers, and let 0 L = 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 a finite subset of F (N). Therefore (l] ∩ J is a finite ideal and thus has a maximal element, so it is 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 that [ Φ: I(L) → P(N); J 7→ J
is an isomorphism: For A ∈ P(N) consider J = {A0 ∈ F (N); A0 ⊆ A} which is clearly an ideal. 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). S S S 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 tells us 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 a more concrete way.
THEOREM 5.4. ([29]) A distributive lattice L is fractionally complete if and only if the following condition holds:
(C) If J ⊆ L is an ideal satisfying
• PJ is join-dense • PJ → J = J
then J is principal.
25 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 is fractionally 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 not fractionally 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 principal since we are dealing with a bounded chain, but J is not principal). These considerations show that 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 shows that 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 a fractionally 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 is fractionally complete.
Proof. It is clear that such a lattice does not possess a proper join-dense ideal. Thus if an ideal J has the property that PJ is join-dense, then we have PJ = L which implies that J is principal.
This gives us an easy recipe to make an arbitrary bounded distributive lattice fractionally complete: 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 L top was a bounded lattice, we get sup(L \{L}) = 1L < L. Therefore, lemma 5.6 implies that Ltop 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 of Priestley spaces. An important tool will be the duality between ideals of bounded distributive lattices and open down-sets of Priestley spaces.
Let X be a Priestley space throughout this chapter; let E(X) be the lattice of all clopen down-sets of X. With I(E(X)) we denote the set of ideals of E(X). Moreover let L be the lattice of open down-sets of X. As noted in [7], we can set up the following connection between I(E(X)) and L.
LEMMA 5.8. ([7]) The map Φ: I(E(X)) → L defined by J 7→ S 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 and Ψ 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-set with U ⊆ S J = Φ(J). Thus U ∈ Ψ(Φ(J)). Conversely, let U ∈ Ψ(Φ(J)). So U is clopen and U ⊆ S J. So U is covered by members of J. Since U is compact, it is covered by finitely many 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 )) = S{U ∈ E(X); U ⊆ W } ⊆ W . In order to prove the other inclusion, it suffices to show that for every w ∈ W there is a clopen down-set Uw such that w ∈ Uw ⊆ W . If W = X, we are done. Suppose W 6= X; since W is a down-set, we have y 6≤ w for every y ∈ X\W . So for every y ∈ X\W there is a clopen up-set Ay that contains y but not w. Now, X\W is covered by the Ay where y 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-isomorphism and whence a lattice isomorphism.
27 5.3 Fractionally complete Priestley spaces
The aim of this chapter is to characterize the Priestley spaces arising from fractionally complete 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) is fractionally 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 have into 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) = S 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 we have 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 every C ∈ 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 exists no clopen down-set D 6= X such that D ⊇ A for all clopen down-sets A that are contained in W .
Proof. Suppose there is such a clopen down-set D. The equation Φ(Ψ(W )) = W tells us that for each w ∈ W there is a clopen down-set A that contains w and is a subset of W . Thus D ⊇ W ; since D is closed we get cl(W ) ⊆ D 6= W . So D is not dense.
28 LEMMA 5.12. An open down-set W of a Priestley space X is dense if and only if it is topologically 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 Priestley topology, there are open down-sets C1,C2 such that ∅ 6= C1 ∩ (X\C2) ⊆ V . Now take 0 0 C = C1 ∩ C2. Since ∅ 6= C1 ∩ (X\C2), we get C 6= C1. Moreover, take any clopen down-set A ⊆ W ∩C1. If there were an a ∈ A∩(X\C2) we would get a ∈ W ∩C1(∩(X\C2)) ⊆ W ∩V = ∅, 0 clearly a contradiction. Thus A ⊆ C = C1 ∩ C2. So we just proved the following: There is 0 0 a clopen down-set C properly contained in C1 such that C ⊇ A for all clopen down-sets A ⊆ 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, fact 5.11 implies that there is no clopen down-set C0 strictly contained in C having the property (P) that C0 ⊇ A for all clopen down-sets A ⊆ C ∩ W . Now if there were a clopen down-set 0 C1 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 C0 ∈ E(X),C0 ⊇ A for all A ∈ E(X) with A ⊆ C ∩ W , then C0 ⊇ 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 suffices to show that W is topologically join-dense. Assume that W is not topologically join-dense. Then there are clopen down-sets C0 6⊇ C such that
C0 ⊇ A for all A ∈ E(X) with A ⊆ C ∩ W (1).
Now let A ∈ J with A ⊆ C. Thus A ⊆ W = S J. So, by (1), A ⊆ C0. Whence C0 ∈ ((C]∩J)u with C0 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 not join-dense. The definition of join-denseness implies that there are clopen down-sets C 6= C0 such that C ⊇ C0 and C0 ∈ ((C] ∩ J)u. Now, U = C\C0 6= ∅ is open. Assume that x ∈ Φ(J) ∩ U. So there is an element D ∈ J with x ∈ D (because Φ(J) = S J). Since J is an ideal, C ∩ D ∈ ((C] ∩ J), but x ∈ (C ∩ D)\C0, so C ∩ D 6⊆ C0 which is a contradiction. So Φ(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.
29 DEFINITION 5.14. Let W be an open down-set of a Priestley space X. Then we define the 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 to the 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] for some D ∈ J. Since Φ is a lattice isomorphism, Φ((C]) ∩ Φ(J) = Φ((D]). Since Φ((C]) = C and 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 clopen down-set, say D. Since Ψ is a lattice isomorphism, Ψ(C) ∩ Ψ(Φ(J)) = Ψ(D) which can be simplified 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 is S clopen 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 to get (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 this equation gives Φ((C]) ∩ Φ(Ψ(W )) = Φ((D]) which in turn yields C ∩ W = D. Therefore C ∈ {A ∈ E(X); A ∩ W clopen}, ie, C ⊆ S{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 two propositions, the proof of statement (2) was taken care of by arguing: ”This follows directly from (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 concerning ideals 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-sets of 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 take any clopen down-set D ⊆ Φ(J1). So Ψ(D) ⊆ Ψ(Φ(J1)) = J1 which shows that D ∈ J1. So S C ∩ 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 all D ∈ E(X) contained in W1). Choose D ∈ J1. It is clear that D ⊆ Φ(J1). Therefore, by our assumption, 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) S with A ⊆ W1 (?). Obviously, C ⊆ x∈C Dx; since C is compact, there are x1, ..., xr such that
C ⊆ Dx1 ∪ ... ∪ Dxr =: D.
0 0 0 0 We claim that D ∈ Ψ(W1) → Ψ(W2). Let D ∈ Ψ(W1). So D∩D = (Dx1 ∩D )∪...∪(Dxr ∩D ) 0 which belongs to Ψ(W2) by (?) and because Ψ(W2) is an ideal. Since D was arbitrary, we conclude that D ∈ Ψ(W1) → Ψ(W2). As C ⊆ D and Ψ(W1) → Ψ(W2) is an ideal, we get C ∈ Ψ(W1) → Ψ(W2).
Conversely, suppose that C ∈ Ψ(W1) → Ψ(W2). This implies that
0 0 C ∈ {C ∈ E(X); C ∩ D ⊆ W2 for all D ∈ E(X) with D ⊆ W1} . | {z } U S 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 and J1 → J2, the relative pseudocomplement for ideals, adequately into the dual category. This enables 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 condition holds:
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 condition holds:
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∈I Li is fractionally complete.
Proof. Assume that all Li are fractionally complete. Let J ⊆ Πi∈I Li 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∈I Li whereas the second P denotes the principal extension in Li.) Let x ∈ πi(PJ). So there exists ζ ∈ Πi∈I Li such that ζ ∈ PJ and πi(ζ) = ζ(i) = x. Thus (ζ] ∩ J = (ξ] for some ξ ∈ J. It is easy to see that 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∈I Li defined by ( x if j = i µ(j) = 0 otherwise ( z if j = i It is then easy to check that (µ] ∩ J = (ν], where ν(j) = . So µ ∈ PJ. Since 0 otherwise πi(µ) = x, we have x ∈ πi(PJ). Claim 2. PK is join-dense. u Proof of Claim 2. Suppose not. So there are x0 > x ∈ L such that x ∈ ((x0] ∩ PK) . Using Claim 1, we see that ξ0, ξ ∈ Πi∈I Li defined by ( ( x0 if j = i x if j = i ξ0(j) = and ξ(j) = 0 otherwise 0 otherwise
32 u satisfy ξ < ξ and ξ ∈ ((ξ0] ∩ PJ) . 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 → ( y if j = i K)\K. Consider µ ∈ Πi∈I Li defined by µ(j) = . With Claim 1, it is easy to 0 otherwise 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 define ξ ∈ Πi∈I Li 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. Now 0 ρ ∈ PJ implies that (ρ] ∩ J = (ρ0] for some ρ0 ∈ J. Pick i ∈ I, define ξ ∈ Πi∈I Li by ( 0 ξ(i) if j = i 0 0 0 ξ (j) = . Now we see that ξ ∈ J. Moreover ρ ∧ ξ ∈ (ρ] ∩ J, so ρ ∧ ξ ≤ ρ0. 0 otherwise 0 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 seen to 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 lattices turns out not to be closed under arbitrary free products. Using our translation of the concept of fractional completeness to the category of Priestley spaces, we will give an example of a fractionally complete Priestley space X such that X × X is not fractionally complete any more.
Consider the topological space βN, that is the Stone-Cechˇ compactification of (N, P(N)). βN is clearly compact, moreover, βN is extremally disconnected, i.e., Hausdorff and the closure of 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} be the 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) is the 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 B is a complete Boolean lattice ([7], Ex. 10.12). By [29] p. 687, B is fractionally complete and X is thus a fractionally complete Priestley space.
33 Consider X × X. Now, the space βN × βN is not extremally disconnected (see [8], 6.3.21); whence by [7], Ex. 10.12, B0 = E(X × X) is not a complete Boolean algebra. By [29], B0 is not 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 whose product with itself is not fractionally complete. This implies that E(X) is a fractionally complete lattice such that the free product of E(X) with itself is not fractionally complete.
34 Chapter 6
Order completeness
6.1 Dualising completeness
Recall that a (distributive) lattice L is called (order-)complete if for every subset A ⊆ L its supremum 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 set U ⊆ 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 and only 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 a basis 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 is open. 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. Consider Φ(I), an open subset of X. The closure C = cl(Φ(I)) is open by assumption and therefore an element of E(X). Under the canonical isomorphism from E(X) to L, C corresponds to the supremum 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). Here we 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 L with itself is complete is equivalent to saying that the product of βN with itself is extremally disconnected – which it is not, by [8], 6.2.28. Therefore coproducts of even finitely many complete distributive lattices need not be complete.
36 Chapter 7
Comparing the three kinds of completeness
7.1 Affine complete vs fractionally complete
When comparing affine complete lattices and fractionally complete lattices, it is easier to start off with lattices that are fractionally complete but not affine complete, for there is a plethora of examples!
EXAMPLE 7.1. Each finite distributive (0,1)-lattice is fractionally complete and not affine complete.
Proof. Let L be finite. Then every ideal in L is principal, so the requirement for fractionally distributive 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 that Q01 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 not affine complete.
Proof. P(X) is not affine complete because it is Boolean. Moreover, example 5.5.4 says that P(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
0 L := {ξ ∈ QN; ξ(n) = 0 for all but finitely many n ∈ N}.
0 0 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 there exists 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 ( q n = m ζ(n) = ξ(n) n 6= m Now it is easy to see that ζ has no complement in [ξ, η]: Suppose ζ ∧ ζ0 = ξ. Then ζ0(m) = ξ(m), but this implies ζ(m)∨ζ0(m) = q 6= η(m), whence ζ∨ζ0 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 L0 = L\{1} is a join-dense ideal of L; the reason is that 1 ( 1 n = k is the supremum of all the functions of the form χk for k ∈ N where χk(n) = . 0 n 6= k Define the ideal 0 J = {ξ ∈ L ; ξ(k) = 0 for all odd k ∈ N}. It is easy to see that J is non-principal.
Claim 1. PJ = L0. Proof of Claim 1. Of course, PJ is contained in L0 (otherwise J would be principal). Now 0 pick α ∈ L . So α(k) 6= 0 for only finitely many k ∈ N. Consider α0 ∈ J defined by ( α(n) if n is even α0(k) = 0 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 = L0, then x ∧ 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 satisfying PJ → 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 of distributive lattices that are complete but not affine complete, abound. We want to exhibit some of them. Remember that a pair (x, y) of elements of a lattice L are called a covering pair iff [x, y] = {x, y}.
EXAMPLE 7.5. Any complete distributive lattice containing a covering pair is complete but not affine complete.
38 Proof. This is clear, since a covering pair is a non-trivial Boolean interval contained in the lattice 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 at least two distinct elements. So, since it is finite, it contains a covering pair which makes it non-affine-complete as we have seen in the last example.
EXAMPLE 7.7. Every complete distributive lattice can be made non-affine-complete by adding a new top element.
There are also plenty of examples of distibutive lattices that are not complete but affine complete. Note that every affine complete contains an order dense chain (as shown in theorem 4.16) and whence it is not surprising that non-complete order dense chains lie at the core of constructing 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. More generally, any product of bounded order dense chains, at least one of them non-complete, is affine 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 bounded chain. 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 continous W W if 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 is fractionally complete. In particular, every algebraic lattice is fractionally complete.
This theorem enables us to construct an example of a bounded complete distributive lattice that is not fractionally complete.
EXAMPLE 7.12. Consider the poset (N, ), where m n if and only if m divides n. Note that 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 of natural numbers not containing 0. It is clear that no positive number divides every number contained in A. So 0 is the only natural number deviding every member of A, whence W A = 0. In order to show that (N, ) is not fractionally complete it suffices to find a non-principal ideal J ⊆ N such that
• PJ is join-dense in (N, ). • PJ → J = J.
n Take J := {2 ; n ∈ N}. This is clearly an ideal, for it is directed since J is a chain, and it is a down-set because any divisor of a power of 2 is also a power of 2 (or 1). Now take any natural s number n ∈ N\{0}. There is a greatest natural number s such that 2 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 not complete. Moreover, one could conjecture that chains are essentially the only examples of lattices that are fractionally complete but not complete. But the conjecture is easily seen to be false, since for each cardinal κ ≥ 2 the product lattice Cκ is not a chain and fractionally complete but not complete.
EXAMPLE 7.13. The distributive lattice (R\Q) × Q with a top and bottom element added is fractinally complete but not complete.
The following lemma provides a partial the converse for theorems 5.19 and 6.3. It shows how to 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 Q 1. If some Li is not complete, then i∈I Li is not complete. Q 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 Y A¯ := {f ∈ Li; f(i) ∈ A} i∈I Q 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. Then consider the ideal Y J¯ := {f ∈ Li; f(i) ∈ A} i∈I Q Q 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 is p0 ∈ J such that (p] ∩ J = (p0]. Whence, setting ρ0 to be p0 in i and 1 everywhere else, we get (ρ ∩ J¯ = ρ0. This shows that Y P J¯ ⊇ {f ∈ Li; f(i) ∈ PJ} i∈I ¯ Q which directly implies that P J is join-dense in i∈I Li. Another easy calculation yields P J¯ → J¯ = J¯.
This shows how to construct a large class of bounded distributive lattices that are fractionally complete 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 the natural numbers ordered by divisibility. Then L×N is complete but not fractionally complete.
41 Part III
Representability and order components
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 bounded distributive lattice?
A poset that is order isomorphic to the poset of prime ideals of some bounded distributive lattice is called representable. By Priestley Duality, a poset is representable if and only if it can 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, for example, the expository article Priestley [26].) I was wondering whether it suffices to consider connected posets for the representability question. 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 if each 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 R0 to be the transitive closure of R. Then R0 is an equivalence relation. An 0 order component of X is an equivalence class [x]R0 of the relation R for some x ∈ X. A poset ist 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 of a representable posets (lemma 8.4), but failed to establish that they must be representable themselves in general. After discussing the problem with M.E. Adams, he came up with a counterexample which is the key ingredient of theorem 8.2:
THEOREM 8.2 (Adams). There exists a representable poset with an order component which 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 compact under 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 then establish, in lemma 8.5, that if every order component of a poset is representable, then so too is the poset. In paragraph 8.3, we define a countably infinite poset which we show to be order-isomorphic to an order component of a representable poset in lemma 8.7, but which, as lemma 8.8 shows, is not itself representable.
For any undefined terms or additional background, we refer the reader to Gr¨atzer [11] and Kelley [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 S S (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.
S Proof. Assume that (i) holds and let U be an open cover of X = k∈K Xk. By Alexander’s subbase 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, UXk has 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 no k ∈ K such that A ∪ B ⊆ Xk. Hence, there are w1, w2 ∈ A ∪ B such that w1 ∈ Xk and 0 w2 ∈ Xk0 for some k 6= k ∈ K. If w1, w2 ∈ A, then {X\(w1],X\(w2]} is a finite subcover of U. If w1 ∈ A, w2 ∈ B, then {X\(w1],X\[w2)} is a finite subcover of U (similarly for
44 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’s subbase lemma we may assume that
U = {Xk\(a] | a ∈ A} ∪ {Xk\[b) | b ∈ B} S 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 compact with its interval topology. Thus {Xk\(a1], ..., Xk\(ar]} ∪ {Xk\[b1), ..., Xk\[bs)} is a finite subcover of Xk.
If (X; ≤) is representable, then, for some topology τ,(X; τ, ≤) is a Priestley space. In particular, (X; τ) is a compact space and, as τi ⊆ τ, so too is (X; τi). Thus, the following is an immediate consequence of 8.3.
LEMMA 8.4. Each order component of a representable poset is compact with respect to its interval topology.
We now go on to show that if the order components of a poset are representable, then so is the poset.
LEMMA 8.5. Let (Xk, ≤k)k∈K be a family of pairwise disjoint nonempty representable S S posets. 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 has more than one element. For any k ∈ K, let τk be a topology making (Xk; τk, ≤k) a Priestley ∗ space. Fix k ∈ K and x ∈ Xk∗ . We now build a subbase for a topology on X in three steps. We set:
S S1 = l∈K\{k∗} τl;
S2 = {U ∈ τk∗ | x∈ / U};
0 ∗ S3 = {U ⊆ X | x ∈ U and U ∩ Xk∗ ∈ τk∗ and, for some k ∈ K\{k },U = [U ∩ Xk∗ ] ∪ S [ l∈K\{k∗,k0} Xl]}.
Then let τ be the topology having S = S1 ∪ S2 ∪ S3 as a subbase. Using Alexander’s subbase lemma we check easily that (X; τ) is compact using the fact that any subbase member containing x is, in some sense, large by virtue of the definition of S3 ⊆ S. Moreover, an inspection by cases shows easily that (X; τ, ≤) is totally order-disconnected.
45 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 an order 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) and y ∈ P (n + 1) \ P (n), where it is the case that x > y and x < y depending on whether n is even or odd, respectively. In particular, ≤ is antisymmetric. Moreover, if n is even, say
n = 2r, then x = pi0,...,i2r−1,k and y = pi0,...,i2r,i providing 0 ≤ k ≤ i2r and 0 ≤ i < ω, and if
n is odd, say n = 2r + 1, then x = pi0,...,i2r,k and y = pi0,...,i2r+1,i providing 0 ≤ k ≤ i2r+1 and 0 ≤ i < ω. In particular, ≤ is transitive and, as claimed, (P ; ≤) is seen to be a countable
connected 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 refer back to later.
46 In order to show that (P ; ≤) is order-isomorphic to an order component of a representable poset, we will define a suitable order on a compact totally disconnected space (C; τ) which itself 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 greatest element 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 some r ∈ Q. Intuitively, (C; ≤) may be thought of as the real interval [0, 1] where every rational element 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, and finite 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 partial order 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, the 1 length of Xi, denoted ln(Xi), is ≤ 2 in the pseudometric obtained from the metric imposed on 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, 1 Xi0,i ⊆ Xi0 , ln(Xi0,i) ≤ 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,
47 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 = ∅ 1 for i 6= j, Xi0,i1,i ⊆ Xi0,i1 , ln(Xi0,i1,i) ≤ 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, 1 Xi0,i1,...,i2r,i ∩ Xi0,i1,...,i2r,j = ∅ for i 6= j, Xi0,i1,...,i2r,i ⊆ Xi0,i1,...,i2r , ln(Xi0,i1,...,i2r,i) ≤ 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) ≤ 1 , and (0, s ), (0, s ] 6∈ X for 0 ≤ i < ω. 22(r+1) 2r+2 2r+2 i0,i1,...,i2r+1,i
Elsewhere on C, let be trivial. Thus, since (X; ) is order-isomorphic to (P ; ≤), (C; ) is
a poset whose order components consist precisely of X = {x} ∪ {xi0,...,in | 0 ≤ n < ω and 0 ≤ ω ij < ω for 0 ≤ j ≤ n} and 2 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 remains to show that, for u, v ∈ C, whenever u 6 v there exists a clopen decreasing set U such that u ∈ 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}
48 and observe that, as [xi0,...,in ) or (xi0,...,in ] is a finite chain depending on whether n is even or odd, respectively, it follows from the choice of elements in X\X(n) that, for 0 ≤ i0, . . . , in < ω, S (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 such that 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. 1 Then, by choice, u is a gap and, since ln(Xi0,...,in ) ≤ n , v 6∈ Xi0,...,in for some 0 ≤ n < ω. 2 S 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,...,jk for 1 any 0 ≤ k < ω, then v is a gap and, since ln(Xjo,...,jm ) ≤ 2m , u 6∈ Xj0,...,jm for some 0 ≤ m < ω. We may assume, again with no loss in generality, that m is odd. Set S U = 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 any
0 ≤ l < ω. Then, providing u 6= xi0,...,in , it is not hard to see that there exists a clopen set U such that u ∈ U, v 6∈ U, and each element of U is incomparable under to any other element of (C; ), whereby U is decreasing. Were it the case that u 6∈ Xl for any 0 ≤ l < ω, then a similar 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 clopen set V such that v ∈ V , u 6∈ V , and each element of V is incomparable under to any other element of (C; ). In this case, set U = C \ V . Likewise, unless v = x, a similar set may be defined 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 , S 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.
49 Assume ik = jk for all 0 ≤ k ≤ n. Then, by hypothesis, m ≥ n + 2 and, since u > v, n is S even. 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, contrary to hypothesis. Thus, we may assume that in < jn. But then it follows that m ≥ n + 2. S Thus, v ∈ V = (Xi ,...,i ,jn,j ,l : 0 ≤ l ≤ jn+2) which is clopen increasing and, since 0 n−1 n+1 S 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, set U = 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 ∈ S U = (Xi ,...,i ,i ,l : 0 ≤ l ≤ ik+1) which is clopen decreasing and, since U ⊆ Xi0,...,i ,i 0 k−1 k S k−1 k 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,jk and u 6∈ Xi0,...,ik−1,jk , u 6∈ V . In this case, set U = 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 is S odd. 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 is S even. 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, S 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 analogous case when n ≤ m.
Since the order components of (C; τ, ) consist of precisely X = {x} ∪ {xi0,...,in | 0 ≤ n < ω ω and 0 ≤ ij < ω for 0 ≤ j ≤ n} and 2 singletons and, by choice, (X; ) is order-isomorphic to (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.
50 Proof. Suppose, contrary to hypothesis, that (P ; ≤) is representable and let (P ; τ, ≤) be a Priestley space for some topology τ.
We claim that, for x ∈ P , there is a sequence (xi : 0 ≤ i < ω) such that either, for 0 ≤ j < i < ω, xi < xj and x is the greatest lower bound of {xi | 0 ≤ i < ω} or, for 0 ≤ 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. In particular, S is not closed. Choose y ∈ cl(S) \ S. If y 6≥ x, then there is a clopen decreasing set U with y ∈ U and x 6∈ U, from which it follows that U ∩ S = ∅, contradicting y ∈ cl(S). If y > 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, as claimed, x is an accumulation point. As similar argument holds in the case that x is the least upper 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 the notation 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 ⊂ B and it is possible to choose x ∈ B \ A. Since x is an accumulation point, there exists a distinct element y ∈ B \ A. Say, without loss of generality, x 6≥ y. Then there exists a clopen decreasing set U with x ∈ U and y 6∈ U. Set C = A ∪ (B ∩ U). Then C is a clopen decreasing set such that A ⊂ C ⊂ B. In particular, C corresponds to an element c ∈ L such that a < 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 denoted by 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+) is uncountable 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 , we were 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 the one-component-compactification yields different topologies on P such that in general we will get non-isomorphic Priestley spaces. So the question remains whether this construction has any 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 not uniquely 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 exactly one topology making it Priestley.
2. A poset (P, ≤) is called uniquely representable in the second sense (ur2) if for any topolo- gies τ, τ 0 making P Priestley, the ordered spaces (P, τ, ≤) and (P, τ 0, ≤) are isomorphic.
It is clear that a ur1 space is ur2, but it is not clear whether there are Priestley spaces that are ur2 but not ur1. Moreover, note that if the interval topology of a representable poset is Hausdorff, then it must be the unique topology on the poset making it Priestley (first, the interval topology is always contained in the Priestley topology, and second, any topology refining a given compact Hausdorff topology is not compact any more). It is natural to ask the converse as has been done 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 categorical constructions like the product. Note that the coproduct (i.e. disjoint union) case has been dealt 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 still ur2?
52 Part IV
Maximal compactness
53 Chapter 9
Joint work with H. P. K¨unzi
9.1 Introduction
Priestley spaces are compact Hausdorff spaces. Every compact Hausdorff space (X, τ) is both maximally compact and minimally Hausdorff, that is, any topology strictly coarser than τ is not Hausdorff and any topology strictly finer that τ is not compact. (This property can be easily 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. K¨unzi 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 maximally compact topology on X containing τ? This question turned out to be open. Observe that Zorn’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 of compact 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 is closed. A topological space is called a US-space provided that each convergent sequence has a unique limit. It is known [36] that each Hausdorff space (= T2-space) is a KC-space, each KC-space is a US-space and each US-space is a T1-space (that is, singletons are closed); and no 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 strictly contained in a compact topology on X. It is known that a topological space is maximal compact if and only if it is a KC-space that is also compact [27]. These spaces will be called compact KC-spaces in the following. Let us note that while there are many maximal compact topologies, minimal noncompact topologies do not exist: Any noncompact space X possesses a strictly increasing open cover
54 {Cα : α < δ} of X where δ is a limit ordinal and C0 can be assumed to be nonempty. Clearly then {∅,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]). A standard example of a maximal compact topology that is not a Hausdorff topology is given by 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 observe by citing an example due to van Douwen (see [35]). In order to discuss that example we first recall some pertinent definitions. A nonempty subspace S of a topological space is called irreducible (see e.g. [13]) if each pair of nonempty open sets of S intersects. Furthermore a topological space X called a Fr´echetspace (see [8, p. 53]) provided that for every A ⊆ X and every x ∈ A there exists a sequence of points of A converging to x. For the convenience of the reader we include a proof of the following observation (compare e.g. Math. Reviews 53#1519 of [28]). LEMMA 9.2. Each Fr´echet US-space X is a KC-space.
Proof. Suppose that x ∈ K where K is a compact subspace of X. Because X is a Fr´echet space, there is a sequence (kn)n∈N of points of K converging to x. Since K is compact, that sequence has a cluster point c in K. Because X is a Fr´echet space, there is a subsequence of (kn)n∈N converging to c (compare [8, Exercise 1.6D]). Hence x = c ∈ K, because X is a US-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 Fr´echet US- space that is anti-Hausdorff. By the preceding lemma that space is a KC-space and hence maximal 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 topology satisfies 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 (just take the supremum of the given topology with the cofinite topology) [33, Theorem 10], the question whether each compact topology is contained in a compact KC-topology (that is, is contained in a maximal compact topology) seems still to be open. Apparently that question was 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 compact topologies need not have a compact supremum: Consider the sequence (τn)n∈N of topologies τn = {∅, 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 only cofinite neighborhoods is clearly contained in a maximal compact topology: Just consider the one-point-compactification Xx of X \{x} where X \{x} is equipped with the discrete topology 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 has a neighborhood base consisting of compact sets. Note that a locally compact KC-space is a regular Hausdorff space.
55 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 compact topology?
In [33, Example 11] it is shown that a compact space need not be the continuous image of a compact T2-space. In fact, a careful analysis of the argument reveals the following general fact (also stated in [14, 3.6]).
PROPOSITION 9.6. A KC-space Y that is the continuous image of a compact T2-space X 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. The conclusion follows, since obviously a closed continuous image of a compact T2-space, is a T2-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 space onto 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 , we see 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 hence Y 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 compact set 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-quotients of maximal compact spaces are not necessarily maximal compact.
PROBLEM 9.8. Are T1-quotient topologies of maximal compact topologies contained in maximal 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 topology τ 0 finer than τ such that (X, τ 0) is a US-space.
Proof. As usual two subsets A and B of X will be called almost disjoint provided that their intersection is finite. Let M = {Ai : i ∈ I} be a maximal (with respect to inclusion) family of pairwise almost disjoint injective sequences in X with a distinct τ-limit (that is, each Ai ∈ 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, let m 0 Ai = {xn : n ∈ N, n ≥ m} ∪ {x}. Let τ be the topology on X which is generated by the m subbase τ ∪ {X \ Ai : i ∈ I, m ∈ N}.
56 n We first show that τ‘ is compact. Let C be a subcollection of Aτ ∪ {Ai : i ∈ I, n ∈ N} with empty intersection. (Here, as in the following, Aτ denotes the set of τ-closed sets.) Denote the intersection of C with Aτ by F. We want to show that there is a finite subcollection of C with an empty intersection. Of course, it will be sufficient to find a finite subcollection of C with finite intersection. If C = F, then such a finite subcollection of C must exist by compactness of (X, τ). So in this case we are finished. If we have in our collection C\F two n m sets Ai and Aj 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 form Am = {x : n ∈ N, n ≥ m} ∪ {a} for some fixed i ∈ I and n ∈ M where M is a nonempty i0 n 0 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 that a 6∈ F0. Since F0 is τ-closed and the injective sequence (xn)n∈N τ-converges to a, we conclude that F0 ∩ {xn : n ∈ N} is finite, since otherwise a ∈ F0. Hence for any m ∈ M we have that F ∩ Am is finite and we are finished again. 0 i0 We deduce from Alexander’s subbase theorem that the topology τ 0 is compact. Next we want to show that (X, τ 0) is a US-space. In order to reach a contradiction, suppose 0 that there is some sequence (xn)n∈N that τ -converges to x and y where x and y are distinct points in X. Replacing (xn)n∈N if necessary by a subsequence, we can and do assume that the 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 the original sequence (xn)n∈N attains each value at most finitely many often, since (X, τ) and 0 0 thus (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 denotes the chosen τ-limit of the sequence (zn)n∈N) belonging to M such that Ai ∩ {xn : n ∈ N} has infinitely many elements. Suppose that there is some p ∈ N such that x or y does not belong p p 0 to Ai . Then X \ Ai 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 both 0 τ -limits of (xn)n∈N. So there is no such p ∈ N and it necessarily follows that x = z = y —a contradiction. We conclude that (X, τ 0) 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 that each compact topology is contained in a compact topology with respect to which each compact countable set is closed.
In order to see this we need the following two auxiliary results. We recall that a topological space is called sequentially compact provided that each of its sequences has a convergent subsequence.
LEMMA 9.12. Let X be a US-space and let {Kn : n ∈ N} be a countable family of sequentially compact sets in X having the finite intersection property. Then ∩n∈NKn is nonempty.
n Proof. For each n ∈ N find xn ∈ ∩i=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
57 m ∈ N such that k 6∈ Km. Since there is a tail of (yn)n∈N belonging to Km and Km is sequentially 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 that k ∈ ∩n∈NKn. LEMMA 9.13. Each compact US-topology is contained in a compact topology with respect to which each compact countable set is closed.
Proof. Let (X, τ) be a compact US-space and let τ 0 be the topology generated by the subbase τ ∪ {X \ K : K ⊆ X is countable and compact} on X. We are going to show that τ 0 is compact. In order to reach a contradiction, assume that C is a subcollection of Aτ ∪ {K ⊆ X : K is countable and compact} having the finite intersection property, but ∩C = ∅. Since τ is compact, we deduce that some compact countable set K belongs to C. Hence by countability of K there must exist a countable subcollection D of C such that ∩D = ∅. Replace in D each member F of D ∩ Aτ by its trace F ∩ K on K to get a countable collection D0 of compact countable sets having the finite intersection property. By a result of [21], each compact countable space is sequentially compact and hence D0 is a countable collection of sequentially compact sets in a US-space. Since ∩D0 = ∅, we have reached a contradiction to the preceding lemma. We conclude that τ 0 is compact by Alexander’s subbase theorem. Evidently each compact countable set in (X, τ 0) is τ-compact and thus τ 0-closed.
PROBLEM 9.14. Given some fixed cardinal κ > ℵ0. Is each compact topology contained in a 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 positively the variant of the main problem (see [6, Question 8-1, p. 56]) formulated for sequential compactness instead of compactness.
THEOREM 9.15. Each sequentially compact topology τ on a set X is contained in a se- quentially compact topology τ 00 that is maximal among the sequential compact topologies on X.
Proof. Since (X, τ) is sequentially compact and any convergent (sub)sequence has a constant or an injective subsequence, it is obvious that any sequence in (X, τ) has a subsequence that converges 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 sequentially compact topology τ on X is a T1-topology. Define now a topology τ 0 on X in exactly the same way as above. We next show that 0 (X, τ ) is sequentially compact provided that (X, τ) is sequentially compact. Let (yn)n∈N be any 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 0). So by choosing an appropriate subsequence of (sn)n∈N if necessary, it suffices to consider the case that (sn)n∈N is injective and that sn 6= a whenever n ∈ N. By maximality of M there is Ai = {zn : n ∈ N} ∪ {z} belonging to M such that {sn : n ∈ N} ∩ Ai is infinite. Hence there is a common injective subsequence 0 of the injective sequences (sn)n∈N and (zn)n∈N in this intersection. By definition of τ that 0 Tn kj subsequence converges to z, because any basic τ -neighborhood G ∩ j=1(X \ Aj ) of z where
58 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 0 0 (yn)n∈N has a τ -convergent subsequence and that (X, τ ) is sequentially compact. As in the preceding proof, one argues that (X, τ 0) is a US-space. We now define a new topology τ 00 on X by declaring A ⊆ X to be τ 00-closed if and only if 0 xn ∈ 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 00 is a topology finer than τ 0 on X with the property that any 0 00 sequence (xn)n∈N that converges to x in (X, τ ) also converges to x in (X, τ ). In particular, it follows that the space (X, τ 00) is sequentially compact, because (X, τ 0) is sequentially compact. 00 Let K be a sequentially compact subset in (X, τ ). Suppose that xn ∈ K whenever n ∈ N and 0 that the sequence (xn)n∈N converges to x in (X, τ ). Then there is a subsequence (yk)k∈N of 0 00 0 (xn)n∈N that converges to r ∈ K in (X, τ ), since K is sequentially compact in (X, τ ) and τ ⊆ τ 00. Thus x = r, since (X, τ 0) is a US-space and hence x ∈ K. By the definition of the topology τ 00 we conclude that K is closed in (X, τ 00). Therefore each sequentially compact subset of (X, τ 00) is τ 00-closed. By [5, Theorem 2.4] we conclude that τ 00 is a maximal sequentially compact 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 maximal compact topologies?
Evidently the cofinite topology on an infinite set X is the infimum of the family of maximal compact Hausdorff topologies of the one-point-compactifications Xx (where x ∈ X) that we have 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). 0 0 Note that if τ and τ are two compact topologies on a set X such that τ ⊆ τ , then Aτ ⊆ Aτ 0 ⊆ Cτ 0 ⊆ 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 unions will be called a ring of sets on X. We consider the set Mτ of all rings G of sets ordered by set-theoretic inclusion on the topological space (X, τ) such that Aτ ⊆ G ⊆ Cτ . Since Aτ is S such 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 of C having the finite intersection property has nonempty intersection. We use this nonstandard convention 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 ∗ 0 that is a compact collection, then G = Aτ 0 where τ is a maximal compact topology finer than τ.
∗ Proof. Suppose that G is a maximal element in Mτ that is compact . Then {X \ K : K ∈ G} 0 ∗ yields the subbase of a topology τ on X. Observe that Aτ ⊆ G ⊆ Aτ 0 . Since G is compact ,
59 τ 0 will be compact, by Alexander’s subbase theorem. Because τ 0 is compact, τ ⊆ τ 0 implies that Aτ 0 ⊆ Cτ . Hence Aτ 0 belongs to Mτ . We conclude that G = Aτ 0 by the maximality of G. It remains to be seen that τ 0 is maximal compact. If τ 00 is a finer topology than τ 0 and 0 00 compact, then Aτ 0 ⊆ Aτ 00 ⊆ Cτ . Hence by maximality of G, Aτ 00 = G = Aτ 0 and so τ = τ . We have shown that τ 0 is maximal compact.
PROPOSITION 9.18. Let (X, τ) be a compact topological space such that each filterbase consisting of compact subsets has a nonempty intersection. Then τ is contained in a maximal compact topology τ 0.
Proof. Let G be any maximal element in Mτ as defined above. Recall that G is closed under finite intersections. Hence any nonempty subcollection G0 of G having the finite intersection property generates a filterbase consisting of compact sets on X. It follows from our hypothesis that G is a compact∗ collection. Furthermore by lemma 4 we conclude that G is equal to the set of closed subsets of a maximal compact topology τ 0 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 application of Proposition 4 we want to show however that the seemingly reasonable conjecture that the product topology of a large family of maximal compact topologies is no longer contained in a maximal compact topology is unfounded. In order to see this we next prove the following result.
LEMMA 9.19. Let (Xi)i∈I be a nonempty family of T1-spaces such that each Xi has the property that every filterbase of compact sets has a nonempty intersection. Then the product Πi∈I Xi also has that property.
Proof. We can (and do) assume that I is equal to some finite ordinal or an infinite limit ordinal . Let F be a filterbase of compact subsets on the product Πγ<Xγ.
For each α < we shall inductively find xα ∈ Xα such that the set Aα = {(yγ)γ< ∈ Πγ<Xγ : yγ = xγ whenever γ ≤ α} satisfies Aα ∩ K 6= ∅ whenever K ∈ F.
Suppose now that for some δ < and all α < δ, xα ∈ Xα have been chosen such that Aα ∩ 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. So let δ be a limit ordinal (possibly equal to 0) and fix K ∈ F. Since for each α < δ, Aα is closed because every space Xα is a T1-space, and since Aα ∩ K 6= ∅ the claim holds by compactness of K and the monotonicity of the sequence {Aα : α < δ}. (For the case that δ = 0 as usual we use the convention that ∩∅ = Πγ<Xγ.) Continuing now with the proof we next consider the filterbase {[pr (∩ A ∩K)] : K ∈ F} Xδ α<δ α of compact sets on Xδ. By our assumption on Xδ there exists some
x ∈ ∩ [pr (∩ A ∩ K)]. δ K∈F Xδ α<δ α
It remains to show that for each K ∈ F, {(yγ)γ< ∈ Πγ<Xγ : yγ = xγ, γ ≤ δ} ∩ K 6= ∅; but this is an immediate consequence of x ∈ pr (∩ A ∩ K). Finally note that ∩ A = δ Xδ α<δ α α< α
60 {(xα)α<} and that — for exactly as in the case of the ordinal δ above — ∩α<Aα ∩ 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-topologies is contained in a maximal compact topology.
Proof. Note first that in a compact KC-topology each filterbase of compact sets has a nonempty intersection. We conclude by the preceding lemma and Proposition 4 that the compact product topology of an arbitrary nonempty family of maximal compact topologies is contained in a maximal compact topology.
COROLLARY 9.21. Let (Xi)i∈I be a nonempty family of spaces each of which is contained in a maximal compact topology. Then also their product topology is contained in a maximal compact topology.
9.4 Sobriety and maximal compactness
Note that the closure of each irreducible subspace of a topological space is irreducible. Recall also that a topological space is called sober (see e.g. [13]) provided that every irreducible closed 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 the intersection 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. T T 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 some maximal compact topology τ 0.
Proof. Since all (compact) sets in a T1-space are saturated, the condition stated in Proposition 4 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 0 obtained in corol- laryollary 3 will be sober, since the only irreducible sets with respect to the coarser topology τ are the singletons. Van Douwen’s example [35] mentioned earlier shows that a maximal compact topology need not be (contained in) a compact sober topology.
EXAMPLE 9.25. Note that the closed irreducible subsets of the one-point-compactification X (of the Hausdorff space) of the rationals are the singletons: Any finite subset of a T1-space with at least two points is discrete and hence not irreducible. Moreover any infinite subset of X contains two distinct rationals and thus cannot be irreducible. We conclude that an arbitrary power of X is a compact, sober T1-space, because products of sober spaces are sober (see e.g. [13, Theorem 1.4]).
61 In the light of the proof of Proposition 4 one wonders which compact sober T1-topologies can be represented as the infimum of a family of maximal compact topologies (compare Problem 5). Our next result provides a partial answer to this question. An interesting space satisfying the hypothesis of Proposition 6 is a T1-space constructed in [18]: It has infinitely many isolated points although each open set is the intersection of two compact open sets. (It was noted in the 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 set A ⊆ 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 or sequential 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 than τ, then K is compact with respect to σ and thus with respect to τ. In order to verify the statement, 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 compact set K0 such that K0 ∩ C is not compact. Suppose first that X is locally compact.
Then there is x ∈ X such that x ∈ clτ C \ C. Let F = {K : K is a compact neighborhood at x in (X, τ)}. Of course, ∩F = {x}, since X is a locally compact T1-space. Suppose that K ∩ C is compact in (X, τ) whenever K ∈ F. Then {K ∩ C : K ∈ F} is a filterbase of compact saturated sets in X. According to the result cited above from [15], we have ∩F ∩ C = T{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 is not compact in (X, τ).
Suppose next that X is sequential. Since C is not closed, there is a sequence (xn)n∈N in C converging 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 is empty, because τ is a T1-topology and (xn)n∈N converges to x—a contradiction. Hence there is 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. 0 Note now that τ ∪ {X \ K0} is a subbase for a compact topology τ on X that is also sober 00 0 and T1. By corollaryollary 3 there is a maximal compact topology τ finer than τ . Observe 00 that X \ C 6∈ τ : Otherwise C ∈ Aτ 00 and, since K0 ∈ Aτ 00 , also K0 ∩ C ∈ Aτ 00 . Therefore 00 K0 ∩ C ∈ Cτ 00 and K0 ∩ C ∈ Cτ —a contradiction. Thus indeed X \ C 6∈ τ . We conclude that τ 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 any two 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 two compact sets is compact is a KC-space.
62 We next give an example of a compact sober T1-topology that is not the infimum of a family of maximal compact topologies.
EXAMPLE 9.29. Let Y be an uncountable set and let −∞ and ∞ be two distinct points not in Y . Set X = Y ∪ {−∞, ∞}. Each point of Y is supposed to be isolated. The neighborhoods of ∞ are the cofinite sets containing ∞ and the neighborhoods of −∞ are the cocountable sets 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 and not closed if and only if A is uncountable, ∞ ∈ A and −∞ 6∈ A : Indeed, if ∞ ∈ A, then A is clearly compact and if A is uncountable and −∞ 6∈ A, then A cannot be closed. In order to prove the converse suppose that A is compact and not closed in (X, τ). Then A is certainly infinite. It therefore follows from compactness of A that ∞ ∈ A. Since A is not closed, we conclude that −∞ ∈ A \ A and hence A is uncountable. 0 0 Of course, if τ is a maximal compact topology such that τ ⊆ τ , then Aτ ⊆ Aτ 0 ⊆ Cτ 0 ⊆ Cτ . Observe that the topology τ 00 generated by the subbase {{−∞}} ∪ τ clearly yields a compact T2-topology finer than τ. Obviously, Cτ \Aτ ⊆ Aτ 00 by the description found above of the 00 nonclosed compact sets in (X, τ). Thus Aτ 00 = Cτ . We conclude that τ is finer than any maximal compact topology containing τ. Hence τ 00 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 limits of each ultrafilter is equal to the closure of some unique singleton. Of course, each compact Hausdorff space satisfies this condition. We finally observe that each locally compact strongly sober topological space (X, τ) possesses a finer compact Hausdorff topology; just take the supremum of τ and its dual topology (see e.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 compact Hausdorff topology.
63 Part V
Outlook: Between topology and order
64 In the category of posets there are different ”measures” of posets; since they seem to have much in common with the cardinal functions used in topology, we will call them poset cardinal functions. The following are examples of poset cardinal functions. Let (P, ≤) be a partially ordered set. Recall that ≤ is called a linear order if for all x, y ∈ P we have x ≤ y or 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 ordering relation ≤. For instance consider u(P ), the so called upper topology which is generated by 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 interval topology 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 are examples of induced topologies on a poset. Recall that also in the category of topological spaces there are certain measures which are called (topological) cardinal functions. Since posets and topological spaces can be linked by means 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 two instances:
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 upper topology (or indeed, any of the induced topologies)?
2. If the ordering relation on a poset P is extended, in what way does this influence the cardinal functions of (P, τi(P ))?
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67 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 (Maturit¨at). 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-finite algebras” (supervisor Prof. Christine Riedtmann). 2001-2004 Doctoral thesis ”Aspects of Priestley Duality” (supervisor Prof. J¨urg Schmid).
Talks:
- ”Representable posets and their order components”, 2003 Summer Conference on Topology and Its Applications, Howard University, Washington, DC, USA. - ”Aspekte der Priestley-Dualit¨at”,Kolloquiumsvortr¨age, 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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