Priestley Rings and Priestley Order- Compactifications Guram

Priestley Rings and Priestley Order- Compactifications

Guram Bezhanishvili & Patrick J. Morandi

Order A Journal on the Theory of Ordered Sets and its Applications

ISSN 0167-8094 Volume 28 Number 3

Order (2011) 28:399-413 DOI 10.1007/s11083-010-9180-2

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Order (2011) 28:399–413 DOI 10.1007/s11083-010-9180-2

Priestley Rings and Priestley Order-Compactifications

Guram Bezhanishvili · Patrick J. Morandi

Received: 14 July 2009 / Accepted: 6 September 2010 / Published online: 18 September 2010 © Springer Science+Business Media B.V. 2010

Abstract We introduce Priestley rings of upsets (of a poset) and prove that inequivalent Priestley ring representations of a bounded distributive lattice L are in 1-1 correspondence with dense subspaces of the Priestley space of L. This generalizes a 1955 result of Bauer that inequivalent reduced field representations of a Boolean algebra B are in 1-1 correspondence with dense subspaces of the Stone space of B. We also introduce Priestley order-compactifications and Priestley bases of an ordered topological space, and show that they are in 1-1 correspondence. This generalizes a 1961 result of Dwinger that zero-dimensional compactifications of a topological space are in 1-1 correspondence with its Boolean bases.

Keywords Ordered topological space · Order-compactification · Priestley space · Stone space · Distributive lattice · Boolean algebra

Mathematics Subject Classifications (2010) Primary 54F05; Secondary 54D35 · 06E15 · 06D05

1 Introduction

It is a celebrated theorem of Stone [19] that the category of Boolean algebras is dually equivalent to the category of zero-dimensional compact Hausdorff spaces, also known as Stone spaces. Soon after, Stone [20] generalized his result to obtain a dual equivalence of the category of bounded distributive lattices and the category of what later became known as spectral spaces. The concept of a spectral space generalizes

G. Bezhanishvili · P. J. Morandi (B) Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM 88003-8001, USA e-mail: [email protected] G. Bezhanishvili e-mail: [email protected] Author's personal copy

400 Order (2011) 28:399–413 that of a Stone space. In particular, spectral spaces are not Hausdorff (nor even T1), and, as a result, they are harder to work with. Another duality for bounded distributive lattices was developed by Priestley [16, 17] some thirty five years later by means of ordered topological spaces, which became known as Priestley spaces. As was shown by Cornish [7, Theorem 2.3], the categories of spectral spaces and Priestley spaces are isomorphic. Thus, it is only a matter of preference whether to work with spectral spaces or Priestley spaces. It follows from Stone’s theorem that every Boolean algebra B is isomorphic to the field Cp(X) of clopen subsets of the Stone space X of B.SinceX is Hausdorff, the field Cp(X) is reduced (that is, any two different points x and y of X can be separated by an element A of Cp(X)), and as X is compact, Cp(X) is perfect (that is, every ultrafilter of Cp(X) is determined by a point of X). Stone’s theorem implies that Cp(X) is a unique up to equivalence reduced and perfect field representation of B. On the other hand, as was shown by Bauer [2], inequivalent reduced field representations of B are in 1-1 correspondence with dense subspaces of X.Anice English exposition of Bauer’s theorem can be found in [8, Theorem 12.1]. Our first goal is to generalize Bauer’s theorem to bounded distributive lattices. We generalize the notion of a reduced field of sets to that of a Priestley ring of upsets (of a poset) and show that inequivalent Priestley ring representations of a bounded distributive lattice L are in 1-1 correspondence with dense subspaces of the Priestley space of L. We recall that a compactif ication of a topological space X is a compact Hausdorff space Y such that X is homeomorphic to a dense subspace of Y. We also recall that a Boolean basis of X is a basis of X that is a field of sets. It was shown by Dwinger [8, Theorem 13.1] (see also [3, 11]) that the poset of zero-dimensional compactifications of X is isomorphic to the poset of Boolean bases of X. Our second goal is to generalize Dwinger’s theorem to the case of ordered topological spaces. We introduce Priestley order-compactifications and Priestley bases of an ordered topological space X, and show that the poset of Priestley order-compactifications of X is isomorphic to the poset of Priestley bases of X. We also give a necessary and sufficient condition for an ordered topological space to have a Priestley order- compactification, and show that among Priestley order-compactifications there exists a largest one, which may be different from the Nachbin order-compactification. Furthermore, we give a necessary and sufficient condition for the Nachbin order- compactification to be a Priestley order-compactification.

2 Priestley Rings and a Generalization of Bauer’s Theorem

For a poset X,letUp(X) denote the bounded distributive lattice of upsets of X.We call a subset R of Up(X) a ring of upsets of X if ∅, X ∈ R and R is closed under finite unions and intersections. In other words, R is a ring of upsets of X if R is a bounded distributive sublattice of Up(X).

Definition 2.1 Let R be a ring of upsets of a poset X.WecallR a Priestley ring if for each x, y ∈ X with x ≤ y,thereisA ∈ R such that x ∈ A and y ∈/ A. Author's personal copy

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The next proposition shows that we can always restrict our attention to Priestley rings.

Proposition 2.2 Let R be a ring of upsets of a poset X. Then there is a Priestley ring P such that P is isomorphic to R.

Proof Define ∼ on X by x ∼ y if for all A ∈ R, we have x ∈ A iff y ∈ A.Then∼ is clearly an equivalence relation on X.Let[x] denote the equivalence class of x ∈ X. We define ≤∼ on X/∼ by [x] ≤∼ y if for all A ∈ R, x ∈ A implies y ∈ A. It is easy to see that ≤∼ is well-defined and that≤∼ is a partial order on X/∼. For A ⊆ X,let A denote {[x] : x ∈ A},andsetP = A : A ∈ R . We show that P is a Priestley ring isomorphic to R. First, it is easy to verify that P⊆ Up(X/∼),that X/∼=[X] is the top of P,that ∅ =[∅] is the bottom of P,that A ∪ [B] = A ∪ B and A ∩ [B] = A ∩ B . Thus, P is a bounded sublattice of Up(X/∼). Next, if [x] ≤∼ y , then there is A ∈ R with x ∈ A and y ∈/ A. Therefore, [A]∈P, [x] ∈ A and y ∈/ A , which implies that P is a Priestley ring. Lastly, the map ϕ : R → P, defined by ϕ(A) = A , is clearly a bounded lattice homomorphism. It is onto by the definition of P . To see that it is 1-1, let A = [B] for some A, B ∈ R.Ifa ∈ A,then[a] ∈ A . Therefore, [a]∈[B] and it follows from the definition of ∼ that a ∈ B. Thus, A ⊆ B. The other inclusion is similar. Consequently A = B,andsoϕ is a lattice isomorphism.

Definition 2.3 Let L be a bounded distributive lattice.

(1) If R is a ring of upsets of a poset X and ξ : L → R is a lattice isomorphism, then we call (R,ξ)a ring representation of L. (2) If (R,ξ) is a ring representation of L and R is a Priestley ring, then we call (R,ξ)a Priestley ring representation of L. (3) Let (R,ξ) and (P, μ) be two ring representations of L over X and Y,respectively. We call (R,ξ) and (P, μ) equivalent if there is an order-isomorphism f : X → Y such that f −1 : P → R is a lattice isomorphism with f −1 ◦ μ = ξ.

Among Priestley ring representations of a bounded distributive lattice, there exists a canonical one, obtained through the Priestley representation theorem for bounded distributive lattices. We recall that an ordered topological space X is a Priestley space if X is compact and satisfies the Priestley separation axiom: x ≤ y implies there exists a clopen upset U of X such that x ∈ U and y ∈/ U. It follows that X is also Hausdorff and zero-dimensional, and so X is a Stone space. In our terminology, all the Priestley separation axiom says is that the ring CpUp(X) of clopen upsets of X is a Priestley ring. If L is a bounded distributive lattice, then it is well known that the Priestley space X of L is constructed as the set of prime filters of L, ordered by set inclusion, with {φ(a) − φ(b) : a, b ∈ L} as a basis for the topology, where φ(a) ={x ∈ X : a ∈ x} is the Stone map. Then φ[L]=CpUp(X) and so φ : L → CpUp(X) is a lattice isomorphism. Clearly (CpUp(X), φ) is a Priestley ring representation of L,whichis the canonical Priestley ring representation of L. Author's personal copy

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Theorem 2.4 Let L be a bounded distributive lattice and let X be its Priestley space. Then there is a 1-1 correspondence between inequivalent Priestley ring representations of L and dense subspaces of X.

Proof Suppose Y is a dense subspace of X. We restrict the order of X to Y,and view Y as a poset. Let R = {U ∩ Y : U ∈ CpUp(X)}.ThenR is a bounded sublattice of Up(Y), thus a ring of upsets of Y. To see that R is a Priestley ring, suppose that x ≤ y in Y.SinceY ⊆ X and X is a Priestley space, there is a clopen upset U of X with x ∈ U and y ∈/ U.ThenU ∩ Y ∈ R, and this set separates x from y. Thus, R is a Priestley ring. Define η : CpUp(X) → R by η(U) = U ∩ Y. It is clear that η is a surjective bounded lattice homomorphism. To see that it is also injective observe that, as Y is dense in X, for each U ∈ CpUp(X) we have U = U ∩ Y. (In fact, U = U ∩ Y for every clopen subset U of X.) Define ξ : L → R by ξ = η ◦ φ,whereφ is the Stone map. Then, since φ : L → CpUp(X) is an isomorphism, we obtain that (R,ξ)is a Priestley ring representation of L. Conversely, let (R,ξ)be a Priestley ring representation of L.ThenR is a Priestley ring of upsets of a poset Y.Definee(R,ξ) : Y → X by e(R,ξ)(y) = {a ∈ L : y ∈ ξ(a)}. It is easy to see that e(R,ξ)(y) is a prime filter of L; thus, e(R,ξ) is well-defined. If x ≤ y and a ∈ e(R,ξ)(x),thenx ∈ ξ(a).Asξ(a) ∈ R is an upset of Y, it follows that y ∈ ξ(a).Soa ∈ e(R,ξ)(y), and thus e(R,ξ)(x) ⊆ e(R,ξ)(y). On the other hand, if x ≤ y,thensinceR is a Priestley ring, there exists A ∈ R with x ∈ A and y ∈/ A. Because ξ : L → R is an isomorphism, there is a ∈ L with x ∈ ξ(a) and y ∈/ ξ(a). But then a ∈ e(R,ξ)(x) and a ∈/ e(R,ξ)(y),andsoe(R,ξ)(x) ⊆ e(R,ξ)(y). Consequently, e(R,ξ) is an order-isomorphism from Y onto e(R,ξ)(Y). To see that e(R,ξ)(Y) is ∈ ξ( ) = −1 (φ( )) dense in X we first show that for each a L we have a e(R,ξ) a . Indeed, ∈ −1 (φ( )) ( ) ∈ φ( ) ∈ ( ) ∈ ξ( ) y e(R,ξ) a iff e(R,ξ) y a iff a e(R,ξ) y iff y a .Now,letU be a nonempty basic open subset of X.ThenU = φ(a) − φ(b) for some a, b ∈ L with a ≤ ∩ ( ) = ∅ −1 ( ) = ∅ ξ( ) − ξ( ) = −1 (φ( )) − b.IfU e(R,ξ) Y ,thene(R,ξ) U . Therefore, a b e(R,ξ) a −1 (φ( )) = −1 (φ( ) − φ( )) = −1 ( ) = ∅ e(R,ξ) b e(R,ξ) a b e(R,ξ) U , which is a contradiction since ξ(a) ⊆ ξ(b). Thus, e(R,ξ)(Y) is dense in X. Hence, to each Priestley ring representation (R,ξ)of L over Y there corresponds a dense subspace e(R,ξ)(Y) of X. To see that this correspondence is onto, let Y be a dense subspace of X.ThenR ={φ(a) ∩ Y : a ∈ L} is a Priestley ring, and ξ : L → R defined by ξ(a) = φ(a) ∩ Y is an isomorphism. Therefore, (R,ξ)is a representation of L over Y and

e(R,ξ)(y) ={a ∈ L : y ∈ φ(a) ∩ Y}={a ∈ L : a ∈ y}=y.

Thus, e(R,ξ)(Y) = Y. To see that this correspondence is 1-1, let (R,ξ)and (P, μ) be two Priestley ring representations of L over Y and Z, respectively. Let e(R,ξ) : Y → : → −1 : ( ) → X and e(P,μ) Z X be the corresponding embeddings. Then e(R,ξ) CpUp X −1 : ( ) → ( ) = R and e(P,μ) CpUp X P are lattice isomorphisms. Therefore, if e(R,ξ) Y ( ) = −1 ◦ e(P,μ) Z , then there is an order-isomorphism f e(P,μ) e(R,ξ) from Y to Z,and −1 so f : P → R is a lattice isomorphism. Moreover, since e(P,μ) ◦ f = e(R,ξ),we −1 = −1 ◦ −1 φ have e(R,ξ) f e(P,μ). Composing both sides with , and noting from above that ξ = −1 ◦ φ = −1 ◦ φ ξ = −1 ◦ ( ,ξ) e(R,ξ) and μ e(P,μ) , we get f μ. Thus, R is equivalent to Author's personal copy

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(P, μ). Consequently, there is a 1-1 correspondence between inequivalent Priestley ring representations of L and dense subspaces of X.

Bauer’s theorem is a particular case of Theorem 2.4:

Corollary 2.5 (Bauer) Let B be a Boolean algebra and X be its Stone space. Then there is a 1-1 correspondence between inequivalent reduced f ield representations of B and dense subspaces of X.

Proof View every set Y as a poset with the trivial partial order =.Thenafield representation F over Y of a Boolean algebra B is reduced iff F is a Priestley ring representation of B. The result then follows from Theorem 2.4.

Remark 2.6 Although a ring representation of a Boolean algebra is in fact a field representation, it does not mean that every (Priestley) ring over a set with trivial partial order is a (reduced) field. For instance, the ring of finite subsets of the natural numbers is a Priestley ring, which is not a field. In fact, that ring and the ring of cofinite subsets of the natural numbers form two non-isomorphic Priestley rings that generate the same field of sets, the field of all finite and cofinite subsets of the natural numbers.

3 Priestley Order-Compactifications and Order-Zero-Dimensional Spaces

Let X be an ordered topological space. For x ∈ X, we recall that N ⊆ X is a neighborhood of x if there exists an open subset U of X such that x ∈ U ⊆ N.

Definition 3.1 Let X be an ordered topological space.

(1) [13, p. 966] We call X order-Hausdorf f if for each x, y ∈ X with x ≤ y,there exists an upset neighborhood N of x and a downset neighborhood M of y such that N ∩ M = ∅. (2) [14, p. 54] We call X completely order-regular if: (i) For each x, y ∈ X with x ≤ y, there exists a continuous order-preserving f : X →[0, 1] such that f (x)> f (y), (ii) For each x ∈ X and each closed set F with x ∈/ F, there exist a continuous order-preserving f : X →[0, 1] and a continuous order-reversing g : X → [0, 1] such that f (x) = 1 = g(x) and F ⊆ f −1(0) ∪ g−1(0).

It is well known that the notion of an order-Hausdorff space generalizes that of a Hausdorff space to ordered topological spaces, and that X is order-Hausdorff iff ≤ is closed in X2 (see [14, Proposition 1] or [13, Theorem 2]). It is also clear that the notion of a completely order-regular space generalizes that of a completely regular space to ordered topological spaces. Author's personal copy

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Definition 3.2 [14, p. 103] Suppose X and Y are ordered topological spaces. We call apair(Y, e) an order-compactif ication of X if Y is compact order-Hausdorff and e : X → Y is an order-homeomorphism onto a dense subspace of Y.

If an ordered topological space X has an order-compactification, then it follows from [14, Theorem 7] that X is completely order-regular. Let X be completely order-regular, and let Y = (Y, eY ) and Z = (Z, eZ ) be two order-compactifications of X.WedefineY ≤ Z if there exists a continuous order-preserving f : Z → Y such that f ◦ eZ = eY . It is easy to verify that ≤ is reflexive and transitive. We say that Y is equivalent to Z if Y ≤ Z and Z ≤ Y. It follows from [14, pp. 103–104] that Y is equivalent to Z iff there is an order-homeomorphism f : Y → Z such that f ◦ eY = eZ .Then≤ induces a partial order on the equivalence classes of order- compactifications of X. One of the fundamental results of Nachbin states that there is an order-compactification n(X) = (n(X), en(X)) of X, which has the following universal mapping property: If f : X → Y is a continuous order-preserving map into a compact order-Hausdorff space Y, then there is a unique continuous order- preserving map nf : n(X) → Y with nf ◦ en(X) = f . As a consequence, the poset of inequivalent order-compactifications of X has n(X) as its largest element. We refer to n(X) as the Nachbin order-compactif ication of X. It generalizes the concept of the Stone–Cechˇ compactification to the setting of ordered topological spaces. We refer to [6, 10, 12, 14, 18] for different constructions of n(X), and point out that the standard construction of n(X) is an analogue of the standard construction of the Stone–Cechˇ compactification. It follows that every completely order-regular space has an order- compactification. Putting all this together gives us the following well-known result:

Theorem 3.3 An ordered topological space X has an order-compactif ication if f X is completely order-regular.

This theorem generalizes a similar theorem about completely regular spaces and compactifications to the setting of ordered topological spaces.

Definition 3.4 Let X and Y be ordered topological spaces. We say that Y is a Priestley order-compactif ication of X if Y is a Priestley space and there exists e : X → Y such that (Y, e) is an order-compactification of X.

Priestley order-compactifications generalize the concept of zero-dimensional compactifications. Let 2 denote the ordered topological space {0, 1} with its usual order and discrete topology. The next theorem gives several equivalent conditions for an ordered topological space to have a Priestley order-compactification. We point out that Condition (4) below is analogous with Condition (2) in Definition 3.1.

Theorem 3.5 Let X be an ordered topological space. Then the following conditions are equivalent:

(1) X has a Priestley order-compactif ication. (2) (i) CpUp(X) is a Priestley ring, and (ii) {U − V : U, V ∈ CpUp(X)} is a basis for the topology. Author's personal copy

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(3) (i) CpUp(X) is a Priestley ring, and (ii) for each x ∈ X and each closed set F with x ∈/ F, there exist a clopen upset U and a clopen downset V with x ∈/ U ∪ Vand F ⊆ U ∪ V. (4) (i) If x ≤ y, then there exists a continuous order-preserving f : X → 2 with f(x) = 1 and f (y) = 0,and(ii) for each x ∈ X and each closed set F with x ∈/ F, there exist a continuous order-preserving f : X → 2 and a continuous order- reversing g : X → 2 such that f(x) = 1 = g(x) and F ⊆ f −1(0) ∪ g−1(0).

Proof (1)⇒(2) Let (Y, e) be a Priestley order-compactification of X.ThenY satisfies the Priestley separation axiom. But then so does X. Consequently, CpUp(X) is a Priestley ring. Now suppose U is an open set in X.Thenthereexists an open set V in Y such that U = e−1(V).SinceY is a Priestley space, {A −B : A, B ∈ CpUp(Y)} is a basis for the topology on Y. Therefore, = ( − ) { , } ⊆ ( ) V i∈I Ai Bi with Ai Bi i∈I CpUp Y .So −1 −1 −1 −1 U = e (V) = e (Ai − Bi) = (e (Ai) − e (Bi)). i∈I i∈I

−1 −1 But e (Ai), e (Bi) ∈ CpUp(X), implying that {C − D : C, D ∈ CpUp(X)} is a basis for the topology on X. (2)⇒(3) Suppose x ∈ X, F ⊆ X is closed, and x ∈/ F.ThenFc is an open neighborhood of x. Therefore, there exist U, W ∈ CpUp(X) such that x ∈ W − U ⊆ Fc.ButthenU is a clopen upset, V = Wc is a clopen downset, x ∈/ U ∪ V and F ⊆ U ∪ V. (3)⇒(4) Suppose x ≤ y.SinceCpUp(X) is a Priestley ring, there exists U ∈ CpUp(X) such that x ∈ U and y ∈/ U.Define f : X → 2 by 1, if z ∈ U f (z) = 0, otherwise. Then f is continuous, order-preserving, f(x) = 1,and f (y) = 0. Next, suppose x ∈ X, F ⊆ X is closed, and x ∈/ F. Then there exist a clopen upset U and a clopen downset V with x ∈/ U ∪ V and F ⊆ U ∪ V.Define f : X → 2 and g : X → 2 as follows: 1, if z ∈/ V 1, if z ∈/ U f (z) = and g(z) = 0, otherwise 0, otherwise. Clearly both f and g are continuous. Also, since Vc is an upset, f is order- preserving, and since U c is a downset, g is order-reversing. Finally, f (x) = 1 = g(x) and F ⊆ f −1(0) ∪ g−1(0). (4)⇒(1) To build a Priestley order-compactification of X we mimic Nachbin’s construction, but replace the unit interval with 2.LetJ be the set of all continuous order-preserving maps f : X → 2 and let P = 2J.Since2 is a Priestley space and a product of Priestley spaces is a Priestley space, we have that P is a Priestley space. Define e : X → P by e(x)( f ) = f (x) for f ∈ J,andletY = e(X).SinceY is a closed subspace of P and a closed subspace of a Priestley space is a Priestley space, we have that Y is also a Priestley space. We claim that (Y, e) is a Priestley order-compactification Author's personal copy

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of X. For this it is sufficient to show that e is an order-homeomorphism from X onto e(X). For g ∈ J and ∈ 2,setMg, ={α ∈ P : α(g) = }. Then {Mg, : g ∈ J, ∈ 2} is the usual subbasis for the product topology on P. (Note that Mg,0 and Mg,1 are set-theoretic complements, Mg,1 is an upset, and Mg,0 is a downset.) We see that e is continuous since −1 −1 e (Mg, ) = g () is open in X for each g,. Next, to see e is open, let U be open in X.Takex ∈ U. By (4)(ii), there exist f ∈ J and a continuous order-reversing g : X → 2 such that f (x) = 1 = g(x) and U c ⊆ f −1(0) ∪ g−1(0). Therefore, f, 1 − g ∈ J, f (x) = 1, (1 − g)(x) = 0, −1 −1 and f (1) ∩ (1 − g) (0) ⊆ U. Consider the open set (M f,1 ∩ M1−g,0) ∩ e(X) in e(X). We have e(x) ∈ M f,1 ∩ M1−g,0. Moreover, for each y ∈ X, −1 if e(y) ∈ M f,1 ∩ M1−g,0,then f(y) = 1 and (1 − g)(y) = 0,soy ∈ f (1) ∩ −1 (1 − g) (0) ⊆ U,andsoe(y) ∈ e(U). Therefore, (M f,1 ∩ M1−g,0) ∩ e(X) is an open neighborhood of e(x) in e(X) contained in e(U). Thus, e is open. It is clear that e is order-preserving. Let x, y ∈ X with e(x) ≤ e(y).Then f(x) ≤ f (y) for each f ∈ J.Ifx ≤ y, then, by (4)(i), there is f ∈ J with f(x) = 1 and f (y) = 0, a contradiction. Therefore, x ≤ y,andsoe is an order-homeomorphism from X onto e(X), which is a dense subspace of Y. Thus, Y is an order-compactification of X,andasY is a Priestley space, Y is a Priestley order-compactification of X.

Remark 3.6 The construction of the Priestley order-compactification (Y, e) of X in the proof above is similar to constructions of both the Nachbin order- compactification and of the Stone–Cechˇ compactification. The space Y is, in fact, the largest Priestley order-compactification of X.Toseethis,letZ be a Priestley space and let f : X → Z be continuous and order-preserving. It is sufficient to show that there exists a continuous order-preserving g : Y → Z such that g ◦ e = f.Let J (resp. K) be the set of all continuous order-preserving maps from X to 2 (resp. Z to 2). The map f induces a continuous order-preserving map ϕ : 2J → 2K, given by ϕ(α)(σ) = α(σ ◦ f) for each α ∈ 2J and σ ∈ K. We recall e : X → 2J and set e : Z → 2K be the analogous map. For x ∈ X we have ϕ(e(x)) = e( f (x)). Therefore, ϕ(e(X)) ⊆ e(Z).SinceZ is compact, e(Z) = e(Z). Thus,

ϕ(Y) = ϕ(e(X)) ⊆ ϕ(e(X)) ⊆ e(Z) = e(Z).

Consequently, g = (e)−1 ◦ ϕ is a continuous order-preserving map Y → Z with g ◦ e = f . In the particular case when the order of X is trivial, this construction produces the largest zero-dimensional (i.e., Stone) compactification of X, first constructed by Banaschewski [1]. In Theorem 5.2 (see also Remark 5.3) we give another construction of the largest Priestley order-compactification of X, and describe the structure of Priestley order-compactifications of X.

Ordered topological spaces satisfying the Priestley separation axiom were called totally order-disconnected in [5, 16]. The concept of a totally order-disconnected ordered topological space generalizes that of a totally disconnected topological space. We recall that a space X is totally disconnected if each pair of distinct points can be separated by a clopen set. These spaces are also called totally separated in the literature. It is known that every zero-dimensional (Hausdorff) space is totally disconnected, but not conversely. Similarly, we have that every ordered topological Author's personal copy

Order (2011) 28:399–413 407 space satisfying one of the equivalent conditions of Theorem 3.5 is totally order- disconnected, but not conversely. (Of course, if X is compact, then the two notions are equivalent to each other and to that of a Priestley space.) This motivates the following definition.

Definition 3.7 We call an ordered topological space X order-zero-dimensional if (i) X satisfies the Priestley separation axiom and (ii) the set {U − V : U, V ∈ CpUp(X)} forms a basis for the topology.

This definition together with Theorem 3.5 immediately give us:

Corollary 3.8 An ordered topological space has a Priestley order-compactif ication if f it is order-zero-dimensional.

As a consequence, we obtain that each order-zero-dimensional space is completely order-regular.

Remark 3.9 The notion of an order-zero-dimensional space coincides with that of a zero-dimensional ordered topological space of Nailana [15, Section 1]. Let X = (X,τ,≤) be an ordered topological space, CpUp(X) be the set of clopen upsets of X,andCpDn(X) be the set of clopen downsets of X.By[15, Theorem 1.3], X is a zero-dimensional ordered space if (i) CpUp(X) ∪ CpDn(X) is a subbasis for τ and (ii) ≤ is the specialization order for the topology generated by CpUp(X). Since U ∈ CpUp(X) iff U c ∈ CpDn(X) and both CpUp(X) and CpDn(X) are closed under finite intersections, the condition that CpUp(X) ∪ CpDn(X) is a subbasis for τ is equivalent to the condition that {U − V : U, V ∈ CpUp(X)} is a basis for τ. Therefore, Condition (i) of Nailana’s definition is equivalent to Condition (ii) of Definition 3.7. In addition, that ≤ is the specialization order for the topology generated by CpUp(X) is equivalent to x ≤ y iff x ∈ U implies y ∈ U for each U ∈ CpUp(X). This last condition is obviously equivalent to CpUp(X) being a Priestley ring. Thus, Condition (ii) of Nailana’s definition is equivalent to Condition (i) of Definition 3.7, and so the two notions of an order-zero-dimensional space and a zero- dimensional ordered space coincide. To call such spaces zero-dimensional ordered spaces is a little ambiguous, because it could be understood as ordered topological spaces which are zero-dimensional. Therefore, we prefer to call them order-zero- dimensional spaces.

Remark 3.10 It is well known that the category of compact Hausdorff spaces and continuous maps is a reflective subcategory of the category of completely regular spaces and continuous maps, where the reflector functor associates with each completely regular space X its Stone–Cechˇ compactification β(X);alsothe category of Stone spaces and continuous maps is a reflective subcategory of the category of zero-dimensional (Hausdorff) spaces and continuous maps, where the reflector associates with each zero-dimensional (Hausdorff) space its largest zero-dimensional compactification. Similarly, the category of compact order- Hausdorff spaces and continuous order-preserving maps is a reflective subcategory of the category of completely order-regular spaces and continuous order-preserving maps, where the reflector functor associates with each completely order-regular Author's personal copy

408 Order (2011) 28:399–413 space X its Nachbin order-compactification n(X); also the category of Priestley spaces and continuous order-preserving maps is a reflective subcategory of the category of order-zero-dimensional spaces and continuous order-preserving maps, where the reflector associates with each order-zero-dimensional space its largest Priestley order-compactification. More generally, let C be a category possessing all inverse limits. If R is a subcategory of C closed under inverse limits in C and containing a co-generator of C,thenR is a reflective subcategory of C. In the case of compact Hausdorff spaces, such a co-generator is the space [0, 1], in the case of Stone spaces it is the discrete space {0, 1}, in the case of compact order-Hausdorff spaces it is the space [0, 1] with its natural order, and in the case of Priestley spaces it is the discrete space {0, 1} with its natural order.

4 Strongly Order-Zero-Dimensional Spaces

Let X be an order-zero-dimensional space. Since X is completely order-regular, the Nachbin order-compactification n(X) of X exists, but n(X) may not be a Priestley space: Dowker’s example (see, e.g., [9, Example 6.2.20]) shows that there is a zero- dimensional normal space X such that the Stone–Cechˇ compactification β(X) of X is not zero-dimensional. If we view X as an ordered topological space (with the trivial order), then X is completely regular, hence completely order-regular, and n(X) = β(X).Butn(X) is not a Priestley space because β(X) is not zero-dimensional. It is well known (see, e.g., [9, Theorem 6.2.12]) that the Stone–Cechˇ compactification of a completely regular space X is zero-dimensional iff X is strongly zero-dimensional. We will generalize this result to the setting of ordered topological spaces.

Definition 4.1 Let X be an ordered topological space, A be an upset of X, B be a downset of X,andA ∩ B = ∅. 1. We say that A and B are completely order-separated if there exists a continuous order-preserving f : X →[0, 1] such that A ⊆ f −1(1) and B ⊆ f −1(0). 2. We say that X satisfies the strong Priestley separation axiom if whenever A and B are completely order-separated, then there exists a clopen upset U with A ⊆ U and B ⊆ U c.

Proposition 4.2 Let X be a completely order-regular space. If X satisf ies the strong Priestley separation axiom, then X satisf ies the Priestley separation axiom.

Proof Let X be completely order-regular, x, y ∈ X,andx ≤ y.Thenthereexists a continuous order-preserving f : X →[0, 1] such that f(x)> f (y).Defineg : [0, 1]→[f(y), f (x)] by ⎧ ⎨ f(x), if f (x) ≤ r ( ) = , ( )< < ( ) g r ⎩ r if f y r f x f(y), if r ≤ f (y). It is easy to verify that g is continuous and order-preserving. Then g ◦ f : X → [ f(y), f (x)] is continuous and order-preserving, and as [ f(y), f (x)] is order- homeomorphic to [0, 1], we obtain that there exists a continuous order-preserving Author's personal copy

Order (2011) 28:399–413 409 h : X →[0, 1] such that x ∈ h−1(1) and y ∈ h−1(0). Thus, ↑x and ↓y are completely order-separated, and by the strong Priestley separation axiom, there exists a clopen upset U of X with ↑x ⊆ U and ↓y ⊆ U c. Therefore, x ∈ U and y ∈/ U,andsoX satisfies the Priestley separation axiom.

The converse of Proposition 4.2 is not necessarily true as we will see shortly.

Definition 4.3 Let X be a completely order-regular space. We say that X is strongly order-zero-dimensional if X satisfies the strong Priestley separation axiom.

Proposition 4.4 For a completely order-regular space X, the Nachbin order-compactif ication n(X) is a Priestley space if f X is strongly order-zero-dimensional.

Proof Suppose n(X) is a Priestley space. We identify X with a dense subspace of n(X).LetA be an upset of X and B a downset of X which are completely order- separated. Then there exists a continuous order-preserving f : X →[0, 1] such that A ⊆ f −1(1) and B ⊆ f −1(0). By the universal mapping property for the Nachbin order-compactification, there is a unique continuous order-preserving nf : n(X) → −1 −1 [0, 1] such that (nf)|X = f . We have (nf) (1) is a closed upset of n(X), (nf) (0) is a closed downset of n(X),and(nf)−1(1) ∩ (nf)−1(0) = ∅.Sincen(X) is a Priestley space, there is a clopen upset V of n(X) with (nf)−1(1) ⊆ V and (nf)−1(0) ⊆ n(X) − V.SetU = X ∩ V.ThenU is a clopen upset of X with A ⊆ U and B ⊆ X − U. Therefore, X is strongly order-zero-dimensional. For the converse, suppose X is strongly order-zero-dimensional. Let x, y ∈ n(X) with x ≤ y.Sincen(X) is completely order-regular, there is a continuous order-preserving g : n(X) →[0, 1] with ↑x ⊆ g−1(1) and ↓y ⊆ g−1(0).DefineA = −1( 2 , ]∩ = −1[ , 1 ) ∩ , g 3 1 X and B g 0 3 X.ThenA B are open in X, A is an upset, and B is a downset. Moreover, ↑x ∩ X ⊆ A and ↓y ∩ X ⊆ B.SinceA and B are completely order-separated, there is a clopen upset U of X with A ⊆ U and B ⊆ − ( ) −1( 2 , ] ( ) = −1( 2 , ]∩ X U. Because X is dense in n X , g 3 1 is open in n X ,andA g 3 1 = −1( 2 , ] −1( 2 , ]⊆ ∈ ∈ X, we have that A g 3 1 . Thus, g 3 1 U,andsox U.Ify U,then −1[ , 1 ) ∩ = ∅ ∩ = ∅ ∈ g 0 3 U . Therefore, B U , which is a contradiction. Thus, y U. We show that U is a clopen upset of n(X).Leth : X → 2 be the characteristic function of U.Thenh is continuous and order-preserving. Let nh : n(X) → 2 be the extension of h.ThenU ⊆ (nh)−1(1),andsoU ⊆ (nh)−1(1).Also,X − U ⊆ (nh)−1(0), and so X − U ⊆ (nh)−1(0). Therefore, U ∩ X − U = ∅. In addition, U ∪ X − U = U ∪ (X − U) = X = n(X). Thus, U = (nh)−1(1),andsoU is a clopen upset of n(X). Consequently, n(X) is a Priestley space.

Remark 4.5 By putting Proposition 4.4 and Dowker’s example together, we obtain that the converse of Proposition 4.2 is not true.

Remark 4.6 Proposition 4.4 is also proved by Nailana [15, Section 2] using a different technique. We note that strongly order-zero-dimensional spaces Nailana calls strongly zero-dimensional ordered spaces. Because of the same reason as in Remark 3.9, we prefer to call such spaces strongly order-zero-dimensional. Author's personal copy

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Since a poset with the discrete topology is automatically strongly order-zero- dimensional, the next corollary, which can be found in [4, Proposition 3.4], is an immediate consequence of Proposition 4.4.

Corollary 4.7 If X is a discretely topologized poset, then n(X) is a Priestley space.

5 Priestley Bases and a Generalization of Dwinger’s Theorem

Definition 5.1 Let X be an ordered topological space. We call a Priestley ring R of clopen upsets of X a Priestley basis if {U − V : U, V ∈ R} is a basis for the topology on X.

Thus, R is a Priestley basis of X if R is a Priestley ring that is a bounded distributive sublattice of CpUp(X),and{U − V : U, V ∈ R} is a basis for the topology on X.Let(PB(X), ⊆) denote the poset of Priestley bases of X,andlet(POC(X), ≤) denote the poset of inequivalent Priestley order-compactifications of X. We note that if X is not order-zero-dimensional, then both POC(X) and PB(X) are empty.

Theorem 5.2 (POC(X), ≤) is isomorphic to (PB(X), ⊆).

Proof We define F : POC(X) → PB(X) as follows. If Y is a Priestley order- compactification of X and e : X → e(X) ⊆ Y is an order-homeomorphism from X −1 onto a dense subspace of Y,thenweletF(Y) = RY = e (U) : U ∈ CpUp(Y) . Since e : X → Y is continuous and order-preserving, it is clear that RY is a bounded distributive sublattice of CpUp(X). Suppose x, y ∈ X with x ≤ y.Thene(x) ≤ e(y). Since Y is a Priestley space, there is U ∈ CpUp(Y) with e(x) ∈ U and e(y)/∈ U. Thus, −1 −1 −1 x ∈ e (U) and y ∈/ e (U). Because e (U) ∈ RY , we see that RY is a Priestley ring. Finally, since Y is a Priestley space, {U − V : U, V ∈ CpUp(Y)} is a basis for the topology on Y.Butthen,asX is order-homeomorphic to e(X), {e−1(U) − e−1(V) : U, V ∈ CpUp(Y)}={A − B : A, B ∈ RY } is a basis for the topology on X. Thus, RY is a Priestley basis of X,andsoF is well-defined. We show that F is order-preserving. Let Y, Z ∈ POC(X) with Y ≤ Z. Let also eY : X → Y and eZ : X → Z be the corresponding order-homeomorphisms from X into Y and Z, respectively. From Y ≤ Z it follows that there is a continuous order- preserving f : Z → Y such that f ◦ eZ = eY . It is sufficient to show that RY ⊆ RZ . ∈ ∈ ( ) = −1( ) = Let A RY . Then there exists U CpUp Y such that A eY U .ButthenA −1 −1( ) −1( ) ∈ ( ) ∈ eZ f U and f U CpUp Z . Thus, A RZ . Conversely, we define G : PB(X) → POC(X) as follows. If R is a Priestley basis of X,weletG(R) = YR be the Priestley space of R.Definee : X → YR by e(x) = {U ∈ R : x ∈ U}. It is clear that e(x) is a prime filter of R, thus e is well- defined. It is also obvious that e is order-preserving, and since R is a Priestley ring, e : X → e(X) is an order-isomorphism. We show that e is a homeomorphism. First observe that for each U ∈ R we have e−1(φ(U)) = U. Indeed, x ∈ e−1(φ(U)) iff e(x) ∈ φ(U) iff U ∈ e(x) iff x ∈ U. It follows that e−1(φ(U)c) = U c. Therefore, e is continuous because {φ(U) − φ(V) : U, V ∈ R} is a basis for the topology on Y. Moreover, e(U) = e(X) ∩ φ(U) for each U ∈ R,sincee(x) ∈ e(U) iff x ∈ U iff U ∈ e(x) iff e(x) ∈ φ(U). Similarly, e(U c) = e(X) − φ(U). Thus, e is an order- Author's personal copy

Order (2011) 28:399–413 411 homeomorphism from X onto e(X). Finally, e(X) is dense in Y, since if φ(U) − φ(V) is a nonempty basic clopen subset of Y,thenU ⊆ V,andso

e(X) ∩ (φ(U) − φ(V)) = (e(X) ∩ φ(U)) ∩ (e(X) − φ(V)) = e(U) ∩ e(Vc) = e(U − V) = ∅ because U − V = ∅. Thus, YR is a Priestley order-compactification of X,andsoG is well-defined. We show that G is order-preserving. Let R, P ∈ PB(X) with R ⊆ P.Itis sufficient to show that YR ≤ YP. By the Priestley duality, f : YP → YR, given by f (∇) =∇∩R, for each prime filter ∇ of P, is continuous and order-preserving. Let eP : X → YP and eR : X → YR be the corresponding order-homeomorphisms from X into YP and YR, respectively. Then f (eP(x)) = eP(x) ∩ R = eR(x). Thus, YR ≤ YP. We finish the proof by showing that F and G are inverses of each other. Let Y ∈ POC(X) and eY : X → Y be the embedding. Then, by Theorem 2.4, h = −1 : ( ) → ( ) eY CpUp Y RY is an isomorphism. If Z is the Priestley space of CpUp Y , = −1 : → then Priestley duality yields the order-homeomorphism f h YRY Z.Let e : X → Y be the embedding and : Y → Z be the canonical isomorphism. YRY RY We show that f ◦ e = ◦ e . For each x ∈ X and U ∈ CpUp(Y) we have U ∈ YRY Y − − f (e (x)) iff e 1(U) ∈ e (x) ={V ∈ R : x ∈ V} iff x ∈ e 1(U) iff e (x) ∈ U YRY Y YRY Y Y Y iff U ∈ (e (x)). Therefore, f ◦ e = ◦ e .Letg = −1 ◦ f.Theng : Y → Y Y YRY Y RY is an order-homeomorphism such that g ◦ e = e . Thus, Y and G(F(Y)) are YRY Y equivalent. Now let R ∈ PB(X),andletYR be the corresponding Priestley order- compactification with the order-homeomorphism e from X onto a dense subspace −1 of YR. Since for each U ∈ R we have U = e (φ(U)), it follows that R = F(G(R)). Thus, (PB(X), ⊆) is isomorphic to (POC(X), ≤).

Remark 5.3 Let X be order-zero-dimensional. By Corollary 3.8, X has a Priestley order-compactification, and so by Theorem 3.5, CpUp(X) is a Priestley basis of X, which is clearly the largest Priestley basis. Thus, by Theorem 5.2, CpUp(X) gives rise to the largest Priestley order-compactification of X. We note that this largest Priestley order-compactification is equivalent to the Priestley order-compactification constructed in Theorem 3.5 (see also Remark 3.6). However, as follows from Proposition 4.4, this largest Priestley order-compactification is not necessarily the Nachbin order-compactification of X.

If X is a discretely topologized poset, then Up(X) is clearly the largest Priestley basis of X. Thus, Corollary 4.7 and Theorem 5.2 immediately give us:

Corollary 5.4 [4, Proposition 3.4] For a discretely topologized poset X, the Nachbin order-compactif ication n(X) of X is the Priestley order-compactif ication of X corresponding to Up(X).

For a topological space X,let(Z(X), ≤) denote the poset of inequivalent zero- dimensional compactifications of X. We recall that a Boolean basis of X is a basis of X that is a field of sets [8, p. 26]. Let (B(X), ⊆) denote the poset of Boolean bases of X. Of course, if X is not zero-dimensional, then both Z(X) and B(X) are empty. Author's personal copy

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If we view X as an ordered topological space, then we have that Z(X) ⊆ POC(X), and that B(X) ⊆ PB(X).Infact,Z(X) consists of exactly those Priestley order- compactifications that have the trivial order, and B(X) consists of exactly those Priestley bases that are fields of sets. Now, for a Priestley basis R of X, the Priestley order-compactification YR corresponding to R has the trivial order iff every prime filter of R is an ultrafilter, which in turn happens iff R is a field of sets. Therefore, Dwinger’s theorem [8, Theorem 13.1] is a corollary to Theorem 5.2.

Corollary 5.5 (Dwinger) Let X be a topological space. Then the poset (Z(X), ≤) is isomorphic to the poset (B(X), ⊆).

We point out that similar to Remark 5.3, where CpUp(X) is the largest element of PB(X),thefieldCp(X) of all clopen subsets of X is the largest element of B(X). Therefore, the largest zero-dimensional compactification of X can alternatively be constructed as the Stone space of Cp(X).

Acknowledgements We would like to thank Leo Esakia and the members of his group at the Razmadze Mathematical Institute for helpful discussions. Some of the main findings of this paper were announced at the international conference “Order, Algebra, and Logics”, Nashville, Tennessee, 2007.

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