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Order (2011) 28:577–592 DOI 10.1007/s11083-010-9193-x Order-Compactifications of Totally Ordered Spaces: Revisited Guram Bezhanishvili · Patrick J. Morandi Received: 9 November 2009 / Accepted: 27 December 2010 / Published online: 18 January 2011 © Springer Science+Business Media B.V. 2011 Abstract Order-compactifications of totally ordered spaces were described by Blatter (J Approx Theory 13:56–65, 1975) and by Kent and Richmond (J Math Math Sci 11(4):683–694, 1988). Their results generalize a similar characterization of order-compactifications of linearly ordered spaces, obtained independently by Fedorčuk (Soviet Math Dokl 7:1011–1014, 1966; Sib Math J 10:124–132, 1969) and Kaufman (Colloq Math 17:35–39, 1967). In this note we give a simple characterization of the topology of a totally ordered space, as well as give a new simplified proof of the main results of Blatter (J Approx Theory 13:56–65, 1975) and Kent and Richmond (J Math Math Sci 11(4):683–694, 1988). Our main tool will be an order-topological modification of the Dedekind-MacNeille completion. In addition, for a zero-dimensional totally ordered space X, we determine which order-compactifications of X are Priestley order-compactifications. Keywords Ordered topological space · Order-compactification · Totally ordered space · Interval topology · Linearly ordered space · Dedekind-MacNeille completion Mathematics Subject Classifications (2010) Primary 54F05; Secondary 54D35 · 06A05 1 Totally Ordered Spaces The study of ordered topological spaces was initiated by Nachbin in the 1940s. Among other things, Nachbin generalized the concept of completely regular space 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] 578 Order (2011) 28:577–592 (resp. compactification) to that of completely order-regular space (resp. ordercompactification), and proved that an ordered topological space has an ordercompactification iff it is completely order-regular. Nachbin also showed that each completely order-regular space X has a greatest order-compactification, which we call the Nachbin order-compactif ication of X, and denote by n(X). We recall that an ordered topological space X is order-Hausdorf f if for each x, y ∈ X with x ≤ y, there exists an upset neighborhood U of x and a downset neighborhood V of y such that U ∩ V = ∅. (Equivalently, the order ≤ is closed in X 2 .) We also recall that X is completely order-regular if the following two conditions are satisfied: (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 subset F of X 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). Obviously each completely order-regular space is order-Hausdorff; however, the converse is not true in general. Let X be an ordered topological space. We recall that a pair (Y, f ) is an ordercompactif ication of X if Y is compact order-Hausdorff and f : X → Y is an orderhomeomorphism onto a dense subspace of Y. The poset of (inequivalent) order-compactifications of a completely order-regular space has been investigated extensively; see, e.g., [2–8, 12–18, 21–23]. It turned out to have a rather complicated structure. However, it becomes tractable in the case of totally ordered spaces. Definition 1.1 (see [4, Sec. 2] or [12, Sec. 1]) Let X be an ordered topological space. 1. The order ≤ on X is total if x ≤ y or y ≤ x for each x, y ∈ X. 2. A subset A of X is convex if x, y ∈ A and x ≤ z ≤ y imply z ∈ A. 3. The topology on X is locally convex if it is generated by open convex subsets of X. 4. X is a totally ordered space if X is order-Hausdorff, the order on X is total, and the topology on X is locally convex. Remark 1.2 Totally ordered spaces coincide with generalized ordered spaces or GOspaces. For a detailed account of GO-spaces we refer to an excellent survey by Bennett and Lutzer [1]. We recall that the interval topology τ≤ on a totally ordered set X is the topology whose closed sets are generated by {↑x : x ∈ X} ∪ {↓x : x ∈ X}, and so the open sets are generated by open intervals (x, y), where x < y (and we allow x = −∞ and y = ∞). We also recall that a linearly ordered space, or a LOTS, is a totally ordered set equipped with the interval topology. Typical examples of totally ordered spaces are linearly ordered spaces. However, not every totally ordered space is a linearly ordered space, as we see from the following simple example. Example 1.3 Let X = [0, 1) ∪ {2}. If we view X as a subspace of the real line R and the order on X as the restriction of the standard order on R, then X is obviously a Order (2011) 28:577–592 579 totally ordered space. However, since 2 is not a limit point of [0, 1), but 2 is a limit point of [0, 1) in the interval topology on X, we obtain that X is not a linearly ordered space. Nevertheless, as was shown by Čech (see, e.g., [1, p. 85]), each totally ordered space is order-homeomorphic to a subspace of a linearly ordered space. We give a simple description of the topology of a totally ordered space. Note that as each totally ordered space (X, τ, ≤) is order-Hausdorff, the interval topology τ≤ is contained in τ . Lemma 1.4 Let X be a totally ordered space and let U be open in X. If U has a minimum (resp. maximum) point x, then ↑x (resp. ↓x) is clopen. Proof Let x be the minimum of U and set C = X − U, a closed set. Then C∩ ↓x is closed and equal to ↓x − {x} = X− ↑x. Thus, ↑x is open. Since it is also closed, ↑x is clopen. The argument when U has a maximum is similar. For a totally ordered space X, let COD(X) denote the set of clopen downsets and COU (X) denote the set of clopen upsets of X. Theorem 1.5 Let (X, τ, ≤) be a totally ordered space. Then τ≤ ∪ COD(X) ∪ COU (X) is a subbasis of τ . Proof Let θ be the topology generated by τ≤ ∪ COD(X) ∪ COU (X). Since τ≤ ⊆ τ , it is clear that θ ⊆ τ . Let U ∈ τ be convex and let x ∈ U. If there are y, z ∈ U with y < x < z, then (y, z) is a θ-neighborhood of x inside U. If x is the least element of U but there is z ∈ U with x < z, then by Lemma 1.4, ↑x ∈ COU (X) and ↑x ∩ (−∞, z) is open in θ, contains x, and is contained in U. Similarly, if x is the greatest element of U but there is y ∈ U with y < x, then we can produce V ∈ θ containing x and contained in U. Finally, if x is both the greatest and least element of U, then U = {x}, both ↑x and ↓x are clopen, and so U = ↑x ∩ ↓x is open in θ . Thus, in all cases we have found an element of θ containing x and contained in U. Therefore, U ∈ θ. Since τ has a basis of convex open sets, this proves θ = τ . 2 Order-Compactifications of Totally Ordered Spaces As follows from [4, Prop. 2.4] (see also [12, p. 689]), an ordered topological space X whose order is total is a totally ordered space iff X is completely order-regular. Therefore, by Nachbin’s Theorem, X has an order-compactification. The structure of all order-compactifications of a totally ordered space X was described by Blatter [4] and Kent and Richmond [12]. This generalizes a similar characterization of all order-compactifications of a linearly ordered space by Fedorčuk [9, 10] and Kaufman [11]. The main goal of the present paper is to give a much simplified description of all order-compactifications of X. At the end of the paper we also describe the structure of Priestley order-compactifications of a zero-dimensional totally ordered space. Our main tool will be an order-topological modification of the DedekindMacNeille completion. 580 Order (2011) 28:577–592 Let (X, ≤) be a totally ordered set and let A ⊆ X. We denote by Au the set of upper bounds of A, and by Al the set of lower bounds of A. It is easy to verify that A ⊆ Aul and A ⊆ Alu . We recall that a downset D of X is normal if D = Dul . Note that if X has a least element 0, then {0} is the least normal downset of X; and if X does not have a least element, then ∅ is the least normal downset of X. It is well known that the set Y of normal downsets of X is a complete totally ordered set, that X embeds in Y by the mapping x →↓x, and that the image of X is both join-dense and meet-dense in Y. Consequently, Y is the Dedekind-MacNeille completion DM(X) of X. We view X and DM(X) as linearly ordered spaces. Since DM(X) is a complete totally ordered set, the interval topology on DM(X) is compact and order-Hausdorff (see, e.g., [4, Sec. 2] or [12, Sec. 1]). Moreover, the map x →↓x is a topological embedding, and the image of X is topologically dense in DM(X). Therefore, the Dedekind-MacNeille completion DM(X) of X is an ordercompactification of X. In fact, DM(X) is the smallest order-compactification of X [11, p. 36]. On the other hand, if X is a totally ordered space, then the Dedekind-MacNeille completion of X may not be an order-compactification of X, as we see in the next example. Example 2.1 Let X = [0, 1) ∪ {2} be the totally ordered space of Example 1.3, where the order and subspace topology on X are inherited from R. Since as a totally ordered set X is isomorphic to [0, 1], the total order on X is complete, and so DM(X) is isomorphic to X. On the other hand, as a totally ordered space X is not compact. In fact, the topology on X is not the interval topology on X, as we saw in Example 1.3. Thus, in order to produce order-compactifications of X, we need to modify the construction of the Dedekind-MacNeille completion of X. Let X be a totally ordered space. We recall that a downset D of X is principal if D = ↓x for some x ∈ X; otherwise D is non-principal. Similarly, an upset F of X is principal if F = ↑x for some x ∈ X; otherwise F is non-principal. For each A ⊆ X, we point out that Au = {↑a : a ∈ A} and Al = {↓a : a ∈ A}. Therefore, Au is a closed upset and Al is a closed downset of X. This implies that each normal downset is a closed downset of X. However, the converse need not hold. Example 2.2 Let X = [0, 1) ∪ {2} be the totally ordered space of Example 1.3. In Example 2.1 we saw that DM(X) is isomorphic to X, hence every normal downset of X is principal. In particular, D = [0, 1) is a closed downset of X which is not normal. It is easy to see that D is the only nonempty closed downset of X which is not normal. As we already saw, X is not compact. In fact, [0, 1] ∪ {2} is an order-compactification of X, which can be obtained as the space of all nonempty closed downsets of X. This example indicates that in order to build order-compactifications of a totally ordered space X, it is reasonable to work with closed downsets instead of normal downsets of X. Of course, if X has a least element, then we need to work with nonempty closed downsets instead of all closed downsets. However, the next example shows that this idea also does not work in general. Example 2.3 Let X = [0, 1) ∪ (2, 3], with order and topology inherited from R. Clearly X is a totally ordered space and D = [0, 1) is the only nonempty nonprincipal closed downset of X. Therefore, the space of nonempty closed downsets Order (2011) 28:577–592 581 of X is isomorphic to [0, 1] ∪ (2, 3], which is not compact (in particular, it is not a linearly ordered space). On the other hand, it is obvious that Y = [0, 1] ∪ [2, 3] is an order-compactification of X, which (up to order-homeomorphism) can be obtained by adding one more point to the space of nonempty closed downsets of X. If we do that, then in the obtained space, we have added a point which is the immediate successor of D, and this point is the infimum of X − D. This example indicates that it is not enough to work just with (nonempty) closed downsets; we may need to add the supremum of a closed downset D and the infimum of its complement X − D. Note that if D is a nonempty closed non-principal downset of X such that F = X − D is a nonempty closed non-principal upset, then the pair (D, F) is a typical example of a non-trivial Dedekind cut. If X is a totally ordered set, then we only need to add D as a new point of the Dedekind-MacNeille completion of X. However, if X is a totally ordered space, then in order to be able to produce an order-compactification of X, we may need to add two points corresponding to this cut. We are able to do this by embedding X inside a sufficiently large complete totally ordered set, as we now do. Definition 2.4 Let X be a totally ordered space. 1. Let CD(X) denote the set of closed downsets of X. 2. Let CD0 (X) = CD(X) − {∅} denote the set of nonempty closed downsets of X. 3. Let CD1 (X) denote the set of closed downsets of X whose complement is a nonempty closed non-principal upset. Note that X ∈ CD1 (X) and if X does not have a least element, then ∅ ∈ CD1 (X). It is also clear that CD1 (X) ⊆ COD(X). Definition 2.5 Let X be a totally ordered space. = (CD0 (X) × {0}) ∪ (CD1 × {1}). 1. Let X by (D, ) ≤ (E, δ) iff D ⊂ E or D = E and ≤ δ. 2. Define ≤ on X We also let Z = CD(X) × {0, 1} and order Z lexicographically; that is, (D, ) ≤ (E, δ) iff D ⊂ E or D = E and ≤ δ. Since CD(X) is totally ordered by the inclusion order, (Z , ≤) is a totally ordered set. It is an immediate consequence of Definition ≤) is a subposet of (Z , ≤). Therefore, ( X, ≤) is a totally ordered set. We 2.5 that ( X, ≤) is in fact complete. show that ( X, is a complete totally ordered set. Lemma 2.6 X is a totally ordered set. Let A ⊆ X. We show that the Proof We already saw that X supremum of A exists in X. Let B = {D ∈ CD0 (X) : (D, ) ∈ A}. Clearly E = B is / B . If E ∈ / B , then we claim that (E, 0) the sup of B in CD0 (X). We have E ∈ B or E ∈ To see this, for each (D, ) ∈ A, we have D ⊂ E. Therefore, is the sup of A in X. (D, ) ≤ (E, 0), and so (E, 0) is an upper bound of A. If (K, δ) is another upper bound of A, then D ⊆ K for each (D, ) ∈ A, so E ⊆ K, and so (E, 0) ≤ (K, δ). Thus, (E, 0) is the sup of A in X. 582 Order (2011) 28:577–592 Now let E ∈ B . First suppose that = 0 for each (D, ) ∈ A with D = E. Then again (E, 0) is the sup of A. Next suppose that there exists (D, 1) ∈ A such that D = E. Then (E, 1) ∈ A, and by the definition of B it is clear that (E, 1) is an upper bound of A. Thus, (E, 1) is the sup of A. Consequently, A has a sup in X. and if Moreover, if X has a least element x, then ({x}, 0) is the least element of X, It follows that X does not have a least element, then (∅, 1) is the least element of X. is a complete totally ordered set. X is a complete totally ordered set, the interval topology on X is compact Since X by e(x) = and order-Hausdorff (see, e.g., [4, Sec. 2] or [12, Sec. 1]). Define e : X → X is (equivalent to) (↓x, 0). It is clear that e is an order-embedding. We show that X the Nachbin order-compactification n(X) of X. def ined above is a topological embedding and Proposition 2.7 The map e : X → X Thus, ( X, e) is an order-compactif ication of X. e(X) is dense in X. is continuous. For this it is sufficient to show Proof First we show that e : X → X First, let α = (D, 0), where that e−1 (↑α) and e−1 (↓α) are closed in X for each α ∈ X. D ∈ CD0 (X). Then e−1 (↓(D, 0)) = {x ∈ X : e(x) ≤ (D, 0)} = {x ∈ X : x ∈ D} = D, which is closed. Also, e−1 (↑(D, 0)) = {x ∈ X : (D, 0) ≤ e(x)} = Du , which is closed by the comments in the paragraph before Example 2.2. Next let α = (D, 1), where D ∈ CD1 (X). Then X − D is a closed non-principal upset. Therefore, e−1 (↓(D, 1)) = D and e−1 (↑(D, 1)) = X − D, both of which are closed. Therefore, e is continuous. Next we show that e is a homeomorphism onto e(X). For this it is sufficient to prove that if C is closed in X, then e(C) is closed in e(X). By Theorem 1.5, we have four cases to consider. First, let C = ↓x. Then e(C) = {e(y) : y ≤ x} = ↓e(x) ∩ e(X). Similarly, if C =↑x, then e(C) = {e(y) : x ≤ y} = ↑e(x) ∩ e(X). Next, let C be a clopen downset of X. If C = ∅, then e(C) = ∅. Otherwise e(C) = {e(y) : y ∈ C} = ↓(C, 0) ∩ e(X). Finally, let C be a clopen upset of X. We have already considered the cases when C is principal or empty. Let C be non-principal and nonempty. Then (X − and C, 1) ∈ X e(C) ={e(y) : y ∈ C} = {e(y) : X − C ⊂ ↓y} = {e(y) : (X − C, 1) ≤ e(y)} = ↑(X − C, 1) ∩ e(X). In all cases, we see that e(C) is closed in e(X). Thus, e is a homeomorphism from X onto e(X). Let U be a nonempty open subset of X. Lastly, we show that e(X) is dense in X. We may assume that U = X− ↑α or U = X− ↓α for some α ∈ X. First, suppose that ↑(D, 0), then there must exist x, y ∈ D α = (D, 0), where D ∈ CD0 (X). If U = X− ↑(D, 0)). On the other with x < y. Therefore, ↓x ⊂ D, and so e(x) ∈ e(X) ∩ ( X− ↓(D, 0), then as (X, 0) is the largest element of X, we have D = X. hand, if U = X− − ↓(D, 0)). Next, suppose Take x ∈ X − D. Then D ⊂ ↓x, and so e(x) ∈ e(X) ∩ ( X ↑(D, 1), then D = ∅. Take x ∈ D. that α = (D, 1), where D ∈ CD1 (X). If U = X− − ↑(D, 1)). Finally, suppose U = X− Then e(x) < (D, 1), and so e(x) ∈ e(X) ∩ ( X ↓ ↓(D, 1). Clearly D = X. Pick x ∈ / D. Then (D, 1) < e(x), and so e(x) ∈ e(X) ∩ ( X− (D, 1)). Therefore, in all cases, we see that e(X) ∩ U is nonempty. Thus, e(X) is dense Consequently, ( X, e) is an order-compactification of X. in X. Order (2011) 28:577–592 583 Remark 2.8 It is easy to see that if D ∈ CD0 (X), then (D, 0) is the join of {e(x) : x ∈ D}. Moreover, if D ∈ CD1 (X), then (D, 1) is the meet of {e(x) : x ∈ X − D}. is either a meet or join of elements of e(X). Note that Therefore, everything in X may be a join of elements of e(X), but not a meet, and vice-versa. elements of X need not be DM(X) in general. This is another way to see that X Theorem 2.9 Let (Y, f ) be an order-compactif ication of X. Then there is a unique → Y with g ◦ e = f . Thus, X is equivalent to continuous order-preserving map g : X the Nachbin order-compactif ication n(X) of X. → Y by Proof Let (Y, f ) be an order-compactification of X. Define g : X sup f (D) if α = (D, 0) with D ∈ CD0 (X) g(α) = . inf f (X − D) if α = (D, 1) with D ∈ CD1 (X) Since Y is compact, hence complete as a poset, the map g is well defined. Note that if x ∈ X, then g(e(x)) = g(↓x, 0) = sup f (↓x) = f (x). Thus, g ◦ e = f . Since the and Y are both interval topologies, it is enough to prove that g topologies on X with α ≤ β. First let is order-preserving to see that it is continuous. Let α, β ∈ X α = (D, 0) with D ∈ CD0 (X). As α ≤ β, we have β = (E, ), where E is a closed downset of X and D ⊆ E. If = 0, then g(D, 0) = sup f (D) ≤ sup f (E) = g(E, 0); and if = 1, then g(D, 0) = sup f (D) ≤ inf f (X − E) = g(E, 1). Next let α = (D, 1), where D ∈ CD1 (X). Let β = (E, ). If D = E, then = 1, so α = β, and so g(α) = g(β). Otherwise, D ⊂ E. If = 0, then g(D, 1) = inf f (X − D) ≤ sup f (E) = g(E, 0); and if = 1, then g(D, 1) = inf f (X − D) ≤ inf f (X − E) = g(E, 1). Thus, g is order-preserving. Finally, if h is another continuous map to Y with h ◦ e = f , then g and h agree on the dense subset e(X) of X. Since from X Y is Hausdorff, this forces g = h. Remark 2.10 In [4] the Nachbin order-compactification of a totally ordered space X is constructed by means of all bounded continuous order-preserving functions f : X → R, and in [12] it is constructed by means of the Wallman order-compactification of X. We find our construction to be much simpler than the ones given in [4] and [12]. Let X be a totally ordered space and let (Y, f ) be an order-compactification of X. → Y such that g ◦ e = f . By Theorem 2.9, there exists a unique g : X → Y be the Lemma 2.11 Let (Y, f ) be an order-compactif ication of X and let g : X unique continuous order-preserving map with g ◦ e = f . 1. If D, E ∈ CD0 (X) with D = E, then g(D, 0) = g(E, 0). 2. If D ∈ CD1 (X) is a principal downset, then g(D, 0) < g(D, 1). Proof (1) It is sufficient to prove that if D ⊂ E, then g(D, 0) < g(E, 0). Let D ⊂ E. Then there is x ∈ E − D, and so D ⊂ ↓x ⊆ E. Since f is an embedding, f (D) is a closed downset of f (X), so f (D) = C ∩ f (X) for some closed subset C of Y. Note that since x is an upper bound of D, we have f (D) = (C ∩ ↓ f (x)) ∩ f (X). Let z = sup(C ∩ ↓ f (x)). Since C ∩ ↓ f (x) is closed in Y, we have z ∈ C ∩ 584 Order (2011) 28:577–592 ↓ f (x). Thus, z ≤ f (x). If z = f (x), then f (x) ∈ f (D), a contradiction since x ∈ D. Thus, z < f (x). Therefore, g(D, 0) = sup( f (D)) ≤ z < f (x) ≤ sup( f (E)) = g(E, 0). (2) Since D ∈ CD1 (X) is principal, there is x ∈ X with D =↓x. Therefore, g(D, 0) = g(e(x)) = f (x). Since D ∈ CD1 (X), we have X − D is a closed subset of X. Therefore, f (X − D) = C ∩ f (X) for some closed subset C of Y. Because x is a lower bound of X − D, we have f (X − D) = (C ∩ ↑ f (x)) ∩ f (X). Let z = inf(C ∩ ↑ f (x)). Then f (x) ≤ z and z ∈ C ∩ ↑ f (x). If z = f (x), then x ∈ X − D, a contradiction. Thus, f (x) < z. Therefore, g(D, 0) = f (x) < z ≤ inf( f (X − D)) = g(D, 1). that may be identified by g are (D, 0) and By Lemma 2.11, the only points of X (D, 1), where D ∈ CD1 (X) is non-principal. Definition 2.12 For a totally ordered space X, let CD2 (X) denote the set of nonempty closed non-principal downsets of X such that X − D is a nonempty closed non-principal upset. Clearly CD2 (X) ⊆ CD1 (X). In our terminology, the main result of [4] and [12] states that the poset of (inequivalent) order-compactifications of X is orderisomorphic to the powerset of CD2 (X). Our final goal in this section is to give a simplified proof of this result. Let OC(X) denote the poset of (inequivalent) order-compactifications of X. For (Y, f ), (Y , f ) ∈ OC(X), we recall that (Y , f ) ≤ (Y, f ) iff there exists a continuous order-preserving h : Y → Y such that h ◦ f = f . Let also P (CD2 (X)) denote the powerset of CD2 (X). We define ϕ : P (CD2 (X)) → OC(X) as follows. Let S ⊆ CD2 (X), and define ∼ S to be the equiva generated by the relations (D, 0) ∼ (D, 1) for each D ∈ lence relation on X / S. Thus, then if X denotes the diagonal of X, / S} ∪ {((D, 1), (D, 0)) : D ∈ / S}. ∼S = X ∪ {((D, 0), (D, 1)) : D ∈ From the definition of ∼ S it is clear that no two points in e(X) are equivalent, and S . Furthermore, it is clear that X/∼ S that we have an induced total ordering on X/∼ S is compact and order-Hausdorff with respect to is a complete poset. Thus, X/∼ → X/∼ S be the canonical projection. It is easy the interval topology. Let π S : X S is a continuous order-embedding. To see that it is to see that π S ◦ e : X → X/∼ a topological embedding, let C be a closed subset of X. Then e(C) is closed in Since [e(x)] = {e(x)} for e(X), so e(C) = F ∩ e(X) for some closed subset F of X. each x ∈ X, it follows that π S (e(C)) = π S (F) ∩ π S (e(X)), and so π S (e(C)) is closed S , π S ◦ e) is an order-compactification of X. We set in π S (e(X)). Therefore, ( X/∼ S , π S ◦ e). Then ϕ : P (CD2 (X)) → OC(X) is well defined. ϕ(S) = ( X/∼ Next, we define ψ : OC(X) → P (CD2 (X)) as follows. Let (Y, f ) be an order → Y be the unique continuous order-preserving compactification of X and let g : X map with g ◦ e = f . We let SY = {D ∈ CD2 (X) : g(D, 0) < g(D, 1)} = {D ∈ CD2 (X) : sup( f (D)) < inf( f (X − D))}, and set ψ(Y, f ) = SY . Then ψ : OC(X) → P (CD2 (X)) is well defined. Order (2011) 28:577–592 585 Theorem 2.13 The maps ϕ and ψ establish an order-isomorphism between (P (CD2 (X)), ⊆) and (OC(X), ≤). Proof It is sufficient to show that both ϕ and ψ are order-preserving, that ψ(ϕ(S)) = S for each S ⊆ CD2 (X), and that ϕ(ψ(Y, f )) is equivalent to (Y, f ) for each (Y, f ) ∈ OC(X). Let S, T ⊆ CD2 (X). If S ⊆ T, then it is easy to see that α ∼T β implies α ∼ S T → X/∼ S by h([α]T ) = [α] S . Clearly h is order-preserving and β. Define h : X/∼ S , π S ◦ e) ≤ ( X/∼ T , πT ◦ e), which means that h ◦ πT ◦ e = π S ◦ e. Therefore, ( X/∼ ϕ(S) ≤ ϕ(T). Thus, ϕ is order-preserving. Next, let (Y, f ), (Y , f ) be order-compactifications of X. If (Y , f ) ≤ (Y, f ), then there exists an order-preserving h : (Y, f ) → (Y , f ) such that h ◦ f = f . Let also → Y and g : X → Y be the unique order-preserving maps with g ◦ e = f g:X and g ◦ e = f , respectively. If D ∈ SY , then g (D, 0) < g (D, 1). Since h ◦ g = g , we have h(g(D, 0)) < h(g(D, 1)). As h is order-preserving and Y is totally ordered, this means that g(D, 0) < g(D, 1), and so D ∈ SY . Thus, SY ⊆ SY , so ψ(Y , f ) ⊆ ψ(Y, f ), and so ψ is order-preserving. Let S ⊆ CD2 (X). Then D ∈ / S iff (D, 0) ∼ S (D, 1), which is equivalent to π S (D, 0) = π S (D, 1). But this means that D ∈ / S X/∼ / S iff D ∈ / S . Therefore, D ∈ ψ(ϕ(S)). Thus, S = ψ(ϕ(S)) for each S ⊆ CD2 (X). → Y be the Finally, let (Y, f ) be an order-compactification of X and let g : X unique order-preserving map with g ◦ e = f . By the description of SY , it is clear that SY → Y, defined by h([D, ]) = g(D, ), is a bijection. From the definition of h : X/∼ h we get h ◦ π SY = g, and from this equation it follows that h is continuous since the SY is the weakest topology making π SY continuous. Thus, since the topology on X/∼ domain is compact and the target Hausdorff, h is a homeomorphism. Thus, (Y, f ) SY , π SY ◦ e) = ϕ(ψ(Y, f )). Consequently, ϕ and ψ establish an is equivalent to ( X/∼ order-isomorphism between (P (CD2 (X)), ⊆) and (OC(X), ≤). The next corollary is an immediate consequence of Theorem 2.13. Parts of it can be found in [4, Sec. 2] and [12, Sec. 2]. Corollary 2.14 Let X be a totally ordered space and let OC(X) be the poset of ordercompactif ications of X. 1. 2. 3. 4. 5. OC(X) is a complete and atomic Boolean algebra. OC(X) always has a least element. OC(X) is never countable. OC(X) is f inite if f CD2 (X) is f inite. X has a unique (up to equivalence) order-compactif ication if f CD2 (X) is empty. In particular, by Corollary 2.14(1), if X is a linearly ordered space, then OC(X) is a complete and atomic Boolean algebra, and so we arrive at the main result of [11, Thm. 8] and [10, Thm. 1]. Remark 2.15 Order-compactifications of sums and products of totally ordered spaces are investigated in [17, 18]. More generally, for results on the structure of posets of order-compactifications of completely order-regular spaces see [13– 15, 22, 23], as well as [7, 8, 21], where posets of order-compactifications of completely 586 Order (2011) 28:577–592 order-regular spaces are investigated by means of intervals in complete lattices of closed quasi-orders on compact Hausdorff spaces. 3 Priestley Order-Compactifications of Totally Ordered Spaces We recall that an ordered topological space X satisfies the Priestley separation axiom if for each x, y ∈ X, whenever x ≤ y, there exists a clopen upset U of X such that x ∈ U and y ∈ / U. Clearly each ordered topological space that satisfies the Priestley separation axiom is order-Hausdorff, but the converse is not true in general. We call an ordered topological space X a Priestley space if X is compact and satisfies the Priestley separation axiom. Clearly each Priestley space is compact and orderHausdorff, but not vice versa. Priestley spaces play an important role in the theory of distributive lattices as they serve as duals of bounded distributive lattices [20]. Definition 3.1 Let X be an ordered topological space. 1. ([2, Def. 3.7] and [19, Sec. 1]) We call X order-zero-dimensional if X satisfies the Priestley separation axiom and the set of clopen convex subsets of X forms a basis for the topology on X. 2. ([2, Def. 3.4]) We call an order-compactification (Y, f ) of X a Priestley ordercompactif ication if Y is a Priestley space. As follows from [2, Thm. 3.5], X has a Priestley order-compactification iff X is order-zero-dimensional. If X is a totally ordered space, then X is order-zerodimensional iff X is zero-dimensional, which happens iff X satisfies the Priestley separation axiom [16, Cor. 3]. The goal of this final section is to describe all Priestley order-compactifications of a zero-dimensional totally ordered space. Definition 3.2 Let X be an ordered topological space. 1. ([2, Def. 4.1.1]) Let A be an upset of X, B be a downset of X, and A ∩ B = ∅. 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. ([2, Def. 4.1.2]) 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 of X such that A ⊆ U and U ∩ B = ∅. 3. ([2, Def. 4.3]) If X is completely order-regular, then we call X strongly orderzero-dimensional if X satisfies the strong Priestley separation axiom. Let X be a completely order-regular space. By [2, Prop. 4.2], if X satisfies the strong Priestley separation axiom, then X satisfies the Priestley separation axiom, and hence is an order-zero-dimensional space. Moreover, by [2, Prop. 4.4] (see also [19, Sec. 2]), X is strongly order-zero-dimensional iff the Nachbin ordercompactification n(X) of X is a Priestley space. We show that each zero-dimensional totally ordered space is strongly order-zero-dimensional. Order (2011) 28:577–592 587 Lemma 3.3 If X is a zero-dimensional totally ordered space, then X is strongly orderzero-dimensional. Proof Let X be a zero-dimensional totally ordered space. Let A be an upset and B be 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). Without loss of generality we may assume that f −1 (1), f −1 (0) = ∅. Then f −1 (1) is a closed upset and f −1 (0) is a closed downset of X. If f −1 (1), f −1 (0) partition X, then f −1 (1) is a clopen upset separating A and B. Else, there exists x ∈ X − ( f −1 (1) ∪ f −1 (0)) which is an upper bound of f −1 (0) and a lower bound of f −1 (1). If x is the minimum of X − f −1 (0), then by Lemma 1.4, ↑x is a clopen upset of X, containing A and missing B. If x is not the minimum, then there is y ∈ X − f −1 (0) with y < x. Then y is also an upper bound of f −1 (0) and a lower bound of f −1 (1). By [16, Cor. 3], X satisfies the Priestley separation axiom. Therefore, there exists a clopen upset U of X such that x ∈ U and y ∈ / U. Thus, A ⊆ f −1 (1) ⊆↑x ⊆ U and −1 U ∩ B ⊆ U ∩ f (0) ⊆ U ∩ ↓y = ∅. Consequently, X satisfies the strong Priestley separation axiom, and so it is strongly order-zero-dimensional. e) is the Let X be a zero-dimensional totally ordered space. By Theorem 2.9, ( X, Nachbin order-compactification of X. Therefore, by Lemma 3.3 and [2, Prop. 4.4], X is a Priestley space. Let (Y, f ) be an order-compactification of X. By Theorem 2.13, S , π S ◦ e). For there is a subset S of CD2 (X) such that (Y, f ) is equivalent to ( X/∼ S ⊆ CD2 (X), let DS = {e(x) : ↑x is clopen in X} ∪ {e(x) : ↓x is clopen in X} ∪ {(D, 1) : D ∈ S}. Theorem 3.4 Let X be a zero-dimensional totally ordered space and S ⊆ CD2 (X). S , π S ◦ e) is a Priestley order-compactif ication of X if f DS is dense in X. Then ( X/∼ S , π S ◦ e) is a Priestley order-compactification of Proof First suppose that ( X/∼ Without loss of generality we X. Let (α, β) be a nonempty open interval in X. Since (α, β) is nonempty, there exists x ∈ X such that may assume that α, β ∈ X. α < e(x) < β. If (α, e(x)) = ∅, then α is an immediate predecessor of e(x), and so Moreover, ↑x = e−1 (↑e(x)) is a clopen upset of X, and ↑e(x) is a clopen upset of X. so {e(x) : ↑x is clopen in X} ∩ (α, β) = ∅. Otherwise, there exists y ∈ X such that α < e(y) < e(x) < β. Therefore, π S (α) ≤ π S (e(y)) ≤ π S (e(x)) ≤ π S (β). As α ∼ S e(y), e(y) ∼ S e(x), and e(x) ∼ S β, we have π S (α) < π S (e(y)) < π S (e(x)) < π S (β). Since S such that π S (e(x)) ∈ S is a Priestley space, there exists a clopen upset U of X/∼ X/∼ U and π S (e(y)) ∈ / U. Therefore, π −1 (X − U), π −1 (U) is a clopen partition of X, S S e(y) ∈ π S−1 (X − U) with π S−1 (X − U) a downset, and e(x) ∈ π S−1 (U) with π S−1 (U) is a compact totally ordered space, there exist ξ, η ∈ X such that an upset. As X η is an immediate successor of ξ , π S−1 (X − U) =↓ξ , and π S−1 (U) =↑η. If η = e(z) for some z ∈ X, then ↑z = e−1 (↑η) is clopen in X. Moreover, α < e(y) < e(z) ≤ e(x) < β. Therefore, {e(z) : ↑z is clopen in X} ∩ (α, β) = ∅. If not, then there exists D ∈ CD1 (X) such that ξ = (D, 0) and η = (D, 1). If D =↓z for some z ∈ X, then ξ = e(z), and so ↓z = e−1 (↓ξ ) is clopen in X. Moreover, α < e(y) ≤ e(z) < e(x) < β, and so {e(z) : ↓z is clopen in X} ∩ (α, β) = ∅. Finally, if D is non-principal, then D ∈ CD2 (X). Since ξ ∼ S η, we have D ∈ S. Moreover, α < e(y) < (D, 0) < (D, 1) < e(x) < β. Thus, {(D, 1) : D ∈ S} ∩ (α, β) = ∅, and so DS is dense in X. 588 Order (2011) 28:577–592 Let α, β ∈ X and let π S (α) < π S (β). Then Next suppose that DS is dense in X. and so ↑π S (β) is a clopen α < β. If (α, β) = ∅, then ↓α and ↑β are clopen in X, upset of X/∼ S which separates π S (α) and π S (β). If (α, β) = ∅, then as DS is dense either there exists y ∈ X such that ↑y is clopen in X and α < e(y) < β, or there in X, exists z ∈ X such that ↓z is clopen in X and α < e(z) < β, or there exists D ∈ S such that α < (D, 1) < β. First suppose that there exists y ∈ X such that ↑y is clopen in X and α < e(y) < β. Then π S (α) < π S (e(y)) < π S (β). Since ↑y is clopen in X, either y has an immediate predecessor z or else E =↓y − {y} is a clopen downset of X. If z is an immediate predecessor of x, then π S (e(z)) is an immediate predecessor of S , and so ↑π S (e(y)) is a clopen upset of X/∼ S separating π S (β) from π S (e(y)) in X/∼ π S (α). Similarly, if E =↓y − {y} is a clopen downset of X, then (E, 0) is an immediate and so π S (E, 0) is an immediate predecessor of π S (e(y)) in predecessor of e(y) in X, S separating π S (β) from π S (α). S . Therefore, ↑π S (e(y)) is a clopen upset of X/∼ X/∼ A dual argument shows that if there exists z ∈ X such that ↓z is clopen in X and α < S − ↓π S (e(z)) is a clopen upset of X/∼ S separating π S (β) from e(z) < β, then X/∼ π S (α). Finally, suppose that there exists D ∈ S such that α < (D, 1) < β. Then α ≤ (D, 0) < (D, 1) < β. Since D ∈ S, we have π S (D, 0) is an immediate predecessor of S . Thus, ↑π S (D, 1) is a clopen upset of X/∼ S separating π S (β) from π S (D, 1) in X/∼ S , π S ◦ e) is a Priestley π S (α). Consequently, X/∼ S is a Priestley space, and so ( X/∼ order-compactification of X. Definition 3.5 ([2, Def. 5.1]) Let X be an ordered topological space. A Priestley basis of X is a bounded sublattice P of the lattice of all clopen upsets of X such that: 1. If x ≤ y, then there is U ∈ P with x ∈ U and y ∈ U. 2. {U − V : U, V ∈ P} is a basis for the topology on X. By [2, Thm. 5.2], the poset POC(X) of Priestley order-compactifications of an order-zero-dimensional space X is isomorphic to the poset (PB(X), ⊆) of Priestley bases of X. In the particular case of a zero-dimensional totally ordered space X, this provides a description of all Priestley order-compactifications of X. Theorem 3.4 provides another description of all Priestley order-compactifications of X. We will show that the two descriptions are very closely related to each other. Lemma 3.6 Let X be a zero-dimensional totally ordered space and let P be a Priestley basis of X. 1. If x ∈ X with ↑x clopen, then ↑x ∈ P. 2. If x ∈ X with ↓x clopen, then X− ↓x ∈ P. Proof (1) Suppose that ↑x is clopen in X. Since P is a Priestley basis of X, we may write ↑x = I (U i − Vi ) for some U i , Vi ∈ P. Therefore, there is j ∈ I with x ∈ U j − V j. As X is totally ordered, we have V j ⊂↑x ⊆ U j. Because U j − V j ⊆↑x, we get U j ⊆↑x ∪ V j =↑x ⊆ U j. Thus, ↑x = U j, and so ↑x ∈ P. (2) Suppose that ↓x is clopen in X. Write ↓x = I (U i − Vi ) with U i , Vi ∈ P. Then there is j ∈ I with x ∈ U j − V j. Therefore, V j ⊂↑x ⊆ U j, and so V j ⊆ X− ↓x =↑x − {x}. If there is y ∈ X− ↓x with y ∈ V j, then x < y, which means Order (2011) 28:577–592 589 y ∈ U j. But then y ∈ (U j − V j) ⊆ ↓x, a contradiction. Thus, X− ↓x = V j, and so X− ↓x ∈ P. For a zero-dimensional totally ordered space X, let PD (CD2 (X)) denote the set Clearly PD (CD2 (X)) ⊆ of those subsets S of CD2 (X) for which DS is dense in X. P (CD2 (X)). Given S ∈ PD (CD2 (X)), we define PS = {↑y : ↑y is clopen in X} ∪ {X− ↓z :↓z is clopen in X} ∪ {X − D : D ∈ S} ∪ {∅, X}. Conversely, if P is a Priestley basis of X, we define S P = {D ∈ CD2 (X) : X− D ∈ P}. Theorem 3.7 Let X be a zero-dimensional totally ordered space. 1. If S ∈ PD (CD2 (X)), then PS ∈ PB(X). 2. If P ∈ PB(X), then S P ∈ PD (CD2 (X)). 3. The maps S → PS and P → S P establish that the poset (PD (CD2 (X)), ⊆) is isomorphic to the poset (PB(X), ⊆). Proof (1) Suppose that S ∈ PD (CD2 (X)). Since X is totally ordered, it follows from the definition of PS that PS is a bounded sublattice of the lattice of all clopen upsets of X. Take x, y ∈ X with x ≤ y. Then y < x. If (y, x) = ∅, then ↑x is clopen. Therefore, by Lemma 3.6(1), ↑x ∈ PS and separates x and y. On the other hand, there is α ∈ DS with e(y) < α < e(x). if (y, x) = ∅, then, since DS is dense in X, If α = e(z) with ↑z clopen, then by Lemma 3.6(1), ↑z ∈ PS and separates x and y. If α = e(z) with ↓z clopen, then by Lemma 3.6(2), X− ↓z ∈ PS and separates x and y. Finally, if α = (D, 1) with D ∈ CD2 (X), then X − D ∈ PS and separates x and y. Thus, in all cases, we can find U ∈ PS separating x and y. Next, to see that {U − V : U, V ∈ PS } is a basis for the topology, we recall the subbasis of the topology on X given in Theorem 1.5. Let W be a subbasic open set and x ∈ W. First, let W ∈ COU (X). If x is the minimum of W, then by Lemma 1.4, W =↑x. By Lemma 3.6(1), ↑x ∈ PS . Therefore, W ∈ PS . If x is not the minimum of W, then there is y ∈ W with y < x. By the previous paragraph, there is U ∈ PS with x ∈ U and y ∈ U. Then U ⊆↑y, so x ∈ U ⊆ W. Next suppose W ∈ COD(X). If x is the maximum of W, then by Lemma 1.4, W =↓x, and by Lemma 3.6(2), X− ↓x ∈ PS . Therefore, X − W ∈ PS . If x is not the maximum of W, there is y ∈ W with x < y. Again, we can find V ∈ PS with y ∈ V and x ∈ V. Then x ∈ X − V ⊆ W. Finally, let W = (y, z). Then y < x < z. Therefore, there is U ∈ PS with x ∈ U and y ∈ U, and there is V ∈ PS with z ∈ V and x ∈ V. Thus, x ∈ U − V ⊆ (y, z). In all cases, we have produced an element of {U − V : U, V ∈ PS } containing x and contained in W. Consequently, PS is a Priestley basis of X. Without (2) Let P ∈ PB(X) and let (α, β) be a nonempty open interval in X. loss of generality we may assume that α, β ∈ X. Take x ∈ X with α < e(x) < β. If (α, e(x)) = ∅, then α is an immediate predecessor of e(x), so ↑e(x) and so ↑x = e−1 (↑e(x)) is clopen in X. Therefore, e(x) ∈ is clopen in X, 590 Order (2011) 28:577–592 DS P . If (α, e(x)) = ∅, then there is y ∈ X with α < e(y) < e(x). Thus, y < x. Take U ∈ P with x ∈ P and y ∈ P. If U = ↑z or U = X− ↓z for some z ∈ X, then e(z) ∈ (α, β) ∩ DS P . If U and X − U are both non-principal, then X − U ∈ S P , and so (X − U, 1) ∈ DS P ∩ (α, β). In all possible cases we see that and so S P ∈ PD (CD2 (X)). (α, β) ∩ DS P is nonempty. Thus, DS P is dense in X, (3) The maps S → PS and P → S P are clearly order-preserving. Recall that if D ∈ CD2 (X), then both D and X − D are proper and non-principal. Thus, S PS = {D ∈ CD2 (X) : X − D ∈ PS } = {D ∈ CD2 (X) : D ∈ S} = S. To see that PS P = P for each Priestley basis P of X, it is clear that both P and PS P contain ∅, X. Also, by Lemma 3.6, both P and PS P contain all the principal clopen upsets and complements of principal clopen downsets of X. If U is a clopen upset for which U and X − U are non-principal, we have U ∈ PS P iff X − U ∈ S P iff U ∈ P. Thus, PS P = P, and so (PD (CD2 (X)), ⊆) is isomorphic to (PB(X), ⊆). It is well known that if X is a completely order-regular space, then the poset OC(X) of order-compactifications of X is closed under upper bounds. In particular, if X is a totally ordered space, then OC(X) is closed under upper bounds. We show that if in addition X is zero-dimensional, then the same is true for the poset POC(X) of Priestley order-compactifications of X. Lemma 3.8 Let X be a zero-dimensional totally ordered space. Then POC(X) is closed under upper bounds. Proof Let {(Yi , fi ) : i ∈ I} be a family of Priestley order-compactificationsof X. By Theorem 2.13, each (Yi , fi ) corresponds to a subset Si of CD2 (X), and I (Yi , fi ) corresponds to S . Clearly if each D is dense in X, then so is Si I i I D Si . Therefore, (Y , f ) ∈ POC (X). i i I More generally, if X is an order-zero-dimensional space and {(Yi , fi ) : i ∈ I} is a family of Priestley order-compactifications ofX, then by [2, Thm. 5.2], each (Yi , fi ) corresponds to a Priestley basis Pi of X, and I (Yi , fi ) corresponds to the Priestley basis I Pi , where I Pi is the bounded distributive sublattice of the lattice of all clopen upsets of X generated by I Pi . Therefore, POC(X) is closed under upper bounds. On the other hand, POC(X) may not have a least element, and hence may not be a lattice in general, even if X is a totally ordered space. Indeed, if {(Yi , fi ) : i ∈ I} is a family of Priestley order-compactifications of X, and Si are the corresponding elements of PD (CD2 (X)), then I (Yi , fi ) corresponds to I Si . Unfor any longer, as we see from the following tunately, D I Si may not be dense in X example. Example 3.9 Let Q be the space of rational numbers with the interval topology, and let Q[0,1] = Q ∩ [0, 1]. Clearly Q[0,1] is a zero-dimensional linearly ordered space. We first describe OC(Q[0,1] ) and POC(Q[0,1] ). Let P denote the set of irrational numbers, and let P[0,1] = P ∩ [0, 1]. Since P[0,1] is in a bijective correspondence with CD2 (Q[0,1] ), it follows from Theorem 2.13 that OC(Q[0,1] ) is isomorphic to the powerset of P[0,1] . Order (2011) 28:577–592 591 In particular, ∅ corresponds to the least order-compactification of Q[0,1] , which is [0, 1] = DM(Q[0,1] ), and P[0,1] corresponds to n(Q[0,1] ), which is obtained from [0, 1] by replacing each irrational number α ∈ [0, 1] by a doubleton α − < α + . Moreover, as {x ∈ Q[0,1] : ↑x is clopen in Q[0,1] } ∪ {x ∈ Q[0,1] : ↓x is clopen in Q[0,1] } = ∅, by Theorem 3.4, POC(Q[0,1] ) is isomorphic to the set D(P[0,1] ) of subsets of P[0,1] which are dense in [0, 1] (which is equivalent to being dense in n(Q[0,1] )). Now, since there exist disjoint subsets of P[0,1] which are dense in [0, 1], we obtain that there exist Priestley order-compactifications of Q[0,1] whose lower bound is [0, 1]. Therefore, POC(Q[0,1] ) is not a lattice. In particular, POC(Q[0,1] ) does not have a least element. Acknowledgements We would like to thank the referees for their detailed suggestions which greatly improved the exposition of the paper. References 1. Bennett, H.R., Lutzer, D.J.: Recent developments in the topology of ordered spaces. Recent Progress in General Topology, II, pp. 83–114. North-Holland, Amsterdam (2002) 2. Bezhanishvili, G., Morandi, P.J.: Priestley rings and Priestley order-compactifications. Order (2011). doi: 10.1007/s11083-010-9180-2 3. 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