T K 1 -ordered Spaces: A New Version of T1-ordered Topological Spaces, Papers of Trigonometry

Download T K 1 -ordered Spaces: A New Version of T1-ordered Topological Spaces and more Papers Trigonometry in PDF only on Docsity! Publ. Math. Debrecen 73/3-4 (2008), 361–377 Ordered separation axioms and the Wallman ordered compactification By HANS-PETER A. KÜNZI (Rondebosch), AISLING E. MCCLUSKEY (Galway) and THOMAS A. RICHMOND (Bowling Green) Abstract. Two constructions have been given previously of the Wallman ordered compactification w0X of a T1-ordered, convex ordered topological space (X, τ,≤). Both of those papers note that w0X is T1, but need not be T1-ordered. Using this as one motivation, we propose a new version of T1-ordered, called T K 1 -ordered, which has the property that the Wallman ordered compactification of a T K1 -ordered topological space is T K1 -ordered. We also discuss the R0-ordered (R K 0 -ordered) property, defined so that an ordered topological space is T1-ordered (T K 1 -ordered) if and only if it is T0-ordered and R0-ordered (R K 0 -ordered). 1. Introduction Given a set X with a topology τ and a partial order ≤, we recall the topology τ ♯ on X as the collection of all τ -open ≤-increasing subsets of X and the topology τ ♭ on X as the corresponding collection of all τ -open ≤-decreasing subsets of X . Thus, we may consider the topological space (X, τ), the ordered topological space (X, τ,≤), or the bitopological space (X, τ ♯, τ ♭). As the study of ordered topological spaces and bitopological spaces devel- oped, important topological properties, including separation axioms, were defined Mathematics Subject Classification: 54F05, 54D10, 54D35, 54A10, 54E55, 18B30, 06F30. Key words and phrases: ordered topological space, T1-ordered, T0-ordered, R0-ordered, T K1 - ordered, RK0 -ordered, Wallman ordered compactification, ordered reflection. Hans-Peter Künzi was partially supported by the South African NRF Grant FA2006022300009 during his stay at the National University of Ireland, Galway, in September 2007. 362 Hans-Peter A. Künzi, Aisling E. McCluskey and Thomas A. Richmond for these new categories. As a general unifying principle, it seems natural that the chain of implications (1) ⇒ (2) ⇒ (3) should hold for the statements (1) (X, τ ♯, τ ♭) satisfies the bitopological (or pairwise) property P , (2) (X, τ,≤) satisfies the ordered property P , and (3) (X, τ) satisfies the (topological) property P . This scheme is borne out in particular by the complete regularity properties when certain other reasonable necessary conditions are assumed. However, as the ordered theory and bitopological theory were often developed independently, there are exceptions and anomalies, which we address here. In particular, the standard definition of the T1-ordered property is seen to fall short of expectations in this regard, as well as in regard to its relation to the Wallman ordered com- pactification. In Section 2, we review the T0-ordered and T1-ordered properties and introduce the R0-ordered property, noting their relation to the corresponding pairwise properties of (X, τ ♯, τ ♭). In Section 3, we introduce the T K1 -ordered and RK0 -ordered properties, and show that the T K 1 -ordered property does not have the shortcomings of the standard T1-ordered property. Continuing in this theme, Section 4 shows that the Wallman ordered compactification behaves more nicely for T K1 -ordered spaces than for T1-ordered spaces. Our notation is that of Nachbin [15]. If (X, τ,≤) is a partially ordered topo- logical space and A ⊆ X , then the increasing hull of A is i(A) = {x ∈ X : ∃ a ∈ A with a ≤ x}. If A = i(A), we say A is an increasing set. The closed increasing hull of A, denoted I(A), is the smallest closed increasing set containing A. The decreasing hull d(A), closed decreasing hull D(A), and decreasing sets are defined dually. If A = {a}, we write i(a) for i({a}). Another useful hull operator, used in the construction of the Wallman ordered compactification, is C(A) = I(A)∩D(A). Following [5] and [6], C(A) is the c-set hull of A, and if A = C(A) we say A is a c-set. We say (X, τ,≤) is convex if τ has a subbase of monotone (i.e., increasing or decreasing) open sets. In other terminology, note that D(A) = clτ♯(A) and I(A) = clτ♭(A) where τ ♯ = {U ∈ τ : U = i(U)} and τ ♭ = {U ∈ τ : U = d(U)}. The T0(-ordered) reflection of a (partially ordered) topological space X is the T0(-ordered) quotient space Y of X such that for any continuous (and increasing) function f from X into any arbitrary T0(-ordered) space Z, there exists a unique continuous (and increasing) function h : Y → Z with f = h ◦ q, where q : X → Y is the quotient map. The construction of the T0-ordered reflection, considered in Section 2, utilizes the equivalence relation defined by x ≈ y if and only if I(x) = I(y) and D(x) = D(y), or equivalently, if and only if C(x) = C(y) (see [8]). Let [x] denote the ≈-equivalence class of x. The closure operator C(·) Ordered separation axioms and the Wallman ordered compactification 365 It is easily seen that (X, τ,≤) being T1-ordered implies (X, τ) is T1, and (X, τ ♯, τ ♭) being pairwise T1 implies (X, τ) is T1. However, (X, τ ♯, τ ♭) being pairwise T1 does not imply (X, τ,≤) is T1-ordered. Indeed, the example after Theorem 15 shows that (X, τ ♯, τ ♭) being middle pairwise T1 (the strongest of these pairwise T1 properties which can possibly be satisfied by ordered spaces other than antichains) does not imply (X, τ,≤) is T1-ordered. That example also shows that (X, τ ♯, τ ♭) being middle pairwise T1 does not imply that (X, τ ♯, τ ♭) is pairwise R0 (for the definition of this concept, see below). The following characterization of the T1-ordered property will be used often. Lemma 3. (X, τ,≤) is T1-ordered if and only if i(x) = ⋂ {U ∈ τ ♯ : x ∈ U} and dually for all x ∈ X . Proof. If X is T1-ordered, then y /∈ i(x) implies x /∈ d(y) = D(y), so i(x) ⊆ X \ D(y) ≡ Hy, and thus i(x) = ⋂ {Hy : y /∈ i(x)} ⊇ ⋂ {U ∈ τ ♯ : x ∈ U}. The reverse inclusion is immediate. Conversely, if y /∈ i(x), then x /∈ d(y) = ⋂ {U ∈ τ ♭ : y ∈ U}, so there exists an open decreasing neighborhood of y disjoint from x and therefore from i(x), so y /∈ cl(i(x)), and thus i(x) is closed.  R0 Properties One motivation for R0 spaces, introduced by Davis [3], is that a space is T1 if and only if it is T0 and R0. The analogous equivalences in the categories of ordered topological spaces and bitopological spaces are used to motivate definitions of the appropriate concepts of R0 in those categories. Note however, as seen in the example mentioned before Lemma 3, that the weak concept of pairwise T1 used in this article does not imply pairwise R0 as defined below. It appears that R0- ordered spaces have not been studied previously, so our discussion here will be more thorough. We start with some theorems providing equivalent definitions of the R0 properties in the three categories in question. Theorem 4. The following are equivalent: (a) (X, τ) is an R0-space. (b) F closed, x /∈ F ⇒ ∃ open U with F ⊆ U , x /∈ U . (c) U open, x ∈ U ⇒ cl{x} ⊆ U . (d) {cl{x} : x ∈ X} is a partition of X . (e) τ is lattice isomorphic to the topology of a T1-space. (f) cl{x} 6= cl{y} ⇒ ∃ neighborhood of x not containing y. (g) F closed, cl{x} ∩ F 6= ∅ ⇒ x ∈ F . (h) ⋂ {U ∈ τ : x ∈ U} = cl{x} for all x ∈ X . 366 Hans-Peter A. Künzi, Aisling E. McCluskey and Thomas A. Richmond (i) The T0-reflection of X is T1. (j) ⋂ {U ∈ τ : A ⊆ U} ⊆ cl A for all A ⊆ X . These equivalences may be found in [3] or [1], except for (j), which is easily shown. (In [1], a T(α,β) space is one whose Tα reflection is already Tβ, so there, R0 spaces are called T(0,1) spaces.) Note that the containment in (j) cannot be strengthened to equality: Let X = [0, 1] with the discrete topology on (0, 1] and the usual neighborhoods of 0. Now (b) holds, but A=(0, 1] = ⋂ {U ∈τ :A ⊆ U} 6= cl A = [0, 1]. It will be interesting to note that many of the characterizations of R0 given above have direct analogs in either the ordered topological setting or the bitopo- logical setting, but not both. The definition given below for an R0-ordered space arises from the necessary and sufficient conditions in [8] for the T0-ordered reflection of an ordered topolog- ical space to be T1-ordered. For an ordered topological space (X, τ,≤), we obtain the T0-ordered reflection (X/ ≈, τ0,≤0) as an ordered quotient of X , modulo the equivalence relation x ≈ y if and only if C(x) = C(y). The order ≤0 on X/ ≈ is the “finite step order” given by [z0] ≤ 0 [zn] ⇐⇒ ∃[z1], [z2], . . . , [zn−1] and ∃z ′ i, z ∗ i ∈ [zi] (i = 0, 1, . . . , n) with z′i ≤ z ∗ i+1 ∀i = 0, 1, . . . , n − 1. Note that any closed or open monotone set S in X is ≈-saturated (that is, x ∈ S implies [x] ⊆ S). We note that Mršević [13] introduced a bitopological “quotient space” which, in the case of (X, τ ♯, τ ♭), is equivalent to the T0-ordered reflection of (X, τ,≤). Lemma 5. Suppose x, y ∈ X , F ⊆ X , and f : X → X/ ≈ is the quotient map from an ordered topological space X to its T0-ordered reflection X/ ≈. (a) If A is closed and increasing in X then f(A) is closed and increasing in X/ ≈, and dually. If A is ≈-saturated, the converse holds. (b) B is closed and increasing in X/ ≈ if and only if f−1(B) is closed and increasing in X , and dually. (c) f(I(x)) = IX/≈([x]) and f(D(x)) = DX/≈([x]). (d) f−1(IX/≈(f(x)) = f −1(IX/≈([x])) = I(x) and dually. (e) If [y] ∈ f(D(F )), and [x] ≤ [y] in X/ ≈, then x ∈ D(F ), and dually. Ordered separation axioms and the Wallman ordered compactification 367 Proof. (a), (b), and (c) were stated and justified in the paragraph before Theorem 3.1 of [8]. See also Corollary 2 and Proposition 5 of [13]. (d) I(x) ⊆ f−1(IX/≈([x])) since the latter set is a closed increasing set con- taining x and the former is the smallest such set. For the reverse inclusion, if z ∈ f−1(IX/≈([x])) but z /∈ I(x), then since I(x) is saturated, applying f (and part (c)) gives f(z) /∈ f(I(x)) = IX/≈([x]), contrary to z ∈ f −1(IX/≈([x])). (e) Suppose [y] ∈ f(D(F )) with [x] ≤ [y] in X/ ≈. Now D(F ) is closed and decreasing, and thus saturated, and contains y. Since X/ ≈ carries the finite step order, it follows that x ∈ D(F ).  Any of the equivalent statements below may be taken as the definition of an R0-ordered space. Theorem 6. For an ordered topological space (X, τ,≤) with T0-ordered reflection X/ ≈, the following are equivalent: (a) (X, τ,≤) is R0-ordered. (b) ⋂ {U ∈ τ ♯ : x ∈ U} = i([x]) and dually for all x ∈ X . (c) The T0-ordered reflection of X is T1-ordered. (d) [x]  [y] in X/ ≈ implies x /∈ D(y) and y /∈ I(x). (e) I(x) = f−1(iX/≈([x])) and D(x) = f −1(dX/≈([x])) for all x ∈ X , where f : X → X/ ≈ is the natural ordered quotient map. Proof. The equivalence of (b) and (c) is Theorem 3.2 of [8], and was the impetus for taking these conditions to be the definition of R0-ordered. (c) ⇒ (d). The proof of this implication is modeled on the proof of the corresponding non-ordered case (Theorem 3.5 (i) ⇒ (ii) in [1]). Suppose (c) holds and [x]  [y] in X/ ≈. Then [y] /∈ iX/≈([x]) = IX/≈([x]). Applying f−1, where f : X → X/ ≈ is the (ordered) quotient map, we have y /∈ f−1(IX/≈([x])) = f−1(f(I(x)) ⊇ I(x), so y /∈ I(x). Similarly, [x] /∈ dX/≈([y]) implies x /∈ D(y). (d) ⇒ (c). Suppose (d) holds and [y] /∈ iX/≈([x]). Then x /∈ D(y) implies that X\D(y) is an open increasing (and thus, saturated) neighborhood of x which does not include y, so [y] /∈ IX/≈([x]). Similarly, y /∈ I(x) shows X \ I(x) is an open decreasing saturated neighborhood of y not containing x. (c) ⇒ (e). Suppose (c) holds. Then iX/≈([x]) is closed and increasing, so f−1(iX/≈([x])) is closed, increasing, and contains x, so I(x) ⊆ f −1(iX/≈([x])). For the reverse inclusion, if z ∈ f−1(iX/≈([x])), then [z] = f(z) ∈ iX/≈([x]) = IX/≈([x]) = f(I(x)), and thus z ∈ I(x) by Lemma 5 (e). Thus, we have I(x) = f−1(iX/≈([x])). The dual argument completes this implication. 370 Hans-Peter A. Künzi, Aisling E. McCluskey and Thomas A. Richmond and e ∈ D(π). It immediately follows that C(x) = {x} for every x ∈ X . Thus, X is a T K1 -ordered space. Consider e and π. There exists no increasing (open) neighborhood of e that does not contain π and no increasing (open) neighborhood of π that does not contain e. Thus, the associated bispace (X, τ ♯, τ ♭) is not a weak pairwise T1 space, and hence is not middle pairwise T1. Noting that π ∈ ⋂ {U ∈ τ ♯ : e∈U} 6= i(e) = {e}, Lemma 3 shows that X is not T1-ordered. This is also easily seen since d(π) is not closed. Now we define RK0 -ordered spaces and subsequently list some of their prop- erties. Definition 12. An ordered topological space (X, τ,≤) is RK0 -ordered if, for all x, y ∈ X , we have y ∈ C(x) implies x ∈ C(y) . Theorem 13. Suppose (X, τ,≤) is an ordered topological space. (a) The following are equivalent: (i) (X, τ,≤) is RK0 -ordered. (ii) C = C−1. (iii) C is an equivalence relation. (iv) x ∈ C(y) if and only if C(x) = C(y) for all x, y ∈ X . (v) x ∈ C(y) if and only if [x] = [y] for all x, y ∈ X . (vi) {C(x) : x ∈ X} is a partition of X . (b) (X, τ ♯, τ ♭) is pairwise R0 ⇒ (X, τ,≤) is RK0 -ordered. (X, τ,≤) is R0-ordered ⇒ (X, τ,≤) is RK0 -ordered. (X, τ,≤) is R K 0 -ordered and convex ⇐⇒ (X, τ) is R0 and (X, τ,≤) is convex. (c) If (X, τ,≤) is a linearly ordered space, then R0-ordered and RK0 -ordered are equivalent properties. Proof. (a) is immediate. (b) The first implication follows from a comparison of part (iv) of (a) above and Theorem 8 (c). X is R0-ordered if and only if X/ ≈ is T1-ordered, which implies X/ ≈ is T K1 -ordered, and hence (by Theorem 15 below) X is R K 0 -ordered. Suppose X is RK0 -ordered and convex. We will show that {cl(x) : x ∈ X} partitions X . Suppose z ∈ cl(x)∩cl(y). Now z ∈ cl(x) ⊆ C(x) implies, by part (v) of (a), that [x] = [z], and similarly [y] = [z] = [x]. Recalling that convexity implies [w] ⊆ cl(w) for any w ∈ X , we have x ∈ [x] = [y] ⊆ cl(y) and y ∈ [y] = [x] ⊆ cl(x). Applying the closure operator now shows cl(x) ⊆ cl(y) ⊆ cl(x), so cl(x) = cl(y), and thus (X, τ) is R0 by Theorem 4 (d). Conversely, suppose (X, τ,≤) is convex Ordered separation axioms and the Wallman ordered compactification 371 and (X, τ) is R0. We will show that z ∈ C(x) implies C(z) = C(x), from which it easily follows that {C(x) : x ∈ X} partitions X and thus X is RK0 -ordered. If z ∈ C(x) and x ∈ C(z), applying the closure operator C shows C(x) = C(z). Thus, suppose z ∈ C(x) and x /∈ C(z). Then x /∈ cl(z), so cl(x) 6= cl(z). Theorem 4 (f) implies the existence of a neighborhood N of z which does not contain x, and by convexity, we may assume N is monotone open. If N is increasing, then X \N is a closed decreasing set containing x and excluding z. This shows z /∈ D(x), giving the contradiction that z /∈ C(x). The dual argument applies if N is decreasing. (c) If X is a linearly ordered RK0 -ordered space, by Theorem 15 below, X/ ≈ is T K1 -ordered. It is easy to see that the finite step order on X/ ≈ is also linear, in which case X/ ≈ is T1-ordered, and therefore X is R0-ordered. The converse follows from part (b).  Observing the appearance of convexity in (b) above, we note that RK0 -ordered need not imply R0 if the topology is not convex. For example, consider X = {⊥, a, b,⊤} where a and b are noncomparable and ⊥ ≤ x ≤ ⊤ for all x ∈ X . Give X the topology having {{⊤,⊥}, {a, b}, {a}} as base of closed sets. It is easy to check that C(x) = X for each x ∈ X , so X is RK0 -ordered. In fact, X is R0-ordered since i([x]) = i(X) = X = ⋂ {U ∈ τ ♯ : x ∈ U} and dually for each x ∈ X . However, {cl(x) : x ∈ X} does not partition X , so X is not R0. The next theorems show that the RK0 -ordered and T K 0 -ordered properties interact as one would hope. Theorem 14. (X, τ,≤) is T K1 -ordered if and only if it is T0-ordered and RK0 -ordered. Proof. If X is T0-ordered and R K 0 -ordered, then for all x ∈ X we have {x} = C−1(x)∩C(x) = C(x)∩C(x) = C(x), where the first equality follows from the T0-ordered property and the second equality from the R K 0 -ordered property. Thus, X is T K1 -ordered. Conversely, if X is T K 1 -ordered, then C(x) = {x} implies C(x) ∩ C−1(x) = {x}, so that X is T0-ordered. Also, y ∈ C(x) = {x} implies x ∈ C(y), so X is RK0 -ordered.  Theorem 15. The T0-ordered reflection X/ ≈ of an ordered topological space X is T K1 -ordered if and only if X is R K 0 -ordered. Proof. If X/ ≈ is T K1 -ordered, then IX/≈([x]) ∩ DX/≈([x]) = {[x]} for all [x] ∈ X/ ≈. Applying f−1 as in Lemma 5 (d) gives C(x) = I(x) ∩ D(x) = [x]. Now y ∈ C(x) = [x] ⇐⇒ C(x) = C(y), so X is RK0 -ordered by Theorem 13 (a). Conversely, suppose X is RK0 -ordered and [y] ∈ CX/≈([x]) = IX/≈([x]) ∩ DX/≈([x]). Applying f −1 as in Lemma 5 (d) gives [y] ⊆ I(x) ∩ D(x) = C(x), 372 Hans-Peter A. Künzi, Aisling E. McCluskey and Thomas A. Richmond and now y ∈ C(x) implies [y] = [x]. It follows that CX/≈([x]) = {[x]} for any [x] ∈ X/ ≈, so X/ ≈ is T K1 -ordered. This direction of the proof also follows from the bitopological quotient construction of Theorem 3.1 of [20], which also appears as Corollary 6 of [13].  (X, τ ♯, τ ♭) (X, τ,≤) (X, τ ) pairwise T0 ⇒ T0-ordered ⇒ T0 weak pairwise T0 ⇐⇒ T0-ordered ⇒ T0 T0-ordered + convex ⇐⇒ ( T0 + (X, τ,≤) convex pairwise R0 6⇒ R0-ordered 6⇒ R0 ⇓ pairwise R0 ⇒ R K 0 -ordered 6⇒ R0 RK0 -ordered + convex ⇐⇒ ( R0 + (X, τ,≤) convex pairwise T1 6⇒ T1-ordered ⇒ T1 ⇓ pairwise T1 ⇐⇒ T K 1 -ordered ⇒ T1 T K1 -ordered + convex ⇐⇒ ( T1 + (X, τ,≤) convex pairwise completely regular + (X, τ,≤) convex + (X, τ,≤) T1-ordered 9>>=>>; ⇒ completely regularlyordered ⇒ completelyregular We note that T1-ordered is a strictly stronger property than T K 1 -ordered. For example, consider the interval [0, 1] ⊆ R with the usual topology. Impose the usual order on (0, 1], with 0 noncomparable to all other points. This ordered topological space is easily seen to be (τ ♯, τ ♭)-pairwise T1, and hence T K 1 -ordered, but since, for example, d(1) = (0, 1] is not closed, it is not T1-ordered. Furthermore, since T K1 -ordered implies T0-ordered, this space is its own T0-ordered reflection. Now the characterizations in Theorem 6 and Theorem 15 show that this space is RK0 - ordered but not R0-ordered. Thus, R0-ordered is a strictly stronger property than RK0 -ordered. The table above summarizes the implications (1) ⇒ (2) ⇒ (3) suggested in the introduction. A bitopological space is pairwise completely regular (see [9]) if and only if it admits a compatible quasi-uniformity U in the sense that τ(U) is the first topology and τ(U−1) is the second topology. An ordered topological space is Ordered separation axioms and the Wallman ordered compactification 375 bases of closed sets, we must use (S(D(x)),S(I(x))) for the construction of the Wallman ordered compactification of a T K1 -ordered space. With this modification, the proof in [2] that (S(d(x)),S(i(x))) is a maximal bifilter for all x ∈ X does not show that the bifilter (S(D(x)),S(I(x))) is maximal if X is T K1 -ordered. We remedy this situation with the following Lemma. Lemma 17. If X is a convex T K1 -ordered topological space and x ∈ X , then (S(D(x)),S(I(x))) is a maximal bifilter. Proof. It is easy to see that (S(D(x)),S(I(x))) is a bifilter. If it is not maximal, then there exists a bifilter (F ,G) ⊇ (S(D(x)),S(I(x))) with either S(D(x)) 6= F or S(I(x)) 6= G. The cases are dual, so we will only consider the case S(D(x)) 6= F . Since F has a base of closed decreasing sets, there exists a closed decreasing set F ∈ F such that F /∈ S(D(x)), or equivalently, D(x) 6⊆ F . Since D(x) ∈ F , we have A = D(x) ∩ F ∈ F . Now A ⊆ F implies D(x) * A, so x /∈ A. We also have I(x) ⊆ X \ A, for if y ∈ I(x) ∩ A ⊆ I(x) ∩ D(x), then we have the contradiction that y = x /∈ A since, by the T K1 -ordered property, I(x)∩D(x) = {x}. Thus, we have X \A ∈ S(I(x)) ⊆ G and A ∈ F , contradicting that F ∨G exists. This shows that the bifilter (S(D(x)),S(I(x))) is maximal.  The Wallman ordered compactification of a T1-ordered space need not be T1- ordered, but the theorem below shows the advantage of the T K1 -ordered property. Theorem 18. If X is any convex T K1 -ordered topological space, the Wallman ordered compactification w0X is T K 1 -ordered. Proof. We will show that w0X satisfies the characterization of T K 1 -ordered given in Theorem 10 (a). Suppose F and G are distinct maximal c-filters on X , that is, F and G are distinct points in w0X . Now either I(G) 6⊆ F or I(G) ⊆ F . In case I(G) 6⊆ F , there exists I(G) ∈ I(G) ⊆ G with I(G) /∈ F . Since G is a filter and F is maximal, it follows that X \ I(G) /∈ G and X \ I(G) ∈ F , so that G /∈ (X \ I(G))∗ and F ∈ (X \ I(G))∗. Now (X \ I(G))∗ is an open decreasing neighborhood of F in w0X which excludes G. In case I(G) ⊆ F , we have D(G) 6⊆ F , for otherwise I(G) ∨ D(G) = C(G) = G ⊆ F , contradicting the maximality of G. Now there exists D(G) ∈ G with D(G) /∈ F , and the dual argument of the previous case shows that (X \ D(G))∗ is an open increasing neighborhood of F in w0X which excludes G. In either case, we have found that G 6= F implies G /∈ ⋂ { open monotone neighborhoods of F}, and it follows that w0X is T K 1 -ordered.  376 Hans-Peter A. Künzi, Aisling E. McCluskey and Thomas A. Richmond We mention that the expected properties of the Wallman ordered compacti- fication remain valid even when the construction is applied to T K1 -ordered topo- logical spaces. For example, if X is convex and T K1 -ordered, ϕ : X → w0X is the natural embedding, and f : X → Y is a continuous increasing function from X into an arbitrary compact T2-ordered space Y (i.e., Y is compact and the graph of its order is closed in Y × Y ), then there exist a unique continuous increasing function f : w0X → Y such that f ◦ ϕ = f . References [1] K. Belaid, O. Echi, and S. 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Math. Debrecen 58 (4) (2001), 693–705. [17] I. Reilly, On Pairwise R0 Spaces, Annales de la Société Scientifique de Bruxelles 88, III (1974), 293–296. Ordered separation axioms and the Wallman ordered compactification 377 [18] M. J. Saegrove, Pairwise completely regularity and compactification in bitopological spaces, J. London Math. Soc. (2) 7 (1973), 286–290. [19] J. Swart, Total disconnectedness in bitopological spaces and product bitopological spaces, Nederl. Akad. Wetensh. Proc. Ser. A 74 Indag. Math. 33 (1971), 135–145. [20] M. R. Žižovic, Some properties of bitopological spaces, Math. Vesnik 11 (26) (1974), 233-237. HANS-PETER A. KÜNZI DEPARTMENT OF MATHEMATICS AND APPLIED MATHEMATICS UNIVERSITY OF CAPE TOWN RONDEBOSCH 7701 SOUTH AFRICA E-mail: hans-peter.kunzi@uct.ac.za AISLING E. MCCLUSKEY DEPARTMENT OF MATHEMATICS NATIONAL UNIVERSITY OF IRELAND, GALWAY GALWAY IRELAND E-mail: aisling.mccluskey@nuigalway.ie THOMAS A. RICHMOND DEPARTMENT OF MATHEMATICS WESTERN KENTUCKY UNIVERSITY BOWLING GREEN, KY 42101 USA E-mail: tom.richmond@wku.edu (Received March 3, 2008; revised June 25, 2008)