Ti-ordered reflections

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 6, No. 2, 2005 pp. 207-216 Ti-ordered reflections Hans-Peter A. K¨ unzi and Thomas A. Richmond ∗ Abstract. We present a construction which shows that the Tiordered reflection (i∈ {0,1,2}) of a partially ordered topological space (X, τ, ≤) exists and is an ordered quotient of (X, τ, ≤). We give an explicit construction of the T0-ordered reflection of an ordered topological space (X, τ, ≤), and characterize ordered topological spaces whose T0-ordered reflection is T1-ordered. 2000 AMS Classification: 54F05, 18B30, 54G20, 54B15, 54C99, 06F30 Keywords: ordered topological space, T2-ordered, T1-ordered, T0-ordered, ordered reflection, ordered quotient 1. Introduction The T0-, T1-, and T2-reflections of a topological space have long been of interest to categorical topologists. Methods of constructions of these in the category TOP are described in the references given in [3] (see p. 302). Here, we consider the corresponding concepts of Ti-ordered reflections in the category ORDTOP of partially ordered topological spaces with continuous increasing functions as morphisms. In Section 2, we construct the Ti-ordered reflection (i= 0,1,2, S2) of a partially ordered topological space (X, τ, ≤). The construction is extrinsic, occurring in the category PREORDTOP of preordered topological spaces with continuous increasing functions as morphisms, which contains ORDTOP as a subcategory. In Section 3, we give an intrinsic construction of the T0-ordered reflection of a partially ordered space (X, τ, ≤) and examine some properties of this reflection. A preordered topological space (X, τ, ) is a set Xwith a topology τand a preorder . Following the notation of Nachbin ([7]), for A⊆X, the increasing hull of Ais i(A) = {y∈X:∃a∈Awith ay}. A set Ais an increasing set ∗The first author would like to thank the South African Research Foundation for partial financial support under Grant Number 2068799. The first version of the article was completed in Germany during the Dagstuhl-Seminar 04351 on Spatial representation: Discrete vs. continuous computational models. 208 H.-P. A. K¨unzi and T. A. Richmond if A=i(A). The closed increasing hull I(A) of A⊆Xis the smallest closed increasing set containing A. Decreasing sets, decreasing hulls d(A), and closed decreasing hulls D(A) are defined dually. A set is monotone if it is increasing or decreasing. There are many compatibility conditions between the topology and order of a preordered topological space which one may stipulate. These include the convex topology condition (τhas a subbase of monotone open sets) or the ordered separation axioms, some of which are defined below. Our preordered spaces need not satisfy any of these compatibility conditions. Suppose (X, τ, ) is a preordered topological space. It is well-known that the preorder induces an equivalence relation ∼=G()∩G(−1) on Xdefined by x∼yif and only if xyand yx. The preordered topological space (X, τ, ) is said to be T0-preordered if any of the following equivalent statements holds. (a) x6∼ y⇒[I(x)6=I(y) OR D(x)6=D(y)] (b) If I(x) = I(y) and D(x) = D(y), then x∼y. (c) If x6∼ y, there exist a monotone open neighborhood of one of the points which does not contain the other point. Observe that if is a partial order, then the relation ∼is equality. A preordered topological space (X, τ, ) is T1-preordered if i(x) and d(x) are closed for every x∈X, or equivalently, if x6 yin Ximplies there exists an open increasing neighborhood of xwhich does not contain yand there exists an open decreasing neighborhood of ywhich does not contain x. A preordered topological space (X, τ, ) is T2-preordered if there is an increasing neighborhood of xdisjoint from some decreasing neighborhood of y whenever x6 y. Equivalently, (X, τ, ) is T2-preordered if the preorder is closed in (X, τ)×(X, τ). A preordered topological space (X, τ, ) is strongly T2-preordered, or for notational convenience, TS2-preordered, if there is an increasing open neighborhood of xdisjoint from some decreasing open neighborhood of ywhenever x6 y. If is a partial order, then (X, τ, ) is a partially ordered topological space, or simply an ordered topological space. If the preorder of a Ti-preordered topological space (X, τ, ≤) is a partial order, we will say (X, τ, ≤) is Ti-ordered. We will typically denote preorders by and partial orders by ≤. To avoid confusion when indicating inclusions, we may represent a preorder ⊑by its graph G(⊑). 2. Existence of Ti-ordered reflections A special quotient. The definition of a T0-preordered topological space (X, τ, ) involved the equivalence relation ∼on Xdefined by a∼bif and only if aband ba. For any preordered topological space (X, τ, ), we obtain a partially ordered topological space by giving X/ ∼the quotient topology τ/ ∼and the order ≤defined by [a]≤[b] if and only if ab. The following properties of this quotient construction are easily verified. Ti-ordered reflections 209 Proposition 2.1. Suppose is a preorder on a set X,∼is the equivalence relation G()∩G(−1)on X, and ≤is the partial order on X/ ∼defined by [a]≤[b]if and only if ab. (a) Any -increasing or -decreasing set is ∼-saturated. (b) The quotient map q:X→X/ ∼carries increasing (decreasing) sets to increasing (decreasing) sets. Specifically, q(i(A)) = i≤(q(A)) for any subset A⊆X, and dually. (c) If ∗is a preorder on Xwith G()⊆G(∗)and ∼∗is defined from ∗ as ∼is defined from , then the ∼∗-equivalence classes are ∼-saturated. Preorders induced by functions. Any continuous increasing function f: (X, τ, ≤)→(Y, γ, ⊑) between two partially ordered topological spaces induces a preorder fon Xdefined by afbif and only if f(a)⊑f(b). Furthermore, G(≤)⊆G(f). Suppose (Y, γ, ⊑) is Ti-ordered for some i∈ {0,1,2, S2}. Now G(f) = (f−1×f−1)(⊑). Noting that f−1and f−1×f−1carry open (respectively, closed, ⊑-increasing, ⊑-decreasing, disjoint) sets to open (respectively, closed, fincreasing, f-decreasing, disjoint) sets and that the Ti-(pre)ordered properties (i∈ {0,1,2, S2}) are defined in terms of open/closed/increasing/decreasing/ disjoint sets, it follows that (X, τ, f) is Ti-preordered. Also, if ∼fis the equivalence relation G(f)∩G(−1 f), observe that the ∼f-equivalence class of a∈Xis [a] = f−1(f(a)), a fiber of the map f. Thus, fis a partial order if and only if fis injective. We now apply the special quotient construction described above to the preorder finduced by a function f. Suppose f: (X, τ, ≤)→(Y, γ, ⊑) is a continuous increasing function between partially ordered spaces (X, τ, ≤) and (Y, γ, ⊑), fis the preorder on Xdefined by afbif and only if f(a)⊑f(b), ∼fis the equivalence relation on Xdefined by a∼fbif and only if afband bfa, and ≤fis the partial order on X/ ∼fdefined by [a]≤f[b] if and only if afb. We now show that if (Y, γ, ⊑) is Ti-ordered for some i∈ {0,1,2, S2}, then the partially ordered space (X/ ∼f, τ/ ∼f,≤f) is also Ti-ordered. Define h: (X/ ∼f, τ/ ∼f,≤f)→(Y, γ, ⊑) by h([a]) = h(f−1f(a)) = f(a), that is, h([a]) = fq−1[a] where q:X→X/ ∼fis the natural quotient map. Now h is continuous, for if Uis open in Y, then h−1(U) = qf−1(U) which is open since fis continuous and the quotient map carries saturated open sets to open sets. The definitions of ≤f,f, h, and hrespectively imply the following implications: [a]≤f[b]⇐⇒ afb⇐⇒ f(a)⊑f(b)⇐⇒ h([a]) ⊑h([b]) ⇐⇒ [a]h[b]. Thus, his increasing and h=≤f. By the remarks of the second paragraph after Proposition 2.1, it follows that (X/ ∼f, τ/ ∼f,≤f) is Ti-ordered. 210 H.-P. A. K¨unzi and T. A. Richmond Proposition 2.2. Suppose (X, τ, )is a preordered topological space, ∼is the equivalence relation on Xdefined by a∼bif and only if aband ba, and ≤is the partial order on X/ ∼defined by [a]≤[b]if and only if ab. Then for i∈ {0,1, S2},(X, τ, )is Ti-preordered if and only if (X/ ∼, τ/ ∼,≤)is Ti-ordered. Proof. Suppose i∈ {0,1, S2}and (X/ ∼, τ/ ∼,≤) is Ti-ordered. The two paragraphs following Proposition 2.1 remain valid for a function ffrom a preordered space to a partially ordered space, and if fis taken to be the quotient map qfrom (X, τ, ) to (X/ ∼, τ/ ∼,≤), then f=. Thus, (X, τ, ) is Ti-preordered. For the converse, first suppose that (X, τ, ) is T0-preordered. If [a]6= [b] in X/ ∼, then there is a -monotone open neighborhood Nof one of the points aor bwhich does not contain the other. Now q(N) is a ≤-monotone open neighborhood of one of the points [a] or [b] in X/ ∼which does not contain the other, so X/ ∼is T0-ordered. Now suppose (X, τ, ) is T1-preordered. For [x]∈ X/ ∼, we have i≤([x]) = i≤(q(x)) = q(i(x)) by Proposition 2.1 (b). Since i(x) is closed and saturated, q(i(x)) will be closed in X/ ∼. With the dual argument, this shows that X/ ∼is T1-ordered. Finally, suppose (X, τ, ) is TS2- preordered. If [a]6≤ [b] in X/ ∼, then a6 bin X, so there exist a -increasing τ-open neighborhood Naof aand a -decreasing τ-open neighborhood Nbof b in Xwhich are disjoint. By Proposition 2.1 (a) and (b), it follows that q(Na) and q(Nb) are the required ≤-monotone τ/ ∼-open neighborhoods separating [a] and [b] in X/ ∼. The result of Proposition 2.2 does not hold for i= 2. While the reasoning of the first paragraph of the proof shows that if (X/ ∼, τ/ ∼,≤) is T2-ordered then (X, τ, ) is T2-preordered, the example below shows that the converse fails. Example 2.3. If (X, τ, ) is a T2-preordered space, ∼is the equivalence relation on Xdefined by a∼bif and only if aband ba, and ≤is the partial order on X/ ∼defined by [a]≤[b] if and only if ab, then (X/ ∼, τ/ ∼,≤) need not be T2-ordered. Let γbe the Euclidean topology on R. Define a topology τon X=Ras follows: Each point of Q\{0}is isolated. For x∈(R\Q)∪{0}, a τ-neighborhood of xis {x} ∪ (U∩Q) where Uis a γ-neighborhood of x. Define a preorder on Xby abif and only if a=bor {a, b} ⊆ R\Q. (In fact, is already an equivalence relation, so ∼=.) The graph of is (R\ Q)2∪∆X. Because each γ-neighborhood of x∈Xcontains a τ-neighborhood of x, it follows that (X, τ) is T2, and thus ∆Xis closed in X×X. Observe that Qis a neighborhood of each of its points, so R\Qis closed in X. It follows that G() = (R\Q)2∪∆Xis closed in X×X, so is T2-preordered. To see that (X/ ∼, τ/ ∼,≤) is not T2-ordered, suppose to the contrary that it is. Now π6 0 in X, so [π]6≤ [0] in X/ ∼, and thus there exist disjoint sets M, N, where Mis a ≤-increasing τ/ ∼-neighborhood of [π] and Nis a ≤-decreasing τ/∼-neighborhood of [0]. If q:X→X/ ∼is the quotient map, Ti-ordered reflections 211 then q−1(N) contains a τ-neighborhood {0}∪(B∩Q) of 0, where Bis a γ-open neighborhood of 0. For any b∈B\Q, we have b∈R\Q= [π]⊆q−1(M). Now q−1(M) is a -increasing τ-neighborhood of b. But a -increasing τneighborhood of b∈R\Qhas form (R\Q)∪Uwhere Uis a γ-neighborhood of b. Now U∩B6=∅⇒U∩(B∩Q)6=∅⇒q−1(M)∩q−1(N)6=∅, contrary to M∩N=∅. Thus, (X/ ∼, τ/ ∼,≤) is not T2-ordered. The existence construction. Suppose (X, τ, ≤) is a given partially ordered topological space. For i∈ {0,1,2, S2}, let Pi={∗:∗is a preorder on X, G(≤)⊆G(∗), and (X/ ∼∗, τ/ ∼∗,≤∗) is Ti−ordered}. Note that Pi6=∅since X×Xbelongs to it. If i∈ {0,1, S2}, by Proposition 2.2, we have Pi={∗:G(≤)⊆G(∗) and (X, τ, ∗) is Ti−preordered}. Let G(i) = TPi. As an intersection of preorders containing G(≤), iis also a preorder containing G(≤). Proposition 2.4. For i∈ {0,1,2, S2},(X/ ∼i, τ/ ∼i,≤i)is Ti-ordered. Proof. i = 0: Suppose [x]6= [y] in (X/ ∼0, τ/ ∼0,≤0). Then x6∼0yin X, so for some ∗∈ P0, there exists a ∗-monotone open neighborhood Nof awhich does not contain b, where {a, b}={x, y}. Applying Proposition 2.1 (b), we obtain a ≤∗-monotone open neighborhood N′of [a]∼∗which does not contain [b]∼∗. Since G(≤0)⊆G(∗), Proposition 2.1 (c) implies the existence of a natural increasing quotient map q:X/ ∼0→X/ ∼∗, and q−1(N′) is a ≤0- monotone open neighborhood of [a]∼0which does not contain [b]∼0. Thus, (X/ ∼0, τ/ ∼0,≤0) is T0-ordered. i = 1: Because the increasing hull of xin 1=TP1is the intersection of the increasing hulls of xin each ∗in P1, and each of these latter increasing hulls is closed, it follows that i1(x) is closed for any x∈X. With the dual argument, we have (X, τ, 1) is T1-preordered. By Proposition 2.2, (X/ ∼1, τ/ ∼1,≤1) is T1-ordered. i = 2: Suppose [a]6≤ [b] in (X/ ∼2, τ/ ∼2,≤2). Then there exists ∗∈ P2 such that [a]∼∗6≤∗[b]∼∗in the T2-ordered space (X/ ∼∗, τ/ ∼∗,≤∗). Let Na and Nbbe disjoint τ/ ∼∗neighborhoods of [a]∼∗and [b]∼∗respectively, with Nabeing ≤∗-increasing and Nbbeing ≤∗-decreasing. Since G(≤2)⊆G(∗), the natural quotient map qfrom X/ ∼2to X/ ∼∗yields q−1(Na) and q−1(Nb) as oppositely directed monotone neighborhoods separating [a] and [b] in X/ ∼2. Thus, (X/ ∼2, τ/ ∼2,≤2) is T2-ordered. Taking Naand Nbabove to be open sets proves the case i=S2.  We are now ready for the main result of this section. Theorem 2.5. Suppose (X, τ, ≤)is a partially ordered topological space and i∈ {0,1,2, S2}. Then the Ti-ordered reflection of (X, τ, ≤)is (X/ ∼i, τ/ ∼i, ≤i). 212 H.-P. A. K¨unzi and T. A. Richmond Proof. Suppose i∈ {0,1,2, S2}is given, (Y, γ, ⊑) is Ti-ordered, and f: (X, τ, ≤)→(Y, γ, ⊑) is continuous and increasing. From the paragraph preceding Proposition 2.2, it follows that (X/ ∼f, τ/ ∼f,≤f) is a Ti-ordered space with G(≤)⊆G(f). From the definition of i, we have G(i)⊆G(f). From Proposition 2.1 (c), there is a natural continuous increasing quotient map q: (X/ ∼i, τ/ ∼i,≤i)→(X/ ∼f, τ/ ∼f,≤f) which carries [a]∼ito [a]∼f. We have shown above that there is a continuous increasing function h: (X/ ∼f, τ/ ∼f, ≤f)→(Y, γ, ⊑). Now hq :X/ ∼i→Yis continuous and increasing. Thus, each continuous increasing function f:X→Ycan be lifted through X/ ∼i, and from the construction, this lifting is unique. Thus, X/ ∼iis the Ti-ordered reflection of X. It is easy to verify that this construction gives the property Q reflection of (X, τ, ≤) as a quotient for any property Q for which (a) (X/ ∼Q, τ/ ∼Q,≤Q) has property Q where a∼Qbif and only if aQb and bQa;G(Q) = T{G(∗) : G(≤)⊆G(∗) and (X/ ∼∗, τ/ ∼∗, ≤∗) is an ordered space with property Q}; and [a]≤Q[b] in X/ ∼Qif and only if aQbin X. (b) If (Y, γ, ⊑) has property Q and f: (X, τ, ≤)→(Y, γ, ⊑) is continuous and increasing, then (X/ ∼f, τ/ ∼f,≤f) has property Q. Furthermore, the methods of this section can be used to find the Tireflection (i= 0,1,2) of a topological space (X, τ) by considering (X, τ) as a discretely ordered topological space (X, τ, =) and taking all preorders to be equivalence relations, so that the resulting quotients are discretely ordered. 3. The T0-ordered reflection The construction of the Ti-ordered reflections in the previous section was an extrinsic construction—working from outside the space (X, τ, ≤)—which produced the Ti-ordered reflection as a quotient based on the intersection of all suitable preorders on Xfor which the indicated quotient construction would yield a Ti-ordered space. In this section, we present an intrinsic construction of the T0-ordered reflection and discuss some other properties of the T0-ordered reflection. Intrinsic constructions of the other Ti-ordered reflections (i > 0) studied in the previous section appear to be much more complicated. In a T0-ordered space, D(x) = D(y) and I(x) = I(y) would imply x=y. If our space is not T0-ordered, then there may be distinct elements xand ywith D(x) = D(y) and I(x) = I(y). Our strategy will be to say two such points are equivalent and mod out by this equivalence relation. Suppose (X, τ, ≤) is an ordered topological space. For x, y ∈X, define x≈yif and only if D(x) = D(y) and I(x) = I(y). Order the set X/ ≈of ≈-equivalence classes by the finite step order: [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. Ti-ordered reflections 213 First note that this is indeed antisymmetric and therefore is a partial order: Suppose [z0]≤0[zn] and [zn]≤0[z0]. Then there exist [zi] (i= 1,...,n,...,m) with [z0] = [zm] and there exist z′ i, z∗ i∈[zi] such that z′ i≤z∗ i+1 for all i= 0,...,m−1. To show [z0] = [zn], suppose not. Then either D(z0)6=D(zn) or I(z0)6=I(zn). Now z′ i≤z∗ i+1 ⇒z∗ i+1 ∈I(z′ i)⇒I(z∗ i+1)⊆I(z′ i)⇒I(zi+1)⊆I(zi). Applying this for i= 0,...,m−1 gives I(z0)⊇I(z1)⊇ · · · ⊇ I(zm) = I(z0). Thus, I(zi) = I(z0)∀i∈ {1,...,m}. Dually, D(zi) = D(z0)∀i∈ {1,...,m}. It follows that zi≈z0∀i∈ {1,...,m}, so [z0] = [zn], and ≤is antisymmetric. In fact, the argument above shows that [x]≤0[y]⇒I(y)⊆I(x) and D(x)⊆D(y). At this point, one can verify that the equivalence relation ≈agrees with ∼0introduced in the previous section and that the finite step order described above agrees with the order ≤0defined in the previous section by [a]≤0[b] if and only if a0bwhere G(0) = TP0, and thus the T0-ordered reflection of (X, τ, ≤) is (X/ ≈, τ/ ≈,≤0). However, we will continue our intrinsic approach and prove this directly. It is easy to show that any closed or open monotone set in Xis ≈-saturated and that the quotient map f:X→X/ ≈carries closed increasing sets to closed increasing sets and open increasing sets to open increasing sets. The dual statement (obtained by replacing “increasing” by “decreasing”) also holds. It follows that fis an ordered quotient map as defined in Definition 6 of [6]. It is easily verified that if D={f−1(y) : y∈X/ ≈} is the decomposition of X associated with the quotient map f:X→X/ ≈, then for each [x]∈ D and each increasing (decreasing) open set Ucontaining [x], there exists a saturated increasing (decreasing) open set containing [x] which is contained in U. We have Ais closed and increasing in Xif and only if f(A) is closed and increasing in X/ ≈, and Bis closed and increasing in X/ ≈if and only if f−1(B) is closed and increasing in X. Furthermore, because I(x) = TCwhere Cis the collection of closed increasing sets containing xand f(TC) = Tf(C) for any collection Cof ≈-saturated sets, it follows that f(I(x)) = IX/≈([x]). Dually, f(D(x)) = DX/≈([x]). Theorem 3.1. Suppose (X, τ, ≤)is a partially ordered topological space, and a≈bif and only if D(a) = D(b)and I(a) = I(b). Then X/ ≈with the quotient topology and the finite-step order is the T0-ordered reflection of X. Proof. First we will show that X/ ≈is T0-ordered. Suppose IX/≈([x]) = IX/≈([y]) and DX/≈([x]) = DX/≈([y]). If f:X→X/ ≈is the natural ordered quotient map, then we have f(I(x)) = f(I(y)) and f(D(x)) = f(D(y)). Applying f−1to the equalities above and recalling that I(x) and D(x) are saturated, we have I(x) = I(y) and D(x) = D(y), which implies [x] = [y]. Thus, X/ ≈is T0-ordered. Now suppose Yis any T0-ordered space and g:X→Yis continuous and increasing. We will show that {g−1(y) : y∈Y}is saturated with respect 214 H.-P. A. K¨unzi and T. A. Richmond to D={f−1([x]) : [x]∈X/ ≈}. Suppose to the contrary that there exists y∈Ysuch that g−1(y) is not D-saturated. Then there exist b∈g−1(y) and a∈X\g−1(y) such that [a] = [b] (that is, f(a) = f(b)). Now g−1(IY(g(b)) is a closed increasing set in Xwhich contains g−1(g(b)) and therefore contains b. But [a] = [b]⇒I(a) = I(b) ⇒ais an element of every closed increasing set containing b ⇒a∈g−1(IY(g(b)) ⇒g(a)∈IY(g(b)) ⇒IY(g(a)) ⊆IY(g(b)). Repeating the argument of the last paragraph with aand binterchanged shows the reverse inclusion, so IY(g(a)) = IY(g(b)). The dual argument shows that DY(g(a)) = DY(g(b)). Since Yis T0-ordered, this implies g(a) = g(b), contrary to a∈X\g−1(y) and b∈g−1(y). Now since {g−1(y) : y∈Y}is D-saturated, there is a natural quotient map hfrom X/D=X/ ≈to Y, and hf =g. From the definition of the finite step order on X/ ≈, it is clear that his increasing, and his clearly unique from the construction. Thus, X/ ≈is the T0-ordered reflection of X. The theorem below characterizes those spaces whose T0-ordered reflections are T1-ordered. Similar results for the non-ordered setting can be found in [1], where a T(i,j)-space is defined to be one whose Ti-reflection satisfies the Tj separation axiom (0 ≤i < j ≤2). Comparing Theorem 3.5(iv) of [1] with Theorem 2(b) of [2], we note that T(0,1)-spaces have been studied by Davis and others subsequently under the name of R0-spaces. Theorem 3.2. The following are equivalent. (a) The T0-ordered reflection X/ ≈of Xis T1-ordered. (b) [x]6≤0[y]in X/ ≈implies there exists an open increasing neighborhood of xnot containing yand there exists an open decreasing neighborhood of ynot containing x. (c) i([x]) = T{N:Nis an open increasing neighborhood of x}for any x∈X, and d([x]) = T{N:Nis an open decreasing neighborhood of x}for any x∈X. Proof. (a)⇒(c): Because closed or open increasing sets are saturated, we have i([x]) ⊆T{N:Nis an open increasing neighborhood of x}. Suppose M=T{N:Nis an open increasing neighborhood of x} 6⊆ i([x]). Then there exists y∈M\i([x]), and since Mis saturated, [y]6⊆ i([x]). In particular, [x]6≤0[y] in the T1-ordered space X/ ≈, so there exists an increasing open neighborhood Jof [x] in X/ ≈disjoint from [y]. Now if f:X→X/ ≈is the quotient map, f−1(J) is an open increasing neighborhood of xdisjoint from y. This contradicts y∈M. This proves that i([x]) = T{N:Nis an open increasing neighborhood of x}for any x∈X. The other statement is proved dually. Ti-ordered reflections 215 (c)⇒(a): Suppose (c). If X/ ≈is not T1-ordered, then there exist [x]6≤0[y] such that either (i) every increasing open neighborhood of [x] in X/ ≈contains [y], or (ii) every decreasing open neighborhood of [y] in X/ ≈contains [x]. If (i) holds, then [y]∈T{N:Nis an open increasing neighborhood of x}=i([x]), contrary to [x]6≤0[y]. If (ii) holds, then [x]∈T{N:Nis an open decreasing neighborhood of y}=d([y]), contrary to [x]6≤0[y]. (a)⇒(b): Suppose (a). Now [x]6≤0[y] in X/ ≈implies there exists an open increasing (respectively, decreasing) neighborhood of [x] (respectively, [y]) not containing [y] (respectively, [x]). Taking f−1of these neighborhoods gives the desired neighborhoods in X. (b)⇒(a): If [x]6≤0[y] in X/ ≈, then by (b) there exists an open increasing neighborhood Nof xnot containing yand there exists an open decreasing neighborhood Mof ynot containing x. Now Mand Nare saturated, and since fis an ordered quotient map, f(M) and f(N) are monotone open neighborhoods of [y] and [x], respectively, which show that X/ ≈is T1-ordered.  A set Awhich satisfies A=I(A)∩D(A) is called a c-set. In [4], maximal filters of c-sets are used to construct the Wallman ordered compactification of an ordered space with convex topology. The Wallman ordered compactification w0Xis a universal compact T1extension. In [5], conditions involving c-sets are given to insure w0Xis T1-ordered. Thus, one might expect c-sets to play a role in the T1-ordered or even T0-ordered reflection. Let C(A) = I(A)∩D(A), that is, let C(A) be the smallest c-set containing A. Proposition 3.3. Suppose (X, τ, ≤)is an ordered topological space and let ≈ be the equivalence relation on Xdefined by x≈yif and only if D(x) = D(y) and I(x) = I(y). Then x≈yif and only if C(x) = C(y). Proof. Suppose C(x) = C(y). Then x∈C(y)⊆I(y), so I(x)⊆I(y). Interchanging xand yshows that I(y)⊆I(x), so I(x) = I(y). Dually, D(x) = D(y), so x≈y. The converse is immediate.  Thus, the equivalence classes of the T0-ordered reflection are determined by the closure operator C(·). If Xhas a convex topology, this closure operator is especially nice. Theorem 3.4. If the ordered topological space (X, τ, ≤)has a convex topology, then the topological space (X′, τ′)underlying its T0-ordered reflection (X′, τ′,≤′) is simply the T0reflection of (X, τ). Proof. Suppose Xhas a convex topology. We will show that C(x) = I(x)∩ D(x) = cl{x}. Clearly y∈cl{x} ⇒ y∈I(x)∩D(x). For the converse, suppose y6∈ cl{x}. Then there exist an increasing open neighborhood Nyof yand a decreasing open neighborhood Myof ywith x6∈ Ny∩My. Thus, either x6∈ Ny or x6∈ My, and taking complements shows that y6∈ D(x) or y6∈ I(x), that is, y6∈ I(x)∩D(x), as needed. By Proposition 3.3, x≈yif and only if cl{x}=cl{y}. It is well-known that the T0reflection of (X, τ) is given by the quotient topology on the quotient set X/ ≃where x≃yif and only if cl{x}=cl{y}.