RichKunAGT.dvi


@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 6, No. 2, 2005

pp. 207-216

Ti-ordered reflections

Hans-Peter A. Künzi and Thomas A. Richmond
∗

Abstract. We present a construction which shows that the Ti-
ordered 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 topolog-
ical 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 inter-
est 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 X with a topology τ and a
preorder �. Following the notation of Nachbin ([7]), for A ⊆ X, the increasing
hull of A is i(A) = {y ∈ X : ∃a ∈ A with a � y}. A set A is 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. Künzi and T. A. Richmond

if A = i(A). The closed increasing hull I(A) of A ⊆ X is 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 (τ has 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 X defined
by x ∼ y if and only if x � y and y � x. The preordered topological space
(X, τ, �) is said to be T0-preordered if any of the following equivalent statements
holds.

(a) x 6∼ 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 x 6∼ 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 x 6� y in X implies there exists an
open increasing neighborhood of x which does not contain y and there exists
an open decreasing neighborhood of y which does not contain x.

A preordered topological space (X, τ, �) is T2-preordered if there is an in-
creasing neighborhood of x disjoint from some decreasing neighborhood of y
whenever x 6� 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 neigh-
borhood of x disjoint from some decreasing open neighborhood of y whenever
x 6� 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 X defined by a ∼ b if and
only if a � b and b � a. For any preordered topological space (X, τ, �), we ob-
tain 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 X with 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 �f on X defined by a �f b if 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−1 and f−1×f−1 carry open (respectively, closed,
⊑-increasing, ⊑-decreasing, disjoint) sets to open (respectively, closed, �f -
increasing, �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 ∼f is the equivalence relation G(�f ) ∩ G(�
−1
f ), observe that the

∼f-equivalence class of a ∈ X is [a] = f
−1(f(a)), a fiber of the map f. Thus,

�f is a partial order if and only if f is injective.

We now apply the special quotient construction described above to the pre-
order �f induced by a function f. Suppose f : (X, τ, ≤) → (Y, γ, ⊑) is a
continuous increasing function between partially ordered spaces (X, τ, ≤) and
(Y, γ, ⊑), �f is the preorder on X defined by a �f b if and only if f(a) ⊑ f(b),
∼f is the equivalence relation on X defined by a ∼f b if and only if a �f b and
b �f a, and ≤f is the partial order on X/ ∼f defined by [a] ≤f [b] if and only
if a �f b. 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/ ∼f is the natural quotient map. Now h
is continuous, for if U is open in Y , then h−1(U) = qf−1(U) which is open
since f is continuous and the quotient map carries saturated open sets to open
sets. The definitions of ≤f, �f , h, and �h respectively imply the following
implications:

[a] ≤f [b] ⇐⇒ a �f b ⇐⇒ f(a) ⊑ f(b) ⇐⇒ h([a]) ⊑ h([b]) ⇐⇒ [a] �h [b].

Thus, h is 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. Künzi and T. A. Richmond

Proposition 2.2. Suppose (X, τ, �) is a preordered topological space, ∼ is the
equivalence relation on X defined by a ∼ b if and only if a � b and 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 f from a pre-
ordered space to a partially ordered space, and if f is taken to be the quotient
map q from (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 N of one of the points
a or b which 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 a 6� b in X, so there exist a �-increasing
τ-open neighborhood Na of a and a �-decreasing τ-open neighborhood Nb of b
in X which 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 rela-
tion on X defined by a ∼ b if and only if a � b and 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 γ be the Euclidean topology on R. Define a topology τ on X = R as
follows: Each point of Q\{0} is isolated. For x ∈ (R\Q)∪{0}, a τ-neighborhood
of x is {x} ∪ (U ∩ Q) where U is a γ-neighborhood of x.

Define a preorder � on X by a � b if and only if a = b or {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 ∈ X contains a τ-neighborhood
of x, it follows that (X, τ) is T2, and thus ∆X is closed in X ×X. Observe that
Q is a neighborhood of each of its points, so R \ Q is closed in X. It follows
that G(�) = (R \ Q)2 ∪ ∆X is 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 π 6� 0 in X, so [π] 6≤ [0] in X/ ∼, and thus there exist disjoint
sets M, N, where M is a ≤-increasing τ/ ∼-neighborhood of [π] and N is 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 B is 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 τ-
neighborhood of b ∈ R \ Q has form (R \ Q) ∪ U where U is a γ-neighborhood
of b. Now U ∩ B 6= ∅ ⇒ 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 Pi 6= ∅ since X×X belongs 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) =
⋂
Pi. As an intersection of preorders containing G(≤), �

i is 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 x 6∼0 y in X, so

for some �∗∈ P0, there exists a �
∗-monotone open neighborhood N of a which

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]∼0 which does not contain [b]∼0. Thus,
(X/ ∼0, τ/ ∼0, ≤

0) is T0-ordered.
i = 1: Because the increasing hull of x in �1=

⋂
P1 is the intersection of the

increasing hulls of x in 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 Nb be disjoint τ/ ∼∗ neighborhoods of [a]∼∗ and [b]∼∗ respectively, with
Na being ≤

∗-increasing and Nb being ≤
∗-decreasing. Since G(≤2) ⊆ G(�∗),

the natural quotient map q from X/ ∼2 to 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 Na and Nb above 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. Künzi 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]∼i to [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→ Y is continuous and increasing. Thus,
each continuous increasing function f : X → Y can be lifted through X/ ∼i,
and from the construction, this lifting is unique. Thus, X/ ∼i is 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 ∼Q b if and only if a �Q b

and b �Q a; G(�Q) =
⋂
{G(�∗) : G(≤) ⊆ G(�∗) and (X/ ∼∗, τ/ ∼∗,

≤∗) is an ordered space with property Q}; and [a] ≤Q [b] in X/ ∼Q if
and only if a �Q b in 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 Ti reflection
(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 X for 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 x and y with
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 ≈ y if 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
∼0 introduced in the previous section and that the finite step order described
above agrees with the order ≤0 defined in the previous section by [a] ≤0 [b] if
and only if a �0 b where G(�0) =

⋂
P0, 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 X is ≈-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 f is 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 U containing [x], there exists a saturated
increasing (decreasing) open set containing [x] which is contained in U. We
have A is closed and increasing in X if and only if f(A) is closed and increasing
in X/ ≈, and B is closed and increasing in X/ ≈ if and only if f−1(B) is
closed and increasing in X. Furthermore, because I(x) =

⋂
C where C is the

collection of closed increasing sets containing x and f(
⋂
C) =

⋂
f(C) for any

collection C of ≈-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 ≈ b if 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 or-
dered quotient map, then we have f(I(x)) = f(I(y)) and f(D(x)) = f(D(y)).
Applying f−1 to the equalities above and recalling that I(x) and D(x) are sat-
urated, we have I(x) = I(y) and D(x) = D(y), which implies [x] = [y]. Thus,
X/ ≈ is T0-ordered.

Now suppose Y is any T0-ordered space and g : X → Y is continuous and
increasing. We will show that {g−1(y) : y ∈ Y } is saturated with respect



214 H.-P. A. Künzi and T. A. Richmond

to D = {f−1([x]) : [x] ∈ X/ ≈}. Suppose to the contrary that there exists
y ∈ Y such 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 X which contains g
−1(g(b))

and therefore contains b. But

[a] = [b] ⇒ I(a) = I(b)

⇒ a is 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 a and b interchanged
shows the reverse inclusion, so IY (g(a)) = IY (g(b)). The dual argument shows
that DY (g(a)) = DY (g(b)). Since Y is 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
h from X/D = X/ ≈ to Y , and hf = g. From the definition of the finite step
order on X/ ≈, it is clear that h is increasing, and h is 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 X is T1-ordered.
(b) [x] 6≤0 [y] in X/ ≈ implies there exists an open increasing neighborhood of
x not containing y and there exists an open decreasing neighborhood of y not
containing x.
(c) i([x]) =

⋂
{N : N is an open increasing neighborhood of x} for any x ∈ X,

and
d([x]) =

⋂
{N : N is 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]) ⊆

⋂
{N : N is an open increasing neighborhood of x}. Suppose

M =
⋂
{N : N is an open increasing neighborhood of x} 6⊆ i([x]). Then there

exists y ∈ M \ i([x]), and since M is saturated, [y] 6⊆ i([x]). In particular,
[x] 6≤0 [y] in the T1-ordered space X/ ≈, so there exists an increasing open
neighborhood J of [x] in X/ ≈ disjoint from [y]. Now if f : X → X/ ≈ is the
quotient map, f−1(J) is an open increasing neighborhood of x disjoint from
y. This contradicts y ∈ M. This proves that i([x]) =

⋂
{N : N is 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] ∈

⋂
{N : N is an open increasing neighborhood of x} = i([x]),

contrary to [x] 6≤0 [y]. If (ii) holds, then [x] ∈
⋂
{N : N is 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−1 of these neighborhoods gives the
desired neighborhoods in X.

(b) ⇒ (a): If [x] 6≤0 [y] in X/ ≈, then by (b) there exists an open increas-
ing neighborhood N of x not containing y and there exists an open decreasing
neighborhood M of y not containing x. Now M and N are saturated, and since
f is an ordered quotient map, f(M) and f(N) are monotone open neighbor-
hoods of [y] and [x], respectively, which show that X/ ≈ is T1-ordered. �

A set A which 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
w0X is a universal compact T1 extension. In [5], conditions involving c-sets are
given to insure w0X is 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 X defined by x ≈ y if and only if D(x) = D(y)
and I(x) = I(y). Then x ≈ y if and only if C(x) = C(y).

Proof. Suppose C(x) = C(y). Then x ∈ C(y) ⊆ I(y), so I(x) ⊆ I(y). Inter-
changing x and y shows 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 X has 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 T0 reflection of (X, τ).

Proof. Suppose X has 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
y 6∈ cl{x}. Then there exist an increasing open neighborhood Ny of y and a
decreasing open neighborhood My of y with x 6∈ Ny ∩My. Thus, either x 6∈ Ny
or x 6∈ My, and taking complements shows that y 6∈ D(x) or y 6∈ I(x), that is,
y 6∈ I(x) ∩ D(x), as needed.

By Proposition 3.3, x ≈ y if and only if cl{x} = cl{y}. It is well-known that
the T0 reflection of (X, τ) is given by the quotient topology on the quotient set
X/ ≃ where x ≃ y if and only if cl{x} = cl{y}. �



216 H.-P. A. Künzi and T. A. Richmond

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Received January 2005

Accepted January 2005

Hans-Peter A. Künzi (kunzi@maths.uct.ac.za)
Department of Mathematics and Applied Mathematics, University of Cape
Town, Rondebosch 7701, South Africa.

Thomas A. Richmond (tom.richmond@wku.edu)
Department of Mathematics, Western Kentucky University, Bowling Green,
KY 42101, USA.