Download Order-Compactifications of Totally Ordered Spaces

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]
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(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
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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.
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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
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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.
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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.
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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 ∩
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↓ 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
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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.
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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.
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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,
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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.
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