Further studies on open well-filtered spaces

The open well-filtered spaces were introduced by Shen, Xi, Xu and Zhao to answer the problem whether every core-compact well-filtered space is sober. In the current paper we explore further properties of open well-filtered spaces. One of the main results is that if a space is open well-filtered, then so is its upper space (the set of all nonempty saturated compact subsets equipped with the upper Vietoris topology). Some other properties on open well-filtered spaces are also studied.


Introduction
The open well-filtered spaces were introduced in [6] and used to give an alternative and more natural proof of the conjectured conclusion that every core compact and well-filtered space is sober.In [6], the authors actually proved a stronger conclusion: every core-compact and open well-filtered space is sober, as every well-filtered space is open well-filtered and the converse conclusion is not true in general.By [8] and [10], one knows that a space is well-filtered if and only if its upper space is well-filtered.It is thus natural to wonder whether a similar result holds for open well-filteredness.In this paper we shall give a partial answer to this problem and prove that if a space is open well-filtered then so is its upper space.A number of new results on open well-filtered spaces will also be presented here.

Preliminaries
We first recall some basic definitions and results that will be used in the paper.
Let P be a poset.A nonempty subset D of P is directed (resp., filtered ) if every two elements of D have an upper (resp., lower) bound in D. A poset P is a directed complete poset, or a dcpo for short, if for any directed subset D ⊆ P , the supremum D exists.
For any subset A of a poset P , ↑A = {y ∈ P : ∃x ∈ A, x ≤ y}, and ↓A = {y ∈ P : ∃x ∈ A, y ≤ x}.
In particular, for each x ∈ X, we write ↑x = ↑{x} and ↓x = ↓{x}.For x, y ∈ P , x is way-below y, denoted by x y, if for any directed subset D of P with D existing, y ≤ D implies x ≤ d for some d ∈ D. Denote x = {y ∈ P : x y} and x = {y ∈ P : y x}.A poset P is continuous, if for any x ∈ P , the set x is directed and x = x.A continuous dcpo is also called a domain.
A subset U of a poset P is Scott open if (i) U = ↑U and (ii) for any directed subset D of P with D existing, D ∈ U implies D ∩ U = ∅.All Scott open subsets of P form a topology on P , called the Scott topology on P and denote by σ(P ).The space ΣP = (P, σ(P )) is called the Scott space of P .
For a T 0 space X, the specialization order ≤ on X is defined by x ≤ y iff x ∈ cl X ({y}), where cl X is the closure operator of X.Clearly, cl X ({y}) = ↓y.
In what follows, if no otherwise specified, the partial order on a T 0 space will mean the specialization order.
For any T 0 space X, we shall often use O(X) to denote the topology of X (the collection of all open sets of X).For any subset A of X, the saturation of A, denoted by Sat X (A), is defined by or equivalently, Sat X (A) = ↑A with respect to the specialization order (see [2, Proposition 4.2.9]).A subset A of X is saturated if A = Sat X (A).
For any U, V ∈ O(X), we write U V for that U is way-below V in the poset (O(X), ⊆).Using a similar proof to that of the Alexander's Subbase Lemma (see [2,Theorem 4.4.29]),we can obtain the following result.Lemma 2.2 Let X be a T 0 space, S be a subbase for O(X), and U, V ∈ O(X).Then U V if and only if one can extract a finite subcover of U from any cover Theorem 2.7 [6] Every core-compact open well-filtered space is sober.
Shen, Xi and Zhao The following result on irreducible sets in product spaces will be used in the sequel.The following lemma will be used in the sequel.
Lemma 2.11 [6, Lemma 2.4] Let X be a T 0 space, {U i : i ∈ I} be a -filtered family of open sets of X, and F be a closed set of X.
In addition, this F 0 is irreducible.

A new characterization for open well-filtered spaces
In [7,9], a characterization for well-filtered spaces by means of KF-sets is obtained.We now prove a similar characterization for open well-filtered spaces.
Definition 3.1 A nonempty subset A of a T 0 space X is called an open well-filtered set, or OWF-set for short, if there exists a -filtered family {U i : i ∈ I} ⊆ O(X) such that cl(A) is a minimal closed set that intersects all U i , i ∈ I.
Denote by OWF(X) the set of all closed OWF-subsets of X.
Remark 3.2 (1) A subset of a topological space is an OWF-set if and only if its closure is an OWF-set.
Theorem 3.3 Let X be a T 0 space.Then the following statements are equivalent: (1) X is open well-filtered; (2) ∀A ∈ OWF(X), there exists a unique x ∈ X such that A = cl({x}). Proof.
(1) ⇒ (2).Let A ∈ OWF(X).Then there exists Then cl({x}) ⊆ A, and it is a closed set that intersects all U i , i ∈ I.By the minimality of A, we have that A = cl({x}).The uniqueness of x is determined by the T 0 separation of X.
(2) ⇒ (1).Let {U i : i ∈ I} ⊆ f lt O(X) and U ∈ O(X) such that i∈I U i ⊆ U .We need to show that U i ⊆ U for some i ∈ I. If, on the contrary, that U i U for all i ∈ I, then by Lemma 2.11, there exists a minimal (irreducible) closed set A ⊆ X \ U that intersects all U i , i ∈ I. Thus A ∈ OWF(X).By the assumption, there exists a unique x ∈ X such that A = cl({x}).For each i ∈ I, since cl({x}) ∩ U i = A ∩ U i = ∅, it follows that x ∈ U i , which implies that x ∈ i∈I U i ⊆ U .Thus x ∈ U ∩ A, which contradicts A ⊆ X \ U .This contradiction completes the proof. 2 Example 3.4 Let N + be the set of all positive integers, N + cof be the space of N + with the co-finite topology (the open sets are ∅ and all the complements of finite subsets of N + ), and let N + α be the Alexandoff space of N + (the open sets are the upper subsets of N + with the usual order of numbers).We have the following claims: (c1) It is trivial to check that each subset of N + is compact in both N + cof and N + α .We then deduce that  From (r1), we have that ↓x (which is exactly the closure of {x} in ΣJ) is not an OWF-set for each x ∈ J.We then deduce that the closure of singletons need not be OWF-sets.From (r2), it follows that the saturated subspace of an open well-filtered spaces need not be open well-filtered.We have the following conclusions.
(2) N + is a Scott closed subset of P , and as a subspace of P , is homeomorphic to N + α , hence is not open well-filtered (by Example 3.4 (2)).
(3) N + , as a subspace of P , is a retract of P .
Since N + is a closed set in ΣP , we have that e is continuous, and note that r −1 (↓n) = ↓(n, ω) is Scott closed for each n ∈ N + , which implies that r is a continuous mapping.It is clear that the composition r • e is the identity mapping on N + .Thus (3) holds.
From above (1)-( 3), we deduce that the closed subspace or the retract of an open well-filtered space need not be open well-filtered.Proposition 3.8 Let X be a core-compact space.Then every irreducible set in X is an OWF-set.
Proof.Suppose that A is an irreducible subset of X.Let Claim 2: cl(A) is a minimal closed set that intersects all members of F. Suppose B is a closed set such that B ⊆ cl(A) and B ∩ U = ∅ for all U ∈ F. We need to prove Therefore Ûi 1 Ûi 3 holds.
For each i ∈ I, let By Proposition 2.4, K is a nonempty saturated compact set, that is K ∈ D(X).
Choose one U i 0 ,t 0 such that 2U i 0 ,t 0 ∩ C 0 = ∅.Then U i 0 ,t 0 ⊆ U i 0 ⊆ E, and it follows that As a matter of fact, if the statement is not true, then there is i 0 ∈ I such that K ∈ Ûi 0 .By the definition of Ûi 0 .Take any U i 0 ,t 0 with 2U i 0 ,t 0 ∩ C 0 = ∅.Then K ⊆ U i 0 ,t 0 .Take an e ∈ K \ U i 0 ,t 0 and let F = ♦cl({e}).Then by Remark 4.1 F ∩ 2U i 0 ,t 0 = ∅.
We show that C 0 ∩ F ∩ U i = ∅ for all i ∈ I.For this, it is enough to show that for any i ∈ I, there is a t i ∈ T i such that 2U i,t i ∩ F ∩ C 0 = ∅.
If not, there exists i 1 ∈ I such that for all t ∈ T i 1 .Choose i 2 such that Ûi 2 Ûi 1 .Then there are By the definition of K, we have that e ∈ K ⊆ U i 2 , and then there is t ∈ T i 2 such that e ∈ U i 2 ,t where 2U i 2 ,t ∩ C 0 = ∅.Thus Sat X ({e}) ∈ F ∩ 2U i 2 ,t = ∅.
for any open sets U, V in N + cof or N + α .(c2) Neither N + cof nor N + α is open well-filtered.For each n ∈ N + , U n = N + \ {1, 2, 3, . . ., n} is a nonempty open set in both N + cof and N + α .From (c1), it follows that the family {U n : n ∈ N + } is -filtered, but n∈N + U n = ∅.By Proposition 2.4, we obtain (c2).The following example shows that a saturated subspace of an open well-filtered space need not be open well-filtered.Example 3.5 Let J = N + ×(N + ∪{ω}) be the Johnstone's dcpo [2,3], which is ordered by (m, n) ≤ (m , n ) iff either m = m and n ≤ n , or n = ω and n ≤ m (refer to Figure 1).

Remark 3 . 6
From Example 3.5, we deduce that if each -filtered family of open sets in a space X contains the empty set, then X must be open well-filtered.The following example shows that neither the closed subspace nor the retract of an open well-filtered space is open well-filtered in general.Example 3.7 [6, Example 4.13] Let P = J ∪ N + , where J is the Johnstone's dcpo.For any x, y ∈ P , define x ≤ y if one of the following conditions holds (refer to Figure2):

Fig. 2 .
Fig. 2. The poset P of Example 3.7 (i) x, y ∈ N + and x ≤ y in N + with the usual ordering; (ii) x, y ∈ J and x ≤ y in J; (iii) x ∈ N + and y = (x, ω).
Proposition 2.10 [2, Proposition 8.4.7]Let {X i : i ∈ I} be a family of topological spaces.The irreducible closed sets in i∈I X i are exactly the sets of the form i∈I C i , where each C i is an irreducible closed set in X i (i ∈ I).