D-completion, well-filterification and sobrification
DD -completion, well-filterification and sobrification (cid:63) Hualin Miao, Longchun Wang, Qingguo Li ∗ School of Mathematics, Hunan University, Changsha, Hunan, 410082, China
Abstract
In this paper, we obtain some sufficient conditions for the D -completion of a T space to be thewell-filterification of this space, the well-filterification of a T space to be the sobrification of thisspace and the D -completion of a T space to be the sobrification, respectively. Moreover, we givean example to show that a tapered closed set may be neither the closure of a directed set nor theclosed KF -set, respectively. Because the tapered closed set is a closed W D -set, the example alsogives a negative answer to a problem proposed by Xu. Meantime, a new direct characterizationof the D -completion of a T space is given by using the notion of pre- c -compact elements. Keywords: D -completion, Well-filterification, Sobrification, Join continuous
1. Introduction D -completion, well-filterification and sobrification of a T space play a fundamental role innon-Hausdorff topological spaces. We know that the D -completion of a T space is contained inthe well-filterification of this space, and the well-filterification is contained in the sobrification.But the converses of them are not necessarily true.In [6], Keimel and Lawson verified that the D -completion of a T space is the sobrification ofthis space if the space is a c -space or qc -space. Lawson, Wu and Xi gave the sufficient conditionthat the space X is core-compact for a well-filtered space X to be sober in [8]. Xu and Shenproved that every first countable well-filtered space is sober in [14]. Xi and Lawson obtained theresult that every monotone convergence space X with the property that ↓ ( K ∩ A ) is a closedsubset of X for any closed subset A and compact saturated set K is well-filtered in [12]. Therefore,naturally, there are some questions in the following:(1) Whether the well-filterification of a first countable T space X coincides with the sobrifi-cation of this space;(2) Whether the well-filterification of a second countable T space agrees with the sobrification;(3) Whether the D -completion of a T space X with the property that ↓ ( K ∩ A ) is a closedsubset of X for any closed KF set A and compact saturated set K coincides with the well- (cid:63) This work is supported by the National Natural Science Foundation of China (No.11771134) and by HunanProvincial Innovation Foundation For Postgraduate (CX20200419) ∗ Corresponding author.
Email addresses: [email protected] (Hualin Miao), [email protected] (Longchun Wang), [email protected] (Qingguo Li) a r X i v : . [ m a t h . GN ] J a n lterification.In [10], Miao and Li proposed the concept of join-continuous poset and showed that every core-compact join continuous dcpo is sober. They also obtain the result that a monotone convergencespace X with the property that ↓ ( A ∩ W ) is a closed subset of X for any irreducible closed subset A of X and upper set W is a sober space. Naturally, some questions are raised below.(4) Whether the D -completion of a locally compact T space X with the property that ↓ ( K ∩ A )is a closed subset of X for any closed KF set A and compact saturated set K agrees with thesobrification;(5) Whether the D -completion of a core-compact join-continuous poset P coincides with thesobrification;(6) Whether the D -completion of a T space X with the property that ↓ ( A ∩ W ) is a closedsubset of X for any irreducible closed subset A of X and upper set W coincides with the sobrifi-cation.In [9], Liu, Li and Wu give a counterexample to show that WD ( X ) may not agree with KF ( X ) for any T space X , which solved the open problem proposed by Xu in [15]. Zhang andLi provided a direct characterization of the D -completion of a T space using the tapered closedsubsets ([16]). We know that every directed set is a tapered set. So, naturally there is a problemin the following. Problem 1.1.
Let X be a T space and A a tapered closed subset of X . Does there exist adirected subset D of X such that A = cl ( D )?In [17], Zhao and Fan gave the D -completion of a poset. Keimel and Lawson presented astandard D -completion of a T topological space in [6]. However, both of their work do not givethe concrete elements of the D -completion of a poset or a T space like the sobrification of thespaces. In [16], Zhang and Li provided a direct description of the elements of the D -completionof a T space by continuous functions. Shall we just characterize the D -completion of a T spaceby the elements of the space itself?The purpose of this paper is to investigate the above questions. We give some sufficientconditions for the D -completion of a T space to be the well-filterification, the well-filterificationof a T space to be the sobrification and the D -completion of a T space to be the sobrification,respectively. For Problem 1.1, we propose a counterexample to reveal that a tapered closedset may not be the closure of a directed set and the closed KF -set, respectively. Meantime, anew direct characterization of the D -completion of a T space is given by using the notion ofpre- c -compact elements.
2. Preliminaries
Without further references, the posets mentioned here are all endowed with the Scott topology.Let X be a topological space. We denote the set of all closed sets of X by Γ( X ) and all opensets by O ( X ). If P is a poset, σ ( P ) denotes the set of all Scott open sets and Γ( P ) the set of all Scott closed sets. A is called d -closed if D ⊆ A implies that sup D ∈ A for any directed subset of A . cl d ( A ) represents A is d -closed. Given a topological space ( X, τ ), we define x ≤ y iff x ∈ cl ( y ).Hence, X with its specialization order is a poset. For any set A ⊆ X , we denote the closure of A by cl ( A ). The irreducible sets of X are denoted by IRR ( X ) and the irreducible closed sets by IRR ( X ), all upper sets of X by up ( X ). 2 efinition 2.1. ([3])(i) Let P be a poset. A subset D of P is directed provided it is nonempty and every finitesubset of D has an upper bound in D . We denote that D is a directed subset of P by D ⊆ ↑ P .(ii) A poset P is a dcpo if every directed subset D has the supremum. Definition 2.2. ([4])(i) A topological space X is locally compact if for every x ∈ X and any open neighborhood U of x , there is a compact saturated subset Q of X such that x ∈ int( Q ) and Q ⊆ U .A set K of a topological space is called saturated if it is the intersection of its open neighbor-hood ( K = ↑ K in its specialization order). If A is any subset of X , the intersection sat A of allits open neighborhood is a saturated set called its saturation .(ii) A topological space X is core-compact if O ( X ) is a continuous lattice.(iii) A topological space X is said to be first countable if each point has a countable neigh-bourhood basis (local base). That is, for each point x in X there exists a sequence N , N , · · · ofthe neighbourhoods of x such that for any neighbourhood N of x , there exists an integer i with N i contained in N .(iv) A topological space is said to be second countable if its topology has a countable base. Definition 2.3. ([4]) A topological space X is sober if it is T and every irreducible closed subsetof X is the closure of a (unique) point. Definition 2.4. ([3]) We shall say that a space X is well-filtered if for each filter basis C ofcompact saturated sets and each open set U with (cid:84) C ⊆ U , there is a K ∈ C with K ⊆ U .For a topological space X , the compact saturated subsets of X are denoted by Q ( X ). Wewrite K ⊆ flt Q ( X ) represents that K is a filtered subfamily of Q ( X ) and F ⊆ fin X representsthat F is a finite subset of X . Definition 2.5. ([11]) Let X be a topological space. A nonempty subset A ∈ KF ( X ) if and onlyif there exists K ⊆ flt Q ( X ) such that cl ( A ) is a minimal closed set that intersects all members of K . The set of all closed KF -subsets of X is denoted by KF ( X ). Definition 2.6. [14] A T space X is called ω -well-filtered , if for any countable filtered family { K i : i < ω } ⊆ Q ( X ) and U ∈ O ( X ), it satisfies (cid:84) i<ω K i ⊆ U ⇒ ∃ i < ω, K i ⊆ U Definition 2.7. [14] Let X be a T space. A nonempty subset A ∈ KF ω ( X ) if and only if thereexists a countable filtered family K ⊆ Q ( X ) such that cl ( A ) is a minimal closed set that intersectsall members of K . The set of all closed KF ω -subsets of X is denoted by KF ω ( X ). Definition 2.8. [14] A subset A of a T space X is called a ω -well-filtered determined set , W D ω set for short, if for any continuous mapping f : X −→ Y to a ω -well-filtered space Y , there existsa unique y A ∈ Y such that cl ( f ( A )) = cl ( { y A } ). The set of all closed ω -well-filtered determinedsubsets of X is denoted by WD ω ( X ). Definition 2.9. ([6]) A T space is a monotone convergence space if and only if the closure ofevery directed sets (in the specialization order) is the closure of a unique point.3 efinition 2.10. ([6]) A D-completion of a T space X is a monotone convergence space Y witha topological embedding j : X −→ Y and cl d ( j ( X )) = Y . Definition 2.11. [10] A poset L is join-continuous if f x is continuous for any x ∈ L where f x : ( L, σ ( L )) → up ( L ) is defined by f x ( y ) = ↑ x ∩ ↑ y for all y ∈ L when up ( L ) is endowed withthe upper Vietoris topology. Definition 2.12. [13] A subset A of a T space X is called a well-filtered determined set , W D set for short, if for any continuous mapping f : X −→ Y to a well-filtered space Y , there existsa unique y A ∈ Y such that cl ( f ( A )) = cl ( { y A } ). The set of all closed well-filtered determinedsubsets of X is denoted by WD ( X ). Lemma 2.13. ([4]) If a topological space X is second countable, then X is first countable. Lemma 2.14. ([16]) The topological space of all tapered closed subsets of a T space X is thestandard D -completion of X . The following construction is due to Ershov [2]. Let topological spaces X and Y x , and x ∈ X ,be given. Let Z = (cid:83) x ∈ X Y x × { x } τ = { U ⊆ Z | ( U ) x ∈ τ ( Y x ) f or any x ∈ X and ( U ) X ∈ τ ( X ) } ,where U x = { y ∈ Y x | ( y, x ) ∈ U } for any x ∈ X and ( U ) X = { x ∈ X | ( U ) x (cid:54) = ∅} . In this paper, the space Z = ( Z, τ ) is denoted by (cid:80) X Y x . For any subset A ⊆ Z , put( A ) x = { y ∈ Y x | ( y, x ) ∈ A } , and ( A ) X = { x ∈ X | ( A ) x (cid:54) = ∅} . Lemma 2.15. [2] Let X be a T space, Y x an irreducible T space for any x ∈ X , and Z = (cid:80) X Y x . For all ( y , x ) , ( y , x ) ∈ Z , we have ( y , x ) ≤ Z ( y , x ) if and only if the following twoalternatives hold: (1) x = x , y ≤ Y x y ; (2) x < x and y = (cid:62) x is the greatest element in Y x with respect to the specializationorder. Lemma 2.16. [9] Let N = ( N, τ cof ) , where τ cof denotes the cofinite topology, and X n an irre-ducible T space for every n ∈ N , such that there are at most finitely many X n ’s that have agreatest element under the specialization order ≤ X n . Let Z = (cid:80) N X n . Then A is a KF -set of Z iff there exists a unique n ∈ N such that A ⊆ X n × { n } and { y ∈ X n | ( y, x ) ∈ A } is a KF -set of X n . Proposition 2.17. ([6]) For a monotone convergence space X . A is a monotone convergencesubspace iff A is a sub-dcpo of X with the specialization order. Theorem 2.18. [15] Every locally compact T space is a Rudin space, that is, IRR ( X ) = KF ( X ) . Theorem 2.19. [18] Let L be a dcpo. Then the following statements are equivalent: (1) σ ( L ) is a continuous lattice; (2) For every dcpo or complete lattice S , one has σ ( S × L ) = σ ( S ) × σ ( L ) . Theorem 2.20. [13] Let X be a T space. Then WD ( X ) with the lower Vietoris topology is thewell-filtered reflection of X . . Sufficient conditionsLemma 3.1. Let X be a T space. Then the following statements are equivalent: (1) If A ∈ KF ( X ) , then ↓ ( A ∩ K ) ∈ Γ( X ) for any K ∈ Q ( X ) ; (2) A is a directed closed subset of X .Proof. (1) ⇒ (2) It suffices to prove that A is directed. Suppose A ∈ KF ( X ). Then there exists { K i } i ∈ I ⊆ flt Q ( X ) such that A is a minimal closed set that intersects all members of { K i } i ∈ I byDefinition 2.5.Claim 1: A = ↓ ( A ∩ K j ) for any j ∈ I .For any i ∈ I , since { K i } i ∈ I ⊆ flt Q ( X ), there exists r ∈ I such that K r ⊆ K i ∩ K j . Thisimplies that ∅ (cid:54) = A ∩ K r ⊆ A ∩ K j ∩ K i ⊆ ↓ ( A ∩ K j ) ∩ K i . Then we have A = ↓ ( A ∩ K j ) by theminimality of A .Claim 2: ↑ x ∩ K i ∩ A (cid:54) = ∅ for any x ∈ A and i ∈ I . x ∈ A = ↓ ( A ∩ K i ) by Claim 1. Thus ↑ x ∩ A ∩ K i (cid:54) = ∅ .Claim 3: A is directed.Let x, y ∈ A . Then we have ↓ ( ↑ x ∩ A ) ∩ K i (cid:54) = ∅ by Claim 2. It follows that ↓ ( ↑ x ∩ A ) = A bythe minimality of A . Note that y ∈ A = ↓ ( ↑ x ∩ A ), thus ↑ y ∩ ↑ x ∩ A (cid:54) = ∅ .(2) ⇒ (1) It remains to prove that ↓ ( A ∩ K ) ∈ Γ( X ).Claim 1: ↓ ( ↑ x ∩ A ) ∈ Γ( X ) for any x ∈ X .If ↑ x ∩ A = ∅ , then we have ↓ ( ↑ x ∩ A ) = ∅ ∈ Γ( X ). Else ↑ x ∩ A (cid:54) = ∅ . Then x ∈ A because A is a lower set. Obviously, ↓ ( ↑ x ∩ A ) ⊆ A . Conversely, ↑ a ∩ ↑ x ∩ A (cid:54) = ∅ for any a ∈ A since A isdirected. So we have ↓ ( ↑ x ∩ A ) = A ∈ Γ( X ).Claim 2: ↓ ( K ∩ A ) ∈ Γ( X ).If K ∩ A = ∅ , then ↓ ( K ∩ A ) = ∅ ∈ Γ( X ). Else K ∩ A (cid:54) = ∅ . Pick k ∈ K ∩ A . This means that A ⊆ ↓ ( ↑ k ∩ A ) ⊆ ↓ ( K ∩ A ) ⊆ A by Claim 1. Hence, ↓ ( K ∩ A ) = A ∈ Γ( X ). Theorem 3.2.
Let X be a locally compact T space with the property ↓ ( K ∩ A ) ∈ Γ( X ) for any A ∈ KF ( X ) and K ∈ Q ( X ) . Then the D -completion of X coincides with the soberification of X .Proof. Let A ∈ IRR ( X ). It suffices to prove that A is directed. By Theorem 2.18, we have IRR ( X ) = KF ( X ). This implies that A is directed by Lemma 3.1.Theorem 3.2 gives a positive answer for the problem (4) in the introduction. Theorem 3.3.
Let X be a T space with the property ↓ ( K ∩ A ) ∈ Γ( X ) for any A ∈ KF ( X ) and K ∈ Q ( X ) . If KF ( X ) endowed with the lower Vietoris topology is the well-filterification of X .Then the D -completion of X agrees with the well-filterification of X .Proof. By Lemma 2.14, we know that all tapered closed subsets of X is the D -completion of X .Note that directed closed subsets of X must be tapered. We need to prove that A is a directedsubset of X for any tapered closed subset A of X . It suffices to prove that D = { cl ( D ) | D ⊆ ↑ X } is a subdcpo of Γ( X ). Let { cl ( D i ) } i ∈ I be any directed subset of Γ( X ) contained in D . Note thatsup i ∈ I cl ( D i ) = cl ( (cid:83) i ∈ I cl ( D i )). We need to verify that sup i ∈ I cl ( D i ) ∈ D . Since D ⊆ KF ( X ),we have cl ( D i ) is directed for any i ∈ I by Lemma 3.1. It suffices to show that (cid:83) i ∈ I cl ( D i ) is adirected subset of X . Now let x, y ∈ (cid:83) i ∈ I cl ( D i ). Then there exists { i x , i y } ⊆ I such that x ∈ l ( D i x ) , y ∈ cl ( D i y ). It follows that there exists i ∈ I such that { x, y } ⊆ cl ( D i x ) ∪ cl ( D i y ) ⊆ cl ( D i )by the directionality of ( cl ( D i )) i ∈ I . Thus there exists z ∈ cl ( D i ) ⊆ (cid:83) i ∈ I cl ( D i ) such that z is anupper bound of x, y because cl ( D i ) is a directed subset of X . So sup i ∈ I cl ( D i ) ∈ D . By Lemma3.1, we have KF ( X ) ⊆ D . Hence, the D -completion of X agrees with the well-filterification of X . We know that a monotone convergence space with the property ↓ ( K ∩ A ) ∈ Γ( X ) for any A ∈ KF ( X ) and K ∈ Q ( X ) is a well-filtered space by [12]. So there is a problem below: Problem 3.4.
Without the condition that KF ( X ) endowed with the lower Vietoris topology isthe well-filterification of X , whether the statement in Theorem 3.3 holds for any T space X withthe property ↓ ( K ∩ A ) ∈ Γ( X ) for any A ∈ KF ( X ) and K ∈ Q ( X ).Note that Problem 3.4 actually is the problem (3) in the introduction. For this problem, wegive a counterexample as follows. Example 3.5.
Let X = ( N , τ cof ) , Y n = ( N , σ ( N )) for any n ∈ X , Z = (cid:80) X Y n , where N is theset of natural numbers. Then ↓ ( K ∩ A ) ∈ Γ( Z ) for any A ∈ KF ( Z ) and K ∈ Q ( Z ), but the D -completion Z d of Z does not agree with the well-filterification Z w of Z . Proof.
By Lemma 2.16 and Lemma 2.15, we have that A is a directed subset of Z for any A ∈ KF ( Z ). This implies that ↓ ( K ∩ A ) ∈ Γ( Z ) for any A ∈ KF ( Z ) and K ∈ Q ( Z ) because ofLemma 3.1. Let N = { Y n × { n } | n ∈ X } . Note that Y n × { n } ⊆ ↑ Z for any n ∈ X . It followsthat N ⊆ Z w .Claim 1: A is compact in Z w for any A ⊆ N .Let { U i } i ∈ I ⊆ O ( Z ) with A ⊆ (cid:83) i ∈ I ♦ U i . Pick Y n × { n } ∈ A ⊆ (cid:83) i ∈ I ♦ U i . So thereexists i n ∈ I such that Y n × { n } ∈ ♦ U i n . ( U i n ) X ∈ τ cof since U i n is open in Z . Let B = X \ ( U i n ) X , C = { n ∈ X | Y n × { n } ∈ A , Y n × { n } ∩ U i n = ∅} . Note that B is finite and C ⊆ B . Then there exists finitely many members of {♦ U i } i ∈ I to cover A .Let A n = N \{ Y i × { i } | i ∈ { , , · · · n }} . Thus {↑ Z w A n } n ∈ X ⊆ flt Q ( Z w ) by Claim 1.Since Z w is well-filtered, we have (cid:84) n ∈ X ↑ Z w A n (cid:54) = ∅ . Choose B ∈ (cid:84) n ∈ X ↑ Z w A n , i.e., B ∈ Z w .Claim 2: ( B ) X is infinite.Suppose not, if ( B ) X is finite. Let n = max ( B ) X . Since B ∈ (cid:84) n ∈ X ↑ Z w A n ⊆ ↑ Z w A n , wehave ↓ Z w B ∩ A n (cid:54) = ∅ . Then there exists n > n such that Y n × { n } ∈ ↓ Z w B ∩ A n , whichcontradicts n / ∈ ( B ) X .By Lemma 2.15, B is not directed. This implies that B is not a tapered closed subset of Z bythe proof of Theorem 3.3. Thus B ∈ Z w \ Z d , that is, the D -completion Z d of Z does not agreewith the well-filterification Z w of Z . Lemma 3.6.
Let L be a poset. Then the following statements are equivalent: (1) σ ( L ) is a continuous lattice; (2) For every dcpo or complete lattice S , one has σ ( S × L ) = σ ( S ) × σ ( L ) .Proof. The proof is similar to Theorem 2.19. 6 heorem 3.7.
Let L be a core-compact and join continuous poset. Then the D -completion of L coincides with the soberification of L .Proof. Let A be a irreducible closed subset of X . Then it suffices to prove that A is directed.Define F : ( L × L, σ ( L × L )) −→ up ( L ) by F ( x, y ) = ↑ x ∩ ↑ y for any ( x, y ) ∈ L × L .Claim 1: F is continuous.Let U ∈ σ ( L ). We need to prove that F − ( (cid:3) U ) is Scott open in L × L . It suffices toprove that F − ( (cid:3) U ) is Scott open when F − ( (cid:3) U ) (cid:54) = ∅ . Obviously, F − ( (cid:3) U ) = ↑ ( F − ( (cid:3) U )).Now let { ( x i , y i ) } i ∈ I be any directed subset of L × L such that sup i ∈ I ( x i , y i ) exists in L × L withsup i ∈ I ( x i , y i ) = (sup i ∈ I x i , sup i ∈ I y i ) ∈ F − ( (cid:3) U ). This implies that ↑ sup i ∈ I x i ∩↑ sup i ∈ I y i ∈ (cid:3) U .So we have f sup i ∈ I x i (sup i ∈ I y i ) ∈ (cid:3) U . It follows that sup i ∈ I y i ∈ f − i ∈ I x i ( (cid:3) U ) is Scott open in L because L is join continuous. This implies that there exists i ∈ I such that y i ∈ f − i ∈ I x i ( (cid:3) U ).Thus ↑ y i ∩ ↑ sup i ∈ I x i ∈ (cid:3) U . Then we have sup i ∈ I x i ∈ f − y i ( (cid:3) U ). By the join continuity of L ,there exists i ∈ I such that x i ∈ f − y i ( (cid:3) U ). Thus there exists i ∈ I such that ( x i , y i ) is an upperbound of { ( x i , y i ) , ( x i , y i ) } because { ( x i , y i ) } i ∈ I is directed. Therefore, ( x i , y i ) ∈ F − ( (cid:3) U ).Hence, F is continuous.Claim 2: A is a directed subset of X .By Lemma 3.6 and Claim 1, we have F : σ ( L ) × σ ( L ) −→ up ( L ) is continuous. Let { x, y } ⊆ A .We need to prove that ↑ x ∩ ↑ y ∩ A (cid:54) = ∅ . Suppose not, if ↑ x ∩ ↑ y ∩ A = ∅ . It follows that ↑ x ∩ ↑ y ⊆ (cid:3) ( X \ A ), that is, F (( x, y )) ∈ (cid:3) ( X \ A ). This implies that there exists { U, V } ⊆ σ ( L )such that ( x, y ) ∈ U × V ⊆ F − ( (cid:3) ( X \ A )) by the continuity of F . Notice that x ∈ A ∩ U , y ∈ A ∩ V . We have A ∩ U ∩ V (cid:54) = ∅ since A is irreducible. Pick z ∈ A ∩ U ∩ V . Hence( z, z ) ∈ U × V ⊆ F − ( (cid:3) ( X \ A )). This means that z ∈ ↑ z ∩ ↑ z ⊆ ( X \ A ), which contradicts z ∈ A .So A is directed.For the problem (5) mentioned in the introduction, we give a positive answer by Theorem3.7. Corollary 3.8.
Let L be a core-compact poset with the property that ↓ ( K ∩ A ) ∈ Γ( L ) for any A ∈ IRR ( L ) and K ∈ Q ( L ) . Then the D -completion of L agrees with the soberification of L .Proof. The proof is similar to Theorem 3.7.In contrast to Theorems 3.2 and Corollary 3.8, we raise the following question.
Problem 3.9.
Whether the statement in Corollary 3.8 holds for arbitrary core-compact topo-logical spaces X with the property that ↓ ( K ∩ A ) ∈ Γ( X ) for any A ∈ IRR ( X ) and K ∈ Q ( X ). Proposition 3.10.
Let X be a second countable space. Then the well-filterification of X coincideswith the soberification of X .Proof. It suffices to prove that the well-filterification of X is second countable by Theorem 4.1in [14]. Let B be a countable basis of X , U = {♦ U | U ∈ B} . Since U is countable, we only needto prove that U is a basis of the well-filterification of X . Let A ∈ WD ( X ), V ∈ O ( X ), A ∈ ♦ V .Then A ∩ V (cid:54) = ∅ . Pick x ∈ A ∩ V . This implies that there exists U ∈ B such that x ∈ U ⊆ V . Itfollows that A ∈ ♦ U ⊆ ♦ V . 7 e give a positive answer for the problem (2) in the introduction by Theorem 3.10. Lemma 3.11.
Let X be a T space, X s the soberification of X . For each ordinal β , define (1) X β +1 = { x ∈ X s | ∃ F ∈ KF ( X β ) , cl X s ( F ) = ↓ X s x } ; (2) X β = (cid:83) γ<β X γ for a limit ordinal β .Then there exists an ordinal α X such that X α X = X α X +1 and the ω -well-filtered reflection is X α X endowed with the lower Vietoris topology.Proof. The proof is similar to Proposition 3.8 in [11].
Theorem 3.12.
Let X be a first countable space. Then the well-filterification of X agrees withthe soberification of X .Proof. By Theorem 4.1 in [14], we only need to prove that the ω -well-filterification of X is firstcountable. There exists an ordinal α X such that X α X = X α X +1 and the ω -well-filtered reflectionis X α X endowed with the lower Vietoris topology by Lemma 3.11. We want to prove that X α endowed with the lower Vietoris topology is first countable for any ordinal α ≤ α X . We useinduction on α . For α = 0, X = {↓ x | x ∈ X } . Then X and X are homeomorphic. This impliesthat X is first countable. Let α be such that α + 1 ≤ α X , and let X α be first countable.Claim 1: X α +1 is first countable.Let A ∈ X α +1 . Then there exists F ∈ KF ( X α ) such that cl X s ( F ) = ↓ X s A . Since F ∈ KF ( X α ), there exists ( K n ) n ∈ N ⊆ Q ( X α ) such that F is a minimal closed set that intersects allmembers of {K n | n ∈ N } . Pick A n ∈ F ∩ K n for any n ∈ N . It follows that F = cl X α ( A ) fromthe minimality of F , where A = { A n | n ∈ N } . We want to prove that cl X s ( F ) = cl X s ( A ). Since cl X s ( F ) ∩ X α = F = cl X α ( A ) = cl X s ( A ) ∩ X α , we have F ⊆ cl X s ( A ). So cl X s ( F ) ⊆ cl X s ( A ).Conversely, note that A n ∈ K n ⊆ X α for any n ∈ N . Hence, A ⊆ cl X s ( A ) ∩ X α = cl X s ( F ) ∩ X α .This implies that A ⊆ cl X s ( F ). Therefore, cl X s ( F ) = cl X s ( A ). Because X α is first countable,there exists a countable neighborhood basis B n = {♦ U ∩ X α | ♦ U ∩ X α ∈ B n } of A n for any n ∈ N .Let B A = {♦ U ∩ X α +1 | ♦ U ∩ X α ∈ (cid:83) n ∈ N B n } . Note that B A is countable. We only need to provethat B A is a countable neighborhood basis of A in X α +1 . Let V ∈ O ( X ) with A ∈ ♦ V ∩ X α +1 .It follows that A ∩ ♦ V (cid:54) = ∅ since ↓ X s A = cl X s ( F ) = cl X s ( A ). Choose A n ∈ A ∩ ♦ V ∩ X α . Thenthere exists ♦ U ∩ X α ∈ B n such that A n ∈ ♦ U ∩ X α ⊆ ♦ V ∩ X α . We need to prove that U ⊆ V .Suppose not, if U (cid:42) V , then there exists u ∈ U ∩ ( X \ V ). This implies that ↓ u ∈ ♦ U ∩ X α ⊆ ♦ V .Thus u ∈ V , which contradicts u ∈ X \ V . So A ∈ ♦ U ∩ X α +1 ⊆ ♦ V ∩ X α +1 .Suppose now that α ≤ α X is a limit ordinal and X β is first countable for any β < α . Bydefinition of X α , we have X α = (cid:83) β<α X α .Claim 2: X α is first countable.For any A ∈ X α , there exists β < α such that A ∈ X β . We have that there exists a countableneighborhood basis B = {♦ U ∩ X β | ♦ U ∩ X β ∈ B} of A in X β since X β is first countable. Itsuffices to prove that B A = {♦ U ∩ X α | ♦ U ∩ X β ∈ B} is a countable neighborhood basis of A in X α . It is obvious that B A is countable. Now let V ∈ O ( X ) with A ∈ X α ∩ ♦ V . Then A ∈ ♦ V ∩ X β . It follows that there exists ♦ U ∩ X β ∈ B such that A ∈ ♦ U ∩ X β ⊆ ♦ V ∩ X β .Similar to the proof of Claim 1, we have U ⊆ V . So A ∈ ♦ U ∩ X α ⊆ ♦ V ∩ X α . This proves that X α endowed with the lower Vietoris topology is first countable for any ordinal α ≤ α X . Hence,the ω -well-filtered reflection X α X of X is first countable.8e give a positive answer by Theorem 3.12 to the problem (1) in the introduction. Theorem 3.13.
Let X be a T space with the property ↓ ( A ∩ W ) ∈ Γ( X ) for any A ∈ IRR ( X ) and W ∈ up ( X ) . Then the D -completion of X coincides with the soberification of X .Proof. Let A ∈ IRR ( X ). It suffices to prove that A is directed. For any x, y ∈ A , we need to provethat ↑ x ∩ ↑ y ∩ A (cid:54) = ∅ . Suppose not, ↑ x ∩ ↑ y ∩ A = ∅ . Let B = ↓ ( ↑ x ∩ A ). Then ↓ ( X \ B ) ∈ Γ( X ). Itfollows that A = ( B ∪ ( X \ B )) ∩ A = ( B ∩ A ) ∪ (( X \ B ) ∩ A ) = B ∪ (( X \ B ) ∩ A ) ⊆ B ∪↓ (( X \ B ) ∩ A ).Thus A ⊆ B or A ⊆ ↓ (( X \ B ) ∩ A ). But y ∈ A ∩ ( X \ B ) and ↑ x ∩ A ∩ ( X \ B ) = ∅ implies that A (cid:42) B and A (cid:42) ↓ (( X \ B ) ∩ A ), which contradicts A ⊆ B or A ⊆ ↓ (( X \ B ) ∩ A ).For the problem (6) in the introduction, we give a positive answer by Theorem 3.13.
4. A counterexample
In the following, we construct an example to give a negative answer to Problem 1.1. It alsocan answer Xu’s problem ([15]).
Figure 1: A tapered non-directed-determined poset.
Example 4.1.
Let L = N × N × ( N ∪ ∞ ), where N is the set of natural numbers. We define anorder ≤ on L as follows:( n , i , j ) ≤ ( n , i , j ) if and only if: • n = n , i = i , j ≤ j ; • n = n , i = j = j , i ≤ i ; • n = n + 1 , j ≤ i , j = ∞ . L can be easily depicted as in Figure 1. Then L is tapered closed in ( L, σ ( L )), but L / ∈ KF ( L )and L / ∈ { cl ( D ) | D ⊆ ↑ L } . Proof.
Claim 1: L is tapered closed. 9et A n = { ( n, i, j ) | ( i, j ) ∈ N × ( N ∪ ∞ ) } . Note that (cid:83) ni =1 A i = cl ( A n \ max A n ) for any n ∈ N and A n \ max A n ⊆ ↑ L . Hence, (cid:83) ni =1 A i is tapered closed. Let B n = (cid:83) ni =1 A i . It followsthat { B n } n ∈ N ⊆ ↑ Γ( L ). Then we have L = sup n ∈ N B n is tapered closed.Claim 2: L / ∈ { cl ( D ) | D ⊆ ↑ L } .Suppose not, if there exists a directed subset D of L such that cl ( D ) = L . We need to provethat ( D ) N = { n ∈ N | A n ∩ D (cid:54) = ∅} is infinite. Assume ( D ) N is finite. Let n = max( D ) N .Then we have D ⊆ (cid:83) i ∈ ( D ) N A i ⊆ (cid:83) n i =1 A i = B n . So L = cl ( D ) ⊆ B n , which contradicts B n (cid:36) L . Thus ( D ) N is infinite. Let n ∈ ( D ) N . Pick ( n , i , j ) ∈ D ∩ A n . By the infinitenessof ( D ) N , there exists m ∈ ( D ) N such that m ≥ n + 2. Choose ( m, i m , j m ) ∈ D ∩ A m . Obviously, ↑ ( n , i , j ) ⊆ A n ∪ A n +1 and ↑ ( m, i m , j m ) ⊆ A m ∪ A m +1 , ( A n ∪ A n +1 ) ∩ ( A m ∪ A m +1 ) = ∅ . Itfollows that ↑ ( n , i , j ) ∩ ↑ ( m, i m , j m ) = ∅ , which contradicts that D is directed.Claim 3: ( K ) N = { n ∈ N | K ∩ A n (cid:54) = ∅} is finite for any K ∈ Q ( L ).Suppose not, if ( K ) N is infinite. For any n ∈ ( K ) N , pick ( n, i n , j ) ∈ K ∩ A n , Then ( n, i n , ∞ ) ∈ K ∩ A n . Let F ⊆ fin ( K ) N , B F = { (cid:83) n ∈ ( K ) N \ F ↓ ( n, i n , ∞ ) } . We need to verify that B F ∈ Γ( L ).Now let D ⊆ ↑ B F . If D is finite, then sup D ∈ D ⊆ B F . Else, there exists a unique n ∈ ( K ) N such that D ⊆ ↓ ( n, i n , ∞ ). This implies that sup D ∈ ↓ ( n, i n , ∞ ) ⊆ B F . ↓ B F = B F is obvious.Note that B F ∩ K (cid:54) = ∅ for any F ⊆ fin ( K ) N . So we have K ∩ (cid:84) F ⊆ fin ( K ) N B F (cid:54) = ∅ . Therefore,there exists ( n, i, j ) ∈ K ∩ (cid:84) F ⊆ fin ( K ) N B F . Since ( K ) N is infinite, we have m ∈ ( K ) N such that m ≥ n + 2.We now distinguish two cases:Case 1, n + 1 ∈ ( K ) N : Then ( n, i, j ) ∈ (cid:84) F ⊆ fin ( K ) N B F ⊆ B { n,n +1 } ∩ A n , which contradicts B { n,n +1 } ∩ A n = ∅ .Case 2, n + 1 / ∈ ( K ) N : Then ( n, i, j ) ∈ (cid:84) F ⊆ fin ( K ) N B F ⊆ B { n } ∩ A n , which contradicts B { n } ∩ A n = ∅ .Claim 4: L / ∈ KF ( L ).Suppose not, L ∈ KF ( L ), then there exists { K i } i ∈ I ⊆ flt Q ( L ) such that L is a minimal closedset that intersects all members of { K i } i ∈ I by Definition 2.5. Pick i ∈ I . ( K i ) N is finite by Claim3. Now let m = max( K i ) N . This implies that K i ⊆ B m . We need to prove that B m ∩ K i (cid:54) = ∅ for any i ∈ I . For any i ∈ I , there exists j ∈ I such that K j ⊆ K i ∩ K i because { K i } i ∈ I is filter.It follows that ∅ (cid:54) = K j = B m ∩ K j ⊆ B m ∩ K i . So L = B m by the minimality of L , whichcontradicts L (cid:54) = B m . Example 4.2.
Let J be Johnstone’s Example. Then J ∈ KF ( J ), but J is not a tapered closedset. Remark 4.3.
By Example 4.1 and 4.2, we know that there is no relation between the taperedclosed subsets and the closed KF -subsets of a T space X .Figure 2 below shows certain relations among some kinds of subsets of a T space.
5. A direct characterization for D -completion Let P be a poset, H = {D ⊆ Γ( P ) \∅ | , ∀ A ∈ D , ∀ a ∈ A, ↓ a ∈ D , cl d ( D ) = D} and A, B ∈ Γ( P ).We say that A is beneath B denoted by A ≺ B , if for every D ∈ H , the relation B ⊆ sup D igure 2: Certain relations among some kinds of subsets of a T space. always implies that A ∈ D .Notice that the relation which we define above is not an auxiliary relation because the elementsof H may not be lower sets. Definition 5.1.
Let P be a poset. An element A of Γ( P ) is called pre-C-compact if A ≺ A . Weuse K (Γ( P )) to denote the set of all the pre- C -compact elements of P .Ho and Zhao introduce C-continuous posets and C -compact elements in [19]. They provedthat the C -compact elements of a poset P are co-primes. They find that the C -compact elementsof Γ( P ) are more than the elements of the D -completion of the posets. Now we want to find away to reduce the C -compact elements of P such that they are equal. Definition 5.2. [19] Let P be a poset and x, y ∈ P . We say that x is beneath y , denoted by x ≺ y , if for every nonempty Scott closed set C ⊆ P for which sup C exists, the relation y ≤ sup C always implies that x ∈ C . An element x of a poset P is called C -compact if x ≺ x .The pre- C -compact elements are C -compact by Definition 5.2. The following example revealsthat C -compact elements may not be pre- C -compact. Example 5.3.
Let J = N × ( N ∪ {∞} ), that is Johnstone’s example. The partial order is( m, n ) ≤ ( a, b ) iff either m = a and n ≤ b or b = ∞ and n ≤ a . That J is C -compact have beenproved in Remark 5.3 of [19]. We now show that J is not pre- C -compact. Let D = {↓ x | x ∈ J } .Note that D ∈ H and sup D = J but J (cid:54)∈ D .The following proposition reveals some order-theoretic properties of the lattice (Γ( P ) , ⊆ ) foran arbitrary poset P . Proposition 5.4.
Let P be a poset. Then for any
D ∈ H , sup Γ( P ) D = (cid:83) D .Proof. Note that each member of H is a Scott-closed subset of P . So to prove the equation, itsuffices to show that (cid:83) D ∈ Γ( P ). Obviously, (cid:83) D is a lower set. Now Let D be any directedsubset of P contained in (cid:83) D such that sup D exists in P . We want to prove that sup D ∈ C forsome C ∈ D . First note that C = {↓ d | d ∈ D } is a directed subset of Γ( P ). Moreover, C ⊆ D by the definition of D . Since D is a d -closed set, so sup C ∈ D . But sup C is precisely ↓ sup D .Hence sup D ∈ C for some C ∈ D . 11 roposition 5.5. Let P be a poset. If A ∈ K (Γ( P )) , then A is an irreducible closed set.Proof. Suppose A is pre- C -compact and B, C ∈ Γ( P ), A ⊆ B ∪ C . Let D = ↓{ B, C } . Then D ∈ H and sup D = B ∪ C . Hence, A ∈ D , so A ⊆ B or A ⊆ C . Lemma 5.6.
Let P be a poset and D be a directed subset of P . Then cl ( D ) ∈ K (Γ( P )) .Proof. Let
D ∈ H . Suppose cl ( D ) ⊆ sup D . Then by Proposition 5.4, D ⊆ sup D and {↓ d | d ∈ D } is a directed subset of D . Hence, cl ( D ) = sup {↓ d | d ∈ D } ∈ D . Corollary 5.7.
Let P be a poset. Then for each x ∈ P , ↓ x ≺ ↓ x holds. Theorem 5.8.
For any poset P , K (Γ( P )) is a dcpo with the inclusion order.Proof. Let ( A i ) i ∈ I be a directed subset in K (Γ( P )). It suffices to show that sup i ∈ I A i ≺ sup i ∈ I A i .Suppose D ∈ H with sup i ∈ I A i ⊆ sup D . Then A i ⊆ sup D for all i ∈ I . Since A i ∈ K (Γ( P )), itfollows that A i ≺ A i and so A i ∈ D . Because D is d -closed, this implies that sup i ∈ I A i ∈ D .We write ˆ P = K (Γ( P )) with the relative topology of Γ( P ). Proposition 5.9.
Let P be a poset. Then ˆ P is a monotone convergence space.Proof. It is easy to prove by Proposition 2.17.
Lemma 5.10.
Let P be a poset. If A ⊆ P is tapered and closed, then A ∈ K (Γ( P )) .Proof. Assume P ∨ is the set of all tapered closed subsets and A ∈ P ∨ , then ( P ∨ , j ) is the D -completion of P by Lemma 2.14. Define f : P → K (Γ( P )) by f ( x ) = ↓ x for all x ∈ P .Then, immediately, f is continuous. As ( P ∨ , j ) is the D -completion of P , it follows that thereexists a unique continuous map ˆ f : P ∨ → K (Γ( P )) s.t. f = ˆ f ◦ j . Thus sup f ( A ) = ˆ f ( A ) byTheorem 3.10 (2) of [19]. We need to prove A ∈ K (Γ( P )). Suppose not, A / ∈ K (Γ( P )). Since ↓ a ≤ A for any a ∈ A and ˆ f is continuous, we have ↓ a = ˆ f ( ↓ a ) ≤ ˆ f ( A ) = sup f ( A ). Hence, A (cid:38) sup f ( A ), i.e. sup f ( A ) ∩ A c (cid:54) = Ø. So we have sup f ( A ) ∈ ♦ ( A c ). Because ˆ f is continuous,this implies ˆ f − ( ♦ ( A c )) is open in P ∨ and A ∈ ˆ f − ( ♦ ( A c )). Hence, there exists U ∈ σ ( P ) suchthat A ∈ ♦ U ⊆ ˆ f − ( ♦ ( A c )). This means that A ∩ U (cid:54) = ∅ , then there exists a ∈ A ∩ U such that ↓ a ∈ ♦ U ⊆ ˆ f − ( ♦ ( A c )), that is ˆ f ( ↓ a ) = ↓ a ∈ ♦ ( A c ) which contradicts a ∈ A . Thus we haveshown the Lemma. Theorem 5.11.
Let P be a poset. Then K (Γ( P )) = cl d ( η ( P )) where η : P → Γ( P ) is defined by η ( x ) = ↓ x for all x ∈ P , i.e. ( ˆ P , η ) is the D -completion of P .Proof. For any x ∈ P , η ( x ) = ↓ x ∈ K (Γ( P )) by Corollary 5.7. Then η ( P ) ⊆ K (Γ( P )). We haveshown that K (Γ( P )) is a subdcpo of Γ( P ), so cl d ( η ( P )) ⊆ K (Γ( P )).Conversely, note that cl d ( η ( P )) ∈ H and sup cl d ( η ( P )) = P . If A ∈ K (Γ( P )), then A ⊆ sup cl d ( η ( P )). This implies A ∈ cl d ( η ( P )) since A ≺ A .From the proof of Theorem 5.11, we have the following: Corollary 5.12.
Let P be a poset. Then the pre-C-compact elements are exactly the taperedclosed sets. efinition 5.13. Let (
X, τ ) be a topological space. X is called determined by its directed subsets if τ = { U ⊆ X | cl ( D ) ∩ U (cid:54) = ∅ implies D ∩ U (cid:54) = ∅} for all directed subsets with respect to theorder of specialization.The following two theorems have the similar proof to that of posets. Theorem 5.14.
Let X be a topological space. Suppose X is determined by its directed subsets,then we can get ( K (Γ( X )) , η ) is the D -completion of X , where η : X → Γ( X ) is defined by η ( x ) = ↓ x for all x ∈ X . Theorem 5.15.
Suppose X is a topological space. Define H = {D ⊆ Γ( X ) | ∀ A ∈ D , ∀ a ∈ A, ↓ a ∈ D , cl d ( D ) = D , cl ( (cid:83) D ) = (cid:83) D} . Then ( K (Γ( X )) , η ) is the D -completion of X . Corollary 5.16.
A topological space X is a monotone convergence space if and only if for any A ∈ K (Γ( X )) , there exists a unique point x in X , such that A = ↓ x .
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