Two Topologise on the Lattice of Scott Closed Subsets?

Two topologise on the lattice of Scott closed subsets?

Yu Chen, Hui Kou∗, Zhenchao Lyu Department of Mathematics, Sichuan University, Chengdu 610064, China

Abstract For a poset P , let σ(P ) and Γ(P ) respectively denote the lattice of its Scott open subsets and Scott closed subsets ordered by inclusion, and set ΣP = (P, σ(P )). In this paper, we discuss the lower Vietoris topology and the Scott topology on Γ(P ) and give some sufficient conditions to make the two topologies equal. We built an adjunction between σ(P ) and σ(Γ(P )) and proved that ΣP is core-compact iff ΣΓ(P ) is core-compact iff ΣΓ(P ) is sober, locally compact and σ(Γ(P )) = υ(Γ(P )) (the lower Vietoris topology). This answers a question in [17]. Brecht and Kawai [2] asked whether the consonance of a topological space X implies the consonance of its lower powerspace, we give a partial answer to this question at the last part of this paper.

Keywords: Scott closed sets; adjunction; distributive continuous lattice; lower powerspace; consonance.

1. Introduction

The lower powerspace over a topological space X is the set of closed subsets of X with the lower Vietoris topology. The lower powerspace coincides with the lower powerdomain for continuous dcpos with the Scott topology, where the latter construction is used in modelling non-deterministic computation (see [1, 19]). There naturally arise a question when the lower Vietoris topology and the Scott topology coincide on the lattice of closed subsets of a topological space. For a poset P , we give some sufficient conditions such that the lower Vietoris topology and the Scott topology coincide on Γ(P ). We observe that there is an adjunction between σ(P ) and σ(Γ(P )), which serves as a useful tool in studying the relation between P and Γ(P ). Particularly, we obtain the conclusion that (P, σ(P )) is core-compact iff (Γ(P ), σ(Γ(P )) is a sober and locally compact space. This positively answers a question in [17]. We show that the two topologies generally don’t coincide on the lattice of closed subsets of a topological space that is not a Scott space. In domain theory, the spectral theory of distributive continuous lattices and the duality theorem have been extensively studied. Hoffmann [12] and Lawson [14] independently proved that a directed complete poset (or dcpo for short) is continuous iff the lattice of its Scott open subsets is completely distributive. Gierz and Lawson [8] proved that quasicontinuous dcpos equipped with the Scott topologies are precisely the spectrum of distributive hypercontinuous lattices. The correspondence between P and Γ(P ) has also been studied (see [7, 13, 14, 21, 11]). A poset P is continuous (resp., quasicontinuous, algebraic, quasialgebraic) if and only if Γ(P ) is continuous (resp., quasicontinuous, algebraic, quasialgebraic). Using the pair of adjoint we can give a unified proof. We see that if L is a continuous lattice such that σ(Lop) = υ(Lop), then (Lop, σ(Lop)) is a sober and locally compact space. We give two equivalent conditions of σ(Lop) = υ(Lop) for continuous lattices. The previous result about core-compact posets implies that if L is a distributive continuous lattice such that the hull-kernel topology arXiv:2103.15139v1 [math.GN] 28 Mar 2021 of SpecL is just the Scott topology then σ(Lop) = υ(Lop). We wonder whether the condition σ(Lop) = υ(Lop) implies that SpecL is a Scott space. If not, what other conditions are needed. Actually, characterize those distributive continuous lattices for which the spectrum is a dcpo equipped with the Scott topology is an open problem [20]. The case of distributive algebraic lattices is simpler. The notion of jointly Scott continuity occurs in our proofs. It is well-known that a complete lattice is sober with respect to the Scott topology if it is jointly Scott continuity [7, Corollary II-1.12]. We give an example of a complete lattice which is sober with the Scott topology but fails to be jointly Scott continuity. Consonance is an important topological property (see [2, 4, 5, 18]). It is proved in [5] that every regular Céch-complete space is consonant. Thus every completely metrizable space is consonant. Bouziad [4] proved

?Research supported by NSF of China (Nos. 11871353, 12001385). ∗Corresponding author Email addresses: [email protected] (Yu Chen), [email protected] (Hui Kou), [email protected] (Zhenchao Lyu)

Preprint submitted to Topology and its applications March 30, 2021 that the set of rational numbers with subspace topology from R is not consonant. A space X is consonant if and only if, for every topological space Y , the compact-open and the Isbell topologies agree on the space of continuous maps from X to Y (see [9]). Brecht and Kawai [2] proved that the consonance of a T0 space X is equivalent to the commutativity of the upper and lower powerspaces of X. There they asked whether the consonance of X implies the consonance of its lower powerspace. We will give a partial answer at the last part of this paper.

2. Preliminaries

We refer to ([1, 7, 9]) for some concepts and notations of domain theory that will be used in the paper. For a poset P and A ⊆ P , let ↓ A = {x ∈ P : ∃a ∈ A, x ≤ a} and ↑ A = {x ∈ P : ∃a ∈ P, a ≤ x}. For x ∈ P , we write ↓x for ↓{x} and ↑x for ↑{x}. A set A is called saturated if and only if A =↑A. If F is a finite subset of P , we denote by F ⊆f P . A poset P is called a directed complete poset (dcpo, for short) if every directed subset D of P has a supremum. For a poset P , the upper topology υ(P ) is the topology generated by taking the collection of sets {P \ ↓x : x ∈ P } as a subbais, the lower topology ω(P ) on P is defined dually. A subset A of P is called Scott closed if A =↓ A and for any directed set D ⊆ A, W D ∈ A whenever the least upper bound W D exists. The Scott topology σ(P ) consists of the complements of all Scott closed sets of P . The topology λ(P ) = σ(P ) ∨ ω(P ) is called the Lawson topology on P . Given a topological space X, let O(X) be the topology on X, then O(X) is a complete lattice ordered by inclusion. We denote by (C(X), υ(C(X))) the lattice of closed subsets of X equipped with the lower Vietoris topology, which is generated by 3U = {A ∈ C(X): A ∩ U 6= ∅} as a subbase, where U ranges over O(X). A topological space is core-compact if and only if O(X) is a continuous lattice. An arbitrary nonempty subset A of a topological space X is irreducible if A ⊆ B ∪ C for closed subsets B and C implies A ⊆ B or A ⊆ C. X is called sober if for every irreducible closed set A, there exists a unique x ∈ X such that {x}− = A. For a poset P and x, y ∈ P , we say that x is way-below y, in symbols x y, if for any directed subset D of P that has a least upper bound in P , y ≤ W D implies x ≤ d for some d ∈ D. For any non-empty subsets F,G of P , we say that F is way-below G and write F G if for any directed subset D of P that has a least upper bound in P , W D ∈↑G implies D ∩ ↑F 6= ∅. P is continuous if the set {d ∈ D : d x} is directed and x = W{d ∈ P : d x} for all x ∈ P . Let K(P ) = {x ∈ P : x x}. P is algebraic if for all x ∈ P the set ↓x ∩ K(P ) is directed and x = W(↓x ∩ K(P )). A poset P is called a quasicontinuous poset (resp., quasialgebraic poset) if for all x ∈ P and U ∈ σ(P ), x ∈ U implies that there is a non-empty finite set F ⊆ P such that x ∈ intσ ↑F ⊆↑F ⊆ U (resp., x ∈ intσ ↑F =↑F ⊆ U). For a complete lattice L and x, y ∈ L, we say that x ≺ y iff whenever the intersection of a non-empty collection of upper sets is contained in ↑y, then the intersection of finitely many is contained in ↑x. x ≺ y ⇔ y ∈ intυ ↑x and x ≺ y implies x y. L is hypercontinuous if {d ∈ L : d ≺ x} is directed and x = W{d ∈ L : d ≺ x} for all x ∈ L. L is hyperalgebraic provided x = W{y ∈ L : y ≺ y ≤ x} for all x ∈ L. Let L be a complete lattice, we say that x is wayway-below y, in notation x ≪ y, if for any subset A of L, W W y ≤ A implies x ≤ a for some a ∈ A. L is prime continuous if x = {d ∈ L : d ≪ x} for all x ∈ L. The next two well-known propositions give various equivalent formulations for completely distributive lattice and hypercontinuous lattice separately. Proposition 2.1. ([7]) Let L be a complete lattice, the following conditions are equivalent: (1) L is prime-continuous, (2) L is a completely distributive lattice, (3) L is distributive and both L and Lop are continuous lattices, Theorem 2.2. ([8]) Let L be a complete lattice, then the following conditions are equivalent: (1) L is a hypercontinuous lattice, (2) L is a continuous lattice and σ(L) = υ(L), (3) L is a continuous lattice, Lop is a quasicontinuous lattice, and the bi-Scott topology agrees with the Lawson topology(λ(L) = σ(L) ∨ σ(Lop)). The following lemma characterizes the binary relation ≺ on the complete lattice of Scott open subsets. It is easy to see that this conclusion is also true for any topological space.

Lemma 2.3. Let P be a poset and U, V ∈ σ(P ). U ≺ V if and only if U ⊆↑F ⊆ V for some F ⊆f P .

2 Tn Proof. (⇐) Let F = {x1, x2, ··· , xn}, then V * P \ ↓xi, for each 1 ≤ i ≤ n. We claim that V ∈ i=1(σ(P )\ ↓ Tn σ(P )(P \ ↓ xi)) ⊆↑ σ(P )U. Indeed, for each W ∈ i=1(σ(P )\ ↓ σ(P )(P \ ↓ xi)), W ∩ ↓ xi 6= ∅ for each 1 ≤ i ≤ n, then U ⊆↑F ⊆ W . Tn (⇒) If U ≺ V , then there exists {Wi ∈ σ(P ) : 1 ≤ i ≤ n} such that V ∈ i=1(σ(P )\ ↓ σ(P )Wi) ⊆↑ σ(P )U, which implies that V * Wi for each i. Let F = {x1, x2, ··· , xn}, where xi ∈ V \ Wi for each 1 ≤ i ≤ n. Now we show that U ⊆↑F . If not, there exists x ∈ U\ ↑F , i.e., F ⊆ P \ ↓x. Then P \ ↓x * Wi for each 1 ≤ i ≤ n, but U * P \ ↓x, which is a contradiction. M. Ern´e[6] and T. Yokoyama [23] respectively give the following Lemma, and prove the result that a spectral Scott space (i.e. a poset on which the Scott topology is spectral) is a quasialgebraic domain.

Lemma 2.4. Let P be a poset. A Scott open subset U of P is Scott compact if and only if there exists F ⊆f P such that U = intσ ↑F =↑F . The correspondence between P and σ(P ) is enumerated below, where (2) and (3) is a direct consequences of Lemma 2.3 and 2.4. Theorem 2.5. ([7, 16, 22]) Let P be a poset, (1) P is continuous (algebraic) iff the lattice σ(P ) of all Scott open sets is a completely distributive lattice (completely distributive algebraic lattice). (2) P is quasicontinuous iff the lattice σ(P ) of all Scott open sets is a hypercontinuous lattice. (3) P is quasialgebraic iff the lattice σ(P ) of all Scott open sets is an algebraic lattice iff σ(P ) is a hyperalgebraic lattice.

3. The correspondent properties between P and Γ(P )

Recently, H. Miao et al [17] proved that for a well-filtered dcpo L,ΣL is locally compact if and only if ΣΓ(L) is a locally compact space. They further asked the following question [17, Problem 5.9]: For a poset P , if ΣP is core-compact, must ΣΓ(P ) be locally compact? In this section, we investigate conditions when the lower Vietoris topology and the Scott topology on Γ(P ) coincide for a poset P and give an positive answer to the above question. Particularly, we show that for a poset L,ΣP is core-compact iff ΣΓ(P ) is core-compact iff ΣΓ(P ) is sober, locally compact and σ(Γ(P )) = υ(Γ(P )) . When we are working on continuous lattices, we notice that the condition σ(Lop) = υ(Lop) is very important. Firstly, several well-known lemmas are needed. Lemma 3.1. ([8, 21]) A complete lattice L is a quasicontinuous (quasialgebraic) lattice iff ω(L) is a continuous (algebraic) lattice. Lemma 3.2. ([7, Theorem V-5.6.]) For a sober space X, the lattice O(X) of open subsets is continuous iff X is locally compact.

The following lemma is easy to verify. Lemma 3.3. For a complete lattice L, both (L, υ(L)) and (L, ω(L)) are sober. Proposition 3.4. Let L be a continuous lattice, if L satisﬁes the condition that υ(Lop) = σ(Lop), then (Lop, σ(Lop)) is a sober and locally compact space.

Proof. By Lemma 3.1, σ(Lop) = υ(Lop) = ω(L) are continuous lattices. Thus (L, σ(Lop)) is a sober and locally compact space by Lemma 3.2 and 3.3 . We give a sufficient condition such that σ(Γ(P )) = υ(Γ(P )) for a poset P , which is crucial for further discussion. n n Q Q Proposition 3.5. Let P be a poset. If Σ( P ) = (ΣP ) for each n ∈ N, then σ(Γ(P )) = υ(Γ(P )). Proof. We only need to prove that σ(Γ(P )) ⊆ υ(Γ(P )). n Q At first, for each n ∈ N, we define a map sn : P → Γ(P ) as follows:

n n Q [ ∀(x1, x2, . . . , xn) ∈ P, sn(x1, x2, . . . , xn) = ↓xk. k=1

3 We claim that sn preseves existing directed sups. Let {(x1i, x2i, . . . , xni): i ∈ I} be a directed subset of n n Q Q W P such that the supremum of {(x1i, x2i, . . . , xni): i ∈ I} exists in P , then (x1i, x2i, . . . , xni) = W W W i∈I ( i∈I x1i, i∈I x2i,..., i∈I xni). We have

n n _ [ _ [ [ _ sn( (x1i, x2i, . . . , xni)) = ↓ ( xki) = ↓ xki = sn(x1i, x2i, . . . , xni). i∈I k=1 i∈I i∈I k=1 i∈I n Q Thus sn is a Scott continuous map from P into Γ(P ). Next, let U be a Scott open subset of Γ(P ) and A ∈ U. With loss of generality, we assume A 6= ∅. Note that S since A = {↓ F : F ⊆f A} and {↓ F : F ⊆f A} is a directed family in Γ(P ), there exists a non-empty finite subset F of A such that ↓F ∈ U. Let F = {x1, x2, . . . , xn}, then sn(x1, x2, . . . , xn) =↓F ∈ U. It follows that n n −1 Q Q (x1, x2, . . . , xn) ∈ sn (U). Since sn is Scott continuous and Σ( P ) = (ΣP ), there exists a family of Scott open subset Uk, k = 1, 2, . . . , n of P such that −1 (x1, x2, . . . , xn) ∈ U1 × U2 × · · · × Un ⊆ sn (U). n T Since xk ∈ A for 1 ≤ k ≤ n, we have A ∈ 3Uk = {B ∈ Γ(P ): B ∩ Uk 6= ∅}. It follows that A ∈ 3Uk ∈ k=1 n T −1 υ(Γ(P )). For any B ∈ 3Uk, there exists yk ∈ B ∩ Uk for 1 ≤ k ≤ n. since (y1, y2, . . . , yn) ∈ sn (U), we have k=1 n n S T ↓yk ∈ U. It follows that B ∈ U, i.e., A ∈ 3Uk ⊆ U. k=1 k=1 Lemma 3.6. ([7, Theorem II-4.13]) Let P be a dcpo, (P, σ(P )) is core-compact if and only if Σ(Q × P ) = ΣQ × ΣP for every dcpo Q. It is not difficult to see that the above lemma is still true when P and Q are posets. Proposition 3.7. Let P be a poset, if (P, σ(P )) is core-compact, then σ(Γ(P )) = υ(Γ(P )). Moreover, (Γ(P ), σ(Γ(P ))) is sober and locally compact. n n Q Q Proof. If (P, σ(P )) is core-compact, then Σ( P ) = (ΣP ) for each n ∈ N. It follows that σ(Γ(P )) = υ(Γ(P )) by Proposition 3.5. Let L = σ(P ) which is a continuous lattice by asuumption and Γ(P ) =∼ Lop. By Proposition 3.4, (Γ(P ), σ(Γ(P ))) is a sober and locally compact space. We have answered Problem 5.9 raised in [17]. The following result appears in J. Goubault-Larrecq’s blog and is given by Matthew de Brecht. Lemma 3.8. ([3]) Let P and Q be two posets. If both (P, σ(P )) and (Q, σ(Q)) are first-countable, then ΣP × ΣQ = Σ(P × Q). Since a countable product of first-countable spaces is first-countable, we have the following result: Proposition 3.9. Let P be a poset, if (P, σ(P )) is first-countable, then σ(Γ(P )) = υ(Γ(P )). We observe that there is an adjunction between σ(P ) and σ(Γ(P )), which serves as a useful tool in studying the relation between P and Γ(P ). For the standard theory of adjunctions or Galois connection we refer to ([7], Section O-3). Proposition 3.10. Let P be a poset. (1) η : P → Γ(P ), ∀x ∈ P, η(x) =↓x. Then η is Scott continuous. (2) A map f : σ(P ) → σ(Γ(P )) is defined by f(U) = 3U = {A ∈ Γ(P ): A ∩ U 6= ∅}. Then f preserves arbitrary sups. −1 −1 −1 (3) Let the map η : σ(Γ(P )) → σ(P ) be defined by η (U) = {x ∈ P : ↓ x ∈ U}. Then η ◦ f = 1σ(P ), −1 −1 f ◦ η ≤ 1σ(Γ(P )), which implys that (η , f) is an adjunction. Proof. (1) Straightforword. S (2) Obviously, f is monotone. Let {Ui : i ∈ I} be arbitrary subset of σ(P ). f( Ui) = {A ∈ Γ(P ): S W i∈I A ∩ i∈I Ui 6= ∅} = {A ∈ Γ(P ): ∃i ∈ I,A ∩ Ui 6= ∅} = i∈I f(Ui). (3) Because η is Scott continuous, η−1 preserves arbitrary sups and finite infs. For any U ∈ σ(P ), x ∈ −1 −1 η (f(U)) ⇔ η(x) ∈ f(U) ⇔↓ x ∩ U 6= ∅ ⇔ x ∈ U, hence η ◦ f = 1σ(P ). For any U ∈ σ(Γ(P )), A ∈ −1 −1 −1 f ◦ η (U) ⇔ A ∩ η (U) 6= ∅ ⇒ A ∈ U, i.e., f ◦ η ≤ 1σ(Γ(P )). n n Q Q For a complete lattice L, the condition that Σ( L) = (ΣL) for each n ∈ N relates to the concept of jointly continuous semilattice. 4 Definition 3.11. Let L be a sup semilattice, the sup operation is jointly continuous with respect to the Scott topology provided that the mapping

(x, y) 7→ x ∨ y :(L, σ(L)) × (L, σ(L)) → (L, σ(L)) is continuous in the product topology. Proposition 3.12. ([7, Corollary II-1.12]) If L is a dcpo and a sup semilattice such that the sup operation is jointly Scott continuous, then (L, σ(L)) is a sober space. Now we complete the conclusion in Proposition 3.7. Theorem 3.13. Let P be a poset, then the following statements are equivalent: (1) (P, σ(P )) is core-compact. (2) (Γ(P ), σ(Γ(P ))) is core-compact. (3) (Γ(P ), σ(Γ(P ))) is sober and locally compact (4) (Γ(P ), σ(Γ(P ))) is sober and locally compact with σ(Γ(P )) = υ(Γ(P )). Proof. (1) ⇒ (4) ⇒ (3). Proposition 3.7. (3) ⇒ (2). The open subsets of a locally compact space form a continuous lattice. (2) ⇒ (1). If (Γ(P ), σ(Γ(P ))) is locally compact, then σ(Γ(P )) is a continuous lattice. By Proposition 3.10, σ(P ) is also a continuous lattice. (2) ⇒ (3). If (Γ(P ), σ(Γ(P ))) is core-compact, then Σ(Γ(P ) × Γ(P )) = ΣΓ(P ) × ΣΓ(P ) by Lemma 3.6. Obviously, the sup operation is jointly Scott continuous. Thus (Γ(P ), σ(Γ(P ))) is sober and locally compact by Proposition 3.12 and Lemma 3.2. It is proved by diﬀerent method in [17] that for a poset P , if ΣΓ(P ) is locally compact then ΣL is core- compact. It seems that the equivalence in Theorem 3.13 only works for posets with the Scott topology. In the following example, we give a topological space X and show that X is a compact Hausdorﬀ space while ΣC(X) is sober but not locally compact. Moreover, σ(C(X)) 6= υ(C(X)).

1 + Example 3.14. Consider the subset X = {0}∪{ n : n ∈ N } of the real line R, with the subspace topology. Let 1 + 1 + C1(X) = {A : A ⊆f { n : n ∈ N }} and C2(X) = {B ∪ {0} : B ⊆ { n : n ∈ N }}, then C(X) = C1(X) ∪ C2(X). Clearly, X is a compact Hausdorff space. We will prove that C(X) is sober but not locally compact with respect to the Scott topology. Hence, σ(C(X)) 6= υ(C(X)) also holds. (1) We show that ↑ C(X)A is Scott open for each A ∈ C1(X). Indeed, for any directed family {Ci}i∈I of closed subsets, if {Ci}i∈I is a finite family of C1(X), then there exists a largest element; if {Ci}i∈I is an infinite W S W S W family of C1(X), then i∈I Ci = {0} ∪ i∈I Ci; otherwise i∈I Ci = i∈I Ci. In all cases, if i∈I Ci ∈↑C(X)A, there exists some j ∈ I such that A ⊆ Cj. Let A be a Scott closed subset of C(X) and MaxA be the set of maximal elements of A. Then A =↓ C(X)MaxA. We claim that A cannot be irreducible whenever MaxA is an infinite subset of C(X), which implies that C(X) is sober with respect to the Scott topology. Case 1. There exist A1,A2 in C1(X) ∩ MaxA such that A1 \ A2 6= ∅ and A2 \ A1 6= ∅. Then ↑ C(X)A1 and ↑ C(X)A2 are two Scott open subsets of C(X). But ↑ C(X)A1∩ ↑ C(X)A2 ∩ A = ∅ by the maximality of A1 and A2. Case 2. There is only one A ∈ C1(X) ∩ MaxA. Then must be some B ∈ C2(X) such that A \ B 6= ∅ and B \ (A ∪ {0}) 6= ∅. Let x ∈ B \ (A ∪ {0}) 6= ∅, then ↑C(X)A and ↑C(X){x} are two Scott open subsets of C(X) which intersect with A. But ↑C(X)A∩ ↑C(X){x} ∩ A = ∅ by the maximality of A. Case 3. MaxA ⊆ C2(X). Let B1 ∪ {0} and B2 ∪ {0} are two different elements in MaxA. Then B1 \ B2 6= ∅ and B2\B1 6= ∅. For any F1 ⊆f B1 and F2 ⊆f B2, then ↑C(X)F1 and ↑C(X)F2 are two Scott open subsets of C(X) which intersect with A. Suppose that A is an irreducible closed subset. We have that ↑C(X)F1∩ ↑C(X)F2∩A= 6 ∅, which implies that F1 ∪ F2 ∪ {0} ∈ A. Since F1,F2 are arbitrary and A is a Scott closed subset, it follows that B1 ∪ B2 ∪ {0} must be in A. This contradicts with the maximality of B1 ∪ {0}. (2) Suppose that (C(X), σ(C(X))) is locally compact. Let U be any Scott open set containing {0} but not ∅. From the assumption, there exists a compact saturated subset K such that {0} ∈ intσ(C(X))K ⊆ K ⊆ U. It 1 + is easy to see that K =↑ C(X)MinK and for each A ∈ MinK, A = {0} or A ⊆f { : n ∈ N }. We claim that K W n W is Scott open. Let {Ci}i∈I be any directed family of closed subsets with i∈I Ci ∈ K, then i∈I Ci ∈↑ C(X)A for some A ∈ MinK. If A = {0}, then there exists some i ∈ I such that Ci ∈ intσ(C(X))K ⊆ K. Otherwise, 1 + A ⊆f { : n ∈ N }, there exists some Cj ⊇ A by previous proof. Then by Lemma 2.4, MinK is finite. But this n S is impossible, since K may contain some A ⊆f X \ MinK. Moreover, according to the previous proof (C(X), υ(C(X))) is core-compact while X is core-compact. By Lemma 3.2 and 3.3, (C(X), υ(C(X))) is sober and locally compact. From this we get that σ(C(X)) 6= υ(C(X)). Indeed, we have the following result: 5 Proposition 3.15. Let Y be a T1 space, denote by C(Y ) the lattice of all closed subsets of Y with inclusion order, if σ(C(Y )) = υ(C(Y )), then Y has the discrete topology. Proof. Suppose not, then there exists an infinite subset A of Y such that A \ A 6= ∅. Let x ∈ A \ A and U = C(Y ) \ ({∅} ∪ {{y} : y ∈ A}), then {x} ∈ U and U ∈ σ(C(Y )). According to the assumptions that σ(C(Y )) = υ(C(Y )), there exists finitely many open neighborhoods {Ui : 1 ≤ i ≤ n} of x such that {x} ∈ 3U1 ∩ 3U2 ∩ · · · ∩ 3Un ⊆ U. Let V = U1 ∩ U2 ∩ · · · ∩ Un. Then {x} ∈ 3V ⊆ 3U1 ∩ 3U2 ∩ · · · ∩ 3Un ⊆ U and C(Y ) \ U ⊆ {H ∈ C(Y ): H ∩ V = ∅}. It follows that A ∩ V = ∅, and then A ∩ V = ∅, a contradiction. Similar to Theorem 2.5, the correspondence between P and Γ(P ) has been studied (see [13, 14, 21, 11]). Here we use the adjunction in Proposition 3.10 to give a unified proof. Lemma 3.16. ([15]) Let P and Q be complete lattices, and let g : P → Q and d : Q → P be maps which preserve arbitrary sups and g ◦ d = 1Q. (1) If P is completely distributive, then so is Q, (2) If P is hypercontinuous, then so is Q. Theorem 3.17. Let P be a poset, (1) P is continuous iff the lattice Γ(P ) is a continuous lattice. (2) P is quasicontinuous iff the lattice Γ(P ) is a quasicontinuous lattice. (3) P is algebraic iff the lattice Γ(P ) is an algebraic lattice. (4) P is quasialgebraic iff the lattice Γ(P ) is a quasialgebraic lattice.

Proof. (1) If P is continuous, then σ(P ) is a completely distributive lattice by Theorem 2.5. Both σ(P ) and Γ(P ) are continuous lattices by Proposition 2.1. Conversely, if Γ(P ) is continuous, then σ(Γ(P )) is completely distributive, and so σ(P ) is a completely distributive lattice by Proposition 3.10 and Lemma 3.16. This shows that P is a continuous poset. (2) If P is quasicontinuous, then σ(P ) is a hypercontinuous lattice by Theorem 2.5. It follows that Γ(P ) is a quasicontinuous lattice by Theorem 2.2. Conversely, if Γ(P ) is quasicontinuous, then σ(Γ(P )) is hypercontinuous, and so σ(P ) is a hypercontinuous lattice by Proposition 3.10 and Lemma 3.16. This shows that P is a quasicontinous poset. (3) For each nonempty B ∈ Γ(P ), it is easy to see that B ≪ B in Γ(P ) if and only if there exists some k ∈ K(P ) such that B =↓k. If P is algebraic, then A = cl(S{↓k : k ∈ A ∩ K(P )}). Thus Γ(P ) is an algebraic lattice. Conversely, if Γ(P ) is an algebraic lattice, P is continuous by (1). Both σ(P ) and Γ(P ) are completely distributive. This implies that P is algebraic. (4) If P is quasialgebraic, then by Theorem 2.5, σ(P ) is a hypercontinuous and algebraic lattice. Let L = σ(P ). Then by Theorem 2.5 and Theorem 2.2, ω(Lop) = υ(L) = σ(L) is an algebraic lattice. This implies that Γ(P ) is a qusialgebraic lattice from Lemma 3.1. Conversely, if Γ(P ) is a quasialgebraic lattice, then by Lemma 3.1, ω(Lop) = υ(L) is an algebraic lattice. By (2) and Theorem 2.5, P is quasicontinuous and L is a hypercontinuous latiice. By Theorem 2.2, σ(L) = υ(L) and is a completely distibutive algebraic lattice. This together implies that σ(P ) is a hyperalgebraic lattice. Thus P is a quasialgebraic poset, by Theorem 2.5.

4. Distributive continuous lattice

Now we return to continuous lattices to consider the condition that σ(Lop) = υ(Lop). At ﬁrst, we list some well-known results.

Fact 4.1. If X and Y are both sober spaces and the open set lattice O(X) is isomorphic to O(Y ), then X is homeomorphic to Y . Lemma 4.2. ([7, Lemma III-5.7]) Let L be a quasicontinuous domain. If A =↑ A is compact in the Scott topology, then every Scott open neighborhood U of A contains a ﬁnite set F such that A ⊆ ↑F ⊆↑ F ⊆ U. Furthermore, A is a directed intersection of all ﬁnitely generated upper sets that contain A in their Scott interior.

Lemma 4.3. ([8]) Let L be a complete lattice.

(1) The Lawson topology (L, λ(L)) is compact and T1. (2) L is quasicontinuous if and only if (L, λ(L)) is Hausdorﬀ. (3) A subset M of L is closed in the lower topology if and only if M =↑M and if for every ultraﬁlter F with M ∈ F , liminfF = W{V F : F ∈ F } ∈ M. (4) A subset M of L is closed in the lower topology if and only if M =↑ M and M is closed in the Lawson topology.

6 Proposition 4.4. Let L be a continuous lattice, the following conditions are equivalent: (1) σ(Lop) = υ(Lop). op (2) The bi-Scott topology σBi(L) = σ(L) ∨ σ(L ) is compact and Hausdorﬀ. (3) Let Q(L) be the set of compact saturated sets of (L, σ(L)), ordered by reverse inclusion, then σ(Lop) =∼ Q(L).

op op Proof. (1) ⇒ (2). The bi-Scott topology σBi(L) = σ(L) ∨ σ(L ) = σ(L) ∨ υ(L ) = σ(L) ∨ ω(L) = λ(L), and is compact and Hausdorff by Lemma 4.3. (2) ⇒ (3). The Lawson topology of L is compact and Hausdorff by the above proposition. If the bi-Scott topology of L is also compact and Hausdorff, then it is equal to the Lawson topology, since λ(L) ⊆ σBi(L). op Let U ∈ σ(L ) and KU = L \ U, then KU is an upper set in L and closed in the bi-Scott topology. This T implies that KU is closed in the lower topology of L and KU = {↑ F : F ⊆f L, KU ⊆↑ F }. We see that T S {↑ F : F ⊆f L, KU ⊆↑ F } is a directed subset of Q(L). For KU ⊆↑ F1,KU ⊆↑ F2, KU ⊆↑ F1 ↑ F2 = {↑ (x∨y): x ∈ F1, y ∈ F2} which is an upper bound for ↑F1, ↑F2 in Q(L). As the supremum of a directed subset of T op Q(L), KU ∈ Q(L). For any A ∈ Q(L), A = {↑F : F ⊆f L, A ⊆ ↑F } by Lemma 4.2, then UA = L\A ∈ σ(L ). It follows that σ(Lop) =∼ Q(L). (3) ⇒ (1). From the assumption and the proof above, we can see that Q(L) =∼ υ(Lop) =∼ σ(Lop). By Lemma 3.1, υ(Lop) = ω(L) is a continuous lattice, since L is a continuous lattice. This implies that σ(Lop) is also a continuous lattice. Then (Lop, σ(Lop)) is a sober space by an argument similar to Theorem 3.13. By hypothesis and Lemma 3.3, (Lop, υ(Lop)) and (Lop, σ(Lop)) are both sober spaces with υ(Lop) =∼ σ(Lop). It follows that (Lop, υ(Lop)) and (Lop, σ(Lop)) are homeomorphic by Fact 4.1. There exists continuous functions op op op op op op op op f :(L , υ(L )) → (L , σ(L )) and g :(L , σ(L )) → (L , υ(L )) such that f ◦g = g◦f = 1Lop . Obviously, f, g preserve arbitrary sups and infs. We finish the proof by showing that any closed set in (Lop, σ(Lop)) is also a closed set in (L, ω(L)). Let A be a closed subset in (Lop, σ(Lop)) and F be an ultrafilter with A ∈ F , then A =↑ A and f(F ) = {f(F ): F ∈ F } is also an ultrafilter with f(A) ∈ f(F ). By (3) of the above lemma, we have liminff(F ) = W{V f(F ): F ∈ F } ∈ f(A), and then g(liminff(F )) = g(W{V f(F ): F ∈ F }) = W{V g ◦ f(F ): F ∈ F } = W{V F : F ∈ F } = liminfF ∈ g ◦ f(A) = A. Thus A is closed in the lower topology of L. We have the following result that is similar to Proposition 3.10. Proposition 4.5. Let L be a distributive continuous lattice such that σ(Lop) = υ(Lop). (1) Define I : L → σ(Lop) by I(x) = L\ ↑x for each x ∈ L, then I preserves arbitrary sups.

op S T (2) For each U ∈ σ(L ), U = {L\ ↑F : F ⊆f L, L\ ↑F ⊆ U} = L \ {↑F : F ⊆f L, L\ ↑F ⊆ U} = L \ KU , T T op where KU = {↑ F : F ⊆f L, L\ ↑ F ⊆ U} = {↑ F : F ⊆f L, F KU }. Define J : σ(L ) → L by W↑ J(U) = ∧F . Then J preserves arbitrary sups, J ◦ I = id and I ◦ J ≤ id op , which implies F KU L σ(L ) that (J, I) is an adjunction and thus J also preserves arbitrary infs. W W S S Proof. (1) I( α∈∆ xα) = L\ ↑ α∈∆ xα = α∈∆ L\ ↑xα = α∈∆ I(xα). op (2) Obviously, for each U ∈ σ(L ), KU = L \U is a compact saturated subset of L and J is a well-defined map. op op Let {Uβ : β ∈ ∇} be an arbitrary family of open subsets in (L , υ(L )). For each Uβ, β ∈ ∇, let L\Uβ = Kβ. S T S W W W Let L \ β∈∇ Uβ = β∈∇ Kβ = K. We will show that J( β∈∇ Uβ) = F K ∧F = β∈∇ F K ∧F = W W W W β β∈∇ J(Uβ), i.e., J preserve arbitrary sups. It is easy to see that F K ∧F ≥ β∈∇ F K ∧F . Indeed, if T β F Kβ for some β ∈ ∇, then F K. In the other direction, if F K = β∈∇ Kβ for some F ⊆f L, then Tn Tn Tn T there exists a finite subset {βi : 1 ≤ i ≤ n} ⊆ ∇, n ∈ N such that F i=1 Kβi . Since i=1 Kβi = i=1{ ↑ T Tn G : G ⊆f L, G Kβi } = { i=1 ↑Gi : Gi Kβi , 1 ≤ i ≤ n}, which is the intersection of a filtered family of Tn finitely generated upper subsets, there exists Gi Kβi , 1 ≤ i ≤ n such that i=1 ↑ Gi ⊆↑ F . Because L is a Tn S Wn Wn V Wn distributive lattice, we have that i=1 ↑ Gi = {↑ i=1 xi : xi ∈ Gi, 1 ≤ i ≤ n} and i=1 ∧Gi = { i=1 xi : Wn xi ∈ Gi, 1 ≤ i ≤ n}, thus ∧F ≤ ∧Gi. W i=1 W For any x ∈ L, J ◦ I(x) = F x ∧F . Since L is a continuous lattice, we have F x ∧F = x. op W↑ For any U ∈ σ(L ), let K = L \U. I ◦ J(U) = L\ ↑ ∧F . We claim that I ◦ J ≤ 1 op . Indeed, U F KU σ(L ) K = T ↑F ⊆↑W↑ ∧F , thus I ◦ J(U) ⊆ L \ K = U. In conclusion, (J, I) is an adjunction between U F KU F KU U op σ(L ) and L. Remark 4.6. There is a one-to-one correspondence between distributive continuous lattices (that is, continuous frames) and locally compact sober spaces in the sense of a duality of categories, namely Hofmann-Lawson duality (cf.[13]). Theorem 3.13 shows that if L is a distributive continuous lattice such that the hull-kernel topology of SpecL is just the Scott topology, then σ(Lop) = υ(Lop) and Σ(Lop) is sober and locally compact. There exists a complete lattice W such that σ(W ) is a continuous lattice, but W is not quasicontinuous (see [7, Theorem 7 VI-4.5.]). By Theorem 2.2, σ(Lop) = υ(Lop) is strictly weaker that σ(L) = υ(L) for a distributive continuous lattice. We wonder whether the condition that σ(Lop) = υ(Lop) in Proposition 3.4 is necessary and whether it implies that SpecL is a Scott space. In the case of distributive algebraic lattices, we have the following result. Proposition 4.7. Let L be a distributive algebraic lattice, then the following conditions are equivalent: (1) L is a hyperalgebraic lattice, (2) σ(Lop) = υ(Lop), (3) The spectrum of L is a quasialgebraic domain equipped with the Scott topology, (4) Lop is a quasialgebraic lattice. Proof. (1) ⇔ (3) ⇔ (4). (cf. [21]) (1) implies (2). Since (1) ⇒ (3), Lop is isomorphic to the lattice of Scott closed subsets of a quasialgebraic domain. Then we have σ(Lop) = υ(Lop) by Proposition 3.7, since any quasialgebraic domain is locally compact with respect to the Scott topology. (2) implies (1). σ(Lop) = υ(Lop) = ω(L), where ω(L) is an algebraic lattice by Lemma 3.1. So we have that Lop is a quasialgebraic lattice from Theorem 2.5 and that σ(Lop) is a hyperalgebraic lattice by Lemma 2.3. There exists an adjunction between σ(Lop) and both preserve arbitrary sups by Proposition 4.5. Thus we get the conclusion that L is a hyperalgebraic lattice. Next, we construct a complete lattice P such that (P, υ(P )) is sober and locally compact, but the Scott topology on the lattice of closed subsets is not equal to the lower Vietoris topology. P endowed with the Scott topology is neither locally compact nor first-countable still we have σ(Γ(P )) = υ(Γ(P )).

Example 4.8. Let P = (N × N) ∪ {>} with a partial order defined as follows: (i) ∀(m, n) ∈ N × N, ⊥ ≤ (m, n) ≤ >; (ii) ∀(m1, n1), (m2, n2) ∈ N × N, (m1, n1) ≤ (m2, n2) iff m1 = m2 and n1 ≤ n2. (1) When considering P with the upper topology and let L = υ(P ), it is easy to see that (P, υ(P )) is a sober space. We also have that (P, υ(P )) is locally compact, indeed L is a distributive algebraic lattice. Obviously, P is isolated in L. For any finite subset F of P , we will show that P \ ↓F P \ ↓F in υ(P ) by the S T Alexander’s Lemma. Let {xi : i ∈ I} be a subset of P such that P \ ↓ F ⊆ P \ ↓ xi, i.e., ↓ xi ⊆↓ F . T i∈I T i∈I If i∈I ↓ xi = {⊥}, then there exists xi, xj, i, j ∈ I such that ↓ xi∩ ↓ xj = {⊥}. If i∈I xi 6= {⊥}, then

{xi : i ∈ I} has a minimal element xi0 6= ⊥. In the ﬁrst case P \ ↓F ⊆ P \ ↓xi ∪ P \ ↓xj and in the second case op op P \ ↓F ⊆ P \ ↓xi0 , which together imply that L is an algebraic lattice. Thus σ(L ) 6= υ(L ) by Proposition 4.7, since P is not quasicontinuous. It is not hard to verify that Σ(Lop) is a sober but not locally compact space.

(2) When considering P with the Scott topology, we have σ(Γ(P )) = υ(Γ(P )). One can easily check that (P, σ(P )) is neither locally compact nor first-countable. For any U ∈ σ(Γ(P )), let A = Γ(P ) \U. We will show that A is a closed subset of (Γ(P ), υ(Γ(P )). Let A∗ = S{A : A ∈ MaxA}, then A∗ ∈ Γ(P ) since A∗ = η−1(A). If > ∈ A∗, then A = Γ(P ). Otherwise, ({i} × N) ∩ A∗ is a finite subset for each i ∈ N. A∗ with the order inherited form P is an algebraic dcpo. We see that A is a closed subset of (Γ(A∗), σ(Γ(A∗))), and is also a closed subset of (Γ(A∗), υ(Γ(A∗))) since σ(Γ(A∗)) = υ(Γ(A∗)). This means that A can be represented as the intersection of a family of finitely generated lower sets in Γ(A∗). Thus A can be represented as the intersection of the family of finitely generated lower sets in Γ(P ), which implies that A is a closed subset of Γ(P ) with the lower Vietoris topology. Hertling [10] constructs a complete lattice that is not jointly Scott continuous. Actually, the construction of Hertling works for any complete lattice that is not core-compact with respect to the Scott topology.

Lemma 4.9. ([7, Theorem II-4.10]) Let X be a T0 space, the set E = {(x, U) ∈ X × O(X): x ∈ U} is open in X × ΣO(X) iﬀ X is core-compact. Proposition 4.10. Let P be a complete lattice and L = P × σ(P ). If (P, σ(P )) is not core-compact, then L is not jointly Scott continuous and then Σ(L × L) 6= ΣL × ΣL. Proof. By the above lemma, the set E = {(x, U) ∈ P ×σ(P ): x ∈ U} is not open in ΣP ×Σ(σ(P )). Suppose that the map sup: (L, σ(L)) × (L, σ(L)) → (L, σ(L)), ((x, U), (y, V )) 7→ (x ∨ y, U ∪ V ) is continuous. Since E is open −1 in ΣL, sup (E) is open in (L, σ(L)) × (L, σ(L)). For any (a, W ) ∈ E, we have (a, ∅) ∨ (⊥P ,W ) = (a, W ) ∈ E. −1 There exists D1,D2 ∈ σ(L) such that ((a, ∅), (⊥P ,W )) ∈ D1 × D2 ⊆ sup (E). Let E1 = {x ∈ P :(x, ∅) ∈ D1} and E2 = {U ∈ σ(P ):(⊥P ,U) ∈ D2}, then a ∈ E1,E1 ∈ σ(P ) and W ∈ E2,E2 ∈ σ(σ(P )). For each b, V ∈ E1 × E2,(b, V ) = (b, ∅) ∨ (⊥P ,V ) ∈ sup(D1 × D2) ⊆ E. Contradiction. Thus L is not jointly Scott continuous and then Σ(L × L) 6= ΣL × ΣL. In the following, we use the complete lattice P in Example 4.8 to construct a complete lattice L such that (L, σ(L)) is sober but not jointly Scott continuous. 8 Figure 1: Complete lattice P in Example 4.8

Example 4.11. Let P be the complete lattice in Example 4.8. Let M = {⊥, >} ∪ (N → N) with a partial order deﬁned as follows:

(i) ∀f ∈ (N → N), ⊥ ≤ f ≤ >; (ii) ∀f, g ∈ (N → N), f ≤ g iﬀ ∀i ∈ N, f(i) ≥ g(i). It is easy to see that M =∼ σ(P ). Let L = P × M, L is not jointly Scott continuous by Proposition 4.10. Let A be a non-empty Scott closed subset of L, then A =↓ MaxA, where MaxA denotes the set of all maximal elements of A. We will show that ΣL is a sober space by proving that A can not be an irreducible closed set when |MaxA| ≥ 2. Case 1. |πM (MaxA)| = 1. πP (A) must be a Scott closed subset of P with |MaxπP (A)| ≥ 2. Since ΣP is a sober space, there exists B,C ∈ Γ(P ) such that πP (A) ⊆ B ∪ C but πP (A) * B and πP (A) * C. It follows that A ⊂ B × πM (A) ∪ C × πM (A) but A does not contained in any of them. Case 2. |πM (MaxA)| ≥ 2 and > ∈ πM (MaxA). Let A0 = {(x, y) ∈ MaxA : y = >} and B0 = {(x, y) ∈ W MaxA : y 6= >}, then A0 6= ∅, B0 6= ∅ and A ⊆↓ A0∪ ↓ B0. We claim that ↓ A0 is a Scott closed subset of W 0 L. Suppose not, there exists a directed subset D ⊆↓A0 such that D/∈↓A0. Let D = {(πP (d), >): d ∈ D}, 0 W 0 W 0 W W 0 W then D is a directed subset in ↓A0. We have that D ∈↓/ A0 since D ≥ D, and that D ∈↓/ B0 since W W 0 W πM ( B0) < πM ( D ) = >. Contradiction. Obviously, A *↓A0 and A *↓ B0. W Case 3. |πM (MaxA)| ≥ 2 and > ∈/ πM (MaxA). Let g = πM ( A), then g ∈ (N → N). Case 3.1. ⊥ ∈/ πM (MaxA). By assumption, there exists some i0 ∈ N such that B0 = {(x, y) ∈ MaxA : W y(i0) > g(i0)} 6= ∅. Let A0 = {(x, y) ∈ MaxA : y(i0) = g(i0)}, then A0 6= ∅ and A ⊆↓ A0∪ ↓ B0. We claim that ↓ A0 is a Scott closed subset of L. Suppose not, there exists some directed subset D ⊆↓ A0 such that W D/∈↓ A0. For each d ∈ D, deﬁne fd : N → N, fd(n) = πM (d)(n) if n 6= i0; fd(n) = g(i0) if n = i0. Let 0 0 W 0 W W 0 D = {(πP (d), fd): d ∈ D}, then D is a directed subset in ↓A0 with D ≥ D. We have that D ∈↓/ A0 W W 0 W W W 0 since D/∈↓A0, and that D ∈↓/ B0 since πM ( B0) < πM ( D ). Contradiction. Obviously, A *↓A0 and W A *↓ B0. Thus A is not irreducible. Case 3.2. ⊥ ∈ πM (MaxA) and |πM (MaxA) ∩ (N → N)| = 1. Let (x, g) ∈ MaxA, then A ⊆↓(x, g) ∪ πP (A) × {⊥}. Obviously, A *↓(x, g) and A * πP (A) × {⊥}. Case 3.3. ⊥ ∈ πM (MaxA) and |πM (MaxA) ∩ (N → N)| = 1. There exists some i0 ∈ N such that B0 = {(x, y) ∈ MaxA : y(i0) > g(i0)}= 6 ∅. Let A0 = {(x, y) ∈ MaxA : y(i0) = g(i0)}, then A0 6= ∅. Let A1 =↓A0 ∪ πP (A) × {⊥} and B1 =↓B0 ∪ πP (A) × {⊥}. Similar to case 3.1, we have that A1 and B1 are both Scott closed subsets of L with A ⊆ A1 ∪ B1. But A is contained in non of them. So we prove that ΣL is a sober space while L is not jointly Scott continuous. And we can see from the proof that for any dcpo Q, if ΣQ is sober that Σ(Q × M) is also a sober space.

5. Consonance of the lower powerspace

Given a topological space X, we compared the Scott topology on C(X) with the lower Vietoris topology on C(X). Where the latter is often called the lower powerspace over X (also called the Hoare powerspace) and is denoted by PH (X). Using Q(X) to denote the set of all compact saturated subsets of X, the upper powerspace over X is Q(X) with the upper Vietoris topology, which is generated by 2U = {K ∈ Q(X): K ⊆ U} as a basis, U ∈ O(X). Q(X) with the upper Vietoris topology is also called the Smyth powerspace over X and is denoted by PS(X). For each K ∈ Q(X), let Φ(K) = {U ∈ O(X): K ⊆ U}. Then Φ(K) is a Scott open ﬁlter of O(X) and K = T Φ(K). The concept of consonance is as follows.

9 Definition 5.1. A topological space X is consonant if and only if for every H ∈ σ(O(X)) and every U ∈ H there exists K ∈ Q(X) such that U ∈ Φ(K) ⊆ H. For a sober space X, by the Hoffmann-Mislove theorem (see [7, Theorem II-1.20]), X is consonant if and only if the Scott topology on O(X) has a basis consisting of Scott open filters. Recently, Brecht and Kawai [2] proved that for a topological space X, the consonance of X is equivalent to the commutativity of the upper ∼ and lower powerspaces in the sense that PH (PS(X)) = PS(PH (X)) under a naturally defined homeomorphism. In that paper, they asked the following question: if X is consonant, is PH (X) also consonant? We will give a partial answer below.

n n Q Q Theorem 5.2. Let X be a consonant topological space. If Σ( O(X)) = (ΣO(X)) for each n ∈ N, then PH (X) is consonant.

n Q Proof. Firstly, for each n ∈ N, we deﬁne a map ϕn : O(X) → υ(C(X)) as follows:

n n Q \ ∀(U1,U2,...,Un) ∈ O(X), ϕn(U1,U2,...,Un) = 3Ui. k=1 S S ϕn preserves arbitrary sups, since 3( i∈I Ui) = i∈I 3Ui for any family {Ui ∈ O(X): i ∈ I}. Thus ϕn is n n n Scott continuous and is a continuous map from Q(ΣO(X)) to Σ(υ(C(X))), since Σ(Q O(X)) = Q(ΣO(X)) by assumption. T Generally, 3U ∩ 3V 6= 3(U ∩ V ) for U, V ∈ O(X). Let B = { U∈F 3U : F ⊆f O(X)}, which serves as a base for PH (X). It is easy to see that B is closed for finite union. For any A ∈ σ(υ(C(X))) and A ∈ A . Then A is equal to the union of a family of elements in B. Without loss of generality, we assume that A= 6 ∅. Here we use a trick that is frequently used. Let fin(A) be the set of all finite unions of elements of the family, then fin(A) is a directed subset of B, since B is closed for S W↑ finite union. And then A = fin(A) = υ(C(X)) fin(A) ∈ A , where A is Scott open. There exists some Tn T {Uj : 1 ≤ j ≤ n} ⊆ O(X) \ {∅} such that j=1 3Uj ∈ fin(A) A , i.e., ϕn(U1,U2,...,Un) ∈ A . Since ϕn is a n Q continuous map from (ΣO(X)) to Σ(υ(C(X))) by previous proof, there exists Hj ∈ σ(O(X)), j = 1, 2, . . . , n such that −1 (U1,U2,...,Un) ∈ H1 × H2 × · · · × Hn ⊆ ϕn (A ).

By assumption, X is consonant. For each 1 ≤ j ≤ n, there is Kj ∈ Q(X) such that Uj ∈ Φ(Kj) ⊆ Hj. Obviously, the function η : x 7→↓x, X → PH (X) is a topological embedding. Tn Tn Claim 1 j=1 ↑ C(X)η(Kj) is compact and saturated in PH (X). j=1 ↑ C(X)η(Kj) is an upper set, we Tn only need to prove the compactness. For any C ∈ C(X), C ∈ j=1 ↑ C(X)η(Kj) if and only if there exists (x1, x2, . . . , xn) ∈ K1 × K2 × · · · × Kn such that ↓x1∪ ↓x2 ∪ · · · ∪ ↓xn ⊆ C. The function (x1, x2, . . . , xn) 7→ n Q Tn ↓x1∪ ↓x2 ∪ · · · ∪ ↓xn from X to PH (X) is continuous. Thus j=1 ↑ C(X)η(Kj) =↑ C(X){↓x1∪ ↓x2 ∪ · · · ∪ ↓ n Q xn :(x1, x2, . . . , xn) ∈ K1 × K2 × · · · × Kn} is compact in PH (X) since K1 × K2 × · · · × Kn is compact in X. Tn Tn Claim 2 j=1 3Uj ∈ Φ( j=1 ↑C(X)η(Kj)) ⊆ A . We have that {↓x1∪ ↓x2 ∪ · · · ∪ ↓xn :(x1, x2, . . . , xn) ∈ Tn Tn Tn K1 × K2 × · · · × Kn} ⊆ j=1 3Uj, since Kj ⊆ Uj for each 1 ≤ j ≤ n. Thus j=1 3Uj ∈ Φ( j=1 ↑C(X)η(Kj)), Tn where Φ( j=1 ↑ C(X)η(Kj)) is a Scott open ﬁlter by deﬁnition. According to the previous proof, in order Tn Tn T to prove Φ( j=1 ↑ C(X)η(Kj)) ⊆ A , we only need to prove Φ( j=1 ↑ C(X)η(Kj)) B ⊆ A . Suppose that Tn T j=1 ↑C(X)η(Kj) ⊆ V ∈F 3V for some F ⊆f O(X). Then for each V ∈ F, there is some 1 ≤ j ≤ n such that Tn Kj ⊆ V . Otherwise, for any 1 ≤ j ≤ n, there is yj ∈ Kj \V . Then ↓y1∪ ↓y2 ∪· · · ∪ ↓yn ∈ j=1 ↑C(X)η(Kj) but ↓y1∪ ↓y2 ∪ · · · ∪ ↓yn ∈/ 3V , which contradicts the assumption. For each 1 ≤ j ≤ n, let Fj = {V ∈ F : Kj ⊆ V } and let V = U ∩ T V . Then V ∈ Φ(K ) ⊆ H for each j, which implies that j j V ∈Fj j j j

−1 (V1,V2,...,Vn) ∈ H1 × H2 × · · · × Hn ⊆ ϕn (A ). Tn Tn T T j=1 3Vj = ϕn(V1,V2,...,Vn) ∈ A and it is obvious that j=1 3Vj ⊆ V ∈F 3V . Thus V ∈F 3V ∈ A , since A is an upper set in the order of set inclusion. Tn Tn Since j=1 ↑C(X)η(Kj) ⊆ j=1 3Uj ⊆ A, we are done. Corollary 5.3. Let X be a consonant topological space. If Σ(O(X)) is core-compact or ﬁrst-countable, then PH (X) is consonant.

A natural question is whether consonance of PH (X) implies that X is consonant. We answer this question negatively by a counterexample. First, we show the following result.

10 Proposition 5.4. Let X be a topological space, X is locally compact if and only if X is core-compact and consonant. Proof. (⇒) If X is a locally compact space, then X is core-compact. For any H ∈ σ(O(X)) and any U ∈ H. ◦ Suppose that U is not empty, then for each x ∈ U there is some Kx ∈ Q(X) such that x ∈ Kx ⊆ Kx ⊆ U. S ◦ W S ◦ S S ◦ { x∈F Kx : F ⊆f U} is a directed subset of O(X) with { x∈F Kx : F ⊆f U} = { x∈F Kx : F ⊆f U} = S ◦ S U ∈ H. There exists some F ⊆f U such that Kx ∈ H, since H is Scott open. Then Kx ∈ Q(X) S x∈F x∈F and U ∈ Φ( x∈F Kx) ⊆ H. (⇐) Let X be a core-compact and consonant space. For any x ∈ X and any open set U containing x, there exists some open set V such that x ∈ V U. We employ the interpolation property to ﬁnd a sequence {Vi}i∈N of open sets such that x ∈ V · · · Vn Vn−1 · · · V1 U (see [7]). Let H = {W ∈ O(X): W ⊇ Vi for some i ∈ N}. It is easy to see that H is a Scott open ﬁlter with U ∈ H. There T T exists some K ∈ Q(X) such that U ∈ Φ(K) ⊆ H. Then x ∈ V ⊆ H ⊆ K = {W ∈ O(X): K ⊆ W } ⊆ U.

Example 5.5. Exercise V-5.25 of [7] gives an example of a T0 space X such that O(X) is a continuous lattice but X itself is not locally compact. By Proposition 5.4, X is not consonant. Let L = O(X), then C(X) =∼ Lop and υ(C(X)) =∼ υ(Lop) = ω(L). By Lemma 3.1, υ(C(X)) is a continuous lattice. According to Lemma 3.2 and 3.3, PH (X) is a sober and locally compact space. We can see from Proposition 5.4 that PH (X) is consonant while X is not.

Acknowledgement

We thank the anonymous referee for improving the presentation of the paper.

References

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