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Some problems about co-consonance of topological spaces

In this paper, we first prove that the retract of a consonant space (or co-consonant space) is consonant (co-consonant). Using this result, some related results have obtained. Simultaneously, we proved that (1) the co-consonance of the Smyth powerspace implies the co-consonance of a topological space under a necessary condition; (2) the co-consonance of a topological implies the co-consonance of the smyth powerspace under some conditions; (3) if the lower powerspace is co-consonant, then the topological space is co-consonant; (4) the co-consonance of implies the co-consonance of the lower powerspace with some sufficient conditions.


Introduction
Given a topological space X, there are two topologies on O(X), let O(X) be the poset of all open sets of X with inclusion order. In this paper, we consider three kinds of topologies on O(X). One is the topology τ that has a base { Q | Q is compact in X}, where Q = {V ∈ O(X) | Q ⊆ V }, the second is the Scott topology σ(O(X)) and the third is the upper topology υ(O(X)). It is not difficult to find that τ ⊆ σ(O(X)). A topological space X is consonant if τ = σ(O(X)). Equivalently, a space X is consonant if the Upper Kuratowski topology and the co-compact on the set of all closed sets of X are equal (see [13]). Consonant spaces have been researched by many scholars and the more conclusions about consonance can see [1,4,13]. Given a poset P , the upper topology υ(P ) on P is a topology that has a subbase {P \ ↓x | x ∈ P }. Clearly, υ(P ) ⊆ σ(P ), where σ(P ) is the Scott topology on P . A topological space is co-consonant if the upper topology and the Scott topology on the open set lattice O(X) coincide (see [2]). Compared with the consonance, the co-consonance is lack of research. In this paper, we focus our interest on co-consonant spaces.
Retract is an ordinary relation of topological spaces. There are many topological properties which are preserved by retraction, such as sobriety, wellfilteredness, d-spaces. Naturally, it arises the following two questions: (1) Is the retract of a consonant space consonant?
(2) Is the retract of a co-consonant space co-consonant?
In section 3, we give a positive answer. Using this result, some related conclusions are discovered.
Given a topological space X, there are two constructions of topological spaces, namely, the lower powerspace P H (X) and smyth powerspace P S (X) (see [5,7]). Recently, M. Brecht and T. Kawai proved that X is consonant iff the lower powerspace and Smyth powerspace on X is commute (see [2]). Furthermore, M. Brecht and T. Kawai asked that whether the consonance is preserved by the Smyth powerspace construction. This question is first answered by Z. Lyu, Y. Chen, and X. Jia (see [11] ). It has proved in [11] that if X is consonant and for natural number n, Σ( n O(X)) = n Σ(O(X)), then P S (X) is consonant. Similarly, Y. Chen, H. Kou and Z. Lyu also showed that if X is consonant and for natural number n, then P H (X) is consonant (see [3]). Naturally, it arises the following questions.
(3) Is X co-consonant when P s (X) is co-consonant?
(4) Is the smyth powerspace P s (X) of a co-consonant space X co-consonant?
(5) Is the lower powerspace P H (X) of a co-consonant space X co-consonant? (6) Is X co-consonant when P H (X) is co-consonant?
In section 4, we will give some answers for these questions.

Preliminaries
Given a poset P and A ⊆ P , let For every x ∈ P , we write ↓x for ↓{x} and ↑x for ↑{x} .
Let P be a poset and x, y ∈ P . We say that x is way below y, in symbols x ≪ y, iff for all directed subsets D ⊆ P for which D exists, y ≤ D implies x ≤ d for some d ∈ D. If ⇓a = {b ∈ P | b ≪ a} is directed and ⇓a = a for all a ∈ P , we call P a continuous poset. A complete lattice L is called a continuous lattice if L is a continuous poset.
Let P be a poset. A subset U of P is Scott open (see [5]) if (i) U= ↑U and (ii) for any directed subset D, D ∈ U implies D ∩ U = ∅ whenever D exists. The Scott open sets on P form the Scott topology σ(P ). The Scott space (P, σ(P )) will be simply written as Σ(P ). The upper topology υ(P ) on poset P is a topology that has {P \ ↓x | x ∈ P } as a subbase.
For a T 0 space (X, τ ), the specialization order ≤ on X is defined by x ≤ y if and only if x ∈ cl({y}) (see [5], p. 42]). We use O(X) (Γ(X)) to denote the lattice of all open(closed) subsets of X. Note that for each subset A ⊆ X, ↑A is equal to the intersection of all the open sets containing A and we say ↑A is the saturation of A.
For a topological space X, we shall use Q(X) to denote the poset of all nonempty compact saturated subsets of X with the reverse inclusion order. The upper Vietoris topology on Q(X) is the topology that has {✷U | U ∈ O(X)} as a base, where ✷U = {K ∈ Q(X) | K ⊆ U}. The upper Vietoris topological space (or called smyth powerspace) is denoted by P S (X). Naturally, {⋄U | U ∈ Γ(X)} is a subbase of the closed sets of P S (X), where ⋄U = {Q ∈ Q(X) | Q ∩ U = ∅}. Then the specialization order of the upper space P S (X) is the reverse inclusion order.
For every x ∈ X, cl({x}) is an irreducible closed set of X. A topological space X is called to be sober if every irreducible closed set of X is the closure of an unique singleton set. A topological space X is well-filtered if for each filtered family F of compact saturated subsets of X and each open set U of X, F ⊆ U implies F ⊆ U for some F ∈ F . It is well known that every sober space is well-filtered (see [5]). A topological space X is coherent if the intersection of two compact saturated subsets is compact. For every complete lattice L, the Scott space Σ(L) is well-filtered and coherent (see [10,14]).
there is an open set V and a compact subset K such that x ∈ V ⊆ K ⊆ U. A topological space X is said to be core-compact if (O(X), ⊆) is a continuous lattice. It has known that every locally compact space is core-compact.
Interestingly, J.Goubault-Larrecq find the following results in his blog which is due to Matthew de Brecht.

The retract of consonant and co-consonant
In the following, we will give positive answers to the question 1 and question 2 proposed in introduction.
(2)A topological space X is locally compact iff X is a core-compact and consonant space (see [3]).
Clearly, X is co-consonant iff the upper topology is agree with the Scott topology on O(X).

Example 3.4 (1)
If X is a quasi-polish space, then P S (X) is a co-consonant space (see [2]).
The Example 3.4 (2) illustrates that the following conclusion holds.
(2) Let X be a T 0 space. Define mappings φ : P S (X) −→ P S (P S (X)) and ϕ : P S (P S (X)) −→ P S (X) as follows: where the mapping ξ : X −→ P S (X) is defined by ξ(x) = ↑x. Then φ and ϕ are continuous such that ϕ • φ = id P S (X) and φ • ϕ ≥ id P S (P S (X)) . So we can conclude that P S (X) is a strong retract of P S (P S (X)).
Remark 3.11 Let L be a complete lattice. Then for any is a retract of Σ(σ(σ(L))).
Claim: f and g are continuous mappings.
Proof. Fix a y 0 ∈ Y , the mapping α y 0 : X −→ X × Y is given by This means that α y 0 is a continuous mapping. Clearly, p • α y 0 = id X , where p : X × Y −→ X is the projection. So we can assert that X is a retract of X × Y . By Theorem 3.6 and Theorem 3.7, X is co-consonant (consonant). ✷ Theorem 3.15 Let X and Y be a pair of consonant spaces. If (i) X and Y are core-compact; Define a mapping f n : Σ( Claim 1: f n is continuous.  N 2 ), · · ·, (M s,i , N s,i ), · · ·, (M n , N n )))) = f n (((M 1 , N 1 ), (M 2 , N 2 ), · · ·, ( i∈I M s,i , i∈I N s,i ), · · ·, (M n , N n ))) = ( , and i∈I f n ((((M 1 , N 1 ), (M 2 , N 2 ), · · ·, (M s,i , N s,i ), · · ·, (M n , N n )))) = i∈I (( By the consonance of X and Y , for every 1 ≤ k ≤ n, there is a Q k ∈ Q(X) and a P k ∈ Q(Y ) such that A k ∈ Q k ⊆ A k and B k ∈ P k ⊆ B k . By (ii), Q k = ↑x k and P k = ↑y k for some x k ∈ X and y k ∈ Y . Let Then C is compact in X × Y .
So (x k , y k ) ∈ R j k × S j k for some j k ∈ J 0 . This means that R j k ∈ ↑x k = Q k and S j k ∈ ↑y k = P k . Then Equivalently, In this section, we will give some answers for those four questions which is proposed in the introduction. Specifically, the answer of question 3 is negative and the partial answers of question 4 and question 5 is given. Furthermore, the answer of question 6 is positive.
Before answering question 3, it is necessary to introduce the concept of strongly compact subsets. A subset K of a topological space X is strongly compact if for each open set U, K ⊆ U implies that there is a finite subset F ⊆ X such that K ⊆ ↑F ⊆ U (see [2]). Lemma 4.1 (see [2]) Let X be a co-consonant T 0 space. Then every compact subset of X is strongly compact.
In the following, we prove that the converse of Lemma 4.1 is true if P S (X) is co-consonant.
Theorem 4.2 Let X be a T 0 space. If P S (X) is co-consonant and every compact subset of X is strongly compact, then X is co-consonant.
Proof. Define a mapping ξ : X −→ P S (X) by ξ(x) = ↑x, for all x ∈ X. Then ξ is continuous. So the mapping f : O(P S (X)) −→ O(X) is well defined, where f (U) = ξ −1 (U) for all U ∈ O(P S (X)). Clearly, f is Scott continuous. For each H ∈ σ(O(X)) and U ∈ H. Then ✷U ∈ f −1 (H) and By the continuity of f , f −1 (H) ∈ σ(O(P S (X))). Since P S (X) is co-consonant, there is a finite subset F ⊆ Γ(P S (X)) such that Since ✷U ∈ ♦F for each F ∈ F , there exists a Q F ∈ Q(X) such that Q F ⊆ U and Q F ∈ F . By assumption, Q is strongly compact. Then there is a finite Clearly, F is a finite subset of Γ(X).
So we have V ∈ H. Hence the claim 2 is proved. By two claims, we can conclude that X is co-consonant. ✷ Note that the co-consonance of P S (X) does not imply that every compact subset of X is strongly compact. Please see the following example. ∀ x ∈ P, x ≤ ∞. Then X = (P, υ(P )) is a quasi-polish space (see [2] Example 3.2). By Example 3.4(1), P s (X) is co-consonant. It is not difficult to verify that X is a compact space. However the compact subset P is not strongly compact in X. So we can assert that X is not co-consonant.
The following conclusion is one of the most important conclusions of this paper. This gives a partial answer of question 4. Proof. Let F ∈ σ(O(P S (X))) and F ∈ F . Then F = i∈I ✷U i for some ✷U k ∈ F . Define a mapping β n : Σ( n (O(X))) −→ Σ(O(P S (X))) as follows: Since β n is Scott continuous for each component V i , β n is continuous. As Σ( n O(X)) = n Σ(O(X)), β n is also continuous form n Σ(O(X)) to Σ(O(P S (X))). Then there are finitely Scott open sets {H k | 1 ≤ k ≤ n} ⊆ σ(O(X)) such that (U 1 , U 2 , · · ·, U n ) ∈ H 1 × H 2 × · · · × H n ⊆ β −1 n (F ). By the co-consonance of X, there is a finite subset E k ⊆ Γ(X) such that This means that W j k ∈ H k for each 1 ≤ k ≤ n. So we have (W j 1 , W j 2 , · · ·, W jn ) ∈ H 1 × H 2 × · · · × H n ⊆ β −1 n (F ).
Whence, 1≤k≤n ✷W j k ∈ F . Since F is an upper set in O(P S (X)), j∈J ✷W j ∈ F .
Those two claims imply that P S (X) is co-consonant. ✷ By Lemma 2.1, Lemma 2.2 and Theorem 4.4, the following conclusion holds immediately.
Corollary 4.5 Let X be a co-consonant space. If X is core-compact or Σ(O(X)) is first-countable, then P S (X) is co-consonant.
The next conclusion demonstrates that co-consonance of original space is necessary for the co-consonance of lower powerspace. Thus we give a positive answer of question 6.
Theorem 4.6 Let X be a T 0 space. If P H (X) is co-consonant, then X is co-consonant.
Proof. Define a mapping ξ : X −→ P H (X) by ξ(x) = cl({x}), for all x ∈ X. Then ξ is a topological embedding. So the mapping η : O(P H (X)) −→ O(X) is well defined, where η(U) = ξ −1 (U) for all U ∈ O(P H (X)). Clearly, η is Scott continuous. Let F ∈ σ(O(X)) and U ∈ F . By the definition of mapping η, we have where J i is finite for each i ∈ I and {U j | j ∈ J i } ⊆ O(X). Then η −1 (F ) ∈ σ(O(P H (X))) and ⋄U ∈ η −1 (F ). By the co-consonance of P H (X), there is a finite subset E ⊆ Γ(P H (X)) such that Then for each A ∈ E, ⋄U ∩ A = ∅. So there exists a nonempty closed Before answering question 5, it is necessary to introduce the following concept. Remark 4.8 If every closed set of a topological space X is irreducible, then X is intersection-compatible. In particular, the Scott spaces of chains are intersection-compatible. Proof. Let F ∈ σ(O(P H (X))) and F ∈ F .
Since β n is Scott continuous for each component V i , β n is continuous. As Σ( n O(X)) = n Σ(O(X)), β n is also continuous form n Σ(O(X)) to Σ(O(P H (X))). Then there are finitely Scott open sets {H k | 1 ≤ k ≤ n} ⊆ σ(O(X)) such that (U 1 , U 2 , · · ·, U n ) ∈ H 1 × H 2 × · · · × H n ⊆ β −1 n (F ). By the co-consonance of X, there is a finite subset F k ⊆ Γ(X) such that U k ∈ {♦A | A ∈ F k } ⊆ H k , for each 1 ≤ k ≤ n. Let F = {✷( n j=1 A j ) | A j ∈ F j , 1 ≤ j ≤ n}. Clearly, F is a finite subset of Γ(P H (X)). A j ) and for every 1 ≤ j ≤ n, A j ∈ F j . Since U j ∈ {♦A | A ∈ F j }, then U j ∩A j = ∅. Take x j ∈ U j ∩A j . Then ⋄V i . Then we have (G 1 , G 2 , · · ·, G n ) ∈ H 1 × H 2 × · · · × H n ⊆ β −1 n (F ). Whence, n k=1 ⋄G k ∈ F . Since F is an upper set, it follows from n k=1 ⋄G k ⊆ m i=1 ⋄V i ⊆ A that A ∈ F . So the claim 2 is proved.
These two claims are enough for the proof. ✷