A complete Heyting algebra whose Scott space is non-sober
aa r X i v : . [ m a t h . GN ] M a r A complete Heyting algebra whose Scottspace is non-sober
Xu Xiaoquan, Xi Xiaoyong, and Zhao Dongsheng
Abstract.
We prove that (1) for any complete lattice L , the set D ( L )of all nonempty saturated compact subsets of the Scott space of L is acomplete Heyting algebra (with the reverse inclusion order); and (2) ifthe Scott space of a complete lattice L is non-sober, then the Scott spaceof D ( L ) is non-sober. Using these results and the Isbell’s example of anon-sober complete lattice, we deduce that there is a complete Heytingalgebra whose Scott space is non-sober, thus give a positive answer to aproblem posed by Jung. We will also prove that a T space is well-filterediff its upper space (the set D ( X ) of all nonempty saturated compactsubsets of X equipped with the upper Vietoris topology) is well-filtered,which answers another open problem. Sobriety and well-filteredness are two of the most important proper-ties for non-Hausdorff topological spaces. The Scott space of every domain(continuous directed complete posets) is sober. Johnstone [ ] gave the firstexample of a dcpo whose Scott space is non-sober. Soon after that, Isbell [ ]constructed a complete lattice whose Scott space is non-sober. The generalproblem in this line is whether each object in a classic class of lattices has asober Scott space. The Isbell’s non-sober complete lattice is not distributive.Thus recently Jung asked whether there is a distributive complete latticewhose Scott space is non-sober. In this paper we shall give a positive answerto Jung’s problem. The main structure we shall use is the poset D ( X ) ofall nonempty saturated compact subsets of a topological space X equippedwith the reverse inclusion order. We first show that for any complete lat-tice L , the poset D ( L ) of all nonempty saturated compact subsets of theScott space of L is a complete Heyting algebra. Then we prove that for acertain type T spaces X , if X is non-sober, then the Scott space of D ( X )is non-sober. An immediate conclusion is that for any complete lattice L , Mathematics Subject Classification.
Key words and phrases.
Scott topology; sober space; well-filtered space; upper space. if the Scott space of L is non-sober, then the Scott space of the completeHeyting algebra D ( L ) is non-sober. Taking L to be the Isbell’s example,we obtain a complete Heyting algebra whose Scott space is non-sober, thusanswer Jung’s problem.Heckmann and Keimel [ ] proved that a space X is sober if and onlyif the upper space D ( X ) of X is sober. In [ ], Xi and Zhao proved thata space X is well-filtered iff its upper space D ( X ) is a d-space. But, it isstill not known, as pointed out in [ ], whether it is true that a space iswell-filtered if and only if the upper space of X is well-filtered. In the lastpart of this paper we will give a positive answer to this problem.
1. Preliminaries A complete Heyting algebra is a complete lattice L satisfying the follow-ing infinite distributive law: a ∧ _ { a i : i ∈ I } = _ { a ∧ a i : i ∈ I } for any a ∈ L and { a i : i ∈ I }⊆ L . Such a complete lattice is also called a frame . Apparently every complete Heyting algebra is a distributive completelattice.An element p of a meet-semilattice S is a prime element if for any a, b ∈ S , a ∧ b ≤ p implies a ≤ p or b ≤ p . A frame (or complete Heyting algebra) A is called spacial if every element of A can be expressed as a meet of primeelements. It is well-known that a complete lattice L is a spacial frame iff itis isomorphic to the lattice of all open subsets of some topological space (cf.[ ]).A subset U of a poset ( P, ≤ ) is Scott open if i) U is an upper set (thatis, U = ↑ U = { x ∈ P : y ≤ x for some y ∈ U } ), and ii) for any directed subset D ⊆ P , W D ∈ U implies D ∩ U = ∅ whenever W D exists. All Scott open setsof a poset P form a topology on P , denoted by σ ( P ) and called the Scotttopology on P . The space ( P, σ ( P )) is denoted by Σ P , called the Scott space of P .A poset is called a directed complete poset ( dcpo , for short) if everydirected subset of the poset has a supremum. For more about Scott topologyand dcpos, see [ ][ ].A subset A of a topological space is called saturated if A equals theintersection of all open sets containing it. A T space X is well-filtered if forany open set U and filtered family F of saturated compact subsets of X , T F ⊆ U implies F ⊆ U for some F ∈F .A subset A of a space X is irreducible if for any closed subsets F , F of X , F ⊆ F ∪ F implies F ⊆ F or F ⊆ F . Obviously, the closure of everysingleton set is irreducible. A space X is called sober if every irreducible COMPLETE HYTING ALGEBRA WHOSE SCOTT SPACE IS NON-SOBER 3 closed set of X is the closure of a unique singleton set. It is well known thatevery sober space is well-filtered. Jonstone [ ] constructed the first exampleof a dcpo whose Scott space is non-sober. Isbell [ ] constructed a completelattice whose Scott space is non-sober. Kou [ ] gave the first example of adcpo whose Scott space is well-filtered but non-sober.The specialization order ≤ τ on a T space ( X, τ ) is defined by x ≤ τ y iff x ∈ cl ( { y } ), where cl ( { y } ) is the closure of set { y } . A space ( X, τ ) iscalled a d-space (or monotone convergence space ) if ( X, ≤ τ ) is a dcpo and τ ⊆ σ (( X, ≤ τ )) (cf. [ ]).
2. The existence of a complete Heyting algebra whose Scottspace is non-sober
For any topological space X , following Heckmann and Keimel [ ] weshall use D ( X ) to denote the set of all nonempty compact saturated subsetsof X . The upper Vietoris topology on D ( X ) is the topology that has { (cid:3) U : U ∈O ( X ) } as a base, where (cid:3) U = { K ∈D ( X ) : K ⊆ U } . The set D ( X )equipped with the upper Vietoris topology is called the Smyth power space or upper space of X (cf. [ ][ ]).The specialization order on the upper space D ( X ) is the reverse inclusionorder ⊇ . In what follows, the partial order on D ( X ) we will concern is justthe reverse inclusion order.For a poset P , we shall use D ( P ) to denote the poset of all nonemptycompact saturated subsets of the Scott space ( P, σ ( P )).A space X is called coherent if the intersection of any two compactsaturated subsets in X is compact. Lemma . For any complete lattice L , D ( L ) is a complete Heyting al-gebra. Proof.
The Scott space Σ L of L is well-filtered by Xi and Lawson[ ],and is coherent by Jia and Jung [ ]. We now show that the poset D ( L ) isa complete Heyting algebra.(i) From that ( L, σ ( L )) is well-filtered, it follows that D ( L ) is closedunder filtered intersections, thus it is a dcpo, in which the infimum of adirected subsets K of D ( L ) is the intersection of K .(ii) Also Σ L is coherent and every member of D ( L ) contains the topelement 1 L of L , so the intersection K ∩ K of any two members K , K of D ( L ) is again a member of D ( L ), which equals the join K ∨ K of K and K in D ( L ). Also D ( L ) has L as the least element, and { L } as the topelement. It follows that D ( L ) is both a dcpo and a join semilattice, and hasa least element. Therefore D ( L ) is a complete lattice. In addition, for any K , K ∈ D ( L ), the meet K ∧ K of K , K in D ( L ) clearly equals K ∪ K . X. XU, X. XI, AND D. ZHAO
Now for any subfamily { K i : i ∈ I }⊆D ( L ), from (i) and (ii) we can deducethat _ { K i : i ∈ I } = \ { K i : i ∈ I } . Then for any K ∈D ( L ) and { K i : i ∈ I }⊆D ( L ), we have K ∧ W i ∈ I K i = K ∪ ( T i ∈ I K i )= T ı ∈ I ( K ∪ K i )= W i ∈ I ( K ∧ K i ) . Hence D ( L ) is a complete Heyting algebra. (cid:3) For any T space ( X, τ ), let ξ X : X → D ( X ) be the canonical mapping given by ξ X ( x ) = ↑ x = { y ∈ X : x ≤ τ y } . It is easy to see that ξ X : ( X, ≤ τ ) −→ ( D ( D ) , ⊇ ) is an order embedding.To emphasize the codomain, we shall use ξ σX to denote the correspondingmapping ξ σX : ( X, τ ) −→ ( D ( X ) , σ ( D ( X ))), where ξ σX ( x ) = ↑ x for each x ∈ X . Theorem . Let X be a T space such that(i) the upper Vietoris topology on D ( X ) is contained in σ ( D ( X )) (thatis, the upper Vietoris topology is weaker than the Scott topology);(ii) the mapping ξ σX : ( X, τ ) −→ ( D ( X ) , σ ( D ( X ))) is continuous; and(iii) ( D ( X ) , σ ( D ( X ))) is sober.Then X is sober. Proof.
Let F be a nonempty closed irreducible subset of X . Then,as ξ σX is continuous, ξ σX ( F ) is an irreducible subset of ( D ( X ) , σ ( D ( X ))).Therefore, there exists K ∈ D ( X ) such that cl σ ( D ( X )) ( ξ σX ( F )) = ↓ D ( X ) K (= { A ∈D ( X ) : K ⊆ A } ) , where cl σ ( D ( X )) is the closure operator in the Scott space ( D ( X ) , σ ( D ( X ))).Claim 1. Every element of K is an upper bound of F in the poset( X, ≤ τ ).In fact, let k ∈ K and x ∈ F . Then ↑ k ∈D ( X ) and ↑ k ⊆ K . In addition, ↑ x = ξ σX ( x ) ∈↓ D ( X ) K , so ↑ x ⊇ K . Hence ↑ x ⊇ K ⊇ ↑ k , which implies x ≤ τ k .Claim 2. K has a least element.If, on the contrary, for each k ∈ K , there is s ( k ) ∈ K such that k τ s ( k ).Then K ⊆ [ { X \ ↓ s ( k ) : k ∈ K } . Thus K ∈ (cid:3) S { X \ ↓ s ( k ) : k ∈ K }∈ σ ( D ( X )) (by the assumption (i) in Theo-rem 1), implying K ∈ cl σ ( D ( X )) ( ξ σX ( F )) ∩ (cid:3) [ { X \ ↓ s ( k ) : k ∈ K } . COMPLETE HYTING ALGEBRA WHOSE SCOTT SPACE IS NON-SOBER 5
Hence cl σ ( D ( X )) ( ξ σX ( F )) ∩ (cid:3) [ { X \ ↓ s ( k ) : k ∈ K }6 = ∅ . Therefore ξ σX ( F ) ∩ (cid:3) S { X \ ↓ s ( k ) : k ∈ K }6 = ∅ . Hence there exists y ∈ F with ↑ y ⊆ S { X \ ↓ s ( k ) : k ∈ K } . It follows that y τ s ( k ) for some s ( k ) ∈ K . Butthis contradicts Claim 1 (every element of K is an upper bound of F ).Therefore K has a least element, say s . Then K = ↑ s .Claim 3. F = cl X ( { s } ). As s is an upper bound of F and F is closed,we only need to confirm that s ∈ F .Assume, on the contrary, that s F . Then K ⊆ X \ F , so K ∈ cl σ ( D ( X )) ( ξ σX ( F )) ∩ (cid:3) ( X \ F )and (cid:3) ( X \ F ) ∈ σ ( D ( X )) . Therefore ξ σX ( F ) ∩ (cid:3) ( X \ F ) = ∅ , which is impossible.Hence s ∈ F , thus F = ↓ s = cl X ( { s } ).All these together show that ( X, τ ) is sober. (cid:3)
Remark . (1) For every T space X , the mapping ξ X : X −→D ( X )(the upper space of X ) is a topological embedding (cf. [ ]).(2) For any poset P , the mapping ξ σP : ( P, σ ( P )) −→ ( D ( P ) , σ ( D ( P )))is continuous, i.e., it preserves all existing suprema.(3) Every well-filtered space is a d-space.(4) A T space X is well-filtered iff D ( X ) is a dcpo and the upper Vietoristopology on D ( X ) is contained in σ ( D ( D )) (equivalently, the upper space D ( X ) is a d-space) (see [ ][ ]). In general, the well-filteredness of X isstronger than the condition that the upper Vietoris topology on D ( X ) iscontained in σ ( D ( D )).For example, consider the poset N of natural numbers with the usualorder. Then every element in N is compact and so N is an algebraic poset.Hence Σ N ( σ ( N ) equals the Alexandroff topology on N ) is locally compactand D ( N ) = {↑ n : n ∈ N } , which is isomorphic to N . Therefore D ( N ) isnot a dcpo. Now we have that the upper Vietoris topology on D ( N ) equalsthe Scott topology σ ( D ( N )) (and also equals the Alexandroff topology on D ( N )). But Σ N is not well-filtered. Example . Let X be any non-countable set and τ be the co-countabletopology on X . Then ( X, τ ) is a T space. Clearly, the nonempty compact(saturated) subsets of ( X, τ ) are exactly the nonempty finite subsets of X ,that is, D ( X ) = { F : F is a nonempty finite subset of X } . Every directedsubset E of D ( X ) has a largest element (which is the intersection of E ), so( X, τ ) is well-filtered but non-sober ( X is an irreducible closed set but notthe closure of any singleton set). Clearly D ( X ) is a dcpo and every element X. XU, X. XI, AND D. ZHAO in D ( X ) is compact. Hence D ( X ) is an algebraic domain and σ ( D ( X ))(which equals the Alexandroff topology on σ ( D ( X ))) is sober. For ( X, τ ),the conditions (i) and (iii) in Theorem 1 are satisfied, but the assumption(ii) does not hold. In this case, the sobriety of ( D ( X ) , σ ( D ( X ))) does notimply the sobriety of ( X, τ ).By Remark 1 (4), Theorem 1 can be restated as the following one.
Theorem . Let ( X, τ ) be a T space such that(i) X is well-filtered;(ii) the mapping ξ σX : ( X, τ ) −→ ( D ( X ) , σ ( D ( X ))) is continuous; and(iii) ( D ( X ) , σ ( D ( X ))) is sober.Then X is sober. By Theorem 2 and Remark 1, we deduce the following.
Corollary . For a dcpo P , if ( P, σ ( P )) is well-filtered and ( D ( P ) , σ ( D ( P ))) is sober (equivalently, the upper space D ( P ) is a d-space and ( D ( P ) , σ ( D ( P ))) is sober), then ( P, σ ( P )) is sober. By Xi and Lawson [ ], for any complete lattice L , ( L, σ ( L )) is well-filtered, hence the upper Vietoris topology on D ( L ) is contained in σ ( L ).Now applying Corollary 1, we obtain the following. Theorem . For any complete lattice L , if ( L, σ ( L )) is non-sober, then ( D ( L ) , σ ( D ( L ))) is non-sober. Now we are ready to answer Jung’s problem mentioned in the introduc-tion.
Example . In [ ], Isbell constructed a complete lattice whose Scotttopology is non-sober, thus answered a question posed by Johnstone in [ ].The Isbell’s complete lattice is not distributive. In one of his recent talk inSingapore, Achim Jung asked whether there is a distributive complete latticewhose Scott topology is non-sober. We now can give a positive answer tothis problem. Take M be the complete lattice constructed by Isbell and let L = D ( M ). Then by Lemma 1, L is a complete Heyting algebra. Sincethe Scott space of M is non-sober, by Theorem 3, the Scott space of L isnon-sober. Hence L is a complete Heyting algebra whose Scott space isnon-sober. Remark . For any T space ( X, τ ), the poset ( D ( X ) , ⊇ ) is a meet-semilattice, where the meet of K , K ∈ D ( X ) equals K ∪ K . Then clearlyevery principle filter ↑ x = { y ∈ X : x ≤ τ y } is a prime element of D ( X ).In addition, for any K ∈ D ( X ), K = ^ {↑ x : x ∈ K } , COMPLETE HYTING ALGEBRA WHOSE SCOTT SPACE IS NON-SOBER 7 showing that every element of D ( X ) can be expressed as a meet of primeelements. Hence by Lemma 1, ( D ( L ) , ⊇ ) is actually a spacial frame for anycomplete lattice L (see [ ] for more about spatial frames). Thus the non-sober complete Heyting algebra L obtained in Example 2 is also a spacialframe.
3. Well-filteredness of upper spaces
In this section, the symbol D ( X ) will denote the upper space of topo-logical space X .In [ ], it is proved that a space X is sober iff the upper space D ( X ) issober. In [ ], it is proved that a T space X is well-filtered if and onlyif its upper space is a d-space. As remarked in [ ], it is still not knownthe answer to the following problem: Must the upper space D ( X ) be well-filtered if X is well-filtered?We now give a positive answer to the above problem. Lemma . ([ ]) Let X be a topological space and A an irreducible subsetof the upper space D ( X ) . Then every closed set C ⊆ X that meets all mem-bers of A contains an irreducible closed subset A that still meets all membersof A . Remark . The irreducible closed set A in Lemma 2 can be take as aminimal one: if A ′ is a closed set, A ′ ⊆ A and meets all members of A , then A ′ = A (see the proof of Lemma 3.1 in [ ]).The following result can be verified straightforwardly (see e.g. [ , page128] or the proof of Lemma 3.1 in [ ]). Lemma . If K⊆D ( X ) is a nonempty compact subset of D ( X ) , then S K∈D ( X ) . Theorem . A topological space X is well-filtered iff its upper space D ( X ) is well-filtered. Proof.
We only need to show that if X is well-filtered, then so is D ( X ).Let {K t : t ∈ T } be a filtered family of saturated compact subsets of D ( X ), U = S { (cid:3) U i : i ∈ I } an open set of D ( X ) such that \ {K t : t ∈ T }⊆U . Suppose that K t for all t , that is, K t ∩ ( D ( X ) \U ) = ∅ . Then as {K t : t ∈ T } is an irreducible subset of the space D ( X ), by Lemma 2 and Remark 3,there is a minimal closed irreducible subset C⊆D ( X ) \U that meets every K t ( t ∈ T ).Let C = T {⋄ C j : j ∈ J } , where each C j is a closed subset of X and ⋄ C j = { F ∈D ( X ) : F ∩ C j = ∅} . For each t ∈ T , let K t = S ( K t ∩C ). As K t ∩C is X. XU, X. XI, AND D. ZHAO nonempty and compact in D ( X ), by Lemma 3 we have that K t ∈D ( X ). Also { K t : t ∈ T } is a filtered family of member of D ( X ), thus K = T { K t : t ∈ T } is a member of C ( X ) because X is well-filtered.Claim 1. K .Assume, on the contrary, that K ∈U . Then K = T { K t : t ∈ T } ⊆ U i forsome i ∈ I . Then, as X is well-filtered, K t ⊆ U i holds for some t ∈ T . Then K t ∩C⊆ (cid:3) U i ⊆U , contradicting C⊆D ( X ) \U .Claim 2. K ∈ T {↑ D ( X ) ( K t ∩C ) : t ∈ T } . Suppose, on the contrary, that K T {↑ D ( X ) ( K t ∩C ) : t ∈ T } . Then thereis t ∈ T such that K
6∈ ↑ D ( X ) ( K t ∩C ). Thus, for any G ∈K t ∩C , there exists e ( G ) ∈ K \ G . Then G ∩ ↓ e ( G ) = ∅ (Note that G is a saturated compactset). Now for any G ∈K t ∩ C and any t ∈ T , since e ( G ) ∈ K (so e ( G ) ∈ K t ),we have e ( G ) ∈ S ( K t ∩C ). Thus there exists H t ∈K t ∩C such that e ( G ) ∈ H t ,implying H t ∈K t ∩C∩ ⋄ ( ↓ e ( G )) . It follows that K t ∩C∩ ⋄ ( ↓ e ( G )) = ∅ , for all t ∈ T. By the minimality of C , we have C∩ ⋄ ( ↓ e ( G )) = C , which implies C⊆ ⋄ ( ↓ e ( G )).Therefore C⊆ T {⋄ ( ↓ e ( G )) : G ∈K t ∩C} . Note that for any G ∈K t ∩C , G ( ↓ e ( G )). Hence ∅ 6 = K t ∩C = K t ∩C ∩ \ {⋄ ( ↓ e ( G )) : G ∈K t ∩C} = ∅ . This contradiction confirms Claim 2.Now K ∈ T {↑ D ( X ) ( K t ∩C ) : t ∈ T }⊆ T {K t : t ∈ T }⊆U , which implies K ∈U .But this contradicts Claim 1.All these together deduce that there must be some t ∈ T such that K t ⊆U .Hence D ( X ) is well-filtered. The proof is completed. (cid:3) The following result collects some of the equivalent conditions for a spaceto be well-filtered.
Theorem . For any T space X , the following statements are equiva-lent: (1) X is well-filtered. (2) The upper space D ( X ) of X is a d-space. (3) The upper space D ( X ) of X is well-filtered. Acknowledgement (1) The first author is sponsored by NSFC (11661057),the Ganpo 555 project for leading talent of Jiangxi Provence and the NaturalScience Foundation of Jiangxi Provence, China (20161BAB2061004).
COMPLETE HYTING ALGEBRA WHOSE SCOTT SPACE IS NON-SOBER 9 (2) The second author is sponsored by NSFC (11361028, 61300153,11671008, 11701500, 11626207) and NSF Project of Jiangsu Province, China(BK20170483).(3) The third author is sponsored by NIE AcRF (RI 3/16 ZDS).
References [1] G. Gierz, K. Hofmann, K. Keimel, J. Lawson, M. Mislove, D. Scott, ContinuousLattices and Domains, Cambridge University Press, 2003.[2] J. Goubault-Larrecq, Non-Hausdorff topology and Domain Theory, Cambridge Uni-versity Press, 2013.[3] R. Heckmann, An upper power domain construction in terms of strongly compactsets, in: Lecture Notes in Computer Science, vol. 598, Springer, Berlin HeidelbergNew York, 1992, pp. 272-293.[4] R. Heckmann, K. Keimel, Quasicontinuous domains and the Smyth powerdomain,Electronic Notes in Theor. Comp. Sci. 298 (2013) 215-232.[5] H. Kou, U k -admitting dcpos need not be sober, in: Domains and Processes, SemanticStructure on Domain Theory, vol. 1, Kluwer, 2001, pp. 41-50.[6] J. Isbell, Completion of a construction of Johnstone, Proc. Amer. Math. Soci. 85(1982) 333-334.[7] X. Jia, A. Jung, A note on coherence of dcpos, Topol. Appl. 209 (2016) 235-238.[8] P. Johnstone, Scott is not always sober, in: Continuous Lattices, Lecture Notes inMath., vol. 871, Springer-Verlag, 1981, pp. 282-283.[9] P. Johnstone, Stone Spaces, Cambridge University Press, 1982.[10] A. Schalk, Algebras for Generalized Power Constructions, PhD Thesis, TechnischeHochschule Darmstadt, 1993.[11] X. Xi, J. Lawson, On well-filtered spaces and ordered sets, Topol. Appl. 228 (2017)139-144.[12] X. Xi, D. Zhao, Well-filtered spaces and their dcpo models, Math. Struct. Comput.Sci. 27 (2017) 507-515. School of Mathematics and Statistics, Minnan Normal University, Fu-jian, Zhangzhou 363000, P.R. China
E-mail address : [email protected] School of mathematics and Statistics, Jiangsu Normal University, Jiangsu,Xuzhou, P.R. China
E-mail address : [email protected] Mathematics and Mathematics Education, National Institute of Edu-cation Singapore, Nanyang Technological University, 1 Nanyang Walk,Singapore 637616
E-mail address ::