Characterization of metrizable Esakia spaces via some forbidden configurations
aa r X i v : . [ m a t h . L O ] S e p CHARACTERIZATION OF METRIZABLE ESAKIA SPACES VIA SOMEFORBIDDEN CONFIGURATIONS
GURAM BEZHANISHVILI AND LUCA CARAI
Abstract.
By Priestley duality, each bounded distributive lattice is represented as thelattice of clopen upsets of a Priestley space, and by Esakia duality, each Heyting algebrais represented as the lattice of clopen upsets of an Esakia space. Esakia spaces are thosePriestley spaces that satisfy the additional condition that the downset of each clopen isclopen. We show that in the metrizable case Esakia spaces can be singled out by forbiddingthree simple configurations. Since metrizability yields that the corresponding lattice ofclopen upsets is countable, this provides a characterization of countable Heyting algebras.We show that this characterization no longer holds in the uncountable case. Our resultshave analogues for co-Heyting algebras and bi-Heyting algebras, and they easily generalizeto the setting of p-algebras. Introduction
Priestley duality [9, 10] provides a dual equivalence between the category
Dist of boundeddistributive lattices and the category
Pries of Priestley spaces; and Esakia duality [5] providesa dual equivalence between the category
Heyt of Heyting algebras and the category
Esa ofEsakia spaces. To make the paper self-contained, we recall main definitions.An ordered topological space is a triple ( X, T , ≤ ) such that ( X, T ) is a topological spaceand ≤ is a partial order on X . When we say that an ordered topological space is compact,metrizable, etc. we mean that the underlying topological space is compact, metrizable, etc.As usual, for A ⊆ X we let ↑ A = { x ∈ X | a ≤ x for some a ∈ A } and ↓ A = { x ∈ X | x ≤ a for some a ∈ A } . If A = { x } , then we write ↑ x and ↓ x , respectively. We call A an upset if ↑ A = A and a downset if ↓ A = A . Definition 1.1. (1) An ordered topological space ( X, T , ≤ ) satisfies the Priestley separation axiom if x (cid:2) y implies that there is a clopen upset U such that x ∈ U and y / ∈ U .(2) A Priestley space is an ordered topological space that is compact and satisfies thePriestley separation axiom.
Mathematics Subject Classification.
Key words and phrases.
Distributive lattice, Heyting algebra, p-algebra, Priestley duality, Esakia duality.
Notation 1.2.
To simplify notation, we will suppress T and ≤ and denote a Priestley spacesimply by X . Remark 1.3.
The following facts about Priestley spaces are well known:(1) Each Priestley space is a Stone space (compact, Hausdorff, zero-dimensional space).(2) If F is closed, then so are ↑ F and ↓ F .(3) There exist Priestley spaces such that the downset or upset of an open set may notbe open.The Priestley space of a bounded distributive lattice L is constructed by taking the set X of prime filters of L , the order on X is the inclusion order, and the topology on X is givenby the basis { α ( a ) \ α ( b ) | a, b ∈ L } where α ( a ) = { x ∈ X | a ∈ x } . Then α is an isomorphism of L onto the lattice of clopen upsets of X . Definition 1.4.
A Priestley space is an
Esakia space if the downset of each open set is open(equivalently, the downset of each clopen set is clopen).
Remark 1.5.
In an Esakia space, the upset of an open set may not be open.Heyting algebras are the bounded distributive lattices L with an additional binary oper-ation → of relative pseudo-complement which satisfies, for all a, b, x ∈ L : a ∧ x ≤ b iff x ≤ a → b. It turns out that the lattice of clopen upsets of a Priestley space X is a Heyting algebra iff itis an Esakia space, where the relative pseudo-complement of two clopen upsets U, V is givenby X \ ↓ ( U \ V ).The three spaces Z , Z , and Z depicted in Figure 1 are probably the simplest examplesof Priestley spaces that are not Esakia spaces. Topologically each of the three spaces ishomeomorphic to the one-point compactification of the countable discrete space { y } ∪ { z n | n ∈ ω } , with x being the limit point of { z n | n ∈ ω } . For each of the three spaces, itis straightforward to check that with the partial order whose Hasse diagram is depicted inFigure 1, the space is a Priestley space. On the other hand, neither of the three spaces is anEsakia space because { y } is open, but ↓ y = { x, y } is no longer open.In this paper we show that a metrizable Priestley space is not an Esakia space exactly whenone of these three spaces can be embedded in it. The embeddings we consider are special inthat the point y plays a special role. We show that this condition on the embeddings, as wellas the metrizability condition, cannot be dropped by presenting some counterexamples. Indoing so, we develop a way to combine two Priestley spaces which has proved to be useful inbuilding Priestley spaces that are not Esakia spaces. An advantage of our characterizationlies in the fact that when a metrizable Priestley space X is presented by a Hasse diagram,it is easy to verify whether or not X contains one of the three “forbidden configurations”. HARACTERIZATION OF METRIZABLE ESAKIA SPACES 3 xyz z z Z xyz z z Z xyz z z Z Figure 1.
The three Priestley spaces Z , Z , and Z .The paper is organized as follows. In Section 2 we present the main result by showing thata metrizable Priestley space is not an Esakia space iff a copy of one of the three forbiddenconfigurations sits inside X in a special way. In Section 3 we translate our main resultinto the dual lattice-theoretic statement, yielding a characterization of countable Heytingalgebras. This characterization easily generalizes to the setting of p-algebras, and also hasanalogues for co-Heyting and bi-Heyting algebras. In Section 4 we present the “down-upsum” of Priestley spaces, and its dual “ideal-filter product” of lattices. Finally, Section 5is devoted to counterexamples. We use the down-up sum to build non-metrizable Priestleyspaces that are not Esakia spaces and yet do not contain a copy of any of the three forbiddenconfigurations. This shows that there is no obvious generalization of our results to the non-metrizable setting. We finish by showing that the additional condition on the embeddingscannot be dropped either. 2. The main theorem
Definition 2.1.
Let X be a Priestley space. We say that Z i ( i = 1 , ,
3) is a forbiddenconfiguration for X if there are a topological and order embedding e : Z i → X and an openneighborhood U of e ( y ) such that e − ( ↓ U ) = { x, y } .The next result shows that whether a metrizable Priestley space is an Esakia space isdetermined by these three forbidden configurations. The key assumption of metrizability isused to show that if x is a limit point of a set, then there is a sequence in the set convergingto x . This can be done already for the Priestley spaces that are sequential spaces (seeRemark 2.3). The necessity of the sequentiality assumption will be discussed in more detailin Section 5. G. BEZHANISHVILI AND L. CARAI
Theorem 2.2.
A metrizable Priestley space X is not an Esakia space iff one of Z , Z , Z is a forbidden configuration for X .Proof. First suppose that one of the Z i is a forbidden configuration for X . Since e : Z i → X is continuous and e − ( ↓ U ) = { x, y } is not open in Z i , we conclude that ↓ U is not open in X . Thus, X is not an Esakia space.Conversely, suppose that X is not an Esakia space. Then there is an open subset U of X such that ↓ U is not open. Therefore, ( ↓ U ) c is not closed. Since X is metrizable, thereis a sequence { x n } ⊆ ( ↓ U ) c such that lim x n = x ∈ ↓ U . As X is Hausdorff, { x n } has tobe infinite, hence we may assume that x n = x m for n = m . Because U is open, we have x ∈ ↓ U \ U . Let y ∈ U be such that x ≤ y . Then x < y .Observe that x n (cid:2) x and x n (cid:2) y for any n because otherwise x n ∈ ↓ U . In addition, ifthere is M such that y ≤ x n for all n ≥ M , then x n ∈ ↑ y for all n ≥ M . Since ↑ y is closedand x = lim x n , this would yield x ∈ ↑ y , a contradiction. Therefore, y (cid:2) x n for some n ≥ M .Thus, we can select a subsequence of { x n } each member of which is not above y . Hence, wemay assume without loss of generality that x n and y are incomparable for all n . We nowhave two cases to consider. Case 1:
There is an infinite subsequence { y n } of { x n } that is totally ordered by ≤ . Since { y n } is an infinite subsequence of { x n } , we have lim y n = x . Consider the closure { y n } . As { y n } is totally ordered, by [2, Lem. 3.1], { y n } is also totally ordered and has max and min.Since x ∈ { y n } which is totally ordered, for each n we have x ≤ y n or y n ≤ x . But, as wealready observed, y n (cid:2) x . Thus, x ≤ y n for each n . Since x / ∈ ( ↓ U ) c , we have x < y n . Wenow define recursively a subsequence { z n } of { y n } such that z > z > z > · · · .Set z = y . If z k = y n k is already defined, then since lim y n = x and x < y n k , thereis a clopen downset V of X such that x ∈ V , y n k = z k / ∈ V , and V contains an infinitesubset of { y n } . So there is y n k +1 ∈ V such that n k +1 > n k . Therefore, y n k +1 < y n k . Set z k +1 = y n k +1 . We thus obtain a sequence z > z > z > · · · such that lim z n = x and each z n is incomparable with y .Let Z = { y, x } ∪ { z n } ⊆ X , and view Z as an ordered topological space with the order andtopology inherited from X . Since Z ∩ U = { y } and { y } is closed in X , we have that { y } isclopen in Z . For each m , we show that the singleton { z m } is clopen in Z . As x < z m , thereis a clopen downset V of X such that x ∈ V and z m / ∈ V , so V c ∩ Z is finite and contains z m .Since X is Hausdorff, so is V c ∩ Z . Because every finite Hausdorff space is discrete, { z m } isclopen in V c ∩ Z , which is clopen in Z . Thus, the singleton { z m } is clopen in Z .Opens in Z containing x are exactly the cofinite subsets of Z because lim z n = x and allthe singletons except { x } are clopen. Therefore, Z is order-isomorphic and homeomorphicto the Priestley space Z . Case 2:
There is no infinite totally ordered subsequence of { x n } . Since every infiniteposet contains either an infinite chain or an infinite antichain (see, e.g., [13, Thm. 1.14]),there is an infinite subsequence { y n } of { x n } that is an antichain. As { y n } is an infinitesubsequence of { x n } , we have that lim y n = x . Our goal is to select a subsequence { z n } of { y n } so that either Z or Z becomes a forbidden configuration. Which of the two becomesa forbidden configuration depends on whether or not ↑ x ∩ { y n } is infinite. HARACTERIZATION OF METRIZABLE ESAKIA SPACES 5
Case 2a: ↑ x ∩ { y n } is infinite. Then { z n } := ↑ x ∩ { y n } is an infinite subsequence of { y n } such that lim z n = x and each z n is incomparable with y . Let Z = { y, x } ∪ { z n } ⊆ X ,and view Z as an ordered topological space with the order and topology inherited from X .Since x < z m for each m , by arguing as in Case 1 we obtain that Z is order-isomorphic andhomeomorphic to the Priestley space Z . Case 2b: ↑ x ∩ { y n } is finite. Then { z n } := ( ↑ x ) c ∩ { y n } is an infinite subsequence of { y n } such that lim z n = x and each z n is incomparable with y . Let Z = { y, x } ∪ { z n } ⊆ X ,and view Z as an ordered topological space with the order and topology inherited from X .Since x and z m are incomparable for each m , by arguing as in Case 1 we obtain that Z isorder-isomorphic and homeomorphic to the Priestley space Z . (cid:3) Remark 2.3.
In the proof of Theorem 2.2 metrizability was used to find in a set that is notclosed a sequence converging outside of it. We recall (see, e.g., [4, p. 53]) that a topologicalspace X is a sequential space provided a set A is closed in X iff together with each sequence A contains all its limits. Thus, Theorem 2.2 holds not only for metrizable Priestley spaces,but more generally, for sequential Priestley spaces.3. Algebraic meaning of the result
Let L , L , and L be the dual lattices of Z , Z , and Z , respectively. Clopen upsetsof Z are the whole space, the empty set, ↑ z n , and ↑ z n ∪ { y } for n ∈ ω . Thus, L can bedepicted as in Figure 2. Note that L is not a Heyting algebra since ¬ c does not exist. { y } = c ∅ ↑ z ∪ { y }↑ z ∪ { y }↑ z ∪ { y }↑ z ↑ z ↑ z Z Figure 2.
The lattice L .Clopen upsets of Z are the whole space, the empty set, and the finite subsets of { y }∪{ z n | n ∈ ω } . Therefore, L is isomorphic to the lattice of finite subsets of ω together with a topelement; see Figure 3. Thus, L is not a Heyting algebra because ¬ F does not exist for anyfinite subset F of ω .Clopen upsets of Z are the whole space, the empty set, finite subsets of { y }∪{ z n | n ∈ ω } ,and { x, y } ∪ C where C is a cofinite subset of { z n | n ∈ ω } . Therefore, if we denote by CF ( ω )the Boolean algebra of finite and cofinite subsets of ω and by the two-element Boolean G. BEZHANISHVILI AND L. CARAI P fin ( ω ) ∅ F Z Figure 3.
The lattice L .algebra, then L is isomorphic to the sublattice of CF ( ω ) × given by the elements of theform ( A, n ) where A is finite or n = 1; see Figure 4. Thus, L is not a Heyting algebrabecause ¬ ( F,
1) does not exist for any finite F . P fin ( ω ) × { } ( F, P fin ( ω ) × { }P cofin ( ω ) × { } Figure 4.
The lattice L . Definition 3.1.
Let L ∈ Dist and let a, b ∈ L . Define I a → b := { c ∈ L | c ∧ a ≤ b } It is easy to check that I a → b is an ideal, and that I a → b is principal iff a → b exists in L , inwhich case I a → b = ↓ ( a → b ).In order to give the dual description of I a → b let X be the Priestley space of L and let α be the isomorphism from L onto the lattice of clopen upsets of X (see the introduction). Itis well known that ideals of L correspond to open upsets of X , and this correspondence isrealized by sending an ideal I of L to α [ I ] := S { α ( a ) | a ∈ I } . On the other hand, filters of HARACTERIZATION OF METRIZABLE ESAKIA SPACES 7 L correspond to closed upsets of X , and this correspondence is realized by sending a filter F of L to α [ F ] := T { α ( a ) | a ∈ F } . Lemma 3.2.
Let L ∈ Dist and let X be its dual Priestley space. If a, b ∈ L , then α [ I a → b ] = X \ ↓ ( α ( a ) \ α ( b )) .Proof. For any c ∈ L we have c ∈ I a → b ⇔ c ∧ a ≤ b ⇔ α ( c ) ∩ α ( a ) ⊆ α ( b ) ⇔ α ( c ) ⊆ X \ ↓ ( α ( a ) \ α ( b ))where the last equivalence follows from the fact that for any upsets U, V, W we have W ∩ U ⊆ V iff W ⊆ X \ ↓ ( U \ V ). Thus, α [ I a → b ] = X \ ↓ ( α ( a ) \ α ( b )). (cid:3) It is a well-known consequence of Stone duality for Boolean algebras that a Boolean algebrais countable iff its Stone space is metrizable (see, e.g., [7, Prop. 7.23]). This fact generalizesto bounded distributive lattices and Priestley spaces (see, e.g., [12, p. 54]). To see this, let L be a bounded distributive lattice and X its Priestley space. The Boolean algebra of clopensof X is isomorphic to the free Boolean extension B ( L ) of L ; see, e.g., [1, Sec. V.4]. Thus,the following three conditions are equivalent: • X is metrizable; • L is countable; • B ( L ) is countable. Theorem 3.3.
Let L be a countable bounded distributive lattice. Then L is not a Heytingalgebra iff one of L i ( i = 1 , , is a homomorphic image of L such that the homomorphism h i : L → L i satisfies the following property: There are a, b ∈ L such that h i [ I a → b ] = I c i → ,where c = c , c = { } , or c = ( ∅ , .Proof. ( ⇒ ) It is sufficient to translate Theorem 2.2 to its dual algebraic statement. Let X bethe Priestley space of L . Then X is a metrizable Priestley space which is not an Esakia space.Thus, by Theorem 2.2, Z i is a forbidden configuration for X for some i = 1 , ,
3. Let e, U beas in Definition 2.1. Then there are a, b ∈ L such that e ( y ) ∈ α ( a ) \ α ( b ) ⊆ U . Therefore, e − ↓ ( α ( a ) \ α ( b )) ⊆ e − ↓ U = { x, y } . On the other hand, since e is order-preserving and e ( y ) ∈ α ( a ) \ α ( b ), we have { e ( x ) , e ( y ) } ⊆ ↓ ( α ( a ) \ α ( b )), so { x, y } ⊆ e − ↓ ( α ( a ) \ α ( b )).Thus, e − ↓ ( α ( a ) \ α ( b )) = { x, y } = ↓ y . We also have that α ( c i ) = { y } ⊆ Z i . By Lemma 3.2, α [ I a → b ] = X \ ↓ ( α ( a ) \ α ( b )) and α [ I c i → ] = Z i \ ↓ y . Let h i : L → L i be the bounded latticehomomorphism corresponding to the embedding e : Z i → X , so h i = e − . Since e is anembedding, h i is onto [9]. Therefore, since e − ( X \ ↓ ( α ( a ) \ α ( b ))) = Z i \ e − ↓ ( α ( a ) \ α ( b )) = Z i \ ↓ y, we conclude that h i [ I a → b ] = I c i → .( ⇐ ) We show that a → b does not exist in L . If a → b exists, then we have I a → b = ↓ ( a → b ). Since h i is an onto lattice homomorphism, I c i → = h i [ I a → b ] = h i [ ↓ ( a → b )] = ↓ h i ( a → b ) . G. BEZHANISHVILI AND L. CARAI
Therefore, c i → h i ( a → b ), and hence c i → L i . The obtained contradictionproves that a → b does not exist in L , and hence L is not a Heyting algebra. (cid:3) Theorem 3.3 yields a characterization of countable Heyting algebras. We conclude thissection by showing that this characterization easily generalizes to countable p-algebras. Werecall (see, e.g., [8]) that a p-algebra is a pseudocomplemented distributive lattice. Priestleyduality for p-algebras was developed in [11]. We call a Priestley space X a p-space providedthe downset of each open upset is open. Then a bounded distributive lattice L is a p-algebraiff its dual Priestley space X is a p-space. Definition 3.4.
Let X be a Priestley space. We say that Z i ( i = 1 , ,
3) is a p-configuration for X if Z i is a forbidden configuration for X and in addition the open neighborhood U of e ( y ) is an upset.We point out that neither of the bounded distributive lattices L , L , L that are dual to Z , Z , Z is a p-algebra. The next result is a direct generalization of Theorems 2.2 and 3.3,so we skip its proof. Corollary 3.5.
Let L be a countable bounded distributive lattice, and let X be its Priestleyspace, which is then a metrizable space. (1) X is not a p-space iff one of Z , Z , Z is a p-configuration for X . (2) L is not a p-algebra iff one of L i ( i = 1 , , is a homomorphic image of L such thatthe homomorphism h i : L → L i satisfies the following property: There is a ∈ L suchthat h i [ I a → ] = I c i → , where c = c , c = { } , or c = ( ∅ , . We recall that co-Heyting algebras are order-duals of Heyting algebras. The Priestleyspaces dual to co-Heyting algebras are the ones with the property that the upset of eachopen is open [6]. Let Z ∗ , Z ∗ , Z ∗ be the Priestley spaces obtained by reversing the order in Z , Z , Z , respectively. Then dualizing Theorem 2.2 yields: Corollary 3.6.
A metrizable Priestley space X is not the dual of a co-Heyting algebra iffthere are a topological and order embedding e from one of Z ∗ , Z ∗ , Z ∗ into X and an openneighborhood U of e ( y ) such that e − ( ↑ U ) = { x, y } . We recall that bi-Heyting algebras are the lattices which are both Heyting algebras andco-Heyting algebras. Priestley spaces dual to bi-Heyting algebras are the ones in which theupset and downset of each open is open. Putting together the results for Heyting algebrasand co-Heyting algebras yields:
Corollary 3.7.
A metrizable Priestley space X is not dual to a bi-Heyting algebra iff one of Z , Z , Z is a forbidden configuration for X or there are a topological and order embedding e from one of Z ∗ , Z ∗ , Z ∗ into X and an open neighborhood U of e ( y ) such that e − ( ↑ U ) = { x, y } . HARACTERIZATION OF METRIZABLE ESAKIA SPACES 9 Ideal-filter product and down-up sum
Definition 4.1.
Let L and M be bounded distributive lattices, I an ideal of L , and F afilter of M . We define the ideal-filter product of L and M as L × FI M := { ( l, m ) ∈ L × M | l ∈ I or m ∈ F } . Lemma 4.2. L × FI M is a bounded sublattice of L × M .Proof. Clearly (0 , ∈ L × FI M since 0 ∈ I , and (1 , ∈ L × FI M because 1 ∈ F . Let( l , m ) , ( l , m ) ∈ L × FI M . If m ∧ m / ∈ F , then m / ∈ F or m / ∈ F , so l ∈ I or l ∈ I , implying that l ∧ l ∈ I . Therefore, ( l ∧ l , m ∧ m ) ∈ L × FI M . If l ∨ l / ∈ I ,then l / ∈ I or l / ∈ I , so m ∈ F or m ∈ F , which implies that m ∨ m ∈ F . Thus,( l ∨ l , m ∨ m ) ∈ L × FI M . (cid:3) L × FI ML I MF
Figure 5. L × FI M as a sublattice of L × M .In order to describe the Priestley space of L × FI M , we recall (see, e.g., [3, p. 17 andp. 269]) the definition of linear sum of two Priestley spaces. Let X, Y be Priestley spaces.For simplicity, we assume for the rest of this section that X and Y are disjoint. If they arenot, then as usual, we can simply replace X with X × { } and Y with Y × { } . Define the linear sum X ⊕ Y as the disjoint union of X and Y with the topology of disjoint union andthe order given by x ≤ y iff ( x, y ∈ X and x ≤ y ) or( x, y ∈ Y and x ≤ y ) or( x ∈ X and y ∈ Y ) . Intuitively, we place X “below” Y . We next modify the definition of the linear sum of X and Y . Definition 4.3.
Let
X, Y be Priestley spaces, D a closed downset of X , and U a closedupset of Y . We define the down-up sum X ⊕ UD Y of X and Y as their disjoint union withthe topology of disjoint union and the order given by x ≤ y iff ( x, y ∈ X and x ≤ y ) or( x, y ∈ Y and x ≤ y ) or( x ∈ D and y ∈ U ) . Intuitively, instead of placing X “below” Y , we are only placing D “below” U (see Fig-ure 6). D UX Y
Figure 6.
The Priestley space X ⊕ UD Y . Lemma 4.4. X ⊕ UD Y is a Priestley space.Proof. Clearly X ⊕ UD Y is compact. That ≤ is reflexive and antisymmetric is obvious, andthat ≤ is transitive follows from D being a downset of X and U an upset of Y . Let x (cid:2) y .First suppose that x, y ∈ X . Then there is a clopen upset A of X containing x and missing y . Therefore, A ∪ Y is a clopen upset of X ⊕ UD Y containing x and missing y .Next suppose that x, y ∈ Y . Then there is a clopen upset B of Y containing x and missing y . But then B is also a clopen upset of X ⊕ UD Y containing x and missing y .If x ∈ Y and y ∈ X , then Y is a clopen upset of X ⊕ UD Y containing x and missing y .Finally, suppose that x ∈ X and y ∈ Y . Since x (cid:2) y , we have x / ∈ D or y / ∈ U . If x / ∈ D ,then since D is a closed downset of X , there is a clopen upset A of X containing x anddisjoint from D . Thus, A is a clopen upset of X ⊕ UD Y containing x and missing y . If y / ∈ U ,then since U is a closed upset of Y , there is a clopen downset B of Y containing y anddisjoint from U . Therefore, A := Y \ B is a clopen upset of Y containing U and missing y .Thus, X ∪ A is a clopen upset of X ⊕ UD Y containing x and missing y . (cid:3) Theorem 4.5.
Let
L, M be bounded distributive lattices, I an ideal of L , and F a filter of M . Let also X be the Priestley space of L , Y the Priestley space of M , V an open upset of X corresponding to the ideal I , D := X \ V , and U the closed upset of Y corresponding tothe filter F . Then X ⊕ UD Y is homeomorphic and order-isomorphic to the Priestley space of L × FI M .Proof. Let α be a lattice isomorphism from L onto the clopen upsets of X and β a latticeisomorphism from M onto the clopen upsets of Y . Define γ from L × FI M to the clopenupsets of X ⊕ UD Y by γ ( l, m ) = α ( l ) ∪ β ( m ). Since l ∈ I , we have α ( l ) ∩ D = ∅ ; and since m ∈ F , we have U ⊆ β ( m ). Thus, γ ( l, m ) is a clopen upset of X ⊕ UD Y , and so γ is welldefined. It is straightforward to see that γ is a one-to-one bounded lattice homomorphism.To see that γ is onto, let A be a clopen upset of X ⊕ UD Y . Let l ∈ L and m ∈ M be suchthat α ( l ) = A ∩ X and β ( m ) = A ∩ Y . If l / ∈ I , then α ( l ) ∩ D = ∅ . So there is d ∈ D ∩ A ,and since A is an upset of X ⊕ UD Y , we have ↑ d ⊆ A . But then, by the definition of the order HARACTERIZATION OF METRIZABLE ESAKIA SPACES 11 on X ⊕ UD Y , we have that U ⊆ ↑ d ⊆ A . Therefore, U ⊆ A ∩ Y = β ( m ), and so m ∈ F .Thus, ( l, m ) ∈ L × FI M , and hence γ is onto. Consequently, L × FI M is isomorphic to theclopen upsets of X ⊕ UD Y , which by Priestley duality yields that X ⊕ UD Y is homeomorphicand order-isomorphic to the Priestley space of L × FI M . (cid:3) Counterexamples
In the definition of X ⊕ UD Y , when the closed upset U coincides with Y , we denote X ⊕ UD Y by X ⊕ D Y . Lemma 5.1.
Let
X, Y be Esakia spaces. Then X ⊕ D Y is an Esakia space iff D is clopenin X .Proof. Without loss of generality we may assume that X and Y are disjoint. First supposethat D is not clopen in X . We have that Y is a clopen upset of X ⊕ D Y and ↓ Y = Y ∪ D .Since ( Y ∪ D ) ∩ X = D , we see that Y ∪ D cannot be clopen in X ⊕ D Y . Therefore, X ⊕ D Y is not an Esakia space.Next suppose that D is clopen in X . Any clopen in X ⊕ D Y can be written as A ∪ B where A is clopen in X and B is clopen in Y . If B = ∅ , then ↓ ( A ∪ B ) = ↓ A is clopen in X , and hence clopen in X ⊕ D Y . If B = ∅ , then ↓ ( A ∪ B ) = ↓ A ∪ ↓ B = ↓ A ∪ ( ↓ B ∩ Y ) ∪ D .Since ↓ A, D are clopen in X and ↓ B ∩ Y is clopen in Y , we conclude that ↓ ( A ∪ B ) is clopenin X ⊕ D Y . (cid:3) Remark 5.2. (1) There is an obvious analogue of Lemma 5.1 for p-spaces: For p-spaces X and Y , X ⊕ D Y is a p-space iff D is clopen in X .(2) If Y = { y } is a singleton, then in the definition of X ⊕ D Y we are adding only onepoint on top of D .We are ready to give examples of non-metrizable (even non-sequential) Priestley spacessuch that they are not Esakia spaces and yet they do not contain the three forbidden con-figurations. Example 5.3.
Let X = βω be the Stone- ˇCech compactification of the discrete space ω . Weview βω as an Esakia space with trivial order. Let D = βω \ ω , Y = { y } , and consider X ⊕ D Y ; see Figure 7.Since D is not clopen, Lemma 5.1 implies that this is an example of a Priestley spacethat is not an Esakia space. It is well known (see, e.g., [4, Cor. 3.6.15]) that there are nonon-trivial convergent sequences in βω . Therefore, there is no sequence in ( ↓ y ) c convergingto an element of ↓ y . Thus, X ⊕ D Y does not contain the three forbidden configurations.The closed downset D corresponds to the ideal I := P fin ( ω ) of finite subsets of P ( ω ), andthe clopen upset Y = { y } corresponds to the filter F = { } of . Thus, the dual lattice of X ⊕ D Y is P ( ω ) × FI ; see Figure 8. Example 5.4.
Let ω be the first uncountable ordinal, and let X be the poset obtained bytaking the dual order of ω + 1. Endow X with the interval topology. It is straightforward yβω \ ω Figure 7.
The space X ⊕ D Y of Example 5.3. P fin ( ω ) × { } P ( ω ) × { } Figure 8.
The dual lattice of the space X ⊕ D Y of Example 5.3.to check that X is an Esakia space. Let D = { ω } ⊆ X , let Y = { y } , and consider X ⊕ D Y ;see Figure 9. ω yX Figure 9.
The space X ⊕ D Y of Example 5.4.Since D is not clopen, Lemma 5.1 implies that X ⊕ D Y is not an Esakia space. On theother hand, there is no sequence in X \ { ω } converging to ω . Thus, X ⊕ D Y does notcontain the three forbidden configurations.The dual lattice of X is ω + 1, the closed downset D corresponds to the ideal I := ω of ω + 1, and the clopen upset Y = { y } corresponds to the filter F = { } of . Thus, the duallattice of X ⊕ D Y is ( ω + 1) × FI ; see Figure 10. HARACTERIZATION OF METRIZABLE ESAKIA SPACES 13 ω Figure 10.
The dual lattice of the space X ⊕ D Y of Example 5.4. Remark 5.5. (1) The space X ⊕ D Y of Example 5.4 can be thought of as a generalization of theforbidden configuration Z , obtained by “stretching” the chain { z n | n ∈ ω } . As aresult, the chain X \ { ω } is “too long” to contain a sequence converging to ω .(2) We can generalize the forbidden configuration Z similarly by “stretching” the an-tichain { z n | n ∈ ω } .(3) The space X ⊕ D Y of Example 5.3 can be thought of as a generalization of theforbidden configuration Z , obtained by “inflating” the point x . As a result, we donot have sequences from ω converging inside βω \ ω .We note that in the definition of the three forbidden configurations, the condition onthe open neighborhood U of e ( y ) cannot be dropped. This can be seen from a generalobservation that every Priestley space embeds into an Esakia space, and hence every boundeddistributive lattice is a homomorphic image of a Heyting algebra. To see this, let L be abounded distributive lattice and X its Priestley space. We let F L be the free boundeddistributive lattice generated by the underlying set of L . The identity on L induces an ontolattice homomorphism h : F L → L . Dually, the onto homomorphism h corresponds to anembedding e : X → L where 2 = { , } is the two-element discrete Priestley space with0 <
1. Since 2 is an Esakia space and products of Esakia spaces are Esakia spaces, 2 L is anEsakia space. Thus, F L is a Heyting algebra. Consequently, we cannot characterize Esakiaspaces by forbidding embeddings of some Priestley spaces. This yields that in the definitionof the three forbidden configurations, the condition on the open neighborhood U of e ( y )cannot be dropped.In most cases, the space 2 L is rather complex. We conclude the paper by presenting muchsimpler examples of Esakia spaces into which the Priestley spaces Z , Z and Z embed. Example 5.6.
Let X be the disjoint union of two copies of the one-point compactificationof the discrete space ω , and let the order on X be defined as in Figure 11.It is straightforward to check that X is a metrizable Esakia space, and yet there is atopological and order embedding of Z into X , described by the white dots in the figure. Figure 11.
The space X of Example 5.6. The white dots represent the imageof Z under the embedding of Z into X .An analogous space for Z can be constructed as follows. Example 5.7.
Let X be the disjoint union of two copies of the one-point compactificationof the discrete space ω , and let the order on X be defined as in Figure 12. Figure 12.
The space X of Example 5.7. The white dots represent the imageof Z under the embedding of Z into X .Then X is a metrizable Esakia space, and the white dots describe a topological and orderembedding of Z into X .We finish by constructing a metrizable Esakia space in which Z is embedded, which ismore complicated than Examples 5.6 and 5.7. Example 5.8.
Let X be a subspace of R as described in Figure 13 with each α mn an isolatedpoint, each sequence { α in | n ∈ ω } converging to α iω , and each sequence { α ni | n ∈ ω } converging to α ωi .Let Y = { α mn | m, n ∈ ω } . Then Y is a discrete subspace of X and X is a compactificationof Y . Clearly X is a compact metrizable space. Each clopen U of X is either a finite unionof subsets of the form • { α mn } for some m, n ∈ ω ; • { α mn | h ≤ m ≤ ω } for some h, n ∈ ω ; • { α mn | k ≤ n ≤ ω } for some k, m ∈ ω ;or the complement of one of these finite unions. From this it is easy to see that X is a Stonespace. HARACTERIZATION OF METRIZABLE ESAKIA SPACES 15 α α α α ω α α α α ω α α α α ω α ω α ω α ω α ωω Figure 13.
The space X .Define a partial order on X by α hk ≤ α mn iff ( h, k ) = ( m, n ) or h, k ≥ m + n where we set m + n = ω if at least one of m, n is ω . α α α ω α ω α ωω ↓ α α ω α ωω ↓ α ω α ωω ↓ α ωω Figure 14.
The principal downsets ↓ α , ↓ α ω , and ↓ α ωω .Figure 14 shows how to calculate principal downsets of points of X . From this and thedescription of clopens of X it is straightforward to check that the downset of each clopen isclopen. α α α α ↑ α α α α α ω ↑ α ω Figure 15.
The principal upsets ↑ α and ↑ α ω .Figure 15 describes how to calculate principal upsets of points of X . From this and thefact that α ωω is the least element of X , it is easy to see that the upset of each point of X isclosed. Thus, X is an Esakia space (see [5]). We can embed Z into X via the map definedby y α ω , z i α ω i +1 , and x α ωω ; see Figure 16. z z z xy Figure 16.
The embedding of Z into X . Acknowledgment
We are grateful to the referee for careful reading and the comments which have improvedthe presentation of the paper.
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