On continuous noncomplete lattices

Algebra univers. 46 (2001) 215 – 230 0002–5240/01/020215 – 16 $ 1.50 + 0.20/0 © Birkhäuser Verlag, Basel, 2001 On continuous noncomplete lattices K. V. Adaricheva, V. A. Gorbunov and M. V. Semenova Abstract. We consider noncomplete continuous and algebraic lattices and prove that finitely generated free lattices are algebraic. We also study the Lawson topology, the second most important topology in the theory of continuous domains, on finitely presented lattices. In particular, we prove that algebraic finitely presented lattices are linked bicontinuous and the Lawson topology on these lattices coincides with the interval topology. Several examples of non-distributive and noncomplete algebraic and continuous lattices are given in the paper. Introduction The notion of the continuous lattice exists in two different worlds. In classical lattice theory by a continuous lattice one usually means a lattice which is both meet and join continuous. In the mathematical theory of computation the notion of continuous lattice was introduced by D. Scott [8] as a model of λ-calculus. In that world only topological aspects of this lattice were essential. What is common for both definitions is that they are applied to a complete lattice, i.e., to a lattice with the operations of join and meet defined for an arbitrary subset of elements of the lattice. In the well-known paper [5], Yu. L. Ershov introduced so-called A-spaces that are a generalized noncomplete version of continuous lattices (or, even wider, continuous domains) in the sense of D. Scott. In the present paper, we are making an attempt to look at noncomplete continuous lattices from the classical lattice theory point of view. In the first section, we consider some facts known for continuous lattices that remain true in noncomplete case. In the second section, we give several examples of noncomplete lattices that are algebraic or continuous. The proof of the central example of this section is given in the third section. In particular, we show that finitely generated free lattices are algebraic. The fourth section is devoted to the study of the Lawson topology on a finitely presented lattice. We show that it coincides with the dual Lawson topology and the interval topology Presented by Professor Wieslaw Dziobiak. Received April 5, 2000; accepted in final form December 12, 2000. 2000 Mathematics Subject Classification: 06B25, 0630, 06B35. Key words and phrases: Lattice, continuous, finitely presented, Lawson topology, interval topology. The research of the first two authors was supported in part by RFBR Grants 96-01-01525 and 96-01-00097, and by DFG Grant 436 113/2670. The research of the third author was supported by RFBR Grants 97-71001 and 99-01-00485, and by the program “Russian Universities: fundamental research”. 215 216 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. provided that the lattice is algebraic. In particular, the interval topology on such a lattice is Hausdorff. 1. Definition and some properties of continuous lattices The content of this section is a re-establishment of some facts from [5] given there in topological setting or a generalization of known facts concerning (complete) continuous lattices [8]. Let us recall some definitions and notation. Let L be an arbitrary lattice and D ⊆ L. By ↓D we denote the set {f ∈ L : f 6 d for some d ∈ D}. Respectively, ↑D denotes the set {f ∈ L : f > d for some d ∈ D}. Next, let x, y ∈ L. We say that x is way-below y, notationally x ≪ y, if W for every up-directed subset D of L that has a least upper bound D in L we have x ∈ ↓D W whenever y 6 D. Also, by ↑x, we denote the set {y ∈ L : x ≪ y}. We say that L is continuous if every x ∈ L is the least upper bound of the set of elements W which are way-below it, i.e., x = {z ∈ L : z ≪ x}. PROPOSITION 1.1. The relation ≪ on any continuous lattice L satisfies the interpolation property: if x ≪ z and x 6= z then there exists y such that x ≪ y ≪ z and x 6 = y. Proof. If L does not have 0, then we add it to L, i.e., we consider L0 = L ∪ {0}, where 0 6 x for any x ∈ L. Let x ≪ z, x 6 = z. Consider Iz = {u ∈ L0 : (∃y) (u ≪ y ≪ z)}. Then 0 ∈ Iz , Iz is an ideal of L0 , and z is an upper bound of Iz . We show that z is the least upper bound of Iz . If it were not the case then there would exist an upper bound z∗ such that z∗ > 6 z. Since L is continuous, there exists y ≪ z with y 6 6 z∗ . The last inequality implies existence of u ≪ y such that u 6 6 z∗ . Then u ∈ Iz but u 6 6 z∗ , a contradiction. W W Since z 6 = 0 and z = Iz , the set Iz′ = Iz \ {0} is a nonempty ideal in L and z = Iz′ . By supposition, x ≪ z, hence x ∈ Iz′ . Then there exists y ∗ such that x ≪ y ∗ ≪ z. Since x 6 = z, there exists u such that u ≪ z, u 6 6 x. Put y ∗∗ = y ∗ ∨ u. Then x ≪ y ∗∗ ≪ z and ∗∗ x 6= y . ¨ Now we introduce the Scott topology on L. A subset U ⊆ L is called Scott-open if U satifies (I) and (II) below. (I) U = ↑U ; W W (II) if D is an up-directed set and D exists then D ∈ U implies D ∩ U 6= ∅. The proofs of two propositions below are similar to those in complete case [8]. Vol. 46, 2001 On continuous noncomplete lattices 217 PROPOSITION 1.2. A subset U of a continuous lattice L is Scott-open if and only if S U = z∈U ↑z. Moreover, the “if” part of this assertion is true for any lattice. PROPOSITION 1.3. For any element u of a continuous lattice L, if U = ↑u then S U = z∈U ↑z. In particular, ↑u is Scott-open. Now we consider Scott-continuous mappings of lattices, i.e., mappings which are continuous in the Scott topology. PROPOSITION 1.4. Let f : L → S be a mapping between lattices L and S. The following are equivalent: (1) f is Scott-continuous, i.e., f −1 (U ) is Scott-open in L whenever U is Scott-open in S; (2) For any up directed set D that has a least upper bound in L, the set f (D) has a W W least upper bound in S and f ( D) = f (D); Moreover, if S and L are continuous, then (1) and (2) are equivalent to either of the following conditions: (3) y ≪ f (x) if and only if y 6 f (w) for some w ≪ x. W (4) f (x) = {f (w) : w ≪ x} for all x ∈ L. Proof. (1) ⇒ (2). First we notice that f is monotone. Indeed, suppose x 6 y but f (x) 6 6 f (y). The set V = S \ ↓f (y) is a Scott-open neighborhood of f (x), hence U = f −1 (V ) is a Scott-open neighborhood of x. Then y ∈ U and f (y) ∈ f (U ) = V , a contradiction. W Now suppose that there exists D = d for some up directed subset D of L. Since f is monotone, f (d) is an upper bound of f (D). Suppose that there exists an upper bound y of f (D) such that y > 6 f (d). The open set V = S \ ↓y contains f (d), hence U = f −1 (V ) is a Scott-open neighborhood of d. By the definition of a Scott-open set we have U ∩ D 6= ∅; let d ′ ∈ U ∩ D. Then f (d ′ ) ∈ V . Since y is an upper bound of f (D), we arrive at a W contradiction. Thus f (d) = f (D). (2) ⇒ (1). A mapping f is Scott-continuous if and only if f −1 (V ) is closed whenever V is. Evidently, V is closed in the Scott topology iff V = ↓V and V is closed with respect to existing least upper bounds of up directed sets. Let V be closed. Since f is monotone, W f −1 (V ) = U = ↓U . Now consider an up directed set D ⊆ U such that D exists. W W W Since f ( D) = f (D) and f (D) is up directed in V , we have f (D) ∈ V and hence W D ∈ U. The equivalence of conditions (1)–(4) for continuous lattices is established as in Theorem II.2.1 in [8]. ¨ 218 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. A lattice L is said to be meet-continuous if for any up-directed subset D ⊆ L such that W D exists in L and any x ∈ L, the set x ∧D = {x ∧d : d ∈ D} has a least upper bound and W W (x ∧ D) = x ∧ D. Notice that in general lattice theory the lattices with this property are called upper continuous (cf. [2, 6]). PROPOSITION 1.5. Any continuous lattice is meet-continuous. W Proof. Evidently, x ∧ D is an upper bound of x ∧ D. Suppose there exists an upper W bound u of x ∧ D such that u > 6 x ∧ D. Since the lattice is continuous, there exists W W y ≪ x ∧ D such that y 6 6 u. Then y 6 x, y ≪ D, whence y 6 x ∧ d 6 u for some W W d ∈ D, a contradiction. Thus (x ∧ D) exists and coincides with x ∧ D. ¨ An element x ∈ L is said to be compact if x ≪ x. A lattice L is algebraic if every element of it is a join of compact elements (cf. [1]). Algebraic lattices are called also compactly generated (see [2]). An element x ∈ L is said to be completely join irreducible W if x = X, X ⊆ L, implies x ∈ X. Completely meet irreducible elements are defined dually. The set of completely join (meet) irreducible elements of L will be denoted by CJ (L) (CM(L), respectively). It is not hard to see that x is completely join irreducible if and only if there exists the unique lower cover x∗ of x, i.e. y < x implies y 6 x∗ . As an easy corollary of definitions we obtain the following PROPOSITION 1.6. Any algebraic lattice is continuous. The following assertion is a generalization of the well-known fact concerning complete algebraic lattices [2]. Recall that a lattice L is weakly atomic if every interval in L contains a cover, i.e., for any a, b ∈ L, a < b, there exist c, d ∈ L such that a 6 c ≺ d 6 b. PROPOSITION 1.7. Any algebraic lattice is weakly atomic. Proof. Consider a, b ∈ L with a < b. Then there exists a compact element c such that c 6 b and c 6 6 a. Hence a < a ∨ c 6 b. Consider the set P = {x ∈ L : a 6 x < a ∨ c}. Since a ∈ P , this set is nonvoid. We show that every chain in P has an upper bound in P . Let D ⊆ P be a chain. If there W exists d = D and d < a ∨ c, then we are done. If d = a ∨ c then by the compactness of c we have c 6 d ′ for some d ′ ∈ D. Since a ∨ c 6 d ′ , we arrive at a contradiction. Finally, suppose that D does not have a least upper bound in L. Then a1 = a ∨ c is an upper bound of D but it is not the least one. Therefore there exists an upper bound a2 of D such that a2 > 6 a1 . Then a 6 a1 ∧ a2 < a1 , hence a1 ∧ a2 is an upper bound of D in P . By Zorn’s lemma, there exists a maximal element d ∈ P , which yields d ≺ a ∨ c. ¨ A homomorphism f : K → L is said to be lower bounded if for every a ∈ L the set f −1 (↑a) = {x ∈ K : f (x) > a} either is empty or has a least element which is denoted Vol. 46, 2001 On continuous noncomplete lattices 219 by β(a). Upper bounded homomorphism is defined dually, and the greatest element of f −1 (↓a) provided it exists is denoted by α(a). A homomorpism is called bounded whenever it is both lower and upper bounded. LEMMA 1.8. Any upper bounded homomorphism is Scott-continuous. Proof. Let f : L → K be an upper bounded homomorphism. According to Proposition 1.4 it suffices to show that f preserves joins of up directed sets whenever W W they exist. Let D be an up directed set in L such that D exists. Obviously, f ( D) is an upper bound of f (D). Let b ∈ L be another upper bound of that set. We have W W W D ⊆ f −1 (↓b), that implies D 6 α(b), i.e., f ( D) 6 b, which means that f ( D) is the least upper bound of f (D). ¨ PROPOSITION 1.9. A bounded homomorphic image of a continuous lattice is continuous. Proof. Let K be a continuous lattice and let f : K → L be bounded surjective homoW morphism. Also, let a ∈ L. Since K is continuous, we have β(a) = {b : b ≪ β(a)}. W The set {b : b ≪ β(a)} is up directed, consequently, by Lemma 1.8 we have a = {f (b) : b ≪ β(a)}. W Let us show that f (b) ≪ a for all b ≪ β(a). Let a 6 D for some up directed set W D. Then β(d) 6 β( D) holds for any d ∈ D, and the set D1 = {β(d) : d ∈ D} is up W W W directed. If β( D) 6 = D1 , then we find z ∈ K, such that β(d) 6 z < β( D) for all W W d ∈ D. Therefore d 6 f (z) for all d ∈ D, i.e., D 6 f (z). Consequently, z > β( D), W W a contradiction with the choice of z. We have b ≪ β(a) 6 β( D) = D1 . By the assumption this implies b 6 β(d) for some d ∈ D. Consequently, f (b) 6 d, which was required. ¨ THEOREM 1.10. A lattice L is continuous if and only if L is a bounded retract of an algebraic lattice. Proof. The “if” part follows from the previous proposition. Conversely, let L be a W I exists}. In fact, continuous lattice. We consider the set I d ∗ (L) = {I ∈ I d(L) : ∗ ∗ ∗ I d (L) is a lattice. Moreover, I d (L) 6 Id(L). Indeed, let I1 , I2 ∈ I d (L) and I = I1 ∨I2 W W W W W be a join in I d(L), then, obviously, I = I1 ∨ I2 . If I = I1 ∩ I2 , then I1 ∧ I2 W W is an upper bound of I . Let y ∈ L and x 6 y hold for every x ∈ I . If z ≪ I1 ∧ I2 , then we can find i1 ∈ I1 , i2 ∈ I2 , such that z 6 i1 ∧ i2 , which means that z ∈ I and, W W W consequently, z 6 y. Now, I = I1 ∧ I2 follows by the continuity of L. Using the natural embedding φ(x) = ↓x, we obtain L 6 Id ∗ (L). W Define a map ψ : I d ∗ (L) → L by ψ(I ) = I . ψ is a homomorphism, as W W W ψ(I1 ∩ I2 ) = (I1 ∩ I2 ) = ( I1 ∧ I2 ) = ψ(I1 ) ∧ ψ(I2 ) 220 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. and ψ(I1 ∨ I2 ) = W W W (I1 ∨ I2 ) = ( I1 ∨ I2 ) = ψ(I1 ) ∨ ψ(I2 ) hold for any I1 , I2 ∈ Id ∗ (L). Moreover, ψ is a retraction, as ψφ(x) = for any x ∈ L. We show that ψ is bounded, i.e., the sets n o _ I >a , ψ −1 (↑a) = I ∈ Id ∗ (L) : n o _ I 6a ψ −1 (↓a) = I ∈ Id ∗ (L) : W ↓x = x holds W have the least and the greatest element, respectively. Since I 6 a implies I ⊆ ↓a, ↓a is T the greatest element in ψ −1 (↓a). We show I ′ = {I : I ∈ ψ −1 (↑a)} ∈ ψ −1 (↑a), i.e., W ′ I = a. We have I ′ ⊆ ↓a as ↓a ∈ ψ −1 (↑a). Let y be an upper bound of I ′ and x ≪ a. The last inequality implies that x ∈ I for all I ∈ ψ −1 (↑a), hence x ∈ I ′ and x 6 y. By the W continuity of L we obtain a 6 y and, consequently, a = I ′ , i.e., I ′ is the least element in ψ −1 (↑a). By Lemma 1.8 and its dual, ψ is Scott-continuous and dually Scott-continuous. Moreover, ψ is complete with respect to existing least upper and greatest lower bounds. Finally, let us show that Id ∗ (L) is algebraic. We prove all principal ideals to be compact. Let {Dk : k ∈ K} be an up-directed subset of Id ∗ (L) and D ∈ Id ∗ (L) be the least upper W bound of this subset, i.e., D exists. We claim that W W (1) d = D = {dk : k ∈ K}, W Dk . Indeed, Dk ⊆ D for all k ∈ K, whence d is an upper bound of where dk = {dk : k ∈ K}. If y is another upper bound then Dk ⊆ ↓y for all k ∈ K. Thus ↓y is an upper bound of {Dk : k ∈ K} in Id ∗ (L) and, consequently, D ⊆ ↓y, which means d 6 y. By (1) W W S we have {Dk : k ∈ K} = {Dk : k ∈ K} in Id ∗ (L). Hence ↓x ⊆ {Dk : k ∈ K} implies x ∈ Dk for some k ∈ K. Thus ↓x ⊆ Dk , i.e., ↓x is compact in Id ∗ (L) for any x ∈ L. ¨ Not all properties of complete algebraic lattices are inherent in noncomplete lattices. For example, any element of a complete continuous (algebraic) lattice is a meet of meet irreducible (completely meet irreducible) elements. This is not true in the general case of noncomplete lattices. EXAMPLE 1.11. There exists an algebraic noncomplete lattice such that every element of it is meet reducible. Consider L = ⊕Q 2, the ordinal sum of two-element chains 2 = {0, 1} over the chain of rational numbers Q. Let 0q and 1q denote the elements of qth summand of this ordinal sum. It is easy to verify that for any q, r ∈ Q the element (1q , 1r ) is compact in the Vol. 46, 2001 On continuous noncomplete lattices 221 lattice L2 . Moreover, any element of L2 is a join of elements of such type. Hence L2 is algebraic. On the other hand, if (a, b) ∈ L2 and a1 > a, b1 > b, where a1 , b1 ∈ L, then (a, b) = (a1 , b) ∧ (a, b1 ). Hence (a, b) is meet reducible. QUESTION 1.12. Which conditions on a noncomplete continuous (algebraic) lattice would suffice for every element to be equal to the meet of meet irreducible (completely meet irreducible) elements? 2. Examples of algebraic and continuous lattices The aim of this section is to give some examples of noncomplete algebraic and continuous lattices which will show the importance of the generalization undertaken in the previous section. 2.1. There is a Boolean noncomplete algebraic lattice. <ω of finite and cofinite subsets of a given (infinite) set X. It is the lattice B<ω Most further examples are not distributive. 2.2. For any finite X, the lattice F L(X) freely generated by X is algebraic. This statement follows from Theorem 3.1 that will be proved in the next section. See also the discussion following that theorem. 2.3. There exist continuous nonalgebraic lattices. (a) Let E n be the Euclidean n-dimensional vector space over the field R of real numbers. A subset X ⊆ E n is said to be convex if for any a, b ∈ X the segment [a, b] is contained in X. A convex body is a compact convex subset in E n . The collection of convex bodies is denoted by CB(E n ). Evidently, CB(E n ) is a lattice that is noncomplete provided that E n is not attached to it as the top element. Also, it contains the lattice CBf (E n ) of finitely generated convex bodies (or polytopes) as a sublattice. The latter is essentially noncomplete because the infinite intersection of polytopes is not necessarily a polytope. The dual lattices of CB(E n ) and CBf (E n ) turn out to be continuous. However, these lattices are not algebraic because neither CBf (E n ) nor CB(E n ) is weakly atomic. For a polytope P , let Pε denote a polytope all whose faces are parallel to the corresponding faces of P with the distance ε between them; obviously, P ⊆ Pε . To show that the dual of CBf (E n ) is continuous, we note that for any polytope T P and any ε > 0, Pε is way-below P . Evidently, P = ε>0 Pε . Hence every element of the dual of CBf (E n ) is the least upper bound of the elements which are way-below it. 222 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. Similarly, if B, C ∈ CB(E n ) and B is a neighborhood of C, then B is way-below C in the dual of CB(E n ). (b) Another example of noncomplete continuous nonalgebraic lattice is given by a dense noncomplete chain. For instance, the chain of rational numbers Q satisfies all these requirements. To give an example of a vast class of noncomplete meet-continuous lattices we recall some definitions from [6]. W A finite subset L of a lattice L is called a join cover of an element a ∈ L if a 6 U and a 6 6 u for any u ∈ U . Meet covers are defined dually. We say that a cover V refines U , notationally V ⊑ U , if for any v ∈ V there exists u ∈ U such that v 6 u. A join cover V of a is minimal if for any join cover T of a, T ⊑ V implies V ⊆ T . We say that L possesses the minimal join cover refinement property if every a ∈ L has only finitely many minimal join covers and every join cover of a refines to a minimal one. Among the lattices satisfying this property are free, projective and finitely presented lattices. LEMMA 2.1. Let L be an arbitrary lattice that possesses the dual minimal join cover refinement property. Let D be an up-directed subset of L with nonempty set of upper bounds U . Then for an arbitrary b ∈ L the set of all upper bounds of b∧D coincides with ↑(b∧U ). Proof. Evidently, any t ∈ ↑(b ∧ U ) is an upper bound of b ∧ D. Now we take any upper bound s of b ∧ D. If b 6 s or d 6 s for some d ∈ D then the conclusion is clear. Otherwise by the assumption, any meet cover b ∧ d, d ∈ D, of s is refined to a maximal one, and there are only finitely many maximal meet covers for s. Hence there exists a maximal meet cover e1 ∧ · · · ∧ ek that refines b ∧ d for all d ∈ D. For each i, if ei > 6 b, then ei > d for all d ∈ D, and hence ei ∈ U . Thus s > e1 ∧ · · · ∧ ek ∈ ↑(b ∧ U ), and we are done. ¨ As a corollary we get the result proven in [6]. COROLLARY 2.2. Every lattice that possesses the dual of minimal join cover refinement property is meet continuous. Proof. Indeed, if D has the least upper bound d then d is the least element in U . Then b ∧ d is the least element in b ∧ U , which implies that it is the least upper bound of b ∧ D. ¨ 2.4. There exists a noncomplete lattice that is meet continuous but not continuous. Let F L(ω) denote a free lattice generated by the set {xi : i ∈ ω}. Since any free lattice possesses the dual minimal join cover refinement property, by Corollary 2.2 we conclude Vol. 46, 2001 On continuous noncomplete lattices 223 that F L(ω) is meet continuous. On the other side, we show that a ≪ b never holds for elements a, b ∈ F L(ω). Indeed, let a, b ∈ F L(ω). Then a and b have canonical representations as terms in variables x1 , x2 , . . . . We may assume that for some n ∈ ω the variables xs , s > n, do not occur in these representations. Consider dm = xn ∨ xn+1 ∨ · · · ∨ xn+m , m ∈ ω. The elements b ∧ dm , m ∈ ω, form a strictly increasing chain with b as an upper bound. We show that there exists no upper bound c > 6 b of this chain. We argue by induction on the length of the canonical representation of c. 6 b imply c = xk > dm for all dm , which If c = xk for some k ∈ ω, then c > b ∧ dm and c > is impossible. Suppose, f 6> b implies that f is not an upper bound of {b ∧ dm : m ∈ ω} whenever the length of the canonical representation of f does not exceed t. Now, if the length of the canonical representation of c is equal to t + 1, then either c = f1 ∧ f2 or c = f1 ∨ f2 , where length of fi , i = 1, 2, does not exceed t. In the first case, if c > b ∧ dm for all m ∈ ω then f1 , f2 > b ∧ dm , which contradicts the inductive hypothesis. If f1 ∨ f2 > b ∧ dm for all m ∈ ω, then either fi > b ∧ dm for some i ∈ {1, 2} or c > dm . By the inductive hypothesis, the first case is possible only for finitely many m. Hence c is an upper bound for all xn+m , m ∈ ω. But F L(ω) does not contain upper bounds of infinitely many generators. Hence c is not an upper bound of {b ∧ dm : m ∈ ω}. W Thus b = m∈ω (b ∧ dm ). Now if a ∈ F L(ω) is a term in variables x1 , x2 , . . . , xn , then it is easy to show that a 6 6 dm . Hence a 6≪ b. In the contrast to this example we can notice that according to Theorem 3.1 any finitely generated sublattice of F L(ω) is algebraic. 3. Algebraic finitely presented lattices In this section we find several properties equivalent to the algebraicity in finitely presented lattices. The main result of this section, Theorem 3.1, combined with the Day’s result that finitely generated free lattices are weakly atomic [3] will justify example 2.2 given in the previous section. For a completely join irreducible element w of a finitely presented lattice we denote by κ(w) the set of elements v maximal with the following property: 6 v. w∗ 6 v, w 6 (2) By Theorem 12 of [7] the set κ(w) is finite for any completely join irreducible element w of a finitely presented lattice. Moreover, every element possessing the property (2) is contained in a maximal one. Evidently, every element v of κ(w) is completely meet irreducible, with unique upper cover w ∨ v. Following [7], we put L(w) = J (w)∨ ∪ 0, 224 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. where J (w) is the smallest closed set containing w, i.e., the smallest set S with the property that for any u ∈ S every join cover of u can be refined to a join cover of u that consists of elements of S. Theorem 9 from [7] states that the set J (w) exists for every element w of a finitely presented lattice L; moreover, if w is completely join irreducible then the map f : L → L(w) defined by W f (u) = {v ∈ J (w) : v 6 u} is a bounded homomorphism (cf. Theorem 14 in [7]). That homomorphism is called standard. A lattice is said to be spatial if every element is a join of completely join irreducible elements. THEOREM 3.1. For a finitely presented lattice L the following conditions are equivalent: 1) 2) 3) 4) 5) 6) L is algebraic; L is weakly atomic; L is spatial; L is approximated by its finite homomorphic images under bounded homomorphisms; L is dually algebraic; L is dually spatial. Proof. The implication 1) ⇒ 2) holds for any lattice and was proved in Proposition 1.7. Since any completely join irreducible element is compact in a meet-continuous lattice, the implication 3) ⇒ 1) holds as well. We show 2) ⇒ 3). Let a ∈ L, a 6= 0. It is enough to prove that for any a1 < a there 6 a1 . exists completely join-irreducible element v1 such that v1 6 a and v1 6 As L is weakly atomic, we have a1 6 u ≺ v 6 a for some u, v ∈ L. Since any finitely presented lattice has the minimal join cover refinement property (cf. [6, 7]), the set V of all elements of L belonging to some nonrefinable join representation of v is finite. As u ≺ v, 6 u. Let v1 be a minimal element in V with such property. Let there exists v1 ∈ V with v1 6 us show that v1 is completely join irreducible. We have v1 ∧ u < v1 . If v1 ∧ u < x < v1 for some x ∈ L, then x ∨ u = v. By the minimal join cover refinement property there exists a nonrefinable join representation v = t1 ∨ · · · ∨ tn , such that {t1 , . . . , tn } ⊑ {x, u}. 6 u for some i 6 n. It follows that ti 6 x < v1 , which contradicts Since u ≺ v, we have ti 6 the minimality of v1 . Therefore we have v1 ∧ u ≺ v1 . In fact v1 is join irreducible, as it belongs to a nonrefinable join cover of v. This together with the last inequality implies that v1 is completely join irreducible. Next, we show 3) ⇔ 4). Let a 6 6 b in L. Then we can find a completely join irreducible element x ∈ L, such that x 6 a and x 6 6 b. By Theorem 14 from [7] the lattice L(x) is a Vol. 46, 2001 On continuous noncomplete lattices 225 bounded homomorphic image of the lattice L, and the standard homomorphism f : L → L(x) separates a and b. Conversely, let the lattice L be approximated by its finite homomorphic images under bounded homomorphisms. For a ∈ L, set A = {x ∈ CJ (L) : x 6 a}. We may assume W a 6 = 0, since otherwise a = ∅. If a is not the least upper bound of A then there is an upper bound a1 of the set A with a1 < a. By the supposition there exists a bounded homomorphism f : L → K, where K is a finite lattice, such that f (a1 ) < f (a). Consequently, x 6 f (a) W and x 66 f (a1 ) for some join irreducible element x ∈ K. Suppose β(x) = X for some W X ⊆ L. Then x = f (β(x)) = f (X) by Lemma 1.8. Since x is join irreducible and the set f (X) is finite, we have x = f (x1 ) for some x1 ∈ X. It follows that x1 > β(x) > x1 and β(x) is completely join irreducible. Finally, we have β(x) 6 β(f (a)) 6 a, but β(x) 66 a1 , W which contradicts the choice of a1 . Consequently, a = A. The equivalences 2) ⇔ 5) ⇔ 6) can be proved by duality. The proof is complete. ¨ In particular, the last theorem can be applied to finitely generated free lattices. The well-known Day’s theorem [3] states that finitely generated free lattices are approximated by finite bounded lattices. Hence, by Theorem 3.1, they are weakly atomic and algebraic. This result is also true for finitely generated projective lattices, and, in particular, for the lattices freely generated by finite ordered sets. Notice, that the question whether finitely generated free lattices are weakly atomic was raised by R. McKenzie in his well-known paper [10]. He proved that this property of finitely generated free lattices is equivalent to the fact that bounded lattices generate the variety of all lattices. Thus Day’s result implied weak atomicity based on the considerations involving lattice varieties. In case of our theorem the weak atomicity follows from the Day’s result as an “inner” property of a finitely generated free lattice. QUESTION 3.2. Does there exist a proof of algebraicity of finitely generated free lattices that does not involve Day’s theorem? We do not know an example of finitely presented lattice that is continuous but not algebraic. As follows from Corollary 2.2, all finitely presented lattices are at least meet continuous. But not all of them are algebraic as an example in [7] shows: there exists a finitely presented lattice that has no covers at all. QUESTION 3.3. Is every finitely presented lattice continuous? Unlike algebraic lattices the weak atomicity does not seem to play a big role in the theory of continuous lattices. Any chain is a continuous lattice but, of course, it is not necessarily weakly atomic. 226 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. On the other side, there exist lattices that are meet continuous and weakly atomic but not continuous (see the example of Jeffrey Leon in [2]). Nevertheless, it would be interesting to learn more about the effect of the weak atomicity in continuous lattices. 4. Lawson topology on a finitely presented lattice In this section, we consider the Lawson topology on an algebraic finitely presented lattice F . Our aim is to show that it coincides with the interval topology on F , which implies that the interval topology on F is Hausdorff in that case. We note also that the same results (and their proofs) remain valid for finitely generated projective lattices (in particular, for finitely generated free lattices) since they are algebraic. We recall that the Lawson topology is defined as a join of the Scott topology and the lower topology, i.e., it has a subbase that consists of Scott-open sets and the sets F \ ↑f , f ∈ F . First we describe the Scott-open sets on F . We introduce a new topology J on F . Let C denote the set CJ (F )∨ = {w1 ∨ · · · ∨ wn : wi ∈ CJ (F ), n ∈ ω}. Let J be the topology with the base of open sets ↑c, c ∈ C. Since any element of F is a join of completely join irreducible elements (see Theorem 3.1), the order on F coincides with the order determined by this topology. With respect to this topology, F is an f -space, i.e., a particular case of an A-space, introduced in [4]. PROPOSITION 4.1. The topology J is the Scott topology on F . In particular, the sets ↑c, c ∈ C, form a base of the Scott topology on F . Proof. Evidently, all sets that are open in J are Scott-open because completely join irreducible elements are compact in an algebraic lattice. Consider a Scott-open set U ⊆ F . W Let CU = ↑(C ∩ U ). Notice that C ∩ U 6= ∅ whenever U 6= ∅. Indeed, 1F = C ∈ U , hence c ∈ U for some c ∈ C as U is Scott-open and C is up directed. Clearly, CU ⊆ U and W CU is open in J . Choose t ∈ U . Then TC = ↓t ∩ C is an up directed set and TC = t. Since U is Scott-open, there exists t ′ ∈ U ∩ TC ⊆ U ∩ C. Since t > t ′ , we have t ∈ CU . Thus U is open in J . ¨ Now we describe the Lawson topology on an algebraic finitely presented lattice F . PROPOSITION 4.2. The Lawson topology on F has a subbase of open sets ↑w and F \↑w, w ∈ CJ (F ). Proof. Let V be a basic Lawson-open set. Then V = U \(↑f1 ∪ · · · ∪ ↑fm ) for some S Scott-open set U and some elements f1 , . . . , fm ∈ F . By Proposition 4.1, U = i∈I ↑ci , S where {ci : i ∈ I } = U ∩ C. Also, for every k 6 m we have F \↑fk = j ∈Jk F \↑wkj , where {wkj : j ∈ Jk } is the set of all completely join irreducible elements below fk . Thus S V = {↑ci \(↑w1j1 ∪ · · · ∪ ↑wmjm ) : i ∈ I, j1 ∈ J1 , . . . , jm ∈ Jm }. Vol. 46, 2001 On continuous noncomplete lattices 227 It remains to notice that ↑ci is a finite intersection of sets ↑w, where w ∈ CJ (F ). Hence V is a union of finite intersections of sets of the form ↑w and F \↑w, where w ∈ CJ (F ). ¨ PROPOSITION 4.3. The Lawson topology on a continuous lattice is Hausdorff. Proof. Let a 6 6 b in L. Then there exists w ≪ a such that w 6 6 b. By Proposition 1.3, ↑w ↑ is a Scott-open set. Thus w is a Lawson-open neighborhood of a and L\↑w is a disjoint Lawson-open neighborhood of b. ¨ By Theorem 3.1, for any algebraic finitely presented lattice F its dual F ∂ is an algebraic finitely presented lattice, too. Hence the dual Lawson topology on F is described as the topology with the subbase of open sets ↓v and F \↓v, v ∈ CM(F ). Similarly to the complete case (cf. [8]), we call a lattice L linked bicontinuous if both L and its dual L∂ are continuous and the Lawson topology coincides with the dual Lawson topology. Now we state the main result of this section. THEOREM 4.4. The Lawson topology and the dual Lawson topology on an algebraic finitely presented lattice agree. In particular, algebraic finitely presented lattices are linked bicontinuous. Proof. First we prove one auxiliary statement that has an interest of its own. LEMMA 4.5. Let w be a completely join irreducible element of a finitely presented lattice F . Then there exist k ∈ ω and completely meet irreducible elements v0 , . . . , vk , such that ↑w ∩ ↓vi = ∅ for all i 6 k and F = ↑w ∪ ↓v0 ∪ · · · ∪ ↓vk . Proof. Let W = ↓w∗ ∩ S {S : S is a minimal join cover of w}. First show that the elements of W , if any, are completely join irreducible. Indeed, let W W w ′ ∈ W and w′ = A. If S is a minimal join cover of w that contains w′ then w 6 (S\ W {w ′ }) ∨ A. As w ∈ CJ (L) and L is meet-continuous, we conclude that w 6 s1 ∨ · · · ∨ sl ∨ a1 ∨ · · · ∨ am for some s1 , . . . , sl ∈ S \{w′ } and a1 , . . . , am ∈ A. Moreover, S ′ = {s1 , . . . , sl } ∪ {a1 , . . . , am } is a join cover of w and S ′ ⊑ S. By the minimality of S we have S ⊆ S ′ which implies w′ = aj for some j . Hence w′ is completely join irreducible. Put S V = {v0 , . . . , vk } = κ(w) ∪ {κ(wi ) : wi ∈ W }. Evidently, V consists of completely meet irreducible elements. Moreover, ↑w ∩ ↓vi = ∅ for all i 6 k. Indeed, if x ∈ ↑w ∩ ↓vi then vi 6∈ κ(w). If vi ∈ κ(w′ ) for some w′ ∈ W , then w ′ 6 w∗ < w 6 x 6 vi , a contradiction with the definition of κ(w′ ). 228 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. Now we show that L = ↑w ∪ ↓v0 ∪ · · · ∪ ↓vk . Let a > 6 w. If w 6 6 w∗ ∨ a, then a 6 w∗ ∨ a 6 vi , for some vi ∈ κ(w). Otherwise, {w∗ , a} is a join cover of w. Let M 0 be a minimal join cover of w that refines {w∗ , a}. Then S 0 = {s ∈ M 0 : s 6 6 a} is nonvoid and S 0 ⊆ W . Hence it consists of completely join irreducible elements. If s 6 a ∨ s∗ for every s ∈ S 0 , then the set S ′ = {a} ∪ S∗0 = {a} ∪ {s∗ : s ∈ S 0 } is a join cover of w. It can be refined to a minimal join cover M 1 with nonvoid subset S 1 = {s ∈ M 1 : s 6 6 a}. Notice that S 1 ⊆ W , S 1 ⊑ S∗0 and, consequently, S 1 ∩ S 0 = ∅. Arguing for S 1 instead of S0 we can either find vi ∈ V that contains a or find a nonvoid subset S 2 in W that satisfies S 2 ⊑ S∗1 = {s∗ : s ∈ S 1 } and S 2 ∩ (S 0 ∪ S 1 ) = ∅. Since F has the minimal join cover refinement property, W is finite. Hence for some 6 a ∨ s∗ will hold. That implies a to be contained in some t 6 ω and s ∈ S t the inequality s 6 ¨ vi ∈ V . We turn to the proof of Theorem 4.4. By Lemma 4.5, for any w ∈ CJ (F ) the set F \↑w is a union of the sets ↓vi , where vi ∈ CM(F ), i ∈ I , for some finite I . Hence T ↑w = i∈I (F \↓vi ) and both ↑w and F \↑w are open in the dual Lawson topology. By the statement dual to Lemma 4.5, any subbasic open set of the dual Lawson topology is open in the Lawson topology. Therefore, these topologies coincide on F . ¨ Recall that the interval topology I on a lattice L is defined by the subbase of open sets of the form L\↓a and L\↑a, a ∈ L. COROLLARY 4.6. The Lawson topology on an algebraic finitely presented lattice F is the interval topology. In particular, the interval topology on F is Hausdorff. Proof. Since the Lawson topology in finer than I, it is sufficient to show that any Lawson-open set is open in the interval topology. By Proposition 4.2 the Lawson topology on F is given by the subbase of open sets ↑w and F \↑w, w ∈ CJ (F ). Since T ↑w = i∈I (F \↓vi ) for some elements vi ∈ F and finite set I , all sets from the subbase of the Lawson topology are open in I. The last statement of the corollary follows from Proposition 4.3. ¨ The following property of finitely presented lattices is a consequence of our topological considerations. COROLLARY 4.7. For any a < b in an algebraic finitely presented lattice F , there exist S S k ∈ ω and f1 , . . . , fk ∈ Fab = F \(↓a ∪ ↑b) such that Fab = i 6 k (↑fi ) ∪ i 6 k (↓fi ). Proof. For any lattice F the statement of the corollary is equivalent to the fact that the interval topology on F is Hausdorff [1]. It remains to refer to Corollary 4.6. ¨ Vol. 46, 2001 On continuous noncomplete lattices 229 The interval topology on a lattice L is known to be compact if and only if L is complete [1]. By a compact topology we mean here a topology with Heine–Borel property. (In [8] such topology is called quasicompact.) Thus, it follows from Corollary 4.6 that algebraic finitely presented lattices are not compact in the Lawson topology. Actually, that assertion is a partial case of more general situation. Similarly to the interval topology, the Lawson topology possesses the following property. PROPOSITION 4.8. A lattice L is compact in Lawson topology if and only if L is complete. Proof. The “if” part is the content of Theorem III.1.9 in [8], so suppose that L is compact in Lawson topology. Let M ⊆ L. For any m ∈ M the set ↓m is Scott- and hence Lawsonclosed. Moreover, ↓m1 ∩ · · · ∩ ↓mk = ↓(m1 ∧ · · · ∧ mk ), whence is nonempty for any T m1 , . . . , mk ∈ M. Since L is compact, we can find x ∈ {↓m : m ∈ M}. It follows, x is a lower bound of M. Consider the set X = {x ∈ L : x 6 m for all m ∈ M}. X is nonvoid and for any x ∈ X the set ↑x is closed in the lower topology and hence is LawsonT T closed. Moreover, i 6 k ↓mi ∩ j 6 l ↑xj = ↓(m1 ∧ · · · ∧ mk ) ∩ ↑(x1 ∨ · · · ∨ xl ), for all m1 , . . . , mk ∈ M, x1 , . . . , ml ∈ X. The last set is obviously nonempty. Therefore, there T T exists y ∈ {↓m : m ∈ M} ∩ {↑x : x ∈ X}. Evidently, y is the greatest lower bound of M. Dual arguments are applied to prove that M has the least upper bound. ¨ Acknowlegments The authors are greatful to Prof. Yu. L. Ershov for many valuable comments to the manuscript, including his suggestions used in the current proof of Lemma 4.5. Also, we would like to thank an anonymous referee for the numerous corrections and helpful suggestions that improved the presentation of our results. The first version of this paper was written while the first two authors were visiting Technische Hochschule Darmstadt in the framework of the joint project on lattices and universal Horn logic. The attention to this work paid by our German colleagues is highly appreciated. REFERENCES [1] Birkhoff, G., Lattice Theory, Amer. Math. Soc., Providence, RI, 1967. [2] Crawley, P. and Dilworth, R. P., Algebraic Theory of Lattices, Prentice-Hall, Englewood Cliffs, NJ, 1973. [3] Day, A., Splitting lattices generate all lattices, Algebra Univers. 7 (1977) no. 2, 163–170. [4] Ershov, Yu. L., Computable functionals of finite type, Algebra and Logic 11 (1974) no. 4, 203–242 (translation from Algebra i Logika 11 (1972) no. 4, 367–437 (Russian)). [5] Ershov, Yu. L., The theory of A-spaces, Algebra and Logic 12 (1975) no. 4, 209–232 (translation from Algebra i Logika 12 (1973) no. 4, 369–416 (Russian)). [6] Freese, R., Ježek, J. and Nation, J. B., Free Lattices, Amer. Math. Soc., Providence, RI, 1995. 230 k. v. adaricheva, v. a. gorbunov and m. v. semenova algebra univers. [7] Freese, R., Finitely presented lattices: canonical forms and the covering relation, Trans. Amer. Math. Soc. 312 (1989) no. 1, 841–860. [8] Gierz, G., Hoffman, K. H., Keimel, K., Lawson, J. D., Mislove, M. and Scott, D. S., A Compendium of Continuous Lattices, Springer Verlag, Berlin — New York, 1980. [9] Gruber, P. M. and Wills, J. M. eds., Handbook of Convex Geometry, North-Holland, 1993. [10] McKenzie, R., Equational bases and nonmodular lattice varieties, Trans. Amer. Math. Soc. 174 (1972) no. 1, 1–43. K. V. Adaricheva 134 South Lombard ave Oak Park, lllinois 60302 USA e-mail: [email protected] M. V. Semenova Sobolev’s Institute of Mathematics Siberian Branch of Russian Academy of Science 4 Prospect of Academician Koptug 630090 Novosibirsk Russia e-mail: [email protected]