MacNeille transferability and stable classes of Heyting algebras

A lattice P is transferable for a class of lattices K\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {K}$$\end{document} if whenever P can be embedded into the ideal lattice IK\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathfrak {I}K$$\end{document} of some K∈K\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K\in \mathcal {K}$$\end{document}, then P can be embedded into K. There is a rich theory of transferability for lattices. Here we introduce the analogous notion of MacNeille transferability, replacing the ideal lattice IK\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathfrak {I}K$$\end{document} with the MacNeille completion K¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\overline{K}$$\end{document}. Basic properties of MacNeille transferability are developed. Particular attention is paid to MacNeille transferability in the class of Heyting algebras where it relates to stable classes of Heyting algebras, and hence to stable intermediate logics.


Introduction
A lattice P is transferable if whenever there is a lattice embedding ϕ : P → IK of P into the ideal lattice of a lattice K, then there is a lattice embedding ϕ : P → K. It is sharply transferable if ϕ can be chosen so that ϕ (x) ∈ ϕ(y) iff x ≤ y. If we restrict K to belong to some class of lattices K, we say P is transferable in K. The notion has a long history, beginning with [16,13], and remains a current area of research [25]. For a thorough account of the subject, see [17], but as a quick account, among the primary results in the area are the following. Theorem 1.1. (see [17, pp. 502-503], [25]) A finite lattice is transferable for the class of all lattices iff it is projective, and in this case it is sharply transferable.

Basic definitions and MacNeille transferability for lattices
For a lattice K, we use K for the MacNeille completion of K. This will be viewed either as the set of normal ideals of K, or abstractly as a complete lattice that contains K as a join and meet dense sublattice. We also consider lattices with one or both bounds as part of their basic type. For τ ⊆ {∧, ∨, 0, 1}, a τ -lattice is a lattice, or lattice with one or both bounds, whose basic operations are of type τ . A τ -homomorphism is a homomorphism with respect to this type, and a τ -embedding is a one-one τ -homomorphism. Definition 2.1. Let τ ⊆ {∧, ∨, 0, 1}, P be a τ -lattice, and K a class of τ -lattices. Then P is τ -MacNeille transferable for K if for any τ -embedding ϕ : P → K where K ∈ K, there is a τ -embedding ϕ : P → K. We say P is sharply τ -transferable for K if ϕ can be chosen so that ϕ (x) ≤ ϕ(y) iff x ≤ y for all x, y ∈ P . We use the terms transferable and sharply transferable when τ = {∧, ∨}.
There are obvious examples of lattices P that are MacNeille transferable for the class of all lattices. Any finite chain, and the 4-element Boolean lattice provide examples. Infinite chains can be problematic, as is seen by a simple cardinality argument for P the chain of real numbers and K the chain of rational numbers. Such difficulties arise also with the traditional study of transferability using ideal lattices. Here we restrict our attention to the case where P is finite. Our first result follows immediately from [18] where it was shown that any lattice can be embedded into the MacNeille completion of a distributive lattice.

Theorem 2.2. A lattice that is MacNeille transferable for the class of all lattices is distributive.
For a bounded lattice K, let K σ be the canonical completion of K [14]. For a lattice K, we let K + be the result of adding a bottom to K if it does not have one, and adding a top to K if it does not have one. Note that for a finite lattice P , there is a lattice embedding of P into K iff there is a lattice embedding of P into K + . Following the convention that ∅ is not an element of the ideal lattice, but can be an element of the MacNeille completion when viewed as normal ideals, it is easily seen that K + = K and I(K + ) = (IK) + .

Theorem 2.3. If a finite lattice P is MacNeille transferable for a class K of lattices that is closed under ultrapowers, then P is transferable for K.
Proof. Suppose that P is isomorphic to a sublattice of the ideal lattice IK of some K ∈ K. Then P is isomorphic to a sublattice of I(K + ). The ideal lattice I(K + ) is isomorphic to the sublattice of open elements of the canonical extension (K + ) σ [14,Lemma 3.3]. In [15] it was shown that the canonical extension (K + ) σ is isomorphic to a sublattice of the MacNeille completion (K + ) * of an ultrapower (K + ) * of K + . Clearly (K + ) * = (K * ) + . So P is isomorphic to a sublattice of (K * ) + = K * . Since K is closed under ultrapowers, K * ∈ K. Then as P is MacNeille transferable for K, there is a sublattice of K * that is isomorphic to P . Therefore, K * satisfies the diagram Δ P of the finite lattice P (see, e.g., [8, pp. 68-69]), hence K satisfies Δ P by Lós' Theorem. Thus, K has a sublattice that is isomorphic to P , and this shows that P is transferable.
Combining the above two results with Theorem 1.1, it follows that if a finite lattice is MacNeille transferable for the class of all lattices, then it is projective in the class of all lattices and distributive. Using the result of Kostinsky [21] that a finite lattice is projective in the class of all lattices iff it is a sublattice of a free lattice, and the result of Galvin and Jónsson [12] characterizing the finite sublattices of free lattices that are distributive as exactly those that do not contain a doubly reducible element (one that is both a join and meet), gives the following. Galvin and Jónsson [12] further characterize all (not just finite) distributive lattices that have no doubly reducible elements. This will be of use in later considerations for us as well. For the definition of linear sum that appears in the following result, see Definition 3.4.
Theorem 2.5. [12] A distributive lattice has no doubly reducible elements iff it is a linear sum of lattices each of which is isomorphic to an eight-element Boolean algebra, a one-element lattice, or 2 × C for a chain C. Problem 1. Is each finite distributive lattice with no doubly reducible elements MacNeille transferable for the class of all lattices?
If we move from the setting of MacNeille transferability with both lattice operations ∧, ∨ involved, to a setting where at most one lattice operation is involved, the situation opens considerably. The key result here relates to the projectivity of a semilattice in the class of all semilattices. A consequence of [20] gives that if P is a semilattice that is the ∧-reduct of a finite distributive lattice, then P is projective in semilattices. We will also need the following slight extension of a result of Baker and Hales [1]. Lemma 2.6. For a lattice K, there is a sublattice S of an ultrapower K * of K and an onto lattice homomorphism μ : S → IK. If K has a 0, then the 0 of K * belongs to S and is the only element of S mapped by μ to 0. Similar comments hold if K has a 1.
Proof. We follow [1]. Let X be the set of all finite subsets of K partially ordered by set inclusion. The principle upsets of X form a filter base. Let U be an ultrafilter extending this filter base. Let M be the set of order preserving maps from X to K. Then M is a sublattice of K X . Let S be the image of M in K * = K X /U. So the elements of S are equivalence classes σ/ U of order preserving functions σ : X → K. Define μ : S → IK by letting μ(σ/ U) be the ideal generated by the image of σ. In [1] it is shown that μ is a well defined onto lattice homomorphism.
If K has a 0, then the constant map from X to K taking value 0 belongs to K X , and the corresponding element of K * is its 0. Since μ is onto, it must preserve 0. If σ : X → K is order preserving and μ(σ/ U) = 0, then the ideal generated by the image of σ is the zero ideal {0}, and that implies that σ is the zero function. So the 0 of S is the only element mapped to 0 in K. Similarly, if K has a 1, then the 1 of K * is the equivalence class of the constant function 1, it belongs to S, and since μ is onto it preserves 1. Suppose μ(σ/ U) = 1. Then 1 is in the ideal of K generated by the image of σ, and since σ is order preserving and X is a lattice, 1 is in the image of σ. Then there is a finite subset A ⊆ K with σ(A) = 1. Since σ is order preserving, σ takes value 1 on the upset generated by A, hence σ is in the equivalence class of the constant function 1. Proof. Assume K ∈ K τ and that ϕ : P → K is a τ -embedding. By symmetry, we may assume that τ does not contain ∨. We consider first the case that τ does not contain ∧. If K is finite, then K = K and there is nothing to show. Otherwise, since P is finite, there is a τ -embedding of P into K. Assume that τ contains ∧. Viewing the MacNeille completion K as a set of normal ideals, the identical embedding ι : K → IK is a τ -embedding, hence so is the composite ι • ϕ : P → IK. Let S ≤ K * and μ : S → IK be as in the proof of the previous lemma. Since P is projective as a semilattice, there is a ∧-semilattice homomorphism γ : P → S with μ • γ = ι • ϕ. Since the composite is an embedding, so is γ. Since the composite preserves the bounds belonging to τ , the result of the previous lemma about the uniqueness of elements of S mapped to bounds of IK shows that γ preserves bounds in τ . Thus, S has a τ -subalgebra isomorphic to P , and hence so does K * . So K * satisfies the diagram of P in the τ -language, and by Lós' Theorem, K satisfies this diagram, hence has a τ -subalgebra isomorphic to P .
We conclude this section with comments related to adding bounds to the type. For a lattice K and k ∈ K, we can consider the principal ideals (↓k) K and (↓k) K . Using the abstract characterization of MacNeille completions (see, e.g., [2, p. 237]), it is easily seen that (↓k) K = (↓k) K . We say that a class K of lattices is closed under principal ideals if for each K ∈ K and each k ∈ K, the lattice (↓k) K belongs to K. Similarly, K is closed under principal filters if each (↑k) K belongs to K. Proof. We prove the result for τ ∪{1}, the result for τ ∪{0} is by symmetry, and the result for τ ∪ {0, 1} follows from these. Suppose K ∈ K and ϕ : P ⊕ 1 → K is a τ ∪ {1}-embedding. Let be the top of P and 1 be the top of P ⊕ 1. Then ϕ( ) = x for some x < 1 in K. Since the MacNeille completion is meet dense, there is k ∈ K with x ≤ k < 1. Then ϕ|P : P → (↓k) K = (↓k) K is a τ -embedding. Since (↓k) K ∈ K and P is τ -MacNeille transferable for K, there is a τ -embedding ψ : P → (↓k) K . Since k < 1, there is a (τ ∪ {1})-embedding of P ⊕ 1 into K.

MacNeille transferability for distributive lattices
One can consider MacNeille transferability for the class of distributive lattices. However, since distributive lattices are not closed under MacNeille completions, one would leave distributive lattices to do so. Instead, we consider Mac-Neille transferability for the class K of lattices whose MacNeille completions are distributive. This class K includes such interesting classes as (the lattice reducts of) Heyting algebras, co-Heyting algebras, and bi-Heyting algebras. Since MacNeille transferability for K holds vacuously for any non-distributive lattice, we consider only the case when a finite distributive lattice is Mac-Neille transferrable for K. In contrast to Theorem 1.2, which says that every finite distributive lattice is transferrable for all distributive lattices, we have the following.
There is a finite distributive lattice P that is not MacNeille transferable for the class K of lattices whose MacNeille completions are distributive.
Proof. Consider the lattices P , in Figure 1 at left, and K, in Before moving to some positive results, we note that a lattice being projective in the class of distributive lattices is not the same as it being distributive and projective in the class of all lattices. The finite lattices that are projective in the class of distributive lattices are characterized in [3] as exactly those where the meet of two join irreducible elements is join irreducible (and 0 is considered join irreducible). Proof. We begin with the result for MacNeille transferability. Let P be a finite distributive lattice that is projective in the class of distributive lattices. Let K be a distributive lattice whose MacNeille completion is distributive, and let ϕ : P → K be a lattice embedding. By Theorem 2.7, there is a ∧-embedding χ : P → K. Since P is projective in distributive lattices, the set of join irreducibles J in P forms a ∧-sub semilattice of P , and χ|J :

Theorem 3.2. Every finite lattice that is projective in the class of distributive lattices is
, and as χ is an embedding a ≤ y. Since x is the join of the members of J beneath it, x ≤ y. So χ : P → K is a lattice embedding, showing P is MacNeille transferable.
Showing that P is {∧, ∨, 0}-MacNeille transferable for the class of lattices with a 0 whose MacNeille completion is distributive is nearly identical. One uses Theorem 2.7 to obtain a {∧, 0}-embedding χ : P → K, and notes that 0 ∈ J. Since the lattice embedding χ : P → K produced agrees with χ on J, it follows that χ preserves 0. The result for {∧, ∨, 1}-MacNeille transferability follows by symmetry. D Figure 2. The lattice D Remark 3.3. The above theorem can be obtained using the transferability of finite distributive lattices in the class of all distributive lattices rather than using Theorem 2.7.
We next provide notation and detail for the linear sums in Theorem 2.5 of Galvin and Jónsson [12]. This theorem will be used to show that there are finite non-projective distributive lattices that are MacNeille transferable for distributive lattices.

Definition 3.4. Let
A be a chain, and for each a ∈ A let (K a , ≤ a ) be a lattice. Then the linear sum A K a is the disjoint union of the K a with the ordering ≤ given by setting x ≤ y iff x ∈ K a and y ∈ K b for some a < b, or x, y ∈ K a for some a and x ≤ a y. Proof. Suppose that K is a distributive lattice that does not contain D as a sublattice. We must show that K does not contain D as a sublattice. By Theorem 2.5, K is a linear sum of lattices A K a with each K a isomorphic to either an 8-element Boolean algebra, a 1-element lattice, or 2 × C for a chain C. We describe the MacNeille completion of K.
Then it is not difficult to see that M S = Mz (S ∩ M z ) with it understood that the join of the emptyset in M z is 0 z . Similar remarks hold for M S. Thus, M is complete and clearly K is a sublattice of M .
We consider the matter of K being join dense in M . Suppose p = 0 z where z ∈ A is the join of the elements of A strictly beneath it. This includes the case when p ∈ M z for any z ∈ A\A since then M z is a singleton lattice.
If ↓p ∩ K p = ∅, then it must be that p = 0 z and 0 z ∈ K z . The first case still covers this situation if z is the join of the elements in A strictly beneath it. Otherwise z ∈ A covers an element y ∈ A, p = 0 z , and K z has no least element. In this case, p is not the join in M of the elements of K beneath it. Similar remarks apply to meet density. Define a special covering pair in M to be an ordered pair formed from elements 1 x , 0 y where x, y ∈ A with x covered by y and either 1 x ∈ K x or 0 y ∈ K y . This implies that 1 x is join irreducible in M , or 0 y is meet irreducible in M , or both. Let θ be the set of all special covering pairs union the diagonal relation on M . It is easy to see that special covering pairs cannot overlap, and hence θ is a congruence relation. The quotient M/θ is the MacNeille completion K.
It remains to show that K = M/θ has no doubly reducible elements, and hence does not have a sublattice that is isomorphic to D. Along the way, we will show that M/θ is distributive. Let z ∈ A. Then M z is either a 1-element lattice, the MacNeille completion of an 8-element Boolean algebra, which is an 8element Boolean algebra, or 2 × C for some chain C. The MacNeille completion 2 × C depends on whether C is bounded. If it is, then 2 × C = 2 × C. If C has a top, but no bottom, then 2 × C = (2 × C)\{(1, 0 C )}, and so forth. But in any case 2 × C is a sublattice of 2 × C. Therefore, each summand M z is distributive and has no doubly reducible elements. Thus, M is distributive and has no doubly reducible elements. In forming the quotient M/θ we only collapse covering pairs where either the lower half is not a proper join, or the upper half is not a proper meet, and hence introduce no doubly reducible elements into the quotient.

Problem 2.
Classify the finite distributive lattices that are MacNeille transferable for the class of lattices whose MacNeille completions are distributive.

MacNeille transferability for Heyting algebras
In this section we consider τ -MacNeille transferability for the class H of Heyting algebras. Here we treat members of H as bounded lattices and note that the notion of MacNeille transferability does not involve the Heyting implication. Since the MacNeille completion of a Heyting algebra is distributive, and in fact is a Heyting algebra, see [2, p. 238] or [19,Theorem 2.3], results of the previous section give the following. We turn attention to {∧, ∨, 0, 1}-MacNeille transferability for H. Using Theorem 3.2, Proposition 2.9, and the fact that for a Heyting algebra K each principle ideal (↓k) K and filter (↑k) K are Heyting algebras, gives the following. We consider a closely related result, but one that to the best of our knowledge requires a completely different approach. Proof. Let K be a Heyting algebra and suppose there is a bounded sublattice of K isomorphic to D⊕1. Then there are normal ideals P, Q, R, S of K situated as in Figure 3. We will show that K has a bounded sublattice isomorphic to D ⊕ 1.
If there are 0 < p 1 < p ∈ P and 0 < q 1 < q ∈ Q, then since P ∧ Q = 0 implies that p ∧ q = 0, it follows from applications of the distributive law that {0, p 1 , q 1 , p 1 ∨ q 1 , p ∨ q 1 , q ∨ p 1 , p ∨ q, 1} is a bounded sublattice of K isomorphic to D ⊕ 1. So we may assume that one of P, Q is an atom of K and hence an atom of K. We assume P = ↓p where p is an atom of K.
Choose r ∈ R\S and s ∈ S\R with p ≤ r, s. Since K is a Heyting subalgebra of K (see, e.g., [19,Section 2] To make further progress with positive results, we consider more restrictive classes of Heyting algebras. We assume the reader is familiar with Esakia duality for Heyting algebras [11,4]. In particular, for the Esakia space X of a Heyting algebra K, the maximum of X is the set of maximal elements of the poset X. maximal cardinality of an antichain in X and that K and X have top width n if n is the cardinality of the maximum of X. Let H w be the class of Heyting algebras of finite width, and H t the class of Heyting algebras of finite top width. Let K be a Heyting algebra. We call S ⊆ K orthogonal if x ∧ y = 0 for any distinct x, y ∈ S. An element x of K is regular if x = x * * where x * is the pseudocomplement of x. It is well known (see, e.g., [24, Section IV.6]) that the regular elements form a Boolean algebra B that is a {∧, 0}-subalgebra of K. So an orthogonal set in B is an orthogonal set in K. The following lemmas are easily proved.  (1) K has top width at most n.
(3) The maximal cardinality of an orthogonal set in K is at most n.
If a finite distributive lattice P is MacNeille transferable for H, then by definition it is MacNeille transferable for H w . The converse holds as well, since P is a sublattice of a finite Boolean algebra, hence of any Heyting algebra that is not of finite width. Similar reasoning shows that P ⊕ 1 is MacNeille transferable for H t iff P ⊕ 1 is MacNeille transferable for H. However, the notion of {∧, ∨, 0, 1}-MacNeille transferability for H w and H t differs from that of H. We begin with the following.

Lemma 4.7. If K is a Heyting algebra of finite top width, then the complemented elements of the MacNeille completion K belong to K.
Proof. Let x be a complemented element of K. For any z ∈ K we have z = (z ∧ x) ∨ (z ∧ x * ). Since K is a Heyting subalgebra of K, for a ∈ K, we have that a * is the pseudocomplement of a in both K and K. Suppose a ≤ x. Then a ≤ a * * ≤ x * * = x. So the normal ideal N = {a ∈ K : a ≤ x} is generated by regular elements of K, and for regular elements a, b ∈ N we have (a∨b) * * ∈ N . Since K has finite top width, by Lemma 4.6, there are finitely many regular elements of K. Therefore, there is a largest regular element in N , so N is a principal ideal of K. Since K is join dense in K, we have x = N , hence x ∈ K.   ↓(1, 0) and ↓(0, 1) are isomorphic to P and Q respectively, while (↓x) K and (↓y) K are isomorphic to (↓x) K and (↓y) K respectively. Since P and Q are {∧, ∨, 0, 1}-MacNeille transferable for H t and the Heyting algebras (↓x) K and (↓y) K belong to H t , there are bounded sublattices of (↓x) K and (↓y) K isomorphic to P and Q respectively. Since K is isomorphic to (↓x) K × (↓y) K , K has a bounded sublattice isomorphic to P × Q. The argument for H w is identical, using that H w ⊆ H t .
As we will see below, the class of finite lattices that are {∧, ∨, 0, 1}-MacNeille transferable for H is not closed under binary products. In fact, Theorem 4.12 shows that the product of any two non-trivial finite distributive lattices is not {∧, ∨, 0, 1}-MacNeille transferable for H. Thus, the results of Theorem 4.9 fail if we replace H t or H w by H. To establish this we require a preliminary definition and lemma.  . . . . . . Figure 6. The Esakia space X Let T be the union of the T i . By [7,Lemma 16], for every finite connected poset F there is a T as above such that F is a p-morphic image of T . Since p-morphisms are order preserving maps, without loss of generality we may assume that F is T = k i=1 T i . We then define recursively Let n = n k and define a map h : S n → F as follows. Let h map y 1 , . . . , y n1−1 bijectively onto T 1 \{r 1 , t r 1 }. The exact nature of this bijection is irrelevant . Let h map x 1 , . . . , x n1 to r 1 and y n1 to t r . , x ni to r i , and map y ni to t r i . Finally, let h map y n k−1 +1 , . . . , y n bijectively onto T k \{r k , t l k } and map x n k−1 +1 , . . . , x n to r k . Then h is the desired map. Proof. Let X be the Esakia space whose domain is {x i , y i , w i , z i | i ≥ 1} ∪ {∞} and that is topologized by the one-point compactification of the discrete topology on {x i , y i , w i , z i | i ≥ 1} with compactification point ∞ and whose ordering is as shown in Figure 6. Let K be the Heyting algebra of clopen upsets of X. Since ∅, X are the only clopen upsets of X that are also downsets, the only complemented elements of K are 0, 1.
For an Esakia space Y , we let Y be the Esakia space of the MacNeille completion of the Heyting algebra corresponding to Y . Since K is in fact a bi-Heyting algebra, by [19,Theorem 3.8], the elements of the MacNeille completion of K are the regular open upsets of X. Therefore, Consequently, X is the disjoint union of two Esakia spaces X 1 and X 2 , each of which carries the topology of the one-point compactification of the discrete topology on {x i , y i | i ≥ 1} and {w i , z i | i ≥ 1}, respectively; see Figure 7.
To show that P is not {∧, ∨, 0, 1}-MacNeille transferable for H, by Esakia duality, it is sufficient to construct an Esakia space Y such that the Esakia dual E of P is a continuous order preserving image of Y but not of Y . Since P  Figure 7. The Esakia space X has a complemented element a = 0, 1, the finite poset E is disconnected. Let E = F 1 ∪ · · · ∪ F m be the decomposition of E into connected components. First suppose that m = 2, so E = F 1 ∪ F 2 . If there were a continuous order preserving onto map f : X → E, then the inverse image of the upset of E corresponding to a would be a complemented element of K different from 0, 1. Since this is a contradiction, there is no continuous order preserving map from X onto E.
We show that there is a continuous order preserving map from X onto E. By Lemma 4.11, there are n 1 and n 2 and order preserving onto maps h 1 : S n1 → F 1 and h 2 : S n2 → F 2 . We have that X is the disjoint union of X 1 and X 2 and we can regard S n1 as a subposet of X 1 and S n2 as a subposet of X 2 in the obvious way. Let m 1 be a minimal element of F 1 that is below h 1 (y n1 ) and let m 2 be a minimal element of F 2 that is below h 2 (y n2 ). Define Then h is the desired continuous order preserving onto map. If m > 2, then define Y to be the disjoint union of X and F 3 , . . . , F m . Then Y is the disjoint union of X and F 3 , . . . , F m . Since E has m components and Y has m − 1 components, there is no continuous order preserving map from Y onto E. But an obvious modification of the above argument provides a continuous order preserving map from Y onto E.
To conclude this section, we give an example to show that even in the setting of Heyting algebras, sharp MacNeille transferability is not the norm.  Figure 2 is not sharply MacNeille transferable for H.
It is routine to verify that K and L are Heyting algebras. The MacNeille completion L of L inserts the "missing" line {(1, y) ∈ R 2 | 0 ≤ y ≤ 1}. We can visualize L and L as follows in Figure 8. The black circles in the picture of L define a lattice embedding ϕ from D into L. Suppose there is a lattice embedding ϕ : D → L that is sharp, meaning that ϕ (x) ≤ ϕ(y) iff x ≤ y. Recall that D = 2 2 ⊕2 2 is built from two 4-element Boolean algebras. Since ϕ is sharp, ϕ and ϕ must agree on the upper copy of 2 2 comprising D. So ϕ maps the doubly reducible element of D to (2, 1). Since ϕ maps the bottom element of D to (1, 0), ϕ must map the bottom element of D to some with a < 1. But there do not exist elements in L whose meet is (a, 0) where a < 1 and whose join is (2,1).
Among many open problems in this area, the following seems to us the most pressing. We expect the answer is negative, but that a counterexample is quite involved.

Bi-Heyting algebras
Restricting attention to the class of bi-Heyting algebras, stronger results of a positive nature are possible. Recall that a Heyting algebra A is a bi-Heyting algebra if the order dual of A is also a Heyting algebra. Note that the property of being a bi-Heyting algebra is preserved under MacNeille completions (see, e.g., [19]). Proof. Suppose the width of A is n. Then the width of X is n, and by [10, p. 3], X can be covered by n maximal chains C 1 , . . . , C n . By [11,Lemma III.2.8], each of these maximal chains is closed in X. To show that U ∪ V is regular open, it is enough to show that IC(U ∪ V ) ⊆ U ∪ V where I and C are the interior and closure operators.
Suppose (A) holds. Then since D := {C i : i ∈ T } is closed and x ∈ D, there is clopen K with x ∈ K ⊆ K ⊆ C(U ) ∪ C(V ) and K disjoint from D. Therefore, K ⊆ (K ∩ C(U )) ∪ C(V ). Since C 1 ∪ · · · ∪ C n covers X and K is disjoint from D, and hence disjoint from each Suppose that (A) does not hold. Then there is i with x ∈ C i and U ∩C i ⊆ C(V ). Let y ∈ U ∩ C i with y ∈ C(V ). Note that if x ∈ U , then there is nothing in showing that x ∈ U ∪ V , and if x ∈ U , then since x, y belong to the chain C i and U is an upset, we must have x < y. Since y ∈ C(V ), there is clopen G with y ∈ G and G disjoint from V . So G is disjoint from C(V ). Since X is an Esakia space and V is an upset, C(V ) is an upset. Therefore, G disjoint from C(V ) implies that the clopen set ↓G is disjoint from C(V ). Thus,

Proposition 5.2. If A is a bi-Heyting algebra of finite width, then A is a bounded sublattice of IA.
Proof. Let X be the Esakia space of A. It is well known that IA is isomorphic to the open upsets of X, and since A is a bi-Heyting algebra, by [19, As a consequence, we obtain the following. Proof. Let P be a finite distributive lattice. If A is a bi-Heyting algebra of finite width and ϕ : P → A is a lattice embedding, then since A is a sublattice of IA, we may consider ϕ : P → IA. Then sharp MacNeille transferability follows from Theorem 1.2. That P is MacNeille transferable for the class of all bi-Heyting algebras follows since P is a sublattice of a finite Boolean algebra, and hence of any bi-Heyting algebra that is not of finite width.

Stable universal classes and stable intermediate logics
In this section we show how to produce stable universal classes of Heyting algebras [5,6]  It is known [5,6] that a universal class K of Heyting algebras is stable iff there is a set P of finite distributive lattices such that K consists of those Heyting algebras K such that no member of P is isomorphic to a bounded sublattice of K. It is convenient to introduce the following notation H(P) = {K ∈ H : K has no bounded sublattice isomorphic to any P ∈ P}.
The definition of τ -MacNeille transferability for τ = {∧, ∨, 0, 1} gives the following.  Remark 6.4. Theorem 6.3(1) is based on Theorem 2.7, which says that the class of lattices that do not contain a {∧, 0, 1}-subalgebra isomorphic to P is closed under MacNeille completions. Using syntactic methods, an analogous result for residuated lattices was given in [9]. These results overlap in the setting of Heyting algebras (see [22]), and can be used to provide examples of stable classes that are closed under MacNeille completions based on omitting finite projective distributive lattices. However, since the lattices D ⊕ 1 and 0 ⊕ D ⊕ 1 are not projective in distributive lattices, the results of Theorem 6.3(2) do not fall in the scope of the results of [9].
There is a related method to create stable classes of Heyting algebras that are closed under MacNeille completions. Note that for a finite lattice P and a bounded lattice K, we have P is isomorphic to a sublattice of K iff at least one of P, 0⊕P, P ⊕1, 0⊕P ⊕1 is a bounded sublattice of K. The following result for a finite lattice P that is MacNeille transferable for H is immediate, and has obvious generalization to a set of such MacNeille transferable P . Proof. Suppose K ∈ H(P ). Then K has no bounded sublattice isomorphic to 2 n ⊕ 1, so by Lemma 4.6, K has finite top width, hence belongs to H t . Since every finite Boolean algebra is projective in distributive lattices [3], Theorem 6.3 provides that 2 n ⊕ 1 is {∧, ∨, 0, 1}-MacNeille transferable for H. Thus, K has no bounded sublattice isomorphic to 2 n ⊕ 1. Also, since each P ∈ P is {∧, ∨, 0, 1}-MacNeille transferable for H t and K ∈ H t , we have that K contains no bounded sublattice isomorphic to any P ∈ P. So K ∈ H(P ). The result for H w is similar, using that each 2 n is MacNeille transferable for H, Lemma 4.5, and Theorem 6.5.

Remark 6.7.
Our primary examples of finite distributive lattices that are {∧, ∨, 0, 1}-MacNeille transferable for H are those of the shape 0 ⊕ P , P ⊕ 1, and 0 ⊕ P ⊕ 1 for some finite projective distributive lattice P , and the lattices 0 ⊕ D ⊕ 1 and D ⊕ 1. We can make stable classes that are closed under MacNeille completions directly from these lattices as in Theorems 6.3 and 6.5.
Since {∧, ∨, 0, 1}-MacNeille transferability for H implies that for H t and H w , by Theorem 4.9, we can take finite products of these lattices, and then use Theorem 6.6 to produce stable classes that are closed under MacNeille completions. This provides a range of examples of stable universal classes that are closed under MacNeille completions that are not within the scope of [9].
Finally, we discuss the impact of our results on the stable intermediate logics from [5,6]. Recall that an intermediate logic is stable if its corresponding variety is generated by a stable universal class of Heyting algebras. Thus, the above results produce stable intermediate logics generated by stable universal classes that are closed under MacNeille completions. Due to [15]   The following provides an infinite family of stable intermediate logics each of which is determined by a stable universal class closed under MacNeille completions. None of these logics is covered by the results of [9]. algebra K validates L n iff none of the algebras D ⊕ 1, 2 n , 0 ⊕ 2 n , 2 n ⊕ 1, or 0 ⊕ 2 n ⊕ 1 is isomorphic to a bounded sublattice of K. If m > n, then 0 ⊕ 2 n ⊕ 1 validates L m but refutes L n , so L m = L n .
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