On poset Boolean algebras

Uri Abraham

Department of Mathematics, Ben Gurion University, Beer-Sheva, Israel

Robert Bonnet

Laboratoire de Mathématiques, Université de Savoie, Le Bourget-du-Lac, France

Wiesław Kubiś

Department of Mathematics, Ben Gurion University, Beer-Sheva, Israel

Matatyahu Rubin

Department of Mathematics, Ben Gurion University, Beer-Sheva, Israel

Abstract

Let ⟨P, ≤⟩ be a partially ordered set. The poset Boolean algebra of P, denoted F (P), is defined as follows: The set of generators of F (P) is {x_{p}: p ∈ P }, and the set of relations is {x_{p} · x_{q} = x_{p}: p ≤ q}. We say that a Boolean algebra B is well-generated, if B has a sublattice G such that G generates B and ⟨G, ≤^{B} ↾G⟩ is well-founded. A well-generated algebra is superatomic. Theorem 1. Let ⟨P, ≤⟩ be a partially ordered set. The following are equivalent. (i) P does not contain an infinite set of pairwise incomparable elements, and P does not contain a subset isomorphic to the chain of rational numbers. (ii) F (P) is superatomic. (iii) F (P) is well-generated.

The equivalence (i) ⇔ (ii) is due to M. Pouzet [P]. A partially ordered set W is well-ordered, if W does not contain a strictly decreasing infinite sequence, and W does not contain an infinite set of pairwise incomparable elements. Theorem 2. Let F (P) be a superatomic poset algebra. Then there are a well-ordered set W and a subalgebra B of F (W), such that F (P) is a homomorphic image of B.

This is similar but weaker than the fact that every interval algebra of a scattered chain is embeddable in an ordinal algebra ([AB1]). Remember that an interval algebra is a special case of a poset algebra.

Keywords: poset algebras, superatomic Boolean algebras, scattered posets, well quasi orderings. Mathematics Subject Classification 2000 (MSC2000) Primary: 03G05, 06A06. Secondary: 06E05, 08A05, 54G12.

1 Introduction

In this paper we investigate poset Boolean algebras and especially superatomic poset Boolean algebras. The construction of a Boolean algebra from a partially ordered set (poset) is natural. Yet, not much is known about the relationship between the properties of the poset and the properties of the Boolean algebra constructed from it. Two of the three theorems proved in this work are of this kind.

A poset ⟨P, ≤^{P} ⟩ is denoted for simplicity by P, and when the context is clear we write ≤ rather than ≤^{P}.

For Boolean algebras we use the notations of [Ko]. Thus +, ·, − and ≤ denote the join, meet, complementation and partial ordering of a Boolean algebra B. The zero and one of B are denoted by 0 and 1.

Poset algebras are defined as follows. Let ⟨P, ≤⟩ be a partially ordered set. The poset Boolean algebra of ⟨P, ≤⟩, denoted by F (⟨P, ≤⟩), is defined by specifying a set of generators together with a set of relations on them. The set of generators for F (⟨P, ≤⟩) is {x_{p}: p ∈ P }. The set of relations is {x_{p} · x_{q} = x_{p}: p, q ∈ P and p ≤ q} union with the set of all identities of the variety of Boolean algebras. We shall need a more formal definition of F (⟨P, ≤⟩).

Definition 1.1. Let Y = {y_{p}: p ∈ P } be a set of distinct variables and T (Y) be the set of all terms in the language {+, ·, −, 0, 1} with variables from Y. Let

R_{P} = {y_{p} · y_{q} = y_{p}: p, q ∈ P and p ≤ q}.

Define an equivalence relation ∼ on T (Y): t ∼ u if u can be obtained from t by a sequence of substitutions using the identities of the variety of Boolean algebras and the relations of R_{P}. For t ∈ T (Y) denote [t] = t/ ∼ and let T [Y] = {[t] : t ∈ T (Y)}. Define [t] + [u] = [t + u], [t] · [u] = [t · u] etc..

It is well-known from universal algebra that such a construction yields a Boolean algebra. The resulting Boolean algebra ⟨T [Y], +, ·, −, 0, 1⟩ is called the poset algebra of P. We denote the set of equivalence classes T [Y] by F (P) and [y_{p}] by x_{p}.

We shall use the following well-known property of T [Y] which is a special case of a more general theorem in universal algebra.

Theorem 1.2. Let B be a Boolean algebra and g : Y → B. Suppose that {g(y) : y ∈ Y } satisfies the relations of R_{P}. That is, for every p ≤ q in P, g(y_{p}) · g(y_{q}) = g(y_{p}). Then there is a homomorphism g̃ : T [Y] → B such that for every y ∈ Y, g̃([y]) = g(y).

There is an alternative description of F (P) using topology. A subset R of P is a final segment if for every p ∈ R and q ∈ P: if p ≤ q then q ∈ R. Let Fs (P) be the set of final segments of P. Viewing a final segment as its characteristic function, it is obvious that Fs (P) is a closed subspace of {0, 1}^{P}. Let F̂(P) be the Boolean algebra of clopen subsets of Fs (P). Consider the map x_{p} ↦ {R ∈ Fs (P) : p ∈ R}. In Theorem 2.3 it will be shown that this map extends to an isomorphism between F (P) and F̂(P). Thus we have a description of F (P) as an algebra of subsets of Fs (P).

Poset algebras can be viewed in yet another way. A topological lattice is a structure of the form ⟨X, τ, ∨, ∧⟩ where ⟨X, τ⟩ is a topological space, ⟨X, ∨, ∧⟩ is a lattice and ∨ and ∧ are continuous. In Fs (P) define x∨y = x∪y and x ∧ y = x ∩ y. This makes Fs (P) into a topological lattice which we denote by L(P). Obviously, L(P) is a distributive lattice.

Our first main theorem (Theorem 2.6) is that every Hausdorff compact 0-dimensional topological distributive lattice is isomorphic to L(P) for some poset P.

We next define the notions used in the statement of the main theorems of Sections 3 and 4.

A poset P is well-founded, if P has no strictly decreasing infinite sequence. Let B be a Boolean algebra. A member a ∈ B is called an atom of B, if a

is a minimal element of B−{0}. A Boolean algebra B is superatomic, if every subalgebra of B has an atom. A Boolean algebra B is called a well-generated algebra, if B has a sublattice G (that is, a subset closed under + and ·) such that:

(1) G generates B;

(2) ⟨G, ≤^{B}↾G⟩ is well-founded.

Every well-generated algebra is superatomic. This is proved in [BR] Proposition 2.7(b). However, the proof is easy, and can be found by the reader.

We now turn to partial orderings. A subset of a poset P consisting of pairwise incomparable elements is called an antichain of P.

A poset P is narrow, if P does not contain infinite antichains.

A poset P is scattered, if P does not contain a subset isomorphic to the chain of rational numbers.

We can now state the second main theorem. It will be proved in Section 3.

Theorem 1.3. Let P be a partially ordered set. Then the following properties are equivalent.

(1) P is narrow and scattered.

(2) F (P) is superatomic.

(3) F (P) is well-generated.

The implication (1) ⇒ (3) is the difficult part of Theorem 1.3. M. Pouzet [P] proved the implication (1) ⇒ (2) (see R. Fraïssé [F] Section 6.7). But now this fact follows easily from the implication (1) ⇒ (3).

The interval algebra of a linearly ordered set ⟨L, <⟩ is the subalgebra of P(L) generated by the set {[a, ∞) : a ∈ L} of left closed rays of L. This algebra is denoted by B(L). The interval algebra of L is isomorphic to the poset algebra of L when L has no minimum, and to the poset algebra of L minus its minimum when L has a minimum. Thus interval algebras are a special case of poset algebras.

In [AB1] Theorem 3.4 it is proved that an interval algebra of a scattered linear ordering is embeddable in the interval algebra of a well-ordering. This result motivates our third main theorem.

A partially ordered set is well-ordered, if it is narrow and well-founded.

Theorem 1.4. Let P be a narrow scattered partially ordered set. Then there is a well-ordered poset W and a subalgebra B of F (W) such that F (P) is a homomorphic image of B.

Theorems 1.3 and 1.4 use a structure theorem for narrow scattered partially ordered sets, which is similar to the theorem of Hausdorff on the structure of scattered linear orderings. This theorem is due to Abraham. It is proved in [AB2] Theorem 3.4. We next quote this theorem.

Suppose that P is a poset and for every a ∈ P, Q_{a} is a poset. We define the lexicographic sum of the indexed family {Q_{a}: a ∈ P }, and denote it by ∑{Q_{a}: a ∈ P }. The universe of ∑{Q_{a}: a ∈ P } is ⋃{{a} × Q_{a}: a ∈ P }. The partial ordering on ∑{Q_{a}: a ∈ P } is defined as follows: ⟨a, q⟩ ≤ ⟨b, r⟩, if either (i) a <^{P} b or (ii) a = b and q ≤^{Q_{a}} r.

We say that a poset ⟨P, ≤⟩ is anti well-ordered, if ⟨P, ≥⟩ is well-ordered. Let W denote the class of posets which are either well-ordered or anti wellordered.

Suppose that ≤^{1} and ≤^{2} are partial orderings of a set P. We say that ⟨P, ≤^{2}⟩ is an augmentation of ⟨P, ≤^{1}⟩ iff ≤^{1} ⊆ ≤^{2} .

Theorem 1.5. ([AB2] Theorem 3.4) Let H_{0} be the class of all posets whose universe is a singleton. Let H be the closure of H_{0} under lexicographic sums indexed by members of W, and under isomorphism. Then for every poset P the following are equivalent.

(1) P is narrow and scattered.

(2) P is an augmentation of a member of H.

Theorem 1.4 follows from Theorem 1.5 and from the following theorem which is of independent interest.

Theorem 1.6. Let P ∈ H. Then there is a well-ordered poset W such that F (P) is embeddable in F (W).

Theorems 1.6 and 1.4 will be proved in Section 4.

We conjecture that Theorem 1.4 cannot be strengthened by requiring that for some well-ordered poset W, F (P) is embeddable in F (W). We thus ask the following question.

Question 1.7. Is there a narrow scattered poset P such that F (P) cannot be embedded in F (W) whenever W is a well-ordered poset?

In [AB1] Theorem 1, it is proved that a superatomic subalgebra of an interval algebra is embeddable in an interval algebra of a scattered linear ordering. We ask whether the analogous fact for partially ordered sets is also true.

Question 1.8. Let B be a superatomic subalgebra of a poset algebra. Does there exist a narrow scattered poset P such that B is embeddable in F (P) ?

2 Topological descriptions of poset algebras

In this section we show that the poset algebra of P is the algebra of clopen sets of the space of final segments of P. We also characterize poset algebras as the algebras of clopen sets of Hausdorff compact 0-dimensional topological distributive lattices.

Referring to Definition 1.1, let P be a poset, Y = {y_{p}: p ∈ P }, B be a Boolean algebra and f : Y → B. We say that f respects the relations of R_{P} if f(y_{p}) · f(y_{q}) = f(y_{p}) whenever p ≤ q. Note that this is equivalent to the fact that the function p ↦ f(y_{p}) is order preserving.

The property of F (P) mentioned in Theorem 1.2 characterizes F (P) up to an isomorphism. This is noted in the following theorem.

Theorem 2.1. Let P be a poset, B be a Boolean algebra and g : Y → B. Suppose that g and B satisfy the following three properties:

(1) g respects the relations of R_{P}.

(2) The image of g generates B.

(3) For every Boolean algebra C and a function h from Y to C: if h respects R_{P}, then there is a homomorphism α : B → C such that α ∘ g = h.

Then F (P) ≅ B and there is such an isomorphism ψ satisfying ψ(x_{p}) = g(y_{p}), for every p ∈ P.

Proof. It is trivial that Clauses (1) - (3) characterize B up to an isomorphism. And by Theorem 1.2, F (P) fulfills these clauses.

Let P(A) denote the powerset of A. P(A) is equipped with its Cantor topology defined as follows. Let σ, τ be finite subsets of A. Denote U_{σ,τ} = {x ∈ P(A) : σ ⊆ x and x ∩ τ = ∅}. The set of all U_{σ,τ}’s is a base for the topology of P(A).

Definition 2.2. A subset R of a poset P is a final segment of P if for every p ∈ R and q ∈ P: if p ≤ q, then q ∈ R. The set of final segments of P is denoted by Fs (P).

For every p ∈ P let V_{p} = {x ∈ Fs (P) : p ∈ x}. Let F̂(P) be the subalgebra of P(Fs (P)) generated by the set {V_{p}: p ∈ P }.

The following theorem provides an alternative (and sometimes more convenient) description of F (P). It will be also used to prove some computational facts about F (P).

Theorem 2.3. Let P be a poset.

(a) Fs (P) is closed in P(P), and F̂(P) is the set of clopen subsets of Fs (P).

(b) There is an isomorphism φ between F (P) and F̂(P) such that for every p ∈ P, φ(x_{p}) = V_{p}.

(c) For every p ≠ q in P, x_{p} ≠ x_{q}.

Proof. (a) It is trivial that Fs (P) is closed. Denote by clop(X) the set of clopen subsets of a space X. If X is a Hausdorff compact 0-dimensional space, and F ⊆ X is closed, then clop(F) = {U ∩ F : U ∈ clop(X)}. It is easy to see that a subset U ⊆ P(P) is clopen iff it is a finite union of sets of the form U_{σ,τ}.

It is also easy to see that for every V, V ∈ F̂(P) iff there is U of the above form such that V = U ∩ Fs (P). It follows that F̂(P) = clop(Fs (P)).

(b) By the definition of F̂(P) for every p, q ∈ P : if p ≤ q, then V_{p} ⊆ V_{q}. So the relation V_{p} ∩ V_{q} = V_{p} holds between the generators V_{p} and V_{q} of F̂(P). Recall that F (P) is T [Y] of Definition 1.1 and x_{p} = [y_{p}]. So by Theorem 1.2, there is a homomorphism φ : F (P) → F̂(P) such that φ([y_{p}]) = V_{p}.

Since {V_{p}: p ∈ P } generates F̂(P), φ is surjective.

Let σ, τ be finite subsets of P. It is easy to see that:

If ⋂{V_{p}: p ∈ σ} ∩ ⋂{−V_{q}: q ∈ τ} = ∅, then for some p ∈ σ
and q ∈ τ, p ≤ q.

(1)

Also, by the definition of the relations of F (P),

If p ≤ q, then in F (P), x_{p} · −x_{q} = 0.  (2)

Now suppose that a ∈ F (P) and φ(a) = ∅. The element a is the sum of elements of the form ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ}. Let us take one of these summands. Then ∅ = φ(∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ}) = ⋂{V_{p}: p ∈ σ} ∩ ⋂{−V_{q}: q ∈ τ}. By (1), there are p ∈ σ and q ∈ τ such that p ≤ q. By (2), ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ} = 0. It follows that a = 0. So φ is injective.

(c) If p ≠ q, then V_{p} ≠ V_{q}. Since φ(x_{p}) = V_{p} and φ(x_{q}) = V_{q}, x_{p} ≠ x_{q}.

Definition 2.4. Let ⟨P, ≤⟩ be a poset.

(a) We denote by ⟨P, ≤⟩^{∗} the inverse ordering of ⟨P, ≤⟩. That is, ⟨P, ≤⟩^{∗} = ⟨P, ≥⟩. We abbreviate ⟨P, ≤⟩^{∗}, and denote it by P^{∗}.

(b) Let a ∈ P. Then P^{≥a} denotes the set {p ∈ P : p ≥ a}. For σ ⊆ P let P^{≥σ} = {p ∈ P : there is a ∈ σ such that p ≥ a}.

(c) Let P and Q be posets and α : P → Q. We say that α is a homomorphism from P to Q, if for every s, t ∈ P: if s ≤ t, then α(s) ≤ α(t).

We say that α is an embedding of P in Q, if for every s, t ∈ P: s ≤ t, iff α(s) ≤ α(t).

The next proposition collects some easy facts needed in the subsequent sections.

Proposition 2.5. Let P be a poset and denote the set of generators of F (P) by X := {x_{p}: p ∈ P }.

(a) We denote the set of generators of F (P^{∗}) by {x_{p}^{∗}: p ∈ P }. Then the function x_{p} ↦ −x_{p}^{∗}, p ∈ P extends to an isomorphism between F (P) and F (P^{∗}). We denote this isomorphism by ψ_{P}.

(b) Let σ and τ be finite subsets of P. The following are equivalent.

(i) ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ} = 0. (ii) There are p ∈ σ and q ∈ τ such that p ≤ q.

(c) Let σ, τ ⊆ P be such that for every p ∈ σ and q ∈ τ, p and q are incomparable. Suppose that b and c belong to the subalgebras of F (P) generated respectively by {x_{r}: r ∈ σ} and {x_{r}: r ∈ τ}. If b·c = 0, then b = 0 or c = 0.

(d) Let Q be a poset and α : P → Q be a homomorphism. Let Z := {z_{q}: q ∈ Q} be the set of generators of F (Q). Define ρ : X → Z as follows: ρ(x_{p}) = z_{α(p)}. Then

(1) ρ extends to a homomorphism ρ_{α} from F (P) to F (Q).

(2) If α is an embedding of P in Q, then ρ_{α} is an embedding of F (Q) in F (P).

Proof. (a) The function ψ′_{P} defined by x_{p} ↦ −x_{p}^{∗}, p ∈ P, is an order preserving function from {x_{p}: p ∈ P } to F (P^{∗}). That is, for every relation x_{p} · x_{q} = x_{p} in the defining set of relations of F (P), the relation (−x_{p}^{∗}) · (−x_{q}^{∗}) = −x_{p}^{∗} holds for the images of the x ’s under ψ′_{P}. So ψ′_{P} extends to a homomorphism ψ_{P} between F (P) and F (P^{∗}). Since Rng(ψ′_{P}) generates F (P^{∗}), ψ_{P} is surjective.

The homomorphism ψ_{P^{∗}} : F (P^{∗}) → F (P) is defined similarly. ψ_{P^{∗}}(x_{p}^{∗}) = −x_{p}. So ψ_{P^{∗}}(−x_{p}^{∗}) = x_{p}. Hence (ψ_{P^{∗}} ∘ ψ_{P}) ↾ {x_{p}: p ∈ P } = Id . It follows that ψ_{P^{∗}} ∘ ψ_{P} = Id . So ψ_{P} is injective.

(b) Suppose that there are p ∈ σ and q ∈ τ such that p ≤ q. Then in F (P), x_{p} · x_{q} = x_{p}. So in F (P), x_{p} − x_{q} = 0. So in F (P), ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ} = 0.

For the second direction we use the isomorphism φ between F (P) and F̂(P) defined in Theorem 2.3(b). Suppose that for every p ∈ σ and q ∈ τ, p ≰ q. The set R:= ⋃{P^{≥p}: p ∈ σ} is a final segment of P. For every p ∈ σ, p ∈ R; and for every q ∈ τ, q ∉ R. So R ∈ ⋂{V_{p}: p ∈ σ} ∩ ⋂{−V_{q}: q ∈ τ}. It follows that

φ(∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ}) = ⋂{V_{p}: p ∈ σ} ∩ ⋂{−V_{q}: q ∈ τ} ≠ ∅. So ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ} ≠ 0.

(c) Part (c) follows easily from Part (b).

(d) (1) A relation between generators of F (P), has the form x_{s} · x_{t} = x_{s}, where s ≤ t in P. Since α is a homomorphism, α(s) ≤ α(t). So z_{α(s)} · z_{α(t)} = z_{α(s)} holds in F (Q). That is, ρ(x_{s}) · ρ(x_{t}) = ρ(x_{s}). This implies (1).

(2) Let σ, τ ⊆ P be finite. Denote x_{σ,τ} = ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ}. It suffices to show that for every finite σ, τ ⊆ P, if ρ_{α}(x_{σ,τ}) = 0, then x_{σ,τ} = 0.

Now, ρ_{α}(x_{σ,τ}) = ∏{z_{α(p)} : p ∈ σ} · ∏{−z_{α(q)} : q ∈ τ}. So if ρ_{α}(x_{σ,τ}) = 0, then by Part (b), there are p ∈ σ and q ∈ τ such that α(p) ≤ α(q). Since α is an embedding, p ≤ q. So by Part (b), x_{σ,τ} = 0.

Recall that the Cantor topology on P(A) was defined by the basic clopen sets

U_{σ,τ} = {x ∈ P(A) : σ ⊆ x and x ∩ τ = ∅}

defined for every σ, τ finite subsets of A. The resulting topology is Hausdorff, compact, and 0-dimensional.

Since the intersection (and unions) of a family of final segments is again a final segment, Fs (P) is a lattice (a sublattice of P(P)) in addition to being a topological space. In fact it is a topological lattice as we shall see. Denote by τ^{P} the Cantor topology on Fs (P) and define

L(P) := ⟨Fs (P), τ^{P} , ∪, ∩⟩.

The main goal of this section is to prove the following theorem.

Theorem 2.6. (a) Let P be a poset. Then L(P) is a Hausdorff compact 0-dimensional topological distributive lattice.

(b) Let ⟨X, τ, ∨, ∧⟩ be a Hausdorff compact 0-dimensional topological distributive lattice. Then there is a poset P such that ⟨X, τ, ∨, ∧⟩ ≅ L(P).

(c) If φ : L(P) ≅ L(Q), then there is α : P ≅ Q such that α induces φ. Actually, it is enough to assume that φ is a lattice isomorphism of Fs (P) and Fs (Q); this already implies that φ is continuous.

We recall first some notions on lattices. Let L be a lattice. A subset F ⊆ L is a filter if (i) F is closed under meet and (ii) for every a ∈ F and b ≥ a, b ∈ F. F is called a proper filter if F ≠ L. I ⊆ L is an ideal if (i) F is closed under join and (ii) for every a ∈ I and b ≤ a, b ∈ I.

An element e of a lattice L is compact if, for every D ⊆ L, e ≤ ⋁ D implies e ≤ ⋁ D_{0} for some finite D_{0} ⊆ D. An element e is join irreducible if e = a ∨ b implies e = a or e = b. An element e of a lattice is completely join prime if e ≤ ⋁ D implies that there exists d ∈ D such that e ≤ d. Observe that the minimal element of a lattice L (if it exists) is not completely join prime. (For if 0 is the minimal element of L then 0 = ⋁ ∅.) Similarly, we say that e is completely meet prime if e ≥ ⋀ D implies that there exists d ∈ D such that e ≥ d. Observe, again, that 1 (if it exists) is not completely meet prime. If e is completely join prime then it is both compact and join irreducible. The converse also holds in distributive lattices.

If a < b and there is no c in the lattice with a < c < b, then we way that b covers a (and b is said to be a cover).

Let e and f be two elements of a complete lattice L. We shall say that ⟨e, f⟩ is a prime pair iff L is a disjoint union of the sets L^{≥e} and L^{≤f}. In this case e is obviously completely join prime, and f is completely meet prime.

If e is any completely join prime in a complete lattice L, then there exists a (uniquely determined) completely meet prime element f such that ⟨e, f⟩ is a prime pair. Indeed, define f = ⋁(L − L^{≥e}). Then e ≰ f, or else the fact that e is completely join prime implies that e ≤ d for some d ∉ L^{≥e} which is impossible. Similarly, if f is completely meet prime and we define e = ⋀(L − L^{≤f}), then ⟨e, f⟩ is a prime pair.

Prime pairs and covers are intimately connected. If ⟨e, f⟩ is a prime pair, then e covers e ∧ f, and e ∨ f covers f. Let L be a distributive, complete lattice, and suppose that b covers a. Define

I = {x ∈ L: x ∧ b ≤ a}

F = {x ∈ L: x ∨ a ≥ b}.

Then I is an ideal, F a filter, I ∩ F = ∅, and I ∪ F = L. To prove this last equation, let x be an arbitrary element of L, and define c = a ∨ (b ∧ x). Then a ≤ c ≤ b, and hence there are two possibilities: c = a which implies that c ∈ I, and c = b which implies that c ∈ F.

Now, since L is complete we can define e = ⋀ F, f = ⋁ I. Then ⟨e, f⟩ is a prime pair. Thus we have proved the following

Lemma 2.7. Suppose that L is a distributive complete lattice, and b covers a. Then there exists a prime pair ⟨e, f⟩ such that e ≤ b and f ≥ a.

A topological lattice is a structure ⟨L, τ, ∨, ∧⟩ such that (L, τ) is a Hausdorff topological space, (L, ∨, ∧) a lattice, and the join and meet functions are continuous (from L × L to L). Observe that the set ≤ (as a subset of L × L) is closed. Observe also that a compact (Hausdorff) topological lattice L is necessarily complete. (In particular, 0^{L} and 1^{L} exist.)

Let P be a poset. We have defined L(P) as the topological lattice of final segments Fs (P) of P. We note that part (a) of Theorem 2.6 is an easy exercise. To prove that L(P) is a Hausdorff, compact, 0-dimensional topological distributive lattice, recall that Fs (P) is closed in P(P) and hence it inherits these topological properties. It is obvious that the set of all sets of the form Fs (P) ∩ U_{σ,τ} is a basis for the topology. It follows now that the lattice operations are continuous.

What are the covers and the completely prime elements of L(P)? The covers of L(P) are those final segments x that possess a minimal element p (that is, p ∈ x and there is no q < p in x). Such x covers the initial segment x−{p} (and possibly other initial segments if it has other minimal elements).

For any p ∈ P, e = P^{≥p} is completely join prime in L(P), f = P − P^{≤p} is completely meet prime, and ⟨e, f⟩ forms a prime pair. For the other direction, every completely join prime member of Fs (P) is some P^{≥p}. To see this, suppose that e ∈ Fs (P) is completely join prime. Since e = ⋁{P^{≥p}: p ∈ e}, e = P^{≥p} for some p ∈ P. Similarly, if f ∈ Fs (P) is completely meet prime, then f = P − P^{≤p} for some p ∈ P, and ⟨P^{≥p}, f⟩ is a prime pair.

Thus e : P → Fs (P) defined by e(p) = P^{≥p} is an order-inversion bijections between P and the set of completely join prime elements of L(P).

We return to the general discussion and prove that prime pairs are connected to clopen filters and ideals.

Lemma 2.8. Let L be a distributive compact topological lattice.

(1) If F is a non-empty closed filter, then e = ⋀ F ∈ F and F = L^{≥e}.

(2) If F is a proper, clopen, prime filter, then e = ⋀ F is completely join prime. Conversely, if e is any completely join prime, then L^{≥e} is a non-empty, non-trivial, clopen, prime filter.

Proof. Given that F is a closed filter, the closed sets F_{a} = {x ∈ F : x ≤ a}, a ∈ F, have the finite intersection property, and hence their non-empty intersection gives the desired e ∈ F.

Assume now that F is a proper, clopen, prime filter. Then e = ⋀ F ∈ F. Define I = L − F. Since F is a proper, prime filter, I is a non-empty ideal, and I is closed, since F is open. It follows, as above, that f = ⋁ I ∈ I exists. Hence e ≰ f, and it follows that e is completely join prime. (If e ≤ ⋁ D then D ∩ F ≠ ∅, or else D ⊆ I and then ⋁ D ≤ f contradicts the assumption that e ≤ ⋁ D.)

We have identified above the completely meet prime elements of Fs (P). So this lemma yields the following characterization.

Lemma 2.9. Let P be a poset. In Fs (P) every clopen, prime, proper filter has the form V_{p} = {X ∈ Fs (P) : p ∈ X} for some p ∈ P.

We are interested in this section in compact, 0-dimensional, distributive lattices, and our first result concerning these lattices is the following.

Lemma 2.10. Let X be a compact, 0-dimensional, distributive topological lattice. Then for every a < b in L there are a′ and b′ such that a ≤ a′ < b′ ≤ b and b′ covers a′.

Proof. Let C be a maximal chain (linearly ordered set) such that a ≤ ⋀ C and ⋁ C ≤ b. Since X is 0-dimensional, there exists a clopen set B such that b ∈ B and a ∉ B. Any maximal chain between a and b is a closed set. Thus B∩C is closed and trivially closed under meets. Hence b′ = ⋀(B∩C) ∈ B∩C. So b′ > a, and we observe that C^{<b′} ∩B = ∅. Thus C^{<b′} = C^{≤b′} −B is closed, and we define a′ = ⋁ C^{<b′}. Then a′ < b′ and b′ covers a′.

Lemma 2.11. Let X be a compact, 0-dimensional, distributive topological lattice. Then:

(1) Every element x ∈ X is the join of the set of completely join prime elements g such that g ≤ x.

(2) The proper, clopen, prime filters of X separate points, and hence generate the algebra of clopen subsets of X.

Proof. Let x ∈ X be an arbitrary element and define x_{0} as the join of all g ≤ x that are completely meet prime. We shall obtain a contradiction from the assumption that x_{0} < x. By the previous lemma there are a′, b′ in X such that x ≤ a′ < b′ ≤ x_{0} and b′ covers a′. By Lemma 2.7, there is a prime pair ⟨e, f⟩ with e ≤ b′ and f ≥ a′. Since e ≤ x and e is completely join prime, e ≤ x_{0}. But then e ≤ f, which contradicts the fact that ⟨e, f⟩ is a prime pair.

Let a, b ∈ X be any two points, and assume that b ≰ a. We argue first that we can assume that a < b. Indeed, a ∧ b < b, and if we find a proper, clopen prime filter F such that a ∧ b ∉ F and b ∈ F, then a ∉ F as well since F is a filter. So we assume that a < b. By the previous lemma there are a′, b′ in X such that x ≤ a′ < b′ ≤ x_{0} and b′ covers a′. By Lemma 2.7, there is a prime pair ⟨e, f⟩ with e ≤ b′ and f ≥ a′. The filter X^{≥e} contains b, excludes a and is as required.

Thus the algebra of clopen subsets of X is generated by the clopen filters of X. (In any compact space a family of clopen sets that separate points generates all clopen sets.)

We are now ready to prove part (b) of Theorem 2.6. So let L = ⟨X, τ, ∨, ∧⟩ be a Hausdorff compact 0-dimensional topological distributive lattice. Let <^{L} denote its lattice ordering. Let P ⊆ X be the set of completely join prime elements of X, and define a <^{P} b iff b <^{L} a. That is reverse the lattice ordering. Then P = ⟨P, <^{P} ⟩ is the required poset for which we prove that L(⟨P, <^{P} ⟩) ≅ L. To define the required isomorphism ψ : L(P) → L, let x ∈ L(P) be any element. Then x ∈ Fs (P) is a final segment in P, and hence it is an initial segment of completely join prime members of X. Anyhow, x ⊆ X and we define

ψ(x) = ⋁ x ∈ X.

It is obvious that x is an order-homomorphism, and Lemma 2.11(1) immediately implies that ψ is onto X. We prove that ψ is an embedding. Suppose that x_{1} ≰ x_{2} and we shall prove that ψ(x_{1}) ≰ ψ(x_{2}). There exists p ∈ x_{1}−x_{2}, and since p is completely join prime, p ≰ ⋁ x_{2}. Thus ψ(x_{1}) ≰ ψ(x_{2}). Hence ψ is a lattice isomorphism. But this implies, as we are going to prove in the following lemma, that ψ preserves the topological structure as well, namely that it is an isomorphism between the topological lattices L(P) and L.

Lemma 2.12. Suppose that L_{1} and L_{2} are Hausdorff compact 0-dimensional topological distributive lattices, and ψ : L_{1} → L_{2} is a lattice isomorphism. Then ψ is continuous.

Proof. Since the notion of being a completely join prime element is latticetheoretic, Lemma 2.8 implies that ψ establish a correspondence between the clopen prime filters of L_{1} and L_{2}. But, by Lemma 2.11, it follows that ψ establishes a correspondence between the clopen sets of L_{1} and L_{2}. Since the topology of our 0-dimensional spaces is generated by the clopen sets, it is clear that ψ is a homemomorphism.

We can now also prove part (c) of Theroem 2.6. Suppose that φ : L(P) ≅ L(Q), namely that φ is a lattice isomorphism (and thus an isomorphism of the topological structures as well). Let J_{L(P)} be the set of completely join prime elements of L(P), and J_{L(Q)} be the corresponding set in L(Q). Then φ is an order isomorphism between J_{L(P)} and J_{L(Q)}. Now, e_{P} : P → J_{L(P)} defined by e_{P} (x) = P^{≥x} is an order anti-iomorphism, and so is e_{Q}. Thus α : P → Q defined by

α(p) = e_{Q}^{−1} ∘ φ ∘ e_{P} (p)

is an order isomorphism between P and Q. We have φ(P^{≥p}) = Q^{≥α(p)}. Hence α induces φ in the sense that for every final segment s ∈ Fs (P), φ(s) = α[s].

Our results reveal that the category of posets with order-preserving maps is (anti) isomorphic to the category of compact, 0-dimensional, distributive topological lattices, with continuous lattice homomorphisms that preserve 0 and 1. For this, we need the following lemma.

Lemma 2.13. Let X and Y be compact 0-dimensional distributive lattices. A map f : X → Y is a continuous lattice homomorphism iff the f-preimage of every clopen prime filter of Y is a clopen prime filter of X.

Proof. Clearly, the condition is necessary: if f is a continuous lattice homomorphism then the preimage of every clopen set is clopen, and the preimage of every prime filter is a prime filter.

Assume that f : X → Y is such that f^{−1}[P] is a clopen prime filter in X, whenever P is a clopen prime filter in Y. Since the complement of a clopen prime ideal is a clopen prime filter, we get that the preimage of any clopen prime ideal is also a clopen prime ideal. Then f is continuous, because by Lemma 2.11 the collection of clopen prime filters and ideals generates the topology.

Suppose x, y ∈ X are such that a = f(x ∧ y) ≠ f(x) ∧ f(y) = b. There exists a clopen prime filter F which separates a, b. In case a ∈ F and b ∉ F, x∧y ∈ f^{−1}[F] (since f^{−1}[F] is a filter) and hence x, y ∈ f^{−1}[F] and therefore f(x), f(y) ∈ F, which is a contradiction. In case a ∉ F and b ∈ F, then we get a contradiction again. Thus f is meet preserving. By a dual argument, f is join preserving.

For every order preserving map f : P → Q there is a continuous homomorphism g : L(Q) → L(P) that maps 0 to 0 and 1 to 1, and satisfies

g^{−1}[V_{p}] = V_{f(p)}  (3)

for every p ∈ P. (Simply define g(X) = f^{−1}[X].)

Lemma 2.13 implies that if f and g are any two functions connected as in (3), then f is order-preserving and g is a continuous homomorphism (preserving 0 and 1).

Now, if g : L(Q) → L(P) is any continuous homomorphism that maps 0 to 0 and 1 to 1, then f that satisfies (3) can be defined by referring to Lemma 2.9.

3 The well-generatedness of poset algebras

In this section we prove Theorem 1.3. It says that the poset algebra of a narrow scattered poset is well-generated.

Recall that W is the class of posets which are either well-ordered or anti well-ordered.

Definition 3.1. We define by induction on the ordinal α the class of posets H_{α}. Let H_{0} be the class of all posets whose universe is a singleton. If α is a limit ordinal, then H_{α} = ⋃{H_{β}: β < α}. If α = β+1, then H_{α} is the class of all posets P such that there is W ∈ W and an indexed family of posets {P_{w}: w ∈ W } ⊆ H_{β} such that P is isomorphic to ∑{P_{w}: w ∈ W }. Define

H = ⋃{H_{α}: α ∈ Ord }.

We shall prove by induction on α that if P ∈ H_{α}, then F (P) is wellgenerated. The rough idea of the proof is as follows. Let P ∈ H_{α+1}, and suppose that P = ∑_{w∈W} P_{w}, where W ∈ W and each P_{w} belongs to H_{α}. For every w ∈ W let L_{w} be a well-founded lattice which generates F (P_{w}). If the L_{w} are appropriately chosen, then ⋃_{w∈W} L_{w} generates a well-founded lattice, and certainly, this lattice generates F (P).

The following key lemma from [BR] is reproved here for the reader’s convenience.

Proposition 3.2. Let B be a well-generated algebra and I be a maximal ideal of B. Then there is a sublattice G ⊆ I such that G generates B, and G is well-founded.

Proof. We prove the following claim. Claim 1. Let G′ be a well-founded sublattice of B and b ∈ B. Let H be the sublattice of B generated by G′ ∪ {b}. Then H is well-founded. Proof. We may assume that 0^{B}, 1^{B} ∈ G′. Then every member of H has the form g + h · b, where g, h ∈ G′.

Let {g_{n}+h_{n}·b : n ∈ ω} be a decreasing sequence in H. Let d_{n} = g_{n}+h_{n}·b and e_{n} = g_{n} + h_{n}. Suppose by contradiction that {d_{n} · b : n ∈ ω} is not eventually constant. We have d_{n} · b = e_{n} · b. So {e_{n} · b : n ∈ ω} is not eventually constant. Then p_{n} := ∏_{i≤n} e_{i}, n ∈ ω, is a decreasing sequence of members of G′ which is not eventually constant. So

(i) {d_{n} · b : n ∈ ω} is eventually constant.

Suppose by contradiction that {d_{n}−b : n ∈ ω} is not eventually constant. d_{n} − b = g_{n} − b. So {g_{n} − b : n ∈ ω} is a decreasing sequence, which is not eventually constant. Then q_{n} := ∏_{i≤n} g_{i}, n ∈ ω, is a decreasing sequence of members of G′ which is not eventually constant. So (ii) {d_{n} − b : n ∈ ω} is eventually constant. It follows from (i) and (ii) that {d_{n} : n ∈ ω} is eventually constant. So H is well-founded. We have proved Claim 1.

We prove the proposition. Let G′ be a well-founded sublattice of B which generates B. If G′ ⊆ I, then there is nothing to prove. Assume that G′ ⊈ I. Let G_{1} := G′ − I. By the maximality of I, G_{1} is a lattice, and since in addition, G_{1} is well-founded and not empty, it has a minimum. Denote the minimum by a. Let G be the sublattice of B generated by (G′ ∩ I) ∪ {b − a : b ∈ G_{1}} ∪ {−a}. Clearly, G is contained in the lattice generated by G′∪{−a}. So by Claim 1, G is well-founded. By the maximality of I, −a ∈ I, so G ⊆ I. Let b ∈ G′. If b ∈ I, then b ∈ G. If b ∉ I, then a ≤ b. So b = (b − a) + a. This means that b belongs to the subalgebra generated by G. So G generates B.

Let ⟨T, ∨, ∧⟩ be a lattice. x ∨ y and x ∧ y denote the join and the meet of x and y in T. The partial ordering of T is denoted by ≤. So x ≤ y if x ∧ y = x.

We shall need the following lemma. It appears in [BR] Lemma 2.8, and the proof is brought here for the readers’s convenience.

Lemma 3.3. Let T be a distributive lattice.

(a) If T is well-founded, then every nonempty subset L of T closed under meet has a minimum m.

(b) Let S be a subset of T such that:

(1) S generates T.

(2) The meet of two elements of S is a finite join of elements of S.

(3) S is well-founded with respect to the partial ordering of T.

Then T is well-founded. In particular, if S ⊆ T is closed under meet, generates T, and is well-founded, then T is well-founded.

Proof. (a) Let m_{1} and m_{2} be minimal elements of L. Then m := m_{1} ∧ m_{2} ∈ L, and thus m = m_{1} = m_{2}.

(b) Clearly, every element of T is a finite join of elements of S. It is easy to check that the following holds. (∗) If w_{0} > w_{1} > ⋯ is a strictly decreasing sequence in T, and w_{0} = ⋁_{i<n} v_{i}, then there is ℓ < n such that {w_{j} ∧ v_{ℓ} : j < ω} contains a strictly decreasing infinite subsequence. The proof uses the distributivity of T.

Next suppose by contradiction that u_{0} > u_{1} > ⋯ is a strictly decreasing sequence in T. We define by induction a strictly decreasing sequence {v_{n}: n < ω} in S. Assume by induction that v_{n} has the following property. (∗∗) There is a strictly decreasing sequence w_{0} > w_{1} > ⋯ in T such that w_{0} < v_{n}. Let U be a finite subset of S such that u_{0} = ⋁ U. By (∗), there is v_{0} ∈ U such that {u_{j} ∧ v_{0} : j < ω} contains a strictly decreasing subsequence. Hence v_{0} satisfies the induction hypothesis. Suppose that v_{n} has been defined, and let v_{n} > w_{0} > w_{1} > ⋯ be as in the induction hypothesis. Let W be a finite subset of S such that ⋁ W = w_{0}. By (∗), there is v_{n+1} ∈ W such that {w_{j} ∧ v_{n+1} : j < ω} contains a strictly decreasing subsequence. So v_{n+1} satisfies the induction hypothesis and v_{n} > w_{0} ≥ v_{n+1}. The sequence {v_{n}: n < ω} ⊆ S is strictly decreasing. This contradicts the well-foundedness of S, so T is well-founded.

Theorem 3.4. Let ⟨T, ∨, ∧⟩ be a distributive lattice. We assume that T has a minimum 0 and a maximum 1. Let ⟨I, ≤⟩ be a poset, and {T_{i}: i ∈ I} be a family of sublattices of T. Suppose that:

(W1) ⟨I, ≤⟩ is a well-ordered poset.

(W2) For every i ∈ I, T_{i} is well-founded.

(W3) ⋃{T_{i}: i ∈ I} generates the lattice T.

(W4) For every i, j ∈ I, s ∈ T_{i} and t ∈ T_{j}: if i < j and s ≠ 1, then either s < t or s ∧ t = 0.

(W5) Let σ ⊆ I, ℓ ∈ I, and for every i ∈ σ ∪ {ℓ}, u_{i} ∈ T_{i}. Suppose that the following properties hold:

(1) σ is finite;

(2) for every i ∈ σ, i ≰ ℓ;

(3) ⋀_{i∈σ} u_{i} ≠ 0 and u_{ℓ} ≠ 1. Then ⋀_{i∈σ∪{ℓ}} u_{i} < ⋀_{i∈σ} u_{i}.

Then T is a well-founded lattice.

Proof. We may assume that for every i ∈ I, 1 ∈ T_{i}. Let T be the set of all functions t⃗ such that:

(1) Dom (t⃗) is a finite subset of I,

(2) for every i ∈ Dom (t⃗), t⃗(i) ∈ T_{i}. Let ⋀ t⃗ := ⋀{t⃗(i) : i ∈ Dom (t⃗)}, and

L := {⋀ t⃗: t⃗ ∈ T }.

Let t⃗ ∈ T and i ∈ Dom (t⃗). We denote t⃗(i) by t_{i}.

Claim 1. Let u ∈ L − {0, 1}. Let t⃗ ∈ T be such that u = ⋀{t_{i}: i ∈ Dom (t⃗)}. Let σ be the set of minimal elements of Dom (t⃗). Then σ is a finite antichain in I and u = ⋀{t_{i}: i ∈ σ}. Proof. For each j ∈ Dom (t⃗) − σ, let m(j) ∈ σ be such that m(j) < j. Since u ≠ 0, t_{m(j)} ∧ t_{j} ≠ 0. By (W4), t_{m(j)} < t_{j}. Hence u = ⋀ t⃗ ≤ ⋀{t_{j}: j ∈ σ} ≤ u. So u = ⋀{t_{i}: i ∈ σ}. Claim 2. For u ∈ L − {0, 1}, there is t⃗^{u} ∈ T satisfying the following properties.

(1) Dom (t⃗^{u}) is an antichain.

(2) u = ⋀ t⃗^{u}.

(3) If i ∈ Dom (t⃗^{u}), then t_{i}^{u} ≠ 0, 1.

(4) For every t⃗ ∈ T such that u = ⋀ t⃗ : Dom (t⃗^{u}) ⊆ Dom (t⃗), and if i ∈ Dom (t⃗^{u}) then t_{i}^{u} ≤ t_{i}. Proof. Let i ∈ I. The set

T_{i}^{u}:= { t′ ∈ T_{i}: there is t⃗∈ T such that i ∈ Dom (t⃗), t_{i} = t′ and u = ⋀ t⃗} is nonempty and closed under ∧. By (W2), T_{i} is well-founded, and it follows from Lemma 3.3(a), that t_{i}^{∗} := min(T_{i}^{u}) exists.

Let r⃗ ∈ T be such that u = ⋀ r⃗. Since u ≠ 1, we may assume that for every k ∈ Dom (r⃗), r_{k} ≠ 1. Let σ be the set of minimal elements of Dom (r⃗) and s⃗ = r⃗ ↾ σ. By Claim 1 and since u ≠ 0, ⋀ s⃗ = u. Also, σ is an antichain. For every k ∈ σ, t_{k}^{∗} ≤ s_{k}, so ⋀_{k∈σ} t_{k}^{∗} ≤ ⋀_{k∈σ} s_{k}. That is,

(i) ⋀{t_{k}^{∗}: k ∈ σ} ≤ u. For every k ∈ σ, let t⃗^{k} ∈ T be such that u = ⋀ t⃗^{k} and t_{k}^{k} = t_{k}^{∗}. Let ρ = {⟨k, j⟩: k ∈ σ, j ∈ Dom (t⃗^{k})}. Obviously,

u = ⋀{⋀ t⃗^{k}: k ∈ σ)} = ⋀{t_{j}^{k}: ⟨k, j⟩ ∈ ρ}.

Since for every k ∈ σ, t_{k}^{∗} ∈ {t_{j}^{k}: ⟨k, j⟩ ∈ ρ}, ⋀{t_{j}^{k}: ⟨k, j⟩ ∈ ρ} ≤ ⋀{t_{k}^{∗}: k ∈ σ}. That is, (ii) u ≤ ⋀{t_{k}^{∗}: k ∈ σ}. It follows from (i) and (ii) that u = ⋀{t_{k}^{∗}: k ∈ σ}.

Let t⃗^{u} be defined as follows: Dom (t⃗^{u}) = σ and for every i ∈ Dom (t⃗^{u}), t⃗^{u}(i) = t_{i}^{∗}, that is t⃗_{i}^{u} = t_{i}^{∗}.

We show that t⃗^{u} is as required.

(1) Dom (t⃗^{u}) = σ is an antichain.

(2) We have proved that u = ⋀ t⃗^{u}.

(3) Since for every i ∈ σ, t_{i}^{∗} ≤ s_{i} < 1, (3) holds.

(4) Let t⃗ ∈ T be such that u = ⋀ t⃗, and suppose by contradiction that ℓ ∈ Dom (t⃗^{u}) − Dom (t⃗). By Claim 1, we may assume that Dom (t⃗) is an antichain. Suppose by contradiction that for some k ∈ Dom (t⃗), k < ℓ. Since Dom (t⃗^{u}) is an antichain and k < ℓ ∈ Dom (t⃗^{u}), for every i ∈ Dom (t⃗^{u}), i ≰ k. Since u ≠ 1, we may assume that for every k ∈ Dom (t⃗), t_{k} ≠ 1. We apply (W5) to Dom (t⃗^{u}), k, {t_{i}^{u}: i ∈ Dom (t⃗^{u})}, and t_{k}, and conclude that

u ∧ t_{k} = ( ⋀{t_{i}^{u}: i ∈ Dom (t⃗^{u})}) ∧ t_{k} < ⋀{t_{i}^{u}: i ∈ Dom (t⃗^{u})} = u.

Hence u ∧ t_{k} < u, and so u ≰ t_{k}. A contradiction. So there is no k ∈ Dom (t⃗) such that k < ℓ.

We now apply (W5) to Dom (t⃗), ℓ, {t_{i}: i ∈ Dom (t⃗)}, and t_{ℓ}^{u}, and conclude that

u ∧ t_{ℓ}^{u} = (⋀{t_{i}: i ∈ Dom (t⃗)}) ∧ t_{ℓ}^{u} < ⋀{t_{i}: i ∈ Dom (t⃗)} = u.

Hence u ∧ t_{ℓ}^{u} < u, and so u ≰ t_{ℓ}^{u}. A contradiction. So Dom (t⃗^{u}) ⊆ Dom (t⃗).

Let i ∈ Dom (t⃗^{u}), and we show that t_{i}^{u} ≤ t_{i}. Since ⋀ t⃗^{u} ∧ ⋀ t⃗ = u, t_{i}^{u} ∧ t_{i} ∈ T_{i}^{u}. Also, t_{i}^{u} = min(T_{i}^{u}). So t_{i}^{u} ≤ t_{i}^{u} ∧ t_{i}. That is, t_{i}^{u} ≤ t_{i}. We have proved Clause (4) of Claim 2. So Claim 2 is proved.

It is obvious that the object t⃗^{u} which satisfies Clauses (1) - (4) of Claim 2 is unique. For every u ∈ L − {0, 1} we denote

σ(u) := Dom (t⃗^{u}).

Claim 3. Let u, v ∈ L − {0, 1} be such that u ≤ v.

(a) For every i ∈ σ(u) ∩ σ(v), t_{i}^{u} ≤ t_{i}^{v}.

(b) For every i ∈ σ(v) there is j ∈ σ(u) such that j ≤ i. Proof. (a) For w ∈ L − {0, 1} and i ∈ I − σ(w) we set t_{i}^{w} = 1. Hence

⋀{t_{i}^{u} : i ∈ σ(u)} = u = u ∧ v = ⋀{t_{i}^{u} : i ∈ σ(u)} ∧ ⋀{t_{i}^{v} : i ∈ σ(v)}
= ⋀{t_{i}^{u} ∧ t_{i}^{v} : i ∈ σ(u) ∪ σ(v)}.

So by Claim 2, for every i ∈ σ(u) ∩ σ(v), t_{i}^{u} ≤ t_{i}^{u} ∧ t_{i}^{v}, and thus t_{i}^{u} ≤ t_{i}^{v}.

(b) Let i ∈ σ(v), and suppose by contradiction that there is no j ∈ σ(u) such that j ≤ i. We apply (W5) to σ(u), i, {t_{j}^{u}: j ∈ σ(u)} and t_{i}^{v}. Then

u ∧ t_{i}^{v} = (⋀{t_{j}^{u}: j ∈ σ(u)}) ∧ t_{i}^{v} < ⋀{t_{j}^{u}: j ∈ σ(u)} = u. So u ≰ t_{i}^{v}. But u ≤ v ≤ t_{i}^{v}. A contradiction.

So there is j ∈ σ(u) such that j ≤ i. Claim 3 is proved.

Let Q be a poset. A subset J of Q is an initial segment of Q if for every q ∈ Q and p ∈ J: if q ≤ p, then q ∈ J. Claim 4. Let Q be a well-ordered poset. Then there is no strictly decreasing sequence of initial segments of Q with respect to set inclusion. Proof. Claim 4 is due to Higman [H]. For completeness, we sketch a proof. Assume by contradiction that {J_{n}: n ∈ ω} is a strictly decreasing sequence of initial segments of Q. For every n ∈ ω, let q_{n} ∈ J_{n} −J_{n+1}. Then {q_{n}: n ∈ ω} has no increasing pair. Hence Q is not well-ordered. A contradiction. We have proved Claim 4.

We define a partial ordering on the set Ant (Q) of antichains of a poset Q: let σ ≤^{m} τ, if for every i ∈ τ there is j ∈ σ such that j ≤ i. Claim 5. If Q is a well-ordered poset, then ⟨ Ant (Q), ≤^{m}⟩ is well-founded. Proof. For σ ∈ Ant (Q) let Q^{≥σ} = {q ∈ Q: (∃p ∈ σ)(p ≤ q)} and I(σ) = Q − Q^{≥σ}. Since Q^{≥σ} is a final segment of Q, its complement I(σ) is an inital segment. The function σ ↦ Q^{≥σ} is one-to-one, and if σ ≤^{m} τ, then Q^{≥σ} ⊇ Q^{≥τ}, and so I(σ) ⊆ I(τ). If {σ_{n}: n ∈ ω} is a strictly decreasing sequence in Ant (Q), then {I(σ_{n}) : n ∈ ω} is a strictly decreasing sequence in the set of initial segments of Q, which by Claim 4 is impossible. So ⟨ Ant (Q), ≤^{m}⟩ is well-founded. Claim 6. L is well-founded. Proof. We prove that L is well-founded. Note that by Claim 3(b), for every u, v ∈ L − {0, 1}: if u ≤ v, then σ(u) ≤^{m} σ(v).

Let {u_{n}: n < ω} be a decreasing sequence in L. Then for every m > n, σ(u_{m}) ≤^{m} σ(u_{n}). So by Claim 5, we may assume that {σ(u_{n}) : n ∈ ω} is a constant sequence. Let σ = σ(u_{0}). By Claim 3(a), if m > n then t_{i}^{m} ≤ t_{i}^{n}. Recall that t_{i}^{n} ∈ T_{i} and that T_{i} is well-founded. Since σ is finite, there is n_{0} ∈ ω such that for every i ∈ σ and m > n_{0}, t_{i}^{m} = t_{i}^{n_{0}}. So u_{m} = u_{n_{0}}. Claim 6 is proved.

We now prove the theorem. By the Claim 6, L is well-founded. It is obvious that L is closed under ∧. Since T is a distributive lattice, and T is generated by ⋃{T_{i}: i ∈ I}, every member of T is a finite sum of members of L. So L generates T. By Lemma 3.3(b), T is well-founded.

Definition 3.5. Let P be a poset. An element b ∈ F (P) is said to be bounded if there is a nonempty finite subset σ of P such that b ≤ ∑{x_{p}: p ∈ σ}. Let I^{bnd} (P) be the set of bounded elements of F (P).

Observe the following trivial fact.

Proposition 3.6. Let F (P) be a poset algebra. Then I^{bnd} (P) is a maximal ideal of F (P).

Proof of Theorem 1.3. (3) ⇒ (2). The easy proof that every wellgenerated Boolean algebra is superatomic appears in [BR] Proposition 2.7(b).

(2) ⇒ (1). Suppose that it is not true that P is narrow and scatterd, and we show that F (P) is not superatomic. If ℚ is embeddable in P, then F (ℚ) (which is atomless) is embeddable in F (P). So F (P) is not superatomic.

If A is an infinite antichain in P, then F (A) (which is an infinite free Boolean algebra) is embeddable in F (P). So F (P) is not superatomic.

(1) ⇒ (3). We prove by induction on α, that for every P ∈ H_{α}, F (P) is well-generated. There is nothing to prove for α = 0 and for limit ordinals. Suppose that the claim is true for α, and let P ∈ H_{α+1} − H_{α}.

Note that if ⟨P, ≤⟩ ∈ H_{α}, then ⟨P, ≥⟩ ∈ H_{α}. Hence by Proposition 2.5(a), we may assume that P = ∑{P_{v}: v ∈ V }, where V is a wellordered poset, and for every v ∈ V, P_{v} ∈ H_{α}. By the induction hypothesis, for every v ∈ V, F (P_{v}) is well-generated. By Proposition 3.6, I^{bnd} (P_{v}) is a maximal ideal in F (P_{v}). So by Proposition 3.2, there is a well-founded sublattice G_{v} of F (P_{v}) such that G_{v} generates F (P_{v}) and G_{v} ⊆ I^{bnd} (P_{v}). Let G be the sublattice of F (P) generated by ⋃{G_{v}: v ∈ V }. We verify that G and {G_{v}: v ∈ V } satisfy the hypotheses of Theorem 3.4.

By the definition, V is well-ordered, and thus (W1) holds.

Since each G_{v} is a well-founded lattice, (W2) holds.

(W3) follows from the definition of G.

(W4) Let v < w in V, g ∈ G_{v} −{1^{F (P)}} and h ∈ G_{w}. We show that either g < h or g ·h = 0. Let γ be a finite subset of P_{v} such that g ≤ ∑{x_{p}: p ∈ γ}. Let b = ∑{x_{p}: p ∈ γ}. It suffices to show that either b < h or b · h = 0. Note that h has a representation of the following form: there is ℓ ∈ ω and for every i < ℓ there are finite disjoint subsets η(i) and τ(i) of P_{w} such that η(i) ≠ ∅ and

h = ∑_{i<ℓ} ( ∏{x_{p}: p ∈ η(i)} · ∏{−x_{q}: q ∈ τ(i)} ).

Case 1. There is i_{0} < ℓ such that τ(i_{0}) = ∅. Hence b < ∏{x_{p}: p ∈ η(i_{0})} ≤ h, and thus b < h. Case 2. For every i < ℓ, τ(i) ≠ ∅. For every i < ℓ, b · ∏{−x_{q}: q ∈ τ(i)} = ∅, and thus b · ∏{x_{p}: p ∈ η(i)} · ∏{−x_{q}: q ∈ τ(i)} = 0. Hence b · h = 0. We have proved that (W4) holds.

(W5) Let ρ ⊆ V, w ∈ V, and {g_{v}: v ∈ ρ ∪ {w}} ⊆ G. We assume that:

(1) ρ is finite; (2) for every v ∈ ρ, v ≰ w; (3) for every v ∈ ρ ∪ {w}, g_{v} ∈ G_{v};

(4) ∏{g_{v}: v ∈ ρ} ≠ 0 and g_{w} ≠ 1.

It needs to be shown that

g_{w} · ∏{g_{v}: v ∈ ρ} < ∏{g_{v}: v ∈ ρ}.

By (W4), arguing as in Claim 1 of Theorem 3.4, we may assume that ρ is an antichain.

We rely on the fact that F (P) and F̂(P) are isomorphic, and argue in F̂(P) rather than in F (P). Recall that the above isomorphism takes x_{p} to V_{p}. Let σ, τ be finite subsets of P. Define

T_{σ,τ} = {x ∈ Fs (P) : σ ⊆ x and x ∩ τ = ∅}.

Clearly, for every T ⊆ Fs (P): T ∈ F̂(P) iff T is a finite union of T_{σ,τ}’s. Also,

(i) T_{σ,τ} ≠ ∅ iff for every p ∈ σ and q ∈ τ, p ≰ q.

Let T = ⋃_{i<n} T_{σ_{i},τ_{i}} and assume that each T_{σ_{i},τ_{i}} is nonempty. Then T is bounded iff for every i < n, σ_{i} ≠ ∅.

We regard the g_{v}’s as members of F̂(P). For every v ∈ ρ ∪ {w} let g_{v} = ⋃_{i<n(v)} T_{σ(v,i),τ(v,i)}, where σ(v, i), τ(v, i) ⊆ P_{v}. We may assume that each T_{σ(v,i),τ(v,i)} is nonempty. Let v ∈ ρ. Since g_{v} ≠ 0, n(v) > 0, Let v ∈ ρ ∪ {w}. Since g_{v} is bounded, σ(v, i) ≠ ∅ for every i < n(v).

Let σ = ⋃_{v∈ρ} σ(v, 0) and x = P^{≥σ}. So x ∈ Fs (P). We shall show that

x ∈ ∏_{v∈ρ} g_{v} and x ∉ g_{w}.

For v ∈ ρ ∪ {w} let η(v) = ⋃_{i<n(v)} σ(v, i) ∪ ⋃_{i<n(v)} τ(v, i). (ii) If u,v ∈ρ are distinct, p∈η(u) and q ∈η(v), then p,q are incomparable. This follows from the fact that ρ is an antichain and η(v) ⊆ P_{v}. We show that for every v ∈ ρ, x ∈ T_{σ(v,0),τ(v,0)}. By the definition of x, σ(v, 0) ⊆ x.

Let q ∈ τ(v, 0). By (i), q ∉ P^{≥σ(v,0)}. By (ii), q ∉ P^{≥σ(u,0)}, for every u ≠ v. So q ∉ x. Hence τ(v, 0) ∩ x = ∅. We have shown that x ∈ T_{σ(v,0),τ(v,0)} ⊆ g_{v}.

Hence x ∈ ∏_{v∈ρ} g_{v}.

Recall that for every v ∈ ρ, v ≰ w. It follows that for every v ∈ ρ, p ∈ η(v) and q ∈ η(w), p ≰ q. That is, denoting η = ⋃_{v∈ρ} η(v),

P^{≥η} ∩ η(w) = ∅.

But σ ⊆ η, and for every i < n(w), σ(w, i) ⊆ η(w). So σ(w, i) ∩ P^{≥σ} = ∅. Recalling that x = P^{≥σ} and that σ(w, i) ≠ ∅, one concludes that for every i < n(w), x ∉ T_{σ(w,i),τ(w,i)}. So x ∉ ⋃_{i<n(w)} T_{σ(w,i),τ(w,i)} = g_{w}.

We have shown that G and {G_{v}: v ∈ V } satisfy the hypotheses (W1) - (W5). So by Theorem 3.4, G is well-founded. For every v ∈ V, G_{v} generates F (P_{v}) and ⋃{F (P_{v}) : v ∈ V } generates F (P). So G = ⋃{G_{v}: v ∈ V } generates F (P).

4 The embeddability of poset algebras in wellordered poset algebras

In this section we prove Theorems 1.6 and 1.4. In fact, we prove the following slight strengthening of 1.6.

Theorem 4.1. Let P ∈ H. Then there is a well-ordered poset W and an embedding φ of F (P) in F (W) such that φ(I^{bnd} (P)) ⊆ I^{bnd} (W).

We shall prove by induction on α that the claim of the theorem is true for every P ∈ H_{α}.

For a poset Q let m_{Q} be an element which does not belong to Q and let Q^{+} = {m_{Q}} + Q. That is, Q^{+} is the lexicographic sum of a singleton and Q over a chain with two elements. Let X := {x_{p}: p ∈ P } be the set of generators of P. Denote by {z_{q} : q ∈ Q} the generators of F (Q) and by {z_{q}^{+} : q ∈ Q^{+}} the generators of F (Q^{+}). Let a ↦ a^{+} be the embedding of F (Q) into F (Q^{+}) which sends z_{q} to z_{q}^{+}. Let φ : F (P) → F (Q) be an embedding. Define φ̂ : X → F (Q^{+}) as follows: φ̂(x_{p}) = x_{m_{Q}}^{+} + (φ(x_{p}))^{+}. The following notation will be used below. If σ, τ are finite subsets of Q, then z_{σ,τ} := ∏_{p∈σ} z_{p} · ∏_{q∈τ} −z_{q}. z_{σ,τ}^{+} is defined similarly.

Lemma 4.2. Let P, Q, φ and φ̂ be as in the preceding paragraph.

(a) Let σ be a finite subset of P. Then:

(1) ∑{φ̂(x_{p}) : p ∈ σ} = z_{m_{Q}}^{+} + ∑{φ(x_{p})^{+} : p ∈ σ}.

(2) ∏{φ̂(x_{p}) : p ∈ σ} = z_{m_{Q}}^{+} + ∏{φ(x_{p})^{+} : p ∈ σ}.

(b) φ̂ can be extended to an embedding of F (P) in F (Q^{+}).

Proof. (a) The proof of is trival.

(b) Let p, q ∈ P and p ≤ q. Then

φ̂(x_{p}) · φ̂(x_{q}) = z_{m_{Q}}^{+} + φ(x_{p})^{+} · φ(x_{q})^{+} = z_{m_{Q}}^{+} + φ(x_{p})^{+} = φ̂(x_{p}). By Theorem 1.2, φ̂ can be extended to a homomorphism φ̃ from F (P) to F (Q).

We show that φ̃ is an embedding. Notice first that for every a ∈ F (P), either φ̃(a) = z_{m_{Q}}^{+} + φ(a)^{+} or φ̃(a) = −z_{m_{Q}}^{+} · φ(a)^{+}. To see this, notice that if φ̃(a) and φ̃(b) have one of the above forms, then so do φ̃(−a) and φ̃(a · b). Let a ∈ F (P) − {0}. We show that φ̃(a) ≠ 0. Write φ(a) as φ(a) = ∑_{i<n} z_{σ_{i},τ_{i}}, where all the summands are different from 0. Then n > 0 and φ(a)^{+} = ∑_{i<n} z_{σ_{i},τ_{i}}^{+}. If φ̃(a) = z_{m_{Q}}^{+} + φ(a)^{+}, then φ̃(a) ≠ 0. Suppose that φ̃(a) = −z_{m_{Q}}^{+} · φ(a)^{+}. Then φ̃(a) = ∑_{i<n} z_{σ_{i},τ_{i}∪{m_{Q}}}^{+}. Since m_{Q} is the minimum of Q^{+} and m_{Q} ∉ σ_{i}, for every p ∈ σ_{i} and q ∈ τ_{i} ∪ {m_{Q}}, p ≰ q. By Proposition 2.5(b), for every i < n, z_{σ_{i},τ_{i}∪{m_{Q}}}^{+} ≠ 0. So φ̃(a) ≠ 0.

Let {x_{i}: i ∈ I} be a set of generators for a Boolean algebra B. Let σ, τ ⊆ I be finite. We denote x_{σ,τ} = ∏{x_{p}: p ∈ σ} · ∏{−x_{q}: q ∈ τ}. If φ is a function from {x_{i}: i ∈ I} to a Boolean algebra C, then we denote x_{σ,τ}^{φ} = ∏{φ(x_{p}) : p ∈ σ} · ∏{−φ(x_{q}) : q ∈ τ}.

The following claim is part of a lemma due to Sikorski. See [Ko] Theorem 5.5.

Proposition 4.3. Let {x_{i}: i ∈ I} be a set of generators for a Boolean algebra B and φ be a homomorphism from B to a Boolean algebra C. Suppose that for every finite σ, τ ⊆ I: if x_{σ,τ}^{φ} = 0, then x_{σ,τ} = 0. Then φ is an embedding of B into C.

Lemma 4.4. Let R = ∑{R_{i}: i ∈ I} be a lexicographic sum of posets. Suppose that σ, τ ⊆ I are finite, and for every i ∈ σ and j ∈ τ, i < j. For every i ∈ σ let a_{i} ∈ I^{bnd} (R_{i}). For every j ∈ τ let r_{j} ∈ R_{j}. Then ∑_{i∈σ} a_{i} < ∏_{j∈τ} x_{r_{j}}.

Proof. We rely on the fact that F (R) and F̂(R) are isomorphic, and argue in F̂(R) rather than in F (R). Recall that the isomorphism between F (R) and F̂(R) takes each x_{r} to V_{r}, where V_{r} = {x ∈ Fs (R) : r ∈ x}. As in the proof Theorem 1.3 we denote T_{σ,τ} = {x ∈ Fs (R) : σ ⊆ x and x ∩ τ = ∅}. Now, each x_{r_{i}} is replaced by V_{r_{i}} and a_{i} has the form a_{i} = ⋃_{ℓ<n(i)} T_{σ(i,ℓ),τ(i,ℓ)}. We may assume that for every i, ℓ, T_{σ(i,ℓ),τ(i,ℓ)} ≠ ∅. By the boundedness of the a_{i}’s, for every i and ℓ, σ(i, ℓ) ≠ ∅. Let s(i, ℓ) ∈ σ(i, ℓ). Clearly, for every i ∈ σ, ℓ < n(i) and j ∈ τ, s(i, ℓ) < r_{j}. Let x ∈ ⋃_{i∈σ} a_{i}. Then for some i and ℓ, s(i, ℓ) ∈ x. Hence for every j ∈ τ, r_{j} ∈ x. So x ∈ ⋂_{j∈τ} V_{r_{j}}. That is, ⋃_{i∈σ} a_{i} ⊆ ⋂_{j∈τ} V_{r_{j}}.

Let x = ⋃_{j∈τ} P^{≥r_{j}}. Then x ∈ ⋂_{j∈τ} V_{r_{j}}. Since for every i ∈ σ and ℓ < n(i) and j ∈ τ, r_{j} ≮ s(i, ℓ), s(i, ℓ) ∉ x. So x ∉ ⋃_{i∈σ} a_{i}. That is, ⋃_{i∈σ} a_{i} ⫋ ⋂_{j∈τ} V_{r_{j}}.

The next lemma contains the main claim in the proof of Theorem 4.1. It will be used in the inductive step.

Lemma 4.5. Let V be a poset. For every v ∈ V let P_{v}, Q_{v} be posets and φ_{v} : F (P_{v}) → F (Q_{v}) be an embedding such that φ_{v}(I^{bnd} (P_{v})) ⊆ I^{bnd} (Q_{v}).

Let P = ∑{P_{v}: v ∈ V }. For every v ∈ V let Q_{v}^{+} be as in Lemma 4.2(b), and let R = ∑{Q_{v}^{+} : v ∈ V }. For every v ∈ V denote m_{Q_{v}} by m(v). We regard each F (P_{v}) as a subalgebra of F (P) and each F (Q_{v}^{+}) as a subalgebra of F (R). Let Z = {z_{r}: r ∈ R} be the set of generators of F (R), and for every v ∈ V let X_{v} = {x_{p}: p ∈ P_{v}} be the set of generators of P_{v}. For v ∈ V define φ̂_{v} : X_{v} → F (R) as follows:

φ̂_{v}(x_{p}) = z_{m(v)} + φ_{v}(x_{p}),

and let

φ^{+} = ⋃_{v∈V} φ̂_{v}.

Then φ^{+} can be extended to an embedding ψ of F (P) in F (R), and ψ(I^{bnd} (P)) ⊆ I^{bnd} (R).

Proof. Claim 1. Let p ∈ P_{v} and q ∈ P_{w}. Suppose that v < w. Then φ^{+}(x_{p}) < x_{m(w)} < φ^{+}(x_{q}). Proof. Claim 1 follows trivially from Lemma 4.4.

We show that φ^{+} can be extended to a homomorphism from F (P) to F (R). By Theorem 1.2 it suffices to show that if p, q ∈ P and p ≤ q, then φ^{+}(x_{p}) ≤ φ^{+}(x_{q}). Suppose that p ∈ P_{v} and q ∈ P_{w}. Case 1 v = w. By Proposition 4.2(b), φ_{v}^{+} can be extended to an embedding χ : F (P_{v}) → F (Q_{v}^{+}). So

φ^{+}(x_{p}) · φ^{+}(x_{q}) = χ(x_{p}) · χ(x_{q}) = χ(x_{p} · x_{q}) = χ(x_{p}) = φ^{+}(x_{p}). Hence φ^{+}(x_{p}) ≤ φ^{+}(x_{q}). Case 2 v ≠ w. Since p ≤ q, v < w. By Claim 1, φ^{+}(x_{p}) < x_{m(w)} < φ^{+}(x_{q}).

So φ^{+} can be extended to a homomorphism.

We prove that the homomorphism extending φ^{+} is an embedding.

By Proposition 4.3, it suffices to show that: (⋆) if x_{σ,τ}^{φ^{+}} = 0, then x_{σ,τ} = 0. Let σ, τ be finite subsets of P. Denote ρ^{+}(σ, τ) = {v ∈ V : σ ∩ P_{v} ≠ ∅}, ρ^{−}(σ, τ) = {v ∈ V : τ ∩ P_{v} ≠ ∅} and ρ(σ, τ) = ρ^{+}(σ, τ) ∪ ρ^{−}(σ, τ). Claim 2. If ρ^{+}(σ, τ) and ρ^{−}(σ, τ) are antichains and x_{σ,τ}^{φ^{+}} = 0, then x_{σ,τ} = 0. Proof. By induction on |ρ(σ, τ)|. Let n = |ρ(σ, τ)|. We show that the induction holds for n = 1. Suppose that x_{σ,τ}^{φ^{+}} = 0 and that |ρ(σ, τ)| = 1. Then there is v ∈ V such that σ ∪ τ ⊆ P_{v}. By Proposition 4.2(b), there is an embedding χ : F (P_{v}) → F (Q_{v}^{+}) which extends φ_{v}^{+}. Then 0 = x_{σ,τ}^{φ^{+}} = x_{σ,τ}^{φ_{v}^{+}} = χ(x_{σ,τ}). Since χ is an embedding, x_{σ,τ} = 0.

Suppose that the induction hypothesis holds for n, and let σ, τ be such that x_{σ,τ}^{φ^{+}} = 0, ρ^{+}(σ, τ) and ρ^{−}(σ, τ) are antichains and |ρ(σ, τ)| = n + 1. Denote ρ^{+} = ρ^{+}(σ, τ), ρ^{−} = ρ^{−}(σ, τ) and ρ = ρ(σ, τ). Case 1 There is v ∈ ρ^{+} and w ∈ ρ^{−} such that v < w. Let p ∈ P_{v} ∩ σ and q ∈ P_{w} ∩ τ. Then p < q. Hence x_{p} · −x_{q} = 0. So x_{σ,τ} = 0.

For the other cases we need some additional notations. For u ∈ P let b_{u}^{+} := b_{u}^{+}(σ, τ) = ∏{φ^{+}(x_{p}) : p ∈ P_{u} ∩ σ}, c_{u}^{+} := c_{u}^{+}(σ, τ) = ∏{φ^{+}(x_{p}) : p ∈ σ − P_{u}}, b_{u}^{−} := b_{u}^{−}(σ, τ) = ∏{−φ^{+}(x_{q}) : q ∈ P_{u} ∩ τ}, c_{u}^{−} := c_{u}^{−}(σ, τ) = ∏{−φ^{+}(x_{q}) : q ∈ τ − P_{u}}. Case 2 ρ^{+} ∩ ρ^{−} ≠ ∅. Let u ∈ ρ^{+} ∩ ρ^{−}. Then 0 = x_{σ,τ}^{φ^{+}} = (b_{u}^{+} · b_{u}^{−}) · (c_{u}^{+} · c_{u}^{−}). Since ρ^{+} and ρ^{−} are antichains, u is incomparable with every member of ρ − {u}. Let η = Q_{u}^{+} and ζ = ⋃_{v∈ρ−{u}} Q_{v}^{+}. Then for every p ∈ η and q ∈ ζ, p, q are incomparable. Also, b_{u}^{+} · b_{u}^{−} belongs to the algebra generated by {z_{r}^{+} : r ∈ η}, and c_{u}^{+} · c_{u}^{−} belongs to the algebra generated by {z_{r}^{+} : r ∈ ζ}. By Proposition 2.5(c), either b_{u}^{+} · b_{u}^{−} = 0, or c_{u}^{+} · c_{u}^{−} = 0.

Suppose that b_{u}^{+} · b_{u}^{−} = 0. Let σ′ = σ ∩ P_{u} and τ′ = τ ∩ P_{u}, then b_{u}^{+} · b_{u}^{−} = x_{σ′,τ′}^{φ^{+}} and ρ(σ′, τ′) = {u}. So by the induction claim for n = 1, x_{σ′,τ′} = 0. Obviously, x_{σ,τ} ≤ x_{σ′,τ′} = 0. Thus x_{σ,τ} = 0.

Suppose next that c_{u}^{+} · c_{u}^{−} = 0. Let σ′ = σ − P_{u} and τ′ = τ − P_{u}. Then c_{u}^{+} · c_{u}^{−} = x_{σ′,τ′}^{φ^{+}} , ρ^{+}(σ′, τ′) = ρ^{+} − {u}, and ρ^{−}(σ′, τ′) = ρ^{−} − {u}. Clearly, |ρ(σ′, τ′)| = |ρ| − 1. So by the induction hypothesis, x_{σ′,τ′} = 0. Obviously, x_{σ,τ} ≤ x_{σ′,τ′} = 0. Thus x_{σ,τ} = 0. Case 3 Cases 1 and 2 do not occur. So

(1) For every v ∈ ρ^{+} and w ∈ ρ^{−}, either w < v or w and v are incomparable. We shall show that in this case x_{σ,τ}^{φ^{+}} ≠ 0. This will contradict our assumption. We shall thus conclude that Case 3 cannot happen.

We compute x_{σ,τ}^{φ^{+}}. For v ∈ ρ^{+} let σ_{v} = σ∩P_{v}. For w ∈ ρ^{−} let τ_{w} = τ ∩P_{w}. Let v ∈ ρ^{+}. Then x_{σ_{v},∅}^{φ^{+}} = ∏{φ_{v}^{+}(x_{p}) : p ∈ σ_{v}}. So by Lemma 4.2(a)(1),

x_{σ_{v},∅}^{φ^{+}} has the form z_{m(v)} + d_{v}, where d_{v} ∈ I^{bnd} (Q_{v}^{+}). So

(2) x_{σ,∅}^{φ^{+}} = ∏{x_{σ_{v},∅}^{φ^{+}}:v ∈ ρ^{+}} = ∏{x_{z(v)} +d_{v}:v ∈ ρ^{+}} ≥ ∏{z_{m(v)}:v ∈ ρ^{+}}.

Let w ∈ ρ^{−}. Then

x_{∅,τ_{w}}^{φ^{+}} = ∏{−φ_{w}^{+}(x_{q}) : q ∈ τ_{w}} = − ∑{φ_{w}^{+}(x_{q}) : q ∈ τ_{w}}.

Then by Lemma 4.2(a)(2), x_{∅,τ_{w}}^{φ^{+}} has the form −(z_{m(w)} + e_{w}), where e_{w} ∈ I^{bnd} (Q_{w}^{+}). So

(3) x_{∅,τ}^{φ^{+}} = ∏{x_{∅,τ_{w}}^{φ^{+}} : w ∈ ρ^{−}} = ∏{−(z_{m(w)} + e_{w}) : w ∈ ρ^{−}} = −(∑{z_{m(w)} + e_{w}: w ∈ ρ^{−}}) .

For every w ∈ ρ^{−} there is a finite set η_{w} ⊆ Q_{w} such that

(4) e_{w} ≤ ∑{z_{q}: q ∈ η_{w}}. By (2), (3) and (4),

(5) x_{σ,τ}^{φ^{+}} = (x_{σ,∅}^{φ^{+}})·(x_{∅,τ}^{φ^{+}}) ≥ ∏{z_{m(v)}:v ∈ ρ^{+}}−∑{z_{m(w)}+e_{w}:w ∈ ρ^{−}} ≥ ∏{z_{m(v)}:v ∈ ρ^{+}}−∑{z_{q}:w ∈ ρ^{−}, q ∈ {m(w)}∪η_{w}}

= ∏{z_{m(v)}:v ∈ ρ^{+}}·∏{−z_{q}:w ∈ ρ^{−}, q ∈ {m(w)}∪η_{w}}.

For every v ∈ ρ^{+} and q ∈ {m(w) : w ∈ ρ^{−}} ∪ ⋃{η_{w}: w ∈ ρ^{−}}, m(v) ≰ q. So by Proposition 2.5(b), the last expression in (5) is different from 0. So x_{σ,τ}^{φ^{+}} ≠ 0. A contradiction, so Case 3 does not happen.

It follows that x_{σ,τ} = 0. So Claim 2 is proved. Claim 3. Let σ, τ be such that x_{σ,τ}^{φ^{+}} = 0. Then there are σ′ ⊆ σ and τ′ ⊆ τ such that x_{σ′,τ′}^{φ^{+}} = 0 and ρ^{+}(σ′, τ′), ρ^{−}(σ′, τ′) are antichains. Proof. Denote ρ^{+} = ρ^{+}(σ, τ) and ρ^{−} = ρ^{−}(σ, τ). Let η^{+} be the set of minimal elements of ρ^{+}, and η^{−} be the set of maximal elements of ρ^{−}. Let σ′ = σ ∩ ⋃{P_{v}: v ∈ η^{+}} and τ′ = τ ∩ ⋃{P_{v}: v ∈ η^{−}}.

Clearly, ρ^{+}(σ′, τ′) = η^{+} and ρ^{−}(σ′, τ′) = η^{−}. So ρ^{+}(σ′, τ′) and ρ^{−}(σ′, τ′) are antichains.

Let p ∈ σ − σ′. Let v ∈ ρ^{+} be such that p ∈ P_{v}. There is w ∈ η^{+} such that w < v. Let q ∈ P_{w} ∩ σ. By the definition of σ′, q ∈ σ′. By Claim 1 φ^{+}(x_{q}) < φ^{+}(x_{p}). It follows that ∏{φ^{+}(x_{p}) : p ∈ σ′} ≤ ∏{φ^{+}(x_{p}) : p ∈ σ}. Since σ′ ⊆ σ, ∏{φ^{+}(x_{p}) : p ∈ σ′} ≥ ∏{φ^{+}(x_{p}) : p ∈ σ}. So x_{σ′,∅}^{φ^{+}} = x_{σ,∅}^{φ^{+}}.

An identical argument shows that x_{∅,η′}^{φ^{+}} = x_{∅,η}^{φ^{+}}. So x_{σ′,η′}^{φ^{+}} = x_{σ,η}^{φ^{+}} = 0. We have proved Claim 3.

We now prove (⋆). Suppose that x_{σ,τ}^{φ^{+}} = 0. Let σ′, τ′ be as assured by Claim 3. By Claim 2, x_{σ′,τ′} = 0. Since σ′ ⊆ σ and τ′ ⊆ τ, x_{σ,τ} = 0. We have proved (⋆).

We have shown that φ^{+} fulfills the condition of Proposition 4.3. So the homomorphism ψ, which extends φ^{+} is an embedding of F (P) into F (Q^{+}). It remains to show that ψ(I^{bnd} (P)) ⊆ I^{bnd} (Q^{+}). Since {x_{p}: p ∈ P } generates I^{bnd} (P), it suffices to show that for every p ∈ P, ψ(x_{p}) ∈ I^{bnd} (R). But this follows from the definition of φ^{+}. Proof of Theorem 4.1. We prove by induction on α, that for every P ∈ H_{α}, F (P) is well-generated. There is nothing to prove for α = 0 and for limit ordinals. Suppose that the claim is true for α, and let P ∈ H_{α+1} −H_{α}.

By Proposition 2.5(a), F (P^{∗}) ≅ F (P). It follows from the definition of H_{α} that if P ∈ H_{α+1} − H_{α}, then P^{∗} ∈ H_{α+1} − H_{α}.

Hence we may assume that P = ∑{P_{v}: v ∈ V }, where V is a wellordered poset, and for every v ∈ V, P_{v} ∈ H_{α}. By the induction hypothesis, for every v ∈ V there is a well-ordered poset Q_{v} and an embedding φ_{v} of F (P_{v}) in F (Q_{v}) such that φ_{v}(I^{bnd} (P_{v})) ⊆ I^{bnd} (Q_{v}).

We may apply Lemma 4.5 to V, {P_{v}: v ∈ V }, {Q_{v}: v ∈ V } and {φ_{v}: v ∈ V }.

Let Q_{v}^{+}, v ∈ V and ψ be as assured by that lemma and R = ∑_{v∈V} Q_{v}^{+}. Since Q_{v}^{+} is obtained from Q_{v} by adding only one element, Q_{v}^{+} is well-ordered. Recall that V is well-ordered.

It is easy to check that if in a lexicographic sum the index poset is wellordered and each summand is well-ordered, then the sum is well-ordered. So R is well-ordered.

Finally, ψ is an embedding of F (P) in F (R), and ψ(I^{bnd} (P)) ⊆ I^{bnd} (R). So R and ψ are as required. Proof of Theorem 1.4. Let ⟨P, ≤⟩ be narrow and scattered. By Theorem 1.5 we may assume that there is ≤′ ⊆ ≤ such that ⟨P, ≤′⟩ ∈ H. By Theorem 1.6, F (⟨P, ≤′⟩) is embeddable in F (W), where W is a well-ordered poset, and obviously F (⟨P, ≤⟩) is a homomorphic image of F (⟨P, ≤′⟩) . Remark. The Boolean algebra F (ω^{∗} · ω_{1}) is not a homomorphic image of a poset algebra of a well-ordered poset. This is proved in [ABK].

References

[AB1] U. Abraham and R. Bonnet: Every superatomic subalgebra of an interval algebra is embeddable in an ordinal algebra. Proc. of Amer. Math. Soc., 115(3), 1992, pp. 585-592.

[AB2] U. Abraham and R. Bonnet: A generalization of Hausdorff theorem on scattered poset, Fund. Math., 159, 1999, pp. 51-69.

[ABK] U. Abraham, R. Bonnet and W. Kubiś: On well-ordered poset algebras, preprint.

[BR] R. Bonnet and M. Rubin: On well-generated Boolean algebras, Ann. Pure Appl. Logic, 105, 2000, pp. 1-50.

[F] R. Fraïssé: Theory of relations. Revised edition. With an appendix by Norbert Sauer. Studies in Logic and the Foundations of Mathematics, North-Holland Publishing Co., Amsterdam, 2000. ii+451 pp.

[H] G. Higman: Ordering by divisibility in abstract algebras, Proc. Lond. Math. Soc., 2, 1952, pp. 326-336.

[Ko] S. Koppelberg: Handbook on Boolean Algebras, Vol. 1, Ed. J.D. Monk, North Holland 1989.

[P] M. Pouzet: On the set of initial intervals of a scattered poset satisfying FAC, 1981, Private communication. Annonced in Ordered sets (Banff, Alta., 1981), Edit. I. Rival, NATO Adv. Sci. Inst., Ser. C Math. Phys. Sci., Reidel, Dordrecht, 83, 1982, p. 847.

E-mails U. Abraham: abraham@math.bgu.ac.il R. Bonnet: bonnet@in2p3.fr M. Rubin: matti@math.bgu.ac.il W. Kubiś: kubis@math.bgu.ac.il