On zero-dimensional continuous images of compact ordered spaces

Lutz Heindorf

February 20, 1995

Throughout this paper X denotes a compact and zero-dimensional Hausdorff space. We shall be concerned with

Theorem 0.1 The following assertions are equivalent.

(A) X is the continuous image of a compact ordered space.

(B) X is the continuous image of a zero-dimensional compact ordered space.

(C) X has a T_{0}-separating cross-free family of clopen sets.

(D) X has a T_{0}-separating non-Archimedian family of clopen subsets.

(E) X can be embedded as a strong T -set into a dendron.

(F) X can be embedded into a dendron.

(G) X has a cross-free closed subbase.

(H) There is a continuous multiplication · on X making it a topological semilattice, i. e., associative, commutative, and idempotent, such that x · y · z ∈ {x · y , x · z} for all x, y, z ∈ X.

Although probably never arranged in single file, the above equivalences are more or less known. The implication G ⇒ F is due to van Mill and Wattel (cf. [7, theorem 6.6]) and does not depend on X being zero-dimensional (but the general case it is much harder to prove).

F ⇒ A follows from the fact that every dendron is a continuous image of a compact ordered space (a result due independently to J. L. Cornette [3] and A. E. Brouwer [2] see also [7, theorem 3.3]).

The implication A ⇒ E is due to J. Nikiel [9]. S. Purisch recently observed E ⇒ D, which he combined with Nikiel’s result to obtain A ⇒ D (cf. [12]), thus solving a problem posed by van Douwen in [4] and in Boolean algebraic form also by Koppelberg and Monk in [6].

The equivalence of F and H does not come out of the blue either. But it is not clear how to get it directly from L. E. Ward’s result in [11].

In this note we offer new proofs of most of the implications according to the following schema.

E ⟹^{t} F ⟹ G ⟹^{3} C ⟹^{2} D ⟹^{5} H ⟹^{2} D ⟹^{5} E

∥ ∨

B ⟹^{t} A ⟹^{2} B ⟹^{4} C,

where trivial implications are marked by ⟹^{t} and ⟹^{i} means that the implication is proved in section i.

The two unmarked implications will not be proved here, because we have no simplifications to offer. In [7] the very first thing van Mill and Wattel prove about dendrons is the fact that they have cross-free closed subbases. From this the implication F ⇒ G is immediate. The second exception is D ⇒ B. Its Boolean algebraic reformulation (see section 4 below) is proved very elegantly in [6, theorem 3.1].

The author wishes to thank J. D. Monk for making [8] available to him.

1 Definitions

Before starting we recall the definitions of the less common notions involved in the statement of the theorem.

A family C of subsets of some set X will be called

• T_{0}–separating if for all x ≠ y in X there is some c ∈ C such that x ∈ c ⇔ y ∉ c.

• cross-free iff for all c, d ∈ C

(cf) c ⊆ d or d ⊆ c or c ∩ d = ∅ or c ∪ d = X.

• non-Archimedian (following [4], the phrase rank 1 family is also popular) if for all c, d ∈ C

(nA) c ⊆ d or d ⊆ c or c ∩ d = ∅.

With the obvious modification in mind we shall use the phrases ‘cross-free’ and ‘non-Archimedian’ also for subfamilies of Boolean algebras.

A dendron is a compact connected Hausdorff space D such that for all distinct x, y ∈ D there is a point z ∈ D which separates x and y, i.e., D \ {z} = U ∪ V for some U and V that are disjoint, open, and such that x ∈ U, y ∈ V.

Finally, a subset X of a dendron D is called a strong T -set if it is closed and each connected component of D \ X is homeomorphic with the open unit interval (0, 1). A good reference for dendrons is the already cited survey [7]. To understand the proofs below, no theory is needed however.

2 The easy implications

Proof of A ⟹ B. Let f : K → X be a continuous surjection, where K is an ordered compact space. By Zorn’s Lemma we can replace K by a minimal closed subspace K′ such that f(K′) is still the whole of X. A well-known elementary result tells us that K′ is also an ordered space.

Let L be a non-empty connected subset of K′. We want to show that L consists of one point only. But otherwise, it would contain a non-empty, densely ordered closed interval, [l_{1}, l_{2}] say. As X is zero-dimensional, the connected set f([l_{1}, l_{2}]) consists of only one point, hence f(K′ \ (l_{1}, l_{2}) ) = f(K′) = X, contradicting minimality. So K′ is totally disconnected, hence zero-dimensional. □

The Proof of C ⟹ D. starts with a

Claim If C is cross-free then C ∪ {X \ c : c ∈ C}is also cross-free.

Indeed, rewriting condition (cf) as

c ∩ (X \ d) = ∅ or d ∩ (X \ c) = ∅ or c ∩ d = ∅ or (X \ c) ∩ (X \ d) = ∅

it becomes clear that the sets c and d and their complements enter symmetrically, from which the claim follows.

Now let C be a cross-free T_{0}-separating family of clopen sets and fix any point x ∈ X. By the claim, T = {c ∈ C : x ∉ c} ∪ {X \ c : x ∈ c ∈ C} is also cross-free and, as x ∉ t for all t ∈ T , even non-Archimedian. A moment’s reflection shows that T remains T_{0} separating. □

Proof of H ⟹ D. Let (X, ·) be a topological semilattice such that x · y · z ∈ {x · y , x · z} for all x, y, z ∈ X. We need a simple fact from topological algebra, which the reader probably knows, at least for groups.

Lemma 2.1 (Numakura’s Theorem [10]) Every compact and zero-dimensional semigroup is profinite.

For semilattices, i.e., commutative and idempotent semigroups, the assertion can be strengthened using the following

Claim For each pair of distinct elements h_{1}, h_{2} of the finite semilattice H there is a homomorphism φ : H → {0, 1} (multplicative) such that φ(h_{1}) ≠ φ(h_{2}).

To see this, assume without loss h_{1} · h_{2} ≠ h_{1} and put

φ(h) = \{ \begin{matrix}1, & h_{1} · h = h_{1} \\ 0, & otherwise \end{matrix}

Then a straight-foreward verification (using associativity, commutativity, and idempotency) shows that φ is as desired.

Combining the claim with Numakura’s Theorem we conclude that any pair of points of X is separated by a continuous homomorphism φ : X → {0, 1}. In other words, the family T consisting of all φ^{−1}(1) is T_{0}-separating. It remains to check that T is non-Archimedian.

Let φ and ψ be continuous homomorphisms X → {0, 1}. Aiming at a contradiction we assume

φ^{−1}(1) ⊈ ψ^{−1}(1) and ψ^{−1}(1) ⊈ φ^{−1}(1) and ψ^{−1}(1) ∩ φ^{−1}(1) ≠ ∅.

This yields points x, y, z ∈ X such that

φ(y) = 1, ψ(y) = 0 and ψ(z) = 1, φ(z) = 0 and φ(x) = ψ(x) = 1.

Then φ(x·y·z) = ψ(x·y·z) = 0, but φ(x·y) = 1 and ψ(x·z) = 1, contradicting x·y·z ∈ {x·y , x·z} □

3 Closed versus clopen

Assume that S is a cross-free closed subbase of X. We show that S contains enough clopen elements to separate the points of X, thus establishing the implication G ⇒ C.

There is no harm in assuming X ∈ S. But then

Claim the finite unions of pairwise disjoint elements of S constitute a closed base of X.

To see this, consider any finite union S_{1} ∪ … ∪ S_{n} of elements of S. If S_{i} ⊆ S_{j} for some i ≠ j, we may drop S_{i}. If S_{i} ∪ S_{j} = X we can replace the whole union by X ∈ S. Having purged it this way, the union has become disjoint.

Now let x ≠ y be given and choose a clopen set A such that x ∈ A ∌ y. As A is closed, it can be represented as an intersection of basic closed sets. As A is also open, a finite intersection

A = ⋂_{i=1}^{n}( S_{1}^{i} ∪ … ∪ S_{m_{i}}^{i} )

is sufficient (by compactness). By the claim, we can further assume that S_{p}^{i} ∩ S_{q}^{i} = ∅ for all i and 1 ≤ p < q ≤ m_{i}. Using the distributivity of the set-theoretic operations we rewrite

A = ⋃_{f} ⋂_{i=1}^{n} S_{f(i)}^{i},

where f runs over all mappings {1, …, n} ⟶ {1, …, max m_{i} } such that f(i) ≤ m_{i} for all i.

The sets ⋂_{i=1}^{n} S_{f(i)}^{i} are, obviously, closed. They are also pairwise disjoint, for f ≠ g means f(j) ≠ g(j) for some j, hence

⋂_{i=1}^{n} S_{f(i)}^{i} ∩ ⋂_{i=1}^{n} S_{g(i)}^{i} ⊆ S_{f(j)}^{j} ∩ S_{g(j)}^{j} = ∅.

As their union A is open, all ⋂_{i=1}^{n} S_{f(i)}^{i} are clopen and one of them separates x and y.

To simplify notation, we now consider a clopen intersection B = S_{1} ∩…∩S_{n} of (without loss) pairwise incomparable elements of S, such that x ∈ B ∌ y There is no loss in further assuming that x ∈ S_{1} ∌ y. Our aim is to show that S_{1} is actually clopen. For j = 2, …, n we must have S_{1} ∪ S_{j} = X, for x ∈ S_{1} ∩ S_{j} ≠ ∅ and comparability is excluded by assumption. It follows that X = S_{1} ∪ (S_{2} ∩ … ∩ S_{n}), hence, denoting (S_{2} ∩ … ∩ S_{n}) by C,

X = S_{1} ∪ C = [ S_{1} \ (S_{1} ∩ C) ] ∪ [ S_{1} ∩ C] ∪ [ C \ (S_{1} ∩ C) ].

Given that S_{1} ∩ C = B is clopen, this is a decomposition of X into three disjoint closed sets. It follows that they are all clopen and so is

S_{1} = [ S_{1} \ (S_{1} ∩ C) ] ∪ [ S_{1} ∩ C],

as was to be shown. □

4 Switching to Boolean algebras

In this section we prove the implication B ⟹ C using the language of Boolean algebras.

If f : K → X is a continuous surjection, then the assignment a ↦ f^{−1}(a) defines an embedding Clop(X) → Clop(K) of the Boolean algebras consisting of all clopen subsets of X, resp. K, under the set-theoretic operations.

The fact that K is an ordered space translates to Clop(K) being a so-called interval algebra. This fact and much background on interval algebras can be found in section 15 of [5].

Each point x ∈ X gives rise to the ultrafilter {a ∈ Clop(X) : x ∈ a} of Clop(X). It follows that in order to establish B ⟹ C it is sufficient to prove

Proposition 4.1 Every subalgebra of an interval algebra contains a cross-free subcollection which separates its ultrafilters.

Before embarking on that proof we introduce some notation.

Let R be a linear ordering with least element 0. The interval algebra on R, to be denoted by Intalg(R), is defined to be the subalgebra of the power-set algebra on R generated by the collection of all half-open intervals [r, s) = {t ∈ R : r ≤ t < s}. As Intalg(R) is an algebra of sets, we use ∩, ∪ and \ to denote the Boolean operations and ∅ resp. R for the zero- resp. unit element.

It is well-known that each non-zero element b of Intalg(R) has a unique representation of the form

b = [r_{0}, r_{1}) ∪ … ∪ [r_{2n}, r_{2n+1}),

where n ≥ 0 and 0 ≤ r_{0} < r_{1} < r_{2} < … < r_{2n} < r_{2n+1} ≤ ∞, with the obvious meaning of [r, ∞). The elements r_{0}, r_{2}, …, r_{2n} will be called the left end-points of b, whereas r_{1}, r_{3}, …, r_{2n+1} are its right end-points. By rel( b) we denote the set {r_{0}, r_{1}, …, r_{2n}, r_{2n+1}} \ {0, ∞} (the relevant points of b). In particular, for the unit element we have rel(R) = rel([0, ∞)) = ∅. By definition, we put rel(∅) = ∅.

We need the following

Observations 4.2 concerning a, b ∈ Intalg(R) and m ∈ R.

(1) rel(R \ b) = rel(b)

(2) If m is a left end-point of a and m ∈ b, then rel(a \ b) ⊆ (rel(a) ∪ rel(b)) \ {m}.

(3) If m is a left end-point of a and m ∉ b, then rel(a ∩ b) ⊆ (rel(a) ∪ rel(b)) \ {m}.

Verifications are very simple and, therefore, omitted.

Now we prove the proposition. Let R be as before and let B be a subalgebra of Intalg(R).

Claim We can find a set M ⊆ R \ {0} such that

(m1) rel(b) ∩ M ≠ ∅ for all b ∈ B \ {∅, R}, and

(m2) for each m ∈ M there exists exactly one b ∈ B such that m ∈ B and rel(b) ∩ M = {m}.

To see this, consider the collection N of all N ⊆ R\{0} for which (m1) holds. Containing R\{0}, the collection is not empty. Moreover, N is closed under intersections of descending chains (as rel(b) is finite for each b). So, by Zorn’s Lemma, there is a minimal element, M say, in N.

Automatically M satisfies (m1). To see that it also satisfies (m2), we fix m ∈ M. By minimality, M \ {m} ∉ N. So rel(b) ∩ (M \ {m}) = ∅, hence rel(b) ∩ M = {m} for some b ∈ B. Switching to R \ b if necessary, we can additionally assume that m is a left end-point of b, i.e. m ∈ b. So, b is as desired in (2). To see that b is unique, consider a second element c with the same properties. By 4.2(2), m is not an end-point of either b \ c or c \ b. It follows that (rel(b \ c) ∪ rel(c \ b)) ∩ M = ∅, which, by (1), implies b \ c = c \ b = ∅, i.e. b = c.

We now fix M as in the claim and denote by c_{m} the unique element of B such that m ∈ c_{m} and rel(c_{m}) ∩ M = {m}. Notice that, by (m2) and 4.2(1),

(m3) R \ c_{m} is the unique element of B such that m ∉ R \ c_{m} and rel(R \ c_{m}) ∩ M = {m}.

In order to see that C_{M} = {c_{m} : m ∈ M} is cross-free, consider m ≠ n ∈ M.

If m ∈ c_{n}, then, by 4.2(2), rel(c_{m} \ c_{n}) ∩ M ⊆ {n}. If rel(c_{m} \ c_{n}) ∩ M = ∅, then, by (m1), c_{m} ⊆ c_{n}.

If rel(c_{m} \ c_{n}) ∩ M = {n}, then, from n ∉ c_{m} \ c_{n} we get c_{m} \ c_{n} = R \ c_{n}, by (m3). It follows that c_{m} ∪ c_{n} = R.

If n ∈ c_{m}, we argue symmetrically.

If m ∉ c_{n} and n ∉ c_{m}, then, by 4.2(3), neither m nor n is in rel(c_{m} ∩ c_{n}), so c_{m} ∩ c_{n} = ∅, by (m1).

Finally, we establish that C_{M} separates the ultrafilters of B. Aiming at a contradiction we assume that p ≠ q are ultrafilters of B that can not be separated by an element of C_{M}. Among the elements of B that separate p and q we choose some b such that the (finite) set rel(b) ∩ M is minimal. Take m ∈ rel(b) ∩ M ( which is possible, by (m1)). Without loss we can assume that m is a left end-point of b (otherwise we replace b by R \ b) and that b ∈ p, b ∉ q (otherwise we interchange p and q). As c_{m} does not separate p and q, we either have c_{m} ∈ p, c_{m} ∈ q or c_{m} ∉ p, c_{m} ∉ q.

In the first case we put d = c_{m} \ b. Then d ∉ p, d ∈ q and, by 4.2(2),

rel(d) ∩ M ⊆ [(rel(c_{m}) ∪ rel(b)) \ {m}] ∩ M ⊆ (rel(b) ∩ M) \ {m},

which contradicts the choice of b. In the second case we get the same contradiction using d = b\c_{m}. □

We close this section with a few further remarks adressed to Boolean algebraists.

The class of pseudo-tree Boolean algebras was introduced by Monk. Many information about these algebras can be found in [6]. The following definition is equivalent (cf. [6, theorem 2.3]) to the original one, but more convenient for our discussion.

Definition 4.3 A Boolean algebra is called a pseudo-tree algebra if it has a non-Archimedian set of generators.

The name can be motivated by the fact that each non-Archimedian subset T of a Boolean algebra is a (downward growing) pseudo-tree under the Boolean partial order, i.e., such that for each t ∈ T the subset {s ∈ T : s ≥ t} is a chain. (To make T a(n upside down) tree, these chains have to be (reversely) well-ordered.)

Recalling the implication C ⇒ D from section 2 and the fact that a subset of Clop(X) is T_{0} separating iff it generates Clop(X) as a Boolean algebra, we can reformulate the above proposition as follows.

Proposition 4.4 Every subalgebra of an interval algebra is a pseudo-tree algebra.

In this form the result answers a question of Koppelberg and Monk in [6]. For the special case of superatomic Boolean algebras (corresponding to scattered spaces) it was earlier obtained by Bonnet [1] with a stronger conclusion.

The converse of 4.4 is also true and this yields the implication D ⟹ B of our main theorem. As we mentioned already in the introduction, we can not improve upon the very nice demonstration given in [6, theorem 3.1]. The fact itself was known to van Douwen (cf. [4, theorem 1.3]) and also proved by M. Bekkali in his thesis.

¿From 4.4 and its converse we immediately get

Corollary 4.5 Every subalgebra of a pseudo-tree algebra is also a pseudo-tree algebra.

5 The embedding in a dendron

Here we prove the implication D ⇒ E. A few additional remarks will then yield D ⇒ H.

Let T be as in D, i.e. a family of clopen subsets of X such that

(t1) for distinct x, y ∈ X there exists t ∈ T such that x ∈ t iff y ∉ t, and

(t2) t ⊆ s or s ⊆ t or s ∩ t = ∅ for all s, t ∈ T

Claim We can assume that T has two additional properties, namely

(t3) ∅, X ∉ T and

(t4) t_{1}△t_{2}△ … △t_{n} ≠ ∅ (symmetric difference) for all pairwise distinct t_{1}, t_{2}, … t_{n} ∈ T ∪ {X}.

There is no problem with (t3), of course. If (t4) is not satisfied by our first choice for T , we replace it by a subset, T′ say, which is maximal with respect to having properties (t3) and (t4). The existence follows from Zorn’s Lemma again. To see that T′ still satisfies (t1), we consider x ≠ y in X. There is some t ∈ T separating x and y. If t ∈ T′, we are done. Otherwise, by maximality,

t△t′_{1}△ … △t′_{n} = ∅ hence t = t′_{1}△ … △t′_{n}

for suitable pairwise distinct t′_{i} ∈ T′ ∪ {X}. It follows that one of the t′_{i} ∈ T′ separates x and y, for, otherwise t would not.

Properties (t3) and (t4) have the consequence that

(t5) for t_{1}, …, t_{n}, s_{1}, …, s_{m} ∈ T ∪ {X} the inclusion t_{1} ∩ … ∩ t_{n} ⊆ s_{1} ∪ … ∪ s_{m} is possible only if either t_{i} ∩ t_{j} = ∅ or t_{i} ⊆ s_{j} for some i, j.

To see this, assume that t_{i} ∩ t_{j} ≠ ∅ for all i, j. Then, by (t2), the t_{i} are pairwise comparable and their intersection reduces to one term, t_{1} say. Among the elements s_{1}, …, s_{m} we first drop those that do not intersect t_{1} and then those that are not maximal. The remaining elements, s_{1}, …, s_{k} say, still cover t_{1} and are pairwise disjoint, by (t2). Therefore, s_{1} ∪ … ∪ s_{k} = s_{1}△ … △s_{k}.

Assuming t_{1} ⊈ s_{j} for all j, we get t_{1} ⊇ s_{j} for all j. Consequently, t_{1} ⊇ s_{1} ∪ … ∪ s_{k} = s_{1}△ … △s_{k} ⊇ t_{1}, contradicting (t4). After these preparations we describe the required dendron. It will be called D and is the subspace of the Tychonoff cube [0, 1]^{T} consiststing of all φ : T → [0, 1] such that (d1) s ⊆ t implies φ(s) ≤ φ(t), (d2) s ∩ t = ∅ implies that either φ(s) = 0 or φ(t) = 0, and (d3) there is at most one value t ∈ T such that φ(t) ∉ {0, 1}. The desired properties will be established in a series of claims. Some routine verifications will be skipped.

(1) D is closed in the cube, hence compact.

(2) X is homeomorphic with the subspace Y of D consisting of all those φ : T → {0, 1} which satisfy (d1) and (d2), i.e., those elements of D that take only the values 0 and 1.

To see this put F (x)(t) = \{ \begin{matrix}1, & x ∈ t \\ 0, & x ∉ t \end{matrix} and verify that F : X → Y is the desired homeomorphism. Condition (t1) yields injectivity, whereas (t5) is needed to get surjectivity. Next we need names for some elements of D. For t ∈ T and θ ∈ [0, 1] we define φ_{t,θ} by setting

φ_{t,θ}(s) = \{ \begin{matrix}1, & t ⊂ s (strong inclusion) \\ θ, & s = t \\ 0, & t ⊈ s. \end{matrix}

and put J_{t} = {φ_{t,θ} : θ ∈ (0, 1)}.

(3) φ_{t,θ} ∈ D

(4) If φ ∈ D and φ(t) = θ ∈ (0, 1), then φ = φ_{t,θ}. It is clear that φ(t) = φ_{t,θ}(t). For s ≠ t, both functions take only values in {0, 1}.

Case 1. t ⊂ s, i.e., φ_{t,θ}(s) = 1.
Then, by (d1), 0 < θ = φ(t) ≤ φ(s), hence φ(s) = 1.
Case 2. t ⊈ s, i.e., φ_{t,θ}(s) = 0.
If s ∩ t = ∅, then φ(s) = 0, for, otherwise, by (d2), 0 = φ(t) = θ > 0.
If s ⊂ t, then φ(s) ≤ φ(t) = θ < 1, which again implies φ(s) = 0.

As an immediate consequence of (4) we get

(5) D \ Y = ⋃_{t∈T} J_{t}. Given that J_{t} is, obviously, homeomorphic to (0, 1), we get the ‘strog T-set’ part of E. In order to prove the connectedness of D we need two preparatory claims.

(6) Every non-empty clopen subset of D intersects Y. Indeed, let ∅ ≠ A ⊆ D be clopen and take φ ∈ A. If φ ∈ Y, we are done. Otherwise, by (5), φ = φ_{t,θ} ∈ J_{t} for some t and θ. As J_{t} is connected, the clopen set A cannot properly intersect its closure \overline{J_{t}}. So A ⊇ \overline{J_{t}} ∋ φ_{t,0} ∈ Y. Let ζ ∈ Y denote the mapping T → {0, 1} constantly taking the value zero.

(7) Every non-empty clopen subset of Y which does not contain ζ can be written as a finite union of sets of the form {φ ∈ Y : φ(t) = 1; φ(s_{1}) = … = φ(s_{m}) = 0}, where t ⊈ s_{i} (and m = 0 is allowed).

Indeed, by (d1) and (d2) the a priori possible more general form

{φ ∈ Y : φ(t_{1}) = … = φ(t_{n}) = 1; φ(s_{1}) = … = φ(s_{m}) = 0}

reduces to the given one. At least one t must be present because of ζ.

(8) D is connected.

Assume not. Then there is a non-empty clopen subset A of D, which does not contain ζ. By (6), its intersection with Y is non-empty. By (7), we get

A ∩ Y = ⋃_{i=1}^{n}{φ ∈ Y : φ(t_{i}) = 1; φ(s_{1}^{i}) = … = φ(s_{m_{i}}^{i}) = 0}.

Let t be maximal among t_{1}, …, t_{n}. Then φ_{t,1} ∈ A ∩ Y but φ_{t,0} ∉ A ∩ Y. It follows that A splits the connected set J_{t}, which is impossible.

(9) D is a dendron

Let α and β be distinct points in D. Without loss, we can assume α(t) < β(t) for some t ∈ T . Choosing θ such that α(t) < θ < β(t), we obtain, by (4),

D \ {φ_{t,θ}} = {φ ∈ D : φ(t) < θ } ∪ {φ ∈ D : φ(t) > θ },

which is the desired decomposition. The implication D ⇒ E is proved. □

To prove D ⇒ H it will be sufficient to introduce the required semilattice structure on the space Y ⊆ {0, 1}^{T} constructed above. The operation · of pointwise multiplication on the cube {0, 1}^{T} is, obviously, associative, commutative, and idempotent.

Moreover, Y is closed under ·. Indeed, if φ, ψ : T → {0, 1} are in Y, i.e., satisfy (d1) and (d2), then the same is true of their pointwise product φ · ψ.

It remains to verify that for all φ, ψ, χ satisfying (d1) and (d2) either φ · ψ · χ = φ · ψ or φ · ψ · χ = φ · χ. If that were not true, we could find s and t in T such that

φ(s) = 1, ψ(s) = 1, χ(s) = 0, and φ(t) = 1, ψ(t) = 0, χ(t) = 1.

But then neither s ∩ t = ∅ (because of φ) nor s ⊆ t (because of ψ) nor t ⊆ s (because of χ). This contradicts T being non-Archimedian. □

References

[1] Bonnet, R. Superatomic subalgebras of an interval algebra. Preprint no. 91- 001, Universität Bielefeld 1991.

[2] Brouwer, A. E. A compact tree-like space is the continuous image of an ordered continuum. Report ZW 33, Math. Centre, Amsterdam 1974.

[3] Cornette, J.L. ‘Image of a Hausdorff arc’ is cyclically extensible and reducible. Trans. Amer. Math. Soc. 199 (1974), 253–267.

[4] van Douwen, E. K. Small tree algebras with nontree subalgebras. Topology and its applications 51 (1993), 173–181.

[5] Koppelberg, S. General theory of Boolean algebras. vol 1. of: J. D. Monk (ed.) Handbook of Boolean algebras. North-Holland, Amsterdam 1989

[6] Koppelberg, S., Monk, J. D. Pseudo-trees and Boolean algebras. Order 8 (1992), 359–374.

[7] van Mill, J., Wattel, E. Dendrons. 59 -82 in: H. R. Bennet, D. J. Lutzer (eds.), Topology and order structures, part I, Mathematical Centre Tracts vol. 142, Amsterdam 1981.

[8] Monk, J. D. Notes on pseudo-tree algebras. informal notes, September 1994.

[9] Nikiel, J. Orderability properties of a zero-dimensional space which is a continuous image of an ordered compactum. Topology and its applications 31 (1989), 269–276.

[10] Numakura, K. Theorems on compact totally disconnected semigroups and lattices. Proceedings Amer. Math. Soc. 8(1957), 623–636.

[11] Mobs, trees and fixed points. Proceedings Amer. Math. Soc. 8(1957), 798–804.

[12] Purisch, S. In Topology proceedings 17(1994), Problem section, p. 412.

Lutz Heindorf Freie Universität Berlin

2. Mathematisches Institut Arnimallee 3 D - 14195 Berlin

heindorf @ math.fu-berlin.de