From the files of KP 1996/11/2

A UNIVERSAL CONTINUUM OF WEIGHT ℵ

KLAAS PIETER HART AND ALAN DOW

Abstract. We prove that every continuum of weight ℵ_{1} is a continuous image of the Čech-Stone-remainder ℝ^{∗} of the real line. It follows that under CH the remainder of the half line [0, ∞) is universal among the continua of weight �.

We complement this result by showing that 1) under MA every continuum of weight less than � is a continuous image of ℝ^{∗}, 2) in the Cohen model the long segment of length ω_{2} +1 is not a continuous image of ℝ^{∗}, and 3) PFA implies that 𝕀_{u} is not a continuous image of ℝ^{∗}, whenever u is a �-saturated ultrafilter.

Introduction

In [13] Parovičenko proved that every compact Hausdorff space of weight ℵ_{1} is a continuous image of the space ω^{∗} (the Čech-Stone remainder of the discrete space ω). It follows that under CH the space ω^{∗} is a universal compact space of weight �, universal in the sense of onto mappings rather than embeddings.

The purpose of this paper is to prove a similar result for ℍ^{∗}, the Čech-Stone remainder of the half line ℍ = [0, ∞). As ℍ^{∗} is compact and connected (a continuum) the following theorem is the proper analogue of Parovičenko’s result.

Theorem 1. Every continuum of weight ℵ_{1} is a continuous image of ℍ^{∗}.

We present proofs of our theorem using each of three quite different approaches to Parovičenko’s theorem from the literature that have proven valuable.

It is clear that new ideas are needed to prove Theorem 1 since the fact that ω^{∗} is zero-dimensional appears to be so essential in each of the proofs of Parovičenko’s theorem.

We now give brief outlines of the proofs of Parovičenko’s theorem and we indicate how our proofs parallel these arguments; more details, including definitions, will appear later in the paper. We shall finish this introduction with some remarks on universal spaces.

Two algebraic proofs. The first two proofs of Parovičenko’s theorem use Stone’s duality, see [15], between compact zero-dimensional spaces and Boolean algebras. One first uses a theorem of Alexandroff, from [3], to find, given the compact space X of weight ℵ_{1}, a compact zero-dimensional space Y of weight ℵ_{1} and a continuous map from Y onto X, thus reducing the problem to the case of zero-dimensional spaces. Stone Duality then tells us that finding a continuous map from ω^{∗} onto Y is equivalent to finding an embedding of the Boolean algebra ℂ of clopen subsets of Y into the Boolean algebra 𝔹 of clopen subsets of ω^{∗}.

This gives rise to two proofs of Parovičenko’s theorem. In the first one constructs the desired embedding by transfinite induction; this is Parovičenko’s original proof.

1991 Mathematics Subject Classification. Primary: 54F15. Secondary: 03E35, 04A30, 54G05.

Key words and phrases. Parovičenko’s theorem, universal continuum, remainder of [0, ∞), ω_{1}-saturated model, elementary equivalence, Continuum Hypothesis, Cohen reals, long segment, Martin’s Axiom, Proper Forcing Axiom, saturated ultrafilter.

The research of the first author was supported by Netherlands Organization for Scientific Research (NWO) — Grant R 61-322.

The second proof recognizes that Parivičenko’s theorem is in fact a special case of a well-known result about ℵ_{1}-saturated models.

Our parallels of these proofs are naturally based on Wallman’s generalization of Stone’s duality to the class of distributive lattices and amount to embedding, for the given continuum, a suitable sublattice of its lattice of closed sets into the lattice of closed subsets of ℍ^{∗}. The key new idea is to extract the hidden model-theoretic aspects of the abovementioned proofs.

A topological proof. The third proof is of a more topological nature, it appears in [5]. It starts by writing the compact space X as an inverse limit \underleftarrow{lim} X_{α} of an ω_{1}-sequence of compact metrizable spaces. The continuous map from ω^{∗} onto X is obtained as the limit of a family of continuous onto maps f_{α} : ω^{∗} ↠ X_{α}. These maps are constructed by transfinite induction and using the following lemma.

Lemma 1. Let M and N be compact metrizable spaces and let f : ω^{∗} ↠ N and g : M ↠ N be continuous surjections. Then there is a continuous surjection h : ω^{∗} ↠ M such that g ∘ h = f.

Our third proof will be along these lines and the main problem is that the above lemma is false if we simply replace ω^{∗} by ℍ^{∗} and assume that M and N are continua — see Example 3.2. Intriguingly, if g is induced by an appropriate elementary embedding then the analogue of Lemma 1 will hold.

Universal spaces. As with Parovičenko’s theorem, Theorem 1 becomes a result about universal spaces if one assumes the Continuum Hypothesis (CH).

Theorem 2 (CH). The space ℍ^{∗} is a universal continuum of weight �.

It is of interest to note that in the class of compact metrizable spaces the situation is somewhat different from what we just found for spaces of weight �. On the one hand the Cantor set is a universal compact metrizable space; this is a theorem of Alexandroff [2] and Hausdorff [8]. On the other hand, in [20] Waraszkiewicz constructed a family of plane continua such that no metric continuum maps onto all of them, hence there is no universal metric continuum.

Part of the balance is restored by the following theorem, due to Aarts and Van Emde Boas [1]; as we shall need this theorem and its proof later on we provide a short argument.

Theorem 3. The space ℍ^{∗} maps onto every metric continuum.

Proof. Consider a metric continuum M and assume it is embedded into the Hilbert cube Q = [0, 1]^{∞}. Choose a countable dense subset A of M and enumerate it as {a_{n} : n ∈ ω}. Next choose, for every n, a finite sequence of points a_{n} = a_{n,0}, a_{n,1}, …, a_{n,k_{n}} = a_{n+1} such that d(a_{n,i}, a_{n,i+1}) < 2^{−n} for all i — this uses the connectivity of M. Finally then let e be map from ℍ to (0, 1] × Q with first coordinate e_{1}(t) = 2^{−t} and whose second coordinate satisfies e_{2}(n + \frac{i}{n}) = d_{n,i} for all n and i and is (piecewise) linear otherwise.

It is clear that e is an embedding and one readily checks that cl e[ℍ] = e[ℍ] ∪ ({0} × M); the Čech-Stone extension βe of e maps ℍ^{∗} onto M.

1. Algebraic Preliminaries

We already indicated that two of our proofs will involve the lattice of closed sets of our continua and that the base for this is Wallman’s generalization, to the class of distributive lattices, of Stone’s representation theorem for Boolean algebras. Wallman’s representation theorem is as follows.

A UNIVERSAL CONTINUUM OF WEIGHT ℵ 3

Theorem 1.1 (Wallman [19]). If L is a distributive lattice then there is a compact T_{1}-space X with a base for its closed sets that is a homomorphic image of L. The homomorphism is an isomorphism iff L is disjunctive, that is if a ≠ b then there is c ∈ L such that c ∧ a = 0 or c ∧ b = 0.

The representing space X is Hausdorff iff the lattice L is normal; this means that given a, b ∈ L with a ∧ b = 0 there are c, d ∈ L such that c ∨ d = 1, a ∧ d = 0 and b ∧ c = 0. Note that this mimics the formulation of normality of topological spaces in terms of closed sets only.

Also, if one starts out with a compact Hausdorff space X and a base C for its closed sets that is closed under finite unions and intersections then this base C is a normal lattice and its representing space is X.

The following theorem shows how one can create onto mappings from maps between lattices. In it we use 2^{X} to denote the family of closed subsets of the space X.

Theorem 1.2. Let X and Y be compact Hausdorff spaces and let C be a base for the closed subsets of Y that is closed under finite unions and intersections. Then Y is a continuous image of X if and only if there is a map φ : C → 2^{X} such that

1. φ(∅) = ∅ and if F ≠ ∅ then φ(F) ≠ ∅;

2. if F ∪ G = Y then φ(F) ∪ φ(G) = X; and

3. if F_{1} ∩ ⋯ ∩ F_{n} = ∅ then φ(F_{1}) ∩ ⋯ ∩ φ(F_{n}) = ∅.

Proof. Necessity is easy: given a continuous onto map f : X → Y let φ(F) = f^{←}[F]. Note that φ is in fact a lattice-embedding.

To prove sufficiency let φ : C → 2^{X} be given and consider for each x ∈ X the family F_{x} = {F ∈ C : x ∈ φ(F)}. We claim that ⋂ F_{x} consists of exactly one point. Indeed, by Condition 3 the family F_{x} has the finite intersection property, so that ⋂ F_{x} is nonempty. Next assume that y_{1} ≠ y_{2} in Y and take F, G ∈ C such that F ∪ G = Y, y_{1} ∉ F and y_{2} ∉ G. Then, by Condition 2, either x ∈ φ(F) and so y_{1} ∉ ⋂ F_{x} or x ∈ φ(G) and so y_{2} ∉ ⋂ F_{x}.

We define f(x) to be the unique point in ⋂ F_{x}.

To demonstrate that f is continuous and onto we show that for every closed subset F of Y we have

f^{←}[F] = ⋂{φ(G) : G ∈ C, F ⊆ int G}.  (∗)

This will show that preimages of closed sets are closed and that, by Condition 1, every fiber f^{←}(y) is nonempty.

To verify (∗), first let x ∈ X \ f^{←}[F]. Then one can find G, H ∈ C such that f(x) ∉ G, hence x ∉ φ(G), and also H ∩ F = ∅ and G ∪ H = Y so that F ⊆ int G.

Second, if F ⊆ int G then one can find H ∈ C such that H ∪ G = X and H ∩ F = ∅. It follows that if x ∉ φ(G) we have x ∈ φ(H), hence f(x) ∈ H and so f(x) ∉ F.

An obvious corollary of this theorem is that Y is a continuous image of X iff there is an embedding of the lattice C into the lattice 2^{X}.

The reason that Theorem 1.2 is formulated the way it is is that we shall apply it in the following way: given the continuum K we shall take a base C for its closed sets and define a map as in the theorem into the canonical base L for the closed sets of ℍ^{∗}, rather than into the family 2^{ℍ^{∗}}. We do not know whether it is actually possible to get a lattice-embedding of C into L.

The canonical base in question is just

L = {A^{∗} : A is closed in ℍ }.

Here, as is common, A^{∗} abbreviates cl A ∩ ℍ^{∗}.

The following lemma will be the first step in the constructions of both embeddings.

Lemma 1.3. Let K be a metric continuum and let x ∈ K. There is a map φ from 2^{K} to 2^{ℍ} such that

1. φ(∅) = ∅ and φ(K) = ℍ;

2. φ(F ∪ G) = φ(F) ∪ φ(G);

3. if F_{1} ∩ ⋯ ∩ F_{n} = ∅ then φ(F_{1}) ∩ ⋯ ∩ φ(F_{n}) is compact; and

4. ℕ ⊆ φ({x}). In addition, if some countable family C of nonempty closed subsets of K is given in advance then we can arrange that for every F in C the set φ(F) is not compact.

Proof. As proved in Theorem 3 there is a map from ℍ^{∗} onto K.

The proof given is flexible enough to allow us to ensure that the embedding e of ℍ into (0, 1] × Q is such that e(n) = ⟨2^{−n}, x⟩ for every n ∈ ℕ and that for every element y of some countable set C the set {t : e_{1}(t) = y} is cofinal in ℍ. In our case we let C be a countable subset of K that meets every element of the family C.

We now identify K and {0} × K and define a map ψ : 2^{K} → 2^{𝕀×Q} by

ψ(F) = {y ∈: 𝕀 × Q : d(y, F) ⩽ d(y, K \ F)}.

In [12, § 21 XI] it is shown that for all F and G we have – ψ(F ∪ G) = ψ(F) ∪ ψ(G);

– ψ(K) = 𝕀 × Q and ψ(∅) = ∅ — by the fact that d(y, ∅) = ∞ for all y; and – ψ(F) ∩ K = F.

Note that for every y ∈ K we have d(⟨t, y⟩, {y}) = d(⟨t, y⟩, K \ {y}) = t and hence 𝕀 × {y} ⊆ ψ({y}).

Now define φ(F) = e^{←}[ψ(F)] or rather, after identifying ℍ and e[ℍ], set φ(F) = ψ(F)∩e[ℍ]. All desired properties are easily verified: 1 and 2 are immediate; to see that 3 holds note that if F_{1} ∩ ⋯ ∩ F_{n} = ∅ then cl φ(F_{1}) ∩ ⋯ ∩ cl φ(F_{n}) ∩ K = ∅, so that cl φ(F_{1}) ∩ ⋯ ∩ cl φ(F_{n}) is a compact subset of ℍ. That 4 holds follows from the way we chose the values e(n) for n ∈ ℕ.

Finally, if F ∈ C and y ∈ C ∩ F then the cofinal set {t : π(e(t)) = y} is a subset of φ(F) so that φ(F) is not compact.

Remark 1.4. If φ is as Lemma 1.3 then the map ψ : C → L defined by

ψ : C ↦ φ(C) ↦ φ(C)^{∗}

is as in Theorem 1.2. Observe that ψ is even a ∪-homomorphism; to get a lattice homomorphism the map φ would have to satisfy 3′ instead of 3, where 3′ reads:

3′. the symmetric difference of φ(F ∩ G) and φ(F) ∩ φ(G) is bounded in ℍ. We do not know how to achieve 3′.

2. Two algebraic proofs

We shall now show how to construct, given a continuum K, a map φ from a base for the closed sets of K into the base L as in Theorem 1.2.

A proof using Model Theory. Our plan is to find the map φ promised above by an appeal to some machinery from Model Theory. The first step would be to show that L is an ℵ_{1}-saturated lattice and hence an ℵ_{2}-universal structure; this latter notion says that every structure of size ℵ_{1} that is elementary equivalent to L is embeddable into L. The second step would then be to show that every lattice of size ℵ_{1} is embeddable into a lattice of size ℵ_{1} that itself is elementary equivalent to L.

There are two problems with this approach: 1) we were not able to show that L is ℵ_{1}-saturated, and 2) Lemma 1.3 does not give a lattice-embedding. We shall deal with these problems in turn.

An ℵ_{1}-saturated structure. As mentioned above we do not know whether L is ℵ_{1}- saturated. We can however find an ℵ_{1}-saturated sublattice:

L′ = {A^{∗} : A is closed in ℍ, and ℕ ⊆ A or ℕ ∩ A = ∅ }.

This lattice is a base for the closed sets of the space H, obtained from ℍ^{∗} by identifying ω^{∗} to a point.

To see that L′ is ℵ_{1}-saturated we introduce another space, namely 𝕄 = ω × 𝕀, where 𝕀 denotes the unit interval. The canonical base M for the closed sets of 𝕄^{∗} is naturally isomorphic to the reduced power (2^{𝕀})^{ω} modulo the cofinite filter. It is wellknown that this structure is ℵ_{1}-saturated — see [9]. The following substructure M′ is ℵ_{1}-saturated as well:

M′ = {A^{∗} : A is closed in 𝕄, and N ⊆ A or N ∩ A = ∅ },

where N = {0, 1} × ω. Indeed, if T is a type over a countable subset of M′ then, because of maximality, either N ⊆ x or N ∩ x = ∅ belongs to T. Any element of M that satisfies T will automatically belong to M′.

We claim that L′ and M′ are isomorphic. To see this consider the map q : 𝕄 → ℍ defined by q(n, x) = n + x. The Čech-Stone extension of q maps 𝕄^{∗} onto ℍ^{∗} and it is readily verified that L ↦ q^{−1}[L] is an isomorphism between L′ and M′. (In topological language: the space H is also obtained from 𝕄^{∗} by identifying N^{∗} to a point.)

A new language. To remedy the problem that Lemma 1.3 does not provide a lattice embedding but rather a map that only partially preserves unions and intersections we introduce a first-order language 𝔏 with equality, with a binary predicate J and for every n ⩾ 2 an n-ary predicate M_{n}. Every lattice is a structure for this language under the following interpretations:

\begin{aligned}J(x, y) & ≡ x ∨ y = 1 \\ M_{n}(x_{1}, …, x_{n}) & ≡ x_{1} ∧ ⋯ ∧ x_{n} = 0. \end{aligned}

A direct consequence of Theorem 1.2 is that compact spaces are homeomorphic iff they have bases for the closed sets that are isomorphic as 𝔏-structures.

The proof. Now let C be a base of size ℵ_{1} for the closed sets of the continuum K. We want to find an 𝔏-structure D of size ℵ_{1} that contains C and that is elementary equivalent to L′. To this end we consider the diagram of C; that is, we add the elements of C to our language 𝔏 and we consider the set T_{C} of all atomic sentences from this expanded language that hold in C. For example if a∩b = ∅ and c∪d = K then M_{2}(a, b) ∧ J(c, d) belongs to T_{C}.

To T_{C} we add the theory T_{L} of L, to get a theory T. Now Lemma 1.3 implies that for every countable subset C′ of C there is an 𝔏-embedding into L′; indeed, Condition 4 says that ω^{∗} may be mapped onto a fixed point x. This implies, by the compactness theorem, that the theory T is consistent and hence has a model D of size ℵ_{1}. The model D is as required: it satisfies the same sentences as L′ and it contains a copy of C, to wit the set of interpretations of the constants from C.

An extension. It is clear from the above reasoning and Lemma 2.2 below that the following theorem holds.

Theorem 2.1. If X is a continuum with an ℵ_{1}-saturated base for its closed sets that maps onto every metric continuum then it maps onto every continuum of weight ℵ_{1}.

Lemma 2.2. Suppose that X is a compact space which has a base C for the closed sets that is ℵ_{1}-saturated as a lattice. Then, if X maps onto a metric space M, there is an embedding of a base D for the closed subsets of M into C such as in Theorem 1.2, that is an 𝔏-embedding.

Proof. Simply fix a map f from X onto M and let D = {D_{n} : n ∈ ω} be a base for the closed subsets of M such that each D_{n} is a regular closed subset of M (that such a base, which in addition is closed under finite unions and intersections, exists is established in [16]). We inductively choose C_{n} ∈ C so that f(C_{n}) = D_{n}. To begin we may assume that D_{0} = ∅ and D_{1} = M, so, of course, we let C_{0} = ∅ and C_{1} = X (note that since C is a lattice we may assume that ∅ ∈ C). Suppose that we have chosen C_{n−1} and we must choose C_{n}. Since we plan to guarantee that f(C_{m}) = D_{m} for each m we will have that ⋂{C_{k} : k ∈ F } is empty for each F ∈ [ω]^{<ω} such that ⋂{D_{k} : k ∈ F } is empty and that C_{m} will be empty whenever D_{m} is empty. Therefore our only concern is that C_{k} ∪ C_{n} should be X whenever D_{k} ∪D_{n} is equal to M. To accomplish we first prove that, for each closed K ⊂ M, there is a C ∈ C such that f^{−1}(int(K)) ⊂ C and f(C) = K. Indeed, let {F_{i} : i ∈ ω} be a family of closed subsets of M whose union is the interior of K. Also let {U_{i} : i ∈ ω} be a neighbourhood base for K in M. For each i, there are J_{i} and Y_{i} in C such that f^{−1}(F_{i}) ⊂ J_{i}, J_{i} ⊂ f^{−1}(int(K)), f^{−1}(K) ⊂ Y_{i}, and Y_{i} ⊂ f^{−1}(U_{i}). Use the ℵ_{1}-saturatedness of C to pick L which contains each J_{i} and is contained in each Y_{i}. Clearly L has the desired properties.

Now for each k < n such that D_{k} ∪D_{n} = M, let J_{k} ∈ C be such that f^{−1}(int(M \ D_{k})) ⊂ J_{k} and f(J_{k}) = M \ int(D_{k}) (recall that D_{k} is regular closed). If D_{k} ∪ D_{n} (for k < n) is not equal to M, then let J_{k} be the empty set. Also let J_{n} ∈ C be any set such that f(J_{n}) = D_{n}. Now define C_{n} to be the union of J_{n} with all J_{k} for k < n. Clearly C_{n} ∪ C_{k} = X for all k < n such that D_{n} ∪ D_{k} = M, so it suffices to show that f(C_{n}) = D_{n}. Since J_{n} ⊂ C_{n}, it follows that D_{n} ⊂ f(C_{n}). Let k < n and assume that D_{k} ∪ D_{n} = M. Then M \ int(D_{k}) ⊂ D_{n} (again using that D_{k} is regular closed). Since f(J_{k}) = M \ int(D_{k}), it follows that f(L) ⊂ D_{n}.

A direct construction. Rather than relying on general model-theoretic results we can also give a direct construction of an 𝔏-embedding of a base for the closed sets of a continuum K into the lattice L. We do this in order to point out what may have seemed to make our main result harder to prove than Parovičenko’s theorem.

We found it necessary to use elementary substructures to guide the induction; as a consequence we shall build the embedding countably many steps at a time, rather than one.

Let K be a continuum of weight ℵ_{1} and fix a base C for the closed sets of K of size ℵ_{1} and closed under finite unions and intersections. We also assume {x} ∈ C for some fixed x ∈ K.

To construct an 𝔏-embedding φ of C into the base L for the closed sets of ℍ^{∗} we construct a map ψ : C → 2^{ℍ} that satisfies

1. if F ≠ ∅ then ψ(F) is not compact;

2. if F ∪ G = K then there is a t in ℍ such that [t, ∞) ⊆ ψ(F) ∪ ψ(G); and 3. if F_{1} ∩ ⋯ ∩ F_{n} = ∅ then ψ(F_{1}) ∩ ⋯ ∩ ψ(F_{n}) is compact.

As in Remark 1.4 we then define φ(F) = ψ(F)^{∗}.

We take a large enough regular cardinal κ such that x, K, ℍ, 2^{K}, 2^{ℍ} and C all belong to H(κ) — we can take κ = (2^{�} )^{+}. We let ⟨M_{α} : α < ω_{1}⟩ be a continuous chain of countable elementary submodels of H(κ) such that

– x, K, ℍ, 2^{K}, 2^{ℍ} and C all belong to M_{0};

– ⟨M_{β} : β ⩽ α⟩ ∈ M_{α+1} for all α; and

– C = ⋃_{α<ω_{1}} C_{α}, where C_{α} = C ∩ M_{α}.

By elementarity C_{0} is a normal distributive lattice and its Wallman representation K_{0} is a metric continuum. We may therefore apply Lemma 1.3 and obtain a map ψ_{0} : C_{0} → 2^{ℍ} such that

– ψ_{0}(∅) = ∅, ψ_{0}(K) = ℍ and if F ≠ ∅ then ψ_{0}(F) is not compact;

– ψ_{0}(F ∪ G) = ψ_{0}(F) ∪ ψ_{0}(G);

– if F_{1} ∩ ⋯ ∩ F_{n} = ∅ then ψ_{0}(F_{1}) ∩ ⋯ ∩ ψ_{0}(F_{n}) is compact; and

– ℕ ⊆ ψ_{0}({x}), where x is the point fixed above.

Note that we can have ψ_{0} ∈ M_{1} by elementarity.

We shall construct for each α a map ψ_{α} : C_{α} → 2^{ℍ} with the desired properties and such that ψ_{β} = ψ_{α} ↾ C_{β} if β < α, and ψ_{α} ∈ M_{α+1} for all α. We already have ψ_{0} and at limit stages we simply take unions so we only have to show how to construct ψ_{α+1} from ψ_{α}.

To this end enumerate C_{α} as {c_{n} : n ∈ ω} and C_{α+1} \C_{α} as {d_{n} : n ∈ ω}. Now for each n we can find, by elementarity, an n-tuple ⟨c_{i}^{n} : i < n⟩ of nonempty elements of C_{α} such that

– for all i, j < n: c_{i}^{n} ∪ c_{j} = K iff d_{i} ∪ c_{j} = K;

– for all i, j < n: c_{i}^{n} ∪ c_{j}^{n} = K iff d_{i} ∪ d_{j} = K;

– for all i < n: x ∈ c_{i}^{n} iff x ∈ d_{i}; and

– for all f, g ⊆ n: ⋂_{i∈f} c_{i}^{n} ∩ ⋂_{j∈g} c_{j} = ∅ iff ⋂_{i∈f} d_{i} ∩ ⋂_{j∈g} c_{j} = ∅.

Given this n-tuple find m_{n} ∈ ℕ such that

– if c_{i}^{n} ∪ c_{j} = K then [m_{n}, ∞) ⊆ ψ_{α}(c_{i}^{n}) ∪ ψ_{α}(c_{j});

– if c_{i}^{n} ∪ c_{j}^{n} = K then [m_{n}, ∞) ⊆ ψ_{α}(c_{i}^{n}) ∪ ψ_{α}(c_{j}^{n});

– if ⋂_{i∈f} c_{i}^{n} ∩ ⋂_{j∈g} c_{j} = ∅ then ⋂_{i∈f} ψ_{α}(c_{i}^{n}) ∩ ⋂_{j∈g} ψ_{α}(c_{j}) ⊆ [0, m_{n});

– if x ∈ c_{i}^{n} then ℕ \ ψ_{α}(c_{i}^{n}) ⊆ [0, m_{n}); and

– if x ∉ c_{i}^{n} then ℕ ∩ ψ_{α}(c_{i}^{n}) ⊆ [0, m_{n}).

We can and will assume in addition that m_{n+1} > m_{n} for all n and also that ψ_{α}(c_{i}^{n}) ∩ [m_{n}, m_{n+1}) ≠ ∅ for all i.

Now we define, for all i,

ψ_{α+1}(d_{i}) = ⋃_{n>i} ψ_{α}(c_{i}^{n}) ∩ [m_{n}, m_{n+1}).

We show that ψ_{α+1} has all the required properties.

The first thing to check is that ψ_{α+1}(d_{i}) is indeed closed. Note that only the points m_{n+1} with n > i can possibly be in cl ψ_{α+1}(d_{i}) but not in ψ_{α+1}(d_{i}). But if m_{n+1} ∈ cl(ψ_{α+1}(d_{i})) then m_{n+1} ∈ cl(ψ_{α}(c_{i}^{n}) ∩ [m_{n}, m_{n+1})) or m_{n+1} ∈ ψ_{α}(c_{i}^{n+1}); in the latter case we are done and in the former case we have m_{n+1} ∈ ψ_{α}(c_{i}^{n}) and hence x ∈ c_{i}^{n}, by the choice of m_{n}, and so x ∈ d_{i}. It then follows that x ∈ c_{i}^{n+1} and so again m_{n+1} ∈ ψ_{α}(c_{i}^{n+1}) ⊆ ψ_{α+1}(d_{i}).

It is now a routine matter to check that ψ_{α+1} is as required. For example, if d_{i} ∪ d_{j} = X them [m_{n}, ∞) ⊆ ψ_{α+1}(d_{i}) ∪ ψ_{α+1}(d_{j}), where n = max(i, j) + 1 and if ⋂_{i∈f} d_{i} ∩ ⋂_{j∈g} c_{j} = ∅ then ⋂_{i∈f} ψ_{α+1}(d_{i}) ∩ ⋂_{j∈g} ψ_{α} ⊆ [0, m_{n}), where n = max(f ∪ g) + 1.

In the end ψ = ⋃_{α<ω_{1}} ψ_{α} is the map that we want.

3. A topological proof

In this section we show how to modify the argument of Błaszczyk and Szymański from [5] to obtain a topological proof of Theorem 1.

As mentioned in the introduction the following statement at first sight seems a reasonable property that ℍ^{∗} should share with ω^{∗}.

False lemma 3.1. If f maps ℍ^{∗} onto the metric continuum X and if g maps the metric continuum Y onto X then there is a map h from ℍ^{∗} onto Y such that f = g ∘ h.

That this lemma is false is witnessed by the following example.

Example 3.2. Define f : ℍ → S^{1} and g : 𝕀 → S^{1} by f(t) = g(t) = e^{2πit}. Now any map from ℍ^{∗} onto 𝕀 is induced by a map h : ℍ → 𝕀 and if we want to have f^{∗} = g^{∗} ∘ h^{∗} then we must have lim_{t→∞}|f(t) − g(h(t))| = 0. Choose N ∈ ℕ such that |f(t) − g(h(t))| < \frac{1}{4} for t ⩾ N and let n > N. The image, under f, of [n − \frac{1}{4}, n + \frac{1}{4}] is the arc from −i, via 1, to i. Therefore the image under g ∘ h of this interval does not contain −1 and so its image under h does not contain \frac{1}{2}; however this last image does contain points above and below \frac{1}{2}. This contradicts the continuity of h.

The interested reader can certainly formulate the appropriate elementary submodel version of Lemma 3.1 by examining the inductive step in our direct algebraic construction and the relation between C_{α} and C_{α+1} exploited therein. We choose a more direct approach.

Let K be the continuum of weight ℵ_{1} and assume that K is a subspace of the Tychonoff cube 𝕀^{ω_{1}}. We shall construct by induction maps f_{α} : ℍ → 𝕀 such that the Čech-Stone extension of the diagonal map f = △_{α} f_{α} maps ℍ^{∗} onto K.

For this it suffices to make sure that the following conditions are met:

1. for every open set U around K there is n ∈ ℕ such that f[[n, ∞)] ⊆ U, and 2. for every open set U that meets K the set of t ∈ ℍ for which f(t) ∈ U is

cofinal in ℍ. To this end we let κ be a reasonably large regular cardinal such that K, ℍ and

𝕀^{ω_{1}} belong to H(κ). We also fix a continuous chain ⟨M_{α} : α < ω_{1}⟩ of countable elementary submodels of H(κ) such that

– K, ℍ and 𝕀^{ω_{1}} belong to M_{0}, and

– for every α the chain ⟨M_{β} : β < α⟩ belongs to M_{α}. Observe that automatically ω_{1} ⊆ ⋃_{α} M_{α}. Denote M_{α} ∩ ω_{1} by δ_{α} and denote the projection of K onto the first δ_{α} coordinates by K_{α}.

The model H(κ) also contains the base B for 𝕀^{ω_{1}} that is built using only open intervals in 𝕀 with rational end points and because this base is uniquely defined it also belongs to M_{0}.

We shall construct the maps f_{β} for β < δ_{α}, by induction on α. To facilitate this we assume that the point 0 with all coordinates zero belongs to K; this entails no loss of generality because 𝕀^{ω_{1}} is homogenous.

We use the proof of Theorem 3 to find a map h : ℍ → 𝕀^{δ_{0}} such that conditions 1 and 2 are met with K replaced by K_{0}. We can and will assume that h(n) = 0 for all n ∈ ℕ and that h ∈ M_{1}. We let f_{β} be the βth coordinate of h of course.

Now let α ∈ ω_{1} and assume that the maps f_{β} have been found for β < δ_{α} such that e_{α} = △_{β<δ_{α}} f_{β} belongs to M_{α+1} and meets conditions 1 and 2 with K replaced by K_{α}, and such that f_{β}(n) = 0 for all β and n.

Denote the family of elements of B that are supported in δ_{α+1} by C and observe that C ⊆ M_{α+1} — every element is defined from finitely many elements of M_{α+1}: finitely many ordinals and rational numbers.

Let {U_{n} : n ∈ ω} enumerate a decreasing neighbourhood base for K_{α+1}, where each U_{n} is a finite union of elements of C and arcwise connected. Furthermore we enumerate the elements of C that meet K_{α+1} as {V_{n} : n ∈ ω}.

Let n ∈ ω and consider U_{n} and the sets V_{i} for i ⩽ n. The set of ordinals determined by these open sets is finite and may be split into two parts: F_{n,0}, those below δ_{α}, and F_{n,1}, the rest. One can express the facts that K_{α+1} ⊆ U_{n} and that each V_{i} meets K_{α+1} in one formula that involves as constants the elements of F_{n,1} and some elements of M_{α}, notably K, the elements of F_{n,0} and some rational numbers. For example if V_{0} = π\overleftarrow{η_{1}}[(q_{1}, q_{2})] ∩ π\overleftarrow{η_{2}}[(q_{3}, q_{4})] then V_{0} ∩ K ≠ ∅ abbreviates

(∃x ∈ K)(q_{1} < x_{η_{1}} < q_{2} ∧ q_{3} < x_{η_{2}} < q_{4})

By elementarity one can find, in the interval (max F_{n,0}, δ_{α}), a finite set of ordinals G_{n} that can replace F_{n,1} without changing the validity of the formula. Using G_{n} we can build open sets U′_{n} and V_{n,i} for i ⩽ n that are reflections of U_{n} and the V_{i}. In the above example this means that if, say η_{1} ∈ M_{α} but η_{2} ∉ M_{α}, there must be an ϵ in M_{α} (larger than η_{1}) such that

(∃x ∈ K)(q_{1} < x_{η_{1}} < q_{2} ∧ q_{3} < x_{ϵ} < q_{4})

If this were the case n = 0 then we would take V_{0,0} = π\overleftarrow{η_{1}}[(q_{1}, q_{2})] ∩ π\overleftarrow{ϵ}[(q_{3}, q_{4})].

We can therefore find, given a natural number N, two natural numbers k and l bigger than N and such that e_{α}[[k, ∞)] ⊆ U′_{n} and every V_{n,i} meets e_{α}[[k, l]].

Use this to find a strictly increasing sequence ⟨k_{n}⟩_{n} in ℕ such that for every n we have e_{α}[[k_{n}, ∞)] ⊆ U′_{n} and V_{n,i} ∩ e_{α}[[k_{n}, k_{n+1}]] ≠ ∅ for i ⩽ n.

Observe that the sets F_{n,1} are increasing and that their union is the interval [δ_{α}, δ_{α+1}). So given β in this interval the set of n with β ∈ F_{n,1} is a final segment of ω. For each such n let β_{n} be the element of G_{n} that corresponds to β. Finally define

f_{β}(t) = \{ \begin{matrix}0 & if t ∈ [k_{n}, k_{n+1}] and β ∉ F_{n,1}, and \\ f_{β_{n}}(t) & if t ∈ [k_{n}, k_{n+1}] and β ∈ F_{n,1}. \end{matrix}

Elementarity considerations will tell us that these f_{β} are as required. We have e_{α+1}[[k_{n}, k_{n+1}]] ⊆ U_{n} because e_{α}[[k_{n}, k_{n+1}]] ⊆ U′_{n} and V_{i} ∩ e_{α+1}[[k_{n}, k_{n+1}]] ≠ ∅ because V_{n,i} ∩ e_{α}[[k_{n}, k_{n+1}]] ≠ ∅ whenever i ⩽ n.

Finally note that that the whole construction can be considered to have taken place in M_{α+2} so that the result is in M_{α+2} as well.

In the end the full diagonal mapping △_{β<ω_{1}} f_{β} is as required.

Remark 3.3. Let us note that in all three proofs it was necessary to map the set ω^{∗} to a single point. In the first proof to get an ℵ_{1}-saturated structure and in the other proofs to make certain inductions work.

4. Other universal continua

We can exhibit several other natural universal continua: for every n ∈ ℕ the remainder (ℝ^{n})^{∗} maps onto ℍ^{∗} and hence onto every continuum of weight ℵ_{1}. Thus we find, under CH, for every n a universal continuum of weight � and of dimension n.

There is one more natural type of continuum that we want to deal with. These we find in the remainder of the space 𝕄 = ω ×𝕀 considered above. For every u ∈ ω^{∗} the set

𝕀_{u} = ⋂_{A∈u} cl(A × 𝕀)

is a component of 𝕄^{∗}. Each of our three proof can be modified to show that every continuum 𝕀_{u} also maps onto every continuum of weight ℵ_{1}.

This is easiest for the model-theoretic approach. Indeed, it is very easy to show that 𝕀_{u} has an ℵ_{1}-saturated base for its closed sets: let I be the family of finite unions of closed intervals in 𝕀 with rational end points. The product I^{ω} is naturally isomorphic to a base for the closed sets of 𝕄 and the ultrapower I^{ω}/u is naturally isomorphic to a base for the closed subsets of 𝕀_{u}. It is well-known that such ultrapowers are ℵ_{1}-saturated.

To show how to adapt the other two proofs we use the map q from 𝕄^{∗} onto ℍ^{∗} defined by q(⟨n, x⟩) = n + x. The modifications will be such that the final map will map each of the continua q[𝕀_{u}] onto the target continuum as well.

The first step is to adapt the proof of Theorem 3. This is quite straightforward: if we make sure that, for every n, the finite set {a_{n,1}, …, a_{n,k_{n}}} contains the points a_{0}, …, a_{n−1} then it follows easily that A ⊆ βe[𝕀_{u}] and hence βe[𝕀_{u}] = M for every u.

The direct algebraic construction. As a result of the adaptation of Theorem 3 we know that we can ensure that for every F ∈ C_{0} there is a natural number n such that ψ_{0}(F) meets [m, m + 1] for every m ⩾ n.

All that is needed to do then is to choose the numbers m_{n} so large that for every i < n and every m ⩾ m_{n} the set ψ_{α}(c_{i}^{n}) meets the interval [m, m + 1].

The topological proof. In this case the fact that h maps every q[𝕀_{u}] onto K_{0} translates to: if V is an open set that meets K_{0} then there is a natural number n such that V meets h[[m, m + 1]] for every m ⩾ n.

The modification needed now is to demand each time that V_{n,i} meet e_{α}[[k, k+1]] for all k > k_{n}.

In [6] it is shown that, under CH, all continua 𝕀_{u} are mutually homeomorphic. This gives us a second one-dimensional universal continuum of weight �. It is different from ℍ^{∗} because ℍ^{∗} is indecomposable and 𝕀_{u} is not.

5. Extensions and limitations

In this section we find analogues of some standard results about continuous images of ω^{∗}.

First we extend our main result by showing that under Martin’s Axiom (MA) every continuum of weight less than � is a continuous image of ℍ^{∗}.

Next we give some limiting results. We show that in the Cohen model the long segment of length ω_{2} is not a continuous image of ℍ^{∗}. This shows that an extra assumption like MA is necessary to push Theorem 1 from ℵ_{1} to ℵ_{2}. The Cohen model result also shows that in Theorem 1 we cannot replace ℵ_{1} by �. We improve this by showing that the Proper Forcing Axiom (PFA) implies that there is a continuum of weight � that is not a continuous image of ℍ^{∗}, so not even MA is strong enough to allow us to prove Theorem 1 for � instead of ℵ_{1}.

Martin’s Axiom. We show that Martin’s Axiom (MA) implies that all continua of weight less than � are continuous images of ℍ^{∗}. We need the following result from [17].

Theorem 5.1 (MA). If X is a compact space of weight less than � then there is a compactification of ω with X as its remainder.

Using this theorem and the ideas from the proof of Theorem 3 it is now quite straightforward to prove the following theorem by an extra application of MA.

Theorem 5.2 (MA). Let K be a continuum of weight less than �. Then K is a continuous image of ℍ^{∗}.

Proof. It suffices to construct a compactification of ℍ with K as its remainder. Let κ be the weight of K and let γω be a compactification of ω with K as its remainder. We embed γω into 𝕀^{κ} and we note the following two facts.

1. If U is a neighbourhood of K in 𝕀^{κ} then ω \ U is finite; and

2. K has arbitrarily small arcwise connected neighbourhoods.

The first fact is immediate; the second follows by noting that every basic open set in 𝕀^{κ} is arcwise connected and that every neighbourhood of K contains a neighbourhood that is both a finite union of basic open sets and connected.

There is a map h : ℍ → 𝕀^{κ} that almost gives the desired compactification of ℍ: the piecewise linear map that connects the natural numbers in their natural order. This map is defined coordinatewise by h_{α}(n + t) = (1 − t)n_{α} + t(n + 1)_{α}. Here n_{α} denotes the αth coordinate of the point of γω that corresponds to n. We need to rectify a few problems: 1) h need not be one-to-one, 2) the intersection h[ℍ] ∩ K may be nonempty, and 3) K may be a proper subset of ⋂_{n∈ω} cl h[[n, ∞)].

The first two may be solved by introducing one more coordinate: the map t ↦ ⟨2^{−t}, h(t)⟩ from ℍ to 𝕀 × 𝕀^{κ} does not suffer from problems 1 and 2.

To remedy 3 we introduce the following partial order ℙ: an element of ℙ is of the form p = ⟨n_{p}, f_{p}, ϵ_{p}, F_{p}, U_{p}⟩, where n_{p} ∈ ω, U_{p} is an arcwise connected neighbourhood of K that contains ω \ n_{p}, f_{p} : [0, n_{p}] → 𝕀^{κ} is such that f_{p}(n) = n for n ⩽ n_{p} — so in particular f_{p}(n_{p}) ∈ U_{p}, ϵ_{p} > 0 and F_{p} ∈ [κ]^{<ω}.

The ordering is defined as follows: p ⩽ q iff n_{p} ⩾ n_{q}, ϵ_{p} ⩽ ϵ_{q}, U_{p} ⊆ U_{q}, F_{p} ⊇ F_{q}, f_{p}[[n_{q}, n_{p}]] ⊆ U_{q} and for all x ∈ [0, n_{q}] and for all α ∈ F_{q} we have |π_{α}(f_{p}(x)) − π_{α}(f_{q}(x))| < ϵ_{q}.

It should be clear that by making F_{p} larger or ϵ_{p} and U_{p} smaller one automatically obtains a stronger condition in ℙ; it follows that the following sets are dense in ℙ:

– D_{1,α} = {p : α ∈ F_{p}}, where α ∈ κ;

– D_{2,n} = {p : ϵ_{p} < 2^{−n}}, where n ∈ ω; and

– D_{3,U} = {p : U_{p} ⊆ U}, where U is a basic neighbourhood of K. We must also consider D_{4,n} = {p : n_{p} ⩾ n}, where n ∈ ω. To see that such a set is dense let n be given. Now if p ∈ ℙ and n_{p} < n then, because U_{p} is arcwise connected and {n_{p}, …, n} ⊆ U_{p}, we can extend f_{p} to g : [0, n] → 𝕀^{κ} with g[[n_{p}, n]] ⊆ U_{p}. It follows that ⟨n, g, ϵ_{p}, F_{p}, U_{p}⟩ is in D_{4,n} and below p.

If G is a filter that meets all of the aforementioned κ many dense sets then the net ⟨f_{p}⟩_{p∈G} converges uniformly on each compact subset of ℍ to a map f, which is continuous. For this map f we do have K = ⋂_{n∈ω} cl f[[n, ∞)]. If we add the extra coordinate to f then we get an embedding f̄: t ↦ ⟨2^{−t}, f(t)⟩ of ℍ into 𝕀 × 𝕀^{κ} that satisfies cl f̄[ℍ] = f̄[ℍ] ∪ K.

It remains to show that ℙ satisfies the ccc. So let Q be an uncountable subset of ℙ. We may and do assume without loss of generality that there are fixed n and ϵ such that n_{p} = n and ϵ_{p} ⩾ ϵ for all p ∈ Q. We also assume that {F_{p} : p ∈ Q} is a Δ-system with root F.

With the uniform metric the function space C([0, n], ℝ^{F} ) is separable, hence the set {π_{F} ∘ f_{p} : p ∈ Q} has a complete accumulation point f in it. We therefore assume, again without loss of generality, that for all p ∈ Q we have d(f, π_{F} ∘f_{p}) < ϵ.

We show that any two elements of Q are compatible. Indeed, given p and q define g : [0, n] → 𝕀^{κ} by

g_{α}(x) = \{ \begin{matrix}f_{α}(x) & if α ∈ F, \\ f_{p,α}(x) & if α ∈ F_{p} \ F, \\ f_{q,α}(x) & if α ∈ F_{q} \ F, and \\ h_{α}(x) & otherwise. \end{matrix}

The quintuple ⟨n, g, ϵ, F_{p} ∪ F_{q}, U_{p} ∩ U_{q}⟩ is below both p and q.

The Cohen Model. In [10, Chapter 12] Kunen investigated for what cardinals κ the σ-algebra R(κ) generated by the family of rectangles equals the full power set of κ^{2}, where a rectangle is a set of the form X × Y with X, Y ⊆ κ.

The following lemma, from [10], gives a convenient sufficient condition for a set to belong to R(κ).

Lemma 5.3. If X ⊆ κ^{2} and if one can find, for all α, subsets x_{α} and y_{α} of ω such that ⟨α, β⟩ ∈ X iff x_{α} ∩ y_{β} is infinite, then X belongs to R(κ).

Proof. Define X_{n} = {α : n ∈ x_{α}} and Y_{n} = {α : n ∈ Y_{n}}. Now observe that ⟨α, β⟩ ∈ X iff (∀m)(∃n ⩾ m)(n ∈ x_{α} ∩ y_{β}). Therefore X = ⋂_{m} ⋃_{n⩾m} X_{n} × Y_{n}.

Using this lemma Kunen showed that R(ω_{1}) = P(ω_{1}^{2}) and that MA implies R(κ) = P(κ^{2}) for every κ < �.

Kunen complemented this by showing that in the model obtained by adding (at least) ℵ_{2} Cohen reals to a model of CH the set L = {⟨α, β⟩ : α < β < ω_{2}} does not belong to R(ω_{2}).

Using the results mentioned above it is nearly immediate that in the Cohen model the ordinal space ω_{2} + 1 is not a continuous image of ω^{∗}; this is very likely the reason that several authors refer to [10] as the source for this result. Indeed, from a continuous onto mapping one immediately obtains a sequence ⟨x_{α} : α < ω_{2}⟩ of subsets of ω such that x_{β} ⊂^{∗} x_{α} whenever α < β. For each α let y_{α} be the complement of x_{α}. It follows that α < β iff x_{α} ∩y_{β} is infinite and so, by Lemma 5.3, L ∈ R(ω_{2}).

We show that a similar result can be formulated a proved for ℍ^{∗}.

Theorem 5.4. In the model obtained by adding (at least) ℵ_{2} Cohen reals to a model of CH the long segment of length ω_{2} is not a continuous image of ℍ^{∗}.

The long segment L_{ω_{2}} of length ω_{2} is the set ω_{2} × [0, 1) ∪ {ω_{2}} ordered lexicographically.

We need the following adaptation of Lemma 5.3.

Lemma 5.5. If there is, with respect to the mod finite order, a strictly increasing ω_{2}-sequence ⟨f_{α} : α < ω_{2}⟩ in ℝ^{ω} then L ∈ R(ω_{2}).

Proof. Much as in the proof of Lemma 5.3 define, for n ∈ ω and q ∈ ℚ, sets X_{n,q} and Y_{n,q} by X_{n,q} = {α : q > f_{α}(n)} and Y_{n,q} = ω_{2} \ X_{n,q}. It is easily seen that L = ⋂_{m} ⋃_{n⩾m,q∈ℚ} X_{n,q} × Y_{n,q}.

To show that in the Cohen model there is no map from ℍ^{∗} onto L_{ω_{2}} we show that the existence of such a map entails the existence of sequence ⟨x_{α} : α < ω_{2}⟩ of subsets of ω such that x_{β} ⊂^{∗} x_{α} whenever α < β or of a sequence ⟨f_{α} : α < ω_{2}⟩ in ℝ^{ω} as in Lemma 5.5.

So assume f : ℍ^{∗} → L_{ω_{2}} is a continuous surjection. Choose, for each α < ω_{2}, a standard open set U_{α} in βℍ that contains f^{←}[[α + 1, ω_{2}]] but whose closure is disjoint from f^{←}[[0, α]]. Let ⟨I_{n} : n ∈ ω⟩ be the sequence of intervals that determines U_{0}.

For each α < ω_{2} we let J_{α} = {n : I_{0,n} ∩ U_{α} ≠ ∅}. Now either ⟨J_{α} : α < ω_{2}⟩ is decreasing with respect to ⊂^{∗}, in which case we are done, or it becomes constant modulo finite sets and then we may as well assume that it is constant and that J_{α} = ω for all α.

Now observe that if β < α we have cl U_{α} ∩ ℍ^{∗} ⊆ U_{β} so there must be an m such that cl U_{α} \ U_{β} ⊆ [0, m]. We can therefore define functions f_{α} : ω → ℝ by f_{α}(n) = inf U_{α} ∩ I_{n}. The sequence ⟨f_{α} : α < ω_{2}⟩ is strictly increasing with respect to <^{∗} and we are done.

The Proper Forcing Axiom. The obvious question whether MA is strong enough to imply that every compact space of weight � is a continuous image of ω^{∗} has a negative answer, see for example the surveys by Baumgartner [4] and Van Mill [18].

We show, by obvious modifications of the arguments from [4], that PFA implies there is a continuum of weight � that is not a continuous image of ℍ^{∗}.

As PFA implies MA+� = ℵ_{2} we know that there is an ℵ_{2}-saturated ultrafilter u on ω, see [7]. Such an ultrafilter has the property that every ultrapower over it of any countable structure is ℵ_{2}-saturated. We apply this to the linearly ordered set ℚ of rational numbers and find an embedding of the lexicographically ordered tree T = 2^{<ω_{2}} into ℚ^{ω}/u and hence into the standard continuum 𝕀_{u}. We show that this standard continuum is not a continuous image of ℍ^{∗}.

Indeed, assume there is a continuous map h from ℍ^{∗} onto 𝕀_{u}. Let x ∈ 2^{ω_{2}} be such that both x^{←}(0) and x^{←}(1) are cofinal in ω_{2}; then x determines a ⟨ω_{2}, ω_{2}^{∗}⟩-gap ⟨l(x), r(x)⟩ in T, where

\begin{aligned}l(x) & = \{s ∈ T : (∃α)(dom s = α + 1, s ↾ α = x ↾ α, s(α) = 0, x(α) = 1)\} \\ r(x) & = \{s ∈ T : (∃α)(dom s = α + 1, s ↾ α = x ↾ α, x(α) = 0, s(α) = 1)\}. \end{aligned}

Via the embedding of T into 𝕀_{u} and the map h we can associate with this gap two sequences {U_{s} : s ∈ l(x)} and {V_{s} : s ∈ r(x)} of standard open sets with the properties that

1. if s ∈ l(x) and t ∈ r(x) then U_{s} ∩ V_{t} = ∅;

2. if s < t in l(x) then cl U_{s} ⊆ U_{t}; and

3. if t < s in r(x) then cl V_{s} ⊆ V_{t}. Now the proof, given in [4], that in the Boolean algebra 𝔹 there are no ⟨ω_{2}, ω_{2}^{∗}⟩-gaps (assuming PFA) works mutadis mutandis to show that no such gaps exist in the lattice L either. It follows that there must be L_{x} ∈ L such that U_{s} ⊆ L_{x} ⊆ ℍ^{∗} \ V_{t} for all s ∈ l(x) and t ∈ r(x).

The assignment x ↦ L_{x} is clearly one-to-one; this is impossible however as there are 2^{ℵ_{2}} possible x’s and only ℵ_{2} possible L_{x}’s.

Yet another universal continuum. In the previous subsection we used a �- saturated ultrafilter u to find a continuum of weight �, to wit 𝕀_{u}, that is not a continuous image of ℍ^{∗}. The methods used in this paper allows us to show that 𝕀_{u} is in fact a universal continuum of weight �.

Indeed, let I be the family of all finite unions of closed intervals in 𝕀 with rational end points. Clearly, I is a base for the closed sets of 𝕀 and so the ultrapower I_{u} = I^{ω}/u is a base for the closed sets of 𝕀_{u}; this ultrapower is �-saturated and hence �^{+}-universal. But now the reasoning leading up to Theorem 2.1 can be used to show that 𝕀_{u} maps onto every continuum of weight � or less.

6. Some questions

Of course every result about continuous images of ω^{∗} suggests the possibility of a corresponding one about continuous images of ℍ^{∗}. The more obvious questions are.

Question 6.1 (compare [14]). Is every perfectly normal continuum a continuous image of ℍ^{∗}?

Question 6.2. Is it consistent with ¬CH (or better still with MA + ¬CH) that every continuum of weight � is a continuous image of ℍ^{∗}?

A more interesting question, raised by G. D. Faulkner and also suggested by the proofs of Theorems 3 and 5.2, is.

Question 6.3. Is every continuum that is a continuous image of ω^{∗} also a continuous image of ℍ^{∗}?

A still more interesting problem is to generalize Parovičenko’s characterization of ω^{∗} to the connected case.

Problem 6.4. Find, assuming CH, topological characterizations of the continua ℍ^{∗} and 𝕀_{u}.

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Faculty of Technical Mathematics and Informatics, TU Delft, Postbus 5031, 2600 GA Delft, the Netherlands

E-mail address: k.p.hart@twi.tudelft.nl

Department of Mathematics and Statistics, York University, 4700 Keele Street, North York, Ontario, Canada M3J 1P3

E-mail address: dowa@mathstat.yorku.ca