EXTENDING REAL-VALUED FUNCTIONS IN βκ

ALAN DOW

Abstract. An Open Coloring Axiom type principle is formulated for uncountable cardinals and is shown to be a consequence of the Proper Forcing Axiom. Several applications are found. We also study dense C^{∗}-embedded subspaces of ω^{∗}, showing that there can be such sets of cardinality 𝔠 and that it is consistent that ω^{∗} \ {p} is C^{∗}-embedded for some but not all p ∈ ω^{∗}.

1. Introduction

We establish a consequence of the proper forcing axiom (PFA) which is a combinatorial principle for uncountable cardinals which is similar to the Open Coloring Axiom of Todorčević [Tod89a] (see also [AS81]), see Definition 4.1. This principle is more general than another similar principle formulated by Todorčević in [Tod89b]. In the second section we show that this principle implies several reflection results concerning extending real-valued continuous functions on subspaces of βκ (for discrete κ) and the reflection of first-countability in spaces of countable tightness. In the third section we show, in response to a question in [vDKvM89], that it is consistent that there is a “small” dense subset of βω \ ω which is C-embedded (i.e. every real-valued continuous function has a continuous extension to all of βω). It is shown in [vDKvM89] that ω^{∗} \ {p} is C-embedded for every p ∈ ω^{∗} and it is asked in [HvM90] if there is a model in which ω^{∗} \ {p} is C-embedded for some but not all p. We show that in the Miller model ω^{∗} \ {p} is not C-embedded if and only if p is a P -point of ω^{∗}. A set X is said to be C^{∗}-embedded in Y if every bounded realvalued continuous function on X has a continuous extension to Y. The distinction is not significant in this article because in each case the space under discussion will not possess any unbounded continous real-valued functions.

2. Uncountable Domains

There are many situations in which one has two families I and J which are subsets of a cardinal κ such that every member of I is almost disjoint from every member of J and one is interested in determining if the two families can be separated by a single set X, i.e. a set X such that every member of I is almost contained in X and every member of J is almost disjoint from X (where almost disjoint means the intersection is in some fixed ideal, e.g. the ideal of finite sets). Of course it is well-known that there are situations where there is no such separating set X. However it is perfectly reasonable to expect that if there is no such X then there should be small subsets (in terms of cardinality) of I and J for which there is also no such separating set X (a useful reflection principle results). It is easily seen that countable subfamilies can always be separated hence ℵ_{1} would be the target. If I ∪ J forms a maximal almost disjoint family of sequences of rationals which converge to reals and I is the set of all those which converge to some member of a Bernstein set, then I and J cannot be separated. However it follows from MA(ω_{1}) that any subsets of I and J of cardinality ℵ_{1} can be separated. We will avoid this problem by assuming that our ideal I is a P -ideal, that is, if I′ ⊂ I is countable, then there is a member I ∈ I such that I′ \ I is finite for each I′ ∈ I′. One other situation must be avoided in order to expect ℵ_{1}-sized reflection (under PFA); this is formulated in the next definition.

Definition 2.1. We will say that two ideals I and J can be countably separated if there are countably many sets X_{n} such that for each pair (I, J) ∈ I ×J, there is an n such that I is almost disjoint from X_{n} and J is almost contained in X_{n} . We will say that the ideals are strongly countably separated if the X_{n}’s have the additional property that X_{n} ∩ I is finite for each I ∈ I and each n ∈ ω (i.e. [X_{n}]^{ω} ∩ I = ∅ for each n).

If we say that a set of pairs {(I_{α}, J_{α}) : α ∈ ω_{1}} are (strongly) countably separated then we mean that the family {I_{α} : α ∈ ω_{1}} is (strongly) countably separated from {J_{α} : α ∈ ω_{1}} .

Notice that the definition of countably separated is symmetric in I and J and is obviously more general than that of strongly countably separated. However we will actually be using the stronger notion in all cases because of the following lemma.

Lemma 2.2. If I and J are countably separated and I is a P -ideal, then they are strongly countably separated.

Proof. Assume that I is a P -ideal and that {X_{n} : n ∈ ω} countably separates I and J . Simply take the family of only those X_{n} such that [X_{n}]^{ω} ∩ I = ∅, this family will also separate the two ideals. Indeed, for each n such that there is an I ∈ I such that X_{n} ∩ I is infinite, let I_{n} be such a member of I. Since I is a P -ideal, there is an I ∈ I such that I_{n} \ I is finite for each n. Therefore I has the property that I ∩ X_{n} is infinite for each n such that there is some I′ ∈ I such that I′ ∩ X_{n} is infinite. Now {X_{n} : I ∩ X_{n} is finite } separates I and J since this family will have to separate the family {I′ ∪ I : I′ ∈ I} and J.

The remarkable thing about the following result is that there is no hypothesis on the ideal J. This result is the combinatorial essence to many reflection results, some of which are discussed in the next section.

Theorem 2.3 (PFA). If I and J are ideals of countable sets with I a P -ideal then either

1. I and J can be countably separated,

or

2. there are ℵ_{1}-generated subideals I′ ⊂ I and J′ ⊂ J which cannot be countably separated.

Stated as above this theorem most resembles the Open Colouring Axiom principle discussed above. However it is useful to explicitly state a more detailed structure theorem that yields Theorem 2.3 as a Corollary. The terms used in this statement are defined below.

Lemma 2.4. If I and J are ideals, with I a P -ideal, which cannot be countably separated then there is a proper poset which introduces a subcollection {(I_{α}, J_{α}) : α ∈ ω_{1}} ⊂ I×J which forms either a Hausdorff-Luzin type family or a finite-to-one graph type family and such that {I_{α} : α ∈ ω_{1}} generates a P -ideal.

Theorem 2.5 (PFA). If I and J are ideals, with I a P -ideal, which cannot be countably separated then there is a subcollection {(I_{α}, J_{α}) : α ∈ ω_{1}} ⊂ I × J which forms either a Hausdorff-Luzin family or a finite-to-one graph family. In addition, {I_{α} : α ∈ ω_{1}} can be chosen to generate a P -ideal.

Clearly what is happening is that the Hausdorff-Luzin families and the finiteto-one graph families are the only families of size ℵ_{1} which are “absolutely” not strongly countably separated – i.e. passing to a larger (ω_{1} preserving) model will not make them strongly countably separated.

The best known example of such absolutely unseparated ideals are the Hausdorff-Luzin type.

Definition 2.6. We say that {(I_{α}, J_{α}) : α ∈ ω_{1}} is a Hausdorff-Luzin type family if for each α < ω_{1} and m ∈ ω, the set of β < α such that I_{α} ∩ J_{β} has cardinality at most m is finite.

Definition 2.7. We say that {(I_{α}, J_{α}) : α ∈ ω_{1}} is a finite-to-one graph type family if the J_{β}’s are pairwise disjoint and there is a sequence of enumerations, {j(β, k) : k ∈ ω}, of the J_{β}’s so that for each α there is a finite-to-one function g_{α} from α to ω such that j(β, g_{α}(β)) ∈ I_{α} for each β < α (roughly speaking, I_{α} contains the graph of a finite-to-one function)

Proposition 2.8. Hausdorff-Luzin type families cannot be strongly countably separated.

Proof. Let {(I_{α}, J_{α}) : α ∈ ω_{1}} be a Hausdorff-Luzin type family. Assume that {X_{n} : n ∈ ω} strongly separates the family {I_{α} : α < ω_{1}} from {J_{α} : α < ω_{1}}. There is an n and an m so that there is an uncountable set S such that for each β ∈ S, |J_{β} \X_{n}| ≤ m . In addition, for each β ∈ S, assume that |I_{β} ∩X_{n}| ≤ m. Now fix α ∈ S such that S∩α is infinite, hence there is a β < α such that |I_{α}∩J_{β}| is bigger than 2m. Consider how I_{α} ∩ J_{β} is split by X_{n}. |(I_{α} ∩ J_{β}) ∩ X_{n}| ≤ |X_{n} ∩ I_{α}| ≤ m and |(I_{α} ∩ J_{β}) \ X_{n}| ≤ |J_{β} \ X_{n}| ≤ m. Therefore |I_{α} ∩ J_{β}| ≤ 2m.

Lemma 2.9. A finite-to-one graph type family cannot be strongly countably separated.

Proof. Fix a family as in Definition 2.7 and assume that {X_{n} : n ∈ ω} strongly separates the I_{α}’s from the J_{α}’s. Certainly there is some X_{n} which contains (mod finite) uncountably many of the J_{β}. Therefore, there is an m so that there is an uncountable set S such that for each β ∈ S, J_{β} \ X_{n} ⊂ {j(β, ℓ) : ℓ < m}. Choose any α such that S ∩ α is infinite. Since g_{α} is finite-to-one g_{α}(β) < m for at most finitely many β, hence I_{α} meets X_{n} in an infinite set.

In the remainder of this section we prove Lemma 2.4. We leave it to the reader to observe that if we fix names for a Hausdorff-Luzin type family or a finite-to-one graph type family then there is a family of ω_{1} dense subsets of the poset such that if a generic filter meets each of these dense sets then the interpretations of the above names will be either a Hausdorff-Luzin type family or a finite-to-one graph type family.

The proof will use the following result of Todorčević [Tod89b]. The proof is available in [Dow92] but since we require a slightly stronger form of it in one part of the proof (and for completeness) we include a proof in the form of Claim 4 of the proof of Theorem 2.3.

Lemma 2.10 ([Tod89b]). If I is an ideal of countable sets then either there are S_{n} (n ∈ ω) such that ⋃ I = ⋃_{n} S_{n} and [S_{n}]^{ω} ⊂ I for each n, or there is a proper poset P which forces an uncountable S ⊂ ⋃ I such that S is almost disjoint from every member of I.

Proof of Lemma 2.4. Assume that I and J are not countably separated. Let κ be a regular cardinal larger than |P(⋃ I)| (without loss of generality ⋃ J = ⋃ I). Let P be the countable condition collapse of |H(κ)| and let G be a P -generic filter. The following claim will be useful: Claim 1: If X ∈ V [G] is such that [X]^{ω} ∩ I is empty, then there is Y in H(κ)^{V} such that X ⊂ Y and [Y]^{ω} ∩ I is empty.

Proof of Claim 1: Let Ẋ be a name for X and let p_{0} ∈ P force that [Ẋ]^{ω} ∩ I is empty. Fix a countable elementary submodel M of a large enough H(θ) with p_{0}, P, Ẋ, I all in M. Fix I ∈ I such that I′ \ I is finite for each I′ ∈ M ∩ I. Recursively choose, if possible, descending p_{n} ∈ P ∩ M, and x_{n} ∈ I ∩ M (n ∈ ω) such that p_{n+1} ⊩ x_{n} ∈ Ẋ and, for n > 0, x_{n} ∉ {x_{0}, …, x_{n−1}}. This process must stop for some n, since P is countably closed and a lower bound would force that Ẋ ∩ I is infinite. Hence there is an n such that p_{n} ⊩ Ẋ ∩ I ∩ M is finite. Let Y be the set of all x such that there is a q < p_{n} which forces x to be in Ẋ. Clearly p_{n} ⊩ Ẋ ⊂ Y. Since Ẋ and p_{n} are both in M, it follows that Y is in M. Since I almost contains every member of I ∩ M it follows that Y meets every member of I ∩ M in a finite set. Therefore Y meets every member of I in a finite set.

It follows by the claim that I and J are not countably separated in V [G] and that the cardinality of I ∪ J is ω_{1}, in fact, H(κ)^{V} has cardinality ω_{1}. Fix an increasing continuous chain {M_{α} : α ∈ ω_{1}} of countable elementary submodels of H(κ)^{V} whose union is all of H(κ)^{V} and such that {M_{β} : β ≤ α} ∈ M_{α+1} for each α < ω_{1} .

For each α < ω_{1}, choose J_{α} ∈ J ∩ M_{α+1} such that J_{α} is not contained (mod finite) in any member of M_{α} which is almost disjoint from every member of I, and so that, if possible J_{α} ⊂ M_{α}. Note that such a set J_{α} will always exist because I and J are not countably separated. Note that if J_{α} cannot be taken to be a subset of M_{α}, then J_{α} \ M_{α} is a suitable choice for J_{α}, hence we assume that either J_{α} ⊂ M_{α} or J_{α} is disjoint from M_{α}. Similarly, we may choose I_{α} ∈ I ∩ M_{α+1} so that I_{α} contains, mod finite, every member of I ∩ M_{α}. We will prove that we can force an uncountable Y ⊂ ω_{1} so that {I_{α} : α ∈ Y } and {J_{α} : α ∈ Y } is either a Hausdorff-Luzin type family or a finite-to-one graph type family.

Claim 2: {I_{α} : α ∈ ω_{1}} and {J_{α} : α ∈ ω_{1}} are not countably separated in V [G] . Proof of Claim 2: Since {I_{α} : α < ω_{1}} is cofinal (mod finite) in I, it suffices to show that I and {J_{α} : α < ω_{1}} are not countably separated. By Claim 1, it suffices to show that there is no countable collection in H(κ)^{V} which witnesses that they are countably separated. However any countable subset of H(κ)^{V} will be contained in M_{α} for some α < ω_{1} and J_{α} will witness that this countable collection does not separate the two collections.

Let K be defined by

K ∈ K ↔ (∃m ∈ ω)(∀α ∈ ω_{1}){β ∈ K : |I ∩ J_{β}| > m} is finite.

The proof splits into two cases according to the application of Lemma 2.10 to the ideal K.

Case 1: There is no cover {Y_{n} : n ∈ ω} of ω_{1} such that [Y_{n}]^{ω} ⊂ K for all n ∈ ω. We will show that there is a proper poset R ∗ Q which introduces an uncountable set Y′ ⊂ ω_{1} such that {(I_{α}, J_{α}) : α ∈ Y′} is a Hausdorff-Luzin type family.

Case 2: There is a cover {Y_{n} : n ∈ ω} of ω_{1} such that [Y_{n}]^{ω} ⊂ K . In this case we fix any indexing {j(β, n) : n ∈ ω} of J_{β} for each β ∈ ω_{1} and we will show that there is an n and a proper poset Q which introduces a Y′ ∈ [Y_{n}]^{ω_{1}} such that {(I_{α}, J_{α}) : α ∈ Y′} is a finite-to-one graph type family. The steps in this case are more involved. We first show that we may assume that the J_{β} are pairwise disjoint (as required in the definition of a finite-to-one graph type family). We define g_{α} : α → ω by

g_{α}(β) = min{m : I_{α} ∩ J_{β} ⊂ {j(β, k) : k ≤ m}} .

We will then show that there is a forcing similar to the one used for the proof of Lemma 2.10 which will introduce Y′ so that g_{α} ↾ (α∩Y′) is finite-to-one as required.

In the sequel, when we refer to one of our ideals in an extension of the model in which it was defined, then we will mean the ideal generated by the ground model ideal. Proof of Case 1: It follows, by Lemma 2.10, that there is a proper poset R such that R introduces an uncountable set Y such that [Y]^{ω} ∩ K is empty. Define the poset Q = [Y]^{<ω} ordered by a < b if a ⊃ b and for each α ∈ b and each β ∈ α ∩ (a \ b) |I_{α} ∩ J_{β}| > |b|. Forcing with Q will introduce a Y′ ⊂ Y such that for each α ∈ Y′ and each m, {β ∈ Y′ ∩ α : |I_{α} ∩ J_{β}| < m} is finite. Showing that Q is ccc (and a simple density argument) establishes that Y′ can be forced to be uncountable.

To see that Q is ccc, let {F_{α} : α ∈ ω_{1}} be a family of finite subsets of Y. We may assume that all the F_{α} have cardinality m and that they form a Δ-system with root F. Let L_{0} be any infinite subset of ⋃_{n} F_{n} \ F. Since L_{0} ⊂ Y, L_{0} ∉ K, hence there is an I′_{0} ∈ I such that there is no bound on the cardinalities of I′_{0} ∩ J_{β} for β ∈ L_{0}. Choose L′_{0} ⊂ L_{0} so that the cardinalities of I′_{0} ∩ J_{β} diverge to infinity for β ∈ L′_{0}. Choose L_{1} an infinite subset of ⋃{F_{n} \ (F ∪ L′_{0}) : L′_{0} ∩ F_{n} ≠ ∅}. Again, find I′_{1} ∈ I and L′_{1} ⊂ L_{1} so that the cardinalities of I′_{1} ∩ J_{β} diverge to infinity for β ∈ L′_{1} . Repeat, obtaining I′_{k} and L′_{k} for k < K ≤ m, until T = {n : F_{n} ⊂ ⋃_{k<K} L′_{k}} is infinite. Let I ∈ I be the union of the I′_{k} and choose α such that I is almost contained in I_{ξ} for each ξ ∈ F_{α}\F. It is easily checked that for each ξ ∈ F_{α}\F there are at most finitely many n ∈ T for which there is a β ∈ F_{n} such that |I_{ξ} ∩J_{β}| < m. Therefore, there is an n ∈ T such that F_{α} is compatible with F_{n}.

Proof of Case 2:

As mentioned above we first show that we can assume that the J_{β} are pairwise disjoint (by restricting to a co-countable set). Claim 3: The set of β such that J_{β} ⊂ M_{β} is countable.

Let S be the set of α such that J_{α} ⊂ M_{α}. We first show, by contradiction, that S is not stationary. Fix n such that S_{0} = Y_{n} ∩ S is stationary and let Z_{0} = ω_{1}. For each γ < ω_{1}, there is a minimal m_{γ} < ω such that, for each α ∈ ω_{1}, there are only finitely many β ∈ Y_{n} ∩ γ for which |I_{α} ∩ J_{β}| is larger than m_{γ}. Since the I_{α} are increasing mod finite, it follows that the m_{γ}’s are bounded by a single m.

Since {I_{γ} : γ ∈ Z_{0}} is cofinal in I it follows that if X_{0} is the set consisting of those x which are members of I_{γ} for uncountably many γ, then X_{0} almost contains each I ∈ I. Fix a countable elementary submodel N_{0}, of some H(θ) which contains everything mentioned, such that N_{0} ∩ ω_{1} = α_{0} ∈ S_{0}. By Claim 1 (dualized), there is X′_{0} ∈ H(κ)^{V} ∩ N_{0} such that X′_{0} ⊂ X_{0} and I \ X′_{0} is finite for each I ∈ I. It follows that X′_{0} ∈ M_{α_{0}} (i.e. N_{0} ⊧ (∃β) X′_{0} ∈ M_{β}, hence there is such a β < α). Choose x_{0} ∈ J_{α_{0}} ∩ X_{0}, which we may do since the complement of X′_{0} contains mod finite every member of I and so cannot contain J_{α_{0}}. Since J_{α_{0}} ⊂ M_{α_{0}}, and α_{0} ∈ S_{1} = {β ∈ S_{0} : {x_{0}} ⊂ J_{β}} ∈ M_{α_{0}}, it follows that S_{1} is stationary. Also, Z_{1} = {γ ∈ Z_{0} : {x_{0}} ⊂ I_{γ}} is cofinal in Z_{0}, hence {I_{γ} : γ ∈ Z_{1}} is cofinal in I and again X_{1} = {x : x ∈ I_{γ} for uncountably many γ ∈ Z_{1}} almost contains every member of I. Therefore, by induction we can find {x_{0}, …, x_{m}} so that S_{m+1} = {β ∈ Y : {x_{0}, …, x_{m}} ⊂ J_{β}} is stationary and Z_{m+1} = {γ : {x_{0}, …, x_{m}} ⊂ I_{γ}} is uncountable. However, if γ ∈ Z_{m+1} is such that S_{m+1} ∩ γ is infinite, then {β ∈ S_{m+1} ∩ γ : |I_{γ} ∩ J_{β}| > m} is infinite contradicting that m “works” for Y_{n} ∩ γ. This completes the proof that S is not stationary.

Now we show that for all but countably many β, J_{β} ∩ M_{β} is empty. Indeed, there must be a C ⊂ [H(κ)^{V}]^{ω} which is closed and unbounded and such that for each M ∈ C and J ∈ J, there is a Y ∈ M such that [Y]^{ω} ∩ I = ∅ and J ∩ M ⊂ Y. Now if γ < ω_{1} is large enough such that C ∈ M_{γ}, then for all β ≥ γ, M_{β} ∈ C, hence β ∉ S.

For each β, fix an enumeration {j(β, m) : m ∈ ω} of J_{β}. Recall that we defined g_{α} : α → ω by

g_{α}(β) = min{m : I_{α} ∩ J_{β} ⊂ {j(β, k) : k ≤ m}} .

Clearly if {I_{α} : α ∈ ω_{1}} and {J_{β} : β ∈ Y_{n}} are countably separated for each n, then so are I and J. Therefore we may fix an n so that {I_{α} : α ∈ ω_{1}} and {J_{β} : β ∈ Y_{n}} are not countably separated. In fact, it follows immediately that we may assume that {I_{β} : β ∈ Y_{n}} and {J_{β} : β ∈ Y_{n}} are not countably separated.

To help us find our desired Y′ we define another auxiliary ideal L.

L ∈ L ↔ (∃m ∈ ω) (∀α ∈ Y_{n}){β ∈ L : I_{α} ∩ J_{β} ⊄ {j(β, ℓ) : ℓ ≤ m}} is finite

The idea is that members L of L are “bad” if {J_{β} : β ∈ L} are to be part of a finite-to-one graph family. Clearly we need to show that we can force a set Y′ ∈ [Y_{n}]^{ω_{1}} such that Y′ ∩ L is finite for each L ∈ L. Just as in Lemma 2.10, we need to show that Y_{n} cannot be written as a countable union, ⋃_{n} Z_{n}, such that [Z_{n}]^{ω} ⊂ L for each n. We show that such a sequence of Z_{n} will contradict that {I_{α} : α ∈ Y_{n}} and {J_{β} : β ∈ Y_{n}} are not countably separated.

Indeed, observe that if Z ⊂ ω_{1} is such that [Z]^{ω} ⊂ L, then there is an m such that

Z′ = ⋃{J_{β} \ {j(β, ℓ) : ℓ < m} : β ∈ Z}

is almost disjoint from every member of I. Indeed, for each γ, there is an m_{γ} witnessing that Z ∩ γ ∈ L and so there is an m so that m_{γ} = m for uncountably many γ. So to see that I ∩ Z′ is finite (for some I ∈ I), choose γ so that m_{γ} = m and I ∩ J_{β} = ∅ for β > γ.

By Lemma 2.10, there is a proper poset R which adds an uncountable Y ⊂ Y_{n} such that [Y]^{ω} ∩L is empty. However as we need a bit more we prove it and Lemma 2.10 now. The proof of Lemma 2.10 follows from Claim 4 by ignoring I since we will not use any properties of L other than the fact that Y_{n} cannot be expressed as a countable union of “L-homogeneous” sets. For each I ∈ I and β ∈ ω_{1}, let g_{I}(β) be the smallest integer such that I ∩ J_{β} ⊂ {j(β, ℓ) : ℓ < g_{I}(β)}.

Claim 4: There is a proper poset R such that R adds an uncountable set Y ⊂ Y_{n} such that [Y]^{ω} ∩ L is empty and, for each α ∈ Y, there is an I_{α} ∈ I such that g_{I_{α}} ↾ α is finite-to-one.

Define L̃ to be all Z ⊂ ω_{1} with the property that [Z]^{ω} ⊂ L. Recall that Y_{n} is not covered by a countable subcollection of L̃. As is usual (see §6 [Dow92]) define p ∈ R iff

p = A_{p} ∪ L_{p} ∪ M_{p} where \{ \begin{matrix}A_{p} ⊂ Y_{n} is finite, \\ L_{p} ⊂ L is finite, \\ M_{p} is a finite ∈-chain, \\ for α < β ∈ A_{p} ∃M ∈ M_{p} s.t. α ∈ M, β ∉ M, \\ α ∉ ⋃(M ∩ L̃) for any M ∈ M_{p} and α ∈ A_{p} \ M, \\ M ∈ M_{p} implies M ≺ H(θ) and |M| = ω . \end{matrix}

and p < q if p ⊃ q and for α ∈ A_{q} and β ∈ α ∩ (A_{p} \ A_{q}), β is not in ⋃ L_{q} and I_{α} ∩ J_{β} \ {j(β, ℓ) : ℓ < |A_{q}|} is not empty. Note that elementarity will guarantee that I_{α} contains mod finite every member of I ∩ M if M ∩ ω_{1} ≤ α.

To show that R is proper, fix a countable M ≺ H(ρ) for a suitably large ρ and let p ∈ M ∩R. We show that p∪{M ∩H(θ)} is (M, R)-generic. Let D ∈ M be a dense open subset of R and let r ∈ D be less than p ∪ {M ∩ H(θ)}. Let {α_{0}, …, α_{n−1}} list A_{r} \ M in increasing order. Also, let r_{0} = r ∩ M; note that r_{0} ∈ R. Let S denote the tree of finite sets s of ordinals in ω_{1} \ (max(A_{r_{0}}) + 1) (ordered by end-extension) for which there is a q ∈ D extending r_{0} such that |A_{q}| = |A_{r}| and A_{q} \ A_{r_{0}} end-extends s. Prune S ( starting from the top) to get T so that for each t ∈ T, T_{t} = {ξ : t^{⌢}ξ ∈ T } is not covered by countably many members of L̃. One uses the fact that M_{r} separates A_{r} and that α ∈ A_{r} \ M′ (for M′ ∈ M_{r}) implies that α is not in any member of L̃ ∩ M′ to show that {α_{0}, …, α_{i}} is in T for each i < n (including i = 0 which yields that ∅ ∈ T). The reason for introducing T is that T can be shown to be a member of M_{α_{0}} while D certainly is not. We must now work in M_{α_{0}} rather than M because we will want to know that the sets I_{α_{i}} are large enough with respect to certain other sets that arise.

Let L = ⋃ L_{r_{0}} and begin a finite recursion. Since ∅ ∈ T, and T ∈ M_{α_{0}}, T_{∅} ∈ M_{α_{0}} and M_{α_{0}} ⊧ T_{∅} ∉ L̃, hence there is a T′ ∈ M_{α_{0}} such that T′ ∈ [T_{∅}]^{ω} \ L. Since T′ ∉ L and I is a P -ideal, there is an I ∈ I ∩ M_{α_{0}} such that, for each m, there is a β_{m} ∈ T′ such that j(β_{m}, ℓ) ∈ I for some ℓ > m. Clearly {β_{m} : m ∈ ω} meets every member of L in a finite set, so we may assume that, in fact, T′ ∩ L is empty. Since I_{α_{i}} almost contains I for each i < n, there is a ξ_{0} ∈ T′ such that I_{α_{i}} ∩ J_{ξ_{0}} \ {j(ξ_{0}, ℓ) : ℓ < |A_{r_{0}}|} ≠ ∅ for each i < n. Continue choosing ξ_{i} ∈ M_{α_{0}} \ L (i < n) so that t = {ξ_{0}, …, ξ_{i}} ∈ T and so that I_{α_{j}} ∩ J_{ξ_{i}} \ {j(ξ_{i}, ℓ) : ℓ < |A_{r_{0}}|} is not empty for each j < n. The argument is the same as above with T_{t} replacing T_{∅}. When we have chosen {ξ_{0}, …, ξ_{n−1}} ∈ T ∩ M_{α_{0}}, we have that there is a q ∈ D ∩ M such that A_{q} = A_{r_{0}} ∪ {ξ_{0}, …, ξ_{n−1}}. One easily checks that q ∪ r is a member of R and by the careful choice of the ξ (i.e. the condition on I_{α_{j}} ∩ J_{ξ_{i}}) we have that q ∪ r is less than both q and r. This completes the proof that R is proper.

If G is R-generic, we let Y′ = ⋃{A_{p} : p ∈ G}. The condition on extension guarantees that g_{α} ↾ (Y′ ∩ α) is finite-to-one for each α ∈ Y. In addition [Y′]^{ω} ∩ L is finite for each L ∈ L since it is easily seen that there is a p ∈ G such that L ∈ L_{p} and then, by the definition of extension, it follows that Y′ ∩ L ⊂ A_{p}. Finally we must show that Y′ is uncountable. This follows directly from the fact that Y_{n} is not covered by countably many members of L̃. Indeed, if p ∈ R is arbitrary, and M is any countable elementary submodel of H(θ) containing p, then Y_{n} \ (L̃ ∩ M) is uncountable.

3. Applications of Theorem 2.3

The results in this section were shown to hold in models in which a supercompact cardinal is collapsed to ℵ_{2} by either the Levy or the Mitchell collapse. In particular Corollary 3.3 is from [DJW89], Corollaries 3.4 and 3.6 are from [DM90], and Corollary 3.5 is from [Dow88]. The notion of a coherent κ-matrix (of integer-valued functions) was introduced in [DJW89] (see Definition 3.1). It was shown that, for a regular uncountable cardinal κ, every space which can be written as an increasing κ-chain of first-countable subspaces would itself be first-countable if and only if every coherent κ-matrix had an extension to a coherent κ + 1-matrix.

Recall that for a cardinal κ, βκ denotes the Stone-Čech compactification of the space κ with the discrete topology. As usual we identify βκ with the Stone space of the Boolean algebra P(κ) in which the fixed ultrafilters are identified with the ordinals in κ. The subspaces κ^{∗} and U(κ) of βκ consist of the free ultrafilters and the uniform ultrafilters, respectively; where an ultrafilter is uniform if each of its members has cardinality κ.

Definition 3.1. For any ordinal number κ, call {f_{α,n} : α ∈ κ, n ∈ ω} a κ-matrix if each f_{α,n} ∈ ^{α}ω and k < n implies f_{α,k} ≤ f_{α,n}. We say that it is a coherent κ-matrix if for each β < α < κ, sup{f_{α,n}(β) : n ∈ ω} = ω and, for each n there is an m such that f_{β,n} ≤ f_{α,m} ↾ β and f_{α,n} ↾ β ≤ f_{β,m}.

Remark 3.1. The condition that f_{α,k} ≤ f_{α,n} for k < n was inadvertently left out of the definition of κ-matrix in [DJW89].

Definition 3.2. For a function g into ω, let g^{↓} denote the set of pairs (x, m) such that m ≤ g(x) (and x is in the domain of g).

Corollary 3.3 (PFA). If {g_{α,n} : α < κ, n ∈ ω} is a coherent κ-matrix of functions for a regular cardinal κ > ω_{1}, then the matrix extends to a κ + 1-matrix.

Proof. We define a pair of ideals I and J and then apply Theorem 2.3. Let I ∈ I if and only if I is a countable subset of κ×ω and such that I ∩g_{α,n}^{↓} is finite for each α < κ and n ∈ ω. It is easily checked that I is a P -ideal. Next, let J ∈ [κ × ω]^{ω} be a member of J if and only if there is an α < κ and n ∈ ω such that J ⊂ g_{α,n}^{↓}.

We first show that any I′ ∈ [I]^{ω_{1}} and J′ ∈ [J]^{ω_{1}} can be countably separated. Indeed, choose λ < κ large enough so that K ⊂ λ×ω for each K ∈ I′ ∪J′. The fact that the matrix is coherent easily implies that {g_{λ,n}^{↓} : n ∈ ω} countably separates I′ and J′.

Therefore by Theorem 2.3 and Lemma 2.2, there are {X_{n} : n ∈ ω} which strongly separate I and J. Without loss of generality X_{n} ⊂ X_{n+1} for each n. For each n, define h_{n} ∈ ^{κ}ω by

h_{n}(α) = min{k : X_{n} ∩ ({α} × [k, ω)) is empty} .

Let K be the set of γ such that the set {h_{n}(γ) : n ∈ ω} is bounded in ω. If we show that K is countable, then by adding only finitely points to each X_{n} we can actually assume that K is empty. Suppose that K is uncountable and fix any α ∈ κ so that there is an m and uncountably many γ ∈ K ∩ α such that h_{n}(γ) is bounded by m. Choose k ∈ ω so that g_{α,k}(γ) > m for uncountably many of these γ. But now this gives rise to a J ∈ J (i.e. a subset of g_{α,k}^{↓}) which is not contained, mod finite, in any X_{n}.

Now we show that we can use {h_{n} : n ∈ ω} to extend the matrix. Fix any α < κ, n ∈ ω and m ∈ M. By the definition of h_{m}, there is a k such that h_{m} ↾ α is bounded by g_{α,k}. We must also show there is a k so that g_{α,n} < h_{k}. This is the same as showing that g_{α,n}^{↓} is contained in some X_{k}. If there were no such k, we could easily define a countable J ⊂ g_{α,n}^{↓} such that J is not contained in any X_{k}. But now J ∈ J hence there is a k such that J is almost contained in X_{k} . Finally we finish by noting that the h_{n}’s are an increasing family and {h_{n}(γ) : n ∈ ω} is unbounded for each γ.

Corollary 3.4 (PFA). For each regular cardinal κ > ω_{1}, the space κ^{∗} \ U(κ) is C^{∗}-embedded in βκ.

Proof. Suppose that Z_{0} and Z_{1} are disjoint zero-set subsets of the space κ^{∗} \ U(κ). Now define I (respectively J) to be the ideal of all those countable subsets, I (respectively J) of κ such that I^{∗} ⊂ Z_{0} (respectively J^{∗} ⊂ Z_{1}). We first note that ℵ_{1}-sized subideals I′, J′ of I and J respectively are countably separated. Indeed, given such subideals there is a λ which contains I ∪ J for each I, J ∈ I′ ∪ J′. Furthermore, there is a Y ⊂ λ such that Y^{∗} ∩ Z_{0} is empty and Y^{∗} ⊃ Z_{1} ∩ λ^{∗}. This Y alone will countably separate I′ and J′.

Therefore there is a family of X_{n} which strongly separates the ideals. In this case J is also a P -ideal so we will show that some X_{n} actually contains, mod finite, every member of J. Otherwise, choose, for each n, a J_{n} ∈ J such that J_{n} \ X_{n} is infinite. Clearly then any J ∈ J which contains, mod finite, each J_{n} will not be contained, mod finite, in any X_{n}. It follows easily that \overline{X_{n}}, for some n, is a clopen subset of βκ which contains Z_{1} and is disjoint from Z_{0}.

Corollary 3.5 (PFA). A countably tight space is first countable if and only if every ≤ ℵ_{1}-sized subspace is first countable. Equivalently, a sequential or Fréchet space is first countable if every ≤ ℵ_{1}-sized subspace is first countable.

Proof. Let Y be a countably tight space and assume that each ≤ ℵ_{1}-sized subspace is first countable. Fix any y ∈ Y. Define I to be the ideal of countable subsets of Y \ {y} which converge to y. Since each countable subspace is first countable, I is a P -ideal. Let J be the ideal of countable subsets of Y which do not have y as a limit point. It is easily seen that if {(I_{α}, J_{α}) : α < ω_{1}} is a family as in (2), then the subspace {y} ∪ ⋃_{α} J_{α} is not first countable. Therefore we assume that {X_{n} : n ∈ ω} is as in (1) and show that Y is first countable at y. Again, without loss of generality, each X_{n} is almost disjoint from every member of I. Suppose that Z has cardinality ℵ_{1} and is a subset of Y \ ⋃_{n} X_{n} (if this set is uncountable). If Z ∪ {y} is first countable, then there is an infinite J ⊂ Z which does not have y as a limit point, i.e. J ∈ J. Since there is no X_{n} which almost contains J, Y \ ⋃_{n} X_{n} must be countable. Therefore we may assume Y \ {y} = ⋃_{n} X_{n}. Since X_{n} is almost disjoint from each member of I and since Y is sequential (this follows from countable tightness and first countable on countable subsets), it follows that y is not a limit point of any X_{n}. Now suppose that y is not in the closure of some Z ⊂ Y. To prove that Y is first countable at y we show that Z is almost contained in some X_{n}. Indeed, if it were not, then Z would have a countable subset J which also was not contained (mod finite) in any of the X_{n}. However y is not a limit of J ⊂ Z, hence J ∈ J. This contradicts that some X_{n} should almost contain J.

Corollary 3.6 (PFA). If {Z_{α} : α < κ} (with cf(κ) > ω_{1}) is a sequence of zerosubsets of κ^{∗} with the property that Z_{α} = Z_{β} ∩ α^{∗} for each α < β < κ, then there is a zero set Z of βκ such that Z ∩ α^{∗} = Z_{α} for each α < κ.

4. More on ω^{∗}

In this section we answer questions that have been raised about the variety of dense C^{∗}-embedded subsets of ω^{∗}. Van Douwen, Kunen and van Mill [vDKvM89] have shown that it is consistent that there are proper dense (even co-dense) subsets of ω^{∗} which are C^{∗}-embedded and have asked if it is consistent that there are dense subsets of size 𝔠 which are C^{∗}-embedded. We show that this follows from PFA (in fact MA plus Todorčević’s strong form of the Open Coloring Axiom). We also show that it holds in the Cohen model. It is well-known that it follows from CH that no proper dense subset is C^{∗}-embedded. We let C(X) denote the ring of real-valued continuous functions on a space X. Let [X]^{2} denote the set of two-element subsets of X (which we identify with the set of ordered pairs (x, y) such that x < y in the case that X is ordered).

Definition 4.1. OCA denotes the statement that if [X]^{2} = K_{0} ∪K_{1} where X ⊂ R and where K_{0} is open in [X]^{2}, then either there is an uncountable 0-homogeneous set or X is the countable union of 1-homogeneous sets. A set Y ⊂ X is 0-homogeneous (respectively 1-homogeneous) if [Y]^{2} ⊂ K_{0} (respectively K_{1}).

Todorčević shows that this form of OCA follows from PFA; hence we may assume MA plus OCA is consistent with ZFC. A much weaker form of OCA was first shown to be consistent in [AS81]. For a family A ⊂ [ω]^{ω}, let ∂^{∞}A denote the set of complete accumulation points of A, i.e. ∂^{∞}A = {x ∈ ω^{∗} : for each u in the ultrafilter x, |A| = |{a ∈ A : u ∩ a ≠^{∗} ∅}|}.

The main theorem of this section follows easily from Lemmas 4.3 and 4.4 by applying simple cardinal arithmetic.

Theorem 4.2. If V is a model of Martin’s Axiom plus OCA or if it is obtained by adding at least 2^{λ} Cohen reals to some ground model satisfying 𝔠 = λ then there are dense C-embedded subspaces of ω^{∗} which have cardinality 𝔠.

Proof. By Lemma 4.3 or 4.4, simply choose an X ∈ [ω^{∗}]^{𝔠} which satisfies the hypotheses of the lemma. That this can be done simply relies on the fact that there are only continuum many A ⊂ [ω]^{ω} which must be considered. In the case that Martin’s Axiom holds we have that 2^{<𝔠} = 𝔠 while in the case that V is obtained by adding at least 2^{λ} Cohen reals, then it suffices to observe that [P(ω)]^{≤λ} has cardinality 𝔠.

Lemma 4.3 (MA+OCA). Assume that X ⊂ ω^{∗} is such that for any A ∪ C ⊂ [ω]^{ω} such that |A ∪ C| < 𝔠 and if ∂^{∞}A ∩ ∂^{∞}C ≠ ∅ then X meets this intersection. Then X is C-embedded in ω^{∗}.

Proof. Let X ⊂ ω^{∗} satisfy the hypotheses of the Lemma and assume that f ∈ C(X). By Lavrentieff’s theorem, f will extend continuously to a G_{δ}-set D with X ⊂ D ⊂ ω^{∗}. Since X meets ∂^{∞}A for each countable A ⊂ [ω]^{ω} and non-empty G_{δ} subsets of ω^{∗} have dense interior, it should be clear that f is a bounded function. Therefore if f does not extend continuously to ω^{∗}, there are reals r < s such that the closure of A_{1} = f^{−1}([s, ∞)) meets the closure of A_{0} = f^{−1}((−∞, r]). Let X denote all disjoint pairs (a, b) of infinite subsets of ω such that a^{∗} ⊂ A_{0} and b^{∗} ⊂ A_{1}. We may think of X as a subset of the square of the Cantor set by identifying a subset of ω with its characteristic function. Define a partition K_{0} ∪ K_{1} of [X]^{2} by ⟨(a, b), (c, d)⟩ ∈ K_{0} if (a ∪ c) ∩ (b ∪ d) is not empty. It is easily seen that K_{0} is open in [X]^{2}. By OCA, there is either

1. an uncountable 0-homogeneous Y ⊂ X, or

2. a countable family of 1-homogeneous sets whose union is X.

We show that both situations lead to a contradiction. In the first instance, assume that {(a_{α}, b_{α}) : α < ω_{1}} is a 0-homogeneous set. We first show that ⋃{a_{α}^{∗} : α < ω_{1}} and ⋃{b_{α}^{∗} : α < ω_{1}} do not have disjoint closures. Indeed, if they did then there would be an a ⊂ ω such that a_{α} \a and b_{α} ∩a are finite for all α ∈ ω_{1}. Then we find an uncountable J and an n ∈ ω so that for all α ∈ J, a_{α} \ a ⊂ n and b_{α} ∩ a ⊂ n. In addition, we may arrange that a_{α} ∩ n and b_{α} ∩ n are the same for all α ∈ J. But now, if α < β are both in J, it follows that (a_{α} ∪ a_{β}) ∩ (b_{α} ∪ b_{β}) is empty, which contradicts that this family was to be 0-homogeneous. Note that each a_{α} is almost disjoint from each b_{β} since a_{α}^{∗} ⊂ A_{0} and b_{β}^{∗} ⊂ A_{1}. By MA (i.e. b > ω_{1}), it follows that ∂^{∞}{a_{α} : α < ω_{1}} meets ∂^{∞}{b_{α} : α < ω_{1}}, and therefore this intersection should meet X. However this would clearly contradict that f is continuous on X.

Now suppose that X is a countable union of 1-homogeneous sets and fix p in \overline{A_{0}} ∩ \overline{A_{1}}. By induction on cardinality it follows that if λ < 𝔠 and \{y_{α} : α < λ\} is a subset of the ultrafilter p, then

D ∩ A_{0} ∩ ⋂{y_{α}^{∗} : α < λ}

has non-empty interior (and similarly with A_{1} replacing A_{0}). Indeed, for each α < λ, choose, by the inductive hypothesis, a_{α} so that a_{α}^{∗} is contained in

D ∩ A_{0} ∩ ⋂{y_{β}^{∗} : β < α} .

Clearly ∂^{∞}{a_{α} : α < λ} meets itself, hence there is an element, x, of X in ∂^{∞}{a_{α} : α < λ}. Since D is a G_{δ}, there are b_{n} ∈ x (n ∈ ω) so that ⋂_{n} b_{n}^{∗} ⊂ D and b_{n}^{∗} ∩ D ⊂ f^{−1}((−∞, r + 1/n)). Observe that x ∈ y_{β}^{∗} for each β < λ, hence, by MA,

⋂{b_{n}^{∗} : n ∈ ω} ∩ ⋂{y_{β}^{∗} : β < λ}

has interior contained in D ∩ A_{0}, as was to be proved.

Now fix an enumeration, {y_{α} : α < c}, of p and choose, for each α < 𝔠, disjoint infinite a_{α} and b_{α} so that

a_{α}^{∗} ⊂ A_{0} ∩ ⋂{y_{β}^{∗} : β < α}

and

b_{α}^{∗} ⊂ A_{1} ∩ ⋂{y_{β}^{∗} : β < α} .

Note that for each J ⊂ 𝔠 with |J| = 𝔠, p is a limit point of both ⋃{a_{α}^{∗} : α ∈ J} and ⋃{b_{α}^{∗} : α ∈ J}, hence ⋃{a_{α} : α ∈ J} is not disjoint from ⋃{b_{α} : α ∈ J}.

However since {(a_{α}, b_{α}) : α ∈ c} is a countable union of 1-homogeneous sets, there is a J ⊂ 𝔠 of cardinality 𝔠, so that {(a_{α}, b_{α}) : α ∈ J} is 1-homogeneous. This is our desired contradiction since 1-homogeneity guarantees that ⋃{a_{α} : α ∈ J} is, in fact, disjoint from ⋃{b_{α} : α ∈ J}.

Lemma 4.4. Let G be Fn(κ, 2)-generic over a model V and let λ = 𝔠^{V}. In V [G], suppose X ⊂ ω^{∗} is such that for any A ∪ C ⊂ [ω]^{ω} such that ∂^{∞}A ∩ ∂^{∞}C ≠ ∅ and |A ∪ C| ≤ λ, then X meets this intersection. Then X is C-embedded in ω^{∗}.

Proof. Let X ⊂ ω^{∗} satisfy the hypotheses of the Lemma and assume that f ∈ C(X). By Lavrentieff’s theorem, f will extend continuously to a G_{δ}-set D with X ⊂ D ⊂ ω^{∗}. Just as in the proof of Lemma 4.3, assume that f has been extended continuously to all of D and it does not extend continuously to ω^{∗}. Therefore there are reals r < s such that the closure of f^{−1}([s, ∞)) meets the closure of f^{−1}((−∞, r]). Let X denote all infinite subsets a of ω such that f[a^{∗}] ⊂ (−∞, r] and similarly Y is all b ⊂ ω such that f[b^{∗}] ⊂ [s, ∞). Since D is a dense G_{δ} and non-empty G_{δ}’s of ω^{∗} have dense interior, it follows that X cannot be separated from Y, i.e. if A ⊂ ω is such that a \ A is finite for each a ∈ X, then there is a b ∈ Y such that A ∩ b is infinite.

Let θ be a sufficiently large regular cardinal in V and let M ≺ H(θ) contain Fn(κ, 2)-names for each of X and Y and suppose that M^{ω} ⊂ M. It follows that M[G] is an elementary submodel of H(θ)[G] and that H(θ)[G] is the H(θ) in the sense of V [G]. In addition, since M^{ω} ⊂ M, it follows that M[G] ∩ [ω]^{ω} is contained in V [G∩M] and, countable subsets of M[G]∩[ω]^{ω} which are members of V [G∩M] are also members of M[G]. For proofs of these facts see §4 of [Dow92].

We will show that the interpretations of A = X ∩ M and C = Y ∩ M also cannot be separated in V [G]. Therefore if |M| ≤ λ, it will follow that X ∩ ∂^{∞}A ∩ ∂^{∞}C ≠ ∅ contradicting that f is continuous on X.

Working in V [G ∩ M], suppose that there is a name Ȧ which is forced to (mod finite) contain every member of X ∩ M[G] and to be (mod finite) disjoint from every member of Y ∩ M[G]. For each condition p (in the countable name Ȧ), let X_{p} denote all those members of X ∩ M[G] which are forced by p to be contained in Ȧ. Clearly, then X_{p} is separated from Y ∩ M[G] by the set {n : p ⊩ n ∈ Ȧ} which is a member of M[G] hence it follows by elementarity that X_{p} is separated from all of Y. It follows then that it suffices to show that if {X_{n} : n ∈ ω} ⊂ P(ω) (in V [G ∩ M]) is such that X_{n} ∩ b is finite for each b ∈ Y and n ∈ ω, then there is an a ∈ X ∩ M[G] such that a \ X_{n} is infinite for each n. As explained above, it follows that {X_{n} : n ∈ ω} ∈ M[G], hence, by elementarity, it suffices to show that there is an a ∈ X such that a \ X_{n} is infinite for each n. Observe that, since X_{n}^{∗} ∩ f^{−1}([s, ∞)) is empty for each n, it follows that f^{−1}((∞, r]) − ⋃_{k<n} X_{k}^{∗} is not empty for each n. For each n, choose a_{n} ∈ X such that a_{n} ∩ X_{k} = ∅ for each k < n. Let x ∈ X be a member of ∂^{∞}{a_{n} : n ∈ ω}. Clearly, f(x) ≤ r. Since D is a G_{δ}, we may choose c_{n} ∈ x (n ∈ ω) so that ⋂_{n∈ω} c_{n}^{∗} is contained in D and, for each n, f(c_{n}^{∗} ∩ D) ⊂ (−∞, r + \frac{1}{n}). Let a ∈ [ω]^{ω} be such that a \ c_{n} is finite for each n. It is easily seen that a ∈ X, and since c_{n} ∩ X_{n} is empty, it follows that a \ X_{n} is infinite for each n.

5. ω^{∗} minus a point in the Miller model

It is shown in [vDKvM89] that it is consistent to suppose that ω^{∗} \ {p} is C^{∗}- embedded for every p ∈ ω^{∗} (e.g. from PFA) and Malyhin has shown that this also holds in the Cohen model. It is asked in [HvM90] if there is a model in which ω^{∗} \ {p} is C^{∗}-embedded for some but not all p ∈ ω^{∗}. We establish that such a model exists: the one obtained by iterating Miller forcing. We obtain that, in this model, ω^{∗} \ {p} is C^{∗}-embedded iff p is not a P -point.

Recall (see [Mil84]) that the Miller poset Q consists of rooted trees T ⊂ ^{<ω}ω with the property that for each t ∈ T, there is an extension t′ ∈ T of t which is branching in the sense that t′ has infinitely many immediate successors in T. The ordering on Q is inclusion: T < T′ if T ⊂ T′. Blass and Shelah [BS87] have shown that the character of every P -point ultrafilter on ω is ω_{1} in the Miller model (see Corollary 5.2 for a very brief sketch). For convenience, we will assume that if T ∈ Q and t ∈ T, then t consists of an increasing function (we can work below any T with this property).

Lemma 5.1. If p ∈ ω^{∗} has a base of cardinality ω_{1}, then ω^{∗} \ {p} is not C^{∗}- embedded in ω^{∗}.

Proof. This is essentially well-known. If p is a P -point then ω^{∗} \ {p} is homeomorphic to ω_{1}^{∗} \ U(ω_{1}) (the so-called sub-uniform ultrafilters on ω_{1}) which, obviously, has a non-trivial compactification, namely ω_{1}^{∗}. Therefore the Stone-Čech compactification of ω^{∗} \ {p} is not ω^{∗}, i.e. the one-point compactification.

If p is not a P -point then fix a pairwise disjoint sequence {b_{n} : n ∈ ω} ⊂ [ω]^{ω} \ p so that every member of p meets infinitely many of the b_{n} in an infinite set. Let {i_{α} : α < ω_{1}} enumerate a generating set for the dual ideal of p such that i_{n} ⊃ b_{n} for each n. We will choose a partition, a_{α} ∪ b_{α}, of each i_{α} so that if β < α, then a_{α} ∩ i_{β} is almost equal to a_{β} ∩ i_{α}. In addition, we must ensure that for each β there is an α so that a_{α} \ i_{β} is infinite. We have already chosen b_{n} and let a_{n} = i_{n} \ b_{n}. Now consider stage α. If i_{α} is in the ideal generated by {i_{β} : β < α} then simply let a_{α} be a suitable finite union of previously chosen a_{β} ∩ i_{α}. Otherwise, fix some infinite j_{α} ⊂ i_{α} such that j_{α} ∩ i_{β} is finite for all β < α. We wish to choose a_{α} ⊂ i_{α} so that

1. j_{α} ⊂ a_{α},

2. for each β < α, a_{β} ∩ i_{α} is almost contained in a_{α},

3. for each β < α, b_{β} ∩ i_{α} is almost disjoint from a_{α}. That is, we are simply asking for a set to separate the family {j_{α}}∪{a_{β} ∩i_{α} : β < α} from the family {b_{β} ∩ i_{α} : β < α}. Since every member of the first family is almost disjoint from every member of the second and both families are countable, this is easily done. We let b_{α} = i_{α} \ a_{α}.

Clearly there are uncountably many α such that i_{α} is not in the ideal generated by {i_{β} : β < ω_{1}} and, for each such α, there was a j_{α} chosen to witness that a_{α} is not in the ideal generated by {i_{β} : β < α}.

Now by the coherence property of the a_{α}’s with respect to the i_{β}’s, it follows that U = ⋃{a_{α}^{∗} : α < ω_{1}} is a clopen subset of ω^{∗} \ {p}. In addition, since no i_{β} contains mod finite all the a_{α}, U is not a compact set (it has p in its closure). Since b_{n}^{∗} ∩ U is empty for each n, it follows that ω^{∗} \ U also has p as a limit. Therefore there is a two-valued function on ω^{∗} \ {p} which does not extend continuously to ω^{∗}.

Corollary 5.2. If G is P_{ω_{2}}-generic over V, where V ⊧ CH and P_{ω_{2}} is the ω_{2}- length countable support iteration of Miller forcing, then ω^{∗}\{p} is not C^{∗}-embedded in ω^{∗} for any P -point ultrafilter p ∈ ω^{∗}.

Proof. Suppose that p is a P -point ultrafilter on ω in V [G]. Fix a name for p and choose an elementary submodel M of H(θ) (in V) for a large enough θ so that M contains the name, M^{ω} ⊂ M and |M| = ω_{1}. Again, standard arguments yield that p ∩ M[G] is a member of V [G ∩ M] and, in that model, is a P -point ultrafilter on ω (see, for example, §4 of [Dow92]). Also, V [G] is obtained from V [G ∩ M] by forcing with a countable support iteration of Miller posets. Miller [Mil84] has shown that single stage Miller forcing preserves P -points and it is shown in Blass and Shelah [BS87], that any countable support iteration of proper forcings that preserve P -points also preserves them. Therefore, p ∩ M[G] generates p in the final model. Since V [G ∩ M] ⊧ CH, it follows that p has a base of cardinality ω_{1}. Now we apply Lemma 5.1.

Theorem 5.3. Let, for λ ≤ ω_{2}, P_{λ} be the λ-length countable support iteration of Miller forcing Q. If G is P_{ω_{2}}-generic over V, a model of CH, and if, in V [G], p ∈ ω^{∗} is not a P -point, then ω^{∗} \ {p} is C^{∗}-embedded in ω^{∗}.

Proof. In V [G], fix a partition {C_{n} : n ∈ ω} of ω so that for each U ∈ p, there are infinitely many n so that U ∩ C_{n} is infinite. Also let f ∈ C^{∗}(ω^{∗} \ {p}) and assume that p is in the closure of f^{−1}(0) and f^{−1}(1). Let C denote the set ⋃_{n} C_{n}^{∗} and define I_{0} to be the ideal of subsets, a, of ω such that a^{∗} ⊂ f^{−1}(0) \ \overline{C} and similarly define I_{1} to be those a such that a^{∗} ⊂ f^{−1}(1) \ \overline{C}. Claim: Each of I_{0} and I_{1} are P -ideals and every member of p contains a member of I_{0} and a member of I_{1}.

By symmetry we just prove the claim for I_{0}. Let u ∈ p. First we find an x ∈ u^{∗} \ C such that f(x) = 0. Since (u \ ⋃_{n<k} C_{n})^{∗} ∩ f^{−1}(0) is not empty for each k, choose x_{k} in this intersection for each k. The sequence {x_{k} : k ∈ ω} has 2^{𝔠} limit points, none of which are in C; let x be such a limit point. Fix a v ⊂ u such that v ∈ x \ p. Now f restricted to v^{∗} is continuous and v^{∗} is homeomorphic to ω^{∗} hence the zero set v^{∗} ∩ f^{−1}(0) \ C has non-empty interior. Now suppose that I is a countable subset of I_{0}. Since each member of I_{0} is almost disjoint from each C_{n}, it follows that there is a set u which is almost disjoint from C_{n} for each n and which contains mod finite each member of I. Again, restricting f to u^{∗} allows us to utilize the fact that the ideal of a ⊂ u such that a^{∗} ⊂ f^{−1}(0) ∩ u^{∗} is a P -ideal. Since I is contained in this ideal this completes the proof of the Claim.

Fix P_{ω_{2}} names for p, the C_{n}’s, and the ideals I_{0} and I_{1}. Since P_{ω_{2}} is proper and has the ℵ_{2}-c.c., there is a λ < ω_{2} of uncountable cofinality so that

1. p′ = p ∩ V [G_{λ}] is an ultrafilter on ω in V [G_{λ}];

2. ⟨C_{n} : n ∈ ω⟩ ∈ V [G_{λ}];

3. I′_{0} = I_{0} ∩V [G_{λ}] and I′_{1} = I_{1} ∩V [G_{λ}] are members of V [G_{λ}] and are P -ideals; and

4. every member of p′ contains an infinite member of each of I′_{0} and I′_{1}.

Since V [G_{λ}] is a model of CH, there are {a_{α} : α < ω_{1}} increasing mod finite and cofinal in I′_{0} and {b_{α} : α < ω_{1}} increasing mod finite and cofinal in I′_{1}. Let g_{λ} be the generic real added by Q_{λ} (recall that g_{λ} is an increasing function). Let C^{↓} = ⋃_{n}(C_{n} ∩ g_{λ}(n)) and for each α, let a_{α}^{↓} = a_{α} ∩ C^{↓} and b_{α}^{↓} = b_{α} ∩ C^{↓}. We finish the proof by showing that, in V [G], the family {a_{α}^{↓} : α < ω_{1}} is not separated from {b_{α}^{↓} : α < ω_{1}}. This will complete the proof since (C^{↓})^{∗} ∩ f^{−1}(0) is contained in a clopen subset of ω^{∗} which is disjoint from (C^{↓})^{∗} ∩ f^{−1}(1) because C^{↓} ∉ p. We proceed in two steps, the first is to show that they are not separated in V [G_{λ+1}] and the second is an easy appeal to results of Shelah [She84] for the preservation by the rest of the iteration. Claim: for every a ∈ [ω]^{ω} ∩ V [G_{λ+1}] such that a_{α}^{↓} ⊂^{∗} a for all α < ω_{1}, there is an α such that a ∩ b_{α}^{↓} is infinite.

To prove this we work in V [G_{λ}] and force with Q. Recall that Q ⊂ P(^{<ω}ω) and that g_{λ} is the union of roots of conditions T which are in the generic filter G such that G_{λ+1} = G_{λ} ∗ G. Let ȧ be the Q-name of a subset of C^{↓} and assume that T ∈ Q is such that T ⊩ |a_{α}^{↓} \ ȧ| < ω for each α < ω_{1}. Let M be a countable elementary submodel of some H(θ) which includes ⟨C_{n} : n ∈ ω⟩, T, ȧ and the sequence {a_{α}, b_{α} : α < ω_{1}}. Since a_{α}^{↓} is simply equal to a_{α} ∩ C^{↓} and C^{↓} is defined in terms of ⟨C_{n} : n ∈ ω⟩ and the canonical generic real it shouldn’t be necessary to specify names for a_{α}^{↓} (and similarly b_{α}^{↓}).

It suffices to show that there is a T′ < T such that T′ ⊩ ȧ ∩ b_{α}^{↓} is infinite where α = M ∩ ω_{1}. This is done by a standard fusion argument in which the following is the main step. Claim: If T̃ < T is a member of M and m ∈ ω, there is a T′ < T̃ (again in M) such that T′ ⊩ (∃k > m) k ∈ ȧ ∩ b_{α}^{↓}.

Let A = {ℓ : (∃T′ < T̃) T′ ⊩ ℓ ∈ ȧ}. Towards a contradiction, assume that there is a β such that a_{β} \ A is infinite. Since a_{β} ∩ C_{n} is finite for each n, there is a J ∈ [ω]^{ω} and a function h ∈ ^{J}ω such that h(n) ∈ a_{β} \ A for each n ∈ J. Let S ∈ Q and m ∈ ω be arbitrary, let s ∈ S be the root and let n be the smallest member of J which is larger than m + |s|. Since s has infinitely many immediate successors in S, there is an s^{⌢}j in S with j > h(n). It follows that S has an extension which forces g_{λ}(n) to be larger than h(n). Since S and m were arbitrary and since h ∈ V [G_{λ}] it follows that, in V [G_{λ+1}], the set {n ∈ J : g_{λ}(n) > h(n)} is infinite. But this is a contradiction since, T̃ forces that a_{β}^{↓} ⊂^{∗} ȧ ⊂ A, while {h(n) : n ∈ J and h(n) < g_{λ}(n)} is an infinite subset of a_{β}^{↓} \ A. So we have established that A is a member of V [G_{λ}] which contains mod finite each a_{β}, hence it follows that there is a β such that A ∩ b_{β} is infinite. Since A is in M and b_{α} contains mod finite each b_{β} ∈ M, it follows that A ∩ b_{α} is infinite. Fix any k > m in this intersection and since k ∈ A there is T′ < T so that T′ ⊩ k ∈ ȧ. Since T ⊩ ȧ ⊂ C^{↓}, it follows that T′ ⊩ k ∈ b_{α}^{↓} = b_{α} ∩ C^{↓}.

Now for the fusion argument. For a condition T ∈ Q and a t ∈ T, the set T_{t} = {s ∈ T : s is comparable with t} is again a member of Q. If A is a maximal set of pairwise incomparable elements of T ∈ Q and for each t ∈ A we are given a T′_{t} ≤ T_{t} in Q, then T (A) = ⋃{T′_{t} : t ∈ A} < T is again a member of Q. Note that the set of branching nodes of T which have precisely n branching predecessors is a maximal set of pairwise incomparable elements. Furthermore, for each branching t ∈ T which is contained in a member of A, t has exactly the same set of immediate successors in T (A) defined as above as it had in T. If, T_{n} = T_{n−1}(A_{n}) is defined recursively, where T_{0} = T and A_{n} is the set of branching nodes of T_{n−1} which have precisely n branching predecessors, then T_{ω} = ⋂ T_{n} is a member of Q. Note that if, for each n, there is a statement φ_{n} of the forcing language such that, at stage n, each T′_{t}, for t ∈ A_{n}, which is used in the construction of T_{n}, forces that φ_{n} holds, then T_{ω} forces that φ_{n} holds for each n. We use this exact process where φ_{n} is the assertion that there is a k > n in ȧ ∩ b_{α}^{↓}. The only inductive assumption necessary is that each T′_{t} is chosen to be a member of M (it is not necessary that T_{n} itself be a member of M). The inductive step is simply as follows: suppose we have chosen T_{n} and t ∈ A_{n+1}. There is a unique t′ ≤ t in A_{n} and, by assumption T′_{t′} is a member of M. By the construction of T_{n}, it follows that (T_{n})_{t} = (T′_{t′})_{t}, is again a member of M. By the above claim, (T_{n})_{t} has an extension T′_{t} as required.

The proof of the theorem is completed by the next lemma.

The following result is also given in [Laf93].

Lemma 5.4. Suppose that A and B are P -ideals in P(ω) which cannot be separated. Then A and B remain unseparated after forcing with a countable support iteration of Miller forcings.

Proof. Obviously we may assume that every member of A is almost disjoint from every member of B. Following the notation of Shelah [She84], unseparated gaps can be coded as a nice pair. For each a ∈ A and b ∈ B, let f_{a,b}(n) = 1 if n ∈ a and f_{a,b}(n) = 2 if n ∈ b \ a and let f_{a,b} = 0 for other values of n. Let F be the set of all f_{a,b} for a ∈ A and b ∈ B. Let R be the following (absolute) two-place relation on ω^{ω}: g R f iff {n : f(n) = 1 ≠ g(n)} is infinite or {n : g(n) = 1 and f(n) = 2} is infinite (i.e. the idea is that f = f_{a,b} and g^{−1}(1) either meets b infinitely or it does not contain a). Recall that F is said to be R-bounding if for every g there is an f ∈ F such that g R f (that is to say that A and B are not separated). To apply the results of [She84] we must show the following: given a countable elementary submodel M there is an f ∈ F so that for every m_{0}, Player II has an absolute winning strategy for the following game. On the k-th move, \underline{player I} chooses g_{k} ∈ ^{ω}ω, f_{k} ∈ F ∩ M, such that g_{k} ↾ m_{ℓ+1} = g_{ℓ} ↾ m_{ℓ+1} for 0 < ℓ < k and g_{k}Rf_{k}, then \underline{player II} chooses m_{k+1} > m_{k}. In the end, player II wins if ⋃_{k} g_{k} ↾ m_{k} R f.

Given M, choose a ∈ A and b ∈ B such that a contains mod finite every member of M ∩ A and b contains mod finite every member of M ∩ B. Let f = f_{a,b}. In a play of the game, we are given g_{k}Rf_{k}. Since f_{k} ∈ F ∩ M, there are a′ ∈ M ∩ A and b′ ∈ B such that f_{k} = f_{a′,b′}. Since {n : f_{a′,b′}(n) = 1 ≠ f_{a,b}(n)} ∪ {n : f_{a′,b′}(n) = 2 ≠ f_{a,b}(n)} is finite, there is an m_{k+1} such that {n < m_{k+1} : f_{a,b}(n) = 1 ≠ g_{k}(n)} ∪ {n : g_{k}(n) = 1 and f_{a,b}(n) = 2} has size at least k. It follows then that ⋃_{k} g_{k} ↾ m_{k} R f.

By the results of [She84] then, it is sufficient to show that Q itself preserves that A cannot be separated from B. This is somewhat easier than the argument in Theorem 5.3 since we do not have to worry about intersection with C^{↓} so we will omit the details.

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