TWO TO ONE IMAGES AND PFA

ALAN DOW

Abstract. We prove that all maps on N^{∗} that are exactly two to one, are trivial if PFA is assumed.

1. Introduction

A map f : X → K is precisely two to one if for each k ∈ K, there are exactly two points of X that map to k. For the remainder of the paper we are assuming that f is a precisely two to one mapping from N^{∗} onto some (compact) space K. The question of whether there are non-trivial two to one maps on N^{∗} is motivated by the papers of van Douwen [vD93] and R. Levy [Lev04]. In particular, Levy asks if every two to one image of N^{∗} is homeomorphic to N^{∗}. In fact there are several questions in [Lev04] that are consistently answered by the results in this paper. The behavior of two to one maps on N^{∗} when CH is assumed is investigated in [DT04]. It is well known that van Douwen has shown that there is a compact separable space which is a ≤2 to one image of N^{∗} and this pathology motivates the current study. R. Levy showed that if f is precisely two to one on N^{∗} then K will have weight equal to 𝔠 and countable discrete subsets of K will have closure homeomorphic to βN.

Proposition 1. If f is locally one to one (every point has a neighborhood on which f is one to one), then N can be partitioned into a ∪ b such that f[a^{∗}] = f[b^{∗}] = K. Since f is two to one, f is then a homeomorphism on each of a^{∗} and b^{∗}.

Proof. If each point of N^{∗} has a neighborhood on which f is one to one, then there is a finite cover by such neighborhoods. Let A be a finite partition of N such that f is one to one on each a^{∗} ∈ A. Enumerate A = {a_{i} : i ≤ n}. We will use induction on n. Consider the compact set B_{0} = f^{−1}(f[a_{0}^{∗}]) \ a_{0}^{∗} and note that f[B_{0}] = f[a_{0}^{∗}]. Since f is two to one, f[B_{0} ∩ a_{1}^{∗}] is disjoint from f[⋃_{1<j} a_{j}^{∗}]. Therefore there is a c_{1} ⊂ a_{1} such that B_{0} ∩ a_{1}^{∗} ⊂ c_{1}^{∗} and f[c_{1}^{∗}] also disjoint from f[⋃_{1<j} a_{j}^{∗}]. Since f is precisely two to one, and is one to one on c_{1}^{∗}, it follows that f[c_{1}^{∗}] ⊂ f[a_{0}^{∗}]. That is, we have shown that B_{0} ∩ a_{1}^{∗} = c_{1}^{∗}. The same argument applies for each i > 0 replacing 1, hence B_{0} is equal to b_{0}^{∗} for some infinite b_{0} ⊂ N \ a_{0}. It follows that the restriction of f to the union of {(a_{1} \ b_{0})^{∗}, (a_{2} \ b_{0})^{∗}, …, (a_{n} \ b_{0})^{∗}} is precisely two to one and is one to one on each piece. □

We will also assume throughout that PFA, OCA and MA hold. The statements of each of these can be found in [Tod89] and some familiarity will be assumed. Basic information about N^{∗} can be found in [Wal74]. Of course it is well known (see [Vel93]) that OCA and MA implies that the mapping, f^{−1} ∘ f from a^{∗} to b^{∗} in the above lemma will actually be a trivial mapping.

2000 Mathematics Subject Classification. Primary 54A25.

Key words and phrases. two to one maps, Stone-Cech, PFA.

Definition 2. Let J be the collection of those sets a ∈ [ω]^{ω} on which f is precisely two to one and locally one to one.

Proposition 3. For each a ∈ J, there will be a permutation h_{a} on a such that h_{a} is never the identity, h_{a}^{2} is the identity, and for each b ⊂ a, f[b^{∗}] = f[h(b)^{∗}].

Proof. By Proposition 1, a can be partitioned as a_{0} ∪ a_{1} such that f[a_{0}^{∗}] = f[a_{1}^{∗}]. It follows easily that f^{−1} ∘ f restricts to a homemorphism from a_{0}^{∗} to a_{1}^{∗}. By OCA and MA, this is a trivial homeomorphism and h_{a} is the witness together with its inverse (with possibly finitely many elements of a removed). □

Naturally the task is to prove that J does not generate a proper ideal. The first step is to prove that J is not empty. We will proceed by first showing that if f is not locally one to one, then there is a point x such that for every countable family {A_{n} : n ∈ ω} ⊂ x, there is an A ∈ J such that A is almost contained in each A_{n}.

There are two main results. The first is Lemma 14 which is critical to establishing that such an x exists. The second, Theorem 17, is to show that this leads to a contradiction.

We will certainly need the following results from [Far00, 3.3.2;3.8.2]

Definition 4. An ideal I ⊂ ℘(N) is ccc over fin, if there is no uncountable family of almost disjoint subsets of N such that none are in I.

The following are consequences of OCA and MA (and therefore of PFA).

Proposition 5. If Φ : ℘(N)/fin → ℘(N)/fin is a homomorphism, then there is an A ⊂ N and an h : A → N, such that {a ⊂ N : Φ(a) = h^{−1}(a)} is ccc over fin.

Proposition 6. [Far00, 3.5.5] If a subset of N^{∗} is a continuous image of N^{∗}, then it is equal to the disjoint union of a clopen set and a nowhere dense set.

It will be useful to explicit record the following corollary combining the previous two results.

Corollary 7. If a nowhere dense set T of N^{∗} is homeomorphic to N^{∗}, then there does not exist an uncountable family of pairwise disjoint clopen subsets of N^{∗} each of which meets T.

2. Basic properties of f and K

We will have to show that K is nowhere ccc (which will just use MA). We fix some closed subset Z_{0} of N^{∗} such that f restricted to Z_{0} is irreducible (meaning no proper closed subset of Z_{0} will map onto). We will show that the image of N^{∗} \ Z_{0} is dense in K (i.e. f restricted to Z_{0} is nowhere precisely two to one).

Lemma 8 (MA). For each clopen subset U of Z_{0}, there is a point x ∉ Z_{0} such that f(x) ∈ f[U].

Proof. Let U be a clopen subset of Z_{0}. If f^{−1}(f[U]) is contained in Z_{0} then there is a b_{0} ∈ [N]^{ω} such that f^{−1}(f[b_{0}^{∗}] ⊂ U. Since f is irreducible, for each clopen a ⊂ Z_{0}, J_{a} = a ∩ f^{−1}(f[Z_{0} \ a]) is a nowhere dense subset of a. Let {b_{α} : α ∈ 𝔠} enumerate [b_{0}]^{ω}. For each α ∈ 𝔠}, J_{α} = J_{b_{α}^{∗}} is a nowhere dense subset of b_{0}^{∗}. It is well known that MA implies that a clopen subset of N^{∗} can not be covered by any family of 𝔠 many nowhere dense sets. Therefore there is a point x ∈ b_{0}^{∗} such that x ∉ ⋃{J_{α} : α ∈ 𝔠}. Since f is precisely two to one, there is a point x′ ≠ x such that f(x′) = f(x). Since x′ ≠ x, there is an α ∈ 𝔠 such that x ∈ b_{α}^{∗} and x′ ∉ b_{α}^{∗}. Since x ∉ J_{α}, it follows that x′ ∉ Z_{0} \ b_{α}^{∗}. Therefore x′ ∉ Z_{0}. □

Lemma 9. For each a ∈ [ω]^{ω}, there is a b ∈ [a]^{ω} such that f ↾ b^{∗} is one-to-one. If f is one to one on b^{∗} and f[b^{∗}] has interior, then f^{−1}(f(b^{∗})) contains c^{∗} for some c ∈ J.

Proof. If a^{∗} is not contained in Z_{0}, then any b ⊂ a such that b^{∗} ∩ Z_{0} will work. Otherwise we have that a^{∗} ⊂ Z_{0}. Since f is irreducible on Z_{0}, K \ f[Z_{0} \ a∗] is not empty. By Lemma 8, there is an x ∈ N^{∗} \ Z_{0} such that f(x) ∈ K \ f[Z_{0} \ a∗]. In fact, there is a element C ∈ x such that C ∩ a = ∅ and f[C^{∗}] is contained in f[a^{∗}]. Since f is two to one, f(x) is not in f[N^{∗} \ (a ∪ C)^{∗}]. Let b be any subset of a such that f[b^{∗}] ∩ f[C^{∗}] and f[b^{∗}] ∩ f[N^{∗} \ (a ∪ C)^{∗}] are empty. It follows that b^{∗} is a clopen subset of Z_{0} with the property that f[b^{∗}] is disjoint from f[N^{∗} \ Z_{0}]. This contradicts Lemma 8

Now assume that f[b^{∗}] has non empty interior in K. If b^{∗} is disjoint from Z_{0}, fix any point z_{0} ∈ Z_{0} such that f(z_{0}) is in the interior of f[b^{∗}]. Find b_{0} ∈ z_{0} such that b_{0} ∩ b is empty and f[b_{0}^{∗}] is a subset of f[b^{∗}]. Clearly f is one to one on b_{0}^{∗} and f^{−1}(f(b^{∗})) contains f^{−1}(f(b_{0}^{∗})), hence it suffices to show that the lemma holds if we assume that b^{∗} ∩ Z_{0} is not empty.

Of course b^{∗} ∩ Z_{0} is a clopen subset of Z_{0}. Since f is irreducible on Z_{0}, there is some c_{0} ⊂ b such that c_{0}^{∗} ∩ Z_{0} is not empty and is such that f[c_{0}^{∗} ∩ Z_{0}] is disjoint from f[Z_{0} \b^{∗}]. Since f is one to one on b^{∗}, it also follows that f[c_{0}^{∗}] is disjoint from f[b^{∗} \ c_{0}^{∗}]. Therefore f[c_{0}^{∗} ∩ Z_{0}] = K \ (f[Z_{0} \ b^{∗}] ∪ f[b^{∗} \ c_{0}^{∗}]) is a clopen subset of K. Let c ⊂ N be chosen so that c^{∗} = f^{−1}(f[c_{0}^{∗} ∩ Z]). Since f is one to one on c_{0}^{∗}, it follows that c_{0}^{∗} ∩ c^{∗} = c^{∗} ∩ Z_{0}. This means that c \ c_{0} and c ∩ c_{0} is the partition of c showing that c ∈ J. □

For b ⊂ N, let f_{+b} denote the mapping f ↾ b^{∗} and f_{−b} = f ↾ (N \ b)^{∗}.

Lemma 10. If f is one to one on b^{∗}, then g_{b} = f_{−b}^{−1} ∘ f_{+b} is an embedding of b^{∗} into (N \ b)^{∗}. If c^{∗} ⊂ g_{b}[b^{∗}], then c ∪ b contains a member of J.

Proof. Since g_{b}^{−1}[c^{∗}] will be a clopen subset of b^{∗}, there is a c_{0} ⊂ b such that g_{b}[c_{0}^{∗}] = c^{∗}. It follows that f[c_{0}^{∗}] = f[c^{∗}] and that f is locally one to one on (c ∪ c_{0})^{∗}. Therefore, c ∪ c_{0} will be in J. □

Lemma 11 (MA). The space K is nowhere ccc.

Proof. It suffices to show that Z_{0} is nowhere ccc since the map is irreducible on Z_{0}. Let A ⊂ N and A^{∗} ∩ Z_{0} ≠ ∅ and assume that A^{∗} ∩ Z_{0} is ccc. Since f[Z_{0} \ A^{∗}] and f[A^{∗} ∩ Z_{0}] meet in a nowhere dense subset of K, there is a clopen subset b_{0} of A^{∗} ∩ Z_{0} such that f[b_{0}] ∩ f[Z_{0} \ A^{∗}] is empty. Since Z_{0} \ A^{∗} is compact, there is a clopen set A_{1}^{∗} of N^{∗} such that Z_{0} \ A^{∗} ⊂ A_{1}^{∗} and f[b_{0}] ∩ f[A_{1}^{∗}] is empty.

Let {b_{α} : α ∈ 𝔠} enumerate the collection of clopen subsets of Z_{0} ∩ A^{∗}. Further let {A_{α} : α ∈ 𝔠} enumerate all the infinite subsets of N with the property that their closures are disjoint from Z_{0}.

For each clopen subset a of b_{0}, let J_{a} = a ∩ f^{−1}[f[Z_{0} \ a]]. Since f is irreducible on Z_{0}, each J_{a} is nowhere dense in Z_{0}. Also let Y_{α} = Z_{0} ∩ A^{∗} ∩ f^{−1}[f[A_{α}^{∗}]] for each α ∈ 𝔠. Since f is one to one on A_{α}^{∗}, f[A_{α}^{∗}] is nowhere ccc. Since we are assuming that Z_{0} ∩ A^{∗} is ccc, it follows that Y_{α} is also nowhere dense in Z_{0}.

We inductively define a family, {a_{β} : β < α} of clopen subsets of A^{∗} ∩ Z_{0}. Let a_{0} = b_{0}. Our inductive hypotheses are that for each α < 𝔠,

(1) {a_{β} : β < α} has the finite intersection property;

(2) a_{β} is contained in one of {b_{β}, b_{0} \ b_{β}};

(3) for each β + 1 < α, a_{β+1} is disjoint from J_{a_{β}} ∪ Y_{β}; and

(4) the family {a_{β} : β < α} has non-empty intersection.

Suppose we have chosen the family {a_{β} : β < α}. Let Z_{α} denote the closed set ⋂{a_{β} : β < α}. For each integer n, the selection of a_{n} is trivial, and since we are assuming that a_{0} is ccc, we can assume that α ≥ ω and that Z_{α} is nowhere dense in Z_{0}. If α is a limit, then simply let a_{α} equal b_{α} if b_{α} meets Z_{α}, otherwise set a_{α} = b_{0} \ b_{α}. Otherwise, α = β + 1 and we must avoid J_{a_{β}} ∪ Y_{β}. It suffices to show that Z_{α} is not contained in J_{a_{β}} ∪ Y_{β} because then we can select a_{α} to meet Z_{α}, miss J_{a_{β}} ∪ Y_{β} and to be contained in one of {b_{α}, b_{0} \ b_{α}}.

Let B_{α} be a Boolean subalgebra of {b_{γ} : γ < 𝔠} which contains {a_{β} : β < α} ∪ {b_{α}}. For each γ < α, we can ensure that there is a countable subset U_{γ} of 𝔅_{α} such that each member of U_{γ} is disjoint from J_{γ} ∪ Y_{γ} and ⋃ U_{γ} is dense in A^{∗} ∩ Z_{0}. Since α < 𝔠, there is such a B_{α} of cardinality less than 𝔠. Since B_{α} is ccc and of cardinality less than 𝔠, its Stone space is separable. Furthermore, U_{α} generates a dense open subset, U_{α}, of the Stone space. Let D be a countable dense subset of U_{α}. The filter base on B_{α} consisting of finite intersections of members of {a_{γ} : γ < α} generates a closed set, F_{α}, in S(B_{α}) with character less than 𝔠. Martin’s axiom implies that there is a countable set {d_{n} : n ∈ ω} ⊂ D which converges to F_{α}. That is, for any finite intersection of members of {a_{γ} : γ < α}, there are only finitely many x_{n} which are not in the intersection. For each n, x_{n} is in fact a filter on B_{α}, hence there is a point z_{n} ∈ Z_{0} which is also in each member of the filter.

For each n, let z′_{n} ∈ N^{∗} be distinct from z_{n} such that f(z′_{n}) = f(z_{n}). Let T = f[Z_{α}], a nowhere dense subset of K. By construction, the image of \overline{\{z_{n} : n ∈ ω\}} is contained in {f(z_{n}) : n ∈ ω} ∪ T. Since K has no isolated points, and {f(z_{n}) : n ∈ ω} is a relatively closed subset of Z_{0} \ T, it is nowhere dense and discrete. Therefore f^{−1}(f(\overline{\{z_{n} : n ∈ ω\}})) meets Z_{0} in a nowhere dense set. Furthermore, this set contains \overline{\{z′_{n} : n ∈ ω\}}.

Let x ∈ Z_{0} be a limit point of {z_{n} : n ∈ ω}, hence x ∈ Z_{α}. There is a point x′ which is a limit point of {z′_{n} : n ∈ ω} such that f(x) = f(x′) since f(\overline{\{z_{n} : n ∈ ω\}}) = f(\overline{\{z′_{n} : n ∈ ω\}}). Clearly x ∈ b_{β} and we claim that x ∉ J_{a_{β}} ∪ Y_{β}. To show this it is sufficient (and necessary) to show that x′ is not in (Z_{0} \ b_{β}) ∪ A_{β}^{∗}. For each n, z_{n} ∉ Y_{β}, hence z′_{n} ∉ A_{β}^{∗}. Since A_{β}^{∗} is clopen and x′ is a limit of {z′_{n} : n ∈ ω} it follows that x′ ∉ A_{β}^{∗}. The collection {z′_{n} : n ∈ ω} ∩ Z_{0} is contained in b_{α}, which is clopen in Z_{0}, hence this closure is disjoint from Z_{0} \ b_{α}. We will be finished if we show that the closure of {z′_{n} : n ∈ I} = {z′_{n} : n ∈ ω} \ Z_{0} is disjoint from Z_{0}. Since {z_{n} : n ∈ ω} ⊂ b_{0} and f[b_{0}] is disjoint from f[A_{1}^{∗}], it follows that {z′_{n} : n ∈ ω} is disjoint from A_{1}^{∗}. Recall that we showed above that f^{−1}(f(\overline{\{z_{n} : n ∈ ω\}})) meets Z_{0} in a nowhere dense set and that it contains \overline{\{z′_{n} : n ∈ ω\}} ∩ Z_{0}. Since it is nowhere dense and b_{0} is ccc, there is a collection {c_{n} : n ∈ ω} of clopen subsets of N^{∗} such that each is disjoint from {z′_{n} : n ∈ ω} and such that b_{0} is contained in the closure of ⋃{c_{n} ∩ b_{0} : n ∈ ω}. For each n ∈ I, let d_{n} be a clopen subset of N^{∗} \A_{1}^{∗} such that z′_{n} ∈ d_{n} and d_{n} ∩(Z_{0} ∪⋃{c_{k} : k ≤ n}) is empty. For each n ∈ ω, shrink c_{n} by removing ⋃{d_{k} : k < n}; note that this does not change c_{n} ∩ Z_{0}. Then we have that ⋃_{n∈I} d_{n} and A_{1}^{∗} ∪ ⋃_{n} c_{n} are disjoint, and as is well-known, they have disjoint closures in N^{∗}. Since the latter closure conains Z_{0} we have finished the proof. □

Proposition 12. If f is one to one on b^{∗}, then b can be partitioned into two, b_{0} and b_{1}, such that f[b_{0}^{∗}] is clopen, and f[b_{1}^{∗}] is nowhere dense.

Proof. Again let g_{b} denote the embedding of b^{∗} into (N \ b)^{∗} given by f_{−b}^{−1} ∘ f_{+b}. By Proposition 6, there is c ⊂ N \ b such that c^{∗} ⊂ g_{b}[b^{∗}] and g_{b}[b^{∗}] \ c^{∗} is nowhere dense. There are b_{0} ⊂ b such that g_{b}[b_{0}^{∗}] = c^{∗} and b_{1} = b \ b_{0}^{∗} will satisfy g_{b}[b_{1}^{∗}] is nowhere dense. It follows that f[b_{0}^{∗}] = f[c^{∗}] and f[b_{0}^{∗}] = K \ f[(N \ (b_{0} ∪ c)^{∗}] is clopen. In addition, f[b_{1}^{∗}] is nowhere dense in K because g_{b}[b_{1}^{∗}] is equal to f_{−b}^{−1}(f[b_{1}^{∗}]) and is nowhere dense in (N \ b)^{∗}. □

3. tree-like families

An embedding of N^{∗} into N^{∗} is said to be trivial, if the embedding lifts to an embedding of βN into βN. It is an open problem to determine if there can be a non-trivial embedding of N^{∗} into N^{∗} under PFA. If there are none, then it is easy to use Levy’s result in [Lev04] to show that the set b_{1} in Proposition 12 would be empty. The main result of this section is used as an alternative approach.

Proposition 13. [Vel93, 2.3] Let A be an uncountable almost disjoint family of infinite subsets of N. Then there is an uncountable B ⊂ A and for each a ∈ B a partitition a = a_{0} ∪ a_{1} such that the family B_{i} = {a_{i} : a ∈ B} is tree-like for each i ∈ {0, 1}.

Lemma 14. Suppose that {a_{α} : α ∈ ω_{1}} is a tree-like family of subsets of ω with the property there is a b_{α} ∈ [a_{α}]^{ω} such that f(b_{α}^{∗}) is disjoint from f[(N \ a_{α})^{∗}]. Then there is an b_{α} = b such that g_{b}[b^{∗}] has interior.

Proof. We may assume that f is one-to-one on b_{α}^{∗} by Lemma 9. For each c ⊂ a_{α} define F (c) ⊂ b_{α} as follows. Since f[(a_{α} \ b_{α})^{∗})] contains f(b_{α}^{∗}) and f is precisely two to one, there will be a subset F (c) of b_{α} such that

f[F (c)^{∗}] = f(b_{α}^{∗}) ∩ f[(c \ b_{α})^{∗}] .

The definition of F on ℘(a_{α}) can also be expressed as F (c)^{∗} is the clopen subset of b^{∗} which is equal to g_{b}^{−1}[c^{∗} ∩ g_{b}(b^{∗})].

It is easily seen that F is a homomorphism from ℘(a_{α})/fin onto ℘(b_{α})/fin. By Corollary 7, we need to find some α and some uncountable family of pairwise disjoint clopen sets each of which meets g_{b_{α}}[b_{α}^{∗}]. Equivalently, by Farah [Far00] (Proposition 5), if the kernel of F is not ccc over fin, then there is a c ⊂ a_{α} \ b_{α} such that F is a trivial isomorphism from ℘(c) to ℘(b_{α}). We proceed as in [Vel93].

Let X denote the set of all pairs ⟨c, d⟩ such that for some α, d ⊂ c ⊂ a_{α} \ b_{α}, and each of F (d) and F (c \ d) are not 0.

We define a set K_{0} ⊂ [X]^{2} according to (⟨c, d⟩, ⟨c̄, d̄⟩) ∈ K_{0} providing

(1) c ⊂ a_{α} and c̄ ⊂ aᾱ implies α ≠ ᾱ

(2) c ∩ F (c̄) and c̄ ∩ F (c) are empty;

(3) c ∩ d̄= c̄ ∩ d;

(4) F (c) ∩ F (d̄) is not equal to F (c̄) ∩ F (d).

The appropriate separable metric topology on X (given by considering it as embedded in ℘(ω)^{4} by the mapping sending ⟨c, d⟩ to ⟨c, d, F (c), F (d)⟩) will result in K_{0} being an open subset of [X]^{2} (see [Vel93]).

Assume that Y is an uncountable subset of X and that [Y]^{2} ⊂ K_{0}. Let I ⊂ ω_{1} be the set of α such that there is ⟨c, d⟩ ∈ Y such that c ⊂ a_{α}. Also let ⟨c_{α}, d_{α}⟩ ∈ Y be chosen for each α ∈ I so that c_{α} ⊂ a_{α}. Since [Y]^{2} ⊂ K_{0} and Y is uncountable, it follows that I is uncountable.

Let C = ⋃{c_{α} : α ∈ I} and D = ⋃{d_{α} : α ∈ I}. Let ⟨c, d⟩ and ⟨c̄, d̄⟩ be an arbitrary distinct pair from Y. Note that c ∩ (F (c) ∪ F (c̄)) is empty. It follows that C is disjoint from ⋃{F (c) : (∃d)⟨c, d⟩ ∈ Y}. Also, D ∩ c will equal d for each ⟨c, d⟩ ∈ Y. Hence (C \ D) ∩ c = c \ d for each ⟨c, d⟩ ∈ Y.

Now consider the two families {F (d_{α}) : α ∈ I} and {F (c_{α} \d_{α}) : α ∈ I}. Assume that E ⊂ ω and that E ∩ F (c_{α}) =^{∗} F (d_{α}) for each α ∈ I. Let n ∈ ω and I′ ∈ [I]^{ω_{1}} such that (E ∩ F (c_{α}))ΔF (d_{α}) is contained in n for all α ∈ I′. Let α ≠ β both be in I′. We may assume that F (c_{α}) ∩ n = F (c_{β}) ∩ n, F (d_{α}) ∩ n = F (d_{β}) ∩ n. Also, we may assume that F (d_{α}) \ F (c_{α}) is n contained in n for all α ∈ I′.

Suppose that j ∈ F (c_{α})∩F (d_{β}) and j ∉ F (c_{β})∩F (d_{α}). Clearly j must be larger than n. Since j ∈ F (d_{β}), it follows that j ∈ F (c_{β}). Therefore j is in E. On the other hand, since j is in E ∩ F (c_{α}), it must follow that j ∈ F (d_{α}), contradicting that j ∉ F (c_{β}) ∩ F (d_{α}). .

It follows then that there is no such E. This means that ⋃{F (d_{α})^{∗} : α ∈ I} and ⋃{F (c_{α} \ d_{α})^{∗} : α ∈ I} do not have disjoint closures in N^{∗}. Fix any x ∈ N^{∗} which is in each of the closures. Notice that x ∉ C^{∗}.

Since f[d_{α}^{∗}] is equal to f[F (d_{α})^{∗}] and f[(c_{α} \ d_{α})^{∗}] = f[(F (c_{α} \ d_{α}))^{∗}], it follows that f(x) is in the image of the closure of ⋃_{α∈I} d_{α}^{∗} and of ⋃_{α∈I}(c_{α} \d_{α})^{∗}. Therefore f(x) is in the image of D^{∗} and of (C \D)^{∗}. However this contradicts that f(x) only has two points mapping to it, we have found points in D^{∗}, (C \ D)^{∗}, and (N \ C)^{∗}.

Therefore by OCA, X can be expressed as a countable union ⋃_{n} Y_{n} such that [Y_{n}]^{2} ∩ K_{0} is empty for each n. For each n, there is a countable Y_{n} ⊂ Y_{n} such that for each integer m and each ⟨c, d⟩ ∈ Y_{n}, there is some ⟨c̄, d̄⟩ ∈ Y_{n} such that c ∩ m = c̄ ∩ m, d ∩ m = d̄∩ m, F (c) ∩ m = F (c̄) ∩ m, and F (d) ∩ m = F (d̄) ∩ m.

Fix any α ∈ ω_{1} such that c̄ ∩ a_{α} is finite for each ⟨c̄, d̄⟩ ∈ ⋃_{n} Y_{n}. Construct an increasing sequence {k_{n} : n ∈ ω} of integers so that for each n and each i ≤ n and each sequence c′, d′, a′, b′ of subsets of k_{n}, if there is a ⟨c, d⟩ ∈ Y_{i} such that c ∩ k_{n} = c′, d ∩ k_{n} = d′, F (c) ∩ k_{n} = a′, and F (d) ∩ k_{n} = b′, then there is a pair ⟨c, d⟩ ∈ Y_{i} that also has this property, and in addition, a_{α} ∩ aᾱ ⊂ k_{n+1} where c ⊂ aᾱ.

Define E_{i} to be ⋃{a_{α} ∩ [k_{3n+i}, k_{3n+i+1}) : n ∈ ω} for i ∈ 3. There is an i ∈ 3 such that F (E_{i}) is not finite. There is a j ∈ 3 such that E_{j} ∩ F (E_{i}) is not finite. Fix any c ⊂ E_{i} such that F (c) is infinite and is contained in E_{j}. Let d_{0}, d_{1}, d_{2} be a partition of c so that F (d_{0}), F (d_{1}), and F (d_{2}) are each infinite. For simplicity we will assume that {i, j} = {1, 2}. Note that for all x ⊂ d_{1}, the pair ⟨c, d_{0} ∪ x⟩ is a member of X (since c \ (d_{0} ∪ x) contains d_{2}).

Under the same (or similar) circumstances it is shown in [Vel93, 2.2] that F ↾ ℘(a_{α}) is Borel. Then one can apply the results in [Far00] to deduce that g_{b_{α}}[b_{α}^{∗}] is not nowhere dense. However, rather than checking that the approach in [Vel93] is applicable, we will directly produce an uncountable family of clopen subsets of N^{∗} each of which will meet g_{b_{α}}[b_{α}^{∗}].

We construct an increasing sequence {m_{i} : i ∈ ω} ⊂ {k_{3ℓ} : ℓ ∈ ω} together with subsets t_{i} ⊂ d_{1} ∩ [m_{i}, m_{i+1}) and possibly infinite sets {J_{i} : i ∈ ω} by induction. We can let m_{0} = 0 and J_{−1} = ∅. For each J ⊂ N, let D(J) = d_{0} ∪(d_{1} ∩⋃{[k_{3j}, k_{3j+3}) : j ∈ J}).

Given that m_{i} and J_{i−1} have been chosen we will construct the set t_{i} also by a finite induction and will then choose m_{i+1}. Fix an enumeration {(e_{ℓ}, n_{ℓ}, n′_{ℓ}) : ℓ < L} of ℘(m_{i})×i×i. We will construct {t(i, ℓ) : ℓ < L} such that t(i+1, ℓ)∩max(t(i, ℓ))+ 1 = t(i, ℓ) and will let t_{i} = ⋃{t(i, ℓ) : ℓ < L}. Our inductive hypotheses on J_{i} are that F (D(N \ J_{i})) \ F (d_{0}) is infinite and that if j ∈ J_{i} \ J_{i−1}, then k_{3j} > m_{i}.

As we define t(i, ℓ), we will also define J(i, ℓ) and will set J_{i} = ⋃{J(i, ℓ) : ℓ < L}. For convenience let t(i, −1) = ∅ and J(i, −1) = J_{i−1}.

Suppose we have chosen t(i, ℓ − 1) and J(i, ℓ − 1) such that F (D(N \ J(i, ℓ − 1))) \ F (d_{0}) is infinite.

First choose, if possible, J′(i, ℓ) ⊃ J(i, ℓ−1) such that J′(i, ℓ)∩max(t(i, ℓ−1)) = J(i, ℓ − 1), ⟨c, D(J′(i, ℓ))⟩ ∈ Y_{n′}, and F (D(N \ J′(i, ℓ))) \ F (d_{0}) is infinite. If there is no such J′(i, ℓ), then let J′(i, ℓ) = J(i, ℓ − 1).

Next choose, if possible, an x ⊂ d_{1} \ max(t(i, ℓ − 1)) such that

F (e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J′(i, ℓ)) ∪ x) \ F (D(J′(i, ℓ − 1)))

is infinite and ⟨c, d_{0} ∪ e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J(i, ℓ − 1)) ∪ x⟩ ∈ Y_{n_{ℓ}}.

If such an x exists, call this Case one, and choose some large enough j′ ∈ x such that there is a k_{3ℓ′} < j′ such that ℓ′ ∉ J′(i, ℓ) and

F (e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J′(i, ℓ)) ∪ x) \ F (D(J′(i, ℓ)))

contains some element of F (c) ∩ [m_{i}, k_{3ℓ′}). We define J(i, ℓ) = J′(i, ℓ), and t(i, ℓ) = t(i, ℓ − 1) ∪ (x ∩ j + 1). Note that ℓ′ will not be in J_{i}.

On the other hand, if no such an x exists, then choose J(ℓ, i) ⊃ J′(ℓ, i) such that each of

F (e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J(i, ℓ))) \ F (D(J′(i, ℓ)))

and

F (D(N \ J(i, ℓ)))

are infinite. Again choose an ℓ′ ∉ J′(i, ℓ) so that k_{3ℓ′} > max(t(i, ℓ − 1)), and ensure that J(i, ℓ) ∩ ℓ′ + 1 equals J′(i, ℓ) ∩ ℓ′. Also ensure that d_{1} ∩ [k_{3ℓ′}, k_{3ℓ′+3}) is not empty and let t(i, ℓ) = t(i, ℓ − 1) ∪ (d_{1} ∩ [k_{3ℓ′}, k_{3ℓ′+3})).

Let m_{i+1} = k_{3n} be chosen so that 3n ∉ J_{i} = ⋃{J(i, ℓ) : ℓ < L} and t_{i} = ⋃{t(i, ℓ) : ℓ < L} ⊂ m_{i+1}.

Let J_{ω} = ⋃_{i} J_{i} and note that, by construction D(J_{ω}) ∩ m_{i+1} = D(J_{i}) ∩ m_{i+1}. For each infinite I ⊂ ω, set e_{I} = ⋃{t_{i} : i ∈ I} ⊂ d_{1}. We finish by showing that the clopen set e_{I}^{∗} meets g_{b_{α}}[b_{α}^{∗}]. To show this, it suffices to show that F (e_{I}) is infinite. Since F (e_{I}) contains F (D(J_{ω}) ∪ e_{I}) \ F (D(J_{ω})) mod finite, it suffices to show this latter set is infinite.

Since ⟨c, D(J_{ω}) ∪ e_{I}⟩ ∈ X, there is an n such that ⟨c, D(J_{ω}) ∪ e_{I}⟩ ∈ Y_{n}. There is also an n′ such that ⟨c, D(J_{ω})⟩ ∈ Y_{n′}. Let i ∈ I be any integer greater than both n and n′. We will show that F (D(J_{ω}) ∪ e_{I}) \ F (D(J_{ω})) is not contained in i.

Set d = D(J_{ω}) ∪ e_{I} and fix ℓ such that at stage i in the construction of the m_{i}’s, e_{ℓ} = d ∩ m_{i} and n_{ℓ} = n, and n′_{ℓ} = n′. We consider the properties of t(i, ℓ). Let Y = D(J′(i, ℓ)). Since ⟨c, D(J_{ω})⟩ is in Y_{n′}, it follows that we were able to ensure that ⟨c, D(J′(i, ℓ))⟩ = ⟨c, Y ⟩ is in Y_{n′}.

Recall that there was an ℓ′ such that m_{i} < k_{3ℓ′} < m_{i+1} and that the maximum of t(i, ℓ) was greater than k_{3ℓ′}. If J(i, ℓ) \ J′(i, ℓ) was infinite, then we know F (e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J(i, ℓ))) \ F (Y) is infinite. Therefore F (d) \ F (Y) is infinite because d ⊃ e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J(i, ℓ)). If we set x̄ = d \ max(t(i, ℓ − 1)), then x̄ would be a witness to the fact that we should have been in case one when defining t(i, ℓ). Therefore J(i, ℓ) = J′(i, ℓ) and at stage ℓ we were able to find some x as in Case one. In addition, t(i, ℓ) was defined as t(i, ℓ − 1) ∪ (x ∩ j + 1) for some j ∈ x \ k_{3ℓ′}.

Since e_{ℓ} = d ∩ m_{i} and i ∈ I, we have that

d ∩ k_{3ℓ′} = (e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J′(i, ℓ)) ∪ x) ∩ k_{3ℓ′}.

That is, if we let d_{x} = e_{ℓ} ∪ t(i, ℓ − 1) ∪ D(J′(i, ℓ)) ∪ x, then ⟨c, d_{x}⟩ ∈ Y_{n} and d_{x} ∩ k_{3ℓ′} = d ∩ k_{3ℓ′}.

Now since ⟨c, d⟩ ∈ Y_{n}, there is a pair ⟨c̄, d̄⟩ ∈ Y_{n} such that c̄ ∩ k_{3ℓ′} = c ∩ k_{3ℓ′}, d̄∩ k_{3ℓ′} = d ∩ k_{3ℓ′}, F (c̄) ∩ k_{3ℓ′} = F (c) ∩ k_{3ℓ′}, and F (d̄) ∩ k_{3ℓ′} = F (d) ∩ k_{3ℓ′}. In addition, if we let ᾱ ∈ ω_{1} such that c̄ ⊂ aᾱ, we have that aᾱ ∩a_{α} ⊂ k_{3ℓ′+1}. Further, recall that (c ∪ F (c)) ∩ [k_{3ℓ′}, k_{3ℓ′+1}) is empty. Observe also that Y ∩ k_{3ℓ′} is equal to D(J_{ω}) ∩ k_{3ℓ′}.

It follows then that each of

(1) c ⊂ a_{α} and c̄ ⊂ aᾱ and α ≠ ᾱ

(2) c ∩ F (c̄) and c̄ ∩ F (c) are empty: because c ∩ F (c̄) ⊂ c ∩ (F (c̄) ∩ a_{α}) ⊂ c∩(F (c̄)∩k_{3ℓ′+1}) ⊂ c∩F (c) and similarly, c̄∩F (c) ⊂ c̄∩F (c)∩k_{3ℓ} ⊂ c∩F (c)

(3) c ∩ d̄= c̄ ∩ d: because each of c ∩ d̄ and c̄ ∩ d are equal to d ∩ k_{3p+ℓ} Therefore, the only reason that (⟨c, d⟩, ⟨c̄, d̄⟩) is not in K_{0} is that F (c) ∩ F (d̄)

must be equal to F (c̄) ∩ F (d). The same is true with d_{x} in place of d, hence F (c) ∩ F (d̄) must be equal to F (c̄) ∩ F (d_{x}). By the choice of x, there is some integer m′ ∈ F (c) ∩ [m_{i}, k_{3ℓ′}) ∩ F (d_{x}) \ F (Y), and therefore m′ ∈ F (c̄) ∩ F (d̄). This then means that m′ ∈ F (d).

By the same reasoning, there is a pair ⟨c′, d′⟩ ∈ Y_{n′} such that c′ ∩ k_{3ℓ′} = c ∩ k_{3ℓ′}, d′∩k_{3ℓ′} = D(J_{ω})∩k_{3ℓ′}, F (c′)∩k_{3ℓ′} = F (c)∩k_{3ℓ′}, and F (d′)∩k_{3ℓ′} = F (D(J_{ω}))∩k_{3ℓ′}. In addition, if we let α′ ∈ ω_{1} such that c′ ⊂ a_{α′}, we have that a_{α′} ∩ a_{α} ⊂ k_{3ℓ′+1}. Further, recall that (c ∪ F (c)) ∩ [k_{3ℓ′}, k_{3ℓ′+1}) is empty. Repeating the argument above with ⟨c, D(J_{ω})⟩ and ⟨c′, d′⟩ in place of ⟨c, d⟩ and ⟨c̄, d̄⟩ respectively, we have that F (D(J_{ω})) ∩ k_{3ℓ′} is equal to F (d′) ∩ k_{3ℓ′}. Furthermore, F (Y) ∩ k_{3ℓ′} will also equal F (d′) ∩ k_{3ℓ′} because Y ∩ k_{3ℓ′} = D(J_{ω}) ∩ k_{3ℓ′}. Since m′ ∈ F (c) ∩ F (c′) \ F (Y), it follows that m′ ∉ F (D(J_{ω})).

This shows that m′ ∈ F (d) \ F (D(J_{ω})) and completes the proof. □

Corollary 15. The space Z_{0} is regular closed in N^{∗}.

Proof. Let A ⊂ N be any infinite set such that A^{∗} ∩ Z_{0} is not empty. Since f is irreducible on Z_{0}, f[A^{∗} ∩ Z_{0}] has interior in K. By Lemma 11, there is an uncountable family {U_{α} : α ∈ ω_{1}} of pairwise disjoint open subsets of f[A^{∗} ∩ Z_{0}]. For each α, let c_{α} be an infinite set such that f[c_{α}^{∗}] is a subset of U_{α} and such that f is one to one on c_{α}^{∗}. If any of the c_{α}’s are such that f[c_{α}^{∗}] has interior then we can apply Lemma 9.

On the other hand, we assume that each c_{α}^{∗} is disjoint from Z_{0}. Fix an a_{α} such that f[a_{α}^{∗}] is a subset of U_{α} and meets f[c_{α}^{∗}]. By Proposition 13, we may assume that {a_{α} : α ∈ ω_{1}} is tree-like. It follows that f[N^{∗} \ (a_{α} ∪ c_{α})^{∗}] is a closed subset of K which does not contain f(c_{α}^{∗}). Let b_{α} be any infinite subset of a_{α} such that f[b_{α}^{∗}] is disjoint from f[N^{∗} \ (a_{α} ∪ c_{α})^{∗}] and the nowhere dense set f[c_{α}^{∗}]. Now we can apply Lemma 14 □

Lemma 16. If f is not locally one to one, then there is a point x such that for every countable family {A_{n} : n ∈ ω} ⊂ x, there is an A ∈ J such that A is almost contained in each A_{n}.

Proof. We will choose a point x and an infinite set b ⊂ N, such that b ∉ x, f(x) ∈ f[b^{∗}], and f is not one to one on any A^{∗} with A ∈ x.

First assume that there is some infinite b ⊂ N such that f[b^{∗}] is nowhere dense in K. Since f[(N \ b)^{∗}] is equal to K, we can fix x ∈ N^{∗}, such that f(x) ∈ f[b^{∗}].

Otherwise, we may assume that f[b^{∗}] has interior for all infinite b ⊂ N. In this case, fix any x ∈ N^{∗} such that f ↾ A^{∗} is not one to one for any A ∈ x. Let b ⊂ N be any infinite set such that f(x) ∈ f[b^{∗}] and b ∉ x.

Assume that {A_{n} : n ∈ ω} ⊂ x. We may assume that for each n, A_{n} ∩b is empty and A_{n+1} ⊂ A_{n}. It follows that f(x) ∉ f[(N \ (b ∪ A_{n}))^{∗}] for each n. By continuity of f, there is a B_{n} ∈ x such that f[B_{n}^{∗}] is disjoint from f[(N \ (b ∪ A_{n}))^{∗}]. Let A be almost contained in B_{n} for each n. It follows that f[A^{∗}] ∩ f[(N \ (b ∪ A_{n}))^{∗}] is empty for each n. We may assume that f is one to one on A^{∗} by Lemma 9. Furthermore, if f[A^{∗}] has interior in K, then by Lemma 9, there is a c ∈ J′ such that f[c^{∗}] ⊂ f[A^{∗}]. Therefore f[c^{∗}] is disjoint from f[(N \ (A_{n} ∪ b))^{∗}] for each n, hence c is almost contained in A_{n} for all n.

Now we consider the case that f[A^{∗}] is nowhere dense for each A almost contained in all the A_{n}’s. Since Z_{0} is regular closed, we can choose, for each n, an infinite b_{n} ⊂ B_{n} such that b_{n}^{∗} ⊂ Z_{0}. We may further suppose that there is a disjoint set c_{n} such that f[b_{n}^{∗}] = f[c_{n}^{∗}]. It follows that there is some m > n such that b_{n}^{∗} ∩ B_{m}^{∗} is empty (otherwise there would be an A ⊂ b_{n} almost contained in each A_{n} such that f[A^{∗}] had interior).

We may also assume that b_{n} is disjoint from b_{m} for each n ≠ m. For each n, let {b(n, α) : α ∈ ω_{1}} be an almost disjoint family of infinite subsets of b_{n}. Inductively define an almost disjoint family {a_{α} : α ∈ ω_{1}} of subsets of ⋃_{n} b_{n} so that a_{α} ∩ b_{n} = b(n, α) for each n. Apply Proposition 13, to find a partition a_{α} as a_{α}^{0} ∪ a_{α}^{1} so that there is an uncountable I ⊂ ω_{1} such that each of the families {a_{α}^{0} : α ∈ I} and {a_{α}^{1} : α ∈ I} are tree-like. For each α ∈ I, at least one of a_{α}^{0} or a_{α}^{1} will meet infinitely many of the b_{n}’s. Therefore we can assume we have an uncountable tree-like family {a_{α} : α ∈ ω_{1}} such that each a_{α} meets infinitely many of the b_{n}’s in an infinite set.

Assume that f[(N \ a_{α})^{∗}] contains f[a_{α}^{∗}] for some α. It then follows that f is one to one on a_{α}^{∗}. By Proposition 12, we can write a_{α} as a_{α}^{0} ∪ a_{α}^{1} so that f[(a_{α}^{1})^{∗}] is nowhere dense and f[(a_{α}^{0})^{∗}] is clopen. It follows then that the clopen set f[(a_{α} ∩ b_{n})^{∗}] is contained in f[(a_{α}^{0})^{∗}] for each n. This however contradicts the current assumption that f[A^{∗}] is nowhere dense for any A ⊂ a_{α} which is almost disjoint from each b_{n}. From this we conclude that there is some infinite b_{α} ⊂ a_{α} such that f[b_{α}^{∗}] is disjoint from f[(N \ a_{α})^{∗}].

Finally, the theorem follows by applying Lemma 14. □

4. locally one to one

Theorem 17 (PFA). The function f is locally one to one.

Proof. Assume that f is not locally one to one and choose an x as in Lemma 16. For each a ∈ J, we have the permutation h_{a} on a such that h_{a} is never the identity and h_{a}^{2} is the identity on a (3).

We define a poset P by p ∈ P if there is a finite set F_{p} ⊂ ω_{1}, a set M_{p} = ⟨M_{α} : α ∈ F_{p}⟩ of elementary submodels, a finite function ⟨a_{α} : α ∈ F_{p}⟩ whose range is a subset of J′, and a finite function s_{p} with domain n_{p} into {0, 1, 2}.

For each α ∈ F_{p}, we require that M_{α} is countable elementary submodel of a suitable H(θ) (containing x, J′, f). We require that a_{α} ∈ J is contained mod finite in each A ∈ x ∩ M_{α}. For β > α also in F_{p}, M_{α} ∈ M_{β}. For convenience, we can require that M_{α} ∩ ω_{1} = α. The final condition is that there must be an i ∈ a_{α} and a j ∈ a_{β} such that s_{p}(i) = 0, s_{p}(j) = 1, h_{a_{α}}(i) = h_{a_{β}}(j), and s_{p}(h_{a_{α}}(i)) = 2.

We show below that P is a proper (Lemma 20) and that for each α ∈ ω_{1}, the set D_{α} = {p ∈ P : F_{p} \ α ≠ ∅} is dense (Lemma 19). Suppose that G ⊂ P is a filter which meets D_{α} for all α ∈ ω_{1}. Let I = ⋃{F_{p} : p ∈ G} and {a_{α} : α ∈ I} be the corresponding elements of J. For each α ∈ I, let h_{α} = h_{a_{α}}. Let A_{0} = {n ∈ ω : (∃p ∈ G)s_{p}(n) = 0}, A_{1} = {n ∈ ω : (∃p ∈ G)s_{p}(n) = 1} and A_{2} = {n : (∃p ∈ G)s_{p}(n) = 2}. For each α ∈ I, let b(α, 0) = A_{0} ∩ a_{α} and b(α, 1) = A_{1} ∩ a_{α}. Let c(α, 0) = A_{2} ∩ h_{α}(b(α, 0)) and c(α, 1) = A_{2} ∩ h_{α}(b(α, 1)).

Assume that d ⊂ A_{2} is such that c(α, 0) ⊂^{∗} d and d ∩ c(α, 1) =^{∗} ∅ for all α ∈ I. There is some m ∈ ω and an uncountable I′ ⊂ I such that c(α, 0) ⊂ d ∪ m and d ∩ c(α, 1) ⊂ m. We can also assume that there is an integer m′ ≥ m, a finite set a′ ⊂ m′ and a function h from a′ into ω such that a_{α} ∩ m′ = a′, h_{α}^{−1}(m) ⊂ a′, and h_{α} ↾ a′ = h for all α ∈ I. Let α < β ∈ I′ and p ∈ G such that {α, β} ⊂ F_{p}. By the definition of P, there is an i ∈ a_{α}, j ∈ a_{β} such that s_{p}(i) = 0, s_{p}(j) = 1, h_{α}(i) = h_{β}(j), and s_{p}(h_{α}(i)) = 2. Clearly i ∈ b(α, 0) and j ∈ b(β, 1). We want to show that each of i, j and k = h_{α}(i) are larger than m. Since h_{α}^{−1}(k) is different than h_{β}^{−1}(k) it follows that k ≥ m. Since h_{α} and h_{β} disagree on i and j, they too are greater than m. Therefore, k ∈ c(α, 0) \ m, hence k ∈ d. However this contradicts that d ∩ c(β, 1) ⊂ m.

It follows that the sets ⋃{c(α, 0)^{∗} : α ∈ I} and ⋃{c(α, 1)^{∗} : α ∈ I} do not have disjoint closures in N^{∗}. Let x be a common limit point. Since f[⋃{b(α, 0)^{∗} : α ∈ I}] contains f[⋃{c(α, 0)^{∗} : α ∈ I}], it follows that there is a point y ∈ A_{0}^{∗} such that f(y) = x. Similarly there is a point z ∈ A_{1}^{∗} such that f(z) = x. Since x ∈ A_{2}^{∗}, we have produced three distinct points mapping to the same element of K. □

For a set A ⊂ ℘(N), and a set a ⊂ N, let a ≺^{∗} A indicate that a ⊂^{∗} A for each A ∈ A.

Lemma 18. Assume that S ⊂ J and that for each countable subset A of x, there is an a ∈ S such that a ≺^{∗} A Then the set S consisting of all k satisfying

(∀A ∈ [x]^{ω})(∃a ∈ S) (a ≺^{∗} A and (k ∈ a)}

is a member of x.

Proof. Let x, S ∈ M be any countable elementary submodel and let S′ denote the elements of S which are mod finite contained in every member of x ∩ M. It follows by elementarity that B = {k : (∃a ∈ S′)k ∈ a} is contained in S. If B ∉ x, then there is some a ∈ S′ such that a ⊂^{∗} N \ B. This is clearly impossible since, by definition, a is a subset of B. Therefore S ∈ x. □

Lemma 19. For each α ∈ ω_{1}, the set D_{α} is a dense subset of P.

Proof. Let p ∈ P be arbitrary and let p, α, J ∈ M for some countable elementary submodel of H(λ). Let δ = M ∩ ω_{1} and set M_{δ} = M.

If F_{p} is empty, then we can simply apply Lemma 9, to find an a_{δ} ∈ J such that a_{δ} ⊂^{∗} A for all A ∈ x ∩ M. and set p′ = (δ, ⟨M⟩, ⟨a_{δ}⟩, s_{p}). It follows easily that p′ ≤ p and p′ ∈ D_{α}.

If F_{p} is not empty, let β be the largest element of F_{p}. Let the domain of s_{p} be contained in some integer m. We must again choose a suitable a_{δ} to be mod finite contained in every member of x ∩ M. If we ensure that a_{δ} ∩ m = a_{β} ∩ m, then for each α′ ∈ F_{p} ∩ β, there are i ∈ a_{α′}, j ∈ m ∩ a_{β} and k = h_{a_{α′}} (i) = h_{a_{β}}(j) < m such that s_{p}(i) = 0, s_{p}(j) = 1, and s_{p}(k) = 2. Therefore we must work to ensure that there is an i ∈ a_{β} \ m and distinct j ∈ a_{δ} \ m such that h_{a_{β}}(i) = h_{a_{δ}}(j) = k. Then we can simply extend s_{p} by sending i to 0, j to 1 and k to 2.

Let a′ = a_{β} ∩ m and set S = {a ∈ J : a ∩ m = a′}. Observe that S ∈ M_{β} where M_{β} denotes the elementary submodel M′ ∈ M_{p} such that M′ ∩ ω_{1} = β. Since a_{β} ∈ S and a_{β} is almost contained in every A ∈ x ∩ M_{β}, it follows by elementarity, that for every A ∈ [x]^{ω}, there is an a ∈ S satisfying a ≺^{∗} A. By Lemma 18, the set S = {k : (∀A ∈ [x]^{ω}) (∃a ∈ S)(k ∈ a)} is a member of x ∩ M_{β}.

Now define B to be the set of all k ∈ S for which there is some countable A ∈ [x]^{ω} and a unique value g(k) such that h_{a}(k) = g(k) for all a ∈ S such that k ∈ a and a ≺^{∗} A. This set B is also a member of M_{β}.

If B is a member of x, then by the three set lemma [Wal74, 6.25], there is a partition B_{0}, B_{1}, B_{2} of B with such that g[B_{i}] ∩ B_{i} is empty for each i. We may assume that B_{0} is a member of x. If B is not a member of x, then let B_{0} = S \ B. Again, we may choose B_{0} to be a member of M_{β} ∩ x.

Therefore a_{β} is contained mod finite in B_{0}. Let k ∈ a_{β} ∩ B_{0} \ m be large enough so that k ∉ g^{−1}({0, …, m}) and i = h_{a_{β}}(k) ∈ a_{β} \ m. If k ∈ B, then there is an a_{δ} = a ∈ S such that a ≺^{∗} M, k ∈ a and j = h_{a}(k) = g(k) ∉ B_{0}. If k ∉ B, then g(k) is not defined. Since k ∈ S, it follows that there is an a_{δ} = a ∈ S such that a ≺^{∗} M, k ∈ a, and h_{a}(k) = j ≠ i.

It follows that p′ < p and p′ ∈ D_{α} where F_{p′} = F_{p} ∪ {δ}, M_{p′} = M_{p} ∪ {M_{δ}}, A_{p′} = A_{p} ∪ {a_{δ}} and define s_{p′} ⊃ s_{p} with s_{p′}(i) = 0, s_{p′}(j) = 1 and s_{p′}(k) = 2. □

Lemma 20. The poset P is proper.

Proof. Let λ be any suitably large cardinal with P ∈ H(λ) and let θ, P ∈ M ≺ H(λ) be a countable elementary submodel. Let p_{0} ∈ M, δ = M ∩ω_{1} and M_{δ} = M ∩H(θ). In the proof of Lemma 19, we have shown that there is a q < p such that M_{δ} ∈ M_{q}.

Let D ∈ M be a dense subset of P, q ∈ D, q < p_{0}, and M_{δ} ∈ M_{q}. To show that P is proper, it suffices to show that there is a q′ ∈ M ∩ D which is compatible with q.

Let m_{0} ∈ ω contain the domain of s_{q}. Let q_{0} = q ∩ M, and F_{q} \ M = {δ_{0}, δ_{1}, …, δ_{n−1}} listed in increasing order. For i < n, let c_{i} = a_{δ_{i}} ∩ m_{0}.

Define T_{n} ⊂ ^{n}(ω_{1} × J) by t = ⟨(γ_{i}, a_{i})⟩_{i<n} ∈ T if there is a p ∈ D such that (1) for each i < n, a_{i} ∩ m_{0} = c_{i};

(2) s_{p} = s_{q};

(3) F_{q_{0}} is an initial segment of F_{p};

(4) F_{p} \ F_{q_{0}} = {γ_{i} : i < n}; and

(5) t ⊂ A_{p}.

Observe that ⟨δ_{i}, a_{δ_{i}}⟩_{i<n} ∈ T_{n} and that T_{n} ∈ H(θ) ∩ M = M_{δ_{0}}.

We define by induction on k < n, a set T_{n−k}. If T_{n−k} has been defined, then let ⟨(γ_{i}, a_{i})⟩_{i<(n−(k+1))} ∈ T_{n−(k+1)} providing that for each A ∈ [x]^{ω} and each γ < ω_{1}, there are γ′ > γ and a′ such that a′ ≺^{∗} A, ⟨(γ_{i}, a_{i})⟩_{i<(n−(k+1))}^{⌢}{(γ′, a′)} ∈ T_{n−k}. The set T_{n−k} ∈ M_{δ} for each k < n.

Since the elements of F_{q} are separated by the elementary submodels in M_{q}, it follows that for each k < n, ⟨(γ_{i}, a_{δ_{i}})⟩_{i<(n−k)} is a member of T_{n−k}. In particular, it follows that ∅ ∈ T_{0}. The approach is based on Todorcevic’s no S-spaces proof [Tod89, 8.9].

The next step is to inductively find a sequence ⟨(β_{i}, a_{i})⟩_{i<n} ∈ M_{δ} so that for each k < n, ⟨(β_{i}, a_{i})⟩_{i<n−k} is in T_{n−k}. This will identify a condition p ∈ M ∩ D but we will have to have chosen the a_{i} more carefully in order to arrange that p is compatible with q. This is similar to the proof of Lemma 19.

Let S_{0} = {a ∈ J : (∃γ)⟨(γ, a)⟩ ∈ T_{1}}. Since a_{δ_{0}} is in S_{0} ∈ M_{δ_{0}}, it follows that S_{0} satisfies the hypothesis of Lemma 18. As in the proof of Lemma 19, there is a k ∈ a_{δ_{0}} \ m_{0} such that with j = h_{a_{δ_{0}}} (k), there is an a_{0} ∈ S_{0} such that k ∈ a_{0} and h_{a_{0}}(k) = i ≠ j. Fix an element ⟨(β_{0}, a_{β_{0}})⟩ ∈ T_{1} ∩ M_{δ} such that k ∈ a_{β_{0}} and h_{a_{β_{0}}} (k) = i. Let m_{1} ∈ ω be larger than each of i, j, k.

Let S_{1} = {a ∈ J : (∃β_{1}) ⟨(β_{0}, a_{β_{0}}), (β_{1}, a)⟩ ∈ T_{2}}. Since ⟨(β_{0}, a_{β_{0}})⟩ ∈ T_{1}, it follows that S_{1} satisfies the hypothesis of Lemma 18. We again find (β_{1}, a_{β_{1}}) ∈ M_{δ} such that ⟨(β_{0}, a_{β_{0}}), (β_{1}, a)⟩ ∈ T_{2}} and there is a k ∈ a_{δ_{1}} ∩a_{β_{1}} such that h_{a_{δ_{1}}} (k) = j is distinct from h_{a_{β_{1}}} (k) = i. Let m_{2} be greater than each of these values {i, j, k}.

Proceeding in this way, we can obtain our desired sequence ⟨(β_{i}, a_{β_{i}})⟩_{i<n} ∈ T_{n} ∩ M together with an increasing sequence m_{0} < m_{1} < ⋯ < m_{n−1} so that for each ℓ < n, there is a k_{ℓ} ∈ a_{β_{ℓ}} ∩ a_{δ_{ℓ}} ∩ [m_{ℓ−1}, m_{ℓ}) such that i_{ℓ} = h_{a_{β_{ℓ}}} (k) ≠ h_{a_{δ_{ℓ}}} (k) = j_{ℓ} are both in [m_{ℓ−1}, m_{ℓ}). Let p ∈ D ∩ M_{δ} be chosen to witness that ⟨(β_{i}, a_{β_{i}})⟩_{i<n} ∈ T_{n} ∩ M.

Define a function s = s_{q′} ⊃ s_{q} so that for each ℓ < n, s(i_{ℓ}) = 0, s(j_{ℓ}) = 1, and s(k_{ℓ}) = 2. Let F_{q′} = F_{p} ∪F_{q}, M_{q′} = M_{p} ∪M_{q}, and A_{q′} = A_{p} ∪A_{q}. The properties of s described above ensure that q′ ∈ P and it is clear that q′ < p, q. □

References

[vD93] Eric K. van Douwen, Applications of maximal topologies, Topology Appl. 51 (1993), no. 2, 125–139.

[DT04] A. Dow and G. Techanie, Two to one images of N^{∗} under CH, in preparation.

[Far00] I. Farah. Analytic quotients: theory of liftings for quotients over analytic ideals on the integers. Number 702 in Memoirs of the American Mathematical Society, vol. 148. 2000. 177 pp.

[Lev04] R. Levy. The weight of certain images of ω^{∗} preprint, 2004

[Tod89] S. Todorcevic, Partition problems in topology, Contemporary Mathematics, 1989.

[Vel93] Boban Velickovic, OCA and automorphisms of P(ω)/fin, Topology and its Applications 49 (1993), 1–13.

[Wal74] Russell C. Walker, The Stone-Čech compactification, Springer-Verlag, New York, 1974, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 83.

Department of Mathematics, UNC-Charlotte, 9201 University City Blvd., Charlotte, NC 28223-0001

E-mail address: adow@uncc.edu

   URL: http://www.math.uncc.edu/∼adow