ON THE WEAK REFLECTION PRINCIPLE

JOHN KRUEGER

Abstract. The Weak Reflection Principle for ω_{2}, or WRP(ω_{2}), is the statement that every stationary subset of P_{ω_{1}}(ω_{2}) reflects to an uncountable ordinal in ω_{2}. The Reflection Principle for ω_{2}, or RP(ω_{2}), is the statement that every stationary subset of P_{ω_{1}}(ω_{2}) reflects to an ordinal in ω_{2} with cofinality ω_{1}. Let κ be a κ^{+}-supercompact cardinal and assume 2^{κ} = κ^{+}. Then there exists a forcing poset ℙ which collapses κ to become ω_{2}, and ⊩_{ℙ} WRP(ω_{2}) ∧ ¬RP(ω_{2}).

In this paper we will be concerned with reflection of stationary subsets of P_{ω_{1}}(ω_{2}). Recall that for an uncountable ordinal α, a set S ⊆ P_{ω_{1}}(α) = {x ⊆ α : |x| < ω_{1}} is stationary if for any function F : [α]^{<ω} → α, there is a set b in S which is closed under F. Let S ⊆ P_{ω_{1}}(ω_{2}) be a stationary set. We say that S reflects to α, where α is an uncountable ordinal in ω_{2}, if S ∩ P_{ω_{1}}(α) is stationary in P_{ω_{1}}(α).

The Weak Reflection Principle for ω_{2}, or WRP(ω_{2}), is the statement that for every stationary set S ⊆ P_{ω_{1}}(ω_{2}), there is an uncountable ordinal α in ω_{2} such that S reflects to α. This principle is a special case of the stronger Weak Reflection Principle introduced in [1]. The statement WRP(ω_{2}) has a number of interesting combinatorial consequences, including 2^{ω} ≤ ω_{2} ([4], [6]), ¬□(ω_{2}) ([7]), and every stationary subset of ω_{2} ∩ cof(ω) reflects to an ordinal in ω_{2} with cofinality ω_{1}.

A related principle is the Reflection Principle for ω_{2}, or RP(ω_{2}), which asserts that every stationary subset of P_{ω_{1}}(ω_{2}) reflects to an ordinal in ω_{2} with cofinality ω_{1}. The standard models for obtaining WRP(ω_{2}) (for example, by Lévy collapsing a large cardinal to become ω_{2}) all satisfy RP(ω_{2}). Also, it tends to be easier to work with RP(ω_{2}) than with WRP(ω_{2}). Thus it is a natural question, and one which has been open for some time, whether WRP(ω_{2}) implies RP(ω_{2}).

A standard argument shows that if ♢_{A} holds for every stationary set A ⊆ ω_{2} ∩ cof(ω), then for any stationary set S ⊆ P_{ω_{1}}(ω_{2}), there is a stationary set T ⊆ S which does not reflect to any uncountable ordinal in ω_{2} with cofinality ω. So assuming the existence of such diamonds, WRP(ω_{2}) implies RP(ω_{2}). Classically, the existence of such diamonds were known to follow from GCH; more recently, Shelah [5] has proven they are a consequence of 2^{ω_{1}} = ω_{2}. Thus assuming 2^{ω_{1}} = ω_{2}, WRP(ω_{2}) implies RP(ω_{2}) (originally, this result was proven in [2] by a different argument).

The Weak Reflection Principle for ω_{2} is equiconsistent with a weakly compact cardinal. However, Sakai [3] has shown that a kind of local reflection is consistent from ZFC. Specifically, assuming GCH and □_{ω_{1}}, there is a generic extension in which there exists a stationary set S ⊆ P_{ω_{1}}(ω_{2}) such that every stationary subset of S reflects to an uncountable ordinal in ω_{2} with cofinality ω.

Date: February 2010.

2000 Mathematics Subject Classification. Primary 03E35; Secondary 03E05.

Key words and phrases. Weak Reflection Principle, Reflection Principle, Generalized Stationarity.

We had the idea to combine parts of Sakai’s argument with the classical methods for constructing models of the Weak Reflection Principle, in an attempt to show that WRP(ω_{2}) does not imply RP(ω_{2}). This paper is the culmination of those efforts. We will prove that if κ is a κ^{+}-supercompact cardinal and 2^{κ} = κ^{+}, then there is a forcing poset ℙ which collapses κ to become ω_{2}, and forces WRP(ω_{2}) ∧ ¬RP(ω_{2}).

1. Background Material

We begin by recording some definitions and results which will be used in the paper.

By a forcing poset, we mean an order which is reflexive and transitive (we do not assume it is antisymmetric or separative). Let ℙ be a forcing poset, and let μ be a regular uncountable cardinal. We say ℙ has the μ-chain condition, or is μ-c.c., if for any sequence ⟨p_{i} : i < μ⟩ of conditions in ℙ, there are i < j such that p_{i} and p_{j} are compatible. We say ℙ is μ-Knaster if for any sequence ⟨p_{i} : i < μ⟩ of conditions in ℙ, there is a set Z ⊆ μ with |Z| = μ such that for all i < j in Z, p_{i} and p_{j} are compatible. Clearly if ℙ is μ-Knaster, then ℙ is μ-c.c.

We say ℙ is μ-distributive if whenever ⟨D_{i} : i < ν⟩ is a sequence of dense open subsets of ℙ, where ν < μ, then ⋂{D_{i} : i < ν} is dense. This property implies that if G is a generic filter on ℙ over V, then for any set x ⊆ V in V [G] such that V [G] ⊧ |x| < μ, x is in V. (Assuming ℙ is separative, these two properties are equivalent, but they are not equivalent in general.)

A map i : ℙ → ℚ, where ℙ and ℚ are forcing posets, is a complete embedding if:

(1) q ≤ p in ℙ implies i(q) ≤ i(p) in ℚ,

(2) if i(p) and i(q) are compatible in ℚ, then p and q are compatible in ℙ,

(3) for all q in ℚ, there is p in ℙ such that whenever t ≤ p in ℙ, i(t) and q are compatible in ℚ.

Suppose i : ℙ → ℚ is a complete embedding. If H is a generic filter on ℚ over V, then i^{−1}(H) is a generic filter on ℙ over V, and V [i^{−1}(H)] ⊆ V [H]. Note that if ℙ ⊆ ℚ and i is the inclusion map, then i^{−1}(H) = {p ∈ ℙ : p ∈ H} = H ∩ ℙ.

A complete embedding i : ℙ → ℚ provides a way to translate ℙ-names to ℚnames. Define i(ẋ), for each ℙ-name ẋ, by recursion, letting i(ẋ) = {(i(p), i(ẏ)) : (p, y) ∈ ẋ}. If H is a generic filter on ℚ over V, then i(ẋ)^{H} = ẋ^{i^{−1}(H)}. Note that if i is the inclusion map, then for any ℙ-name ẋ, i(ẋ) = ẋ. So if H is a generic filter on ℚ over V, then ẋ^{H} = ẋ^{H∩ℙ}.

A map i : ℙ → ℚ is a dense embedding if

(1) q ≤ p in ℙ implies i(q) ≤ i(p) in ℚ,

(2) if i(p) and i(q) are compatible in ℚ, then p and q are compatible in ℙ,

(3) i[ℙ] is dense in ℚ. Every dense embedding is a complete embedding. If i is a dense embedding, then ℙ and ℚ are forcing equivalent, in the sense that they have the same generic extensions. We will consider a particular kind of dense embedding, as follows. Say a map i : ℙ → ℚ is an isomorphism if i is a bijection, and q ≤ p in ℙ iff i(q) ≤ i(p) in ℚ. If i : ℙ → ℚ is an isomorphism of ℙ onto a dense subset of ℚ, then i is a dense embedding.

Let ℙ and ℚ be forcing posets with maximum elements 1_{ℙ} and 1_{ℚ} respectively. A map π : ℚ → ℙ is a projection mapping if:

(1) π(1_{ℚ}) = 1_{ℙ}, (2) q ≤ p in ℚ implies π(q) ≤ π(p) in ℙ,

(3) if t ≤ π(q) in ℙ, then there is r ≤ q in ℚ such that π(r) ≤ t in ℙ.

Suppose π : ℚ → ℙ is a projection mapping. If H is a generic filter on ℚ over V, then π[H] generates a generic filter on ℙ over V. Furthermore, the projection mapping π allows a factorization of ℚ into a two-step iteration ℙ∗(ℚ/ℙ) as follows. Let G be a generic filter on ℙ over V. In V [G], define ℚ/ℙ as the forcing poset with underlying set {q ∈ ℚ : π(q) ∈ G}, ordered as in ℚ. Then the map i : ℚ → ℙ∗(ℚ/ℙ) defined by i(q) = (π(q), q̌) is an isomorphism of ℚ onto a dense subset of ℙ ∗ (ℚ/ℙ). Moreover, if K is a generic filter on ℚ over V, and J is the filter on ℙ generated by π[K], then K is a generic filter on ℚ/ℙ over V [J], and V [K] = V [J][K].

Lemma 1.1. Suppose π : ℚ → ℙ is a projection mapping. If D is a dense subset of ℚ, then ℙ forces D ∩ (ℚ/ℙ) is a dense subset of ℚ/ℙ.

Proof. Let G be a generic filter on ℙ over V. Then ℚ/ℙ = {q ∈ ℚ : π(q) ∈ G}. Let p be in ℚ/ℙ, and we find r ≤ p in D ∩ (ℚ/ℙ). Note that π(p) ∈ G.

Let D′ = {u ≤ π(p) : ∃r ≤ p (r ∈ D ∧ π(r) = u)}. We claim D′ is dense below π(p). So let t ≤ π(p) in ℙ be given. By the definition of a projection mapping, fix q ≤ p in ℚ such that π(q) ≤ t. Since D is dense, choose r ≤ q in D. Let u = π(r). Then u = π(r) ≤ π(q) ≤ t, so u ≤ t. Also r ≤ q ≤ p, so r ≤ p, r ∈ D, and π(r) = u. So u ≤ t and u is in D′.

Since π(p) is in G and D′ is dense below π(p), fix u in D′ ∩ G. Then there is r ≤ p in D such that π(r) = u. Since π(r) = u is in G, r is in D ∩ (ℚ/ℙ). □

Suppose ℙ and ℚ are forcing posets. The Product Lemma says that for any set G ⊆ ℙ × ℚ, G is a generic filter on ℙ × ℚ over V iff there is a generic filter G_{ℙ} on ℙ over V, and a generic filter G_{ℚ} on ℚ over V [G_{ℙ}], such that G = G_{ℙ} × G_{ℚ}. Moreover, V [G] = V [G_{ℙ}][G_{ℚ}], where G = G_{ℙ} × G_{ℚ}. Since ℙ × ℚ is isomorphic to ℚ × ℙ by the map (p, q) ↦ (q, p), it follows that if G_{ℙ} is a generic filter on ℙ over V, and G_{ℚ} is a generic filter on ℚ over V [G_{ℙ}], then G_{ℙ} is a generic filter on ℙ over V [G_{ℚ}], and V [G_{ℙ}][G_{ℚ}] = V [G_{ℚ}][G_{ℙ}].

We will apply the Product Lemma in the following way. Consider a generic extension V [G][H], where G is a generic filter on ℙ over V, and H is a generic filter on ℚ over V [G]. If it happens that the forcing poset ℚ is already in V, then by the Product Lemma, V [G][H] = V [G × H] = V [H × G] = V [H][G]. So V [G][H] = V [H][G].

Let X be a set. A nice ℙ-name for a subset of X is a ℙ-name of the form ⋃{A_{x} × {x̌} : x ∈ X}, where for each x in X, A_{x} is an antichain of ℙ. A basic fact about ℙ-names is that if p ⊩ Ȧ ⊆ X, then there is a nice ℙ-name Ḃ for a subset of X such that p ⊩ Ȧ = Ḃ. Nice names are useful for controlling the continuum function. Let Z be the collection of all nice ℙ-names for a subset of X. If G is a generic filter on ℙ over V, then in V [G], the map Ȧ ↦ Ȧ^{G} is a surjective function from Z onto P (X). Hence V [G] ⊧ |P (X)| ≤ |Z|. Usually the chain condition and the size of ℙ are used to put an upper bound on the size of Z.

Suppose f : X → Y is a function and a is a subset of X. If we say a is closed under f, this implicitly assumes that a is non-empty. If f is a function and A ⊆ dom(f), we write f[A] for the pointwise image of A, {f(a) : a ∈ A}.

For any set A, let P_{ω_{1}}(A) = {a ⊆ A : |a| < ω_{1}}. Suppose ω_{1} ⊆ A. A set C ⊆ P_{ω_{1}}(A) is club if (1) whenever ⟨a_{n} : n < ω⟩ is a sequence of sets in C such that a_{n} ⊆ a_{n+1} for all n < ω, then ⋃{a_{n} : n < ω} ∈ C, and (2) for all b in P_{ω_{1}}(A), there is c in C with b ⊆ c. A set S ⊆ P_{ω_{1}}(A) is stationary if for every club set C ⊆ P_{ω_{1}}(A), S ∩ C is non-empty.

A useful characterization of stationarity is given in terms of functions. A set S ⊆ P_{ω_{1}}(A) is stationary iff for any function F : [A]^{<ω} → A, there is a set a in S which is closed under F. Fodor’s Lemma says that whenever S ⊆ P_{ω_{1}}(A) is stationary, and F : S → A satisfies that F (b) ∈ b for all b in S, then there is a stationary set S′ ⊆ S and x in A such that F (a) = x for all a ∈ S′.

When we say a cardinal θ is much larger than a set X, we mean that P (P (P (X))) is in H(θ). We will frequently consider countable elementary substructures of H(θ), where θ ≥ ω_{2} is a regular cardinal. Let N ≺ H(θ) be countable. We claim that for any uncountable ordinal ν in N, N ∩ ν is not in N. For if N ∩ ν is in N, then since ν ⊈ N ∩ ν, let ξ be the least ordinal in ν \ (N ∩ ν). Then by elementarity, ξ is in N, which is a contradiction.

Let N ≺ H(θ) be countable. Suppose C ∈ N is a club subset of P_{ω_{1}}(A). Then N ∩ A ∈ C. Suppose ℙ is a forcing poset in N. If G if a generic filter on ℙ over V, then N[G] = {ẋ^{G} : ẋ is a ℙ-name in N} is an elementary substructure of H(θ)^{V [G]}. Assume ℙ preserves ω_{1}, Ċ is a ℙ-name in N, and p forces Ċ is club in P_{ω_{1}}(A). If G is a generic filter on ℙ over V and p ∈ G, then C = Ċ^{G} is in N[G], and hence N[G] ∩ A ∈ C.

A condition q is N-generic (for ℙ) if q ⊩ N[Ġ]∩V = N, where Ġ is the canonical ℙ-name for the generic filter. Suppose p ⊩ Ċ is club in P_{ω_{1}}(A), where Ċ ∈ N, and p is N-generic. Then by the last paragraph, p ⊩ N[Ġ] ∩ A ∈ Ċ. But since p is N-generic, p ⊩ N[Ġ] ∩ A = N ∩ A. So p ⊩ N ∩ A ∈ Ċ. A descending sequence of conditions ⟨p_{n} : n < ω⟩ in ℙ ∩ N is N-generic if for every dense open set D ⊆ ℙ in N, there is n < ω such that p_{n} ∈ D ∩ N. It is always possible to construct an N-generic sequence, by simply enumerating the dense open subsets of ℙ which are in N in order type ω, and defining a descending sequence of conditions in N ∩ ℙ by recursion, meeting each dense open set in the enumeration. If ⟨p_{n} : n < ω⟩ is an N-generic sequence, and q ≤ p_{n} for all n < ω, then q is an N-generic condition.

We will use the following special case of the Δ-System Lemma. Assume CH. Then for any sequence ⟨x_{i} : i < ω_{2}⟩ of countable sets, there is a set Z ⊆ ω_{2} with |Z| = ω_{2}, and a countable set b, such that for all i < j in Z, x_{i} ∩ x_{j} = b.

A cardinal κ is λ-supercompact, where κ ≤ λ, if there exists an elementary embedding j : V → M with critical point κ such that j(κ) > λ and M^{λ} ⊆ M.

Suppose j : N → M is an elementary embedding, where N and M are transitive inner models of ZFC. Suppose ℙ is a forcing poset in N, G is a generic filter on ℙ over N, and H is a generic filter on j(ℙ) over M. Then j can be extended to j : N[G] → M[H] such that j(G) = H iff j[G] ⊆ H (namely, let j(ẋ^{G}) = j(ẋ)^{H}).

Let M ⊆ N be inner models of ZFC, suppose λ is a regular uncountable cardinal of N, and assume M^{<λ} ∩ N ⊆ M. Let ℙ be a forcing poset in M, and assume N models that ℙ is λ-c.c. If G is a generic filter on ℙ over N, then M[G]^{<λ} ∩ N[G] ⊆ M[G].

2. Outline of Proof

The objective of this paper is to define a forcing poset ℙ such that

⊩_{ℙ} WRP(ω_{2}) ∧ ¬RP(ω_{2}).

Specifically, we will define a forcing poset ℙ which satisfies that whenever K is a generic filter on ℙ over V, then in V [K] there is a set S satisfying:

(1) S is a stationary subset of P_{ω_{1}}(ω_{2}),

(2) S does not reflect to any ordinal in ω_{2} ∩ cof(ω_{1}),

(3) every stationary subset of S reflects to an uncountable ordinal in ω_{2}∩cof(ω),

(4) every stationary subset of P_{ω_{1}}(ω_{2}) \ S reflects to an ordinal in ω_{2} ∩ cof(ω_{1}). Clearly statements (1) and (2) imply that ¬RP(ω_{2}) holds in V [K]. Statements (3) and (4) imply WRP(ω_{2}) holds in V [K]. For if U is a stationary subset of P_{ω_{1}}(ω_{2}), then either U ∩ S is stationary, or U \ S is stationary. In the first case, (3) implies there is an uncountable ordinal α in ω_{2} ∩ cof(ω) such that U ∩ S reflects to α. In the second case, (4) implies there is an ordinal α in ω_{2} ∩ cof(ω_{1}) such that U \ S reflects to α. In either case, there is an uncountable ordinal α in ω_{2} such that U reflects to α.

We will define ℙ in a model V in which there is a cardinal κ which is κ^{+}- supercompact and 2^{κ} = κ^{+}. The forcing poset ℙ will be of the form

Coll(ω_{1}, <κ) ∗ ℙ̇_{ω_{2}} ∗ ℚ̇.

So in the first step, we Lévy collapse κ to become ω_{2}. Let G be a generic filter on Coll(ω_{1}, <κ) over V. Note that in V [G], 2^{ω} = ω_{1}, 2^{ω_{1}} = ω_{2}, and 2^{ω_{2}} = ω_{3}.

In V [G] we define a forcing poset ℙ_{ω_{2}} which adds a generic stationary set S ⊆ P_{ω_{1}}(ω_{2}) which does not reflect to any ordinal in ω_{2} ∩ cof(ω_{1}).

Definition 2.1. Let ℙ_{ω_{2}} be the forcing poset consisting of pairs (X, F) satisfying: (1) X is a countable subset of P_{ω_{1}}(ω_{2}) \ {∅},

(2) F is a function whose domain is a countable subset of ω_{2} ∩ cof(ω_{1}),

(3) for all ξ in dom(F), F (ξ) is a ⊆-increasing and continuous sequence ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩ of countable subsets of ξ, for some γ_{ξ} < ω_{1},

(4) for all ξ in dom(F) and i ≤ γ_{ξ}, a_{i}^{ξ} is not in X,

(5) for all ξ in dom(F), a_{γ_{ξ}}^{ξ} ⊆ ⋃ X.

Define an order on ℙ_{ω_{2}} by letting (Y, K) ≤ (X, F) if:

(a) X ⊆ Y,

(b) for all b in Y \ X, b ⊈ ⋃ X,

(c) dom(F) ⊆ dom(K),

(d) for all ξ in dom(F), F (ξ) is an initial segment of K(ξ).

The forcing poset ℙ_{ω_{2}} is ω_{1}-distributive. Since 2^{ω} = ω_{1} holds in V [G], ℙ_{ω_{2}} is ω_{2}-c.c. It is easy to see that ℙ_{ω_{2}} forces 2^{ω} = ω_{1}, 2^{ω_{1}} = ω_{2}, and 2^{ω_{2}} = ω_{3}.

Let H be a generic filter on ℙ_{ω_{2}} over V. Define S = ⋃{X : ∃F (X, F) ∈ H}. Then S is a stationary subset of P_{ω_{1}}(ω_{2}). For each ν in ω_{2} ∩ cof(ω_{1}), the union of the collection of sequences {F (ν) : ∃X (X, F) ∈ H, ν ∈ dom(F)} is an increasing and continuous sequence ⟨a_{i}^{ν} : i < ω_{1}⟩, with union equal to ν, which is disjoint from S. It follows that S does not reflect to any ordinal in ω_{2} ∩ cof(ω_{1}). Moreover, by upwards absoluteness, S does not reflect to any ordinal in ω_{2} ∩ cof(ω_{1}) in any outer model of V [G ∗ H] with the same ω_{1} and ω_{2}.

In V [G ∗ H], we define a forcing iteration ℚ, which is designed to destroy the stationarity of any subset of S which does not reflect to an uncountable ordinal in ω_{2} ∩ cof(ω). This forcing poset is ω_{1}-distributive and ω_{2}-c.c. in V [G ∗ H]. Let I be a generic filter on ℚ over V [G∗H]. Then in V [G∗H ∗I], S is a stationary subset of P_{ω_{1}}(ω_{2}) which does not reflect to any ordinal in ω_{2} ∩ cof(ω_{1}), and moreover, every stationary subset of S reflects to an uncountable ordinal in ω_{2} ∩ cof(ω).

Finally, let K = G ∗ H ∗ I. Statements (1), (2), and (3) hold in V [K], as we have already observed. It remains to show (4), that every stationary subset of P_{ω_{1}}(ω_{2})\S reflects to an ordinal in ω_{2} ∩ cof(ω_{1}).

In the ground model V, let j : V → M be an elementary embedding with critical point κ such that j(κ) > κ^{+} and M^{κ^{+}} ⊆ M. Let \overline{G} ∗ \overline{H} ∗ \overline{I} be a generic filter on j(Coll(ω_{1}, <κ) ∗ ℙ̇_{ω_{2}} ∗ ℚ̇) over V. Let G = \overline{G} ∩ Coll(ω_{1}, <κ) and H = \overline{H} ∩ ℙ_{ω_{2}}. Then G is a generic filter on Coll(ω_{1}, < κ) over V, and H is a generic filter on ℙ_{ω_{2}} over V [G]. In V [\overline{G} ∗ \overline{H}], we can extend j to j : V [G ∗ H] → M[\overline{G} ∗ \overline{H}] such that j(G ∗ H) = \overline{G} ∗ \overline{H}. Also let G′ = \overline{G} ∩ Coll(ω_{1}, [κ, j(κ))). Note that M[\overline{G}] = M[G][G′].

In the model M[\overline{G}], j(ℙ_{ω_{2}}) factors as a two-step iteration

ℙ_{ω_{2}} ∗ ℙ(κ, S, j(κ)).

(The definition of ℙ(κ, S, j(κ)) will be given in the next section; it is a generalization of Definition 2.1.) Since j(ℙ_{ω_{2}}) is forcing equivalent to ℙ_{ω_{2}} ∗ ℙ(κ, S, j(κ)), there exists a generic filter H′ on ℙ(κ, S, j(κ)) over M[\overline{G}][H] such that M[\overline{G}][\overline{H}] = M[\overline{G}][H][H′].

In the model M[\overline{G} ∗ \overline{H}], there exists a projection mapping π : j(ℚ) → ℚ. This allows us to factor j(ℚ) as ℚ ∗ (j(ℚ)/ℚ). Let I be the filter on ℚ generated by π[\overline{I}]. Then M[\overline{G} ∗ \overline{H}][\overline{I}] = M[\overline{G} ∗ \overline{H}][I][\overline{I}]. In the model V [\overline{G} ∗ \overline{H} ∗ \overline{I}], we extend j to j : V [G ∗ H ∗ I] → M[\overline{G} ∗ \overline{H} ∗ \overline{I}] such that j(G ∗ H ∗ I) = \overline{G} ∗ \overline{H} ∗ \overline{I}. By several applications of the Product Lemma, the model M[\overline{G} ∗ \overline{H} ∗ \overline{I}] can be shown to be equal to M[G ∗ H ∗ I][G′][H′][\overline{I}].

Suppose T is a stationary subset of P_{ω_{1}}(ω_{2}) \ S in the model V [G ∗ H ∗ I]. Since the forcing poset Coll(ω_{1}, <κ) ∗ ℙ̇_{ω_{2}} ∗ ℚ̇ is κ-c.c., M[G ∗ H ∗ I]^{κ^{+}} ∩ V [G ∗ H ∗ I] ⊆ M[G ∗ H ∗ I]. So in M[G ∗ H ∗ I], T is a stationary subset of P_{ω_{1}}(κ). The model M[\overline{G} ∗ \overline{H} ∗ \overline{I}] is a generic extension of M[G ∗ H ∗ I] by the forcing poset Coll(ω_{1}, [κ, j(κ))) ∗ ℙ(κ, S, j(κ)) ∗ (j(ℚ)/ℚ). It turns out that this forcing poset preserves the stationarity of any stationary subset of P_{ω_{1}}(κ) in M[G ∗ H ∗ I] which is disjoint from S. Thus, T remains stationary in M[\overline{G} ∗ \overline{H} ∗ \overline{I}].

In the model M[\overline{G} ∗ \overline{H} ∗ \overline{I}], j(T) reflects to κ, since j(T) ∩ P_{ω_{1}}(κ) = T and T is stationary in P_{ω_{1}}(κ). In this model, κ is an ordinal in j(κ) = ω_{2} with cofinality ω_{1}. By the elementarity of j, in V [G ∗ H ∗ I] the set T reflects to an ordinal in ω_{2} with cofinality ω_{1}.

3. The forcing poset ℙ(α, A, β)

In the previous section we defined the forcing poset ℙ_{ω_{2}}, which adds a stationary subset of P_{ω_{1}}(ω_{2}) which does not reflect to any ordinal in ω_{2} with cofinality ω_{1}. For the purpose of extending an elementary embedding after forcing with ℙ_{ω_{2}}, we will need to consider a more general version of this forcing poset.

Definition 3.1. Let α < β ≤ ω_{2} be ordinals, where β has uncountable cofinality. Let A be a subset of P_{ω_{1}}(α). If cf(α) = ω_{1}, also assume P_{ω_{1}}(α) \ A is stationary in P_{ω_{1}}(α). Define a forcing poset ℙ(α, A, β) as follows. A condition in ℙ(α, A, β) is a pair (X, F) satisfying:

(1) X is a countable subset of P_{ω_{1}}(β), (2) for all b in X, b is not a subset of α,

(3) F is a function whose domain is a countable subset of [α, β) ∩ cof(ω_{1}),

(4) for all ξ in dom(F), F (ξ) is a ⊆-increasing and continuous sequence ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩ of countable subsets of ξ, for some γ_{ξ} < ω_{1},

(5) for all ξ in dom(F) and i ≤ γ_{ξ}, a_{i}^{ξ} is not in X ∪ A,

(6) for all ξ in dom(F), a_{γ_{ξ}}^{ξ} ⊆ ⋃ X.

Define an order on ℙ(α, A, β) by letting (Y, K) ≤ (X, F) if:

(a) X ⊆ Y,

(b) for all b in Y \ X, b ⊈ ⋃ X,

(c) dom(F) ⊆ dom(K),

(d) for all ξ in dom(F), F (ξ) is an initial segment of K(ξ).

Note that property (2) implies every set in X is non-empty. Also, for any condition (X, F) in ℙ(α, A, β), X ∩ A is empty. Namely, if b is in X, then b is not a subset of α by (2). But A is a subset of P_{ω_{1}}(α).

If (X, F) is a condition, then X is an approximation of a generic stationary set S ⊆ P_{ω_{1}}(β). The function F in (X, F) approximates an array of clubs which witnesses that S does not reflect to any ordinal in [α, β) with cofinality ω_{1}.

We will only be interested in two particular cases of the definition. The first case is when α = 0 and A = ∅. In this case, we abbreviate ℙ(α, A, β) as ℙ_{β}. For example, the forcing poset ℙ_{ω_{2}} described in the previous section is equal to ℙ(0, ∅, ω_{2}).

The second case of interest occurs in the following context. Let ν < ω_{2} be an ordinal with cofinality ω_{1}. Let H be a generic filter on ℙ_{ν} over V. In V [H], define A = ⋃{X : ∃F (X, F) ∈ H}. So A is a subset of P_{ω_{1}}(ν). As we will prove later, P_{ω_{1}}(ν) \ A is stationary in P_{ω_{1}}(ν). So it is meaningful to define ℙ(ν, A, ω_{2}) in the model V [H]. As it turns out, ℙ_{ω_{2}} is forcing equivalent to the two-step iteration ℙ_{ν} ∗ ℙ(ν, A, ω_{2}). This factorization is the motivation for Definition 3.1.

Notation 3.2. Let β ≤ ω_{2} be an ordinal with uncountable cofinality. Then ℙ_{β} denotes ℙ(0, ∅, β).

The proof of the next lemma is straightforward.

Lemma 3.3. Let β_{1} < β_{2} ≤ ω_{2} be ordinals with uncountable cofinality. Then ℙ_{β_{1}} ⊆ ℙ_{β_{2}}. If p and q are in ℙ_{β_{1}}, then q ≤ p in ℙ_{β_{1}} iff q ≤ p in ℙ_{β_{2}}.

Lemma 3.4. Let p = (X, F) be a condition in ℙ(α, A, β), and let d be a set in P_{ω_{1}}(β) such that d is not a subset of α and d is not a subset of ⋃ X. Then q = (X ∪ {d}, F) is a condition in ℙ(α, A, β) and q ≤ p.

Proof. Properties (1), (2), (3), (4), and (6) of Definition 3.1 are immediate. For (5), let ξ be in dom(F) and let F (ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Let i ≤ γ_{ξ}. Then a_{i}^{ξ} is not in X ∪ A. Also a_{i}^{ξ} is a subset of ⋃ X. Since d is not a subset of ⋃ X, a_{i}^{ξ} is not equal to d. So a_{i}^{ξ} is not in (X ∪ {d}) ∪ A. It is immediate that q ≤ p. □

Lemma 3.5. Let b be a set in P_{ω_{1}}(β). Then there are densely many conditions (Y, K) in ℙ(α, A, β) such that b ⊆ ⋃ Y.

Proof. Let (X, F) be a condition. Choose a countable set d ⊆ β such that b ∪ (⋃ X) ∪ {α} ⫋ d. By Lemma 3.4, (X ∪ {d}, F) is a condition in ℙ(α, A, β) which is below (X, F), and clearly b ⊆ ⋃(X ∪ {d}). □

Lemma 3.6. Let ξ be an ordinal in [α, β) ∩ cof(ω_{1}), let π < ω_{1}, and let b be a countable subset of ξ. Then there are densely many conditions in ℙ(α, A, β) of the form (Y, K), where ξ is in dom(K), and K(ξ) = ⟨b_{i}^{ξ} : i ≤ γ_{ξ}⟩, where π ≤ γ_{ξ} and b ⊆ b_{γ_{ξ}}^{ξ}.

Proof. Let (X, F) be a condition in ℙ(α, A, β). We choose sets c and d as follows, depending on whether ξ > α or ξ = α. First suppose ξ > α. Since ξ is uncountable and ⋃ X is countable, we can choose countable sets c and d such that ((⋃ X) ∩ ξ) ∪ b ∪ {α} ⫋ c ⫋ d ⊆ ξ. Now suppose ξ = α. Then cf(α) = ω_{1}, so P_{ω_{1}}(α) \ A is stationary in P_{ω_{1}}(α). So we can choose c in P_{ω_{1}}(α)\A such that ((⋃ X)∩α)∪b ⫋ c. Now let d = c ∪ {α}.

Define Y = X ∪ {d}. By the choice of d, α ∈ d and d ⊈ ⋃ X. So by Lemma 3.4, (Y, F) is a condition in ℙ(α, A, β) which is below (X, F).

Now we define a function K. Let K have domain equal to dom(F) ∪ {ξ}. For all ν in dom(K) different from ξ, let K(ν) = F (ν). To define K(ξ), we consider the two cases of whether ξ is in dom(F) or not. Suppose ξ is not in dom(F). Then let K(ξ) be the sequence ⟨b_{i}^{ξ} : i ≤ π⟩, where b_{i}^{ξ} = c for all i ≤ π. Suppose ξ is in dom(F). Let F (ξ) = ⟨a_{i}^{ξ} : i ≤ \overline{γ}_{ξ}⟩. Let γ_{ξ} = max(\overline{γ}_{ξ}, π) + 1. Now define K(ξ) = ⟨b_{i}^{ξ} : i ≤ γ_{ξ}⟩, where b_{i}^{ξ} = a_{i}^{ξ} for i ≤ \overline{γ}_{ξ}, and b_{j}^{ξ} = c for \overline{γ}_{ξ} < j ≤ γ_{ξ}. It is routine to verify that (Y, K) is a condition in ℙ(α, A, β) which is below (X, F), and (Y, K) clearly satisfies the conclusion of the lemma. □

We will frequently consider descending sequences of conditions in ℙ(α, A, β). So it will be useful to establish some notation and basic facts about such sequences.

Suppose ⟨(X_{n}, F_{n}) : n < ω⟩ is a descending sequence of conditions in ℙ(α, A, β). Let us define what we will call the infinum of the sequence of functions ⟨F_{n} : n < ω⟩. Let K have domain equal to ⋃{dom(F_{n}) : n < ω}, which is a countable subset of [α, β) ∩ cof(ω_{1}). Consider ξ in the domain of K. If n < m and ξ is in the domain of F_{n}, then since (X_{m}, F_{m}) ≤ (X_{n}, F_{n}), F_{n}(ξ) is an initial segment of F_{m}(ξ). So if we take the union of all sequences F_{n}(ξ), where ξ ∈ dom(F_{n}), we obtain an increasing and continuous sequence of the form ⟨a_{i}^{ξ} : i < γ_{ξ}⟩, where γ_{ξ} < ω_{1}. If γ_{ξ} is a successor ordinal, let K(ξ) be the sequence ⟨a_{i}^{ξ} : i ≤ γ_{ξ} − 1⟩. If γ_{ξ} is a limit ordinal, define a_{γ_{ξ}}^{ξ} = ⋃{a_{i}^{ξ} : i < γ_{ξ}}, and let K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Then K(ξ) is an increasing and continuous sequence of countable subsets of ξ. We refer to K as the infinum of the sequence ⟨F_{n} : n < ω⟩.

Lemma 3.7. Let θ be a regular cardinal much larger than ℙ(α, A, β), and suppose N is a countable elementary substructure of H(θ) such that ℙ(α, A, β) is in N. Let ⟨(X_{n}, F_{n}) : n < ω⟩ be an N-generic sequence for ℙ(α, A, β).

Let X′ = ⋃{X_{n} : n < ω} and let K be the infinum of ⟨F_{n} : n < ω⟩. For each ξ in dom(K), write K(ξ) as ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Then:

(1) ⋃ X′ = N ∩ β,

(2) dom(K) = N ∩ [α, β) ∩ cof(ω_{1}),
(3) for all ξ in dom(K), γ_{ξ} = N ∩ ω_{1} and a_{γ_{ξ}}^{ξ} = N ∩ ξ.

Proof. For each n < ω, X_{n} is in N. Since X_{n} is countable, X_{n} ⊆ N. It follows that X′ = ⋃{X_{n} : n < ω} ⊆ N. So if b is in X′, then b is in N, and therefore since b is countable, b ⊆ N ∩ β. It follows that ⋃ X′ ⊆ N ∩ β. On the other hand, if ν is in N ∩ β, then by Lemma 3.5 there is a dense set D of conditions (X, F) such that ν is in ⋃ X. By elementarity, D is in N. So fix n < ω such that (X_{n}, F_{n}) is in D. Then ν is in ⋃ X_{n} ⊆ ⋃ X′, so ν is in ⋃ X′. Thus ⋃ X′ = N ∩ β. Statements (2) and (3) can be proven by similar arguments, using Lemma 3.6. □

The next result is our main tool for constructing lower bounds of descending sequences.

Proposition 3.8. Let θ be a regular cardinal much larger than ℙ(α, A, β). Let N be a countable elementary substructure of H(θ) such that ℙ(α, A, β) is in N, and moreover, if cf(α) = ω_{1}, then N ∩ α is not in A. Let ⟨(X_{n}, F_{n}) : n < ω⟩ be an N-generic sequence for ℙ(α, A, β).

Let X′ = ⋃{X_{n} : n < ω} and let K be the infinum of ⟨F_{n} : n < ω⟩. Suppose that Y is a countable subset of P_{ω_{1}}(β) such that X′ ⊆ Y, and for all b in Y \ X′, either b = N ∩ β, or there is an uncountable ordinal ν in (α, β) ∩ N ∩ cof(ω) such that b = N ∩ ν. Then (Y, K) is a condition in ℙ(α, A, β) and (Y, K) ≤ (X_{n}, F_{n}) for all n < ω.

Proof. We prove that (Y, K) is a condition in ℙ(α, A, β). Note that properties (1), (3), and (4) of Definition 3.1 are obvious, and (2) follows from the fact that α is in N. To prove (5) and (6), consider ξ in dom(K). Let K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. By Lemma 3.7, ⋃ Y = N ∩ β, ξ ∈ N, and a_{γ_{ξ}}^{ξ} = N ∩ ξ. So a_{γ_{ξ}}^{ξ} = N ∩ ξ ⊆ N ∩ β = ⋃ Y, proving (6).

To prove (5), let i ≤ γ_{ξ}, and we show a_{i}^{ξ} is not in Y ∪ A. First suppose i < γ_{ξ}. Then a_{i}^{ξ} is in N. Note that every set in Y \ X′ is not in N. So if a_{i}^{ξ} is in Y, then it is in X′. But if a_{i}^{ξ} is in X′, then we can find n < ω such that a_{i}^{ξ} is in X_{n} and also appears on the sequence F_{n}(ξ), contradicting that (X_{n}, F_{n}) is a condition. So a_{i}^{ξ} is not in Y. For some n < ω, a_{i}^{ξ} is on the sequence F_{n}(ξ), which implies by definition that a_{i}^{ξ} is not in A. So a_{i}^{ξ} is not in Y ∪ A.

Now assume i = γ_{ξ}. Then by Lemma 3.7, a_{γ_{ξ}}^{ξ} = N ∩ ξ. In particular, a_{γ_{ξ}}^{ξ} is not in N. So a_{γ_{ξ}}^{ξ} is not in X′. If ξ = α, then α has cofinality ω_{1}. By the assumptions on N, N ∩ ξ = N ∩ α = a_{γ_{ξ}}^{ξ} is not in A. If ξ > α, then α ∈ N ∩ ξ = a_{γ_{ξ}}^{ξ}, and therefore a_{γ_{ξ}}^{ξ} is not in A. So we have proven that a_{γ_{ξ}}^{ξ} is not in X′ ∪ A.

Let b be in Y \ X′, and we show b is not equal to a_{γ_{ξ}}^{ξ}. First suppose b = N ∩ β. Then ξ ∈ N ∩ β = b, whereas a_{γ_{ξ}}^{ξ} is a subset of ξ. So b is not equal to a_{γ_{ξ}}^{ξ}. Now suppose b = N ∩ ν, for some uncountable ordinal ν in (α, β) ∩ N ∩ cof(ω). Since ν has cofinality ω and ξ has cofinality ω_{1}, either ξ < ν or ν < ξ. If ξ < ν, then ξ is in N ∩ ν = b, whereas a_{γ_{ξ}}^{ξ} ⊆ ξ, so a_{γ_{ξ}}^{ξ} is not equal to b. If ν < ξ, then ν is in N ∩ ξ = a_{γ_{ξ}}^{ξ}, whereas b ⊆ ν, so a_{γ_{ξ}}^{ξ} is not equal to b. This completes the proof that (Y, K) is in ℙ(α, A, β).

Let n < ω be given, and we show (Y, K) ≤ (X_{n}, F_{n}). Properties (a), (c), and (d) of Definition 3.1 are immediate. Suppose b is in Y \X_{n}, and we show b ⊈ ⋃ X_{n}. If b is in X′, then there is m > n such that b is in X_{m} \X_{n}. Since (X_{m}, F_{m}) ≤ (X_{n}, F_{n}), b ⊈ ⋃ X_{n}. Now suppose b is not in X′. Then b = N ∩ ν, where either ν = β, or ν is an uncountable ordinal in (α, β) ∩ N ∩ cof(ω). Suppose for a contradiction that b ⊆ ⋃ X_{n}. Then b = b ∩ ν ⊆ (⋃ X_{n}) ∩ ν ⊆ N ∩ ν = b. So (⋃ X_{n}) ∩ ν = N ∩ ν. But ⋃ X_{n} and ν are in N, so by elementarity, (⋃ X_{n}) ∩ ν is in N. So N ∩ ν is in N, which is impossible. So indeed b ⊈ ⋃ X_{n}. □

Theorem 3.9. The forcing poset ℙ(α, A, β) is ω_{1}-distributive.

Proof. Let {D_{n} : n < ω} be a collection of dense open subsets of ℙ(α, A, β). Let (X, F) be a condition. We find a condition (Y, K) ≤ (X, F) in ⋂{D_{n} : n < ω}.

Fix a regular cardinal θ much larger than ℙ(α, A, β). Choose a countable elementary substructure N of H(θ) which contains as members the objects ℙ(α, A, β), (X, F), and {D_{n} : n < ω}, and moreover, if cf(α) = ω_{1}, then N ∩ α is not in A. This is possible since if cf(α) = ω_{1}, then P_{ω_{1}}(α) \ A is stationary in P_{ω_{1}}(α).

Choose an N-generic sequence ⟨(X_{n}, F_{n}) : n < ω⟩ with (X_{0}, F_{0}) = (X, F). Let Y = ⋃{X_{n} : n < ω} and let K be the infinum of the sequence ⟨F_{n} : n < ω⟩. By Proposition 3.8, (Y, K) is a condition and (Y, K) ≤ (X_{n}, F_{n}) for all n < ω. In particular, (Y, K) ≤ (X, F). Let n < ω be given. Then D_{n} is in N. So for some m, (X_{m}, F_{m}) is in D_{n}. Since D_{n} is open, (Y, K) is in D_{n}. Thus (Y, K) is in ⋂{D_{n} : n < ω}. □

Theorem 3.10. If 2^{ω} = ω_{1}, then ℙ(α, A, β) is ω_{2}-c.c.

Proof. We will prove that ℙ(α, A, β) is ω_{2}-Knaster. Let ⟨(X_{j}, F_{j}) : j < ω_{2}⟩ be a sequence of conditions in ℙ(α, A, β). For each j < ω_{2} and ξ ∈ dom(F_{j}), let F_{j}(ξ) = ⟨a(j)_{i}^{ξ} : i ≤ γ_{ξ}^{j}⟩. We will thin out this sequence of conditions in five steps.

First, apply the Δ-System Lemma to the sequence of countable sets ⟨dom(F_{j}) : j < ω_{2}⟩ to obtain an unbounded set Z_{0} ⊆ ω_{2} and a countable set B such that for all j < k in Z_{0}, dom(F_{j}) ∩ dom(F_{k}) = B.

Secondly, an easy counting argument using CH shows there are at most ω_{1} many possibilities for a sequence of the form ⟨γ_{ξ}^{j} : ξ ∈ B⟩, for j in Z_{0}. So we can fix an unbounded set Z_{1} ⊆ Z_{0} and a sequence ⟨γ_{ξ} : ξ ∈ B⟩ such that for all j ∈ Z_{1} and ξ ∈ B, γ_{ξ}^{j} = γ_{ξ}.

Thirdly, another straightforward counting argument shows there are at most ω_{1} many possibilities for a sequence of the form ⟨a(j)_{i}^{ξ} : ξ ∈ B, i ≤ γ_{ξ}⟩, for j ∈ Z_{1}. Fix an unbounded set Z_{2} ⊆ Z_{1} and a sequence ⟨a_{i}^{ξ} : ξ ∈ B, i ≤ γ_{ξ}⟩ such that for all j ∈ Z_{2}, ξ ∈ B, and i ≤ γ_{ξ}, a(j)_{i}^{ξ} = a_{i}^{ξ}.

Fourthly, apply the Δ-system lemma to the sequence of countable sets ⟨⋃ X_{j} : j ∈ Z_{2}⟩ to obtain an unbounded set Z_{3} ⊆ Z_{2} and a countable set E such that for all j < k in Z_{3}, (⋃ X_{j}) ∩ (⋃ X_{k}) = E.

Fifthly, a routine counting argument shows there are at most ω_{1} possibilities for a set of the form P (E) ∩ X_{j}, for j ∈ Z_{3}. So we can choose an unbounded set Z_{4} ⊆ Z_{3} so that for all j < k in Z_{4}, P (E) ∩ X_{j} = P (E) ∩ X_{k}.

Claim 3.11. Let j and k be distinct ordinals in Z_{4}, and suppose b is in X_{j} \ X_{k}. Then b is not a subset of ⋃ X_{k}.

Proof. Suppose for a contradiction b is in X_{j} \ X_{k} and b ⊆ ⋃ X_{k}. Since b is in X_{j}, b ⊆ ⋃ X_{j}. So b ⊆ (⋃ X_{j}) ∩ (⋃ X_{k}) = E. Therefore b is in P (E) ∩ X_{j}. But P (E) ∩ X_{j} = P (E) ∩ X_{k}. So b is in X_{k}, which is a contradiction. □

Let j < k be in Z_{4}, and we prove (X_{j}, F_{j}) and (X_{k}, F_{k}) are compatible. Define (Y, K) as follows. Let Y = X_{j} ∪ X_{k}. Let dom(K) = dom(F_{j}) ∪ dom(F_{k}). If ξ is in dom(F_{j}) \ dom(F_{k}), let K(ξ) = F_{j}(ξ), and if ξ is in dom(F_{k}) \ dom(F_{j}), let K(ξ) = F_{k}(ξ). If ξ is dom(F_{j}) ∩ dom(F_{k}) = B, then F_{j}(ξ) and F_{k}(ξ) are both equal to ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩, and we define K(ξ) to be this sequence. This completes the definition of (Y, K).

We prove that (Y, K) is a condition. Properties (1), (2), (3), and (4) of Definition 3.1 are immediate. For (5) and (6), suppose ξ is in dom(K). Let K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Consider i ≤ γ_{ξ}. Since a_{i}^{ξ} appears on either the sequence F_{j}(ξ) or F_{k}(ξ), a_{i}^{ξ} is either a subset of ⋃ X_{j} or ⋃ X_{k}. These sets are in turn subsets of ⋃ Y. So a_{i}^{ξ} ⊆ ⋃ Y. This verifies (6).

It remains to show that a_{i}^{ξ} is not in Y ∪ A. Since a_{i}^{ξ} appears on either F_{j}(ξ) or F_{k}(ξ), it is not in A. Now we show a_{i}^{ξ} is not in Y. First assume ξ is in dom(F_{j}) \ dom(F_{k}). Then K(ξ) = F_{j}(ξ). So a_{i}^{ξ} appears on the sequence F_{j}(ξ). Since (X_{j}, F_{j}) is a condition, a_{i}^{ξ} is not in X_{j} and a_{i}^{ξ} ⊆ ⋃ X_{j}. If for a contradiction a_{i}^{ξ} is in Y, then a_{i}^{ξ} is in X_{k} \ X_{j}. By Claim 3.11, a_{i}^{ξ} is not a subset of ⋃ X_{j}, which is a contradiction. The case when ξ is in dom(F_{k}) \ dom(F_{j}) is the same, with the roles of j and k reversed. Now assume ξ is in dom(F_{j}) ∩ dom(F_{k}) = B. Then a_{i}^{ξ} appears on the sequence F_{j}(ξ) = F_{k}(ξ). Since (X_{j}, F_{j}) and (X_{k}, F_{k}) are conditions, a_{i}^{ξ} is neither in X_{j} nor X_{k}. Therefore a_{i}^{ξ} is not in Y = X_{j} ∪ X_{k}. This completes the proof that (Y, K) is in ℙ(α, A, β).

For proving that (Y, K) ≤ (X_{j}, F_{j}), (X_{k}, F_{k}), properties (a), (c), and (d) of Definition 3.1 are immediate, and (b) follows from Claim 3.11. □

4. The Set S(α, A, β)

Fix for the remainder of the section α, β, and A as described in the assumptions of Definition 3.1. Specifically, α < β ≤ ω_{2}, β has uncountable cofinality, A is a subset of P_{ω_{1}}(α), and if cf(α) = ω_{1}, then P_{ω_{1}}(α) \ A is stationary in P_{ω_{1}}(α). Also assume 2^{ω} = ω_{1}.

Notation 4.1. Let Ṡ(α, A, β) be a ℙ(α, A, β)-name such that

⊩ Ṡ(α, A, β) = ⋃{X : ∃F (X, F) ∈ Ḣ},

where Ḣ is the canonical name for the generic filter.

Notation 4.2. In the case α = 0 and A = ∅, let Ṡ_{β} denote Ṡ(0, ∅, β).

Consider a generic filter H on ℙ(α, A, β) over V. Since ℙ(α, A, β) is ω_{1}-distributive and ω_{2}-c.c., all cardinals and cofinalities are preserved in V [H]. Also by distributivity, P_{ω_{1}}(β) is computed the same in V and V [H]. The set S(α, A, β) is a subset of P_{ω_{1}}(β) in V [H]. Let ξ be in [α, β)∩cof(ω_{1}). By Lemma 3.6, if we take the union of all sequences of the form F (ξ), where (X, F) ∈ H for some X, and ξ ∈ dom(F), then we obtain an increasing and continuous sequence ⟨a_{i}^{ξ} : i < ω_{1}⟩ cofinal in P_{ω_{1}}(ξ).

Lemma 4.3. In V [H], for all ξ in [α, β) ∩ cof(ω_{1}), the set {a_{i}^{ξ} : i < ω_{1}} is a club subset of P_{ω_{1}}(ξ) which is disjoint from S(α, A, β). So S(α, A, β) does not reflect to any ordinal in [α, β) with cofinality ω_{1}.

Proof. If b is in S(α, A, β), fix (X, F) in H such that b ∈ X. Suppose for a contradiction b is equal to a_{i}^{ξ} for some ξ in [α, β) ∩ cof(ω_{1}) and i < ω_{1}. Then there is (Y, K) ≤ (X, F) such that b appears on the sequence F (ξ). But then b is in Y and on the sequence F (ξ), which contradicts property (5) of Definition 3.1. □

Note that the statement that S(α, A, β) does not reflect to any ordinal in [α, β)∩ cof(ω_{1}) is upwards absolute between V [H] and any outer model W of V [H] with the same ω_{1}.

Proposition 4.4. The forcing poset ℙ(α, A, β) forces that Ṡ(α, A, β) is a stationary subset of P_{ω_{1}}(β).

Proof. Let (X, F) be a condition in ℙ(α, A, β) and suppose (X, F) forces Ċ is a club subset of P_{ω_{1}}(β). We will find (Y, K) ≤ (X, F) and a set z such that (Y, K) ⊩ z ∈ Ṡ(α, A, β) ∩ Ċ.

Fix a regular cardinal θ much larger than ℙ(α, A, β) and Ċ. Let N be a countable elementary substructure of H(θ) which contains as members the objects ℙ(α, A, β), (X, F), and Ċ, and moreover, if cf(α) = ω_{1}, then N ∩ α is in P_{ω_{1}}(α) \ A. Let ⟨(X_{n}, F_{n}) : n < ω⟩ be an N-generic sequence with (X_{0}, F_{0}) = (X, F).

Let X′ = ⋃{X_{n} : n < ω}. Define Y = X′ ∪ {N ∩ β}. Let K be the infinum of the sequence ⟨F_{n} : n < ω⟩. By Proposition 3.8, (Y, K) is a condition in ℙ(α, A, β), and (Y, K) ≤ (X_{n}, F_{n}) for all n < ω. In particular, (Y, K) ≤ (X, F) and (Y, K) is an N-generic condition. Since N ∩ β ∈ Y, (Y, K) ⊩ N ∩ β ∈ Ṡ(α, A, β).

Since Ċ is in N, ℙ(α, A, β) forces that N[Ḣ] ∩ β ∈ Ċ, where Ḣ is the canonical ℙ(α, A, β)-name for the generic filter. As (Y, K) is N-generic, (Y, K) ⊩ N[Ḣ] ∩ β = N ∩ β. So (Y, K) ⊩ N ∩ β ∈ Ṡ(α, A, β) ∩ Ċ. □

The next easy lemma will be used frequently throughout the paper.

Lemma 4.5. Let (X, F) be a condition in ℙ(α, A, β). Then

(X, F) ⊩ Ṡ(α, A, β) ∩ P (⋃ X) = X.

Proof. Clearly (X, F) ⊩ X ⊆ Ṡ(α, A, β) ∩ P (⋃ X). Let H be a generic filter on ℙ(α, A, β) with (X, F) in H. Suppose for a contradiction there is x in S(α, A, β) ∩ P (⋃ X) which is not in X. Fix a condition (Y, K) in H such that x is in Y. Choose (Z, L) in H which is below (X, F) and (Y, K). Then x is in Z \ X and x ⊆ ⋃ X, which contradicts that (Z, L) ≤ (X, F). □

Lemma 4.6. The forcing poset ℙ(α, A, β) forces that for any countable set B ⊆ β,

|Ṡ(α, A, β) ∩ P (B)| ≤ ℵ_{0}.

Proof. If B is a countable subset of β in a generic extension by ℙ(α, A, β), then since ℙ(α, A, β) is ω_{1}-distributive, B is in the ground model. There are densely many conditions of the form (X, F), where B ⊆ ⋃ X. If (X, F) is such a condition, then by Lemma 4.5, (X, F) forces that any subset of B which is in Ṡ(α, A, β) is in Ṡ(α, A, β) ∩ P (⋃ X) = X. So (X, F) ⊩ Ṡ(α, A, β) ∩ P (B) ⊆ X. Since X is countable, (X, F) ⊩ |Ṡ(α, A, β) ∩ P (B)| ≤ ℵ_{0}. □

The next technical, unmotivated lemma will be crucial to several arguments in later sections.

Lemma 4.7. Let θ ≥ ω_{2} be a regular cardinal.

(1) ℙ(α, A, β) forces there are stationarily many N in P_{ω_{1}}(H(θ)) such that Ṡ(α, A, β) ∩ P (N ∩ β) ⊆ N.

(2) Assume cf(α) = ω_{1}, and let T be a stationary subset of P_{ω_{1}}(α) which is disjoint from A. Then ℙ(α, A, β) forces there are stationarily many N in P_{ω_{1}}(H(θ)) such that Ṡ(α, A, β) ∩ P (N ∩ β) ⊆ N and N ∩ α ∈ T.

Proof. Let (X, F) be a condition in ℙ(α, A, β), and suppose (X, F) forces Ċ is a club subset of P_{ω_{1}}(H(θ)). (By H(θ), we mean H(θ) as computed in the generic extension.) Let χ be a regular cardinal much larger than ℙ(α, A, β), θ, and Ċ. Let M be a countable elementary substructure of H(χ) which contains as members the objects ℙ(α, A, β), (X, F), θ, and Ċ, and if cf(α) = ω_{1}, then M ∩ α ∈ P_{ω_{1}}(α) \ A. In addition, in case (2), also assume M ∩ α is in T.

Let ⟨(X_{n}, F_{n}) : n < ω⟩ be an M-generic sequence with (X_{0}, F_{0}) = (X, F). Let Y = ⋃{X_{n} : n < ω} and let K be the infinum of the sequence ⟨F_{n} : n < ω⟩. By Proposition 3.8, (Y, K) is in ℙ(α, A, β), and (Y, K) ≤ (X_{n}, F_{n}) for all n < ω. In particular, (Y, K) ≤ (X, F) and (Y, K) is M-generic. Since ⋃ Y = M ∩ β, Lemma 4.5 implies (Y, K) ⊩ Ṡ(α, A, β) ∩ P (M ∩ β) = Y.

Let H be a generic filter on ℙ(α, A, β) with (Y, K) ∈ H. Since (Y, K) is Mgeneric, M[H] ∩ V = M. Also S(α, A, β) ∩ P (M ∩ β) = Y. Let N = M[H] ∩ H(θ)^{V [H]}. We show that N is as desired.

Since θ is in M, H(θ)^{V [H]} is in M[H], so by standard arguments, N is an elementary substructure of H(θ)^{V [H]}. As Ċ is in M, N is in C. And since (Y, K) is M-generic, N ∩ β = M[H] ∩ β = M ∩ β, and N ∩ α = M[H] ∩ α = M ∩ α. If cf(α) = ω_{1}, then N ∩ α = M ∩ α is in T.

It suffices to show S(α, A, β) ∩ P (N ∩ β) ⊆ N. Let b be in S(α, A, β) ∩ P (N ∩ β). Since N ∩ β = M ∩ β, b is in S(α, A, β) ∩ P (M ∩ β) = Y. So b is in Y. But Y = ⋃{X_{n} : n < ω}. So fix n < ω such that b is in X_{n}. Then since X_{n} ∈ M and X_{n} is countable, X_{n} ⊆ M, so b is in M. Hence b is in M[H]. Since β ≤ θ, b is in M[H] ∩ H(θ) = N. □

Corollary 4.8. Assume that cf(α) = ω_{1}, and let T be a stationary subset of P_{ω_{1}}(α) which is disjoint from A. Then ℙ(α, A, β) forces that T is stationary in P_{ω_{1}}(α).

Proof. Let H be a generic filter on ℙ(α, A, β) over V. In V [H] let C be a club subset of P_{ω_{1}}(α). Fix a regular cardinal θ in V [H] much larger than ω_{2} and C. Note that in V, θ ≥ ω_{2} is regular. By Lemma 4.7, we can find a countable elementary substructure M of H(θ) with C in M such that M ∩ α ∈ T. Since C is in M, M ∩ α ∈ C ∩ T. □

Corollary 4.9. The forcing poset ℙ(α, A, β) forces that P_{ω_{1}}(β) \ Ṡ(α, A, β) is stationary in P_{ω_{1}}(β).

Proof. Let H be a generic filter on ℙ(α, A, β) over V. In V [H] consider a function F : [β]^{<ω} → β. Fix a regular cardinal θ much larger than β. By Lemma 4.7, we can choose a countable elementary substructure M of H(θ) with F in M such that S(α, A, β)∩P (M ∩β) ⊆ M. By elementarity, M ∩β is closed under F. Since M ∩β is in P (M ∩ β) but not in M, M ∩ β is not in S(α, A, β). Thus P_{ω_{1}}(β) \ S(α, A, β) is stationary in P_{ω_{1}}(β). □

We end this section by proving a general lemma regarding subsets of P_{ω_{1}}(β) which do not reflect to any ordinal of cofinality ω_{1}. We will use this result in Section 12.

Lemma 4.10. Let β ≤ ω_{2} be an ordinal with uncountable cofinality. Suppose S is a subset of P_{ω_{1}}(β) which satisfies the property that for all ν in β ∩ cof(ω_{1}), S does not reflect to ν. Then there is a club set C ⊆ P_{ω_{1}}(β) such that for all b in C, for all ν in b ∩ cof(ω_{1}), b ∩ ν is not in S.

Proof. Suppose for a contradiction there is a stationary set U ⊆ P_{ω_{1}}(β) such that for all b in U, there is an ordinal ν in b ∩ cof(ω_{1}) such that b ∩ ν is in S. By Fodor’s Lemma, fix a stationary set U′ ⊆ U and an ordinal ν in β ∩ cof(ω_{1}) such that for all b in U′, ν is in b and b ∩ ν is in S.

We prove that S ∩ P_{ω_{1}}(ν) is stationary in P_{ω_{1}}(ν), which contradicts our assumption on S. So consider a function F : [ν]^{<ω} → ν. Since U′ is stationary, fix b in U′ such that b is closed under F. Then b ∩ ν is closed under F. Since b is in U′, b ∩ ν is in S. □

5. Factorization

Fix for the remainder of the section ordinals ν < β ≤ ω_{2} with uncountable cofinality. Assume 2^{ω} = ω_{1}. Recall that ℙ_{ν} and Ṡ_{ν} abbreviate ℙ(0, ∅, ν) and Ṡ(0, ∅, ν). In this section we prove that ℙ_{β} factors as the two-step iteration ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β). This factorization will be important for the purpose of extending an elementary embedding.

Notation 5.1. Let (Y, K) be a condition in ℙ_{β}. For each ξ in dom(K), let K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Define:

(1) Y_{ν} = {b ∈ Y : b ⊆ ν},

(2) K_{ν} = K ↾ ν,

(3) Y^{ν} = {b ∈ Y : b ⊈ ν},

(4) K^{ν} = K ↾ [ν, β).

We define a dense suborder ℙ_{ν,β} of ℙ_{β}. The definition is designed so that if (Y, K) is in ℙ_{ν,β}, then (Y_{ν}, K_{ν}) is in ℙ_{ν}, and (Y_{ν}, K_{ν}) ⊩_{ℙ_{ν}} (Y^{ν}, K^{ν}) ∈ ℙ(ν, Ṡ_{ν}, β).

Definition 5.2. Let ℙ_{ν,β} be the suborder of ℙ_{β} consisting of conditions (Y, K) such that ⋃ Y_{ν} = (⋃ Y^{ν}) ∩ ν.

Lemma 5.3. The suborder ℙ_{ν,β} is dense in ℙ_{β}.

Proof. Let (X, F) be a condition in ℙ_{β}. Since ⋃ X is countable and ν is uncountable, we can choose a countable set e such that (⋃ X) ∩ ν ⫋ e ⊆ ν. Now choose f, a countable subset of β, such that f ∩ ν = e, ν ∈ f, ⋃ X ⊆ f, and f \ ν ⊈ ⋃ X.

Let Y = X ∪ {e, f}. By two applications of Lemma 3.4, (Y, F) is a condition in ℙ_{β} which is below (X, F). We show (Y, F) is in ℙ_{ν,β}.

The set e is in Y_{ν}, and e is the maximum element of Y_{ν}. So ⋃ Y_{ν} = e. Also the set f is the maximum element of Y, and since f is in Y^{ν}, f is the maximum element of Y^{ν}. It follows that ⋃ Y^{ν} = f. So ⋃ Y_{ν} = e = f ∩ ν = (⋃ Y^{ν}) ∩ ν. □

Lemma 5.4. Suppose (Y, K) is in ℙ_{ν,β}. For each ξ in dom(K), let K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Then:

(1) ⋃ Y = ⋃ Y^{ν},

(2) ⋃ Y_{ν} = (⋃ Y) ∩ ν,
(3) for all ξ in dom(K), a_{γ_{ξ}}^{ξ} ⊆ ⋃ Y^{ν},
(4) for all ξ in dom(K) and i ≤ γ_{ξ}, if a_{i}^{ξ} ⊆ ν, then a_{i}^{ξ} ⊆ ⋃ Y_{ν}.

Proof. (1) Clearly ⋃ Y^{ν} ⊆ ⋃ Y. Suppose α is in b, where b is in Y. If b is in Y^{ν}, then α is in ⋃ Y^{ν}. Otherwise b is in Y_{ν}, so α is in ⋃ Y_{ν}, which is equal to (⋃ Y^{ν})∩ν. So α is in ⋃ Y^{ν}. (2) ⋃ Y_{ν} = (⋃ Y^{ν}) ∩ ν by definition of ℙ_{ν,β}, which is equal to (⋃ Y) ∩ ν by (1). (3) By Definition 3.1, a_{γ_{ξ}}^{ξ} ⊆ ⋃ Y, and ⋃ Y = ⋃ Y^{ν} by (1). (4) If a_{i}^{ξ} ⊆ ν, then by (3), a_{i}^{ξ} ⊆ (⋃ Y^{ν}) ∩ ν = ⋃ Y_{ν}. □

Proposition 5.5. (I) If (Y, K) is in ℙ_{ν,β}, then (Y_{ν}, K_{ν}) is in ℙ_{ν}.

(II) Suppose (Z, L) ≤ (Y, K) in ℙ_{β}. If (Z, L) and (Y, K) are both in ℙ_{ν,β}, then

(Z_{ν}, L_{ν}) ≤ (Y_{ν}, K_{ν}) in ℙ_{ν}. (III) If (Y, K) is in ℙ_{ν,β}, then (Y_{ν}, K_{ν}) ⊩_{ℙ_{ν}} (Y^{ν}, K^{ν}) ∈ ℙ(ν, Ṡ_{ν}, β). (IV) Suppose (Z, L) ≤ (Y, K) in ℙ_{β}. If (Z, L) and (Y, K) are both in ℙ_{ν,β}, then

(Z_{ν}, L_{ν}) ⊩_{ℙ_{ν}} (Z^{ν}, L^{ν}) ≤ (Y^{ν}, K^{ν}) in ℙ(ν, Ṡ_{ν}, β).

Proof. (I) We verify the properties of Definition 3.1 in the case of ℙ_{ν} = ℙ(0, ∅, ν). Properties (1), (2), (3), and (4) are immediate. To show (5) and (6), consider ξ in dom(K_{ν}) = dom(K) ∩ ν. Let K_{ν}(ξ) = K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Let i ≤ γ_{ξ}. Since (Y, K) is in ℙ_{β}, a_{i}^{ξ} is not in Y. But Y_{ν} ⊆ Y, so a_{i}^{ξ} is not in Y_{ν}. Also a_{i}^{ξ} is obviously not in ∅. This proves (5). For (6), we know that a_{γ_{ξ}}^{ξ} ⊆ ⋃ Y, and a_{γ_{ξ}}^{ξ} ⊆ ξ ⊆ ν. So by Lemma 5.4(2), a_{γ_{ξ}}^{ξ} ⊆ (⋃ Y) ∩ ν = ⋃ Y_{ν}.

(II) Properties (a), (c), and (d) of Definition 3.1 are immediate. For (b), suppose b is in Z_{ν} \ Y_{ν}. Then b is in Z and b ⊆ ν. Since b is not in Y_{ν}, b is not in Y. So b is in Z \ Y. As (Z, L) ≤ (Y, K), b ⊈ ⋃ Y. But ⋃ Y_{ν} ⊆ ⋃ Y, so b ⊈ ⋃ Y_{ν}.

(III) It is immediate that ℙ_{ν} forces that (Y^{ν}, K^{ν}) satisfies properties (1), (2), (3), and (4) of Definition 3.1. For (5) and (6), consider ξ in dom(K^{ν}) = dom(K)∩[ν, β). Let K^{ν}(ξ) = K(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. Let i ≤ γ_{ξ}. Since (Y, K) is in ℙ_{β}, a_{i}^{ξ} is not in Y, so a_{i}^{ξ} is not in Y^{ν}. We need to show (Y_{ν}, K_{ν}) ⊩_{ℙ_{ν}} a_{i}^{ξ} ∉ Ṡ_{ν}. If a_{i}^{ξ} is not a subset of ν, then this is clear. Suppose a_{i}^{ξ} ⊆ ν. Then by Lemma 5.4(4), a_{i}^{ξ} ⊆ ⋃ Y_{ν}. By Lemma 4.5, (Y_{ν}, K_{ν}) ⊩_{ℙ_{ν}} Ṡ_{ν} ∩ P (⋃ Y_{ν}) = Y_{ν}. But a_{i}^{ξ} is not in Y, and hence is not in Y_{ν}, and a_{i}^{ξ} is in P (⋃ Y_{ν}). So (Y_{ν}, K_{ν}) ⊩ a_{i}^{ξ} ∉ Ṡ_{ν}. This proves (5). By Lemma 5.4(3), a_{γ_{ξ}}^{ξ} ⊆ ⋃ Y^{ν}, proving (6).

(IV) Since properties (a), (b), (c), and (d) of Definition 3.1 are absolute between V and a generic extension by ℙ_{ν}, it suffices to show these properties hold in V. Properties (a), (c), and (d) are immediate. For (b), suppose b is in Z^{ν} \ Y^{ν}. Then b is Z and b ⊈ ν. But then b is not in Y, since otherwise b would be in Y^{ν}. So b is in Z \ Y. Since (Z, L) ≤ (Y, K), b ⊈ ⋃ Y. As ⋃ Y^{ν} ⊆ ⋃ Y, b ⊈ ⋃ Y^{ν}. □

If (Y, K) is in ℙ_{ν,β}, then (Y_{ν}, K_{ν}) is in ℙ_{ν}. By Lemma 3.3, ℙ_{ν} ⊆ ℙ_{β}, so (Y_{ν}, K_{ν}) is in ℙ_{β} as well.

Lemma 5.6. Let (Y, K) be in ℙ_{ν,β}. Then (Y, K) ≤ (Y_{ν}, K_{ν}) in ℙ_{β}.

Proof. Properties (a), (c), and (d) of Definition 3.1 are immediate. For (b), if b is in Y \ Y_{ν}, then b ⊈ ν. But ⋃ Y_{ν} ⊆ ν. So b ⊈ ⋃ Y_{ν}. □

Lemma 5.7. Let (X, F) be a condition in ℙ_{ν}. Suppose that Y, K are sets such that (X, F) ⊩_{ℙ_{ν}} (Y̌ , Ǩ) ∈ ℙ(ν, Ṡ_{ν}, β). Let Z = X ∪ Y and let L = F ∪ K. Then (Z, L) is in ℙ_{β}. Moreover, Z_{ν} = X, L_{ν} = F, Z^{ν} = Y, and L^{ν} = K.

Proof. Properties (1), (2), (3), and (4) of Definition 3.1 are immediate. For (5) and (6), let ξ be in dom(L). Let L(ξ) = ⟨a_{i}^{ξ} : i ≤ γ_{ξ}⟩. If ξ is in dom(F), then a_{γ_{ξ}}^{ξ} ⊆ ⋃ X ⊆ ⋃ Z, and if ξ is in dom(K), then a_{γ_{ξ}}^{ξ} ⊆ ⋃ Y ⊆ ⋃ Z. This proves (6).

To show (5), let i ≤ γ_{ξ}, and we show a_{i}^{ξ} is not in Z ∪ ∅ = Z = X ∪ Y. First assume ξ is in dom(K). Since (X, F) ⊩_{ℙ_{ν}} (Y̌ , Ǩ) ∈ ℙ(ν, Ṡ_{ν}, β), a_{i}^{ξ} is not in Y. By Definition 3.1(5), (X, F) ⊩_{ℙ_{ν}} a_{i}^{ξ} ∉ Ṡ_{ν}, which implies a_{i}^{ξ} ∉ X. So a_{i}^{ξ} is not in Z. Now suppose ξ is in dom(F). Then clearly a_{i}^{ξ} is not in X. Since dom(F) ⊆ ν, a_{i}^{ξ} is a subset of ν. But for all b in Y, b ⊈ ν. So a_{i}^{ξ} is not in Y.

So indeed (Z, L) is a condition in ℙ_{β}. The last statement of the lemma is immediate. □

Lemma 5.8. There is a dense set of conditions in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β) of the form (X, F) ∗ (Y̌ , Ǩ) such that ⋃ X = (⋃ Y) ∩ ν. Suppose (X, F) ∗ (Y̌ , Ǩ) is such a condition. Let Z = X ∪ Y and L = F ∪ K. Then (Z, L) is in ℙ_{ν,β}, and (Z_{ν}, L_{ν}) ∗ (Ž^{ν}, Ľ^{ν}) is equal to (X, F) ∗ (Y̌ , Ǩ).

Proof. Let (X_{0}, F_{0}) ∗ (Ẏ_{0}, K̇_{0}) be in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β). Since ℙ_{ν} is ω_{1}-distributive, we can fix (X_{1}, F_{1}) ≤ (X_{0}, F_{0}) and sets Y_{1} and K_{1} such that (X_{1}, F_{1}) forces Ẏ_{0} = Y̌_{1} and K̇_{0} = Ǩ_{1}.

Choose a countable set e such that ⋃ X_{1} ⫋ e ⊆ ν and (⋃ Y_{1}) ∩ ν ⊆ e. Let X_{2} = X_{1} ∪ {e}. By Lemma 3.4, (X_{2}, F_{1}) is in ℙ_{ν} and (X_{2}, F_{1}) ≤ (X_{1}, F_{1}). Now choose a countable set f ⊆ β such that f ∩ ν = e and (⋃ Y_{1}) ∪ {ν} ⫋ f. Let Y_{2} = Y_{1} ∪ {f}. By Lemma 3.4, (X_{2}, F_{1}) forces (Y̌_{2}, Ǩ_{1}) is in ℙ(ν, Ṡ_{ν}, β) and (Y̌_{2}, Ǩ_{1}) ≤ (Y̌_{1}, Ǩ_{1}).

It follows that (X_{2}, F_{1}) ∗ (Y̌_{2}, Ǩ_{1}) ≤ (X_{0}, F_{0}) ∗ (Ẏ_{0}, K̇_{0}). Note that e is the maximum element of X_{2}, so ⋃ X_{2} = e. Also f is the maximum element of Y_{2}, so ⋃ Y_{2} = f. So ⋃ X_{2} = e = f ∩ ν = (⋃ Y_{2}) ∩ ν, proving the first part of the lemma.

Now suppose (X, F) ∗ (Y̌ , Ǩ) is a condition in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β) such that ⋃ X = (⋃ Y)∩ν. Let Z = X ∪Y and L = F ∪K. By Lemma 5.7, (Z, L) is in ℙ_{β}, Z_{ν} = X, L_{ν} = F, Z^{ν} = Y, and L^{ν} = K. Also ⋃ Z_{ν} = ⋃ X = (⋃ Y) ∩ ν = (⋃ Z^{ν}) ∩ ν, so (Z, L) is in ℙ_{ν,β}. □

Theorem 5.9. Define i : ℙ_{ν,β} → ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β) by letting i(Y, K) = (Y_{ν}, K_{ν}) ∗ (Y̌^{ν}, Ǩ^{ν}). Then i is an isomorphism of ℙ_{ν,β} onto a dense subset of ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β).

Proof. By (I) and (III) of Proposition 5.5, if (Y, K) is in ℙ_{ν,β}, then i(Y, K) is in ℙ_{ν} ∗ℙ(ν, Ṡ_{ν}, β). By (II) and (IV) of Proposition 5.5, if (Z, L) and (Y, K) are in ℙ_{ν,β} and (Z, L) ≤ (Y, K) in ℙ_{β}, then i(Z, L) ≤ i(Y, K) in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β). It is easy to see that i is an injection, and the range of i is dense in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β) by Lemma 5.8. To complete the proof, it remains to show that if (Z, L) and (Y, K) are in ℙ_{ν,β} and i(Z, L) ≤ i(Y, K) in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β), then (Z, L) ≤ (Y, K) in ℙ_{β}.

Let (Z, L) and (Y, K) be in ℙ_{ν,β}, and assume i(Z, L) ≤ i(Y, K) in ℙ_{ν} ∗ℙ(ν, Ṡ_{ν}, β). This means that (Z_{ν}, L_{ν}) ≤ (Y_{ν}, K_{ν}) in ℙ_{ν}, and (Z_{ν}, L_{ν}) ⊩_{ℙ_{ν}} (Z^{ν}, L^{ν}) ≤ (Y^{ν}, K^{ν}) in ℙ(ν, Ṡ_{ν}, β). We prove that (Z, L) ≤ (Y, K) in ℙ_{β}.

Since (Z_{ν}, L_{ν}) ≤ (Y_{ν}, K_{ν}) in ℙ_{ν}, we have: (1) Y_{ν} ⊆ Z_{ν}, (2) for all b in Z_{ν} \ Y_{ν}, b ⊈ ⋃ Y_{ν}, (3) dom(K_{ν}) ⊆ dom(L_{ν}), and (4) for all ξ in dom(K_{ν}), K(ξ) is an initial segment of L(ξ). Fix a generic filter H on ℙ_{ν} with (Z_{ν}, L_{ν}) in H. Then in V [H], (Z^{ν}, L^{ν}) ≤ (Y^{ν}, K^{ν}) in ℙ(ν, S_{ν}, β), which means: (5) Y^{ν} ⊆ Z^{ν}, (6) for all b in Z^{ν} \ Y^{ν}, b ⊈ ⋃ Y^{ν}, (7) dom(K^{ν}) ⊆ dom(L^{ν}), and (8) for all ξ in dom(K^{ν}), K(ξ) is an initial segment of L(ξ). Note that properties (5 – 8) are absolute between V and V [H], so in fact they hold in V.

To show that (Z, L) ≤ (Y, K) in ℙ_{β}, we show (a) Y ⊆ Z, (b) for all b in Z \ Y, b ⊈ ⋃ Y, (c) dom(K) ⊆ dom(L), and (d) for all ξ in dom(K), K(ξ) is an initial segment of L(ξ). Since Y = Y_{ν} ∪ Y^{ν} and Z = Z_{ν} ∪ Z^{ν}, (a) follows from (1) and (5). Similarly, K = K_{ν} ∪ K^{ν} and L = L_{ν} ∪ L^{ν}, so (c) and (d) follow from (3), (4), (7), and (8).

It remains to prove (b). So let b be in Z \Y, and we show b ⊈ ⋃ Y. First assume that b ⊆ ν. Then b is in Z_{ν} \ Y_{ν}. By (2), b ⊈ ⋃ Y_{ν}. Since (Y, K) is in ℙ_{ν,β}, by Lemma 5.4(2), (⋃ Y) ∩ ν = ⋃ Y_{ν}. So if b ⊆ ⋃ Y, then b ⊆ (⋃ Y) ∩ ν = ⋃ Y_{ν}, which is a contradiction. Now assume that b ⊈ ν. Then b is in Z^{ν} \Y^{ν}. By (6), b ⊈ ⋃ Y^{ν}. As (Y, K) is in ℙ_{ν,β}, by Lemma 5.4(1), ⋃ Y = ⋃ Y^{ν}. Since b ⊈ ⋃ Y^{ν}, b ⊈ ⋃ Y. □

As ℙ_{ν,β} is a dense suborder of ℙ_{β}, it follows that ℙ_{β} is forcing equivalent to ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β).

The following is a technical lemma which we will need in the next section.

Lemma 5.10. Let i : ℙ_{ν,β} → ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β) be the map i(Y, K) = (Y_{ν}, K_{ν}) ∗ (Y̌^{ν}, Ǩ^{ν}). Suppose (X, F) is a condition in ℙ_{ν} (and hence (X, F) is in ℙ_{β} as well). Let (Y, K) be in ℙ_{ν,β}, and assume (Y, K) ≤ (X, F) in ℙ_{β}. Then i(Y, K) ≤ (X, F)∗1 in ℙ_{ν} ∗ ℙ(ν, Ṡ_{ν}, β).

Proof. For (X, F) and (Y, K) as above, i(Y, K) ≤ (X, F) ∗ 1 means (Y_{ν}, K_{ν}) ∗ (Y̌^{ν}, Ǩ^{ν}) ≤ (X, F)∗1, which is equivalent to (Y_{ν}, K_{ν}) ≤ (X, F) in ℙ_{ν}. So it suffices to show that if (X, F) is in ℙ_{ν}, (Y, K) is in ℙ_{ν,β}, and (Y, K) ≤ (X, F) in ℙ_{β}, then (Y_{ν}, K_{ν}) ≤ (X, F) in ℙ_{ν}.

Since (Y, K) ≤ (X, F) in ℙ_{β}, X ⊆ Y. If b is in X, then b is in Y, and since (X, F) is in ℙ_{ν}, b is a subset of ν. So b is in Y_{ν}. Hence X ⊆ Y_{ν}. Suppose b is in Y_{ν} \ X. Then b is in Y \ X, so b ⊈ ⋃ X. As (Y, K) ≤ (X, F) in ℙ_{β}, dom(F) ⊆ dom(K). But also dom(F) ⊆ ν. So dom(F) ⊆ dom(K) ∩ ν = dom(K_{ν}). If ξ is in dom(F), then as (X, F) ≤ (Y, K), F (ξ) is an initial segment of K(ξ) = K_{ν}(ξ). □

6. Extending an Elementary Embedding, Part 1

We now deal with the problem of extending an elementary embedding after Lévy collapsing a large cardinal to become ω_{2}, and then forcing with ℙ_{ω_{2}}. Fix for the remainder of the section a cardinal κ which is κ^{+}-supercompact, and an elementary embedding j : V → M with critical point κ such that j(κ) > κ^{+} and M^{κ^{+}} ⊆ M. Assume 2^{κ} = κ^{+}.

Recall that the Lévy collapse Coll(ω_{1}, < κ) is ω_{1}-closed, κ-c.c., has size κ, and forces that κ is equal to ω_{2}. It is also easy to see that Coll(ω_{1}, < κ) forces 2^{ω} = ω_{1} and 2^{ω_{1}} = ω_{2}. Since we are assuming 2^{κ} = κ^{+} in V, Coll(ω_{1}, < κ) forces 2^{ω_{2}} = ω_{3}. We will be interested in the two-step iteration Coll(ω_{1}, <κ) ∗ ℙ̇_{κ}. (Note that ℙ̇_{κ} is the same as ℙ̇_{ω_{2}}.) Since CH holds after forcing with the Lévy collapse, this two-step iteration is κ-c.c. and ω_{1}-distributive. Occasionally we will write j(Coll(ω_{1}, < κ) ∗ ℙ̇_{κ}) as Coll(ω_{1}, < j(κ)) ∗ ℙ̇_{j(κ)}. To reduce superscripts, we will write κ^{+} to denote the successor of κ in V, even when working in a generic extension in which this ordinal is no longer a cardinal.

Theorem 6.1. Let \overline{G} ∗ \overline{H} be a generic filter on j(Coll(ω_{1}, < κ) ∗ ℙ̇_{κ}) over V. Let G = \overline{G} ∩ Coll(ω_{1}, < κ) and H = \overline{H} ∩ ℙ_{κ}. Then G is a generic filter on Coll(ω_{1}, <κ) over V, H is a generic filter on ℙ_{κ} over V [G], and j can be extended in V [\overline{G} ∗ \overline{H}] to j : V [G ∗ H] → M[\overline{G} ∗ \overline{H}] such that j(G ∗ H) = \overline{G} ∗ \overline{H}. Moreover:

(1) M[G ∗ H]^{κ^{+}} ∩ V [G ∗ H] ⊆ M[G ∗ H], (2) M[\overline{G}∗\overline{H}] is equal to M[\overline{G}][H][H′], where H′ is a generic filter on ℙ(κ, S_{κ}, j(κ))

over M[\overline{G}][H],

(3) j(S_{κ}) = S_{j(κ)}, S_{j(κ)} = S_{κ} ∪ S(κ, S_{κ}, j(κ)), S_{κ} = {b ∈ S_{j(κ)} : b ⊆ κ}, and

S(κ, S_{κ}, j(κ)) = {b ∈ S_{j(κ)} : b ⊈ κ}.

Proof. Since the inclusion map from Coll(ω_{1}, < κ) to Coll(ω_{1}, < j(κ)) is a complete embedding, G = \overline{G} ∩ Coll(ω_{1}, < κ) is a generic filter on Coll(ω_{1}, < κ) over V. As j ↾ Coll(ω_{1}, < κ) is the identity, j[G] = G ⊆ \overline{G}. So j can be extended to j : V [G] → M[\overline{G}] such that j(G) = \overline{G}.

As Coll(ω_{1}, <κ) is κ-c.c., M[G]^{κ^{+}}∩V [G] ⊆ M[G]. In particular, ℙ_{κ} is computed the same in V [G] and M[G]. By standard facts, M[\overline{G}] = M[G][G′], where G′ = \overline{G} ∩ Coll(ω_{1}, [κ, j(κ))). Since Coll(ω_{1}, [κ, j(κ))) is ω_{1}-closed, ℙ_{κ} is computed the same in M[G] and M[\overline{G}].

In M[\overline{G}] consider the map i : ℙ_{κ,j(κ)} → ℙ_{κ} ∗ ℙ(κ, Ṡ_{κ}, j(κ)) given by i(Z, K) = (Z_{κ}, K_{κ}) ∗ (Ž^{κ}, Ǩ^{κ}). By Theorem 5.9, i is an isomorphism of ℙ_{κ,j(κ)} onto a dense subset of ℙ_{κ} ∗ ℙ(κ, Ṡ_{κ}, j(κ)). But being an isomorphism is absolute. It follows that i[\overline{H} ∩ ℙ_{κ,j(κ)}] generates a generic filter F on ℙ_{κ} ∗ ℙ(κ, Ṡ_{κ}, j(κ)) over V [\overline{G}], and M[\overline{G}][\overline{H}] = M[\overline{G}][F].

Let k : ℙ_{κ} → ℙ_{κ} ∗ ℙ(κ, Ṡ_{κ}, j(κ)) be the map k(p) = p ∗ 1. Then k is a complete embedding. It follows that k^{−1}(F) = {p ∈ ℙ_{κ} : p ∗ 1 ∈ F } is a generic filter on ℙ_{κ} over V [\overline{G}]. Let H = k^{−1}(F).

We claim that H is equal to \overline{H}∩ℙ_{κ}. So let p be in H. Then p is in ℙ_{κ}, and p∗1 is in F, the filter generated by i[\overline{H} ∩ℙ_{κ,j(κ)}]. So there is a condition (Y, K) in \overline{H} ∩ℙ_{κ,j(κ)} such that i(Y, K) = (Y_{κ}, K_{κ}) ∗ (Y̌^{κ}, Ǩ^{κ}) ≤ p ∗ 1. This implies (Y_{κ}, K_{κ}) ≤ p in ℙ_{κ}. By Lemma 3.3, (Y_{κ}, K_{κ}) ≤ p in ℙ_{j(κ)}. By Lemma 5.6, (Y, K) ≤ (Y_{κ}, K_{κ}) in ℙ_{j(κ)}. So (Y, K) ≤ p in ℙ_{j(κ)}. Since (Y, K) is in \overline{H}, p is in \overline{H}. This shows H ⊆ \overline{H} ∩ ℙ_{κ}.

On the other hand, let q be in \overline{H} ∩ ℙ_{κ}, and we show q is in H. In other words, we show q ∗ 1 is in F. Since ℙ_{κ,j(κ)} is dense in ℙ_{j(κ)} and q is in \overline{H}, choose q′ ≤ q in \overline{H} ∩ ℙ_{κ,j(κ)}. Then i(q′) ∈ i[\overline{H} ∩ ℙ_{κ,j(κ)}], so i(q′) is in F. Since q′ ≤ q in ℙ_{j(κ)} and q is in ℙ_{κ}, by Lemma 5.10, i(q′) ≤ q ∗ 1 in ℙ_{κ} ∗ ℙ(κ, Ṡ_{κ}, j(κ)). As i(q′) is in F, q ∗ 1 is in F. This completes the proof that H = \overline{H} ∩ ℙ_{κ}.

Now j ↾ ℙ_{κ} is the identity, as can be easily seen, so j[H] = H = \overline{H} ∩ ℙ_{κ} ⊆ \overline{H}. Therefore we can extend j to j : V [G][H] → M[\overline{G}][\overline{H}] such that j(H) = \overline{H}. So j(G ∗ H) = j(G) ∗ j(H) = \overline{G} ∗ \overline{H}. Since Coll(ω_{1}, <κ) ∗ ℙ_{κ} is κ-c.c., M[G ∗ H]^{κ^{+}} ∩ V [G ∗ H] ⊆ M[G ∗ H].

Recall that M[\overline{G}][\overline{H}] = M[\overline{G}][F], where F is a generic filter on ℙ_{κ} ∗ℙ(κ, Ṡ_{κ}, j(κ)) over M[\overline{G}]. By standard facts about two-step iterations, if we let H′ = \{ṡ^{H} : ∃p p ∗ ṡ ∈ F \}, then H′ is a generic filter on ℙ(κ, S_{κ}, j(κ)) over M[\overline{G}][H], and M[\overline{G}][F] = M[\overline{G}][H][H′]. So M[\overline{G}][\overline{H}] = M[\overline{G}][H][H′].

It is immediate by the elementarity of j and the fact that j(H) = \overline{H} that j(S_{κ}) = S_{j(κ)}. If b is in S_{κ}, then by elementarity, j(b) = b is in j(S_{κ}) = S_{j(κ)}. So S_{κ} ⊆ S_{j(κ)}.

By definition, S(κ, S_{κ}, j(κ)) is equal to ⋃{Y : ∃K (Y, K) ∈ H′}. As a subclaim, let us prove that S(κ, S_{κ}, j(κ)) is equal to ⋃\{Z^{κ} : ∃L (Z, L) ∈ \overline{H} ∩ ℙ_{κ,j(κ)}\}.

On the one hand, suppose b is in Z^{κ}, where for some L, (Z, L) is in \overline{H} ∩ ℙ_{κ,j(κ)}. Then i(Z, L) = (Z_{κ}, L_{κ}) ∗ (Ž^{κ}, Ľ^{κ}) is in F. By the definition of H′, (Z^{κ}, L^{κ}) is in H′. So Z^{κ} ⊆ S(κ, S_{κ}, j(κ)). Hence b is in S(κ, S_{κ}, j(κ)).

On the other hand, suppose b is in S(κ, S_{κ}, j(κ)). Then there is (Y, K) in H′ such that b is in Y. By the definition of H′, fix p ∗ ṡ in F such that ṡ^{H} = (Y, K). Since p ∗ ṡ is in F, there is (Z, L) in \overline{H} ∩ ℙ_{κ,j(κ)} such that i(Z, L) = (Z_{κ}, L_{κ}) ∗ (Ž^{κ}, Ľ^{κ}) ≤ p ∗ ṡ. By Lemma 5.6, (Z, L) ≤ (Z_{κ}, L_{κ}) in ℙ_{j(κ)}, so (Z_{κ}, L_{κ}) is in \overline{H} ∩ ℙ_{κ} = H. Since (Z_{κ}, L_{κ}) ∗ (Ž^{κ}, Ľ^{κ}) ≤ p ∗ ṡ and (Z_{κ}, L_{κ}) ∈ H, (Z^{κ}, L^{κ}) ≤ ṡ^{H} = (Y, K). In particular, Y ⊆ Z^{κ}. So b is in Z^{κ}. This proves S(κ, S_{κ}, j(κ)) ⊆ ⋃{Z^{κ} : ∃L (Z, L) ∈ \overline{H} ∩ ℙ_{κ,j(κ)}\}, finishing the proof of the subclaim.

Let us prove that S_{j(κ)} = S_{κ} ∪ S(κ, S_{κ}, j(κ)). We already observed S_{κ} ⊆ S_{j(κ)}. Let b be in S(κ, S_{κ}, j(κ)). By the subclaim, there is (Z, L) in \overline{H} ∩ ℙ_{κ,j(κ)} such that b is in Z^{κ}. Since (Z, L) is in \overline{H}, Z ⊆ S_{j(κ)}. But b is in Z^{κ} ⊆ Z, so b is in S_{j(κ)}.

Now let b be in S_{j(κ)}, and we show b is in S_{κ} ∪ S(κ, S_{κ}, j(κ)). Fix (Y, K) in \overline{H} such that b is in Y. Since ℙ_{κ,j(κ)} is dense in ℙ_{j(κ)}, fix (Z, L) ≤ (Y, K) in \overline{H} ∩ℙ_{κ,j(κ)}. Then b is in Z. By the subclaim, Z^{κ} ⊆ S(κ, S_{κ}, j(κ)). So if b is in Z^{κ}, then b is in S(κ, S_{κ}, j(κ)). Otherwise b ⊆ κ, so b is in Z_{κ}. By Lemma 5.6, (Z, L) ≤ (Z_{κ}, L_{κ}) in ℙ_{j(κ)}, so (Z_{κ}, L_{κ}) is in \overline{H} ∩ ℙ_{κ} = H. This implies Z_{κ} ⊆ S_{κ}, and hence b is in S_{κ}.

We have proven S_{j(κ)} = S_{κ} ∪ S(κ, S_{κ}, j(κ)). Now every set in S_{κ} is a subset of κ, and every set in S(κ, S_{κ}, j(κ)) is not a subset of κ. Therefore it is immediate that S_{κ} = {b ∈ S_{j(κ)} : b ⊆ κ} and S(κ, S_{κ}, j(κ)) = {b ∈ S_{j(κ)} : b ⊈ κ}. □

7. Adding a Club

We briefly describe a forcing poset due to Sakai [3] for adding a club subset of P_{ω_{1}}(ω_{2}). Specifically, for a given set T ⊆ P_{ω_{1}}(ω_{2}), we define a forcing poset ℚ(T) which adds a generic function F : ω_{2} ×ω_{2} → ω_{1} for which there is no set in T closed under F. In the next section we describe an iteration of this forcing poset.

Definition 7.1. Let T be a subset of P_{ω_{1}}(ω_{2}). Define ℚ(T) as the forcing poset consisting of conditions p satisfying:

• p is a function of the form p : a^{p} × a^{p} → ω_{1}, where a^{p} is a countable subset of ω_{2},

• for all x in T, if x ⊆ a^{p}, then x is not closed under p. Let q ≤ p if q extends p as a function, that is, if a^{p} ⊆ a^{q} and q ↾ (a^{p} × a^{p}) = p.

Notation 7.2. For a countable set a ⊆ ω_{2}, we write a^{2} for a × a.

Lemma 7.3. Let p and q be conditions in ℚ(T), and assume p ↾ (a^{p} ∩ a^{q})^{2} = q ↾ (a^{p} ∩ a^{q})^{2}. Then p and q are compatible.

In particular, if a^{p} ∩ a^{q} = ∅, then p and q are compatible.

Proof. We define a condition r which is below p and q. Let a^{r} = a^{p} ∪ a^{q}. As a^{r} is countable, fix an ordinal ζ in ω_{1} \ a^{r}. Define r : (a^{r})^{2} → ω_{1} as follows. Let (i, j) be in (a^{r})^{2}. If i and j are both in a^{p}, let r(i, j) = p(i, j). If i and j are both in a^{q}, let r(i, j) = q(i, j). Note this is well-defined, since if i and j are both in a^{p} ∩ a^{q}, then p(i, j) = q(i, j) by assumption. The remaining possibility is that i is in a^{p} \ a^{q} and j is in a^{q} \ a^{p}, or vice versa. In this case, let r(i, j) = ζ.

So r is a function of the form r : (a^{r})^{2} → ω_{1}, where a^{r} is a countable subset of ω_{2}. Note that r extends both p and q. To show r is a condition, suppose x is in T and x ⊆ a^{r}. If x is a subset of a^{p}, then x is not closed under p, since p is a condition. So x is not closed under r. Likewise, if x is a subset of a^{q}, then x is not closed under q, so x is not closed under r. If neither of the above holds, then there is i in a^{r} \ a^{q} and j in a^{r} \ a^{p}. But then i is in a^{p} \ a^{q} and j is in a^{q} \ a^{p}. By the definition of r, r(i, j) = ζ. Since ζ is not in a^{r}, ζ is not in x, so x is not closed under r. □

Lemma 7.4. Let b be a countable subset of ω_{2}. Then there are densely many conditions t in ℚ(T) such that b ⊆ a^{t}.

Proof. Let p : (a^{p})^{2} → ω_{1} be a given condition. If b ⊆ a^{p}, then we are done. Otherwise b \ a^{p} is non-empty. As b is countable, fix an ordinal ζ in ω_{1} \ b. Define s by letting a^{s} = b \ a^{p}, and for all (i, j) in (a^{s})^{2}, let s(i, j) = ζ. If x is in T and is a subset of a^{s}, then ζ is not in x, so x is not closed under s. This shows s is in ℚ(T). Now a^{s} ∩ a^{p} = ∅, so by Lemma 7.3, s and p are compatible. Let t ≤ s, p. Then b ⊆ a^{s} ∪ a^{p} ⊆ a^{t}. □

Suppose I is a generic filter on ℚ(T) over V. Then by the last lemma, it is easy to see that F = ⋃ I is a total function F : ω_{2}^{V} × ω_{2}^{V} → ω_{1}^{V}, and if b is in T, then b is not closed under F. So if ω_{1} and ω_{2} are preserved, then ℚ(T) destroys the stationarity of T.

Depending on the assumptions of the ground model and on the choice of T, ℚ(T) may or may not preserve cardinals. In the sections which follow, we will define and analyze an iteration of forcing posets of the form ℚ(T). In our context, the forcing posets ℚ(T) which we use, together with the iteration itself, are ω_{1}-distributive and ω_{2}-c.c.

8. Iteration of Club Adding

In this section we define an iteration of forcing posets of the form ℚ(T) from the previous section. We work in a model V [H], where H is a generic filter on ℙ_{ω_{2}} over V. As usual, let S_{ω_{2}} = ⋃{X : ∃F (X, F) ∈ H}. So S_{ω_{2}} is a stationary subset of P_{ω_{1}}(ω_{2}) in V [H] which does not reflect to any ordinal in ω_{2} with cofinality ω_{1}.

Our goal is to define a forcing poset ℚ which forces that every stationary subset of S_{ω_{2}} reflects to an uncountable ordinal in ω_{2} with cofinality ω. The forcing poset ℚ will be an iteration of forcing posets of the form ℚ(T), where T is a subset of S_{ω_{2}} which does not reflect to any uncountable ordinal in ω_{2} of cofinality ω.

The iteration we define is an adaptation of an iteration developed by Sakai, which he used to prove the following theorem:

Theorem 8.1 (Sakai [3]). Assume GCH and □_{ω_{1}}. Then there exists a forcing poset which forces that there is a stationary set S ⊆ P_{ω_{1}}(ω_{2}) such that every stationary subset of S reflects to an uncountable ordinal in ω_{2} with cofinality ω.

Let us describe in more detail the context in which we will define ℚ. We start with a ground model V which satisfies 2^{ω} = ω_{1}, 2^{ω_{1}} = ω_{2}, and 2^{ω_{2}} = ω_{3}. Let H be a generic filter on ℙ_{ω_{2}} over V. Let S_{ω_{2}} = ⋃{X : ∃F (X, F) ∈ H}.

We verify that in V [H], it is still the case that 2^{ω} = ω_{1}, 2^{ω_{1}} = ω_{2}, and 2^{ω_{2}} = ω_{3}. Since ℙ_{ω_{2}} is ω_{1}-distributive, ℙ_{ω_{2}} does not add any new subsets of ω. So 2^{ω} = ω_{1} in V [H]. As CH holds in V, it is easy to see that ℙ_{ω_{2}} has cardinality ω_{2}. Since ℙ_{ω_{2}} is ω_{2}-c.c., it has at most ω_{2}^{ω_{1}} = 2^{ω_{1}} = ω_{2} many antichains. So the number of nice ℙ_{ω_{2}}-names in V for a subset of ω_{1} is bounded by ω_{2}^{ω_{1}} = 2^{ω_{1}} = ω_{2}. It follows that 2^{ω_{1}} = ω_{2} in V [H]. Similarly, the number of nice ℙ_{ω_{2}}-names in V for a subset of ω_{2} is ω_{2}^{ω_{2}} = 2^{ω_{2}} = ω_{3}. So 2^{ω_{2}} = ω_{3} in V [H].

We will work in V [H] to define a forcing iteration ℚ which destroys the stationarity of any subset of S_{ω_{2}} which does not reflect to an uncountable ordinal in ω_{2} with cofinality ω. Actually, we will find it convenient to define ℚ in a way that is not literally a forcing iteration, although it is forcing equivalent to one.

Before beginning the definition of ℚ, we fix several auxiliary objects which will be needed in the definition. Since we will remind the reader of these objects later in the paper, we provide them with numbers for reference.

Notation 8.2. For the purposes of bookkeeping, fix a surjective function f : ω_{3} → ω_{3} × ω_{3} satisfying that whenever f(α) = (i, j), then i ≤ α.

It will be convenient to have at hand an auxiliary poset ℝ, which will be useful for notational purposes. However, we will not be interested in forcing with ℝ.

Notation 8.3. Let ℝ denote the set of all non-empty partial functions s of the form s : (a^{s})^{2} → ω_{1}, where a^{s} is a countable subset of ω_{2}. Let t ≤_{ℝ} s if t extends s as a function, that is, a^{s} ⊆ a^{t} and t ↾ (a^{s})^{2} = s.

Note that by CH, ℝ has cardinality ω_{2}.

Notation 8.4. For each ordinal α in ω_{2}, fix a surjective function g_{α} : ω_{1} → α. Let E be the set of b in P_{ω_{1}}(ω_{2}) such that ot(b) is a limit ordinal and for all α in b, b is closed under g_{α}. Note that E is a club subset of P_{ω_{1}}(ω_{2}).

We are now ready to define by recursion the forcing poset ℚ in the model V [H]. Specifically, we define by recursion sequences ⟨ℚ_{α} : α ≤ ω_{3}⟩ and ⟨Ṫ_{j}^{i} : i, j ≤ ω_{3}⟩. For each α < ω_{3}, we let Ṫ (α) be Ṫ_{j}^{i}, where f(α) = (i, j). Finally, the forcing poset ℚ is defined as ℚ_{ω_{3}}.

We will maintain several recursion hypotheses.

Recursion Hypotheses 8.5. For all α ≤ ω_{3}:

(1) If p is in ℚ_{α}, then p is a partial function p : α → ℝ with a countable domain, and for all p and q in ℚ_{α}, q ≤ p iff dom(p) ⊆ dom(q) and for all ν in dom(p), q(ν) ≤_{ℝ} p(ν).

(2) Let β < α. Then: (a) for all q in ℚ_{α}, q ↾ β is in ℚ_{β}, (b) ℚ_{β} ⊆ ℚ_{α}, (c) if q is in ℚ_{α} and s ≤ q ↾ β in ℚ_{β}, then letting t = s ∪ (q ↾ [β, α)), t is in ℚ_{α} and t ≤ s, q in ℚ_{α}, and (d) the inclusion map ℚ_{β} → ℚ_{α} is a complete embedding.

(3) ℚ_{α} is ω_{1}-distributive and ω_{2}-c.c.

(4) If α < ω_{3}, then the sequence ⟨Ṫ_{i}^{α} : i < ω_{3}⟩ is an enumeration of all nice ℚ_{α}-names Ṫ for a subset of P_{ω_{1}}(ω_{2}) satisfying the property that ℚ_{α} forces Ṫ ⊆ S_{ω_{2}} ∩ E and Ṫ does not reflect to any uncountable ordinal in ω_{2} with cofinality ω.

Let α ≤ ω_{3} be given, and assume the sequences ⟨ℚ_{i} : i < α⟩ and ⟨Ṫ_{i}^{β} : β < α, i < ω_{3}⟩ are defined and satisfy the recursion hypotheses.

If α < ω_{3} and we have defined ℚ_{α} which satisfies recursion hypotheses (1), (2), and (3), then recursion hypothesis (4) is straightforward. This is clear if α = 0. Suppose α > 0. By CH the poset ℝ has cardinality ω_{2}. So the collection of all partial functions p : α → ℝ with a countable domain has cardinality ω_{2}. Since ℚ_{α} is a subset of this collection by recursion hypothesis (1), ℚ_{α} has cardinality at most ω_{2}. By recursion hypothesis (3), ℚ_{α} is ω_{2}-c.c., so the number of antichains of ℚ_{α} is at most ω_{2}^{ω_{1}} = ω_{2}. Hence the number of nice ℚ_{α}-names for a subset of P_{ω_{1}}(ω_{2}) is at most ω_{2}^{ω_{2}} = ω_{3}. Consequently, it is possible to choose an enumeration ⟨Ṫ_{i}^{α} : i < ω_{3}⟩ satisfying recursion hypothesis (4).

Now we define ℚ_{α}. The definition splits into three cases.

α is equal to 0

Let ℚ_{0} be the trivial forcing poset consisting of just the empty function. Note that recursion hypotheses (1), (2), and (3) are trivially true.

α is a limit ordinal

Let ℚ_{α} consist of all partial functions p : α → ℝ with a countable domain which satisfy the property that for all β < α, p ↾ β is in ℚ_{β}. Let q ≤ p in ℚ_{α} if dom(p) ⊆ dom(q) and for all ν in dom(p), q(ν) ≤_{ℝ} p(ν).

Note that recursion hypothesis (1) is immediate.

α is a successor ordinal

Let α = α_{0} + 1. Recall our bookkeeping function f from Notation 8.2. Let f(α_{0}) = (i, j). Then i ≤ α_{0} and j < ω_{3}, so Ṫ_{j}^{i} is defined and is equal to Ṫ (α_{0}).

By recursion hypothesis (4), Ṫ (α_{0}) is a nice ℚ_{i}-name for a subset of P_{ω_{1}}(ω_{2}). Since the inclusion map ℚ_{i} → ℚ_{α_{0}} is a complete embedding by recursion hypothesis (2), Ṫ (α_{0}) is a nice ℚ_{α_{0}}-name for a subset of P_{ω_{1}}(ω_{2}). Also by (4), ℚ_{i} forces that Ṫ (α_{0}) is a subset of S_{ω_{2}} ∩ E which does not reflect to any uncountable ordinal in ω_{2} with cofinality ω. But this property of Ṫ (α_{0}) is upwards absolute between a generic extension by ℚ_{i} and a generic extension by ℚ_{α_{0}}. Therefore ℚ_{α_{0}} forces that Ṫ (α_{0}) is a subset of S_{ω_{2}} ∩ E which does not reflect to any uncountable ordinal in ω_{2} with cofinality ω.

Let ℚ_{α} be the set of all partial functions p : α → ℝ with a countable domain which satisfy that p ↾ α_{0} is in ℚ_{α_{0}}, and if α_{0} is in dom(p), then p ↾ α_{0} ⊩_{ℚ_{α_{0}}} p(α_{0}) ∈ ℚ(Ṫ (α_{0})). Let q ≤ p if dom(p) ⊆ dom(q) and for all ν in dom(p), q(ν) ≤_{ℝ} p(ν).

Note that recursion hypothesis (1) is immediate.

Lemma 8.6. Let p and q be in ℚ_{α_{0}+1}. Then q ≤ p iff q ↾ α_{0} ≤ p ↾ α_{0} in ℚ_{α_{0}} and if α_{0} is in dom(p), then α_{0} is in dom(q) and q(α_{0}) ≤_{ℝ} p(α_{0}).

The proof is straightforward, using recursion hypothesis (1) applied to ℚ_{α_{0}}.

Lemma 8.7. The forcing poset ℚ_{α_{0}+1} is isomorphic to a dense subset of ℚ_{α_{0}} ∗ ℚ(Ṫ (α_{0})). So ℚ_{α_{0}+1} and ℚ_{α_{0}} ∗ ℚ(Ṫ (α_{0})) are forcing equivalent.

Proof. Define i : ℚ_{α_{0}+1} → ℚ_{α_{0}} ∗ ℚ(Ṫ (α_{0})) by letting i(p) = (p ↾ α_{0}) ∗ ˙1 if α_{0} is not in dom(p), and i(p) = (p ↾ α_{0}) ∗ p(α_{0}) if α_{0} is in dom(p). By the definition of ℚ_{α_{0}+1}, it is immediate that i(p) is in ℚ_{α_{0}} ∗ ℚ(Ṫ (α_{0})).

Clearly i is an injection. Note that ℚ_{α_{0}} forces that for all a and b in ℚ(Ṫ (α_{0})), a and b are in ℝ, and b ≤ a in ℚ(Ṫ (α_{0})) iff b ≤_{ℝ} a. Using this observation and Lemma 8.6, it is routine to verify that for all p and q in ℚ_{α_{0}+1}, q ≤ p iff i(q) ≤ i(p).

It remains to show that the range of i is dense. So let p ∗ ṡ be in ℚ_{α_{0}} ∗ ℚ(Ṫ (α_{0})). By recursion hypothesis (3), ℚ_{α_{0}} is ω_{1}-distributive, so ℚ_{α_{0}} forces ṡ is in the ground model. Hence we can find a set s and a condition q ≤ p in ℚ_{α_{0}} such that q ⊩ š = ṡ. Let r = q ∪ {(α_{0}, s)} if s is non-empty, and r = q if s is empty. Then r is in ℚ_{α_{0}+1} and i(r) = q ∗ š ≤ p ∗ ṡ. □

This completes the recursive definition.

We have taken care of recursion hypotheses (1) and (4). Proving that ℚ_{α} is ω_{1}-distributive and ω_{2}-c.c. will take some work, and we postpone it until the next section. The proof of recursion hypothesis (2) is routine, but we include it for completeness.

Lemma 8.8. Let β < α. Then:

(a) for all q in ℚ_{α}, q ↾ β ∈ ℚ_{β},

(b) ℚ_{β} ⊆ ℚ_{α},

(c) if q is in ℚ_{α} and s ≤ q ↾ β in ℚ_{β}, then letting t = s ∪ (q ↾ [β, α)), t is in ℚ_{α} and t ≤ s, q in ℚ_{α},

(d) the inclusion map ℚ_{β} → ℚ_{α} is a complete embedding.

Proof. (a) Let q be in ℚ_{α}. If α is a limit ordinal, then q ↾ β is in ℚ_{β} by the definition of ℚ_{α}. Otherwise α is a successor ordinal. Let α = α_{0} + 1. If β = α_{0}, then q ↾ α_{0} is in ℚ_{α_{0}} by the definition of ℚ_{α}. If β < α_{0}, then q ↾ β is equal to (q ↾ α_{0}) ↾ β, which is in ℚ_{β} by recursion hypothesis (2a) applied to ℚ_{α_{0}}.

(b) First suppose α is a successor ordinal. Let α = α_{0} + 1. Note that by the definition of ℚ_{α}, ℚ_{α_{0}} ⊆ ℚ_{α}. By recursion hypothesis (2b) applied to ℚ_{α_{0}}, ℚ_{β} ⊆ ℚ_{α_{0}}. So ℚ_{β} ⊆ ℚ_{α}. Now assume α is a limit ordinal, and let p be in ℚ_{β}. To show p is in ℚ_{α}, we show that p is a partial function p : α → ℝ with a countable domain, and for all γ < α, p ↾ γ is in ℚ_{γ}. The first statement follows from recursion hypothesis (1) applied to ℚ_{β}. Now let γ < α and we show p ↾ γ is in ℚ_{γ}. If γ < β, then p ↾ γ is in ℚ_{γ} by recursion hypothesis (2a) applied to ℚ_{β}. If β ≤ γ, then p ↾ γ = p, and by recursion hypothesis (2b) applied to ℚ_{γ}, p is in ℚ_{γ}.

(c) Let q be in ℚ_{α}, and suppose s ≤ q ↾ β in ℚ_{β}. Let t = s ∪ (q ↾ [β, α)). We will prove t is in ℚ_{α}. It is then easy to verify that t ≤ s, q in ℚ_{α}. First assume α is a limit ordinal. Clearly t is a partial function t : α → ℝ with a countable domain. Let γ < α and we show t ↾ γ is in ℚ_{γ}. If γ ≤ β, then t ↾ γ = s ↾ γ, which is in ℚ_{γ} by recursion hypothesis (2a) applied to ℚ_{β}. Suppose β < γ. Let \overline{q} = q ↾ γ. Then by (a), \overline{q} is in ℚ_{γ}, and s ≤ q ↾ β = \overline{q} ↾ β. By recursion hypothesis (2c) applied to ℚ_{γ}, s ∪ (\overline{q} ↾ [β, γ)) is in ℚ_{γ}. But s ∪ (\overline{q} ↾ [β, γ)) is equal to t ↾ γ.

Now assume α is a successor ordinal. Let α = α_{0} + 1. First consider the case when β = α_{0}. If α_{0} is not in dom(q), then t = s, which is in ℚ_{α} by (b). Suppose α_{0} is in dom(q). Then t = s ∪ {(α_{0}, q(α_{0}))}. So t ↾ α_{0} = s, which is in ℚ_{α_{0}}. By the definition of ℚ_{α}, q ↾ α_{0} ⊩ q(α_{0}) ∈ ℚ(Ṫ (α_{0})). As s ≤ q ↾ α_{0}, s = t ↾ α_{0}, and t(α_{0}) = q(α_{0}), t ↾ α_{0} ⊩ t(α_{0}) ∈ ℚ(Ṫ (α_{0})). So t is in ℚ_{α}.

Now suppose β < α_{0}. Then s ≤ q ↾ β = (q ↾ α_{0}) ↾ β. By recursion hypothesis (2c) applied to ℚ_{α_{0}}, letting t′ = s ∪ (q ↾ [β, α_{0})), t′ is in ℚ_{α_{0}} and t′ ≤ s, q ↾ α_{0} in ℚ_{α_{0}}. But t′ = t ↾ α_{0}, so t ↾ α_{0} is in ℚ_{α_{0}}. If α_{0} is not in dom(q), then t′ = t and we are done by (b). Otherwise since t′ ≤ q ↾ α_{0}, t′ forces q(α_{0}) = t(α_{0}) is in ℚ(Ṫ (α_{0})). So t is in ℚ_{α}.

(d) We show that the inclusion map from ℚ_{β} to ℚ_{α} is a complete embedding. Let q ≤ p in ℚ_{β}. Then dom(p) ⊆ dom(q), and for all ν in dom(p), q(ν) ≤_{ℝ} p(ν).

But this is also the definition of the ordering on ℚ_{α}. So q ≤ p in ℚ_{α}. Suppose p_{1} and p_{2} are in ℚ_{β} and r ≤ p_{1}, p_{2} in ℚ_{α}. By (a), r ↾ β is in ℚ_{β}, and clearly r ↾ β ≤ p_{1}, p_{2} in ℚ_{β}. Finally, let q be in ℚ_{α}. By (c), if p ≤ q ↾ β in ℚ_{β}, then p ∪ (q ↾ [β, α)) is in ℚ_{α} and is below p and q in ℚ_{α}. Thus p and q are compatible in ℚ_{α}. □

9. Distributivity and the Chain Condition

Our next goal is to prove that ℚ_{α} is ω_{1}-distributive and ω_{2}-c.c., which will complete the verification of the recursion hypotheses. For this purpose, it will be useful to first analyze the iteration in more detail. The proofs in this section are independent of whether α is a successor ordinal or a limit ordinal.

Notation 9.1. Suppose that p is a partial function p : α → ℝ with a countable domain (for example, if p is in ℚ_{α}). For each ξ in dom(p), let a_{ξ}^{p} be a^{p(ξ)}, that is, a_{ξ}^{p} is the non-empty countable set a such that dom(p(ξ)) = a^{2}.

Occasionally we will consider a partial function p : α → ℝ, and write a_{ξ}^{p} and p(ξ), even though we do not know whether ξ is in dom(p). In this case, a_{ξ}^{p} and p(ξ) both denote the empty set if ξ is not in dom(p).

The next observation is obvious, but we record it for later reference.

Lemma 9.2. Let p be in ℚ_{α}. Suppose β is in dom(p) and x is a countable subset of a_{β}^{p} which is closed under p(β). Then p ↾ β ⊩_{ℚ_{β}} x ∉ Ṫ (β).

Proof. By Lemma 8.8(a), p ↾ (β + 1) is in ℚ_{β+1}. By the definition of ℚ_{β+1}, p ↾ β ⊩_{ℚ_{β}} p(β) ∈ ℚ(Ṫ (β)). The conclusion of the lemma now follows by the definition of ℚ(Ṫ (β)) as given in Definition 7.1. □

The next lemma will be useful throughout the remainder of the paper. The proof is straightforward, but we include it for completeness.

Lemma 9.3. Suppose p is a partial function p : α → ℝ with a countable domain. If p is not in ℚ_{α}, then there exist β, t, and x such that:

• β is in dom(p),

• p ↾ β is in ℚ_{β},

• t ≤ p ↾ β in ℚ_{β},

• x is a countable subset of a_{β}^{p} closed under p(β),

• t ⊩_{ℚ_{β}} x ∈ Ṫ (β).

Proof. We assume the statement holds for ℚ_{ν}, for all ν < α, and verify that it also holds for ℚ_{α}. It is clear when α = 0.

Assume α is a successor ordinal α_{0} + 1. Let p be a partial function p : α → ℝ with a countable domain, and suppose p is not in ℚ_{α}. There are two ways this could happen. The first possibility is that p ↾ α_{0} is not in ℚ_{α_{0}}. In this case, the conclusion of the lemma follows from the induction hypothesis applied to p ↾ α_{0} in ℚ_{α_{0}}.

The second possibility is that p ↾ α_{0} is in ℚ_{α_{0}}, α_{0} is in dom(p), and p ↾ α_{0} does not force that p(α_{0}) is in ℚ(Ṫ (α_{0})). So there is t_{0} ≤ p ↾ α_{0} such that t_{0} forces p(α_{0}) is not in ℚ(Ṫ (α_{0})). Since p(α_{0}) is in ℝ, t_{0} forces there exists a countable set x ⊆ a^{p(α_{0})} such that x is closed under p(α_{0}) and x is in Ṫ (α_{0}). By recursion hypothesis (3) applied to ℙ_{α_{0}}, ℙ_{α_{0}} is ω_{1}-distributive. So t_{0} forces there is such a set x in the ground model. Hence we can choose t ≤ t_{0} and x such that t forces x satisfies the above properties. Then x is a countable subset of a_{α_{0}}^{p} closed under p(α_{0}) and t forces x is in Ṫ (α_{0}).

Now assume α is a limit ordinal. If p is not in ℚ_{α}, then by the definition of ℚ_{α}, there is γ < α such that p ↾ γ is not in ℚ_{γ}. The conclusion of the lemma now follows by applying the induction hypothesis to p ↾ γ in ℚ_{γ}. □

Recall that ⊩_{ℚ_{β}} Ṫ (β) ⊆ S_{ω_{2}} ∩ E. So if t ⊩_{ℚ_{β}} x ∈ Ṫ (β), as in the conclusion of Lemma 9.3, then x is in S_{ω_{2}} ∩ E.

Lemma 9.4. Suppose p and q are conditions in ℚ_{α} such that for all β in dom(p)∩ dom(q), p(β) ↾ (a_{β}^{p} ∩ a_{β}^{q})^{2} = q(β) ↾ (a_{β}^{p} ∩ a_{β}^{q})^{2}. Then p and q are compatible. Moreover, there is r ≤ p, q such that dom(r) = dom(p) ∪ dom(q), for all β in dom(p) \ dom(q), r(β) = p(β), and for all β in dom(q) \ dom(p), r(β) = q(β).

In particular, if p and q are conditions in ℚ_{α} such that for all β in dom(p) ∩ dom(q), a_{β}^{p} ∩ a_{β}^{q} is empty, then p and q are compatible.

Proof. Choose an ordinal ζ in ω_{1} which is not in a_{β}^{p}∪a_{β}^{q} for any β in dom(p)∩dom(q). Define r as follows. The domain of r is equal to dom(p) ∪ dom(q). Let β be in dom(r), and we define r(β). If β ∈ dom(p) \ dom(q), let r(β) = p(β). If β ∈ dom(q) \ dom(p), let r(β) = q(β). Suppose β is in dom(p) ∩ dom(q). Define a_{β}^{r} = a_{β}^{p} ∪ a_{β}^{q}. Let (i, j) be in (a_{β}^{r})^{2}, and we define r(β)(i, j). If (i, j) is in (a_{β}^{p})^{2}, let r(β)(i, j) = p(β)(i, j). If (i, j) is in (a_{β}^{q})^{2}, let r(β)(i, j) = q(β)(i, j). Note this is well-defined, since if i and j are both in a_{β}^{p} ∩ a_{β}^{q}, then p(β)(i, j) = q(β)(i, j). In any other case, let r(β)(i, j) = ζ.

It is easy to see that if r is a condition in ℚ_{α}, then r ≤ p, q. Suppose for a contradiction r is not in ℚ_{α}. By Lemma 9.3, fix β, t, and x such that β is in dom(r), r ↾ β is in ℚ_{β}, t ≤ r ↾ β in ℚ_{β}, x is a countable subset of a_{β}^{r} closed under r(β), and t ⊩_{ℚ_{β}} x ∈ Ṫ (β). Note that r ↾ β is below p ↾ β and q ↾ β in ℚ_{β}. Therefore t ≤ p ↾ β, q ↾ β.

We consider different cases regarding β, and derive a contradiction in each case. First consider the case that β is in dom(p) \ dom(q). Then by the definition of r, r(β) = p(β). Therefore x is a countable subset of a_{β}^{p} closed under p(β). By Lemma 9.2, p ↾ β ⊩_{ℚ_{β}} x ∉ Ṫ (β). But this contradicts that t ≤ p ↾ β. A similar argument handles the case when β is in dom(q) \ dom(p).

Now consider the case when β is in dom(p) ∩ dom(q). If x is a subset of a_{β}^{p} or x is a subset of a_{β}^{q}, then the same argument as in the previous paragraph yields a contradiction. Otherwise there is i in x \ a_{β}^{p} and j in x \ a_{β}^{q}. Then by the definition of r, r(β)(i, j) = ζ, which is not in x. So x is not closed under r(β), contradicting the choice of x. □

Lemma 9.5. Let z be a countable subset of α and let b be a countable subset of ω_{2}. Then there are densely many conditions t in ℚ_{α} such that z ⊆ dom(t) and for all β in z, b ⊆ a_{β}^{t}.

Proof. Let p be a condition in ℚ_{α}. We will find a condition t ≤ p such that z ⊆ dom(t) and for all β in z, b ⊆ a_{β}^{t}. Fix ζ in ω_{1} \ b. Let X be the set of β in dom(p) ∩ z such that b ⊆ a_{β}^{p}.

Define s in ℚ_{α} as follows. The domain of s is equal to z \ X. Consider β in dom(s). If β is in z \ dom(p), let s(β) : b^{2} → ω_{1} satisfy s(β)(i, j) = ζ for all (i, j) ∈ b^{2}. Suppose β is in z ∩ dom(p). Since β is not in X, b \ a_{β}^{p} is non-empty. Define s(β) : (b \ a_{β}^{p})^{2} → ω_{1} by letting s(β)(i, j) = ζ for all (i, j) in (b \ a_{β}^{p})^{2}.

The set s is a partial function s : α → ℝ with a countable domain. Note that s satisfies the property that for all β in dom(s), there are no subsets of a_{β}^{s} which are closed under s(β). It follows easily from Lemma 9.3 that s is in ℚ_{α}. Also note that if β is in dom(s) ∩ dom(p), then a_{β}^{s} ∩ a_{β}^{p} is empty. By Lemma 9.4, s and p are compatible. Fix t ≤ s, p in ℚ_{α}. Then clearly t is as desired. □

It will be useful to consider a certain kind of condition in ℚ_{α}, which we call a square condition.

Notation 9.6. A condition p in ℚ_{α} is square if there is a set a such that for all β in dom(p), a_{β}^{p} = a. In this case, we let a^{p} denote the set a as above.

Proposition 9.7. The forcing poset ℚ_{α} is ω_{1}-distributive. The set of square conditions is dense in ℚ_{α}.

Proof. We prove both statements simultaneously. So let {D_{n} : n < ω} be a family of dense open subsets of ℚ_{α}, and let p be a condition in ℚ_{α}. We prove there exists a square condition q ≤ p which is in ⋂{D_{n} : n < ω}.

Fix a regular cardinal θ much larger than ℚ_{α}. Recall that we are working in the model V [H], where H is a generic filter on ℙ_{ω_{2}} over V. We will apply Lemma 4.7 to the forcing poset ℙ_{ω_{2}} = ℙ(∅, 0, ω_{2}). By Lemma 4.7, in V [H] there are stationarily many N in P_{ω_{1}}(H(θ)) such that S_{ω_{2}} ∩ P (N ∩ ω_{2}) ⊆ N. Choose N in P_{ω_{1}}(H(θ)) such that N ≺ H(θ), the objects ℚ_{α}, p, and {D_{n} : n < ω} are in N, and S_{ω_{2}} ∩ P (N ∩ ω_{2}) ⊆ N.

Choose an N-generic sequence ⟨p_{n} : n < ω⟩ for ℚ_{α}, with p_{0} = p. By Lemma 9.5 and N-genericity, it is easy to see that ⋃{dom(p_{n}) : n < ω} = N ∩ α, and for each β in N ∩ α, ⋃{a_{β}^{p_{n}} : n < ω} = N ∩ ω_{2}.

Define q by letting dom(q) = ⋃{dom(p_{n}) : n < ω}, and for each β in dom(q), letting q(β) be the union of the set of functions {p_{n}(β) : n < ω}. Thus q is a partial function q : α → ℝ with a countable domain. We will prove that q is in ℚ_{α}. It is then easy to see that q ≤ p_{n} for all n < ω, and hence q is in ⋂{D_{n} : n < ω}. Also, for all β in dom(q), a_{β}^{q} = N ∩ ω_{2}, so q is a square condition.

Suppose for a contradiction q is not in ℚ_{α}. By Lemma 9.3, fix β, t, and x such that β is in dom(q) = N ∩ α, q ↾ β is in ℚ_{β}, t ≤ q ↾ β in ℚ_{β}, x is a countable subset of a_{β}^{q} = N ∩ ω_{2} closed under q(β), and t ⊩_{ℚ_{β}} x ∈ Ṫ (β). Note that q ↾ β ≤ p_{n} ↾ β for all n < ω. Since t ⊩_{ℚ_{β}} x ∈ Ṫ (β), x is in S_{ω_{2}}. Also x ⊆ N ∩ ω_{2}. So x is in S_{ω_{2}} ∩ P (N ∩ ω_{2}). By the choice of N, x is in N.

Let D be the set of conditions s in ℚ_{α} such that β is in dom(s) and x ⊆ a_{β}^{s}. By Lemma 9.5, D is dense open, and by elementarity, D is in N. So fix n < ω such that p_{n} is in D. Since x is closed under q(β) and q(β) ↾ (a_{β}^{p_{n}})^{2} = p_{n}(β), x is closed under p_{n}(β). By Lemma 9.2, p_{n} ↾ β ⊩_{ℚ_{β}} x ∉ Ṫ (β). As t ⊩_{ℚ_{β}} x ∈ Ṫ (β), p_{n} ↾ β and t are incompatible. But t ≤ q ↾ β ≤ p_{n} ↾ β, so we have a contradiction. □

Proposition 9.8. The forcing poset ℚ_{α} is ω_{2}-c.c.

Proof. Recall that CH holds in V [H]. We will show that ℚ_{α} is ω_{2}-Knaster. Let ⟨p_{i} : i < ω_{2}⟩ be a sequence of conditions in ℚ_{α}. Without loss of generality we may assume each p_{i} is a square condition.

Using CH, apply the Δ-System Lemma to the sequence of countable sets ⟨dom(p_{i}) : i < ω_{2}⟩ to get an unbounded set Z_{0} ⊆ ω_{2} and a countable set b such that for all i < j in Z_{0}, dom(p_{i}) ∩ dom(p_{j}) = b. Now apply the Δ-System Lemma again to the sequence of countable sets ⟨a^{p_{i}} : i ∈ Z_{0}⟩ to get an unbounded set Z_{1} ⊆ Z_{0} and a countable set c such that for all i < j in Z_{1}, a^{p_{i}} ∩ a^{p_{j}} = c.

An easy counting argument using CH shows there are at most ω_{1} many possibilities for a sequence ⟨p_{i}(β) ↾ c^{2} : β ∈ b⟩, for i in Z_{1}. So we can fix an unbounded set Z_{2} ⊆ Z_{1} such that for all i < j in Z_{2}, for all β in b, p_{i}(β) ↾ c^{2} = p_{j}(β) ↾ c^{2}.

Consider i < j in Z_{2}. Then for all β in dom(p_{i}) ∩ dom(p_{j}) = b, p_{i}(β) ↾ (a^{p_{i}} ∩ a^{p_{j}})^{2} = p_{i}(β) ↾ c^{2} = p_{j}(β) ↾ c^{2} = p_{j}(β) ↾ (a^{p_{i}} ∩ a^{p_{j}})^{2}. By Lemma 9.4, p_{i} and p_{j} are compatible. □

This completes the verification of the recursion hypotheses.

We conclude this section with several technical lemmas which will be needed later in the paper. Recall that ℚ is ℚ_{ω_{3}}.

Lemma 9.9. Let p and q be conditions in ℚ, and assume dom(p) ⊆ dom(q), and for all β in dom(p), a_{β}^{p} ⊆ a_{β}^{q}. If p and q are compatible, then q ≤ p.

Proof. To show q ≤ p, it suffices to show that whenever β is in dom(p), q(β) ↾ (a_{β}^{p})^{2} = p(β). Fix r ≤ p, q. Then q(β) ↾ (a_{β}^{p})^{2} = r(β) ↾ (a_{β}^{p})^{2} = p(β). □

Lemma 9.10. Suppose I is a generic filter on ℚ over V [H]. Let {p_{n} : n < ω} be a countable subset of I. Then there is a condition r in I such that r ≤ p_{n} for all n < ω.

Proof. Since ℚ is ω_{1}-distributive, the set {p_{n} : n < ω} is in the ground model V [H]. Let z = ⋃{dom(p_{n}) : n < ω} and let b = ⋃{a_{β}^{p_{n}} : n < ω, β ∈ dom(p_{n})}. By Lemma 9.5, there are densely many conditions t in ℚ such that z ⊆ dom(t), and for all β in z, b ⊆ a_{β}^{t}. Fix such a condition t in I. Let n < ω, and we show t ≤ p_{n}. By the choice of t, dom(p_{n}) ⊆ z ⊆ dom(t), and for all β in dom(p_{n}), a_{β}^{p_{n}} ⊆ b ⊆ a_{β}^{t}. Also p_{n} and t are both in I, so they are compatible. By Lemma 9.9, t ≤ p_{n}. □

The next easy lemma is technical and unmotivated, but we will need it in a construction later in the paper.

Lemma 9.11. Suppose p is a square condition in ℚ. Let ν be an ordinal in ω_{2} \ a^{p} and let ζ be an ordinal in ω_{1} which is different from ν. Let b be a subset of dom(p). Then there is a condition t ≤ p in ℚ such that dom(t) = dom(p), for all β in b, t(β) = p(β), and for all β in dom(t) \ b, ν ∈ a_{β}^{t} and t(β)(ν, ν) = ζ.

Proof. Define s with domain equal to dom(p)\b so that for all β in dom(s), a_{β}^{s} = {ν} and s(β)(ν, ν) = ζ. Since for each β in dom(s), there do not exist any subsets of a_{β}^{s} which are closed under s(β), it follows easily from Lemma 9.3 that s is a condition in ℚ. Now for all β in dom(p)∩dom(s), a_{β}^{p} ∩a_{β}^{s} is empty. So by Lemma 9.4, we can choose t ≤ s, p such that dom(t) = dom(s) ∪ dom(p) (which is equal to dom(p)), and for all β in dom(p) \ dom(s), t(β) = p(β).

Let us verify that t is as desired. We already noted that dom(t) = dom(p). If β is in b, then by the choice of s, β is in dom(p) \ dom(s), so t(β) = p(β). If β is in dom(t) \ b, then β is in dom(p) \ b = dom(s). By the choice of s, ν is in a_{β}^{s}. Since t ≤ s, ν is in a_{β}^{t}, and t(β)(ν, ν) = s(β)(ν, ν) = ζ. □

10. Preserving the Stationarity of S_{ω_{2}}

In this section we will prove that ℚ preserves the stationarity of S_{ω_{2}}, and forces that every stationary subset of S_{ω_{2}} reflects to an uncountable ordinal in ω_{2} ∩cof(ω).

Proposition 10.1. The forcing poset ℚ forces that S_{ω_{2}} is stationary in P_{ω_{1}}(ω_{2}).

Proof. Recall that ℚ is defined in the model V [H], where H is a generic filter on ℙ_{ω_{2}} over V. So it suffices to show that in V, the forcing poset ℙ_{ω_{2}} ∗ ℚ̇ forces Ṡ_{ω_{2}} is stationary.

Let (X, F)∗ṗ be a condition in ℙ_{ω_{2}} ∗ℚ̇, and suppose (X, F)∗ṗ forces ḣ : [ω_{2}]^{<ω} → ω_{2} is a function. We will find (Y, K) ∗ q̇ ≤ (X, F) ∗ ṗ and a set z such that (Y, K) ∗ q̇ forces z is in Ṡ_{ω_{2}} and z is closed under ḣ. Note that (X, F) ∗ ṗ forces there are club many ordinals ν < ω_{2} which are closed under ḣ. Choose (X′, F′) ∗ ṗ′ ≤ (X, F) ∗ ṗ and an uncountable ordinal ν in ω_{2} ∩cof(ω) such that (X′, F′)∗ṗ′ forces ν is closed under ḣ.

For each α < ω_{3}, ℙ_{ω_{2}} ∗ℚ̇_{α} forces that Ṫ (α) is a subset of P_{ω_{1}}(ω_{2}) which does not reflect to any uncountable ordinal in ω_{2} with cofinality ω. In particular, ℙ_{ω_{2}} ∗ ℚ̇_{α} forces that Ṫ (α) does not reflect to ν. So choose a ℙ_{ω_{2}} ∗ ℚ̇_{α}-name ḟ_{α} for a function ḟ_{α} : [ν]^{<ω} → ν such that ℙ_{ω_{2}} ∗ ℚ̇_{α} forces there is no set in Ṫ (α) ∩ P_{ω_{1}}(ν) which is closed under ḟ_{α}.

Fix a regular cardinal θ much larger than ℙ_{ω_{2}} ∗ ℚ̇, ḣ, and ⟨ḟ_{α} : α < ω_{3}⟩. Let N be a countable elementary substructure of H(θ) which contains as elements the objects ℙ_{ω_{2}} ∗ ℚ̇, (X′, F′) ∗ ṗ′, ḣ, ν, and ⟨ḟ_{α} : α < ω_{3}⟩. We attempt to define a condition (Y, K) ∗ q̇ below (X′, F′) ∗ ṗ′ which is N-generic for ℙ_{ω_{2}} ∗ ℚ.

Choose an N-generic sequence ⟨(X_{n}, F_{n}) ∗ ṗ_{n} : n < ω⟩ such that (X_{0}, F_{0}) ∗ ṗ_{0} = (X′, F′) ∗ ṗ′. Let K be the infinum of the sequence ⟨F_{n} : n < ω⟩. Let Y = ⋃{X_{n} : n < ω} ∪ {N ∩ ν}. By Proposition 3.8, (Y, K) is a condition in ℙ_{ω_{2}} and (Y, K) ≤ (X_{n}, F_{n}) for all n < ω. In particular, (Y, K) is N-generic.

Since N ∩ ν is in Y, (Y, K) ⊩ N ∩ ν ∈ Ṡ_{ω_{2}}. Using Lemma 3.7, it is easy to see that ⋃ Y = N ∩ ω_{2}. So by Lemma 4.5, (Y, K) ⊩ Ṡ_{ω_{2}} ∩ P (N ∩ ω_{2}) = Y.

Let H be a generic filter on ℙ_{ω_{2}} over V which contains the condition (Y, K). Since (Y, K) is N-generic, N[H] ∩ V = N. For each n < ω, let p_{n} = ṗ_{n}^{H}. Then ⟨p_{n} : n < ω⟩ is an N[H]-generic sequence for ℚ. By Lemma 9.5 and N[H]-genericity, ⋃{dom(p_{n}) : n < ω} = N[H]∩ω_{3} = N ∩ω_{3}, and for each β in ⋃{dom(p_{n}) : n < ω}, ⋃{a_{β}^{p_{n}} : n < ω} = N[H] ∩ ω_{2} = N ∩ ω_{2}.

We define a lower bound q of the sequence ⟨p_{n} : n < ω⟩. Let the domain of q be equal to ⋃{dom(p_{n}) : n < ω}, and for each β in dom(q), let q(β) be the union of the set of functions {p_{n}(β) : n < ω}. Clearly q is a partial function q : ω_{3} → ℝ with a countable domain. Also, dom(q) = N ∩ ω_{3}, and for each β in dom(q), a_{β}^{q} = N ∩ ω_{2}.

We show that q is a condition in ℚ. It is then clear that q ≤ p_{n} for all n < ω. Suppose for a contradiction q is not a condition. By Lemma 9.3, fix β, t, and x such that β is in dom(q), q ↾ β is in ℚ_{β}, t ≤ q ↾ β in ℚ_{β}, x is a countable subset of a_{β}^{q} which is closed under q(β), and t ⊩_{ℚ_{β}} x ∈ Ṫ (β). Note that β is in N ∩ ω_{3} and x is in S_{ω_{2}} ∩ P (N ∩ ω_{2}) = Y. Clearly q ↾ β ≤ p_{n} ↾ β for all n < ω, so in particular, q ↾ β is N[H]-generic for ℚ_{β}.

Now Y = ⋃{X_{n} : n < ω} ∪ {N ∩ ν}, and ⋃{X_{n} : n < ω} ⊆ N. So either x is in N, or x = N ∩ν. First suppose x is in N. Let D be the dense set of conditions s in ℚ such that β ∈ dom(s) and x ⊆ a_{β}^{s}. By elementarity, D is in N. Fix n < ω such that p_{n} ∈ D. Then x ⊆ a_{β}^{p_{n}}. Since x is closed under q(β) and q(β) ↾ (a_{β}^{p_{n}})^{2} = p_{n}(β), x is closed under p_{n}(β). By Lemma 9.2, p_{n} ↾ β ⊩_{ℚ_{β}} x ∉ Ṫ (β). Since t ⊩_{ℚ_{β}} x ∈ Ṫ (β), p_{n} ↾ β and t are incompatible. But t ≤ q ↾ β ≤ p_{n} ↾ β, and we have a contradiction.

Now assume x = N ∩ ν. Recall that ℙ_{ω_{2}} ∗ ℚ̇_{β} forces that ḟ_{β} : [ν]^{<ω} → ν is a function such that there is no set in Ṫ (β) ∩ P_{ω_{1}}(ν) which is closed under ḟ_{β}. Since β is in N, ḟ_{β} is in N. Let I_{β} be a generic filter on ℚ_{β} over V [H] such that t is in I_{β}. Since t ≤ q ↾ β and q ↾ β is N[H]-generic, N[H][I_{β}] ∩ V [H] = N[H]. In particular, N[H][I_{β}] ∩ ν = N[H] ∩ ν = N ∩ ν. Let f_{β} = ḟ_{β}^{H∗I_{β}}. Since ḟ_{β} is in N, f_{β} is in N[H][I_{β}]. By the elementarity of N[H][I_{β}], N[H][I_{β}] ∩ ν = N ∩ ν is closed under f_{β}. On the other hand, as t is in I_{β}, x = N ∩ ν is in T (β) = Ṫ (β)^{I_{β}}. So x is a member of T (β) ∩ P_{ω_{1}}(ν) which is closed under f_{β}, contradicting the choice of f_{β}. This completes the proof that q is in ℚ.

Let q̇ be a ℙ_{ω_{2}}-name such that (Y, K) forces q̇ is in ℚ̇ and q̇ ≤ ṗ_{n} for all n < ω. Then (Y, K) ∗ q̇ is N-generic for ℙ_{ω_{2}} ∗ ℚ̇, and (Y, K) ∗ q̇ ≤ (X, F) ∗ ṗ. We claim (Y, K) ∗ q̇ forces that N ∩ ν is in Ṡ_{ω_{2}} and is closed under ḣ, which completes the proof. Since (Y, K) ∗ q̇ ≤ (X′, F′) ∗ ṗ′, (Y, K) ∗ q̇ forces ν is closed under ḣ. As ḣ is in N, (Y, K) ∗ q̇ forces N[Ḣ ∗ İ] ∩ ω_{2} is closed under ḣ, where Ḣ ∗ İ is the canonical ℙ_{ω_{2}} ∗ ℚ̇-name for the generic filter. So (Y, K) ∗ q̇ forces N[Ḣ ∗ İ] ∩ ν is closed under ḣ. As (Y, K) ∗ q̇ is N-generic, (Y, K) ∗ q̇ ⊩ N[Ḣ ∗ İ] ∩ ν = N ∩ ν. But N ∩ ν is in Y, so (Y, K) ∗ q̇ forces N ∩ ν is in Ṡ_{ω_{2}} and is closed under ḣ. □

Theorem 10.2. The forcing poset ℚ forces that S_{ω_{2}} is a stationary subset of P_{ω_{1}}(ω_{2}) which does not reflect to any ordinal in ω_{2} with cofinality ω_{1}, and for every stationary set T ⊆ S_{ω_{2}}, there is an uncountable ordinal ν in ω_{2} with cofinality ω such that T reflects to ν.

Proof. Let p be a condition in ℚ and suppose p forces U̇ is a subset of S_{ω_{2}} which does not reflect to any uncountable ordinal in ω_{2} with cofinality ω. We will prove that p forces U̇ is non-stationary. Since ℚ is ω_{1}-distributive, it is easy to see that the set E from Notation 8.4 is still club in any generic extension by ℚ. So it suffices to show p forces U̇ ∩ E is non-stationary.

Fix a nice ℚ-name Ṫ for a subset of P_{ω_{1}}(ω_{2}) such that p ⊩_{ℚ} Ṫ = U̇ ∩ E, and ℚ forces if p is not in the generic filter, then Ṫ is the empty set. For some sequence of antichains ⟨A_{x} : x ∈ P_{ω_{1}}(ω_{2})⟩ of ℚ, Ṫ is equal to ⋃{A_{x} × {x̌} : x ∈ P_{ω_{1}}(ω_{2})}. As ℚ is ω_{2}-c.c., each A_{x} has cardinality at most ω_{1}. Since P_{ω_{1}}(ω_{2}) has cardinality ω_{2} and ℚ = ⋃{ℚ_{α} : α < ω_{3}}, we can fix β < ω_{3} such that for all x in P_{ω_{1}}(ω_{2}), A_{x} ⊆ ℚ_{β}. Then Ṫ is a nice ℚ_{β}-name.

For each ν in ω_{2} with cofinality ω, ⊩_{ℚ} Ṫ does not reflect to ν. So we can choose a nice ℚ-name Ċ_{ν} for a club subset of P_{ω_{1}}(ν) which is disjoint from Ṫ. As above, choose ξ < ω_{3} larger than β such that for all such ν, Ċ_{ν} is a ℚ_{ξ}-name. Now it is easy to see that ℚ_{ξ} forces that Ċ_{ν} ∩ Ṫ is empty, for all uncountable ν in ω_{2} with cofinality ω.

In summary, Ṫ is a nice ℚ_{ξ}-name for a subset of P_{ω_{1}}(ω_{2}) such that ℚ_{ξ} forces that Ṫ ⊆ S_{ω_{2}} ∩ E, and for all uncountable ν in ω_{2} with cofinality ω, Ṫ does not reflect to ν. By recursion hypothesis (4), there is i < ω_{3} such that Ṫ = Ṫ_{i}^{ξ}.

Let γ < ω_{3} be an ordinal such that f(γ) = (ξ, i). Then Ṫ (γ) = Ṫ_{i}^{ξ} = Ṫ. By Lemma 8.7, ℚ_{γ+1} adds a club subset of P_{ω_{1}}(ω_{2}) which is disjoint from Ṫ. Since the inclusion map ℚ_{γ+1} → ℚ is a complete embedding, ℚ adds a club which is disjoint from Ṫ. As p ⊩ Ṫ = U̇ ∩ E, p forces U̇ ∩ E is non-stationary. □

11. Extending an Elementary Embedding, Part 2

Fix for the remainder of the section a cardinal κ which is κ^{+}-supercompact, and assume 2^{κ} = κ^{+}. Let j : V → M be an elementary embedding with critical point κ such that j(κ) > κ^{+} and M^{κ^{+}} ⊆ M. In Section 6 we showed how to extend the elementary embedding j after forcing with Coll(ω_{1}, <κ)∗ℙ̇_{κ}. Specifically, let \overline{G}∗\overline{H} be a generic filter on j(Coll(ω_{1}, <κ)∗ℙ̇_{κ}) over V. Let G = \overline{G}∩Coll(ω_{1}, <κ) and H = \overline{H} ∩ ℙ_{κ}. Then G ∗ H is a generic filter on Coll(ω_{1}, < κ) ∗ ℙ̇_{κ} over V, and in V [\overline{G} ∗ \overline{H}] we can extend j to j : V [G ∗ H] → M[\overline{G} ∗ \overline{H}] such that j(G ∗ H) = \overline{G} ∗ \overline{H}.

In this section we will prove that in M[\overline{G} ∗ \overline{H}], there is a projection mapping π : j(ℚ) → ℚ. As a consequence, if \overline{I} is a generic filter for j(ℚ) over V [\overline{G} ∗ \overline{H}], then in V [\overline{G} ∗ \overline{H} ∗ \overline{I}] we can extend j to j : V [G ∗ H ∗ I] → M[\overline{G} ∗ \overline{H} ∗ \overline{I}] such that j(G ∗ H ∗ I) = \overline{G} ∗ \overline{H} ∗ \overline{I}, where I is the filter on ℚ generated by π[\overline{I}].

As in Section 6, we write κ^{+} to indicate the ordinal which is the successor of κ in V, even when working in a model in which this ordinal is no longer a cardinal. In the models M[G ∗ H] and V [G ∗ H], κ is equal to ω_{2} and κ^{+} is equal to ω_{3}. However, in the model M[\overline{G}∗\overline{H}], j(κ) is equal to ω_{2}, and κ and κ^{+} are both ordinals of cofinality ω_{1} strictly between ω_{1} and ω_{2}.

By Theorem 6.1(1), M[G ∗ H]^{κ^{+}} ∩ V [G ∗ H] ⊆ M[G ∗ H]. It easily follows from this fact and from Recursion Hypotheses 8.5 (1) and (4) that the sequences ⟨ℚ_{α} : α ≤ κ^{+}⟩ and ⟨Ṫ (α) : α < κ^{+}⟩ are in M[G ∗ H].

Recall that in V [G ∗ H], ℝ is the poset consisting of all non-empty functions s : (a^{s})^{2} → ω_{1}, where a^{s} is a countable subset of ω_{2}, ordered by extension of functions.

In the next lemma and in what follows, ℝ means (ℝ)^{V [G∗H]}.

Lemma 11.1. In M[\overline{G} ∗ \overline{H}], ℝ = \{s ∈ j(ℝ) : a^{s} ⊆ κ\}. Hence if t is in j(ℝ), then t ↾ κ^{2} is in ℝ.

Proof. Let s be in ℝ. Since s is a countable subset of κ^{2} × ω_{1}, j(s) = s. So by the elementarity of j, s is in j(ℝ), and obviously a^{s} ⊆ κ. On the other hand, suppose s is in j(ℝ) and a^{s} ⊆ κ. Since s is a countable subset of κ^{2} × ω_{1} and M[\overline{G} ∗ \overline{H}] is a generic extension of M by an ω_{1}-distributive forcing poset, s is in M. So s is in V, and thus s is in V [G ∗ H]. As above, j(s) = s. Since j(s) is in j(ℝ), s is in ℝ. □

We will define for each α ≤ κ^{+} a map π_{α} : j(ℚ_{α}) → ℚ_{α} in M[\overline{G} ∗ \overline{H}]. So fix α ≤ κ^{+}, and consider a condition q in j(ℚ_{α}). Then q is a partial function q : j(α) → j(ℝ) with a countable domain. So dom(q) ∩ j[α] is a countable subset of j[α], and the set j^{−1}(dom(q) ∩ j[α]) is a countable subset of α. Since M[\overline{G} ∗ \overline{H}] is a generic extension of M by an ω_{1}-distributive forcing poset, j^{−1}(dom(q) ∩ j[α]) is in M, and hence is in V [G ∗ H].

For each ordinal γ in j^{−1}(dom(q) ∩ j[α]), q(j(γ)) is in j(ℝ). So q(j(γ)) ↾ κ^{2} is in ℝ by Lemma 11.1. Define π_{α}(q) as the function with domain equal to j^{−1}(dom(q)∩ j[α]) such that for all γ in this domain, π_{α}(q)(γ) = q(j(γ)) ↾ κ^{2}. Clearly π_{α}(q) is a partial function π_{α}(q) : α → ℝ with a countable domain, and again by distributivity, π_{α}(q) is in V [G ∗ H]. It will take some work to prove that π_{α}(q) is in ℚ_{α}.

We record our definition of π_{α} for easy reference.

Definition 11.2. Let α ≤ κ^{+}. In M[\overline{G}∗\overline{H}] define a map π_{α} with domain j(ℚ_{α}) as follows. Given q in j(ℚ_{α}), let the domain of π_{α}(q) be equal to j^{−1}(dom(q) ∩ j[α]), and for each γ in this domain, let π_{α}(q)(γ) = q(j(γ)) ↾ κ^{2}.

The next lemma is immediate from the definition.

Lemma 11.3. Let α ≤ κ^{+}. For all q in j(ℚ_{α}), j[dom(π_{α}(q))] = dom(q) ∩ j[α], and for all γ in the domain of π_{α}(q), a_{γ}^{π_{α}(q)} = a_{j(γ)}^{q} ∩ κ.

Lemma 11.4. Let α_{1} ≤ α_{2} ≤ κ^{+} and let q be in j(ℚ_{α_{2}}). Then π_{α_{1}}(q ↾ j(α_{1})) = π_{α_{2}}(q) ↾ α_{1}.

Proof. The domain of π_{α_{1}}(q ↾ j(α_{1})) is equal to j^{−1}(dom(q ↾ j(α_{1}))∩j[α_{1}]), whereas the domain of π_{α_{2}}(q) ↾ α_{1} is j^{−1}(dom(q) ∩ j[α_{2}]) ∩ α_{1}, and it is easy to see these sets are equal. Let γ be in the domain of π_{α_{1}}(q ↾ j(α_{1})). Then π_{α_{1}}(q ↾ j(α_{1}))(γ) is equal to (q ↾ j(α_{1}))(j(γ)) ↾ κ^{2} = q(j(γ)) ↾ κ^{2} = π_{α_{2}}(q)(γ) = (π_{α_{2}}(q) ↾ α_{1})(γ). □

Lemma 11.5. Let α ≤ κ^{+} and let q be in j(ℚ_{α}). Suppose t is in ℚ_{α}, dom(π_{α}(q)) ⊆ dom(t), and for all γ in dom(π_{α}(q)), t(γ) ≤_{ℝ} π_{α}(q)(γ). Then j(t) and q are compatible in j(ℚ_{α}). In particular, if π_{α}(q) is in ℚ_{α} and t ≤ π_{α}(q) in ℚ_{α}, then j(t) and q are compatible in j(ℚ_{α}).

Proof. To show that j(t) and q are compatible in j(ℚ_{α}), by Lemma 9.4 it suffices to show that for all β in dom(j(t)) ∩ dom(q),

j(t)(β) ↾ (a_{β}^{j(t)} ∩ a_{β}^{q})^{2} = q(β) ↾ (a_{β}^{j(t)} ∩ a_{β}^{q})^{2}.

First we claim that

dom(j(t)) ∩ dom(q) = dom(q) ∩ j[α].

The domain of t is a countable subset of α, so dom(j(t)) = j(dom(t)) = j[dom(t)], which is a subset of j[α]. So dom(j(t)) ∩ dom(q) ⊆ dom(q) ∩ j[α]. Suppose on the other hand β is in dom(q) ∩ j[α]. Recall that dom(π_{α}(q)) = j^{−1}(dom(q) ∩ j[α]). So letting γ = j^{−1}(β), γ is in the domain of π_{α}(q). But dom(π_{α}(q)) ⊆ dom(t) by assumption, so γ is in the domain of t. Hence j(γ) = β is in dom(j(t)) ∩ dom(q).

Next we claim that for all β in the domain of j(t), letting γ = j^{−1}(β),

a_{β}^{j(t)} = a_{γ}^{t}

and

j(t)(β) = t(γ).

So let β = j(γ) be in the domain of j(t). Then t(γ) : (a_{γ}^{t})^{2} → ω_{1}, where a_{γ}^{t} is a countable subset of κ. So a_{β}^{j(t)} = j(a_{γ}^{t}) = a_{γ}^{t}, proving the first claim. Now

for any pair (i, k) in (a_{β}^{j(t)})^{2}, j(t)(β)(i, k) = j(t)(j(γ))(j(i), j(k)) = j(t(γ)(i, k)) = t(γ)(i, k). So j(t)(β) = t(γ).

Now we prove that for all β in dom(j(t)) ∩ dom(q), letting γ = j^{−1}(β),

a_{β}^{j(t)} ∩ a_{β}^{q} = a_{γ}^{π_{α}(q)}.

This makes sense since if β is in dom(j(t)) ∩ dom(q), then β is in dom(q) ∩ j[α], so γ is in j^{−1}(dom(q) ∩ j[α]) = dom(π_{α}(q)). Since a_{β}^{j(t)} = a_{γ}^{t}, which is a subset of κ,

a_{β}^{j(t)} ∩ a_{β}^{q} ⊆ a_{β}^{q} ∩ κ = a_{γ}^{π_{α}(q)}. On the other hand, t(γ) ≤_{ℝ} π_{α}(q)(γ) by assumption.

In particular, a_{γ}^{π_{α}(q)} ⊆ a_{γ}^{t} = a_{β}^{j(t)}. Also a_{γ}^{π_{α}(q)} = a_{β}^{q} ∩κ ⊆ a_{β}^{q}. So a_{γ}^{π_{α}(q)} ⊆ a_{β}^{j(t)} ∩a_{β}^{q}.

We finish the proof. Let β be in dom(j(t)) ∩ dom(q). Then β is in dom(q) ∩ j[α]. Let γ = j^{−1}(β). Then a_{β}^{j(t)} ∩ a_{β}^{q} = a_{γ}^{π_{α}(q)} and j(t)(β) = t(γ). Since t(γ) ≤_{ℝ}

π_{α}(q)(γ), j(t)(β) ↾ (a_{β}^{j(t)} ∩ a_{β}^{q})^{2} = t(γ) ↾ (a_{γ}^{π_{α}(q)})^{2} = π_{α}(q)(γ) = q(β) ↾ (a_{γ}^{π_{α}(q)})^{2} =

q(β) ↾ (a_{β}^{j(t)} ∩ a_{β}^{q})^{2}. □

Lemma 11.6. Let α ≤ κ^{+} and let t be in ℚ_{α}. Suppose u ≤ j(t) in j(ℚ_{α}). Then dom(t) ⊆ dom(π_{α}(u)), and for all γ in dom(t), π_{α}(u)(γ) ≤_{ℝ} t(γ). In particular, if π_{α}(u) is in ℚ_{α}, then π_{α}(u) ≤ t in ℚ_{α}.

Proof. Since dom(t) is a countable subset of α, dom(j(t)) = j(dom(t)) = j[dom(t)]. As u ≤ j(t) in j(ℚ_{α}), j[dom(t)] ⊆ dom(u). So if γ is in dom(t), j(γ) is in dom(u) ∩ j[α], so γ is in j^{−1}(dom(u)∩j[α]) = dom(π_{α}(u)). This shows dom(t) ⊆ dom(π_{α}(u)).

Now let γ be in dom(t), and we show π_{α}(u)(γ) ≤_{ℝ} t(γ). Note that j(t)(j(γ)) = j(t(γ)) = t(γ), because t(γ) is a countable subset of κ^{2} × ω_{1}. Since u ≤ j(t) in j(ℚ_{α}), u(j(γ)) ≤_{j(ℝ)} j(t)(j(γ)) = t(γ). In other words, u(j(γ)) extends t(γ) as a function. But a_{γ}^{t} is a subset of κ, so u(j(γ)) ↾ κ^{2} = π_{α}(u)(γ) extends t(γ) as a function. So π_{α}(u)(γ) ≤_{ℝ} t(γ). □

Now we are ready to show π_{α} : j(ℚ_{α}) → ℚ_{α} is a projection mapping.

Proposition 11.7. For all α ≤ κ^{+}, π_{α} is a projection mapping π_{α} : j(ℚ_{α}) → ℚ_{α}. Moreover, if t ≤ π_{α}(q) in ℚ_{α}, then j(t) and q are compatible in j(ℚ_{α}).

Proof. The proof is by induction on α ≤ κ^{+}. So let α ≤ κ^{+} be given, and assume for all β < α, π_{β} satisfies the statement of the proposition.

First we prove that if q is in j(ℚ_{α}), then π_{α}(q) is in ℚ_{α}. We observed above that π_{α}(q) is in V [G ∗ H], and π_{α}(q) is a partial function π_{α}(q) : α → ℝ with a countable domain. So if π_{α}(q) is not in ℚ_{α}, then by Lemma 9.3, we can fix β, u, and x such that β is in dom(π_{α}(q)), π_{α}(q) ↾ β is in ℚ_{β}, u ≤ π_{α}(q) ↾ β in ℚ_{β}, x is a countable subset of a_{β}^{π_{α}(q)} which is closed under π_{α}(q)(β), and u ⊩_{ℚ_{β}} x ∈ Ṫ (β).

The set x is a countable subset of κ, so j(x) = x. By Lemma 11.4, π_{β}(q ↾ j(β)) = π_{α}(q) ↾ β. So u ≤ π_{β}(q ↾ j(β)) in ℚ_{β}. Applying Lemma 11.5 to u and q ↾ j(β), we get that j(u) and q ↾ j(β) are compatible in j(ℚ_{β}). Fix s ≤ j(u), q ↾ j(β) in j(ℚ_{β}).

Recall that u ⊩_{ℚ_{β}} x ∈ Ṫ (β). Since j(x) = x and s ≤ j(u), by the elementarity of j, s ⊩_{j(ℚ_{β})} x ∈ j(Ṫ (β)). The set x is a subset of a_{β}^{π_{α}(q)} = a_{j(β)}^{q} ∩ κ, and x is closed under π_{α}(q)(β) = q(j(β)) ↾ κ^{2}. Therefore x is a countable subset of a_{j(β)}^{q} which is closed under q(j(β)). By Lemma 9.2, q ↾ j(β) ⊩_{j(ℚ_{β})} x ∉ j(Ṫ (β)). But s ≤ q ↾ j(β), and s ⊩_{j(ℚ_{β})} x ∈ j(Ṫ (β)). This is a contradiction. So indeed π_{α}(q) is in ℚ_{α}.

Now we prove that π_{α} : j(ℚ_{α}) → ℚ_{α} is a projection mapping. Clearly π_{α} sends the empty condition in j(ℚ_{α}) to the empty condition in ℚ_{α}. Suppose r ≤ q in j(ℚ_{α}), and we show π_{α}(r) ≤ π_{α}(q) in ℚ_{α}. Since dom(q) ⊆ dom(r), dom(π_{α}(q)) = j^{−1}(dom(q) ∩ j[α]) ⊆ j^{−1}(dom(r) ∩ j[α]) = dom(π_{α}(r)). Let γ be in dom(π_{α}(q)). Then j(γ) is in dom(q), and since r ≤ q, r(j(γ)) extends q(j(γ)) as a function. It follows that r(j(γ)) ↾ κ^{2} extends q(j(γ)) ↾ κ^{2} as a function, that is, π_{α}(r)(γ) ≤_{ℝ} π_{α}(q)(γ). Finally, let q be in j(ℚ_{α}), and suppose v ≤ π_{α}(q) in ℚ_{α}. We show there is u ≤ q in j(ℚ_{α}) such that π_{α}(u) ≤ v in ℚ_{α}. By Lemma 11.5, j(v) and q are compatible in j(ℚ_{α}). So fix u ≤ j(v), q. Then by Lemma 11.6, π_{α}(u) ≤ v in ℚ_{α}.

We have established that π_{α} : j(ℚ_{α}) → ℚ_{α} is a projection mapping. If t ≤ π_{α}(q) in ℚ_{α}, then j(t) and q are compatible in j(ℚ_{α}) by Lemma 11.5. □

Let π : j(ℚ) → ℚ be the map π_{κ^{+}}.

We now turn to showing that we can extend the elementary embedding j after forcing with ℚ. First we need several easy technical lemmas.

Lemma 11.8. Let α ≤ κ^{+}. If q is in j(ℚ_{α}) and π_{α}(q) ≤ p in ℚ_{α}, then q ≤ j(p) in j(ℚ_{α}).

Proof. The domain of j(p) is equal to j(dom(p)) = j[dom(p)], and for all j(γ) in dom(j(p)), j(p)(j(γ)) = j(p(γ)) = p(γ), since p(γ) is a countable subset of κ^{2} × ω_{1}.

First we show dom(j(p)) ⊆ dom(q). So let β be in dom(j(p)). Then β = j(γ) for some γ in dom(p). Since π_{α}(q) ≤ p, γ is in dom(π_{α}(q)) = j^{−1}(dom(q) ∩ j[α]). So j(γ) = β is in dom(q).

Now let β = j(γ) be in dom(j(p)). Then q(β) ↾ κ^{2} = π_{α}(q)(γ) ≤_{ℝ} p(γ) = j(p)(β), that is, q(β) ↾ κ^{2} extends j(p)(β) as a function. Therefore q(β) extends j(p)(β) as a function, that is, q(β) ≤_{j(ℝ)} j(p)(β). □

Lemma 11.9. Let α ≤ κ^{+}. If p is in ℚ_{α}, then π_{α}(j(p)) = p.

Proof. The domain of π_{α}(j(p)) is equal to j^{−1}(dom(j(p)) ∩ j[α]). But dom(j(p)) = j(dom(p)) = j[dom(p)], which is a subset of j[α]. So the domain of π_{α}(j(p)) is equal to j^{−1}(j[dom(p)]) = dom(p). For each γ in the domain of p, π_{α}(j(p))(γ) = j(p)(j(γ)) ↾ κ^{2} = j(p(γ)) ↾ κ^{2}. But p(γ) is a countable subset of κ^{2} × ω_{1}, so j(p(γ)) = p(γ). Since the domain of p(γ) is a subset of κ^{2}, π_{α}(j(p))(γ) = j(p(γ)) ↾ κ^{2} = p(γ) ↾ κ^{2} = p(γ). So π_{α}(j(p)) = p. □

Since π : j(ℚ) → ℚ is a projection mapping, if \overline{I} is a generic filter on j(ℚ) over V [\overline{G} ∗ \overline{H}], then π[\overline{I}] generates a generic filter I on ℚ over V [\overline{G} ∗ \overline{H}].

Lemma 11.10. π[\overline{I}] = I.

Proof. It suffices to show π[\overline{I}] is upwards closed. So let q be in \overline{I} and suppose π(q) ≤ p in ℚ. By Lemma 11.8, q ≤ j(p) in j(ℚ). Therefore j(p) is in \overline{I}. By Lemma 11.9, π(j(p)) = p, so p is in π[\overline{I}]. □

The next theorem is an extension of Theorem 6.1.

Theorem 11.11. Let \overline{I} be a generic filter on j(ℚ) over V [\overline{G} ∗ \overline{H}]. Let I = π[\overline{I}]. Then j can be lifted to j : V [G∗H∗I] → M[\overline{G}∗\overline{H}∗\overline{I}] such that j(G∗H∗I) = \overline{G}∗\overline{H}∗\overline{I}.

Proof. It suffices to show j[I] ⊆ \overline{I}. So let p be in I, and we show j(p) is in \overline{I}. Since p is in I = π[\overline{I}], fix q in \overline{I} such that π(q) = p. In particular, π(q) ≤ p. By Lemma 11.8, q ≤ j(p) in j(ℚ). As q is in \overline{I}, j(p) is in \overline{I}. □

12. Preserving Stationary Sets

As usual, fix a cardinal κ which is κ^{+}-supercompact, and assume 2^{κ} = κ^{+}. Let j : V → M be an elementary embedding with critical point κ such that j(κ) > κ^{+} and M^{κ^{+}} ⊆ M. Let \overline{G} ∗ \overline{H} ∗ \overline{I} be a generic filter on j(Coll(ω_{1}, < κ) ∗ ℙ̇_{κ} ∗ ℚ̇) over V. Let G = \overline{G} ∩ Coll(ω_{1}, < κ), H = \overline{H} ∩ ℙ_{κ}, and I = π[\overline{I}]. Extend j to j : V [G ∗ H ∗ I] → M[\overline{G} ∗ \overline{H} ∗ \overline{I}].

Our goal in this section is to prove the following theorem. After we complete the proof of this result, we will have all the tools necessary to prove the main theorem of the paper.

Theorem 12.1. In V [G ∗ H ∗ I], let T be a stationary subset of P_{ω_{1}}(κ) which is disjoint from S_{κ}. Then T is a stationary subset of P_{ω_{1}}(κ) in the model M[\overline{G}∗\overline{H} ∗\overline{I}].

By Theorem 6.1, M[\overline{G} ∗ \overline{H}] is equal to M[\overline{G}][H][H′], where H′ is a generic filter on ℙ(κ, S_{κ}, j(κ)) over M[\overline{G}][H]. Let G′ = G ∩ Coll(ω_{1}, [κ, j(κ))). Then M[\overline{G}] = M[G][G′]. So M[\overline{G} ∗ \overline{H}] = M[G][G′][H][H′].

Lemma 12.2. M[\overline{G} ∗ \overline{H} ∗ \overline{I}] = M[G ∗ H ∗ I][G′][H′][\overline{I}].

Proof. We already know

M[\overline{G} ∗ \overline{H} ∗ \overline{I}] = M[G][G′][H][H′][\overline{I}].

Since π : j(ℚ) → ℚ is a projection mapping and I = π[\overline{I}], we have

M[G][G′][H][H′][\overline{I}] = M[G][G′][H][H′][I][\overline{I}].

So

M[\overline{G} ∗ \overline{H} ∗ \overline{I}] = M[G][G′][H][H′][I][\overline{I}].

Now we repeatedly apply the Product Lemma. Since ℙ_{κ} is in M[G],

M[G][G′][H] = M[G][H][G′].

So

M[\overline{G} ∗ \overline{H} ∗ \overline{I}] = M[G][H][G′][H′][I][\overline{I}].

Since ℚ is in M[G][H],

M[G][H][G′][H′][I] = M[G][H][I][G′][H′].

Therefore

M[\overline{G} ∗ \overline{H} ∗ \overline{I}] = M[G ∗ H ∗ I][G′][H′][\overline{I}].
□

Now we begin the (long) proof of Theorem 12.1.

Proof. Let T be a stationary subset of P_{ω_{1}}(κ) in V [G ∗ H ∗ I] which is disjoint from S_{κ}. Since Coll(ω_{1}, <κ)∗ℙ̇_{κ}∗ℚ̇ is κ-c.c., M[G∗H∗I]^{κ^{+}} ∩V [G∗H∗I] ⊆ M[G∗H∗I]. So T is in M[G ∗ H ∗ I].

By the preceding lemma, M[\overline{G} ∗ \overline{H} ∗ \overline{I}] is equal to M[G ∗ H ∗ I][G′][H′][\overline{I}]. Since Coll(ω_{1}, [κ, j(κ))) is ω_{1}-closed, and hence proper, T remains a stationary subset of P_{ω_{1}}(κ) in the model M[G ∗ H ∗ I][G′].

By Corollary 4.8, letting α = κ, A = S_{κ}, and β = j(κ), we get that ℙ(κ, S_{κ}, j(κ)) forces over M[G ∗ H ∗ I][G′] that T is stationary in P_{ω_{1}}(κ). Thus T is stationary in the model M[G ∗ H ∗ I][G′][H′]. So to show that T is stationary in M[G ∗ H ∗ I][G′][H′][\overline{I}], it suffices to show that in the model M[G ∗ H ∗ I][G′][H′], the forcing poset j(ℚ)/ℚ = {q ∈ j(ℚ) : π(q) ∈ I} forces that T is stationary in P_{ω_{1}}(κ).

Goal: To prove that in the model M[G ∗ H ∗ I][G′][H′] = M[\overline{G} ∗ \overline{H}][I], the forcing poset j(ℚ)/ℚ forces that T is a stationary subset of P_{ω_{1}}(κ).

Before beginning the proof, let us recall some of the notation used in the definition of ℚ, and fix some additional parameters.

The forcing poset ℚ is defined in the model V [G ∗ H], and in this model, κ = ω_{2} and κ^{+} = ω_{3}. (As before, we will always write κ^{+} to denote the ordinal which is the successor of κ in the ground model V.) In M[\overline{G}∗\overline{H}], both κ and κ^{+} are ordinals with cofinality ω_{1} strictly between ω_{1} and j(κ) = ω_{2}.

Recall Notation 8.4, where we fixed, in V [G∗H], a surjective function g_{α} : ω_{1} → α for each α in ω_{2} = κ. We let E be the set of b in P_{ω_{1}}(κ) such that ot(b) is a limit ordinal and for all α in b, b is closed under g_{α}. Note that the sequence ⟨g_{α} : α ∈ κ⟩ is in M[G ∗ H].

The definition of ℚ involved the sequence of nice names ⟨Ṫ (α) : α < κ^{+}⟩. Recall that for all α < κ^{+}, ℚ_{α} forces that Ṫ (α) is a subset of S_{κ} ∩ E which does not reflect to any ordinal in κ with cofinality ω. By elementarity, j(ℚ) is defined in M[\overline{G} ∗ \overline{H}] by the sequence of nice names j(⟨Ṫ (α) : α < κ^{+}⟩), which we denote by ⟨Ṫ^{j}(β) : β < j(κ^{+})⟩.

By upwards absoluteness, in the model M[\overline{G} ∗ \overline{H}][I], S_{κ} is a subset of P_{ω_{1}}(κ) which does not reflect to any ordinal in κ with cofinality ω_{1}. Applying Lemma 4.10, fix a club set D ⊆ P_{ω_{1}}(κ) as follows.

Notation 12.3. Let D be a club subset of P_{ω_{1}}(κ) in M[\overline{G} ∗ \overline{H}][I] such that for all b in D, for any ν in b ∩ cof(ω_{1}), b ∩ ν is not in S_{κ}.

For the remainder of the proof, we work in the model M[\overline{G}∗\overline{H}][I], unless specified otherwise. We will prove that in this model, the forcing poset j(ℚ)/ℚ forces that T is a stationary subset of P_{ω_{1}}(κ). To prove this, it suffices to show that whenever p is a condition in j(ℚ)/ℚ and p ⊩ Ċ ⊆ P_{ω_{1}}(κ) is club, then there is q ≤ p in j(ℚ)/ℚ and a set z in T such that q ⊩ z ∈ Ċ. So let p and Ċ be given, and assume p ⊩ Ċ ⊆ P_{ω_{1}}(κ) is club.

Fix a regular cardinal θ much larger than j(ℚ)/ℚ and Ċ. We use Lemma 4.7 to choose an appropriate elementary substructure of H(θ). The model we are working in, namely, M[\overline{G} ∗ \overline{H}][I], is equal to M[G ∗ H ∗ I][G′][H′], which is a generic extension of M[G ∗ H ∗ I][G′] by the forcing poset ℙ(κ, S_{κ}, j(κ)). In M[G ∗ H ∗ I][G′], θ is a regular cardinal larger than ω_{2} = j(κ), κ has cofinality ω_{1}, and T is a stationary subset of P_{ω_{1}}(κ) which is disjoint from S_{κ}. By Lemma 4.7, in M[G ∗ H ∗ I][G′][H′] there are stationarily many N in P_{ω_{1}}(H(θ)) such that S(κ, S_{κ}, j(κ)) ∩ P (N ∩ j(κ)) ⊆ N and N ∩ κ ∈ T.

Notation 12.4. Let N be a countable subset of H(θ) in M[\overline{G} ∗ \overline{H}][I] satisfying: • N ≺ H(θ),

• N ∩ κ ∈ T ∩ D,

• S(κ, S_{κ}, j(κ)) ∩ P (N ∩ j(κ)) ⊆ N,

• N contains as members the objects j(ℚ), ⟨Ṫ^{j}(β) : β < j(κ^{+})⟩, I, j(ℚ)/ℚ,

p, Ċ, j ↾ κ^{+}, and ⟨g_{α} : α ∈ κ⟩.

Since N ∩ κ is in T, it suffices to find q ≤ p in j(ℚ)/ℚ such that q ⊩ N ∩ κ ∈ Ċ. If q ≤ p is N-generic, then since Ċ is in N, q ⊩ N[K̇] ∩ κ = N ∩ κ ∈ Ċ, where K̇ is the canonical j(ℚ)/ℚ-name for the generic filter. So it suffices to define an N-generic sequence ⟨p_{n} : n < ω⟩ for j(ℚ)/ℚ, with p_{0} = p, and find a lower bound q of this sequence.

We will choose the N-generic sequence carefully. Enumerate all dense open subsets of j(ℚ)/ℚ in N as ⟨D_{n} : n < ω⟩. By the ω_{1}-distributivity of the forcing posets, the set N ∩ κ is in M[G ∗ H]. So by Lemma 4.6, S_{κ} ∩ P (N ∩ κ) is countable in M[G ∗ H], and hence is countable in M[\overline{G} ∗ \overline{H}][I]. Let ⟨x_{n} : n < ω⟩ be an enumeration of S_{κ} ∩ P (N ∩ κ) in which each set appears infinitely often.

Now we define the N-generic sequence ⟨p_{n} : n < ω⟩. Let p_{0} = p. Suppose n < ω and p_{n} is defined. Choose a condition p_{n}^{∗} ≤ p_{n} which is in D_{n} ∩ N. Since the set of square conditions in j(ℚ) is dense in j(ℚ), the set of square conditions in j(ℚ)/ℚ is dense in j(ℚ)/ℚ by Lemma 1.1. Choose a square condition p′_{n} ≤ p_{n}^{∗} in (j(ℚ)/ℚ) ∩ N. We split the definition of p_{n+1} into three cases.

Case 1: x_{n} ⊆ a^{p′_{n}}.

In this case, let p_{n+1} = p′_{n}.

Case 2: Case 1 fails and N ∩ ω_{1} ⊆ x_{n}.

In this case, again let p_{n+1} = p′_{n}.

Case 3: Case 1 and Case 2 fail.

By the failure of Case 1, x_{n} is not a subset of a^{p′_{n}}. Fix an ordinal ν_{n} which is in x_{n} \ a^{p′_{n}}. By the failure of Case 2, N ∩ ω_{1} is not a subset of x_{n}. Fix an ordinal ζ_{n} in (N ∩ ω_{1}) \ x_{n}. Since ν_{n} is in x_{n} and ζ_{n} is not in x_{n}, ν_{n} is not equal to ζ_{n}.

To define p_{n+1}, we apply Lemma 9.11. The condition p′_{n} is a square condition in j(ℚ). The ordinal ν_{n} is in j(κ) \ a^{p′_{n}} and ζ_{n} is an ordinal in ω_{1} different from ν_{n}. Let b = dom(p′_{n}) ∩ j[κ^{+}]. By Lemma 9.11, there is a condition p_{n+1} ≤ p′_{n} in j(ℚ) such that dom(p_{n+1}) = dom(p′_{n}), for all β in b, p_{n+1}(β) = p′_{n}(β), and for all β in dom(p_{n+1}) \ b, ν_{n} is in a_{β}^{p_{n+1}} and p_{n+1}(β)(ν_{n}, ν_{n}) = ζ_{n}. Since j(ℚ), p′_{n}, b, ν_{n}, and ζ_{n} are all in N, by elementarity we can choose p_{n+1} in N.

We claim that p_{n+1} is in j(ℚ)/ℚ, that is, π(p_{n+1}) ∈ I. Now dom(p_{n+1}) = dom(p′_{n}) and p_{n+1}(β) = p′_{n}(β) for all β in dom(p′_{n})∩j[κ^{+}]. It follows that π(p_{n+1}) = π(p′_{n}). For dom(π(p_{n+1})) = j^{−1}(dom(p_{n+1}) ∩ j[κ^{+}]) = j^{−1}(dom(p′_{n}) ∩ j[κ^{+}]) = dom(π(p′_{n})), and for all γ in dom(π(p_{n+1})), π(p_{n+1})(γ) = p_{n+1}(j(γ)) ↾ κ^{+} = p′_{n}(j(γ)) ↾ κ^{2} = π(p′_{n})(γ). Since π(p′_{n}) is in I, π(p_{n+1}) is in I. Therefore p_{n+1} is in j(ℚ)/ℚ.

This completes the definition of the N-generic sequence ⟨p_{n} : n < ω⟩.

We define q, a lower bound of the sequence ⟨p_{n} : n < ω⟩, as follows. Let the domain of q be equal to ⋃{dom(p_{n}) : n < ω}. If β is in the domain of q, define q(β) as the union of the set of functions {p_{n}(β) : n < ω}. Clearly q is a partial function q : j(κ^{+}) → j(ℝ) with a countable domain. Furthermore, by the N-genericity of the sequence ⟨p_{n} : n < ω⟩, dom(q) = N ∩ j(κ^{+}), and for each β in dom(q), a_{β}^{q} = N ∩ j(κ).

We will prove that q is in j(ℚ)/ℚ. It then follows easily that q ≤ p_{n} for all n < ω, which completes the proof. The proof that q is in j(ℚ)/ℚ splits into two parts: (1) proving that q is in j(ℚ), and (2) proving that π(q) is in I. The second part can be handled quickly, so we take care of it first.

Claim 12.5. Suppose q is in j(ℚ). Then π(q) ∈ I, so q is in j(ℚ)/ℚ.

Proof. Assume q is in j(ℚ). Then π(q) is in ℚ. As a subclaim, we prove

dom(π(q)) = ⋃{dom(π(p_{n})) : n < ω},

and for all γ in dom(π(q)),

π(q)(γ) = ⋃{π(p_{n})(γ) : n < ω}.

We have that dom(π(q)) = j^{−1}(dom(q) ∩ j[κ^{+}]) = j^{−1}((⋃{dom(p_{n}) : n < ω}) ∩ j[κ^{+}]) = j^{−1}(⋃{dom(p_{n}) ∩ j[κ^{+}] : n < ω}) = ⋃{j^{−1}(dom(p_{n}) ∩ j[κ^{+}]) : n < ω} = ⋃{dom(π(p_{n})) : n < ω}. Now let γ be in dom(π(q)). Then π(q)(γ) = q(j(γ)) ↾ κ^{2} = (⋃{p_{n}(j(γ)) : n < ω}) ↾ κ^{2} = ⋃{p_{n}(j(γ)) ↾ κ^{2} : n < ω} = ⋃{π(p_{n})(γ) : n < ω}.

For each n < ω, p_{n} is in j(ℚ)/ℚ, and so π(p_{n}) is in I. Hence ⟨π(p_{n}) : n < ω⟩ is a descending sequence of conditions in I. By Lemma 9.10, there is r in I such that r ≤ π(p_{n}) for all n < ω. It follows from the subclaim that dom(π(q)) ⊆ dom(r), and for all γ in dom(π(q)), r(γ) ↾ a_{γ}^{π(q)} = π(q)(γ). Thus r ≤ π(q) in ℚ. Since r is in I and I is a filter, π(q) is in I. □

It remains to show that q is in j(ℚ). We defined the condition q in the model M[\overline{G} ∗ \overline{H}][I], whereas the forcing poset j(ℚ) is defined in M[\overline{G} ∗ \overline{H}]. But ℚ is ω_{1}-distributive in M[\overline{G} ∗ \overline{H}]. This is true since j(ℚ) is ω_{1}-distributive in M[\overline{G} ∗ \overline{H}], and π : j(ℚ) → ℚ is a projection mapping. It follows that q is in M[\overline{G} ∗ \overline{H}]. We will prove in the model M[\overline{G} ∗ \overline{H}] that q is in j(ℚ).

Suppose for a contradiction q is not in j(ℚ). Applying Lemma 9.3 in the model M[\overline{G} ∗ \overline{H}], we can fix β, t, and x such that:

• β is in dom(q),

• q ↾ β is in j(ℚ)_{β},

• t ≤ q ↾ β in j(ℚ)_{β},

• x is a countable subset of a_{β}^{q} closed under q(β),

• t ⊩_{j(ℚ)_{β}} x ∈ Ṫ^{j}(β).

Recall that ⊩_{j(ℚ)_{β}} Ṫ^{j}(β) ⊆ j(S_{κ}) ∩ j(E). Therefore, since t ⊩_{j(ℚ)_{β}} x ∈ Ṫ^{j}(β), x is in j(S_{κ}) ∩ j(E). By Theorem 6.1, j(S_{κ}) = S_{j(κ)} = S_{κ} ∪ S(κ, S_{κ}, j(κ)). So there are two possibilities concerning x: either x ⊆ κ and x ∈ S_{κ}, or x ⊈ κ and x ∈ S(κ, S_{κ}, j(κ)).

Note that β is in N, since dom(q) = N ∩j(κ^{+}). Since q ↾ β is in j(ℚ)_{β}, it is easy to see that q ↾ β ≤ p_{n} ↾ β for all n < ω. In particular, q ↾ β is N-generic for j(ℚ)_{β}.

Claim 12.6. For all n < ω, if β ∈ dom(p_{n}), then x is not a subset of a_{β}^{p_{n}}.

Proof. Suppose x is a subset of a_{β}^{p_{n}}. Since q(β) extends p_{n}(β) as a function, x is closed under p_{n}(β). By Lemma 9.2, p_{n} ↾ β ⊩_{j(ℚ)_{β}} x ∉ Ṫ^{j}(β). But t ≤ q ↾ β ≤ p_{n} ↾ β, so t ⊩_{j(ℚ)_{β}} x ∉ Ṫ^{j}(β), contradicting the choice of t. □

Claim 12.7. The set x is not in N.

Proof. Let D be the dense open set of conditions s in j(ℚ)/ℚ such that β is in dom(s) and x ⊆ a_{β}^{s}. If x is in N, then by elementarity, D is in N. By N-genericity, fix n < ω such that p_{n} is in D. Then x ⊆ a_{β}^{p_{n}}, contradicting the last claim. □

Claim 12.8. The set x is in S_{κ}.

Proof. If x is not in S_{κ}, then x is in S(κ, S_{κ}, j(κ)). Since q(β) = ⋃{p_{n}(β) : n < ω}, a_{β}^{q} = ⋃{a_{β}^{p_{n}} : n < ω}, which by N-genericity is equal to N ∩ j(κ). Since x ⊆ a_{β}^{q}, x is a subset of N ∩ j(κ). Therefore x is in S(κ, S_{κ}, j(κ)) ∩ P (N ∩ j(κ)). By the choice of N, as described in Notation 12.4, S(κ, S_{κ}, j(κ)) ∩ P (N ∩ j(κ)) ⊆ N. Thus x is in N. But this contradicts Claim 12.7. □

So x is a subset of κ. Also x is a subset of a_{β}^{q}, which is equal to N ∩ j(κ). So x is a subset of N ∩ κ.

Let j(⟨g_{α} : α < κ⟩) = ⟨g_{δ}^{j} : δ < j(κ)⟩. For all α < κ, g_{α}^{j} = g_{j(α)}^{j} = j(g_{α}), which is equal to g_{α}, since g_{α} is a subset of ω_{1} × α. Since x is in j(E), it follows that for all α in x, x is closed under g_{α}.

Claim 12.9. The ordinal β is not in j[κ^{+}].

Proof. Suppose for a contradiction β is in j[κ^{+}]. Let γ = j^{−1}(β), so j(γ) = β. Then γ is in j^{−1}(dom(q) ∩ j[κ^{+}]) = dom(π(q)).

By Claim 12.8, x ⊆ κ. Since q(β) ↾ κ^{2} = π(q)(γ), x is a countable subset of a_{γ}^{π(q)} which is closed under π(q)(γ). Since M[\overline{G} ∗ \overline{H}] is a generic extension of M by an ω_{1}-distributive forcing poset, x is in M, and hence is in V [G ∗ H]. Applying Lemma 9.2 in V [G ∗ H], π(q) ↾ γ ⊩_{ℚ_{γ}} x ∉ Ṫ (γ).

By Lemma 11.4, π_{γ}(q ↾ β) = π(q) ↾ γ. So π_{γ}(q ↾ β) ⊩_{ℚ_{γ}} x ∉ Ṫ (γ). Now t ≤ q ↾ β, so π_{γ}(t) ≤ π_{γ}(q ↾ β). Hence π_{γ}(t) ⊩_{ℚ_{γ}} x ∉ Ṫ (γ). Since j(x) = x, j(ℚ_{γ}) = j(ℚ)_{β}, and j(Ṫ (γ)) = Ṫ^{j}(β), by the elementarity of j we have j(π_{γ}(t)) ⊩_{j(ℚ)_{β}} x ∉ Ṫ^{j}(β). On the other hand, t ⊩_{j(ℚ)_{β}} x ∈ Ṫ^{j}(β). But by Proposition 11.7, j(π_{γ}(t)) and t are compatible in j(ℚ)_{β}, which is a contradiction. □

We split the remainder of the proof into two cases. The first case is that x is an initial segment of N ∩ κ, and the second case is that x is not an initial segment of N ∩ κ. We will derive a contradiction in both cases. This shows that q is indeed a condition in j(ℚ), which completes the proof.

Case a: x is an initial segment of N ∩ κ.

Note that x cannot be equal to N ∩ κ itself. For by the choice of N, as described in Notation 12.4, N ∩ κ ∈ T. But x is in S_{κ}, and T is disjoint from S_{κ}.

Let ν be the least ordinal in N ∩ κ which is not in x. Then x = N ∩ ν. Recall that since t ⊩_{j(ℚ)_{β}} x ∈ Ṫ^{j}(β), x is in j(S_{κ}) ∩ j(E). By the definition of E, this implies ot(x) is a limit ordinal. It follows that ν is a limit ordinal.

Recall by Notation 12.3, D is a club subset of P_{ω_{1}}(κ) in M[\overline{G} ∗ \overline{H}][I] such that for all b in D, if γ is in b ∩ cof(ω_{1}), then b ∩ γ is not in S_{κ}. By the choice of N, as described in Notation 12.4, N ∩ κ is in D. Since N ∩ ν = x is in S_{κ}, ν is not in (N ∩ κ) ∩ cof(ω_{1}). So cf(ν) = ω.

In the model M[\overline{G}∗\overline{H}], the forcing poset j(ℚ)_{β} forces that Ṫ^{j}(β) does not reflect to any ordinal in j(κ) with cofinality ω. As ν has cofinality ω, we can fix a j(ℚ)_{β}- name Ḟ_{ν} for a function Ḟ_{ν} : [ν]^{<ω} → ν such that any b in Ṫ^{j}(β) which is a subset of ν is not closed under Ḟ_{ν}. Since j(ℚ), β, Ṫ^{j}(β), and ν are in N, by elementarity we can choose such a name Ḟ_{ν} in N.

As Ḟ_{ν} is in N, j(ℚ)_{β} forces that N[K̇] ∩ ν is closed under Ḟ_{ν}, where K̇ is the canonical j(ℚ)_{β}-name for a generic filter. Recall that q ↾ β ≤ p_{n} ↾ β for all n < ω, and so q ↾ β is N-generic for j(ℚ)_{β}. Since t ≤ q ↾ β, t is also N-generic. So t forces N[K̇] ∩ ν = N ∩ ν = x is closed under Ḟ_{ν}. But t ⊩_{j(ℚ)_{β}} x ∈ Ṫ^{j}(β), contradicting the choice of Ḟ_{ν}.

Case b: x is not an initial segment of N ∩ κ.

Fix ordinals γ < α < κ such that γ is in N \ x and α is in x. Recall that g_{α} is a surjection of ω_{1} onto α. Since α is in N, g_{α} is in N. As γ is in N ∩ α, by elementarity fix ξ in N ∩ ω_{1} such that g_{α}(ξ) = γ. Since α is in x, x is closed under g_{α}. Therefore ξ is not in x, since otherwise g_{α}(ξ) = γ would be in x.

Recall that ⟨x_{n} : n < ω⟩ is an enumeration of S_{κ} ∩ P (N ∩ κ) in which each set appears infinitely often. But x is in S_{κ}∩P (N∩κ). Since β is in ⋃{dom(p_{n}) : n < ω}, choose n large enough so that β is in dom(p_{n}) and x_{n} = x.

Now recall the three cases in the definition of p_{n+1}. The first case is that x_{n} ⊆ a_{β}^{p′_{n}}. This is false by Claim 12.6. The second case is that N ∩ ω_{1} ⊆ x_{n}. This is false, since ξ is in N ∩ ω_{1} but not in x. So we are in the third case in the definition of p_{n+1}. In that case we fixed an ordinal ν_{n} which is in x_{n} \ a^{p′_{n}}, and an ordinal ζ_{n} which is in (N ∩ ω_{1}) \ x_{n}. Then we chose p_{n+1} so that for all δ in dom(p_{n+1}) \ j[κ^{+}], ν_{n} is in a_{δ}^{p_{n+1}} and p_{n+1}(δ)(ν_{n}, ν_{n}) = ζ_{n}. By Claim 12.9, β is in dom(p_{n+1}) \ j[κ^{+}]. Therefore ν_{n} is in a_{β}^{p_{n+1}} and p_{n+1}(β)(ν_{n}, ν_{n}) = ζ_{n}. It follows that q(β)(ν_{n}, ν_{n}) = ζ_{n}. Since ν_{n} is in x_{n} and ζ_{n} is not in x_{n}, x_{n} is not closed under q(β). But x = x_{n} and x was assumed to be closed under q(β). Thus we have reached our final contradiction. □

13. The Main Theorem

We now have all the tools necessary to complete the proof of the main theorem of the paper.

Theorem 13.1. Suppose κ is a κ^{+}-supercompact cardinal and 2^{κ} = κ^{+}. Then there is a forcing poset ℙ, namely,

Coll(ω_{1}, <κ) ∗ ℙ̇_{κ} ∗ ℚ̇,

such that

⊩_{ℙ} WRP(ω_{2}) ∧ ¬RP(ω_{2}).

Proof. Let ℙ denote Coll(ω_{1}, < κ) ∗ ℙ̇_{κ} ∗ ℚ̇. Then ℙ is ω_{1}-distributive and ω_{2}- c.c. Let j : V → M be an elementary embedding with critical point κ such that j(κ) > κ^{+} and M^{κ^{+}} ⊆ M.

Let φ be the statement WRP(ω_{2}) ∧ ¬RP(ω_{2}). To show ⊩_{ℙ} φ, it suffices to show whenever \overline{G} ∗ \overline{H} ∗ \overline{I} is a generic filter on j(ℙ) over V, then letting G = \overline{G} ∩ Coll(ω_{1}, < κ), H = \overline{H} ∩ ℙ_{κ}, and I = π[\overline{I}], V [G ∗ H ∗ I] ⊧ φ. For if ⊩_{ℙ} φ is false, then there is a condition p ∗ q̇ ∗ ṙ in ℙ such that p ∗ q̇ ∗ ṙ ⊩_{ℙ} ¬φ. Choose \overline{G} ∗ \overline{H} ∗ \overline{I} which contains the condition j(p ∗ q̇ ∗ ṙ). By Theorems 6.1 and 11.11, extend j to j : V [G ∗ H ∗ I] → M[\overline{G} ∗ \overline{H} ∗ \overline{I}] such that j(G ∗ H ∗ I) = \overline{G} ∗ \overline{H} ∗ \overline{I}. Since j(p) = p is in \overline{G} = j(G), p is in G. As j(q) = q is in \overline{H} = j(H), q is in H. By Lemma 11.9, π(j(r)) = r is in π[\overline{I}] = I. So p ∗ q̇ ∗ ṙ is in G ∗ H ∗ I, and thus V [G ∗ H ∗ I] ⊧ ¬φ.

So let \overline{G} ∗ \overline{H} ∗ \overline{I} be a generic filter on j(ℙ) over V, and let G, H, and I be as above. We prove V [G ∗ H ∗ I] ⊧ φ. By Theorems 6.1 and 11.11, we can extend j to j : V [G ∗ H ∗ I] → M[\overline{G} ∗ \overline{H} ∗ \overline{I}] such that j(G ∗ H ∗ I) = \overline{G} ∗ \overline{H} ∗ \overline{I}. By Theorem 10.2, S_{κ} is a stationary subset of P_{ω_{1}}(ω_{2}) in V [G ∗ H ∗ I] which does not reflect to any ordinal in ω_{2} with cofinality ω_{1}. Therefore V [G ∗ H ∗ I] ⊧ ¬RP(ω_{2}).

It remains to show that V [G ∗ H ∗ I] ⊧ WRP(ω_{2}). So let U be a stationary subset of P_{ω_{1}}(ω_{2}) in V [G ∗ H ∗ I]. First assume U ∩ S_{κ} is stationary. Then by Theorem 10.2, there is an uncountable ordinal ν in ω_{2} with cofinality ω such that (U ∩ S_{κ}) ∩ P_{ω_{1}}(ν) is stationary in P_{ω_{1}}(ν). So U reflects to ν.

Now assume U ∩ S_{κ} is non-stationary. Let T = U \ S_{κ}. Then T is a stationary subset of P_{ω_{1}}(ω_{2}) in V [G ∗ H ∗ I] which is disjoint from S_{κ}. By Theorem 12.1, T is a stationary subset of P_{ω_{1}}(κ) in M[\overline{G} ∗ \overline{H} ∗ \overline{I}]. For any b in T, j(b) = b is in j(T) by the elementarity of j. So T ⊆ j(T) ∩ P_{ω_{1}}(κ) in M[\overline{G} ∗ \overline{H} ∗ \overline{I}] (in fact, these sets are equal). It follows that j(T) ∩ P_{ω_{1}}(κ) is stationary in P_{ω_{1}}(κ) in M[\overline{G} ∗ \overline{H} ∗ \overline{I}]. So M[\overline{G} ∗ \overline{H} ∗ \overline{I}] models the statement that j(T) reflects to an ordinal in ω_{2} with cofinality ω_{1}. By the elementarity of j, V [G ∗ H ∗ I] models the statement that T reflects to an ordinal in ω_{2} with cofinality ω_{1}. Since T ⊆ U, U reflects to an uncountable ordinal in ω_{2}. □

References

[1] M. Foreman, M. Magidor, and S. Shelah. Martin’s maximum, saturated ideals and non-regular ultrafilters I. Ann. of Math., 127:1–47, 1988.

[2] B. Koenig, P. Larson, and Y. Yoshinobu. Guessing clubs in the generalized club-filter. Fund. Math., 195(2):177–191, 2007.

[3] H. Sakai. Partial stationary reflection principle in P_{ω_{1}}(ω_{2}). In Axiomatic Set Theory and Set-Theoretic Topology, number 1595 in RIMS Kokyuroku, pages 47–62, 2008.

[4] S. Shelah. Around Classification Theory of Models, volume 1182 of Lecture Notes of Mathematics. Springer-Verlag, Berlin, 1986.

[5] S. Shelah. Diamonds. Preprint, 2008.

[6] S. Todorcevic. Reflecting stationary sets. Handwritten Notes, 1985.

[7] B. Veličković. Forcing axioms and stationary sets. Adv. Math., 94(2):256–284, 1992.

Department of Mathematics, University of North Texas, 1155 Union Circle #311430, Denton, TX 76203, USA

   E-mail address: jkrueger@unt.edu