RADIN FORCING AND ITS ITERATIONS

JOHN KRUEGER

Abstract. We provide an exposition of supercompact Radin forcing and present several methods for iterating Radin forcing.

In this paper we give an exposition of supercompact Radin forcing using coherent sequences of ultrafilters. This version of Radin forcing includes as special cases the Prikry forcing and Magidor forcing, both the measurable and supercompact versions. We also introduce some methods for iterating Radin forcing. First we show how to iterate Radin forcing over the same cardinal infinitely many times. Secondly we show that Magidor’s method of iterating Prikry forcing over different cardinals can be extended to iterate Radin forcing.

Radin forcing was introduced in [8]. Mitchell [7] presented a version of Radin forcing which uses coherent sequences of ultrafilters in place of a measure sequence. Foreman and Woodin [5] developed a supercompact version of Radin forcing using measure sequences in the context of a proof that GCH can fail for every cardinal. See [2] for a more recent exposition of Radin forcing on a measurable cardinal.

In Section 1 we review notation and prove some technical lemmas we need in the paper. Part I, consisting of Sections 2 to 7, is an exposition of supercompact Radin forcing using coherent sequences. Part II, consisting of Sections 8 and 9, presents two methods for iterating Radin forcing. Section 8 covers iterations of Radin forcing over the same cardinal. Section 9 extends Magidor’s method of iterating Prikry forcing over different cardinals to Radin forcing.

1. Notation and Background

We assume that the reader is familiar with forcing, Prikry forcing, and supercompact cardinals; see [3] or [4].

For cardinals κ ≤ λ with κ regular, let P_{κ}λ denote the set of a in [λ]^{<κ} such that a ∩ κ is an ordinal. Then P_{κ}λ is a club subset of [λ]^{<κ}. For a, b in P_{κ}λ, let a ⊂ ∼ b if a ⊆ b and |a| < b ∩ κ.

In this paper, normal ultrafilter means a normal, fine, non-principal ultrafilter on some P_{κ}λ. By fineness we mean that for all i < λ, the set {a ∈ P_{κ}λ : i ∈ a} is in the ultrafilter. Normality is the property that for any function F : P_{κ}λ → λ such that F (a) ∈ a for all a, there is β < λ such that the set {a : F (a) = β} is in the ultrafilter. If U is an ultrafilter on P_{κ}λ and A ⊆ κ, we say that U concentrates on A if the set {a ∈ P_{κ}λ : a ∩ κ ∈ A} is in U; equivalently, κ ∈ j(A) where j : V → Ult(V, U).

Date: January 2007.

This work was partially supported by FWF project number P16790-N04.

If U_{0} and U_{1} are normal ultrafilters on P_{κ}λ, let U_{0} ◁ U_{1} if U_{0} is in Ult(V, U_{1}) (i.e. U_{0} is less than U_{1} in the Mitchell ordering). This relation is transitive and well-founded.

The following lemma is straightforward.

Lemma 1.1. Suppose U is a normal ultrafilter on P_{κ}λ and j : V → Ult(V, U). If D is a subset of P_{κ}λ, then j(D) ∩ P_{κ}j“λ = j“D.

Suppose κ ≤ λ_{0} ≤ λ and U is an ultrafilter on P_{κ}λ. Let U ↾ λ_{0} be the ultrafilter on P_{κ}λ_{0} defined by letting X be in U ↾ λ_{0} iff X ⊆ P_{κ}λ_{0} and {a ∈ P_{κ}λ : a∩λ_{0} ∈ X} is in U. If U is normal and j : V → Ult(V, U), then X ∈ U ↾ λ_{0} iff j“λ_{0} ∈ j(X).

Suppose that κ is λ-supercompact. Then by a theorem of Solovay, λ^{<κ} is equal to λ^{+} if cf(λ) < κ and is equal to λ if cf(λ) ≥ κ. See [3] for a proof.

If M is an inner model, we say that M is α-closed if ^{α}M ⊆ M.

If ℙ is a forcing poset and p is in ℙ, let ℙ/p denote the poset consisting of conditions q ≤ p, ordered the same as in ℙ.

We will often use the following style of proof. Suppose ℙ is a forcing poset, G is generic for ℙ over V, and φ is a statement in the forcing language. Then V [G] ⊧ φ iff for all p in G, there is q ≤ p such that q forces φ.

A Prikry type forcing poset is a triple ⟨ℙ, ≤, ≤^{∗}⟩ such that ⟨ℙ, ≤⟩ and ⟨ℙ, ≤^{∗}⟩ are both forcing posets, q ≤^{∗} p implies q ≤ p, and the triple satisfies the Prikry property: if φ is a statement in the forcing language for ⟨ℙ, ≤⟩ and p is in ℙ, then there is q ≤^{∗} p such that q decides φ.

If φ is a statement in the forcing language for some forcing poset, let ^{0}φ denote φ and ^{1}φ denote ¬φ.

We will use the following theorem of Cummings and Shelah [1]. Suppose that λ is a cardinal in V and ℙ is a λ^{+}-c.c. forcing poset. Then ℙ forces that cf(λ) = cf(|λ|).

The following lemma generalizes a well-known characterization of the Mitchell ordering on a measurable cardinal.

Lemma 1.2. Let U_{0} and U_{1} be normal ultrafilters on P_{κ}λ. For each a in P_{κ}λ let π_{a} : a → o.t.(a) be the unique order preserving bijection. Then U_{0} ◁ U_{1} iff there exists a function f : P_{κ}λ → V_{κ} such that:

(1) {a ∈ P_{κ}λ : f(a) is a normal ultrafilter on P_{(a∩κ)} o.t.(a)} is in U_{1},

(2) For every X ⊆ P_{κ}λ, X ∈ U_{0} iff the set of a in P_{κ}λ such that

X_{a} = {π_{a}“c : c ∈ X ∩ P_{(a∩κ)}a} ∈ f(a)

is in U_{1}.

Proof. Write j : V → M = Ult(V, U_{1}). In M, [a ↦ a] = j“λ, [a ↦ a ∩ κ] = κ, and [a ↦ o.t.(a)] = λ. Let X be a subset of P_{κ}λ. We show that j(a ↦ X_{a})(j“λ) = X. Note that π_{j“λ} = j^{−1} ↾ j“λ and j(X) ∩ P_{κ}j“λ = j“X. So j(a ↦ X_{a})(j“λ) = X_{j“λ} = {j^{−1}“c : c ∈ j“X} = X. Therefore [a ↦ X_{a}] = X in M.

Suppose that U_{0} ◁ U_{1}. Let [f] = U_{0} in M. Clearly (1) holds. For (2), X is in U_{0} iff [a ↦ X_{a}] is in [f]. On the other hand suppose (1) and (2) are true. Then (2) implies that [f] = U_{0} in M, so U_{0} ◁ U_{1}. □

Suppose that M is a transitive inner model of set theory which is λ^{<κ}-closed. Then by Lemma 1.2, if U_{0} and U_{1} are normal ultrafilters on P_{κ}λ which are in M, then U_{0} ◁ U_{1} iff M ⊧ U_{0} ◁ U_{1}.

The next lemma is a standard result about supercompact cardinals.

Lemma 1.3. Suppose that κ ≤ λ are cardinals. Assume j : V → M is an elementary embedding such that M is a 2^{(λ^{<κ})}-closed inner model, crit(j) = κ, and j(κ) > λ. Let U^{∗} be the normal ultrafilter on P_{κ}λ defined by letting X be in U^{∗} if X ⊆ P_{κ}λ and j“λ ∈ j(X). Write i : V → N = Ult(V, U^{∗}). Then there is an elementary embedding k : N → M such that j = k ∘ i and crit(k) = (2^{(λ^{<κ})})^{+N}.

Proof. Define k as follows. Let a be in N and let f : P_{κ}λ → V be a function such that [f]_{N} = a. Define k(a) = j(f)(j“λ). It is straightforward to check that k is a well-defined elementary embedding and j = k ∘ i.

If β ≤ λ, then β is represented by the function a ↦ o.t.(a ∩ β) in N. By the definition of k, k(β) = o.t.(j“λ ∩ j(β)) = o.t.(j“β) = β in M. So crit(k) > λ.

Since N and M are λ^{<κ}-closed and crit(k) > λ, k(P_{κ}λ) = P_{κ}λ and k(P(P_{κ}λ)) = P(P_{κ}λ). If a is in P_{κ}λ, then k(a) is in k(P_{κ}λ) = P_{κ}λ, and α ∈ k(a) iff k(α) ∈ k(a) iff α ∈ a. Therefore k ↾ P_{κ}λ is the identity. By this fact and a similar argument, k ↾ P(P_{κ}λ) is the identity.

We prove by induction that for all β less than (2^{(λ^{<κ})})^{+N}, k(β) = β. Fix β and suppose k(α) = α for all α less than β. Since β is less than (2^{(λ^{<κ})})^{+N}, there is a surjective function s : P(P_{κ}λ) → β in N. By elementarity, k(s) is a surjection of k(P(P_{κ}λ)) = P(P_{κ}λ) onto k(β). If A is in P(P_{κ}λ), then k(s)(A) = k(s)(k(A)) = k(s(A)) = s(A), so k(s) = s. Therefore k(β) = β.

Now (2^{(λ^{<κ})})^{+N} < i(κ) < (2^{(λ^{<κ})})^{+} = (2^{(λ^{<κ})})^{+M}, and k maps (2^{(λ^{<κ})})^{+N} to (2^{(λ^{<κ})})^{+M}. So crit(k) = (2^{(λ^{<κ})})^{+N}. □

Lemma 1.4. Suppose that κ ≤ λ are cardinals, M is a λ^{<κ}-closed inner model, and U is a normal ultrafilter on P_{κ}λ in M. Let j_{U} : V → N_{U} = Ult(V, U) and j_{U}^{M} : M → M_{U} = Ult(M, U).

(1) If f is in ^{(P_{κ}λ)}M, then [f]_{N_{U}} = [f]_{M_{U}}. Therefore M_{U} ⊆ N_{U}.

(2) j_{U} ↾ M = j_{U}^{M}.

Proof. We prove (1) by induction on the rank of [f]_{M_{U}}. Suppose that f is in (P_{κ}λ)M and (1) holds for all functions g such that [g]_{M_{U}} has rank less than the rank of [f]_{M_{U}}. By the closure of M, f is in M so [f]_{M_{U}} is defined. We show [f]_{M_{U}} = [f]_{N_{U}}. Suppose that x is in [f]_{N_{U}}. Then there is g in ^{(P_{κ}λ)}M such that [g]_{N_{U}} = x and g(a) ∈ f(a) for almost all a. Clearly [g]_{M_{U}} is in [f]_{M_{U}}, and by induction [g]_{M_{U}} = x. The other direction is similar.

To prove (2), let x be in M. Then j_{U}(x) = [f]_{N_{U}}, where f(a) = x for all a in P_{κ}λ. Since f is in ^{(P_{κ}λ)}M, by (1) we have [f]_{N_{U}} = [f]_{M_{U}}, and clearly [f]_{M_{U}} = j_{U}^{M}(x).

□

Suppose that κ is a supercompact cardinal. Then there exists a function f : κ → V_{κ}, known as a Laver function, such that whenever x is a set and λ ≥ κ is a cardinal such that λ ≥ |tr(x)|, there is a normal ultrafilter U on P_{κ}λ with elementary embedding j : V → Ult(V, U) such that j(f)(κ) = x. We will also need the following variant of this fact.

Lemma 1.5. Suppose that κ ≤ λ are cardinals, X is a subset of κ, and there is a normal ultrafilter on P_{κ}2^{2^{(λ^{<κ})}} which concentrates on X. Then there is a function f : κ → κ and an ultrafilter on P_{κ}2^{(λ^{<κ})} which concentrates on X with elementary embedding j : V → M such that j(f)(κ) = λ.

Proof. Define f : κ → κ by recursion. Suppose that f ↾ α is defined. Assume there is a cardinal β with α ≤ β < κ such that for every normal ultrafilter on P_{α}2^{(β^{<α})} concentrating on X ∩ α with elementary embedding j : V → M, j(f ↾ α)(α) is not equal to β. Let f(α) be the least such β. Otherwise let f(α) = 0.

Suppose for a contradiction that the statement of the lemma fails for f. Let λ_{κ} be the least cardinal such that for every normal ultrafilter on P_{κ}2^{(λ_{κ}^{<κ})} concentrating on X with elementary embedding i : V → M, i(f)(κ) is not equal to λ_{κ}. By assumption, λ_{κ} exists and λ_{κ} ≤ λ.

Let U be a normal ultrafilter on P_{κ}2^{2^{(λ^{<κ})}} concentrating on X and write j : V → M = Ult(V, U). By the closure of M, the definition of f, and Lemma 1.4(2), j(f)(κ) = λ_{κ}. Let U^{∗} = U ↾ 2^{(λ^{<κ})} and write i : V → N = Ult(V, U^{∗}). Note that U^{∗} concentrates on X. Apply Lemma 1.3 to obtain k : N → M such that j = k ∘ i and crit(k) = (2^{2^{(λ^{<κ})}})^{+N}. Then k(i(f)(κ)) = j(f)(κ) = λ_{κ}, so i(f)(κ) must be equal to λ_{κ}. This contradicts the definition of λ_{κ}. □

PART I: SUPERCOMPACT RADIN FORCING

2. Coherent Sequences of Supercompact Ultrafilters

Let κ ≤ λ be cardinals. A coherent sequence of ultrafilters on P_{κ}λ is a sequence

U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩,

for some partial function oU⃗ : κ + 1 → On satisfying the following properties. Let g : κ → V be the map g(α) = ⟨U(α, i) : i < oU⃗(α)⟩.

(1) There is a map β ↦ λ_{β} defined on the domain of oU⃗ such that λ_{κ} = λ, each λ_{β} ≥ β is a cardinal, and whenever oU⃗(α) is defined, α is closed under this map. Moreover, for α in the domain of oU⃗ and i < oU⃗(α), the set {a ∈ P_{α}λ_{α} : o.t.(a) = λ_{a∩α}} is in U(α, i).

(2) Each g(α) = ⟨U(α, i) : i < oU⃗(α)⟩ is a sequence of normal ultrafilters on P_{α}λ_{α}.

(3) Each U(α, i) concentrates on α ∩ dom(oU⃗), and every α in the domain of oU⃗ is strongly inaccessible.

(4) Fix β < oU⃗(α) and let j_{β}^{α} : V → Ult(V, U(α, β)). Then j_{β}^{α}(g ↾ α)(α) = ⟨U(α, i) : i < β⟩.

In Part II of the paper we will need to know that coherent sequences exist under certain conditions. The next two lemmas provide this information.

Proposition 2.1. Suppose that ⟨U_{i} : i < β⟩ is a sequence of normal ultrafilters on P_{κ}λ for some β ≤ κ such that:

(1) U_{i} ↾ κ ≠ U_{j} ↾ κ for i < j < β,

(2) For all α_{0} < α_{1} < β there exists a function f_{α_{0}}^{α_{1}} : κ → V_{κ} such that, letting j_{α_{1}} : V → Ult(V, U_{α_{1}}), j_{α_{1}}(f_{α_{0}}^{α_{1}})(κ) = U_{α_{0}},

(3) There is a function h_{0} : κ → κ such that j_{0}(h_{0})(κ) = λ. Then there is a coherent sequence U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ of ultrafilters on P_{κ}λ such that oU⃗(κ) = β and U(κ, i) = U_{i} for i < β.

Moreover, suppose X is a subset of κ such that each U_{i} concentrates on X. Then we can choose U⃗ so that each U(α, i) concentrates on X ∩ α, and the domain of oU⃗ is a subset of X.

Proof. For α_{0} < α_{1} < β define a partial function h_{α_{0}}^{α_{1}} : κ → κ as follows. The domain of h_{α_{0}}^{α_{1}} is the set of γ in X such that f_{α_{0}}^{α_{1}}(γ) is a normal ultrafilter on P_{γ}μ for some cardinal γ ≤ μ < κ. If γ is such an ordinal then let h_{α_{0}}^{α_{1}}(γ) = μ.

We construct a sequence ⟨X_{i} : i < β⟩ of subsets of κ such that for i < j < β, X_{i} ∩ X_{j} = ∅ and X_{i} is in U_{i} ↾ κ. By (1), for distinct i and j below β we can choose X_{i}(j) in (U_{i} ↾ κ) \ (U_{j} ↾ κ). Let X_{i}(i) = κ. Define X_{i}^{∗} = △{X_{i}(j) : j < β}. Finally let X_{i} = X_{i}^{∗} \ ⋃{X_{j}^{∗} : j < i}.

Now define a sequence ⟨A_{i} : i < β⟩ of disjoint subsets of κ using the following recursive definition: α is in A_{i} if

(a) α is in X_{i} ∩ X, α > i, and α is strongly inaccessible,

(b) if ξ < min({α, β}), then α is closed under the mapping h_{0}^{ξ} ↾ α,

(c) for all β_{0} < β_{1} < i, h_{β_{0}}^{i}(α) = h_{β_{1}}^{i}(α), and if i = 0 then h_{0}(α) ≥ α is a cardinal,

(d) for ξ < i, f_{ξ}^{i}(α) is a normal ultrafilter on P_{α}h_{ξ}^{i}(α) which concentrates on the sets A_{ξ} ∩ α and X ∩ α,

(e) for β_{0} < β_{1} < i, letting k : V → Ult(V, f_{β_{1}}^{i}(α)), k(f_{β_{0}}^{β_{1}} ↾ α)(α) = f_{β_{0}}^{i}(α). By recursion it is straightforward to check that each A_{i} is in U_{i} ↾ κ; in particular,

(e) follows from Lemma 1.4(2).

Now define U⃗ as follows. For α < κ let oU⃗(α) = i iff α is in A_{i}, and let oU⃗(κ) = β.

For α < κ and ξ < oU⃗(α), let U(α, ξ) = f_{ξ}^{oU⃗ (α)}(α), and let U(κ, ξ) = U_{ξ}. For α ≤ κ

let g(α) = ⟨U(α, i) : i < oU⃗(α)⟩. Define λ_{α} = h_{0}^{oU⃗ (α)}(α) if oU⃗(α) > 0 and let λ_{α} = h_{0}(α) if oU⃗(α) = 0.

Fix α < κ and ξ < oU⃗(α). Let k : V → Ult(V, U(α, ξ)). By (d), α is in k(A_{ξ}), so k(oU⃗)(α) = k(ξ) = ξ. By the definition of U⃗ and (e), k(g ↾ α)(α) =

⟨k(f_{i}^{ξ})(α) : i < ξ⟩ = ⟨f_{i}^{oU⃗ (α)}(α) : i < ξ⟩ = ⟨U(α, i) : i < ξ⟩. Similarly, if ξ < β and j : V → Ult(V, U(κ, ξ)), then j(g)(κ) = ⟨U_{i} : i < ξ⟩. □

In particular, if there is a ◁-increasing sequence ⟨U_{i} : i < β⟩ of normal ultrafilters on a cardinal κ, where β ≤ κ, then there is a coherent sequence U⃗ on κ with oU⃗(κ) = β. We do not know whether, for example, the existence of a ◁-increasing sequence of ultrafilters on κ with length κ^{+} implies the existence of a coherent sequence U⃗ with oU⃗(κ) = κ^{+}.

If U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ is a coherent sequence, then let U⃗ ↾ (β + 1) denote the coherent sequence ⟨U(α, i) : α ≤ β, i < oU⃗(α)⟩. The next result gives a sufficient condition for a coherent sequence with the maximum possible length.

Proposition 2.2. Suppose that κ ≤ λ are cardinals. Assume that there is an elementary embedding j : V → M, where M is a 2^{(λ^{<κ})}-closed inner model, such that the critical point of j is κ, j(κ) > λ, and there is a function f : κ → κ such that j(f)(κ) = λ. Then there is a coherent sequence U⃗ of ultrafilters on P_{κ}λ such that oU⃗(κ) = (2^{(λ^{<κ})})^{+} and j(U⃗) ↾ (κ + 1) = U⃗.

Assume moreover that X is a subset of κ such that κ is in j(X). Then we can choose U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ so that each U(α, i) concentrates on X ∩ α and the domain of oU⃗ is a subset of X.

Proof. Without loss of generality assume that for all α, α ≤ f(α) and f(α) is a cardinal. Write f(α) = λ_{α}.

Define g : κ → V_{κ} by induction. Suppose that α < κ and g ↾ α is defined. If α is not a strongly inaccessible cardinal in X closed under f, then oU⃗(α) and g(α) are not defined. Otherwise let X_{α} denote the set of increasing sequences ⟨U_{i} : i < γ⟩ of normal ultrafilters on P_{α}λ_{α}, with each U_{i} concentrating on X ∩ α, such that for all β < γ, letting j_{β} : V → Ult(V, U_{β}), j_{β}(g ↾ α)(α) = ⟨U_{i} : i < β⟩. Note that X_{α} is closed under unions of chains and contains the empty set. By Zorn’s Lemma, let g(α) be a maximal element of X_{α}. Write g(α) = ⟨U(α, i) : i < oU⃗(α)⟩.

Now define oU⃗(κ) = j(oU⃗ ↾ κ)(κ). Write

j(g)(κ) = ⟨U(κ, i) : i < oU⃗(κ)⟩.

Note that each U(κ, i) concentrates on X. By the closure of M and Lemma 1.4(2), U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ is a coherent sequence of ultrafilters on P_{κ}λ.

We show that oU⃗(κ) = (2^{(λ^{<κ})})^{+}. For each i < oU⃗(κ), any ultrafilter on P_{κ}λ in Ult(V, U(κ, i)) is represented by a function P_{κ}λ → V_{κ}. So there are at most 2^{(λ^{<κ})} many ultrafilters less than U(κ, i) in the Mitchell ordering. Therefore oU⃗(κ) ≤ (2^{(λ^{<κ})})^{+}.

Suppose for a contradiction that oU⃗(κ) < (2^{(λ^{<κ})})^{+}, and we will show that j(g)(κ) = ⟨U(κ, i) : i < oU⃗(κ)⟩ is not maximal in M. So we will find a normal ultrafilter U^{∗} on P_{κ}λ in M concentrating on X such that, letting j^{∗} : M → Ult(M, U^{∗}), j^{∗}(j(g) ↾ κ)(κ) = j^{∗}(g)(κ) = ⟨U(κ, i) : i < oU⃗(κ)⟩. Define U^{∗} by letting Y be in U^{∗} iff Y ⊆ P_{κ}λ and j“λ ∈ j(Y). Then U^{∗} is a normal ultrafilter on P_{κ}λ which concentrates on X. By the closure of M, U^{∗} is in M. Let i : V → N = Ult(V, U^{∗}). By Lemma 1.3 there is an elementary embedding k : N → M with critical point (2^{(λ^{<κ})})^{+N} such that j = k ∘ i.

Write i(g)(κ) = ⟨U_{α}^{∗} : α < γ⟩. Note that k(U_{α}^{∗}) = U_{α}^{∗} for each α. Now k maps i(g)(κ) to ⟨U(κ, i) : i < oU⃗(κ)⟩, which by assumption has length less than (2^{(λ^{<κ})})^{+}. So i(g)(κ) has length less than (2^{(λ^{<κ})})^{+N} = crit(k). It follows that i(g)(κ) has length oU⃗(κ). Therefore i(g)(κ) = ⟨U(κ, α) : α < oU⃗(κ)⟩. If j^{∗} : M → Ult(M, U^{∗}), then by Lemma 1.4(2), j^{∗}(g) = i(g), so j^{∗}(g)(κ) = ⟨U(κ, α) : α < oU⃗(κ)⟩. □

3. Supercompact Radin Forcing

Let κ ≤ λ be cardinals and suppose that we have a coherent sequence

U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩

of ultrafilters on P_{κ}λ. Let α ↦ λ_{α} be the map such that each U(α, i) is a normal ultrafilter on P_{α}λ_{α}. We define a Radin forcing ℝ(U⃗) = ℝ.

For each a in P_{κ}λ let π_{a} : a → o.t.(a) denote the unique order preserving bijection. Define AU⃗ as the set of a in P_{κ}λ such that oU⃗(a ∩ κ) is defined and o.t.(a) = λ_{a∩κ}. Note that if a ⊂ ∼ b and a and b are in AU⃗, then λ_{a∩κ} < b ∩ κ.

A condition in ℝ is a pair

⟨d⃗, A⟩ = ⟨⟨d_{1}, …, d_{n}⟩, A⟩

such that:

(1) A is in ⋂{U(κ, i) : i < oU⃗(κ)} and A ⊆ AU⃗,

(2) d⃗ is a finite sequence, possibly empty,

(3) For each 0 < i ≤ n, either d_{i} is a set a_{i} in AU⃗ with oU⃗(a_{i} ∩ κ) = 0, or d_{i} is a pair ⟨a_{i}, A_{i}⟩ where a_{i} is in AU⃗, oU⃗(a_{i} ∩ κ) > 0, A_{i} ⊆ AU⃗ ∩ P_{(a_{i}∩κ)}a_{i}, and the set {π_{a_{i}}“c : c ∈ A_{i}} is in ⋂{U(a_{i} ∩ κ, β) : β < oU⃗(a_{i} ∩ κ)},

(4) For 0 < i < j ≤ n, a_{i} ⊂ ∼ a_{j}, and if d_{j} = ⟨a_{j}, A_{j}⟩, then for all a in A_{j}, a_{i} ⊂ ∼ a;

also, for all a in A, a_{n} ⊂ ∼ a.

We will abbreviate a condition ⟨∅, A⟩ as ⟨A⟩ and ⟨⟨d_{1}, …, d_{n}⟩, A⟩ as ⟨d_{1}, …, d_{n}, A⟩. Note that in the definition of ℝ, n is the size of d⃗. We will sometimes write a condition ⟨d⃗, A⟩ = ⟨d_{1}, …, d_{n}, A⟩ even when d⃗ might be empty, in which case n = 0.

Now we define the ordering on ℝ. Suppose that

p = ⟨d_{1}, …, d_{n}, A⟩

and

q = ⟨e_{1}, …, e_{m}, B⟩

are conditions in ℝ. Let q ≤ p if:

(1) m ≥ n,

(2) B ⊆ A,

(3) if n > 0, then there are i_{1} < … < i_{n} ≤ m such that for each 0 < k ≤ n, (a) if d_{k} is a set in P_{κ}λ, then e_{i_{k}} = d_{k},

(b) if d_{k} = ⟨a_{k}, A_{k}⟩, then e_{i_{k}} = ⟨a_{k}, B_{k}⟩, where B_{k} ⊆ A_{k},

(4) for each 0 < l ≤ m not equal to any i_{k},

(a) if n = 0 or if l > i_{n}, then either e_{l} is in A, or e_{l} = ⟨b_{l}, B_{l}⟩ where b_{l} is in A and B_{l} is a subset of A,

(b) if n > 0 and k is least such that l < i_{k}, then d_{k} is of the form ⟨a_{k}, A_{k}⟩, and either e_{l} is in A_{k}, or e_{l} = ⟨b_{l}, B_{l}⟩ where b_{l} is in A_{k} and B_{l} is a subset of A_{k}. If q ≤ p and n = m, then we let q ≤^{∗} p and say that q is a direct extension of p.

We will prove in Section 5 that ⟨ℝ, ≤, ≤^{∗}⟩ is a Prikry type forcing poset.

Note that ⟨AU⃗⟩ is the maximum element of ℝ, which we will denote by 1_{ℝ}. If p ≤^{∗} 1_{ℝ} then p = ⟨B⟩ for some B.

The following lemma is immediate from the completeness of the ultrafilters.

Lemma 3.1. Let p = ⟨d_{1}, … d_{n}, A⟩ be a condition in ℝ. Suppose that ξ < κ is an ordinal such that every d_{m} of the form ⟨a_{m}, A_{m}⟩ satisfies that ξ < a_{m} ∩ κ. Then for any collection {p_{i} : i < ξ} of direct extensions of p, there is q such that q ≤^{∗} p_{i} for i < ξ.

Similarly, if p and q are two conditions with the same finite sequence of elements from P_{κ}λ, then p and q are compatible. The next lemma follows from this observation.

Proposition 3.2. The poset ℝ is (λ^{<κ})^{+}-c.c.

The following lemma says in effect that if p is a condition in which A appears as a measure one set, then there are many ways to extend p to include elements from A.

Lemma 3.3. Suppose that A is in ⋂{U(κ, i) : i < oU⃗(κ)}. Let X be the set of a in A such that the set

{π_{a}“c : c ∈ A ∩ P_{(a∩κ)}a}

is in ⋂{U(a ∩ κ, i) : i < oU⃗(a ∩ κ)}. Then X is in ⋂{U(κ, i) : i < oU⃗(κ)}

Suppose that ⟨b, B⟩ appears in p for some p in ℝ. Let Y be the set of a in P_{(b∩κ)}b such that the set {π_{a}“c : c ∈ B ∩ P_{(a∩κ)}a} is in ⋂{U(a ∩ κ, i) : i < oU⃗(a ∩ κ)}. Then the set {π_{b}“c : c ∈ Y } is in ⋂{U(b ∩ κ, i) : i < oU⃗(b ∩ κ)}.

Proof. Suppose that β < oU⃗(κ) and write j_{β} : V → Ult(V, U(κ, β)). We prove that j_{β}“λ ∈ j_{β}(X). Note that π_{j_{β}“λ} = j_{β}^{−1} ↾ j_{β}“λ. By Lemma 1.1, j_{β}(A) ∩ P_{κ}(j_{β}“λ) = j_{β}“A, and for all j_{β}(b) in this set, (π_{j“λ})“j_{β}(b) = j_{β}^{−1}“(j_{β}“b) = b. Therefore

{π_{(j_{β}“λ)}“c : c ∈ j_{β}(A) ∩ P_{κ}(j_{β}“λ)} = A,

which is in ⋂{U(κ, i) : i < β} = ⋂{U(j_{β}“λ ∩ j_{β}(κ), i) : i < oU⃗(j_{β}“λ ∩ j_{β}(κ))}. So j_{β}“λ is in j_{β}(X).

Fix ⟨b, B⟩ as in the second statement. Let β < oU⃗(b ∩ κ) and write j_{β} : V → Ult(V, U(b ∩ κ, β)). We show that j_{β}“(λ_{(b∩κ)}) is in the set j_{β}({π_{b}“c : c ∈ Y }). Let d = j_{β}“b. We claim that j_{β}(π_{b})“d = j_{β}“(λ_{(b∩κ)}). For ξ is in j_{β}(π_{b})“d iff there is γ in b such that ξ = j_{β}(π_{b})(j_{β}(γ)) = j_{β}(π_{b}(γ)) iff ξ is in j_{β}“(λ_{(b∩κ)}). So it suffices to show that d is in j_{β}(Y). Note that π_{d} = (π_{b} ∘ j_{β}^{−1}) ↾ j_{β}“b and j_{β}(B) ∩ P_{(d∩j_{β}(κ))}d = j_{β}“B. So

{π_{d}“c : c ∈ j_{β}(B) ∩ P_{(d∩j_{β}(κ))}d} = {(π_{b} ∘ j_{β}^{−1})“c : c ∈ j_{β}“B} = {π_{b}“e : e ∈ B},

which is in ⋂{U(b ∩ κ, i) : i < β} = ⋂{U(d ∩ j_{β}(κ), i) : i < j_{β}(oU⃗)(d ∩ j_{β}(κ))}. □

4. Factorization

Let p = ⟨d_{1}, …, d_{n}, A⟩ be a condition in ℝ. Fix 0 < m ≤ n and suppose d_{m} = ⟨a_{m}, A_{m}⟩ (so in particular, oU⃗(a_{m} ∩κ) > 0). Define p^{>m} = ⟨d_{m+1}, …, d_{n}, A⟩. Clearly p^{>m} is in ℝ. Define p^{≤m} = ⟨e_{1}, …, e_{m−1}, B⟩ as follows. Let B = {π_{a_{m}}“c : c ∈ A_{m}}, which is in ⋂{U(a_{m}∩κ, i) : i < oU⃗(a_{m}∩κ)}. Fix 0 < i < m. If d_{i} is in P_{κ}λ, then oU⃗(d_{i} ∩κ) = 0 and d_{i} is in P_{(a_{m}∩κ)}a_{m}. In this case let e_{i} = π_{a_{m}}“d_{i}. Otherwise d_{i} = ⟨a_{i}, A_{i}⟩. Then a_{i} is in P_{(a_{m}∩κ)}a_{m}. Let e_{i} = ⟨b_{i}, B_{i}⟩, where b_{i} = π_{a_{m}}“a_{i} and B_{i} = {π_{a_{m}}“c : c ∈ A_{i}}. The reader can check that p^{≤m} is a condition in the Radin forcing ℝ(U⃗ ↾ ((a_{m} ∩ κ) + 1)). We will abbreviate ℝ(U⃗ ↾ ((a_{m} ∩ κ) + 1))/p^{≤m} as ℝ/p^{≤m}.

Lemma 4.1 (Factorization Lemma I). Suppose that p = ⟨d_{1}, …, d_{n}, A⟩ is a condition in ℝ, 0 < m ≤ n, and d_{m} = ⟨a_{m}, A_{m}⟩. Then there is

i : ℝ/p → ℝ/p^{≤m} × ℝ/p^{>m}.

which is an isomorphism with respect to both ≤ and ≤^{∗}.

Suppose that q = ⟨e_{1}, …, e_{m}, B⟩ is a condition in ℝ such that for all 0 < i ≤ m, e_{i} is in P_{κ}λ. Then there is

j : ℝ/q → ℝ/⟨B⟩

which is an isomorphism with respect to both ≤ and ≤^{∗}.

Proof. Define i as follows. Let u = ⟨f_{1}, …, f_{l}, C⟩ be a condition in ℝ below p. Fix j ≤ l with f_{j} = ⟨a_{m}, C_{j}⟩. Define i(u) = ⟨u^{≤j}, u^{>j}⟩.

Now we define j. Suppose that s = ⟨g_{1}, …, g_{k}, D⟩ is below q. Then the sequence g_{1}, …, g_{m} is exactly e_{1}, …, e_{m}. Let j(s) = ⟨g_{m+1}, … g_{k}, D⟩. □

5. The Prikry Property

Now we turn to proving that ℝ satisfies the Prikry property. First we show that it suffices to consider only direct extensions of 1_{ℝ}.

Lemma 5.1. Suppose that for any coherent sequence U⃗, if φ is a statement in the forcing language for ℝ(U⃗) and p ≤^{∗} 1_{ℝ(U⃗)}, then there exists q ≤^{∗} p which decides

φ. Then for any coherent sequence U⃗, ℝ(U⃗) satisfies the Prikry property.

Proof. We prove the lemma by induction. Suppose that for all μ < κ, if U⃗ is a coherent sequence on P_{μ}λ for some λ, then ℝ(U⃗) satisfies the Prikry property. Let U⃗ be a coherent sequence on P_{κ}λ for some λ and let ℝ = ℝ(U⃗). Consider a condition p = ⟨d_{1}, …, d_{n}, A⟩ in ℝ with n > 0.

Suppose first that each d_{i} is in P_{κ}λ. Then by Lemma 4.1 there is an isomorphism j : ℝ/p ≅ ℝ/⟨A⟩. Note that j(p) = ⟨A⟩, which is a direct extension of 1_{ℝ}. By assumption ⟨A⟩ can be directly extended to decide any particular forcing statement, so p can as well.

Otherwise fix m ≤ n maximal such that d_{m} = ⟨a_{m}, A_{m}⟩. Combining the maps i and j in Lemma 4.1, there is an isomorphism k : ℝ/p ≅ ℝ/p^{≤m} × ℝ/⟨A⟩. Let φ be a statement in the forcing language for ℝ.

Sublemma 5.2. There is a dense set D of r in ℝ/p^{≤m} for which there exists t ≤^{∗} ⟨A⟩ such that ⟨r, t⟩ decides φ.

Proof. Let q in ℝ/p^{≤m} be given. Reverse the order of the product to ℝ/⟨A⟩ × ℝ/p^{≤m}. Let ṡ be an ℝ/⟨A⟩-name for a condition below q which decides φ over the generic extension by ℝ/⟨A⟩. For each r ≤ q let ⟨A(r)⟩ be a direct extension of ⟨A⟩ which decides whether ṡ = r. Since |ℝ/p^{≤m}| < κ, t = ⟨⋂{A(r) : r ≤ q}⟩ is a condition. Clearly there must exist r ≤ q such that t forces ṡ = r. Fix t′ ≤^{∗} t which decides for some l < 2 that r forces ^{l}φ. Then ⟨t′, r⟩ forces ^{l}φ, and therefore ⟨r, t′⟩ forces ^{l}φ. □

For each r in the dense set D described in the sublemma, choose ⟨B(r)⟩ directly below ⟨A⟩ such that ⟨r, ⟨B(r)⟩⟩ decides φ. Let B = ⋂{B(r) : r ∈ D}. Then ℝ/p^{≤m} forces that B decides φ. By induction the poset ℝ/p^{≤m} satisfies the Prikry property. So find q ≤^{∗} p^{≤m} which decides which way that B decides φ. Then ⟨q, ⟨B⟩⟩ decides φ. It follows that k^{−1}(⟨q, ⟨B⟩⟩) is a direct extension of p which decides φ. □

We will need the next two lemmas to prove the Prikry property. For the remainder of this section let I denote the set of sequences d⃗ = ⟨d_{1}, …, d_{n}⟩ such that for some B, ⟨d_{1}, …, d_{n}, B⟩ is in ℝ.

Lemma 5.3. Suppose that U is a normal ultrafilter on P_{κ}λ. Let F : P_{κ}λ → I be a function such that, letting a_{1}, …, a_{n} denote the finite sequence from P_{κ}λ appearing in F (a), a_{n} ⊂ ∼ a. Then there is X in U such that F is constant on X.

Proof. Write j : V → M = Ult(V, U) and let j(F)(j“λ) = ⟨d_{1}, …, d_{n}⟩. For 0 < m ≤ n write d_{m} = a_{m} or d_{m} = ⟨a_{m}, A_{m}⟩. Then each a_{m} is in P_{κ}j“λ and each A_{m} is a subset of P_{κ}j“λ with size less than κ. Let b_{m} = j^{−1}“a_{m}, and if A_{m} is defined let B_{m} = {j^{−1}“b : b ∈ A_{m}}. Define ⟨e_{1}, …, e_{n}⟩ by letting e_{m} = ⟨b_{m}, B_{m}⟩ if B_{m} is defined, otherwise e_{m} = b_{m}. Then j(⟨e_{1}, …, e_{n}⟩) = ⟨d_{1}, …, d_{n}⟩ = j(F)(j“λ), and by elementarity ⟨e_{1}, …, e_{n}⟩ is in I. Let X be the set of a such that F (a) = ⟨e_{1}, …, e_{n}⟩. □

Lemma 5.4. Suppose that U is a normal ultrafilter on P_{κ}λ. Let ⟨X(d⃗) : d⃗ ∈ I⟩ be a sequence such that each X(d⃗) is in U. Define the diagonal intersection △{X(d⃗) : d⃗ ∈ I} by letting a be in △{X(d⃗) : d⃗ ∈ I} iff a is in X(d⃗) for any d⃗ in I such that every b in P_{κ}λ appearing in d⃗ satisfies that b ⊂ ∼ a. Then △{X(d⃗) : d ∈ I} is in U.

Proof. Suppose for a contradiction that the diagonal intersection is not in U, and let Y be its complement. Define F : Y → I by letting F (a) be some member of I below a such that a is not in X(F (a)). By Lemma 5.3 there is a set X ⊆ Y in U and d⃗ in I such that F (a) = d⃗ for all a in X. But then X ∩ X(d⃗) is empty, which is impossible. □

Proposition 5.5. The poset ℝ satisfies the Prikry Property.

Proof. Let p = ⟨A⟩ be a condition in ℝ which is a direct extension of 1_{ℝ} and let φ be a statement in the forcing language. We prove that there is a direct extension of p which decides φ. The proof will follow closely the same argument from [2].

Associate to each d⃗ in I and a in A an integer l(d⃗, a) < 2 if there exists a condition below ⟨A⟩ which forces ^{l(d⃗,a)}φ whose finite sequence is either d⃗ ⟨a, C⟩ for some C, or d⃗ a if oU⃗(a ∩ κ) = 0. Otherwise let l(d⃗, a) = 2. For each d⃗ in I and i < oU⃗(κ), let A(d⃗, i) ⊆ A be a set in U(κ, i) such that for all a and b in A(d⃗, i), l(d⃗, a) = l(d⃗, b). Define A(i)^{∗} = △{A(d⃗, i) : d⃗ ∈ I}, which is in U(κ, i) by Lemma 5.4. Define A(0) = A(0)^{∗} ∩ {a ∈ P_{κ}λ : oU⃗(a ∩ κ) = 0}. For i > 0 let A(i) = A(i)^{∗} ∩ {a ∈ P_{κ}λ : oU⃗(a ∩ κ) > 0}. Finally, let B = ⋃{A(i) : i < oU⃗(κ)}. Then B is in ⋂{U(κ, i) : i < oU⃗(κ)}.

Suppose for a contradiction that ⟨B⟩ does not have a direct extension which decides φ. Let q = ⟨d_{1}, …, d_{n}, C⟩ be an extension of ⟨B⟩ which decides φ such that n is minimal. By assumption, n > 0. Fix l < 2 such that q forces ^{l}φ. We will define a set D ⊆ C such that ⟨d_{1}, …, d_{n−1}, D⟩ forces ^{l}φ, which contradicts the minimality of n. Let a_{1}, …, a_{n} be the finite sequence from P_{κ}λ which appears in d_{1}, …, d_{n}. Let d⃗ = ⟨d_{1}, …, d_{n−1}⟩. Fix some i such that a_{n} is in A(i).

Since q forces ^{l}φ, l(d⃗, a_{n}) = l, and so for all a in A(i) ∩ C, l(d⃗, a) = l. So for each a in A(i) ∩ C fix a condition r_{a} = ⟨d⃗, f_{a}, B_{a}⟩ below ⟨A⟩ which forces ^{l}φ whose finite sequence from P_{κ}λ is a_{1}, …, a_{n−1}, a.

First suppose that i = 0. Define D = C ∩ △{B_{a} : a ∈ C ∩ A(i)}. Suppose for a contradiction that q′ = ⟨d_{1}, …, d_{n−1}, D⟩ does not force ^{l}φ. Then there is a condition r = ⟨e_{1}, …, e_{m}, E⟩ below q′ which forces ¬(^{l}φ). If n > 1 then fix k ≤ m such that e_{k} and d_{n−1} have the same element from P_{κ}λ; if n = 1 then let k = 0 and e_{k} = ∅. If the least element a of r above e_{k} satisfies that oU⃗(a ∩ κ) = 0, then a is in C ∩ A(0). Then r_{a} and r are compatible, which is impossible. Otherwise oU⃗(a ∩ κ) > 0, so we can extend r to r′ by adding an element a from A(0) ∩ C immediately above e_{k}. Then r_{a} and r′ are compatible, which is a contradiction.

Now assume that i > 0. For each a in A(i) ∩ C, write f_{a} = ⟨a, b_{a}⟩, so r_{a} = ⟨d⃗, ⟨a, b_{a}⟩, B_{a}⟩.

We define three sets D(< i), D(i), and D(> i). Let D(< i) be the set represented by the function a ↦ {π_{a}“c : c ∈ b_{a}} in Ult(V, U(κ, i)). For all a in A(i) ∩ C, {π_{a}“c : c ∈ b_{a}} is in ⋂{U(a ∩ κ, β) : β < oU⃗(a ∩ κ)}. Since A(i) ∩ C is in U(κ, i), D(< i) is in ⋂{U(κ, β) : β < i}. Also note that if j : V → Ult(V, U(κ, i)), a ↦ b_{a} represents j“D(< i).

Let D(i) be the set of a in A(i) ∩ C such that D(< i) ∩ P_{(a∩κ)}a = b_{a}.

Claim 5.6. The set D(i) is in U(κ, i).

Proof. Write j : V → Ult(V, U(κ, i)). Then by Lemma 1.1, j(D(< i)) ∩ P_{κ}j“λ = j“D(< i). But j“D(< i) is represented by the function a ↦ b_{a} and P_{κ}j“λ is represented by the function a ↦ P_{(a∩κ)}a. □

Finally, if i is not the maximum ordinal below oU⃗(κ), then let D(> i) be the set of a in ⋃{A(j) : i < j < oU⃗(κ)} for which there exists β < oU⃗(a ∩ κ) such that the set {π_{a}“c : c ∈ D(i) ∩ P_{(a∩κ)}a} is in U(a ∩ κ, β). If i is the maximum ordinal below oU⃗(κ) then let D(> i) be the empty set.

Claim 5.7. The set D(> i) is in ⋂{U(κ, l) : i < l < oU⃗(κ)}.

Proof. Fix l and let j : V → Ult(V, U(κ, l)). The set D(i) is in U(κ, i). But D(i) is equal to the set {π_{j“λ}“c : c ∈ j(D(i)) ∩ P_{κ}j“λ}. □

Let D(< i)^{∗} be the set of a in D(< i) such that {π_{a}“c : c ∈ D(< i) ∩ P_{(a∩κ)}a} is in ⋂{U(a ∩ κ, l) : l < oU⃗(a ∩ κ)}. Note that D(< i)^{∗} is in ⋂{U(κ, l) : l < i}.

Now define

D = C ∩ △{B_{a} : a ∈ C ∩ A(i)} ∩ (D(< i)^{∗} ∪ D(i) ∪ D(> i)).

Suppose for a contradiction that q′ = ⟨d_{1}, …, d_{n−1}, D⟩ does not force ^{l}φ. Then there is a condition r = ⟨e_{1}, …, e_{m}, E⟩ below q′ which forces ¬(^{l}φ). If n > 1 then fix k ≤ m such that e_{k} and d_{n−1} have the same element from P_{κ}λ; if n = 1 then let k = 0 and e_{k} = ∅. Recall that for each a in A(i)∩C, r_{a} = ⟨d_{1}, …, d_{n−1}, ⟨a, b_{a}⟩, B_{a}⟩ is a condition which forces ^{l}φ. We will get a contradiction by showing that there is a such that r and r_{a} are compatible. First note that we can extend each r_{a} to r′_{a} whose finite sequence below ⟨a, b_{a}⟩ has length k and has the same finite sequence from P_{κ}λ as does e_{1}, …, e_{k}.

We consider three cases. First suppose k = m. Extend r to r′ = ⟨e_{1}, …, e_{m}, e, E′⟩ such that the element a of e from P_{κ}λ is in A(i). Then r′ and r′_{a} have the same finite sequence of elements from P_{κ}λ and so are compatible.

Now assume k < m. Suppose that for all k < j ≤ m, the element of e_{j} in P_{κ}λ is in D(< i)^{∗}. Fix a in D(i) such that r′ = ⟨e_{1}, …, e_{m}, ⟨a, b_{a} ∩ E⟩, F ⟩ ≤ r. By the definition of D(i), b_{a} = D(< i)∩P_{(a∩κ)}a. So the elements of P_{κ}λ from e_{k+1}, …, e_{m} can be added to r′_{a} to obtain a condition with the same finite sequence as r′. Then r′_{a} and r′ are compatible.

Finally, suppose there is a minimum j with k < j ≤ m such that e_{j} = ⟨a_{j}, B_{j}⟩ where a_{j} is in D(i) ∪ D(> i). First assume that a_{j} is in D(i). For k < l < j, define e_{l}^{∗} by intersecting the measure one set appearing in e_{l} (if it exists) with D(< i). Then

⟨e_{1}, …, e_{k}, e_{k+1}^{∗}, …, e_{j−1}^{∗}, ⟨a_{j}, B_{j} ∩ b_{a_{j}}⟩, e_{j+1}, …, e_{m}, E ∩ B_{a_{j}}⟩

is below r′_{a_{j}} and r. Otherwise the element a_{j} of e_{j} is in D(> i). By the definition of D(> i), we can extend r by adding an element of D(i) immediately below e_{j}. This condition satisfies that the least member a of the finite sequence above e_{k} which is in D(i) ∪ D(> i) is actually in D(i), and we just proved that any such condition is compatible with r′_{a}. □

Now we can strengthen Sublemma 5.2.

Lemma 5.8. Suppose that p = ⟨d_{1}, …, d_{n}, A⟩ is a condition in ℝ and m ≤ n with d_{m} = ⟨a_{m}, A_{m}⟩. Let G be generic for ℝ/p^{≤m}. Then in V [G]:

(1) for any φ in the forcing language for ℝ and any condition s in ℝ/p^{>m}, there is t ≤^{∗} s which decides φ.

(2) Suppose that ξ < κ is an ordinal such that for all l with m < l ≤ n, if d_{l} = ⟨a_{l}, A_{l}⟩ where oU⃗(a_{l} ∩ κ) > 0 then ξ < a_{l} ∩ κ. Then for any collection {p_{i} : i < ξ} of direct extensions of p^{>m}, there is q such that q ≤^{∗} p_{i} for i < ξ.

(3) Suppose that ξ < κ is an ordinal such that any a in P_{κ}λ with oU⃗(a ∩ κ) > 0 which appears either in the finite sequence of p^{>m} or in a measure one set of p^{>m} satisfies that a ∩ κ > ξ. Then p^{>m} forces that no new subsets of ξ are added over V [G].

Proof. The proof of (1) is similar to the proof of Sublemma 5.2, using Lemma 3.1 and the Prikry property for ℝ in place of the assumption of Lemma 5.1. The proof of (2) is similar, using Lemma 3.1.

For (3), every r ≤ p^{>m} satisfies the assumptions of (2). Suppose q ≤ p^{>m} and q forces ḟ : ξ → 2. For each i < ξ let q_{i} ≤^{∗} q decides the value of ḟ(i). Apply (2) and obtain r such that r ≤^{∗} q_{i} for each i. Then r forces that ḟ is in the ground model. □

6. The Radin Club

Suppose that p is a condition in ℝ and a is in P_{κ}λ. We say that a appears in p if either a is in the finite sequence of p, or there is d in the finite sequence of p such that d = ⟨a, A⟩ for some A.

Suppose that G is generic for ℝ over V. Define the Radin club

C_{G} = {a ∈ (P_{κ}λ)^{V} : ∃ p ∈ G (a appears in p)}.

By the compatibility of conditions in G, if a and b are in C_{G}, then either a ⊂ ∼ b or

b ⊂ ∼ a. Therefore either a ∩ κ < b ∩ κ or vice versa. So C_{G} is a well ordering and

o.t.(C_{G}) ≤ κ.

Proposition 6.1. The set C_{G} is a closed and unbounded subset of (P_{κ}λ)^{V}. Furthermore, for all a in C_{G}:

(1) a is a successor point of C_{G} iff oU⃗(a ∩ κ) = 0 and a is not equal to min(C_{G}).

(2) a is a limit point of C_{G} iff oU⃗(a ∩ κ) > 0.

Proof. Suppose a is in C_{G} and oU⃗(a ∩ κ) > 0. Fix p in G such that a appears in p. Let c be in P_{(a∩κ)}a. Then there is a dense set of r below p in which there appears some b with c ⊂ ∼ b ⊂ ∼ a. For if q ≤ p, then ⟨a, A⟩ is in q for some A, and it is possible

to add elements b from A to q with c ⊂ ∼ b ⊂ ∼ a. So a is a limit point of C_{G}.

Suppose that a is in C_{G}, oU⃗(a ∩ κ) = 0, and a is not equal to min(C_{G}). Fix p in G such that a appears in p. It is easy to show that since a is not minimal and oU⃗(a ∩ κ) = 0, there is b ⊂ ∼ a appearing in p. Let b be the maximal b ⊂ ∼ a appearing

in p. Then by the definition of the ordering, whenever c ⊂ ∼ a is in C_{G}, c ⊂ ∼ b. So a

is the successor of b in C_{G}.

The other directions of (1) and (2) now follow immediately.

An easy density argument shows that C_{G} is unbounded in (P_{κ}λ)^{V}. Suppose for a contradiction that there is an initial segment C of C_{G} such that ⋃ C = A is not in C_{G}. Let a be the least element of C_{G} such that A ⊂ a. Then C_{G} is not unbounded in P_{(a∩κ)}a, so a is not a limit point of C_{G}. By (2) it follows that oU⃗(a ∩ κ) = 0. By (1) there is b in C_{G} such that a is the successor of b. But then every element of C is a subset of b, so A ⊂ b. This contradicts the minimality of a. □

The generic filter G can be recovered from C_{G}. Let p = ⟨d_{1}, …, d_{n}, A⟩ be a condition in ℝ and let a_{1}, …, a_{n} be its finite sequence from P_{κ}λ. Then p is in G iff

(1) each a_{i} is in C_{G},

(2) for each a in C_{G} with a_{n} ⊂ ∼ a, a is in A and the set {π_{a}“c : c ∈ A ∩ P_{(a∩κ)}a}

is in ⋂{U(a ∩ κ, i) : i < oU⃗(a ∩ κ)}, and

(3) write a_{0} = ∅; then for each i ≤ n, if d_{i} = ⟨a_{i}, A_{i}⟩, then for all a in C_{G} with a_{i−1} ⊂ ∼ a ⊂ ∼ a_{i}, a is in A_{i} and the set {π_{a}“c : c ∈ A_{i} ∩ P_{(a∩κ)}a} is in ⋂{U(a ∩ κ, i) :

i < oU⃗(a ∩ κ)}. For any condition p in G satisfies these properties, and any condition which satisfies these conditions is compatible with each member of G, and so is in G.

We will abuse terminology and say that C_{G} itself is generic for ℝ.

Lemma 6.2 (Factorization Lemma II). Suppose that p = ⟨d_{1}, …, d_{n}, A⟩ is a condition in ℝ, 0 < m ≤ n, and d_{m} = ⟨a_{m}, A_{m}⟩.

Let G be generic for ℝ containing p. Define C^{≤m} = {π_{a_{m}}“c : c ∈ C_{G}, c ⊂ ∼ a_{m}}

and C^{>m} = {c ∈ C_{G} : a_{m} ⊂ ∼ c}. Then C^{≤m} is generic for ℝ/p^{≤m} over V and C^{>m}

is generic for ℝ/p^{>m} over V [C^{≤m}].

On the other hand, if C_{0} is generic for ℝ/p^{≤m} over V and C_{1} is generic for ℝ/p^{>m} over V [C_{0}], then C = {π_{a_{m}}^{−1}“c : c ∈ C_{0}} ∪ {a_{m}} ∪ C_{1} is generic for ℝ/p over V.

Proof. Let i : ℝ/p → ℝ/p^{≤m} × p^{>m} be the isomorphism described in the proof of Lemma 4.1. Write i“G = G_{0} × G_{1}. Then C_{G_{0}} = C^{≤m} and C_{G_{1}} = C^{>m}.

For the second statement, let G = i^{−1}“(G_{C_{0}} × G_{C_{1}}). Then C_{G} = C. □

Lemma 6.3. Suppose that p = ⟨d_{1}, …, d_{n}, A⟩ is a condition in ℝ, 0 < m ≤ n, and d_{m} = ⟨a_{m}, A_{m}⟩. Let G be generic for ℝ containing p and factor C_{G} as C^{≤m} and C^{>m}. Then every subset of λ_{(a_{m}∩κ)} in V [G] is in V [C^{≤m}].

Proof. The condition p^{>m} satisfies the property that for any a in P_{κ}λ appearing either in the finite sequence of p^{>m} or in a measure one set of p^{>m}, a ∩ κ > λ_{a_{m}∩κ}. Now apply Lemma 5.8(3). □

If 0 < oU⃗(κ) < κ then ℝ is very similar to Magidor forcing, as we see next.

Lemma 6.4. (1) If the set {a ∈ C_{G} : oU⃗(a ∩ κ) ≥ oU⃗(κ)} is non-empty, then it has a maximal element.

(2) For any b in C_{G}, if the set {a ∈ C_{G} : a ⊂ ∼ b, oU⃗(a ∩ κ) ≥ oU⃗(b ∩ κ)} is

non-empty, then it has a maximal element.

Proof. Let C = {a ∈ C_{G} : oU⃗(a ∩ κ) ≥ oU⃗(κ)}. If C is empty then we are done. Suppose that C is non-empty. Fix p = ⟨d_{1}, …, d_{n}, A⟩ in G such that there is 0 < m ≤ n with d_{m} = ⟨a_{m}, A_{m}⟩ and a_{m} is the largest element of C appearing in p. Define q ≤ p by replacing A with A′ = {a ∈ A : oU⃗(a ∩ κ) < oU⃗(κ)}. Then q forces that any c in C_{G} with oU⃗(c ∩ κ) ≥ oU⃗(κ) is a subset of b. Therefore q forces that a is the maximal element of C.

Statement (2) follows from the first statement applied to the Radin forcing ℝ(U⃗ ↾ ((b ∩ κ) + 1)). □

Theorem 6.5. Suppose that 0 < oU⃗(κ) < κ. If there is a in C_{G} such that oU⃗(a ∩ κ) ≥ oU⃗(κ), then let a′ be the maximal such set. Otherwise let a′ = ∅. Then the set {b ∈ C_{G} : a′ ⊂ ∼ b} is a closed and unbounded subset of (P_{κ}λ)^{V} with order type

ω^{oU⃗ (κ)}.

Proof. We prove the theorem by induction on oU⃗(κ). Let C′_{G} = {a ∈ C_{G} : a′ ⊂ ∼ a}.

Suppose that oU⃗(κ) = 1. Then for all b in C′_{G}, oU⃗(b ∩ κ) = 0. By Proposition 6.1(2), C′_{G} has no limits points. So the order type of C′_{G} is ω.

Suppose that oU⃗(κ) = ξ + 1 and the statement holds for ordinals less than ξ + 1. Then for all b in C′_{G}, oU⃗(b ∩ κ) ≤ ξ. By a density argument the set C_{ξ} = {a ∈ C′_{G} : oU⃗(a ∩ κ) = ξ} is unbounded in C′_{G}. By Proposition 6.4(2), C_{ξ} has no limit points in C′_{G} and so has order type ω. Let ⟨a_{n} : 0 < n < ω⟩ enumerate the elements of C_{ξ} in increasing order and let a_{0} = a′. By the the induction hypothesis and Lemma 6.2, for each n ≥ 1 the set C_{n} = {b ∈ C′_{G} : a_{n−1} ⊂ ∼ b ⊂ ∼ a_{n}} has order

type ω^{ξ}. Thus C′_{G} has order type ω^{ξ} · ω = ω^{ξ+1}.

Now suppose that oU⃗(κ) is a limit ordinal and the statement holds for all δ < oU⃗(κ). Fix δ < oU⃗(κ). By a density argument, there is a set b in C′_{G} with oU⃗(b∩κ) = δ. So by the induction hypothesis and Lemma 6.2, C′_{G} has order type at least ω^{δ}. Therefore C′_{G} has order type at least sup{ω^{δ} : δ < oU⃗(κ)} = ω^{oU⃗ (κ)}. Suppose for a contradiction that the order type of C′_{G} is larger than ω^{oU⃗ (κ)}. Let b be the ω^{oU⃗ (κ)}-th element of C′_{G}. By the induction hypothesis and Lemma 6.2, there is a tail of C′_{G} ∩ P_{(b∩κ)}b with order type ω^{oU⃗ (b∩κ)}. But every tail of C′_{G} ∩ P_{(b∩κ)}b has

order type ω^{oU⃗ (κ)}, which is a contradiction. □

Corollary 6.6. Suppose that 0 < oU⃗(κ) < κ. Then there is p ≤^{∗} 1 which forces that the order type of C_{G} is ω^{oU⃗ (κ)}.

Proof. Let p = ⟨A⟩, where A is the set of a in AU⃗ with oU⃗(a ∩ κ) < oU⃗(κ). □

Note that if oU⃗(α) < oU⃗(κ) < κ for all α < κ, then the order type of C_{G} is ω^{oU⃗ (κ)}.

7. Preservation of Cardinals

When λ = κ then ℝ preserves all cardinals. When λ > κ the situation is a bit more complicated, but we have a description of which cardinals are collapsed.

For each a in P_{κ}λ write κ_{a} = a ∩ κ. Let G be generic for ℝ over V and let C_{G} be the Radin club described in the last section.

Recall that since κ is λ-supercompact, λ^{<κ} is equal to λ if cf(λ) ≥ κ and is equal to λ^{+} if cf(λ) < κ.

Proposition 7.1. Let μ ≥ κ be a cardinal.

(1) All cardinals greater than λ^{<κ} are preserved.

(2) If μ ≤ λ and cf(μ) ≥ κ, then in V [G], cf(μ) = cf(o.t.(C_{G})).

(3) If κ < μ ≤ λ^{<κ} then μ is collapsed in V [G].

Proof. Statement (1) follows from the fact that ℝ is (λ^{<κ})^{+}-c.c. Suppose that κ ≤ μ ≤ λ and cf(μ) ≥ κ in V. By a density argument the set {sup(a∩μ) : a ∈ C_{G}} is closed and unbounded in μ with order type o.t.(C_{G}). So cf(μ) = cf(o.t.(C_{G})). Since o.t.(C_{G}) ≤ κ, every successor cardinal in the interval (κ, λ] is collapsed. It follows that every cardinal in this interval is collapsed.

If cf(λ) ≥ κ, then λ^{<κ} = λ and the proof is complete. If cf(λ) < κ then λ^{<κ} = λ^{+}. Every member of (P_{κ}λ)^{V} is a subset of some a in C_{G}, and for any a in C_{G} there are fewer than κ many subsets of a in V. It follows that in V [G] the ordinal (λ^{<κ})^{V} has size no larger than |C_{G}| · κ = κ. □

Proposition 7.2. Let μ < κ be a cardinal.

(1) The cardinal μ is collapsed iff there is a in C_{G} with oU⃗(κ_{a}) > 0 such that κ_{a} < μ ≤ λ_{κ_{a}}^{<κ_{a}}.

(2) Suppose there is a in C_{G} with oU⃗(κ_{a}) > 0 such that κ_{a} ≤ μ ≤ λ_{κ_{a}} and cf(μ) ≥ κ_{a}. Then in V [G], cf(μ) = cf(o.t.(C_{G} ∩ P_{κ_{a}}a)).

(3) If μ is regular and μ changes its cofinality, then there is a in C_{G} with oU⃗(κ_{a}) > 0 such that κ_{a} ≤ μ ≤ λ_{κ_{a}}^{<κ_{a}}.

Proof. Suppose that a is in C_{G} with oU⃗(κ_{a}) > 0 and κ_{a} < μ ≤ λ_{κ_{a}}^{<κ_{a}}. Then we can factor C_{G} as C_{0} and C_{1} such that C_{0} is generic for ℝ(U⃗ ↾ (κ_{a} + 1)) over V. By Proposition 7.1(3), μ is collapsed in V [C_{0}] and so also in V [G]. By Proposition 7.1(2), if μ ≤ λ_{κ_{a}} and cf(μ) ≥ κ_{a}, then in V [C_{0}], cf(μ) = cf(o.t.(C_{0})) = cf(o.t.(C_{G}∩ P_{κ_{a}}a)). By Lemma 6.3, the same is true in V [G]. This proves (2) and the “if” direction of (1).

(3) Suppose that μ is regular, β < μ, p = ⟨d_{1}, …, d_{n}, A⟩ is in G, and p forces that ḟ : β → μ is cofinal. Assume for a contradiction that there is no q ≤ p which forces the conclusion of (3). Then in particular, if 0 < i ≤ n and d_{i} = ⟨a_{i}, A_{i}⟩, then μ is not in the interval [κ_{a_{i}}, λ_{a_{i}}^{<κ_{a_{i}}}]. Define V_{1} and p^{∗} as follows. If there is a largest m ≤ n such that oU⃗(κ_{a_{m}}) > 0 and μ > κ_{a_{m}}, let V_{1} be V [G^{≤m}], where G^{≤m} is the generic for ℝ/p^{≤m} given by G, and let p^{∗} = p^{>m}; otherwise let V_{1} = V and p^{∗} = p. In either case, μ is regular in V_{1}, and for any a appearing in p^{∗} with oU⃗(κ_{a}) > 0, μ < κ_{a}. For each pair ⟨i, α⟩ in β × μ, choose q(i, α) ≤^{∗} p^{∗} which decides the statement ḟ(i) = α. By Lemmas 3.1 and 5.8(2), there is q such that q ≤^{∗} q(i, α) for each i and α. Define g : β → μ in V_{1} by letting g(i) be such that q forces ḟ(i) = g(i). Then g is cofinal in μ and therefore μ is singular in V_{1}, which is a contradiction.

“Only if” direction of (1): First we show that if μ is collapsed, then there is a in C_{G} with oU⃗(κ_{a}) > 0 such that κ_{a} ≤ μ ≤ λ_{κ_{a}}^{<κ_{a}}. If μ is regular then this follows from (3). Suppose μ is singular. Then there is a sequence ⟨α_{i} : i < cf(μ)⟩ of regular cardinals unbounded in μ in V which are all collapsed in V [G]. For each i fix a_{i} in C_{G} such that κ_{a_{i}} ≤ α_{i} ≤ λ_{κ_{a_{i}}}^{<κ_{a_{i}}}. If the map i ↦ a_{i} is not eventually constant, then μ is a limit point of the club set {a ∩ κ : a ∈ C_{G}}, and so is in this set. This is a contradiction since μ is singular in V, and if a is a limit point of C_{G} then a ∩ κ is measurable in V. If a is in C_{G} such that a_{i} = a for a tail of i’s, then κ_{a} ≤ μ ≤ λ_{κ_{a}}^{<κ_{a}}.

Now suppose that μ = κ_{a} for some a in C_{G} with oU⃗(κ_{a}) > 0. We show that μ is not collapsed. In V [G] the cardinal μ is a limit of point of {κ_{b} : b ∈ C_{G}, oU⃗(b∩κ) = 0}, and by the last paragraph every ordinal in this set is a cardinal in V [G]. □

By Proposition 7.2(1), the set {κ_{a} : a ∈ C_{G}} is a club subset of κ consisting of cardinals. So κ remains a cardinal in V [G].

By the theorem of Cummings and Shelah about changes of cofinality (see Section 1) and the fact that ℝ is (λ^{<κ})^{+}-c.c., the cofinality of λ^{<κ} in V [G] is equal to cf(|(λ^{<κ})^{V} |) = cf(κ) = cf(o.t.(C_{G})). Similarly, if a is in C_{G} then in V [G] the cofinality of λ_{κ_{a}}^{<κ_{a}} is equal to cf(o.t.(C_{G} ∩ P_{(a∩κ)}a)).

PART II: ITERATED RADIN FORCING

8. Iterating Radin forcing over the same cardinal

In this section we show how to iterate Radin forcing infinitely many times over the same cardinal. Suppose that κ is a supercompact cardinal. Let ⟨κ_{n} : n < ω⟩ be a sequence of cardinals such that either (1) each κ_{n} is equal to κ, or (2) κ_{0} ≥ κ and κ_{n} < (κ_{n}^{<κ})^{+} ≤ κ_{n+1} for each n. We will define a forcing iteration ℝ^{ω} with length ω such that at each stage n we force with a Radin forcing defined from a carefully chosen coherent sequence of ultrafilters on P_{κ}κ_{n}.

Define by recursion sequences ⟨ℝ_{n} : n < ω⟩ and ⟨ℝ^{n} : n < ω⟩ as follows. We let ℝ^{0} be the trivial poset and for n > 0 let ℝ^{n} = ℝ_{0} ∗ … ∗ ℝ_{n−1}. We maintain the following for each n ≥ 0:

Recursion Hypotheses (1) ℝ^{n} forces that κ is supercompact, and

(2) if n > 0 then |ℝ^{n}| ≤ 2^{(κ_{n−1}^{<κ})} and ℝ^{n} is (κ_{n−1}^{<κ})^{+}-c.c.

Obviously ℝ^{0} forces that κ is supercompact. Suppose that n ≥ 0 and ℝ^{n} is defined and satisfies the recursion hypotheses. Let G^{n} be generic for ℝ^{n} over V and write V_{n} = V [G^{n}].

We would like to define a Radin forcing ℝ_{n} on P_{κ}κ_{n} in V_{n} which forces that κ remains supercompact. In order to achieve this we must choose our coherent sequence carefully. Until further notice we are working in the model V_{n}. We begin with a variation of Proposition 2.2.

Proposition 8.1. Suppose that λ ≥ 2^{(κ_{n}^{<κ})}, U is a normal ultrafilter on P_{κ}λ with j : V → M = Ult(V, U), and there is a function f : κ → κ such that j(f)(κ) = κ_{n}. Then there exists a coherent sequence of ultrafilters U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ on P_{κ}κ_{n} such that:

(1) 0 < oU⃗(κ) < (2^{(κ_{n}^{<κ})})^{+},

(2) j(U⃗) ↾ (κ + 1) = U⃗, and

(3) ⋂{U(κ, i) : i < oU⃗(κ)} is a subset of U ↾ κ_{n}.

Proof. Define g : κ → V_{κ} recursively as follows. The domain of g is the set of strongly inaccessible cardinals α such that f(α) ≥ α is a cardinal and α is closed under f. Suppose that α is in the domain of g and g ↾ α is defined. Let X_{α} be the collection of sequences ⟨U_{i} : i < γ⟩ of normal ultrafilters on P_{α}f(α) such that for all β < γ, letting j_{β} : V → Ult(V, U_{β}), j_{β}(g ↾ α)(α) = ⟨U_{i} : i < β⟩; moreover, for each β < γ, U_{β} \ ⋃{U_{i} : i < β} is non-empty. Let g(α) = ⟨U(α, i) : i < oU⃗(α)⟩ be some maximal element in X_{α}, which exists by Zorn’s Lemma.

Let ⟨U(κ, i) : i < oU⃗(κ)⟩ be equal to j(g)(κ). This defines a coherent sequence U⃗ on P_{κ}κ_{n} such that j(U⃗) ↾ (κ + 1) = U⃗.

Note that oU⃗(κ) < (2^{(κ_{n}^{<κ})})^{+}. For otherwise choose X_{i} in U(κ, i) \ ⋃{U(κ, j) : j < i} for each i < (2^{(κ_{n}^{<κ})})^{+}. Then X_{i} ≠ X_{j} for i < j, since X_{j} is not in U(κ, i). This is a contradiction since there are only 2^{(κ_{n}^{<κ})} many subsets of P_{κ}κ_{n}.

Let U′ = U ↾ κ_{n} and write i : V → N = Ult(V_{n}, U′). Let k : N → M be the elementary embedding from Lemma 1.3 such that j = k∘i and crit(k) = (2^{(κ_{n}^{<κ})})^{+N}. Now k maps the critical point of k to (2^{(κ_{n}^{<κ})})^{+}, which is larger than oU⃗(κ), so i(g)(κ) has length oU⃗(κ) and in fact is equal to ⟨U(κ, i) : i < oU⃗(κ)⟩. Since U′ is in M, by the maximality of j(g)(κ), U′ \ ⋃{U(κ, i) : i < oU⃗(κ)} is empty. Suppose that A is in ⋂{U(κ, i) : i < oU⃗(κ)}. If A is not in U′, then (P_{κ}κ_{n} \ A) is in U′. So there is i < oU⃗(κ) such that (P_{κ}κ_{n} \ A) and A are both in U(κ, i), which is impossible. □

Lemma 8.2. There exists a coherent sequence U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ of ultrafilters on P_{κ}κ_{n} in V_{n} with 0 < oU⃗(κ) < (2^{(κ_{n}^{<κ})})^{+} satisfying the following property: for every λ ≥ 2^{(κ_{n}^{<κ})} there is a normal ultrafilter U_{λ} on P_{κ}λ with j_{λ} : V_{n} → M_{λ} = Ult(V, U_{λ}) such that j_{λ}(U⃗) ↾ (κ+1) = U⃗ and ⋂{U(κ, i) : i < oU⃗(κ)} ⊆ U_{λ} ↾ κ_{n}.

Proof. By Lemma 1.3 it suffices to prove there are unboundedly many regular cardinals λ satisfying the required property. Suppose for a contradiction that for each coherent sequence W⃗ on P_{κ}κ_{n} with oW⃗ (κ) < (2^{(κ_{n}^{<κ})})^{+}, there is μ(W⃗) ≥ 2^{(κ_{n}^{<κ})} such that for every regular λ ≥ μ(W⃗), there is no U_{λ} on P_{κ}λ as desired. Choose a regular cardinal λ which is larger than μ(W⃗) for each coherent sequence W⃗.

Let f : κ → V_{κ} be a Laver function for κ. Then there exists a normal ultrafilter U on P_{κ}λ with elementary embedding j : V_{n} → M = Ult(V_{n}, U) such that j(f)(κ) = κ_{n}. Apply Proposition 8.1 to obtain U⃗, which contradicts that λ is greater than μ(U⃗). □

Now let ℝ_{n} be an ℝ^{n}-name for a Radin forcing defined from a coherent sequence on P_{κ}κ_{n} satisfying the statement of Lemma 8.2. Define ℝ^{n+1} = ℝ^{n} ∗ ℝ_{n}.

Proposition 8.3. The poset ℝ^{n+1} forces that κ is supercompact.

Proof. Suppose that G^{n} ∗ G_{n} is generic for ℝ^{n+1} = ℝ^{n} ∗ ℝ_{n} over V and write V_{n} = V [G^{n}]. Let U⃗ = ⟨U(α, i) : α ≤ κ, i < oU⃗(α)⟩ be the coherent sequence on P_{κ}κ_{n} in V_{n} such that ℝ_{n} = ℝ(U⃗). Fix λ ≥ 2^{(κ_{n}^{<κ})}. Let U be a normal ultrafilter on P_{κ}λ with j : V_{n} → M = Ult(V, U) such that j(U⃗) ↾ (κ + 1) = U⃗ and ⋂{U(κ, i) : i < oU⃗(κ)} ⊆ U ↾ κ_{n}.

Define A^{∗} to be the set of a in j(AU⃗) such that j“κ_{n} ⊂ ∼ a. Recall that in M,

π_{(j“κ_{n})} = j^{−1} ↾ j“κ_{n}. Let s be the condition

⟨ ⟨j“κ_{n}, j“AU⃗⟩, A^{∗} ⟩

in j(ℝ_{n}). Then j(ℝ_{n})/s factors as ℝ_{n}×j(ℝ_{n})/⟨A^{∗}⟩ with isomorphism i as described in the proof of Lemma 4.1.

For any p = ⟨d_{1}, …, d_{n}, A⟩ in ℝ_{n}, j(p) = ⟨j(d_{1}), …, j(d_{n}), j(A)⟩. Let

p′ = ⟨j(d_{1}), …, j(d_{n}), ⟨j“κ_{n}, j“A⟩, j(A) ∩ A^{∗}⟩.

Since A is in ⋂{U(κ, i) : i < oU⃗(κ)⟩, A is in U ↾ κ_{n}. Therefore j“κ_{n} ∈ j(A). It follows that p′ is below j(p). Clearly p′ is below s as well. Moreover, i(p′) = ⟨ p, ⟨j(A) ∩ A^{∗}⟩ ⟩.

Define U^{∗} in V_{n}[G_{n}] by letting X be in U^{∗} iff X ⊆ P_{κ}λ, there is a name Ẋ for X, and there is t ≤^{∗} ⟨A^{∗}⟩ in j(ℝ_{n})/⟨A^{∗}⟩ such that t forces over M[G_{n}] that j“λ is in j(Ẋ).

First we show that the definition does not depend on the choice of the name Ẋ. Suppose that X = Ẋ^{G_{n}} = Ẏ^{G_{n}} in V_{n}[G_{n}]. Then there is p = ⟨d_{1}, …, d_{n}, A⟩ in G_{n} which forces that Ẋ = Ẏ. Since p′ ≤ j(p), p′ forces over M that j(Ẋ) = j(Ẏ). But i(p′) = ⟨p, ⟨j(A) ∩ A^{∗}⟩⟩, so ⟨j(A) ∩ A^{∗}⟩ forces over M[G_{n}] that j(Ẋ) = j(Ẏ). Applying the Prikry property of ℝ/⟨A^{∗}⟩ in M[G_{n}], let tẊ (respectively, tẎ) be a direct extension of ⟨A^{∗}⟩ which decides whether j“λ is in j(Ẋ) (respectively, in j(Ẏ)). But tẊ, tẎ, and ⟨j(A) ∩ A^{∗}⟩ are compatible, so tẊ and tẎ must decide their respective statements the same way.

Similar arguments show that U^{∗} is a non-principal, fine ultrafilter on P_{κ}λ which extends U. To prove normality, suppose that p = ⟨d_{1}, …, d_{n}, A⟩ is in G_{n} and p forces that ḟ : P_{κ}λ → λ is a regressive function. Then ⟨j(A) ∩ A^{∗}⟩ forces over M[G_{n}] that j(ḟ) : j(P_{κ}λ) → j(λ) is a regressive function. For each i less than λ let t_{i} be a direct extension of ⟨j(A) ∩ A^{∗}⟩ which decides whether j(ḟ)(j“λ) = j(i).

By Lemma 5.8 there is t such that t ≤^{∗} t_{i} for all i. Clearly there must exist i such that t forces that j(ḟ)(j“λ) = j(i). Then the set {a ∈ P_{κ}λ : f(a) = i} is in U^{∗}. □

This proves the first recursion hypothesis. If n = 0 then ℝ^{1} = ℝ_{0}, which has size at most 2^{(κ_{0}^{<κ})} and is (κ_{0}^{<κ})^{+}-c.c. Suppose that n > 0. Then by recursion ℝ^{n} has size at most 2^{(κ_{n−1}^{<κ})} and is (κ_{n−1}^{<κ})^{+}-c.c.

Lemma 8.4. Suppose that G^{n} is generic for ℝ^{n} over V. Then the cardinals κ_{n}^{<κ} and 2^{(κ_{n}^{<κ})} are computed the same in both V and V [G^{n}].

Proof. First assume that cf(κ_{n}) < κ in V. Since κ is supercompact in V and V [G^{n}], κ_{n}^{<κ} = κ_{n}^{+} in both models. But ℝ^{n} is κ_{n}-c.c. Now suppose that cf(κ_{n}) ≥ κ in V. Then κ_{n}^{<κ} = κ_{n} in V. If cf(κ_{n})^{V} changes its cofinality in V [G^{n}], then by Proposition 7.1 it must change its cofinality to κ. So κ_{n}^{<κ} = κ_{n} in V [G^{n}] as well.

The poset ℝ^{n} has at most 2^{(κ_{n−1}^{<κ})} many antichains. So there are at most 2^{(κ_{n}^{<κ})} many canonical names for subsets of κ_{n}^{<κ}. □

Since ℝ^{n} is (κ_{n−1}^{<κ})^{+}-c.c. and ℝ^{n} forces that ℝ_{n} is (κ_{n}^{<κ})^{+}-c.c., ℝ^{n+1} is (κ_{n}^{<κ})^{+}- c.c. The chain condition and sizes of ℝ^{n} and ℝ_{n}, along with Lemma 8.4, imply that |ℝ^{n+1}| ≤ 2^{(κ_{n}^{<κ})}.

Now we define the iteration ℝ^{ω}. The poset ℝ^{ω} is basically a Magidor iteration, except instead of iterating Prikry type forcing posets over different cardinals, we iterate Radin forcing over a fixed cardinal.

A condition in ℝ^{ω} is a function p with domain ω such that for all n < ω, p ↾ n is in ℝ^{n}; moreover, there is a finite set a_{p} ⊆ ω such that for all n < ω, n is in ω \ a_{p} iff p ↾ n forces that p(n) ≤^{∗} 1. Let q ≤ p if q ↾ n ≤ p ↾ n in ℝ^{n} for all n. Let q ≤^{∗} p if for all n, q ↾ n ⊩ q(n) ≤^{∗} p(n).

For each n < ω, ℝ_{ω} factors as ℝ^{n} ∗ℝ_{n}^{ω} in the obvious way, where ℝ_{n}^{ω} is a name for the iteration defined using the sequence ⟨ℝ_{m} : n ≤ m < ω⟩ in the generic extension by ℝ^{n}.

Proposition 8.5. The poset ℝ^{ω} satisfies the Prikry property.

Proof. Let φ be a statement in the forcing language for ℝ^{ω} and let p be a condition in ℝ^{ω}. Suppose for a contradiction that there is no direct extension of p which decides φ. We inductively define a ≤^{∗}-decreasing sequence ⟨q_{n} : n < ω⟩ in ℝ^{ω} as follows. Let q_{0} = p. Suppose q_{n} is defined for some n ≥ 0. Let q_{n+1} ↾ [n + 1, ω) be a sequence of names such that q_{n} ↾ (n + 1) forces the following: if there is a direct extension of q_{n} ↾ [n + 1, ω) which decides φ, then q_{n+1} ↾ [n + 1, ω) is such a direct extension; otherwise q_{n+1} ↾ [n + 1, ω) = q_{n} ↾ [n + 1, ω). Define q_{n+1} ↾ n = q_{n} ↾ n. Let q_{n+1}(n) denote a name for a direct extension of q_{n}(n) which decides whether q_{n+1} ↾ [n + 1, ω) decides φ, and if so, which way it decides φ.

Now define q = ⟨q_{(n+1)}(n) : n < ω⟩. Clearly q ≤^{∗} p. Also note that for all n < ω, q ↾ n = q_{n} ↾ n. By assumption there is no direct extension of q which decides φ. For each r ≤ q let b_{r} be the set of m < ω such that r ↾ m does not force that r(m) ≤^{∗} q(m). Then b_{r} ⊆ a_{r}, so b_{r} is finite. Choose r ≤ q which decides φ such that |b_{r}| is minimal. Let n = max(b_{r}), which exists since r is not a direct extension of q. Fix l < 2 such that r forces ^{l}φ. Then

r ↾ n ⊩ r(n) ⊩ ( r ↾ [n + 1, ω) ≤^{∗} q_{n} ↾ [n + 1, ω) ) & ( r ↾ [n + 1, ω) ⊩ ^{l}φ ).

Since r ↾ n ≤ q_{n} ↾ n and q ↾ [n + 1, ω) ≤^{∗} q_{n+1} ↾ [n + 1, ω), by the definition of q(n) we have

r ↾ n ⊩ q(n) ⊩ q ↾ [n + 1, ω) ⊩ ^{l}φ.

It follows that s = (r ↾ n) q(n) (q ↾ [n + 1, ω)) forces ^{l}φ. But b_{s} = b_{r} \ {n}, which contradicts the minimality of |b_{r}|. □

Now we analyze preservation of cardinals and changes in cofinality. We need the following simple lemma.

Lemma 8.6. Suppose ξ is less than κ and p is a condition in ℝ^{ω}. Assume that for each n < ω, p ↾ n forces that if a is in P_{κ}κ_{n} and appears in p(n), and the order of a ∩ κ is greater than 0, then a ∩ κ > ξ. If {p_{i} : i < ξ} is a collection of direct extensions of p, then there is q such that q ≤^{∗} p_{i} for i < ξ.

Proof. Define q inductively. Suppose q ↾ n is defined so that q ↾ n ≤^{∗} p_{i} ↾ n for i < ξ. Apply Lemma 3.1 to find a name q(n) such that q ↾ n forces that q(n) ≤^{∗} p_{i}(n) for i < ξ. □

Let G be generic for ℝ^{ω} over V. For n < ω let G ↾ n denote the set {p ↾ n : p ∈ G}. Then G ↾ n is generic for ℝ^{n} over V.

Lemma 8.7. Suppose that μ < κ. Then there is m < ω such that P^{V [G]}(μ) is in V [G ↾ m].

Proof. Suppose that p is a condition in G. Let n = max(a_{p}). For each m > n write p(m) = ⟨Ȧ_{m}⟩, where Ȧ_{m} is a name for a subset of P_{κ}κ_{m}. Define q by letting q ↾ (n + 1) = p ↾ (n + 1) and for m > n, let q(m) be a name for ⟨ {a ∈ Ȧ_{m} : a ∩ κ > μ} ⟩.

We claim that q forces that P(μ) is in V [G ↾ (n + 1)]. Suppose that r ≤ q and r forces ḟ : μ → 2. For each i < μ let s_{i} be a name for a condition in ℝ_{n+1}^{ω} such that r ↾ (n + 1) forces that s_{i} ≤^{∗} r ↾ [n + 1, ω) and s_{i} decides the statement ḟ(i) = 0. Apply Lemma 8.6 to ℝ_{n+1}^{ω} to find a name s such that r ↾ (n + 1) forces that s ≤^{∗} s_{i} for i < μ. Then r ↾ (n + 1) forces that s decides all the values ḟ(i) for i < μ. Therefore (r ↾ (n + 1)) s is a refinement of r which forces that ḟ is in V [G ↾ (n + 1)]. □

In particular, if μ < κ is a cardinal in V which is collapsed in V [G], then there is m < ω such that μ is collapsed in V [G ↾ m]. Also if μ < κ and cf(μ) = δ in V [G], then there is m < ω such that cf(μ) = δ in V [G ↾ m].

Proposition 8.8. In V [G], κ is a strong limit singular cardinal with cofinality ω.

Proof. For each n let C_{n} be the Radin club on P_{κ}κ_{n} in V [G]. Let D = {min(C_{n})∩κ : n < ω}. We claim that D is unbounded in κ, and therefore cf(κ) = ω in V [G]. Fix ξ < κ. Suppose that p is in G and let n = max(a_{p}). For each m > n write p(m) = ⟨Ȧ_{m}⟩. Define q ≤ p by letting q ↾ (n + 1) = p ↾ (n + 1), and for m > n let q(m) be a name for ⟨{a ∈ Ȧ_{m} : a ∩ κ > ξ}⟩. Then q forces that for all m > n, min(C_{m}) ∩ κ > ξ. So D is unbounded in κ.

To show that κ is a strong limit cardinal in V [G] it suffices to show that for all μ < κ, 2^{μ} < κ. Fix μ < κ. By Lemma 8.7 there is m < ω such that P^{V [G]}(μ) = P^{V [G↾m]}(μ). But κ is supercompact in V [G ↾ m]. □

Suppose we are in the case that κ_{n} < κ_{n+1} for n < ω. Let κ_{ω} = sup({κ_{n} : n < ω}).

Proposition 8.9. Suppose that κ ≤ μ < κ_{ω} and cf(μ) ≥ κ in V. Then cf(μ) = ω in V [G]. Therefore all cardinals in the interval (κ, κ_{ω}] are collapsed to κ.

Proof. Let C_{n} be the Radin club on P_{κ}κ_{n} in V [G]. Define D_{μ} = {sup(min(C_{n})∩μ) : n < ω}. As in the proof of Proposition 8.8, D_{μ} is an unbounded subset of μ. □

Proposition 8.10. The poset ℝ^{ω} is κ_{ω}^{+}-c.c.

Proof. Suppose for a contradiction that A is an antichain in ℝ^{ω} with size κ_{ω}^{+}. Then there is B ⊆ A with size κ_{ω}^{+} and a finite set a ⊆ ω such that for all p in B, a_{p} = a. Let n = max(a). We claim that the set {p ↾ (n+1) : p ∈ B} is an antichain in ℝ^{n+1}, which will contradict the fact that ℝ^{n+1} is (κ_{n}^{<κ})^{+}-c.c. Suppose that p and q are in B and r is below both p ↾ (n + 1) and q ↾ (n + 1). Then r forces that p ↾ [n + 1, ω) and q ↾ [n + 1, ω) are both direct extensions of 1 and hence are compatible. Let u be a name such that r forces that u is below both q ↾ [n + 1, ω) and p ↾ [n + 1, ω). Then r u is below both p and q, which is impossible. □

Let us summarize the situation. In V [G], κ is a strong limit singular cardinal with cofinality ω. If μ is a cardinal in V such that κ < μ ≤ κ_{ω}, then μ is collapsed to κ in V [G]. The successor of κ_{ω} in V is preserved and becomes the successor of κ in V [G].

9. Iterating Radin forcing over different cardinals

Magidor [6] introduced a method for iterating Prikry forcing over different cardinals. In this section we show that the same method works more generally for Radin forcing.

First we show how to iterate Radin forcing using coherent sequences of arbitrary length. At each stage we use a coherent sequence derived from a supercompact ultrafilter. This works as we will see because the iteration preserves supercompactness. Afterwards we refine this construction, iterating Radin forcing defined from coherent sequences without appealing to a larger supercompact ultrafilter.

Suppose that A is a set of cardinals satisfying the following properties:

(1) There is a map α ↦ λ_{α} defined on A, and every β in A is closed under this map. Let A^{∗} denote the set ⋃{[α, λ_{α}^{+}] : α ∈ A}.

(2) For each α in A, there is a normal ultrafilter U_{α} on P_{α}(2^{2^{(λ{<α \atop α})}}) with elementary embedding j_{α} : V → Ult(V, U_{α}) such that min(j(A^{∗}) \ α) > λ_{α}. In particular, U_{α} concentrates on α \ A^{∗}. We would like to add a Radin club to P_{α}λ_{α} for each α in A.

Define by recursion an iterated forcing ⟨𝕄_{α}, ℝ_{α} : α ≤ sup(A)⟩, maintaining the following recursion hypotheses:

Recursion Hypotheses:

(1) If α is a strong limit cardinal closed under the map β ↦ λ_{β}, then |𝕄_{α}| ≤ 2^{α} and 𝕄_{α} is α^{+}-c.c.

Suppose that α is in A.

(2) The poset 𝕄_{α} forces there is a normal ultrafilter U_{α}^{∗} which extends U_{α}.

(3) The cardinals λ_{α}^{<α}, 2^{(λ_{α}^{<α})}, and 2^{2^{(λ{<α \atop α})}} are computed the same in both V and any generic extension by 𝕄_{α}.

(4) |𝕄_{α+1}| ≤ 2^{(λ_{α}^{<α})}.

Let α ≤ sup(A) and assume that 𝕄_{α} is defined and satisfies the recursion hypotheses. If α is not in A then let ℝ_{α} be a name for the trivial poset and define 𝕄_{α+1} = 𝕄_{α} ∗ ℝ_{α}.

Suppose that α is in A. Let G_{α} be generic for 𝕄_{α} over V. By the recursion hypotheses there is a normal ultrafilter U_{α}^{∗} in V [G_{α}] which extends U_{α}. By Lemma 1.5 and Proposition 2.2, for each ξ ≤ (2^{(λ_{α}^{<α})})^{+} there is a coherent sequence of ultrafilters U⃗ = ⟨U(β, i) : β ≤ α, i < oU⃗(β)⟩ on P_{α}λ_{α} such that each U(β, i) concentrates on β \ A^{∗}, the domain of oU⃗ is disjoint from A^{∗} ∩ α, and oU⃗(κ) = ξ. Choose some such coherent sequence on P_{α}λ_{α} and let ℝ_{α} be the Radin forcing defined from this coherent sequence. Let 𝕄_{α+1} = 𝕄_{α} ∗ ℝ_{α}. The poset 𝕄_{α} is α^{+}- c.c., has size no larger than 2^{α}, and forces that |ℝ_{α}| ≤ 2^{(λ_{α}^{<α})}. So |𝕄_{α+1}| ≤ 2^{(λ_{α}^{<α})}.

Suppose that α is a limit ordinal and 𝕄_{β} is defined for β < α. Define 𝕄_{α} as the set of functions p with domain α such that for all β < α, p ↾ β is in 𝕄_{β}; moreover, there exists a finite set a_{p} ⊆ α such that for β < α, β is in α \ a_{p} iff p ↾ β forces that p(β) is a direct extension 1 in ℝ_{β}. Let q ≤ p if q ↾ β ≤ p ↾ β for β < α. Let q ≤^{∗} p if for all β < α, q ↾ β forces that q(β) ≤^{∗} p(β).

The iteration 𝕄_{α} satisfies the Prikry property; for a proof see [2] or [6].

We verify the recursion hypotheses. Suppose that α is a strong limit cardinal closed under the map β ↦ λ_{β}. First assume sup(A) = γ < α. Then γ is a strong limit cardinal closed under the map β ↦ λ_{β}, so by recursion |𝕄_{α}| = |𝕄_{γ+1}| < α. Otherwise sup(A) = α. Then for each γ < α, |𝕄_{γ}| < α. So ⋃{𝕄_{γ} : γ < α} has size α, and therefore 𝕄_{α} has size at most 2^{α}.

Suppose that B ⊆ 𝕄_{α} with |B| = α^{+}. Since there are only α many finite subsets of α, there is a finite set a and C ⊆ B with size α^{+} such that a_{p} = a for all p in C. But there are fewer than α many possibilities for p ↾ a_{p}. Let p and q be in C such that p ↾ a = q ↾ a. Then p and q are compatible.

Lemma 9.1. Suppose that p is a condition in 𝕄_{α}, ξ < min(A), and at least one of the following two statements holds:

(1) p is a direct extension of 1,

(2) for all β in A ∩ α, p ↾ β forces that any a in P_{β}λ_{β} appearing in p(β) satisfies that a ∩ β > ξ. If {p_{i} : i < ξ} is a family of direct extensions of p, there is q such that q ≤^{∗} p_{i} for i < ξ.

Proof. For (1), write p_{i}(β) = ⟨Ȧ_{i}(β)⟩ for β in A ∩ α, where Ȧ_{i}(β) is a name for a subset of P_{β}λ_{β}. Define q by letting q(β) = ⟨ ⋂{Ȧ_{i}(β) : i < ξ} ⟩. Since ξ < min(A), the completeness of the ultrafilters implies that q is a condition in 𝕄_{α}. The proof of (2) is similar, using Lemma 3.1. □

Lemma 9.2. Suppose that p is a condition in 𝕄_{α} and ξ < min(A). Assume that for all β in A ∩ α, p ↾ β forces that whenever a in P_{β}λ_{β} is either in the finite sequence of p(β) or in a measure one set appearing in p(β), then a ∩ β > ξ. Then p forces that no new subsets of ξ are added.

Proof. Any q ≤ p satisfies (2) of Lemma 9.1. Suppose that q ≤ p and q forces ḟ : ξ → 2. For each i < ξ let q_{i} ≤^{∗} q decide the value of ḟ(i). Then there is r such that r ≤^{∗} q_{i} for i < ξ. Clearly r forces that ḟ is in the ground model. □

Now we verify the second recursion hypothesis.

Proposition 9.3. Suppose that α is in A. Then the poset 𝕄_{α} forces there is a normal ultrafilter U_{α}^{∗} which extends U_{α}.

Proof. If α is not a limit point of A, then |𝕄_{α}| < α, so U_{α} can be lifted. Suppose that α is a limit point of A.

Write j : V → M = Ult(V, U_{α}). Let β = min(j(A) \ α), which by assumption is larger than λ_{α}. Factor j(𝕄_{α}) = 𝕄_{α} ∗ 𝕄_{β,j(α)}. Write μ_{α} = 2^{2^{(λ{<α \atop α})}}. Note that β > μ_{α}.

Let G_{α} be generic for 𝕄_{α} over V. Define U_{α}^{∗} in V [G_{α}] by letting X be in U_{α}^{∗} iff X ⊆ P_{α}μ_{α} and there is a name Ẋ for X and s ≤^{∗} 1 in 𝕄_{β,j(β)} such that s forces over M[G_{α}] that j“μ_{α} ∈ j(Ẋ).

We show that the definition does not depend on the choice of Ẋ. Suppose that X = Ẋ^{G_{α}} = Ẏ^{G_{α}}. By the Prikry property for 𝕄_{β,j(α)}, there is sẊ (respectively sẎ) directly extending 1 which decides whether j“μ_{α} is in j(Ẋ) (respectively, in j(Ẏ)). Fix p in G_{α} which forces that Ẋ = Ẏ. Then j(a_{p}) = a_{p}, so j(p) = p r where r is a direct extension of 1 in 𝕄_{β,j(α)}. Moreover, r forces over M[G_{α}] that j(Ẋ) = j(Ẏ). Since r, sẊ, and sẎ are direct extensions of 1, they are compatible. So sẊ and sẎ must decide their respective statements the same way.

Similar arguments show that U_{α}^{∗} is an ultrafilter which extends U_{α}. We prove that U_{α}^{∗} is normal. Suppose that f : P_{α}μ_{α} → μ_{α} is regressive. Let ḟ be a name for f and fix p in G_{α} which forces that ḟ is regressive. Then j(p) = p r, where r is a direct extension of 1 and r forces over M[G_{α}] that j(ḟ) is regressive. For each i < μ_{α} choose s_{i} ≤^{∗} 1 in 𝕄_{β,j(α)} which decides the statement j(ḟ)(j“μ_{α}) = j(i). By Lemma 9.1 there is s such that s ≤^{∗} s_{i} for all i. Clearly there must exist i such that s forces j(ḟ)(j“μ_{α}) = j(i). But then {a ∈ P_{α}μ_{α} : f(a) = i} is in U_{α}^{∗}. □

The proof of the third recursion hypothesis is similar to the proof of Lemma 8.4. This completes the construction of the iteration. Suppose that α is in A and α′ is the least element of A greater than α. Write

𝕄_{sup(A)} = 𝕄_{α+1} ∗ 𝕄_{tail}. Unlike the case of an Easton support iteration of strategically closed posets or even a Magidor iteration of Prikry forcing, the poset 𝕄_{tail} will add bounded subsets to α′ over the generic extension by 𝕄_{α+1}. One might worry that this tail iteration will damage the universe below α′ in such a way that it ruins the effect achieved in forcing with 𝕄_{α+1}. It turns out this is not true, due to a nice factorization of the tail iteration.

Theorem 9.4. Suppose that α is in A and α′ = min(A\(α+1)). Write 𝕄_{sup(A)} = 𝕄_{α+1} ∗ 𝕄_{tail}. Let G_{α+1} be generic for 𝕄_{α+1} over V. Then in V [G_{α+1}], there is a dense set of q in 𝕄_{tail} such that 𝕄_{tail}/q factors into a finite iteration

ℙ_{0} ∗ ⋯ ∗ ℙ_{n}

such that for each i ≤ n, either ℙ_{i} does not add subsets to λ_{α}^{+} or ℙ_{i} is a Radin forcing on some P_{γ}γ′ where γ′ < α.

Proof. Working in V [G_{α+1}], let p be in 𝕄_{tail}. We define q ≤ p as follows. Fix ξ in A \ (α + 1). Let U⃗ be a name for the coherent sequence such that ℝ_{ξ} = ℝ(U⃗). Then p ↾ ξ forces that for all a in P_{ξ}λ_{ξ} appearing in the finite sequence of p(ξ), either a ∩ ξ < α or a ∩ ξ > λ_{α}^{+}. This is true since α is in A and therefore the interval [α, λ_{α}^{+}] is disjoint from the domain of oU⃗. Write p(ξ) = ⟨d⃗_{ξ}, Ȧ_{ξ}⟩. Define q(ξ) = ⟨d⃗_{ξ}, Ḃ_{ξ}⟩, where Ḃ_{ξ} = {a ∈ Ȧ_{ξ} : a ∩ ξ > λ_{α}^{+}}.

Write a_{q} \ {α′, sup(A)} = {ξ_{1}, …, ξ_{n−1}}, and let ξ_{0} = α′ and ξ_{n} = sup(A). For i ≤ n write s_{i} = q ↾ (ξ_{i}, ξ_{i+1}). Then 𝕄_{tail}/q factors as

(ℝ_{ξ_{0}}/q(ξ_{0})) ∗ (𝕄_{(ξ_{0},ξ_{1})}/s_{0}) ∗ … ∗ (ℝ_{ξ_{n−1}}/q(ξ_{n−1})) ∗ (𝕄_{(ξ_{n−1},ξ_{n})}/s_{n−1}) ∗ (ℝ_{ξ_{n}}/q(ξ_{n})).

By Lemma 9.2 and the definition of q, each 𝕄_{(ξ_{i},ξ_{i+1})}/s_{i} does not add any subsets to λ_{α}^{+}. By the definition of q, the Factorization Lemma, and Lemma 5.8(3), each ℝ_{ξ_{i}}/q(ξ_{i}) factors into a product of a Radin forcing below α by a Radin forcing which adds no subsets of λ_{α}^{+}. □

We show now that the stationarity of A is always destroyed. Let α be a limit point of A with cofinality greater than ω. Suppose that G_{α} is generic for 𝕄_{α} over V. For each ξ in A ∩ α let c_{ξ} denote the Radin club added by a generic for the Radin forcing on P_{ξ}λ_{ξ}.

Proposition 9.5. Define C_{𝕄_{α}} in V [G_{α}] as the set of β < α such that for all ξ ≥ β in A ∩ α, min(c_{ξ}) ∩ ξ ≥ β. Then C_{𝕄_{α}} is a club subset of α disjoint from A ∩ α.

Proof. If ξ is in A ∩ α then min(c_{ξ}) ∩ ξ < ξ, so ξ is not in C_{𝕄_{α}}. Suppose that C is a bounded subset of C_{𝕄_{α}} and let β = sup(C). We show that β is in C_{𝕄_{α}}. Consider ξ ≥ β in A∩α. For all β′ in C, min(c_{ξ})∩ξ ≥ β′. Therefore min(c_{ξ})∩ξ ≥ sup(C) = β.

To show that C_{𝕄_{α}} is unbounded in α, fix γ < α and let p be in G_{α}. Fix an ordinal β < α not in A ∩ α such that β is larger than max(a_{p}) and γ. Define q as follows. Let q ↾ (β + 1) = p ↾ (β + 1). Fix β < ξ < α. If ξ is not in A ∩ α then let q(ξ) = p(ξ). Otherwise write p(ξ) = ⟨Ȧ_{ξ}⟩. Let q(ξ) be a name for the condition ⟨ {a ∈ Ȧ_{ξ} : a ∩ ξ > β} ⟩. Then q ≤ p and q forces that β is in C_{𝕄_{α}}. □

Now we will refine this construction somewhat. Ideally we would like to do something like the following. Start with a set of cardinals A and a coherent sequence of ultrafilters on each α in A. Define the Magidor iteration as above. Maintain the recursion hypothesis that for each α in A, the coherent sequence on α can be lifted after forcing with 𝕄_{α}. This idea can be made to work when the coherent sequences are short using Proposition 2.1. But we do not know whether a long coherent sequence on α can be lifted after forcing with 𝕄_{α} without making further assumptions.

Suppose that A is a set of cardinals satisfying the following properties:

(1) For each α in A there is a ◁-increasing sequence ⟨U_{i}^{α} : i < β_{α}⟩ of normal ultrafilters on α, for some β_{α} ≤ α.

(3) Each U_{i}^{α} contains α \ A. Note that the hypotheses of Proposition 2.1 are satisfied.

Define by recursion an iterated forcing ⟨𝕄_{α}, ℝ_{α} : α ≤ sup(A)⟩ using the same definition as above. The only difference is that we need to prove for all α in A, 𝕄_{α} forces that the sequence ⟨U_{i}^{α} : i < β_{α}⟩ can be lifted in such a way that it still satisfies the hypotheses of Proposition 2.1. Clearly it suffices to prove that the lifted sequence is ◁-increasing. Then we apply Proposition 2.1 to obtain a coherent sequence U⃗ = ⟨U(β, i) : β ≤ α, i < oU⃗(β)⟩ on α in the generic extension by 𝕄_{α} such that oU⃗(α) = β_{α}, each U(β, i) contains β \ A, and the domain of oU⃗ is disjoint from A ∩ α. Let ℝ_{α} be the Radin forcing defined from this coherent sequence.

To prove that the sequence on α can be lifted, we need a more complicated recursion hypothesis. In addition to recursion hypotheses 1, 3, and 4 in the previous construction, we maintain the following.

Recursion Hypothesis: Suppose that α ≤ sup(A) is a limit point of A. Assume U is a normal ultrafilter on α which contains α\A. Write j_{U}(𝕄_{α}) = 𝕄_{α} ∗𝕄_{α,j_{U} (α)}.

Let G_{α} be generic for 𝕄_{α} over V. Define U^{∗} in V [G_{α}] by letting X be in U^{∗} iff X ⊆ α, there is a name Ẋ for X, and there is s ≤^{∗} 1 in 𝕄_{α,j_{U} (α)} such that s forces over M_{U}[G_{α}] that α ∈ j_{U}(Ẋ). Then U^{∗} is a normal ultrafilter on α which extends U. Moreover, if U ◁ U_{1} and j_{1} : V → M_{1} = Ult(V, U_{1}), then U^{∗} is in M_{1}[G_{α}] and satisfies the same definition in M_{1}[G_{α}] as it does in V [G_{α}].

Suppose that 𝕄_{α} is defined and α is a limit point of A. We verify the recursion hypothesis. Let U be a normal ultrafilter on α which contains α \ A. Define U^{∗} as in the statement of the recursion hypothesis. The proof that U^{∗} is a normal ultrafilter on α is similar to the proof of Proposition 9.3. Suppose that U ◁ U_{1} and j_{1} : V → M_{1} = Ult(V, U_{1}). Write k : M_{1} → N = Ult(M_{1}, U). Then j_{U} ↾ M_{1} = k by Lemma 1.4(2), so k(𝕄_{α}) = 𝕄_{α} ∗ 𝕄_{α,j_{U} (α)}. Define U^{∗∗} extending U in M_{1}[G_{α}] exactly as U^{∗} was defined in V [G_{α}], using k in place of j_{U}. Since 𝕄_{α} is α^{+}-c.c., M_{1}[G_{α}] is α-closed in V [G_{α}]. So U^{∗∗} really is an ultrafilter. We show that U^{∗∗} and U^{∗} are equal by proving U^{∗∗} ⊆ U^{∗}. Suppose Ẋ is a name in M_{1}[G_{α}], s is a direct extension of 1 in 𝕄_{α,j_{U} (α)}, and s forces over N[G_{α}] that α ∈ k(Ẋ). Then s forces over N[G_{α}] that α ∈ j_{U}(Ẋ). Since N[G_{α}] ⊆ M_{U}[G_{α}] by Lemma 1.4(1) and the fact that the statement α ∈ j_{U}(Ẋ) is upwards absolute, s forces over M[G_{α}] that α ∈ j_{U}(Ẋ). So Ẋ^{G_{α}} is in U^{∗}.

This completes the proof of the main recursion hypothesis. We omit the proofs of the other recursion hypotheses since they are the same as above.

Now we prove that when α is in A, 𝕄_{α} forces that the sequence of ultrafilters on α can be lifted as desired. Suppose that α is in A and 𝕄_{α} is defined. Let G_{α} be generic for 𝕄_{α} over V.

First assume that α is not a limit point of A. Then |𝕄_{α}| < α. So each U_{i}^{α} generates a normal ultrafilter (U_{i}^{α})^{∗} in the generic extension. We prove that ⟨(U_{i}^{α})^{∗} : i < β_{α}⟩ is ◁-increasing.

Fix α_{0} < α_{1} < β_{α}. We define g_{α_{0}}^{α_{1}} as follows. The domain of g_{α_{0}}^{α_{1}} is the set of ξ < α larger than |𝕄_{α}| such that f_{α_{0}}^{α_{1}}(ξ) is a normal ultrafilter in V on ξ. Let g_{α_{0}}^{α_{1}}(ξ) denote the ultrafilter generated by f_{α_{0}}^{α_{1}}(ξ) in V [G_{α}].

Write j_{α_{1}}^{∗} : V [G_{α}] → M_{α_{1}}^{∗} = Ult(V [G_{α}], (U_{α_{1}}^{α})^{∗}). We want to show that j_{α_{1}}^{∗}(g_{α_{0}}^{α_{1}})(α) = (U_{α_{0}}^{α})^{∗}. Write j_{α_{1}} : V → M_{α_{1}} = Ult(V, U_{α_{1}}^{α}). Since |𝕄_{α}| < α, j_{α_{1}}“G_{α} is generic for j_{α_{1}}(𝕄_{α}) over M_{α_{1}}, and j_{α_{1}}^{∗} is equal to the lift of j_{α_{1}} to V [G_{α}] → M_{α_{1}}[j_{α_{1}}“G_{α}] = M_{α_{1}}^{∗}. Therefore j_{α_{1}}^{∗}(f_{α_{0}}^{α_{1}})(α) = U_{α_{0}}^{α} and so j_{α_{1}}^{∗}(g_{α_{0}}^{α_{1}})(α) is the ultrafilter generated by U_{α_{0}}^{α} in M_{α_{1}}^{∗}, which is (U_{α_{0}}^{α})^{∗}.

Now assume that α is a limit point of A. For each i < β_{α} define (U_{i}^{α})^{∗} as in the statement of the recursion hypothesis. We prove that the sequence ⟨(U_{i}^{α})^{∗} : i < β_{α}⟩ is ◁-increasing.

Fix α_{0} < α_{1} < β_{α} and define g_{α_{0}}^{α_{1}} as follows. The domain of g_{α_{0}}^{α_{1}} is the set of ξ < α such that ξ is a limit point of A and f_{α_{0}}^{α_{1}}(ξ) is a normal ultrafilter in V on ξ which contains ξ \ A. Note that the domain of g_{α_{0}}^{α_{1}} is in U_{α_{1}}^{α}. Let g_{α_{0}}^{α_{1}}(ξ) be equal to (f_{α_{0}}^{α_{1}}(ξ))^{∗} as defined in the statement of the recursion hypothesis in the model V [G_{α} ↾ β]. Let ġ_{α_{0}}^{α_{1}} be a name such that 𝕄_{α} forces that ġ_{α_{0}}^{α_{1}} satisfies the definition of g_{α_{0}}^{α_{1}} just given.

Write j_{α_{1}} : V → M_{α_{1}} = Ult(V, U_{α_{1}}^{α}) and j_{α_{1}}^{∗} : V [G_{α}] → M_{α_{1}}^{∗} = Ult(V [G_{α}], (U_{α_{1}}^{α})^{∗}). Let U′ = j_{α_{1}}^{∗}(g_{α_{0}}^{α_{1}})(α). We want to show that U′ = (U_{α_{0}}^{α})^{∗}.

Lemma 9.6. The poset 𝕄_{α,j_{α_{1}} (α)} forces over M_{α_{1}}[G_{α}] that j_{α_{1}}(ġ_{α_{0}}^{α_{1}})(α) is equal to (U_{α_{0}}^{α})^{∗}.

Proof. By the recursion hypothesis, M_{α_{1}}[G_{α}] correctly computes (U_{α_{0}}^{α})^{∗}. Since j_{α_{1}}(f_{α_{0}}^{α_{1}})(α) = U_{α_{0}}^{α}, the lemma follows. □

Lemma 9.7. The set of ξ < α such that g_{α_{0}}^{α_{1}}(ξ) is an ultrafilter in V [G_{α}] is in (U_{α_{1}}^{α})^{∗}.

Proof. Let Ḃ be a name such that 𝕄_{α} forces that Ḃ is the set of ξ < α in the domain of ġ_{α_{0}}^{α_{1}} such that ġ_{α_{0}}^{α_{1}}(ξ) is an ultrafilter in V [G_{α}]. To show that Ḃ^{G_{α}} is in (U_{α_{1}}^{α})^{∗} it suffices to find s ≤^{∗} 1 in 𝕄_{α,j_{α_{1}} (α)} which forces over M_{α_{1}}[G_{α}] that j_{α_{1}}(g_{α_{0}}^{α_{1}})(α) is an ultrafilter. Since j_{α_{1}}(g_{α_{0}}^{α_{1}})(α) is in ultrafilter in M_{α_{1}}[G_{α}], it suffices to find s ≤^{∗} 1 which forces that no new subsets of α are added over M_{α_{1}}[G_{α}].

Define s as follows. For ξ in j_{α_{1}}(A) \ α let s(ξ) be a name for the maximal condition intersected with ξ \ (α + 1). By Lemma 9.2, s forces that every subset of α is in M_{α_{1}}[G_{α}]. □

It follows that U′ is an ultrafilter in M_{α_{1}}^{∗} and hence in V [G_{α}]. So to prove that U′ = (U_{α_{0}}^{α})^{∗}, it suffices to prove (U_{α_{0}}^{α})^{∗} ⊆ U′.

Lemma 9.8. If X ⊆ α in V [G_{α}], then X is represented in M_{α_{1}}^{∗} by the function β ↦ X ∩ β. So X is in U′ iff the set

{β < α : X ∩ β ∈ g_{α_{0}}^{α_{1}}(β)}

is in (U_{α_{1}}^{α})^{∗}.

The proof is standard. Suppose that X is in (U_{α_{0}}^{α})^{∗}. Let Ẋ be a name for X such that 𝕄_{α} forces that Ẋ is a subset of α. Let Ẏ be a name such that 𝕄_{α} forces

Ẏ = {β < α : Ẋ ∩ β ∈ ġ_{α_{0}}^{α_{1}}(β)}.

By Lemma 9.8, if Ẏ^{G_{α}} is in (U_{α_{1}}^{α})^{∗}, then X is in U′. By the definition of (U_{α_{1}}^{α})^{∗}, it suffices to prove there is s which directly extends 1 in 𝕄_{α,j_{α_{1}} (α)} such that s forces over M_{α_{1}}[G_{α}] that α is in j_{α_{1}}(Ẏ). In other words, s must force

j_{α_{1}}(Ẋ) ∩ α ∈ j_{α_{1}}(ġ_{α_{0}}^{α_{1}})(α)

By Lemma 9.6. it suffices to find s ≤^{∗} 1 which forces that

j_{α_{1}}(Ẋ) ∩ α ∈ (U_{α_{0}}^{α})^{∗}.

Since X is in (U_{α_{0}}^{α})^{∗}, the following lemma completes the proof.

Lemma 9.9. There is s ≤^{∗} 1 in 𝕄_{α,j_{α_{1}} (α)} which forces over M_{α_{1}}[G_{α}] that X is a subset of j_{α_{1}}(Ẋ) ∩ α.

Proof. Fix β in X. Let p_{β} be a condition in G_{α} which forces that β is in Ẋ. Write j_{α_{1}}(p_{β}) = p_{β} s_{β} where s_{β} ≤^{∗} 1 in 𝕄_{α,j_{α_{1}} (α)}. Then s_{β} forces over M_{α_{1}}[G_{α}] that j_{α_{1}}(β) = β is in j_{α_{1}}(Ẋ) ∩ α. So s_{β} forces that β is in the desired set. By Lemma 9.1 there is s such that s ≤^{∗} s_{β} for all such β. □

By Proposition 2.1, 𝕄_{α} forces that there is a coherent sequence on α as desired. Let ℝ_{α} be a name for the Radin forcing defined from such a sequence. This completes the construction of the iteration.

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Department of Mathematics, University of California, Berkeley

   E-mail address: jkrueger@math.berkeley.edu
   URL: http://www.math.berkeley.edu/∼jkrueger