Sharps and the Σ_{3}^{1} correctness of K

Ralf Schindler

Institut für Formale Logik, Universität Wien, 1090 Wien, Austria

                    rds@logic.univie.ac.at
              http://www.logic.univie.ac.at/^{∼}rds/

The purpose of the present paper is to present a new, simple, and purely combinatorial proof of the following result.

Theorem 0.1 (Steel-Welch 1993, [6, Theorem 4.1] ) Let A ⊂ ℝ be Π_{2}^{1}. Suppose that there is some sequence (x_{n}: n < ω) such that x_{0} ∈ A and for all n < ω, x_{n+1} = x_{n}^{#}. Suppose also that there is some N < ω such that there is no inner model with N strong cardinals. Then A ∩ K ≠ ∅.

Here, K denotes the core model; cf. the remark right after Definition 0.2. It is open whether Theorem 0.1 still holds if we replace the second sentence in its statement by “Suppose that there is some x ∈ A such that x^{#} exists.” It is also open whether Theorem 0.1 still holds if we replace the third sentence in its statement by “Suppose also that there is no inner model with a Woodin cardinal, but K exists” (cf. Definition 0.2).

We refer the reader to [4]. The current argument will exploit, among other things, the argument of [4].

Definition 0.2 Let A be a transitive model of ZFC. Then by K^{A} we denote the model which is recursively constructed inside A in the manner of [5, §6], if it exists (otherwise we let K^{A} undefined). If K^{A} ↓ then we say that K^{A} exists. If K^{V} ↓ then we write K = K^{V} and say that K exists.

It is shown in [3] that K exists if 0^{|^{•}} doesn’t exist. It is also shown in [5] that K^{M} exists if M = V_{Ω}^{H}, where H is a transitive model of “ZFC^{−} + Ω is measurable + there is no inner model with a Woodin cardinal” (in this case we’ll sometimes also write K^{H} for K^{M}).

We shall prove Theorem 0.1 with the third sentence in its statement being replaced by “Suppose also that 0^{¶} doesn’t exist.” We’ll leave a proof of Theorem 0.1 as stated as an exercise to the reader.

Definition 0.3 Let (♣) denote the following assertion. Let x ∈ ℝ be such that x^{#} exists. If K^{L[x]} and K^{L[x^{#}]} both exist and are coiterable then there is some α ∈ OR such that K^{L[x^{#}]}||α iterates past K^{L[x]}.

Lemma 0.4 Suppose that 0^{¶} doesn’t exist. Then (♣) holds.

Proof. Suppose that x ∈ ℝ witnesses the failure of (♣). It is fairly easy too see that then K^{L[x^{#}]} ⊧ “there is a strong cardinal.” (Cf. the proof of [6, Lemma 3.3].) Let us now assume that 0^{¶} doesn’t exist. We aim to derive a contradiction. Let us work in L[x^{#}].

Claim 1. cf(κ^{+K^{L[x]}}) = ω for all κ.

Proof. It is true that cf(κ^{+L[x]}) = ω for all κ. Let us thus fix some K^{L[x]}-cardinal κ such that κ^{+K^{L[x]}} < κ^{+L[x]}. Let λ = Card^{L[x]}(κ) ≤ κ. Then λ is neither an x-indiscernible nor singular in L[x]. Let η < λ be the largest x-indiscernible which is smaller than λ.

Let τ_{n} enumerate the Skolem terms of L[x]. The sequence (λ_{n}: n < ω), where

λ_{n} = sup({τ_{n}(ξ⃗): ξ⃗ < η} ∩ λ) < λ

witnesses that cf(λ) = ω. But we’ll have that cf^{L[x]}(κ^{+K^{L[x]}}) = λ by weak covering applied inside L[x] (cf. [2]).

□ (Claim 1)

Claim 2. K^{L[x]} doesn’t move in the comparison with K^{L[x^{#}]}.

Proof sketch. The point is that by our assumption K^{L[x]} absorbs all coiterable set-sized premice which exist in L[x^{#}]. Jensen’s argument yielding that below 0^{¶} any universal weasel is an iterate of K then gives this Claim.

□ (Claim 2)

It is now easy to see that Claims 1 and 2, combined with an application of weak covering applied inside L[x^{#}] (cf. [2]) yields the Lemma.

□ (Lemma 0.4)

The proof of Lemma 0.4 is certainly more interesting than its result. If we had assumed the existence of x^{##} then we could have just cited [6, Lemma 3.3]. We conjecture that (♣) still holds under much weaker assuptions than the non-existence of 0^{¶} (cf. [6, p. 188, Question3]).

We are now going to prove the following result, which will immediately give Theorem 0.1 (the third sentence in its statement being replaced by “Suppose also that 0^{¶} doesn’t exist”) via Lemma 0.4.

We emphasize that Theorem 0.5 is not given by the results of [6]; the proof of [6, Theorem 4.1] which is given in [6] heavily uses universal iterations which are not known to exist significantly above 0^{¶}.

Theorem 0.5 Let A ⊂ ℝ be Π_{2}^{1}. Suppose that there is some sequence x⃗ = (x_{n}: n < ω) such that x_{0} ∈ A and for all n < ω, x_{n+1} = x_{n}^{#}. Suppose also that there is no inner model with a Woodin cardinal, that K^{L[x⃗]} exists, and that (♣) holds. Then A ∩ K^{L[x⃗]} ≠ ∅.

Proof of Theorem 0.5. Let A = {z ∈ ℝ: Φ(z)} where Φ(−) is Π_{2}^{1}. There is a tree T ∈ K^{L[x⃗]} searching for a quadruple (y⃗, M⃗, T⃗ , σ⃗) such that the following hold true.

• y⃗ = (y_{n}: n < ω) ∈ ^{ω}ℝ,

• M⃗ = (M_{n}: n < ω) such that for all n < ω do we have the following: (a) M_{n} = (J_{α_{n}}[y_{n}]; ∈, y_{n}, U_{n}) for some α_{n}, U_{n},

(b) M_{n+1} ⊧ “M_{n} = y_{n}^{#}” (in particular, M_{n+1} thinks that M_{n} is iterable),

(c) y_{n+1} is the master code of M_{n},

(d) M_{0} ⊧ Φ(y_{0}), and

(e) setting κ = crit(U_{n+1}), there is an initial segment of K^{M_{n+1}} which iterates past K^{L_{κ}[y_{n}]}, and

• (T⃗ , σ⃗) witnesses that each individual K^{M_{n}} (n < ω) is iterable (cf. [4]), i.e., T⃗ = (T_{n}: n < ω), σ⃗ = (σ_{n}: n < ω), and for all n < ω do we have the following:

(a) T_{n} is a countable tree of successor length on K^{L[x⃗]}, and

(b) σ_{n}: K^{M_{n}} → M_{∞}^{T_{n}}||β_{n}, some β_{n} ≤ M_{∞}^{T_{n}} ∩ OR, is elementary.

We are now going to prove that

∅ ≠ p[T] = {y_{0}: ∃(y_{1}, y_{2}, ...)∃M⃗∃T⃗ ∃σ⃗((y_{0}, y_{1}, ...), M⃗, T⃗ , σ⃗) ∈ [T]} ⊂ A.

We may well leave the verification of p[T] ≠ ∅ as an exercise to the reader.

Now fix (y⃗, M⃗, T⃗ , σ⃗) ∈ T. Let y⃗ = (y_{n}: n < ω) and M⃗ = (M_{n}: n < ω). Let us prove that y_{0} ∈ A. Let (_{n}^{α}) denote the assertion that the α^{th} iterate of M_{n} is well-founded. It clearly suffices to prove the following.

Main Claim. For all α, for all n, (_{n}^{α}) holds.

Definition 0.6 Let n < ω. We write (M_{n}^{i}, π_{n}^{ij}: i ≤ j ≤ α) for the putative iteration of M_{n} of length α + 1, if it exists; and if so then for i < α we write κ_{n}^{i} for the critical point of π_{n}^{0i}(U_{n}), i.e., of the top extender of M_{n}^{i}. We call α a uniform indiscernible provided that for all n < ω, the putative iteration (M_{n}^{i}, π_{n}^{ij}: i ≤ j ≤ α) of M_{n} of length α + 1 exists and {κ_{i}: i < α} is (closed and) unbounded in α.

Proof of the Main Claim. We’ll prove the Main Claim by induction on α.

Case 1. α is not a uniform indiscernible.

Let n < ω. By our case assumption, there are some m > n and β < α such that α ∈ M_{m}^{β}. But M_{m}^{β} ⊧ “M_{n} = y_{n}^{#},” so that we may argue inside M_{m}^{β} and deduce that the α^{th} iterate of M_{n}, viz. M_{n}^{α}, is well-founded.

Case 2. α is a uniform indiscernible.

Let n < ω. Let κ = crit(U_{n+1}) = κ_{n+1}^{0}, and let P be the proper initial segment of K^{M_{n+1}} = K^{L_{κ}[y_{n+1}]} which iterates past K^{L_{κ}[y_{n}]}. Let (T , U) be the coiteration of K^{L_{κ}[y_{n}]} with K^{L_{κ}[y_{n+1}]}. (M_{n}^{i}, π_{n}^{ij}: i ≤ j < α) is the putative iteration of M_{n} of length α + 1. Let

σ: M_{n+1} →_{U_{n+1}} M′,

i.e., σ = π_{n+1}^{01} and M′ = M_{n+1}^{1}.

Let X ∈ P(κ) ∩ M_{∞}^{U}. Then X = π_{i∞}^{U}(X̄), some i < κ, X̄, and σ(X) = π_{i∞}^{σ(U)}(X̄) = π_{κ∞}^{σ(U)}(X). Therefore, π_{κ∞}^{σ(U)} ↾ κ^{+M_{∞}^{U}} = σ ↾ κ^{+M_{∞}^{U}}. The same argument shows that π_{n}^{κσ(κ)} ↾ κ^{+M_{n}^{κ}} = σ ↾ κ^{+M_{n}^{κ}}; we construe M_{n} in such a way that crit(U_{n})^{+M_{n}} = M_{n} ∩ OR, so that this latter equality means that π_{n}^{κσ(κ)} ↾ M_{n}^{κ} ∩ OR = σ ↾ M_{n}^{κ} ∩ OR. Let us write λ = σ(κ) = κ_{n+1}^{1}.

Now (σ(T ), σ(U)) is the coiteration of K^{L_{λ}[y_{n}]} with K^{L_{λ}[y_{n+1}]}. We’ll have that κ^{+M_{∞}^{U}} = κ^{+M_{κ}^{σ(U)}} = κ^{+M_{κ}^{σ(T )}} ≥ κ^{+K^{L_{λ}[yn]}} = κ^{+L_{λ}[y_{n}]} = κ^{+M_{n}^{κ}}, so that we get that

π_{n}^{κλ} ↾ M_{n}^{κ} ∩ OR = π_{κ∞}^{σ(U)} ↾ M_{n}^{κ} ∩ OR.

This now buys us that if we let (T^{∗}, U^{∗}) denote the coiteration of K^{L_{α}[y_{n}]} with K^{L_{α}[y_{n+1}]} then for typical i ≤ j < α (namely, for all i ≤ j ∈ {κ_{n+1}^{β}: β < α}) we’ll have that

π_{n}^{ij} ↾ M_{n}^{i} ∩ OR = π_{ij}^{U^{∗}} ↾ M_{n}^{i} ∩ OR.

Moreover, U^{∗} may be construed as an iteration of P. As P is iterable, we may thus conclude that the α^{th} iterate of M_{n}, viz. M_{n}^{α}, is well-founded (cf. the argument of [4]).

□ (Main Claim) □ (Theorem 0.5)

Using [1] it can be verified that Theorem 0.5 still holds if the assumption that K^{L[x⃗]} is being crossed out and the conclusion is being replaced by “Then there is some lightface iterable premouse M with A ∩ M ≠ ∅.

References

[1] Bill Mitchell, A universal extender model without large cardinals, preprint.

[2] Bill Mitchell and E.Schimmerling, Covering without countable closure, Math.Research Letters 2 (1995), pp. 595-609

[3] Ralf Schindler, The core model for almost linear iterations, Annals of Pure and Appl. Logic 116 (2002), pp. 207 - 274.

[4] Ralf Schindler, A simple proof of Σ_{3}^{1} correcntess of K, preprint.

[5] John Steel, The core model iterability problem, Lecture Notes in Logic ♯8, Springer-Verlag 1996.

[6] John Steel and Philip Welch, Σ_{3}^{1} absoluteness and the second uniform indiscernible, Israel Journal of Mathematics 104 (1998), pp. 157-190.