Rank-to-rank hypotheses

and the failure of GCH

Vincenzo Dimonte

Introduction

Regular Cardinals

λ

Open Questions

Rank-to-rank hypotheses and the failure of GCH

Vincenzo Dimonte

20 June 2012

Rank-to-rank hypotheses

and the failure of GCH

Vincenzo Dimonte

Introduction

Regular Cardinals

λ

Open Questions

Question

Is I0 consistent with GCH?

Answer Yes. Question How can GCH possibly fail with I3, I2, I1, I0.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction Regular Cardinals λ Open Questions

Theorem (Kunen, 1971) If j : V ≺ M, then M ≠ V . The critical sequence has an important role in the proof: Definition κ_{0} = crit(j), κ_{n+1} = j(κ_{n}), λ = sup_{n∈ω} κ_{n}. Kunen’s proof uses a choice function that is in V_{λ+2}. So Corollary There is no j : V_{η} ≺ V_{η}, with η ≥ λ + 2.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Remark: we are going to use a test case, i.e. 2^{κ} = κ^{++}. In all the following theorems, this can be extended to any 2^{κ} = E(κ) with E Easton function such that E ↾ λ is definable over V_{λ}. Moreover, we can extend the theorems to other combinatorial properties, like diamonds, V = HOD, etc.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

This leaves room for a new breed of large cardinal hypotheses: Definition

I3 iff there exists λ s.t. ∃j : V_{λ} ≺ V_{λ};

I2 iff there exists λ s.t. ∃j : V ≺ M, with V_{λ} ⊆ M and λ is the supremum of the critical sequence;

I1 iff there exists λ s.t. ∃j : V_{λ+1} ≺ V_{λ+1}.

I0 For some λ there exists a j : L(V_{λ+1}) ≺ L(V_{λ+1}), with crt(j) < λ

In what sense they are large cardinals? The critical point of j is measurable, n-huge, strongly compact in V_{λ}. λ is a strong limit cardinal (in fact, Rowbottom).

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Strategy:

Start with a model of I^{∗};

force the failure of GCH at regulars; take care of not destroying I^{∗}.

Theorem (Corazza, D., Friedman, Hamkins) Suppose I^{∗}. Then there exists a generic extension in which I^{∗} holds and 2^{κ} = κ^{++} for any regular cardinal.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Easy cases:

(V_{λ+1})^{V [G]} = V_{λ+1}: this case is trivial, j is still a witness in V [G]; ℙ ∈ V_{crt(j)}: define the extension k(τ_{G}) = k(τ)_{G}.

Note that crt(j) can be unboundedly large under λ. Therefore any cardinal different from λ can be Cohen-ized. What about λ?

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Strong limitation: Theorem (Solovay, 1974) Let κ be a strongly compact cardinal. Let λ be a singular strong limit cardinal greater than κ. Then 2^{λ} = λ^{+}.

Easy case: Theorem (Gitik, 2002) Suppose λ is a cardinal of cofinality ω such that {κ < λ : o(κ) ≥ κ^{+n}} is cofinal in λ for every n ∈ ω. Then there exists a cardinal preserving extension having the same bounded subsets of λ and satisfying 2^{λ} = λ^{++}.

Corollary Suppose I3(λ). Then there exists a cardinal preserving extension satisfying I3(λ) and 2^{λ} = λ^{++}.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Strong limitation: Theorem (Solovay, 1974) Let κ be a strongly compact cardinal. Let λ be a singular strong limit cardinal greater than κ. Then 2^{λ} = λ^{+}.

Easy case: Theorem (Gitik, 2002) Suppose λ is a cardinal of cofinality ω such that {κ < λ : o(κ) ≥ κ^{+n}} is cofinal in λ for every n ∈ ω. Then there exists a cardinal preserving extension having the same bounded subsets of λ and satisfying 2^{λ} = λ^{++}.

Corollary Suppose I3(λ). Then there exists a cardinal preserving extension satisfying I3(λ) and 2^{λ} = λ^{++}.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Change of strategy:

Start with a model of the failure of GCH at regulars; force I1 on a regular; take care of not destroying 2^{κ} = κ^{++}.

Theorem (D., Friedman) Suppose I 0(λ), witnessed by j. Let ℙ be the Prikry forcing on the critical point of j. Then there exists a generic extension in which I 1(crt(j)) and 2^{crt(j)} = crt(j)^{++} hold.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Theorem (D., Friedman) Suppose I 0(λ), witnessed by j with critical point κ. Let δ be the minimum such that J_{δ}(V_{λ+1}) ≺_{Σ_{1}} J_{δ+1}(V_{λ+1}) and α < δ. Let ℙ be the Prikry forcing on the critical point of j. Then there exists a generic extension in which ∃k : L_{α}(V_{κ+1}) ≺ L_{α}(V_{κ+1}) and 2^{κ} = κ^{++} hold.

Corollary (Woodin) Suppose I 0(λ) and that the U(j)-conjecture is true. Let ℙ be the Prikry forcing on the critical point of j. Then there exists a generic extension in which I 0(κ) and 2^{κ} = κ^{++} hold.

Rank-to-rank hypotheses

and the failure of GCH Vincenzo Dimonte

Introduction

Regular Cardinals λ

Open Questions

Does (ZFC + I 1) imply (ZFC + ∃λ (I 1(λ) ∧ 2^{λ} > λ^{+}))? Is ZFC + ∃λ (I 1(λ) ∧ 2^{λ} > λ^{+} ∧ ∀κ < λ 2^{κ} = κ^{+}) consistent? If so, what is its consistency strength? Is ZFC + ∃λ (I 0(λ) ∧ 2^{λ} > λ^{+}) consistent? If so, what is its consistency strength?