I0 and Rank-Into-Rank Axioms

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2 Introduction

R In 1984 Woodin showed [... ] that ADL( ) follows from a proposition stronger than I1 and straining the limits of consistency [... ] More- over, Woodin’s investigations of I0 was to produce a detailed and coherent structural theory for L(Vδ+1).

With attention shifted upward to the strongest large cardinal hypotheses, Woodin boldly formulated I0 just at the edge of the Kunen inconsistency and ampliﬁed Martin’s embeddings proof to establish the following in early 1984:

R Theorem 2.1 (Woodin). Assume I0. Then ADL( ). This sparse lines in Kanamori’s The Higher Infinite [21] were probably for years the only published trace of the mysterious I0. This fact is even more surprising as [21] was written in 1994, so ten years after the striking result and even more from its first formulation. What happened? Why all this mystery? I0 was just one of the many large cardinals that were created in a short span of time: strong cardinals, supercompact cardinals, huge cardinals, Woodin cardinals, rank-into-rank axioms (its smaller siblings), ... were introduced in a rapid fire, shifting the paradigms of set theory. But I0 could not keep up R the pace with all of them: its main victory was short-termed, as ADL( ) was proven to be of much lower consistency strength, there were no other examples of important propositions implied by I0, and it was definitely too large for inner model theory (in fact, the larger cardinal axiom of all times). So, before even a publication, it quietly disappeared, and with it the proof of Theorem 2.1. Still, under the surface there was some activity: the bound between I0 and R ADL( ) ran deeper than just Theorem 2.1, it involved the whole structure of a model of the form L(Vλ+1), that under I0 had properties very similar than those in L(R) under AD. This peculiarity was explored by Woodin, and the first output was Kafkoulis’ lecture notes An AD-like axiom [39] (at that time I0 was called Axiom A), a transcription of Woodin’s lectures on the argument, that started to be copied and distributed among mathematicians at the beginning of the Nineties. This underground distribution was, of course, very limited (Internet was at its infancy). In the meantime the rank-into-rank axioms were enjoying a period of pop- ularity, thanks to their connections with algebra, and Laver, in 2001, tried to push some of his results on rank-into-rank (in particular reflection) towards I0. In [28] we can find therefore the first officially published fragments of the theory of I0, taken from Kafkoulis notes, and in 2004 Kafkoulis himself published a paper on Coding Lemmata that built on such notes [20]. After that, still silence for many years. The rhythm of publications changed abruptly in 2011. The reason is Wood- ins’ Suitable extender models magnum opus. The aim of such paper (published in two parts) was to push inner model theory up to all large cardinals, so it was

2 necessary to know as much as possible as the top of the hierarchy. Therefore in the sprawling 561 pages there are whole sections dedicated to I0, the structure of L(Vλ+1) under it (in a much more developed way respect to Kafkoulis notes) and even stronger axioms. There was a new exciting realm to explore, finally mature, and with connections to other fields of set theory. The people that were Woodin’s students in that period (the author included) continued the work, expanding the research on I0 in different directions. Such a rapid expansion, after a long silence, carries the risk of excessive fragmentation. It is then time to take stock of the situation and pose solid bases on such a venture, not based on underground notes, folklore, and oral tradition. The objective of this paper is therefore to collect all the results on rank-into-rank axioms and I0, with a stress on the latest one, both from the I0 folklore and from the published papers, so that anybody can have all the tools for reading the latest papers or doing research on her own. Section 3 glances briefly at some important large cardinals, up until the first inconsistency in the large cardinal hierarchy, i.e., the (non-)existence of j : V ≺ V . Starting from this, Section 4 is about rank-into-rank axioms, i.e., axioms of the form j : Vλ ≺ Vλ (I3) and j : Vλ+1 ≺ Vλ+1 (I1), with a particular attention to the application operator. In Section 5 I0 is introduced, and with it all the structural properties of L(Vλ+1) that we can have for free. Much of this part is taken from [39]. Section 6 is a detailed analysis of many (all?) axioms from I3 to I0, with all the proofs of strong implications among them. The techniques of inverse limits and square roots are introduced, that are now a focal centre of the investigations around I0 (see [3]). Section 7 uses again [39] as main source, introducing all the results about R similarities between I0 and ADL( ), including the newest developments: measurability of λ+, Coding Lemma, large cardinal properties of Θ, perfect set property, an analogous of weakly homogenous Suslin-ness. In Section 8 all these axioms are treated in relation to forcing, with many results of indestructibility and independence results. In Section 9 it is noticed how such independence result can actually destroy R the similarities between I0 and ADL( ): the examples treated are the Ultrafilter Axiom, Wadge Lemma, a partition property and Turing Determinacy. Section 10 casts a quick glance on axioms stronger than I0, introduced in [41]: the Icarus sets. Section 11 describes highly speculatively some possible future lines of inquiry.

3 Kunen’s Theorem

To reach the top of the large cardinal hierarchy, it is worth to have a glance at the lower levels before. In the bottom, large cardinal axioms are very combinatoric in nature: an inaccessible cardinal is a cardinal that is regular and closed under exponentiation, a Ramsey cardinal is a cardinal that satisﬁes a particular partition property, a Rowbottom cardinal is a cardinal that satisﬁes another partition property, and so on2. The pivotal moment was when Scott proved

2Admittedly, this is a very rough classiﬁcation, as for example indescribable cardinals are diﬃcult to insert in this narrative

3 that a combinatoric property on a cardinal is equivalent to a global property of the universe: Definition 3.1 (Ulam, 1930). A cardinal κ is measurable iff there exists a κ-complete ultrafilter on κ. Definition 3.2. Let M, N be sets or classes. Then j : M → N is an elementary embedding and we write j : M ≺ N iff it is injective and for any formula ϕ and a1,...,an ∈ M, M ϕ(a1,...,an) iff N ϕ(j(a1),...,j(an)). Fom now on, if we write j : M ≺ N we mean that j is not the identity. So an elementary embedding is a homeomorphism of the logical structure. In particular, M and N must satisfy the same sentences, so for example M ZF iff N ZF. It is easy to see that for all ordinals α, j(α) ≥ α. If M AC or N ⊆ M, than it is possible to prove that there must exist an ordinal α such that j(α) > α. Definition 3.3. Let M, N be sets or classes such that M AC or N ⊆ M. Then the critical point of j, crt(j), is the least α such that j(α) > α. Theorem 3.4 (Scott, 1961). A cardinal κ is measurable iff there exist an inner model M and a j : V ≺ M such that crt(j)= κ. This was the start of an intense research on new breed of large cardinal hypotheses, defined more requirements for M. For example, one can ask how much of the original universe is in M: Definition 3.5 (Gaifman, 1974). A cardinal κ is γ-strong iff there exists j : V ≺ M, crt(j)= κ, γ < j(κ) and Vκ+γ ⊆ M. Or one can ask how close M is. The starting point is to notice that if j : V ≺ M witnesses that κ is measurable, then M κ ⊆ M. Definition 3.6 (Solovay, Reinhardt). A cardinal κ is γ-supercompact iff there exists j : V ≺ M, crt(j)= κ, γ < j(κ) and M γ ⊆ M. It is immediate to see that if κ is γ-supercompact then it is γ-strong. All these definitions fix j(κ) as the limit of the closeness of M. Overcoming this means to define stronger large cardinals: Definition 3.7 (Gaifman, 1974). A cardinal κ is superstrong iff there exists j : V ≺ M, crt(j)= κ and Vj(κ) ⊆ M. Definition 3.8 (Kunen, 1972). A cardinal κ is huge iff there exists j : V ≺ M, crt(j)= κ and M j(κ) ⊆ M. We can go on, defining stronger and stronger large cardinals: Definition 3.9. Let M, N be sets or classes such that M AC or N ⊆ M. Then the critical sequence of j : M ≺ N, hκn : n ∈ ωi, is defined as:

• κ0 = crt(j);

• κn+1 = j(κn). Deﬁnition 3.10. A cardinal κ is n-superstrong iﬀ there exists j : V ≺ M,

crt(j)= κ and Vj(κn) ⊆ M.

4 Definition 3.11. A cardinal κ is n-huge iff there exists j : V ≺ M, crt(j)= κ and M j(κn) ⊆ M. This ascent eventually must stop. In 1969 Reinhardt, in his thesis, asked whether it is possible to have directly j : V ≺ V . Kunen, few months later, proved that it is inconsistent. Such result must be approached cautiously: if we work in ZFC, it is not surprising: Theorem 3.12 (Naive Kunen’s Theorem). (ZFC) There is no j : V ≺ V . Proof by Suzuki, [38]. As we are working in ZFC, it means that j must be a class that is definable with some first order formula. Let ϕ be a formula. Let a be a parameter so that ϕ(a) defines an elementary embedding j from V to itself with least critical point κ among all the possible choices of the parameter. Then κ is definable in V , without parameters. So j(κ) satisfies the same definition, and it is the least critical point of embeddings definable with ϕ. But j(κ) >κ, contradiction. Kunen’s Theorem is stronger than this, as it involves also classes that are not definable: in the original paper, [24], the proof is in MK, Morse-Kelley theory. It is possible to weaken the theory to NBG, or to ZFC(j), i.e., ZFC with j as a predicate that can be inserted in replacement formulas. We leave the interested reader to [23], where all such problems are solved in great detail, and we state Kunen’s Theorem in vague terms, with the only provisos that j is not necessarily definable in V , replacement for j holds and AC holds in V . Theorem 3.13 (Kunen’s Theorem, [24]). (AC) There is no j : V ≺ V . There are many proofs of this result. Instead of the original proof by Kunen, we show the proof by Woodin ([21], Theorem 23.12, second proof), as it gives information that will be useful in Section 5. It builds on the following classical result by Solovay: Theorem 3.14 (Solovay). Let κ>ω be a regular cardinal, and X ⊆ κ stationary. Then X can be partitioned in κ disjoint stationary sets, i.e., there is hXη : η<κi such that X = η<κ Xη and all the Xη’s are pairwise disjoint. It will be useful this technicalS remark, that will be used constantly in the paper without comment: Remark 3.15. Let j : M ≺ N, M AC or N ⊆ M. Then for any γ < crt(j) and any sequence haα : α<γi ∈ M, j(haα : α<γi) = hj(aα): α<γi. So if A ∈ M is well-orderable (for example M AC) and |A| < crt(j), then j(A)= {j(a): a ∈ A}. Proof. Consider the sequence as a function f ∈ M with domain γ. As γ < crt(j), j(f)(α)= j(f)(j(α)) = j(f(α)). Proof of 3.13 by Woodin, [39]. Let λ be the supremum of the critical sequence of j. Then λ must be a fixed point for j, because

j(λ)= j(suphκn : n ∈ ωi) = sup j(hκn : n ∈ ωi) = suphj(κn): n ∈ ωi = λ.

λ+ + λ+ Let Sω = {α < λ : cof(α) = ω}. Note that Sω is stationary: if C is a club in λ+, than any ω-limit of elements of C must be in C, as λ+ is regular, and

5 3 λ+ has coﬁnality ω . We prove that Sω cannot be partitioned in crt(j) disjoint stationary sets, in contradiction with Solovay’s result. λ+ Suppose that hSα : α < crt(j)i is a partition in stationary subsets of Sω . Then j(hSα : α< crt(j)i)= hTβ : β < j(crt(j))i

λ+ + + + must be a partition in stationary subsets of Sω (because j(λ )= j(λ) = λ ) of length j(crt(j)). Let C = {α < λ+ : j(α)= α}. Then C is an ω-club in λ+: If hαn : n ∈ ωi is a sequence of ordinals in C, then

j(suphαn : n ∈ ωi) = sup j(hαn : n ∈ ωi)=

= suphj(αn): n ∈ ωi = suphαn : n ∈ ωi.

+ To see that it is unbounded, let α < λ , and consider the sequence α0 = α, αn+1 = j(αn). Then suphαn : n ∈ ωi ∈ C. As Tcrt(j) is stationary, it must intersect the closure of C in some η. But as η ∈ Tcrt(j), it has cofinality ω, so it must be in C, therefore j(η) = η. Since hSα : α< crt(j)i was a partition, there must be some α such that η ∈ Sα. But then η = j(η) ∈ j(Sα)= Tj(α). But crt(j) cannot be in the image of j (it is not a fixed point, and below it they are all fixed points), so Tcrt(j) 6= Tj(α), so they should have been disjoint, but they have η in common. Contradiction. We can now analyze such proof, to understand what it really needs of the hypotheses, with the objective of defining new axioms for which the proof does not work, so that there is a hope that they are not inconsistent. The first remark is that Solovay’s Theorem uses AC in an essential way, so the proof would not stand without AC.

Definition 3.16. (ZF)[Reinhardt] A cardinal κ is Reinhardt iff there exists j : V ≺ V , with crt(j)= κ. It is not immediate to see, but a Reinhardt cardinal is consistently stronger than all the hypotheses we will introduce in this paper. It is still open whether a Reinhardt cardinal would imply a contradiction, and the study of Reinhardt and even larger cardinals is still going on. Another approach would be to weaken the ground theory, for example not permitting replacement with j, so that the critical sequence is not in V and λ is not definable. This is the “Wholeness Axiom”, and it was studied by Paul Corazza ([1]). A third way would be to consider something smaller than V (but large enough so that the eventual large cardinal would sit at the top of the hierarchy). This will be implemented in the next section.

4 Rank-into-rank embeddings

Analyzing the proof of 3.13, one realizes that the key object is hSα : α< crt(j)i. Now, if β < λ+, there is a well-order of λ that has length β, and such well-order can be coded as a subset of λ. Therefore every ordinal less then λ+ can be

3 λ+ In fact, Sω is ω-stationary, that is, it intersects all the ω-clubs, i.e., the sets that are coﬁnal in λ+ and closed under supremum of ω-sequences

6 + coded as an element of Vλ+1, and a subset of λ can be coded as an element of Vλ+2. So the whole proof can be carried on for:

Corollary 4.1. For any η, there is no j : Vη+2 ≺ Vη+2.

Proof. Suppose that there is j : Vη+2 ≺ Vη+2. Let λ be the supremum of its critical sequence. Then λ ≤ η, because λ is limit. Then Vλ+2 ⊆ Vη+2, and the proof of 3.13 gives the contradiction. Therefore the proof of 3.13 does not exclude the following cases:

I3 There exists j : Vλ ≺ Vλ;

I1 There exists j : Vλ+1 ≺ Vλ+1.

There is a small ambiguity in the deﬁnition of I3. On one hand λ must be at least the supremum of the critical sequence (because ran(j) ⊆ Vλ), on the other hand λ must be less then the supremum of the critical sequence +2, because of the Corollary, but, as it is written, it can be the supremum of the critical sequence +1. In general we will assume, without loss of generality, that λ is the supremum of the critical sequence:

I3 There exists j : Vλ ≺ Vλ, where λ is the supremum of its critical sequence.

Admittedly, not being excluded by a proof is not the most solid of foundations for new axioms. In fact, the “I” in their name stands for “Inconsistency”, as this was the ﬁrst thought of the community when they were introduced. Yet, they survived all attempts of proving their inconsistency, and they are now in the center of a rich and complex theory, that has many interesting results, some of which will be shown in this paper. Let j : Vλ ≺ Vλ. The mental picture is the following:

λ λ . . . .

κ3 . κ3 .

κ2 κ2

κ1 κ1

κ0 κ0

so Vλ is divided in two: under a certain rank, all sets are ﬁxed, above a certain rank, all sets are moved. There are ω “stripes”, and all the sets on one stripe are moved into the next. There is one stripe, then, that contains no image of j. The members of the critical sequence are large cardinals:

7 Remark 4.2. Let hκn : n ∈ ωi be the critical sequence of j : Vλ ≺ Vλ. Then all the κn’s are measurable.

Proof. Consider the ultraﬁlter Uj = {X ∈P(κ0): κ1 ∈ j(X)}. As usual, it is a κ0-complete measure, so κ0 is measurable. Then by elementarity j(κ0)= κ1 is measurable, and so on.

Therefore λ is strong limit: if α < λ, then there is a κn such that α<κn, α and 2 < κn < λ. So |Vλ| = λ, and we can code every pair of sets in Vλ (in fact every ﬁnite sequence of sets) with a single set of Vλ. Therefore we can code every λ-sequence of elements in Vλ+1 as a single element in Vλ+1 (coding the sequence haη : η < λi with {(η, x): x ∈ aη}). The “stripes” structure permits to extend j beyond its scope:

+ Deﬁnition 4.3. Let j : Vλ ≺ Vλ. Deﬁne j : Vλ+1 → Vλ+1 as

+ ∀A ⊂ Vλ j (A)= j(A ∩ Vβ ). β<λ [ While it is not clear whether j+ is an elementary embedding, every elementary embedding from Vλ+1 to itself is the ‘plus’ of its restriction to Vλ:

+ Lemma 4.4. If j : Vλ+1 ≺ Vλ+1, then (j ↾ Vλ) = j. Thus every j : Vλ+1 ≺ Vλ+1 is deﬁned by its behaviour on Vλ, i.e., for every j, k : Vλ+1 ≺ Vλ+1,

j = k iﬀ j ↾ Vλ = k ↾ Vλ.

Proof. The critical sequence hκn : n ∈ ωi is a subset of Vλ, so it belongs to

Vλ+1. But then for every A ⊆ Vλ, {A ∩ Vκn : n ∈ ω} ∈ Vλ+1, so

+ (j ↾ Vλ) (A)= (j ↾ Vλ)(A ∩ Vκn ) n∈ω [

= j(A ∩ Vκn ) n∈ω [

= j( (A ∩ Vκn )) n∈ω [ = j(A).

It is worth noting that there is a strong connection between ﬁrst-order formulas in Vλ+1 and second-order formulas in Vλ. In fact, all the elements of Vλ+1 are subsets of Vλ, so they can be replaced with predicate symbols:

Lemma 4.5. Let A ∈ Vλ+1 \ Vλ and ϕ(v0, v1,...,vn) be a formula. Fix Aˆ ∗ a predicate symbol, and deﬁne ϕ (v1,...,vn) in the language of LST expanded with Aˆ as following:

• for every occurrence of v0, substitute Aˆ; • for every non-bounded quantiﬁed variable x, substitute every occurrence of x with X, a second-order variable.

8 Then for every a1,...,an ∈ Vλ

∗ Vλ+1 ϕ(A, a1,...,an) iﬀ (Vλ, A) ϕ (a1,...,an).

Proof. The proof is by induction on the complexity of ϕ. The previous Lemma is key in clarifying the relationship between elementary + embeddings in Vλ and Vλ+1, and to ﬁnally prove that j is a Σ0 elementary 1 embedding from Vλ+1 to itself (or, alternatively that j is a Σ0 elementary embedding from Vλ to itself).

+ Theorem 4.6. Let j : Vλ ≺ Vλ. Then j : Vλ+1 → Vλ+1 is a ∆0-elementary embedding.

Proof. Let Aˆ be a symbol for an 1-ary relation,a ˆ1,..., aˆn symbols for constants and let ϕ be a formula in LST∗, the language of LST expanded with Aˆ and aˆ1,..., aˆn. We want to prove that for any A ⊆ Vλ:

+ (Vλ,A,a1,...,an) ϕ iﬀ (Vλ, j (A), j(a1),...,j(an)) ϕ.

Actually we will prove only one direction for every ϕ, the other one following by considering ¬ϕ. mi First of all, we skolemize ϕ, so we ﬁnd f1,...,fm functions, fi : (Vλ) → Vλ, ∗ ∗ such that ϕ is equivalent to a formula ϕ in LST expanded with fˆ1,..., fˆm, ∗ ′ ′ where ϕ is ∀x1∀x2 ... ∀xm ϕ , with ϕ a ∆0 formula. ′ Let t0(x1,...,xm),...,tp(x1,...,xm) be all the terms that are in ϕ . We can express as a logical formula the phrase “ti(x1,...,xm) exists“: we work by induction on the tree of the term, writing a conjuction of formulas in this way

′ • every time that there is an occurrence of a function, i.e. fˆi(t ), we add to ′ the formula ∃y fi(t )= y;

• every time that there is an occurrence of a constant, i.e.a ˆi, we add to the formulaa ˆi 6= ∅, if ai 6= ∅; otherwise we don’t write anything. Then we have that

∗ (Vλ,A,a1,...,an,f1,...,fm) ϕ iﬀ ′ ′ ∀δ < λ (Vδ , A ∩ Vδ,a1,...,an,f1 ∩ Vδ,...,fm ∩ Vδ) ′ ∀x1,..., ∀xm( ti(x1,...,xm) exists → ϕ ), i

9 So

(Vλ,A,a1,...,an) ϕ ∗ ↔ (Vλ,A,a1,...,an,f1,...,fm) ϕ ′ ′ ↔ ∀δ < λ (Vδ, A ∩ Vδ,a1,...,an,f1 ∩ Vδ,...,fm ∩ Vδ) ′ ∀x1,..., ∀xm ( ti(x1,...,xm) exists → ϕ ) i

The only thing left to prove is the right direction of the fifth line. The bump + is in the fact that if a j (fi) is not total, then it would be possible to have the equation in the fourth line satisfied because the relevant terms do not exist, + but the fifth line not satisfied on the elements not in the domain of j (fi). We + prove that actually the j (fi) are total, in fact even more: + Claim 4.7. If j : Vλ ≺ Vλ and f : Vλ → Vλ, even partial, then j (dom(f)) = dom(j+(f)). Proof of claim. By definition,

+ j (dom(f)) = j(dom(f) ∩ Vκm ), m∈ω [ while

+ dom(j (f)) = dom( j(f ∩ Vκm )) = m∈ω [

= dom(j(f ∩ Vκm )) = j(dom(f ∩ Vκm )). m∈ω m∈ω [ [ It is not immediate to see why the two are the same, because it is possible that dom(f) ∩ Vκm and dom(f ∩ Vκm ) are diﬀerent, as on the left we are cutting the dominion, on the right both the dominion and codominion. The solution is to divide everything in even smaller pieces, that would be the same on the left and on the right side.

dom(f) ∩ Vκm = dom(f ∩ Vκn ) ∩ Vκm , n∈ω [ and

dom(f ∩ Vκm )= dom(f ∩ Vκm ) ∩ Vκn . n∈ω [ Therefore:

10 + j (dom(f)) = j(dom(f) ∩ Vκm )= j( dom(f ∩ Vκn ) ∩ Vκm )= m∈ω m∈ω n∈ω [ [ [

= j(dom(f ∩ Vκn ) ∩ Vκm )= j(dom(f ∩ Vκn ) ∩ Vκm )= m∈ω n∈ω n∈ω m∈ω [ [ [ [ + = j(dom(f ∩ Vκn )) = dom(j (f)). n∈ω [

The third and ﬁfth equivalence are a delicate point. In the third equivalence,

the ω-sequence hdom(f ∩ Vκn ) ∩ Vκm : n ∈ ωi ∈ Vλ, even if it is defined with κn, therefore j behaves properly. In the fifth equivalence, what seems an infinite union is in fact finite.

+ Therefore every j : Vλ ≺ Vλ can be extended to a unique j : Vλ+1 → Vλ+1 that is at least a ∆0-elementary embedding. Moreover it is possible to prove

• I3 holds iﬀ for some λ there exists a j : Vλ+1 → Vλ+1 that is a ∆0- elementary embedding;

• I1 holds iﬀ for some λ there exists a j : Vλ+1 ≺ Vλ+1 that is a Σn- elementary embedding for every n

+ Theorem 4.6 implies that if j, k : Vλ ≺ Vλ, then j (k): Vλ ≺ Vλ. This operation between elementary embeddings is called application, and we write j+(k) = j · k. Coupled with composition (not to be mistaken to!) they create an interesting algebra.

Theorem 4.8 (Laver, 1992 [25]). Fix λ and let Eλ = {j : Vλ ≺ Vλ}. For all j ∈Eλ the closure of {j} in (Eλ, ·) is the free algebra generated by the law

(Left Distributive Law) i · (j · k) = (i · j) · (i · k).

Moreover Laver proved that the free algebra generated by the laws above satisﬁes the word problem. His proof used extensively I3, but later Dehornoy ([6]) managed to prove the same thing in ZFC. A proof of this can be found in the Handbook of Set Theory [16], Chapter 11. There are still open problems at the I3 level:

Question 4.9. Are there j, k : Vλ ≺ Vλ such that the algebra generated by them is free? Another interesting result in I3 regards a function on the integers. Consider j : Vλ ≺ Vλ with critical point κ and let Aj be the closure of j in (Ej , ·). Then deﬁne f(n)= |{crt(k): k ∈ Aj , κn < crt(k) <κn+1}|.

Then f(0) = f(1) = 0 and f(2) = 1, because the simplest element of Aj that has a critical point not in the critical sequence of j is ((j · j) · j) · (j · j). However f(3) is very large. Laver ([26]) proved that for any n, f(n) is ﬁnite, but f dominates the Ackermann function, so f cannot be primitive recursive ([13]).

11 A standard way of using application is to consider j, j(j), j(j(j)). . . , and 0 n+1 n we write j = j and j = j(j ). If hκn : n ∈ ωi is the critical sequence of j, then by elementarity, crt(j(j)) = j(crt(j)) = κ1 and hκn : n ≥ 1,n ∈ ωi 2 n is the critical sequence of j . By induction, crt(j ) = κn. This is the reason why, when talking about I3 or I1, the critical point is never mentioned, when usually it is the main object of research (see strongness, supercompactness. . . ): if there exists a λ that witnesses I3, then there are inﬁnite critical points tied to the same λ, so it is really λ the relevant cardinal, while the critical points are interchangeable and unbounded in λ. Still, they carry a lot of strength. The following result shows that I3 really sits on top of the large cardinal hierarchy:

Proposition 4.10. Let hκi : i<ωi be the critical sequence of j. Then for every i, κi is γ-supercompact for any γ < λ and n-huge for every n. We use the characterization via ultrafilter: Proposition 4.11 (Solovay, Reinhardt). A cardinal κ is γ-supercompact iff there exists a fine, normal ultrafilter on Pκ(λ). It is n-huge iff there is a κ- complete normal ultrafilter U over some P(η) and κ = η0 < η1 < ··· < ηn = η such that for any i

Proof of 4.10. Let κ0 ≤ γ < λ, and suppose γ<κn; then it is routine to check that γ-supercompactness of κ0 is witnessed by

n ′′ n U = {X ⊆Pκ0 (γ) : (j ) γ ∈ j (X)} and that n-hugeness of κ0 is witnessed by

′′ U = {X ⊆P(κn): j κn ∈ j(X)} and κ0,...,κn.

Then κn is γ-supercompact and n-huge by elementarity. On the other hand λ (the “true” rank-into-rank cardinal) does not enjoy many large cardinal properties, as it is of coﬁnality ω, and usually large cardinals are regular. The most one can say is that it is a Rowbottom cardinal. We end the section with some interesting reﬂection properties of Vλ:

Lemma 4.12. Suppose j : Vλ ≺ Vλ and let hκn : n ∈ ωi be its critical sequence.

Then for every n ∈ ω, Vκn ≺ Vλ.

Proof. Since j is the identity on Vκ0 , it is easy to see that Vκ0 ≺ Vκ1 . Considering j(j), since crt(j(j)) = κ1 and

j(j)(κ1)= j(j)(j(κ0)) = j(j(κ0)) = κ2,

we have also that Vκ1 ≺ Vκ2 . We can generalize this to prove that for every n ∈ ω, Vκn ≺ Vκn+1 . But then h(Vκn , idVκn ),n ∈ ωi forms a direct system, whose direct limit is Vλ.

In particular, as κ0 is inaccessible, Vλ ZFC.

12 5 L(Vλ+1) and I0

It would seem that one could not go any further than I1, as already j : Vλ+2 ≺ Vλ+2 is inconsistent. But there is a way to enlarge the domain of j without stumbling in Kunen’s inconsistency: Deﬁnition 5.1. Let A be a set. Then the class of sets constructible relative to A is:

• L0(A)= the transitive closure of A;

• Lα+1(A) = Def(Lα(A));

• Lγ(A)= α<γ Lα(A);

• L(A)= Sα∈Ord Lα(A). This is theS smallest ZF-model that contains A. We can now introduce I0:

Deﬁnition 5.2 (Woodin). I0 There exists j : L(Vλ+1) ≺ L(Vλ+1) with critical point less then λ.

As L(Vλ+1)) V = L(Vλ+1), it must be that j(Vλ+1)= Vλ+1. Thanks to the added assumption that the critical point must be less than λ, then, I0 is stronger than I1, as j ↾ Vλ+1 : Vλ+1 ≺ Vλ+1 (otherwise it could have been the identity). This implies that λ must be the supremum of the critical sequence. If one wants to avoid the menace of Kunen’s inconsistency, it must be that hSα : α< crt(j)i is not in L(Vλ+1). This is constructed via the Axiom of Choice on Vλ+1. So, for I0 to hold, it must be that L(Vλ+1) 2 AC, and in particular Vλ+1 must not R be well-orderable. This is the first affinity with ADL( ), as in that case we have L(R) 2 AC. More affinities will be introduced in section 7. At a superficial glance I0 seems a Reinhardt cardinal, as it is an embedding in a “universe” of ZF. But the situation is completely different: our ground model is V , we assume that V ZFC, and we construct L(Vλ+1) inside V . So j is a definable class of V , and we will see that L(Vλ+1) does not satisfy replacement for j, so such a j is not a witness for Reinhardt, as in that case the theory of the ground model must be richer. Before delving in I0, we analyze the structure of L(Vλ+1) itself, without further assumptions, that still carry similarities with L(R). As L(Vλ+1) is L relativized to Vλ+1, we can relativize in the same way OD and HOD:

Definition 5.3. ODVλ+1 , the class of the sets that are ordinal-definable over Vλ+1, is the Gödel closure of {Vα : α ∈ Ord} ∪ Vλ+1.

HODVλ+1 , the class of the sets that are hereditarily ordinal-deﬁnable over

Vλ+1, is the class of sets whose transitive closure is contained in ODVλ+1 .

Lemma 5.4. There exists a deﬁnable surjection Φ: Ord × Vλ+1 ։ L(Vλ+1), therefore L(Vλ+1) V = HODVλ+1 Proof. This is immediate from the theory of relative constructibility.

13 The ﬁrst application of this Lemma comes in form of partial Skolem functions. Since L(Vλ+1) 2 AC, we possibly could not have Skolem functions. But with the previous Lemma we can deﬁne for every formula ϕ(x, x1,...,xn), a ∈ Vλ+1, a1,...,an ∈ L(Vλ+1):

L(Vλ+1) hϕ,a(a1,...,an)= y where y is the minimum in (ODa)

such that L(Vλ+1) ϕ(y,a1,...,an). These are partial Skolem functions, and the Skolem Hull of a set is its closure under such Skolem functions. We indicate with ։ the surjectivity of a function. Deﬁnition 5.5. ΘL(Vλ+1) = sup{γ : ∃f : V ։ γ,f ∈ L(V )}. Vλ+1 λ+1 λ+1

In the following, we will call ΘL(Vλ+1) = Θ, because it is a lighter notation Vλ+1 and there is no possibility of misinterpretation. The role of Θ in L(Vλ+1) is exactly the same of its correspondent in L(R). In the usual setting, under AC, to measure the largeness of a set we fix a bijection from this set to a cardinal or, equivalently, the ordertype of a well-ordering of the set. Since there is no Axiom of Choice in L(Vλ+1), it is not always possible to define cardinality for sets that are not in Vλ+1 in the usual way, so to quantify the “largeness” of a subset of Vλ+1 we will not use bijections, but surjections, or, equivalently, not well-orders, but prewellorderings (pwo for short). An order is a pwo if it satisfies antireflexivity, transitivity, and every subset has a least element; in other words, it is a well-order without the antisymmetric property. It is easy to see that the counterimage of a surjective function is a pwo. One can think of a pwo as an order whose equivalence classes are well- ordered, or a well-ordered partition. This creates a strong connection between subsets of Vλ+1 and ordinals in Θ:

Lemma 5.6. 1. For every α < Θ, there exists in L(Vλ+1) a pwo in Vλ+1 with ordertype α, that is codeable as a subset of Vλ+1;

2. for every Z ⊆ Vλ+1, Z ∈ L(Vλ+1) there exists α < Θ such that Z ∈ Lα(Vλ+1).

Proof. 1. Let ρ : Vλ+1 ։ α. Then

Rα = {(a,b) ∈ Vλ+1 × Vλ+1 : ρ(a) ≤ ρ(b)}

is a pwo in Vλ+1. Moreover, Vλ+1 × Vλ+1 can be codiﬁed as a subset of Vλ+1, so also Rα can.

Lγ (Vλ+1) 2. Let γ be such that Z ∈ Lγ(Vλ+1) and consider H (Vλ+1,Z) the Lγ (Vλ+1) Skolem Hull in Lγ(Vλ+1) of Vλ+1 and Z. Then, since H (Vλ+1,Z) =∼ Lγ(Vλ+1), by condensation its collapse X = Lα(Vλ+1) for some α. But Lγ (Vλ+1) H (Vλ+1,Z) is the closure under the Skolem functions, and since there is a surjection from Vλ+1 to the Skolem functions, this surjection Lγ (Vλ+1) transfers to H (Vλ+1,Z) and to Lα(Vλ+1), so α< Θ. Since Z and Lγ (Vλ+1) all its elements are in H (Vλ+1,Z), Z is not collapsed and then Z ∈ Lα(Vλ+1).

14 We can also have a “local” version of Θ: Deﬁnition 5.7.

ΘLα(Vλ+1) = sup{γ : ∃f : V ։ γ,f ∈ L(V ), Vλ+1 λ+1 λ+1

{(a,b): f(a)

The reason why we do not just relativize Θ to Lα(Vλ+1) is because there can be pwos in L(Vλ+1) that are not in Lα(Vλ+1), but can be coded in Lα(Vλ+1), so in this way ΘLα(Vλ+1) is more “truthful” to indicate the ordinals that are “hidden” inside Vλ+1. The notion of “stable” also neatly generalizes. Let δ be the minimum ordinal such that Lδ(Vλ+1) ≺1 LΘ(Vλ+1). Then Lδ(Vλ+1) is the collapse of the closure of Vλ + 1 under the Skolem functions hϕ,a such that ϕ is Σ1. The map that associates to any (ϕ, a, x) the collapse of hϕ,a(x) is therefore a Σ1 partial surjection from Vλ+1 to Lδ(Vλ+1). A study of the ﬁne structure of Lδ(Vλ+1) has the following results:

Lemma 5.8. • For any β<δ, there exists a total surjection from Vλ+1 to Lδ(Vλ+1) Lβ(Vλ+1) that is in Lδ(Vλ+1). Therefore Θ = δ.

2 2 L(Vλ+1) • δ is the supremum of the ∆1 pwos of Vλ+1 and (∆1) = Lδ(Vλ+1) ∩ Vλ+2.

• Let ρ : Vλ+1 → Lδ(Vλ+1) be a partial Σ1 map. Then for any Z ⊆ dom(ρ), Z ∈ Lδ(Vλ+1), ρ ↾ Z is bounded. The proofs are the same as in [37]. Deﬁnition 5.9.

<λ DCλ : ∀X ∀F : (X) →P(X) \ ∅ ∃g : λ → X ∀γ<λg(γ) ∈ F (g ↾ γ).

Note that this is a generalization of DC, since DC = DCω: we use directly a function on < λ-sequences instead of considering a binary relation because binary relations cannot handle the limit stages.

Lemma 5.10. In L(Vλ+1) the following hold: 1. Θ is regular ([39]);

2. DCλ.

Proof. 1. We fix a definable surjection Φ : Ord×Vλ+1 ։ L(Vλ+1). For every ξ < Θ there is a surjection h : Vλ+1 ։ ξ. First of all, we choose one surjection for each ξ: we define t :Θ \ ∅ → Ord, where for every ξ < Θ, t(ξ) is the least γ such that there exists x ∈ Vλ+1 such that Φ(γ, x) is a surjection from Vλ+1 to ξ. Then we define

Φ(t(ξ), x)(y) if Φ(t(ξ), x) is a map in ξ; hξ(hx, yi)= (∅ else.

We have that hξ : Vλ+1 → ξ is well deﬁned, because hx, yi is codeable in Vλ+1, and it is indeed a surjection: by deﬁnition there exists x ∈ Vλ+1

15 such that Φ(t(ξ), x) is a surjection from Vλ+1 to ξ, so for every β < ξ there exists y ∈ Vλ+1 such that Φ(t(ξ), x)(y) = hξ(hx, yi) = β. Moreover hξ is deﬁnable in L(Vλ+1). Now, suppose that Θ is not regular, i.e., there exists π : α → Θ coﬁnal in

Θ with α< Θ. Then we claim that H(hx, yi)= hπ◦hα(x)(y) is a surjection from Vλ+1 to Θ: let β < Θ; then there exists γ < α such that π(γ) >β and there must exist x ∈ Vλ+1 such that hα(x)= γ; so π(hα(x)) >β, and there

exists y ∈ Vλ+1 such that H(hx, yi)= hπ◦hα(x)(y)= β. Contradiction. 2. We have to prove that

∀X ∀F : (X)<λ →P(X) \ ∅ ∃g : λ → X ∀γ<λg(γ) ∈ F (g ↾ γ).

The proof is through several steps. First of all, DCλ(Vλ+1), that is DCλ only for X = Vλ+1, is quite obvious, because for every F as above by AC there exists a g as above in V , but since g is a λ-sequence of elements in Vλ+1 we have that g is codeable in Vλ+1, so g ∈ L(Vλ+1).

Then we prove DCλ(α × Vλ+1) for every ordinal α. The idea is roughly to divide F in two parts, and to define g using the minimum operator for the ordinal part, and DCλ(Vλ+1) for the other part. For every s ∈ <λ (α × Vλ+1) , we define m(s) as the minimum γ such that there exists <λ <λ x ∈ Vλ+1 such that (γ, x) ∈ F (s). We call π2 : (α × Vλ+1) → (Vλ+1) <λ the projection. For every t ∈ (Vλ+1) , say t = hxξ : ξ<νi, we define <λ c(t) by induction as a sequence in (α × Vλ+1) such that π2(c(t)) ⊆ t: c(t)= h(γξ, xξ): ξ < ν¯i, where

γξ = min{γ : (γ, xξ) ∈ F (c(t ↾ ξ))},

so thatν ¯ is the smallest one such that there is no γ such that (γ, xν¯) ∈ <λ F (c(t) ↾ ν¯). Let G : (Vλ+1) →P(Vλ+1) \∅ deﬁned as

G(t)= {x ∈ Vλ+1 : (m(c(t)), x) ∈ F (c(t))},

then by DCλ(Vλ+1) there exists g : λ → Vλ+1 such that for every β < λ g(β) ∈ G(g ↾ β). Now let f : λ → (α × Vλ+1) be deﬁned by induction, f(β) = (m(f ↾ β),g(β)). We want to prove that f(β) ∈ F (f ↾ β) for every β < λ. We prove by induction that f ↾ β = c(g ↾ β). Suppose that for every ξ<β, f ↾ ξ = c(g ↾ ξ). By deﬁnition c(g ↾ β)= h(γξ,g(ξ)) : ξ < β¯i, with

γξ = min{γ : (γ,g(ξ)) ∈ F (c(g ↾ ξ))},

so (γξ,g(ξ)) ∈ F (c(g ↾ ξ)). But g(ξ) ∈ G(g ↾ ξ), so by deﬁnition of G,

(m(c(g ↾ ξ)), g(ξ)) ∈ F (c(g ↾ ξ)),

therefore γξ = m(c(g ↾ ξ)) and

f(ξ) = (m(f ↾ ξ),g(ξ)) = (m(c(g ↾ ξ)),g(ξ)) = (γξ,g(ξ)) = c(g ↾ β)(ξ).

So f ↾ β = c(g ↾ β) and, since for every ξ, (γξ,g(ξ)) ∈ F (c(g ↾ ξ)), f(β) ∈ F (f ↾ β).

16 ′′ Finally, let X ∈ L(Vλ+1). Let α be such that Φ (α × Vλ+1) ⊇ X, and let <λ <λ F : (X) → P(X) \∅. For every t = h(γξ, xξ): ξ<νi ∈ (α × Vλ+1) , <λ we call c(t) = hΦ(γξ, xξ): ξ < ν¯i ∈ X , whereν ¯ is the minimum such <λ that Φ(γν¯, xν¯) ∈/ X. Then we deﬁne G : (α × Vλ+1) →P(α × Vλ+1) \∅,

G(t)= {(γ, x):Φ(γ, x) ∈ F (c(t))},

and by DCλ(α×Vλ+1) we ﬁnd g : λ → (α×Vλ+1) such that g(β) ∈ G(g ↾ β) for every β < λ. Then f =Φ ◦ g is as we wanted, because for every β < λ, Φ(g(β)) ∈ F (c(g ↾ β)), and as above we can prove that c(g ↾ β)= f ↾ β.

So we have DCλ(X) for every X ∈ L(Vλ+1), that is exactly DCλ.

Suppose now for the rest of the section that j : L(Vλ+1) ≺ L(Vλ+1), with crt(j) < λ. Deﬁne

Uj = {X ⊆ Vλ+1 : X ∈ L(Vλ+1), j ↾ Vλ ∈ j(X)}.

Then Uj is a normal non-principal L(Vλ+1)-ultraﬁlter in Vλ+1, and we can construct the ultraproduct Ult(L(Vλ+1),Uj ). Note that for every f,g : Vλ+1 → L(Vλ+1)

[f] = [g] iﬀ {x ∈ Vλ+1 : f(x)= g(x)} ∈ Uj

iﬀ j ↾ Vλ ∈ j({x ∈ Vλ+1 : f(x)= g(x)})=

= {x ∈ Vλ+1 : j(f)(x)= j(g)(x)}

iﬀ j(f)(j ↾ Vλ)= j(g)(j ↾ Vλ), and in the same way [f] ∈ [g] iﬀ j(f)(j ↾ Vλ) ∈ j(g)(j ↾ Vλ), so

Ult(L(Vλ+1),Uj ) =∼ {j(f)(j ↾ Vλ) : dom f = Vλ+1}.

Let i : L(Vλ+1) → Ult(L(Vλ+1),Uj ) be the natural embedding of the ultraproduct, then for every a ∈ L(Vλ+1), i(a) = [ca] corresponds in the equivalence to j(ca)(j ↾ Vλ) = j(a), so we can suppose i = j. Is j an elementary embedding from L(Vλ+1) to Ult(L(Vλ+1),Uj )? Since we do not have AC, the answer is not immediate because we possibly do not haveLos’ Theorem. We will proveLos’ Theorem for this case, and this will imply that j is an elementary embedding. It is clear that the only real obstacle is to prove that for every formula ϕ and f1,...,fn ∈ L(Vλ+1) such that dom fi = Vλ+1

Ult(L(Vλ+1),Uj ) ∃x ϕ([f1],..., [fn])

iﬀ {x ∈ Vλ+1 : L(Vλ+1) ∃y ϕ(f1(x),...,fn(x))} ∈ Uj.

The direction from left to right is immediate: if [g] witness the left side, then g(x) witness the right side. For the opposite direction, we need a sort of Uj - choice, i.e., we need to ﬁnd a function g such that

{x ∈ Vλ+1 : L(Vλ+1) ϕ(g(x),f1(x),...,fn(x))} ∈ Uj .

We re-formulate this considering f : Vλ+1 → L(Vλ+1) \∅,

f(x)= {y ∈ L(Vλ+1): L(Vλ+1) ϕ(y,f1(x),...,fn(x))}.

17 Lemma 5.11 (Los’ Theorem for I0, Woodin, [39]). Let j : L(Vλ+1) ≺ L(Vλ+1) and Uj as above. Then for every F : Vλ+1 → L(Vλ+1) \ {∅} there exists H : Vλ+1 → L(Vλ+1) \ {∅} such that {x ∈ Vλ+1 : H(x) ∈ F (x)} ∈ Uj .

Proof. First we consider the case ∀a ∈ Vλ, F (a) ⊆ Vλ+1. We have to deﬁne H such that j(H)(j ↾ Vλ) ∈ j(F )(j ↾ Vλ). Fix a b ∈ j(F )(j ↾ Vλ), and deﬁne

c if k : V ≺ V , k(k)= j and k(c)= b H(k)= λ λ (∅ otherwise

Note that

Kb := {k ∈ Vλ+1 | k : Vλ ≺ Vλ, k(k)= j, ∃c k(c)= b} ∈ Uj ,

because

j(Kb)= {k ∈ Vλ+1 | k : Vλ ≺ Vλ, k(k)= j(j), ∃c k(c)= j(b)}

and j ↾ Vλ ∈ j(Kb) (with c = b), so {x ∈ Vλ+1 : H(x) 6= ∅} ∈ Uj. Then

c if k : V ≺ V , k(k)= j(j) and k(c)= j(b) j(H)(k)= λ λ (0 otherwise

so j(H)(j ↾ Vλ)= b ∈ j(F )(j ↾ Vλ). For the more general case ∀a ∈ Vλ F (a) ⊆ L(Vλ+1) fix Φ : Ord × Vλ+1 ։ L(Vλ+1) definable and define

Fˆ(a)= {x ∈ Vλ+1 : ∃γ Φ(γ, x) ∈ F (a)}.

Then there exists Hˆ such that {a ∈ Vλ+1 : Hˆ (a) ∈ Fˆ(a)} ∈ Uj . Let γa = min{γ : Φ(γ, Hˆ (a)) ∈ F (a)}. Therefore H(a)=Φ(γa, Hˆ (a)) is as desired. Therefore, calling

Z = {j(f)(j ↾ Vλ): f ∈ L(Vλ+1), dom(F )= Vλ+1},

j : L(Vλ+1) → Z is an elementary embedding, and Z =∼ L(Vλ+1). Let kU be the inverse of the collapse of Z. We’ve seen in the proof of the previous Lemma that for every b ∈ Vλ+1 there exists h such that j(h)(j ↾ Vλ), so Vλ+1 ⊆ Z and kU : L(Vλ+1) ≺ Z. Moreover, if R is a pwo in Vλ+1, then R = {a ∈ Vλ+1 : j(a) ∈ j(R)}, and since j(a), j(R) ∈ Z and Vλ+1 ⊆ Z we have that R is not collapsed, so Θ ⊆ Z and crt(kU ) > Θ.

Theorem 5.12 (Woodin, [39]). For every j : L(Vλ+1) ≺ L(Vλ+1) there exist a L(Vλ+1)-ultraﬁlter U in Vλ+1 and jU , kU : L(Vλ+1) ≺ L(Vλ+1) such that jU is the elementary embedding from U, j = jU ◦ kU and j ↾ LΘ(Vλ+1) = jU ↾ LΘ(Vλ+1).

Deﬁnition 5.13. Let j : L(Vλ+1) ≺ L(Vλ+1). We say that j is weakly proper if j = jU .

18 Note that U/∈ L(Vλ+1), otherwise we could run the naive Kunen’s Theorem argument, so L(Vλ+1) does not satisfy replacement for j, and this proves that I0 is not a “L(Vλ+1)-Reinhardt”. In fact, one cannot even have j ↾ LΘ(Vλ+1) ∈ L(Vλ+1), otherwise one could construct U from that, so j is not amenable. In a certain sense weakly proper embeddings are the “right” embeddings for I0, and the most natural extension of I1, if one considers that LΘ(Vλ+1) is, loosely speaking, the Vλ+2 of L(Vλ+1), i.e., the set of subsets of Vλ+1. Every embedding has therefore a core part, that comes from an ultraproduct, it is unique and depends solely on its behaviour on LΘ(Vλ+1), and a spurious part (note that kU is not an I0-embedding, as it has critical point > λ) that moves things up above without really changing the important part of the embedding. For example, if there are indiscernibles, it could be a simple shift of indiscernibles, so nothing that carries real information. In fact, the behaviour of an embedding that is weakly proper depends only on its Vλ part:

Lemma 5.14 (Woodin, [39]). For every j1, j2 : L(Vλ+1) ≺ L(Vλ+1), if j1 ↾ Vλ = j2 ↾ Vλ then j1 ↾ LΘ(Vλ+1)= j2 ↾ LΘ(Vλ+1).

Proof. We can suppose that j1 and j2 are weakly proper. By the usual analysis of the ultraproduct, we have that every strong limit cardinal with cofinality bigger than Θ is a fixed point for j1 and j2, so I = {η : j1(η) = j2(η) = η} is L(Vλ+1) ∗ a proper class. Let M = H (I ∪ Vλ+1). Since Vλ+1 ⊆ M k , the inverse ∗ of the transitive collapse of M, has domain L(Vλ+1). If k is not the identity then, as Θ ⊆ M by Lemma 5.6, crt(k∗) > Θ. But in that case crt(k∗) is a strong limit cardinal with cofinality bigger than Θ, so crt(k∗) ∈ I, and this is a contradiction, because I ⊆ ran(k∗) and crt(k∗) ∈/ ran(k∗). So k∗ is the identity L(Vλ+1) and L(Vλ+1)= H (I ∪ Vλ+1). Therefore every element of LΘ(Vλ+1) is definable with parameters in I∪Vλ+1. Let A ∈ L(Vλ+1) ∩ Vλ+2, A = {x ∈ Vλ+1 : L(Vλ+1) ϕ(η,a)} with η ∈ I and a ∈ Vλ+1. Since j1 ↾ Vλ = j2 ↾ Vλ, we have that j1(a)= j2(a). Then

j1(A)= {x ∈ Vλ+1 : L(Vλ+1) ϕ(η, j1(a))} =

= {x ∈ Vλ+1 : L(Vλ+1) ϕ(η, j2(a))} = j2(A).

But every element of LΘ(Vλ+1) is deﬁnable from an ordinal α < Θ and an element of Vλ+1, α is deﬁnable from some pwo in L(Vλ+1) ∩ Vλ+2, so j1 ↾ LΘ(Vλ+1)= j2 ↾ LΘ(Vλ+1).

Corollary 5.15 (Woodin, [39]). Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j). Then C = {α< Θ: j(α)= α} is unbounded in Θ. Proof. We can suppose that j is weakly proper. Then every ordinal β < Θ is deﬁnable from some elements of I ∪ Vλ+1, say Iβ = {i0,...,in} ∈ I. Let α0 < Θ and consider the transitive collapse of the Skolem closure Z of

β<α0 Iβ ∪Vλ+1. Then by condensation it must be some Lγ(Vλ+1). As α0 < Θ, there exists π : Vλ+1 ։ Z, therefore γ < Θ. As α0 ⊆ Z, we have α0 < γ. As S j(Z) = Z, we have j(γ) = γ. Therefore for any α0 < Θ there exists a γ ∈ C such that α0 <γ< Θ.

19 6 Strong implications and inverse limits

When introducing new axioms, it is good practice to prove that they really are adding something, i.e., that their consistency strength is strictly larger (or smaller) than the consistency of the already known ones. Deﬁnition 6.1. Let A(x) and B(x) two large cardinal properties. We say that A(x) strongly implies B(x) iﬀ for any κ, A(κ) implies B(κ) and the smallest κ such that A(κ) holds is bigger than the smallest η such that B(η) holds. This is even stronger then just the strict implications, cfr the note after Lemma 6.22.

Proposition 6.2. If j : Vλ ≺ Vλ with crt(j) = κ, then there is a normal ultraﬁlter U on κ such that {α<κ : α is n-huge for every n} ∈ U. Therefore I3 strongly implies being n-huge for every n. Proof. By 4.10 κ is n-huge for every n ∈ ω. Let U = {X ⊆ κ : κ ∈ j(X)} and let X0 = {α<κ : α is n-huge for every n}. Then j(X0)= {α < j(κ): α is n-huge for every n},

and κ ∈ j(X0), so X0 ∈ U. Note that “ω-huge”, i.e, j : V ≺ M with M λ ⊆ M 4, is inconsistent by Kunen’s original proof of Kunen’s Theorem, so this is the best we can do.

α Deﬁnition 6.3. Let j : M0 ≺ M1. Then we say that (Mα, j ) is the α-th iterate if: • j0 = j; • jα+1 = jα(jα);

α • Mα+1 = j (Mα) is well-founded;

α α−1 γ α • jγ,α+1 = j ◦ j ◦···◦ j . So jα,α+1 = j ;

α • if α is limit, then (Mα, j , jn,α) is the direct limit of the system (Mβ, jβ,γ) with β,γ <α, if it is well-founded.

α We say that j is α-iterable if (Mα, j ) exists. We say that j is iterable if it is α-iterable for any α ordinal.

4This is not a standard definition. For different authors, “ω-huge” can mean different things.

20 j0,α

jn,α

j2,n j0 j1 jω M0 M1 M2 ··· Mn ··· Mω Mω+1 ··· Mα

j0,2 j1,ω jω+1,α j0,ω

α α α It is possible for the iterate to not exists, whether because j (j ) or j (Mα) are not defined, or because the model Mα is not well-founded. The direct limit of the system (Mβ, jβ,γ) is defined as an equivalence relation on the set of (β,a) such that a ∈ Mβ: if β<γ then (β,a) is equivalent to (γ,b) iff jβ,γ(a)= b, and [(β,a)] is in [(γ,b)] iff jβ,γ(a) ∈ b. If γ<β, viceversa. Then jβ,α(a) is defined α as [(β,a)] and j = j0,α(j). Ambiguously, we call Mα both the domain of the model defined via the equivalence class and the ∈-model equivalent to it. So if M1 ⊆ M0, then for all α such that Mα is defined we have Mα ⊆ M0, but it is possible that M0 ⊆ Mα if seen as ∈-models. It will be clear from context which of the two will be used. Note that if a ∈ Mα, with α limit, then it must be in the range of some jβ,α, with β < α. n So let j : Vλ ≺ Vλ. We have already defined j , but

jn(jn)= j(jn−1)(j(jn−1)) = j(jn−1(jn−1)) = j(jn)= jn+1,

by left distributive law and induction, therefore the jn’s are iterates of j. More- n over, j(Vλ)= Vλ, so the first ω-iterates of j are (Vλ, j ) and j is finitely iterable. Could we push it any further? As all the Mα ⊆ Vλ, technically there is no problem in defining it, the only decisive factor being whether Mα is well-founded.

Proposition 6.4. “There exists j : Vλ ≺ Vλ ω-iterable” strongly implies I3(λ).

Proof. Suppose j : Vλ ≺ Vλ is ω-iterable, let hκn : n ∈ ωi be its critical sequence. Then Mω is the set of equivalence classes of (n,a) such that a ∈ Vλ, where if nn, jn,m(β) = β, therefore [(n,β)] = [(m,β)]. So for any m ∈ ω,β <λ, [(m,β)] < [(n, α)] iﬀ [(n,β)] < [(n, α)]: If m ≤ n, then jm,n(β) < α<κn = jm,n(κm) therefore by elementarity β<κm and [(m,β)] = [(n,β)]. If m>n, then β < jn,m(α)= α, therefore also [(n,β)] < [(n, α)]. So jn,ω(α)= [(n, α)] = {[(n,β)] : β < α}, that is by induction α. We want to calculate now j0,ω(κ0). If α < j0,ω(κn), then there exists n ∈ ω such that α = jn,ω(β) for some β < λ. But then jn,ω(β) < j0,ω(κ0) =

21 jn,ω(j0,n(κ0)) = κn, therefore α = jn,ω(β)= β < λ. On the other hand, if α < λ then there exists n ∈ ω such that α<κn, so jn,ω(α)= α < jn,ω(κn)= j0,ω(κ0). Therefore j0,ω(κ0)= λ.

If a ∈ Vλ, then there exists n ∈ ω such that a ∈ Vκn . So jn,ω(a)= a ∈ Mω. Therefore Vλ ⊆ Mω. Let

T = {he0,...,eni : ∃hαi : i ≤ n +1i, ∀i ≤ n +1 αi < λ,

∀i ≤ n∃ei : Vαi ≺ Vαi+1 , ∀i

the tree of approximations of an I3 elementary embedding. Note that T = T Mω . As j is a branch of T , T is ill-founded, and by absoluteness of well-foundedness it is ill-founded also in Mω, with the diﬀerence that a branch cannot go up to λ, as λ is regular in Mω by elementarity. So Mω ∃γ < λ ∃k : Vγ ≺ Vγ . So by elementarity ∃γ<κ0 ∃k : Vγ ≺ Vγ .

Corollary 6.5. Let α be a countable ordinal. Then “There exists j : Vλ ≺ Vλ ω · (α + 1)-iterable” strongly implies “There exists j : Vλ ≺ Vλ ω · α-iterable”.

Proof. By induction, Vλ ⊆ Mω·α ⊆ Mω·(α+1). We build again a tree of approximations, but this time we want to add rank functions, so that the ω·α-th iterate

of n∈ω en will be well-founded. So a branch of T is (he0,...,eni, hf0,...,fni), where fi is a rank function on {(β,a): β <ω · α, a ∈ Vα } that is coherent S i with the equivalence on Mω·α. A branch of T will give an embedding that is ω · α-iterable, and the proof is as before.

Proposition 6.6. If j : Vλ ≺ Vλ is ω1-iterable, then it is iterable. Proof. This is a standard result in the theory of the iterations of elementary embeddings. Suppose that j is not iterable, and let ν be the least such that Mν is ill-founded. Let N be a countable substructure of Vλ, let h : N ≺ Vλ indicate the embedding resulting form the collapse. Let h(M¯ 0)= Vλ, h(¯)= j, and build ¯ −1 the iterate sequence in N. Then h(Mβ)= Mh(β) and letν ¯ = h (ν). In partic- ′ ularν ¯ is countable. For any β ≤ ν deﬁne hβ so that the diagram commutes:

jα,β jβ,ν¯ Vλ ··· Mα Mβ Mν¯ ··· Mν

′ ′ ′ ′ h h0 hα hβ hν¯ h ¯ ¯ α,β ¯ β,ν¯ M¯ 0 ··· M¯ α Mβ M¯ ν¯

′ ¯ ¯ ¯ Let h0 = h ↾ M0. If β is a limit ordinal, then every x ∈ Mβ is some jα,β(y), ¯ ′ ′ with α<β and y ∈ Mα. Then hβ(x) = jα,β(hα(y)). If β = γ + 1, then by ¯ ¯ ′ ′ elementarity, as j : Vλ ≺ Vλ, Mβ = Mγ and Mβ = Mγ , so hβ = hγ. ′ ¯ But then hν¯ : Mν¯ ≺ Mν¯. As Mν is illfounded, then by elementarity also Mν¯ is illfounded, and since ν was the least andν ¯ is countable, ν =ν ¯ is countable and j is not ω1-iterable. While the “ω-hugeness” is inconsistent, it is not immediately clear what happens with “ω-superstrongness”:

22 Definition 6.7. I2 There exists j : V ≺ M, with M ⊆ V , such that Vλ ⊆ M for some λ = j(λ) > crt(j). Proposition 6.8. Suppose j witnesses I2. Then λ is the supremum of its critical sequence, and j ↾ Vλ : Vλ ≺ Vλ. Proof. Let η be the supremum of the critical sequence of j. Then it must be η ≤ λ, as η is the first fixed point of j above crt(j). If η +2 ≤ λ then there is a contradiction with Kunen’s Theorem, so we only have to check that η +1 6= λ, i.e., not all subsets of Vη are in M. Suppose that j′′η ∈ M, then define U = {X ⊆ P(η): j′′η ∈ j(X)} and let i : V ≺ N be the ultrapower via U. Then i(η)= η, and

i′′P(η)= { i(X ∩ α): X ∈P(η)} ∈ N. α<η [ Now, coding λ+ as sets in P(λ), i′′λ+ ∈ N, and Woodin’s proof of Kunen’s theorem goes smoothly (because C is deﬁnable from i′′λ+). Contradiction. ′′ So j η∈ / M, and η must be λ. In particular j ↾ Vλ : Vλ ≺ Vλ. The previous Proposition says that I2 implies I3. We can do more:

Proposition 6.9. If j : V ≺ M witnesses I2(λ) then j ↾ Vλ is iterable.

Mν Proof. Consider an iteration of j : V ≺ M with Vλ ⊆ M. Suppose that (Vλ) is ill-founded, with ν the least. Let α0 be the minimum such that j0,ν (α0) is ill-founded. In other words, Vλ “α0 is the least α such that ∃µ such that j0,µ(α) is ill-founded”. Let β < j0,ν (α0) ill-founded. Let η<ν and α1 such that jη,ν (α1) = β. Consider j0,η(α0). As jη,ν (α1) < j0,ν (α0), we must have α1 < j0,η(α0). But by elementarity

Mη (Vλ) “j0,η(α0) is the least α such that ′′ ∃µ such that [j0,η(j)]0,µ(α) is ill-founded .

Note that [j0,η(j)]0,µ = [(jη,η+1)]0,µ = jη,η+µ. But then α1 is such that, contradiction because j0,η(α0) should have been the minimum. Another way to extend I3 comes from the way we characterized I3 and I1, respectively as Σ0- and Σω-elementary embeddings from Vλ+1 to itself. We can consider all the intermediate steps:

1 1 Definition 6.10 (Σn Elementary Embedding). Let j : Vλ ≺ Vλ. Then j is Σn + 1 iff j is a Σn-elementary embedding, i.e., iff for every Σn-formula ϕ(X), for every A ⊆ Vλ, + Vλ ϕ(A) ↔ ϕ(j (A)). It is not immediate, however, to see if these hypothesis are really different, and what is their relationship with I2. In fact, they are not always different: 1 1 the following theorems will prove that if n is odd, then j is Σn iff it is Σn+1. 1 During the proof it will be clear that I2 is the same as j is Σ1, therefore proving that I1 implies I2. To prove this we need what is often called the ‘descriptive set theory’ on Vλ. The fact that λ has cofinality ω and is a strong limit, makes possible to

23 develop a description of the closed subsets of Vλ as projection of trees similar 1 1 to the classic one in descriptive set theory. The fact that both Σ1 and Σ2 sets can be represented with trees will give the ﬁrst equivalence, in a way similar to Schoenﬁeld Absoluteness Lemma. Fix λ and a critical sequence hκ0,κ1,... i. 1 Let ϕ(X, Y )aΣ0-formula, namely

∀x0∃y0∀x1∃y1 ····n ψ(X,Y,x0,y0,... ).

We deﬁne the tree Tϕ(X,Y ). The m-th level of Tϕ(X,Y ) is the set of (a,b,F,P ) such that:

• a,b ⊆ Vκm ;

≤n+1 • F is a partial function : (Vκm ) → Vκm such that for all d0,...,dn where F is deﬁned then

ψ(a,b,d0, F (d0), d1, F (d0, d1),... );

≤n+1 • P : ((Vκm ) \ dom(F )) → (ω \ (m + 1)). We say that (a,b,F,P ) < (a′,b′, F ′, P ′), with the ﬁrst term in the m-th level ′ ′ ′ ′ ′ and the second in the m -th level, when a ⊆ a , b ⊆ b , a ∩Vκm = a, b ∩Vκm = b, F ⊆ F ′, and if P (d~)

ϕ, and P is booking the level in which the elements of Vκm that are not yet in the dominion of F will be in its extension. It is clear that if Tϕ(X,Y ) has an inﬁnite branch, the union of the a’s and b’s will give suitable X and Y , and the F ’s will construct a Skolem function, that is total thanks to the P ’s.

1 Lemma 6.11. Let ϕ(X, Y ) a Σ0-formula and B ⊆ Vλ. Then Vλ ∃X ϕ(X,B) iﬀ Tϕ(X,B) has an inﬁnite branch. 1 The Σ2 case is a bit more complex. Fix a B ⊆ Vλ. For every a ⊆ Vκm let

Gm(a)= {(c,F,P ) : (a,c,F,P ) ∈ m-th level of T¬ϕ(X,B,Y )}.

Then the m-th level of Tϕ(X,B) is

+ {(a,H): a ⊆ Vκm ,H : Gm(a) → λ }.

′ ′ ′ ′ ′ ′ ′ We say that (a,H) < (a ,H ) when a ∩ Vκm and if (a,c,F,P ) < (a ,c , F , P ), then H(c,F,P ) >H′(c′, F ′, P ′).

1 Lemma 6.12. Let ϕ(X, Y ) a Σ0-formula and B ⊆ Vλ. Then Vλ ∃X∀Y ϕ(X,B) iﬀ Tϕ(X,B) has an inﬁnite branch.

If Tϕ(X,B) has an inﬁnite branch, with A the union of the a’s in the branch, then the H’s assure that there are no possible inﬁnite branches in T¬ϕ(A,B,Y ) for every Y ⊂ Vλ, because otherwise it would be possible to build a descending chain in λ+.

1 + Lemma 6.13. Let j : Vλ ≺ Vλ. Then j is Σ1 iﬀ j preserves the well-founded relations.

24 1 1 Proof. ‘Being well-founded’ is a ∆1 relation, so if j is Σ1 it preserves well- 1 founded relations. Vice versa, let ϕ(X, Y )aΣ0 formula, B ⊆ Vλ and suppose that Vλ ∃Xϕ(X,B). Fix an X ⊆ Vλ such that Vλ ϕ(X,B). Therefore + + + by Theorem 4.6 Vλ ϕ(j (X), j (B)) and Vλ ∃X ϕ(X, j (B)). If Vλ ∀Xϕ(X,B), then T¬ϕ(X,B) is without inﬁnite branches. Fix a well-ordering R of Vλ, and deﬁne the relative Kleene-Brouwer order. Therefore the Kleene- Brouwer order of T¬ϕ(X,B) is well-founded. But then the Kleene-Brouwer order + + relative to j (R) on T¬ϕ(X,j+(B)) is well-founded. So Vλ ∀Xϕ(X, j (B)).

1 Corollary 6.14. If j : V ≺ M witnesses I2(λ), then j ↾ Vλ is Σ1. + Proof. If R is well-founded, then (j ↾ Vλ) (R) = j(R) is well founded in M. But by absoluteness of well-foundedness, j(R) is well-founded in V .

Definition 6.15 (λ-Ultrafilters Tower). Fix λ and hκn : n ∈ ωi a cofinal sequence of regular cardinals in λ. Then U~ = hUn : n ∈ ωi is a λ-ultrafilters tower κn iff for every n ∈ ω, Un is a normal, fine, κ-complete ultrafilter on [κn+1] , and if m

κn {x ∈ [κn+1] : x ∩ κm+1 ∈ A}∈Un.

A λ-ultraﬁlters tower is complete if for every sequence hAi ∈ i ∈ ωi with Ai ∈Ui there exists X ⊆ λ such that for every i ∈ ω, X ∩ κi+1 ∈ Ai.

Lemma 6.16 (Gaifman ([15]), Powell ([34]), 1974). If U~ is a complete λ-tower, then the direct limit of the ultrapowers Ult(V, Un) is well-founded.

When j : Vλ ≺ Vλ, it is natural to consider the sequence U~j = hUn : n ∈ ωi deﬁned as κn Un = {A ⊆ [κn+1] : j”κn+1 ∈ j(A)}.

1 ~ Lemma 6.17. If j is Σ1, then Uj is a complete λ-tower.

Proof. Let hAi : i ∈ ωi such that Ai ∈Ui. Let X = j”λ. Then for every i ∈ ω, X ∩ κi+2 = j”κi+1 ∈ j(Ai), so

∃X ⊂ λ∀i ∈ ω X ∩ κi+2 ∈ j(Ai).

1 1 This is a Σ1 statement, so by Σ1-elementarity ∃X ⊂ λ∀i ∈ ω X ∩κi+1 ∈ Ai. 1 Deﬁnition 6.18. Let M be a transitive class, with Vλ ⊂ M. Then M is Σn- 1 correct in λ if for every Σn-formula ϕ(X), for every A ⊆ Vλ and A ∈ M, M (Vλ ϕ(A)) iﬀ Vλ ϕ(A).

1 Theorem 6.19 (Martin ([29])). Let j : Vλ ≺ Vλ. If j is Σ1 then

• Vλ ⊆ Ult(V, U~j ), so we have I2;

~ 1 1 • Ult(V, Uj ) is Σ2-correct, therefore j is Σ2.

Proof. Consider the direct limit Ult(V, U~j ) of the ultrapowers Ult(V, Un). By Lemma 6.16 and Lemma 6.17 Ult(V, U~j ) is well founded, and it is possible to collapse it on a model M.

25 j 0 ∼ i0 V M0 = Ult(V, U0) M =∼ Ult(V, U~j )

j01 j1 i1

M1 =∼ Ult(V, U1)

j j12 n i n . .

Mn =∼ Ult(V, Un)

. .

By the usual theory of normal ultraﬁlters, it is possible to prove that the

above diagram commute, that crt(jn) = κ, crt(jn,n+1) = κn, so Vκn ⊂ Mn, crt(in)= κn and therefore Vλ ⊂ M. Let i = in ◦jn for every n. Then i ↾ Vλ = j, i(λ)= λ and i((λ+)V ) = (λ+)V . M Observe that for every B ⊆ Vλ, B ∈ M we have that (Tϕ(B)) = Tϕ(B), so 1 1 M is Σ1-correct in λ. Unfortunately, the same cannot be said to the Σ2 case, M since it is possible that (Tϕ(X,B,Y )) 6= Tϕ(X,B,Y ). The reason for this is that M while it is true that (Gm(a)) = Gm(a) for every m ∈ ω and a ∈ Vκm , it is + possible that there exists H : Gm(a) → λ with H/∈ M.

Claim 6.20. If c ∈ Vλ and F : c → Ord, then i ◦ F ∈ M.

Proof. Let c ∈ Vα, with α < λ, and pick n ∈ ω such that crt(in) > α. Then i ◦ F = in ◦ jn ◦ F = in(jn ◦ F ) ◦ in ↾ c = in(jn ◦ F ) ∈ M.

1 By Σ1 correctness in λ, M (Vλ ∃X∀Y ϕ(X,B,Y )) → Vλ ∃X∀Y ϕ(X,B,Y ), so it suffices to prove the other direction. M M Suppose that (Vλ ∀X∃Y ¬ϕ(X,B,Y )) . Then (Tϕ(X,B,Y )) is well- M ˜ founded. Let L : (Tϕ(X,B,Y )) → Ord that witnesses it. Define L : Tϕ(X,B,Y ) → Ord as L˜((a,H)) = L((a,i ◦ H)). By the previous claim i ◦ H ∈ M, and more- + + M over i ◦ H is a function from Gm(a) to i(λ )= λ , so(a,i ◦ H) ∈ (Tϕ(X,B,Y )) ˜ ˜ and L is well-defined. It remains to prove that L witnesses that Tϕ(X,B,Y ) is ′ ′ well-founded. Suppose that (a,H) < (a ,H ). If in Tϕ(X,B,Y ) we have that (a,c,F,P ) < (a′,c′, F ′, P ′) then H(c,F,P ) >H′(c′, F ′, P ′), so

i ◦ H(c,F,P ) >i ◦ H′(c′, F ′, P ′), therefore (a,i ◦ H) < (a′,i ◦ H′). It follows that

L((a,i ◦ H)) >L((a′,i ◦ H′)),

26 so L˜((a,H)) > L˜((a′,H′). Then Tϕ(X,B,Y ) cannot have an inﬁnite branch and Vλ ∀X∃Y ¬ϕ(X,B,Y ), 1 i.e., M is Σ2-correct in λ. 1 1 This proves that i ↾ Vλ is Σ2: if ϕ is Σ2 in Vλ and V ϕ(B), then by ~ 1 elementarity Ult(V, Uj ) ϕ(i(B)), therefore by Σ2-correctness also V ϕ(i(B)). Since i ↾ Vλ = j we’re done.

1 1 This theorem proves that Σ1 and Σ2 elementary embeddings in Vλ are the same, and they are equivalent to I2(λ). It is possible, after a pair of technical 1 1 lemmas, to generalize this for Σn and Σn+1. 1 1 Lemma 6.21 (Laver, [27]). Let n be odd. Then “j is Σn” is a Πn+1 formula in Vλ, with j as a parameter. 1 Proof. By deﬁnition j is Σn iﬀ

1 ∀B ⊆ Vλ ∀Σ0 formula ϕ(X1,...,Xn, Y ) + ∃A1 ∀A2 ... ∃An Vλ ϕ(A,~ j (B)) → ∃A1 ∀A2 ... ∃An Vλ ϕ(A,B~ ). ~ But for every D ⊂ Vλ, Vλ ϕ(A,D) iﬀ there exists a branch in Tϕ(A~) whose projection is D. Using this is easy to check the Lemma.

1 Lemma 6.22 (Square root Lemma for Σn, Laver, [27]). Let n be odd, n > 1. 1 Let j be Σn, crt(j)= κ. Let β<κ, A, B ⊆ Vλ. Then there exists a k : Vλ ≺ Vλ 1 ′ ′ that is Σn−1 such that k(k) = j, k(B) = j(B), k(A ) = A for some A ⊆ Vλ and β < crt(k) <κ. 1 Proof. The formula “∃k,Y, k is Σn−2, k(k) = j, k(B) = j(B), k(Y ) = A,β < 1 crt(k) <κ” is Σn by Lemma 6.21. The following formula is clearly true:

1 j is Σn−2, j(j)= j(j), j(j(B)) = j(j(B)), j(A)= j(A), j(β)= β < crt(j) < j(κ) but then, with a smart quantiﬁcation of some of the parameters of the formula above, we have

1 ∃k,Y k is Σn−2, k(k)= j(j), k(j(B)) = j(j(B)), k(Y )= j(A), j(β) < crt(k) < j(κ) By elementarity the lemma is proved. The lemma therefore provides a “square root” for j, with certain desirable 1 characteristics, but with the catch that for a square root of a Σn-elementary 1 embedding we can only assure that it is Σn−2. It is not possible to do better: 1 suppose that κ is the minimum critical point of j : Vλ ≺ Vλ that is Σn. Then the critical point of the square root k of j is less then κ, therefore k cannot be 1 Σn. Also note that the square root of an iterable embedding is iterable, as a square root of j is, in a certain sense, j−1. Note that this would be enough to prove strong implication if we considered the existence of the embeddings as a property of the critical point, and we would have also that I2(κ) strongly implies iterable I3(κ). But we already argued that this seems not the right way to look at this properties. Moreover, it is possible to prove the strong implications of such axioms respect to λ:

27 Corollary 6.23. I2 strongly implies iterable I3.

1 Proof. Note that “j is ω1-iterable”, and therefore “j is iterable”, is Σ1 (there exist α<ω1, M0,...,Mα), j0,...,jα such that. . . ). Therefore one can put “j is iterable” in 6.22 and the proof works.

1 1 Theorem 6.24 (Laver, [27]). Let n be odd. Then if j is Σn, it is also Σn+1. Proof. Theorem 6.19 is the case n = 1. We proceed by induction on n. Let B ⊆ Vλ. We have to prove that for every ψ Σn+1 formula and B ⊆ Vλ+1,

+ Vλ ψ(B) iﬀ Vλ ψ(j (B)).

By induction, the direction from left to right is immediate, so it suﬃces to show the other direction. Suppose

+ Vλ ∃X1 ∀X2 ∃X3 ... ∀Xn+1 ϕ(X1,...,Xn+1, j (B)).

1 Let X1 = A a witness. By Lemma 6.22, we can pick a k that is Σn−2 such that ′ ′ k(B)= j(B) and there exists A ⊆ Vλ+1 such that k(A )= A. By induction, k 1 is also Σn−1, and since

′ Vλ ∀X2 ∃X3 ... ∀Xn+1 ϕ(k(A ),X2,...,Xn+1, k(B)), then ′ Vλ ∀X2 ∃X3 ... ∀Xn+1 ϕ(A ,X2,...,Xn+1,B). Therefore A′ is a witness for ψ(B). Theorem 6.24 shows a peculiar asimmetry. What problems arise when n is even? The key is in the proof of Lemma 6.21. When the number of the quantifiers of a Σn formula is odd, then the last one is an existential quantifier. Since 1 1 by Lemma 6.12 the satisfaction relation for Σ1 formulae is Σ1, this quantifier is absorbed by the satisfaction formula. When n is even, however, the last quan- 1 1 tifier is a universal one, and “being Σn“ is still Πn+2. This seems an ineffectual detail, but in fact Laver proved that it is an essential one, since he showed that 1 1 being a Σn+1-embedding, for n even, is strictly stronger than being a Σn one. 1 Theorem 6.25 (Laver, [27]). Let h and k be Σn embeddings. Then h · k is a 1 Σn-embedding.

1 Proof. By Theorem 6.24 we can suppose n odd. Let ϕ be a Σ0 formula with n + 1 variables. Then

Vλ ∀Y ∃X1 ∀X2 ... ∃Xn ϕ(X1,...,Xn, k(Y )) → ∃X1 ∀X2 ... ∃Xn ϕ(X1,...,Xn, Y ) . 1 1 This formula is Πn+1. Since by Theorem 6.24 h is Σn+1, we have

Vλ ∀Y ∃X1 ∀X2 ... ∃Xn ϕ(X1,...,Xn,h(k)(Y )) → ∃X1 ∀X2 ... ∃Xn ϕ(X1,...,Xn, Y ) . 1 so h(k) is Σn.

28 Note that the converse is not true: minimizing the critical point, we can ﬁnd 1 1 a k that is Σn such that any h so that h(h) = k cannot be Σn (see comment after Lemma 6.22). However, if we switch application with composition, then also the converse is true.

1 1 1 Lemma 6.26. If h and k are Σm and h ◦ k is Σm+1, then k is Σm+1. Proof. Suppose that

Vλ ∀X1 ∃X2 ... ∀Xm+1 ϕ(X1,...,Xm+1,B) with B ⊆ Vλ. Then

Vλ ∀X1 ∃X2 ... ∀Xm+1 ϕ(X1,...,Xm+1,h ◦ k(B)).

Let A ⊆ Vλ. Then in particular

Vλ ∃X2 ... ∀Xm+1 ϕ(h(A),X2,...,Xm+1,h ◦ k(B)).

By elementarity

Vλ ∃X2 ... ∀Xm+1 ϕ(A, X2,...,Xm+1, k(B)).

Since this is true for every A ⊆ Vλ, we have

Vλ ∀X1 ∃X2 ... ∀Xm+1 ϕ(X1,...,Xm+1, k(B)).

1 1 Theorem 6.27 (Laver, [27]). Let h, k ∈Eλ. Then h, k are Σn iﬀ h ◦ k is Σn.

1 1 Proof. If h and k are Σn, then obviously h ◦ k is Σn. We prove by induction on 1 m ≤ n that h and k are Σm. The case m = 0 is by hypothesis. Suppose it is true for m. Then by Theorem 1 1 6.25 h(k) is Σm. It is easy to calculate that h ◦ k = h(k) ◦ h, and this is Σm+1 1 by hypothesis. By using Lemma 6.26 in the left side, we have that k is Σm+1, 1 and using it on the right side we have that h is Σm+1. 1 Theorem 6.27 is promising for our objective, that is proving that beingΣn+2 1 is strictly stronger than being Σn for an elementary embedding. The most natural idea for doing this is using some sort of reﬂection, to prove that if there 1 ′ 1 ′ is a j ∈ Eλ Σn+2, then there is a k ∈ Eλ that is Σn for some λ < λ. It would be tempting to use, as in the case of iterability, direct limit of elementary embeddings, but unfortunately in this setting there are counterexamples:

1 Theorem 6.28 (Laver, [26]). There exists a j that is Σn that has a stabilizing 1 direct limit of members of Aj that is not Σ1, So we will consider inverse limits instead. Let hj0, j1,... i be a sequence of elements of Eλ, and let J = j0 ◦ j1 ◦ ... be the inverse limit of the sequence. By deﬁnition the dominion of J is {x ∈ Vλ :

∃nx ∀i>nx ji(x) = x}. But we know that ji(x) = x iﬀ x ∈ Vcrt(ji), so this is

{x ∈ Vλ : ∃nx ∀i>nx x ∈ Vcrt(ji)}. That is, x ∈ dom J depends only on the

29 rank of x, and this implies that dom J = Vα for some α. It is also possible to calculate α, since β < α iﬀ ∃n ∀m ≥ nβ< crt(jm):

α = sup inf crt(jm) = lim inf crt(jn). n≥0 m≥n n∈ω

With some cosmetic change, we can also suppose that α as the supremum of the critical points, not only the limit inferior. This will also simplify the following proofs and notations. So let λn = infm≥n crt(jm). Then α = supn∈ω λn, and λn is increasing in n. If the supremum is also a maximum, we incur in the trivial case, where J is in fact just a finite composition of elementary embeddings: if n is the first one such that λn = α, then all crt(jm) with m>n are bigger than α, so they are constant in the domain of J and they don’t change anything. Suppose then that α is a proper supremum of the λn sequence. We can suppose crt(jn) = λn by aggregating multiple elementary embeddings in just one: consider the largest n such that crt(jn) = λ0 (there will be a largest one because α is a proper supremum of the λn sequence), define the new k0 as the old j0◦···◦jn, and repeat this for every λn. The following is a graphical example:

k0 k1 k2 k3

The columns represent the behaviours of each jn on λ, where the column on the left represent j0, and the horizontal lines indicate the critical point.

Deﬁnition 6.29. Let J = j0 ◦ j1 ◦ ... . Then we deﬁne Jn = jn ◦ jn+1 ◦ ... and J0(n−1) = j0 ◦ j1 ◦···◦ jn−1.

Lemma 6.30. Let jm ∈Eλ for every m ∈ ω, deﬁne αm = crt(jm) and suppose that for every m ∈ ω, αm < αm+1. Let α = supm∈ω αm and J = j0 ◦ j1 ◦ ... . Then • J ′′α is unbounded in λ;

• J : Vα ≺ Vλ is elementary.

Proof. • Let δm = sup Jm“α. We prove that when δm < λ, then δm 6= δm+1. Obviously δm ≥ α, because otherwise Jm(δm) ∈ Jm“α and Jm(δm) <δm. In particular δm is above αm, i.e., the critical point of jm, so δm is moved by jm and there exists a µ<δm such that jm(µ) >δm. By contradiction, suppose that δm = δm+1. Then µ is also less than δm+1, so by deﬁnition there exists an i ∈ ω such that Jm+1(αi) ≥ µ. Therefore

Jm(αi)= jm ◦ Jm+1(αi) ≥ jm(µ) ≥ δm,

contradiction.

30 Suppose then that δ0 < λ. Therefore for any m ∈ ω δm+1 <δm, but this creates a strictly descending sequence of ordinals, contradiction.

• Fix n ∈ ω and let k = j0(j1(...jn(jn) ... )). Then

crt(k) = crt(j0(j1(...jn(jn) ... ))) =

= j0(j1(...jn(crt(jn)) ... )) = j0 ◦ j1 ◦···◦ jn(αn)= J(αn).

By Lemma 4.12, then, VJ(αn) ≺ Vλ. But J ↾ Vαn : Vαn → VJ(αn) is an

elementary embedding, because J ↾ Vαn = J0,n ↾ Vαn , so for every n ∈ ω,

J ↾ Vαn ≺ Vλ. With methods similar to those in the proof of Theorem 4.6, it is possible to prove that this implies J : Vα ≺ Vλ.

Like in the Vλ case, we can extend J to Vα+1 in the expected way: when A ⊆ Vα, J(A) = β<α J(A ∩ Vβ). Now we want to prove an equivalent of Lemma 6.22, but for inverse limits. S 1 Lemma 6.31 (Square root for inverse limits of Σn). Let J : Vα ≺ Vλ an 1 inverse limit of Σn+1 elementary embeddings. Then for all A, B ⊆ Vα there 1 ′ exist K : Vα ≺ Vλ inverse limit of Σn elementary embeddings and A ⊆ Vα such that k(A′)= A and k(B)= J(B).

Proof. We deﬁne km and Am by induction, with repeated uses of Lemma 6.22. At the end, K will be the inverse limit of the km’s, and the Am’s will be the images of A′ through the inverse limit. 1 Let A0 = A, k0 and A1 such that k0 is a Σn elementary embedding, A1 ⊆ Vλ, k0(A1)= A = A0, k0(J1(B)) = j0(J1(B)) = J(B) and crt(k0) < crt(j0). 1 More generally, km+1 and Am+2 are such that km+1 is a Σn elementary embedding, Am+2 ⊆ Vλ, km+1(Am+2)= Am+1,

km+1(Jm+2(B)) = jm+1(Jm+2(B)) and crt(jm) < crt(km+1) < crt(jm+1). So km and Am satisfy:

1 • km is Σn;

• km(Am+1)= Am;

• km(Jm+1(B)) = Jm(B);

• crt(k0) < crt(j0) < crt(k1) < ··· < crt(jm) < crt(km) < crt(jm+1) < . . .

Let K be the inverse limit of the k’s. Then supm∈ω crt(km) = α, and by Lemma 6.30 K : Vα ≺ Vλ is an elementary embedding. Note that

K(B ∩ Vcrt(km))= k0 ◦ k1 ◦···◦ km(Km+1(B ∩ Vcrt(km))=

= k0 ◦···◦ km(B ∩ Vcrt(km))= k0 ◦···◦ km(Jm+1(B ∩ Vcrt(km))=

= J(B ∩ Vcrt(km)).

31 So K(B)= J(B).

Finally, consider Am+1 ∩ Vcrt(km):

Am+2 ∩ Vcrt(km) = km+1(Am+2 ∩ Vcrt(km))= Am+1 ∩ Vcrt(km), so

K(Am+1 ∩ Vcrt(km))= k0 ◦···◦ km(Am+1 ∩ Vcrt(km))= A ∩ VK(crt(km)). ′ ′ Deﬁne A = m∈ω(Am+1 ∩ Vcrt(km)). Then K(A )= A. We use LemmaS 6.31 to calculate the strength of an inverse limit: 1 Theorem 6.32 (Laver, [27]). If J : Vα ≺ Vλ is an inverse limit of Σn elemen- 1 tary embeddings, then J is Σn. Proof. The case n = 0 is Lemma 6.30 combined with an obvious generalization of Theorem 4.6, so we proceed by induction on n. 1 1 Suppose that J is Σn−1, we need to prove that for every ϕ Πn−1-formula, and any B ⊆ Vλ,

Vλ ∃Xϕ(X,J(B)) → Vα ∃Xϕ(X,B).

Suppose Vλ ∃Xϕ(X,J(B)), and ﬁx A a witness. Using Lemma 6.31, we 1 ′ ﬁnd K inverse limit of Σn−2 elementary embeddings such that K(A )= A and ′ ′ K(B)= J(B) for some A ⊆ Vα. So Vλ ϕ(K(A ),K(B)), and by elementarity ′ Vα ϕ(A ,B), that is Vα ∃X ϕ(X,B).

1 Finally, we can prove that the existence of a Σn+2 elementary embedding 1 strongly implies the existence of a Σn elementary embedding. 1 Theorem 6.33 (Laver, [27]). Let j : Vλ ≺ Vλ be Σn+2. Then

• for every B ⊆ Vλ, there exist an α < λ and a kα : Vα ≺ Vλ such that kα(Bα)= B for some Bα ⊆ Vα. In fact, we can ﬁnd an ω-club C ⊆ λ of such α’s.

1 • there exist an α < λ and a jα : Vα ≺ Vα that is Σn. Moreover, we can ﬁnd 1 an ω-club C ⊆ λ of such α’s. Therefore “there exists j : Vλ ≺ Vλ Σn+2” 1 strongly implies “j : Vλ ≺ Vλ Σn+2”. In particular I2 strongly implies I3. Proof. • Let

1 G = {hl0,...,lmi : li : Vλ ≺ Vλ is Σn, crt l0 < crt l1 < ··· < crt lm <κ0,

∃B0,...,Bm l0(B0)= B, ∀i li+1(Bi+1)= Bi}. By Lemma 6.22 the set

{θ<κ0 : ∃l hl0,...,lm,li ∈ G, crt(l)= θ}

is unbounded in κ0. Pick an infinite branch hl0,l1,... i of G, and let α = supi∈ω crt(li). Let kα be the inverse limit of the li’s. Then by 1 ′ Theorem 6.32 kα is Σn. Define B = m∈ω(Bm+1 ∩ Vcrt(lm)) as in Lemma ′ 6.31 to have kα(B )= B. To prove the existence of the ω-club C, note that we could have used any infinite branchS of T , and the set of the ordinals that are the supremum of the critical points of the elementary embeddings appearing in an infinite branch of T (like α) contains an ω-club.

32 • Let the B above be j. Then there exists α (again, any α ∈ C works), kα : Vα ≺ Vλ, and jα ⊆ Vα such that kα(jα) = j. Suppose, by Theorem 1 1 6.24, that n is odd. Then by Lemma 6.21 ”j is Σn” is Πn+1. Again by 1 Theorem 6.24 and Lemma 6.32 kα is Σn+1, so by elementarity jα : Vα ≺ Vα 1 is Σn.

The situation for I1 has some peculiarities. First of all, if j : Vλ+1 ≺ Vλ+1, then j is deﬁnable from j ↾ Vλ, so it is actually a deﬁnable class of Vλ+1 This is not possible in the I3 case, as Vλ ZFC, so it would contradict the naive Kunen’s Theorem. We can reformulate Lemma 6.21 in this way:

Remark 6.34. Suppose that j : Vλ+1 → Vλ+1. Then there are Πn+1 formulas ϕn(x) such that for any n ∈ ω, Vλ+1 ϕn(j ↾ Vλ) iﬀ j is an elementary embedding.

− − − So let j : Vλ+1 ≺ Vλ+1. Consider j = j ↾ Vλ. We can define j (j ): Vλ ≺ − − + Vλ. By the remark above [j (j )] is an elementary embedding, and we write it simply as j(j). So we can iterate also the I1 embeddings. 1 As j : Vλ+1 ≺ Vλ+1 is Σn for every n, by Lemma 6.22 for any n it is possible 1 to find a square root of j that is Σn. Therefore for any n is it possible to find 1 α < λ and K : Vα ≺ Vλ inverse limit that is Σn, so I1 strongly implies any 1 “There exists j : Vλ ≺ Vλ that is Σn”. We can do more: 1 Corollary 6.35. I1(λ) strongly implies ∀n ∃j : Vλ ≺ Vλ that is Σn Proof. Run the proof of Theorem 6.33 n times, each time considering that j is 1 Σn. So there are Cn ω-clubs in λ such that for any α ∈ Cn there exists k : Vα ≺ 1 Vα that is Σn. The intersection of ω ω-clubs is an ω-club, so α ∈ n∈ω Cn is as desired. T It is possible to push this kind of proof even more:

Lemma 6.36 (Square root Lemma for L(Vλ+1), Laver, [27]). Let j : L1(Vλ+1) ≺ L1(Vλ+1). For every A, B ⊆ Vλ, β < crt(j) there exists k : Vλ+1 ≺ Vλ+1, with ′ β < crt(k) < crt(j), such that k(B)= j(B) and there exists A ⊆ Vλ such that k(A′)= A.

Proof. The only detail we should care of is the fact that “j : Vλ+1 ≺ Vλ+1” must be deﬁnable in L1(Vλ+1). But this is true, because the satisfaction relation in Vλ+1 is deﬁnable in L1(Vλ+1), so the proof is, mutatis mutandis, the same as Lemma 6.22.

Theorem 6.37 (Laver, [27]). “There exists j : L1(Vλ+1) ≺ L1(Vλ+1)” strongly implies “there exists j : Vλ+1 ≺ Vλ+1”. Proof. The proof follows the same method of Theorem 6.33.

1 Note that in this case there is a fundamental diﬀerence between being Σn for every n and being an elementary embedding from Vλ+1 to itself. Suppose that we have ji : Vλ+1 ≺ Vλ+1. Then the inverse limit J of hji : i ∈ ωi is not an elementary embedding from Vλ′+1 to Vλ+1, because the domain of J is just not Vλ′+1. Consider again the deﬁnition of the domain of J:

H = {x ∈ Vλ+1 : ∃n ∈ ω ∀m ≥ n jm(x)= x}.

33 ′ Following this definition, for example, λ ∈/ H, because it is moved by all ji, while on the other hand λ ∈ H, because it is never moved. What we can prove with the methods provided, is that the unique extension of the inverse limits of the ji ↾ Vλ to Vλ′+1 is an elementary embedding. As I0 is equivalent to j : LΘ(Vλ+1) ≺ LΘ(Vλ+1), we consider now hypotheses like “there exists j : Lα(Vλ+1) ≺ Lα(Vλ+1)”, α < Θ, with the aim of finding strong implications among them. Note that with such aim not any α will be appropriate. We defined strong implications only between properties of λ, while “there exists j : Lα(Vλ+1) ≺ Lα(Vλ+1)” is a property of α and λ. We should consider therefore only α’s that depend on λ, for example α’s that are definable from λ (e.g., λ+, λ+ + ω, ...) or just absolutely definable (e.g., ω, ωω, . . . ). This kind of ordinals have also the characteristic of being a fixed point for any large enough embedding, so we already have that I0 implies “there exists j : Lα(Vλ+1) ≺ Lα(Vλ+1)”. Our objective is now, given j : Lβ(Vλ+1) ≺ Lβ(Vλ+1) to find a α<β and ¯ α¯, λ < λ such that Lα¯(Vλ¯+1) ≡ Lα(Vλ+1) and there exists k : Lα¯(Vλ¯+1) ≺ Lα¯(Vλ¯+1). With β and α as above, this will imply that “exists j : Lβ(Vλ+1) ≺ Lβ(Vλ+1)” strongly implies “exists j : Lα(Vλ+1) ≺ Lα(Vλ+1)”. The strategy is to follow Laver’s proof for strong implications as much as possible. The first step is therefore to prove some square root principle like Lemma 6.22. The best possible outcome would be to find for any j : Lα+1(Vλ+1) ≺ Lα+1(Vλ+1) a square root k that is k : Lα(Vλ+1) ≺ Lα(Vλ+1). For this, it is needed that Lα+1(Vλ+1) ∃k : Lα(Vλ+1) ≺ Lα(Vλ+1),

i.e., that j ↾ Lα(Vλ+1) ∈ Lα+1(Vλ+1). Can we assume that? The next step would be to define an inverse limit of embeddings at the Lα(Vλ+1) level. Here another problem arises following the remarks after The- orem 6.37: the domain inverse limit of the ki : Lα(Vλ+1) ≺ Lα(Vλ+1) is not some Lα¯(Vλ¯+1), in fact it does not even contain Vλ¯+1, therefore we should do the inverse limit of ki ↾ Vλ, and then extend it to some Lα¯(Vλ¯+1). How to do it? Both problems were solved with the study of the structure of the sets Lα(Vλ+1) for α< Θ made by Laver in [28]. Definition 6.38 (Laver, [28]). Let λ be a cardinal and let α < Θ. Then α is good iff every element of Lα(Vλ+1) is definable in Lα(Vλ+1) from an element in Vλ+1. The successor of a good ordinal is a good ordinal: let α be good; the largest ordinal in Lα+1(Vλ+1) is λ +1+ α, therefore α is definable in Lα+1(Vλ+1) with λ as a parameter, so Lα(Vλ+1) is definable in Lα+1(Vλ+1) with λ as a parameter. But every element in Lα+1(Vλ+1) is definable using Lα(Vλ+1) and elements of Lα(Vλ+1) (which in turn are definable with parameters from Vλ+1), and therefore α + 1 is good. This proves that the natural numbers are good. But also ω is good: every element of Lω(Vλ+1) is in some Ln(Vλ+1), n ∈ Vλ+1 and n is good for every n. Following the same line of reasoning, every ordinal up to λ is good, and considering that all ordinals less than λ+ are coded as + subsets of λ, and therefore in Vλ+1, every ordinal up to λ is good. On the other side, non-good ordinals exist. The definition of good ordinal is restricted to ordinals strictly less than Θ because larger ones are trivially not

34 good: if x is deﬁnable with a parameter, then it is uniquely determined by its deﬁnition, therefore for any α there exists in L(Vλ+1),

π : Vλ+1 ։ Gα = {x ∈ Lα(Vλ+1):

Lα(Vλ+1) x is deﬁnable from an element in Vλ+1};

then if Θ ⊆ Lα(Vλ+1), Gα must be strictly contained in Lα(Vλ+1), by definition of Θ. But non-good ordinals exist also below Θ: deﬁne Lγ(Vλ+1) as the collapse of the Skolem closure of Vλ+1 in LΘ(Vλ+1); as LΘ(Vλ+1) ∃x x not deﬁnable from an element in Vλ+1, by elementarity the same must be true in Lγ(Vλ+1), as the collapse does not collapse Vλ+1, therefore γ is not good. One can ask how many good ordinals there are. Lemma 6.39 (Laver, [28]). Let λ be a strong limit cardinal. Then the good ordinals are unbounded in Θ. Therefore assuming α good is in most cases a reasonable choice. By the usual condensation argument, if α< Θ and i : Lα(Vλ+1) → Lα(Vλ+1) then i ∈ LΘ(Vλ+1). If α is good, however, it is possible to be much more precise: Lemma 6.40 (Laver, Woodin, [39],[28]). Let λ and α be such that α is good and there exists i : Lα(Vλ+1) ≺ Lα(Vλ+1) with crt(i) < λ. Then i is induced by i ↾ Vλ, and therefore i ∈ Lα+1(Vλ+1). Finally:

Lemma 6.41 (Square root Lemma for Lα(Vλ+1)). Let α be good, and let j : Lα+1(Vλ+1) ≺ Lα+1(Vλ+1). Then for any A, B ∈ Lα(Vλ+1), for any β < crt(j) there exists k : Lα(Vλ+1) ≺ Lα(Vλ+1) such that β < crt(k) < crt(j), k(k ↾ Vλ)= j ↾ Vλ, k(B)= j(B) and there exists A ∈ ran(k). Proof. Exactly as Lemma 6.22. The key point is that as α is good, “∃k k : Lα(Vλ+1) ≺ Lα(Vλ+1)” is expressible in Lα+1(Vλ+1) and it is satisfied by j ↾ 1 Lα(Vλ+1), so it can be put inside the formula instead of “j is Σn−2”. Note that if α is not good, then we can have a similar result: for j : Lβ(Vλ+1) ≺ Lβ(Vλ+1), let α be maximum such that j ↾ Lα(Vλ+1) ∈ Lβ(Vλ+1). Then we can find a square root that extend to Lα(Vλ+1). The difference is that in this case the gap can be large, and it is somewhat arbitrary, while for the good case is just one. Moreover, I0 implies “∃j : Lα(Vλ+1) ≺ Lα(Vλ+1)”:

Lemma 6.42 (Woodin, [39]). Let λ be a cardinal. If there exists j : LΘ(Vλ+1) ≺ LΘ(Vλ+1), then for any α< Θ there exists an i : Lα(Vλ+1) ≺ Lα(Vλ+1). Proof. Suppose it is false. Then there is a counterexample α such that every i : Lα(Vλ+1) → Lα(Vλ+1) is not an elementary embedding. All such i’s are in LΘ(Vλ+1) (see remark before Lemma 6.40), therefore LΘ(Vλ+1) ∃α (α is a counterexample). Let α0 be the least counterexample. Then α0 is

deﬁnable in LΘ(Vλ+1) and j(α0)= α0. Then j ↾ Lα0 (Vλ+1) is as in the lemma, contradiction.

35 Compare this lemma with the fact that Uj ∈/ L(Vλ+1), and therefore j ↾ LΘ(Vλ+1) ∈/ L(Vλ+1). Then if j(α)= α and

j ↾ Lα(Vλ+1): Lα(Vλ+1) ≺ Lα(Vλ+1), for any β > α there must be a j1 elementary embedding with domain Lβ(Vλ+1) and such that j1 6= j ↾ Lβ(Vλ+1), otherwise one could build j ↾ LΘ(Vλ+1) inside L(Vλ+1). In other words, the tree of embeddings with domain of the type Lα(Vλ+1) has levels of “cardinality” < Θ (by 7.7), has height Θ and there is no cofinal branch in L(Vλ+1). In a certain sense, it is a Θ-Aronszajn tree in L(Vλ+1). Now we can do the inverse limit like in Theorem 6.31, and so we have J : Vλ¯ ≺ Vλ, inverse limit of embeddings that can be extended to Lα(Vλ+1). Withα ¯ fixed, we can extend J to Lα¯(Vλ¯+1). But how to find aα ¯ so that J is an elementary embedding? This point needs more work, and has been solved by Scott Cramer in his PhD Thesis. The idea is always to use 6.41 repeatedly, to build inverse limits that have in the range witnesses for existential formulas, so that in the end the inverse limit is elementary. The difference is that we need λ different witnesses to be considered every time, so the plan is to build a forcing that at the same time constructs a limit of inverse limits and collapses λ to ω. We follow with a sketch of the ideas behind such a construction, for the details we refer the reader to [3]. The structure of the proof of Theorem 6.31 is like this:

j ↾ Vλ j ↾ Vλ j ↾ Vλ j ↾ Vλ ...

K : k0 k1 k2 k3 ...

where each embedding below is a square root of the embedding above. We have some freedom for the critical points in the embedding belows, in that for each one we can say that it must be bigger than a certain ﬁxed ordinal, and (optionally) less than crt(j). In this way, we can push the limit of the critical points of the ki’s to be λ, so that the inverse limit will be from Vλ to itself, or less than crt(j), as in Lemma 6.31, so to have J : Vλ¯ ≺ Vλ. So, by Lemma 6.41, the existence of j : Lα+1(Vλ+1) ≺ Lα+1(Vλ+1) implies the existence of an inverse limit J of square roots of j ↾ Vλ such that J : Vλ¯ ≺ Vλ (possibly with λ¯ = λ), and all the square roots can be extended to Lα(Vλ+1). Instead of j, we can start already with an inverse limit:

J : j0 j1 j2 j3 ...

K : k0 k1 k2 k3 ...

We say that K (the inverse limit of the ki’s) is a limit root of J (the inverse limit of the ji’s) iﬀ there exists an n ∈ ω such that ki = ji for i

36 The next step is to build a limit of inverse limits. Let j be an embedding that extends to Lα+ω(Vλ+1). Then consider something like this:

j j j j ...

J : j0 j1 j2 j3 ...

1 1 1 K1 j0 k1 k2 k3 ...

1 1 2 K2 j0 k1 k2 k3 ...

......

where J is an inverse limit of square roots of j, and every Ki+1 is a limit n root of Ki. We can choose the km’s so that they all extend to Lα(Vλ+1): Let ni be an increasing sequence of integers. Then choose ji to be a square root m of j extendible to Lα+ni (Vλ+1). Once deﬁned ki , if it is extendible to some m+1 m Lα+n(Vλ+1) with n > 0, then choose ki to be a square root of ki that is m+1 m extendible to Lα+n−1(Vλ+1), otherwise ki = ki . As ni was increasing, the last case can happen only a ﬁnite amount of times, so Km+1 will always be a limit square root of Km. Also, at each stage, for any Am ⊆ Vλ¯ (where Vλ¯ ) m+1 m is the domain of J) and any Bm ⊆ Vλ, we can have K (A) = K (A) and m+1 m+1 m m+1 B ∈ ran(ki ◦ ki+1 ◦ ... ), where i is the minimum such that ki 6= ki . Note that this is less than a Square root Lemma for inverse limits for Lα(Vλ+1), as we ask only B to be in the range of a tail of the limit, and not of K. But this will be enough.

Theorem 6.43 (Cramer, 2015 [3]). Let α be good, and let j : Lα+ω(Vλ+1) ≺ Lα+ω(Vλ+1). Let J be as the ﬁrst step of the construction above, with Vλ¯ its ¯ domain. Then there exist α,¯ λ < λ such that Lα¯(Vλ¯+1) ≡ Lα(Vλ+1).

Proof. First of all, via a definable surjection ρ : Vλ+1 ։ Lα(Vλ+1), we can think of any element of Lα(Vλ+1) as a subset of Vλ. We work in V [G], where G is the generic of the collapse of λ on ω. Therefore Vλ¯+1 is countable: let han : n ∈ ωi be an enumeration of it, and let hϕi : i ∈ ωi be an enumeration of all formulas. We continue the construction of the limit of limit roots as above. The idea is that we gradually add to Am all the elements of Vλ¯ , each time adding just a finite amount, while Bm will be the set of witnesses of existential formulas with ′′ ′′ parameters in ρ KmAm. So suppose that we have defined Km and that there exists an nm such that m Am = hai : i

Lα(Vλ+1) ∃x ψ(x, ρ(Km(ai0 )),...ρ(Km(ail ))), with i0,...,il < nm. We are following the construction of the limit of limit roots, so at this step there exists a p such that Km+1 should be a limit square m+1 m m+1 m m+1 m root of Km such that k0 = k0 , ... , kp = kp , kp+1 6= kp+1, etc... Call m m (Km)p = kp ◦ kp+1 ◦ ... . Then by elementarity (and the deﬁnability of ρ)

Lα(Vλ+1) ∃x ψ(x, ρ((Km)p(ai0 )),...ρ((Km)p(ail ))).

37 Let b that witnesses this, and let Bm be such that ρ(Bm) = b. Finally let Km+1 be the limit root of Km such that Km+1(Am) = Km(Am) and Bm ∈ ran((Km+1)p). By the remarks above Km+1 exists, and there exists a ∈ Vλ¯+1 such that ρ((Km+1)p(a)) = b, so

Lα(Vλ+1) ψ(ρ((Km+1)p(a)),ρ((Km+1)p(ai0 )),...ρ((Km+1)p(ail )))

(note that (Km+1)p(ai) = (Km)p(ai)). By elementarity

Lα(Vλ+1) ψ(ρ((Km+1)(a)),ρ((Km+1)(ai0 )),...ρ((Km+1)(ail ))).

Let a = at, and ﬁx Am+1 = han : n ≤ ti. If a = an ∈ Vλ¯ , then for any m such that nm >nKm(a)= Km+1(a). Deﬁne m J∗ : Vλ¯+1 ≺ Vλ+1 as J∗(an) = K (an) with nm > n. It is called the common ′′ ′′ part of the sequence hKi : i<ωi. Let M = ρ J∗ Vλ+1¯ . Then M ≺ Lα(Vλ+1): suppose that

Lα(Vλ+1) ∃x ψ(x, ρ(J∗(ai0 )),...ρ(J∗(ail ))), then there exists m such that nm >ni,0,...,nil and

Lα(Vλ+1) ∃x ψ(x, ρ(Km(ai0 )),...ρ(Km(ail ))).

But we built Km+1 so that

Lα(Vλ+1) ψ(ρ((Km+1)(a)),ρ((Km+1)(ai0 )),...ρ((Km+1)(ail ))), and Km+1(a) = J∗(a). By condensation, the collapse of M is a Lα¯(Vλ¯+1) for someα ¯. By absoluteness, this is true also in V . Theorem 6.43 gives us finally the rightα ¯ where to extend J, the inverse limit of embeddings that can be extended to Lα(Vλ+1) in the most natural ¯ way: asα ¯ is (λ−) good, then every element b ∈ Lα¯(Vλ¯+1) is defined from some parameter a ∈ Vλ¯+1, and J(b) is the element of Lα(Vλ+1) that is defined with the same formula and parameter J(a). It remains to see that J is actually an elementary embedding. Theorem 6.43 gives us an elementary embedding, J∗, but it is in V [G]. But if we consider the partial order of all possible sequences of Ki’s, that start with J, coupled with the collapse, then J∗ will be the common part of the generic of such forcing. Considering J as a sequence of length 1, ∗ for any b ∈ Lα¯(Vλ¯+1) we can always extend J so that J (b) = J(b), and then elementarity comes from absoluteness. All the details are in [3].

Theorem 6.44 (Cramer, 2015, [3]). If j : Lα+ω(Vλ+1) ≺ Lα+ω(Vλ+1), then ¯ there exist α¯, λ < λ and k : Lα¯(Vλ¯ ) ≺ Lα¯(Vλ¯ ). In particular, if α depends only on λ, then “∃j : Lα+ω(Vλ+1) ≺ Lα+ω(Vλ+1)” strongly implies “∃j : Lα(Vλ+1) ≺ Lα(Vλ+1)”. So all the following are strong implications:

I0(λ) ⇒···⇒∃j : Lα+ω(Vλ+1) ≺ Lα+ω(Vλ+1) ⇒ ∃j : Lα(Vλ+1) ≺ Lα(Vλ+1) ⇒

⇒···⇒∃j : Lω(Vλ+1) ≺ Lω(Vλ+1) ⇒ ∃j : L1(Vλ+1) ≺ L1(Vλ+1 ⇒ 1 1 ⇒ I1(λ) ≡ ∃j : Vλ ≺ Vλ Σω ⇒ ∀n ∃j : Vλ ≺ VλΣn ⇒ ... 1 1 ···⇒∃j : Vλ ≺ Vλ Σn ≡ ∃j : Vλ ≺ Vλ Σn+1 (n is odd) ⇒ ... 1 1 ···⇒∃j : Vλ ≺ Vλ Σ2 ≡ j : Vλ ≺ VλΣ1 ≡ I2(λ) ⇒ iterable I3(λ) ⇒ ... ···⇒ ω · α-iterable I3(λ) ⇒ α-iterable I3(λ) ⇒ I3(λ).

38 7 Similarities with ADL(R)

We already noticed that for I0(λ) to hold it is necessary that L(Vλ+1) 2 AC, just like for AD to hold in L(R) it is necessary that L(R) 2 AC. This is the first sign of something bigger. The research is still ongoing on this respect, but there are many results that indicate a strong similarity between L(Vλ+1) under I0 and L(R) under AD. On the other hand, there are also striking differences, so there need to be further work on how much the two models are similar and why (and why they are not too similar). The first similarity will be the following:

Theorem 7.1 (Solovay). Suppose L(R) AD. Then L(R) ω1 is measurable.

Lemma 7.2 (Woodin, [39]). Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Then for every δ < λ regular, deﬁne

λ+ + Sδ = {η < λ : cof(η)= δ},

the stationary set of ordinals with the same cofinality and let F be the club filter + λ+ on λ . Then there exists η < λ and hSα : α < ηi ∈ L(Vλ+1) a partition of Sδ + such that for every α < η F ↾ Sα is a L(Vλ+1)-ultrafilter in λ . In particular λ+ is measurable.

Proof. Define κ0 = crt(j). Suppose that there is a δ such that this is false, and pick the minimum one. Then δ is definable (using only λ as a parameter, that is a fixed point), so j(δ)= δ, and this means that δ<κ0. Moreover, note that + + + j ↾ λ is in L(Vλ+1), because the elements of λ are well-orders of λ, so j ↾ λ depends on j ↾ Vλ+1, that in turn is defined by j ↾ Vλ, that is in Vλ+1. λ+ As in the proof of 3.13, there is no partition of Sδ in κ0 stationary sets. The difference is that here we need that C = {α < λ+ : j(α) = α} is a δ-club. But it is: let T~ be a δ-sequence of elements of D. Then T~ can be coded in Vλ+1, and since j(η) = η for every η ∈ ~t and j(δ) = δ we have j(~t) = ~t, so j(sup~t) = sup j(~t) = sup~t. Let I be the non-stationary ideal, i.e. the ideal of the non-stationary sets, λ+ and consider B = P(Sδ )/I. We prove that B is atomic, and its atoms are dense. Looking for a contradiction suppose that there exists a set S ∈ I+ that contains no atom, so that

+ + ∀T ⊂ S (T ∈ I → (∃T0,T1 ∈ I (T0 ∪ T1 = T ∧ T0 ∩ T1 = ∅))).

We construct by induction a tree T of subsets of S that refines reverse inclusion. Indicating Levγ(T ) as the γ-th level of T we define Lev0(T ) = {S} and every + T ∈ T has exactly two successors T0,T1 ∈ I such that T0∪T1 = T and T0∩T1 = DC ∅, chosen with λ. At a limit α, we put the intersections T = β<α Tβ, with T ∈ Lev (T ), such that T ∈ I+. We call Res(α) = S \ Lev (T ) the set of β β αT the points in S that don’t belong to any T in Levα(T ). Now, let α ≤ κ0 be a limit ordinal, we want to prove by induction that Res(α) ∈ I and Levα(T ) 6= ∅. Let x ∈ S \ β<α Res(β), then for every β < α there exists T ∈ Levβ(T ) such that x ∈ T , so we can define bx = {Z ∈ Levβ(T ): x ∈ Z} a S β<α branch of T of length α and x ∈ bx. So S is the union of Res(β), S β<α the points already skimmed by the process, and all of the bx. By induction T S T 39 every Res(β) ∈ I, and since I is the dual of the club filter, I is λ+-complete DC by λ, so β<α Res(β) ∈ I. We consider { bx : x ∈ S \ β<α Res(β)}: it is a partition of S \ Res(β) ∈ I+ and since λ is a strong limit we have S β<α T S that there cannot be more than 2|α| < λ branches, so by λ+-completeness there S + must be at least one set T ∈ I such that T = bx for some branch bx, i.e. + there must exist T ∈ Levα(T ). As for the branches such that bx ∈/ I , their union, again by completeness, must be in I, so T T + Res(α)= Res(β) ∪ { bx : bx ∈/ I } ∈ I, β<α [ [ \ \ and we can carry on the induction. If we consider a branch in T of length κ0, say hTα : α<κ0i, then

{Res(κ0) ∪ (T0 \ T1)}∪{Tβ \ Tβ+1 : β<κ0} is a partition of S in κ0 stationary sets, and we’ve just seen that this cannot be, so we’ve reached a contradiction. λ+ We’ve proved that the atoms of B = P(Sδ )/I are dense, so for every λ+ + S ⊆ Sδ there exists T ⊆ S atom for B, in other words there exists T ⊆ S in I + such that F ↾ T is an ultraﬁlter. Since by DCλ we have that F is λ -complete, F ↾ T is a measure. By induction (using DCλ) we deﬁne hSα : α < ηi:

λ+ + • Let S = Sδ . Then there exists a T ∈ I such that F ↾ T is a measure, and we choose S0 = T . λ+ • Let α ordinal, and suppose that for every β < α, Sβ is defined. If Sδ \ β<α Sβ ∈ I, then we stop the sequence and η = α. Otherwise there + λ + ↾ Sexists Sα ⊆ Sδ \ β<α Sβ such that Sα ∈ I is stationary and F Sα is a measure. S + In particular, F ↾ S0 is a measure for λ . It is remarkable that the proof is completely different, yet in both theorems the measure is not just any measure, but it comes from the club filter. Woodin calls the cardinals that are measurable for this reason “ω-strongly measurable cardinals”, and they have a key role in the development of the inner model descriptive set theory. R The fact that λ+ is measurable has a lot of AD -like consequences:

Theorem 7.3 (Shi, Trang, 2017 [36]). Suppose I0(λ) holds. Then in L(Vλ+1): • there is no λ+-Aronszajn tree;

• λ fails;

• there is no scale at λ in Vλ+1;

• ♦λ+ fails; • there is no λ+-sequence of distinct members of P(λ). In fact Theorem 7.1 was just the ﬁrst ω-strongly measurable cardinal found in L(R), when there is a rich structure of them under AD. The same is true in L(Vλ+1) under I0. Moschovakis’ Coding Lemma is the way to ﬁnd them:

40 Theorem 7.4 (Coding Lemma, Moschovakis). Suppose L(R) AD. Let η< Θ and ρ : R ։ η, ρ ∈ L(R). Then there exists γρ < Θ such that for every A ⊆ R × R, A ∈ L(R), there exists B ⊆ R × R such that: • B ⊆ A;

• B ∈ Lγρ (R); • for every α < η if ∃(a,b) ∈ A ρ(a)= α then ∃(a,b) ∈ B ρ(a)= α.

It is a sort of very weak choice principle for subsets of R: Consider Aξ = {b : (a,b) ∈ A, ρ(a) = ξ}. Thus A can be seen as an η-sequence of subsets of R. If we had the Axiom of Choice, we could have a selector for such a sequence.

Instead the Coding Lemma gives us a sequence Bξ ⊆ Aξ that is in some Lγρ (R). But however “complex” (i.e., high in the constructive hierarchy) we choose A, γρ is the same, so the “selector” is actually simpler than A. We want to prove the Coding Lemma in L(Vλ+1). First we prove a weakening of the Coding Lemma, that will be used in the proof of the Coding Lemma itself. Lemma 7.5 (Weak Coding Lemma, Woodin, [39]). Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Let η < Θ and ρ : Vλ+1 ։ η, ρ ∈ L(Vλ+1). Then there exists γρ < Θ such that for every A ⊆ Vλ+1 × Vλ+1, A ∈ L(Vλ+1), there exists B ⊆ Vλ+1 × Vλ+1 such that: • B ⊆ A;

• B ∈ Lγρ (Vλ+1); • for cofinally α < η if ∃(a,b) ∈ A ρ(a)= α then ∃(a,b) ∈ B ρ(a)= α. Proof. Since this lemma is quite rich in terms of quantifiers, we will use some abbreviation, calling WCL the sentence of the Lemma, WCLη the same, but with fixed η, WCLη,ρ the same, but with fixed η and ρ, and so on... Looking for a contradiction, suppose that WCL is false and let η be the least such that ¬WCLη. Then η is definable, so j(η) = η, and η is a limit: in fact, WCLη → WCLη+1, since for cofinally α < η +1 means for α = η, and in that case we can just choose an element in {(a,b) ∈ A : ρ(a)= η}. Let ρ be such that ¬WCLη,ρ. We define by induction, using DCλ, h(γξ,Zξ): ξ < λi:

• Suppose that we have deﬁned (γξ,Zξ). Since ¬WCLη,ρ, for every γ < Θ

¬WCLη,ρ,γ; so by ¬WCLη,ρ,γξ there exists Zξ+1 ⊆ Vλ+1 ×Vλ+1 such that

– {ρ(a): ∃b (a,b) ∈ Zξ+1} is unbounded in η;

– for every B ⊆ Zξ+1, B ∈ Lγξ (Vλ+1), {ρ(a): ∃b (a,b) ∈ B} is bounded in B.

Let γξ+1 be such that Zξ+1 ∈ Lγξ+1 (Vλ+1) and γξ + ω<γξ+1. By Lemma 5.6 we can suppose that γξ+1 < Θ.

• In the limit case, we choose γξ such that γξ >γζ + ω for every ζ < ξ and

a Zξ ⊆ Vλ+1 × Vλ+1 such that Zξ ∈ Lγξ (Vλ+1).

So, for every ξ1 < ξ2 < λ we have:

41 • γξ1 + ω<γξ2 ;

• Zξ1 ∈ Lγξ1 (Vλ+1);

•{ρ(a): ∃b (a,b) ∈ Zξ+1} is unbounded in η and for every B ⊆ Zξ+1,

B ∈ Lγξ (Vλ+1), {ρ(a): ∃b (a,b) ∈ B} is bounded in B.

Let ρ0 = ρ, ρn+1 = j(ρn),

0 0 h(γξ ,Zξ ): ξ < λi = h(γξ,Zξ): ξ < λi and n+1 n+1 n n h(γξ ,Zξ ): ξ < λi = j(h(γξ ,Zξ ): ξ < λi). n n+1 Let n be the minimum such that γκ0+1 ≤ γκ0+1 (there must exist, otherwise n hγκ0 : n ∈ ωi would be a descendent chain of ordinals). By elementarity we have that:

n • Z ∈ Lγn (Vλ+1); κ0+1 κ0+1

n+1 n+1 •{ρn+1(a): ∃b (a,b) ∈ Zκ1+1} is unbounded in η and for every B ⊆ Zκ1+1, B ∈ L n+1 (Vλ+1) we have that {ρn+1(a): ∃b (a,b) ∈ B} is bounded in η; γκ1

n n+1 • j(Zκ0+1)= Zκ1+1. n Let B = {(j(a), j(b)) : (a,b) ∈ Zκ0+1}. The parameters used in the deﬁnition n of B are j ↾ Vλ+1 (that in turn is deﬁned by j ↾ Vλ) and Zκ0+1, so B ∈ n Lγn +1(Vλ+1) ⊆ Lγn (Vλ+1). If (a,b) ∈ Z , then κ0+1 κ1 κ0+1

n n+1 j((a,b)) = (j(a), j(b)) ∈ j(Zκ0+1)= Zκ1+1,

n+1 n so B ⊆ Zκ1+1. Finally, for every α < η there exists (a,b) ∈ Zκ0+1 such that α< ρn(a), so α ≤ j(α) <ρn+1(j(a)), and {ρn+1(a): ∃b (a,b) ∈ B} is unbounded in η. Contradiction. Theorem 7.6 (Coding Lemma, Woodin, [39]). Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Let η< Θ and ρ : Vλ+1 ։ η, ρ ∈ L(Vλ+1). Then there exists γρ < Θ such that for every A ⊆ Vλ+1 × Vλ+1, A ∈ L(Vλ+1), there exists B ⊆ Vλ+1 × Vλ+1 such that: • B ⊆ A;

• B ∈ Lγρ (Vλ+1); • for every α < η if ∃(a,b) ∈ A ρ(a)= α then ∃(a,b) ∈ B ρ(a)= α.

Proof. The proof is by induction on η. Let ρ : Vλ+1 ։ η. For every α < η we deﬁne ρα : Vλ+1 ։ α:

ρ(a) if ρ(a) < α; ρα(a)= (0 otherwise.

By induction, for every α < η there exists γρα < Θ that satisﬁes the Coding

Lemma for η,ρα. Let β0 = sup{γρα : α < η}. Since Θ is regular, we have that

42 β0 < Θ. Let γρ be the ordinal that witnesses the Weak Coding Lemma for η,ρ, and let β1 = sup{β0,γρ}. Since β1 < Θ, there exists π : Vλ+1 ։ Lβ1 (Vλ+1), and this π can be codiﬁed as a subset of Vλ+1. We call β the ordinal < Θ such that the code of π is in Lβ(Vλ+1). We prove that β + 1 witnesses the Coding Lemma for η,ρ. Fix A and deﬁne A<α = {(a,b) ∈ A : ρ(a) < α}. We can suppose that for every α < η there exists (a,b) ∈ A such that ρ(a)= α. We want to code the set {(A<α,B): α<η, B witnesses the Coding Lemma for η,ρα, A<α} as a subset of Vλ+1 × Vλ+1:

∗ A = {(a,b) ∈ Vλ+1 × Vλ+1 : ρ(a) > 0, π(b) ⊆ A<ρ(a), ∀ξ<ρ(a) ∃(x, y) ∈ π(b) ρ(x)= ξ}.

Since β witnesses the Coding Lemma for all the ρα, we have that {ρ(a): ∃b (a,b) ∈ A∗} = η, and since the Weak Coding Lemma holds for η,ρ,β, ∗ ∗ ∗ ∗ there exists B ⊆ A such that B ∈ Lβ(Vλ+1) and {ρ(a): ∃b (a,b) ∈ B } is coﬁnal in η. ∗ Let B = {π(b) : (a,b) ∈ B }. Then B ⊆ A and B ∈ Lβ+1(Vλ+1). More- ∗ over, for every α < η, there exists (a,b) ∈ B such that ρ(a) > α, π(b) ∈ A<ρ(a) and there existsS (x, y) ∈ π(b) such that ρ(x) = α. But then (x, y) ∈ B, and B witnesses the Coding Lemma for η,ρ,β +1, A.

The Coding Lemma is used to prove a sort of inaccessibility of Θ in L(Vλ+1): γ as L(Vλ+1) 2 AC, we cannot say 2 < Θ for all γ < Θ; but we have the following:

Lemma 7.7 (Woodin, [39]). Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Then in L(Vλ+1) for every α < Θ there exists a surjection π : Vλ+1 ։ P(α).

Proof. Fix α< Θ, ρ : Vλ+1 ։ α, and let γ be the maximum between the least β < Θ such that ρ ∈ Lβ(Vλ+1) and the witness for the Coding Lemma for α, ρ. ∗ ∗ For every A ⊆ α deﬁne A = {(a, 0) ∈ Vλ+1 : ρ(a) ∈ A}. Then A ∈ L(Vλ+1). ∗ ∗ ∗ ∗ So there exists B ⊆ A , B ∈ Lγ(Vλ+1) such that {ρ(a): ∃b (a,b) ∈ B } = A. But then A ∈ Lγ+1(Vλ+1). This means that P(α) ⊆ Lγ+1(Vλ+1), and using π : Vλ+1 ։ γ + 1 we’re done. The following is the strongest theorem related to the Coding Lemma that has been proved until the writing of this paper, and it is akin to a similar one R in ADL( ) (cfr. with [32]):

Theorem 7.8 (Woodin, [39]). Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j)= κ0 < λ. Then Θ is a limit of γ such that:

1. γ is weakly inaccessible in L(Vλ+1);

2. γ =ΘLγ (Vλ+1) and j(γ)= γ;

3. for all β<γ, P(β) ∩ L(Vλ+1) ∈ Lγ(Vλ+1);

4. for coﬁnally κ<γ, κ is an ω-strongly measurable cardinal in L(Vλ+1);

5. Lγ(Vλ+1) ≺ LΘ(Vλ+1).

43 Proof. Fix α < Θ, j(α) = α and let γ < Θ be the minimum such that Lγ(Vλ+1) ≺1 LΘ(Vλ+1). As also j(Θ) = Θ, we have j(γ) = γ. This will be the γ as in the Theorem. By Corollary 5.15 we can have γ as large as we want below Θ. It is tempting to stop here, and say that γ has the same properties of Θ by elementarity, but we do not have this directly, as Θ ∈/ LΘ(Vλ+1). 2 Lγ (Vλ+1) Note that, by 5.8 relativized to Σ1(α) formulas, ∆1(α)sets, etc..., Θ = γ, and therefore (2). Also, cof(δ) > λ: Otherwise let hβξ : ξ<µi be a sequence 2 DC cofinal in δ with µ < λ. Then any βξ is the length of a ∆1(α) pwo. With λ choose one pwo for any ξ, and then gluing this pwos together it is possible to 2 find a ∆1(α) pwo of γ, and this contradicts 5.8. Now the Weak Coding Lemma Lγ (Vλ+1) can be re-proved in Lγ(Vλ+1): the key points are that since Θ = γ, γξ+1 <γ and since cof(γ) > λ the sequence of the Zξ is longer than λ, and we can find the contradiction. With the Weak Coding Lemma we prove that γ is regular. It is similar to the proof that its cofinality is larger than λ, using the Weak Coding Lemma instead of DCλ: Let β<γ and f : β → γ cofinal in γ. Let π be aΣ1(α) partial Lγ (Vλ+1) bounded surjection from Vλ+1 to Lγ(Vλ+1), as in Lemma 5.8. As Θ = γ, there exists ρ : Vλ+1 ։ β with ρ ∈ Lγ(Vλ+1). Let δ witness the Weak Coding Lemma for ρ in Lγ(Vλ+1), and consider

A = {(a,b): b ∈ dom(π) ∧ π(b)= f(ρ(a))}.

Then there exists B ⊆ A, B ∈ Lδ(Vλ+1) such that for cofinally ξ<β if there exists (a,b) ∈ A ρ(a)= ξ then there exists (a,b) ∈ B such that ρ(a)= ξ. 0 As for every ξ, Aξ 6= ∅, this means that for cofinally ξ, Bξ 6= ∅. Let B = {a : ∃b (a,b) ∈ B}. Then π′′B0 is cofinal in β, so (f ◦ π)′′B0 is cofinal in γ. ′′ ′′ 0 Let Z = {b : ∃a ∈ Vλ+1 (a,b) ∈ B}. As B ⊆ A, ρ Z = (f ◦ π) B , but that contradicts the boundedness of ρ. Since γ is regular, then the Coding Lemma can be re-proved relativized to Lγ(Vλ+1), so also Lemma 7.7 holds in Lγ(Vλ+1), therefore (3) is proved. So γ is weakly inaccessible because of regularity and (3). γ γ Let Sω = {α<γ : cof(α) = ω}. Let F be the club filter on Sω. Instead of proving that F is an ultrafilter on a stationary set, we define another filter F0 ⊆ F, that is an ultrafilter on a stationary set, and that coincides with F there. γ Let E = {k : Lγ(Vλ+1) ≺ Lγ(Vλ+1)}. For each k ∈ E, let Ck = {η ∈ Sω : k(η)= η}. As γ is regular, Ck is an ω-club. For each σ ⊆E such that |σ|≤ λ, let γ γ Cσ = ∩{Ck : k ∈ σ}. Let F0 be the filter on Sω generated by the sets Cσ ∩ Sω. + DC γ It is λ -complete by λ. All its elements are clubs on Sω, so F0 ⊆ F. Also, j(F0)= F0. As in 7.2 S cannot be partitioned in κ0 many F0-positive sets, since j ↾ Lγ(Vλ+1) ∈E and therefore Cj ∈F0. So there must exists a T ⊆ S stationary such that F0 ↾ T is an ultrafilter. As it is a maximal filter on T , it must be that F ↾ T = F0 ↾ T . It remains to show that F on T is γ-complete. Let β<γ and hAξ : ξ<βi a sequence of subsets of T of F-measure one. Let ρ : Vλ+1 ։ β, with ρ ∈ Lγ(Vλ+1). Let η to witness the Coding Lemma in Lγ(Vλ+1) for ρ. Ideally, we would like to use the Coding Lemma on

A = {(Aξ, Cσ): σ ⊆E, |σ|≤ λ, ξ<β, Cσ ⊆ Aξ},

44 to “choose” some Cσ inside every Aξ, so that we can intersect them in the following way: Suppose we can ﬁnd B ⊆ A “small” enough, so that, if

EB = {k ∈E : ∃ξ ∃σξ<β, σ ⊆E, |σ|≤ λ, (Aξ, Cσ) ∈ B k ∈ σ}, for any ξ<γ, sup{k(ξ): k ∈ EB} < γ. Then pick ξ0 > α, ξn+1 = sup{k(ξn): k ∈EB}. Then ξω = supn∈ω ξn ∈ T and for any k ∈EB k(ξω)= ξω. This means that for any Aξ, ξω ∈ Aξ, so we found an intersection. Therefore we have to code A as a subset of Vλ+1 × Vλ+1: Note that γ is good, therefore by 6.40 for any k1, k2 ∈E, k1 = k2 iﬀ k1 ↾ Vλ = k2 ↾ Vλ, so there is a bijection between E and E0 = {k ↾ Vλ : k ∈E}⊆ Vλ+1. Moreover let F be λ any bijection between Vλ+1 and Vλ+1: for a ∈ Vλ+1 we can deﬁne Ca = Cσ(a) with σ(a)= {k : k ↾ Vλ ∈ F (a)}. Let A = {(a,b) ∈ Vλ+1 × Vλ+1 : F (b) ⊆E0, Cb ⊆ Aρ(a)}.

Let B ⊆ A, B ∈ Lη(Vλ+1) given by the Coding Lemma. Let

EB = ∪{F (b): ∃a (a,b) ∈ B} ∈ Lη+1(Vλ+1), a set of codes for embeddings k such that Ck ⊆ Aξ for some ξ. Let a ∈ dom(π). Note that, as π is Σ1(α)-deﬁnable, if a ∈ dom(π) then k(a) ∈ dom(π) for any k embedding, so

Wa = {k(a): k ∈E, k ↾ Vλ ∈EB}⊆ dom(π), and Wa ∈ Lη+2(Vλ+1). So

′′ π Wa = {π(k(a)) : k ∈E, k ↾ Vλ ∈EB} = {k(π(a)) : k ∈E, k ↾ Vλ ∈EB}.

′′ Let ξ<γ, and a ∈ Vλ+1 such that π(a)= ξ. So π Wa = {k(ξ): k ∈E, k ↾ Vλ ∈ ′′ EB}. As Wa is bounded, then π Wa is bounded, so sup{k(ξ): k ∈E, k ↾ Vλ ∈ EB} <γ. Now we can ﬁnd a ﬁxed point for all the extension of embeddings in EB, and we are done. For the point (5) we just sketch the proof. The idea is to consider P , the theory of LΘ(Vλ+1), and relativize the previous proof for Lγ(Vλ+1)[P ]. Then

(Lγ (Vλ+1)[P ], P ∩ Lγ(Vλ+1)) ≺1 (LΘ(Vλ+1), P ) implies full elementarity. There are still open questions about the combinatorial nature of cardinals R between λ and Θ, following the similarities with ADL( ):

++ +++ Question 7.9. Suppose I0(λ), are λ , λ regular or singular in L(Vλ+1)?

Question 7.10. Suppose I0(λ). Is it true that in L(Vλ+1) every regular cardinal below Θ is measurable? Another point in favour of AD is the regularity properties that it entails, and this is partially true also for I0. We consider on Vλ+2 the topology with basic open sets O(α,a) = {x ⊆ Vλ : x∩Vα = a}, with α < λ and a ⊆ Vα. Then a set X ⊆ Vλ+1 is perfect iﬀ Πn∈ωκn can be continuously embedded into X. A set has the perfect set property iﬀ it has cardinality ≤ λ or it contains a perfect subset.

45 Theorem 7.11 (Cramer, 2015 [3]). Suppose there exists j : L(Vλ+1) ≺ L(Vλ+1), with crt(j) < λ. Then every set in L(Vλ+1) ∩ Vλ+2 has the perfect set property in L(Vλ+1). The proof uses heavily all the inverse limit tools developed in Section 6. It exploits the fact that there is a multiplicity of possible square roots: Consider J : Lα¯+1(Vλ¯+1) ≺ Lα+1(Vλ+1), inverse limit, and let X ∈ Lα+1(Vλ+1) ∩ Vλ+2 ¯ ¯ and X ∈ Lα¯+1(Vλ¯+1) such that J(X)= X. Then if |X| > λ we can ﬁnd for any a ∈ X¯ many (> λ) limit roots of J, all with diﬀerent images on a. Repeated uses of this kind of pigeonhole principle will be used to construct a perfect set inside X.

Question 7.12. Is there an equivalent in L(Vλ+1) for Baire property or Lebesgue measurability? Some key properties of subsets of R are proven thanks to the fact that under AD all such sets have a nice tree-structure, namely they are weakly homogeneously Suslin. Woodin introduced a deﬁnition that wants to develop a similarity with this concept: U(j)-representability. If x ∈ Vλ+1 and hλi : i ∈ ωi is coﬁnal in λ, then we can see x as the ω-

sequence hx ∩ Vλi : i ∈ ωi, just like any real can be seen as an ω-sequence of natural numbers. So any set X ⊆ Vλ+1 can be seen as a set of ω-sequences. In R, homogeneous Suslin-ness consists in labeling any finite sequence of natural numbers with an ultrafilter, so that if a sequence labeled with U2 extends another labeled with U1, then U2 projects to U1. If U2 projects to U1, then there is a related elementary embedding from Ult(V,U1), to Ult(V,U2), therefore for any infinite sequence one can define the direct limit of all the ultrapowers of the ultrafilters that label its finite initial segments. Then a set Z is homogeneously Suslin iff there is such labeling so that a branch is in Z iff such direct limit is well-founded.

. . . .

U5 Ult(V,U5)

U4 Ult(V,U4)

U3 Ult(V,U3) U3

U2 Ult(V,U2) U2

U1 Ult(V,U1) U1

46 Weakly homogeneous Suslin-ness adds another dimension, considering instead the couples of finite sequences. So the first step is to decide which ultrafilters are going to label our sequences:

Deﬁnition 7.13. Let U(j) be the set of U ∈ L(Vλ+1) such that in L(Vλ+1) the following hold: • U is a λ+-complete ultraﬁlter;

• for some γ < Θ, U ⊆P(Lγ(Vλ+1)); • for some A ∈ U and all suﬃciently large n ∈ ω, jn(U)= U and {a ∈ A : jn(a)= a} ∈ U.

Lκ(V +1) Let κ < Θ, κ ≤ Θ λ . Let hai : i<ωi be such that for all i ∈ ω there n exists n ∈ ω such that j (ai) = ai. Then U(j, κ, hai : i ∈ ωi) is the set of U ∈ U(j) such that there exists an n ∈ ω such that for all k : Lκ(Vλ+1) ≺ Lκ(Vλ+1) m l with the property that there are m,l ∈ ω, k = j ↾ Lκ(Vλ+1), if for all i ≤ n k(ai)= ai, then {a ∈ Lκ(Vλ+1): k(a)= a} ∈ U. Now we can introduce U(j) representability:

Deﬁnition 7.14 (Woodin, 2011 [41]). Let Z ∈ L(Vλ+1)∩Vλ+2. Then Z is U(j)- representable if there are κ, hai : i ∈ ωi, an increasing sequence hλi : i ∈ ωi, and a function π : {Vλi+1 × Vλi+1 ×{i} : i ∈ ω} → U(j, κ, hai : i ∈ ωi) such that the following hold: S i • for all i ∈ ω and (a,b,i) ∈ dom(π) there exists A ⊆ (L(Vλ+1)) such that A ∈ π(a,b,i);

• for all i ∈ ω and (a,b,i) ∈ dom(π), if m

dom(π) and π(a,b,i) projects to π(a∩Vλm ,b∩Vλm ,m) ∈ dom(π), i.e., for

every A ∈ π(a ∩ Vλm ,b ∩ Vλm ,m)

i {x ∈ (L(Vλ+1)) : x ↾ m ∈ A} ∈ π(a,b,i);

• for all x ⊆ Vλ, x ∈ Z iﬀ there exists y ⊆ Vλ such that the tower hπ(x ∩

Vλm ,y ∩ Vλm ,m : m ∈ ωi is well-founded.

The question is now whether there are sets in L(Vλ+1) ∩ Vλ+2 that are U(j)- representable, and particularly which ones are. The desired result would be, in similarity with AD, that they are all U(j)-representable. There has been some twists and turns on the still ongoing quest towards this result, and it is worth to see them step by step, with some advice for the reader that wants to know all the details. The main source for U(j)-representability is [41]. There the ﬁrst closure properties are introduced, reminiscent of the closure properties for weakly homogeneously Suslin sets: Theorem 7.15 (Woodin, Lemma 114, Lemma 115 and Theorem 134 in [41]). Suppose j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Then • the λ-union of U(j)-representable sets is U(j)-representable

47 • the projection of a U(j)-representable set is U(j)-representable

• if Z ∈ Lλ(Vλ+1) is U(j)-representable, then its complement is U(j)-representable.

Therefore all sets in Lλ(Vλ+1) ∩ Vλ+2 are U(j)-representable. This was the best result for U(j)-representability in [41]. Further research was blocked by the Tower Condition. Remember by 6.16 that a tower of ultrafilters hUn : n ∈ ωi is well-founded iff there exists an X such that X ∩ dom(Un) ∈ Un for all n ∈ ω. The Tower Condition holds for U(j) iff for any U(j)- representation, it is possible to find such an X for any well-founded tower uniformly: Definition 7.16 (Woodin, 2011 [41]). The Tower Condition holds for U(j) iff for any A ⊆ U(j), A ∈ L(Vλ+1), |A|≤ λ, there is a function F : A → L(Vλ+1) such that, for all hUi : i ∈ ωi ∈ L(Vλ+1) that satisfies:

i • for all i ∈ ω there exists Z ∈ Ui such that Z ⊆ L(Vλ+1) ;

• for all i ∈ ω Ui ∈ A and Ui+1 projects to Ui.

Then the tower is wellfounded iﬀ there exists a function f : ω → L(Vλ+1) such that for all i ∈ ω, f ↾ i ∈ F (Ui). Note that if π is a U(j) representation and A = ran(π), then all towers in A satisfy the conditions. Woodin immediately noticed that the Tower Condition was central for the study of U(j)-representable sets: Lemma 7.17 (Woodin, Lemma 127 in [41]). If the Tower Condition holds for U(j), then the complement of a U(j)-representable set is U(j)-representable, therefore all the sets in Lλ+ (Vλ+1) are U(j)-representable. For much of Section 7 of [41], the Tower Condition is a constant and rather cumbersome presence, as at the time of the publication it was still not proven. In [3] Cramer actually managed to prove the Tower Condition, therefore unlocking much of the content of Section 7 of [41]. The suggestion for the completist reader is therefore ﬁrst to read [3], and reading Section 7 of [41] after, so that the Tower Condition provides the most results. This approach would also make some proofs in [41] more direct and easier to read. For example:

Lemma 7.18 (Woodin, Lemma 121 of [41]). Suppose Z ∈ L(Vλ+1) is U(j)- representable and |Z| > λ. Then Z contains a R-perfect set5.

The proof of such Lemma starts by ﬁxing an elementary embedding j0 : λ+ V → M0 such that (M0) ⊆ M0, A0 = {j0(U): U ∈ ran(π)}. All of this ′′ is to prove the Tower Condition for j0 A0. Now that we know that the Tower Condition holds, we can ignore this trick and just consider M0 = V , j0 = id and A0 = ran(π). It is also possible to improve 7.17:

5With R-perfect set we intend that 2ω can be continuously embedded in Z, a weaker property than to be a perfect set. Therefore this Lemma is now incorporated in 7.11

48 Lemma 7.19 (Woodin, Corollary 148 of [41]). Suppose there is j : L(Vλ+1) ≺ + + L[A] L(Vλ+1) with crt(j) < λ. Let κ = λ and η = sup{(κ ) : A ⊆ λ}. Then every set in Lη(Vλ+1) is U(j)-representable. So it is still open:

Question 7.20. Are all sets in L(Vλ+1) ∩ Vλ+2 U(j)-representable?

8 Rank-into-rank and forcing

A recurrent image for describing the difference of the extension of the universe of sets via large cardinals or via forcing is that large cardinals extend the universe vertically, while forcing extends the universe horizontally. Such image has definitively its shortcomings, first because stronger large cardinals do not always mean larger large cardinals (we will see an example of this), second because it implies that large cardinals and forcing are orthogonal, while there are cases of interdependence. It is therefore natural, facing new axioms, to ask what is their relationship with forcing. The ideal results are two:

• A forcing always destructs the large cardinal axiom. This would be a strong indication that the propositions forced could be inconsistent with the large cardinal. • A forcing never destructs the large cardinal axiom. This means that the proposition forced is independent from the large cardinal, and we are actually in a situation of orthogonality. The test case has always been Cohen forcing: any large cardinal cannot be destructed by Cohen forcing, therefore the Continuum Hypotheses is independent from any large cardinal. The main lemma for proving the indestructibility of large cardinals via forcing is the following: Lemma 8.1 (Lifting Lemma). Let j : M ≺ N. If P ∈ M is a forcing notion, and G is P-generic, then for any H j(P)-generic such that j′′G ⊆ H there exists j∗ : M[G] ≺ N[H] such that j∗ ↾ M = j. Proof. The key point is that the forcing relation is definable. For any P-name ∗ τ, define j (τG) = (j(τ))H . Then M[G] ϕ(τG) iff there exists p ∈ G such that p ϕ(τ). So by elementarity j(p) ϕ(j(τ)). But j(p) ∈ H, i.e., N[H] ∗ ϕ((j(τ))H ), i.e., N[H] ϕ(j (τG)). The opposite direction holds considering ¬ϕ.

This gives immediately a result of lifting: Let j : Vλ ≺ Vλ, P ∈ Vcrt(j). Then j(P) = P, so G = H in the Lifting Lemma works, as j′′G = G. Therefore j ∗ extends to j : Vλ[G] ≺ Vλ[G]. But this is still not V [G] I3(λ), because that V [G] is ∃k : V [G]λ ≺ V [G]λ, where V [G]λ = (Vλ) . The diﬀerence is subtle, but fundamental: Vλ[G] is the set of τG such that τ ∈ Vλ, V [G]λ is the set of τG with rank of τG less than λ in V [G]. As the rank of the interpretation of a name is always less or equal than the rank of the name, clearly Vλ[G] ⊆ V [G]λ, but it is not immediate to see whether the opposite direction is true. This is solved by the nice names lemma:

49 Lemma 8.2 (Name Rank Lemma). If τ is a P-name, P ∈ Vγ+1 and τG ∈ V [G]β for any G P-generic, then there is a name σ ∈ Vγ+3·β such that σG = τG for any G P-generic.

In particular, if P ∈ Vcrt(j), any element of V [G]λ will be in some V [G]κn , and so it will have a name in V3·κn ⊆ Vλ, therefore Vλ[G]= V [G]λ. This is peculiar to rank-into-rank axioms: the indestructibility proof must pass two phases: ﬁrst we have to extend the embedding, then we have to make sure that the domain of the embedding is exactly what we want. By the discussion before Proposition 4.10, crt(j) can be as large as we want below λ, therefore we have the following:

Lemma 8.3. Suppose I3(λ), let P ∈ Vλ be a forcing notion and G be P-generic. Then V [G] I3(λ), So I3 is not destructed by “small” forcings.

The Lifting Lemma holds easily also for j : Vλ+1 ≺ Vλ+1 and P ∈ Vcrt(j), so ∗ there exists j : Vλ+1[G] ≺ Vλ+1[G]. We have to prove that Vλ+1[G]= V [G]λ+1.

If τG ∈ V [G]λ+1, then τG ⊆ V [G]λ, so τG = n∈ω(τG ∩ V [G]κn ). As Vλ[G] = V [G]λ, any τG ∩ V [G]κ has a name in Vλ. The name of a union of sets is the n S union of the names of the sets, therefore there exists a name for τG that is the ω-union of sets in Vλ, so it must be in Vλ+1. Again, crt(j) can be unbounded below λ, so: 1 Lemma 8.4. Suppose I1(λ) (or anything slightly weaker, like I2 or Σn), let P ∈ Vλ be a forcing notion and G be P-generic. Then V [G] I1(λ) (or the corresponding weaker axiom). So I1 is not destructed by “small” forcings.

Finally we consider I0 and P ∈ Vcrt(j). Again, the Lifting Lemma holds, ∗ ∗ so there exists j : L(Vλ+1)[G] ≺ L(Vλ+1)[G] with crt(j ) < λ. This time we cannot always prove that L(Vλ+1)[G] = L(V [G]λ+1), but it actually does not matter: as a constructible set will have a constructible name, we have L(V [G]λ+1) ⊆ L(Vλ+1)[G]. Also, L(V [G]λ+1) is a deﬁnable (with parameter λ) subclass of L(Vλ+1)[G], therefore ∗ j ↾ L(V [G]λ+1): L(V [G]λ+1) ≺ L(V [G]λ+1) and we have I0(λ) in the extension. Can we say that crt(j) can be unbounded below λ? For this, we have to prove that a j that witnesses I0 is at least ﬁnitely iterable. If j is not weakly proper, then it is not clear whether this is possible. But if j is weakly proper, then it is:

Lemma 8.5 (Woodin, Lemma 16 of [41]). Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ and weakly proper. Then j is finitely iterable. Proof. Let I be the class of fixed points of ξ (it is a class by the proof of <ω L(Vλ+1) 5.14). For each s ∈ [I] , let Zs = Hull (Vλ+1 ∪{Vλ+1} ∪ s). Then <ω hZs : s ∈ [I] i is a directed system. Let Z be its limit. Then Z ≺ L(Vλ+1). <ω For each s ∈ [I] , j ↾ Zs is definable from {j ↾ Vλ,s} ∈ L(Vλ+1), therefore ks = j(j ↾ Zs) is well-defined. As j(Zs)= Zs, we have that ks : Zs ≺ Zs. Then we can define k∗ : Z ≺ Z, its direct limit. Composing with the collapse of Z, we have k : L(Vλ+1) ≺ L(Vλ+1), and k ↾ Vλ = j(j ↾ Vλ). If k is not weakly proper, then we consider kU its weakly proper part, and we call it still k. This proves the existence of j2. As j2 is weakly proper, we can prove by induction that jn exists and is weakly proper for every n ∈ ω.

50 n n n In particular j ↾ Vλ = (j ↾ Vλ) , therefore crt(j )= κn, and so:

Lemma 8.6. Suppose I0(λ), let P ∈ Vλ be a forcing notion and G be P-generic. Then V [G] I0(λ), So I0 is not destructed by “small” forcings.

The reason why we do not know whether L(Vλ+1)[G]= L(V [G]λ+1) is quite deep:

Theorem 8.7 (Woodin, Theorem 175 in [41]). Suppose j : L(Vλ+1) ≺ L(Vλ+1) ω ω V [G] with crt(j) < λ. Let P ∈ Vλ, G be P-generic and λ 6= (λ ) . Then Vλ+1 ∈/ 6 L(V [G]λ+1), so L(Vλ+1)[G] 6= L(V [G]λ+1) .

Proof. Let hκi : i ∈ ωi be the critical sequence of j. As P ∈ Vλ, we can suppose ∗ that P ∈ Vκ0 , and then there exists j : L(V [G]λ+1) ≺ L(V [G]λ+1) that extends j. Suppose that Vλ+1 ∈ L(V [G]λ+1). Then by Theorem 7.11 in L(V [G]λ+1) it ω V [G] is possible to embed continuously (λ ) in Vλ+1. We do some cosmetic changes so that the proof is easier to read: we ﬁx ρn bijections between Vκn+1 and |Vκn+1|, rename |Vκn+1| as κn, and modify the V [G] continuous embedding so that its domain is (Πn∈ωκi) . So there exists π that V [G] V V [G] continuously embed (Πn∈ωκi) to (Πn∈ωκi) . We prove that (Πn∈ωκi) ⊆ V , and therefore (λω)V [G] = λω, contradiction.

Let π be such embedding. We deﬁne for every s ∈ n∈ω Πi

• Fix n = lh(s). Let α ∈ Ss. Then let Pα = {β<κn : ∃p ∈ p p β = v min{π˙ (a)(n): a ∈ (Πn∈ωκn) , π(a) ↾ n = s, π˙ (a)(n) > α}}. Then |Pα| < κ0, therefore Pα is bounded in κn. As for any s ∈ n∈ω Πi

Note that not only Ss ∈ V , but {Ss : s ∈ n∈ω Πi

• Applying (*) with n = 0, a anything and α = S∅(a(0)), there exists b0 such that S∅(a(0)) < π(b0)(0)

• Suppose that bn is already deﬁned. Applying (*) with a = bn and α =

Sπ(bn)↾n(a(n + 1)), there exists bn+1 such that π(bn+1) ↾ n = π(bn) ↾ n

and Sπ(bn)↾n(a(n + 1)) < π(bn+1)(n + 1)

As π is continuous, the bn’s are more and more closer. Let b be their limit. Then we deﬁne π′(a)= π(b).

6 The proof in [41] is more convoluted, as it proves that Vλ+1, if in L(V [G]λ+1) and U(j)- representable, is a perfect set, and this brings a contradiction. By 7.11 we immediately have that Vλ+1 is perfect.

51 ′ Note that, if α and α are two consecutive members of Ss, then |{π(b)(n): V [G] ′ ′ ′ ∃a ∈ (Πn∈ωκn) , π (a) = π(b), π(b) ↾ n = s} ∩ [α, α ]| = 1. Therefore π is well-deﬁned and injective. But then the inverse of π′ is deﬁnable in V , as for V [G] ′ any a ∈ (Πn∈ωκn) , π (a)(n)= |Sπ′(a)↾n ∩ a(n)|.

On the other hand, if the forcing is ω-closed in Vλ then we have equality:

Theorem 8.8. Suppose j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Let P ∈ Vλ be ω ω V [G] ω-closed in Vλ, i.e., (Vλ) = ((Vλ) ) , G be P-generic. Then L(Vλ+1)[G]= L(V [G]λ+1).

Proof. Fix a bijection in V between Vλ and λ, this extends naturally to a bijection from Vλ+1 and P(λ), so it suﬃces to deﬁne P(λ) inside L(V [G]λ+1):

V [G] P(λ)= {a ∈P (λ): ∀n ∈ ω a ∩ κn ∈ V }.

This is because if a ∩ κn = an ∈ V , then han : n ∈ ωi ∈ V and a = n∈ω an ∈ V . S Corollary 8.9. Suppose j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Let P ∈ Vλ be ω a forcing notion, G be P-generic. Then L(Vλ+1)[G] = L(V [G]λ+1) iﬀ (Vλ) = ω V [G] ((Vλ) ) . So I3, I2, I1 and I0 are not changed by forcing notions bounded in λ. Also, trivially, if a forcing notion is λ-closed, then V [G]λ+1 = Vλ+1, therefore I3, I2, I1 and I0 are witnessed by the exact same embedding, and are not destructed. In other words, they are not changed by forcings that are above λ. This case can give some very nice results: Theorem 8.10 (Shi, Trang, 2017 [36]). Suppose I0(λ) holds. Then all the following statements are true in a generic extension where I0(λ) still holds: • special λ+-Aronszajn trees exist; • λ+-Suslin tree exist; • there is a very good scale at λ+; • Stationary Reﬂections fails at λ+;

• ♦λ+ : This is particularly striking if compared with Theorem 7.3. It remains to see what happens when the forcing has size λ. The ﬁrst case is when the forcing is unbounded in λ (the second case will be when the forcing is exactly at λ). The typical case is when the forcing is a λ-iteration of small forcings. In such case, the reverse Easton iteration has been proven in the past to behave very well with large cardinals:

Deﬁnition 8.11. Let Pα be a forcing iteration of length α, where α is either a strong limit cardinal or is equal to ∞, the class of all ordinals. Then Pα is a reverse Easton iteration if nontrivial forcing is done only at inﬁnite cardinal stages, direct limits are taken at all inaccessible cardinal limit stages, and inverse limits are taken at all other limit stages; moreover, Pα is the direct limit of hPδ : δ < αi if α is regular or ∞, the inverse limit otherwise.

52 Such iteration was introduced by Easton to force different behaviours of the function κ 7→ 2κ, the power function. There are two known rules for the power function: if κ < η then 2κ ≤ 2η, and cof(2κ) >κ for κ regular. Easton proved that any function that satisfies this two rules can be the power function on the regular cardinals: Definition 8.12. Let E : Reg → Card be a class function. Then E is an Easton function iff: • α<β → E(α) ≤ E(β); • cof(E(α)) > α for all α ∈ Reg. Then Easton proved: Theorem 8.13 (Easton, 1970 [14]). Let E be an Easton function. Then there exists a generic extension V [G] of V such that V [G] ∀κ(κ ∈ Reg → 2κ = E(κ)). The forcing used was a reverse Easton iteration. The easiest application is γ + to force GCH: in that case for any γ < α Qγ forces that 2 = γ if γ is a cardinal, it is trivial otherwise, Pγ+1 = Pγ ∗ Q˙ γ and G will the generic filter of the reverse Easton iteration P∞. The problem is when one, for example, starts with a model of GCH and wants to increase everywhere on the regulars the ω ω1 power function: If we force with Qω 2 = ω2 and after that we force 2 = ω3 ω via Cohen forcing, then in the final model 2 will collapse to ω1 again. The idea is then that Qδ is defined only on δ closed under E, and Qδ is the Easton product7 of the forcings that make 2γ = E(γ) for all γ ∈ [δ, δ∗), where δ∗ is the smallest cardinal closed under E bigger than δ (this is still in [14]). κ So let E be an Easton function, and Pλ that forces 2 = E(κ) for any regular κ, with the stages indicated with Qδ for δ < λ. We want to know for which E’s the forcing Pλ does not destroy I3, ... , I0. ∗ α If for some δ < λ |Qδ|≥ λ, then there is a α<δ such that in V [G]2 ≥ λ, therefore λ is not strong limit anymore in V [G] and the embedding is destroyed. Therefore we must have |Qδ| < λ for any δ < λ. α j(α) If for some δ < λ we have that j(Pδ) 6= Pj(δ), then possibly j(2 ) 6=2 for some α<δ, therefore we should ask that j(Pδ) = Pj(δ). If E is definable then this is trivial. Also note that Qδ is δ-directed closed. This is enough to lift I3, ... , I0 in the generic extension:

Theorem 8.14 ([9]). Suppose j witnesses I3(λ) (or I1(λ) ... ). Let Pλ be a forcing iteration of length λ that is reverse Easton, directed closed (i.e., for all δ < λ, Qδ is δ-directed closed), λ-bounded (i.e., for all δ < λ, |Qδ| < λ) and j-coherent (i.e., for all δ < λ, j(Pδ) = Pj(δ)). Then j lifts to any generic extension via Pλ. Sketch of proof. The proof is in two stages: lifting and right domain. First notice that if γ<κn, then |Pγ| <κn, by elementarity, λ-boundedness and n applications of j. Fixing an n<ω, consider the image of Qκn under j: it

7It is the product with support the Easton ideal, i.e., the sets that are bounded below every inaccessible cardinal.

53 is a subset of Qκn+1 of cardinality less than κn+1, therefore by directed closeness

there is an element Qκn+1 that extends all the j(p) with p ∈ Qκn . Doing the same for all n and also for Pκn,κn+1 , one gets a condition of Pλ that, if in a generic G, implies that j′′G ⊆ G. Therefore we can apply the Lifting Lemma. As for the right domain, since the forcing is directed closed every element in

the generic extension is actually in a generic extension for some Pκn , therefore the previous proofs still hold. Theorem 8.14 is just the final step of a series of theorems that goes in that direction: in [22] Hamkins has proved such theorem for I1 and reverse Easton δ directed closed j-coherent iterations such that |Qδ| < 2 ; in [2] Corazza has proved it for I3 and reverse Easton directed closed j-coherent iterations such that |Qδ| < i(δ), where i(δ) is the smallest inaccessible larger than δ; in [17] Friedman has proved it for I2 (called there ω-superstrong) and GCH. The hypotheses in Theorem 8.14 are very reasonable, and most of the reverse Easton forcing usually employed satisfy them. So, for example: Corollary 8.15. Suppose I3 (or I1, ... ). Then in a generic extension can hold: • I3 + GCH; • I3 + 2κ = κ+++ for any κ regular; • I3 + V = HOD; • I3 + ♦ everywhere; • I3 + every κ supercompact is Laver indestructible (see [35]); • ... Finally the last case is when the forcing is changing λ+. This is the hardest case, and at the moment the indestructibility results are very scarce. For example, Gitik short extenders Prikry forcing construction in [18] can be carried +m on under I3 (it needs o(κn) ≥ (κn) for any n,m ∈ ω): such construction λ +δ+1 push 2 to λ for any desired δ < ℵ1, but does not add bounded subsets of λ, therefore V [G]λ = Vλ and I3 is not destructed. Yet, this construction for I1 does not work, as Vλ+1 6= V [G]λ+1. In [5] Cummings and Foreman, using something more than I2, manage to prove that I2(λ) is consistent with 2λ = λ++. The most successful strategy has been instead to start with I0(λ), and using some sophisticated tool to prove that, adding a Prikry sequence to the critical point κ0, in the extension I1(κ0) holds. Definition 8.16. Let κ be a measurable cardinal, and U a normal measure on κ. Then P, the Prikry forcing on κ via U, is the set of (s, A) such that s ∈ [κ]<ω, A ∈ U and min A> max s. We say that (s, A) < (t,B) if s ⊒ t, A ⊆ B and for any n ∈ lh(s) \ lh(t), s(n) ∈ B. If (s, A) and (t,B) are in the generic set G, then, s and t must be compatible. Therefore by density {s : ∃A (s, A) ∈ G} is an ω-sequence cofinal in κ. So (cof(κ)= ω)V [G], but it is a very delicate forcing, as it does not add any bounded subset of κ. S There is a convenient condition for an ω-sequence cofinal in κ to be generic:

54 Theorem 8.17 (Mathias Condition, or geometric condition). Let κ be a measurable cardinal, P the Prikry forcing on κ via the normal ultrafilter U and let hαn : n ∈ ωi be a cofinal sequence in κ. Then hαn : n ∈ ωi is generic for P iff for any A ∈ U the set hαn : n ∈ ωi\ A is finite. In Theorem 8.5 we noticed that if j witnesses I0(λ) is weakly proper, then it is finitely iterable, and its iterates are still weakly proper embeddings from L(Vλ+1) to itself. Woodin proved that iterability can go further:

Theorem 8.18 (Woodin, Lemma 21 in [41]). Suppose that j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ is weakly proper. Then j is iterable.

From the proof of Lemma 6.4 j0,ω(κ0) = λ, therefore Mω is different than L(Vλ+1): in it, by elementarity, λ is regular. Let U be the ultrafilter from j. Then j0,ω(U) is a normal ultrafilter on λ, therefore λ is measurable in Mω. If P is the Prikry forcing on κ0 via U, then j0,ω(P) is the Prikry forcing on λ via j0,ω(U).

Remark 8.19. If hκn : n ∈ ωi is the critical sequence of j, then hκn : n ∈ ωi is j0,ω(P)-generic in Mω. Proof. Note that Mathias’ characterization needs the Axiom of Choice, but by Mω elementarity (Vλ) AC, so we can use it. If A ∈ j0,n(U), then by the n deﬁnition of U κn ∈ jn,ω(A), as A ∈ j0,n(U) iﬀ κn ∈ j (A) and

n κn = jn+1,ω(κn) ∈ jn+1,ω(j (A)) = jn,ω(A).

So if A ∈ j0,ω(U), there exist n ∈ ω and A¯ ∈ L(Vλ+1) such that A = jn,ω(A¯). By elementarity A¯ ∈ j0,n(U) and jn,n+i(A¯) ∈ j0,n+i(U). So for any i ∈ ω, κn+i ∈ jn+i,ω(jn,n+i(A¯)) = A.

So it make sense to consider M[hκi : i ∈ ωi].

Theorem 8.20 (Generic Absoluteness, Woodin, [41]). Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) be a weakly proper elementary embedding. Let hκn : n ∈ ωi be its critical sequence, and Mω its ω-iterated model. Let δ be such that all the sets in Lδ(Vλ+1) ∩ Vλ+2 are U(j)-representable. Then there exists an elementary embedding

π : Lδ(Mω[hκn : n ∈ ωi] ∩ Vλ+1) ≺ Lδ(Vλ+1)

such that π ↾ Vλ+1 is the identity. Two examples to understand better the theorem:

• As hκn : n ∈ ωi ∈ Vλ+1, clearly

Mω [hκn:n∈ωi] Mω[hκn : n ∈ ωi] ∩ Vλ+1 = (Vλ+1) ⊆ Vλ+1.

Since π ↾ Vλ+1 = id, the Theorem says that Mω[hκn : n ∈ ωi] ∩ Vλ+1 ≺ Vλ+1.

• If A is deﬁnable on Mω[hκn : n ∈ ωi], then π(A) will be the set deﬁned in the same way on Vλ+1, and π(A) ∩ Mω[hκn : n ∈ ωi]= A.

55 The proof exploits the fact that in the deﬁnition of U(j)-representability we had jn(U) = U for the ultraﬁlters used in a representation, therefore given a set in Lδ(Vλ+1), its representation is also in Mω, in a certain sense is the set encrypted. When we add the critical sequence to λ, then we can rebuild something similar to the set (if not the set itself) using such representation. The Theorem uses the theory of Lδ(Vλ+1), as a set, to build an embedding. This essential tool has an immediate corollary:

Corollary 8.21. Suppose that there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j)= κ0 < λ. Let P be the Prikry forcing on κ0. Then there exists G P-generic such that there exists k : V [G]κ0+1 ≺ V [G]κ0+1.

Proof. We have seen that j ↾ Vλ+1 is deﬁnable from j ↾ Vλ, therefore it is in L1(Vλ+1). Then

−1 Mω [hκn:n∈ωi] Mω [hκn:n∈ωi] π (j)= j ∩ Mω[hκn : n ∈ ωi] : (Vλ+1) ≺ (Vλ+1) .

So in Mω there is a condition in j0,ω(P) that forces the existence of a I1(λ) embedding. By elementarity, in V there is a condition on P that forces the existence of a I1(κ0) embedding. This corollary, coupled with Theorem 8.14, can expand the possibilities of κ0 + consistency results: for example one can force 2 >κ0 with the reverse Easton forcing, and then force with Prikry to have I1(κ0). Prikry will not change κ0 + λ + 2 >κ0 , and therefore we have a model for I1(λ)+2 > λ .

Theorem 8.22. Suppose there exists j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ. Then for every Easton function E such that E ↾ λ is definable over Vλ, there is a γ generic extension V [G] of V that satisfies ∃γ < λ ∃k : Vγ+1 ≺ Vγ+1 ∧2 = E(γ). In particular this proves that it is consistent with I1(λ) that there are no strongly compact or supercompact cardinals below λ, since λ is singular and Solovay’s Theorem says that above a strongly compact or a supercompact cardinal we have 2λ = λ+ on singular cardinals. Therefore this is one of the cases where “stronger” does not mean “larger”. The first way to improve such results is to note that Theorem 8.20 can express more than just I1, therefore the same proof works for stronger hypotheses (for example ∃j : Lλ(Vλ+1) ≺ Lλ(Vλ+1)), with the only bound given by which sets are U(j)-representable, making Question 7.20 even more relevant. Another way is to notice that in the proof of Theorem 8.20 it is not necessary to limit ourselves to Prikry forcing and to hκn : n ∈ ωi: the important thing is that there is a G ∈ V j0,ω(P)-generic and there is an ω-sequence cofinal in λ in Mω[G]. If G/∈ V then there are easy counterexamples: if CH holds in V , then it holds also in Mω, and a forcing that adds a Cohen real to Mω would make Mω [G] Vλ+1 not elementary in Vλ+1 (not even contained, in fact). There are numerous variations on Prikry forcing that add ω-sequences to a large cardinal, and as λ is a very large cardinal in Mω (see 4.10) they are usually applicable in this context. For example, it is clear in Theorem 8.22 that the reverse Easton forcing forces 2η = E(η) for all the regular cardinals η<γ, not only for γ, therefore with the construction of Theorem 8.22, based on classical Prikry forcing, γ is not the first cardinal on which GCH fails. Being the first one is a stronger property, that is forced usually with Gitik-Magidor long extender

56 Prikry forcing (see [?]). If P is therefore the Gitik-Magidor long extender Prikry forcing on κ0 and it could be possible to find a G ∈ V j0,ω(P)-generic in Mω, then by Generic Absoluteness we would have a generic extension in which I1(κ0) κ0 + GCH + 2 >κ0 and below κ0. To check if this and other forcing notions work with Generic Absoluteness there is the need of a criterion that, if satisfied, implies that there is a G ∈ V , j0,ω(P)-generic in Mω. Definition 8.23. Let λ be an infinite cardinal. A partially ordered set P is λ-good if it adds no bounded subsets of λ and for every P-generic G and every A ∈ V [G], A ⊂ Ord, |A| < λ, there is a non-⊂-decreasing ω-sequence hAi : i< ωi⊆ V such that A = i∈ω Ai.

Proposition 8.24 (Shi,S Proposition 3.20 in [35]). Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j)= κ0 < λ and let P ∈ Vλ be a κ0-good forcing. Then if (Mω, j0,ω) is the ω-iterate of j, there is a G ∈ V j0,ω(P)-generic on Mω. The original definition of λ-goodness involved families of dense sets, and this is a more convenient way to prove that various Prikry forcings are λ-good. The following is a sufficient condition: Definition 8.25 ([10]). Let λ be an infinite cardinal. Let P be a forcing notion equipped with a length measure of the conditions l : P → ω such that l(1P)=0 and for p, q ∈ P, if p ≤ q then l(p) ≥ l(q) (for example if P is the Prikry forcing, then l(s, A) = lh(s)). We say that P is λ-geometric if it does not add bounded subsets of λ and for any α < λ, any hDβ : β < αi collection of open dense sets and any p ∈ P there exists a condition q ≤ p such that whenever a filter contains q and meets all the dense open sets En = {p : l(p) >n}, it also meets all the Dβ’s. Remark 8.26. Let λ be an infinite cardinal. Let P be a λ-geometric forcing notion. Then P is λ-good. Proof. Let G be P-generic. Let A ⊆ Ord, |A| = µ < λ, A ∈ V [G]. Let f : µ → Ord such that ran(f) = A, f ∈ V [G]. For any α<µ, let Dα = {p ∈ P : ∃β p f˙(α) = β}, the set of the conditions that decide the α-th element of A. By λ-geometricity, the set of q’s such that every filter that contains q and intersects En intersects also all Dα’s, is dense in P, therefore there exists q0 ∈ G with such property. Define by induction qn+1 that extends qn such that qn+1 ∈ G ∩ En. Let Bn = {α<µ : qn ∈ Dα} ∈ V be the set of α’s such that qn

decides f(α). Then since F = n∈ω Fqn , by λ-geometricity, intersects all Dα’s (Fq is the ﬁlter generated by qn), for any α<µ there exists n ∈ ω such that n S qn ∈ Dα, therefore µ = n∈ω Bn. Deﬁne S An = {β : ∃α ∈ Bn qn f˙(α)= β} ∈ V.

If β ∈ A, then there exists α<µ such that f(α) = µ, and there exists n ∈ ω such that α ∈ Bn, i.e., qn decides f(α). But since qn ∈ G, it must be that ˙ qn f(α)= β, so β ∈ An, therefore A = n∈ω An. The key point is that many variationsS of Prikry forcing satisfy a variation of the Prikry condition: Let p ≤∗ q be p ≤ q and l(p)= l(q). Then such variation says that for any dense set D and every p ∈ P there is a q ≤∗ p, such that

57 for any r ≤ q sith l(r) = l(q)+ n, r ∈ D. In other words every condition can be “changed” so that any extension above a certain length is in the dense set. This condition appears frequently in literature, for example in [30] for Prikry forcing, or [31] for Gitik-Magidor long extender Prikry forcing. In [11] there are other examples, and in both [36] and [11] it is exploited to prove the λ- goodness of Gitik-Magidor long extender Prikry forcing, in different ways. If ≤∗ is sufficiently closed, it is easy to see that with repeated uses of this variation of Prikry condition one can find a q that witnesses λ-geometricity, and therefore λ-goodness. It is possible also to prove it directly:

Remark 8.27 ([10]). Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j)= κ0 < λ and let P ∈ Vλ be a κ0-geometric forcing. Then if (Mω, j0,ω) is the ω-iterate of j, there is a G ∈ V j0,ω(P)-generic on Mω.

Proof. Note that in V there are only λ open dense sets of j0,ω(P): as P ∈ Vλ there exists n ∈ ω such that P ∈ Vκn , so j0,ω(P) ∈ Mω ∩ Vj0,ω (κn). But Mω ∩ Vj0,ω (λ) is the range of all jn,ω ↾ Vλ, therefore |j0,ω(P)| < |Mω ∩ Vj0,ω (λ)| = |Vλ| = λ. Let hDα : α < λi be an enumeration of the dense sets of j0,ω(P) in V . For every n ∈ ω, fix qn that witnesses λ-geometricity for hDα : α<κni. Then for ′ ′ every m there exists a qn,m < qn such that qn,m ∈ Em. Let H be the filter ′ n,m∈ω Fqn,m , with Fq the filter generated by q. Then H ∈ V is generic. S Therefore, thanks to the work that in literature has already been done with the structure of Prikry-like forcing, for many variations of Prikry forcing it is possible to prove Generic Absoluteness. Interestingly, for many applications such variations need a forcing preparation that is actually a reverse Easton iteration, therefore Theorem 8.22 is necessary. In [10] are collected some results: Theorem 8.28. Suppose I0(λ). Then the following are consistent: • I1(γ) + 2γ >γ+ + ∀κ<γ 2κ = κ+ (see [19], preparation forcing to force GCH, then Gitik-Magidor long extender Prikry forcing); • I1(γ) + Tree property at γ+ (see [33], preparation forcings that make supercompact cardinals Laver indestructible and add many subsets to all regular cardinals, then diagonal supercompact Prikry forcing); • I1(γ) + Tree property at γ++ (see [12], preparation forcing is reverse Easton iteration of Sacks forcings, then Prikry forcing); • ...

9 Dissimilarities with ADL(R)

R While there are many similarities between I0 and ADL( ), not everything re- R ally matches. The ﬁrst thing that stands out is that while L(R) ADL( ), L(Vλ+1) 2 I0(λ). But the big problem is Theorem 8.6: the fact that many structural properties below λ are independent from I0 in fact implies that I0 cannot possibly prove them, and these properties can be the similarities we are R interested in. In the ADL( ) case this situation is avoided as there are no forc- + ings below ω. For example, we have proven that λ is measurable in L(Vλ+1)

58 under I0, just like ω1 was measurable in L(R) under AD. But in the L(R) case we have that the club filter itself is an ultrafilter, while Theorem 7.2 stops short and proves only that the club filter on a stationary set is measurable. So the question is: how close can we get to the original case? λ+ λ+ λ+ As λ>ω, for any α < λ regular Sα is stationary, and for α 6= βSα ∩Sβ = ∅, therefore it is not possible to have the club filter to be an ultrafilter. The best outcome would be then to have the club filter to be an ultrafilter on every λ+ Sα ; this is called Ultrafilter Axiom at λ. But this is consistently false because of Theorem 8.6:

Proposition 9.1. Let j : L(Vλ+1) ≺ L(Vλ+1) with crt(j) < λ, let ω<γ<λ regular. Then in a generic extension the following is true: I0(λ) still holds and λ+ in L(Vλ+1) there are S1 and S2 disjoint stationary sets such that S1 ∪S2 = Sγ . Proof. First of all, notice that λ+ is coded by subsets of λ, so it can be considered a subset of Vλ+1. As Vλ+1 is closed by λ-sequences, the coﬁnality of a cardinal + less than λ is the same in V and in L(Vλ+1) (because is witnessed by an element + + λ V λ L(Vλ+1) of Vλ+1), therefore (Sη ) = (Sη ) for any η regular. Let P ∈ L(Vλ+1) be Coll(γ+,γ). Then by 8.6 I0(λ) holds in V [G]. As γ>ω, P is ω-closed. Let λ+ V λ+ V G be P generic. Consider (Sγ+ ) and (Sγ ) , both in L(Vλ+1) and therefore in + λ+ V λ+ V L(Vλ+1)[G]. Since P does not add unbounded sets in λ , (Sγ+ ) and (Sγ ) are + + λ V [G] λ L(Vλ+1)[G] still stationary in L(Vλ+1)[G], are both subsets of (Sγ ) = (Sγ ) and of course are disjoint. Since P is ω-closed, by 8.8 L(Vλ+1)[G]= L(V [G]λ+1), λ+ V λ+ V so (Sγ+ ) , (Sγ ) ∈ L(V [G]λ+1) are S1 and S2.

Note that if γ = ω then the proof breaks down: then L(Vλ+1)[G] 6= L(V [G]λ+1) and it is not clear why S1 and S2 would be in L(V [G]λ+1). The following problems are still open: Question 9.2. Is the Ultrafilter Axiom consistent with I0? Question 9.3. Is it consistent with I0 to have the club filter an ultrafilter on cofinality ω? Could it be a consequence of I0? Towards a solution of the second problem, there are the following results: Theorem 9.4 (Woodin, Theorem 176 in [41]). Suppose I0(λ). Then there is λ+ no partition of Sω into two stationary sets such that all the sets definable from them are U(j)-representable8.

♯ Theorem 9.5 (Cramer, Corollary 4.6 in [3]). Suppose there exists j : Lω(Vλ+1, Vλ+1) ≺ ♯ λ+ Lω(Vλ+1, Vλ+1). Then there is no partition of Sω into two stationary (in V ) sets such that S1, S2 ∈ L(Vλ+1).

Note that if S is stationary in V , then it is also stationary in L(Vλ+1), but not necessarily viceversa, therefore it is still open whether there can be two

8This rather clumsy statement comes from the fact that the proof uses the fact that some ZS defined from the stationary set S is U(j)-representable. If all the sets in some Lδ(Vλ+1), with δ limit, are U(j)-representable, this implies that we cannot find two disjoint stationary sets in Lδ(Vλ+1). If all the sets are U(j)-representable, this means that the club filter is an ultrafilter on cofinality ω.

59 λ+ disjoint subsets of Sω that are in L(Vλ+1), stationary there, but not stationary in V . Hypotheses like in Theorem 9.5 are investigated further in Section 10. It is proven using inverse limits. Towards a solution for the first problem, we have a weakening of the Ultra- λ+ filter Axiom. Theorem 7.2 says that we can split Sγ in hSα : α < ηγi disjoint + stationary sets for any γ < λ regular. The Ultrafilter Axiom says that ηγ =1 + + for any γ < λ regular. The Weak Ultrafilter Axiom states that ηγ ≤ γ for any γ < λ+ regular. Theorem 9.6 (Woodin, Theorem 16 in [42]). Suppose I0(λ). Then in a generic extension I0(λ) + the Weak Ultrafilter Axiom hold. Another problematic point is the Wadge hierarchy. Wadge’s Lemma states that under AD for any two subsets A, B of R either there is a continuous function f with A = f −1[B] or there is one with B = f −1[R \ A]. Therefore any two subsets of R are reducible one to another. Under I0 this is consistently false: Theorem 9.7 (Woodin, Theorem 174 in [41]). Suppose I0(λ). Suppose c is a generic Cohen real of V . Then there exist X, Y ∈ V [c]λ+2 such that X/∈ Lω(Y, V [c]λ+1) and Y/∈ Lω(X, V [c]λ+1).

As by 8.6 in L(V [c]λ+1) I0(λ) holds, this proves the consistency of I0 with a negation of Wadge Lemma. Another aspect one can consider is partition combinatorics. In L(R), under AD α9 , for all α<ω1, ω1 → (ω1)ω . A direct generalization always fails:

Remark 9.8 (ZF+DC). Suppose there exists an ω1-sequence of distinct reals. ω1 Then there is no κ such that κ → (κ)2 . + + ω So the best attempt could only be λ → (λ )λ . It is still an open question ω even whether κ → (κ)2 is consistent at all. In [41] Lemma 212 proves that a + + ω universal strong version of λ → (λ )λ is consistently false with I0. But there is a partial result: Proposition 9.9 (Lemma 180 in [41]). Suppose I0(λ). Then for all α < λ, + + α + L(Vλ+1) λ → (λ )(λ,<λ), i.e., for all π : {σ ⊆ λ : ot(σ) = α} → λ there exists H ⊆ λ+ such that |{π(σ): σ ⊆ H, ot(σ)= α}| < λ. Finally, we consider the measure on Turing degrees. A standard result of AD is that the ﬁlter U = {A : A contains a cone} on sets of Turing degrees is an ultraﬁlter (Martin), i.e., every set of degrees either contains or is disjoint from a cone. This result is usually called Turing Determinacy, and it is at the base of many results, like Silver Dichotomy for every set in L(R) under AD. In the context of I0 Turing degrees can be substituted with Zermelo degrees: given a ⊆ λ, let M(a) be the smallest model of Zermelo set theory that contains a. We say that a is equivalent with b if M(a) = M(b), and a is reducible to b if M(a) ⊆ M(b). Then a Zermelo degree is a class of such equivalence relation. Unfortunately Zermelo Degree Determinacy is consistently false under I0:

9 α Recall that κ → (κ)β is that if π : {σ ⊆ κ : ot(σ) = α} → β then there is a set H ⊆ κ of cardinality κ homogeneous for π, i.e., |{π(σ): σ ⊆ H, ot(σ) = α}| = 1.

60 Theorem 9.10 (Shi, 2015, [35]). Suppose I0(λ) and that all the λ-supercompact cardinals under λ are Laver indestructible. Then there is a set of Zermelo degrees in L(Vλ+1) that neither contains nor is disjoint from a cone. This result is significantly more problematic respect the previous results: in 9.1 and Lemma 212 in [41] we had that small changes would destroy that R ADL( )-like properties, but in this case, modulo a reasonable indestructibility preparation, it seems that it is intrinsic from I0, and for deep reasons. But there is one ambiguity still to settle: for now, there is no proof that there are λ singular cardinals of cofinality ω such that in L(Vλ+1) Zermelo Degree Determinacy holds, so maybe its failure is not something peculiar to I0, but a general property of such λ’s, just like it is not possible to have a club filter an ultrafilter everywhere or the full partition property. Question 9.11. Is there a λ such that Zermelo Degree Determinacy holds in L(Vλ+1)?

10 Icarus sets

Is it possible to go beyond I0? The question is very interesting: if I0 holds, and since Reinhardt is inconsistent in ZFC, there should be something in between, or maybe even a hierarchy between the two, where at a certain point the inconsistency comes up, so it is worth to investigate exactly where it is. But the question is not only speculative: results like 8.22 are reﬂection results, from I0 to I1, so to ﬁnd combinatorial properties that are independent from I0 one should have a starting point in a higher ground. It seems that there is not much space above I0: already an elementary embedding from L(Vλ+2) to itself is inconsistent. We consider then models that are strictly between the two, i.e., L(Vλ+1) ⊂ M ⊂ L(Vλ+2). They should be models of ZF, so something like L(N), with Vλ+1 ⊂ N ⊂ Vλ+2, i.e., models L(R) built from collections of subsets of Vλ+1. In similarity with AD , the most important of such models are of the form L(X, Vλ+1), with X ⊂ Vλ+1:

Deﬁnition 10.1. A set X ⊆ Vλ+1 is an Icarus set iﬀ there is an elementary embedding j : L(X, Vλ+1) ≺ L(X, Vλ+1) with crt(j) < λ. The name comes from the fact that we are trying to go as high as possible before reaching an inconsistency. Trivially, if I0(λ) holds, Vλ+1 (and in fact every set in L(Vλ+1) ∩ Vλ+2) is Icarus. On the other hand, a well-ordering of Vλ+1 cannot be Icarus, because of Kunen’s theorem. In this kind of models many structural properties of L(Vλ+1) still hold:

Deﬁnition 10.2. Let X ⊆ Vλ+1. Then in L(X, Vλ+1) DCλ holds and Θ = ΘL(X,Vλ+1) is regular. Vλ+1 The structure of such embeddings is also very similar to the I0 ones:

Theorem 10.3 (Woodin, Lemma 5 in [41]). Let X ⊆ Vλ+1 be an Icarus set. For every j : L(X, Vλ+1) ≺ L(X, Vλ+1) there exist a L(X, Vλ+1)-ultraﬁlter U in Vλ+1 and jU , kU : L(X, Vλ+1) ≺ L(X, Vλ+1) such that jU is the elementary embedding from U, j = jU ◦ kU and j ↾ LΘ(X, Vλ+1)= jU ↾ LΘ(X, Vλ+1).

61 Proof. The proof is the same as in 5.12. The only delicate point is to prove that the collapse of the ultrapower is indeed L(X, Vλ+1), as by elementarity Z ≡ L(j(X), Vλ+1). But j(X) ∈ L(X, Vλ+1), and X = {a ∈ Vλ+1 : j(a) ∈ j(X)}, therefore L(j(X), Vλ+1)= L(X, Vλ+1). There is one difference: in the I0 casewe had that a weakly proper embedding was nicely amenable, and for every a ∈ L(Vλ+1) if we defined a0 = a and an+1 = j(an), then han : n ∈ ωi ∈ L(Vλ+1). This is because everything is definable from Vλ+1 and fixed points, and Vλ+1 is morally closed under ω- sequences. But in the Icarus case it is not clear whether it is necessary that hXn : n ∈ ωi ∈ L(X, Vλ+1).

Deﬁnition 10.4 (Woodin, [41]). Let X ⊆ Vλ+1 be an Icarus set, and let j : L(X, Vλ+1) ≺ L(X, Vλ+1) with crt(j) < λ. Then:

• j is weakly proper iﬀ j = jU ;

• j is proper iﬀ it is weakly proper and hXn : n ∈ ωi ∈ L(X, Vλ+1), with X0 = X and Xn+1 = j(Xn).

In the particular case of X = Vλ+1, then, weakly proper and proper are the same concept. One could ask whether it is a good definition, as there can be a different Z ⊆ Vλ+1 such that L(X, Vλ+1) = L(Z, Vλ+1). For this, we have to study the fixed points of j under Θ:

Proposition 10.5. Let X ⊆ Vλ+1 be an Icarus set and let j : L(X, Vλ+1) ≺ L(X, Vλ+1) with crt(j) < λ. Then hXn : n ∈ ωi ∈ L(X, Vλ+1) iff the fixed points of j are cofinal in Θ.

Proof. Suppose that hXn : n ∈ ωi = Y ∈ L(X, Vλ+1). We can suppose that j is weakly proper. Then the proof is similar to the proof of Corollary 5.15: In this case every ordinal β < Θ is definable from some elements of I ∪ Vλ+1 ∪ {X}. As Y ∈ L(X, Vλ+1), L(Y, Vλ+1) = L(X, Vλ+1), and j(Lω(Y, Vλ+1)) = Lω(Y, Vλ+1). Let α0 < Θ, and suppose that α0 is definable from {i0,...,in} ∪ Lω(Y, Vλ+1). Let α be the supremum of the ordinals that are definable from {i0,...,in}∪Lω(Y, Vλ+1). Since Θ is regular, α< Θ, and since j(Lω(Y, Vλ+1)) = Lω(Y, Vλ+1), j(α)= α > α0. Now suppose that the fixed points of j are cofinal in Θ. Let j(X) ∈ Lγ(X, Vλ+1) and α>γ such that j(α)= α. Then j(Lα(X, Vλ+1)) = Lα(X, Vλ+1) 2 and Lα(X, Vλ+1) is closed by ω-sequences, therefore hj(X), j (X),... i ∈ Lα(Vλ+1).

Proposition 10.6 (Woodin, Lemma 6 of [41]). Let X ⊆ Vλ+1 and j : L(X, Vλ+1) ≺ L(X, Vλ+1) with crt(j) < λ. Then if hXn : n ∈ ωi ∈ L(X, Vλ+1), for all Y ∈ L(X, Vλ+1) ∩ Vλ+2, hYn : n ∈ ωi ∈ L(X, Vλ+1), with Y0 = Y and Yn+1 = j(Yn).

Proof. Again, let Y ∈ Lγ(X, Vλ+1) and α>γ such that j(α) = α. Then j(Lα(X, Vλ+1)) = Lα(X, Vλ+1) and Lα(X, Vλ+1) is closed by ω-sequences, therefore hYn : n ∈ ωi ∈ Lα(Vλ+1). Not surprisingly, given that being proper implies some degree of amenability, properness is bound to iterability:

62 Theorem 10.7 (Woodin, Lemma 16 and Lemma 21 in [41]). Let X ⊆ Vλ+1 and j : L(X, Vλ+1) ≺ L(X, Vλ+1) with crt(j) < λ. If j is weakly proper, then it is ﬁnitely iterable. If j is proper, then it is iterable. The Coding Lemma holds for any embedding, but the equivalent of 7.8 is proven in Lemma 207 of [41] only for the case where the embedding is proper. It is not known whether properness is necessary. In fact it is not even known whether it is a concept that is distinct from weak properness:

Question 10.8. Is there an Icarus set X ⊆ Vλ+1 and an elementary embedding j : L(X, Vλ+1) ≺ L(X, Vλ+1) that is weakly proper but not proper? We have seen that if we limit ourselves to I0 the answer is negative. Towards a solution for this problem in [7] and [8] it is proven that there are N1 and N2, Vλ+1 ⊆ N1,N2 ⊆ Vλ+2, such that there are embeddings weakly proper but 10 non proper from L(N1) to itself, and such that all weakly proper embeddings from L(N2) to itself are not proper. But it is also proven that there cannot be X ⊆ Vλ+1 such that L(N1)= L(X, Vλ+1) or L(N2)= L(X, Vλ+1). The existence of an Icarus set outside L(Vλ+1) ∩ Vλ+2 is not automatically an axiom strictly stronger than I0. We end this section introducing canonical Icarus sets, whose existence is strictly stronger than I0, and that form a hierarchy above it.

+ Deﬁnition 10.9. Let A be a transitive set. Deﬁne LA as LST, the language of set theory, extended with the constants {C}∪{ca}a∈A ∪{di}i∈ω. Let X ⊆ Vλ+1. Suppose that in L(X, Vλ+1) there is a class of indiscernibles ♯ + for formulas with parameters in {X} ∪ Vλ+1. Then X is the theory in LA of

(L(X, Vλ+1), {X} ∪ Vλ+1, {a}a∈{X}∪Vλ+1 , {di}i∈ω), where di is the i-th indiscernible.

♯ ♯ We can consider therefore (Vλ+1) . Contrary to the 0 case, we do not know ♯ whether an I0 embedding implies the existence of (Vλ+1) : it is an inner model ♯ question, see Remark 28 in [41]. On the other hand, if (Vλ+1) exists, with a shift of indiscernibles there is surely an embedding k : L(Vλ+1) ≺ L(Vλ+1), but it is not an I0 embedding, as crt(k) > Θ. ♯ So we suppose that (Vλ+1) exists.

Lemma 10.10. Suppose that (V )♯ exists. Then cof(ΘL(Vλ+1))= ω. λ+1 Vλ+1

♯ Proof. By the theory of the sharp, if (Vλ+1) exists every element of L(Vλ+1) is defined from a finite subset of Vλ+1 and finite indiscernibles. Let αn be the supremum of the ordinals definable from Vλ+1 and the first n indiscernibles. Then the supremum of the α ’s is ΘL(Vλ+1). n Vλ+1

♯ Therefore, if (Vλ+1) exists, L(Vλ+1) is very far from V (cfr. with Theorem 7.8). ♯ What if (Vλ+1) is an Icarus set? Note then that I0 is the same as ∃j : ♯ ♯ ♯ (Vλ+1, (Vλ+1) ) ≺ (Vλ+1, (Vλ+1) ), therefore not only (Vλ+1) is Icarus is strictly ♯ stronger then I0(λ), but any I0-embedding must be deﬁnable in L((Vλ+1) , Vλ+1).

10 We say that j : L(N) ≺ L(N) is proper if for any X ∈ N, hXn : n ∈ ωi ∈ L(N).

63 ♯ By the same proof of 6.37, therefore, (Vλ+1) is Icarus strongly implies I0(λ) ♯ ♯ (in fact, a j : Lω((Vλ+1) , Vλ+1) ≺ Lω((Vλ+1) , Vλ+1) suﬃces to carry on the construction of the inverse limits). If such construction is carefully analyzed, one can realize that the λ¯ such that I0(λ¯) holds constructed in this way has more properties, as it is limit of critical points of embeddings:

♯ Theorem 10.11 (Shi, Trang, 2017 [36]). Suppose that (Vλ+1) is Icarus and let λ¯ < λ be such that I0(λ¯) holds because of the inverse limit construction. Then λ¯ is limit of λ¯+-supercompact cardinals. Therefore there is no λ¯+-Aronszajn tree, there is no good scale at λ¯ and Stationary Reﬂection at λ¯+ is true. This is interesting if compared to Theorem 8.10. The following is the key to the analysis of Icarus sets:

Theorem 10.12 (Woodin, Lemma 213 in [41]). Suppose that X ⊆ Vλ+1 is an L(Y,Vλ+1) L(X,Vλ+1) Icarus set, and let Y ∈ L(X, Vλ+1) ∩ Vλ+2 such that Θ < Θ . ♯ ♯ Then Y exists and Y ∈ L(X, Vλ+1).

L(X,Vλ+1) L(Vλ+1) So let X ⊆ Vλ+1 be an Icarus set such that Θ > Θ . Then ♯ L(X,Vλ+1) (Vλ+1) exists and, since it is deﬁnable, it is an Icarus set. Moreover Θ ≥ ♯ L((Vλ+1) ,Vλ+1) ♯ Θ . In this sense (Vλ+1) is a “canonical” Icarus set. If X ⊆ Vλ+1 ♯ L((Vλ+1) ,Vλ+1) ♯♯ is any Icarus set such that ΘL(X, Vλ+1) > Θ , then (Vλ+1) exists, ♯♯ it is an Icarus set and ΘL(X,Vλ+1) ≥ ΘL((Vλ+1) ,Vλ+1). L(X,Vλ+1) In other words, if X ⊆ Vλ+1 is an Icarus set then the largeness of Θ “measures” how strong it actually is, as an axiom, and the Θ’s of the models ♯ ♯♯ built from the sharps (therefore ΘL((Vλ+1) ,Vλ+1 ,ΘL((Vλ+1) ,Vλ+1 , ... ) are the reference for such measure. Also, the construction for strong implication still holds. Therefore we have a new hierarchy:

Proposition 10.13. Let λ be a cardinal. Then Vλ+1 is Icarus is strongly im- ♯ ♯♯ plied by (Vλ+1) is Icarus, that is strongly implied by (Vλ+1) is Icarus, etc. (n+1)♯ n♯ In other words, (Vλ+1) is Icarus strongly implies that (Vλ+1) is Icarus. L(X,Vλ+1) Moreover, if X ⊆ Vλ+1 is Icarus and there exists n ∈ ω such that Θ < n♯ m♯ ΘL((Vλ+1) ,Vλ+1), then there exists an m

11 Further Developments

The main line of research right now is ﬁnding an answer for Question 7.20. The research is ongoing and very promising: if the answer is yes, then not only many already known results will get unlocked (for example the Ultraﬁlter Axiom at ω or the full power of Generic Absoluteness), but it would give an elegant tree

64 representation to all subsets of Vλ+1. The potential would be huge, just like R what happened for ADL( ). Another approach would be to investigate the inner model theory of I0. Here the main result is this: Theorem 11.1 (Woodin, 2011, [40], [43]). Let δ be an extendible cardinal. Assume that N is a weak extender model for δ supercompact and γ >δ is a cardinal in N. Let j : H(γ+)N → H(j(γ)+)N be an elementary embedding with δ ≤ crt(j) and j 6= id. Then j ∈ N. In other words, if there is a (reasonable) canonical inner model for supercompactness and I0 holds, then I0 must hold in the inner model. There is no hope therefore to reach results of consistency equivalence to I0 via inner models, and for this reason Woodin conceived Ultimate-L, a canonical inner model for all large cardinals. If such hypothesis will prove solid, then we would have an ac- tual canonical model, a preferred V , to work with. This would overcome all the problems in the Section 9, and moreover it would be interesting to understand the role of I0 in this setting.

Question 11.2. Is it true that Ultimate-L I0(λ) iﬀ Ultimate-L L(Vλ+1) 2 AC? One can ask if there are other axioms above I0, still in ZFC. The most natural approach would be to add to L(Vλ+1) the measures that deﬁne j, just like for L[µ], but this is still open:

Question 11.3. Let j : L(Vλ+1) ≺ L(Vλ+1) weakly proper, and Uj that defines it. Is V 6= L(Vλ+1)[Uj ]? Finally, there is the Reinhardt cardinal approach, i.e., large cardinals in ZF without AC. There is a whole hierarchy above, like super-Reinhardt cardinals and Berkeley cardinals, and it goes frontally against the Ultimate-L approach. Yet, the situation about Reinhardt cardinals is similar to the I0 one: very few published results (for example [?]), and a lot of underground activity, via lecture notes, seminar slides, Wikipedia pages, MathOverflow questions, ... . The possible intersections with inner model theory, though, promise to make this topic a catalyst of set theorists’ attention in the near future, and the hope is that this will bring more publications, and maybe even a survey, of this exciting topic. Acknowledgments. The paper was written under the Italian program “Rita Levi Montalcini 2013”, but it is a collection of notes, thoughts, sketches and ideas that started in the Summer Semester of 2008 at Berkeley, propelled by a course on “Beyond I0” by Woodin. With nine years in the making it is therefore a hard task to thank all the people that helped me writing this paper, as almost all the people I interacted with in my academic career has been an inspiration for this, directly or indirectly. Knowing that this is just a small percentage of the whole picture, I would like to thank first of all Hugh Woodin, the demiurge of the I0 world, that welcomed me in Berkeley when I was still a student and continued to be a creative influence during the years; Alessandro Andretta, for his steady support in all this years, and because of his (decisive) insistence for this survey to be written. I would also like to thank Sy Friedman and Liuzhen Wu, that helped to push my research in an unexpected direction, and Scott

65 Cramer and Xianghui Shi, for the many, many discussions on I0 and similars. Special thanks also for Luca Motto Ros, that helped me in shaping this paper.

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