On the ∆ 1 n Problem of Harvey Friedman †

In this paper, we prove the following. If n ≥ 3, then there is a generic extension of L, the constructible universe, in which it is true that the set P (ω) ∩ L of all constructible reals (here—subsets of ω ) is equal to the set P (ω) ∩ ∆n of all (lightface) ∆n reals. The result was announced long ago by Leo Harrington, but its proof has never been published. Our methods are based on almost-disjoint forcing. To obtain a generic extension as required, we make use of a forcing notion of the form Q = C ×∏ν Qν in L, where C adds a generic collapse surjection b from ω onto P (ω) ∩ L, whereas each Qν , ν < ωL 2 , is an almost-disjoint forcing notion in the ω1-version, that adjoins a subset Sν of ωL 1 . The forcing notions involved are independent in the sense that no Qν -generic object can be added by the product of C and all Qξ , ξ 6= ν . This allows the definition of each constructible real by a Σn formula in a suitably constructed subextension of the Q -generic extension. The subextension is generated by the surjection b , sets Sω·k+j with j ∈ b(k) , and sets Sξ with ξ ≥ ω ·ω . A special character of the construction of forcing notions Qν is L, which depends on a given n ≥ 3, obscures things with definability in the subextension enough for vice versa any ∆n real to be constructible; here the method of hidden invariance is applied. A discussion of possible further applications is added in the conclusive section.


Introduction
Problem 87 in Harvey Friedman's treatise One hundred and two problems in mathematical logic [1] requires proof that for each n in the domain 2 < n ≤ ω there is a model of ZFC + "the constructible reals are precisely the ∆ 1 n reals".
(1) (For n ≤ 2 this is definitely impossible e.g., by the Shoenfield's absoluteness theorem.) This problem was generally known in the early years of forcing, see, e.g., problems 3110, 3111, 3112 in an early survey [2] (the original preprint of 1968) by Mathias. At the very end of [1], it is noted that Leo Harrington had solved this problem affirmatively. For a similar remark, see [2] (p. 166), a comment to P 3110. And indeed, Harrington's handwritten notes [3] (pp. 1-4) contain a sketch of a generic extension of L, based on the almost-disjoint forcing of Jensen and Solovay [4], in which it is true that ω ω ∩ L = ∆ 1 3 . Then a few sentences are added on page 5 of [3], which explain, as how Harrington planned to get a model in which ω ω ∩ L = ∆ 1 n holds for a given (arbitrary) natural index n ≥ 3, and a model in which where ∆ 1 ∞ = n ∆ 1 n (all analytically definable reals). This positively solves Problem 87, including the case n = ∞. Different cases of higher order definability are observed in [3] as well.
Yet no detailed proofs have ever emerged in Harrington's published works. An article by Harrington, entitled "Consistency and independence results in descriptive set theory", which apparently might have contained these results among others, was announced in the References list in Peter Hinman's book [5] (p. 462) to appear in Ann. of Math., 1978, but in fact, this or similar article has never been published by Harrington. One may note that finding a model for (1) belongs to the "definability of definable" type of mathematical problems, introduced by Alfred Tarski in [6], where the definability properties of the set D 1M , of all sets x ⊆ ω definable by a parameter-free type-theoretic formula with quantifiers bounded by type M , are discussed for different values of M < ω . In this context, case n = ∞ in (1) is equivalent to case M = 1 in the Tarski problem, whereas cases n < ∞ in (1) can be seen as refinements of case m = 1 in the Tarski problem, because classes ∆ 1 n are well-defined subclasses of D 11 = n<ω ∆ 1 n . The goal of this paper is to present a complete proof of the following part of Harrington's statement that solves the mentioned Friedman's problem. No such proof has been known so far in mathematical publications, and this is the motivation for our research.
Theorem 1 (Harrington). If 2 ≤ n < ∞ then there is a generic extension of L in which it is true that the constructible reals are precisely the ∆ 1 n+1 reals.
The ∆ 1 ∞ case of Harrington's result, as well as different results related to Tarski's problems in [6], will be subject of a forthcoming publication.
This paper is dedicated to the proof of Theorem 1. This will be another application of the technique introduced in our previous paper [7] in this Journal, and in that sense this paper is a continuation and development of the research started in [7]. However, the problem considered here, i.e., getting a model for (1), is different from and irreducible to the problems considered in [7] and related to definability and constructability of individual reals. Subsequently the technique applied in [7] is considerably modified and developed here for the purposes of this new application. In particular, as the models involved here by necessity satisfy ω 1 L < ω 1 (unlike the models considered in [7], which satisfy the equality ω 1 L = ω 1 ), the almost-disjoint forcing is combined with a cardinal collapse forcing in this paper. And hence we will have to substantially deviate from the layout in [7], towards a modification that shifts the whole almost-disjoint machinery from ω to ω 1 . Section 2: we set up the almost-disjoint forcing in the ω 1 -version. That is, we consider the sets SEQ = (ω 1 ) <ω 1 and FUN = (ω 1 ) ω 1 in L, the constructible universe, and, given u ⊆ FUN , we define a forcing notion Q[u] which adds a set G ⊆ SEQ such that if f ∈ FUN in L then G covers f iff f / ∈ u, where covering means that f ξ ∈ G for unbounded-many ξ < ω 1 L . We also consider two types of transformations related to forcing notions of the form Q[u]. Section 3. We let I = ω 2 L be the index set. Arguing in L, we consider systems U = {U(ν)} ν∈I , where each U(ν) ⊆ FUN is dense. Given such U , the product almost-disjoint forcing Q[U] = C × ∏ ν∈I + Q[U(ν)] (with finite support) is defined in L, where C = (P (ω)) <ω is a version of Cohen's collapse forcing. Such a forcing notion adjoins a generic map b G : ω onto −→ P (ω) ∩ L to L, and adds an array of sets G(ν) ⊆ SEQ (where ν ∈ I ) as well, so that each G(ν) is a Q[U(ν)]-generic set over L. We also investigate the structure of related product-generic extensions and their subextensions, and transformations of forcing notions of the form Q[U].
Section 4. Given n ≥ 2, we define a system U ∈ L as above, which has some remarkable properties, in particular, (1) being Q[U(ν)]-generic is essentially a Π 1 n property in all suitable generic extensions, (2) if ν ∈ I and G ⊆ Q[U] is generic over L, then the extension L[b G , {G(ν )} ν =ν ] contains no Q[U(ν)]-generic reals, and (3) all submodels of L[G] of certain kind are elementarily equivalent w. r. t. Σ 1 n formulas. The latter property is summarized in the key technical instrument, Theorem 4 (the elementary equivalence theorem), whose proof is placed in a separate Section 6.
To prove Theorem 1, we make use of a related generic extension L[b G , {G(ν)} ν∈W [G] ], where We introduce forcing approximations in Section 5, a forcing-like relation used to prove the elementary equivalence theorem. Its key advantage is the invariance under some transformations, including the permutations of the index set I , see Section 5. 4. The actual forcing notion Q = Q[U] is absolutely not invariant under permutations, but the n-completeness property, maintained through the inductive construction of U in L, allows us to prove that the auxiliary forcing relation is connected to the truth in Q -generic extensions exactly as the true Q -forcing relation does-up to the level Σ 1 n of the projective hierarchy (Lemma 33). We call this construction the hidden invariance technique (see Section 6.1).
Finally, Section 6 presents the proof of the elementary equivalence theorem, with the help of forcing approximations, and hence completes the proof of Theorem 1.
The flowchart can be seen in Figure 1 on page 3. And we added the index and contents as Supplementary Materials for easy reading.

Almost-Disjoint Forcing
Almost-disjoint forcing as a set theoretic tool was invented by Jensen and Solovay [4]. It has been applied in many research directions in modern set theory, in particular, in our paper [7] in this Journal. Here we make use of a considerably different version of the almost-disjoint forcing technique, which, comparably to [7], (1) considers countable cardinality instead of finite cardinalities in some key positions, (2) accordingly considers cardinality ω 1 instead of countable cardinality. In particular, sequences of finite length change to those of length < ω 1 . And so on.
Assumption 1. During arguments in this section, we assume that the ground set universe is L, the constructible universe. Recall that in L, HC = L ω 1 and Hω 2 = L ω 2 .
For the sake of brevity, we call ω 1 -size sets those X satisfying card X ≤ ω 1 .

Almost-Disjoint Forcing: ω 1 -Version
This subsection contains a review the basic notation related to almost-disjoint forcing in the ω 1 -version. Arguing in L, we put FUN = ω 1 ω 1 = all ω 1 -sequences of ordinals < ω 1 .
{Λ}, the set of all non-empty sequences s of ordinals < ω 1 , of length lh s = dom s < ω 1 . We underline that Λ, the empty sequence, does not belong to SEQ .
If S/ f is unbounded in ω 1 then say that S covers f , otherwise S does not cover f .
The following or very similar version of the almost-disjoint forcing was defined by Jensen and Solovay in [4] ([ § 5]). Its goal can be formulated as follows: given a set u ⊆ FUN in the ground universe, find a generic set S ⊆ SEQ such that the equivalence f ∈ u ⇐⇒ S does not cover f holds for each f ∈ FUN in the ground universe.
Definition 1 (in L). Q * is the set of all pairs p = S p ; F p of finite sets F p ⊆ FUN , S p ⊆ SEQ . Elements of Q * will be called (forcing) conditions. If p ∈ Q * then put a tree in SEQ . If p, q ∈ Q * then let p ∧ q = S p ∪ S q ; F p ∪ F q ; a condition in Q * . Let p, q ∈ Q * . Define q p (that is, q is stronger as a forcing condition) iff S p ⊆ S q , F p ⊆ F q , and the difference S q S p does not intersect Lemma 1 (in L). Conditions p, q ∈ Q * are compatible in Q * iff (1) S q S p does not intersect F ∨ p , and (2) S p S q does not intersect F ∨ q . Therefore any p, q ∈ P * are compatible in P * iff p ∧ q p and p ∧ q q.
Proof. If (1), (2) hold then p ∧ q p and p ∧ q q, thus p, q are compatible.
Any conditions p, q ∈ Q[u] are compatible in Q[u] iff they are compatible in Q * iff the condition p ∧ q = S p ∪ S q ; F p ∪ F q ∈ Q[u] satisfies both p ∧ q p and p ∧ q q. Therefore, we can say that conditions p, q ∈ Q * are compatible (or incompatible) without an explicit indication which forcing notion Q[u] containing p, q is considered. Lemma 2 (in L). If u ⊆ FUN and A ⊆ Q[u] is an antichain then card A ≤ ω 1 .

Proof.
Suppose towards the contrary that card A > ω 1 . If p = q in A are incompatible then obviously S p = S q . Yet {S p : p ∈ Q * } = all finite subsets of SEQ , is a set of cardinality ω 1 , a contradiction.

Almost-Disjoint Generic Extensions
To work with L-sets FUN and SEQ in generic extensions of L, possibly in those obtained by means of cardinal collapse, we let -in other words, FUN L and SEQ L are just FUN and SEQ defined in L.
Define a condition p by F p = F q and S p = S q ∪ { f m}; we have p ∈ D f l and p q.) Pick, by the density, any p ∈ D f l ∩ G . Then sup(S G / f ) > l . We conclude that S G / f is infinite because l is arbitrary.
(ii) Let p ∈ G . Then obviously s p ⊆ S G . If there exists s ∈ (S G S p ) ∩ F ∨ p then s ∈ S q for some q ∈ G . Then conditions p, q are incompatible by Lemma 1, which is a contradiction. Now assume that p ∈ P[u] G . There is a condition q ∈ G incompatible with p. We have two cases by Lemma 1. First, there is some s ∈ (S q S p ) ∩ F ∨ p . Then s ∈ S G S p , so p is not compatible with S G . Second, there is some s ∈ (S p S q ) ∩ F ∨ q . In this case, s / ∈ S r holds for any condition r ≤ q. It follows that s / ∈ S G , hence S p ⊆ S G , and p cannot be compatible with S G .
is an immediate corollary of (iv) since ω L 2 remains a cardinal in L[G] by Lemma 2. Finally, to prove (iv) let f ∈ FUN L u and λ < ω L 1 . The set D f λ of all conditions p ∈ Q[u], such that f λ ⊂ g for some g ∈ S p , is dense in Q[u]. Therefore G contains some p ∈ D f λ . Let this be witnessed by some g ∈ S p . Now, if ξ < λ belongs to X f , so that s = f ξ ∈ S G , then s must belong to S p by (ii), therefore ξ belongs to the finite set {lh s : Now we consider two types of transformations related to the forcing notion Q * .

Lipschitz Transformations
We argue in L. Let LIP be the group of all ⊆-automorphisms of SEQ , called Lipschitz transformations. Any λ ∈ LIP preserves the length lh of sequences, i.e., lh s = lh (λ · s) for all s ∈ SEQ . Any transformation λ ∈ LIP acts on: sequences s ∈ SEQ : by λ · s = λ(s); Lemma 4 (routine). The action of any λ ∈ LIP is an order-preserving automorphism of Q * . If u ⊆ FUN and p ∈ Q[u] then λ · p ∈ Q[λ · u].
We proceed with an important existence lemma. If f = g belongs to FUN then let β( f , g) be equal to the least ordinal β < ω 1 such that f (β) = g(β) (or, similarly, the largest ordinal β with f β = g β). Say that sets X, Y ⊆ FUN are intersection-similar, or i-similar for brevity, if there is a bijection b : ) for all f = g in X -such a bijection b will be called an i-similarity bijection.
(ii) there exists a transformation λ ∈ LIP such that λ · u = v and λ · X = Y .

Proof.
The key argument is that if A ⊆ u, B ⊆ v are at most countable, b : A onto −→ B is an i-similarity bijection, and f ∈ u A, then by the density of v there is g ∈ v B such that the extended is still an i-similarity bijection. This allows proof of (i), iteratively extending an initial i-similarity bijection b 0 : X onto −→ Y by a ω 1 -step back-and-forth argument involving eventually all elements f ∈ u and g ∈ v, to an i-similarity bijection u onto −→ v required. See the proof of Lemma 5 in [7] for more detail.

Substitution Transformations
We continue to argue in L. Assume that conditions p, q ∈ Q * satisfy We define a transformation h pq acting as follows.
If p = q then define h pq (r) = r for all r ∈ Q * , the identity. Suppose that p = q.
Then p, q are incompatible by (4) and Lemma 1. Define d pq = {r ∈ Q * : r p ∨ r q}, the domain of h pq . Let r ∈ d pq . We put h pq (r) = r := S r , F r , where F r = F r and S r =    (S r S p ) ∪ S q in case r p , (S r S q ) ∪ S p in case r q .
Thus, assuming (4), the difference between S r and S r lies entirely within the set X = F ∨ p = F ∨ q , so that if r p then S r ∩ X = S p but S r ∩ X = S q , while if r q then S r ∩ X = S q but S r ∩ X = S p . Lemma 6. (i) If u ⊆ FUN is dense and p 0 , q 0 ∈ Q[u] then there exist conditions p, q ∈ Q[u] with p p 0 , q q 0 , satisfying (4).
(ii) Let p, q ∈ Q * satisfy (4). If p = q then h pq is the identity transformation. If p = q then h pq an order automorphism of d pq = {r ∈ Q * : r p ∨ r q}, satisfying h pq (p) = q and h pq = (h pq ) −1 = h qp .
(iii) If u ⊆ FUN and p, q ∈ Q[u] satisfy (4) then h pq maps the set Q[u] ∩ d pq onto itself order-preserving.
Please note that unlike the Lipschitz transformations above, transformations of the form h pq , called substitutions in this paper, act within any given forcing notion of the form Q[u] by claim (iii) of the lemma, and hence the forcing notions of the form Q[u] considered are sufficiently homogeneous.

Almost-Disjoint Product Forcing
Here we review the structure and basic properties of product almost-disjoint forcing and corresponding generic extensions in the ω 1 -version. There is an important issue here: a forcing C, which collapses ω 1 to ω , enters as a factor in the product forcing notions considered.

Product Forcing
In L, we define C = P (ω) <ω , the set of all finite sequences of subsets of ω , an ordinary forcing to collapse P (ω) ∩ L down to ω . We will make use of an ω 2 -product of Q * with C as an extra factor. (In fact, C can be eliminated since Q * collapses ω L 1 anyway by Lemma 3 (v). Yet the presence of C somehow facilitates the arguments since C has a more transparent forcing structure.) Technically, we put I = ω 2 (in L) and consider the index set I + = I ∪ {−1}. Let Q * be the finite-support product of C and I copies of Q * (Definition 1 in Section 2.1), ordered componentwise. That is, Q * consists of all maps p defined on a finite set dom p = |p| + ⊆ I + so that p(ν) ∈ Q * for all ν ∈ |p| := |p| + {−1}, and if −1 ∈ |p| + then b p := p(−1) ∈ C. If p ∈ Q * then put F p (ν) = F p(ν) and S p (ν) = S p(ν) for all ν ∈ |p|, so that p(ν) = S p (ν) ; F p (ν) . We order Q * componentwise: p q ( p is stronger as a forcing condition) iff |q| + ⊆ |p| + , b q ⊆ b p in case −1 ∈ |q| + , and p(ν) q(ν) in Q * for all ν ∈ |q|. Put In particular, Q * contains the empty condition ∈ Q * satisfying | | + = ∅; obviously is the -least (and weakest as a forcing condition) element of Q * . Because of the factor C, it takes some effort to define p ∧ q for p, q ∈ Q * , and only assuming that b p , b q are compatible, i.e., b p ⊆ b q or b q ⊆ b p . In such a case define p ∧ q ∈ Q * as follows. First, |p ∧ q| + = |p| + ∪ |q| + . If ν ∈ |p| + |q| + then put (p ∧ q)(ν) = p(ν), and similarly if ν ∈ |q| + |p| + then (p ∧ q)(ν) = q(ν). Now suppose that ν ∈ |p| + ∩ |q| + .

Systems
Arguing in L, we consider certain subforcings of the total product forcing notion Q * .
is the finite-support product of C and sets Q[U(ν)], ν ∈ |U|, i.e., Suppose that c ⊆ I + . If p ∈ Q * then define p = p c ∈ Q * so that |p | + = c ∩ |p| + and Writing p c, U c etc., it is not assumed that c ⊆ |p| + .

Lemma 8 (in L). If U is a system and A ⊆ Q[U] is an antichain then card
It follows by the ∆-system lemma that there is a set A ⊆ A of the same cardinality card A = card A > ω 1 , and a finite set d ⊆ I + , such that |p| + = d for all p ∈ A . Then we have S p = S q for all p = q in A , easily leading to a contradiction, as in the proof of Lemma 2.

Outline of Product Extensions
We consider sets of the form Q[U], U being a system in L, as forcing notions over L. Accordingly, we'll study Q[U]-generic extensions L[G] of the ground universe L. Define some elements of these extensions. Suppose that G ⊆ Q * . Put |G| = p∈G |p|; |G| ⊆ I . Let Thus, S G (ν) ⊆ SEQ L , and S G (ν) = ∅ for any ν / ∈ |G|. By the way, this defines a sequence Lemma 9. Let U be a system in L, and G ⊆ Q[U] be a set Q[U]-generic over L. Then:

Names for Sets in Product Extensions
The next definition introduces names for elements of product-generic extensions of L considered. Assume that in L, K ⊆ Q * , e.g., K = Q[U], where U is a system, and X is any set. By N X (K) ( K-names for subsets of X ) we denote the set of all sets Proof. It follows from general forcing theory that there is a name σ ∈ N X (Q[U]), not necessarily an ω 1 -size name, such that where A x ⊆ Q x is a maximal antichain for any x . We observe that card A x ≤ ω 1 in L for all x by Lemma 8, hence τ ∈ SN X (Q[U]). And on the other hand, we have To prove the additional claim, note that by the product forcing theorem if Y ∈ L[G c] then the original name σ can be chosen in N X (Q[U] c), and repeat the argument.

Names for Reals in Product Extensions
Now we introduce names for reals (elements of ω ω ) in generic extensions of L considered. This is an important particular case of the content of Section 3. 4.
Define the restrictions SN ω iff all sets τ "j are pre-dense in K below p 0 , i.e., any condition q ∈ K , q p 0 , is compatible with some r ∈ τ j (and this holds for all j < ω ).
Suppose that τ ∈ SN ω ω (Q * ). A set G ⊆ K is minimally τ-generic iff it is compatible in itself (if p, q ∈ G then there is r ∈ G with r p, r q), and intersects each set τ "x , x ∈ X . In this case, put

Equivalent names.
Names τ, µ ∈ SN ω ω (Q * ) are equivalent iff conditions q, r are incompatible whenever q ∈ τ " j, k and r ∈ µ " j, k for some j and k = k . Names τ, µ are equivalent below some p ∈ Q * iff the triple of conditions p, q, r is incompatible (that is, no common strengthening) whenever q ∈ τ " j, k and r ∈ µ " j, k for some j and k = k .

Lemma 12.
Suppose that in L, p ∈ Q * , and names µ, τ ∈ SN ω ω (Q * ) are equivalent (resp., equivalent below p ). If G ⊆ Q * is minimally µ-generic and minimally τ-generic (resp., and containing p ), then µ Proof. Suppose that this is not the case. Then by definition there exist numbers j and k = k and conditions q ∈ G ∩ (τ " j, k ) and r ∈ G ∩ (µ " j, k ). Then p, q, r are compatible (as elements of the same generic set), contradiction.
The next lemma provides a useful transformation of names. Recall that p ∧ p is defined in Section 3.1.

Permutations
We continue to argue in L. There are three important families of transformations of the whole system of objects related to product forcing, considered in this Subsection and the two following ones.

Lemma 15 (routine).
If λ ∈ LIP I then p −→ π · p is an order-preserving bijection of Q * onto Q * , and if U is a system then p Proof. Apply Lemma 5 componentwise for every ν ∈ I .

Multi-Substitutions
Assume that conditions p, q ∈ Q * satisfy the following: In particular, (4) of Section 2.4 holds for all ν. We define a transformation H pq acting as follows. First, we let D pq , the domain of H pq , contain all conditions r ∈ Q * such that (b) if ν ∈ |r| ∩ |p| and p(ν) = q(ν), then r(ν) p(ν) or r(ν) q(ν), thus, in other words, r(ν) ∈ d p(ν)p(ν) in the sense of Section 2. 4.
Please note that all conditions r p and all r p belong to D pq . On the other hand, if r ∈ Q * satisfies |r| ∩ |p| = ∅ and (a), then r belongs to D pq as well. In particular, ∈ D pq .
If r ∈ D pq , then define r = H pq (r) ∈ Q * so that |r | + = |r| + and: Transformations of the form H pq will be called multi-substitutions.
(ii) If conditions p, q ∈ Q * satisfy (6), then H pq is an order automorphism of D pq = D qp , and we have H pq = (H pq ) −1 = H qp and H pq (p) = q.
(iii) If U is a system, and p, q ∈ Q[U] satisfy (6), then H pq maps the set Q[U] ∩ D pq onto itself order-preserving.

The Basic Forcing Notion and the Model
In this paper, we let ZFC − be ZFC minus the Power Set axiom, with the schema of Collection instead of Replacement, with AC is assumed in the form of well-orderability of every set, and with the axiom: "ω 1 exists". See [8] on versions of ZFC sans the Power Set axiom in detail.
Let ZFC − 2 be ZFC − plus the axioms: V = L, and the axiom "every set x satisfies card x ≤ ω 1 ".

Jensen-Solovay Sequences
Arguing in L, let U, V be systems. Suppose that M is any transitive model of ZFC − 2 . Define U M U iff U U and the following holds: Proof. Let d = |U| ∪ z. By definition SEQ is ω-closed as a forcing: any ⊆-increasing sequence {s n } n<ω of s n ∈ SEQ has the least upper bound in SEQ , equal to the union of all s n . It follows that the countable-support product SEQ (d×ω 1 ) is ω-closed, too. Therefore, as card M ≤ ω 1 , there Proof. We claim that F = ν∈|U| (U (ν) U(ν)) is multiply SEQ -generic over M . Suppose, for the sake of brevity, that (ii) If moreover λ < ω 2 and M |= ZFC − 2 is a transitive model containing { M ξ , U ξ } ξ<λ then M, U ∈ sJS and M ξ , U ξ ≺ M, U , ∀ ξ .
(iii) The same is true in case λ = ω 2 , but then the model M is not necessarily a ω 1 -sizemodel, and we require M, U ∈ JS rather than sJS, of course.

Proof.
The same arguments work as in the proof of Lemma 19.

Stability of Dense Sets
If U is a system, D is a pre-dense subset of P[U], and U is another system extending U , then in principle D does not necessarily remain maximal in P[U ], a bigger set. This is where the genericity requirement (a) in Section 4.1 plays its role to seal the pre-density of sets in M w. r. t. further extensions. This is the content of the following key theorem. Moreover, the product forcing arguments will allow us to extend the stability result in pre-dense sets not necessarily in M , as in items (ii), (iii) of the theorem.
Theorem 2 (stability of dense sets). Assume that, in L, M, U ∈ sJS, U is a system, and U M U . If D is a pre-dense subset of Q[U] (resp., pre-dense below some p ∈ Q[U] ) then D remains pre-dense in Q[U ] (resp., pre-dense below p ) in each of the following three cases: , where G ⊆ P is P-generic over L, and P ∈ M is a PO set; Proof. Arguing in L, we consider only the case of sets D pre-dense in Q[U] itself; the case of pre-density below some p ∈ Q[U] is treated similarly.
(i) Suppose, towards the contrary, that a condition p ∈ Q[U ] is incompatible with each q ∈ D . As D ⊆ P[U], we can w. l. o. g. assume that |p| ⊆ |U|.
We are going to define a condition p ∈ Q[U], also incompatible with each q ∈ D , contrary to the pre-density. To maintain the construction, consider the finite sequence f = f 1 , . . . , f m of all elements f ∈ FUN occurring in ν∈|p| F p (ν) but not in U . It follows from U M U that f is SEQ m -generic over M . Moreover, p being incompatible with D is implied by the fact that f meets a certain family of dense sets in SEQ m , of cardinality ≤ ω 1 in M . Therefore, we will be able to simulate this in M , getting a sequence g ∈ M which meets the same dense sets, and hence yields a condition p ∈ Q[U], also incompatible with each q ∈ D .
To present the key idea in sufficient detail in a rather simplified subcase, we assume that Then To make the argument even more transparent, we suppose that (The general case follows the same idea and can be found in [4]; we leave it to the reader.) The plan is to replace the functions f , g by some functions f , g ∈ U(ν 0 ) so that the incompatibility of p with conditions in D will be preserved.
It holds by the choice of p and Lemma 1 that and D 1 depends on f , g via F p (ν 0 ). The equality D = D 1 ( f , g) ∪ D 2 can be rewritten as Please note that each A ∈ A is a finite subset of SEQ , so we can re-enumerate A = { A κ : κ < ω 1 } in M and rewrite (10) as follows: As the pair f , g is SEQ -generic, there is an index µ 0 < ω 1 such that (11) is forced over M by σ 0 , τ 0 , where σ 0 = f µ 0 and τ 0 = g µ 0 . In other words, A κ ∩ S( f , g ) = ∅ holds for all κ < ω 1 whenever f , g is SEQ -generic over M and σ 0 ⊂ f , τ 0 ⊂ g . It follows that for any κ < ω 1 and sequences σ, τ ∈ SEQ extending resp. σ 0 , τ 0 there are sequences σ , τ ∈ SEQ extending resp. σ, τ , at least one of which extends one of sequences w ∈ A κ . This allows us to define, in M , a pair of sequences f , g ∈ FUN , such that σ 0 ⊂ f , τ 0 ⊂ g , and for any κ < ω 1 at least one of f , g extends one of w ∈ A k . In other words, we have It follows that the condition p defined by |p | (9)), and further D = D 1 ( f , g ) ∪ D 2 , thus p is incompatible with each q ∈ D . Yet p ∈ M since f , g ∈ M , which contradicts the pre-density of D .

Complete Sequences and the Basic Forcing Notion
In L, we say that a pair M, U ∈ sJS solves a set D ⊆ sJS iff either M, U ∈ D or there is no pair M , U ∈ D that extends M, U . Let D solv be the set of all pairs M, U ∈ sJS which solve a given set D ⊆ sJS. A sequence { M ξ , U ξ } ξ<ω 2 ∈ − → JS ω 2 is called n-complete (n ≥ 3) iff it intersects every set of the form D solv , where D ⊆ sJS is a Σ Hω 2 n−2 (Hω 2 ) set. Recall that Hω 2 is the collection of all sets x whose transitive closure TC(x) has cardinality card (TC(x)) < ω 2 . Furthermore, Σ Hω 2 n−2 (Hω 2 ) means definability by a Σ n−2 formula of the ∈language, in which any definability parameters in Hω 2 are allowed, while Σ Hω 2 n−2 means parameter-free definability. Similarly, ∆ Hω 2 n−1 ({ω 1 }) in the next theorem means that ω 1 is allowed as a sole parameter. It is a simple exercise that sets {SEQ} and SEQ are ∆ Hω 2 1 ({ω 1 }) under V = L. Generally, we refer to e.g., ([9] (Part B, 5.4)), or ([10] (Chapter 13)) on the Lévy hierarchy of ∈-formulas and definability classes Σ H n , Π H n , ∆ H n for any transitive set H .

Proof.
To account forω 1 as a parameter, note that the set ω 1 is Σ Hω 2 1 , and hence the singleton {ω 1 } is ∆ Hω 2 2 . Indeed "being ω 1 " is equivalent to the conjunction of "being uncountable"-which is Π Hω 2 1 , and "every smaller ordinal is countable"-which is Σ Hω 2 1 since the quantifier "for all smaller ordinals" is bounded, hence, it does not increase the complexity.
To check the definability property, make use of the well-known fact that the restriction < L Hω 2 is a ∆ Hω 2 1 relation, and if n ≥ 1, p ∈ ω ω is any parameter, and R(x, y, z, . . . ) is a finitary ∆ Hω 2 n (p) relation on HC then the relations ∃ x < L y R(x, y, z, . . . ) and ∀ x < L y R(x, y, z, . . . ) (with arguments y, z, . . . ) are ∆ Hω 2 n (p) as well.

Definition 2 (in L)
. Fix a number n ≥ 2 during the proof of Theorem 1.
We define Q = Q[U] (the basic forcing notion), and Q ξ = Q[U ξ ] for ξ < ω 2 . Thus, Q is the finite-support product of the set C and sets Q(ν) = Q[U(ν)], i ∈ I ; so that Q ∈ L.
(ii) If G ⊆ Q is a set Q-generic over L then the set G ξ = G ∩ Q ξ is Q ξ -generic over M ξ .

Proof.
Make use of Lemma 20 and Corollary 2 in Section 4.2.

Proof.
The sequence { M ξ , U ξ } ξ<ω 1 is ∆ Hω 2 n−1 by definition, hence the relation f ∈ U(ν) is Σ Hω 2 n−1 . On the other hand, if f ∈ Fun belongs to some M ξ then f ∈ U(ν) obviously implies f ∈ U ξ (ν), leading to a Π HC n−1 definition of the relation f ∈ U(ν). To prove the last claim, note that by Corollary 3 if a name τ ∈ SN ω ω (P ξ ) ∩ M ξ is P ξ -full then it remains P-full.

Basic Generic Extension
The proof of Theorem 1 makes use of a generic extension of the form L[G z], where G ⊆ Q is a set Q-generic over L, and z ⊆ I + , z / ∈ L. The following two theorems will play the key role in the proof. Define formulas Γ ν (ν ∈ I ) as follows:

Lemma 22.
Suppose that a set G ⊆ Q is Q-generic over L, and ν ∈ I , c ∈ L[G], ∅ = c ⊆ I + . Then , and generally, there are no sets S ⊆ SEQ L in L[G =ν ] satisfying Γ ν (S).
To prove (iv), we need more work. Let X = SEQ L . Suppose towards the contrary that some S ∈ L[G =ν ], S ⊆ X = SEQ L satisfies Γ ν (S). It follows from Lemma 10 (with U = U and There is an ordinal , it follows that there is a sequence s ∈ SEQ L , s ⊂ f , such that S contains no subsequences of f extending s. Take any µ ∈ I , µ = ν. By Corollary 4 (i), there exists a function g ∈ U(µ) U ξ (µ), g / ∈ U(ν), satisfying s ⊂ g. Then, S covers g by Γ ν (S). However, this is absurd by the choice of s.
The proof of the next important elementary equivalence theorem will be given below in Section 6. 3.

Theorem 4 (elementary equivalence theorem).
Assume that in L, −1 ∈ d ⊆ I + , sets Z , Z ⊆ I d satisfy card (I Z) ≤ ω 1 and card (I Z ) ≤ ω 1 , the symmetric difference Z ∆ Z is at most countable, and the complementary set Let G ⊆ Q be Q-generic over L, and x 0 ∈ L[G d] be any real. Then any closed Σ 1 n formula ϕ, with real parameters in L[x 0 ], is simultaneously true in L[x 0 , G Z] and in L[x 0 , G Z ]. 4. 5

. The Main Theorem Modulo the Elementary Equivalence Theorem: The Model
Here we begin the proof of Theorem 1 on the base of Theorem 4 of Section 4. 4. We fix a number n ≥ 2 during the proof. The goal is to define a generic extension of L in which for any set x ⊆ ω the following is true: x ∈ L iff x ∈ ∆ 1 n+1 . The model is a part of the basic generic extension defined in Section 4. 4. In the notation of Definition 2 in Section 4.3, consider a set G ⊆ Q , Q-generic over L. Then b G = G(−1) is a C-generic map from ω onto P (ω) ∩ L by Lemma 9 (i). We define and We also define, for any m < ω , of w[G] is necessarily infinite and coinfinite, and it codes the target set b G (k) since It will be important below that definition (12) is monotone w.

Proof of the Key Claim of Lemma 23
The proof of Lemma 23 (ii) is based on several intermediate lemmas.
Lemma 24 (compare with Lemma 33 in [7]). Suppose that G ⊆ Q is Q-generic over L, and m < ω . Let c ⊆ w <m [G] be any set in L. Then any closed Σ 1 Proof (Lemma 24). There is an ordinal ξ < ω 2 such that all parameters in ϕ belong to To prepare for Theorem 4 of Section 4.4 In continuation of the proof of Lemma 23 (ii), suppose that ( †) ϕ(·) and ψ(·) are parameter-free Σ 1 n+1 formulas that provide a ∆ 1 n+1 definition for a set x ⊆ ω , Here, G is the canonical Q-name for the generic set G ⊆ Q , as usual, whileW is a name for W ∈ L.
Proof. Suppose, for the sake of simplicity, that p 0 is the empty condition (i.e., |p 0 | + = ∅); the general case does not differ much.
Proof (Proposition). Suppose on the contrary that q does not force "L[G (w + [G] ∪W)] |= ϕ( )". As p, q satisfy (6), the associated transformation H pq maps the set Q p = { p ∈ Q : p p} onto Q q = {q ∈ Q : q q} order-preserving by Lemma 17 (with U = U). By the choice of q, there is a set G q ⊆ Q q , generic over L, containing q, and such that ϕ ] by ( †) (and the assumption that p 0 = ).
The set G p = {(H pq ) −1 (q ) : q ∈ G q ∧ q q} ⊆ Q p is Q-generic over L as well (as H pq is an order isomorphism), and contains p, and hence ϕ( ) is true and Case 1: (I) holds, i.e., b p = b q . Then by definition On the other hand, the sets G p and G q are equi-constructible by means of the application of H pq , and hence G p (w + [G p ] ∪ W) and G q (w + [G q ] ∪ W) are equi-constructible, that is, the classes L[G p (w + [G p ] ∪ W)] and L[G q (w + [G q ] ∪ W)] coincide. However, ϕ( ) is true in one of them and false in the other one, a contradiction.
, and also implies w + ≥m [G p ] = w + ≥m [G q ], while the difference between the sets w <m [G p ], w <m [G q ] is that for any k < m and any j, Moreover, (II) implies G p =−1 = G q =−1 , and hence S G p (ν) = S G q (ν) for all ν ∈ I via H pq . We conclude that L[G p Z] = L[G q Z] for any set Z ∈ L[b G p ], Z ⊆ I + , in particular, However, ϕ( ) holds in L[G p (w + [G p ] ∪ W)], see above. It follows by Lemma 24 that ϕ( ) holds in , which is a contradiction to the above. If (II) (b) holds, then argue similarly using the formula ψ( ).
(Proposition 1) Coming back to Lemma 25, suppose towards the contrary that "L[G (w + [G] ∪W)] |= ϕ( )" is not Q-decided by p 0 = . There are two conditions p, q ∈ Q such that p Q-forces "L[G (w + [G] ∪W)] |= ϕ( )" while q Q-forces the negation. We may w. l. o. g. assume, by Lemma 17 (i), that p, q satisfy (6) of Section 3. 8. We claim that p, q can be connected by a finite chain of conditions in Q in which each two consecutive terms are close neighbours in the sense above, satisfying (6) in Section 3.8-then Proposition 1 implies a contradiction and concludes the proof of Lemma 25.
Thus, it remains to prove the connection claim. Let p ∈ Q be defined by b p = b p and p =−1 = q =−1 . Then p, p are close neighbours and (6) holds for this pair as it holds for p, q.
Still r is a close neighbour to both p and q, and (6) holds for p , r and q, r . Thus, the chain p-p -r -q proves the connection claim.

Forcing Approximation
To prove Theorem 4 of Section 4.4 and thus complete the proof of Theorem 1 in the next Section 6, we define here a forcing-like relation forc, and exploit certain symmetries of objects related to forc. This similarity will allow us to only outline really analogous issues but concentrate on several things which bear some difference.
We argue under Blanket Assumption 1.
Recall that ZFC − is ZFC minus the Power Set axiom, with the schema of Collection instead of Replacement, with the axiom "ω 1 exists", and with AC in the form of wellorderability of every set, and ZFC − 2 is ZFC − plus the axioms: V = L, and "every set x satisfies card x ≤ ω 1 ".

Formulas
Here we introduce a language that will help us to study analytic definability in Q[U]-generic extensions, for different systems U , and their submodels.
Let L be the 2nd order Peano language, with variables of type 1 over ω ω . If K ⊆ Q * then an L(K) formula is any formula of L, with some free variables of types 0, 1 replaced by resp. numbers in ω and names in SN ω ω (K), and some type 1 quantifiers are allowed to have bounding indices B (i.e., ∃ B , ∀ B ) such that B ⊆ I + satisfies either card B ≤ ω 1 or card(I B) ≤ ω 1 (in L). In particular, I + itself can serve as an index, and the absence If ϕ is a L(Q * ) formula, then let NAM ϕ = the set of all names τ that occur in ϕ; IND ϕ = the set of all quantifier indices B which occur in ϕ; |ϕ| + = τ∈NAM ϕ |τ| + (a set of ω 1 -size); ||ϕ|| = |ϕ| + ∪ IND ϕ − so that |ϕ| + ⊆ ||ϕ|| ⊆ I + .
If a set G ⊆ Q * is minimally ϕ-generic (that is, minimally τ-generic w. r. t. every name τ ∈ NAM ϕ, in the sense of Section 3.5), then the valuation ϕ[G] is the result of substitution of τ[G] for any name τ ∈ NAM ϕ, and changing each quantifier ∃ B x , ∀ B x to resp. ∃ (∀ ) x ∈ ω ω ∩ L[G B], while index-free type 1 quantifiers are relativized to ω ω ; ϕ[G] is a formula of L with real parameters, and some quantifiers of type 1 relativized to certain submodels of L[G].
An arithmetic formula in L(K) is a formula with no quantifiers of type 1 (names in SN ω ω (K) are allowed). If n < ω then let a LΣ 1 n (K), resp., LΠ 1 n (K) formula be a formula of the form Define LΠ 1 n (K, M) similarly.

Forcing Approximation
We introduce a convenient forcing-type relation p forc M U ϕ for pairs M, U in sJS and formulas ϕ in L(K), associated with the truth in K-generic extensions of L, where K = Q[U] ⊆ Q * and U ∈ L is a system. The next theorem classifies the complexity of forc in terms of projective hierarchy. Please note that if M, U ∈ sJS and k ≥ 1 then any formula ϕ in LΠ 1 , M) belongs to M if we somehow "label" any large index B ∈ IND ϕ (such that card(I B) ≤ ω 1 ) by its small complement I B ∈ M . Therefore, the sets and Forc(Σ 1 k ) similarly defined, are subsets of Hω 2 (in L).
Proof (sketch). Suppose that ϕ is LΠ 1 1 . Under the assumptions of the theorem, items (F1)a, (F1)b of (F1) are ∆ Hω 2 1 relations, while (F2) is reducible to a forcing relation over M that we can relativize to M . The inductive step goes on straightforwardly using (F3), (F4). Please note that the quantifier over names in (F3) is a bounded quantifier (bounded by M ), hence it does not add any extra complexity.

Further Properties of Forcing Approximations
The notion of names ν, τ ∈ SN ω ω (Q * ) being equivalent below some p ∈ Q * , is introduced in Subsection 3. 5. We continue with a couple of routine lemmas. The induction steps LΠ 1 k → LΣ 1 k+1 and LΣ 1 k → LΠ 1 k are carried out by an easy reduction to items (F3), (F4) of Section 5.2.

Elementary Equivalence Theorem
The goal of this section is to prove Theorem 4 of Section 4.4, and accomplish the proof of Theorem 1. We make use of the relation forc defined above, and exploit certain symmetries in forc studied in Section 5.4.

Hidden Invariance
To explain the idea, one may note first that elementary equivalence of subextensions of a given generic extension is usually a corollary of the fact that the forcing notion considered is enough homogeneous, or in different words, invariant w. r. t. a sufficiently large system of order-preserving transformations. The forcing notion Q = Q[U] we consider, as well as basically any Q[U], is invariant w. r. t. multi-substitutions by Lemma 17. However, for a straightaway proof of Theorem 4 we would naturally need the invariance under permutations of Section 3.6-to interchange the domains Z and Z , whereas Q is definitely not invariant w. r. t. permutations.
On the other hand, the relation forc is invariant w. r. t. both permutations (Lemma 29) and multi-Lipschitz (Lemma 30), as well as still w. r. t. multi-substitutions by Lemma 31. To bridge the gap between forc (not explicitly connected with Q in any way) and Q -generic extensions, we prove Lemma 33, which ensures that forc admits a forcing-style association with the truth in Q -generic extensions, bounded to formulas of type Σ 1 n and below. This key result will be based on the n-completeness property (Definition 2 in Section 4.3). Speaking loosely, one may say that some transformations, i.e., permutations and multi-Lipschitz, are hidden in construction of Q , so that they do not act per se, but their influence up to nth level, is preserved.
This method of hidden invariance, i.e., invariance properties (of an auxiliary forcing-type relationship like forc) hidden in Q by a suitable generic-style construction of Q , was introduced in Harrington's notes [3] in a somewhat different terminology. We may note that the hidden invariance technique is well known in some other fields of mathematics, including more applied fields, see e.g., [12,13].

Approximations of the n-Complete Forcing Notion
We return to the forcing notion Q = Q[U] defined in L as in Definition 2 in Section 4.3 for a given number n ≥ 2 of Theorem 1. Arguing in L, we let the pairs M ξ , U ξ , ξ < ω 2 , also be as in Definition 2.
The next lemma shows that forc ∞ satisfies a key property of forcing relations up to the level of Π 1 n−1 formulas.

Lemma 32.
If ϕ is a closed formula in LΠ 1 k (Q), 2 ≤ k < n, p ∈ Q , and all names in ϕ are Q-full below p, then there is a condition q ∈ Q , q p, such that either q forc ∞ ϕ, or q forc ∞ ϕ ¬ .

Proof.
As the names considered are ω 1 -sizeobjects, there is an ordinal η < ω 2 such that p ∈ Q η , and all names in ϕ belong to M η ∩ SN ω ω (Q η ); then all names in ϕ are Q η -full below p, of course. As k < n, the set D of all pairs M, U ∈ sJS that extend M η , U η and there is a condition q ∈ Q[U], q p, satisfying q forc M U ϕ ¬ , belongs to Σ HC n−2 by Lemma 26. Therefore, by the n-completeness of the We have two cases.
Then there is a condition q ∈ K[U ζ ], q p, satisfying q forc Now we prove another key lemma which connects, in a forcing-style way, the relation forc ∞ and the truth in Q -generic extensions of L, up to the level of Σ 1 n formulas.

Lemma 33.
Suppose that ϕ is a formula in LΠ 1 k (Q) ∪ LΣ 1 k+1 (Q), 1 ≤ k < n, and all names in ϕ are Q-full. Let G ⊆ Q be Q-generic over L. Then ϕ [G] is true in L[G] iff there is a condition p ∈ G such that p forc ∞ ϕ.

Proof.
We proceed by induction and begin with the case of LΠ 1 1 formulas. Consider a closed formula ϕ in LΠ 1 1 (Q). As names in the formulas considered are ω 1 -sizenames in SN ω ω (Q), there is an ordinal ξ < ω 2 such that ϕ is a LΠ 1 1 (Q ξ ) formula. Please note that since G ⊆ P is Q-generic over L, the smaller set G ξ = G ∩ Q ξ is Q ξ -generic over M ξ by Corollary 2 in Section 4. iff there is p ∈ G ξ which Q ξ -forces ϕ over M ξ , iff ∃ p ∈ G ξ (p forc ξ ϕ) by (F2) in Section 5.2, easily getting the result required since ξ is arbitrary.
The step from LΣ 1 k to LΠ 1 k , k ≥ 2. Prove the theorem for a LΠ 1 k (Q) formula ϕ, assuming that the result holds for ϕ ¬ . Suppose that ϕ [G] is false in L[G]. Then ϕ ¬ [G] is true, and hence by the inductive hypothesis, there is a condition p ∈ G c such that p forc ∞ ϕ ¬ . Then it follows from (I) and (II) above that q forc ∞ ϕ fails for all q ∈ G .
Conversely let p forc ∞ ϕ fail for all p ∈ G . Then by Lemma 32 there exists q ∈ G satisfying q forc ∞ ϕ ¬ . It follows that ϕ ¬ [G] is true by the inductive hypothesis, therefore ϕ[G] is false.
If conversely (∃ B x ϕ(x))[G] is true, then by Lemma 11 there is a Q-full name τ ∈ SN ω ω (Q) B such that ϕ(τ)[G] is true. Then, by the inductive hypothesis, there is a condition p ∈ G such that p forc ∞ ϕ(τ). Therefore p forc ∞ ∃ B x ϕ(x) by the choice of τ .
The case of ∃ x ϕ(x) is treated similarly.

The Elementary Equivalence Theorem
We begin the proof of Theorem 4 of Section 4.4, so let d, Z, Z , x 0 be as in the theorem.
We also assume w. l. o. g. that the sets Z, Z satisfy the requirement that Z Z and Z Z are infinite (countable) sets. Indeed, otherwise, under the assumptions of Theorem 4, one easily defines a third set Z such that each of the pairs Z, Z and Z , Z still satisfies the assumptions of the theorem, and in addition, all four sets Z Z , Z Z , Z Z and Z Z are infinite. Please note that this argument necessarily requires that the complementary set I (d ∪ Z ∪ Z ) is infinite.
Step 2. We are going to reorganize the quantifier prefix of ϕ, in particular, by assigning the indices Z and Z to certain quantifiers, to reflect the relativization to classes L[x 0 , G Z] and L[x 0 , G Z ]. This is not an easy task because generally speaking there is no set Z 0 ⊆ I in L satisfying L[x 0 ] = L[G Z 0 ]. However, nevertheless we will define an LΣ 1 n formula, say ψ Z (v), and then ψ Z (v) by the substitution of Z for Z , such that the following will hold: To explain this transformation, assume that n = 4 for the sake of brevity, and hence ϕ(v) has the form ∃ x ∀ y ϑ(v, x, y), where ϑ is a Σ 1 2 formula. To begin with, we define and define ψ Z 1 (v) accordingly.
Thus, the formulas ψ Z 1 , ψ Z 1 do satisfy (A), but they are not LΣ 1 n formulas as defined in Section 5.1, of course. It will take some effort to convert them to a LΣ 1 n form. We must recall some instrumentarium known in Gödel's theory of constructability of reals.
is unknown whether this result generalizes to higher classes ∆ 1 n , n ≥ 4, or ∆ 1 ∞ , and whether it can be strengthened towards ⇐⇒ instead of =⇒ . This is a very interesting and perhaps difficult question. 2. Another question to be mentioned here is the following. Please note that in any extension of L satisfying Theorem 1, it is true that every universal Σ 1 n+1 set u ⊆ ω × ω is by necessity Σ 1 n+1 but non-∆ 1 n+1 , and hence nonconstructible. This gives another proof of Theorem 3 in [7]. (It claims, for any n ≥ 2, the existence of a generic extension of L in which there is a nonconstructible Σ 1 n+1 set a ⊆ ω whereas all ∆ 1 n+1 sets are constructible.) And the problem is, given n ≥ 2, to find a model in which all ∆ 1 n+1 reals are constructible, but there exists a Σ 1 n+1 nonconstructible real u ⊆ ω , which satisfies V = L[u].