HISTORIC FORCING FOR Depth ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH

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1 HISTORIC FORCING FOR Depth ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH Abstract. We show that, consistently, for some regular cardinals θ <, there exists a Boolean algebra B such that B = + and for every subalgebra B B of size + we have Depth(B ) = θ. 0. Introduction The present paper is concerned with forcing a Boolean algebra which has some prescribed properties of Depth. Let us recall that, for a Boolean algebra B, its depth is defined as follows: Depth(B) = sup{ X : X B is well-ordered by the Boolean ordering }, Depth + (B) = sup{ X + : X B is well-ordered by the Boolean ordering }. (Depth + (B)isusedtodealwithattainmentpropertiesinthedefinitionofDepth(B), see e.g. [5, 1].) The depth (of Boolean algebras) is among cardinal functions that have more algebraic origins, and their relations to topological fellows is often indirect, though sometimes very surprising. For example, if we define Depth H+ (B) = sup{depth(b/i) : I is an ideal in B }, then for any (infinite) Boolean algebra B we will have that Depth H+ (B) is the tightness t(b) of the algebra B (or the tightness of the topological space Ult(B) of ultrafiltersonb), see[3, Theorem4.21]. AsomewhatsimilarfunctiontoDepth H+ is obtained by taking sup{depth(b ) : B is a subalgebraofb }, but clearlythis brings nothing new: it is the old Depth. But if one wants to understand the behaviour of the depth for subalgebras of the considered Boolean algebra, then looking at the following subalgebra Depth relation may be very appropriate: Depth Sr (B) = {(κ,µ) : there is an infinite subalgebra B of B such that B = µ and Depth(B ) = κ }. A number of results related to this relation is presented by Monk in [3, Chapter 4]. There he asks if there are a Boolean algebra B and an infinite cardinal θ such that (θ,(2 θ ) + ) Depth Sr (B), while (ω,(2 θ ) + ) / Depth Sr (B) (see Monk [3, Problem 14]; we refer the reader to Chapter 4 of Monk s book [3] for the motivation and background of this problem). Here we will partially answer this question, showing that it is consistent that there is such B and θ. The question if that can be done in ZFC remains open Mathematics Subject Classification. Primary 03E35, 03G05; Secondary 03E05, 06Exx. Key words and phrases. Boolean algebras, depth, historic forcing. The first author thanks the KBN (Polish Committee of Scientific Research) for partial support through grant 2 P03 A The research of the second author was partially supported by the Israel Science Foundation. Publication

2 2 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH Our consistency result is obtained by forcing, and the construction of the required forcing notion is interesting per se. We use the method of historic forcing which was first applied in Shelah and Stanley [9]. The reader familiar with [9] will notice several correspondences between the construction here and the method used there. However, we do not relay on that paper and our presentation here is self-contained. Let us describe how our historic forcing notion is built. So, we fix two (regular) cardinals θ, and our aim is to force a Boolean algebra Ḃθ such that Ḃθ = + and for every subalgebra B Ḃθ of size + we have Depth(B) = θ. The algebra Ḃ θ will be generated by x i : i U for some set U +. A condition p will be an approximation to the algebra Ḃθ, it will carry the information on what is the subalgebra B p = x i : i u p Ḃθ for some u p +. A natural way to describe algebras in this context is by listing ultrafilters (or: homomorphisms into {0, 1}): Definition 1. For a set w and a family F 2 w we define cl(f) = {g 2 w : ( u [w] < ω )( f F)(f u = g u)}, B (w,f) is the Boolean algebra generated freely by {x α : α w} except that if u 0,u 1 [w] < ω and there is no f F such that f u 0 0, f u 1 1 then x α ( x α ) = 0. α u 1 α u 0 This description of algebras is easy to handle, for example: Proposition 2 (see [6, 2.6]). Let F 2 w. Then: (1) Each f F extends (uniquely) to a homomorphism from B (w,f) to {0,1} (i.e. it preserves the equalities from the definition of B (w,f) ). If F is closed, then every homomorphism from B (w,f) to {0,1}extends exactly one element of F. (2) If τ(y 0,...,y l ) is a Boolean term and α 0,...,α l w are distinct then B (w,f) = τ(x α0,...,x αl ) 0 if and only if ( f F)({0,1} = τ(f(α 0 ),...,f(α k )) = 1). (3) If w w, F 2 w and ( f F)( g F )(f g) and ( g F )(g w cl(f)) then B (w,f) is a subalgebra of B (w,f ). So each condition p in our forcing notion P θ will have a set up [ + ] < and a closedsetf p 2 up (andtherespectivealgebrawillbe B p = B (u p,f p )). Buttomake the forcing notion work, we will have to put more restrictions on our conditions, and wewill be takingonlythoseconditionsthat haveto be takentomakethe arguments work. For example, we want that cardinals are not collapsed by our forcing, and demanding that P θ is + -cc (and somewhat (<) closed) is natural in this context. How do we arguethat a forcing notion is + cc? Typically we start with a sequence of + distinctconditions, wecarryoutsome cleaningprocedure (usuallyinvolving the lemma etc), and we end up with (at least two) conditions that can be put together. Putting together two (or more) conditions that are approximations to a Boolean algebra means amalgamating them. There are various ways to amalgamate conditions - we will pick one that will work for several purposes. Then, once we declare that some conditions forming a clean sequence of length θ are in P θ, we will be bound to declare that the amalgamation is in our forcing notion. The

3 HISTORIC FORCING FOR Depth 3 amalgamation (and natural limits) will be the only way to build new conditions from the old ones, but the description above still misses an important factor. So far, a condition does not have to know what are the reasons for it to be called to P θ. This information is the history of the condition and it will be encoded by two functions h p,g p. (Actually, these functions will give histories of all elements of u p describing why and how those points were incorporated to u p. Thus both functions will be defined on u p ht(p), were ht(p) is the height of the condition p, that is the step in our construction at which the condition p is created.) We will also want that our forcing is suitably closed, and getting (<) strategically closed would be fine. To make that happen we will have to deal with two relations on P θ : pr and. The first ( pure ) is (<) closed and it will help in getting the strategic closure of the second (main) one. In some sense, the relation pr represents the official line in history, and sometimes we will have to rewrite that official history, see Definition 6 and Lemma 7 (on changing history see also Orwell [4]). The forcing notion P θ has some other interesting features. (For example, conditions are very much like fractals, they contain many self-similar pieces (see Definition 10 and Lemma 11).) The method of historic forcing notions could be applicable to more problems, and this is why in our presentation we separated several observations of general character (presented in the first section) from the problem specific arguments (section 2) Notation: Our notation is standard and compatible with that of classical textbooks on set theory (like Jech [1]) and Boolean algebras (like Monk [2], [3]). However in forcing considerations we keep the older tradition that the stronger condition is the greater one. Let us list some of our notation and conventions. (1) Throughout the paper, θ, are fixed regular infinite cardinals, θ <. (2) A name for an object in a forcing extension is denoted with a dot above (like Ẋ) with one exception: the canonical name for a generic filter in a forcing notion P will be called Γ P. For a P name Ẋ and a P generic filter G over V, the interpretation of the name Ẋ by G is denoted by ẊG. (3) i,j,α,β,γ,δ,... will denote ordinals. (4) For a set X and a cardinal, [X] < stands for the family of all subsets of X of size less than. The family of all functions from Y to X is called X Y. If X is a set of ordinals then its order type is denoted by otp(x). (5) In Boolean algebras we use (and ), (and ) and for the Boolean operations. If B is a Boolean algebra, x B then x 0 = x, x 1 = x. (6) For a subset Y of an algebra B, the subalgebra of B generated by Y is denoted by Y B. We would like to thank the referee for valuable com- Acknowledgements: ments and suggestions. 1. The forcing and its basic properties Let us start with the definition of the forcing notion P θ. By induction on α < we will define sets of conditions Pα θ,, and for each p Pθ, α we will define u p,f p,ht(p),h p and g p. Also we will define relations α and α pr on Pα θ,. Our inductive requirements are:

4 4 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH (i) α for each p Pα θ,: u p [ + ] <, ht(p) α, F p 2 up is a non-empty closed set, g p is a function with domain dom(g p ) = u p ht(p) and values of the form (l,τ), where l < 2 and τ is a Boolean term, and h p : u p ht(p) θ + 2 is a function, (ii) α α, α pr are transitive and reflexive relations on Pθ, α, and α extends α pr, (iii) α if p,q Pα θ,, p α q, then u p u q, ht(p) ht(q), and F p = {f u p : f F q }, and if p α pr q, then for every i u p and ξ < ht(p) we have h p (i,ξ) = h q (i,ξ) and g p (i,ξ) = g q (i,ξ), (iv) α if β < α then P θ, β P θ, α, and α pr extends β pr, and α extends β. For a condition p P θ, α, we will also declare that B p = B (u p,f p ) (the Boolean algebra defined in Definition 1). We define P θ, 0 = { ξ : ξ < + } and for p = ξ we let F p = 2 {ξ}, ht(p) = 0 and h p = = g p. The relations 0 pr and 0 both are the equality. [Clearly these objects are as declared, i.e, clauses (i) 0 (iv) 0 hold true.] If γ < is a limit ordinal, then we put Pγ = { p ξ : ξ < γ : ( ξ < ζ < γ)(p ξ P θ, ξ & ht(p ξ ) = ξ & p ξ ζ pr p ζ) }, Pγ θ, = Pα θ, Pγ, α<γ and for p = p ξ : ξ < γ P γ we let u p = ξ<γu p ξ, F p = {f 2 up : ( ξ < γ)(f u p ξ F p ξ )}, ht(p) = γ and h p = ξ<γ h p ξ and g p = ξ<γ g p ξ. We define γ and γ pr by: p γ pr q if and only if either p,q Pα θ,, α < γ and p α pr q, or q = q ξ : ξ < γ Pγ, p Pα θ, and p α pr q α for some α < γ, or p = q; p γ q if and only if either p,q Pα θ,, α < γ and p α q, or q = q ξ : ξ < γ, p Pθ, α and p α q α for some α < γ, or p = p ξ : ξ < γ Pγ, q = q ξ : ξ < γ Pγ and ( δ < γ)( ξ < γ)(δ ξ p ξ ξ q ξ ). [It is straightforward to show that clauses (i) γ (iv) γ hold true.] Suppose now that α <. Let Pα+1 consist of all tuples such that for each ξ 0 < ξ 1 < θ: ζ,τ,n,u, p ξ,v ξ : ξ < θ (α) ζ < θ, n < ω, τ = τ (y 1,...,y n ) is a Boolean term, u [ + ] <, (β) p ξ0 Pα θ,, ht(p) = α, v ξ0 [u p ξ 0 ] n, (γ) the family {u p ξ : ξ < θ} forms a system with heart u and u p ξ 0 \u and sup(u ) < min(u p ξ 0 \u ) sup(u p ξ 0 \u ) < min(u p ξ 1 \u ),

5 HISTORIC FORCING FOR Depth 5 (δ) otp(u p ξ 0 ) = otp(u p ξ 1 ) and if H : u p ξ 0 u p ξ 1 is the order isomorphism then H u is the identity on u, F p ξ 0 = {f H : f F p ξ 1 }, H[v ξ0 ] = v ξ1 and ( j u p ξ 0)( β < α)(h p ξ0(j,β) = h p ξ1(h(j),β) & g p ξ0(j,β) = g p ξ1(h(j),β)). We put P θ, α+1 = Pθ, α P α+1 and for p = ζ,τ,n,u, p ξ,v ξ : ξ < θ Pα+1 we let u p = u p ξ and ξ<θ F p = {f 2 up : ( ξ < θ)(f u p ξ F p ξ ) and for all ξ < ζ < θ f(σ maj (τ 3 ξ,τ 3 ξ+1,τ 3 ξ+2 )) f(σ maj (τ 3 ζ,τ 3 ζ+1,τ 3 ζ+2 ))}, where τ ξ = τ (x i : i v ξ ) for ξ < θ (so τ ξ is an element of the algebra B p ξ = B (u p ξ,f p ξ) ), and σ maj (y 0,y 1,y 2 ) = (y 0 y 1 ) (y 0 y 2 ) (y 1 y 2 ). Next we let ht(p) = α+1 and we define functions h p,g p on u p (α+1) by h p (j,β) = g p (j,β) = Next we define the relations α+1 pr h p ξ (j,β) if j u p ξ, ξ < θ, β < α, θ if j u, β = α, θ +1 if j u p ζ \u, β = α, ξ if j u p ξ \u, ξ < θ, ξ ζ, β = α, g p ξ (j,β) if j u p ξ, ξ < θ, β < α, (1,τ ) if j v ξ, ξ < θ, β = α, (0,τ ) if j u p ξ \v ξ, ξ < θ, β = α. and α+1 by: p α+1 pr q if and only if either p,q Pα θ, and p α pr q, or q = ζ,τ,n,u, q ξ,v ξ : ξ < θ Pα+1, p Pθ, α, and p α pr q ζ, or p = q; p α+1 q if and only if either p,q Pα θ, and p α q, or q = ζ,τ,n,u, q ξ,v ξ : ξ < θ Pα+1, p Pθ, α, and p α q ξ for some ξ < θ, or p = ζ,τ,n,u, p ξ,v ξ : ξ < θ, q = ζ,τ,n,u, q ξ,v ξ : ξ < θ are from Pα+1 and ( ξ < θ)(p ξ α q ξ & u p ξ = u q ξ ). [Again, it is easy to show that clauses (i) α+1 (iv) α+1 are satisfied.] After the construction is carried out we let P θ = Pα θ, and pr = α pr and = α< α< α< α. One easily checks that pr is a partial order on P θ and that the relation is transitive and reflexive, and that pr. Lemma 3. Let p,q P θ. (1) If p q then ht(p) ht(q), u p u q and F p = {f u p : f F q } (so B p is a subalgebra of B q ). If p q and ht(p) = ht(q), then q p. (2) For each j u p, the set {β < ht(p) : h p (j,β) < θ} is finite. (3) If p pr q and i u p, then h q (i,β) θ for all β such that ht(p) β < ht(q).

6 6 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH (4) If i,j u p are distinct, then there is β < ht(p) such that θ h p (i,β) h p (j,β) θ. (5) For each finite set X ht(p) there is i u p such that {β < ht(p) : h p (i,β) < θ} = X. (6) If p pr q then there is a pr increasing sequence p ξ : ξ ht(p) P θ such that p ht(p) = p, p ht(q) = q and ht(p ξ ) = ξ (for ξ ht(p)). (In particular, if p pr q and ht(p) = ht(q) then p = q.) (7) If ht(p) = γ is a limit ordinal, p = p ξ : ξ < γ, then for each i u p and ξ < γ: i u p ξ if and only if ( ζ < γ)(ξ ζ h p (i,ζ) θ). Proof. 1) Should be clear (an easy induction). 2) Suppose that p P θ and j up are a counterexample with the minimal possible value of ht(p). Necessarily ht(p) is a limit ordinal, p = p ξ : ξ < ht(p), ht(p ξ ) = ξ and ζ < ξ < ht(p) p ζ pr p ξ. Let ξ < ht(p) be the first ordinal such that j u p ξ. By the choice of p, the set {β ξ : h p (j,β) < θ} is finite, but clearly h p (j,β) θ for all β (ξ,ht(p)). 3) An easy induction on ht(q) (with fixed p). 4) We show this by induction on ht(p). Suppose that ht(p) = α + 1, so p = ζ,τ,n,u, p ξ,v ξ : ξ < θ, and i,j u p are distinct. If i,j u p ξ for some ξ < θ, then by the inductive hypothesis we find β < α such that θ h p (i,β) = h p ξ (i,β) h p ξ (j,β) = h p (j,β) θ. If i u p ξ \ u, j u p ζ \ u and ξ,ζ < θ are distinct, then look at the definition of h p (i,α), h p (j,α) these two values cannot be equal (and both are distinct from θ). Finally suppose that ht(p) is limit, so p = p ξ : ξ < ht(p). Take ξ < ht(p) such that i,j u p ξ and apply the inductive hypothesis to p ξ getting β < ξ such that h p (i,β) h p (j,β) (and both are not θ). 5) Again, it goes by induction on ht(p). First consider a limit stage, and suppose that ht(p) = γ is a limit ordinal, X [γ] <ω and p = p ξ : ξ < γ. Let ξ < γ be such that X ξ. By the inductive hypothesis we find i u p ξ such that {β < ξ : h p (i,β) < θ} = X. Applying clause (3) we may conclude that this i is as required. Nowconsiderasuccessorcaseht(p) = α+1. Let p = ζ,τ,n,u, p ξ,v ξ : ξ < θ, andletξ < θ beζ ifα X,andbeζ +1otherwise. Applytheinductivehypothesis to p ξ and X α to get suitable i u p ξ, and note that this i works for p and X too. 6), 7) Straightforward. Definition 4. We say that conditions p,q P θ are isomorphic if ht(p) = ht(q), otp(u p ) = otp(u q ), and if H : u p u q is the order isomorphism, then for every β < ht(p) ( j u p )(h p (j,β) = h q (H(j),β) & g p (j,β) = g p (H(j),β)). [In this situation we may say that H is the isomorphism from p to q.] Lemma 5. Suppose that q 0,q 1 P θ are isomorphic conditions and H is the isomorphism from q 0 to q 1.

7 HISTORIC FORCING FOR Depth 7 (1) If ht(q 0 ) = ht(q 1 ) = γ is a limit ordinal, q l = qξ l : ξ < γ (for l < 2), then H u q0 ξ is an isomorphism from q 0 ξ to qξ 1. (2) If ht(q 0 ) = ht(q 1 ) = α+1, α <, and q l = ζl,τ l,n l,u l, ql ξ,vl ξ : ξ < θ (for l < 2), then ζ0 = ζ1, τ0 = τ1, n 0 = n 1, H u q0 ξ is an isomorphism from qξ 0 to q1 ξ and H[v0 ξ ] = v1 ξ (for ξ < θ). (3) F q0 = {f H : f F q1 }. (4) Assume p 0 q 0. Then there is a unique condition p 1 q 1 such that H u p0 is the isomorphism from p 0 to p 1. [The condition p 1 will be called H(p 0 ).] Proof. 1), 2) Straightforward (for (1) use Lemma 3(7)). 3), 4) Easy inductions on ht(q 0 ) using (1), (2) above. Definition 6. By induction on α <, for conditions p,q P θ, α such that p α q, we define the p transformation T p (q) of q. If α = 0 (so necessarily p = q) then T p (q) = p. Assume that ht(q) = α+1, q = ζ,τ,n,u, q ξ,v ξ : ξ < θ. If p q ξ for some ξ < θ, then let ξ be such that p q ξ. Next for ξ < θ let q ξ = T H ξ,ξ (p)(q ξ ), where H ξ,ξ is the isomorphism from q ξ to q ξ. Define T p (q) = ξ,τ,n,u, q ξ,v ξ : ξ < θ. Suppose now that p = ζ,τ,n,u, p ξ,v ξ : ξ < θ and u p ξ = u q ξ, p ξ q ξ (forξ < θ). Letq ξ = T p ξ (q ξ )andputt p (q) = ζ,τ,n,u, q ξ,v ξ : ξ < θ. Assume now that ht(q) is a limit ordinal and q = q ξ : ξ < ht(q). If ht(p) < ht(q) then p q ε for some ε < ht(q), and we may choose q ξ (for ξ < ht(q)) such that ht(q ξ ) = ξ, ξ < ξ < ht(q) q ξ pr q ξ, and q ζ = T p(q ζ ) for ζ [ε,ht(q)). Next we let T p (q) = q ζ : ζ < θ. If ht(p) = ht(q), p = p ξ : ξ < ht(p) and p ξ q ξ for ξ > δ (for some δ < ht(p)) then we define T p (q) = p. To show that the definition of T p (q) is correct one proves inductively (parallely to the definition of the p transformation of q) the following facts. Lemma 7. Assume p,q P θ, p q. Then: (1) T p (q) P θ, utp(q) = u q, ht(t p (q)) = ht(q), (2) p pr T p (q) q T p (q), (3) ht(p) = ht(q) T p (q) = p, (4) if q P θ is isomorphic to q and H : uq u q is the isomorphism from q to q, then H is the isomorphism from T p (q) to T H(p) (q ), (5) if q pr q then T p (q) pr T p (q ). Proposition 8. Every pr increasing chain in P θ of length < has a pr upper bound, that is the partial order (P θ, pr) is (< ) closed. Let us recall that a forcing notion(q, ) is (<) strategically closed if the second player has a winning strategy in the following game (Q). The game (Q) lasts moves. The first player starts with choosing a condition p Q. Later, in her i th move, the first player chooses an open dense subset D i of Q. The second player (in his i th move) picks a condition p i Q so that p 0 p,

8 8 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH p i D i and p i p j for all j < i. The second player looses the play if for some i < he has no legal move. It should be clear that (<) strategically closed forcing notions do not add sequences of ordinals of length less than. The reader interested in this kind of properties of forcing notions and iterating them is referred to [7], [8]. Proposition 9. Assume that θ < are regular cardinals, < =. Then (P θ, ) is a (< ) strategically closed + cc forcing notion. Proof. It follows from Lemma 7(2) that if D P θ is an open dense set, p Pθ, then there is a condition q D such that p pr q. Therefore, to win the game (P θ ), the second player can play so that the conditions p i that he chooses are pr increasing,andthustherearenoproblemswithfinding pr bounds(remember Proposition 8). Now, to show that P θ is + cc, suppose that p δ : δ < + is a sequence of distinct conditions from P θ. We may find a set A [+ ] + such that conditions {p δ : δ A} are pairwise isomorphic, the family {u p δ : δ A} forms a system with heart u, if δ 0 < δ 1 are from A then sup(u ) < min(u p δ 0 \u ) sup(u p δ 0 \u ) < min(u p δ 0 \u ). Take an increasing sequence δ ξ : ξ < θ of elements of A, let τ = 1, v ξ = (for ξ < θ), and look at p = 0,τ,0,u, p δξ,v ξ : ξ < θ. It is a condition in P θ stronger than all p δξ s. Definition 10. By induction on ht(p) we define α components of p (for p P θ, α ht(p)). First we declare that the only ht(p) component of p is the p itself. If ht(p) = β + 1, p = ζ,τ,n,u, p ξ,v ξ : ξ < θ and α = β, then α components of p are p ξ (for ξ < θ); if α < β, then α components of p are those q which are α components of p ξ for some ξ < θ. If ht(p) is a limit ordinal, p = p ξ : ξ < ht(p) and α < ht(p), then α components of p are α components of p ξ for ξ [α,ht(p)). Lemma 11. Assume p P θ and α < ht(p). (1) If q is an α component of p then q p, ht(q) = α, and for all j 0,j 1 u q and every β [α,ht(p)): h p (j 0,β) θ & h p (j 1,β) θ h p (j 0,β) = h p (j 1,β). Moreover, for each i u p there is a unique α component q of p such that i u q and ( j u q )( β [α,ht(p)))(h p (i,β) θ h p (j,β) θ). (2) If H is an isomorphism from p onto p P θ, and q is an α component of p, then H(q) is an α component of p. If q 0,q 1 are α components of p then q 0,q 1 are isomorphic. (3) There is a unique α component q of p such that q pr p. Proof. Easy inductions on ht(p). Definition 12. By induction on ht(p) we define when a set Z is p closed for a condition p P θ.

9 HISTORIC FORCING FOR Depth 9 If ht(p) = 0 then every Z is p closed; if ht(p) is limit, p = p ξ : ξ < ht(p), then Z is p closed provided it is p ξ closed for each ξ < ht(p); if ht(p) = α+1, p = ζ,τ,n,u, p ξ,v ξ : ξ < θ and α / Z, then Z is p closed whenever it is p ζ closed; if ht(p) = α+1, p = ζ,τ,n,u, p ξ,v ξ : ξ < θ and α Z, then Z is p closed provided it is p ζ closed and {β < α : ( j v ζ {min(u p ζ \u )})(h p ζ (j,β) < θ)} Z. Lemma 13. (1) If p P θ and w [ht(p)]< ω, then there is a finite p closed set Z ht(p) such that w Z. (2) If p,q P θ are isomorphic and Z is p closed, then Z is q closed. If Z is p closed, α < ht(p) and p is an α component of p, then Z α is p closed. Proof. Easy inductions on ht(p) (remember Lemma 3(2)). Definition 14. Suppose that p P θ and Z ht(p) is a finite p closed set. Let Z = {α 0,...,α k 1 } be the increasing enumeration. (1) We define (2) We let U[p,Z] def = {j u p : ( β < ht(p))(h p (j,β) < θ β Z)}. Υ p (Z) = ζ l,τ l,n l, g l,h l 0,...,h l n l 1 : l < k, where, for l < k, ζ l is an ordinal below θ, τ l is a Boolean term, n l < ω and g l,h l 0,...,h l n l 1 : l 2, andthey allaresuchthatforevery(equivalently: some) α l +1 component q = ζ,τ,n,u, q ξ,v ξ : ξ < θ of p we have: ζ l = ζ, τ l = τ, n l = n and if v ξ = {j 0,...,j nl 1} (the increasing enumeration) then ( m < n l )( l < l)(h l m(l ) = h q (j m,α l )), and if i 0 = min(u q ζ \u ) then ( l < l)(g l (l ) = h q (i 0,α l )). (Note that ζ l,τ l,n l, g l,h l 0,...,h l n l 1 are well-defined by Lemma 11. Necessarily, for all m < n l and β α l \Z we have h q (i 0,β),h q (j m,β) θ; remember that Z is p closed.) Note that if Z ht(p) is a finite p closed set, α = max(z) and p is the α+1 component of p satisfying p pr p (see 11(3)), then U[p,Z] u p. Lemma 15. Suppose that p P θ and Z 0,Z 1 ht(p) are finite p closed sets such that Υ p (Z 0 ) = Υ p (Z 1 ). Then otp(u[p,z 0 ]) = otp(u[p,z 1 ]), and the order preserving isomorphism π : U[p,Z 0 ] U[p,Z 1 ] satisfies ( ) ( l < k)(h p (i,α 0 l ) = hp (π(i),α 1 l )), where {α x 0,...,αx k 1 } is the increasing enumeration of Z x (for x = 0,1). Proof. We prove this by induction on Z 0 = Z 1 (for all p,z 0,Z 1 satisfying the assumptions). Step Z 0 = Z 1 = 1; Z 0 = {α 0 0}, Z 1 = {α 1 0}. Take the α x component q x of p such that q x pr p. Then, for x = 0,1, q x = ζ,τ,n,u x, q x ξ,vx ξ : ξ < θ, and for each i vx ξ, β < αx 0 we have hqx ξ (i,β) θ.

10 10 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH Also, if i x 0 = min(uqx ζ \ u x ) and β < α x 0, then hqx ζ (i x 0,β) θ. Consequently, n = vξ x 1, and if n = 1 then {ix 0} = vζ x (remember Lemma 3(4)). Moreover, U[p,Z x ] = U[q x,z x ] = {H x ξ,ζ (ix 0 ) : ξ < θ}, where Hξ,ζ x is the isomorphism from qx ζ to qx ξ. Now it should be clear that the mapping π : Hξ,ζ 0 (i0 0 ) H1 ξ,ζ (i1 0 ) : U[p,Z 0] U[p,Z 1 ] is the order preserving isomorphism(rememberclause(γ)ofthedefinitionofpα+1), andithastheproperty described in ( ). Step Z 0 = Z 1 = k +1; Z 0 = {α 0 0,...,α0 k }, Z 1 = {α 1 0,...,α1 k }. Let Υ p (Z 0 ) = Υ p (Z 1 ) = ζ l,τ l,n l, g l,h l 0,...,hl n l 1 : l k. For x = 0,1, let q x = ζ,τ,n,u x, qξ x,vx ξ : ξ < θ be the αx k + 1 component of p such that q x pr p. The sets Z x α x k (for x = 0,1) are qx ξ closed for every ξ < θ, and clearly Υ p (Z 0 α 0 k ) = Υ p(z 1 α 1 k ). Hence, by the inductive hypothesis, otp(u[qξ 0,Z 0 \ {α 0 k }]) = otp(u[q1 ξ,z 1 \ {α 1 k }]) (for each ξ < θ), and the order preservingmappings π ξ : U[qξ 0,Z 0\{α 0 k }] U[q1 ξ,z 1\{α 1 k }] satisfy the demand in ( ). Leti x ξ = min(uqx ξ \u x ). Then, asqξ x andqx ζ areisomorphicandtheisomorphism is the identity on u x, we have ( l < k)(h p (i x ξ,αx l ) = g k(l)). Hence π ξ (i 0 ξ ) = i1 ξ, and therefore π ξ [u 0 U[qξ 0,Z 0\{α 0 k }]] = u1 U[qξ 1,Z 1\{α 1 k }]. But since the mappings π ξ are order preserving, the last equality implies that π ξ (u 0 U[qξ 0,Z 0\{α 0 k }]) = π ζ (u 0 U[qζ 0,Z 0 \{α 0 k }]), and hence π = π ξ is a function, and it is an order isomorphism from U[q 0,Z 0 ] = U[p,Z 0 ] onto U[q 1,Z 1 ] = U[p,Z 1 ] satisfying ( ). ξ<θ 2. The algebra and why it is OK (in V Pθ ) Let Ḃθ and U be P θ names such that P θ Ḃθ = {B p : p Γ P θ } and P θ U = {u p : p Γ P θ }. Note that U is (a name for) a subset of +. Let F be a P θ name such that P θ F = {f 2 U : ( p ΓP θ )(f u p F p )}. Proposition 16. Assume θ < are regular, < =. Then in V Pθ : (1) F is a non-empty closed subset of 2 U, and Ḃ θ is the Boolean algebra generated B ( U,Ḟ) (see Definition 1); (2) U = Ḃθ = + ; (3) For every subalgebra B Ḃθ of size + we have Depth + (B) > θ. Proof. 2) Note that if p P θ, sup(up ) < j < + then there is a condition q p such that j u q. Hence U = +. To show that, in V Pθ, the algebra Ḃ θ is of size + it is enough to prove the following claim. Claim Let p P θ, j up. Then x j / x i : i j u p B p. Proof of the claim. Suppose not, and let p,j be a counterexamplewith the smallest possible ht(p). Necessarily, ht(p) is a successor ordinal, say ht(p) = α +1. So let p = ζ,τ,n,u, p ξ,v ξ : ξ < θ and suppose that v [u p j] < ω is such that x j x i : i v B p. If j u then v u and we immediately get a

11 HISTORIC FORCING FOR Depth 11 contradiction (applying the inductive hypothesis to p ζ ). So let ξ < θ be such that j u p ξ \ u. We know that x j / x i : i u (v u p ξ ) B p ξ (remember clause (γ) of the definition of Pα+1 ), so we may take functions f 0,f 1 F p ξ such that f 0 (u (v u p ξ )) = f 1 (u (v u p ξ )), f 0 (j) = 0, f 1 (j) = 1. Let g 0,g 1 : u p 2 be such that g l u p ξ = f l, g l u p ζ = f 0 H ζ,ξ for ζ ξ (where H ζ,ξ is the order isomorphism from u p ζ to u p ξ ). Now one easily checks that g 0,g 1 F p (remember the definition of the term σ maj ). By our choices, g 0 (i) = g 1 (i) for all i v, and g 0 (j) g 1 (j), and this is a clear contradiction with the choice of i and v. 3) Supposethat ȧ ξ : ξ < + isap θ namefora+ sequenceofdistinct members of Ḃθ and let p Pθ. Applying standard cleaning procedures we find a set A + of the order type θ, an ordinal α < and τ,n,u and p ξ,v ξ : ξ A such that p p ξ, ht(p ξ ) = α, p ξ ȧ ξ = τ (x i : i v ξ ) and q def = 0,τ,n,u, p ξ,v ξ : ξ A P α+1, where A is identified with θ by the increasing enumeration (so we will think A = θ). For ξ < θ let τ ξ = τ (x i : i v ξ ) B p ξ. Since ȧ ξ were (forced to be) distinct we know that B q = τ ξ τ ζ for distinct ξ,ζ. Hence τ ξ / x i : i u B p ξ (for each ξ) and therefore we may find functions fξ 0,f1 ξ Fp ξ such that fξ 0 u = fξ 1 u, and fξ 0(τ ξ) = 0, fξ 1(τ ξ) = 1, and if ξ < ζ < θ, and H ξ,ζ is the isomorphism from p ξ to p ζ, then fξ l = fl ζ H ξ,ζ. Now fix ξ < ζ < θ and let g def = fα 0 fα 1. α 3 ξ+2 3 ξ+2<α<θ It should be clear that g is a function from u q to 2, and moreover g F q. Also easily g(σ maj (τ 3 ξ,τ 3 ξ+1,τ 3 ξ+2 )) = 0 and g(σ maj (τ 3 ζ,τ 3 ζ+1,τ 3 ζ+2 ))} = 1. Hence we may conclude that B q = σ maj (τ 3 ξ,τ 3 ξ+1,τ 3 ξ+2 ) < σ maj (τ 3 ζ,τ 3 ζ+1,τ 3 ζ+2 ) for ξ < ζ < θ (remember the definition of F q and Proposition 2). Consequently we get q Depth + ( ȧ ξ : ξ < + Ḃθ ) > θ, finishing the proof. Theorem 17. Assume θ < are regular, = <. Then P θ Depth(Ḃθ ) = θ. Proof. By Proposition 16 we know that Depth + (Ḃθ ) > θ, so what we have to show is that there are no increasing sequences of length θ + of elements of Ḃθ. We will show this under an additional assumption that θ + < (after the proof is carried out, it will be clear how one modifies it to deal with the case = θ + ). Due to this additional assumption, and since the forcing notion P θ is (< ) strategically closed (by Proposition 9), it is enough to show that Depth(B p ) θ for each p P θ. So suppose that p P θ is such that Depth(Bp ) θ +. Then we find a Boolean term τ, an integer n and sets w ρ [u p ] n (for ρ < θ + ) such that ρ 0 < < θ + B p = τ(x i : i w ρ0 ) < τ(x i : i w ρ1 ). For each ρ < θ + use Lemma 13 to choose a finite p closed set Z ρ ht(p) containing the set {β < ht(p) : ( j w ρ )(h p (j,β) < θ)}.

12 12 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH Look at Υ p (Z ρ ) (see Definition 14). There are only θ possibilities for the values of Υ p (Z ρ ), so we find ρ 0 < < θ + such that (i) Z ρ0 = Z ρ1, Υ p (Z ρ0 ) = Υ p (Z ρ1 ) = ζ l,τ l,n l, g l,h l 0,...,h l n l 1 : l < k, (ii) if π : Z ρ0 Z ρ1 is the order isomorphism then π Z ρ0 Z ρ1 is the identity on Z ρ0 Z ρ1, (iii) if π : U[p,Z ρ0 ] U[p,Z ρ1 ] is the order isomorphism, then π[w ρ0 ] = w ρ1. Note that, by Lemma 15, otp(u[p,z ρ0 ]) = otp(u[p,z ρ1 ]) and the order isomorphism π satisfies ( j U[p,Z ρ0 ])( β Z ρ0 )(h p (j,β) = h p (π(j),π (β))), and hence π is the identity on U[p,Z ρ0 ] U[p,Z ρ1 ] (remember Lemma 3). For a function f F p let G ρ0 (f) : u p 2 be defined by G ρ0 (f)(j) = Claim For each f F p, G ρ0 (f) F p. f(π(j)) if j U[p,Z ρ0 ], f(π 1 (j)) if j U[p,Z ρ1 ]\U[p,ρ 0 ], 0 otherwise. Proof of the claim. By induction on α ht(p) we show that for each α component q of p, the restriction G ρ0 (f) u q is in F q. If α is limit, we may easily use the inductive hypothesis to show that, for any α component q of p, G ρ0 (f) u q F q. Assume α = β+1 and let q = ζ,τ,n,u, q ξ,v ξ : ξ < θ be an α component of p. We will consider four cases. Case 1: β / Z ρ0 Z ρ1. Then (U[p,Z ρ0 ] U[p,Z ρ1 ]) u q u q ζ and G ρ0 (f) (u q ξ \u ) 0 for each ξ ζ. Since, by the inductive hypothesis, G ρ0 (f) u q ξ F q ξ for each ξ < θ, we may use the definition of Pβ+1 and conclude that Gρ0 (f) u q F q (remember the definition of the term σ maj ). Case 2: β Z ρ0 \Z ρ1. Let Z ρ0 = {α 0,...,α k 1 } be the increasing enumeration. Then β = α l for some l < k and ζ = ζ l, τ = τ l, n = n l. Moreover, if v ξ = {j ξ 0,...,jξ n l 1 } (the increasing enumeration), ξ < θ, then for m < n l : ( l < l)(h l m(α l ) = h q (j ξ m,α l )) and ( γ β \Z ρ0 )(h q (j ξ m,γ) θ). Note that U[p,Z ρ1 ] u q u q ζ, so if U[p,Z ρ0 ] u q =, then we may proceed as in the previous case. Therefore we may assume that U[p,Z ρ0 ] u q. So, for each γ Z ρ0 \α we may choose i γ U[p,Z ρ0 ] u q such that ( i U[p,Z ρ0 ] u q )(h p (i,γ) θ h p (i,γ) = h p (i γ,γ)) (remember Lemma 11(1)). Let i = max{i γ : γ Z ρ0 \α} (if β = max(z ρ0 ), then let i be any element of U[p,Z ρ0 ] u q ). Note that then ( i U[p,Z ρ0 ] u q )( γ Z ρ0 \α)(h p (i,γ) θ h p (i,γ) = h p (i,γ)) [Why? Remember Lemma 11(1) and the clause (γ) of the definition of Pβ+1.] By Lemma 11, we find a (π (β)+1) component q = ζ,τ,n,u, q ε,v ε : ε < θ of p such that π(i ) u q and ( j u q )( γ (π (β),ht(p)))(h p (π(i ),γ) θ h p (j,γ) θ).

13 HISTORIC FORCING FOR Depth 13 We claim that then ( ) ( j U[p,Z ρ0 ] u q )(π(j) u q U[p,Z ρ1 ]). Why? Fix j U[p,Z ρ0 ] u q. Let r,r be components of p such that r pr p, r pr p, ht(r) = β +1, ht(r ) = π (β)+1 (so r and q, and r,q, are isomorphic). The sets Z ρ0 (β + 1) and Z ρ1 (π (β) + 1) are p closed, and they have the same values of Υ, and therefore U[p,Z ρ0 (β +1)] and U[p,Z ρ1 (π (β)+1)] are (order) isomorphic. Also, these two sets are included in u r and u r, respectively. So looking back at our j, we may successively choose j 0 u r U[p,Z ρ0 (β +1)], j 1 u r U[p,Z ρ1 (π (β)+1)], and j u q such that ( γ β)(h q (j,γ) = h r (j 0,γ)), ( l l)(h r (j 0,α l ) = h r (j 1,π (α l ))), and ( γ π (β))(h r (j,γ) = h q (j,γ)). Then we have ( l l)(h q (j,α l ) = h q (j,π (α l )) and ( γ π (β)\z ρ1 )(h q (j,γ) θ). To conclude ( ) it is enough to show that π(j) = j. If this equality fails, then there is γ < ht(p) such that θ h p (π(j),γ) h p (j,γ) θ. If γ π (β), then necessarily γ Z ρ1, and this is impossible (remember h p (j,α l ) = h p (π(j),π (α l )) for l l). So γ > π (β). If h p (π(j),γ) = θ +1, then h p (j,γ) < θ and (by the choice of q ) h p (π(i ),γ) < θ. Then γ Z ρ1 and h p (i,(π ) 1 (γ)) < θ, and also h p (i,(π ) 1 (γ)) = h p (j,(π ) 1 (γ)) = θ +1 (by the choice of i ), a contradiction. Thus necessarily h p (π(j),γ) < θ (so γ Z ρ1 ) and therefore θ > h p (j,(π ) 1 (γ)) = h p (i,(π ) 1 (γ)) = h p (π(i ),γ) = h p (j,γ) (as the last is not θ), again a contradiction. Thus the statement in ( ) is proven. Now we may finish considering the current case. By the definition of the function Υ (and by the choice of ρ 0, ) we have ζ = ζ l, τ = τ l, n = n l, and π[v ξ ] = v ξ for ξ < θ (and π v ξ is order preserving). Therefore G ρ0 (f)(τ (x i : i v ξ )) = f(τ (x i : i v ξ )) (for every ξ < θ). By the inductive hypothesis, G ρ0 (f) u q ξ F q ξ (for ξ < θ), so as f F p (and hence f u q F q ) we may conclude now that G ρ0 (f) u q F q. Case 3: β Z ρ1 \Z ρ0 Similar. Case 3: β Z ρ0 Z ρ1 If U[p,Z ρ0 ] u q = = U[p,Z ρ1 ] u q, then G ρ0 (f) u q 0 and we are easily done. If one of the intersections is non-empty, then we may follow exactly as in the respective case (2 or 3). Now we may conclude the proof of the theorem. Since B p = τ(x i : i w ρ0 ) < τ(x i : i w ρ1 ), we find f F p such that f(τ(x i : i w ρ0 )) = 0 and f(τ(x i : i w ρ1 )) = 1. It should be clear from the definition of the function G ρ0 (f) (and the choice of ρ 0, ) that G ρ0 (f)(τ(x i : i w ρ0 )) = 1 and G ρ0 (f)(τ(x i : i w ρ1 )) = 0.

14 14 ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH But it follows from Claim 17.1 that G ρ0 (f) F p, a contradiction. Conclusion 18. It is consistent that for some uncountable cardinal θ there is a Boolean algebra B of size (2 θ ) + such that Depth(B) = θ but (ω,(2 θ ) + ) / Depth Sr (B). Problem 19. Assume θ < = < are regular cardinals. Does there exist a Boolean algebra B such that B = + and for every subalgebra B B of size + we have Depth(B ) = θ? References [1] Thomas Jech. Set theory. Springer Monographs in Mathematics. Springer-Verlag, Berlin, The third millennium edition, revised and expanded. [2] J. Donald Monk. Cardinal Invariants of Boolean Algebras. Lectures in Mathematics. ETH Zurich, Birkhauser Verlag, Basel Boston Berlin, [3] J. Donald Monk. Cardinal Invariants of Boolean Algebras, volume 142 of Progress in Mathematics. Birkhäuser Verlag, Basel Boston Berlin, [4] George Orwell Harcourt Brace Jovanovich, San Diego, [5] Andrzej Roslanowski and Saharon Shelah. Cardinal invariants of ultrapoducts of Boolean algebras. Fundamenta Mathematicae, 155: , arxiv:math.lo/ [6] Saharon Shelah. On Monk s questions. Fundamenta Mathematicae, 151:1 19, arxiv:math.lo/ [7] Saharon Shelah. Not collapsing cardinals κ in (< κ) support iterations. Israel Journal of Mathematics, 136:29 115, arxiv:math.lo/ [8] Saharon Shelah. Successor of singulars: combinatorics and not collapsing cardinals κ in (< κ)-support iterations. Israel Journal of Mathematics, 134: , arxiv:math.lo/ [9] Saharon Shelah and Lee Stanley. A theorem and some consistency results in partition calculus. Annals of Pure and Applied Logic, 36: , Department of Mathematics, University of Nebraska at Omaha, Omaha, NE , USA, and Mathematical Institute of Wroclaw University, Wroclaw, Poland address: roslanowski@unomaha.edu URL: aroslano Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, Israel, and Department of Mathematics, Rutgers University, New Brunswick, NJ 08854, USA address: shelah@math.huji.ac.il URL: shelah