PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 143, Number 4, April 2015, Pages 1771–1784 S 0002-9939(2014)12360-4 Article electronically published on December 4, 2014
FORCING WITH COPIES OF COUNTABLE ORDINALS
MILOSˇ S. KURILIC´
(Communicated by Mirna Dˇzamonja)
Abstract. Let α be a countable ordinal and P(α) the collection of its subsets isomorphic to α. We show that the separative quotient of the poset P(α), ⊂ is isomorphic to a forcing product of iterated reduced products of Boolean γ algebras of the form P (ω )/Iωγ ,whereγ ∈ Lim ∪{1} and Iωγ is the corre- sponding ordinal ideal. Moreover, the poset P(α), ⊂ is forcing equivalent to + a two-step iteration of the form (P (ω)/Fin) ∗ π,where[ω] “π is an ω1- + closed separative pre-order” and, if h = ω1,to(P (ω)/Fin) . Also we analyze δ I the quotients over ordinal ideals P (ω )/ ωδ and the corresponding cardinal invariants hωδ and tωδ .
1. Introduction The posets of the form P(X), ⊂,whereX is a relational structure and P(X) the set of (the domains of) its isomorphic substructures, were considered in [7], where a classification of the relations on countable sets related to the forcing-related properties of the corresponding posets of copies is described. So, defining two structures to be equivalent if the corresponding posets of copies produce the same generic extensions, we obtain a rough classification of structures which, in general, depends on the properties of the model of set theory in which we work. For example, under CH all countable linear orders are partitioned in only two classes. Namely, by [6], CH implies that for a non-scattered countable linear order L the poset P(L), ⊂ is forcing equivalent to the iteration S ∗ (P (ˇω)/ Fin)+,where S is the Sacks forcing. Otherwise, for scattered orders, by [8] we have
Theorem 1.1. For each countable scattered linear order L the separative quotient of the poset P(L), ⊂ is ω1-closed and atomless. Under CH, it is forcing equivalent to the poset (P (ω)/ Fin)+.
The aim of this paper is to get a sharper picture of countable scattered linear orders in this context and we concentrate our attention on ordinals α<ω1.So,in Section 3 we describe the separative quotient of the poset P(α), ⊂ and, in Section + 5, factorize it as a two-step iteration (P (ω)/ Fin) ∗π,where[ω] “π is an ω1-closed separative pre-order” (which implies that the equality h = ω1 implies that all posets
Received by the editors April 29, 2013 and, in revised form, September 6, 2013. 2010 Mathematics Subject Classification. Primary 03E40, 03E10, 03C15; Secondary 03E35, 03E17, 06A06. This research was supported by the Ministry of Education and Science of the Republic of Serbia (Project 174006).
c 2014 American Mathematical Society Reverts to public domain 28 years from publication 1771
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1772 MILOSS.KURILIˇ C´
P(α), ⊂ are forcing equivalent to (P (ω)/ Fin)+ again). In Section 4 we factorize γ the quotients P (ω )/Iωγ ,forγ ∈ Lim, and, in Section 6, consider the quotients over δ the ordinal ideals P (ω )/Iωδ ,0<δ<ω1, and analyze the corresponding cardinal invariants hωδ and tωδ . We note that, while the results of the present paper are obtained using the Cantor normal form theorem for ordinals, the corresponding results concerning countable scattered linear orders given in [8] are obtained from its analogue, Laver’s theorem, stating that each countable scattered linear order is a finite sum of hereditarily additively indecomposable linear orders.
2. Preliminaries In this section we recall some definitions and basic facts used in the paper. If X is a relational structure, X its domain and A ⊂ X,thenA will denote ∼ the corresponding substructure of X.LetP(X)={A ⊂ X : A = X} and let IX = {A ⊂ X : X → A}. It is easy to check that X is an indivisible structure (that is, for each partition X = A ∪ B we have X → A,orX → B)iffIX is an ideal. We will use the following elementary fact.
Fact 2.1. Let X and Y be relational structures and f : X −→iso Y.Then (a) A ∈IX ⇔ f[A] ∈IY,foreachA ⊂ X; ∼ (b) P (X) \IX, ⊂ = P (Y ) \IY, ⊂.
A linear order L is said to be scattered iff it does not contain a dense suborder or, equivalently, iff the rational line, Q,doesnotembedinL.ByS we denote the class of all countable scattered linear orders. A linear order L is said to be additively indecomposable iff for each decomposition L = L0 +L1 we have L→ L0 or L→ L1. The class H of hereditarily additively indecomposable (or ha-indecomposable) linear orders is the smallest class of order types of countable linear orders containing the ∗ one element order type, 1, and containing the ω-sum, ω Li,andtheω -sum, ∈ H ω∗ Li, for each sequence Li : i ω in satisfying
(2.1) ∀i ∈ ω |{j ∈ ω : Li → Lj }| = ℵ0.
Fact 2.2 (Laver, [10]). H⊂S.IfL ∈S,thenL ∈Hiff L is additively indecom- posable (see also [11], p. 196 and p. 201). ∈H ∈ Fact 2.3 (See [8]). (a) Let L = ω Li ,where Li : i ω is a sequence in H satisfying (2.1). Then A ⊂ L contains a copy of L iff for each i, m ∈ ω there is ⊂ \ → ∩ finite K ω msuch that Li j∈K Lj A. ∈H H (b) Let L = i≤n Li,whereLi are ω-sums of sequences in satisfying ∈H P ⊂ ∼ P ⊂ (2.1) and Li + Li+1 ,fori License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1773 Fact 2.4. For a countable limit ordinal α the following conditions are equivalent: (a) α is indecomposable (i.e. α is not a sum of two smaller ordinals); (b) β + γ<α,foreachβ,γ < α; (c) A ∈ P(α)orα \ A ∈ P(α), for each A ⊂ α; (d) α = ωδ, for some countable ordinal δ>0; (e) α ∈H; (f) α is an indivisible structure; (g) Iα = {I ⊂ α : α → I} is an ideal in P (α). Proof. For the equivalence of (b), (c) and (d) see [5], p. 43. For (a) ⇔ (d) see 1.3.6 of [2]. By [11], p. 176, (d) holds iff α is additively indecomposable which is, by Fact 2.2, equivalent to (e). (a) ⇔ (f) is 6.8.1 of [2]. (f) ⇔ (g)isevident. δ + Fact 2.5. For each ordinal α we have P(α)=P (α) \Iα.ThusP(ω )=(Iωδ ) . Proof. The inclusion “⊂”istrivial.Ifα→ A ⊂ α, then, using the fact that for each increasing function f : α → α we have β ≤ f(β), for each β ∈ α, we easily show that type(A)=α, which means that A ∈ P(α). A partial order P = P, ≤ is called separative iff for each p, q ∈ P satisfying p ≤ q there is r ≤ p such that r ⊥ q.Theseparative modification of P is the separative pre-order sm(P)= P, ≤∗,wherep ≤∗ q ⇔∀r ≤ p ∃s ≤ rs≤ q.The separative quotient of P is the separative partial order sq(P)= P/=∗, ,where p =∗ q ⇔ p ≤∗ q ∧ q ≤∗ p and [p] [q] ⇔ p ≤∗ q (see [4]). Fact 2.6. Let P, Q and Pi, i ∈ I, be partial orderings. Then (a) P,sm(P)andsq(P) are forcing equivalent forcing notions; ∼ ∼ ∼ (b) P =Q implies that sm P = sm Q and sq P = sqQ; P P P ∼ P (c) sm( i∈I i)= i∈I sm i and sq( i∈I i) = i∈I sq i. Let X be an infinite set, I P (X) an ideal and [X]<ω ⊂I.Then (d) sm P (X) \I, ⊂ = P (X) \I, ⊂I ,whereA ⊂I B ⇔ A \ B ∈I. + (e) sq P (X) \I, ⊂ = (P (X)/ =I ) , ≤I,whereA =I B ⇔ A B ∈Iand + [A] ≤I [B] ⇔ A \ B ∈I. Usually this poset is denoted by (P (X)/I) . Let κ be a regular cardinal. A pre-order P, ≤ is κ-closed iff for each γ<κand each sequence pα : α<γ in P, such that α<β⇒ pβ ≤ pα,thereisp ∈ P such that p ≤ pα, for all α<γ. Fact 2.7. Let κ be a regular cardinal and λ an infinite cardinal. Then P ∈ P (a) If i, i I,areκ-closed pre-orders, then the product i∈I i is κ-closed. (b) If c = ω1, then each atomless separative ω1-closed pre-order of size ω1 is + forcing equivalent to (P (ω)/ Fin) (and to the collapsing algebra Coll(ω1,ω1)). (c) If λ<κ = λ, then each atomless separative κ-closed pre-order P of size λ,such that 1P |λˇ| =ˇκ, is forcing equivalent to the collapsing algebra Coll(κ, λ). 3. The separative quotient of P(α) ⊂ For a Boolean lattice B = B,≤,byrp(B) we will denote the reduced power ω ω B / ≡, ≤≡,wherefor bi, ci∈B , bi≡ ci (resp. [ bi]≡ ≤≡ [ ci]≡)iff bi = ci (resp. bi ≤ ci), for all but finitely many i ∈ ω.Forn ∈ ω we define the set rpn(B)by:rp0(B)=B and rpn+1(B)=rp(rpn(B)). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1774 MILOSS.KURILIˇ C´ The aim of this section is to prove the following statement. γn+rn γ0+r0 Theorem 3.1. If α = ω sn +···+ω s0 +k is a countable ordinal presented in the Cantor normal form, where k ∈ ω, ri ∈ ω, si ∈ N, γi ∈ Lim ∪{1} and γn + rn > ···>γ0 + r0,then n + si ∼ ri γi sq P(α), ⊂ = rp (P (ω )/Iωγi ) . i=0 A proof of Theorem 3.1 is given at the end of the section. We remind the reader that, if I and J are ideals on the sets X and Y respectively, then their Fubini product I×J is the ideal on the set X × Y defined by I×J = + {A ⊂ X × Y : {x ∈ X : πY [A ∩ ({x}×Y )] ∈J }∈I},whereπY : X × Y → Y is the projection. In particular, if X = ω, I = Fin and Li = {i}×Y ,fori ∈ ω,then for A ⊂ ω × Y we have (3.1) A ∈ Fin ×J ⇔ ∃j ∈ ω ∀i ≥ jπY [A ∩ Li] ∈J. For convenience let us define the sets ωn × Y , n ∈ ω, recursively by ω0 × Y = Y and ωn+1 × Y = ω × (ωn × Y ). Also we define the ideal Finn ×J on the set ωn × Y by: Fin0 ×J = J and Finn+1 ×J =Fin×(Finn ×J ). Some parts of the following lemma are folklore but, for completeness, we include their proofs. Lemma 3.2. For each ordinal 1 ≤ β<ω1 and each n ∈ ω we have: β+n ∼ n β n (a) P(ω ), ⊂ = P (ω × ω ) \ (Fin ×Iωβ ), ⊂; ∼ n (b) Iωβ+n = Fin ×Iωβ ; β+n ∼ n β n + (c) sq P(ω ), ⊂ = (P (ω × ω )/(Fin ×Iωβ )) ; n β n ∼ n β (d) P (ω × ω )/(Fin ×Iωβ ) = rp (P (ω )/Iωβ ); β+n ∼ n β + (e) sq(P(ω ), ⊂) = (rp (P (ω )/Iωβ )) . Proof. For n = 0 the statement follows from Fact 2.5. So, in the sequel we prove the statement for n ∈ N. Using induction we prove (a) and (b) simultaneously. First we show that β+1 ∼ + (3.2) P(ω ), ⊂ = (Fin ×Iωβ ) , ⊂. β+1 By the properties of ordinal multiplication and exponentiation we have ω , ∈ β ∈ ∼ × β L L L ∈ L = ω ω, = ω ω , iso β (3.3) fi = πωβ | Li : Li, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1775 (ii) By (3.1), A ∈ Fin ×Iωβ iff for each j ∈ ω there exists i ≥ j such that πωβ [Li ∩ A] ∈Iωβ . But, by (3.3) and Fact 2.1(a) we have: πωβ [Li ∩ A] ∈Iωβ iff β ∩ ∈I β ∩ ∈I L → ∩ → ∩ fi[Li A] ω iff Li A Li iff i Li A iff ω Li A. ⊃ ⊂ ∈I+ By Claim 3.3, the inclusion “ ” in (3.4) is satisfied and we prove “ ”. If A L ∈ ∈ \ <ω β → ∩ and j ω, then, by Claim 3.3(i), there are K [ω j] and g : ω i∈K Li A. Let i =max{i ∈ K : g[ωβ] ∩ L ∩ A = ∅}.ThenF = g[ωβ] ∩ L ∩ A is a final part 0 ∼ i i0 of the linear order g[ωβ] = ωβ and, since type(g[ωβ] \ F ) <ωβ,byFact2.4(c)we β β → ∩ ≥ have type(F )=ω and, hence ω Li0 A and i0 j. By (ii) of Claim 3.3 we + have A ∈ (Fin ×Iωβ ) and (3.4) is proved. So (3.2) is true. β+1 ∼ By (3.4) we have IL =Fin×I β .Since ω , ∈ = L, by Fact 2.1(a) we have ∼ ω Iωβ+1 = IL and, hence, ∼ (3.5) Iωβ+1 = Fin ×Iωβ . Let us assume that the statements (a) and (b) are true for n. By (3.5) we have ∼ ∼ n n+1 Iωβ+n+1 = Fin ×Iωβ+n = Fin ×(Fin ×Iωβ )=Fin ×Iωβ .ByFact2.5wehave β+n+1 + ∼ n+1 + P(ω ), ⊂ = (Iωβ+n+1 ) , ⊂ = (Fin ×Iωβ ) , ⊂. (c) follows from (a) and Fact 2.6(b) and (e). (d) We use induction. For a proof of (d) for n = 1 we show that the mapping β β ω F : P (ω × ω )/ = ×I , ≤ ×I → P (ω )/ =I , ≤I / ≡, ≤≡, given Fin ωβ Fin ωβ ωβ ωβ β ∩ ∈ by F ([A]=Fin ×I )=[[πω [A Li]]=I : i ω ]≡, is an isomorphism. ωβ ωβ β Claim 3.4. For A, B ⊂ ω × ω we have: A = ×I B if and only if Fin ωβ ∩ ∈ ∩ ∈ (3.6) [ [πωβ [A Li]]=I : i ω ]≡ =[[πωβ [B Li]]=I : i ω ]≡. ωβ ωβ Proof. First, by (3.1) we have (3.7) A = ×I B ⇔∃j ∈ ω ∀i ≥ jπβ [(A B) ∩ L ] ∈I β . Fin ωβ ω i ω On the other hand, (3.6) holds iff there is j ∈ ω such that for all i ≥ j we have πωβ [A∩Li]πωβ [B ∩Li] ∈Iωβ , that is, since the restriction πωβ | Li is a bijection, (πωβ | Li)[(A ∩ Li) (B ∩ Li)] = πωβ [(A B) ∩ Li)] ∈Iωβ . By Claim 3.4, F is a well-defined injection. ∈ ∈ β ω ≡ For [ [Xi]=I : i ω ]≡ (P (ω )/ =I ) / we have F ([A]= ×I )= β ωβ Fin β ∈ω { }× ω [ [X ] I : i ω ]≡,whereA = i X ,soF is a surjection. i = β i∈I i ω ≤ By (3.1) we have [A]=Fin ×I Fin ×I β [B]=Fin ×I iff ωβ ω ωβ (3.8) ∃j ∈ ω ∀i ≥ jπωβ [A \ B ∩ Li] ∈Iωβ ∩ ∈ ≤ ∩ ∈ ∈ and [ [πωβ [A Li]]=I : i ω ]≡ ≡ [ [πωβ [B Li]]=I : i ω ]≡ iff there is j ω ωβ ωβ such that for all i ≥ j we have πωβ [A ∩ Li] \ πωβ [B ∩ Li] ∈Iωβ , that is, since the restriction πωβ | Li is a bijection, (πωβ | Li)[(A∩Li)\(B ∩Li)] = πωβ [A\B ∩Li)] ∈ Iωβ .ThusF is an isomorphism. Assuming that the statement is true for n, by (b) and (d) for n =1wehave n+1 β n+1 ∼ n β n ∼ P (ω × ω )/(Fin ×Iωβ ) = P (ω × (ω × ω ))/(Fin ×(Fin ×Iωβ )) = β+n ∼ β+n ∼ n β n ∼ P (ω ×ω )/(Fin ×Iωβ+n ))=rp(P (ω )/Iωβ+n ) =rp(P (ω ×ω )/(Fin ×Iωβ ) = n β ∼ n+1 β rp(rp (P (ω )/Iωβ )) = rp (P (ω )/Iωβ ). (e) follows from (c) and (d). For n ∈ N,lettheidealFinn on the set ωn = ω × (ω ×···×(ω × ω) ...)(n-many factors) be defined by: Finn =Fin×(Fin ×···×(Fin × Fin) ...)(n-many factors). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1776 MILOSS.KURILIˇ C´ Then, by Lemma 3.2 we have Corollary 3.5. For each n ∈ N we have: n ∼ n n ∼ n (a) P(ω ), ⊂ = P (ω ) \ Fin , ⊂ and Iωn = Fin ; ∼ − (b) sq(P(ωn), ⊂) = (rpn 1(P (ω)/ Fin))+. ∼ Lemma 3.6. P(γ + k), ⊂ = P(γ), ⊂, for each limit ordinal γ and each k ∈ N. Proof. First we prove P(γ + k)={C ∪{γ,γ +1,...,γ + k − 1} : C ∈ P(γ)}. The inclusion “⊃”isevident.IfA ∈ P(γ + k)andf : γ + k→ γ + k,where A = f[γ + k], then, since f is an increasing function, we have f(β) ≥ β,foreach β ∈ γ + k, which implies f(γ + i)=γ + i,fori Proof of Theorem 3.1. By Lemma 3.6 we can assume that k =0.So,wehaveα = γn+rn γn+rn γ0+r0 γ0+r0 ω +···+ω +···+ω +···+ω = n L . By Lemma 3.7(a) j< si j ∈H i=0 H for each j the order Lj and it is an ω-sum of elements of satisfying (2.1). By ∼ Lemma 3.7(b) Lj + Lj+1 ∈Hso, by Fact 2.3(b), P(α), ⊂ = n P(Lj ), ⊂ j< i=0 si n P γi+ri ⊂si = i=0 (ω ), , which, together with Fact 2.6(b),(c) and Lemma 3.2(e), n n P ⊂ ∼ P γi+ri ⊂si ∼ P γi+ri ⊂ si ∼ gives sq (α), = sq i=0 (ω ), = i=0(sq (ω ), ) = n ri γi I + si i=0((rp (P (ω )/ ωγi )) ) . ∼ Corollary 3.8. sq P(ωn), ⊂ = ((P (ω)/ Fin)+)n,foreachn ∈ N. γ 4. Forcing with the quotient P (ω )/Iωγ By Theorem 3.1, the poset P(α), ⊂ is forcing equivalent to a forcing product γ of iterated reduced products of Boolean algebras of the form P (ω )/Iωγ .Inthis section we consider such algebras and assume that γ ≥ ω is a countable limit ∈ \{ } L ordinal, δn : n ω a fixed increasing cofinal sequence in γ 0 and = L, < ∼ δn ∈ ∈ ∩ ∅ = n∈ω Ln, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1777 Theorem 4.1. For each countable limit ordinal γ we have: γ γ + (a) The partial orders P(ω ), ⊂ and (P (ω )/Iωγ ) are forcing equivalent to + + ∗ ˇ Iˇ −1 the two-step iteration (P (ω)/ Fin) (P (L)/ qˇ [Γ1]) ; + ˇ Iˇ −1 (b) [ω] “(P (L)/ qˇ [Γ1]) is an ω1-closed, separative and atomless poset”. Fact 4.2. Let f : ω → ω be an increasing function. Then δ δ δ δ (a) ωf(0) + ω f(1) + ···+ ω f(m) = ω f(m) ,foreachm ∈ ω; (b) ωδf(n) = ωγ ; ∼n∈ω (c) L = ωγ . Proof. We prove (a) by induction. Assuming that (a) is true for m ∈ ω we have − ωδf(0) + ···+ ωδf(m+1) = ωδf(m) + ωδf(m+1) = ωδf(m) · 1+ωδf(m)+(δf(m+1) δf(m)) = δ δ −δ δ δ −δ δ ω f(m) (1 + ω f(m+1) f(m) )=ω f(m) ω f(m+1) f(m) = ω f(m+1) . δf(n) (b) By (a) and basic properties of ordinal arithmetic we have n∈ω ω = { δf(n) ∈ } { δf(m) ∈ } γ sup n≤m ω : m ω =sup ω : m ω = ω . L ∼ δn γ (c) By (b) we have = n∈ω ω = ω . Lemma 4.3. For A ⊂ L and m ∈ ω we have: m ⊂ \ (a) SA supp A m; ⇒ m1 ⊃ m2 (b) m1 Proof. (a) If A1,A2 ∈IG and supp(A1 \ I1), supp(A2 \ I2) ∈ G,whereI1,I2 ∈I, then, since (A1 ∪A2)\(I1 ∪I2) ⊂ (A1 \I1)∪(A2 \I2) and supp(X ∪Y ) = supp(X)∪ supp(Y ), we have supp((A1 ∪ A2) \ (I1 ∪ I2)) ⊂ supp(A1 \ I1) ∪ supp(A2 \ I2) ∈ G and, since I1 ∪ I2 ∈I,wehaveA1 ∪ A2 ∈IG. ⊂ ∈ \I ⊂ (b) Let A IG B and C P (L) G,whereC I A. Then, since C = (C \ A) ∪ (C ∩ A \ B) ∪ (C ∩ A ∩ B), A \ B ∈IG and C \ A ∈I⊂IG,wehave ∗ D = C ∩ A ∩ B ∈ P (L) \IG and D ⊂I C, B.ThusA ⊂I B. Conversely, suppose License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1778 MILOSS.KURILIˇ C´ ∗ that A ⊂I B and A \ B ∈IG. Then, for C = A \ B there is D ∈ P (L) \IG such that D ⊂I A \ B and D ⊂I B, which implies D ∈I. A contradiction. We remind the reader that, if P ≤P, 1P and Q ≤Q, 1Q are pre-orders, then a mapping f : P → Q is a complete embedding, in notation f : P →c Q iff (ce1) p1 ≤P p2 ⇒ f(p1) ≤Q f(p2), (ce2) p1 ⊥P p2 ⇔ f(p1) ⊥Q f(p2), (ce3) ∀q ∈ Q ∃p ∈ P ∀p ≤P pf(p ) ⊥Q q. Then, for q ∈ Q the set red(q)={p ∈ P : ∀p ≤P pf(p ) ⊥Q q} is the set of reductions of q to P. The following fact is folklore (see [5]). Fact 4.5. If f : P →c Q,thenQ is forcing equivalent to the two-step iteration P ∗ π, ≤π, 1ˇQ,where1P P π ⊂ Qˇ and for each p ∈ P and q, q1,q2 ∈ Q (a) p qˇ ∈ π iff p ∈ red(q); (b) p qˇ1 ≤π qˇ2 iff q1 ≤Q q2 and p ∈ red(q1). Proposition 4.6. The following pre-orders are forcing equivalent: 1. P(ωγ ), ⊂, γ + 2. (P (ω )/Iωγ ) , 3. P (L) \I, ⊂I , 4. [ω]ω, ⊂∗∗ P (ˇL) \ Iˇ , ⊂ , Γ IˇΓ + + ∗ ˇ Iˇ −1 5. (P (ω)/ Fin) (P (L)/ qˇ [Γ1]) . Proof. By Facts 2.5, 2.6(a) and (e) the posets 1 and 2 are forcing equivalent. By γ γ Facts 2.5, 2.1(b) and 2.6(a),(d) the poset P(ω ), ⊂ = P (ω ) \Iωγ , ⊂ is isomor- phic to the poset P (L) \I, ⊂, forcing equivalent to P (L) \I, ⊂I . The forcing + equivalence of the posets 4 and 5 is evident – note that G1 is a (P (ω)/ Fin) -generic −1 ω ⊂∗ \I ⊂ filter iff G = q [G1]isan [ω] , -generic filter over V and that sq P (L) G, IG + =(P (L)/IG) . Thus the forcing equivalence of the posets 3 and 4 remains to be proved. ω ∗ Claim 4.7. The mapping f : [ω] , ⊂ → P (L) \I, ⊂I defined by f(S)= \I ⊂ ≤ n∈S Ln is a complete embedding. In addition, t( P (L) , I ) t. ∈ ω ∼ δn γ ∼ L Proof. By Fact 4.2(b) and (c), for S [ω] we have f(S) = n∈S ω = ω = , ω thus f(S) ∈ P (L) \I.LetS, T ∈ [ω] . ⊂∗ | \ | | | | \ | (ce1) If S T ,then supp(f(S) f(T )) = supp( n∈S\T Ln) = S T <ω and, by Lemma 4.3(f), f(S) \ f(T ) ∈I,thatis,f(S) ⊂I f(T ). ⊥ | ∩ | | | | ∩ | (ce2) If S T ,then supp(f(S) f(T )) = supp( n∈S∩T Ln) = S T <ω and, by Lemma 4.3(f), f(S) ∩ f(T) ∈I,thatis,f(S) ⊥I f(T ). If S ⊥ T ,then ∩ ∈ ω ∩ ∩ ∈ \I S T [ω] and f(S) f(T )= n∈S∩T Ln = f(S T ) P (L) and, hence, f(S) ⊥I f(T ). (ce3) First we show that for S ∈ [ω]ω and A ∈ P (L) \I we have ∈ ⊂∗ m ∈ (4.1) S red(A)iffS SA , for each m ω. ∈ \ m ∈ ω ∈ Suppose that S red(A)andthatT = S SA [ω] ,forsomem ω. Then there is B ∈ P (L) \I such that B ⊂I f(T ),A. Now we use Lemma 4.3. By (h), there ∈ m1 m2 ∅ ∗ { } are m1,m2 ω such that SB\f(T ) = SB\A = .By(b),form =max m, m1,m2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1779 m∗ m∗ ∅ m∗ m∗ we have SB\f(T ) = SB\A = and, by (g), SB = S(B∩A∩f(T ))∪(B\f(T ))∪(B\A) = m∗ ∪ m∗ ∪ m∗ m∗ m∗ ⊂ SB∩A∩f(T ) SB\f(T ) SB\A = SB∩A∩f(T ).But,by(a),(b)and(c),SB∩A∩f(T ) m∗ ∩ m∗ ⊂ ∩ m ∅ m∗ ∅ ∈I Sf(T ) SA T SA = ,thatis,SB = ,which,by(e),impliesB .A contradiction. ⊂∗ m ∈ ω ⊂∗ Let S SA ,foreachm ω,andlet[ω] T S. In order to find a set B ∈ P (L) \I such that B ⊂I f(T ),A by recursion we construct a sequence nk : k ∈ ω such that for each k ∈ ω we have: (i) nk ∈ T , (ii) nk By the previous claim, Fact 4.5 and (4.1), the pre-order P (L) \I, ⊂I is forcing ω ∗ equivalent to the iteration [ω] , ⊂ ∗ π, ≤π, Lˇ,whereω π ⊂ (P (L) \I)ˇ and for each S ∈ [ω]ω and A, B ∈ P (L) \I we have ˇ ∈ ⇔∀ ∈ ⊂∗ m (4.2) S A π m ωS SA ; (4.3) S Aˇ ≤π Bˇ ⇔ S Aˇ ∈ π ∧ A ⊂I B. ˇ Claim 4.8. (a) ω “Iˇ and IˇΓ are ideals in P (Lˇ)=P (L) ; ˇ (b) ω π = P (L) \ IˇΓ; (c) ω ≤π=⊂Iˇ ∩(π × π). Proof. Let G be an [ω]ω, ⊂∗-generic filter over V . (a) Since the forcing [ω]ω, ⊂∗ is ω-distributive, in V [G]wehaveP V [G](L)= P V (L) and, for the same reason, I remains to be an ideal in P V [G](L). By Lemma V [G] 4.4(a), the set IG = {A ⊂ L : ∃I ∈I supp(A \ I) ∈ G} is an ideal in P (L). (b) We show that πG = {A ∈ P (L) \I : ∀I ∈I supp(A \ I) ∈ G}.Let A ∈ P (L) \I.IfA ∈ πG, then there is S ∈ G such that S Aˇ ∈ π.ForI ∈Iwe ∩ ∈I ∗ ∈ m∗ ∅ have A I and, by Lemma 4.3(e), there is m ω such that SA∩I = .Thus, ⊂∗ m∗ m∗ ∪ m∗ m∗ ⊂ \ by (4.2) and Lemma 4.3(g) we have S SA = SA∩I SA\I = SA\I supp(A I), which implies that supp(A \ I) ∈ G.SoA ∈ P (L) \IG. ∈ ∈ ¬ ˇ ∈ | ∩ m| If A πG, then there is S G such that S A π. Suppose that S SA = ω, ∈ ∩ m ∈ for each m ω. Then, by Lemma 4.3(b), S SA , m ω, would be a decreasing ω ∈ ω ⊂∗ ∩ m sequence in [ω] and, hence, there would be T [ω] such that T S SA ,for ˇ ∗ each m ∈ ω,which,by(4.2),impliesT A ∈ π. But this is impossible since T ⊂ S ¬ ˇ ∈ | ∩ m∗ | ∗ ∈ ∩ and S A π.Thus S SA <ω,forsomem ω.LetI = n∈S A Ln.By m∗ ⊂ ∩ m∗ | m∗ | Lemma 4.3(c) we have SI S SA and, hence, SI <ω,which,byLemma 4.3(d), implies I ∈I. Since supp(A\I)∩S = ∅ and S ∈ G,wehavesupp(A\I) ∈ G. So A ∈ P (L) \IG. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1780 MILOSS.KURILIˇ C´ 2 (c) We show that (≤π)G =⊂I ∩(P (L) \IG) .LetA, B ∈ πG and let S ∈ G where S A,ˇ Bˇ ∈ π.IfA(≤π)GB, then there is T ∈ G such that T Aˇ ≤π Bˇ, which, by (4.3), implies A ⊂I B.IfA ⊂I B, then, since S Aˇ ∈ π, by (4.3) we have S Aˇ ≤π Bˇ and, hence, A(≤π)GB. Thus, the pre-order P (L) \I, ⊂I is forcing equivalent to the two-step iteration ω ∗ ˇ [ω] , ⊂ ∗ P (L) \ IˇΓ, ⊂ˇI and, by Lemma 4.4(b) applied in V [G], to the iteration ω ⊂∗∗ ˇ \ Iˇ ⊂ [ω] , P (L) Γ, IΓ . Proposition 4.9. According to the notation of Proposition 4.6 we have (a) ω “ P (ˇL) \ Iˇ , ⊂ is a separative, ω -closed and atomless pre-order”. Γ IˇΓ 1 + ˇ Iˇ −1 (b) [ω] “(P (L)/ qˇ [Γ1]) is a separative, ω1-closed and atomless poset.” Proof. (a) The separativity follows from Fact 2.6(d) and we prove ω1-closure. We easily show that for S ∈ [ω]ω and A, B ∈ P (L) satisfying S A,ˇ Bˇ ∈ π we have: (4.4) S Aˇ ⊂ Bˇ ⇔∀T ⊂∗ S ∃I ∈I |T \ supp(A \ B) \ I))| = ω. IˇΓ ωˇ Since the forcing [ω]ω, ⊂∗ is ω-distributive we have ω P (ˇL) =((P (L)ω)V )ˇ ˇ ω ω and, clearly, ω π ⊂ P (L). So, assuming that An : n ∈ ω∈P (L) , S ∈ [ω] and S ∀n ∈ ωˇ (Aˇ ∈ π ∧∀m ≥ n Aˇ ⊂ Aˇ ), that is, by (4.2) and (4.4), n m IˇΓ n ∀ ∈ ⊂∗ m (4.5) m, n ωS SAn , ∗ (4.6) ∀R ⊂ S ∃I ∈I |R \ supp((An+1 \ An) \ I)| = ω, it is sufficient to find T ∈ [ω]ω and A ∈ P (L) such that T ⊂∗ S, T Aˇ ∈ π and T Aˇ ⊂Iˇ Aˇn, for all n ∈ ω. Γ Claim 4.10. For r ∈ ω,letS = S ∩ Sm and B = A ∩ L .Then r m,n≤r An r r k∈Sr k (a) Br ∈ P (L) \I; (b) Br+1 ⊂I Br. Proof. (a) If m ∈ ω,thenk ∈ Sm iff k ∈ S and ωδm → B ∩ L = A ∩ L ;thus Br r r k r k m ∩ m | m | ∈ \I SB = Sr SA and, by (4.5), SB = ω. Now, by Lemma 4.3(d), Br P (L) . r r ⊂ r ⊂ (b) Suppose that Br+1 IBr. Then, since Sr+1 Sr,wewouldhaveC = B \ B =(A \ A ) ∩ L ∈ P (L) \I and, by Lemma 4.3(b) and r+1 r r+1 r k∈Sr+1 k ∈ ω ⊂∗ m ∈ m ⊂ (d), there would be R [ω] such that R SC , for all m ω.SinceSC ∗ supp(C) ⊂ Sr+1 ⊂ S we would have R ⊂ S and, by (4.6), there would be I ∈I ∗ such that R ⊂ supp((Ar+1 \ Ar) \ I)). Since (Ar+1 \ Ar) ∩ I ∈I, by Lemma 4.3(e) ∗ ∗ there is m∗ ∈ ω such that Sm = ∅ and, by Lemma 4.3(g), Sm = (Ar+1\Ar )∩I Ar+1\Ar ∗ ∗ ∗ Sm ⊂ supp((A \A )\I). But, by Lemma 4.3(c), R ⊂∗ Sm ⊂ Sm , (Ar+1\Ar )\I r+1 r C Ar+1\Ar ∗ thus R ⊂ supp((Ar+1 \ Ar) \ I). A contradiction. By Theorem 1.1 the pre-order P (L) \I, ⊂I is ω1-closed so, by Claim 4.10, there is A ∈ P (L) \I such that (4.7) ∀n ∈ ωA⊂I Bn ⊂ An. By Lemma 4.3(b) and (d) there is T ∈ [ω]ω such that ∀ ∈ ⊂∗ m (4.8) m ωT SA . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1781 ∗ By (4.7) we have A \ B ∈I, by Lemma 4.3(e) there is m∗ ∈ ω such that Sm = n A\Bn m∗ m∗ m∗ m∗ m∗ ∅ and, by Lemma 4.3(g) we have S = S ∩ ∪ S = S ∩ ⊂ S ⊂ A A Bn A\Bn A Bn Bn ⊂ ⊂ ⊂∗ m∗ ⊂∗ supp(Bn) Sn S. By (4.8) we have T SA and, hence, T S. By (4.8) and (4.2) we have T Aˇ ∈ π. By (4.7), for each n ∈ ω we have A ⊂I An and, hence, T Aˇ ⊂ Aˇ . IˇΓ n Taking an [ω]ω, ⊂∗-generic filter G we prove that the pre-order π , ⊂ is G IˇG ∈ ∈ ⊂∗ m atomless. If A πG, then, by (4.2), there is S G such that S SA ,foreach ∈ 0 ⊃ 1 ⊃ m ∅ m ω. By Lemma 4.3(b) we have SA SA ... and, clearly, m∈ω SA = . ⊂ 0 ∩ m \ m+1 ∈ W.l.o.g. suppose that S SA.ThenS = m∈ω S (SA SA ) and, for n ∩ m \ m+1 δm → ∩ δm ∪˙ S (SA SA )thereisϕn: ω A Ln.Let ϕn[ω ]=Bn Cn,where ∼ δm m ∩ m Bn,Cn = ω and let B = m∈S Bn and C = m∈S Cn.ThenSB = S SA ⊂∗ m ∈ ∈ ∈ and, hence, S SB , for all m ω, which implies B πG and, similarly, C πG. ⊂ ⊂ ∩ ∅ Since B,C A we have B,C IG A and B C = implies that B and C are ⊂Iˇ -incompatible. G \I ⊂ The proof of (b) is similar to the proof of (a). Note that P (L) G, IG is + ω1-closed (atomless) iff (P (L)/IG) is ω1-closed (atomless). 5. Forcing with P(α), ⊂ ∼ ∼ If P, Q and R are pre-orders, then, clearly, P × Q = Q × P and (P × Q) × R = P×(Q×R), that is, concerning the forcing equivalence of pre-orders, direct product is a commutative and associative operation. The following lemma generalizes the associativity law. Lemma 5.1. Let P and Q be pre-orders and π, ≤π, 1π a P-name for a pre-order. Then there is a P-name for a pre-order π , ≤ , 1 such that ∼ 1 π1 π1 (a) (P ∗ π) × Q = P ∗ π1. (b) If P is ω-distributive, 1P P “π is ω1-closed” and Q is ω1-closed, then 1P P “π1 is ω1-closed”. (c) If 1P P “π is separative” and Q is separative, then 1P P “π1 is separative”. (d) If 1P P “π is atomless” or Q is atomless, then 1P P “π1 is atomless”. ≤ Proof. It is easy to show that the triple π1, π1 , 1π1 works, where π1 = { τ,qˇ,p : p ∈ P ∧ τ ∈ dom π ∧ q ∈ Q ∧ p P τ ∈ π}, ≤ { ∈ ∧ ≤ ∧ ≤ } π1 = τ0,q0 , τ1,q1 ˇ,p : p P τ0,τ1 π τ0 π τ1 q0 Q q1 , 1π1 = 1π, 1Q ˇ. Fact 5.2. Let B be a non-trivial Boolean algebra, U⊂P (ω) a non-principal ultra- filter and Bω/U the corresponding ultrapower. Then ω + (a) The poset (B /U) is ω1-closed and separative (folklore). (b) If the algebra B is atomless, then (Bω/U)+ is an atomless poset (folklore). (c) (See [3].) The poset (rp(B))+ is forcing equivalent to the two-step iteration + ω + (P (ω)/ Fin) ∗ (B /Γ1) . Theorem 5.3. For each countable ordinal α ≥ ω + ω the partial order P(α), ⊂ is forcing equivalent to a two-step iteration of the form (P (ω)/ Fin)+ ∗ π,where[ω] “π is an ω1-closed, separative atomless forcing”. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1782 MILOSS.KURILIˇ C´ γn+rn ··· γ0+r0 Proof. Using the notation of Theorem 3.1, for α = ω sn + + ω s0 + k n P ⊂ ∼ ri γi I + si we have sq (α), = i=0((rp (P (ω )/ ωγi )) ) . ≤ γn ··· γ0 ∈ If ri = 0, for all i n,then α = ω sn + + ω s0 + k,whereγn Lim or n P ⊂ ∼ γi I + si ≥ γn =1,andsq (α), = i=0((P (ω )/ ωγi ) ) .So,ifγn ω, then, by the ∼ γn + associativity of direct products, sq P(α), ⊂ = (P (ω )/Iωγn ) ∗ Q,whereQ is an ω1-closed, separative and atomless poset (see Theorem 1.1 and Facts 2.5 and 2.7(a)). Thus, by Theorem 4.1, the poset sq P(α), ⊂ is forcing equivalent to the product + R =((P (ω)/ Fin) ∗ π) × Q,where[ω] “π is an ω1-closed, separative atomless + forcing” and, by Lemma 5.1, R forcing equivalent to an iteration (P (ω)/ Fin) ∗π1, where [ω] “π1 is an ω1-closed, separative atomless forcing”. If γn =1,then ∼ + α = ω · sn and, by the assumption, sn ≥ 2. Thus sq P(α), ⊂ = (P (ω)/ Fin) × − − ((P (ω)/ Fin)+)sn 1 =(P (ω)/ Fin)+ × π,whereπ = (((P (ω)/ Fin)+)sn 1)ˇ. ≤ If ri0 > 0, for some i0 n, then, by the associativity and commutativity of ∼ − P ⊂ ri0 1 γi0 I γ + × Q Q direct products, sq (α), = (rp(rp (P (ω )/ ω i0 ))) ,where is an ω1-closed, separative and atomless poset (see Theorem 1.1, Lemma 3.2 and Fact − B ri0 1 γi0 I γ P ⊂ 2.7(a)). If =rp (P (ω )/ ω i0 ), then, by Fact 5.2(c), sq (α), is forcing + ω + equivalent to the product ((P (ω)/ Fin) ∗π)×Q,where[ω] π =(Bˇ /Γ1) , and, by + Fact 5.2(a) and (b) applied in extensions by (P (ω)/ Fin) ,[ω] “π is an ω1-closed, separative atomless forcing”. By Lemma 5.1, sq P(α), ⊂ is forcing equivalent to + an iteration (P (ω)/ Fin) ∗ π1,where[ω] “π1 is an ω1-closed, separative atomless forcing”. Theorem 5.4. If h = ω1, then for each countable ordinal α ≥ ω the partial order P(α), ⊂ is forcing equivalent to (P (ω)/ Fin)+. ∼ Proof. If α<ω+ ω, then, by Theorem 3.1, sq P(α), ⊂ = (P (ω)/ Fin)+. Otherwise, by Theorem 5.3, P(α), ⊂ is forcing equivalent to a two-step iteration + (P (ω)/ Fin) ∗π,where[ω] “π is an ω1-closed, separative atomless forcing”. Now, V |= h = ω1 implies that CH holds in each generic extension V(P (ω)/ Fin)+ [G] and, by Fact 2.7(b) applied in V(P (ω)/ Fin)+ [G], the pre-order πG is forcing equivalent to ((P (ω)/ Fin)+)V [G]. But, since forcing by (P (ω)/ Fin)+ does not produce reals, ((P (ω)/ Fin)+)V [G] =((P (ω)/ Fin)+)V and, hence, P(α), ⊂ is forcing equivalent to (P (ω)/ Fin)+ × (P (ω)/ Fin)+.Now,inV we have c<ω1 = c and the posets + + + (P (ω)/ Fin) and (P (ω)/ Fin) ×(P (ω)/ Fin) are ω1-closed of size c. In addition, h = ω1 implies that they collapse c to ω1 and, by Fact 2.7(c) they are forcing equivalent (to Coll(ω1, c)). + n Example 5.5. If hn denotes the distributivity number of the poset ((P (ω)/ Fin) ) then, clearly, h ≥ h2 ≥ h3 ≥···≥ω1 and, by Corollary 3.8, h(sq P(ωn), ⊂)=hn. By a result of Shelah and Spinas [12], for each n ∈ N there is a model of ZFC in which hn+1 < hn and, hence, the posets P(ωn), ⊂ and P(ω(n +1)), ⊂ are not forcing equivalent. 6. Forcing with quotients over ordinal ideals δ δ The ideals Iωδ = {I ⊂ ω : ω → I},where0<δ<ω1, are called ordinal or indecomposable ideals.Ifδ = γ + r,whereγ ∈ Lim ∪{1} and r ∈ ω, then, by Facts 2.5, 2.6 and Theorem 3.1, we have δ γ+r + ∼ r γ + (6.1) sq P(ω ), ⊂ =(P (ω )/Iωγ+r ) = (rp (P (ω )/Iωγ )) . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FORCING WITH COPIES OF COUNTABLE ORDINALS 1783 δ + δ + Let hωδ = h((P (ω )/Iωδ ) )andtωδ = t((P (ω )/Iωδ ) ). Then we have Theorem 6.1. For each γ ∈ Lim ∪{1} we have (a) h ≥ hωγ ≥ hωγ+1 ≥ ··· ≥ hωγ+r ≥ ··· ≥ ω1 and, hence, there is r0 ∈ ω such ≥ that hωγ+r = hωγ+r0 ,foreachr r0; (b) t ≥ tωγ ≥ tωγ+1 ≥ ··· ≥ tωγ+r ≥ ··· ≥ ω1 and, hence, there is r0 ∈ ω such ≥ that tωγ+r = tωγ+r0 ,foreachr r0. δ Proof. (a) By Theorem 1.1, for each δ<ω1 the poset sq P(ω ), ⊂ is ω1-closed + δ and, by Theorem 5.3, (P (ω)/ Fin) →c sq P(ω ), ⊂.Thusω1 ≤ tωδ ≤ hωδ ≤ h. Itisknown(see[9])thath((rp(B))+) ≤ h(B+), for each Boolean algebra + r+1 γ + B satisfying h(B ) ≥ ω1,so,by(6.1),hωγ+r+1 = h((rp (P (ω )/Iωγ )) )= r γ + r γ + γ+r + h((rp(rp (P (ω )/Iωγ ))) ) ≤ h((rp (P (ω )/Iωγ )) )=h((P (ω )/Iωγ+r ) )= hωγ+r . γ (b) First we prove that tωγ ≤ t,forγ ∈ Lim. By Proposition 4.6, P (ω )\Iωγ , ⊂ ∼ γ + ∼ + = P (L) \I, ⊂ which implies (P (ω )/Iωγ ) = (P (L)/I) . Thus, by Claim 4.7, γ + + tωγ = t((P (ω )/Iωγ ) )=t((P (L)/I) )=t(P (L)\I, ⊂I ) ≤ t. The rest of the proof is similar to the proof of (a). We use the fact (see [9]) that t((rp(B))+) ≤ t(B+), + for each Boolean algebra B satisfying t(B ) ≥ ω1. ∼ Example 6.2. By Corollary 3.5(a) we have Iω2 = Fin × Fin and, hence, hω2 = h((P (ω × ω)/(Fin × Fin))+). In [3] Hern´andez-Hern´andez proved that in the Math- + ias model h((P (ω × ω)/(Fin × Fin)) )=ω1, while h = c = ω2.So,byTheorem 6.1, in this model we have ω2 = c = h = hω1 > hω2 = hω3 = ···= ω1. By a result of Szyma´nski and Zhou [13] the poset (P (ω × ω)/(Fin × Fin))+ is not ω2-closed. Thus, by Theorem 6.1(b), tω2 = tω3 = ···= ω1 holds in ZFC. We remark that the generic ultrafilter of (P (ω × ω)/(Fin × Fin))+ was analyzed by Blass, Dobrinen and Raghavan in [1] and similar analysis may be possible for the other quotient algebras appearing in this paper. References [1] A. Blass, N. Dobrinen, D. Raghavan, The next best thing to a P -point, submitted, http://arxiv.org/abs/1308.3790 [2] Roland Fra¨ıss´e, Theory of relations, Revised edition, Studies in Logic and the Foundations of Mathematics, vol. 145, North-Holland Publishing Co., Amsterdam, 2000. With an appendix by Norbert Sauer. MR1808172 (2002d:03084) [3] Fernando Hern´andez-Hern´andez, Distributivity of quotients of countable products of Boolean algebras, Rend. Istit. Mat. Univ. Trieste 41 (2009), 27–33 (2010). 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Department of Mathematics and Informatics, Faculty of Science, University of Novi Sad, Trg Dositeja Obradovica´ 4, 21000 Novi Sad, Serbia E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use