Arch. Math. Logic 42, 695–710 (2003) Digital Object Identifier (DOI): 10.1007/s00153-003-0174-2

Mathematical Logic

Teruyuki Yorioka

Forcings with the countable chain condition and the covering number of the Marczewski ideal Received: 1 May 2002 / Revised version: 18 October 2002 / Published online: 13 June 2003 – © Springer-Verlag 2003 Abstract. We prove that the covering number of the Marczewski ideal is equal to ℵ1 in the extension with the iteration of Hechler forcing.

1. Introduction and notation The Marczewski ideal s 0 has been introduced in 1935. We call a subset X of the reals a Marczewski null set, or an s 0 -set, if any perfect set of the reals has a perfect subset which does not meet X([JMS]). We denote the family of the Marczewski null subsets of the reals by s 0 , and call it the Marczewski ideal. In this paper, we deal with the covering number cov(s 0 ) of the Marczewski ideal, which is the smallest cardinality of a family of members of s 0 whose union covers all reals, and forcings with the countable chain condition: Cohen, random and Hechler forcings. In [Sp] Spinas has proved that cov(m0 ) = ℵ1 holds in the Cohen extension, where m0 is the Miller ideal (the collection of subsets X of ωω such that any superperfect set on ωω has a superperfect subset which does not meet X). He proved this by using the following lemma: If T is a C-name for a superperfect tree and π is a C-name for a member of ωω , then in V there exists a superperfect tree S such that
C “S and T are compatible and π ∈ [S].” It’s folklore (see Lemma 3.2) that in an extension which adds a new real, any perfect subset of the real line coded in the ground model has a perfect subset which does not meet the old reals. In section 2, we show that in extensions with Cohen, random and Hechler forcings, any perfect subset of the real line has a perfect subset which does not meet the old reals. That is, in these extensions, the set of old reals is Marczewski null. In [Bl1], Blass has proved that in any model obtained by adjoining uncountably many Cohen or random reals to any model of ZFC, g = ℵ1 holds, where g is the groupwise density number. This implies Spinas’ result by the above (because cov(m0 ) ≤ g holds in ZFC). The point of this proof is that the universe in the forcing extension is the strictly increasing union of an ℵ1 -chain of subuniverses. In section 3, we prove that cov(s 0 ) = ℵ1 in the extension with the iteration of Hechler forcing by an argument similar to Blass’ argument. And we also prove that g = ℵ1 T. Yorioka: Graduate School of Science and Technology, Kobe University, Rokkodai, Nada-ku, Kobe 657-8501, Japan. e-mail: [email protected]

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holds in the Hechler model (which is a conjecture of Blass, see the comments on this in section 11.6 of [Bl2]). In general, any finite support iteration of absolutely definable partial orders with the countable chain condition (for example the forcing notion adding an eventually different real or the localization forcing) should satisfy the first statement of lemma 3.10 (always true for the last statement of lemma 3.10). For a forcing notion P, let κ(P) be the least cardinal κ such that forcing with P adds a bijection from κ onto c and λ(P) the least cardinal λ such that forcing with P adds a cofinal function from λ into c (see also [GRSS]). The additivity add(s 0 ) of the Marczewski ideal is the least cardinal of a family of members of s 0 whose union is not in s 0 . Simon has proved that the Sacks forcing S collapses the continuum c to b, that is, κ(S) ≤ b holds ([Si]). By this result, we can easily see that if b < c then S collapses c. It is known (see [V] and [CL]) that it is consistent that b = c and S does not collapse cardinals. In [JMS], it has been proved that MA(+¬ CH) implies add(s 0 ) = λ(S). From this proof, we note that if κ(S) is regular, then add(s 0 ) ≤ κ(S) holds, and if m(C) = c, then add(s 0 ) = κ(S) holds. In [JMS], it’s also proved that“ MA +cov(s 0 ) = ℵ1 < ℵ2 = c” is consistent with ZFC, therefore so is “add(s 0 ) = κ(S) < b”. Thus it is relatively consistent with ZFC that b = c and S collapses the continuum. We show this by a different approach. In this paper we use Cohen, random and Hechler forcings which are defined as follows. Definition 1.1. (Cohen) For sets I, J and an infinite cardinal κ, let Fn(I, J, κ) be the partial order {f ⊆ I × J ; |f | < κ ∧ f is a function} ordered by reverse inclusion. Then define C = Fn(ω, 2, ω) and CI = Fn(I, 2, ω) for I ⊆ κ. (random) For an infinite cardinal κ, let µ be the standard product measure on 2κ , B(κ) the family of all Baire sets on 2κ and Nκ = {X ⊆ 2κ ; µ(X) = 0}, ordered by inclusion. Then define B = B(ω)/Nω and BI = B(I )/NI for I ⊆ κ. (Hechler) Define D = {s, f  ∈ ω<ω × ωω ; s ⊆ f }, for members s, f  and t, g of D, s, f  ≤ t, g if s ⊇ t and ∀n ∈
ω(f (n)  ≥ g(n)). In this paper, for p ∈ D we write p = sp , fp . For a forcing notion P, we denote the order by ≤P , and for conditions p, q, “p P q” means that p and q are compatible, and “p⊥P q” means that p and q are incompatible. We sometimes omit the subscript if it is clear. Cohen and random forcing satisfy the product lemmata. We use them to prove cov(s 0 ) = ℵ1 in the Cohen and random extension. Lemma 1.2 ([K2]). Let κ be an uncountable cardinal and X, Y subsets of κ with X ∩ Y = ∅ and X ∪ Y = κ. Let f ∈ 2κ . Then 1. f is Cκ -generic over V iff f  X is CX -generic over V and f  Y is CY -generic over V[f  X]. 2. f is Bκ -generic over V iff f  X is BX -generic over V and f  Y is BY -generic over V[f  X].

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In this paper, we consider the reals as functions in 2ω which is the set of functions from ω to 2. We denote 2<ω to be the set of finite sequences of 0’s and 1’s. For a subset T of 2<ω , T is called a tree if it is closed under taking initial segments. For a tree T ⊆ 2<ω and σ ∈ 2<ω , let Tσ := {t ∈ T ; t ⊆ σ ∧ σ ⊆ t}, [σ ] := {x ∈ 2ω ; σ ⊆ x} and [T ] := {f ∈ 2ω ; ∀n ∈ ω(f  n ∈ T )}, where f  n = f ∩ (n × ω). We note that X ⊆ 2ω is closed iff X = [T ] for some tree T ⊆ 2<ω . A tree T is called perfect if ∀σ ∈ T ∃τ, ν ∈ T (σ ⊆ τ, ν ∧ τ ⊥ ν), where τ ⊥ ν means that τ ⊆ ν and ν ⊆ τ , that is, τ and ν are incomparable. We note that X ⊆ 2ω is a perfect set iff X = [T ] for some perfect tree T ⊆ 2<ω . For a tree T , σ ∈ T is called a splitting node if both σ  0 and σ  1 are in T . We write splitn (T ) as the set of n-th splitting nodes. 2. Perfect sets in extensions In this section, we show that in extensions with Cohen, random and Hechler forcings, any perfect subset of the real line has a perfect subset which does not meet the old reals. In other words, Cohen, random and Hechler forcings force the set of old reals to a Marczewski null set. 2.1. Cohen extension It’s folklore that Cohen forcing adds a perfect set of Cohen reals, so this set does not meet the old reals. The point of this proof is that every countable forcing notion is equivalent to Cohen forcing. More explicitly, we may consider the partial ordered set [2<ω ]<ω (see also [BJ] etc.), ordered by end-extension. We show the following lemma using such a countable forcing notion for each C-name for a perfect tree. Lemma 2.1. In the Cohen extension, any perfect subset of the real line has a perfect subset which does not meet the old reals. Formally, letting G be C-generic over V, V[G] |= “∀ perfect A ⊆ R (A \ (R ∩ V) has a perfect subset )”, that is, V[G] |=“R ∩ V ∈ s 0 ”. Proof. We consider C = 2<ω , ordered by reverse inclusion. Let T˙ be a C-name for a perfect tree with [T˙ [G]] = A. We define a partial order Q(T˙ ) as follows: Q(T˙ ) = {f ; f is a function from 2n into [2<ω ]<ω for some n ∈ ω ∧ ∀σ ∈ n 2 (f (σ ) is a finite tree ∧ σ
“f (σ ) ⊆ T˙ ”) } and for conditions f, g ∈ Q(T˙ ), letting nf be such that dom(f ) = 2nf for a condition f ∈ Q(T˙ ), f ≤ g if nf ≥ ng and for every σ ∈ 2nf , f (σ ) is an end extension of g(σ  ng ). Since Q(T˙ ) is countable, Q(T˙ )  C holds, so there is a C-generic filter over V iff there is a Q(T˙ )-generic filter over V. Let H ∈ V[G] be Q(T˙ )-generic over V. We note that for every n ∈ ω there is a condition fn in H with dom(fn ) = 2n ˙ ˙ ˙ (by the genericity  of H ). For a Q(T )-name r˙ , put a Q(T )-name S(˙r ) as follows: ˙ ˙
“S(˙r ) := n∈ω fn (˙r  n).” Then in V[G],  1. S( G) is a perfect tree, 2. [S( G)] ∩ V = ∅, and 3. S( G) ⊆ T˙ [G].

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Check 1. For σ ∈ 2<ω and t ∈ 2<ω , let Dσ,t = {f ∈ Q(T˙ ); either ∀τ ∈ with σ ⊆ τ, ∃n < |t|(t  n ∈ f (τ ) ∧ (t  n) 1 − t (n) ∈ f (τ )) or ∃τ ∈ n f 2 with σ ⊆ τ, t ∈ f (τ ) ∧ ∃u, v ∈ f (τ )(t ⊆ u, v ∧ u⊥v)}, which is dense. 2nf

Given f ∈ Q(T˙ ) and τ ∈ 2nf , assuming t ∈ f (τ ), τ
“t ∈ f (τ ) ⊆ T˙ ∧ ∃v0 , v1 ∈ 2<ω (v0 ⊥v1 ∧ v0 , v1 ∈ T˙ ∧ t ⊆ v0 , v1 ).” Thus there are µ ≤ τ and a0 , a1 ∈ 2<ω such that a0 ⊥a1 , t ⊆ a0 , a1 and µ
“aˇ0 , aˇ1 ∈ T˙ .” And then we define f (τ )+ := f (τ ) ∪ {s ∈ 2<ω ; s ⊆ a0 ∨ s ⊆ a1 }. Let ng := |µ| and define a function g from 2ng into [2<ω ]<ω by: for ν ∈ 2ng , g(ν) = f (τ )+ if ν ⊇ µ, otherwise g(ν) = f (ν  nf ). Then g is an extension of f and is in Dσ,t . Similarly {f ∈ Q(T˙ ); ∃τ ∈ dom(f )(σ ⊆ τ ∧ every terminal node of f (τ ) has the length larger thann} is also dense for all σ ∈ 2<ω and n ∈ ω. Therefore S( G) is a perfect tree in 2ω . Check 2. For σ ∈ 2<ω and x ∈ 2ω ∩ V, {f ∈ Q(T˙ ); ∃τ ∈ dom(f )∃m ∈ ω(σ ⊆ τ ∧ x  m ∈ f (τ ))} is also dense. Check 3. It’s trivial by the definition of Q(T˙ ).   2.2. Random extension Neither B nor B × B add a perfect set of random reals, but there are many forcing notions that add a perfect set of random reals (see [BJ] or [BrJ]). The proof of the next lemma differs from the previous one. The point of the difference is that for names for perfect trees, we do not make forcing notions but names for its perfect subtrees. Lemma 2.2. Letting G be B-generic over V, V[G] |= “∀ perfect A ⊆ R (A \ (R ∩ V) has a perfect subset )”, that is, V[G] |=“R ∩ V ∈ s 0 ”. Proof. It’s enough to show that for a condition b ∈ B and a B-name T˙ such that b
“ T˙ ⊆ 2<ω is a perfect tree ”, there are an extension c ≤ b and a B-name S˙ ˙ ∩ V = ∅ ”. such that c
“ S˙ ⊆ T˙ , S˙ is a perfect tree, and [S] To show this, we use the following two claims. Claim 2.3. Assume that b ∈ B and a B-name T˙ such that b
“T˙ ⊆ 2<ω is a perfect tree ”. Then ∀ε > 0∀n ∈ ω∃k ∈ ω(µ([[|T˙ ∩ 2k | ≥ n]] · b) > µ(b)(1 − ε)). Proof. It’s true that b
“∀s ∈ T˙ ∩ 2<ω ∃t = u ∈ T˙ ∩ 2<ω (s ⊆ t, u)”. For a natural number n, there are c ≤ b and ti ; i < 2n  such that c
“{ti ; i < 2n } is the set of all n-th splitting level nodes of T˙ ". Thus for every n ∈ ω, there is c ≤ b such that c
“|T˙ ∩2k | ≥ n ”. Since if k ≤ l then [[|T˙ ∩ 2k | ≥ n]] ≤ [[|T˙ ∩ 2l | ≥ n]], k∈ω [[|T˙ ∩ 2k | ≥ n]] · b = b. Therefore for a positive real ε and a natural number n, there is k ∈ ω such that µ([[|T˙ ∩ 2k | ≥ n]] · b) > µ(b)(1 − ε).  

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Claim 2.4. Assume that b ∈ B, m, n ∈ ω and a B-name T˙ such that b
“T˙ ⊆ 2<ω is a perfect tree ∧ |T˙ ∩ 2n | ≥ 2m ”. Then there is a name S˙ such that 1. b
“S˙ ⊆ T˙ ∩ 2n ∧ |S˙ ∩ 2n | ≥ 2”, ˙ · b) ≤ µ(b)/m), and 2. ∀s ∈ 2n (µ([[s ∈ S]] 3. ∀c ∈ B(c ⊥ b ⇒ c
“S˙ ∩ 2<ω = ∅”).    Proof. Let S := {A ∈ [2n ]2m ; µ( s∈A [[s ∈ T˙ ]]) > 0}. Then A∈S s∈A [[s ∈ T˙]] ≥ b. Let S  ⊆ S and dA ∈ B for A ∈ S  be such that for all A ∈ S  , ˙ dA ≤ s∈A [[s ∈ T ]] ·b, A∈S  dA = b and for all distinct A and B in S  , dA ⊥ dB . For every A ∈ S  , we take {ai ; i < 2m} ⊆ [A]2 such that |{i < 2m; t ∈ ai }| = 2 for all t ∈ A. We define dAi ∈ B for A ∈ S  and i < 2m such that  1. i<2m dAi = dA , j 2. ∀i = j < 2m(dAi ⊥dA ), and 3. ∀i < 2m(µ(dAi ) = µ(dA )/2m). Then we define a name S˙ such that for all A ∈ S  and i < 2m, [[S˙ = ai ]] · dA = i dA and for any condition c ∈ B incompatible with b, c
“S˙ is not a set of members ˙ ≤ b.) of 2<ω ”. (So for all s ∈ 2n , [[s ∈ S]]   Suppose that b
“T˙ ⊆ 2<ω is a perfect tree”. By induction on m ∈ ω, we construct conditions bm ∈ B, numbers km ∈ ω and B-names S˙m such that • b = b0 ≥ b1 ≥ · · · and k0 ≤ k1 ≤ k2 ≤ · · · , • ∀m ∈ ω(bm
“S˙m ⊆ T˙ ∩2km ∧|S˙m ∩2km | ≥ 2m ∧∀s ∈ S˙m ∃t, u ∈ S˙m+1 (t ⊥ u∧ s ⊆ t, u)”), • ∀m ∈ ω∀t ∈ 2km (µ([[t ∈ S˙m ]] · bm ) ≤ µ(bm )/m), • ∀m ∈ ω(µ(bm+1 ) ≥ µ(bm ) · (1 − 2−m−3 )). ˇ Basic Step m = 0, 1 Let b0 = b, k0 = 0 and S˙0 be such that b
“S˙0 = {∅}”. By Claim 2.3, there is k1 ∈ ω such that µ([[|T˙ ∩ 2k1 | ≥ 2]] · b) ≥ µ(b) · (1 −

1 ). 24

Let b1 = [[|T˙ ∩ 2k1 | ≥ 2]] · b. By Claim 2.4, there is a B-name S˙1 such that • b1
“S˙1 ∩ 2k1 ⊆ T˙ ∩ 2k1 ∧ |S˙1 ∩ 2k1 | ≥ 2”, and • ∀t ∈ 2k1 (µ([[t ∈ S˙1 ]] · b1 ) ≤ µ(b1 )). Induction Step m → m+1 Suppose that bm
“S˙m = {t˙i ; i < 2m } ⊆ T˙ ∩2km ”. By Claim 2.3, there is km+1 ∈ ω such that for all i < 2m , µ([[|T˙t˙i ∩ 2km+1 | ≥ 2m + 2]] · bm ) ≥ µ(bm ) · (1 − 

1 1 · ). 2m+4 2i+1

[[|T˙t˙i ∩ 2km+1 | ≥ 2m + 2]] · bm . Then  1 1 1 · i+1 ) ≥ µ(bm ) · (1 − m+4 ). µ(bm+1 ) ≥ µ(bm ) · (1 − m+4 2 2 2 m

Let bm+1 =

i<2m

i<2

By Claim 2.4, for all i <

2m

i there are B-names S˙m+1 such that

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i i • bm+1
“S˙m+1 ∩ 2km+1 ⊆ T˙t˙i ∩ 2km+1 ∧ |S˙m+1 ∩ 2km+1 | ≥ 2”, and i k • ∀t ∈ 2 m+1 (µ([[t ∈ S˙m+1 ]] · bm+1 ) ≤ µ(bm+1 )/(m + 1)).  i Let S˙m+1 be such that bm+1
“S˙m+1 = i<2m S˙m+1 ”, which completes the construction.  Let c = m∈ω bm . Since for all m ≥ 1, 1 µ(bm ) = µ(bm−1 ) · (1 − 2m+2 ) = ···  1 = µ(b0 ) · i
Since bm ; m ∈ ω is a decreasing sequence and Lebesgue measure µ is σ -additive, µ(c) > 0 holds, hence c is a condition of B. Let S˙ be such that c
“S˙ = {σ ∈ 2ω ; ∃m ∈ ω∃τ ∈ S˙m (σ ⊆ τ )}”. By the construction, c
“S˙ ⊆ 2<ω is a perfect ˙ ∩ V = ∅”, which completes the subtree of T˙ ”. From now on we show that c
“[S] proof. ˙ Then Assume not, then there are c ≤ c and x ∈ 2ω ∩V such that c
“x ∈ [S]”.  ˙ Therefore for every n ∈ ω, µ([[x  n ∈ S]]) ˙ ≥ µ(c ). c
“∀n ∈ ω(x  n ∈ S)”. Now we choose m ∈ ω such that µ(c ) > µ(b)/m. We take t ∈ 2km such that t = x  km . Then ˙ µ(c ) ≤ µ([[x  kn ∈ S]]) ˙ = µ([[t ∈ S]]) = µ([[t ∈ S˙m ]]) = µ([[t ∈ S˙m ]] · bm ) ≤ m1 µ(bm ) ≤ m1 µ(b) < µ(c ), which is a contradiction.

 

2.3. Hechler extension The following lemma is proved by an argument similar to the previous one, that is, for names for perfect trees we make names for its perfect subtrees. Lemma 2.5. Letting G be D-generic over V, V[G] |= “∀ perfect A ⊆ R (A \ (R ∩ V) has a perfect subset )”, that is, V[G] |=“R ∩ V ∈ s 0 ”. Proof. Let T˙ be a D-name for a perfect tree on 2<ω (, i.e.
“T˙ ⊆ 2<ω is perfect”). By induction on n ∈ ω, we construct D-names L˙ n , n ∈ ω, for members of n [2<ω ]2 and maximal antichains An , n ∈ ω, on D such that  (i)
“ n∈ω L˙ n ⊆ T˙ ∧ ∀n ∈ ω∀σ ∈ L˙ n ∃τ, ρ ∈ L˙ n+1 (σ ⊆ τ, ρ ∧ τ ⊥ ρ)”, (ii) ∀n ∈ ω(An+1 refines An ),

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(iii) ∀n ∈ ω∀p ∈ An (n ≤ |sp | ∧ p decides the value of L˙ n ), (define v(L˙ n , p) to be the value of L˙ n decided by p) (iv) ∀n ∈ ω∀p ∈ An (p
“L˙ n is pairwise incompatible”), (v) ∀p ∈ D∃k ∈ ω∃q, r ≤ p(q ⊥ r ∧ both q and r decide the value of L˙ k ∧ ∀σ ∈ v(L˙ k , q)∀τ ∈ v(L˙ k , r)(σ ⊥ τ )).  ˇ 1 } and A0 = {1}, where 1= ∅, 0, 0, · · · ∈ D. Basic Step n = 0 Let L˙ 0 = { ∅, Induction Step n → n + 1 Assume that we have already constructed L˙ n and An . Now we consider a fixed p ∈ An . Since
“T˙ is perfect”, p
“∃K ∈ n+2 [2<ω ]2 (L˙ n ⊆ K ⊆ T˙ ∧ ∀σ ∈ L˙ n ∃g ∈ K 4 (∀i < 4(σ ⊆ g(i)) ∧ ran(g) is p pairwise incompatible))”. By the maximal principle, there is a D-name K˙ n for a n+2 p p ˙ <ω 2 4 member of [2 ] such that p
“L˙ n ⊆ K˙ n ⊆ T˙ ∧ ∀σ ∈ L˙ n ∃g ∈ (Kn ) (∀i < 4(σ ⊆ g(i))∧ran(g) is pairwise incompatible )”. Then there is a maximal antichain p p p Bn below p such that all members of Bn decide the value of K˙ n . 
In this proof, for a condition p of D and j ∈ ω, we define p(j ) = sp(j ) , fp(j ) such that sp(j ) = sp  j  and fp(j ) (n) = j if n = |sp |, otherwise fp(j ) (n) = fp (n) p for any n ∈ ω. We note that for all p ∈ An , {q (j ) ; q ∈ Bn ∧ j ≥ fq (|sq |)} is a maximal antichain below p. 4 p p,q,σ p ∈ v(K˙ n , q) be such that for all For σ ∈ v(L˙ n , p) and q ∈ Bn , let gn p,q,σ p,q,σ i < 4, σ ⊆ gn (i) and ran(gn ) is pairwise incompatible. We define L˙ n+1 and An+1 such that p

• An+1 = {q (j ) ; ∃p ∈ An (q ∈ Bn ∧ j ≥ fq (|sq |))}, and p • ∀p ∈ An ∀q ∈ Bn ∀j ≥ fq (|sq |), p,q,σ p,q,σ (0), gn (1); σ ∈ v(L˙ n , p)},” – j : even ⇒ q (j )
“L˙ n+1 = {gn p,q,σ p,q,σ (j ) ˙ – j : odd ⇒ q
“Ln+1 = {gn (2), gn (3); σ ∈ v(L˙ n , p)}.” By the construction, (i) through (iv) clearly holds. We check the statement (v) now. Assuming that p is a condition of D, we take k ≥ |sp |. Then there is p  ∈ Ak which is compatible with p. Since |sp | ≥ k ≥ |sp |, sp ⊇ sp holds. We choose p

p ∈ Bk which is compatible with p. Since p  is an extension of p  , sp ⊇ sp ⊇ sp holds. We define conditions q and r such that, taking an large enough j ∈ ω, sq = sp  j , sr = sp  j + 1 and for all n ∈ ω,

fq (n) =

fr (n) =

if n < |sq | sq (n) , max{fp (n), fp (n)} otherwise if n < |sr | sr (n) , max{fp (n), fp (n)} otherwise

which are as desired. Let S˙ be a D-name such that
“S˙ = {σ ∈ 2<ω ; ∃n ∈ ω∃τ ∈ L˙ n (σ ⊆ τ )}”. By the condition (i),
“S˙ is a perfect subtree of T˙ ”. From now on we show that ˙ ∩ V = ∅”, which completes the proof.
“[S]

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˙ So p
If not, there are p ∈ D and x ∈ 2ω ∩ V such that p
“x ∈ [S]”. ˙ “∀n ∈ ω∃σ ∈ Ln (σ ⊆ x)”. We take k ∈ ω and q, r ≤ p satisfying (v) for p. Then there are σ ∈ v(L˙ k , q) and τ ∈ v(L˙ k , r) such that σ, τ ⊆ x. Thus σ and τ are compatible, which contradicts (v).   3. The covering number of the Marczewski ideal in extensions In this section, we show that in extensions with Cohen forcing, random forcing and an iteration of Hechler forcing, cov(s 0 ) = ℵ1 holds. We prove these by an argument similar to Blass’ argument mentioned in the introduction. Main Lemma 3.1. Suppose that W is a universe and Vα (α < ω1 ) are subuniverses of W such that W ∩ R = α<ω1 Vα ∩ R and ∀α < β < ω1 (Vα ∩ R
Vβ ∩ R). If for every perfect set A ∈ W, ∀α < ω1 (W |= “A \ (Vα ∩ R) has a perfect subset "), then W |=“cov(s 0 ) = ℵ1 .”

  ∩ R are s 0 -sets in W. If a universe satisfies the

This is clear because then all Vα assumption of the lemma, then g = ℵ1 holds in the universe (see [Bl1] pp 22–25). The results in cases of Cohen and random extensions are folklore. The points of the proof are the product lemmata and the following well-known lemma.

Lemma 3.2 (Folklore). Let V and W be models of ZFC such that V is a subuniverse of W and there is a real in W \ V. If P ⊆ R is coded in V and W |= “P is perfect ”, then W |= “∃Q ⊆ P (Q is perfect ∧ Q ∩ V = ∅)”. Proof. In this proof, we consider R as 2ω . Let r ∈ (W \ V) ∩ 2ω . Let T ⊆ 2<ω be a perfect tree coded in V. By recursion on n ∈ ω, we define Sn := {s ∈ T ; ∃t  ∈ Sn−1 ∃t ∈ split2n (T )(t  ⊆ t ⊆ s ∧ |s| = |t| + 1 ∧ s(|t|) = r(n))}. Let S := {s ∈ T ; ∃k ∈ ω∃t ∈ Sk (s ⊆ t)}. Then S is also perfect and [S] ∩ V = ∅. (If x ∈ [S] ∩ V, since T is coded in V, there is ki ; i ∈ ω ∈ ωω ∩ V such that x  ki ∈ splitki (T ) for all i ∈ ω. Then r(n) = x(k2n ) for all n ∈ ω, by the definition of S, which is a contradiction.)   Theorem 3.3 (Folklore). Let κ be an uncountable regular cardinal. In the extensions with Cκ and Bκ , cov(s 0 ) = ℵ1 holds.  Proof. Let Iα ; α < ω1  be an increasing sequence of subsets of κ such that α<ω1 Iα = κ and Iα
Iβ for any α < β < ω1 and Vα = VCIα . Let W = VCκ . Vα ; α < ω1  satisfies the assumption of Main lemma 3.1. We now show that Vα ; α < ω1  satisfies the conclusion of Main lemma 3.1. We fix α < ω1 and let T˙ be a Cκ -name for a perfect tree (the case of Bκ is similar). Let β < ω1 be such that T˙ is a CIβ -name. By the product lemma 1.2, T˙ [G] is a perfect tree coded in Vβ (where G is Cκ -generic over V). So by Lemma C 3.2, there is a perfect subtree S ⊆ T˙ [G] in W such that [S] ∩ V Iβ is empty, hence   [S] ∩ VCIα is also empty.

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3.1. Iteration of Hechler forcing One must be careful when showing that the value of cov(s 0 ) is equal to ℵ1 in the iteration of Hechler forcing, because there is no product lemmata, as for Cohen or random forcings, for iterations Dκ of Hechler forcing. But we can consider Dκ as a limit of ℵ1 many iterations of Hechler forcing. Throughout this section, let κ be a cardinal with an uncountable cofinality and Iα ; α < ω1  such that • ∀α < β < ω1 (Iβ \ Iα is cofinal in κ), •  ∀α < β < ω1 (Iα
Iβ ), and • α<ω1 Iα = κ. For α < ω1 , we define iterations Pα as follows: γ γ +1 γ γ By induction on κ, we define Pα , γ ≤ κ, such that Pα = Pα ∗ Q˙α , where

˙ if γ ∈ Iα {1} γ Q˙α = ˙ . D otherwise
 γ is a direct limit of Pνα ; ν < γ . Then we let Pα = Pκα for all If γ < κ is limit, Pα  α < ω1 and Pω1 = α<ω1 Pα . Then Pω1 is a finite support iteration of Hechler forcing Dκ . Suppose that P and Q are forcing notions with P ⊆ Q. Then we write P ◦ Q and say P is completely embedded in Q iff

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1. ∀p, q ∈ P(p P q ⇐⇒ p Q q), and 2. ∀A ⊆ P(A is a maximal antichain in P ⇒ A is a maximal antichain in Q).

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Lemma 3.4 (Folklore). P ◦ Q iff 1. ∀p, q ∈ P(p ≤P q ⇐⇒ p ≤Q q), and 2. ∀q ∈ Q∃p ∈ P∀r ∈ P(r ≤P p ⇒ r Q q).

 

The following is well-known. It is an application of Suslin forcing (see also [BJ]).

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Lemma 3.5. ∀α < β ≤ ω1 (Pα ◦ Pβ ). Proof. It’s enough to show that ˙ ◦ Q ∗ D, ˙ and 1. P ◦ Q ⇒ P ∗ D γ γ 2. if γ is a limit ordinal and Pνα ◦ Pνβ for all ν < γ , then Pα ◦ Pβ .

<

<

<

<

˙ Without loss of generality, we may assume that for s ∈ ω<ω 1. Let q ∈ Q ∗ D.
 and Q-name f˙, q(0)
Q “q(1) = sˇ , f˙ ”. Since P ◦ Q, we can get a reduced Q-name g˙ satisfying the condition 2 of Lemma 3.4 (see also [Br2]): For simplicity, we may assume that P and Q are the complete Boolean algebras. For t ∈ ω<ω with s ⊆ t, let bt := [[t ⊆ f˙ ]] Q · q(0) and at∗ ∈ P a projection of bt . (Then bs = q(0) and for n > |s|, q(0) = t⊇s&|t|=n bt and {bt ; t ⊇ s&|t| = n} is a maximal antichain below q(0).) We define at by recursion on n = |t| and j ∈ ω such that as := as∗ and at := at
(n−1) · (at∗ \ j
<

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 at = j ∈ω at j  and {at j  ; j ∈ ω} is a maximal  antichain below at . Therefore by induction on n ∈ ω, we can show that as∗ = t∈ωn &t⊇s at .) We define a P-name ˙ ˙ Q ). (Then as∗
“ s, g ˙ ∈ D”.) g˙ such that as∗
“g˙ ∈ ωω ” and ∀t ⊇ s(at = [[t ⊆ g]] ˙ Let p = as , ˇs , g ˙ ∈ P ∗ D which is a reduction of q. γ 2. We have nothing to do because for p ∈ Pβ , there exists ν < γ such that p is also in Pνβ . By inductive hypothesis, we can find a reduction q ∈ Pνα of p in Pνβ , γ   which is also a reduction of p in Pβ . We now show that for α < β ≤ ω1 , the extension with Pα is a proper subuniverses of the extension with Pβ . To prove this, we must note that Claim 3.6. (See [K1] page 244, D4). Assume P of a forcing notion that is a subset of Q,

<◦ Q and let Q/P be the P-name

ˇ q is compatible with every p ∈ G}” ˙
P“Q/P = {q ∈ Q; ˙ = {p, (where G ˇ p; p ∈ P}). Let G be a P-generic filter over V and H a (Q/P)[G]generic filter over V[G]. Then 1. H is Q-generic over V, and 2. G = H ∩ P.

 

Let α < β ≤ ω1 , G a Pα -generic filter overV and H a (Pβ /Pα )[G]-generic ˙ ) [H  γ ] for γ ∈ Iβ , filter over V[G]. In V[H ], we define dγ := p∈H sp(γ (H  γ := {p ∈ H ; supp(p) ⊆ γ }) which is considered as a Hechler real added at  stage γ . By the above claim, for all γ ∈ Iα , dγ = p∈G sp(γ ˙ ) [G  γ ] is in V[G]. We prove that dγ ∈ V[G] for all γ ∈ Iβ \ Iα . To show this, we prepare a few lemmata. Lemma 3.7 (Folklore). For α < ω1 and m ∈ ω, let D(Pα , m) := {p ∈ Pα ; ∃ 
˙ ) = sˇγ ” ∧|sγ | ≥ m)}. sγ ; γ ∈ supp(p) ⊆ ω<ω ∀γ ∈ supp(p)(p  γ
“sp(γ Every D(Pα , m) is dense in Pα . Proof. Take a condition p ∈ Pα . By recursion on n ∈ ω, we choose pn ∈ Pα and sn ∈ ω<ω such that • • • •

γ−1 := κ, γn = max(supp(pn )), γn+1 < γn and |sn | ≥ m for all n ∈ ω, p0 = p, pn ∈ Pα  γn−1 , pn+1  γn ≤ pn  γn and pn+1
γn “spn˙(γn ) = sˇn ” for all n ∈ ω, if supp(pn ) = {0}, let pk = pn and γk = 0 for all k > n.

Since γn ; n ∈ ω is a decreasing sequence of ordinals, there is n ∈ ω such that γn = 0. Let n be the least such number. We define p  ∈ Pα such that supp(p  ) = {γi ; i < n}, p (0) = pn−0 (0) and p  γn−i
“p  (γn−i ) = p(γn−i )” for all i ≤ n. Then p   γn−(i+1) ≤ pn−i holds for i ≤ n. Therefore p   γ−1 (= p  ) ≤ p0 = p   and p ∈ D(Pα , m).

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Therefore from now on we may consider all conditions of Pβ as members of D(Pβ , 0). Now assume that α < β ≤ ω1 and 0 ∈ Iβ \ Iα ( by passing to an intermediate model). Let t, g ∈ D and j := g(|t|). (We consider a condition t, g in D as a condition p in Pβ  γ for every γ < κ whose support is {0} and p(0) = t, g.) Recall that for a condition p of D and j ∈ ω, p(j ) = sp(j ) , fp(j )  is a condition of D such that sp(j ) = sp  j  and fp(j ) (n) = j if n = |sp |, otherwise fp(j ) (n) = fp (n) for any n ∈ ω. For all i ≥ j , we define an embedding ei [t, g]  1(= ei ) from D[t, g(i) ] := {p ∈ D; p ≤ t, g(i) } into D[t, g(j ) ] as follows: For all p ∈ D[t, g(i) ],

s (n) if n = |t| for all n < |sp |, • |sei (p) | = |sp | and sei (p) (n) = p j if n = |t| • fei (p)  [ |sp | + 1, ω) = fp  [ |sp | + 1, ω). It is easy to see that ei [t, g]  1 is an isomorphism. By recursion on γ < κ, we define an isomorphism ei [t, g]  γ (= ei ) from Pβ [t, g(i) ]  γ := {p ∈ Pβ  γ ; p(0) ≤ t, g(i) } onto Pβ [t, g(j ) ]  γ as follow:
 ˙ ) ∈ Let p ∈ Pβ [t, g(i) ]  (γ + 1), then p  γ
“p(γ ) = sp(γ ˇ ) , fp(γ ˙ Without loss of generality, we may consider fp(γ ˙ ) as a (Pβ [t, g(i) ]  γ )D”. name. Define ei (p) such that ei (p)  γ = ei (p  γ ), sei (p)(γ ˙ ) = sp(γ ˇ ) and ˙ ) = (ei )∗ (fp(γ ˙ ) ), where for an embedding e : P → Q and a P-name fei (p)(γ σ e∗ (σ ) is a Q-name such that e∗ (σ ) = {e∗ (τ ), e(p) ; τ, p ∈ σ } (see [K1] 7.12. & 13.). We note that if e : P → Q is a dense embedding, and if φ(x1 , · · · , xn ) is a formula and σ1 , · · · , σn are P-names, then p
P“φ(σ1 , · · · , σn )” ⇐⇒ e(p)
Q“φ(e∗ (σ1 ), · · · , e∗ (σn ))”.  If γ is a limit  ordinal, then ei  γ = δ<γ ei  δ, i.e.  ei (p) = (ei  ( supp(p) + 1))(p) (we note that p ∈ Pβ  ( supp(p) + 1)). We check that ei is really an isomorphism. γ → γ + 1 For p, q ∈ Pβ [t, g(i) ]  (γ + 1), p ≤ q ⇔ p  γ ≤ q  γ ∧ p  γ
“p(γ ) ≤ q(γ )” ˙ ) ≤ fp(γ ˙ )” ˇ ) ∧ fq(γ ⇔ p  γ ≤ q  γ ∧ p  γ
“sp(γ ˇ ) ⊇ sq(γ ˇ ) ⇔ ei (p)  γ ≤ ei (q)  γ ∧ ei (p)  γ
“sei (p)(γ ˙ ) ≤ fei (p)(γ ˙ )” ⊇ sei (q)(γ ˇ ) ∧ fei (q)(γ (see also [K1] 7.13) ⇔ ei (p) ≤ ei (q). It’s possible to prove that ei is injective (using the above argument). Show that ei is onto. Let p ∈ Pβ [t, g(j ) ]  (γ + 1). By inductive hypothesis, ˙ )∈ there is q  γ ∈ Pβ [t, g(i) ]  γ such that ei (q)  γ = p  γ . Since p  γ
“fp(γ (j ) ω ω ”, for any k ∈ ω there is a maximal antichain Ak ⊆ Pβ [t, g ]  γ and a

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˙ ) (k) ˇ = Fkˇ(r)” for all r ∈ Ak . We function Fk from Ak into ω such that r
“fp(γ (j ) ˙ define a Pβ [t, g ]  γ -name fq(γ ) for a member of D such that ˙ ) (k) ˇ = Fkˇ(r)”), ∀k ∈ ω∀r ∈ Ak ∀r  ∈ Pβ [t, g(i) ]  γ (ei (r  ) ≤ r ⇒ r 
“fq(γ  ˇ r  ∈ fq(γ ˙ ) . Then
“fq(γ ˙ ) ∈ ωω ”, hence q  (γ + 1) ∈ i.e. (k, Fk (r) ), ˙ ) = fp(γ ˙ ) ” also holds. Pβ [t, g(i) ]  (γ + 1) holds. And ei (q)  γ
“fei (q)(γ  Assume not, there is q ≤ ei (q)  γ and numbers k, n0 and n1 ∈ ω such that ˙ ) (k) ˇ = nˇ0 = nˇ1 = fp(γ ˙ ) (k) ˇ ”, hence n0 = n1 . Let r ∈ Ak and q 
“fei (q)(γ (i)   r ∈ Pβ [t, g ]  γ such that ei (r ) is a common extension of r and q  . Then ˇ r ∈ ˙ ) (k) ˇ = Fkˇ(r)”, hence Fk (r) = n1 . Since (k, n1  ), ei (r  )
“nˇ1 = fp(γ  ˇ ei (r  ) ∈ fe (q)(γ ˙ ) , (k, n1  ), ˙ ) , i.e. ei (r  )
“nˇ1 = fei (q)(γ ˙ ) (k) ˇ = nˇ0 ”, hence fq(γ i n0 = n1 which is a contradiction. γ : limit In this case, it’s trivial that ei is an isomorphism by inductive hypothesis because for every p ∈ Pβ [t, g(i) ]  γ , there is γ  < γ with p ∈ Pβ [t, g(i) ]  γ  . Lemma 3.8. Let α < β ≤ ω1 , 0 ∈ Iβ \ Iα , t, g ∈ D and i ≥ g(|t|). For any p ∈ Pα and q ∈ Pβ [t, g(i) ], q ≤ p iff ei (q) ≤ p. ˇ = n” ˇ = n” ˇ for r ∈ Proof. We prove that r
“f˙(k) ˇ ⇐⇒ ei (r)
“f˙(k) Pβ [t, g(i) ]  γ , a Pα  γ -name f˙ for a function in ωω and k, n ∈ ω, by induction on γ < κ. Then we can show the lemma by induction on γ < κ as follows: For p ∈ Pα  γ and q ∈ Pβ [t, g(i) ]  γ , q ≤ p ⇔ q  γ ≤ q  γ ∧ p  γ
“p(γ ) ≤ q(γ )” ˙ ) ≤ fq(γ ˙ )” ⇔ q  γ ≤ p  γ ∧ q  γ
“sq(γ ˇ ) ⊇ sp(γ ˇ ) ∧ fp(γ ˙ ) ≤ fei (q)(γ ˙ )” ⇔ ei (q)  γ ≤ p  γ ∧ ei (q)  γ
“sei (q)(γ ˇ ) ⊇ sp(γ ˇ ) ∧ fp(γ ⇔ ei (q) ≤ p. It’s easy to show in case γ = 1 and γ limit, because p(0) = 1 and for every p ∈ Pβ [t, g(i) ]  γ , there is γ  < γ with p ∈ Pβ [t, g(i) ]  γ  . Assume that r ∈ Pβ [t, g(i) ]  γ , f˙ is a Pα  γ - name for a function in ˇ = n” ˇ = n”. ωω , k, n ∈ ω, r
“f˙(k) ˇ and ei (r) 
“f˙(k) ˇ Let B ⊆ Pα  γ be a maximal antichain and a function F from B into ω such that for all u ∈ B, ˇ ˇ = F (u)”. ˇ = m”. u
“f˙(k) There exist r  ≤ r and m = n such that ei (r  )
“f˙(k) ˇ   Then there is u ∈ B which is compatible with r . Let r be a common extension of ˇ = f˙(k) ˇ = f˙(k) ˇ = n” ˇ = m”. r  and u. Then r 
“F (u) ˇ and ei (r  )
“F (u) ˇ Hence n = F (u) = m, which is a contradiction.   For a condition p ∈ Pβ and i ≥ fp(0) (|sp(0) |), we define an extension p i of p such that pi (0) = p(0)(i) and p i (γ ) = p(γ ) for all γ = 0 < κ. We note that if i ≥ fp(0) (|sp(0) |), p i ≤ p holds. Lemma 3.9. Let α < β ≤ ω1 , 0 ∈ Iβ \ Iα , t, g ∈ D and i ≥ g(|t|). For any p ∈ Pβ there is p+ ≤ p such that p+ (0) = p(0) and for all q ∈ Pβ and i ≥ j := fp(0) (|sp(0) |), q ≤ p+ i implies ei (q) ≤ p+ j .

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˙ ) ∈ ωω ”, so we can conProof. For γ ∈ supp(p), p j  γ
“fpj˙ (γ ) = fp(γ (j ) sider fpj˙ (γ ) as a (Pβ [t, g ]  γ )-name. Therefore for any k ∈ ω, there is a maximal antichain Ak in Pβ [t, g(j ) ]  γ and a function Fk from Ak into ω ˇ = Fkˇ(r)” for every r ∈ Ak . For i ≥ j , we define a such that r
“fpj˙ (γ ) (k) (i) (Pβ [t, g ]  γ )-name a˙i for a function in ωω such that γ

ˇ = Fkˇ(r)”). ∀k ∈ ω∀r ∈ Ak ∀r ∈ Pβ [t, g(i) ]  γ (ei (r  ) ≤ r ⇒ r 
“a˙γi (k) 

We note that pi  γ
“a˙γi ∈ ωω ”. We define a condition p+ ∈ Pβ such that • supp(p+ ) = supp(p) and p+ (0) = p(0), • for all γ ∈ supp(p) \ {0}, – p+  γ
“sp+˙(γ ) = sp(γ ˇ ) ”, ˙ ) ”, – p+ j  γ
“fp+˙ (γ ) = fp(γ ˙ )  |sp(γ ˇ ) | = fp(γ ˇ ) |(= sp(γ ˇ ) |(fp+˙ (γ ) – p+ i  γ
“fp+˙ (γ )  |sp(γ ˇ ) ) ∧ ∀k ≥ |sp(γ ˙ i ˙ (k) = fp(γ ) (k) + aγ (k))”, for all i > j . By the definition, p+ ≤ p. From now on we check that p+ works. Assume that q ≤ p+ i . By induction on γ ∈ supp(q), show that ei (q)  (γ + 1) ≤ p+ j  (γ + 1). Basic Step(γ = 0) Since q(0) ≤ p+ i (0) = p+ (i) (0) and ei is an isomorphism, ei (q)(0) ≤ ei (p+ (i) )(0) = p+ (j ) (0) = p+ j (0). Induction Step Assume that γ ∈ supp(q) and ei (q)  γ ≤ p+ j  γ . Since q  γ
“q(γ ) ≤ p+ i (γ )”, which is implied from q ≤ p+ i , sei (q)(γ ) = sq(γ ) ⊇ ˙ ) (k) + ˇ ) |(fp i˙ (γ ) (k) = fp(γ sp+ i (γ ) = sp(γ ) = sp+ j (γ ) and q  γ
“∀k ≥ |sp(γ + ˙ i ˙ ˙ ˙ aγ (k) ≤ fq(γ ) (k))”. If ei (q)  γ 
“fp+ j (γ ) ≤ fei (q)(γ ) ”, then there are k ≥ ˇ = nˇ0 > nˇ1 = |sp(γ ) |, n0 , n1 ∈ ω and r ≤ ei (q)  γ such that r
“fp+ j˙ (γ ) (k) (i)   ˙ ) (k)”. ˇ We take r ∈ Pβ [t, g ]  γ with ei (r ) = r. Then r  ≤ q  γ . fei (q)(γ ˇ = nˇ0 ”. Then r 
Strengthening r  if need, we may assume that r 
“a˙γi (k) ˇ = fp(γ ˙ ) (k) ˇ + nˇ0 > nˇ1 = fq(γ ˙ ) (k)”, ˇ which is a contradiction.   “fp i˙ (γ ) (k) +

We note that for i ≥ fp(0) (|sp(0) (0)|), it is true that p+ i ≤ pi ≤ p.  Lemma 3.10. ∀α < β ≤ ω1 (VPα ∩2ω
VPβ ∩2ω ) and VDκ ∩2ω = α<ω1 VPα ∩ 2ω . Proof. Let γ ∈ Iβ \ Iα . Without loss of generality, we may assume that γ = 0 ∈ Iβ \ Iα . (We work in V[H  γ ].) Let x˙ be a Pα -name for a real in 2ω (so is a Pβ -name). It’s enough to show ˇ modulo 2 ⊆ x˙ ” } is dense in Pβ , because then d0 that Dx˙ := {p ∈ Pβ ; p
“sp(0) modulo 2, hence d0 itself, are not in the extension with Pα . For all m ∈ ω, there are a maximal antichain Am in D(Pα , m) and a function Gm from Am into ωm such that r
“x˙  m = Gmˇ(r)” for all r ∈ Am . Let p ∈ Pβ and we put j := fp(0) (|sp(0) |). We take m ∈ ω with m > |sp(0) |.

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In this proof, we write e = ej +1 [p(0)] and take p+ in Lemma 3.9 for p. Then there is r ∈ Am which is compatible with p+ j +1 . We take q ∈ Pβ which is a j common extension of r and p+ j +1 . By Lemmata 3.8 and 3.9, e(q) ≤ r, p+ also holds. Therefore q
“x˙  m = Gmˇ(r)”, e(q)
“x˙  m = Gmˇ(r)”, and sq(0) (|sp(0) |) = j + 1 = j = se(q)(0) (|sp(0) |). Since |sp(0) | < m, either q or e(q) is in Dx˙ and both q and e(q) are extensions of p, which completes the proof of ω. VPα ∩ 2ω
VPβ ∩ 2 For the proof of α<ω1 VPα ∩ ωω ⊇ VDκ ∩ ωω , given a Dκ -name f˙ for an element of ωω , for any n ∈ ω there are a maximal antichain Bn ⊆ Dκ and a function Fn : Bn→ ωsuch that ∀p ∈ Bn (p
Dκ “f˙(n) = Fn ˇ(p)”). Then there is α < κ with n∈ω p∈Bn supp(p) ⊆ Iα . Then we can consider f˙ as a Pα -name.  Therefore α<ω1 VPα ∩ ωω ⊇ VDκ ∩ ωω .   In the proof, we have shown that for all γ ∈ Iβ \ Iα the Hechler real added at stage γ is in VPβ ∩ 2ω \ VPα ∩ 2ω . Corollary 3.11. VDκ
“g = ℵ1 ” for any cardinal κ with uncountable cofinality. Proof. It follows from the comment below main lemma 3.1 and lemma 3.10.

 

Theorem 3.12. VDκ
“cov(s 0 ) = ℵ1” for any cardinal κ with uncountable cofinality.
 Proof. It’s sufficient to prove that VPα ; α < ω satisfies the condition of Main Lemma 3.1, i.e. for α < ω1 , any perfect subset of 2ω in VPω1 (= VDκ ) has a perfect subset which does not meet the reals in VPα . Let T˙ be a Pω1 -name for a perfect tree in 2<ω . Then there is α < ω1 such that T˙ is a Pα -name. For σ ∈ 2<ω , there is a maximal antichain Aσ in Pω1 all members of which decide “σˇ ∈ T˙ ”. Then there is α < κ such that for all σ ∈ 2<ω and p ∈ Aσ , we have supp(p) ⊆ Iα . And then there is λ < κ such that T˙ is a (Pα  λ)-name. For all σ ∈ 2<ω and p ∈ Aσ , we have supp(p) ⊆ λ. Without loss of generality, we may assume that λ = 0 ∈ Iα . (We work in VPα
λ .) Then we can consider T˙ as a member T of the ground model. The following argument is very similar to the proof of lemma 3.2. Let F be a isomorphism from split(T ) onto 2<ω . For n ∈ ω, we define a Pω1 -name S˙n such that
“S˙n = {t ∈ split2n (Tˇ ); ∀k < n(Fˇ (t)(2k) = d˙0 (k) modulo 2)}” (recall that do is the Hechler real added at the first stage of Pω1 ). Then we define a Pω1 -name S˙ such that
“S˙ = {σ ∈ 2<ω ; ∃n ∈ ω∃τ ∈ S˙n (σ ⊆ τ )}”. Then 1.
“S˙ is a perfect subtree of Tˇ ”, and ˙ ∩ V[G˙P ] = ∅ ”, where G˙P = {p, ˇ p; p ∈ Pα }. 2.
“[S] α α

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(1) It’s enough to show that for any n ∈ ω, p ∈ Pω1 and t ∈ 2<ω if p
“tˇ ∈ S˙n ”, ˙ ∧ uˇ ⊥ vˇ ∧ tˇ ⊆ u, ˇ vˇ ∈ Sn+1 ˇ v”. ˇ then there are q ≤ p and u, v ∈ 2<ω such that q
“u, ˙ ˇ Assume p
“t ∈ Sn ”, then we take any condition q ≤ p such that |sq(0) | ≥ n + 1. Let i = sq(0) (n) modulo 2, u = F −1 (F (t) 0, i) and v = F −1 (F (t) 1, i), which are witnesses. (2) Let G be Pα -generic over V and H (Pω1 /Pα )[G]-generic over V[G]. Assume ˙ ]] ∩ V[G]. Then there is a strictly increasing chain that there is a real x in [S[H σk ; k ∈ ω of members of 2<ω in V[G] such that for all k ∈ ω, |σk | = k and F −1 (σk ) ⊆ x. We define c ∈ 2<ω such that c(k) = σ2k+2 (2k) for all k ∈ ω. We prove that c = d0 modulo 2, thus d0 modulo 2 is in V[G], which is a contradiction to lemma 3.10 (which says that d0 modulo 2 is not in V[G]). Then for k ∈ ω, c(k) = σ2k+2 (2k) = F (F −1 (σ2k+2 ))(2k) ˙ [H ]). = d0 (k) modulo 2 (because F −1 (σ2k+2 ) ∈ Sk+1   Corollary 3.13. It is relatively consistent with ZFC that κ(S) = cov(s 0 ) < b = c. Proof. In VDω2 , b = m(C) = c = ℵ2 and add(s 0 ) = cov(s 0 ) = ℵ1 hold. Since m(C) = c, κ(S) = add(s 0 ) also holds in the extension (see introduction).   Acknowledgements. While carrying out the research for this paper, I discussed with J¨org Brendle and he pointed out many articles to me. I greatly appreciate his help. And I would like to thank: Masaru Kada, who read the first version of the paper carefully, Hiroaki Minami, who pointed out a gap of the proof of Lemma 2.1 in the first version, and the referee, who gave me many useful comments and suggestions.

References [BJ]

Bartoszy´nski, T., Judah, H.: Set Theory: On the structure of the real line. A.K. Peters, Wellesley, Massachusetts, 1995 [Bl1] Blass, A.: Applications of superperfect forcing and its relatives. Set theory and its applications 1401, 18–40 (1989) [Bl2] Blass, A.: Combinatorial cardinal characteristics of the continuum. Handbook of Set Theory, to appear. [Br1] Brendle, J.: Strolling through Paradise. Fundamenta Mathematicae 148, 1–25 (1995) [Br2] Brendle, J.: Mad families and iteration theory. Logic and Algebra, Yi Zhang (editor), Contemporary Mathematics, vol. 302, American Mathematical Society, 2003 [BrJ] Brendle, J., Judah, H.: Perfect sets of random reals. Israel J. Math. 83, 153–176 (1993) [CL] Carlson, T., Laver, R.: Sacks reals and Martin’s axiom. Fundamenta Mathematicae 133(2), 161–168 (1989) [GRSS] Goldstern, M., Repick´y, M., Shelah, S., Spinas, O.: On tree ideals. Proc. American Math. Society 123(5), 1573–1581 (1995) [JMS] Judah, H., Miller, A., Shelah, S.: Sacks forcing, Laver forcing, and Martin’s Axiom. Arch. Math. Logic 31, 145–162 (1992)

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T. Yorioka Kunen, K.: Set Theory: An Introduction to Independence Proofs, volume 102 of Studies in Logic, North Holland, 1980 Kunen, K.: Random and Cohen reals. Handbook of Set-Theoretic Topology, chapter 20, 887–911 Miller, A.: Special subsets of the real line. Handbook of set-theoretic topology, K. Kunen, J.E. Vaughan (editors), North Holland, Amsterdam, 1984, pp. 201–233 Simon, P.: Sacks forcing collapses c to b. Comment. Math. Univ. Carolin. 34(4), 707–710 (1993) Spinas, O.: Generic trees. J. Symbolic Logic 60, 705–726 (1995) Velickovic, B.: CCC posets of perfect trees. Compositio Mathematica 79(3), 279– 294 (1991)