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Bar Ilan University and U.C. Chile

arXiv:math/9306214v1 [math.LO] 15 Jun 1993STRONG MEASURE ZERO SETSWITHOUT COHEN REALSJanuary 1991Martin Goldstern1Bar Ilan UniversityHaim Judah1Bar Ilan University and U.C. ChileSaharon Shelah1,2Hebrew University of JerusalemABSTRACT.If ZFC is consistent, then each of the followingare consistent with ZFC + 2ℵ0 = ℵ2:1.

X ⊆IR is of strong measure zero iff|X| ≤ℵ1 + thereis a generalized Sierpinski set.2. The union of ℵ1 many strong measure zero sets is astrong measure zero set + there is a strong measurezero set of size ℵ2.1 The authors thank the Israel Foundation for Basic Research, Israel Academy of Science.2 Publication 438

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals§0. IntroductionIn this paper we continue the study of the structure of strong measure zero sets.

Strongmeasure zero sets have been studied from the beginning of this century. They were dis-covered by E. Borel, and Luzin, Sierpinski, Rothberger and others turned their attentionto the structure of these sets and proved very interesting mathematical theorems aboutthem.

Most of the constructions of strong measure zero sets involve Luzin sets, whichhave a strong connection with Cohen reals (see [6]). In this paper we will show that thisconnection is only apparent; namely, we will build models where there are strong measurezero sets of size c without adding Cohen reals over the ground model.Throughout this work we will investigate questions about strong measure zero sets underthe assumption that c = 2ℵ0 = ℵ2.

The reason is that CH makes many of the questionswe investigate trivial, and there is no good technology available to deal with most of ourproblems when 2ℵ0 > ℵ2.0.1 Definition: A set X ⊆IR of reals has strong measure zero if for every sequence⟨εi : i < ω⟩of positive real numbers there is a sequence ⟨xi : i < ω⟩of real numbers suchthatX ⊆[i<ω(xi −εi, xi + εi)We let S ⊂P(IR) be the ideal of strong measure zero sets.0.2 Remark: (a) if we work in ω2 then X ⊆ω2 has strong measure zero if(∀h ∈ωω)(∃g ∈Ynh(n)2)(∀x ∈X)(∃∞n)(g(n) = x|h(n))or equivalently,(∗)(∀h ∈ωω)(∃g ∈Ynh(n)2)(∀x ∈X)(∃n)(g(n) = x|h(n))(b) To every question about strong measure zero sets in IR there is a corresponding ques-tion about a strong measure zero set of ω2, and for all the questions we consider thecorresponding answers are the same. So we will work sometimes in IR, sometimes in ω2.0.3 Definition: Assume that H ⊆ωω.

We say that ¯ν has index H, if ¯ν = ⟨νh : h ∈H⟩and for all h ∈H, νh is a function with domain ω and ∀n νh(n) ∈h(n)2. We letX¯ν :=\h∈H[k∈ω[νh(k)](where we let [η] := {f ∈ω2 : η ⊆f}).We say that X¯ν is the set “defined” by ¯ν.4381January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals0.4 Fact: Assume H ⊆ωω is a dominating family, i.e., for all f ∈ωω there is h ∈H suchthat ∀n f(n) < h(n). Then:(1) If ¯ν has index H, then X¯ν is a strong measure zero set.

(2) If X is a strong measure zero set, then there is a sequence ¯ν with index Hsuch that X ⊆X¯ν.0.5 Definition: A set of reals X ⊆IR is a GLuzin (generalized Luzin) set if for everymeager set M ⊆IR, X ∩M has cardinality less than c. X is a generalized Sierpinski setif set if for every set M ⊆IR of Lebesgue measure 0, X ∩M has cardinality less than c.0.6 Fact: (a) If c is regular, and X is GLuzin, then X has strong measure zero. (b) A set of mutually independent Cohen reals over a model M is a GLuzin set.

(c) If c > ℵ1 is regular, and X is a GLuzin set, then X contains Cohen reals over L.Proof: See [6].0.7 Theorem: [6] Con(ZF) implies Con(ZFC + there is a GLuzin set which is not strongmeasure zero).0.8 Theorem: [6] Con(ZF) implies Con(ZFC + c > ℵ1 + ∃X ∈[IR]c, X a strong measurezero set + there are no GLuzin sets).In theorem 0.16 we will show a stronger form of 0.8.0.9 Definition:(1) Let Unif(S) be the following statement: “Every set of reals of size lessthan c is a strong measure zero set.”(2) We say that the ideal of strong measure zero sets is c-additive, or Add(S),if for every κ < c the union of κ many strong measure zero sets is a strongmeasure zero set. (So Add(S) ⇒Unif(S).

)0.10 Remark: Rothberger ([13] and [12]) proved that the following are equivalent:(i) Unif(S)(ii) for every h : ω →ω, for every F ∈[Qn h(n)]

)In this spirit, we first prove the following result:4382January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals0.11 Theorem: If Unif(S) and d = c, then there exists a strong measure zero set of sizec.We start the proof by proving the following0.12 Fact: If d = c, then there is a set {fi : i < c} of functions in ωω such that for everyg ∈ωω, the set{i < c : fi ≤∗g}has cardinality less than c.Proof of the fact: We build ⟨fi : i < c⟩by transfinite induction. Let ωω = {gj : j < c}.We will ensure that for j < i, fi ̸<∗gj.

This will be sufficient.But this is easy to achieve, as for any i, the family {gj : j < i} is not dominating, so thereexists a function fi such that for all j < i, for infinitely many n, fi(n) > gj(n).This completes the proof of 0.12.0.13 Proof of 0.11: Using d = c, let ⟨fi : i < c⟩be a sequence as in 0.12. Let F : ωω →[0, 1] −Q be a homeomorphism.

(Q is the set of rational numbers.) We will show thatX := {F(fi) : i < c} is a strong measure zero set.Let ⟨εn : n < ω⟩be a sequence of positive numbers.Let {rn : n ∈ω} = Q.ThenU1 := Sn∈ω(rn −ε2n, rn+ε2n) is an open set.

So K := [0, 1]−U1 is closed, hence compact.As K ⊆rng(F), also F −1(K) ⊆ωω is a compact set. So for all n the projection of F −1(K)to the nth coordinate is a compact (hence bounded) subset of ω, say ⊆g(n).

SoF −1K ⊆{f ∈ωω : f ≤∗g}Let Y := X −U1 ⊆K. Then Y ⊆F(F −1(K)) ⊆{F(fi) : fi ≤∗g} is (by assumption on⟨fi : i < c⟩) a set of size < c, hence has strong measure zero.

So there exists a sequence ofreal numbers ⟨xn : n < ω⟩such that Y ⊆U2, whereU2 :=[n∈ω(xn −ε2n+1, xn + ε2n+1)and X ⊆U1 ∪U2. So X is indeed a strong measure zero set.In section 2 we will build models where Add(S) holds and the continuum is bigger thanℵ1 without adding Cohen reals.

First we will show in 3.4:0.14 Theorem: If ZFC is consistent, thenZFC + c = ℵ2 + S = [IR]≤ℵ1 + there are no Cohen reals over Lis consistent.Note that c = ℵ2 and S = [IR]≤ℵ1 implies(1) Add(S). (Trivially)(2) b = d = ℵ1.

(By 0.11)4383January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsThe same result was previously obtained by Corazza[3]. In his model the nonexistence ofstrong measure zero sets of size c is shown by proving that every set of size c can be mappeduniformly continuously onto the unit interval (which is impossible for a strong measurezero set).

Thus, the question arises whether is possible to get a model of S = [IR]

)By a remark of Miller [8, §2] a generalized Sierpinski set cannot be mapped continuouslyonto [0, 1] (not even with a Borel function).Pawlikowski [11] showed that the additivity of the ideal S of strong measure zero sets doesnot imply the additivity of the ideal M of meager sets. For this he built a model satisfyingAdd(S) + c = ℵ2 + b = ℵ1.

He used a finite support iteration of length ω2. So he addsmany Cohen reals, and in the final model Cov(M) holds (i.e., IR can not be written as theunion of less than c many meager sets).

We will improve his result in the next theorem:0.16 Theorem: If ZFC is consistent, thenZFC + c = d = ℵ2 > b + Add(S) + no real is Cohen over Lis consistent. (Note that by 0.11, d = c + Add(S) implies that there is a strong measure zero set ofsize c.)0.17 Notation: We use standard set-theoretical notation.

We identify natural numbersn with their set of predecessors, n = {0, . .

., n −1}. AB is the set of functions from A intoB, A<ω := Sn<ωnA.

|A| denotes the cardinality of a set A. P(A) is the power set of aset A, A ⊂B means A ⊆B & A ̸= B. A −B is the set-theoretic difference of A and B.

[A]κ := {X ⊆A : |X| = κ}. [A]<κ and [A]≤κ are defined similarly.

(We write A := B orB =: A to mean: the expression B defines the term (or constant) A. )Ord is the set of ordinals.

cf(α) is the cofinality of an ordinal α. Sαβ := {δ ∈ωβ : cf(δ) =ωα}. In particular, S12 is the set of all ordinals < ω2 of uncountable cofinality.IR is the set of real numbers.

c = |IR| is the size of the continuum. For f, g ∈ωω we letf < g ifffor all n f(n) < g(n), and f <∗g if for some n0 ∈ω, ∀n ≥n0 f(n) < g(n).

The“bounding number” b and the “dominating number” d are defined asb := min{|H| : H ⊆ωω, ∀g ∈ωω ∃h ∈H ¬(h <∗g)}d := min{|H| : H ⊆ωω, ∀g ∈ωω ∃h ∈H g < h}= min{|H| : H ⊆ωω, ∀g ∈ωω ∃h ∈H g <∗h}(It is easy to see ω1 ≤b ≤d ≤c. )4384January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsWe call a set H ⊆ωω dominating, if ∀g ∈ωω ∃h ∈H g < h.M is the ideal of meager subsets of IR (or of ω2). S is the ideal of strong measure zero sets.For any ideal J ⊂P(IR), Add(J ) abbreviates the statement: “The union of less than cmany sets in J is in J .” Cov(J ) means that the reals cannot be covered by less than cmany sets in J .If f is a function, dom(f) is the domain of f, and rng(f) is the range of f. For A ⊆dom(f),f|A is the restriction of f to A.

For η ∈2<ω, [η] := {f ∈ω2 : η ⊆f}.0.18 More Notation: If Q is a forcing notion, GQ is the canonical name for the genericfilter on Q. We interpret p ≤q as q is stronger (or “has more information”) than p. (Sop ≤q ⇒q|⊢p ∈GQ.

)When we deal with a (countable support) iteration ⟨Pα, Qα : α < ε⟩, we write Gα for thecanonical name of the generic filter on α, and G(α) for the generic filter on Qα. If thereis a natural way to associate a “generic” real to the generic filter on Qα, we write gα forthe real given by G(α).

We write |⊢α for the forcing relation of Pα. If β < α, Gβ alwaysstands for Gα ∩Pβ.

V = V0 is the ground model, and Vα = V [Gα]. Pε is the countablesupport limit of ⟨Pα : α < ε⟩.

Pε/Gα is the Pα-name for {p ∈Pε : p|α ∈Gα} (with thesame ≤relation as Pε). The forcing relation with respect to Pε/Gα (in Vα) is denoted by|⊢αε.There is a natural dense embedding from Pε into Pα ∗Pε/Gα.

Thus we always identifyPα-names for Pε/Gα-names with the corresponding Pε-names.∅α is the weakest condition of Pα, and ∅α|⊢αϕ is usually abbreviated to |⊢αϕ. (So|⊢α(|⊢αδϕ) iff|⊢δϕ).0.19 Even more Notation: The following notation is used when we deal with trees offinite sequences:For η ∈V <ω, i ∈V , η⌢i is the function η ∪{⟨|η|, i⟩} ∈V <ω.p ⊆ω<ω is a tree if p ̸= ∅, and for all η ∈p, all k < |η|, η|k ∈p.

Elements of a tree areoften called “nodes”. We call |η| the “length” of η.

We reserve the word “height” forthe notion defined in 2.2.For p ⊆ω<ω, η ∈p, we let succp(η) := {i : η⌢i ∈p}.If p is a tree, η ∈p, let p[η] := {ν ∈p : η ⊆ν or ν ⊆η}.If p ⊆ω<ω is a tree, b ⊆p is called a branch, if b is a maximal subset of p that is linearlyordered by ⊆.Clearly, if ∀η ∈p succp(η) ̸= ∅, then a subset b ⊆p is a branch iffb is of the formb = {f|n : n ∈ω} for some f ∈ωω.We let stem(p) be the intersection of all branches of p.4385January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals§1. A few well known factsWe collect a few more or less well known facts about forcing, for later reference.1.1 Definition: An ultrafilter U on ω is called a P-Point, if for any sequence ⟨An : n ∈ω⟩of sets in U there is a set A in U that is almost contained in every An (i.e., ∀n A −An isfinite).1.2 Definition: For any ultrafilter U on ω, we define the P-point game G(U) as follows:There are two players, “IN” and “NOTIN”.

The game consistsof ω many moves.In the n-th move, player NOTIN picks a set An ∈U, and playerIN picks a finite set an ⊆An.Player IN wins if after ω many moves, Sn an ∈U.We write a play (or run) of G(U) as⟨A0; a0 →A1; a1 →A2; . .

.⟩.It is well known that an ultrafilter U is a P-point iffplayer NOTIN does not have a winningstrategy in the P-point game.For the sake of completeness, we give a proof of the nontrivial implication “⇒” (which isall we will need later):Let U be a P-point, and let σ be a strategy for player NOTIN. We will construct a run ofthe game in which player NOTIN followed σ, but IN won.Let A0 be the first move according to σ.

For each n, let An be the set of all responses ofplayer notin according to σ in an initial segment of a play of length ≤n in which playerIN has played only subsets of n:An := {Ak : k ≤n, ⟨A0; a0 →A1; . .

. ; ak−1 →Ak⟩is aninitial segment of a play in which NOTINobeyed σ, and a0, .

. ., ak−1 ⊆n}Note that A0 = {A0}, and for all n, An is a finite subset of U.As U is a P-point, there is a set X ∈U such that for all A ∈Sn An, X −A is finite.Let X ⊆A0 ∪n0, and for k > 0 let nk satisfynk > nk−1and∀A ∈Ank−1 X ⊆A ∪nkEither Sk∈ω[n2k, n2k+1) ∈U, or Sk∈ω[n2k+1, n2k+2) ∈U.Without loss of generality we assume Sk∈ω[n2k, n2k+1) ∈U.Now define a play ⟨A0; a0 →A1; a1 →A2; .

. .⟩of the game G(U) by induction as follows:A0 is given.Given Aj, let aj := Aj ∩[n2j, n2j+1) and let Aj+1 be σ’s response to aj.4386January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsThen as a0, . .

. , aj−1 ⊆n2j, we have X ⊆Aj ∪n2j for all j.

Therefore for all j wehave X ∩[n2j, n2j+1) ⊆(Aj ∪n2j) ∩[n2j, n2j+1) = Aj ∩[n2j, n2j+1) = aj. So Sj∈ω aj ⊇X ∩Sj∈ω[n2j, n2j+1) ∈U.Thus player IN wins the play ⟨A0; a0 →A1; a1 →A2; .

. .⟩in which player NOTIN obeyedσ.1.3 Definition: We say that a forcing notion Q preserves P-points, if for every P-pointultrafilter U on ω, |⊢Q“U generates an ultrafilter”, i.e.|⊢Q ∀x ∈P(ω) ∃u ∈U (u ⊆x or u ⊆ω −x).”[9] defined the following forcing notion:1.4 Definition: “Rational perfect set forcing”, RP is defined as the set of trees p ⊆ω<ωsatisfying(1) for all η ∈p, |succp(η)| ∈{1, ℵ0} (See 0.19)(2) for all η ∈p there is ν ∈p with η ⊆ν and |succp(η)| = ℵ0.We let p ≥q iffp ⊆q.Then the following hold:1.5 Lemma:(1) RP preserves P-points.

([9, 4.1])(2) RP adds an unbounded function. ([9, §2])(3) RP is proper.

(This is implicit in [9]. See also 2.16)The next lemma can be found, e.g., in [7, VII ?

? and Exercise H2]:1.6 Fact: If Q is a forcing notion satisfying the ℵ2-cc, then(1) If |⊢Q c∼: ωV2 →ωV2 , then there is a function c : ω2 →ω2 such that|⊢Q∀α < ω2 : c∼(α) < c(α).

(2) |⊢QℵV2 = ℵ2. (3) For every stationary S ⊆ℵ2, |⊢Q “S is stationary on ℵ2”.The following fact is from [14, V 4.4]:1.7 Fact: Assume ⟨Pα, Qα : α < ω2⟩is an iteration of proper forcing notions Qα.

Thenfor every δ ≤ω2 of cofinality > ω, |⊢δωω ∩Vδ = ωω ∩Sα<δ Vα, or in other words: “no newreals appear in limit stages of cofinality > ω”.As a consequence, |⊢ω2“If X ⊆ωω, |X| ≤ℵ1, then there is δ < ω2 such that X ∈Vδ.”We also recall the following facts about iteration of proper forcing notions:1.8 Lemma: Assume CH, and let ⟨Pα, Qα : α < ω2⟩be a countable support iterationsuch that for all α < ω2, |⊢α“Qα is a proper forcing notion of size ≤c.”Then(1) ∀α < ω2: |⊢αc = ℵ1. (see [14, III 4.1])4387January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals(2) |⊢ω2c ≤ℵ2. (This follows from 1.7 and (1))(3) For all α ≤ω2, Pα is proper [14, III 3.2] and satisfies the ℵ2-cc.

(See [14,III 4.1])(4) |⊢ω2ℵV1 = ℵ1. (See [14, III 1.6])In [2, 4.1] the following is proved:1.9 Lemma: Assume ⟨Pα, Qα : α < ω2⟩is as in 1.8, and for all α < ω2:|⊢α“Qα preserves P-points.”Then for all α ≤ω2, Pα preserves P-points.1.10 Definition: We say that a forcing notion Q is ωω-bounding, if the set of “old”functions is a dominating family in the generic extension by Q, or equivalently,|⊢Q∀f ∈ωω ∃g ∈ωω ∩V ∀n f(n) < g(n)[14, V 4.3] proves:1.11 Lemma: Assume ⟨Pα, Qα : α < ω2⟩is as in 1.8, and for all α < ω2:|⊢α“Qα is ωω-bounding and ω-proper.”Then for all α ≤ω2, Pα is ωω-bounding.

(We may even replace ω-proper by “proper”, see [14], [4])The following is trivial to check:1.12 Fact: Assume Q is a forcing notion that preserves P-points or is ωω-bounding. Then|⊢Q“There are no Cohen reals over V ”1.13 Definition: A forcing notion P is strongly ωω-bounding, if there is a sequence⟨≤n : n ∈ω⟩of binary reflexive relations on P such that for all n ∈ω:(1) p ≤n q ⇒p ≤q.

(2) p ≤n+1 q ⇒p ≤n q. (3) If p0 ≤0 p1 ≤1 p2 ≤3 · · · , then there is a q such that ∀n pn+1 ≤n q.

(4) If p |⊢“α∼is an ordinal,” and n ∈ω, then there exists q ≥n p and a finiteset A ⊆Ord such that Q|⊢α∼∈A.1.14 Definition: (1) If ⟨Pα, Qα : α < ε⟩is an iteration of strongly ωω-bounding forcingnotions, F ⊆ε finite, n ∈ω, p, q ∈Pε, we say that p ≤F,n q iffp ≤q and ∀α ∈F q|α|⊢p(α) ≤n q(α). (2) A sequence ⟨⟨pn, Fn⟩: n ∈ω⟩is called a fusion sequence if ⟨Fn : n ∈ω⟩is an increas-ing family of finite subsets of ε, ⟨pn : n ∈ω⟩is an increasing family of conditions in Pε,∀n pn ≤n,Fn pn+1 and Sn dom(pn) ⊆Sn Fn.Note that 1.13 is not a literally a strengthening of Baumgarter’s “Axiom A” (see [1]), aswe do not require that the relations ≤n are transitive, and in (2) we only require pn+1 ≤n qrather than pn+1 ≤n+1 q.

Nevertheless, the same proof as in [1] shows the following fact:4388January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals1.15 Fact:(1) If the sequence ⟨⟨pn, Fn⟩: n ∈ω⟩is a fusion sequence, then there exists acondition q ∈Pε such that for all n ∈ω, pn+1 ≥Fn,n q. (2) If α∼is a Pε-name of an ordinal, n ∈ω, F ⊆Pε finite, then for all p thereexists a condition q ≥n,F p and a finite set A of ordinals such that q|⊢α∼∈A.

(3) If X∼is a Pε-name of a countable set of ordinals, n ∈ω, F ⊆Pε finite,then for all p there exists a condition q ≥n,F p and a countable set A ofordinals such that q|⊢X∼⊆A.The next fact is also well known:1.16 Fact: Let B be the random real forcing. Then B is strongly ωω-bounding.

[Proof: Conditions in B are Borel subsets of [0, 1] of positive measure, p ≤q iffp ⊇q.We let p ≤n q iffp ≤q and µ(p −q) ≤10−n−1µ(p), where µ is the Lebesgue measure.Then if p0 ≥0 p1 ≥1 · · ·, letting q := Tn pn we have for all n, all k ≥n, µ(pk −pk+1) ≤10−k−1µ(pk) ≤10−k−1µ(pn), so µ(pn −q) ≤10−n−1 + 10−n−2 + · · · ≤2 ∗10−n−1µ(pn).In particular, µ(q) ≥0.8 ∗µ(p0), so q is a condition, and q ≥n−1 pn for all n > 0.Given a name α∼, an integer n and a condition p such that p|⊢“α∼is an ordinal,” let A bethe set of all ordinals β such that [[α∼= β]] ∩p has positive measure ([[ϕ]] is the booleanvalue of the statement ϕ, i.e. the union of all conditions forcing ϕ).

Since Pβ∈A µ([[α∼=β]] ∩p) = µ(p) there is a finite subset F ⊆A such that letting q := p ∩Sβ∈A[[α∼= β]] wehave µ(q) ≥(1 −10−n−1)µ(p). So q ≥n p and q|⊢α∼∈F.

]We will also need the following lemma from [17, §5, Theorem 9]:1.17 Lemma: Every stationary S ⊆ℵ2 can be written as a union of ℵ2 many disjointstationary sets.Finally, we will need the following easy fact (which is true for any forcing notion Q)1.18 Fact: If f∼is a Q-name for a function from ω to ω, |⊢Q f∼/∈V , and r0, r1 are any twoconditions in Q, then there are l ∈ω, j0 ̸= j1, r′0 ≥r0, r′1 ≥r1 such that r′0|⊢f∼(l) = j0,r′1|⊢f∼(l) = j1. [Proof: There are a function f0 and a sequence r0 = r0 ≤r1 ≤· · · of conditions in Q suchthat for all n, rn|⊢f∼|n = f0|n.

Since r1|⊢f∼/∈V , r1|⊢∃l f∼(l) ̸= f0(l). There is a conditionr′1 ≥r1 such that for some l ∈ω and some j1 ̸= f0(l), r′1|⊢f∼(l) = j1.

Let j0 := f0(l), andlet r′0 := rl+1. ]4389January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals§2H-perfect treesIn this section we describe a forcing notion PTH that we will use in an iteration in thenext section. We will prove the following properties of PTH:(a) PTH is proper and ωω-bounding.

(b) PTH preserves P-points. (c) PTH does not “increase” strong measure zero sets defined in the groundmodel.

(d) PTH makes the reals of the ground model (and hence, by (c), the union ofall strong measure zero sets defined in the ground model) a strong measurezero set.2.1 Definition: For each function H with domain ω satisfying ∀n ∈ω 1 < |H(n)| < ω,we define the forcing PTH, the set of H-perfect trees to be the set of all p satisfying(A) p ⊆ω<ω is a tree. (B) ∀η ∈p ∀l ∈dom(η) : η(l) ∈H(l).

(C) ∀η ∈p : |succp(η)| ∈{1, |H (|η|)|}. (D) ∀η ∈p ∃ν ∈p : η ⊆ν, |succp(ν)| = |H(|ν|)|.2.2 Definition:(1) For p ∈PTH, we let the set of “splitting nodes” of p besplit(p) := {η ∈p : |succp(η)| > 1}(2) The height of a node η ∈p ∈PTH is the number of splitting nodes strictlybelow η:htp(η) := |{ν ⊂η : ν ∈split(p)}|(Note that htp(η) ≤|η|.

)(3) For p ∈PTH, k ∈ω, we let the kth splitting level of p be the set of splittingnodes of height k.splitk(p) := {η ∈split(p) : htp(η) = k}(Note that split0(p) = {stem(p)}. )(4) For u ⊆ω, we letsplitu(p) :=[k∈usplitk(p)2.3 Remarks:(i) Since H(n) is finite, (3) just means that either η has a unique successorη⌢i, or succp(η) = H(|η|).

)(ii) Letting H′(n) = |H(n)|, clearly PTH is isomorphic to PTH′ (and theobvious isomorphism respects the functions η 7→htp(η), ⟨p, k⟩7→splitk(p),etc)43810January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals2.4 Remark: If we let H(n) = ω for all n, then 2.1(A)–(D) define RP, rational perfectset forcing. The definitions in 2.2 make sense also for this forcing.

Since we will not usethe fact that H(n) is finite before 2.12, 2.5–2.11 will be true also for RP.2.5 Fact: Let p, q ∈PTH, n ∈ω, η, ν ∈ω<ω. Then(1) If η ⊂ν ∈p, then htp(η) ≤htp(ν).

If moreover η ∈split(p), thenhtp(η) < htp(ν). (2) If b ⊆p is a branch, then b ∩splitn(p) ̸= ∅.

(3) If p ⊇q, then for all n, q ∩splitn(p) ̸= ∅. (4) If η ∈p and htp(η) ≤n then ∃ν ∈p, η ⊆ν and ν ∈splitn(p).

(5) If η0 ̸= η1 are elements of splitn(p), then η0 ̸⊆η1, and η1 ̸⊆η0.Proof: (1) is immediate form the definition of ht.For (2), it is enough to see that b ∩split(p) is infinite. (Then ordering b by inclusion, thenth element of b ∩split(p) will be in splitn−1(p).

)So assume that b∩split(p) is finite. Recall that each η ∈b−split(p) has a unique successorin p. By 2.1(C), b cannot have a last element, so b is infinite.

Hence there is η0 ∈b suchthat∀ν ∈b : η0 ⊆ν ⇒|succp(ν)| = 1.A trivial induction on |ν| shows that this implies∀ν ∈p : η0 ⊆ν ⇒ν ∈b.Hence∀ν ∈p : η0 ⊆ν ⇒|succp(ν)| = 1.This contradicts 2.1(D).To prove (3), let b be any branch of q. b is also a branch of p, so (2) shows that q∩splitn(p) ⊇b ∩splitn(p) ̸= ∅.Proof of (4): Let b be a branch of p containing η. By (2) there is ν ∈b ∩splitn(p).

Ifν ⊂η, then htp(η) > htp(ν) = n, which is impossible. Hence η ⊆ν.

(5) follows easily from (1).2.6 Definition: For p, q ∈PTH, n ∈ω, we let(1) p ≤q (“q is stronger than p”) iffq ⊆p. (2) p ≤n q iffp ≤q and splitn(p) ⊆q.

(So also splitk(p) ⊆q for all k < n.)2.7 Fact: If p ≤n q, n > 0, then stem(p) = stem(q).2.8 Fact: Assume p, q ∈PTH, n ∈ω, p ≤n q. (0) For all η ∈q, htq(η) ≤htp(η).

(1) For all k ≤n, splitk(p) ⊆split(q). (2) For all k < n, splitk(p) = splitk(q).43811January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals(3) If p ≤n q ≤n r, then p ≤n r.Proof: (0) is clear. (1): Let η ∈splitk(p) for some k < n, then by 2.5(4) there is a ν, η ⊆ν ∈splitn(p) ⊆q,so η ∈q.

(2): Let η ∈splitk(p), then η ∈split(q). Clearly htq(η) ≤htp(η) = k. Using (1) induc-tively, we also get htq(η) ≥k.

(3): Let η ∈splitn(p). So η ∈q, htq(η) ≤htp(η) = n. By 2.5(4), there is ν ∈splitn(q),η ⊆ν.

As ν ∈r, η ∈r.2.9 Definition and Fact: If p0 ≤1 p1 ≤2 p2 ≤3 · · · are conditions in PTH, then we callthe sequence ⟨pn : n < ω⟩a “fusion sequence”. If ⟨pn : n < ω⟩is a fusion sequence, then(1) p∞:= Tn∈ω pn is in PTH(2) For all n: pn ≤n+1 p∞.2.10 Fact:(1) If η ∈p ∈PTH, then p[η] ∈PTH, and p ≤p[η].

(See 0.19. )(2) If p ≤q are conditions in PTH, η ∈q, then p[η] ≤q[η].2.11 Fact: If for all η ∈splitn(p), qη ≥p[η] is a condition in PTH, then(1) q :=Sη∈splitn(p)qη is in PTH,(2) q ≥n p(3) for all η ∈splitn(p), q[η] = qη.2.12 Fact: If n ∈ω, p ∈PTH, then splitn(p) is finite.Proof: This is the first time that we use the fact that each H(n) is a finite set: Assumethat the conclusion is not true, so for some n and p, splitn(p) is infinite.

Then alsoT := {η|k : η ∈splitn(p), k ≤|η|} ⊆pis infinite. As T is a finitely splitting tree, there has to be an infinite branch b ⊆T.

By2.5(2), there is ν ∈b ⊆T, htp(ν) > n. This is a contradiction to 2.5(1).2.13 Fact: PTH is strongly ωω-bounding, i.e. :If α∼is a PTH-name for an ordinal, p ∈PTH, n ∈ω, then there exists a finite set A ofordinals and a condition q ∈PTH, p ≤n q, and q|⊢α ∈A.Proof: Let C := splitn(p).

C is finite. For each node η ∈C, let qη ≥p[η] be a conditionsuch that for some ordinal αη qη|⊢α∼= αη.

Now letq :=[η∈CqηandA := {αη : η ∈C}Since any extension of q must be compatible with some q[η] (for some η ∈C), q|⊢α∼∈A.43812January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals2.14 Corollary: PTH is proper (and indeed satisfies axiom A, so is α-proper for anyα < ω1) and ωω-bounding. Moreover, if n ∈ω, p ∈PTH, τ∼a name for a set of ordinals,then there exists a condition q ≥n p such that(1) If p |⊢“τ∼is finite”, then there is a finite set A such that q|⊢“τ∼⊆A”.

(2) If p |⊢“τ∼is countable”, then there is a countable set A such that q|⊢“τ∼⊆A”.Proof: Use 2.13 and 2.9.Similarly to 2.13 we can show:2.15 Fact: Assume that α∼is a RP-name for an ordinal, p ∈RP, n ∈ω.Then there exists a countable set A of ordinals and a condition q ∈PTH, p ≤n q, andq|⊢α ∈A.Proof: Same as the proof of 2.13, except that now the set C and hence also the set A maybe countable.2.16 Fact: RP is proper (and satifies axiom A). Proof: By 2.15 and 2.9.2.17 Definition: Let G ⊆PTH be a V -generic filter.

Then we let g∼be the PTH-namedefined byg∼:=\p∈GpWe may write g∼H or g∼P TH for this name g∼. If PTH is the αth iterand Qα in an iteration,we write g∼α for g∼H.2.18 Fact: ∅P TH forces that(0) g∼is a function with domain ω,(1) ∀n g∼(n) ∈H(n).

(2) For all f ∈V , if ∀n f(n) ∈H(n) then ∃∞n f(n) = g∼(n).Furthermore, for all p ∈PTH,(3) p|⊢“{g∼|n : n ∈ω} is a branch through p.(4) p|⊢∀k∃ng∼|n ∈splitk(p)Proof: (0) and (2) are straightforward density arguments. (1) and (3) follow immedaitelyfrom the definition of g∼.

(4) follows from (3) and 2.5(2), applied in V P TH.2.19 Remark: Since Unif(S) is equivalent tofor every H : ω →ω, for every F ∈[Qn H(n)]

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsWe will show a stronger result in 3.3: If P := Pω2 is the limit of a countable supportiteration ⟨Pα, Qα : α < ω2⟩, where “many” Qα are of the form PTHα for some Hα, thensome bookkeeping argument can ensure that V P |= Add(S).Since PTH is ωω-bounding, it does not add Cohen reals. The same is true for a countablesupport iteration of forcings of the form PTH.

However, in 3.9 we will have to consider aforcing iteration in which some forcing notions are of the form PTH, but others do add anunbounded real. To establish that even these iterations do not add Cohen reals, we willneed the fact that the forcing notion PTH preserves many ultrafilters.2.20 Definition: Let Q be a forcing notion, x∼a Q-name, p ∈Q, p|⊢x∼⊆ω.

We say thatx∗⊆ω is an interpretation of x∼(above p), if for all n there is a condition pn ≥p suchthat pn|⊢x∼∩n = x∗∩n.2.21 Fact: Assume Q, p, x∼are as in 2.20. Then(1) There exists x∗⊆ω such that x∗is an interpretation of x∼above p.(2) If p ≤p′ and x∗is an interpretation of x∼above p′, then x∗is an interpretationof x∼above p.2.22 Lemma: PTH preserves P-points, i.e.

: If U ∈V is a P-point ultrafilter on ω, then|⊢P TH“U generates an ultrafilter.”Proof: Assume that the conclusion is false. Then there is a PTH-name τ∼for a subset ofω and a condition p0 such thatp0|⊢P TH∀x ∈U : |x ∩τ∼| = |(ω −x) ∩τ∼| = ℵ0.For each p ∈PTH we choose a set τ(p) such that· τ(p) is an interpretation of τ∼above p.· If there is an interpretation of τ∼above p that is an element of U, thenτ(p) ∈U.Note that if τ(p) ∈U and p ≥p′, then also τ(p′) ∈U, since (by 2.21(2)) we could havechosen τ(p′) := τ(p).

Hence either for all p τ(p) ∈U, or for some p1 ≥p0, all p ≥p1,τ(p) /∈U.In the second case we let σ∼be a name for the complement of τ∼, and letσ(p) = ω −τ(p). Then σ(p) ∈U for all p ≥p1.

Also, σ(p) is an interpretation of σ∼abovep.So wlog for all p ≥p1, τ(p) ∈U for some p1 ∈PTH, p1 ≥p0.We will show that there is a condition q ≥p1 and a set a ∈U such that q|⊢a ⊆τ∼.Recall that as U is a P-point, player NOTIN does not have a winning strategy in theP-point game for U (see 1.2).We now define a strategy for player NOTIN. On the side, player NOTIN will construct afusion sequence ⟨pn : n < ω⟩and a sequence ⟨mn : n < ω⟩of natural numbers.43814January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsp0 is given.Given pn, we letAn =\η∈splitn+1(pn)τ(pn[η])This set is in U. Player IN responds with a finite set an ⊆An.

Let mn := 1 + max(an).For each η ∈splitn+1(pn) there is a condition qη ≥pn[η] forcing τ∼∩mn = τ(pn[η]) ∩mn,so in particularqη|⊢an ⊆τ∼∩mnLet pn+1 =Sη∈splitn+1(pn)qη.Then(∗)pn+1 ≥n+1 pnandpn+1|⊢an ⊆τ∼This is a well-defined strategy for player NOTIN. As it is not a winning strategy, thereis a play in which IN wins.During this play, we have constructed a fusion sequence⟨pn : n < ω⟩.

Letting a := Sn an, q := Tn pn, we have that a ∈U, p0 ≤q ∈PTH (by 2.9),and q|⊢a ⊆τ∼(by (∗)), a contradiction to our assumption.The following facts will be needed for the proof that if we iterate forcing notions PTH withcarefully chosen functions H, then we will get a model where the ideal of strong measurezero sets is c-additive.2.23 Fact and Definition: Assume p ∈PTH, u ⊆ω is infinite, v = ω −u.Thenwe can define a stronger condition q by “trimming” p at each node in splitv(p).(See2.2(4).) Formally, let ⃗ı = ⟨iη : η ∈splitv(p)⟩be a sequence satisfying iη ∈H(|η|) for allη ∈splitv(p).Thenp⃗ı := {η ∈p : ∀n ∈dom(η) : If η|n ∈splitv(p), then η(n) = iη|n}is a condition in PTHProof: Let q := p⃗ı.

q satisfies (A)–(B) of the definition 2.1 of PTH. The definition of p⃗ıimmediately implies:(1) If η ∈splitv(p) ∩q, then succq(η) = {iη}.

(2) If η ∈splitu(p) ∩q, then succq(η) = succp(η) = H(|η|). (3) If η ∈q −split(p), then η ∈p−split(p), so succq(η) = succp(η) is a singleton.Note that split(p) = splitu(p) ∪splitv(p), so (1)–(3) cover all possible cases for η ∈q.So q also satisfies 2.1(C).From (1)–(3) we can also conclude:(4) For all η ∈q: succq(η) ̸= ∅.43815January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsTo show that q ∈PTH, we still have to check condition 2.1(D). So let η ∈q.

Since u isinfinite, there is k ∈u, k > |η|. By (4), there is an infinite branch b ⊆q containing η. By2.5(2) there is ν ∈b, htp(ν) = k. Then η ⊆ν, and ν ∈split(q).2.24 Fact: p⃗ı |⊢“η ⊆g∼& η ∈splitv(p)⇒g∼(|η|) = iη” (where g∼is a name for thegeneric branch defined in 2.18).Proof: p⃗ı|⊢g∼⊆p⃗ı and succp⃗ı(η) = {iη}.To simplify notation, we will now assume that for all n, H(n) ∈ω.

(If H(n) are justarbitrary finite sets as in 2.1, then we could prove analogous statements, replacing 0 and1 by any two elements 0n ̸= 1n of H(n). )2.25 Definition: Let f∼be a PTH-name for a function from ω to ω.

We say that f∼splitson p, k if for all η ∈splitk(p) there are l and j1 ̸= j0 such thatp[η⌢0]|⊢f∼(l) = j0p[η⌢1]|⊢f∼(l) = j12.26 Remark: If f∼splits on p, k, and q ≥k+1 p, then f∼splits on q, k.(Proof: splitk(p) = splitk(q), and for η ∈splitk(p), p[η⌢i] ≤q[η⌢i]. )2.27 Lemma: If p|⊢f∼/∈V , k ∈ω, then there is q ≥k+1 p such that f∼splits on q, k.Proof: For η ∈splitk(p), i ∈{0, 1} we let ηi be the unique element of splitk+1(p) satisfyingη⌢i ⊆ηi.By 1.18, for each η ∈splitk(p) we can find conditions qη0 ≥p[η0], qη1 ≥p[η1] and integerslη, jη,0 ̸= jη,1 such that qη0|⊢f∼(lη) = j0, qη1|⊢f∼(lη) = j1.

If ν ∈splitk+1(p) is not of theform η0 or η1 for any η ∈splitk(p), then let qν := p[ν].By 2.11, q :=Sν∈splitk+1(p)qν is a condition, q ≥k+1 p, and qν = q[ν] for all ν ∈splitk+1(p).We finish the proof of 2.27 by showing that f∼splits on q, k: Let η ∈splitk(p) = splitk(q).Then q[η⌢0] = q[η0] = qη0, so q[η⌢0]|⊢f∼(lη) = jη,0. Similarly, q[η⌢1]|⊢f∼(lη) = jη,1.2.28 Lemma: If p|⊢f∼/∈V , then there is q ≥p, f∼splits on q, k for all k.Proof: By 2.27, 2.26 and 2.9 (using a fusion argument).2.29 Lemma: Assume Q is a strongly ωω-bounding forcing notion.

Let f∼be a Q-namefor a function, p a condition, n ∈ω, p|⊢f∼/∈V . Then there exists a natural numer k suchthat(∗)for all η ∈k2 there is a condition q ≥n p, q|⊢f∼/∈[η].We will write kp,n or kf∼,p,n for the least such k. Note that for any k ≥kp,n, (∗) will alsohold.Proof: Assume that this is false.

So for some f∼, n0, p0,(⋆)∀k ∈ω ∃ηk ∈k2 : ¬(∃q ≥n0 p0 q|⊢f∼/∈[ηk])43816January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsLetT := {ηk|l : l ≤k, k ∈ω}T is a finitely branching tree (⊆ω2) of infinite height, so it must have an infinite branch.Let f ∗∈ω2 be such that {f ∗|j : j ∈ω} ⊆T.Since f ∗∈V but p0|⊢Qf∼/∈V , there exists a name m∼of a natural number such thatp0|⊢f ∗|m≁= f∼|m∼. Let q ≥n0 p0 be such that for some m∗∈ω, q|⊢m∼< m∗.Claim: For some k, q|⊢f∼/∈[ηk].

This will contradict (⋆).Proof of the claim: We have q|⊢f∼|m∗̸= f ∗|m∗. Since f ∗|m∗∈T, there is a k ≥m∗suchthat f ∗|m∗= ηk|m∗.

Hence q|⊢f∼|m∗̸= f ∗|m∗= ηk|m∗, so q|⊢f∼/∈[ηk|m∗]. But then alsoq|⊢f∼/∈[ηk].This finishes the proof of the claim and hence of the lemma.2.30 Lemma: Assume that Q is a strongly ωω-bounding forcing notion, H is a dominatingfamily in V , and ¯ν = ⟨νh : h ∈H⟩has index H. Then|⊢Q\h∈H[k∈ω[νh(k)] ⊆VProof: Assume that for some condition p and some Q-name f∼,p|⊢f∼/∈V & f∼∈\h∈H[n∈ω[νh(n)].We will define a tree of conditions such that along every branch we have a fusion sequence.Specifically, we will define an infinite sequence ⟨ln : n ∈ω⟩of natural numbers, and foreach n a finite sequence⟨p(η0, .

. ., ηn−1) : η0 ∈η02, .

. ., ηn−1 ∈ln−12⟩of conditions satisfying(0) p() = p(1) For all n: ∀η0 ∈η02, .

. ., ηn−1 ∈ln−12 : ln ≥kp(η0,...,ηn−1),n.(2) For all n: ∀η0 ∈η02, .

. ., ηn−1 ∈ln−12 ∀ηn ∈ln2(a) p(η0, .

. ., ηn−1) ≤n p(η0, .

. ., ηn−1, ηn).

(b) p(η0, . .

., ηn−1, ηn)|⊢f∼/∈[ηn].Given p(η0, . .

., ηn−1) for all η0 ∈η02, . .

., ηn−1 ∈ln−12, we can find ln satisfying condition(1). The by the definition of kp(η0,...,ηn−1),n we can (for all ηn ∈ln2) find p(η0, .

. ., ηn−1, ηn).Now let h ∈H be a function such that for all n, h(n) > ln.

Define a sequence ⟨ηn : n ∈ω⟩by ηn := νh(n)|ln, and let pn := p(η0, . .

., ηn).Then p ≤p0 ≤0 p1 ≤1 · · ·, so thereexists a condition q extending all pn. So for all n, q|⊢f∼/∈[ηn].

But then also for all n,q|⊢f∼/∈[νh(n)], a contradiction.43817January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsLemma 2.30 will be needed later to show that if we iterate focings of the form PTH togetherwith random real forcing, after ω2 many steps we obtain no strong measure zero sets ofsize ℵ2. The proof (in 3.4) would be much easier if we could omit “strongly” from thehypothesis of 2.30, i.e., if we could answer the following question positively:2.31 Open Problem: Assume H ⊆ωω is a dominating family (or even wlog H = ωω),and ¯ν has index H. Let Q be an ωω-bounding forcing notion.

Does this imply|⊢Q\h∈H[n[νh(n)] ⊆V ?2.32 Fact:Assume h∗: ω →ω −{0}, H∗(n) = h∗(n)2. Let H ⊆ωω be a dominatingfamily, and let ¯ν have index H. Let g∼∗be the name of the generic function added byPTH∗.Then|⊢P TH∗∃h ∈H[k∈ω[νh(k)] ⊆[n∈ω[g∼∗(n)]Proof: Assume not, then there is a condition p such that(∗)p|⊢∀h ∈H[k∈ω[νh(k)] ̸⊆[n∈ω[g∼∗(n)]Let h ∈H be a function such that ∀k ∈ω∀η ∈split2k+1(p) h∗(|η|) ≤h(k).

This functionh will be a witness contradicting (∗).For η ∈split2k+1(p) let iη ∈succp(η) = H∗(|η|) = h∗(|η|)2 be defined by iη := νh(k)|h∗(|η|). (Note that νh(k) ∈h(k)2 and h(k) ≥h∗(|η|).

)Let ⃗ı := ⟨iη : η ∈split2k+1(p), k ∈ω⟩and let q := p⃗ı.Then q|⊢∀n∀k (g∼|n ∈split2k+1(p) ⇒g∼(n) = ig∼|n ⊆νh(k)) by 2.24.Since also q|⊢∀k∃n g∼|n ∈split2k+1(p), we get q|⊢∀k∃n [νk(k)] ⊆[g∼(n)]. This contra-dicts (∗).§3Two models of Add(S).Recall that S12 := {δ < ω2 : cf(δ) = ω1}.3.1 Lemma: Let ⟨Pα, Qα : α < ω2⟩be an iteration of proper forcing noitions as in 1.8,p ∈Pω2, A∼a Pω2-name.

If p|⊢“A∼is a strong measure zero set,” then there is a closedunbounded set C ⊆ω2 and a sequence ⟨¯νδ : δ ∈C ∩S12⟩such that each ¯νδ is a Pδ-name,andp|⊢ω2 ¯νδ has index ωω ∩Vδ and A∼⊆\h∈ωω∩Vδ[n∈ω[νh(n)]Proof: Let c∼be a Pω2-name for a function from ω2 to ω2 such that for all α < ω2,|⊢ω2∀h ∈ωω ∩Vα ∃νh ∈V c∼(α) : ∀n νh(n) ∈h(n)2 & A∼⊆[n[νh(n)]43818January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals(Why does c∼exist? Working in V [Gω2], note that there are only ℵ1 many functions inωω ∩Vα, and for each such h there is a νh as required in Sβ<ω2 Vβ, by 1.7.

)As Pω2 satisfies the ℵ2-cc, by 1.6(1) we can find a function c ∈V such that |⊢ω2∀α c∼(α)

In V , we can assign to each Pα-name h∼(for α < δ ∈C)a Pδ-name ν∼h∼such that|⊢ω2∀n ν∼h∼(n) ∈h∼(n)2 & A∼⊆[n[ν∼h∼(n)]Now in V [Gδ] we can choose for each h ∈ωω an α < δ and a Pα-name h∼such thath = h∼[Gδ]. Then we let νh := (ν∼h∼)[Gδ].

Thus we found a sequence ¯ν = ⟨νh : h ∈Vδ⟩∈Vδas required.3.2 Lemma:Assume ⟨Pα, Qα : α < ω2⟩is a countable support iteration of proper forcing notions, wherefor each ordinal δ ∈S12 |⊢δQδ = PTHδ for some Pδ-name Hδ. We will write gδ for thegeneric function added by Qδ.Assume H∼is a name for a dominating family (⊆ω(ω −{0})) in Vω2, and|⊢ω2“For all h ∈H∼, Sh := {δ < ω2 :cf(δ) = ω1, Qδ = PTHVδ}is stationary (where H(n) = h(n)2).”Let Gω2 ⊆Pω2 be V -generic, then in V [Gω2], a set A ⊆IR is a strong measure zero set iffthere is a closed unbounded set C ⊆ω2 such that for every δ ∈C ∩S12, A ⊆Sn[gδ(n)].Proof: First we prove the easy direction.

Assume that for some club C, for all δ ∈C ∩S12,A ⊆Sn[gδ(n)]. Then for every h ∈Vω2 ∩ω(ω −{0}) there is a δ = δh ∈C ∩Sh ⊆S12.So Qδh = (PTH)Vδh, where H(n) = h(n)2.

Since gδh(n) ∈h(n)2, and A ⊆S[gδh(n)] forarbitrary h, A is a strong measure zero set.Now for the reverse implication: In Vω2, let A be a strong measure zero set. By the previouslemma, there is a club set C ⊆ω2 and a sequence ⟨¯νδ : δ ∈C ∩S12⟩such that each ¯ν ∈Vδis a sequence with index ωω ∩Vδ and Vω2 |= A ⊆X¯νδ.

By 2.32 we have for all δ ∈C ∩S12:Vδ+1 |= ∃h ∈Vδ[n[νhδ (n)] ⊆[n[gδ(n)]So fix h0 ∈Vδ witnessing this. This inclusion is absolute, so alsoVω2 |=[n[νh0δ (n)] ⊆[n[gδ(n)]43819January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsThusVω2 |= A ⊆X¯νδ ⊆[n[νh0δ (n)] ⊆[n[gδ(n)]and we are done.3.3 Corollary: Assume Pω2 is as above. Then |⊢Pω2 Add(S).Proof: Let ⟨Ai : i ∈ω1⟩be a family of strong measure zero sets in Vω2.

To each i we canassociate a closed unbounded set Ci as in 3.2. Let C :=\i∈ω1Ci, then also C is closedunbounded, and for all δ ∈C ∩S12,[i∈ω1Ai ⊆[n∈ω[gδ(n)].

Again by 3.2,[i∈ω1Ai is a strongmeasure zero set.Our first goal is to show that Unif(S) does not guarantee the existence of a strong measurezero set of size c. Clearly the model for this should satisfy d = ℵ1 (if c = ℵ2), so we willconstruct a countable support iteration of ωω-bounding forcing notions.3.4 Theorem: If ZFC is consistent, thenZFC + c = ℵ2 + S = [IR]≤ℵ1 + no real is Cohen over L+ there is a generalized Sierpinski setis consistent.Proof: We will start with a ground model V0 satisfying V = L. Let H := ω(ω −{0}) ∩L ={hα : α < ω1}, and let Hα(n) = hα(n)2.Let ⟨Sα : α < ω1⟩be a family of disjoint stationary sets ⊆{δ < ω2 : cf(δ) = ω1}.Construct a countable support iteration ⟨Pα, Qα : α < ω2⟩satisfying(1) For all even α < ω2:|⊢PαFor some h : ω →ω −{0}, letting H(n) = h(n)2, Qα = PTH. (2) If δ ∈Sα, then |⊢δQδ = PTHα.

(3) For all odd α < ω2:|⊢Pα Qα = random real forcing.By 1.11 (or as a consequence of 1.15), Pω2 is ωω-bounding, so |⊢ω2“H is a dominatingfamily.” By 1.8(3) and 1.6 the assumptions of 3.3 are satisfied, so |⊢ω2Add(S). Also,|⊢ω2“c = ℵ2 and there are no Cohen reals over L.” Letting X be the set of random realsadded at odd stages, X is a generalized Sierpinski set: Any null set H ∈Vω2 is coveredby some Gδ null set H′ that coded in some intermediate model.

As coboundedly manyelements of X are random over this model, |H ∩X| ≤|H′ ∩X| ≤ℵ1.To conclude the proof of 3.4, we have to showVω2 |= “If X ⊆IR is of strong measure zero, then |X| < c.”43820January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsSince H is a dominating family, by 0.4 it is enough to show that in Vω2 the following holds:If ¯ν has index H, then |X¯ν| ≤ℵ1.We will show: If ¯ν ∈Vα has index H, then X¯ν ⊆ωω ∩Vα. (This is sufficient, by 1.7.

)Assume to the contrary that Gω2 is a generic filter, ¯ν ∈Vα, and in V [Gω2] there is δ > α,f ∈Vδ −Sγ<δ Vγ, f ∈X¯ν. So alsoV [Gδ] |= f ∈Vδ −[γ<δVγ and f ∈X¯νLet f∼be a Pδ-name, ¯ν∼a Pα-name, and let p ∈Pδ be a condition forcing all this.

δ cannotbe a successor ordinal, by 2.30. So δ is a limit ordinal, and cf(δ) must be ω, otherwise wewould have ω2 ∩Vδ = ω2 ∩Sγ<δ Vγ.So we have reduced the problem to the following lemma:3.5 Lemma: Let ⟨Pα, Qα : α < ε⟩be a countable support iteration of forcings where eachQα (for even α) is of the form PTHα for some (Pα-name) Hα, and Qα is random realforcing for odd α.

Let δ ≤ε be a limit ordinal of countable cofinality, and let f∼be aPδ-name of a function in ω2 such that |⊢δ∀α < δ f∼/∈Vα.Let H ∈V0 be a dominating family of functions, and assume that ¯ν has index H.Then |⊢δ f∼/∈Th∈HSn∈ω[νh(n)].For notational simplicity, we again assume that for all even α, |⊢α“Hα : ω →ω (ratherthan Hα : ω →2<ω).”Before we prove this lemma, we need the following two definitions (which make sense forany countable support iteration ⟨Pα, Qα : α < ω2⟩).3.6 Definition and Fact: For p ∈Pε, α < ε, p|α|⊢p(α) ≤r∼∈Qα, we define p ∧r∼asfollows: (p ∧r∼)(γ) = p(γ) for γ ̸= α, and (p ∧r∼)(α) = r∼.Then p ∧r∼∈Pε, p ∧r∼≥p, and (p ∧r∼)|α = p|α, so in particular p|α|⊢p ∧r∼∈Pε/Gα.Furthermore, p|α|⊢(p ∧r∼)(α) = r∼.Also, if p(α) = r∼, then p ∧r∼= p.3.7 Definition and Fact: If p ∈Pα, A a countable subset of ε, and p|⊢r∼∈Pε/Gα & r∼≥p & dom( r∼) ⊆α ∪A, then we define p ∧r∼as follows:For γ < α, (p ∧r∼)(γ) = p(γ). For γ ≥α and γ ∈A, (p ∧r∼)(γ) = r(γ).Again, p ∧r∼∈Pε, p ∧r∼≥p, and (p ∧r∼)|α = p|α, so in particular p|α|⊢p ∧r∼∈Pε/Gα.Also, if p1 ≤p2, then p1 ∧r∼≤p2 ∧r∼.3.8 Proof of 3.5: cf(δ) = ω, so we can find an increasing sequence ⟨δn : n < ω⟩of evenordinals converging to δ.

Assume there is a condition p forcing that f∼∈Th∈HSn∈ω[νh(n)].43821January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsWe will define sequences ⟨pn : n < ω⟩,⟨Fn : n < ω⟩,⟨ℓn : n < ω⟩,⟨s∼n : n ∈ω⟩⟨pin : n ∈ω, i ∈{0, 1}⟩,such that the following hold: For each n, pn, p0n, p1n are conditions in Pδ, δn is an evenordinal < δ, Fn is a finite subset of δn, ℓn is an integer, and s∼n is a Pδn-name for anelement of ω<ω. (We let p0 = p, F0 = ∅, ℓ0 = 0, p10 = p00 = p0, s0 = ∅).

For all n > 0 wewill have:(1) pn−1 ≤Fn,n pn. (2) Fn ⊆δn, Fn−1 ⊆Fn+1, Sk dom(pk) ⊆Sk Fk.

(3) δn−1 ∈Fn. (4) pn|δn|⊢s∼n = stem(pn(δn)) = stem(pn−1(δn))(5) For i ∈{0, 1}, pin = pn ∧pn(δn)[sn⌢i].

(6) pn|δn|⊢δn“∃l < ℓn ∃j0 ̸= j1 ∀i ∈{0, 1} : pin|⊢δn,δ f∼(l) = ji.”Note that (5) implies:(5’) pn|δn|⊢pin ∈Pδ/Gδn,and (6) implies(6’) For all η ∈ℓn2: pn|δn|⊢∃i ∈{0, 1} : pin|⊢δn,δ f∼|ℓn ̸= η[Proof of (6) ⇒(6’): In Vδn, let i ∈{0, 1} be such that η(l) ̸= ji, where l is as in (6). ]Finally, let q = Sn pn.

Then q|δn|⊢stem(pn(δn)) = stem(q(δn)) = s∼n by (1), (3) and (4)and 2.7. Let h∗∈H be a function such that for all n, ℓn < h∗(n).

So for all n, νh∗(n)|ℓnis a well-defined member of ℓn2.For each n, let i∼n be a Pδn-name of an element of {0, 1} such that(6”) pn|δn|⊢p i∼nn |⊢f∼|ℓn ̸= νh∗(n)|ℓnNow define a condition q′ as follows: For α /∈{δn : n ∈ω}, q′(α) = q(α), andq′(δn) = q(δ)[sn⌢i∼n](This is a Pδn-name. )Claim: q′ ≥q ≥p (this is clear) and q′|⊢f∼/∈Sn∈ω[νh∗(n)].To prove this claim, let Gδ ⊆Pδ be a generic filter containing q′, and assume f := f∼[Gδ] isin [νh∗(n)].

Let in := i∼n[Gδn]. Now q ∈Gδ implies pn ∈Gδ, so in particular pn|δn ∈Gδn.Note that stem(q(δn)) = stem(pn(δn)) = sn, so q′ ∈Gδ & pn ∈Gδ implies pinn ∈Gδ.

Also,by (6”) we have q′|⊢f∼/∈Sn[sf ∗(n)], a contradiction.This finishes the proof of 3.5 modulo the construction of the sequences pn, Fn, etc.First we fix enumerations dom(r) = {αmr : m ∈ω} for all r ∈Pδ. We will write αmn forαmpn.43822January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsAssume pn−1 is given. Let Fn := δn ∩(Fn−1 ∪{αmk : k < n, m < n} ∪{δn−1}).

This willtake care of (2) and (3).To define pn, first work in V [Gδn], where pn−1|δn ∈Gδn.We let sn := stem(pn−1(δn)).We let r0 := pn−1 ∧pn−1(δn)[sn⌢0], and r1 := pn−1 ∧pn−1(δn)[sn⌢1].By 1.18, we can find l, j0 ̸= j1 and r′0, r′1 such that r′i ≥ri, and r′i|⊢f∼(l) = ji.We now define a condition r ∈Pδ/Gδn as follows:· r|δn = pn−1|δn.· r(δn) = r′0(δn) ∪r′1(δn) ∪S{pn−1(δn)[sn⌢i] : i ∈succpn−1(δn)(sn) −{0, 1}}. (So stem(r(δn)) = sn.

)· If γ ∈dom(pn−1)∪dom(r′0)∪dom(r′1) and γ > δn, we let r(γ) be a Pγ-namesuch thatpn−1|δn |⊢δn|⊢δn,γ“For i in {0, 1}: If s∼n⌢i ⊆gδn, then r(γ) = r′i(γ),and if gδn extends neither s∼n⌢0 nor s∼n⌢1,then r(γ) = pn−1(γ)(We write gδn for gQδn, the branch added by the forcing Qδn. )This is a condition in Pδ/Gδn.

Note that we have the following:(i) stem(pn(δn)) = sn = stem(pn−1(δn))(ii) For i ∈{0, 1}, r ∧r′i(δn) ≥r′i. (iii) r ≥δnδ pn−1.Coming back to V , we can find names r∼, .

. .

, such that the above is forced by pn−1|δn.Now let ¯r be a condition in Pδn satisfying the following:(a) ¯r ≥Fn,n pn−1|δn. (b) For some countable set A ⊆δ, ¯r|⊢dom( r∼) ⊆A.

(c) For some ℓn ∈ω, ¯r|⊢l∼< ℓn.We can find a condition ¯r satisfying (a)–(c) by 1.15.Finally, let pn := ¯r ∧r∼. So pn|δn = ¯r.And let pin be defined by (5).Why does this work?First we check (1): pn−1|δn ≤Fn,n pn|δn by (a), and pn−1 ≤pn, because pn|δn|⊢pn =¯r ∧r∼≥r∼≥pn−1 (by (iii)).

So pn−1 ≤Fn,n pn. (2) and (3) are clear.Proof of (4): pn|δn|⊢stem(pn(δn)) = stem((¯r ∧r∼)(δn)) = stem( r∼(δn)) = s∼n.

(6): Let Gδn be a generic filter containing pn|δn. Work in V [Gδn].

We write r for r∼[Gδn],etc.43823January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realsWe want to show p0n|⊢δnδ f∼(l) = j0. (p1n|⊢δnδ f∼(l) = j1 is similar.) As r′0|⊢δnδ f∼(l) = j0, itis enough to see p0n ≥r′0.First we note that p0n ≥pn ≥pn−1.

Also p0n = pn ∧pn(δn)[sn⌢0] ≥pn = ¯r ∧r ≥r.Finally, p0n(δn) = r(δn)[sn⌢0] = r′0(δn).So p0n = p0n ∧r′0(δn) ≥r ∧r′0(δn) ≥r′0, and we are done.Our next model will satisfy(∗)Unif(S) + d = c = ℵ2.This in itself is very easy, as it is achieved by adding ℵ2 Cohen reals to L. (Also Miller[10] showed that Unif(S) + c = ℵ2 + b = ℵ1 is consistent. )Our result says that we can obtain a model for (∗) (and indeed, satisfying Add(S)) withoutadding Cohen reals.

In particular, (∗) does not imply Cov(M).3.9 Theorem: Con(ZFC) impliesCon(ZFC + c = d = ℵ2 > b + Add(S) + no real is Cohen over L)Proof (sketch): We will build our model by a countable support iteration of length ω2 whereat each stage we either use a forcing of the form PTH, or rational perfect set forcing. Astandard bookkeeping argument ensures that the hypothesis of 3.3 is satisfied, so we get|⊢ω2Add(S).

Using rational perfect set forcing on a cofinal set yields |⊢ω2d = c = ℵ2.Since all P-point ultrafilters from V0 are preserved, no Cohen reals are added.Proof (detailed version): Let {δ < ω2 : cf(δ) = ω1} ⊇Sγ<ω2 Sγ, where ⟨Sγ : γ < ω2⟩is afamily of disjoint stationary sets. Let Γ : ω2 × ω1 →ω2 be a bijection.

We may assumethat δ ∈SΓ(α,β) ⇒δ > α.First we claim that there is a countable support iteration ⟨Pα, Qα : α < ω2⟩and a sequenceof names ⟨⟨H∼βα : α < ω2⟩: β < ω1⟩such that(1) For all α < ω2, all β < ω1, H∼βα is a Pα-name. (2) For all α < ω2, |⊢α{Hβα : β < ω1} = ω(ω −{0, 1}).

(3) For all α < ω2: If α /∈Sγ<ω2 Sγ, then |⊢αQα = RP. (4) For all α < ω2, all β < ω1, all δ ∈SΓ(α,β): |⊢δQδ = PTH∼βα.Proof of the first claim: By induction on α we can first define Pα, then ⟨H∼βα : β < ω1⟩(by1.8(1)), then Qα (by (3) or (4), depending on whether α ∈Sγ<ω2 Sγ or not).Our second claim is that letting H∼be a name for all functions from ω to ω −{0, 1} inV [Gω2], the assumptions of 3.3 are satisfied, namely:(a) |⊢ω2“∀H ∈H∼∃γ < ω2 Sγ ⊆SH.”(b) |⊢ω2“∀γ < ω2 Sγ is stationary.”(b) follows from 1.8(3) and 1.6, and (a) follows from|⊢ω2“For all H ∈H∼there is α < ω2 and β < ω1 such that H = H∼βα.”43824January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen realswhich in turn is a consequence of 1.7.So by 3.3, Vω2 |= Add(S).Let Gω2 ⊆Pω2 be a generic filter, Vω2 = V [Gω2].Again by 1.7, every H ⊆ωω ∩Vω2 of size ≤ℵ1 is a subset of some Vα, α < ω2, so Hcannot be a dominating family, as rational perfect set forcing Qα+1 will introduce a realnot bounded by any function in H ⊆Vα ⊆Vα+1. Hence d = c = ℵ2.Finally, any P-point ultrafilter from V is generates an ultrafilter in Vω2, so there are noCohen reals over V .This ends the proof of 3.9.REFERENCES.

[1] J. Baumgartner, Iterated forcing,in: Surveys in set theory (A. R. D. Mathias,editor),London Mathematical Society Lecture Note Series, No. 8, CambridgeUniversity Press, Cambridge, 1983.

[2] A. Blass and S. Shelah, There may be simple Pℵ1- and Pℵ2-points, and the Rudin-Keisler order may be downward directed,Annals of pure and applied logic, 33(1987), pp.213–243. [3] P. Corazza, The generalized Borel Conjecture and strongly proper orders, Transac-tions of the American Mathematical Society, 316 (1989), pp.115–140.

. [4] M. Goldstern, H. Judah, On Shelah’s preservation theorem, preprint.

[5] H. Judah, S Shelah, H. Woodin, The Borel Conjecture, Annals of pure and appliedlogic, to appear. [6] H. Judah, Strong measure zero sets and rapid filters, Journal of Symbolic logic, 53(1988), pp.393–402.

[7] K. Kunen, Set Theory: An Introduction to Independence Proofs,[8] A. Miller, Mapping a set of reals onto the reals, Journal of Symbolic logic, 48 (1983),pp.575–584. [9] A. Miller, Rational Perfect Set Forcing, in: Axiomatic Set Theory, Boulder, Co1983, 143–159, Contemporary Math 31, AMS, Providence RI 1984.

[10] A. Miller, Some properties of measure and category, Transactions of the AmericanMathematical Society, 266,1 (1981), pp.93–114. [11] J. Pawlikowski, Power of transitive bases of measure and category, Proceedings ofthe American Mathematical Society, 93 (1985), pp.719–729.43825January 1991

Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals[12] Rothberger, Sur des families indenombrables de suites de nombres naturels etles probl`emes concernant la propriete´e C, Proc. Cambr.

Philos. Soc., 37 (1941),pp.109–126.

[13] Rothberger, Eine Versch¨arfung der Eigenschaft C, Fundamenta Mathematicae, 30(1938), pp.50–55. [14] S. Shelah, Proper Forcing, Lecture Notes in Mathematics Vol.

942, Springer Verlag. [15] S. Shelah, Proper and Improper Forcing, to appear in Lecture Notes in Mathematics,Springer Verlag.

[16] R. Solovay and S. Tennenbaum, Iterated Cohen extensions and Souslin’s problem,Annals of Mathematics, 94 (1971), pp.201–245. [17] R. Solovay, Real valued measurable cardinals,in: Axiomatic Set Theory,Proc.Symp.

Pure Math. 13 I (D. Scott, ed.

), pp.397–428, AMS, Providence RI, 1971.Martin GOLDSTERNgoldstrn@bimacs.cs.biu.ac.ilHaim JUDAHjudah@bimacs.cs.biu.ac.ilDepartment of MathematicsBar Ilan University52900 Ramat Gan, IsraelSaharon SHELAHshelah@shum.huji.ac.ilDepartment of MathematicsGivat RamHebrew University of Jerusalem43826January 1991

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