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Perfect sets of random reals

arXiv:math/9209205v1 [math.LO] 15 Sep 1992Perfect sets of random realsJ¨org Brendle∗and Haim Judah∗∗Abraham Fraenkel Center for Mathematical LogicDepartment of MathematicsBar–Ilan University52900 Ramat–Gan, IsraelAbstractWe show that the existence of a perfect set of random reals over a model M of ZFC doesnot imply the existence of a dominating real over M, thus answering a well-known openquestion (see [BJ 1] and [JS 2]). We also prove that B × B (the product of two copies ofthe random algebra) neither adds a dominating real nor adds a perfect set of random reals(this answers a question that A. Miller asked during the logic year at MSRI).∗The first author would like to thank the MINERVA-foundation for supporting him∗∗The second author would like to thank the Basic Research Foundation (the IsraelAcademy of Sciences and Humanities) for supporting him1

IntroductionThe goal of this work is to give several results concerning the relationship betweenperfect sets of random reals, dominating reals, and the product of two copies of the randomalgebra B. Recall that B is the algebra of Borel sets of 2ω modulo the null sets.

Also,given two models M ⊆N of ZFC, we say that g ∈ωω ∩N is a dominating real over Miff∀f ∈ωω ∩M ∃m ∈ω ∀n ≥m (g(n) > f(n)); and r ∈2ω ∩N is random over M iffravoids all Borel null sets coded in M iffr is the real determined by some filter which isB-generic over M (see [Je 1, section 42] for details).A tree T ⊆2<ω is perfect iff∀t ∈T ∃s ⊇t (sˆ⟨0⟩∈T ∧sˆ⟨1⟩∈T). For a perfecttree T we let [T] := {f ∈2ω; ∀n (f↾n ∈T)} denote the set of its branches.

Then [T]is a perfect set (in the topology of 2ω). Conversely, given a perfect set S ⊆2ω there isperfect tree T ⊆2<ω such that [T] = S. This allows us to confuse perfect sets and perfecttrees in the sequel; in particular, we shall use the symbol T for both the tree and the setof its branches.

— As a perfect tree is (essentially) a real, the statement there is a perfectset of reals random over M in N (where M ⊆N are again models of ZFC) asserts theexistence of a certain kind of real in N over M; and thus we may ask how it is related tothe existence of other kinds of reals (like dominating reals). This will be our main topic.— We recall that the existence of a random real does not imply the existence of a perfectset of random reals; in fact Cicho´n showed that B does not add a perfect set of randomreals [BJ 1, Theorem 2.1].

(Here, we say that a p.o. P adds a perfect set of random realsiffthere is a perfect set of reals random over M in M[G], where G is P-generic over M; asimilar definition applies to dominating reals etc.

)We note that being a perfect set of random reals over some model M of ZFC isabsolute in the following sense: if M ⊆N0 ⊆N1 are models of ZFC, T ∈(2<ω)ω ∩N0is a perfect tree so that [T] ∩N0 consists only of reals random over M, then every real in[T] ∩N1 is random over M as well (see [Je 1, Lemma 42.3]).We now state the main results of our work, and explain how they will be presented in§§ 1 – 3; then we will give some further motivation for the study of perfect sets of randomreals, and close with some notation.The main results and the organization of the paper.Using techniques of [Ba], Bar-2

toszy´nski and Judah proved in [BJ 1, Theorem 2.7] that(*) given models of ZFC M ⊆N such that N contains a dominating real over M, N[r]contains a perfect set of random reals over M, where r is random over N.Our first result shows that the converse does not hold (1.4 – 1.7).Theorem 1. There is a p.o.

which adds a perfect set of random reals and does notadd dominating reals.The framework for proving Theorem 1 (developed in § 1) will enable us to give generalpreservation results for not adding dominating reals in both finite support iterations andfinite support products of ccc forcing notions (1.8). The former will be exploited in 1.9to discuss cardinal invariants closely related to our subject.

As a special instance of thelatter we shall show (1.10)Theorem 2. B × B does not add dominating reals.

(This result was proved earlier by Shelah but never published.) The algebra B×B is ratherdifferent from B; e.g.

it is well-known that B × B adds Cohen reals whereas B does not(see [Je 2, part I, 5.9]). As it is known that some other forcing notions adding both Cohenand random reals (like B × C ∼= B ∗C and C ∗B, where C is the Cohen algebra, and ∗denotes iteration) do not add perfect sets of random reals (see [JS 2, 2.3] for B × C and[BJ 1, Theorem 2.13] for C ∗B), we may ask whether B × B does.

We shall show in § 2that the answer is again negative.Theorem 3. B × B does not add a perfect set of random reals.The argument for this proof (which uses ideas from the proof that B does not add a perfectset of random reals — see [BJ 1, 2.1 – 2.4]) is rather long and technical; and one mightget a shorter proof if the following question has a positive answer.Question 1.

Is B × B a complete subalgebra of C ∗B?We note here that all other embeddability relations between these three algebras addingboth Cohen and random reals (namely, B × C, B × B, C ∗B) are known. We shall sketchthe arguments which cannot be found in literature in 3.1.Two further open problems are closely tied up with the Bartoszy´nski–Judah Theorem(*) and our Theorem 1, respectively.Question 2.

Given models of ZFC M ⊆N such that N contains both a dominatingreal and a random real over M, is there a perfect set of random reals over M in N?3

We shall show in 3.2 that to answer Question 2 it suffices to consider the problem whetherB × D ∼= B ∗D adds a perfect set of random reals, where D is Hechler forcing. We notethat for many p.o.s P adding a dominating real (e.g.

Mathias forcing) it is true that B×Padds a perfect set of random reals (3.3).Question 3. Given models of ZFC M ⊆N, does the existence of a perfect set ofrandom reals over M in N imply the existence of an unbounded real over M in N?Here we say that g ∈ωω ∩N is an unbounded real over M iff∀f ∈ωω ∩M ∃∞n (g(n) >f(n)), where ∃∞n means there are infinitely many n (dually, ∀∞n abbreviates for all butfinitely many n).Motivation.

One of the reasons for studying perfect sets of random reals concernsfinite support iterations of ccc forcing notions. Namely, let ⟨Pn, ˘Qn; n ∈ω⟩be an ω-stagefinite support iteration such that for all n ∈ω, ∥−P n ” ˘Qn is ccc”.

Then the following areequivalent [JS 2, Theorem 2.1]:(i) There exists r ∈V [Gω] \ Sn V [Gn] random over V ,(ii) there exists n ∈ω and T ∈V [Gn] a perfect set of random reals over V ,where ⟨Gi; i ≤ω⟩is a chain of Pi-generic filters. So adding a random real in the ω-th stageis stronger than just adding a random real in an initial step (on the other hand, Pω adds adominating real is simply equivalent to there is an n ∈ω such that Pn adds a dominatingreal [JS 1, Theorem 2.2]).

— Also perfect sets of random reals seem to play an importantrole in the investigation of the problem, posed by Fremlin, whether the smallest coveringof the real line by measure zero sets can have cofinality ω. To build a model of ZFC wherethis is true we suggest an iterated forcing construction (with finite supports) which firstlyadds ωω many Cohen reals over L to produce a family of ωω null sets which will still coverthe real line in the final extension, and then goes through every subalgebra of the randomalgebra which is the random algebra restricted to some small inner model (in which thecontinuum has size < ωω) in ωω+1 steps (see the introduction of [JS 3] for details).

Byconstruction, we destroy all small covering families. So the main problem is to show thatwe do not add a real which does not belong to the family of ωω null sets added in theintermediate stage.

To do this, it suffices (essentially) to prove that the whole iterationdoes not add a perfect set of random reals over the ground model L. We think that ourTheorem 3 is a small but important step in this direction, and we hope that the ideas4

involved can be generalized to give a positive answer toQuestion 4. Let A be a complete subalgebra of B.

Let ˘BA be an A-name for B. Is ittrue that B ∗˘BA does not add a perfect set of random reals?Cicho´n’s Theorem [BJ 1, Theorem 2.1] says that this is true if A = B, and our Theorem 3gives a positive answer in case A is trivial.Notation.

Our notation is fairly standard. We refer the reader to [Je 1] for set theoryand to [Ox] for measure theory.

Most of the cited material will appear in the forthcomingbook [BJ 2]. We now explain some notions which might be less familiar.Given a finite sequence s (i.e.

either s ∈2<ω or s ∈ω<ω), we let lh(s) := dom(s)denote the length of s; for ℓ∈lh(s), s↾ℓis the restriction of s to ℓ.ˆ is used forconcatenation of sequences; and ⟨⟩is the empty sequence.Furthermore, for s ∈2<ω,[s] := {f ∈2ω; f↾lh(s) = s} is the set of branches through s (the open subset of 2ωdetermined by s).Given a perfect tree T ⊆2<ω and s ∈T, we let Ts := {t ∈T; t ⊆s or s ⊆t}; andstem(T) := ∪{s ∈T; Ts = T} is the stem of T. For ℓ∈ω, we let T↾ℓ:= {s ∈T; lh(s) ≤ℓ}, the finite initial part of T of height ℓ. We will confuse finite trees T with all branchesof fixed length ℓwith the set of branches [T] := {s ∈T; lh(s) = ℓ}.We assume the reader to be familiar with forcing and Boolean-valued models (see [Je1], [Je 2]).

We suppose that all our p.o.s (forcing notions) have a largest element 1 . Givena p.o.

P ∈V , we shall denote P-names by symbols like ˘f, ˘T, ... and their interpretationin V [G] (where G is P-generic over V ) by ˘f[G], ˘T[G], ... If φ is a sentence of the P-forcinglanguage, we let ∥φ ∥be the Boolean value of φ; i.e.

the maximal element forcing φ in thecomplete Boolean algebra r.o. (P) associated with P. We shall often confuse P and r.o.

(P).We equip B × B := {(p, q); p, q ∈B \ {0}} ∪{0} with the product measure (i.e.µ(p, q) = µ(p) · µ(q)). Then µ : B × B →[0, 1] is finitely additive and strictly positive (anynon-zero condition has positive measure).

By [Ka, Proposition 2.1], µ can be extended toa finitely additive, strictly positive measure on r.o. (B × B).

This will be used in 2.5. Notethat this measure is not σ-additive.5

§ 1 Not adding dominating reals1.1 We shall now introduce the framework needed to prove Theorem 1. Besides givingthe latter result this framework will also provide us with preservation results for not addingdominating reals in finite support products and finite support iterations.Let P be an arbitrary p.o.

A function h : P →ω is a height function iffp ≤q impliesh(p) ≥h(q). A pair (P, h) is soft iffP is a p.o., h is a height function on P, and thefollowing two conditions are satisfied:(I) (decreasing chain property) if {pn; n ∈ω} is decreasing and ∃m ∈ω ∀n ∈ω (h(pn) ≤m), then ∃p ∈P ∀n ∈ω (p ≤pn);(II) (weak finite cover property) given m ∈ω and {pi; i ∈n} ⊆P there is {qj; j ∈k} ⊆Pso that(i) ∀i ∈n, j ∈k, qj is incompatible with pi;(ii) whenever q is incompatible with all pi and h(q) ≤m then there exists j ∈k sothat q ≤qj.We also consider the following property of pairs (P, h) — where P is a p.o.

and h a heightfunction on P:(*) given a maximal antichain {pn; n ∈ω} ⊆P and m ∈ω there exists n ∈ω such that:whenever p is incompatible with {pj; j ∈n} then h(p) > m.1.2 Lemma. If (P, h) is soft, then (P, h) has property (*).Proof.

Suppose not and let {pn; n ∈ω} and m ∈ω witness the contrary. For eachn ∈ω let {qnj ; j ∈kn} be a weak finite cover with respect to {pi; i ∈n}, m accordingto (II).

By assumption none of these sets can be empty and we can assume that each qnjhas height ≤m. By the cover property (II) (ii) they form an ω-tree with finite levelswith respect to ”≤”.

By K¨onig’s Lemma this tree has an infinite branch. By (I) thereis a condition below this branch, contradicting the fact that {pn; n ∈ω} is a maximalantichain.1.3 Theorem.

Suppose P is a ccc p.o., h is a height function on P, and (P, h) satisfiesproperty (*). Then any unbounded family of functions in ωω∩V is still unbounded in V [G],where G is P-generic over V .6

Proof. Let F be unbounded in ωω ∩V .

Suppose ∥−P ˘f ∈ωω. For each m ∈ω let{pmn ; n ∈ω} be a maximal antichain deciding the value ˘f(m).

Choose nm according to (*)so that: whenever p is incompatible with {pmj ; j ∈nm}, then h(p) > m. Define f : ω →ωby setting f(m) := the maximum of the values of ˘f(m) decided by {pmj ; j ∈nm}. Letg ∈F be a function which is not dominated by f. We claim that ∥−P ” ˘f does not dominateg”.For suppose there is a p ∈P and a k ∈ω so thatp ∥−∀m ≥k ˘f(m) > g(m).Choose m ≥k so that h(p) ≤m and g(m) ≥f(m).

Then p must be compatible with pmjfor some j ∈nm. But if q is a common extension, thenq ∥−˘f(m) > g(m) ≥f(m) ≥˘f(m),a contradiction.1.4 Towards the proof of Theorem 1.

We think of B as consisting of sets B ⊆2ω ofpositive measure so that for all t ∈2<ω, if [t] ∩B ̸= ∅then µ([t] ∩B) > 0; for m ∈ω letB ∩2m := {t ∈2m; [t] ∩B ̸= ∅}. Then we define the following p.o.

(P, ≤):(B, n) ∈P ⇐⇒B ∈B ∧n ∈ω;(B, n) ≤(C, m) ⇐⇒B ⊆C ∧n ≥m ∧B ∩2m = C ∩2m.It follows from the ccc-ness of any product of finitely many copies [Je 2, part I, 5.7] of Bthat P is ccc, too. Clearly, P generically adds a perfect set of random reals, and we haveto show that it does not add dominating reals.

To this end, we will introduce a heightfunction on P.In fact, let P′ ⊆P be the set of conditions (B, n) ∈P so that ∀s ∈2n ∩B µ([s]∩B) ≥2−(lh(s)+1). P′ is dense in P (by the Lebesgue density Theorem [Ox, Theorem 3.20]).

Wedefine h : P′ →ω by h((B, n)) = n and work with P′ from now on.1.5 Lemma. (P′, h) is soft.Remark.

By 1.2 and 1.3 the proof of this Lemma finishes the proof of Theorem 1.Proof. (I) is clear (for if {(Bn, m); n ∈ω} is decreasing then (∩Bn, m) is a lowerbound because we took our conditions from P′).

For (II) we use:7

1.6 Main Claim.Given (B, n), (C, m) ∈P′ and k ∈ω there are finitely manyconditions {qi; i ∈j} below (C, m) so that(i) each qi is incompatible with (B, n);(ii) if q is incompatible with (B, n), h(q) ≤k, and q ≤(C, m), then ∃i ∈j (q ≤qi).Proof. Without loss k ≥m, n. Assume n ≥m.

Let ℓbe such that m ≤ℓ≤n. Wenow describe which conditions of height ℓwe put into our finite set.

(i) For each T ⊆2ℓwith T↾m = C ∩2m and T ⊆C ∩2ℓand T ̸= B ∩2ℓlet CT ∈Bbe such that CT ∩2ℓ= T and CT ∩[t] = C ∩[t] for each t ∈T. If (CT , ℓ) ∈P′, then put(CT , ℓ) into the set.

(ii) For each T ⊆2n with T↾m = C ∩2m and T ⊆C ∩2n and T↾ℓ= B ∩2ℓandT ̸⊇B ∩2n let CT ∈B be such that CT ∩2n = T and CT ∩[t] = C ∩[t] for each t ∈T. If(CT , ℓ) ∈P′, then put (CT , ℓ) into the set.

(iii) For each T ⊆2n with T↾m = C ∩2m and T ⊆C ∩2n and T↾ℓ= B ∩2ℓand T ⊇B ∩2n and for each t ∈B ∩2n let CT,t ∈B be such that CT,t ∩2n = T andCT,t ∩[s] = C ∩[s] for each s ∈T \ {t} and CT,t ∩[t] = (C ∩[t]) \ B. If (CT,t, ℓ) ∈P′, thenput (CT,t, ℓ) into the set.It is easy to see that any condition of height ℓbelow (C, m) which is incompatiblewith (B, n) lies below one the conditions defined in (i) – (iii) above.Next suppose that n ≤ℓ≤k.

Then we can again find a finite set of conditions ofheight ℓsatisfying the requirements of the main claim for conditions of height ℓby anargument similar to the one in (i) – (iii) above.This takes care of the case when n ≥m. So assume now n ≤m.

Then we get our setof conditions as in the preceding paragraph.1.7 Proof of (II) of Lemma 1.5 from the main claim 1.6. We make induction usingthe main claim repeatedly.

I.e. let (B, n) = p0 and (C, m) = (2ω, 0) and apply the mainclaim to them to get {qi; i ∈j}.

Then let (B, n) = p1 and (C, m) = qi (i ∈j) and applythe main claim j times to get a new family. Etc.This finishes the proof of Lemma 1.5 and of Theorem 1.1.8 Theorem.

(i) Suppose ⟨Pα, ( ˘Qα, ˘hα); α < κ⟩is a finite support iteration ofarbitrary length κ (κ limit) such that∥−P α ” ˘Qα is ccc, ˘hα is a height function on ˘Qα and ( ˘Qα, ˘hα) has property (*)”.8

Then Pκ = limα<κ Pα does not add dominating reals. (ii) Suppose ⟨(Pα, hα); α < κ⟩is a sequence of soft ccc p.o.s of arbitrary length κ.Then there is a height function h on the finite support product P of the Pα (α < κ) so that(P, h) is soft.Remark.

In particular, both the finite support iteration and the finite support productof an arbitrary number of ccc soft p.o.s does not add dominating reals (cf 1.2, 1.3).Proof. (i) It suffices to show by induction on α that∥−P αωω ∩V is unbounded in ωω.If α is a limit ordinal, this follows from [JS 1, Theorem 2.2].

So suppose α is a successor.Then the result follows from Theorem 1.3 and the induction hypothesis. (ii) We make again induction on α.

Let Qα be the finite support product of the Pβwhere β < α. We shall recursively construct height functions gα : Qα →ω such that(a) for α < β, gα ⊆gβ;(b) gα(q) ≥maxβ<α hβ(q↾Pβ);(c) gα(q) ≥|supp(q)|;(d) (Qα, gα) is soft.Clearly, a gα satisfying (b) and (c) will satisfy the decreasing chain property (I) as well.

(We assume without loss that ∀β < κ, hβ(1 ) = 0. )We first deal with the case when α is a successor ordinal, α = β + 1.

Then Qα =Qβ × Pβ. Let m := max{gβ(q), hβ(p)} and define gα : Qα →ω bygα(q, p) :=n m + 1if |supp(q)| = m and p ̸= 1motherwisefor (q, p) ∈Qβ × Pβ.

gα is a height function on Qα which is easily seen to satisfy (a) —(c) above.To show that (Qα, gα) satisfies the weak finite cover property (II), let {(qi, pi); i ∈n}be a finite subset of Qα and let m ∈ω. For each A ⊆n let {qAj ; j ∈kA} be a weak finitecover with respect to {qi; i ∈A} and m in Qβ (i.e.

(i) ∀i ∈A, j ∈kA, qAj is incompatiblewith qi; and (ii) whenever q is incompatible with all qi (i ∈A) and h(q) ≤m then thereexists j ∈kA so that q ≤qAj ), and let {pAj ; j ∈ℓA} be a weak finite cover with respect to{pi; i ∈n\A} and m. We claim that the family F := {(qAi , pAj ); A ⊆n ∧i ∈kA ∧j ∈ℓA}is a weak finite cover with respect to {(qi, pi); i ∈n} and m.9

Clearly, F satisfies (i) of the definition of the weak finite cover property (II). Fur-thermore, if (q, p) is incompatible with all (qi, pi) (i ∈n) there exists A ⊆n such thatq is incompatible with all qi for i ∈A and p is incompatible with all pi for i ∈n \ A.So if gα(q, p) ≤m (in particular, gβ(q) ≤m and hβ(p) ≤m) then we can find j ∈kAand j′ ∈ℓA such that q ≤qAj and p ≤pAj′; i.e.

(q, p) ≤(qAj , pAj′). This shows (ii) in thedefinition of the weak finite cover property (II).Now suppose α is a limit ordinal.

Then let gα := Sβ<α gβ. gα clearly satisfies (a)— (c), and the weak finite cover property (II) for (Qα, gα) follows from the weak finitecover properties of the (Qβ, gβ) for β < α (because (II) talks only about finitely manyconditions).1.9 We note here that the notions discussed so far are closely tied up with somecardinal invariants of the continuum.

Namely, we let N denote the ideal of null sets andadd(N ) := the least κ such that ∃F ∈[N ]κ (S F ̸∈N );wcov(N ) := the least κ such that ∃F ∈[N ]κ (2ω \ S F does not contain a perfect set);cov(N ) := the least κ such that ∃F ∈[N ]κ (S F = 2ω);unif(N ) := the least κ such that [2ω]κ \ N ̸= ∅;wunif(N ) := the least κ such that there is a family F ∈[[2<ω]ω]κ of perfect sets with ∀N ∈N ∃T ∈F (N ∩T = ∅);cof(N ) := the least κ such that ∃F ∈[N ]κ ∀A ∈N ∃B ∈F (A ⊆B);b := the least κ such that ∃F ∈[ωω]κ ∀f ∈ωω ∃g ∈F ∃∞n (g(n) > f(n));d := the least κ such that ∃F ∈[ωω]κ ∀f ∈ωω ∃g ∈F ∀∞n (g(n) > f(n)).Then we can arrange these cardinals in the following diagram.put diagram 1 hereHere the invariants get larger as one moves up in the diagram. b ≥add(N ) (and duallyd ≤cof(N )) is due to Miller [Mi].

The dotted line says that wcov(N ) ≥min{cov(N ), b}(and dually, wunif(N ) ≤max{unif(N ), d}). This can be seen from the Bartoszy´nski–Judah result (*) in the Introduction as follows.

Suppose λ := wcov(N ) < min{cov(N ), b}.Let M be a model of enough ZFC of size λ containing a weak covering family. As λ < bthere is a real f ∈ωω dominating all reals in M. Let N be a model of enough ZFC of sizeλ containing M and f. As λ < cov(N ), there is a real r ∈2ω random over N. By (*) this10

implies that there is a perfect set of random reals over M, a contradiction. — Iteratingthe p.o.

from Theorem 1 we get:Theorem 1’. For any regular cardinal κ, it is consistent that wcov(N ) = κ whileb = ω1; Dually, it is consistent that wunif(N ) = ω1 while d = κ.Proof.

(a) Assume CH. We make a finite support iteration of length κ of the p.o.

Pdescribed in 1.4. In the generic extension we have wcov(N ) = κ because we added κ manyperfect sets of random reals; and b = ω1 by 1.5, 1.2 and 1.8 (i).

(b) Assume MA + 2ω = κ; and make a finite support iteration of length ω1 of P.Again standard arguments show that wunif(N ) = ω1 and d = κ in the generic extension.The most interesting open question concerning the relationship between these cardi-nals is connected with Question 3 in the Introduction.Question 3’. Is it consistent that wcov(N ) > d?

Dually, is it consistent that wunif(N )

§ 2 Not adding perfect sets of random reals2.1 This whole section is devoted to the proof of Theorem 3. Lemmata 2.2 and 2.3below which we single out from the principal argument bear the imprint of the proof ofCicho´n’s Theorem in [BJ 1, 2.1 – 2.4].

The main new idea comes in in 2.4. The rest (2.5– 2.9) is mostly technical.Given k′, k ∈ω, k′ < k, we let ǫk,k′ := 21−k · (1 +k1+ ... +kk′−1).

Clearly, givenany k′ ∈ω, we can find k > k′ so that ǫk,k′ is arbitrarily small.2.2 Lemma. Given n, k, k′ ∈ω (k′ ≤k and k ≤2n) and Z ⊆2n and real num-bers aT for each T ⊆2n with |T ∩Z| ≥k there exists Z′ ⊆Z of size ≤|Z|2such thatP|T ∩Z′|≥k′ aT ≥PT aT · (1 −ǫk,k′).Proof.

Let a := PT aT . For any λ close to 1 (λ < 1) we can choose ℓ∈ω and{Ti; i ∈ℓ} so thataT · λ < |{i; Ti = T}|ℓ· a < aT · λ−1for all T. For i ∈ℓlet Zi := {Z′ ⊆Z; |Z′ ∩Ti| ≥k′ and |(Z \ Z′) ∩Ti| ≥k′}.

Then|Zi| · 2−|Z| ≥1 −ǫk,k′ for all i ∈ℓ. We claim that there is an X ⊆ℓof size ≥ℓ· (1 −ǫk,k′)so that Ti∈X Zi ̸= ∅.For suppose not.

Then for each Z′ ⊆Z, |{i ∈ℓ; Z′ ∈Zi}| < ℓ· (1 −ǫk,k′). Hence2|Z| · ℓ· (1 −ǫk,k′) ≤Pi∈ℓ|Zi| = PZ′⊆Z |{i ∈ℓ; Z′ ∈Zi}| < 2|Z| · ℓ· (1 −ǫk,k′), acontradiction.Now choose Z′ ∈Ti∈X Zi.

Then either |Z′| ≤|Z|2 or |Z \ Z′| ≤|Z|2 . Assume withoutloss that |Z′| ≤|Z|2 .

FurthermoreX|T ∩Z′|≥k′aT · λ−1 >X|T ∩Z′|≥k′|{i; Ti = T}|ℓ· a ≥|X|ℓ· a ≥a · (1 −ǫk,k′).Because there are only finitely many possibilities for the sum on the lefthand side, we canchoose λ so small that for the Z′ chosen according to this λ we haveX|T ∩Z′|≥k′aT ≥a · (1 −ǫk,k′).12

This finishes the proof of the Lemma.2.3 Lemma. Given a real ǫ > 0 and m ∈ω the following is true for large enoughk, n ∈ω: given real numbers aT for each T ⊆2n with |T| ≥k there exists a Z ⊆2n of size≤2n−m such that PT ∩Z̸=∅aT ≥PT aT · (1 −ǫ).Remark.

We say that a statement is true for large enough n iff∃k ∈ω so that ∀n ≥kthe statement is true.Proof.Construct recursively a sequence ⟨ki; i ≤m⟩of natural numbers so thatQi∈m(1 −ǫki+1,ki) ≥(1 −ǫ) where k0 = 1. Let k ≥km and n so large that k ≤2n.

Nowapply Lemma 2.2 m times to get Z.2.4 Diagonal chains. It turns out that a detailed investigation of antichains in B × Bis necessary for the proof of Theorem 3.

We say (p, q) ∈B × B is quadratic iffµ(p) = µ(q).Clearly the quadratic conditions are dense in B × B so that it suffices (essentially) toconsider them. — More generally, given (p, q) ∈B × B, (p′, q′) is quadratic in (p, q) iffp′ ≤p, q′ ≤q and µ(p′)µ(p) = µ(q′)µ(q) .

{(pn, qn); n ∈ω} is said to be a first order diagonal chain in B × B iff(1) each (pn, qn) is quadratic;(2) both {pn; n ∈ω} and {qn; n ∈ω} are maximal antichains in B.More generally we say that C = {(pστn , qστn ); n ∈ω, σ, τ ∈ω

2.5 Towards the proof of Theorem 3. Let ˘T be a B × B-name so that∥−B×B ” ˘T is perfect”.We want to construct a null set N in the ground model so that∥−B×B ”N ∩˘T ̸= ∅”.More explicitly, using Lemma 2.3, we shall construct sequences ⟨nm; m ∈ω⟩and ⟨Zm; m ∈ω⟩so that Zm ⊆2nm, |Zm| ≤2nm−m and∥−B×B∃x ∈˘T ∃∞m (x↾nm ∈Zm).This will imply the required result for N := {x ∈2ω; ∃∞m (x↾nm ∈Zm)} is a null set(see below in 2.9).

— We set n0 := 0 and Z0 := ∅. Now let m > 0 and assume that nm−1and Zm−1 have been defined.

We shall describe the construction of nm and Zm.Let δ > 0 be very small; let ⟨zj; j ∈m⟩, ⟨yj; j ∈m −1⟩be sequences of naturalnumbers so that z0 > 1, yj = 4·zj andyjzj+1 is very small; let ǫ > 0 such that δ−2·ǫ·m·zm−1is very small (in fact, we want that ζm = ζ = 2 · m · (ǫ + δ−2 · ǫ · m · zm−1 + δ) + Pm−2j=0yjzj+1is – say – smaller than14m – cf 2.6); let v ∈ω be such that zm−1 ≤v · (1 −ǫ)2. Choose kaccording to Lemma 2.3 for ǫ and m+2·nm−1.

Let {(pστi , qστi ); i ∈ω, σ, τ ∈ω

⟨⟩⟨⟩is relevant. Choose j⟨⟩⟨⟩∈ω such that a⟨⟩⟨⟩:= Pi∈j⟨⟩⟨⟩µ(p⟨⟩⟨⟩i) ≥1 −ǫ.Suppose aστ and jστ are defined for relevant pairs στ of length ℓ(0 ≤ℓ< m −1).

Then14

σˆ⟨i⟩τˆ⟨j⟩is relevant iffi, j ∈jστ and i ̸= j.Furthermore, for each such i, j, choosejσˆ⟨i⟩τˆ⟨j⟩∈ω such thataσˆ⟨i⟩τˆ⟨j⟩:=Xi′∈jσˆ⟨i⟩τˆ⟨j⟩µ(qσˆ⟨i⟩τˆ⟨j⟩i′) · µ(pστi ) ≥µ(pστi ) · µ(qστj ) · (1 −ǫ). (5.2)Now letnm := n :=maxστ relevant,i∈jστ nστi .Fix s ∈2≤nm−1.

For T ⊆2n with |T| ≥k and s ⊆stem(T) and for relevant tuplesστ letaστT :=Xi∈jστµ(∥s ∈˘T ∧˘Ts↾n = T ∥∩(pστi , qστi ))µ(pστi )· µ(pσ↾(ℓ−1)τ↾(ℓ−1)σ(ℓ−1)). (5.3)And letaστs:=Xi∈jστµ(∥s ̸∈˘T ∥∩(pστi , qστi ))µ(pστi )· µ(pσ↾(ℓ−1)τ↾(ℓ−1)σ(ℓ−1)).Then PT aστT + aστs= aστ (this uses the finite additivity of the measure on r.o.

(B × B)– see Introduction). Let ajT := Plh(σ)=lh(τ)=j aστT , ajs := Plh(σ)=lh(τ)=j aστsand aj :=PT ajT + ajs = Plh(σ)=lh(τ)=j aστ.Let aT := Pj∈majTaj−ajs .Apply Lemma 2.3 to getZsm := Zs ⊆2n of size ≤2n−m−2·nm−1 such thatXT ∩Zs̸=∅aT ≥XTaT · (1 −ǫ) = m · (1 −ǫ).

(5.4)Finally, set Zm := Z := Ss∈2≤nm−1 Zs. This completes the construction of nm = nand Zm = Z.2.6 Main Claim.Let s ∈2≤nm−1.Suppose (p, q) is a quadratic condition suchthat (p, q) ∥−”s ∈˘T ∧Zsm ∩˘Ts = ∅”.Then µ(p, q) <14m + ζm (where ζm = ζ =2 · m · (ǫ + δ−2 · m · zm−1 + δ) + Pm−2j=0yjzj+1 as in 2.5).Proof.

The proof of the main claim will take some time (up to 2.8); to make ourargument (which is essentially one big estimation) go through smoothly we need to makesome conventions and introduce a few more notions.15

If σ is a sequence of length ℓ≥1, ˆσ = σ↾(ℓ−1) will be the sequence with the lastvalue deleted. For ℓ= 0, pˆ⟨⟩ˆ⟨⟩⟨⟩(ℓ−1) = qˆ⟨⟩ˆ⟨⟩⟨⟩(ℓ−1) = 1 , the maximal element of the Booleanalgebra B. Pij will always stand for Pi,j∈jστ,i̸=j, where στ is clear from the context;similarly, Pστ means that the sum runs over all relevant στ of some fixed length ℓ(whereℓis again clear from the context).

— For a relevant pair στ we letAστ := {i ∈jστ; µ(p ∩pστi ) < δ · µ(pστi )}(6.1)Bστ := {i ∈jστ; µ(q ∩qστi ) < δ · µ(qστi )} \ Aστ(6.2)Cστ := jστ \ (Aστ ∪Bστ)(6.3)Let lh(σ) = lh(τ) = m −1. We say the relevant pair στ is nice iffXi∈Cστµ(qστi ) · µ(pˆσˆτσ(m−2)) ≤δ−2 · aστ · ǫ · m · zm−1.

(6.4)More generally, if lh(σ) = lh(τ) = ℓ(where 0 ≤ℓ< m −1), we say the pair στ is nice iff(I)Xi∈Cστµ(qστi ) · µ(pˆσˆτσ(ℓ−1)) ≤δ−2 · aστ · ǫ · m · zm−1;(6.5)(II)Xij,σˆ⟨i⟩τˆ⟨j⟩niceaσˆ⟨i⟩τˆ⟨j⟩≥(1 −yℓzℓ+1) ·Xijaσˆ⟨i⟩τˆ⟨j⟩. (6.6)(Note that this is a definition by backwards recursion on ℓ.

)2.7 Claim. for any ℓ(0 ≤ℓ≤m −1), Pστ nice aστ ≥(1 −1zℓ) · aℓ.(7.1)Proof.

We first show that for any ℓwe haveδ−2 · aℓ· ǫ · m ≥XστXi∈Cστµ(qστi ) · µ(pˆσˆτσ(ℓ−1)). (7.2)By construction (5.4), PT ∩Zs̸=∅aT ≥m · (1 −ǫ); i.e.

PT ∩Zs̸=∅Pj∈majTaj−ajs ≥m ·(1 −ǫ). Hence PT ∩Zs̸=∅aℓT ≥(aℓ−aℓs)(1 −ǫ · m).

As (p, q) ∥−”s ∈˘T ∧˘Ts↾n ∩Zs = ∅”we get (using (5.3) and also the definition of Cστ ((6.1) — (6.3)))aℓ· ǫ · m ≥(aℓ−aℓs) · ǫ · m ≥XστXi∈Cστµ(pˆσˆτσ(ℓ−1)) · µ(p ∩pστi , q ∩qστi )µ(pστi )≥δ2 ·XστXi∈Cστµ(qστi ) · µ(pˆσˆτσ(ℓ−1)).16

This shows that formula (7.2) holds.Next, we prove the claim by backwards induction. So assume ℓ= m −1.

In thatcase, it follows immediately from formula (7.2) and the definition of niceness (6.4) thatPστ nice aστ ≥(1 −1zm−1 ) · am−1.So let ℓ< m −1 and assume the claim has been proved for ℓ+ 1. We let Σ(I) :={στ; lh(σ) = lh(τ) = ℓand στ satisfies (I) of the definition of niceness } and Σ(II) :={στ; lh(σ) = lh(τ) = ℓand στ satisfies (II) of the definition of niceness }.

By the argumentof the preceding paragraph we know that Pστ∈Σ(I) aστ ≥(1 −1zm−1 ) · aℓ. We claim thatXστ∈Σ(II)Xijaσˆ⟨i⟩τˆ⟨j⟩≥(1 −1yℓ) · aℓ+1.

(7.3)For suppose not. Then Pστ̸∈Σ(II)Pij aσˆ⟨i⟩τˆ⟨j⟩>1yℓ·aℓ+1.

But if στ does not satisfy(II) thenXij,σˆ⟨i⟩τˆ⟨j⟩not niceaσˆ⟨i⟩τˆ⟨j⟩>yℓzℓ+1·Xijaσˆ⟨i⟩τˆ⟨j⟩.HenceXστ̸∈Σ(II)Xij,σˆ⟨i⟩τˆ⟨j⟩not niceaσˆ⟨i⟩τˆ⟨j⟩>1zℓ+1· aℓ+1,contradicting the induction hypothesis.As (1 −ǫ) · aℓ≤11−ǫ · aℓ+1 + PστPi∈jστ µ(pστi , qστi ) <11−ǫ · aℓ+1 + 1v ·11−ǫ · aℓ((5.1)and (5.2)), we have1(1−ǫ)2 · aℓ+1 ≥(1 −1v ·1(1−ǫ)2 ) · aℓ. Hence by (7.3)Xστ∈Σ(II)aστ ≥Xστ∈Σ(II)Xijaσˆ⟨i⟩τˆ⟨j⟩≥(1 −1yℓ) · aℓ+1 ≥(1 −1yℓ−1v ·1(1 −ǫ)2 −2 · ǫ) · aℓ.Putting everything together we get thatXστ niceaστ ≥(1 −1zm−1−1yℓ−1v ·1(1 −ǫ)2 −2 · ǫ) · aℓ≥(1 −1zℓ) · aℓ.This shows in particular that the pair ⟨⟩⟨⟩is nice.17

2.8 Claim. If στ is nice of length ℓ(0 ≤ℓ≤m −1) thenµ(p ∩pˆσˆτσ(ℓ−1), q ∩qˆσˆττ(ℓ−1)) < (14m−ℓ+ ζℓ) · µ(pˆσˆτσ(ℓ−1), qˆσˆττ(ℓ−1)),(8.1)where ζℓ:= 2 · (m −ℓ) · (ǫ + δ−2 · ǫ · m · zm−1 + δ) + Pm−2j=ℓyjzj+1 (in particular ζ0 = ζ).Proof.

We know that for arbitrary nice στ,δ−2 · aστ · ǫ · m · zm−1 ≥Xi∈Cστµ(qστi ) · µ(pˆσˆτσ(ℓ−1))(cf (6.4) and (6.5)), and, by symmetry (because our conditions are relatively quadratic),δ−2 · aστ · ǫ · m · zm−1 ≥Xi∈Cστµ(pστi ) · µ(qˆσˆττ(ℓ−1)).Also δ · Pi∈jστ µ(pστi ) > Pi∈Aστ µ(pστi∩p) and δ · Pi∈jστ µ(qστi ) > Pi∈Bστ µ(qστi∩q).So it suffices to calculate Pi∈Bστ,j∈Aστ µ(pστi∩p, qστj∩q).For this, we make again backwards induction on ℓ. Assume ℓ= m −1.

Then thedisjointness of Bστ and Aστ (see (6.1) — (6.3)) implies thatXi∈Bστ,j∈Aστµ(p ∩pστi , q ∩qστj ) < 14 · µ(pˆσˆτσ(m−2), qˆσˆττ(m−2)).Now it follows from the discussion in the preceding paragraph (and (5.2)) that formula(8.1) holds for ℓ= m −1.So assume the claim has been proved for ℓ+ 1 ≤m −1. Let στ be nice of length ℓ.Then (6.6)Xij,σˆ⟨i⟩τˆ⟨j⟩not niceaσˆ⟨i⟩τˆ⟨j⟩≤yℓzℓ+1· µ(pˆσˆτσ(ℓ−1), qˆσˆττ(ℓ−1)).

(8.2)And by induction (and the disjointness of Bστ and Aστ) we haveXi∈Bστ ,j∈Aστ,σˆ⟨i⟩τˆ⟨j⟩niceµ(p ∩pστi ,q ∩qστi ) < (14m−ℓ−1 + ζℓ+1) ·Xi∈Bστ ,j∈Aστµ(pστi , qστi )< 14 · (14m−ℓ−1 + ζℓ+1) · µ(pˆσˆτσ(ℓ−1), qˆσˆττ(ℓ−1)). (8.3)18

Putting everything ((5.2), the first paragraph of this proof, (8.2), (8.3)) together we getagain that formula (8.1) holds.The main claim 2.6 now follows from claims 2.7 and 2.8.2.9 Proof of Theorem 3 from the Main Claim 2.6. As remarked in 2.5 we let N :={x ∈2ω; ∃∞m (x↾nm ∈Zm)}.

Then Pm|Zm|2nm ≤Pm 2−m < ∞. Hence N is a null setcoded in V .

We claim that∥−B×B ˘T ∩N ̸= ∅.We first note that it suffices to prove∥−B×B∀ℓ∈ω ∀s ∈˘T (lh(s) = nℓ⇒∃∞m ≥ℓ∃t (lh(t) = nm ∧s ⊆t ∧t ∈˘T ∩Zm)).For if the latter holds then we can recursively construct an x ∈˘T[G] ∩N in the genericextension V [G].So assume that there is a (p, q) ∈B × B, an ℓ∈ω and an s of length nℓsuch that(p, q) ∥−s ∈˘T ∧∀m > ℓ∀t ((lh(t) = nm ∧s ⊆t) ⇒t ̸∈˘T ∩Zm);i.e. (p, q) ∥−s ∈˘T ∧∀m > ℓ(Zsm ∩˘T = ∅).Without loss (p, q) is quadratic.

Choose m ≥ℓso large that µ(p, q) ≥14m + ζm. Then(p, q) ∥−s ∈˘T ∧Zsm ∩˘T = ∅contradicts the main claim 2.6.19

§ 3 Final remarks3.1 We discuss the relationship between C ∗B, B ∗C ∼= B × C and B × B. Truss [T2] proved that C ∗B cannot be completely embedded in B ∗C by showing that the formeradds a new uncountable subset of ω1 containing no old countable subset whereas the latterdoes not. In fact, he proved [T 2, Theorem 3.1] that any uncountable subset of ω1 in V [r],where r is random over V , contains a countable subset in V ; the rest follows from the factthat C has a countable dense subset.

It is easy to see that Truss’ argument for B can begeneralized to B × B so that C ∗B cannot be completely embedded in B × B either.Another argument for showing that C ∗B cannot be completely embedded in B ∗Cis by remarking that the former produces two random reals the sum of which is Cohen(namely, let c be Cohen over V and r random over V [c]; then both r and c −r are randomover V ) whereas the latter does not (by [JS 2, 2.3], if we force with C over V [r], r randomover V , then no new real is random over V ; so the sum of two random reals must lie inV [r] and cannot be Cohen). B × B also produces two random reals the sum of which isCohen (by [Je 2, part I, 5.9], if r0, r1 are the random reals added by B × B, then r0 + r1is Cohen).

So B × B cannot be completely embedded in B ∗C.On the other hand, Pawlikowski (see the last paragraph of § 3 in [Pa]) proved thatB ∗C can be completely embedded into any algebra adding both Cohen and random reals;in particular B ∗C

Namely, suppose there aremodels of ZFC M ⊆N such that N contains both a dominating and a random real overM, but does not contain a perfect set of random reals over M. Without loss, N = M[r][d],where r is random over M, and d is dominating over M. By the ωω-bounding property ofrandom forcing [Je 2, part I, 3.3 (a)], d is dominating over M[r]. Let c be Cohen over N.A result of Truss [T 1, Lemma 6.1] says that d + c is Hechler over M[r].

(Recall that Hechler forcing D is defined as follows. D := {(n, f); n ∈ω ∧f ∈ωω},(n, g) ≤(m, f) iffn ≥m and ∀ℓ∈ω (g(ℓ) ≥f(ℓ)) and f↾m = g↾m.

D generically adds adominating real. )By [JS 2, 2.3] there is no new real random over M in N[c], in particular, there is no20

perfect set of random reals over M in N[c], thus showing that B ∗D ∼= B × D does not adda perfect set of random reals. Hence Question 2 is equivalent toQuestion 2’.

Does B × D add a perfect set of random reals?We will now see that for many forcing notions P adding a dominating real it is true thatB × P adds a perfect set of random reals.3.3 Proposition. Let M be Mathias forcing.

Then B×M adds a perfect set of randomreals.Remark. Mathias forcing is defined as follows.

M := {(s, S); s ∈ω<ω ∧S ∈[ω]ω ∧max s < min S}, (t, T) ≤(s, S) ifft ⊇s and T ⊆S and ∀n ∈dom(t) \ dom(s) (t(n) ∈S).Sketch of proof. In V [G], where G is B × M-generic over V , let r be the randomreal and d the Mathias real (which is dominating).

We claim that T := {f ∈2ω; ∀n ∈ω (f↾[d(n), d(n + 1)) = r↾[d(n), d(n + 1)) ∨f↾[d(n), d(n+ 1)) = (1 −r)↾[d(n), d(n + 1)))}is a perfect set of reals random over V in V [G].We show that given a null set N ∈V and a condition (B, (s, S)) ∈B × M, there is a(B′, (s, S′)) ≤(B, (s, S)) such that(B′, (s, S′)) ∥−˘T ∩N = ∅,where ˘T is a name for the perfect set defined above. First note that by [Ba] there arepartitions {Ii; i ∈ω} and {I′i; i ∈ω} of ω into finite intervals with max(Ii) < min(Ij),max(I′i) < min(I′j) for i < j, sequences ⟨Ji; i ∈ω⟩and ⟨J′i; i ∈ω⟩such that Ji ⊆2Ii,J′i ⊆2I′i, Pi∈ω|Ji|2|Ii| < ∞, Pi∈ω|J′i|2|I′i| < ∞and N ⊆{f ∈2ω; ∃∞n (f↾In ∈Jn)} ∪{f ∈2ω; ∃∞n (f↾I′n ∈J′n)}.Now find S′ ⊆S such that for all n ∈ω, |S′ ∩In| ≤1 and |S′ ∩I′n| ≤1.

Let in be theunique element of S′ ∩In, and i′n the unique element of S′ ∩I′n (if it exists — if not, letin ∈In and i′n ∈I′n be arbitrary). Set Kn := {s ∈2In; s ∈Jn ∨1 −s ∈Jn ∨(s↾in) ∪(1 −s↾[in, ∞)) ∈Jn ∨(1 −s↾in) ∪(s↾[in, ∞)) ∈Jn}; similarly we define K′n.

We chooseB′ ≤B such that B′ ∩({f ∈2ω; ∃∞n (f↾In ∈Kn)} ∪{f ∈2ω; ∃∞n (f↾I′n ∈K′n)}) = ∅.We leave it to the reader to verify that this works.We note that a similar argument works for Laver forcing, for Mathias forcing with aq-point ultrafilter etc. But it is unclear whether B ∗M adds a perfect set of random reals.21

References[Ba] T. Bartoszy´nski, On covering of real line by null sets, Pacific Journal of Math-ematics, vol. 131 (1988), pp.

1-12. [BJ 1] T. Bartoszy´nski and H. Judah, Jumping with random reals, Annals of Pureand Applied Logic, vol.

48 (1990), pp. 197-213.

[BJ 2] T. Bartoszy´nski and H. Judah, Measure and category: the asymmetry, forth-coming book. [Je 1] T. Jech, Set theory, Academic Press, San Diego, 1978.

[Je 2] T. Jech, Multiple forcing, Cambridge University Press, Cambridge, 1986. [JS 1] H. Judah and S. Shelah, The Kunen-Miller chart (Lebesgue measure, the Baireproperty, Laver reals and preservation theorems for forcing), Journal of SymbolicLogic, vol.

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129-138. [JS 3] H. Judah and S. Shelah, Adding dominating reals with random algebra, toappear.

[Ka] A. Kamburelis, Iterations of Boolean algebras with measure, Archive for Math-ematical Logic, vol. 29 (1989), pp.

21-28. [Mi] A. Miller, Additivity of measure implies dominating reals, Proceedings of theAmerican Mathematical Society, vol.

91 (1984), pp. 111-117.

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51 (1986), pp. 957-968.

[T 1] J. Truss, Sets having calibre ℵ1, Logic Colloquium 76, North-Holland, Amster-dam, 1977, pp. 595-612.

[T 2] J. Truss, The noncommutativity of random and generic extensions, Journal ofSymbolic Logic, vol. 48 (1983), pp.

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