ADDING MANY RANDOM REALS MAY ADD MANY COHEN REALS 7 MOHAMMADGOLSHANI 1 0 2 Abstract. Letκbeaninfinitecardinal. Then,forcingwithR(κ)×R(κ)addsageneric n filter for C(κ); whereR(κ)and C(κ) arethe forcingnotions for addingκ-manyrandom a J realsandaddingκ-manyCohenrealsrespectively. 6 1 ] O L 1. Introduction . h t For a cardinal κ>0 let R(κ) be the forcing notion for adding κ-many random reals and a m let C(κ) be the Cohen forcing for adding κ-many Cohen reals1. [ 1 Itisawell-knowfactthatforcingwithR(1)×R(1)addsaCohenreal;infact,ifr1,r2 are v the added random reals, then c =r +r is Cohen [1]. This in turn implies all reals c+a, 6 1 2 5 where a∈RV, are Cohen, and so, we have continuum many Cohen reals over V. However, 1 4 the sequence hc+a : a ∈ RVi fails to be C((2ℵ0)V)-generic over V. In fact, there is no 0 . 1 sequence hc :i<ω i∈V[r ,r ] of Cohen reals which is C(ω )-generic over V. i 1 1 2 1 0 7 In this paper, we extend the above mentioned result by showing that if we force with 1 v: R(κ)×R(κ), then in the resulting extension, we can find a sequence hci : i < κi of reals i X which is C(κ)-generic over the ground model: r a Theorem 1.1. Let κ be an infinite cardinal. Then, forcing with R(κ)×R(κ)adds a generic filter for C(κ). In Section 2, we briefly review the forcing notions C(κ) and R(κ). Then in Section 3, we state some results from analysis which are needed for the proofof abovetheorem and in Section 4, we give a proof of Theorem 1.1. The author’s research has been supported by a grant from IPM (No. 95030417). He also thanks Moti Gitikforhisusefulcomments andsuggestions. 1SeeSection2forthedefinitionoftheforcingnotions R(κ)andC(κ). 1 2 M.GOLSHANI 2. Cohen and Random forcings In this section we briefly review the forcing notions C(κ) and R(κ), and present some of their properties. 2.1. Cohen forcing. LetI beanon-emptyset. TheforcingnotionC(I),theCohenforcing for adding |I|-many Cohen reals is defined by C(I)={p:I ×ω →2:|p|<ℵ }, 0 which is ordered by reverse inclusion. Lemma 2.1. C(I) is c.c.c. Assume G is C(I)-generic over V, and set F = SG : I ×ω → 2. For each i ∈ I set c :ω →2 be defined by c (n)=F(i,n). Then: i i Lemma 2.2. For each i ∈ I,c ∈ 2ω is a new real and for i 6= j in I,c 6= c . Further, i i j V[G]=V[hc :i∈Ii]. i The reals c are called Cohen reals. By κ-Cohen reals over V, we mean a sequence i hc :i<κi which is C(κ)-generic over V. i 2.2. Random forcing. In this subsection we briefly review random forcing. Suppose I is a non-empty set and consider the product measure space 2I×ω with the standard product measure µ on it. Let B(I) denote the class of Borel subsets of 2I×ω. Recall that B(I) is I the σ-algebra generated by the basic open sets [s ]={x∈2I×ω :x⊇p}, p where p∈C(I). Also µ ([s ])=2−|p|. I p For Borel sets S,T ∈B(I) set S ∼T ⇐⇒ S△T is null, where S △T denotes the symmetric difference of S and T. The relation ∼ is easily seen to be an equivalence relation on B(I). Then R(I), the forcing for adding |I|-many random reals, is defined as ADDING MANY RANDOM REALS MAY ADD MANY COHEN REALS 3 R(I)=B(I)/∼. Thus elements of R(I) are equivalent classes [S] of Borel sets modulo null sets. The order relation is defined by [S]≤[T] ⇐⇒ µ(S\T)=0. The following fact is standard. Lemma 2.3. R(I) is c.c.c. Using the above lemma, we can easily show that R(I) is in fact a complete Boolean algebra. LetF beanR(I)-nameforafunctionfromI×ωto2suchthatforeachi∈I,n∈ω ∼ and k <2, kF∼(i,n)=kkR(I)=pik,n, where pi,n =[x∈2I×ω :x(i,n)=k]. k This defines R(I)-names r ∈2ω,i∈I, such that ∼i k∀n<ω, ∼ri(n)=F∼(i,n)kR(I)=1R(I) =[2I×ω]. Lemma 2.4. Assume G is R(I)-generic over V, and for each i ∈ I set r = r [G]. Then i ∼i each r ∈2ω is a new real and for i6=j in I,r 6=r . Further, V[G]=V[hr :i∈Ii]. i i j i The reals r are called random reals. By κ-random reals over V, we mean a sequence i hr :i<κi which is R(κ)-generic over V. i 3. Some results from analysis A famous theorem of Steinhaus [2] from 1920 asserts that if A,B ⊆ Rn are measurable sets with positive Lebesgue measure, then A+B has an interior point; see also [3]. Here, we need a version of Steinhaus theorem for the space 2κ×ω. For S,T ⊆ 2κ×ω, set S +T = {x+y : x ∈ S and y ∈ T}, where x+y : κ×ω → 2 is defined by (x+y)(α,n)=x(α,n)+y(α,n) (mod 2). Note that the above addition is continuous. 4 M.GOLSHANI Lemma 3.1. Suppose S ⊆ 2κ×ω is Borel and non-null. Then S−S contains an open set around the zero function 0. Proof. We follow [3]. Set µ = µ be the product measure on 2κ×ω. As S is Borel and κ non-null, there is a compactsubset of S of positive µ-measure,so may suppose that S itself is compact. Let U ⊇ S be an open set with µ(U)<2·µ(S). By continuity of addition, we can find an open set V containing the zero function 0 such that V +S ⊆U. We show that V ⊆ S−S. Thus suppose x ∈ V. Then (x+S)∩S 6= ∅, as otherwise we will have (x+S)∪S ⊆ U, and hence µ(U) ≥ 2·µ(S), which is in contradiction with our choice of U. Thus let y ,y ∈ S be such that x+y = y . Then x = y −y ∈ S −S as 1 2 1 2 2 1 required. (cid:3) Similarly, we have the following: Lemma 3.2. Suppose S,T ⊆ 2κ×ω are Borel and non-null. Then S+T contains an open set. Suppose S,T ⊆ 2κ×ω are Borel and non-null. It follows from Lemma 3.2 that for some p ∈ C(κ), [s ] ⊆ S +T. Thus, by continuity of the addition, we can find x ∈ S and y ∈ T p such that: • (x+y)↾dom(p)=p. • The sets S∩[s ] and T ∩[s ] are Borel and non-null. x↾dom(p) y↾dom(p) 4. Proof of Theorem 1.1 In this section, we complete the proof of Theorem 1.1. Thus force with R(κ)×R(κ) and let G×H be generic over V. Let hhr : α < κi,hs : α < κii be the sequence of random α α reals added by G×H. For α<κ set c =r +s . The following completes the proof: α α α Lemma 4.1. The sequence hc :α<κi is a sequence of κ-Cohen reals over V. α Proof. It suffices to prove the following: For every ([S],[T])∈R(κ)×R(κ), and every open dense subset D ∈V (∗) of C(κ), there is ([S¯],[T¯])≤([S],[T]) such that ([S¯],[T¯])k−“hc :α∈κi ∼α ADDING MANY RANDOM REALS MAY ADD MANY COHEN REALS 5 extends some element of D”. Thus fix ([S],[T])∈R(κ)×R(κ) and D ∈V as above, where S,T ⊆2κ×ω are Borel and non-null. By Lemma 3.2 and the remarks after it, we can find p∈C(κ) and (x,y)∈S×T such that: (1) [s ]⊆S+T. p (2) (x+y)↾dom(p)=p. (3) The sets S∩[s ] and T ∩[s ] are Borel and non-null. x↾dom(p) y↾dom(p) Now let q ∈D be such that ([S∩[sx↾dom(p)]],[T ∩[sy↾dom(p)]])(cid:13)“q≤C(κ) p”. UsingcontinuityoftheadditionandfurtherapplicationofLemma3.2andtheremarksafter it, we can find x′,y′ such that: (4) x′ ∈S∩[s ] and y′ ∈T ∩[s ]. x↾dom(p) y↾dom(p) (5) (x′+y′)↾dom(q)=q. (6) The sets S∩[sx′↾dom(q)] and T ∩[sy′↾dom(q)] are Borel and non-null. It is now clear that ([S∩[sx′↾dom(q)]],[T ∩[sy′↾dom(q)]])(cid:13)“h∼cα :α∈κi extends q”. The result follows. (cid:3) References [1] Bartoszynski, Tomek; Judah, Haim, Set theory. On the structure of the real line. A K Peters, Ltd., Wellesley,MA,1995.xii+546pp.ISBN:1-56881-044-X. [2] Steinhaus, Hugo, Sur les distances des points dans les ensembles de mesure positive, Fund. Math. 1 (1920), 93-104. [3] Stromberg,Karl,AnelementaryproofofSteinhaus’stheorem.Proc.Amer.Math.Soc.36(1972), 308. Mohammad Golshani, School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746,Tehran-Iran. E-mail address: [email protected] http://math.ipm.ac.ir/golshani/