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How Special are Cohen and Random Forcings

arXiv:math/9303208v1 [math.LO] 15 Mar 1993November 8, 2018How Special are Cohen and Random Forcingsi.e. Boolean Algebras of the family of subsetsof reals modulo meagre or nullSaharon ShelahInstitute of Mathematics, The Hebrew University, Jerusalem, IsraelDepartment of Mathematics, Rutgers University, New Brunswick, N.J., U.S.A.The feeling that those two forcing notions -Cohen and Random- (equivalently thecorresponding Boolean algebras P(R)/(meagre sets), P(R)/(null sets)) are special, wasprobably old and widespread.A reasonable interpretation is to show them unique, or“minimal” or at least characteristic in a family of “nice forcing” like Borel.We shallinterpret “nice” as Souslin as suggested by Judah Shelah [JdSh 292]; (discussed below).We divide the family of Souslin forcing to two, and expect that: among the first part, i.e.those adding some non-dominated real, Cohen is minimal (=is below every one), whileamong the rest random is quite characteristic even unique.

Concerning the second classwe have weak results, concerning the first class, our results look satisfactory.Related is von Neumann’s problem which in our language is:(∗) is there a ωω-bounding c.c.c. forcing notion adding reals which is not equivalent tothe measure algebra (i.e.

control measure problem)?dn 2/92, Publication 480, partially supported by the basic research fund, IsraeliAcademy, and partially sponsored by the Edmund Landau Center for researchin Mathematical Analysis, supported by the Minerva Foundation (Germany) .1

Velickovic (and , as I have lately learnt, also Fremlin) suggests another problem (itsays less on forcings which are ωω-bounding but it says also much on the others). (∗∗) is there a c.c.c.forcing notion P which adds new reals and such that for everyf ∈ωω ∩V P there is h ∈V such that (∀n)|h(n)| ≤2n, and f(n) ∈h(n) for all n.The version of it for Souslin forcing was our starting point.We have two main results: one (1.14) says that Cohen forcing is “minimal” in thefirst class, the other (1.10) says that all c.c.c.

Souslin forcing have a property shared byCohen forcing and Random real forcing (this is the answer to (∗∗) for Souslin forcing), soit gives a weak answer to the problem on how special is random forcing, but says much onall c.c.c. Souslin forcing.

Earlier by Gitik Shelah [Sh 412] , any σ-centered Souslin forcingnotion add a Cohen real. We thank Andrzej Roslanowski for proof reading the paper verycarefully correcting many and pointing out a flawed proof.§1 A Souslin forcing which adds an unbounded real add a Cohen real1.1 Notation: 0) ℓg(η) is the length of η.1) T denotes subtrees of ω>ω, i.e., T ⊆ω>ω is non empty, [ν ⊳η & η ∈T ⇒ν ∈T] and[ν ∈T ⇒(∃η ∈T)(ν ⊳η)].

For η ∈T let T [η] def= {ν ∈T : ν ⊴η or η ⊴ν} and letlim T = {η ∈ωω : Vn η↾n ∈T}2) sp(T) = {η ∈T : (∃≥2k)[ηb⟨k⟩∈T]},ℓsp(T) = {ℓg(η) : η ∈sp(T)}.3) [A]µ = {B ⊆A : |B| = µ}, [A]<µ = S0≤κ<µ[A]κ4) We say T is u-large if: u ∈[ω]ℵ0 and for some n∗< ω: if n∗< n < m < ω, n ∈u, m ∈uthen [n, m) ∩ℓsp(T) ̸= ∅.5) We say T is strongly u−large if: u ∈[ω]ℵ0, and for some n∗< ω, if n∗< n < m < ω,2

n ∈u, m ∈u then (∀η ∈T ∩n2)(∃ν)[η ⊴ν ∈spT & ℓgν < m].6) Ok is a sequence of length k of zeroes.7) (∀∞n) means: for every large enough n < ω. (∃∞n) means for infinitely many n < ω.8) We say T is (u, ¯h)-large if : u ∈[ω]ℵ0, hk : ω →ω \ {0, 1}, ¯h = ⟨hk : k < ω⟩and forevery k < ω, T is (u, hk)-large which means: for infinitely many n ∈u we have:n ≤m ∈u & |u ∩m \ n| < hk(n) ⇒Min (ℓsp(T) \ m) < Min (u \ (m + 1)).Note(∗) if hn = n for every n < ω this is equivalent to: for every k < ω, for someconsequtive members i0 < i1 < .

. .

< ik of u , for every ℓ< k we have [iℓ, iℓ+1) ∩ℓsp(T) isnot empty9) We say ⟨Tℓ: ℓ< n⟩is (u, ¯h)-large if: u ∈[ω]ℵ0, ¯h = ⟨hk : k < ω⟩, hk : ω →ω \ {0, 1} andfor every k < ω for infinitely many n ∈u we haven ≤m ∈u & |u ∩m \ n| < hk(n) ⇒VℓMin (ℓspTℓ\ m) < Min (u \ (n + 1)).10) If hk = h for k < ω we write h instead of ¯h.11) We use forcing notions with the convention that larger means with more information.12) In a partial order (=forcing notion), incompatible means have no common upper bound.1.2 Definition: A statement ϕ(x) on reals is absolute if for every model M extending Vwith the same ordinals (mainly M = V or a generic extension) and N, a model of ZFC−(which is a transitive set or class of M ) with ωM1⊆N and a ∈N, we have N |= ϕ[a] iffM |= ϕ[a].1.3 Definition: 1) P is a c.c.c. Souslin forcing notion if: P = (P, ≤) is such that:(a) there is a P11-definition ϕa of the set P (which is ⊆R)3

(b) there is a P11-definition ϕb of a partial order ≤on P(c) there is a P11-definition ϕc of the relation “p, q incompatible in P”, see (1.1(12))(hence it is ∆11, as by the above it is Σ11, now use Definition 1.3(1) (a)+(b), it impliesbeing compatible is P11 hence being incompatible is Q11). (d) (P, ≤) satisfies the c.c.c..2) Note: we do not distinguish strictly between P and the three P11-formulas ϕa, ϕb, ϕcrespectively appearing in the definition.3) P is a Souslin forcing notion if (a) + (b) holds.1.3A Remark: On (c.c.c.) Souslin forcing see Judah Shelah [JuSh 292] e.g.1.4 Claim: 1) “ϕa, ϕb, ϕc are P11-formulas as in 1.3(1)” is absolute.2) For P a c.c.c.

Souslin forcing notion, “{rn : n < ω} is a maximal antichain of P” is aconjunction of a P11 and a Q11 statements.3) Being a maximal antichain is absolute (even conjunction of Q11 and P11) hence so is“being a P-name of a member of ω2 (or ωω)”.4) If P is a c.c.c. Souslin forcing notion , p0 ∈P and P ∗=df {p ∈P : P |= ”p0 ≤p} (withthe inherited order) then P ∗is a c.c.c.

Souslin forcing notion too .Proof: E.g. (2) The P11 part is to say “rn ∈P”, so if “x ∈P” is also Q11 then this statement is Q11; theQ11 part is to say (∀x)hx /∈P ∨Wn<ω(x, rn compatible)i(by Definition 1.3(1)((a)+(c)));4

a third part is Vn

Souslin forcing,˜r a P-name of a new member of ω2.Then for some infinite u ⊆ω, for every p ∈P, the tree Tp[˜r] (see Definition 1.6 below) isu-large (see 1.1(5)).1.6 Definition: Tp[˜r] = {η ∈ω>2 : p /⊢P “η ̸=˜r↾ℓgη”} (clearly it is a tree).Before we turn to proving Lemma 1.5, we prove:1.7 Claim: 1) For a given c.c.c. Souslin forcing notion P (i.e.

as in Definition 1.3(1))and P-name˜r of a member of ω2, the conclusion of 1.5 is an absolute statement (actuallyP12).2) The statement on u, p (and also on˜r) that they is as required in 1.5, is a P11 statement.3) Also “˜r is a P-name of a new real” is absolute in fact a Q11-statement ..4) If P is c.c.c. Souslin forcing notion, above every p ∈P there are two incompatibleconditions then forcing with P add a new real.Proof: 1) Let˜r be represented byD⟨(pηi , tηi ) : i < ω⟩: η ∈ω>2Ewhere {pηi : i < ω} ⊆P isa maximal antichain of P, tηi a truth value and pηi ⊢P “η ⊳˜r ifftηi ”.

For 1.5, the failureof the statement can be expressed by:(∗) (∀u)(∃p)hu ⊆ω finite or p ∈P & (∃∞n ∈u)(∀η)[η ∈n2 & η ∈Tp(˜r) ⇒¬(∃ν)[η ⊴ν & ℓgν < Min (u \ (n + 1)) & νb⟨0⟩∈Tp(˜r) & νb⟨1⟩∈Tp(˜r)]i.5

Now the statement “ρ ∈Tp(˜r)” is equivalent to “p /⊢P [ρ /⊳˜r]” which is equivalent to(∗∗) Wi<ω(tρi = truth & p, pρi compatible. )It is enough to show that (∗) is a Q12-statement hence it is enough to show that insidethe large parenthesis there is a P11-statement.

In (∗) inside the large parenthesis, ignoringquantifications over ω, we note that “p ∈P” is P11, and then we have to consider (∗∗),on which it is enough to prove that it is a ∆11 statement [actually we have three instancesof it - all negatives]. By Definition 1.3(1)(c) it is Q11 and by Definition 1.3(1)(b) (and thecompatible meaning having a common upper bound) it is P11.2) The proof is included in the proofs of parts (1) and (3).3) Easy.hWhy?

the statement is (∀p)[p /∈P ∨Wη∈ω>2 [ηb⟨0⟩∈Tp(˜r) & ηb⟨1⟩∈Tp(˜r)]]. Now inside the parenthesis we have p /∈P which is Q11 and two instances of (∗∗)which, as shown above, is a Q11-statement.i4) Easy, e.g.

in V Levy(ℵ0,2ℵ0) we ask: is there p ∈P such that Gp =df {q : q ∈P V , q ≤p} is a directed subset of P V , generic over V , i.e. not disjoint to any maximalantichain of P V from V ?

By the assumption if such p exist , necessarily Gp /∈V ,and bythe homogeneity we can find Levy-names˜p,˜G˜p of such objects so in V Levy we can find aperfect set of such Gp’s , so the p’s form an antichain of size continuum but this is absolute. So there is no such p, letting {{pi,j : i < ω} : j < ω} list the maximal antichains of P Vfrom P V (the list in V Levy ), and we define a p-name˜η ∈ωω : (in V Levy):˜η(n) =theunique m such that pn,m ∈GP , the generic subset of P V Levy .

This is a P-name of a newreal (all in V Levy(ℵ0,2ℵ0) and by part (3) +1.4(2) its existence is absolute .□1.7Remark: The use of QD below can be replaced. QD is called Mathias forcing.

See on it[Sh-b].6

1.8 Proof of Lemma 1.5: Assume that the conclusion fails (for˜r, a P-name of a newmember of ω2, which will be fixed until the end of the proof of Lemma 1.5). For D a filteron ω (containing the co-bounded subsets of ω) let QD = {(w, A) : w ⊆ω finite, A ∈Dand max(w) < Min A (when w ̸= ∅)} (and if w ⊆ω is finite A ⊆ω we identify (w, A) with(w, A ∩(max w, ω))); the order is defined by (w1, A1) ≤(w2, A2) iffw1 ⊆w2 ⊆w1 ∪A1,A1 ⊇A2.

Let (w1, A1) ≤pr (ω2, A2) (pure extension) iffw1 = w2, A1 ⊇A2. Clearly QDis a partial order satisfying the c.c.c.

and {q : q0 ≤pr q} is directed for each q0 ∈QD. Let˜w = ∪{w : (w, A) ∈˜GQD}, clearly˜w is a QD-name and any G ⊆QD generic over V canbe reconstructed from˜w[G] : G = {(v, A) ∈QD : v ⊆˜w[G] ⊆v ∪A}.

Without loss ofgenerality CH holds (by claim 1.7(1), e.g. force with Levy(ℵ1, 2ℵ0)), hence we can chooseD as a Ramsey ultrafilter on ω.

So as is well known that:⊗1 if˜ℓ< 2 is a QD-name and q ∈Q then for some q′, q ≤pr q′ ∈QD, q′ forces a value to˜ℓ.So after forcing with QD, the conclusion of 1.5 still fails (by claim 1.7(1)). Hence forsome q∗∈QD and QD-names˜p,˜T , (remember that˜r remains a P-name ) we have q∗⊢QD “˜p ∈P,˜r of ω2,˜T = T˜p[˜r] is not˜w-large, such that: for arbitrarily large n ∈˜w,(˜w-the QD-name) the intervalhn, Min (˜w \ (n + 1))is disjoint to lsp(˜T) ” ; also we canassume that⊢QD “˜p ∈P ,˜r remains a P-name of a new member of ω2 and˜T = T˜p[˜r] ”.

For q ∈QD let S[q] =: {η ∈ω>2 : for some q′, q ≤pr q′ and q′ ⊢QD “η ∈˜T”}; note thatS[q] is also equal to {η ∈ω>2 : for no q′, q ≤pr q′ ∈QD, q′ ⊢QD “η /∈˜T”} (just apply ⊗1).Now note⊗2 S[q] is a subtree (of ω>2) and if q1 ≤pr q2 (in QD) then S[q1] ⊇S[q2], (in fact theyare equal).⊗3 if q∗≤q ∈QD then for some q1 ≥q and m we have: S[q1] has no splitting in any7

level ≥m.Why? let n = max wq; so q forces that: for some m, m ∈˜w, m ≥n, and Min [(ℓspT˜p[˜r] \m] ≥Min [˜w \ (m + 1)].

Before proving ⊗3, repeatedly using ⊗1 we can assume⊗4 if m ∈Aq, v ⊆m ∩Aq, η ∈m2 then the condition (wq ∪v ∪{m}, A \ (m + 1)) ∈QDforces ( ⊢QD) a truth values to the following:(α) η ∈T˜p(˜r)(β) (∃ν)hη ⊴ν ∈ω>2 & ℓgν < Min (˜w \ (m + 1)) & ν ∈sp(T˜p(˜r))i. (Recall the definition of a Ramsey ultrafilter by game.

)By the sentence before the last, for some m ∈Aq and v ⊆Aq ∩m for every η∗∈m2,if we get a positive answer for (α) then we get a negative answer for (β); let q′ = (wq ∪v ∪{m}, Aq \ (m + 1)); so q′ forces those two statements. Let for k ∈Aq \ (m + 1),qk = (wq ∪v ∪{m}, Aq \ k) (so q′ ≤pr qk) and it forces ( ⊢QD) “every η ∈m2 ∩T˜p[˜r] hasa unique extension in T˜p(˜r) ∩k2”, as required in ⊗3.The rest of the argument will be used again so just note that proving Claim 1.9 belowenough for finishing the proof of 1.5.1.9 Claim: Assume P is a c.c.c.

Souslin forcing,˜r a P-name of a new real, QD, S[q] (forq ∈QD) chosen as above. Then ⊗3 above is impossible.Proof: So assume ⊗3 holds and we shall get, eventually a contradiction.For this end we define a forcing notion Q∗= Q∗D, Q∗D = {( ¯m, ¯q): for some n =n(¯q) = n( ¯m, ¯q) we have ¯q = ⟨qη : η ∈n2⟩, q∗≤qη ∈QD, for each η ∈n2 the sequence⟨|S[qη] ∩k2| : k < ω⟩is bounded, ¯m = ⟨mν : ν ∈n>2⟩, mν < ω and if νb⟨ℓ⟩⊴ηℓ∈n2 forℓ= 0, 1 and k < ω then |S[qη0] ∩S[qη1] ∩k2| ≤mν}.8

The order is defined by ( ¯m1, ¯q1) ≤( ¯m2, ¯q2) iffn(¯q1) ≤n(¯q2), ¯m1 = ¯m2↾n(¯q1)>2 andfor η ∈n(¯q2)2 we have q1η↾n(¯q1) ≤q2η.Clearly Q∗D satisfies the c.c.c. as for any ( ¯m∗, ¯w∗) the set {( ¯m, ¯q) ∈Q∗D : ¯m = ¯m∗, wqη = w∗ηfor every η ∈n(¯q)2} is directed (in Q∗D).Also,⊗5 for every n, {( ¯m, ¯q) ∈Q∗D: n(¯q) ≥n} is a dense (and open) subset of Q∗D.hWhy?

it is enough to prove that for any given ( ¯m0, ¯q0) ∈Q∗D with n(0)def= n(¯q0),there is ( ¯m1, ¯q1) such that ( ¯m0, ¯q0) ≤( ¯m1, ¯q1) ∈Q∗D with n(¯q1) = n(0) + 1; letq1η = q0η↾n(0), and m1ν is: m0ν if ν ∈n(0)>2 , and max{|S[q0ν] ∩k2| : k < ω} if ν ∈n(0)2.Checki.For G∗⊆Q∗D generic over V , let for η ∈ω>2,˜wη[G∗] be S{wr : there is ( ¯m, ¯q) ∈G∗,r = qη↾n and n = n(¯q) ≤ℓg(η)} it is well defined).If η ∈(ω2)V [G∗] let˜wη[G∗] beSk<ω wη↾k[G∗].Also Q∗D adds a perfect set of generics for QD, moreover: in V [G∗] for every η ∈(ω2)V [G∗],˜wη[G∗] defined above is generic for QD over V , which means Gηdef= {(v, A) : v ⊆˜wη[G∗] ⊆v ∪A} is a generic subset of QD over V ; this holds by ⊗6 , ⊗7 below .⊗6 if ( ¯m, ¯q) ∈Q∗D and˜τ is a QD-name of an ordinal then we can find ¯q1 such that( ¯m, ¯q) ≤( ¯m, ¯q1), n(¯q1) = n(¯q) and for every η ∈n(¯q1)2, the condition q1η forces a valueto˜τ.hWhy? let ⟨ηk : k < 2n(¯q)⟩list n(¯q)2.

We now define by induction on k ≤2n(¯q), asequence ¯rk = ⟨rkη : η ∈n(¯q)2⟩such that:(a) ( ¯m, ¯rk) ∈Q∗D(b) ( ¯m, ¯rk) ≤( ¯m, ¯rk+1) (i.e. QD |= “rkη ≤rk+1η” for η ∈n(¯q)2)9

(c) ¯r0 = ¯q(d) rk+1ηkforces a value to˜τ (for the forcing notion QD).If we succeed then ¯q1 def= ¯r(2n(¯q)) is as required; as ¯r0 is as required the only problemis to find ¯rk+1 being given ¯rk. First we can find mk < ω, such that no S[rkη] (for η ∈n(¯q)2)has a splitting node of level ≥mk, and mk > n(¯q).

Second we find rk,∗ηk ∈QD such that:QD |=“rkηk ≤pr rk,∗ηk ” and rk,∗ηk forces a truth value to each statement of the form “ν ∈T˜p[˜r]”for ν ∈mk≥2. By the definition of S[rkηk] necessarily rk,∗ηk⊢QD“Tp[˜r] ∩mk≥2 ⊆S[rkηk]”.Thirdly choose rk+1ηk∈QD, such that QD |= “ rk,∗ηk ≤rk+1ηk ” and rk+1ηkforces a value to˜τ(possible by density) and S[rk+1ηk ] has no splitting above some level (use ⊗3).

Fourth, letrk+1η= rkη for η ∈n(¯q)2 \ {ηk}; we still have to check |S[rk+1ν0 ] ∩S[rk+1ν1 ] ∩m2| ≤mν whenνb⟨ℓ⟩⊴νℓ∈n(¯q)2; by the induction hypothesis without loss of generality ηk ∈{ν0, ν1}, solet ηk = νℓ(∗). If m ≤mk then S[rk+1νℓ(∗)] ∩m2 ⊆S[rkνℓ(∗)] and we are done by the inductionhypothesis.

If m > mk, by the choice of mk, S[rk+1ν1−ℓ(∗)] has no splitting nodes of level ≥mkhence |S[rk+1νℓ(∗)] ∩S[rk+1ν1−ℓ(∗)] ∩m2| ≤|S[rk+1νℓ(∗)] ∩S[rk+1ν1−ℓ(∗)] ∩(mk)2|, and use the previoussentence. So we can carry the induction and ¯r(2n(¯q)) is as required in ⊗6i.⊗7 if ( ¯m, ¯q) ∈Q∗D and k < ω then we can find ¯q1 such that ( ¯m, ¯q) ≤( ¯m, ¯q1), n(¯q1) = n(¯q)and for every η ∈n(¯q1)2, the condition q1η forces some mη to be in wη \ {k}[Why?

proof similar to that of ⊗6 . ]Now for every η ∈(ω2)V [G∗] we know that V [Gη] |= “˜p[Gη] ∈P,˜r is still a P-name ofa member of (2ω)V [Gη][˜GP ]and T˜p[Gη][˜r] is not˜w[Gη]-large”, by 1.7 this holds in V [G∗] too.A closer look shows that for η ̸= ν (from (ω2)V [G∗]) the tree T˜p[Gη][˜r]∩T˜p[Gν][˜r] has finitelymany splittings.

So the conditions˜p[Gη],˜p[Gν] are incompatible in P V [G∗]. Contradiction10

to “P is c.c.c. Souslin and this is absolute (1.4(1))”.□1.9□1.5Now we can answer Velickovic’s question for Souslin forcings.1.10 Conclusion: Let P be a c.c.c.

Souslin forcing, adding a new real. (1) The following is impossible: for every P-name of a new˜r ∈ωω for some tree T ⊆ω>ω,T ∈V and p ∈P we have p ⊢P “˜r ∈lim T”, and Vn<ω |T ∩nω| ≤2n, we can replace:“for every n” by “for infinitely many n”.

(2) The following is impossible: for some P-name of a new˜r ∈ωω for every strictlyincreasing {ni : i < ω} ⊆ω from V for some tree T ⊆ω>ω, T ∈V and p ∈P wehave p ⊢P “˜r ∈lim T” and Vi<ω |T ∩niω| ≤2i, we can replace: “for every i” by “forinifinitely many i”. (3) The following is impossible: for some P-name˜r of a new member of ω2 for everystrictly increasing {ni : i < ω} ⊆ω from V for some tree T ⊆ω>2 and q we haveq ∈P and q ⊢P “˜r ∈lim T” and Vi<ω |T ∩ni2| ≤2i.

(4) The following is impossible: for some r ∈(ω2)V P \ ω2 for every strictly increasingsequence ⟨ni : i < ω⟩∈V of natural numbers, for some tree T ⊆ω>2 from V we have:r ∈lim T, and (∃∞i)|T ∩ni2| ≤i2 or at least (∃∞i)|T ∩(ni+1)2| ≤2ni.Proof: (1), (2) follow by part (3) . Suppose that˜r is a P-name of a new member of ωω .

Let˜ηn be (the P-name ) 0˜r(n)+1b⟨1⟩and let˜r∗be the following P-name : the concatenationof˜η0,˜η1,˜η2, . .

. .

By part (3) there is a strictly increasing sequence ⟨ni : i < ω⟩of naturalnumbers such that for no q ∈P and T , does q ⊢“˜r ∈Lim(T) ” and for inifinitely many11

i < ω we have (2) and replacing˜r will give (1) too . (3) Follows from part (4).

(4) Let˜r be a P-name of a new real.By Lemma 1.5 for some infinite u ⊆ω we have(∗) for every p ∈P, Tp[˜r] is u-large (see 1.1(4)).We now choose by induction on i, ni < ω, such that ni > sup{nj : j < i} and |(ni, ni+1) ∩u| > 2ni + 2. If (4) fails for˜r we apply the statement to the sequence ⟨ni : i < ω⟩, so forsome p ∈P and subtree T of ω>2 from V , we have:(a) p ⊢P “˜r ∈lim T”(b) for infinitely many i < ω, we have |T ∩(ni+1) 2| ≤2ni.By the choice of u, for some j∗< ω, we know that Tp[˜r] has a splitting of level ∈[j0, j1)for each j0, j1 ∈u, j∗< j0 < j1.So if i > j∗, then |Tp[˜r] ∩(ni+1)2| is at least the number of levels < ni+1 of splittingnodes of Tp[˜r] which is ≥|(ni, ni+1) ∩u| which is > 2ni.

But p ⊢“˜r ∈lim T” impliesTp[˜r] ⊆T so |T ∩(ηi+1)2| is > 2ni, ( for every i < ω such that i > j∗), this contradicts thechoice of T hence we finish.□1.101.11 Remark: This means that any c.c.c. Souslin forcing which is ωω-bounding is quitesimilar to the Random real forcing in some sense.

More exactly every c.c.c. Souslin forcinghas a property shared by the Random real forcing and the Cohen forcing.1.12 Claim: 1) Assume(a) P is a forcing notion(b)˜r is a P-name of a member of ω2.12

(c) ¯h = ⟨hn : n < ω⟩, hn = n (i.e. hn(i) = n for every i < ω) and u ⊆ω is infinite(d) for every p ∈P, for some η ∈Tp[˜r] the set {k : ηbOk−ℓgηb⟨1⟩∈Tp[˜r]} is(u, ¯h)-large (see (*) of 1.1(8)).Then forcing with P add a Cohen real.2) We can weaken (d) to(d)−for every p ∈P for some n < ω, and η0, .

. ., ηn−1 ∈Tp[˜r] the set {k : for some ℓ< n,ηℓbOk−ℓgηℓb⟨1⟩∈Tp[˜r]} is (u, ¯h)-large.Proof: 1), 2) Let u \ {0} = {ni : 1 ≤i < ω}, n0def= 0 < n1 < n2 < .

. ., let ⟨k(i, ℓ) :ℓ< ω⟩be such that i = Pℓk(i, ℓ)2ℓ, k(i, ℓ) ∈{0, 1}, so k(i, ℓ) = 0 when 2ℓ> i. Letρ∗m = ⟨k(i, ℓ) : ℓ≤[log2(i + 1)]⟩where i = iu(m) is the unique i such that ni ≤m < ni+1.We define a P-name˜s (of a member of (ω2)V P ) : let {˜ki : i < ω} list in increasing order{k < ω :˜r(k) = 1} and˜s be ρ∗k0 bρ∗k1 bρ∗k2 b .

. ..Clearly by condition (d)−, for every p ∈P and n < ω we have p /⊢“˜r(k) = 0 for everyk ≥n”.

Hence⊢P “{k < ω :˜r(k) = 1} is infinite, hence⊢P “˜s ∈ω2”. It is enough toprove that⊢P “˜s is a Cohen real over V ”.

So let T ∈V be a given subtree of ω>2 which isnowhere dense, i.e. (∀η ∈T)(∃ν)[η ⊳ν ∈ω>2 \ T], and we should prove⊢P “˜s /∈lim T”.So assume p ∈P, p ⊢P “˜s ∈lim T” and we shall get a contradiction.

Having our p ∈P wecan apply (d)−(or (d) , which is stronger), so we can find n < ω and η0, . .

. , ηn−1 ∈Tp[˜r]as there such that A = {k < ω: for some ℓ< n, ηℓbOk−ℓgηℓb⟨1⟩∈Tp[˜r]} is (u, ¯h)-large.Let for each ℓ< n, {kℓj : j < jℓ} list in increasing order {k < ℓg(ηℓ) : ηℓ(k) = 1}and let ρℓ= ρ∗kℓ0 bρ∗kℓ1 b .

. .

ρ∗kℓjℓ−1. Now we can choose by induction on ℓ≤n, a sequenceνℓ∈ω>2 such that: ν0 = ⟨⟩, νℓ⊴νℓ+1 and ρℓbνℓ+1 /∈T (each time use “T is nowheredense”).13

Next we choose m(∗) ∈A such that νn ⊴ρ∗m(∗); possible as A is (n, ¯h)-large (checkDefinition 1.1(8): the set {iu(m) : m ∈A} contains an interval of length > 2ℓg(νn), so bythe definition of ρ∗m, some m(∗) in this interval is as required). Now we can find p1 ∈Psuch that p ≤p1 and p1 ⊢P “for some ℓ< n, ηℓbOm(∗)−ℓgηℓb⟨1⟩⊴˜r” hence p1 ⊢P“for some ℓ< n, ρℓbρ∗m(∗) ⊴˜s”, so by the choice of νℓ+1, and as νℓ+1 ⊴νn ⊴ρ∗m(∗) weget p1 ⊢P “˜s /∈lim T” hence we get contradiction to: p ⊢P “˜s ∈lim T”, hence we finishproving⊢P “˜s is a Cohen real over V .”□1.121.13 Claim: Let P be a c.c.c.

Souslin forcing1) “P add a Cohen real” is absolute (as well as “x is a P-name of a Cohen real”).2) “x is a P-name of a dominating real” is absolute.3) “P add a non dominated real” is absolute (as well as “x is a P-name of a non dominatedreal”).4) for a given ¯h, “ there is u ∈[ω]ℵ0 such that (d) of Claim 1.12(1) holds” is absolute;similarly (d)−of 1.12(2).5) “x is a P-name of a member of ωω , dominating η1 ∈ωω and not dominating η2 /∈ωω” is absolute (in fact , conjunction of Q11 and P11 statements )Proof: 1) Let ϕ(x) say:(a) x = ⟨⟨pηi , tηi : i < ω⟩: η ∈ω>ω⟩, pηi ∈P, tηi a truth value, ⟨pηi : i < ω⟩a maximalantichain, ( for each η ∈ω≥ω )(b) if η, ν ∈ω>ω and pηi , pνj are compatible then: [η ⊴ν ∧tνj truth ⇒tηi =truth] and[η, ν are ⊳-incomparable V &tηi =truth⇒tνj =false]. (c) for every p ∈P for some η ∈ω>ω for every ν, η ⊴ν ∈ω>ω, we have Wi<ω(p, pνicompatible ∧tνi =truth).14

Now by 1.4(1)+(2) part (a) is a conjunction of Q11 and P11 statements, part (b) is bothQ11 and P11 and part (c) is Q11 (we use: compatibility is both Q11 and P11 and (∀p ∈P) [. .

. ]means (∀p)[p /∈P ∨.

. .

]). So ϕ(x) is a conjunction of Q11 and P11 statements.

Now ϕ(x)says “x represents a P-name of a Cohen real” so (∃x)ϕ(x) which is a P12 statement, expressthe statement “forcing with P add a Cohen real.”2) We repeat the proof of part (1) but clause (c) is replaced by:(c)′ for every p ∈P and f ∈(ωω)V there are q ∈P and n∗such that p ≤q and:(c)q,f,n∗if q, pηi are compatible, tηi truth and n∗≤n < ℓg(η) then f(n) ≤η(n).Now (c)q,f,n∗is a Q11 and P11, so (c)′ has the form (∀p, f)[p /∈∨(∃q, n∗)[q ∈P&p ≤q&(c)q,f,n∗] which is Q12 hence “x is a P-name of a dominating real” is an absolute state-ment.3) Use the proof of Part (1) but clause (c) is replaced by:(c)′′ for p ∈P for infinitely many n the set {η(n) : η ∈ω>ω, ℓgη > n, i < ω, tni =truth, pηi , p compatible} is infinite.Now (c)′′ is Q11 and we can finish as there.4) The statement (d) and (d)−from 1.12 for given p,˜r, ¯h, u is a Q11 statement (as by theproof of 1.7(1) ν ∈Tp[˜r] is a Q11-statement and a P11-statement . )5) Easier than the proof of (3)□1.141.14 Conclusion: If P is a c.c.c.

Souslin forcing notion adding˜g ∈ωω not dominated byany f ∈(ωω)V then forcing with P add a Cohen real.Proof: Without loss of generality,˜g is strictly increasing. Let˜r = {˜g(i) : i < ω}, it is asubset of ω identified with its characteristic function.

We imitate the proof of Lemma 1.5(using here 1.13(3) instead of 1.7 there) so as there without loss of generality there is a15

Ramsey ultrafilter D on ω and let the forcing notion QD be as there. Let ¯h be as in Claimin 1.12 condition (c); we ask:⊗1 is there an infinite u ⊆ω such that condition (d)−of claim 1.12 holds?If yes we are done by claim 1.12 .

So from now on we asssume not.Let G ⊆QD be generic over V , condition (d)−fails also in V [G] (using absolutenesswhich holds by claim 1.13(4)), in particular for u =df˜w[G]. For p ∈P V [G] and η ∈ω>2we letC[η, p] =df {k : ηb0k−ℓg(η)b⟨1⟩∈Tp[˜r]}.Hence for some p∗∈P V [G], (remeber (*) of 1.1(8)) :(∗)1 V [G] |= “ for every j < ω and η∗ℓ∈Tp∗[˜r] for ℓ< j for some n∗letting C =dfS{C[η∗ℓ, p∗] : ℓ< j} , for no n∗+ 1 consequtive members i0 < i1 < .... < in∗of u do wehave : for every m < n∗the sets [im, im+1) , C are not disjoint ”.Let for n < ω , h0(n) be like n∗in (∗)1 for {η∗ℓ: ℓ< j} =df T˜p∗[˜r] ∩n≥2.

Let forn < ω , h(n) =df Pi≤n(h0(i) + 1). Note that we have: h is strictly increasing .So for some q∗∈G (which is a subset of QD ) , and QD-name˜p∗of a member of P ,we have: q∗forces that˜p∗,˜h,˜C[η,˜p∗] (for η ∈ω>2 , ) are as above.Wlog 0 ∈wq∗.Let˜w =˜w ∪{0} = {˜ni : i < ω} be strictly increasing , note˜w,˜ni are QD-names.Wlog⊗2 for every k ∈Aq∗and subset v of Aq∗∩k the condition (wq∗∪v∪{k}, Aq∗\(k+1))forces a value to the following:(A) T˜p∗[˜r] ∩k≥2, say t(B) truth value to [˜ni,˜ni+1) ∩C[η,˜p∗] = ∅for η ∈t and i ≤k + 1.So16

⊗3 Assume q1 ≥q∗, n < ω and q1 forces˜h(n) = n∗and T˜p∗[˜r] ∩n≥2 = t.Ifi0 < .... < in∗are the first n∗+ 1 members of Aq1, then for some m < n∗the conditionq2 =df (wq1 ∪{i0, ..., im−1}, Aq1 \ im) forces that:for no η ∈t and j do we have: im−1 ≤j < Min(˜w \ (im−1 + 1)) and j ∈C[η,˜p∗].Now we want to imitate the proof of 1.9 . We define a forcing notion Q∗∗= Q∗∗D asfollows :a member of Q∗∗D has the form ( ¯m, ¯q) such that:(1) ¯q = ⟨qν : ν ∈n2⟩, q∗≤qν ∈QD, n = n(¯q) ,(2) for each ν ∈n2 the number i[qν] =df Max(wqν) is well defined,(3) qν forces (in QD) a value tν to T˜p∗[˜r] ∩n≥2 and a value hν(j) to˜h(j) for j ≤nand a value cν(η,˜p∗) to˜C(η,˜p∗) ∩i[qν](4) qν ⊢QD “ if η ∈tν then for no k do we have: i[qν] ≤k < Min(˜w \ (i[qν] + 1)) andk ∈C[η,˜p∗] ”(5) ¯m = ⟨mi : i ≤n⟩, mi < ω and(6) if νb⟨j⟩⊴νj ∈n2 for j = 0, 1 and η ∈tν0 ∩tν1 then cν0[η, p∗] ∩cν1[η, p∗] isbounded by max{mℓg(η), mℓg(ν)}The order is defined by : ( ¯m1, ¯q1) ≤( ¯m2, ¯q2) iffn(¯q1) ≤n(¯q2), ¯m1 = ¯m2↾(n(¯q1) + 1)and for η ∈n(¯q2)2 we have q1η↾n(¯q1) ≤q2η.We now continue as in the proof of 1.9 .Why are the˜p∗[Gν] for ν ∈ω2 pairwise incompatible?

If ν1, ν2 are not equal , and˜p∗[Gν1],˜p∗[Gν2] are compatible in P let p be a common upper bound. We know that forsome η ∈Tp[˜r] we have C[η, p] is infinite as otherwise from Tp[˜r] we can define a functionf ∈ωω such that q ⊢P “˜g ≤f” contradicting the assumption .17

□1.14∗∗∗The question is whether a forcing adding half Cohen real (see below) adds a Cohenreal is due to Bartoszy´nski and Fremlin, appears in [B].1.15 Definition: If V ⊆V 1, r ∈(ωω)V 1 we say that r is a half Cohen real over V if forevery η ∈(ωω)V , for infinitely many n < ω, r(n) = η(n).1.16 Conclusion: If a ccc Souslin forcing add a half Cohen real then it adds a Cohenreal.Proof: If˜r is a P-name a member of ωω which is (forced to be) half Cohen over V , then⊢P “˜r ∈ωω is not dominated by any “old” h ∈ωω (i.e. h ∈(ωω)V ” trivially-you can usethe definition on h + 1 to get strict inequality.

Now use 1.14.1.17 Claim :Assume Q is a ccc Souslin forcing and˜r is a Q-name of a memberof ωω . Assume further that for some η ∈ωω , a Cohen real over V and G , a subsetof QV [η] generic over V [η] we have˜r[G] dominate η i.e.

for every n < ω large enoughη(n) ≤˜r[G](n) .Then ( in V )˜r is forced by some q ∈QV to be a dominating real i.e. for every G , asubset of QV generic over V to which p belongs and ρ ∈(ωω)V [G] , for every n < ω largeenough ρ(n) ≤˜r[G](n)Proof : Assume that not ; then some p ∈QV [η] force the negation , i.e˜r is as abovebut the conclusion fail .

By the homogeneity of Cohen forcing there is a Cohen name˜p ofsuch a condition .18

Also in V there is a maximal antichain J of Q and sequence ⟨ρq : q ∈J⟩such that foreach q ∈J either q forces that˜r does not dominate ρq and ρq ∈(ω(ω \ {0}) OR q forcesthat˜r dominate every ρ ∈(ωω)V and ρq(n) = 0 for every n < ω (all in V ) . Now we canfind ρ∗∈ωω dominating all ρq for q ∈J .Also wlog˜p is above some some q∗∈J so necessarily ρq∗(n) ̸= 0 .We define a forcing notion R as follows: members has the form (f, g) where for somen = n(f) < ω, f is a function from n>2 to ω and g is a function form ω to ω .

The order is:(f1, g2) ≤(f2, g2) iffn(f1) ≤n(f2) , f2 extend f1, for every n < ω we have g1(n) ≤g2(n)and for every k satisfying n(f1) ≤k < n(f2) , for all except at most one ν ∈k2 we haveg1(k) ≤f2(ν) .subclaim Let G be a subset of R generic over V . (1) In V [G] , for every ν ∈ω2 , ην =df f ◦ν = ⟨f(ν↾ℓ) : ℓ< ω⟩is a Cohen real over V , so(2) pην =df˜p[ην] is a member of QV [η] which by absoluteness is a subset of QV [G] .Now in V [ην] clearly pην forces that˜r dominate ην but not ρ∗.

By absoluteness this holdsalso in V [G] . Now if ν1, ν2 are distinct members of (ω2)V [G] and pην1 , pην2 are compatiblein QV [G] , let p∗∈QV [G] be a common upper bound then it forces that˜r dominates ρ∗,but it is also above a member q∗of J such that ρq∗is not a sequence of zeroes.By absoluteness we get a contradiction .□1.17References[JdSh 292] H. Judah and S. Shelah, Souslin-forcing, J. Symb.

Logic, Vol. 53 no.

4 (1988):1188-1207. [Sh b] Proper forcing, Springer Lecture Notes, 940 (1982) 496 + xxix.19

[GiSh 412] M. Gitik and S. Shelah, More on simple forcing notions, and forcings with idealsAnnals of Pure and Applied Logic. [B] T. Bartoszy´nski, Combinatorial aspects of measure and category, Fundamenta Math-ematicae, Vol.

127(1987):225–239.20

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