Examples

We are now ready to prove that some specific forcings are subcomplete. Since these forcings will be presented as sets of conditions rather than Boolean algebras, we set:

Definition LetPbe a set of conditions.

V^P=DfV^BA(P), δ(P) =Dfδ(BA(P))

where BA(P) is the canonical Boolean algebra overPas defined in Chapter 0.

We may refer to the elements ofV^P as ’P-names’. We note:

Fact 1 LetN =L^A_τ be a ZFC⁻ model. Let BA(P)∈Hθ⊂N. Letδ⊂C ≺N, where BA(P)∈Candδ=δ(P). Then for eachp∈Pthere isq∈C∩Ps.t. [q]⊂[p].

(Hence every set predense inC∩Pis predense inP.)

Proof. LetB= BA(P). By definition there aref,∆∈Hθ s.t. ∆ is dense inBand f :δ ↔∆. Hence there are suchf,∆∈C. But ∆⊂C, sinceδ⊂C. Let p∈P. There is b ∈ ∆ s.t. b ⊂[p]. Hence there is q∈ C∩P s.t. [q] ⊂b, since C ≺N.

Hence [q]⊂[p]. QED(Fact 1)

Our first example is Prikry forcing.

Lemma 6.1 Prikry forcing is subcomplete.

Proof. Let U be a normal ultrafilter on a measurable cardinal κ. We define the Prikry forcing determined byU to be the setP=P_U consisting of all pairshs, Xi

s.t. X∈U and s⊂κis finite. The extension relation≤P is defined by:

hs, Xi ≤ ht, Yi iff X ⊂Y, s⊃t, and t= lub(t)∩s.

P does not collapse cardinals or add new bounded subsets of κ IfG is P-generic, theP-sequence added byGis

S=SG=[ {s|W

Xhs, Xi ∈G}.

ThenSis unbounded inκand has order typeω. Gis, in turn, definable fromSby:

G=GS={hs, Xi ∈P|s=S∩lub(s)∧S\s⊂X}.

Definition We call S⊂κa P-sequence (or Prikry sequence) iffS=SG for some P-genericG.

The following characterization of Prikry sequences is well known:

Fact 2 S is a Prikry sequence iffS⊂κhas order typeω and is almost contained in everyX ∈U (i.e. W

ν < κ S\ν⊂X).

We now prove thatPis subcomplete. To this end we letθ >2^2κ and letN =L^A_τ be a ZFC⁻ model s.t. τ > θ andHθ⊂N. Furthermore we assume thatσ:N ≺N whereN is countable and full. We also suppose that

σ(θ, U ,P, s) =θ, U,P, s.

Hence σ(B) =B, where B= BA(P) andB= BA(P) inN. We must show:

Main Claim There is p ∈ P s.t. whenever G ∋ p is P-generic. Then there is σ^′ ∈V[G] s.t.

(a) σ^′:N≺N,

(b) σ^′(θ, U ,P, s) =θ, U,P, s,

(c) C_δ^N(rngσ^′) =C_δ^N(rngσ), whereδ=δ(P).

(d) σ^{′ ′′}G⊂G.

Note that if we set: S=SG andS=S_G inN[G], then (d) becomes equivalent to:

(d^′) σ^{′ ′′}S=S.

LetC=C_δ^N(rngσ). Using Fact 1 we get:

(1) Let X ∈U. Then there isY ∈C∩U s.t. Y ⊂X.

Proof. Suppose not. Then for eachν < κthe set ∆ν is dense inP∩C where

∆ν ={hs, Yi ∈P∩C|s\ν 6⊂X}.

Hence ∆ν is predense inP by Fact 1. Let Gbe P-generic. ThenG∩∆ν 6=∅ for ν < κ. HenceSGis not almost contained inX. Contradiction! by Fact 2. QED(1)

Hence:

(2) Sis aP-generic sequence iffShas order typeωand is almost contained in every X∈C∩U.

(3) δ≥κ,

since otherwiseC < κ andC∩U would have a minimal elementY =T

(C∩U).

Definition We define N0, k0, σ0 by: k0 : N0∼

↔C, whereN0 is transitiveσ0 = k⁻¹₀ ◦σ. We also set: Θ0,P₀, U0, s0=σ0(θ,P, U , s).

By Chapter 3.2, however, we have an alternative characterization:

(4) Let σ0(δ) =δ, ν =δ^+N, H =H_ν^N. Then hN0, σ0i is the liftup ofhN , σ↾ Hi.

Moreoverk0 is defined by the condition:

k0:N0≺N, k0σ0=σ, k0↾ν0= id, whereν0= supσ^′′ν.

Sinceν is regular inN, we conclude:

(5) σ0is aν-cofinal map andσ0(ν) =ν0.

Definition α0=δN0= the leastαs.t. Lα(N0) is admissible.

Our Main Claim will reduce to the assertion that a certain language L0 on Lα0(N0) is consistent. We define:

Definition L0 is the language onLα0(N0) with:

Predicate: ∈ Constants:

◦

S, σ,^◦ x (x∈Lα0(N0)) Axioms: •Basic axioms and ZFC⁻

•

S◦isP

0-sequence overN0

• ◦

σ:N ≺N₀ κ-cofinally, where σ(κ) =κ

• ◦

σ(θ,P, U , s) =θ₀,P₀, U₀, s₀

• ◦

σ^′′S=S^◦.

We first show thatL0 is consistent. To this end we define:

Definition hN1, σ1i= the liftup ofhN , σ↾H_κ^Ni.

k1= the uniquek:N1≺N0 s.t. kσ1=σ0 andk↾κ1= id, whereκ1= supσ^′′κ.

θ1,P₁, U1, s1=σ1(θ,P, U , s),S1=σ1′′S.

Note thatκ1=σ1(κ), sinceσ1 isκ-cofinal intoN1andκis regular inN. Then:

(6) (a) S1is aP₁-sequence overN1, (b) σ1:N ≺N1 κ-cofinally, (c) σ1(θ,P, U , s) =θ1,P₁, U1, s1, (d) σ1′′S =S1.

Proof. (b)–(d) are immediate. (a) follows by:

Claim LetX ∈U1. ThenS1 is almost contained inX. Proof. LetX ∈σ1(w), wherew < κinN. ThenY =T

(U∩w) is almost contained in every z ∈ U ∩w and Y ∈ U. Hence Y = σ1(Y) is almost contained in every Y ∈ U1∩σ1(w). In particular, Y is almost contained in X. But S1 is almost

contained inY. QED(6)

Now let:

α1=δN1 = the leastαs.t. Lα(N1) is admissible.

Let L1 be the language Lα1(N1) which is defined as L0 was defined on Lα0(N0), substitutingθ1,P₁,U1,s1,κ1forθ0,P₀,U0,s0,κ0. Then

(7) L1 is consistent.

Proof. hHκ, S1, σ1imodelsL1by (6). QED(7)

Note, however, that:

N1

k1

//N0

N σ1

OO

σ0

>>

||

||

||

||

where all maps are cofinal and all models are almost full.

ThenL0 is Σ1(Lα0[N0]) in N0and the parameters:

κ, P₀, κ, N , θ, P, U , s, θ0, P₀, U0, s0.

ButL1 is Σ1(Lα1[N1]) inN1and thek1-preimages of these parameters by the same Σ1-formula. Since N1 is almost full and k1 : N1 ≺ N0 cofinally, we conclude by Lemma 4.5:

(8) L0 is consistent.

From this we now derive the Main Claim: Work in a generic extensionV[F] ofV in whichLα0[N] is countable. ThenL0has a solid model

A=h|A|,S^◦A,σ^◦Ai.

SetS=S^◦A,σ^′ =k0◦σ^◦A. Then (8) (a) σ^′ :N ≺N,

(b) S isP-generic overV, (c) σ^′(θ,P, U , s) =θ,P, U, s, (d) C_δ^N(rngσ^′) =C, (e) S =σ^{′ ′′}S.

Proof. (a), (c) are immediate. To see (d) note thatN0=C_δ^N0(rngσ^◦A), since σ^◦Ais κ-cofinal andδ≥κ=σ^◦A(κ). Hence:

C=k0′′N0=C_δ^N(rngk0◦σ^◦A).

Since k0 ↾ (κ+ 1) = id we haveU ∩N0 =U ∩C. Hence (b) follows by (2). (e)

follows byσ^′↾κ=σ^◦A↾κ. QED(9)

We have almost proven the Main Claim, the only problem being thatσ^′ is not necessarily an element ofV[S]. We now show:

(10) There isσ^′∈V[S] satisfying (9).

Proof. Work inV[S]. Letµbe regular inV[S] s.t. N ∈Hµ. Set:

M =hHµ, N, S, θ,P, U, s, σi. We define a languageL2 on the admissible structure M as follows:

Definition L2 is the language onM with Predicate: ∈

Constants: σ,^◦ x, (x∈M)

Axioms: • ZFC⁻ and basic axioms

• ◦

σ:N ≺N

• ◦

σ(θ,P, U , s) =θ,P, U , s

• C_δ^N(rngσ) =^◦ C

• S=σ^◦′′S

L2 is clearly consistent, sincehM, σ^′iis a model ofL2 inV[F], whereσ^′ is defined as above.

Now letπ: ˜M ≺M, where ˜M is countable and transitive. Let ˜L2 be the language on ˜M with the same Σ1 definition, replacing all parameters by their preimages under π. Then ˜L2 is consistent. Since ˜M ∈Hω^V1^[S]=H_ω^V₁, it follows that ˜L2 has a solid model ˜AinV. Let ˜σ=σ^◦A˜ and set: σ^′=π◦σ. The verification of (9) is then˜

straightforward. QED(10)

But, sinceS is a Prikry sequence, there must bep∈GS which forces the existence of such aσ^′. This proves the Main Claim. QED(Lemma 6.1) Lemma 6.2 AssumeCH. Then Namba forcing is subcomplete.

Proof. We first define Namba forcing. The setω₂^<ωof monotone finite sequences in ω2 is a tree ordered by inclusion. The setNof Namba conditions is the collection of all subtreesT 6=∅of ω^<ω₂ s.t. T is downward closed in ω₂^<ω and for eachs∈T the set{t|s≤T t} has cardinalityω2. The extension relation≤Nis defined by:

T ≤T^′←→DfT ⊂T^′. IfGis N-generic, thenS =S T

Gis a cofinal map of ω into ω^V₂. We rite S =SG

and call any such S aNamba sequence. Gis then recoverable fromS by:

G=GS ={T ∈N|V

n < ω S ↾n∈T}.

It is known that, if CH holds, then Namba forcing adds no reals.

We shall also make use of the following fact, which is proven in the Appendix to [DSF]:

Fact Let S be a Namba sequence. Let S^′ ∈V[S] be a cofinal ω-sequence in ω^V₂. ThenS^′ is a Namba sequence andV[S^′] =V[S].

Note thatδ(N)≥ω2, since otherwise ω2would not be collapsed.

We now turn to the proof. Letθ >2^2ω2. LetN =L^A_τ be a ZFC⁻ model s.t. τ > θ andHθ⊂N. Letσ:N ≺N whereN is countable and full. Letσ(θ,N, s) =θ,N, s.

LetGbeN-generic overN. It suffices to show:

Main Claim There is p ∈ N s.t. whenever G ∋ p is N-generic, then there is σ^′ ∈V[G] with:

(a) σ^′:N≺N, (b) σ^′(θ,N, s) =θ,N, s,

(c) C_δ^N(rngσ^′) =C_δ^N(rngσ) whereδ=δ(N), (d) σ^{′ ′′}G⊂G.

Note We shall actually prove a stronger form of (c):

C_ω^N₂(rngσ^′) =C_ω^N₂(rngσ).

Note (d) can equivalently be replaced by:

σ^{′ ′′}S=S, where S=S_G, S=SG. Definition SetC=C_ω^N₂(rngσ). Definek0 by:

k0:N0∼

↔C, where N0is transitive, σ0=k₀⁻¹◦σ, θ0,N₀, s0=σ0(θN, s).

Just as before we get:

(1) hN0, σ0iis the liftup ofhN , σ↾H_ω^N₃i,

k0 is the unique k:N0≺N s.t. k0σ0=σandk0↾ω₃^N0 = id, (whereω₃^N0 = supσ0′′ω^N₃ ).

Now let α0 be the least αs.t. Lα(N0) is admissible. We define a languageL0 on Lα0(N0) as follows:

Definition L0 is the language onLα0(N0) with:

Predicate: ∈

Constants: σ,^◦ x (x∈Lα0(N0)) Axioms: • Basic axioms and ZFC⁻

• ◦

σ:N ≺N₀ ω^N₂-cofinally

• ◦

σ(θ, N , s) =θ0,N₀, s0. (2) L0 is consistent.

Proof. LethN1, σ1ibe the liftup ofhN , σ↾H_ω^N₂i. Definek1:N1≺N0 by:

k1σ1=σ0, k1↾γ1= id, where γ1= supσ^′ω₂^N =σ1(ω₂^N).

LetL1 be the corresponding language onLα1(N1), whereα1=δN1. Just as before it suffices to show that L1 is consistent. This clear, however, since hHω2, σ1i is a

model. QED(2)

Now let S^′ be a Namba sequence. Work in V[S^′]. Letµ be a regular cardinal in V[S^′] withN ∈Hµ. Set:

M =hHµ, N, σ,N, si.

Let π : ˜M ≺ M, where ˜M is transitive and countable. Then ˜M ∈ Hω1 ⊂V in V[S^′]. Let

π( ˜N,σ,˜ N ,˜ L,˜ ˜k,C) =˜ N, σ, N,L0, k0, C0.

LetA∈Vbe a solid model of ˜L. Set ˜σ= ˜k◦σ^◦A; σ^′=π◦˜σ. It follows easily that:

(3) (a) σ^′ :N ≺N

(b) σ^′(θ,N, s) =θ,N, s (c) C_ω^N₁(rngσ^′) =C

Now letS=S_G and set: S=σ^{′ ′′}S. ThenS∈V[S^′] is a cofinalω-sequence inω^V₂; hence:

(4) S is a Namba sequence andV[S] =V[S^′]. (Henceσ^′∈V[S].) But we know:

(5) S=σ^{′ ′′}S.

Let G = GS. There is then a p ∈ G which forces the existence of a σ^′ ∈ V[S]

satisfying (3), (5). This proves the Main Claim. QED(Lemma 6.2) Now letκ > ω1be a regular cardinal. LetA⊂κbe a stationary set ofω-cofinal ordinals. Our final example is the forcingP_A which is designed to shoot a cofinal normal sequence of order type ω1 throughA:

Definition P_A is the set of normal functions p:ν+ 1→A, where ν < ω1. The extension relation is defined by:

p≤q in P_A←→Df q⊂p.

Clearly, ifG isP_A-generic, then S

G:ω1 →A is normal and cofinal inκ. P_A adds no new countable subsets of the ground model. If, however,{λ < κ|cf(λ) = ω∧λ /∈A}is stationary, thenP_Awill not be a complete forcing.

Lemma 6.3 P_A is subcomplete.

Proof. Clearlyδ(P_A)≥κ, since otherwiseκwould remain regular. Now letθ >2^2κ. LetN =L^B_τ be a ZFC⁻ model s.t. τ > θ andHθ⊂N. Letσ:N ≺N where N is countable and full. Letσ(θ,P, A, κ, s) =θ,P_A, A, κ, s. LetGbe P-generic overN. It suffices to show:

Main Claim There isp∈Ps.t. wheneverG∋pisP_A-generic, there isσ^′ ∈V[G]

s.t.

(a) σ^′:N≺N,

(b) σ^′(θ,P, κ, A, s) =θ,P, κ, A, s,

where all maps are cofinal and all structures are almost full. L1 is trivially consis-tent, however, since hHκ, σ1imodelsL1. QED(2) Now letM =hHκ, N0, κ0, A0, σ0i. Let π: ˜M ≺M s.t. ˜M is countable and transi-tive. Let π( ˜L) =L0. Then ˜Lis a consistent language on Lα˜( ˜N) =π⁻¹(Lα0(N0)).

Hence ˜Lhas a solid modelA. Set:

σ^′ =k0◦π◦σ^◦A.

Thenσ^′ satisfies (a), (b), (c), (e) of (1). QED(Lemma 6.3)

Chapter 4

Iterating subcomplete forcing

Thetwo step iteration theoremfor subcomplete forcing says that ifAis subcomplete and

A

◦

B is subcomplete, thenA∗B^◦is subcomplete. Equivalently:

Theorem 1 Let A⊆BwhereAis subcomplete and ABˇ/G^◦ is subcomplete.

Then Bis subcomplete.

(Note The definitions ofA∗B^◦, ˇB/G^◦and other Boolean conventions employed here can be found in Chapter 0. G^◦is the canonical generic name – i.e. A

◦

Gis ˇA-generic over ˇV, and [[ˇa∈G]] =^◦ afora∈A.)

Proof of Theorem 1. Letθ be big enough thatθ verifies the subcompleteness ofA and:

Aθˇ verifies the subcompleteness of ˇB/G.^◦

LetN =L^A_τ be a ZFC⁻ model s.t. Hθ⊂N andτ > θ. Letσ:N ≺N where N is countable and sound. Let:

σ(θ,A,B, s) =θ,A,B, s

where s∈N. Let Gbe B-generic overN. We must findb∈B\ {0}s.t. whenever G ∋b is B-generic, there is σ^′ ∈ V[G] satisfying (a)–(d) in the definition of sub-completeness.

LetG0=G∩A. ThenG0isA-generic overN. Sinceθverifies the subcompleteness of A, there exist a ∈ A\ {0}, σ^◦0 ∈ V^A s.t. whenever G0 ∋ a is A-generic and σ0=σ^◦G₀⁰, then (a)–(d) hold withA,G0,A,G0, σ0 in place ofB,G,B,G,σ^′. Let B^∗ =B/G0. LetG0 ∋a beA-generic. Set: B^∗=B/G0. Clearly,σ0 extends toσ^∗₀ s.t

σ^∗₀ :N[G0]≺N[G0] and σ₀^∗(G0) =G0 51

In other words,σ₀^∗:N^∗≺N^∗where: N =L^A_τ,N^∗=L^A,G_τ ⁰,N =L^A_τ,N^∗=L^A,G_τ ⁰. Note thatH_θ^V[G0]=H_θ^V[G0]⊂N^∗. MoreoverG^∗ isB^∗-generic overN^∗ where:

G^∗=G/G0={b/G0|b∈G}.

Clearly

σ₀^∗(θ,A,B,B^∗, s) =θ,A,B,B^∗, s.

Sinceθverifies the subcompleteness ofB^∗inV[G0], we conclude that there isb^∗∈B^∗ s.t. whenever G^∗ ∋ b^∗ is B^∗-generic over V[G0], then there is σ^∗ ∈ V[G0][G^∗] with:

(a^∗) σ^∗:N^∗≺N^∗,

(b^∗) σ^∗(θ,A,B,B^∗, s) =θ,A,B,B^∗, s

(c^∗) C_δ^N∗^∗(rng(σ^∗)) =C_δ^N∗^∗(rng(σ₀^∗)), whereδ^∗=δ(B^∗).

(d^∗) σ^{∗ ′′}G^∗⊂G^∗.

Note thatG=G0∗G^∗=Df{b∈B|b/G0∈G^∗}. SetG=G0∗G^∗. Setσ^′=σ^∗↾N. Thenσ^′ ∈V[G] =V[G0][G^∗]. We show:

Claim σ^′ satisfies:

(a) σ^′:N ≺N

(b) σ^′(θ,A,B, s) =θ,A,B, s

(c) C_δ^N(rng(σ^′)) =C_δ^N(rng(σ)), whereδ=δ(B).

(d) σ^{′ ′′}G⊂G.

We note first that the claim proves the theorem, since G is B-generic and there must, therefore, be ab∈Gwhich forces the existence of such aσ^′. We now prove the claim. (a), (b), (d) are immediate. We prove (c). Note thatδ≥δ^∗, we have:

C_δ^N∗(rng(σ^∗)) =C_δ^N∗(rng(σ₀^∗)).

SinceC_δ^N(rng(σ0)) =C_δ^N(rng(σ)), it suffices to show:

(1) C_δ^N(rng(σ^′)) =N∩C_δ^N∗(rng(σ^∗)) (2) C_δ^N(rng(σ0)) =N∩C_δ^N∗(rng(σ^∗₀)).

We proof (1), the proof of (2) being virtually identical. (⊂) is trivial. We prove (⊃).

Letx∈N∩C_δ^N∗(rng(σ^∗)). ThenxisN[G0]-definable inξ,σ^∗(z),G0, whereξ < δ, z∈N. But, lettingt∈N^As.t. hz, G0i=t^G0, we have:

hσ^∗(z), G0i=σ^∗(hz, G0i) =σ^∗(t^G0) =σ^′(t)^G0. Hence:

x= that x s.t. N[G0]ϕ[x, ξ, σ^′(t)^G0].

But since σ^′(B) = B, we have: σ^′(δ) = δ, where δ = δ(B). Since δ ≥ δ(A), there is f ∈ N mapping δ onto a dense subset of A. Hence σ^′(f) maps δ onto

a dense subset of A. Hence there is ν < δ s.t. σ^′(f)(ν) ∈ G0 and σ^′(f)(ν) forces ϕ(ˇx,ξ, σˇ ^′(t)). Thus: x = that x s.t. σ^′(f)(ν) ^NA ϕ(ˇx,ξ, σˇ ^′(t)). Hence

x∈C_δ^N(rng(σ^′)). QED(Theorem 1)

The proof of Theorem 1 shows more than we have stated. We can omit the assumption that Ais subcomplete and omit the map σ, assuming, however, that ABˇ/G^◦0 is subcomplete, where G^◦0 is the canonicalA-generic name. Letθ be big enough that

Aθˇ verifies the subcompleteness of ˇB/G^◦0.

LetN be as before and letN be countable and full. Suppose thata∈A\ {0}and σ◦∈V^A are given s.t. wheneverG0∋aisA-generic andσ0=σ^◦G₀⁰, then

• σ0:N ≺N

• σ0(θ,A,B, s) =θ,A,B, s

• σ0′′G0⊂G0.

Our proof then yields a b^∗ ∈B^∗ = (B/G0)\ {0} s.t. if G ⊃G0 is B-generic and b^∗∈G^∗ =G/G0={c/G0|c∈G}, then there is σ^′∈V[G] s.t. (a), (b), (d) and:

(c^′) C_δ^N(rng(σ^′)) =C_δ^N(rng(σ0))

hold, where δ=δ(B). We can improve on this still further. Suppose thatt ∈V^A s.t. and at∈Nˇ. This means thatt^G0 ∈N wheneverG0 ∋ais A-generic. We can then select ourb^∗so as to force:

(e*) σ^∗(t^G0) =σ0(t^G0).

in addition to (a*)–(d*). It then follows that:

(e^′) σ^′(t^G0) =σ0(t^G0).

Since, wheneverG0∋aisA-generic, we can find ab^∗∈B/G0 forcing (a), (b), (d), (c^′), (e^′), we conclude that there is b⁰ ∈ V^A s.t. aforces b^∗ = b^◦G0 to have these properties. We may assume w.l.o.g. that A

◦

b ∈ Bˇ/G^◦0 and [[b^◦ 6= 0]]A =a. By Chapter 0, Fact 4 there is then a uniqueb∈Bs.t. Aˇb/G^◦0= b^◦. Lettingh=hA,B

be defined as in Chapter 0 by h(c) =T

{a∈A|c ⊂a} fora∈B, we conclude by Chapter 0, Fact 3 that:

h(b) = [[ˇb/G^◦06= 0]] = [[b^◦6= 0]] =a.

Clearly, if G∋b is B-generic, thenG0 ∋ais A-generic, whereG0 =G∩A. Thus b/G0= b^◦G0 =b^∗ has the above properties and (a), (b), (d), (c^′),(e^′) hold.

Putting all of this together, we get a very useful technical lemma:

Lemma 1.1 Let A⊆Band let: ABˇ/G^◦is subscomplete. Letθbe big enough that B ∈Hθ and: Aθˇverifies the subcompleteness of Bˇ/G. Let^◦ N =L^A_τ be a ZFC⁻

model s.t. Hθ⊂N andθ < τ. LetN be countable and full. LetA⊆BinN, where Gis B-generic over N. Set: G0=G∩A. Suppose that a∈A\ {0}, σ^◦0∈V^A s.t.

wheneverG0∋aisA-generic and σ0=σ^◦G₀⁰, then:

(i) σ0:N≺N

(ii) σ0(θ,A,B, s) =θ,A,B, s (iii) σ0′′G0⊂G0

(iv) t^G0∈N.

Leth=hA,B. Then there areb∈B\ {0}, σ^◦∈V^B s.t. a=h(b)and wheneverG∋b isB-generic,σ=σ^◦G, andG0=G∩A, then

(a) σ:N ≺N

(b) σ(θ,A,B, s) =θ,A,B, s

(c) C_δ^N(rng(σ)) =C_δ^N(rng(σ0)) (δ=δ(B)) (d) σ^′′G⊂G

(e) σ(t^G0) =σ0(t^G0).

⋆ ⋆ ⋆ ⋆ ⋆

We now prove a theorem about iterations of lengthω.

Theorem 2 Let hB_i | i < ωi be s.t. B₀ = 2, B_i ⊆ B_i+1 and B_i (ˇB_i+1/G^◦ is subcomplete)for i < ω. LetB_ω be the inverse limit ofhB_i|i < ωi. Then B_ω is subcomplete.

Proof. Letθ be big enough thatB_i θˇverifies the subcompleteness of ˇB_i+1/G^◦for i < ω. Let N =L^A_τ s.t. Hθ⊂N, θ < τ, andN is a ZFC⁻ model. Letσ:N ≺N s.t. σ(θ,hB_i | i ≤ ωi, s) = θ,hB_i | i ≤ ωi, s, and N is countable and full. Let G =Gω be B_ω-generic over N and set Gi = G∩B_i. Then Gi is B_i-generic over N. We claim that there isb∈B_ω\ {0} s.t. wheneverG∋b isB_ω-generic, there is σ^′ ∈V[G] s.t.

(a) σ^′:N≺N

(b) σ^′(θ,hB_i|i < ωi, s) =θ,hB_i|i < ωi, s

(c) C_δ^N(rng(σ^′)) =C_δ^N(rng(σ)), whereδ=δ(B_ω).

(d) σ^{′ ′′}G⊂G.

We first construct a sequence bi, σ^◦i (i < ω) s.t. bi ∈ B_i, hi(bi+1) = bi (where hi=hB_i,B_i+1 and wheneverGi∋bi isB_i-generic, then, lettingσi=σ_i^Gi, we have:

(a^′) σi :N ≺N

(b^′) σi(θ,hB_i|i≤ωi, s) =θ,hB_i|i≤ωi, s (c^′) C_δ^N(rng(σi)) =C_δ^N(rng(σ))

(d^′) σ_i^′′G⊂Gi.

Now let hxi|i < ωienumerateN. Set: ui= theN-leastus.t. σ(xi)∈σi(u) and u≤δ=Df δ(B) inN. (This exists, since rng(σ) ⊂C_δ^N(rng(σi)) =S

{σi(u)| u≤ δin N} by Chapter 3, Lemma 5.5.) σi will satisfy the additional requirements:

(e^′) σ0=σ

(f^′) σi(xh) =σh(xh) forh < i, whereσh=Df

σ◦h(Gi∩B_h).

(Note Then σh = σ^◦G_hⁱ, since we assume: V^Bh ⊆ V^Bi (i.e. the identity is the natural injection ofV^Bh into V^Bi). Thus t^Gi∩Bh=t^Gi fort∈V^Bh,h < i.)

(g^′) σi(uh) =σh(uh) forh < i.

Note that ui = u^◦G_iⁱ for a u^◦i ∈ V^Bi. We set: b0 = 1, σ^◦0 = ˇσ. Given bi, σ^◦i, Lemma 1.1 then gives us bi+1, σ^◦i+1. (Take σi+1(t^Gi) = σi(t^Gi) where B_i t = hˇx0, . . . ,ˇxi,u^◦0, . . . ,u^◦ii.) Since hi(bi+1) = bi, the sequence~b = hbi | i < ωi is a thread inhB_i|i < ωi. Henceb=T

i

bi6= 0 inB_ω, sinceB_ωis the inverse limit. Now letG∋bbe B_ω-generic. Set Gi=G∩B_i, σi=σ^◦G_i =σ^◦G_i ⁱ. Then (a^′)–(g^′) hold for i < ω. By (f^′) we can defineσ^′ :N ≺N by: σ^′(x) =σi(x) foris.t. σi(x) =σj(x) fori≤j. (a), (b) are then trivial. We prove:

(c) C_δ^N(rng(σ^′)) =C_δ^N(rng(σ)).

Proof. SetCi =C_δ^N(rng(σi)) for i < ω. (HenceC0=C_δ^N(rng(σ)).) (⊂) It suffices to show rng(σ^′)⊂C0. Butσ^′(xi) =σi(xi)∈Ci=C0. (⊃) We show rng(σ)⊂C_δ^N(rng(σ^′)).

σ(xi)∈σi(ui) =σ^′(ui)⊂S

{σ^′(u)|u≤δin N}=C_δ^N(rng(σ^′)). QED(c) Finally we show:

(d) σ^{′ ′′}G⊂G.

Proof. We first note that σ^{′ ′′}Gi ⊂ G for i < ω, since if a ∈ Gi, then σ^′(a) = σj(a)∈Gj ⊂G for some j ≥i. Now let a∈ G. Since B_ω is the inverse limit of hB_i |i < ωi, we may assume w.l.o.g. thata= T

i<ω

ai where ha|i < ωi is a thread in hB_i |i < ωi. Let σ^′(hai | i < ωi) =hai |i < ωi. Then hai | i < ωiis a thread in hB_i |i < ωiandσ^′(a) =σ^′(T

i

ai) =T

i

ai ∈Gby the completeness ofGwrt. V,

sinceai∈Gfori < ω. QED(Theorem 2)

Note Theorem 2 can be generalized to countable support iterations of length< ω2. Atω2it can fail, however, since in a countable support iteration we are required to take a direct limit atω2. If some earlier stage changed the cofinality ofω2toω(e.g.

ifB₁were Namba forcing), then the direct limit would not be subcomplete. Hence for longer iterations we must employrevised countable support iterations, which we discuss in the next section.

Revised countable support iterations

In dealing with an iteration we shall employ obvious notational simplifications, writing e.g. i for B_i, [[ϕ]]i for [[ϕ]]B_i etc. We also write: hi(b) = hB_i(b) =Df

Our definition of “iteration” permits great leeway in definingB_λ at limitλ. In practice people usually employ one of a number of standard limiting procedures, such as finite support (FS), countable support (CS) or revised countable support (RCS) iterations. RCS iterations are particulary suited to subcomplete forcing.

The definition of RCS iteration is given in Chapter 0. For present purposes all we need to know is:

Proof. Set: δi=δ(B_i). Then (1) δi ≤δj fori≤j < α,

since if X is dense inB_j, then{hi(a)|a∈X} is dense inB_i. (2) ν≤δν forν < α.

Proof. Suppose not. Let ν be the least counterexample. Thenν >0 is a cardinal.

If ν < ω, then δν < ω and hence B_ν is atomic with δν the number of atoms. Let ν = n+ 1. Then δn < δν < n+ 1 by (a). Hence δn < n. Contradiction! Hence ν ≥ω is a cardinal. If ν is a limit cardinal, thenδν ≥sup

i<ν

δν ≥ν. Contradiction!

Thusν is a successor cardinal. LetX ⊂B_ν be dense inB_ν withX=δν< ν. Then X ⊂B_η for anη < ν by the regularity ofν. HenceB_η=B_ν, contradicting (a).

QED(2) By induction oniwe prove:

Claim LetG be B_h-generic, h≤ i. Then B_i/G is subcomplete in V[G]. (Hence B_i ≃B_i/{1} is subcomplete in V, taking h= 0.) The case h=i is trivial, since thenB_i/G≃2. Hencei= 0 is trivial. Now leti=j+ 1. Then B_j/G⊂B_i/G. Let G˜beB_j/G-generic overV[G]. ThenG^′=G∗G˜ ={b∈B_j|b/G∈G}˜ isB_j-generic over V. But then (B_i/G)/G˜ ≃B_i/G^′ is subcomplete in V[G^′] = V[G][ ˜G] by (b).

Hence we have shown: B_j/G(( ˇB_i/G)/G^◦ is subcomplete).ButB_j/Gis subcomplete inV[G] by the induction hypothesis, so it follows by the two step theorem thatB_i/G is subcomplete in V[G]. There remains the case thati =λ is a limit ordinal. By our induction hypothesisB_j/Gh is subcomplete inV[Gh] forh≤j < λ. But then hB_h+i/Gh|i < λ−hisatisfies the same induction hypothesis, since ifi≤k < λ−h and ˆGisB_h+i/Gh-generic overV[Gh], thenG=Gh∗G˜isB_h+i-generic overVand (B_h+k/Gh)/G˜ ≃B_h+k/Gis subcomplete inV.

Case 1 cf(λ)≤δi for ani < λ.

Then cf(λ)≤ω1 in V[Gj] fori < j < λwhenever Gj is B_j-generic. It suffices to prove the claim for such j, since ifh < j and Gh is B_h-generic, we can then use the two step theorem to show – exactly as in the successor case – that B_λ/Gh is subcomplete in V[Gh]. Hence it will suffice to prove:

ClaimAssume cf(λ)≤ω1 inV. ThenB_λ is subcomplete,

since the same proof can then be carried out in V[Gj] to show that B_λ/Gj is subcomplete. Fixf :ω1→λs.t. supf^′′ω1=λ. Letθ > λbe a cardinal s.t. B< θ andθ is big enough that:

i(ˇθwitnesses the subcompleteness of ˇB_j/G)^◦

fori≤j < λ. Let N =L^A_τ be a ZFC⁻ model s.t. Hθ⊂N,θ < τ. Letσ:N ≺N s.t. N is countable and full. Suppose also that: σ(θ,B, λ, f , s) =θ,B, λ, f, s.

ClaimThere is b∈B_λ\ {0}s.t. wheneverG∋bisB_λ-generic, there is σ^′∈V[G]

s.t.

(a) σ^′:N≺N

We now closely imitate the proof of Theorem 2, constructing a sequence bi, σ^◦i

(i < ω) s.t. bi ∈ B_ξ

(hxℓ|ℓ < ωibeing an arbitrarily chosen enumeration ofN.)

(g^′) σi(uh) =σh(uh) (h≤i), whereui= the N-leastus.t. σ(xi)∈σi(u) and u < δ=Dfσ⁻¹(δ) inN.

The construction is exactly as before using that σi(B

ξ_j) = B_ξ

j for all j and that B_ξj (ˇB_ξ

j+1/G^◦is subcomplete). As before set: σ^′(x) =σi(x), whereiis big enough that σi(x) = σj(x) for i ≤j. The verification of (a)–(c) is exactly as before. To verify (d), we first note that, as before,

(2) σ^{′ ′′}Gi⊂Gfori < ω.

We then consider two cases: If cf(λ) =ω inN, then cf(λ) =ωinN and ˜λ=λ. B

λ

is then the inverse limit ofhB

ξ_i |i < ωiandB_λ is the inverse limit ofhB_ξ

i |i < ωi.

We then proceed exactly as before. If cf(λ) =ω1,B_λis the direct limit – i.e. S

i<ω

B_i is dense inB_λ. The conclusion then follows by (2). QED(Case 1) Case 2 Case 1 fails.

(c) C_λ^N(rng(σ^′)) =C_λ^N(rng(σ)) intention is, again, that ifc=T

i

ci∈GandGisB_λ-generic, then we can define the embedding σ^′ ∈V[G] from the sequence σi = σ^◦G_i (i < ω). However, since we no longer have the functionf available in defininghξi|i < ωi, we shallnot be able to enforce: σi(ξ_j) =ξjforj < ω. Nonetheless we can enforce: supσi′′λ= ˜λ, and shall have to make do with that. We inductively constructci ∈B_ξ

i, σ^◦i ∈V^Bξi with the

It remains only to constructci,σ^◦iand verify (I), (II). This will be somewhat trickier than the construction in Theorem 2. We shall also have to add further induction hypotheses to (I), (II). Before definingci we define abi∈B_ξ

i s.t.

(III) (a) b0= 1,σ^◦0= ˇσ

(b) hξj(bi) =cj ifi=j+ 1

(c) (II)(a)–(g) hold wheneverbi∈G.

σ◦i will be defined simultaneously with bi, before defining ci. Our next induction hypothesis states an important porperty ofbi:

Definition Letν≤ξi< µ <˜λs.t. ξh< ν forh < i,

a^jνµ=Dfbi∩[[σ^◦i(ˇξ_j) = ˇν∧σ^◦i(ˇξ_j+1) = ˇµ]]ξi. It follows easily that:

(4) a^jνµ∩a^{j′ν′µ′} = 0 if hj, ν, µi 6=hj^′, ν^′, µ^′i Proof. Suppose a^jνµ ∩a^{j′ν′µ′} ∈ G where G is B_ξ

i-generic. Then j = j^′, since if e.g. j < j^′, then µ ≤ σi(ξ_j+1) ≤ σi(ξ_j′) = ν^′ ≤ ξi Contradiction! But then ν =σi(ξj) =ν^′,µ=σi(ξ_j+1) =µ^′. Contradiction! QED(1) Our final induction hypothesis reads:

(IV) a^jνµ∩[[σ^◦i(ˇx) = ˇy]]ξi ∈B_ν if sup

h<i

ξh< ν ≤ξi< µ.

Hence a^jνµ=a^jνµ∩[[σ^◦i(ˇ0) = ˇ0]]∈B_ν.

Definition A=Ai = the set of alla^jνµ6= 0 s.t. sup

h<i

ξh< ν≤ξi< µ.

By (IV) we see that for eacha=a^jνµ∈Athere is σ^◦a ∈V^Bν s.t.

(5) σ^◦G_a^ν =σ_i^G forB_ξ

i-genericG∋a.

But:

(6) If G∋aisB_ν-generic, thenGextends to aB_ξ

i-genericG^′ s.t. G=G^′∩B_ν. Hence: σ^◦G_a =σ^◦G_a^′ =σ^◦G_i^′.

Thus we have:

(7) Let G ∋ a be B_ν-generic, where a = a^jνµ ∈ Ai. Then (II) holds with σa= σ^◦G_a in place of σi, σh =σ^◦G_h =σ^◦G_h^ξh forh < i, whereGη =DfG∩B_η (η≤ν).

Note Sincea∈G, (d) then reduces to: σa′′G_ξ

j ⊂G.

Note We then have: σa(xh) =σh(xh),σa(uh) =σh(uh) forh < i.

Whenever ν < µ < λ andG isB_ν-generic, we know that B_µ/G is subcomplete in V[G]. Then, using (7), Lemma 1.1, and repeating the construction of bi+1, σ^◦i+1

frombi, σ^◦i in the proof of Theorem 2, we get:

(8) Let a ∈ Ai, a = a^jνµ. There are ˜a ∈ B_µ, σ^◦′_a ∈ V^Bµ s.t. hν(˜a) = a and wheneverG∋a˜isB_µ-generic,σa=σ^◦G, and σ_a^′ = σ^◦′G_a , then we have:

(a) σ^′_a:N ≺N

(b) τ_a^′(θ,B, λ, s) =θ,B, λ, s (c) C_λ^N(rng(σ_a^′)) =C_λ^N(rng(σa)) (d) σ^′_a^′′G_ξ

j+1⊂G

(e) Letrbe least s.t. µ≤ξr. Thenσ^′_a(xh) =σa(xh) forh < r.

(f) Let r be as above. Then σ_a^′(u^a_h) = σa(u^a_h) for h < r, where u^a_h = the N-leastu∈N s.t. u≤λin N andσ(xh)∈σa(uh).

(g) σ^′_a(ξ_ℓ) =σa(ξ_ℓ) forℓ≤j+ 1.

For each a∈Ai fix such a pair ˜a, σ^◦′_a, which can be regarded as an instruction to be used later in forming br, where r is least s.t. µ ≤ξr. If Gis B_ξr-generic and a∩br ∈G, we shall want: ˜a∈Gand σr = σ^◦′G_a (whereσr = σ^◦G_r). In particular, we want: a∩br = ˜a. But we shall also require: hξi(br) = ci. Hence we need:

a∩ci=hξ(a∩br) =hξ(˜a). This is whybimust be “shrunk” toci. Accordingly we defineci as follows:

Definition Letbi be given. Setb=bi\S

Ai. Then:

ci=Dfb∪ [

a∈Ai

hξi(˜a).

We are working by induction oni. We assume (I)–(IV) to hold below i and (III), (IV) to hold at i. We must now verify (I), (II) ati. (II) is immediate by (III)(c), sinceci⊂bi. (I)(b) holds, sincehξjhξi(˜a) =hξj(˜a) =hξjhν(˜a) =hξj(a) fori=j+1 anda=a^ℓµν ∈Ai. Hence:

hξj(ci) =hξj(b)∪ [

a∈A

hξj(˜a) =hξj(b∪ [

a∈A

hξi(˜a)) =hξj(bi) =cj.

For (I)(a) note thatA0={a}wherea=a^0,0,ξ1= 1, sinceσ0=σby (III)(a). Hence c0=h0(˜a) = 1. This completes the verification of (I)–(IV) ati, given (III), (IV) at i and (I)–(IV) belowi. Now let (I)–(IV) hold below i. We must definebi, σ^◦i and verify (III), (IV) ati. For i= 0 set: b0= 1, σ^◦0= ˇσ. The verifications are trivial.

Now let i=j+ 1. Note that Aℓ, h˜a|a∈Aℓi, hσ_a^′ |a∈Aℓihave been defined for ℓ≤j. Set:

Definition Aˆj = the set ofa=a^hνµ∈ S

ℓ≤j

Aℓ s.t. ξj< µ.

(9) Leta, a^′∈Aˆj,a=a^hνµ,a^′=a^{h′ν′µ′}. Thena∩a^′= 0 ifhh, ν, µi 6=hh^′, ν^′, µ^′i.

Proof. Suppose not. Let a∈Aℓ,a^′ ∈Aℓ^′. Thenℓ6=ℓ^′ by (4). Let e.g. ℓ < ℓ^′ Let a∩a^′∈G, whereGisB_j-generic. Setσℓ=σ^◦G_ℓ forℓ≤j. Then:

σℓ(ξ_h) =ν ≤ξℓ< ν^′≤ξℓ^′ ≤ξj< µ=σℓ(ξ_h+1).

Hence σℓ^′ = σℓ by (II)(g). Hence h < h^′, since σℓ(ξ_h′) = ν^′ > σℓ(ξ_h). Hence

σℓ(ξ_h+1)≤ν^′ < µ. Contradiction! QED(9)

We now define:

Definition bi=S

{hξi(˜a)|a∈Aˆj}fori=j+ 1.

To defineσi we set: ˜A= the set ofa^iνµ∈Aˆj s.t. µ≤ξi. σ^◦i∈V^Bi is then a name s.t. [[σ^◦i=σ^◦′_a]] = ˜aifa∈A, [[˜ σ^◦i=σ^◦j]]∩bi=bi\SA.˜

Then:

(10) (III)(c) holds at i.

Proof. LetG∋bi beB_ξi-generic.

Case 1 a˜∈Gfor ana∈A.˜

Leta=a^hνµ∈Aℓ, µ≤ξi (hence ξj < µ≤ξi). Thusσi =σ^′_a. (II)(a)–(d) hold by (8)(a)–(d). Note that therin (8)(e), (f) isr=i. But, ifa∈Aℓ,ℓ≤j, thenσℓ =σa. Hence σℓ(ξ_h) =ν ≤ξℓ ≤ξℓ^′ < σℓ(ξ_h+1) =µ forℓ≤ℓ^′ ≤j. Hence: σa =σℓ^′ for ℓ≤ℓ^′≤j. But then (II)(e), (f) hold by (8)(e), (f). Finally (II)(g) holds vacuously, sinceξj < µ=σi(ξ_h+1)≤ξi, henceξj < σi(ξ_m) whereσi(ξ_m)≤ξi< σi(ξ_m+1).

Case 2 Case 1 fails.

Thenσi=σj. (II)(a)–(g) then follow trivially. QED(10) (III)(a) holds vacuously ati=j+ 1. We prove:

(11) (III)(b) holds ati.

Proof. Clearlyhξ_j(bi) = S

a∈Aˆj

hξ_j(˜a). Hence we need:

Claim cj= S

a∈Aˆj

hξj(˜a).

For j = 0 this is trivial, so let j =ℓ+ 1. Recall that cj =b∪ S

a∈Aj

hξj(˜a), where b=bj\ S

a∈Aj

a, so it suffices to show:

Claim b= S

a∈A^′

hξj(˜a) whereA^′= ˆAj\Aj. (⊃) Leta^′ ∈A^′. Thena^′ ∈Aˆℓ. Hencehξj(˜a^′)⊂ S

a∈Aˆℓ

hj(˜a) =bj. But for alla∈Aj

we havea∩a^′ = 0 by (10). Hencehξj(˜a)∩hξj(˜a^′) =a∩hξj(˜a^′) =hξj(a∩a˜^′) = 0, sincea∩a˜^′ ⊂a∩a^′= 0. Hence hξj(˜a^′)⊂b.

(⊂) Suppose not. Then there isa∈Aˆj\A^′ s.t. b∩hξj(˜a)6= 0. But thena∈Aj

andhξj(˜a) =a. Hencea∩b= 0 by the definition ofb. QED(11) It remains only to show:

(12) (IV) holds ati.

Proof. Leta =a^h,ν,µ ∈Ai. Then ξj < ν ≤ξi. a∩[[σ^◦i(ˇx) = ˇy]] =bi∩d, where d = [[σ^◦i(ˇξ_h) = ˇν∧σ^◦i(ξ_h+1) = ˇµ∧σ^◦i(ˇx) = ˇy]]B_ξi = [[ϕ(σ^◦i)]], where the formula ϕ(v) is Σ0 in the parameters ˇξ_h, ˇξ_h+1, ˇν, ˇµ, ˇx, ˇy, all of which lie inV². Recall that we are assumingV^Bη ⊆V^Bτ forη ≤τ (i.e. B_η is completely contained in B_τ and the identity is the natural embedding ofV^Bη inV^Bτ). As mentioned in Chapter 0, this has the consequence that ifψis a Σ0formula andt1, . . . , tm∈V^Bη, then:

aB_τ ψ(~t)←→aB_ηψ(~t) for a∈B_η, or in other words: [[ψ(~t)]]B_τ = [[ψ(~t)]]B_η ∈B_η.

We shall make strong use of this. We know: bi = S

e∈Aˆj

hξi(˜e). Hence it suffices to assign to eache∈Aˆj ane^∗∈B_ν s.t.

hξi(˜e)∩d=e^∗, since then we have:

bi∩d= [

e∈Aˆj

e^∗∈B_ν.

Forhξi(˜e)∩d= 0 we, of course, sete^∗= 0. Now lethξi(˜e)∩d6= 0. Lete=a^h,ν,µ∈ Aˆj. LetG∋hξi(˜e)∩dbeB_ξi-generic. Set: σi =σ^◦G, σ^◦j=σ^◦G_j.

Case 1 µ≤ξi.

Then ˜e = hξi(˜e) ∈ B_µ ∧G. Hence σi = σe^′ =Df

Then ˜e = hξi(˜e) ∈ B_µ ∧G. Hence σi = σe^′ =Df

Im Dokument Subcomplete Forcing and L -Forcing (Seite 45-0)