Frédéric Wang Subscribe   About Me   Blog Archive   Mathematics   Computer Science

Set theory - Chapter 16: Iterated Forcing and Martin’s Axiom

16.1

Let PP be a partial order, Q˙VP\dot{Q}\in V^{P} a name for a partial order. Let BB be the completion of PP and C˙VP\dot{C}\in V^{P} a name for the completion of Q˙\dot{Q}. Let D=B*CD=B*C. Any pPp\in P can be viewed as an element of B(P)B(P) and is mapped to some π(p)D\pi(p)\in D by the embedding of lemma 16.3. For any q˙VP\dot{q}\in V^{P} such that PqQ˙\Vdash_{P}q\in\dot{Q}, we can find a name c˙VP\dot{c}\in V^{P} such that Pc˙C˙\Vdash_{P}\dot{c}\in\dot{C} and P``c˙ is the image of q˙ by the embedding Q˙C˙′′\Vdash_{P}``\dot{c}\text{ is the image of }\dot{q}\text{ by the embedding }% \dot{Q}\rightarrow\dot{C}^{{\prime\prime}}. Again, we can associate the corresponding element ρ(q˙)D\rho(\dot{q})\in D. Consequently, we can define a mapping P*Q˙DP*\dot{Q}\rightarrow D by (p,q˙)π(p)ρ(q˙)(p,\dot{q})\mapsto\pi(p)\cdot\rho(\dot{q}).

Let (p1,q˙1),(p2,q˙2)P*Q˙(p_{1},\dot{q}_{1}),(p_{2},\dot{q}_{2})\in P*\dot{Q}. Suppose (p1,q˙1)(p2,q˙2)(p_{1},\dot{q}_{1})\leq(p_{2},\dot{q}_{2}). Then p1p2p_{1}\leq p_{2} and p1q˙1q˙2p_{1}\Vdash\dot{q}_{1}\leq\dot{q}_{2}. But p1π(p1)=π(p2)=1C˙p_{1}\Vdash\pi(p_{1})=\pi(p_{2})=1_{{\dot{C}}} and so p1π(p1)ρ(q˙1)C˙π(p2)ρ(q˙2)p_{1}\Vdash\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{\dot{C}}}\pi(p_{2})\cdot\rho% (\dot{q}_{2}). This inequality is clearly forced by -p1-p_{1} since -p1π(p1)=0C˙-p_{1}\Vdash\pi(p_{1})=0_{{\dot{C}}} and so Pπ(p1)ρ(q˙1)C˙π(p2)ρ(q˙2)\Vdash_{P}\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{\dot{C}}}\pi(p_{2})\cdot\rho(% \dot{q}_{2}) that is π(p1)ρ(q˙1)Dπ(p2)ρ(q˙2)\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{D}}\pi(p_{2})\cdot\rho(\dot{q}_{2}).

Conversely, if π(p1)ρ(q˙1)Dπ(p2)ρ(q˙2)\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{D}}\pi(p_{2})\cdot\rho(\dot{q}_{2}) then in particular p1π(p1)ρ(q˙1)C˙ρ(q˙2)p_{1}\Vdash\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{\dot{C}}}\rho(\dot{q}_{2}). Since p1π(p1)=1C˙p_{1}\Vdash\pi(p_{1})=1_{{\dot{C}}}, we get p1ρ(q˙1)C˙ρ(q˙2)p_{1}\Vdash\rho(\dot{q}_{1})\leq_{{\dot{C}}}\rho(\dot{q}_{2}) and so p1q˙1q˙2p_{1}\Vdash\dot{q}_{1}\leq\dot{q}_{2}. Moreover, π(p1)ρ(q˙1)C˙π(p2)=1\|\pi(p_{1})\cdot\rho(\dot{q}_{1})\leq_{{\dot{C}}}\pi(p_{2})\|=1. But we have p1-p2=π(p1-p2)=1C˙=π(p1)-π(p2)=1C˙π(p1)ρ(q˙1)-π(p2)=0C˙ρ(q˙1)=0C˙p_{1}\cdot-p_{2}=\|\pi(p_{1}\cdot-p_{2})=1_{{\dot{C}}}\|=\|\pi(p_{1})\cdot-\pi% (p_{2})=1_{{\dot{C}}}\|\cdot\|\pi(p_{1})\cdot\rho(\dot{q}_{1})\cdot-\pi(p_{2})% =0_{{\dot{C}}}\|\leq\|\rho(\dot{q}_{1})=0_{{\dot{C}}}\|. By definition of the embedding Q˙C˙\dot{Q}\rightarrow\dot{C} we have (in V[G]V[G]), ρ(q˙1)0C˙\rho(\dot{q}_{1})\neq 0_{{\dot{C}}} and so p1-p2Gp_{1}\cdot-p_{2}\notin G. This means that -(p1-p2)G-(p_{1}\cdot-p_{2})\in G. But -(p1-p2)π(p1-p2)=0C˙-(p_{1}\cdot-p_{2})\Vdash\pi(p_{1}\cdot-p_{2})=0_{{\dot{C}}} that is -(p1-p2)π(p1)π(p2)-(p_{1}\cdot-p_{2})\Vdash\pi(p_{1})\leq\pi(p_{2}) and finally -(p1-p2)p1p2-(p_{1}\cdot-p_{2})\Vdash p_{1}\leq p_{2}. Hence in V[G]V[G], and a fortiori in VV, we have p1p2p_{1}\leq p_{2}. Finally (p1,q˙1)(p2,q˙2)(p_{1},\dot{q}_{1})\leq(p_{2},\dot{q}_{2}).

By the previous paragraph, if π(p1)ρ(q˙1)=Dπ(p2)ρ(q˙2)\pi(p_{1})\cdot\rho(\dot{q}_{1})=_{{D}}\pi(p_{2})\cdot\rho(\dot{q}_{2}) then (p1,q˙1)(p2,q˙2)(p_{1},\dot{q}_{1})\leq(p_{2},\dot{q}_{2}) and (p2,q˙2)(p1,q˙1)(p_{2},\dot{q}_{2})\leq(p_{1},\dot{q}_{1}). Hence (p2,q˙2)(p1,q˙1)(p_{2},\dot{q}_{2})\equiv(p_{1},\dot{q}_{1}) by the usual identification (see for example Akihiro Kanamori’s book ”The Higher Infinite”, p 123) and so the mapping P*Q˙DP*\dot{Q}\rightarrow D is one-to-one.

It remains to show that the embedding is dense. If c˙D\dot{c}\in D and then we can find q˙VB\dot{q}\in V^{B} such that Pρ(q˙)c˙\Vdash_{P}\rho(\dot{q})\leq\dot{c}. Let pPp\in P. We have Pπ(p)ρ(q˙)C˙c˙\Vdash_{P}\pi(p)\cdot\rho(\dot{q})\leq_{{\dot{C}}}\dot{c}. Hence π(p)ρ(q˙)Dc˙\pi(p)\cdot\rho(\dot{q})\leq_{D}\dot{c}. Finally, since DD is a complete boolean algebra, we have B(P*Q˙)=B(D)=D=B(P)*B(Q˙)B(P*\dot{Q})=B(D)=D=B(P)*B(\dot{Q}).

16.2

Consider the application from P×QP\times Q to P*QˇP*\check{Q} defined by (p,q)(p,qˇ)(p,q)\mapsto(p,\check{q}). We have qˇQˇ=1\|\check{q}\in\check{Q}\|=1 that is PqˇQˇ\Vdash_{P}\check{q}\in\check{Q} and so the application is well-defined. If (p1,q1ˇ)=(p2,q2ˇ)(p_{1},\check{q_{1}})=(p_{2},\check{q_{2}}) then p1=p2p_{1}=p_{2} and q1=q1ˇG=q2ˇG=q1q_{1}=\check{q_{1}}^{G}=\check{q_{2}}^{G}=q_{1} so the application is one-to-one. If (p1,q1)(p2,q2)(p_{1},q_{1})\leq(p_{2},q_{2}) then p1p2p_{1}\leq p_{2} and q1ˇq1ˇ=1\|\check{q_{1}}\leq\check{q_{1}}\|=1 so p1q1ˇq1ˇp_{1}\Vdash\check{q_{1}}\leq\check{q_{1}} and (p1,q1ˇ)(p2,q2ˇ)(p_{1},\check{q_{1}})\leq(p_{2},\check{q_{2}}). Conversely, if this latter equality holds, then p1p2p_{1}\leq p_{2} and p1q1ˇq1ˇp_{1}\Vdash\check{q_{1}}\leq\check{q_{1}}. But q1ˇq1ˇ{0,1}\|\check{q_{1}}\leq\check{q_{1}}\|\in\{0,1\} and so actually q1ˇq1ˇ=1\|\check{q_{1}}\leq\check{q_{1}}\|=1 and consequently q1q2q_{1}\leq q_{2}. So we get an embedding from P×QP\times Q to P*QˇP*\check{Q}.

To show that the embedding is dense, consider (p,q˙)P*Qˇ(p,\dot{q})\in P*\check{Q} that is pPp\in P, q˙VP\dot{q}\in V^{P} and Pq˙Qˇ\Vdash_{P}\dot{q}\in\check{Q}. In particular, we have psQˇ,sq˙p\Vdash\exists s\in\check{Q},s\leq\dot{q}. Hence there is rpr\leq p and sQs\in Q such that rsˇq˙r\Vdash\check{s}\leq\dot{q}. Thus (r,sˇ)(p,q˙)(r,\check{s})\leq(p,\dot{q}).

16.3

For any dDd\in D, we have

dI˙G\displaystyle\|d\in\dot{I}\|\in G {bB:bd=0}G\displaystyle\Leftrightarrow\sum\{b\in B:b\cdot d=0\}\in G
bG,bd=0\displaystyle\Leftrightarrow\exists b\in G,b\cdot d=0
bG,b-d\displaystyle\Leftrightarrow\exists b\in G,b\leq-d
-dF˙\displaystyle\Leftrightarrow-d\in\dot{F}

So D:B=Bˇ/G˙=Bˇ/I˙D:B=\check{B}/\dot{G}=\check{B}/\dot{I}.

16.4

Let AA be a name for a subset of D:BD:B and e={d:d/IA=1}e=\sum\{d:\|d/I\in A\|=1\}. Let aa is such that a/IA=1\|a/I\in A\|=1. Then aDea\leq_{D}e and so a/ID:Be/Ia/I\leq_{{D:B}}e/I. Let dDd\in D such that a,a/IA=1a/ID:Bd/I\forall a,\|a/I\in A\|=1\implies a/I\leq_{{D:B}}d/I. Then for such a aa we have a-dI=1\|a\cdot-d\in I\|=1 and so 1(a-d)=01\cdot(a\cdot-d)=0 that is aDda\leq_{D}d. Hence eDde\leq_{D}d and e/ID:Bd/Ie/I\leq_{{D:B}}d/I. Hence e/I=A=1\|e/I=\sum A\|=1 and D:BD:B is complete.

Let c1,c2VBc_{1},c_{2}\in V^{B} such that c1,c2D:B=1\|c_{1},c_{2}\in D:B\|=1. We have c1=(D:B)c2=c1c2I={bB:b(c1c2)=0}\|c_{1}=_{{(D:B)}}c_{2}\|=\|c_{1}\triangle c_{2}\in I\|=\sum\{b\in B:b\cdot(c_% {1}\triangle c_{2})=0\}. If c1=Dc2c_{1}=_{D}c_{2} then c1c2=0c_{1}\triangle c_{2}=0 and so c1=(D:B)c2=1\|c_{1}=_{{(D:B)}}c_{2}\|=1. Conversely, if this later equality is true, c1c2=0c_{1}\triangle c_{2}=0. Otherwise, -(c1c2)<1-(c_{1}\triangle c_{2})<1 and we could find bBb\in B with both b(c1c2)=0b\cdot(c_{1}\triangle c_{2})=0 and b-(c1c2)b\nleq-(c_{1}\triangle c_{2}). Hence c1=Dc2c_{1}=_{D}c_{2}. As a consequence c1=(D:B)c2=1c1=Dc2\|c_{1}=_{{(D:B)}}c_{2}\|=1\Leftrightarrow c_{1}=_{D}c_{2} and so D=B*(B:D)D=B*(B:D).

16.5

Let h:P*Q˙Ph:P*\dot{Q}\rightarrow P be defined by h(p,q˙)=ph(p,\dot{q})=p. If (p1,q1˙)(p2,q2˙)(p_{1},\dot{q_{1}})\leq(p_{2},\dot{q_{2}}) then p1p2p_{1}\leq p_{2} and so h(p1,q1˙)h(p2,q2˙)h(p_{1},\dot{q_{1}})\leq h(p_{2},\dot{q_{2}}). If (p,q˙)P*Q˙(p,\dot{q})\in P*\dot{Q} and ph(p,q˙)=pp^{{\prime}}\leq h(p,\dot{q})=p then e(p)1=q˙=q˙q˙q˙e(p^{{\prime}})\leq 1=\|\dot{q}=\dot{q}\|\leq\|\dot{q}\leq\dot{q}\| so pq˙q˙p^{{\prime}}\Vdash\dot{q}\leq\dot{q} and so (p,q˙)(p,q˙)(p^{{\prime}},\dot{q})\leq(p,\dot{q}). Hence (p,q˙)(p^{{\prime}},\dot{q}) is compatible with (p,q˙)(p,\dot{q}) and h(p,q˙)=pph(p^{{\prime}},\dot{q})=p^{{\prime}}\leq p^{{\prime}}. So hh satisfies the hypothesis of Lemma 15.45 and VPVP*Q˙V^{{P}}\subseteq V^{{P*\dot{Q}}}.

16.6

Suppose PP has property (K) and PQ˙ has property (K)\Vdash_{P}\dot{Q}\text{ has property (K)}. Let {(pi,qi˙):iI}\{(p_{i},\dot{q_{i}}):i\in I\} be an uncountable set of pairwise distinct elements of P*Q˙P*\dot{Q} that is ijpipjpiqi˙qj˙i\neq j\implies p_{i}\neq p_{j}\vee p_{i}\Vdash\dot{q_{i}}\neq\dot{q_{j}}.

Assume there is at most countably many distinct pip_{i}. Then by the pigeon principle, there is I1II_{1}\subseteq I uncountable on which all the pip_{i} are equal. Otherwise, since PP has property (K), we can still find I1II_{1}\subseteq I uncountable on which the pip_{i} are pairwise compatible (but pairwise distinct).

Suppose any subset of I1I_{1} on which ijPqi˙qj˙i\neq j\implies\Vdash_{P}\dot{q_{i}}\neq\dot{q_{j}} is at most countable. Then by the pigeon principle, there is I2I1I_{2}\subseteq I_{1} uncountable and q˙VB\dot{q}\in V^{B} such that Pqi˙=q˙\Vdash_{P}\dot{q_{i}}=\dot{q}. In particular, we are in the case where the pip_{i} are pairwise distinct or otherwise for any distinct i,jI2i,j\in I_{2}, we get pi=p=pjp_{i}=p=p_{j} and pqi˙=q˙=qj˙p\Vdash\dot{q_{i}}=\dot{q}=\dot{q_{j}}. Then {(pi,qi˙):iI2}\{(p_{i},\dot{q_{i}}):i\in I_{2}\} is uncountable set of pairwise distinct and compatible elements of P*Q˙P*\dot{Q}.

Otherwise, let I2I1I_{2}\subseteq I_{1} uncountable on which ijPqi˙qj˙i\neq j\implies\Vdash_{P}\dot{q_{i}}\neq\dot{q_{j}}. Since PP satisfies the ccc, I2I_{2} is still uncountable in V[G]V[G] and since PQ˙ has property (K)\Vdash_{P}\dot{Q}\text{ has property (K)}, there is I3I2I_{3}\subseteq I_{2} uncountable such that i,jI3\forall i,j\in I_{3}, PqQ˙,qqi˙,qj˙\Vdash_{P}\exists q\in\dot{Q},q\leq\dot{q_{i}},\dot{q_{j}}. Then the set {(pi,qi˙):iI3}\{(p_{i},\dot{q_{i}}):i\in I_{3}\} is uncountable set of pairwise distinct (because I3I2I_{3}\subseteq I_{2}) of P*Q˙P*\dot{Q}. For any i,jI3i,j\in I_{3}, we can find ppi,pjp\leq p_{i},p_{j} (because I3I1I_{3}\subseteq I_{1}) and q˙VB\dot{q}\in V^{B} such that Pq˙qi˙,qj˙\Vdash_{P}\dot{q}\leq\dot{q_{i}},\dot{q_{j}} (because VBV^{B} is full). Then (p,q˙)(pi,qi˙),(pj,qj˙)(p,\dot{q})\leq(p_{i},\dot{q_{i}}),(p_{j},\dot{q_{j}}) and so the (pi,qi˙)(p_{i},\dot{q_{i}}) are pairwise compatible.

Finally, P*Q˙P*\dot{Q} has property (K).

16.7

16.8

16.9

Let PαP_{\alpha} be a finite support iteration. The first part is a particular case of exercise 16.17. So let’s show by induction on α\alpha that for all β<α\beta<\alpha, Gβ={p|β:pG}G_{\beta}=\{p_{{|\beta}}:p\in G\}. The successor step is deduced from theorem 16.2 so suppose α\alpha is limit.

Let pβ,qβPβp_{\beta},q_{\beta}\in P_{\beta}. Suppose pβGβp_{\beta}\in G_{\beta} and pβqβp_{\beta}\leq q_{\beta}. Let pGp\in G such that p|β=pβp_{{|\beta}}=p_{\beta}. Define qPαq\in P_{\alpha} by q(ξ)=qβ(ξ)q(\xi)=q_{\beta}(\xi) if ξ<β\xi<\beta and q(ξ)=p(ξ)q(\xi)=p(\xi) if βξ<α\beta\leq\xi<\alpha. We have pαqp\leq_{\alpha}q and so qGq\in G. Finally qβ=q|βGβq_{\beta}=q_{{|\beta}}\in G_{\beta}. Suppose instead that pβ,qβGβp_{\beta},q_{\beta}\in G_{\beta} and that p,qGp,q\in G are such that p|β=pβp_{{|\beta}}=p_{\beta} and q|β=qβq_{{|\beta}}=q_{\beta}. Let rαp,qr\leq_{\alpha}p,q such that rGr\in G. Then r|βpβ,qβr_{{|\beta}}\leq p_{\beta},q_{\beta} and r|βGβr_{{|\beta}}\in G_{\beta}.

Finally, if DPβD\subseteq P_{\beta} is dense then D={dPα:d|βD}D^{{\prime}}=\{d\in P_{\alpha}:d_{{|\beta}}\in D\} is dense in PαP_{\alpha}. So DGD^{{\prime}}\cap G\neq\emptyset and a fortiori DGβD\cap G_{\beta}\neq\emptyset. Hence GβG_{\beta} is a generic filter on PβP_{\beta}.

16.10

Let PP be the notion of forcing producing a Cohen generic real. PP satisfies the countable chain condition. For each g:ω{0,1}g:\omega\rightarrow\{0,1\} define Dg={pP:pg}D_{g}=\{p\in P:p\nsubseteq g\}. Each DgD_{g} is dense since any function pPp\in P can be extended to some qPq\in P such that q(n)g(n)q(n)\neq g(n) for some nωdom(p)n\in\omega\setminus\operatorname{dom}(p). If g1g2g_{1}\neq g_{2} then there is nωn\in\omega such that g1(n)g2(n)g_{1}(n)\neq g_{2}(n). Then {(n,g1(n))}Dg2Dg1\{(n,g_{1}(n))\}\in D_{{g_{2}}}\setminus D_{{g_{1}}} so Dg2Dg1D_{{g_{2}}}\neq D_{{g_{1}}}. Hence {Dg:g2ω}\{D_{g}:g\in 2^{\omega}\} is a family of dense subset of size 202^{{\aleph_{0}}}. We adjoin to this family, the countable family of dense subsets Dn={pP:ndom(p)D_{n}=\{p\in P:n\in\operatorname{dom}(p).

If GG is generic for the final family then f=Gf=\bigcup G is a function and there is pDfGp\in D_{f}\cap G. f|dom(p)=pff_{{|\operatorname{dom}(p)}}=p\nsubseteq f. A contradiction.

16.11

Let PP be the notion of forcing that collapses ω1\omega_{1}. In particular, it does not satisfy the countable chain condition. For each α<ω1\alpha<\omega_{1}, let Dα={pP:αran(p)}D_{\alpha}=\{p\in P:\alpha\in\operatorname{ran}(p)\}. We can always extend a function pp arbitrarily at a point α\alpha and so DαD_{\alpha} is dense. If αβ\alpha\neq\beta then {(0,α)}DαDβ\{(0,\alpha)\}\in D_{\alpha}\setminus D_{\beta} so DαDβD_{\alpha}\neq D_{\beta}. Hence {Dα:α<ω1}\{D_{\alpha}:\alpha<\omega_{1}\} is a family of dense subset of size 1\aleph_{1}. We adjoin to this family, the countable family of dense subsets Dn={pP:ndom(p)D_{n}=\{p\in P:n\in\operatorname{dom}(p).

If GG is generic fot the final family then f=Gf=\bigcup G is a function from ω\omega onto ω1\omega_{1}. A contradiction.

16.12

Let κ<20\kappa<2^{{\aleph_{0}}} and suppose that MAκ\text{MA}_{\kappa} holds for complete Boolean Algebra. Let PP be a partial ordering and e:PB(P)e:P\rightarrow B(P). Let 𝒟\mathcal{D} be a family of at most κ\kappa dense subsets of PP. Then its image by ee is a family of at most κ\kappa dense subsets of B(P)B(P). If HH is a generic filter for that family then e-1(H)e_{{-1}}(H) is a 𝒟\mathcal{D}-generic filter.

16.13

This is the same as Lemma 16.12. The set QQ mentioned in the lemma can be defined by Q=nQnQ=\bigcup_{{n\in\mathbb{N}}}Q_{n} where QnQ_{n} is defined by induction: Q0=D𝒟WDQ_{0}=\bigcup_{{D\in\mathcal{D}}}W_{D} and Qn+1=Qn{rp,q:p,qQn}Q_{{n+1}}=Q_{n}\bigcup\{r_{{p,q}}:p,q\in Q_{n}\} where rp,qr_{{p,q}} is a refinement of p,qp,q if there is one (and an arbitrary element of QnQ_{n}, say pp, otherwise). Each WDW_{D} is at most countable and 𝒟\mathcal{D} is of size at most κ\kappa so Q0Q_{0} is of size at most κ\kappa. We deduce that each QnQ_{n} is is of size at most κ\kappa and finally |Q|κ|Q|\leq\kappa. The rest of the proof is not changed.

16.14

Let TT be a Suslin tree, PTP_{T} the associated forcing notion and let PP be the notion of forcing that adjoins κ\kappa Cohen generic reals. PP and PTP_{T} satisfy ccc. In particular, PTP_{T} preserves cardinal and the definition of PTP_{T} remains the same in any generic extension: PTPˇ satisfies ccc\Vdash_{{P_{T}}}\check{P}\text{ satisfies ccc}. We apply corollary 16.6 in the two directions to deduce that PT×PP_{T}\times P satisfy ccc first and then that PPTˇ satisfies ccc\Vdash_{{P}}\check{P_{T}}\text{ satisfies ccc}.

Suppose that TT is actually a normal Suslin tree. Then since PP preserves ω1\omega_{1} and since o(x)o(x) and other concepts involved in (9.9) are absolute, TT remains a normal ω1\omega_{1}-tree in V[G]V[G]. By the previous paragraph, TT satisfy the countable chain condition in V[G]V[G] and by exercise 9.7, we deduce that TT is a Suslin tree in V[G]V[G].

We have V[G]20=κV[G]\models 2^{{\aleph_{0}}}=\kappa and so the existence of Suslin tree is consistent with 1<20\aleph_{1}<2^{{\aleph_{0}}}.

16.15

Let GG be a generic filter for the forcing notion that collapses ω2\omega_{2} to ω1\omega_{1} and f=Gf=\bigcup G. In V[f]V[f], ω1\omega_{1} is preserved and ω2\omega_{2} is collapsed. ff is of size 1\aleph_{1} and so is its transitive closure. Let (ω1,E)(\omega_{1},E) be isomorphic to (TC({f}), )(\operatorname{TC}(\{f\}),\in) and define X=Γ(E)ω1X=\Gamma(E)\subseteq\omega_{1}. We have XV[f]X\in V[f] and so V[X]V[f]V[X]\subseteq V[f]. Conversely, if MM is the transitive collapse of (ω1,E)(\omega_{1},E) in V[X]V[X] it is also the transitive collapse of (ω1,E)(\omega_{1},E) in V[f]V[f] and so fTC({f})=MV[X]f\in\operatorname{TC}(\{f\})=M\in V[X]. Finally V[f]=V[X]V[f]=V[X]. In particular ω1V[X]=ω1\omega_{1}^{{V[X]}}=\omega_{1}.

We can apply the almost disjoint forcing method given p 277 to find AωA\subseteq\omega such that ω1V[A]=ω1\omega_{1}^{{V[A]}}=\omega_{1} and XV[A]X\in V[A]. In particular fV[f]=V[X]V[A]f\in V[f]=V[X]\subseteq V[A] and so ω2\omega_{2} is collapsed in V[A]V[A].

16.16

Assume MAκ\text{MA}_{\kappa} and let {Xα:α<κ}\{X_{\alpha}:\alpha<\kappa\} be a sequence of infinite subsets of ω\omega such that XβXαX_{\beta}\setminus X_{\alpha} is finite for α<β\alpha<\beta. We consider the forcing notion {(s,F):s[ω]<ω,F[κ]<ω}\{(s,F):s\in{[\omega]}^{{<\omega}},F\in{[\kappa]}^{{<\omega}}\} and (s,F)(s,F)(s^{{\prime}},F^{{\prime}})\leq(s,F) iff sss\subseteq s^{{\prime}}, FFF\subseteq F^{{\prime}} and ssXαs^{{\prime}}\setminus s\subseteq X_{\alpha} for all αF\alpha\in F. It is clear that the relation is reflexive and antisymmetric. The transitivity is almost obvious, just note that if (s3,F3)(s2,F2)(s_{3},F_{3})\leq(s_{2},F_{2}) and (s2,F2)(s1,F1)(s_{2},F_{2})\leq(s_{1},F_{1}) then for all αF1F2\alpha\in F_{1}\subseteq F_{2} we have s3s1s3s2s2s1Xαs_{3}\setminus s_{1}\subseteq s_{3}\setminus s_{2}\cup s_{2}\setminus s_{1}% \subseteq X_{\alpha}.

The forcing notion satisfies ccc: since [ω]<ω{[\omega]}^{{<\omega}} is countable, for any uncountable subset WW there is t[ω]<ωt\in{[\omega]}^{{<\omega}} such that Z={(s,F)W:s=t}Z=\{(s,F)\in W:s=t\} is uncountable. Then any (t,F1),(t,F2)Z(t,F_{1}),(t,F_{2})\in Z have a common refinement (t,F1F2)(t,F_{1}\cup F_{2}).

For all n<ωn<\omega, define Dn={(s,F):|s|n}D_{n}=\{(s,F):|s|\geq n\}. Let α1>α2>>αk\alpha_{1}>\alpha_{2}>...>\alpha_{k} the elements of FF. We show by induction on 1mk1\leq m\leq k that i=1mXαi\bigcap_{{i=1}}^{m}X_{{\alpha_{i}}} is infinite. This is true for m=1m=1 by assumption. If it is true for m-1m-1 then

i=1m-1Xαi=i=1mXαii=1mXαiXαm\bigcap_{{i=1}}^{{m-1}}X_{{\alpha_{i}}}=\bigcap_{{i=1}}^{m}X_{{\alpha_{i}}}% \cup\bigcap_{{i=1}}^{m}X_{{\alpha_{i}}}\setminus X_{{\alpha_{m}}}

The left hand side is infinite by induction hypothesis. The second term of the right hand side is included in X1XαmX_{1}\setminus X_{{\alpha_{m}}} and thus is finite. Hence the first term is infinite and the result is true for mm. Finally, for m=km=k, we get that αFXα\bigcap_{{\alpha\in F}}X_{\alpha} is infinite. Pick x1,x2,,xnx_{1},x_{2},...,x_{n} distinct elements from that set and define (s,F)=(s{x1,xn},F)(s^{{\prime}},F^{{\prime}})=(s\cup\{x_{1},...x_{n}\},F). We have (s,F)Dn(s^{{\prime}},F^{{\prime}})\in D_{n}, sss\subseteq s^{{\prime}}, FFF\subseteq F^{{\prime}} and for all αF\alpha\in F, ss={x1,xn}Xαs^{{\prime}}\setminus s=\{x_{1},...x_{n}\}\subseteq X_{\alpha}. This shows that DnD_{n} is dense. For each α<κ\alpha<\kappa, the set Eα={(s,F):αF}E_{\alpha}=\{(s,F):\alpha\in F\} is also dense: for any (s,F)(s,F) consider (s,F)=(s,F{α})(s^{{\prime}},F^{{\prime}})=(s,F\cup\{\alpha\}).

By MAκ\text{MA}_{\kappa} there is a generic filter GG for the previous family of dense sets. Let X={n<ω:(s,F)G,ns}X=\{n<\omega:\exists(s,F)\in G,n\in s\}. For all n<ωn<\omega, there is (s,F)GDn(s,F)\in G\cap D_{n} and so |X||s|n|X|\geq|s|\geq n. Hence XX is infinite. Let α<κ\alpha<\kappa and (s1,F1)GEα(s_{1},F_{1})\in G\cap E_{\alpha}. For any xXx\in X, there is (s2,F2)G(s_{2},F_{2})\in G such that xs2x\in s_{2}. Hence there is (s3,F3)G(s_{3},F_{3})\in G a refinement of (s1,F2),(s2,F2)(s_{1},F_{2}),(s_{2},F_{2}). We have xs2s3x\in s_{2}\subseteq s_{3} and s3s1Xαs_{3}\subseteq s_{1}\subseteq X_{\alpha}. Hence XXαs1X\setminus X_{\alpha}\subseteq s_{1} is finite.

16.17

Let PαP_{\alpha} be an iteration. To prove VPβVPαV^{{P_{\beta}}}\subseteq V^{{P_{\alpha}}} it suffices to prove that the conditions of Lemma 15.45 hold. We do that by induction on α\alpha and the successor step is essentially exercise 16.5. So suppose α\alpha is limit and let β<α\beta<\alpha. If pαqp\leq_{\alpha}q the pβqp\leq_{\beta}q. Let pPαp\in P_{\alpha} and qPβq\in P_{\beta} such that qβp|βq\leq_{\beta}p_{{|\beta}}. Define rr by r(ξ)=q(ξ)r(\xi)=q(\xi) if ξ<β\xi<\beta and r(ξ)=p(ξ)r(\xi)=p(\xi) if βξ<α\beta\leq\xi<\alpha. By definition 16.29, (iii) (b), we have rPαr\in P_{\alpha}, r|β=qqr_{{|\beta}}=q\leq q and rr is compatible with pp (actually rpr\leq p).

16.18

16.19

Let II be a κ\kappa-closed ideal on α\alpha and PαP_{\alpha} and iteration of {Q˙β}β{\{{\dot{Q}}_{{\beta}}\}}_{{\beta}} with II-support. Suppose that β<α,βQ˙β is κ-closed\forall\beta<\alpha,\Vdash_{\beta}{\dot{Q}}_{\beta}\text{ is }\kappa\text{-closed}. Let λ<κ\lambda<\kappa and p1p2pμp^{1}\geq p^{2}\geq...\geq p^{\mu}\geq... (μ<λ\mu<\lambda) a descending sequence in PαP_{\alpha}. By definition for all βα\beta\leq\alpha and μ<λ\mu<\lambda, pβμPβp^{\mu}_{{\restriction\beta}}\in P_{\beta} and the pβμp^{\mu}_{{\restriction\beta}} form a descending sequence in PβP_{\beta}. We shall construct by induction on γα\gamma\leq\alpha an α\alpha-sequence pp such that pβPβp_{{\restriction\beta}}\in P_{\beta} and s(pβ)Ss(p_{{\restriction\beta}})\subseteq S where S=μ<λs(pμ)IS=\bigcup_{{\mu<\lambda}}s(p^{\mu})\in I. In particular, s(pβ)Is(p_{{\restriction\beta}})\in I.

Since Q0Q_{0} is κ\kappa-closed, we let p(0)p(0) be a lower bound of PμP_{\mu} and even p(0)=1p(0)=1 if 0μ<λs(pμ)0\notin\bigcup_{{\mu<\lambda}}s(p^{\mu})). Then s(p1)Ss(p_{{\restriction 1}})\subseteq S. Clearly p1P1p_{{\restriction 1}}\in P_{1} and μ<λ,p11p1μ\forall\mu<\lambda,p_{{\restriction 1}}\leq_{1}p^{\mu}_{{\restriction 1}}.

If pβp_{{\restriction\beta}} is constructed then βxQ˙β,μ<λ,xpμ(β)\Vdash_{\beta}\exists x\in{\dot{Q}}_{\beta},\forall\mu<\lambda,x\leq p^{\mu}(\beta) and since VPβV^{{P_{\beta}}} is full there is xx such that βxQ˙β\Vdash_{\beta}x\in{\dot{Q}}_{\beta} and μ<λ\forall\mu<\lambda βxpμ(β)\Vdash_{\beta}x\leq p^{\mu}(\beta). We let p(β)=xp(\beta)=x and can take p(β)=1p(\beta)=1 if βμ<λs(pμ)\beta\notin\bigcup_{{\mu<\lambda}}s(p^{\mu})). Again, s(pβ+1)Ss(p_{{\restriction\beta+1}})\subseteq S. Then pβ+1Pβ+1p_{{\restriction{\beta+1}}}\in P_{{\beta+1}} and μ<λ\forall\mu<\lambda, pβ+1β+1pβ+1μp_{{\restriction{\beta+1}}}\leq_{{\beta+1}}p^{\mu}_{{\restriction{\beta+1}}}.

Now suppose β\beta is limit and that pγp_{{\restriction\gamma}} has been constructed for all γ<β\gamma<\beta. We just let pβ=γ<βpγp_{{\restriction\beta}}=\bigcup_{{\gamma<\beta}}p_{{\restriction\gamma}}. By induction hypothesis, each γ<β,pγPγ\forall\gamma<\beta,p_{{\restriction\gamma}}\in P_{\gamma} and s(pβ)γ<βs(pγ)Ss(p_{{\restriction\beta}})\subseteq\bigcup_{{\gamma<\beta}}s(p_{{\restriction% \gamma}})\subseteq S. Hence pβPβp_{{\restriction\beta}}\in P_{{\beta}} Moreover for all γ<β\gamma<\beta and μ<κ\mu<\kappa we have pγγpγμp_{{\restriction\gamma}}\leq_{\gamma}p^{\mu}_{{\restriction\gamma}} and so pβγpβμp_{{\restriction\beta}}\leq_{\gamma}p^{\mu}_{{\restriction\beta}}.

Finally at stage β=α\beta=\alpha, we obtain pPαp\in P_{\alpha} such that μ<κ,pαpμ\forall\mu<\kappa,p\leq_{\alpha}p^{\mu}. Hence PαP_{\alpha} is <κ<\kappa-closed.

16.20

Let κ2\kappa\geq\aleph_{2} be a regular cardinal. Let PP be a countable support iteration of length κ\kappa such that for all β<κ\beta<\kappa, PβP_{\beta} has a dense subset DβD_{\beta} of size <κ<\kappa.

If WPβW\subseteq P_{{\restriction\beta}} is an antichain, then for each wWw\in W we can find wDDβw_{D}\in D_{\beta} such that wwDw\leq w_{D}. If wD=wDw,ww_{D}=w^{{\prime}}_{D}\leq w,w^{{\prime}} then w=ww=w^{{\prime}} since WW is an antichain. Hence |W||Dβ|<κ|W|\leq|D_{\beta}|<\kappa. Hence PβP_{\beta} satisfies the κ\kappa-chain condition.

PκP_{\kappa} is a countable support iteration and cf(κ)=κ>ω\operatorname{cf}(\kappa)=\kappa>\omega so PκP_{\kappa} is a direct limit. The set S={β<κ:cf(β)=1}S=\{\beta<\kappa:\operatorname{cf}(\beta)=\aleph_{1}\} is stationary in κ\kappa and for each βS\beta\in S we have cf(β)>ω\operatorname{cf}(\beta)>\omega and so PβP_{\beta} is a direct limit.

Finally, by theorem 16.30, PκP\kappa satisfies the κ\kappa-chain conditon.