Gitik’s Long Extender Forcing – Main Results

This post continues the previous one. As before, I’m following the presentation from Gitik’s  Handbook chapter, none of the results here are mine, and any mistakes are my fault.

Now that we’ve defined the long extender forcing, we want to prove the following theorem:

Theorem (Gitik) : Assume GCH holds in V, and let \langle \kappa_n : n\in \omega\rangle be an increasing sequence of \lambda-strong cardinals with \sup_{n<\omega}\kappa_n=\kappa and \lambda\geq \kappa^{++}, and let (\mathbb{P},\leq) be the corresponding long extender forcing. If G\subseteq \mathbb{P} is V-generic for (\mathbb{P}, \leq), then the following hold in V[G]:

  1. No New bounded subsets of \kappa are added;
  2. All cardinals are preserved;
  3. \kappa^{\omega}\geq\lambda;
  4. In particular, since GCH holds in the ground model, we have that V[G]\models GCH first fails at \kappa.

We will proceed by first showing that (\mathbb{P},\leq) satisfies the \kappa^{++} chain condition. From there, we will black box a rather technical lemma for later, but use this technical lemma to show that forcing with (\mathbb{P},\leq) preserves \kappa^+ and adds no new bounded subsets of \kappa. Finally, we will show that \lambda-many sequences are added to \prod_{n<\omega}\kappa_n. In order to show that \mathbb{P} satisfies the \kappa^{++}-chain condition, we will heavily abuse the fact that the forcings Q_{n1} are basically Cohen forcing.

We begin by fixing some terminology. Letting p=\langle p_n : n<\omega\rangle\in \mathbb{P}, we call p\upharpoonright \ell(p) the stem of p and the rest of the condition the tail of p. Next recall the three lemmas about the extender ordering (as they will be abused quite a bit) in the form that we’ll be using them:
Lemma 1: (GCH) Fix n<\omega, and \tau<\kappa_n. If \{ \alpha_\nu : \nu<\tau\} is a collection of ordinals below \lambda, then there are \lambda-many \alpha<\lambda such that \alpha\geq_{E_n}\alpha_\nu for each \nu<\tau. In particular, \leq_{E_n} is \kappa_n-directed.

Lemma 2: Suppose that \{ \alpha_{\eta} : \eta< \gamma <\kappa_n\} is a collection of ordinals with \alpha_\eta\leq \alpha<\lambda for each \eta<\gamma. If \alpha_\eta\leq_{E_n}\alpha for each \eta<\gamma, then \{\nu<\kappa : (\forall \eta<\gamma)( \pi_{\alpha,\alpha_\eta}(\nu)<\pi_{\alpha,\alpha_\eta}(\nu))\}\in E_{n\alpha}.

Lemma 3: Suppose that\{ \alpha_{\eta} : \eta< \gamma <\kappa_n\} is a collection of ordinals with such that \alpha_\eta\leq_{E_n}\alpha_{\eta'}\leq_{E_n}\alpha for each \eta,\eta'<\gamma, then there is some A\in E_{n\alpha} such that, for each \nu\in A, \pi_{\alpha,\alpha_{\eta}}(\nu)=\pi_{\alpha_{\eta'}\alpha_\eta}(\pi_{\alpha,\alpha_{\eta'}}(\nu)).

Lemma : (\mathbb{P},\leq) satisfies the \kappa^{++} chain condition.

Proof: Let \{ p^\alpha : \alpha<\kappa^{++}\} be a collection of conditions from \mathbb{P} with p^\alpha=\langle p^\alpha_n : n<\omega\rangle and p^\alpha_n=\langle a^\alpha_n, A^\alpha_n,f^\alpha_n\rangle for n\geq\ell(p). We begin by noting that we can refine our collection of conditions to a collection of size \kappa^{++} with the property that \ell=\ell(p^\alpha)=\ell(p^\beta) for each \alpha,\beta<\kappa^{++}. Furthermore, since Q_{n1} is a collection of partial functions from \lambda to \kappa of size <\kappa^+ and GCH holds in the ground model, we can further refine to a collection \{ p^\alpha : \alpha<\kappa^{++}\} with the following properties:

  1. For each n<\ell, the collection \{ dom(p^\alpha_n) : \alpha<\kappa^{++}\} forms a \Delta-system with root r such that p^\alpha_n\upharpoonright r=p^\beta_n\upharpoonright r for each \alpha,\beta<\kappa^{++}.
  2. For each n\geq \ell, the collection \{dom(f^\alpha_n)\cup a^\alpha_{n} : \alpha<\kappa^{++}\} forms a \Delta-system with root r such that f^\alpha_n\upharpoonright r=f^\beta_n\upharpoonright r, and f^\alpha_n\cap a^\beta_n=\emptyset for each \alpha,\beta<\kappa^{++}.

Our next step is to show that, given any \alpha,\beta<\kappa^{++}, there is a condition q=\langle q_n : n<\omega\rangle\in \mathbb{P} (directly) extending both p^\alpha and p^\beta. That the extension is direct is more an artifact of the proof, and not in and of itself important. We begin by fixing such \alpha,\beta<\kappa^{++}, and note that we can easily construct q_n for n<\ell by setting q_n=p^\alpha_n\cup p^\beta_n since they both agree on the root of the \Delta-system. Constructing q_n=\langle b_n, B_n, g_n\rangle for n>\ell is only slightly harder, as we need to make sure that each portion of the condition agrees with the others.

We construct g_n=f^\alpha_n\cup f^\beta_n as before. We would like to construct b_n in the same fashion, but we need to make sure that b_n has a \leq_{E_n}-maximal element, and that said maximal element doesn’t get into the domain of g_n. This is easy though, as Lemma 1 guarantees that there is some \rho > \sup (\bigcup dom(g_n)) with \rho such that it \leq_{E_n}-dominates every element of a^\alpha_n\cup a^\beta_n. With this in mind, let b_n=a^\alpha_n\cup a^\beta_n\cup\{\rho\}, and note that our choice of B_n is completely decided by our choice of b_n. That is, let B'_n=\pi_{\rho\max(a^\alpha_n)}^{-1}{}''A_n^\alpha\cap\pi_{\rho\max(a^\beta_n)}^{-1}{}''A_n^\beta, and use Lemma 2 and Lemma 3 to shrink B'_n to B_n satisfying the conclusions of said lemma for b_n. Then by construction q_n=\langle b_n, B_n, g_n\rangle\in Q_{n0}, and note that we can slightly change each b_n by adding finitely many elements to make sure that b_m\subseteq b_n for \ell\leq m\leq n <\omega. Thus q=\langle q_n : n<\omega\rangle is a condition directly extending both p^\alpha and p^\beta and hence we have shown that (\mathbb{P},\leq) has no anti-chains of length \kappa^{++}.

At this point, we would like to show that (\mathbb{P},\leq,\leq^*) satisfies the Prikry condition. In order to do this, we will go through a version of the strong Prikry lemma for this forcing, as it will allow us to show that \kappa^{+} is preserved in the extension. Before doing this, we need to figure out what it means to take an n-step extension of the stem of a condition. So, let p=\langle p_n : n<\omega\rangle\in \mathbb{P} with p_m=\langle a_m, A_m, f_m\rangle for m\geq \ell(p). For \langle \nu_{\ell(p)},\ldots,\nu_m\rangle\in\prod_{k=\ell(p)}^m A_k, we define the m+1-step extension of the stem of p by  \langle \nu_{\ell(p)},\ldots,\nu_m\rangle to be the condition p\frown^*\langle \nu_{\ell(p)},\ldots,\nu_m\rangle defined as

\langle p_n : n<\ell(p)\rangle \frown \langle p_n\frown^* \nu_n : \ell(p)\leq n\leq m\rangle \frown \langle p_n : m<n<\omega \rangle.

As before, p_n=\langle a_n,A_n,f_n\rangle \frown^*\nu_n= f_{n}\cup \{\langle \beta, \pi_{\max_{a_n}\beta}(\nu_n)\rangle : \beta\in a_n\}. From our prior discussion, we’re basically replacing the first m+1 terms of p past the stem with the minimal extensions (in the Q_n order) by Cohen conditions agreeing with \nu_n. With this in hand we have an idea of what the strong Prikry lemma should state:

Lemma: Let p\in \mathbb{P}, and let D\subseteq \mathbb{P} be dense open in (\mathbb{P}\leq). Then there is some p^*\leq^* p and n_p<\omega such that every n_p-step extension of p^* lands in D. That is, for every \langle \nu_0,\ldots, \nu_{n_p-1}\rangle\in \prod_{m=\ell(p)}^{\ell(p)+n_p-1}A_m, we have that p^*\frown^*\langle n_0,\ldots, \nu_{n_p-1}\rangle\in D.

The proof of this lemma is rather technical, so I’ll hold off on the proof of this until the next entry. For now, the plan is to use the above lemma to finish showing that no new bounded subsets of \kappa are added, and that \kappa^+ is preserved in the extension.

Lemma: (\mathbb{P},\leq,\leq^*) satisfies the Prikry condition. That is, for every condition p\in \mathbb{P}, and every sentence \phi in the forcing language$, there is some q\leq^* p such that q decides \phi.

Proof: Let p\in \mathbb{P} be a condition and \phi a sentence in the forcing language, and let D_{\phi}=\{ q\in\mathbb{P} : q\|\phi\}. Using the strong Prikry lemma, we can find a p^*\leq^* p and n_p<\omega be such that every n_p-step extension of p^* lands in D_\phi. Let p^*=\langle p_n^* : n<\omega\rangle with p_n^*=\langle a_n, A_n, f_n\rangle for the sake of simplifying notation. Our goal is to show that we can refine p^* to a further direct extension q that lands in D_\phi. In order to do this, we will simply refine the measure one sets in the tail of p^* so that it only allows for n_p step extensions that all decide \phi in the same way. But then, given any n_p-1-step extension of q, we see that all non-direct extensions will decide \phi in the same way. So, each n_p-1-step extension will decide \phi in the same way, and we can continue going backwards until we see that q itself actually decided \phi. For the sake of simplicity we will assume that n_p=2, since this case is demonstrative of how the refining process works, and the indexing gets way too annoying for arbitrary n.

So by our assumption, we have for any pair of ordinals \nu_{0},\nu_1 with \nu_0\in A_{\ell(p)} and \nu_1\in A_{\ell(p)+1}, that p^*\frown^*\langle \nu_0,\nu_1\rangle\in D_\phi. Fix a faithful enumeration \{\nu_\eta : \eta<\kappa_{\ell(p)}\} of A_{\ell(p)}, and for each \eta<\kappa_n, set

A_{\eta}^0=\{\nu\in A_{\ell(p)+1} : p^*\frown^*\langle \nu_\eta,\nu\rangle\Vdash\phi\}, and

  A_{\eta}^1=\{\nu\in A_{\ell(p)+1} : p^*\frown^*\langle\nu_\eta,\nu\rangle\Vdash\neg\phi\}.

Let A^*_\eta denote whichever one of the above sets ends up in E_{\ell(p)+1,\max(a_{\ell(p)+1})}. Next, let  A^*_{\ell(p),0}=\{\nu\in A_{\ell(p)}: A^*_\eta=A^0_\eta\}, and A^*_{\ell(p),1}=\{\nu\in A_{\ell(p)}: A^*_\eta=A^1_\eta\}. Since one of these above sets has measure one, let A^*_n be whichever one of those two sets ending up in E_{\ell(p),\max(a_{\ell(p)})}. Finally, set

A^*_{\ell(p)+1}=\bigcap \{A^*_\eta :\nu_\eta\in A^*_{\ell(p)}\} .

Noting that shrinking the measure one sets in our Q_{n0} conditions doesn’t change the fact that they’re still Q_{n0} condition, let q denote the condition resulting from starting with p^*, and replacing p^*_n=\langle a_n,A_n,f_n\rangle with \langle a_n, A^*_n, F_n\rangle for n=\ell(p),\ell(p+1). Then by construction we have, for every \nu_0\in A^*_{\ell(p)} and every \nu_1\in A^*_{\ell(p)+1}, that q\frown^*\langle \nu_0,\nu_1\rangle decides \phi in the same way. Hence, we see that our reasoning from earlier allows us to conclude that every non-direct extension of q decides \phi in the same way and thus so does q. This completes the proof.

As usual, now that we’ve seen that \mathbb{P} satisfies the Prikry condition, we can implement the closure of the tail under the direct extension ordering and show that forcing with (\mathbb{P}, \leq) adds no new bounded subsets of \kappa.

Lemma: Forcing with (\mathbb{P}, \leq) adds no new bounded subsets of \kappa

Proof: We’re going to show that any bounded subset of \kappa in the extension is completely decided by one condition, so let \delta<\kappa, \dot{B} be a name, and p\in \mathbb{P} be such that \kappa_{\ell(p)}>\delta and p\Vdash\dot{B}\subseteq\check{\delta}. We will inductively construct a \leq^*-decreasing sequence of conditions sequence \langle p^\alpha : \alpha<\delta\rangle as follows:

  1. Let p_0=p;
  2. At successor stages, assume that p^\alpha has been defined, and use the Prikry condition to find p^{\alpha+1}\leq^*p^\alpha be such that p^{\alpha+1}\|\check{\alpha}\in\dot{B}.
  3. At limit stages, assume that p^\beta has been defined for each \beta<\alpha<\delta, and define p^\alpha=\langle p^\alpha_n : n<\omega\rangle in the obvious way. For n<\ell(p), we simply set p^\alpha_n=\bigcup_{\beta<\alpha}p^\beta_n and note that p^\alpha_n\in Q_{n1} since each it still had size at most \kappa. For n>\ell(p), set f^\alpha_n=\bigcup_{\beta<\alpha}f^\beta_n. Now set a^\alpha_n=\bigcup_{\beta<\alpha}a^\beta_n\cup\{\rho\} where \rho is above the domain of f^\alpha_n and \leq_{E_n}-dominates every member of \bigcup_{\beta<\alpha}a^\beta_n (use the fact that \kappa_{\ell(p)}>\delta combined with Lemma 1). Finally set B^\alpha_n=\bigcap_{\beta<\alpha} \pi_{\rho,\max(a^\beta_n)}^{-1}{}''A^\beta_n using the fact that E_{n\rho} is \kappa_n complete for n>\ell(p), and use Lemma 2 and Lemma 3 (and the fact that \delta<\kappa_{\ell(p)}) to shrink B^\alpha_n to A^\alpha_n satisfying the conclusions of those lemmas for the collection a^\alpha_n which has size <\kappa_n by construction. Finally set p^\alpha_n=\langle a^\alpha_n,A^\alpha_n,f^\alpha_n\rangle.

Note that, by construction, we have that p^\alpha \| \check{\beta}\in\dot{B} for each \beta<\alpha at stage \alpha<\delta. Now that we have our desired sequence, we can use the limit stage construction on \langle p^\alpha : \alpha<\delta\rangle to construct a condition p^\delta that is a \leq^* lower bound for the sequence. But by construction, we have that, for all \alpha<\delta,  p^\delta\|\check{\alpha}\in\dot{B}. Hence \dot{B} is completely decided by a single condition, and so B\in V. Since \delta<\kappa and \dot{B} were arbitrary, it follows that forcing with (\mathbb{P},\leq) adds no new bounded subsets of \kappa.

All we have left to do is use the strong Prikry lemma to show that \kappa^+ is preserved in the extension. One thing to note is that the argument actually generalizes to other Prikry forcings that satisfy the strong Prikry condition. In fact, all of the previous lemmas depended on the fact that the strong Prikry lemma was satisfied, and the tail is sufficiently closed under the direct extension ordering. In other words, once we have the strong Prikry lemma and the \kappa^{++} chain condition for a Prikry forcing over \kappa, we’ve successfully shown that cardinals are preserved in the extension, and no new bounded subsets of \kappa are added. This observation allows for replacing the (\kappa_n,\lambda+1) extenders that we’re using with shorter ones.

So why try to use shorter extenders? On one hand, we would weaken the consistency strength of the assumptions used, but that’s not a huge deal for us. The problem with using long extenders is that it prevents us from collapsing \kappa to any sufficiently small cardinal such as \aleph_{\omega}. What we would like to do is weave collapses into our Prikry forcing in order to simultaneously blow up the power of \kappa while collapsing it to \aleph_{\omega}. This turns out to be difficult since Prikry sequences eventually end up in the measures we have lying around (this follows form standard density arguments) and are therefore form a collection of indiscernibles for V. So in order to collapse \kappa, we want our collapses to miss the indiscernibles, which doesn’t allow for much collapsing when using long extenders. So, if we shorten the length of the extenders, we will actually shrink the intervals in which indiscernibles appear. If we can sufficiently shrink these intervals, we can collapse \kappa down as far as \aleph_\omega. Anyway, we now show that forcing with (\mathbb{P},\leq) preserves \kappa^+.

Lemma : Forcing with (\mathbb{P},\leq) preserves \kappa^+.

Proof : Suppose otherwise, and note that in the extension, cf((\kappa^+)^V) will be below \kappa. Work in V, and let \delta<\kappa, \dot{g} be a name, and p\in \mathbb{P} be a condition with \kappa_{\ell(p)}>\delta and such that p\Vdash\dot{g} : \check{\delta}\to\check{\kappa^+} has unbounded range. We will use the strong Prikry lemma, along with the fact that |[\kappa_n]^{<\omega}|=\kappa_n  to manufacture a bound on the range of g. For each \tau<\delta, let

D_\tau=\{ q\leq p : (\exists \alpha<\kappa^+) (q\Vdash \dot{g}(\check{\tau})=\check{\alpha})\}.

We now use the strong Prikry lemma to inductively define a \leq^*-decreasing  sequence \langle p^\tau : \alpha<\delta\rangle as follows:

  1. Set p^0=p;
  2. At successor stages, assume that p^\tau has been defined and use the strong Prikry lemma to find p^{\tau+1}\leq^*p^\tau and n_\tau such that every n_\tau-step extension of the stem of p^{\tau}+1 ends up in D_\tau.
  3. At limit stages, use the construction from the previous lemma to obtain a \leq^*  lower bound p^\tau for \langle p^\eta : \eta<\tau\rangle.

Let p^\delta be a \leq^* lower bound for the sequence langle p^\tau : \tau<\delta\rangle, and note that for each \tau<\delta, there is some n_\tau<\omega such that every n_\tau-step extension of the stem of p^\delta decides the value of \dot{g}(\check{\tau}). For each \tau<\delta, define

\alpha_\tau=\sup\{\alpha_a : (\exists a\in\prod_{m=\ell(p)}^{\ell(p) + n_\tau-1} A_m)(p^\delta\frown^* a\Vdash \dot{g}(\check{\tau})=\check{\alpha_a})\}.

Since |[\kappa_n]^{<\omega}|=\kappa_n, we see that the above supremum is over at most \kappa_{\ell(p)+n_\tau-1} ordinals \alpha_a. Thus, since \kappa^+ is regular in the ground model, \alpha_\tau< \kappa^+, and thus p^\delta\Vdash \dot{g}(\check{\tau})<\check{\alpha_\tau}<\check{\kappa^+}. Finally, set \alpha=\sup_{\tau<\delta}\alpha_\tau, and note that we have by construction

p^\delta \Vdash (\forall\tau<\delta)(\dot{g}(\check{\tau})<\check{\alpha}).

But p^\delta \leq p and p forces that the range of g is unbounded in \kappa, giving us a contradiction.

We are now a few density arguments away from proving the following:

Theorem (Gitik) : Assume GCH holds in V, and let \langle \kappa_n : n\in \omega\rangle be an increasing sequence of \lambda-strong cardinals with \sup_{n<\omega}\kappa_n=\kappa and \lambda\geq \kappa^{++}, and let (\mathbb{P},\leq) be the corresponding long extender forcing. If G\subseteq \mathbb{P} is V-generic for (\mathbb{P}, \leq), then the following hold in V[G]:

  1. No New bounded subsets of \kappa are added;
  2. All cardinals are preserved;
  3. \kappa^{\omega}\geq\lambda;
  4. In particular, since GCH holds in the ground model, we have that V[G]\models GCH first fails at \kappa.

Proof: We’ve already shown 1 and 2, and 4 will follow immediately once we prove 3. So, let G\subseteq \mathbb{P} be V-generic for (\mathbb{P},\leq). Let n<\omega, and define a function F_n:\lambda \to \kappa_n by setting F_n(\alpha)=\nu if and only if there is some p\in G with \ell(p)>n and such that p_n(\alpha)=\nu. By genericity, F_n is defined on all of \lambda. Using these functions, we define for each \alpha<\lambda the sequence t_\alpha=\langle F_n(\alpha) : n<\omega\rangle, noting that t_\alpha\in\prod_{n<\omega}\kappa_n for each \alpha<\lambda. Now, it might turn out that a good chunk of these sequences come from the ground model, since they could be that they are determined completely by a single condition. However, the following claim will give us that the set \{ t_\alpha : \alpha<\lambda\} has size \lambda:

Claim: If \beta<\lambda, then there is some \alpha with \beta<\alpha<\lambda such that t_\gamma<_{Fin}t_\alpha for each \gamma<\alpha.

Proof (of claim): What we want to show is that for a large enough \alpha>\beta, such that for every \gamma<\alpha, we can find a condition r\in G such that, for every n\geq \ell(r) where r_n=\langle c_n,C_n,h_n\rangle, both \alpha and \gamma are in c_n. If we can do this it follows that for each n\geq \ell(r), we have \pi_{\max(c_n)\alpha}(\nu)>\pi_{\max(c_n)\gamma}(\nu). Thus, it follows that for all n>\ell(r), F_n(\gamma)<F_n(\alpha) by definition of the ordering.

So, we work in V and let p=\langle p_n : n<\omega\rangle\in G with p_n=\langle a_n, B_n, f_n\rangle for each n\geq\ell(p). We will attempt to extend p to a condition as above. We begin by picking some  \alpha \in \lambda \setminus (\sup(\bigcup_{n<\omega}dom(f_n)) \cup (\bigcup_{n\geq\ell(p)} a_n)\cup \beta), and noting that we can generically extend p to a condition q\in G with \alpha\in b_n for each n\geq\ell(q) where q_n=\langle a_n,A_n,g_n\rangle. Do the same to extend q to r\in G with \alpha replaced by \gamma to get the desired condition. This completes the proof of the lemma.

The only thing left to prove is the strong Prikry lemma, and we will have proved that the long extender forcing does what it was supposed to do. As I mentioned earlier, the proof is rather technical, so I will save it for the next entry by itself since I should work through it fully anyway.


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