Gitik’s Long Extender Forcing – The Definition

This post is a continuation of the previous one.

As usual, none of the results and proofs are due to me, and I’m still following the presentation from Gitik’s handbook chapter, and Spencer Unger’s notes form the Appalachian Set Theory workshop.

In this post, I will only give the definition of the forcing, and discuss what the forcing is trying to accomplish, and how it’s doing that. Since the forcing itself is rather complicated, I will follow Gitik’s approach and define it in pieces.  We begin by fixing n<\omega, and for each \alpha,\beta<\lambda such that \beta\leq_{E_n}\alpha, fix a witnessing function \pi_{\alpha\beta} with \pi_{\alpha\alpha}=id. We now define the nth cell Q_n as follows:

Definition:

Set Q_{n1}=\{ f is a partial function from \lambda to \kappa : |f|\leq \lambda\}, and say that f\leq_{Q_{n1}}g if f\supseteq g.

Define Q_{n0} to be the collection of conditions of the form \langle a, A, f\rangle such that

  1. f\in Q_{n1};
  2. a\in [\lambda]^{<\kappa_n}, such that a contains a \leq_{E_n}-maximal element \max(a), and a\cap dom(f)=\emptyset;
  3. A\in E_{n\max(a)};
  4. For \alpha>\beta if \alpha,\beta\in a, then for every \nu\in A, \pi_{\max(a)\alpha}(\nu)>\pi_{\max(a)\beta}(\nu);
  5. For \alpha,\beta,\gamma\in a, if \alpha\geq_{E_n}\beta\geq_{E_n}\gamma, then for every \nu\in\pi_{\max(a)\alpha}{}''A, \pi_{\alpha\gamma}(\nu)=\pi_{\beta\gamma}(\pi_{\alpha\beta}(\nu)).

We say that \langle a,A,f\rangle \leq_{Q_{n0}}\langle b,B,g\rangle if

  1. f\supseteq g;
  2. a\supseteq b;
  3. \pi_{\max(a)\max(b)}{}''A\supseteq B.

Set Q_n=Q_{n1}\cup Q_{n0}, and define the direct extension ordering by \leq_n^*=\leq_{Q_{n1}}\cup\leq_{Q_{n0}}. For p,q\in Q_n, we say that p\leq_n q if either p\leq^* q or q=\langle a,A,f\rangle and p\in Q_{n1} is such that:

  1. p\supseteq f;
  2. dom(p)\supseteq a;
  3. p(\max(a))\in A;
  4. For every \beta\in a, p(\beta)=\pi_{\max(a)\beta}(p(\max(a)).

We may think of the ordering (Q_n, \leq_n,\leq^*_n) as the nth coordinate of the forcing that we actually want to define when we string all of these blocks together. We should note that, since every condition bottoms out in a Cohen condition, the forcing (Q_n,\leq_n) is actually equivalent to Cohen forcing. Outside of that though, this forcing seems like it will do what we want it to (given the discussion in the previous post). The a,A portions of our conditions give us a set of ordinals of size <\kappa_n that have a maximal element in the extender ordering, and cohere well with respect to it, as witnessed by A\in E_{n\max(a)}. This will allow us to perform the diagonal Prikry forcing construction at the top elements of these sequences (when we glue everything together) in a nice way. In particular, the Cohen conditions give us a nice way of assigning each element of the sequence a to an element of its associated ultrafilter that agrees with the extender ordering and witnessing projections. Finally, rolling up the Cohen conditions along with the a,A parts allows us a way of “predicting” extensions of a Q_{n0} condition to a Cohen condition. We will see what is meant by this in the proof of the upcoming lemma.

Of course, for any Cohen condition extending a condition from Q_{n0}, its values on the sequence a are fully determined by the value at \max(a). It turns out that each block is actually a Prikry forcing.

Lemma: (Q_n,\leq_n^*,\leq_n) satisfies the Prikry condition. That is, for condition p\in Q_n and any sentence \phi in the forcing language, there exists some q\leq_n^* p deciding \phi.

Proof : We begin by noting that it suffices to consider extensions of the form p=\langle a, A, f\rangle since the non-direct and direct extension orderings are the same if p is a Cohen condition. The basic idea here is that we will diagonalize against all non-direct extensions deciding \phi, and construct a direct extension that predicts these non-direct extensions. In order to do this, we look at a sort of minimal extension of a Q_{n0} condition to a Cohen condition. We first fix a condition p=\langle a,A,f\rangle and a sentence \phi in the forcing language. Now, for a given \nu\in A, we set

\langle a,A,f\rangle\frown\nu=f\cup\{\langle \beta, \pi_{\max(a)\beta}(\nu) \rangle : \beta\in a \}.

Basically, we take the minimal extension of \langle a,A,f\rangle by a Cohen condition with values decided by \nu\in A. We will now inductively build a \leq_n^* decreasing chain of conditions from Q_{n0} of length \kappa_n. At each stage, we will find some non-direct extension deciding \phi, and alter the Cohen part of our previous condition so that it predicts this non-direct extension while leaving the first two coordinates of our condition fixed.

We begin by fixing an increasing enumeration \langle \nu_\eta : \eta<\kappa_n\rangle of A, and set p_0=\langle a,A,f_0\rangle. We first describe what to do at the successor stage of this construction. So suppose that p_\eta=\langle a, A, f_\eta \rangle has been defined for some \eta<\kappa_n. As we mentioned earlier, we will only focus on altering the Cohen parts of our condition, while keeping the a and A parts fixed. Let q_{\eta}\leq_n p_\eta\frown \nu_\eta be a non-direct extension of p_\eta deciding \phi, and set f_{\eta+1}=q_\eta\upharpoonleft(dom(q_\eta\setminus a). Define p_{\eta+1}=\langle a, A, f_{\eta+1}\rangle, and note that p_{\eta+1}\frown \nu_\eta=q_\eta. Furthermore, note by definition that f_{\eta+1}\supseteq f_\eta since q_\eta\supseteq f_{\eta}, and dom(f_\eta)\cap a=\emptyset.

Next, we describe how to proceed at limit stages. So, let \gamma<\kappa_n be limit and suppose that p_\eta has been defined for all eta<\gamma. Then, we simply set f_{\gamma}=\cup_{\eta<\gamma}f_{\eta}, and note that |dom(f_\gamma)|\leq \kappa. Then we define p_\gamma=\langle a, A, f_{\eta} \rangle. Now we have a sequence \langle p_\eta : \eta<\kappa_n \rangle with p_\eta=\langle a,A,f_\eta\rangle\in Q_{n0} and such that f_{\eta}\supseteq f_{\eta'} for \eta'\leq\eta<\kappa_n. Now define g=\cup_{\eta<\kappa_n}f_\eta, and note that \langle a, A, g\rangle is still a condition in Q_{n0}. Furthermore, we have by construction that \langle a, A, g\rangle\frown \nu_\eta=q_\eta and hence \langle a,A,g\rangle \frown\nu decides \phi for each \nu\in A.

The only thing left to do is shrink down A so that every extension by \nu\in A decides \phi in the same way. So, we set

A_0=\{\nu\in A : \langle a,A,g\rangle\frown\nu\Vdash\phi\} and A_1=\{\nu\in A : \langle a,A,g\rangle\frown\nu\Vdash\neg\phi\}.

Since A\in E_{n\max(a)}, it follows that one of these above sets is in E_{n\max(a)}, so assume without loss of generality that A_0\in E_{n\max(a)}. Then, every non-direct extension of \langle a,A_0,g\rangle is below some \langle a, A_0,g\rangle\frown\nu for some \nu\in A_0 and hence forces \phi. Thus, every non-direct extension of \langle a,A_0,g\rangle forces \phi and hence so must \langle a,A_0,g\rangle. This completes the proof.

Definition : We now define the forcing (\mathbb{P}, \leq,\leq^*). Conditions are of the form p=\langle p_n : n\in \omega\rangle such that:

  1. Each p_n is a condition in Q_n;
  2. There is some \ell(p)<\omega such that p_n\in Q_{n1} for each n<\ell(p), and p_n\in Q_{n0} for each n\geq\ell(p);
  3. If n\geq\ell(p), and p_n=\langle a_n,A_n,f_n\rangle, then m\geq n\implies a_m\supseteq a_n.

We say that p\leq q if p_n\leq_n q_n for each n<\omega, and that p\leq^* q if p_n\leq^*_n q_n for each n<\omega.

In the next entry, I want to show that (\mathbb{P},\leq,\leq^*) satisfies all of the necessary properties provided that it satisfies the strong Prikry condition. The proof of the strong Prikry condition is rather technical, but I would like to go through it in the last entry in the sequence of entries on the long extender forcing.

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