There are some exercises which we solved during problem sessions during the day; I will preserve these in the text and can post solutions as well if there is interest.
We start off with Monroe Eskew's lectures on the duality theorem.
Basic facts: This section is meant to be supplemented by my earlier postings about ideal properties. The basic situation is that I is an ideal on Z\subseteq P(X), G is a generic for the forcing P(Z)/I, and j:V\rightarrow M=V^Z/G is the generic ultrapower by G.
Definition: An ideal I has the disjointing property if every antichain A in P(Z)/I has a pairwise disjoint system of representatives.
Exercise: If I is a \kappa-complete, \kappa^+-saturated ideal, then I has the disjointing property. If I is a normal ideal on Z\subseteq P(X), and I is |X|^+-saturated, then I has the disjointing property.
Lemma: Suppose A\in I^+ and I has the disjointing property, and [A]\Vdash \dot{\tau} \in V^Z/\dot{G}. Then there is f:Z\rightarrow V such that A\Vdash \tau = [f]_G.
Proof: Exercise.
Lemma: Suppose I is a countably closed ideal with the disjointing property. Let \kappa=\mathrm{crit}(j). Then I is precipitous, and the generic ultrapower M is closed under \kappa-sequences from V[G].
Proof: Precipitousness follows from the combinatorial characterization of precipitousness. Suppose that \Vdash \langle \tau_\alpha:\alpha<\kappa\rangle \subseteq V^Z/G. By the previous lemma, for all \alpha<\kappa, there is f_\alpha:Z\rightarrow V in V such that \Vdash \tau_\alpha=[f_\alpha]_G and k:Z\rightarrow \mathrm{ON} in V such that \Vdash \kappa=[k]_G. Finally, take f:Z\rightarrow V given by f(z)=\langle f_\alpha(z):\alpha<k(z)\rangle. In M, [f]_G=\langle \tau_\alpha:\alpha<\kappa\rangle. \Box
Definition: I is (\lambda,\kappa)-presaturated if for any sequence of antichains \langle A_\alpha:\alpha<\gamma<\lambda\rangle, the set \{X:\forall\alpha<\gamma\, |\{a\in A_\alpha:a\cap x\in I^+\}|<\kappa\} is dense.
Exercise: (\lambda^+, \kappa^+)-presaturation and \kappa-completeness imply that the generic ultrapower M is closed under \lambda-sequences from V[G]. Also show this for normal and (\lambda^+,|X|^+)-presaturated ideals.
We will use the notation \mathcal{B}(\mathbb{P}) to denote Boolean completion of a poset. As we have seen in previous posts, this is very handy when dealing with generic ultrapowers.
Exercise: \mathbb{P} and \mathbb{Q} are separative posets. The following are equivalent:
- \mathcal{B}(\mathbb{P})=\mathcal{B}(\mathbb{Q}).
- There is a \mathbb{P}-name \dot{h} and a \mathbb{Q}-name \dot{g} so that \Vdash_\mathbb{P} \dot{h} \textrm{ is }\mathbb{Q}-\textrm{generic over }V, \Vdash_\mathbb{Q} \dot{g} \textrm{ is }\mathbb{P}-\textrm{generic over }V, and\Vdash_\mathbb{P} \dot{g}^{\dot{h}^\dot{G}}=\dot{G} \textrm{ and } \Vdash_\mathbb{Q} \dot{h}^{\dot{g}^\dot{H}}=\dot{H}.
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