1

Suppose \({\mathbb{P}}\) is a poset, \(\phi(\overline{x})\) is a formula with free variables as shown and \(\overline{\tau}\) are \({\mathbb{P}}\)-names. Show that TFAE

1. \(p\Vdash^* \phi(\overline{\tau})\),
2. \(\left(\forall q\leq p\right)\left(q\Vdash^* \phi(\overline{\tau})\right)\),
3. \(\left\{q\leq p: q\Vdash^*\phi(\overline{\tau})\right\}\) is dense below \(p\).

It is immediate that (b) implies (a) and (c). We induct on the complexity of \(\phi\) to show the other implications.

Case \(\phi \equiv x_1 = x_2\).

\((a)\Rightarrow(b)\). We just need to observe that when a set is dense below \(p\), for every \(q\leq p\), it is also dense below \(q\).

\((c)\Rightarrow(a)\). To show \(p\Vdash^* \tau_1 = \tau_2\), let \((\pi_1, s_1)\in \tau_1\), we want \[ \left\{r\leq p: r\leq s_1 \implies \left(\exists (\pi_2, s_2)\in \tau_2\right)\left(r\leq s_2 \land r\Vdash^*\pi_1 = \pi_2\right)\right\} \] to be dense below \(p\). Let \(p' \leq p\) and by (c) let \(q \leq p'\) with \(q\Vdash^* \tau_1 = \tau_2\), then there exists \(r \leq q \leq p' \leq p\) in the set as required. The symmetric case is similar.

Case \(\phi \equiv x_1\in x_2\).

Same argument as \(\phi \equiv x_1 = x_2\).

Case \(\phi(\overline{x},\overline{y}) \equiv \psi_1(\overline{x}) \land \psi_2(\overline{y})\).

Trivially follows from induction hypothesis.

Case \(\phi(\overline{x}) \equiv \lnot\psi(\overline{x})\).

\((a)\Rightarrow(c)\). Assume (a), then \(\lnot\left(\exists q\leq p\right)\left(q\Vdash^*\psi(\overline{\tau})\right)\). So for all \(r\leq p\), \(\lnot\left(\exists q\leq r\right)\left(q\Vdash^*\psi(\overline{\tau})\right)\) and (c) follows.

\((c)\Rightarrow(a)\). Assume (a) fails and we show (c) fails, let \(q\leq p\) witness \(q\Vdash^*\psi(\overline{\tau})\), by induction hypothesis \(\left(\forall r\leq q\right)\left(r\Vdash^*\psi(\overline{\tau})\right)\) so the set \(\left\{q'\leq p: q'\Vdash^*\lnot\psi(\overline{\tau})\right\}\) is empty below \(q\) and (c) fails.

Case \(\phi(\overline{x}) \equiv \left(\exists y\right)\left(\psi(y,\overline{x})\right)\).

Same argument as \(\phi \equiv x_1 = x_2\).

2

Suppose \(\mathbf{M}\) is a ctn of ZFC and \({\mathbb{P}}\) is a poset in \(\mathbf{M}\) in which every condition has two incompatible extensions. Show that there exists \(\left\langle\left\langle M_n, G_n\right\rangle: n<\omega\right\rangle\) such that the following hold.

1. \(\mathbf{M}_0 = \mathbf{M}\)
2. For every \(n<\omega\), \(G_n\) is a \({\mathbb{P}}\)-generic filter over \(\mathbf{M}_n\) and \(\mathbf{M}_{n+1} = \mathbf{M}_n[G_n]\).
3. There is no ctm \(\mathbf{N}\) of ZFC such that \(o(\mathbf{N}) = o(\mathbf{M}), \mathbf{M}\subseteq\mathbf{N}\) and \(\left\langle G_n:n<\omega\right\rangle\in\mathbf{N}\).

Let \(z\in 2^\omega\) be a real that encodes an ordinal above \(o(\mathbf{M})\). The idea is to make \(G_n\) encode the \(n\)-th bit of \(z\), then no ctm \(\mathbf{N}\) that satisfies (c) can exist since from \(\left\langle G_n:n<\omega\right\rangle\) we can recover \(z\) and the ordinal that \(z\) encodes.

Let \(p_0,p_1\in{\mathbb{P}}\) be two incompatible extensions of \(1_{\mathbb{P}}\). Let \(\mathbf{M}_0 = \mathbf{M}\) and suppose \(\mathbf{M}_n\) has already been defined, if \(z(n) = 0\), we repeat the construction of Lemma 16.5 but starting from \(p_0\), this gives us a generic filter \(G_n\) which contains \(p_0\), proceed similarly with \(p_1\) if \(z(n) = 1\). Define \(\mathbf{M}_{n+1} = \mathbf{M}_n[G_n]\).

3

Suppose \(\mathbf{M}\) is a ctm of ZFC, \({\mathbb{P}}\) a poset in \(\mathbf{M}\) and \(G,H\) are \({\mathbb{P}}\)-generic filters over \(\mathbf{M}\). Show that either \(G = H\) or there exist \(p\in G\) and \(q\in H\) such that \(p\perp q\).

Assume \(G\ne H\), without loss of generality let \(p \in G\setminus H\). By choice in \(\mathbf{M}\) let \(A\in\mathbf{M}\) be a maximal antichain containing \(p\). Since \(H\) is generic \(H\cap A\) is nonempty, let \(q\in H\cap A\) and note that \(q \ne p\), the result follows.

4

Suppose \(\mathbf{M}\) is a ctm of ZFC and \(\left(\kappa\text{ is a cardinal}\right)^\mathbf{M}\). Show the following

1. \(\operatorname{Fn}(\omega,\kappa)\in\mathbf{M}\).

As \(\left\lvert X\right\rvert<\omega\) is absolute, and \[\operatorname{Fn}(X,Y) = \left\{p:\left\lvert p\right\rvert<\omega \land p\text{ is a function} \land \operatorname{dom}(p)\subseteq X\land \operatorname{range}(p)\subseteq Y\right\},\] then for any \(X,Y\in\mathbf{M}\) (including \(\omega, \kappa\)), \(\operatorname{Fn}(X,Y) = \operatorname{Fn}(X,Y)^\mathbf{M}\in \mathbf{M}\).

2. Let \(G\) be a \(\operatorname{Fn}(\omega,\kappa)\)-generic filter over \(\mathbf{M}\). Show that \(\left(\kappa\text{ is countable}\right)^{\mathbf{M}[G]}\).

Since \(G\in \mathbf{M}[G]\), let \(f = \bigcup G \in \mathbf{M}[G]\). It is enough to show that \(f:\omega\to\kappa\) is onto. As \(G\) is a filter, \(f\) is a function with \(\operatorname{dom}(f)\subseteq \omega\) and \(\operatorname{range}(f)\subseteq \kappa\). To show \(\operatorname{dom}(f) = \omega\) let \(n\in\omega\), consider this set \[D_n = \left\{p\in\operatorname{Fn}(\omega,\kappa):n\in \operatorname{dom}(p)\right\}\] which is dense. By absoluteness \(D_n\in\mathbf{M}\) also, so \(G\cap D_n \ne 0\) and \(n\in\operatorname{dom}(f)\).

To see that \(\operatorname{range}(f) = \kappa\) let \(\lambda\in\kappa\) and similarly consider \[E_\lambda = \left\{p\in\operatorname{Fn}(\omega,\kappa):\lambda\in\operatorname{range}(p)\right\}\] which is also dense (as \(\omega\) is infinite) and in \(\mathbf{M}\) by absoluteness, so \(G\cap E_\lambda \ne 0\) and \(\lambda \in \operatorname{range}(f)\).

5

Let \({\mathbb{P}}\) be a poset in which every condition has two incompatible extensions. Define \({\mathsf{cc}}({\mathbb{P}})\) to be the least cardinal \(\kappa\) such that every antichain in \({\mathbb{P}}\) has cardinality \(<\kappa\). For \(p\in {\mathbb{P}}\), let \({\mathbb{P}}_{\leq p} = \left\{q\in {\mathbb{P}}: q\leq p\right\}\). Prove the following.

1. \(\left(\forall p,q\in{\mathbb{P}}\right)\left(p\leq q\implies {\mathsf{cc}}({\mathbb{P}}_{\leq p}) \leq {\mathsf{cc}}({\mathbb{P}}_{\leq q})\right)\).

Let \(p,q\in{\mathbb{P}}\) with \(p\leq q\). Since \({\mathbb{P}}_{\leq p} \subseteq {\mathbb{P}}_{\leq q}\) every antichain in \({\mathbb{P}}_{\leq p}\) is also an antichain in \({\mathbb{P}}_{\leq q}\) and the result follows.

2. For every \(p\in{\mathbb{P}}\), there exists \(q\leq p\) such that \[\left(\forall r,s\in{\mathbb{P}}\right)\left[\left(r\leq p\land s\leq q\right)\implies {\mathsf{cc}}({\mathbb{P}}_{\leq r}) = {\mathsf{cc}}({\mathbb{P}}_{\leq s})\right]\]

Fix \(p\in{\mathbb{P}}\) and let \(q_0 = p\). Suppose \(\left(\forall r,s\in {\mathbb{P}}_{\leq q_0}\right)\left({\mathsf{cc}}({\mathbb{P}}_{\leq r}) = {\mathsf{cc}}({\mathbb{P}}_{\leq s})\right)\), then we are done. In the other case let \(r,s\leq q_0\) witness \({\mathsf{cc}}({\mathbb{P}}_{\leq r}) < {\mathsf{cc}}({\mathbb{P}}_{\leq s})\), then define \(q_1 = r\). We can repeat this construction to get \(q_2, q_3,\dots\) and so on but this construction terminates in finitely many stages. Otherwise by part (a) we will get an infinite strictly decreasing sequence of ordinals \({\mathsf{cc}}({\mathbb{P}}_{q_0}) > {\mathsf{cc}}({\mathbb{P}}_{q_1}) > \dots\). Now define \(q\) to be \(q_n\) where this construction stops and the result follows.

3. Show that \({\mathsf{cc}}({\mathbb{P}})\) is a regular cardinal.

Suppose not, let \(\kappa = {\mathsf{cc}}({\mathbb{P}})\) and \(\lambda = \operatorname{cf}(\kappa) < \kappa\). Let \(a(X)\) denote \(\sup\left\{\left\lvert A\right\rvert: A\subseteq X \text{ is an antichain}\right\}\).

Fix \(M\) a maximal antichain in \({\mathbb{P}}\). We first claim that for any regular \(\mu < \kappa\), whenever \(\mu > \left\lvert M\right\rvert\) there is a \(p\in M\) with \({\mathsf{cc}}({\mathbb{P}}_{\leq p}) = \mu\). By minimality of \(\kappa = {\mathsf{cc}}({\mathbb{P}})\) let \(A\) be an antichain with \(\left\lvert A\right\rvert = \mu\). Regularity of \(\mu\) gives us a \(p\in M\) that is compatible with \(\mu\) many members of \(A\). Let \(A_p\) denote everyone in \(A\) compatible with \(p\). For each \(q\in A_p\) choose \(r_q \in {\mathbb{P}}_{\leq p}\cap{\mathbb{P}}_{\leq q}\), then \(\left\{r_q:q\in A_p\right\}\subseteq {\mathbb{P}}_{\leq p}\) is an antichain with size \(\mu\). This together with part (a) gives \(\sup\left\{a({\mathbb{P}}_{\leq p}):p\in M\right\} = \kappa\).

Now let \(\left\langle p_\xi: \xi < \lambda\right\rangle\) be a sequence in \(M\) such that \(a({\mathbb{P}}_{\leq p_\xi})\) is increasing and \(\sup_{\xi<\lambda} a({\mathbb{P}}_{\leq p_\xi}) = \kappa\). At each \(\xi < \lambda\) choose \(A_\xi\) an antichain witnessing \(a({\mathbb{P}}_{\leq p_\xi})\), then \(A = \bigcup_{\xi<\lambda} A_\xi\) is still an antichain in \({\mathbb{P}}\) but \(\left\lvert A\right\rvert = \kappa\) which contradicts \({\mathsf{cc}}({\mathbb{P}})\).