In this lecture, at around 15-16 minutes in, Velickovic makes the following claim:
Suppose $\mathbb Q$ is proper. Let $\theta > 2^{|\mathbb Q|}$. Let $\mathbb M(\mathbb Q)$ be the partial order consisting of pairs $(s,q)$ such that $s$ is a finite $\in$-chain of countable elementary $M \prec H_\theta$, and $q\in\mathbb Q$ is $M$-generic for all $M \in s$. $(s’,q’) \leq (s,q)$ when $s’ \supseteq s$ and $q’ \leq q$.
Claim: Suppose $\theta^* > 2^{|\mathbb M(\mathbb Q)|}$, $M^* \prec H_{\theta^*}$ is countable, $(s,q) \in M^*$, and $M = H_\theta \cap M^*$, and $q’ \leq q$ is $M$-generic. Then $(s \cup \{M\},q’)$ is $M^*$-generic.
Can someone give a proof of the claim? Thanks.
Let $D \in M^*$ be dense in $\mathbb M(\mathbb Q)$, and let $(s_0,q_0) \leq (s \cup \{ M \}, q’)$ be an arbitrary extension. It suffices to show that $(s_0,q_0)$ is compatible with some $p \in D \cap M^*$.
Let $t_0 = s_0 \cap M^*$. Then $t_0 \in M^*$ is a finite $\in$-chain of elementary submodels of $H_\theta$. Let $E = \{ q \in \mathbb Q :$ either $q$ cannot be extended to a $t_0$-generic condition, or there is $t$ such that $t \supseteq t_0$ and $(t,q) \in D\}$. $E$ is definable from paramters in $M^*$. $E$ is dense in $\mathbb Q$ because if $q$ can be extended to a $t_0$-generic condition $r$, then $(t_0,r) \in \mathbb M(\mathbb Q)$, so we can further extend to some $(t’,r’) \in D$.
Since $q_0$ is $M$-generic, it is compatible with some $q_1 \in E \cap M$. Since $q_0$ is also $t_0$-generic, $q_1$ can be extended to a $t_0$-generic condition, so membership in $E$ is witnessed by the second clause. By the elementarity of $M^*$, there is $t_1 \in M^*$ such that $t_1 \supseteq t_0$ and $(t_1,q_1) \in D$.
Since $t_1 \in M^*$, $s_2 = t_1 \cup s_0$ is a finite $\in$-chain. Let $q_2 \leq q_0,q_1$. Then $q_2$ is $s_2$-generic, so $(s_2,q_2) \in \mathbb M(\mathbb Q)$. $(s_2,q_2) \leq (s_0,q_0),(t_1,q_1)$, and the latter is in $M^* \cap D$.