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Suppose that I want to force to add a "single" new subset of $\omega$ and not much else. For example, consider the Cohen forcing consisting of finite partial functions from $\omega$ to 2. The forcing I am interested in is different (in fact not CCC, but Proper), but still the generic set codes a subset of $\omega$.

Question 1: Suppose that $G$ is $P$-generic over $M$, and there is at least one $r\in(\mathscr{P}(\omega)\cap M[G]) \setminus M$. Is there a condition $(*)$ such that

$P$ satisfies $(*)$ if and only if $\forall t\in M[G]\cap \mathscr{P}(\omega) \exists a, b\in M\cap \mathscr{P}(\omega) [ t = (a \cap r) \cup (b\setminus r)] $ ?

Question 2 (Iterability): Consider a model $M_\delta$ resulting from an iteration of forcings $P_\alpha : \alpha<\delta$. Say that each $P_\alpha$ adds a real $r_\alpha$. Then $ \{r_\alpha : \alpha<\delta\} \subseteq \mathscr{P}(\omega)\cap M_\delta$. Is there a model $M_\delta'\subseteq M_\delta$ such that the reals of $M_\delta'$ are just $(\mathscr{P}(\omega)\cap M)\cup \{ (a\cap r_\alpha)\cup (b\setminus r_\beta) : a,b\in \mathscr{P}(\omega)\cap M; \ \alpha, \beta<\delta\} $ ?

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    $\begingroup$ My first guess would be that there is no forcing that satisfies property (*). Or do you have an example with this property? $\endgroup$
    – Goldstern
    Commented Jan 25, 2013 at 9:44
  • $\begingroup$ Hello. Yes you are probably right, it seems I am probably being too hopeful. I don't have an example in mind although I thought I did. $\endgroup$
    – Kiochi
    Commented Jan 27, 2013 at 2:39

1 Answer 1

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There is no forcing that satisfies $(*)$. Let $r$ be a new real, $r \in M[G]\cap \mathcal{P}(\omega) \setminus M$. Take $t = \{n < \omega \mid n + 1\in r\}$, and let $a, b\in M$ such that $t = (a\cap r)\cup (b\setminus r)$.

I will show that it is possible to reconstruct $r$ from $a, b$ and the bit $0\in r$. This implies that $r\in M$.

Let $n < \omega$ and assume that we know whether $n \in r$ or not. We want to check if $n + 1\in r$. Split into four cases:

  1. If $n \in a \cap b$ then $n\in t$ and therefore $n + 1\in r$.

  2. If $n \notin a \cup b$, $n\notin t\implies n+1\notin r$.

  3. If $n \in a\setminus b$ then $n\in t \iff n\in r$ and therefore $n + 1\in r \iff n\in r$.

  4. If $n \in b\setminus a$ then $n\in t \iff n\notin r$ and therefore $n + 1\in r \iff n\notin r$.

So we can reconstruct $r$, bit by bit, in $M$ (up to two possibilities) from $a, b$.

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