9
$\begingroup$

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\} $ ?

$\endgroup$
  • 1
    $\begingroup$ My first guess would be that there is no forcing that satisfies property (*). Or do you have an example with this property? $\endgroup$ – Goldstern Jan 25 '13 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 Jan 27 '13 at 2:39
6
$\begingroup$

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 $a, b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.