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Working in $\sf ZFC$, is it provable, or at least consistent (say, over $L$), that you have $\aleph_1$ forcings, $\Bbb P_\alpha$ such that:

  1. $\Bbb P_\alpha$ is c.c.c.
  2. $\Bbb P_\alpha$ adds a real which determines the generic.
  3. For every countable $A\subseteq\omega_1$, and $\alpha\notin A$ the finite support product $\prod_{\beta\in A}\Bbb P_\beta$ does not add a $V$-generic real for $\Bbb P_\alpha$?
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    $\begingroup$ What about the following: First note that by Judah, Shelah, if $CH$ holds, then there exists a $c.c.c.$ forcing adding a minimal real. Now let $\mathbb{P}_\alpha$ be the finite support iteration of length $\alpha+1$, where at each step we force with Judah-Shelah forcing. Parts (1) and (2) are clearly satisfied, and for part (3), consider the $\alpha$-th real added by $\mathbb{P}_\alpha$. It is not produced by the finite support product of $\prod_{\beta<\alpha}\mathbb{P}_\beta.$ $\endgroup$
    – Mohammad Golshani
    Commented Oct 13, 2015 at 13:10
  • $\begingroup$ This might work, but will be useless for the thing I want to do. I'll edit the third condition when I am next to a proper keyboard. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 13, 2015 at 13:24
  • $\begingroup$ Asaf, your conditions don't seem to require that the generic filter for $\mathbb{P}_\alpha$ is determined by a real, but is that what you had in mind? For example, we could make each $\mathbb{P}_\alpha$ to be adding a Cohen real (in order to satisfy 2) and also forcing over a Suslin tree $T_\alpha$, where the Suslin trees are independent in the sense of 3. $\endgroup$
    – Joel David Hamkins
    Commented Oct 13, 2015 at 13:30
  • $\begingroup$ @Joel: You're absolutely right. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 13, 2015 at 16:30
  • $\begingroup$ @Mohammad: I changed it so your suggestion doesn't work anymore. (And that you, and Joel, for the helpful suggestions!) $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 13, 2015 at 16:30

1 Answer 1

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First add $\omega_1$ Cohen reals, then partition this set of Cohen reals into $\omega_1$ disjoint sets $A_i$ each of size $\omega_1$. Let $P_i$ be a sigma centered forcing whose generic is a real coding a meager set covering $A_i$. Then, it is easy to check that the family $\{P_i : i < \omega_1\}$ is as required.

Some details: Let $p \in Q_i$ iff $p = (F, \overline{n} = \langle n_k : k \leq N \rangle, \overline{\sigma} = \langle \sigma_k : k < N \rangle)$ where $F$ is a finite subset of $A_i$, $n_0 = 0, n_k \in \omega$ are increasing and $\sigma_k \in 2^{[n_k, n_{k+1})}$, $q \leq p$ iff $F_p \subseteq F_q$, $\overline{\sigma}_q, \overline{n}_q$ extend $\overline{\sigma}_p, \overline{n}_p$, and for every $x \in F_p$, every new string $\sigma$ from $\overline{\sigma}_q$, $x$ disagrees with $\sigma$ somewhere. $Q_i$ is sigma centered and adds a real, namely $\bigcup \{\overline{\sigma}_p : p \in G_{Q_i}\}$, coding a meager set covering $A_i$. Let $P_i$ be the complete subalgebra of $Q_i$ generated by this real. Now note that if $y$ is any real in $\prod \{P_i : i \neq i_0\}$, then the meager set coded by $y$ cannot cover $A_{i_0}$ - in fact none of the reals in $A_{i_0}$ is covered. Clauses 1 and 2 are obvious.

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