Given a countable model $M$ of set theory and an atomless, separative partial order $\mathbb{P} \in M$, can we construct (in the real universe) $2^\omega$ many pairwise mutually $\mathbb{P}$-generic filters $\{ G_r : r \in \mathbb{R} \}$?

If CH holds, then the answer is yes. Recursively on the countable ordinals, we construct models $M_\alpha = M[G_\alpha]$, where $G_\alpha \subseteq \mathbb{P}$ meets all dense sets in $\bigcup_{\beta<\alpha} M_\beta$.

We can also construct $2^\omega$ many distinct generic extensions. We can build (externally) a tree $T \subseteq \mathbb{P}$, where $T = \{ p_s : s \in 2^{<\omega} \}$. Make $T$ such that: (a) If $|s| = n$, then $p_s \in D_n$, where $\{ D_n : n \in \omega \}$ lists the dense subsets of $M$. (b) For $s$ a prefix of $t$, $p_s \geq p_t$. (c) For all $s$, $p_{s0} \perp p_{s1}$. Every real $r$ determines a distinct branch through $T$ and an associated $\mathbb{P}$-generic filter $G_r$. So we have $2^\omega$ many distinct filters, and since the models are countable, a pigeonhole argument shows that we must have $2^\omega$ distinct models. But this argument does not show that the filters are mutually generic.


I think you can handle this by slightly modifying the construction in the last paragraph of your question. In addition to what you did there, enumerate all the dense subsets of $\mathbb P\times\mathbb P$, say as $E_n$ ($n\in\omega$). Do this in such a way that every dense open set occurs infinitely often in the enumeration. Then at the stage of your recursion where you handle nodes $s$ of length $n$, extend your $p_s$'s so that every two of them, considered as a pair, are in $E_n$. That's only finitely much additional work at stage $n$, because there are only $2^n$ $s$'s to take care of. At the end of the construction, if $G$ and $G'$ are two of the generic filters you've produced, corresponding to paths $r$ and $r'$ through $T$, these will be mutually generic because any dense subset of $\mathbb P\times\mathbb P$ occurs as $E_n$ for infinitely many $n$, in particular for some $n$ beyond the level where $r$ and $r'$ branched apart.

  • $\begingroup$ Very nice, thanks. 2 minor remarks. It seems clearer if the $E_n$'s are dense open. Also, I don't see why you need to repeat them infinitely often. For any $n$, $(p_{r \restriction n}, p_{r' \restriction n}) \in E_n \cap G \times G'$, showing mutual genericity because $n$ is arbitrary, right? $\endgroup$ – Monroe Eskew Aug 1 '14 at 21:25
  • $\begingroup$ @MonroeEskew Yes, I intended "dense open" and I actually wrote that in the sentence about infinite repetition, but I neglected to do so when first introducing the $E_n$'s. About your second observation: I think some more work would be needed to avoid the infinite repetition. The problem is that you could have $r\neq r'$ yet $r\upharpoonright n=r'\upharpoonright n$ for a particular $n$. The infinite repetition ensures that $E_n$ will be treated again, for larger values of $n$, where this problem no longer occurs. $\endgroup$ – Andreas Blass Aug 1 '14 at 21:30
  • $\begingroup$ Oh I see, it's only when you have a pair of incompatible elements that you extend to a pair in $E_n$. $\endgroup$ – Monroe Eskew Aug 1 '14 at 21:52

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