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Suppose $\kappa$ is measurable with a witnessing normal ultrafilter $U$ and $P$ is a ccc forcing. Let $\langle p_i: i < \kappa \rangle$ be a sequence of conditions in $P$ such that for every $X \in U$, $\{p_i: i \in X\}$ is predense in $P$. Must there exist an infinite $X \subseteq \kappa$ such that for every infinite $Y \subseteq X$, $\{p_i: i \in Y \}$ is predense in $P$?

$D$ is predense in $P$ means: Every condition in $P$ is compatible with some condition in $D$.

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  • $\begingroup$ This is the case for separative c.c.c. partial orders with greatest element that satisfy $\vert \mathbb{P} \vert < \kappa$ (in which case a sequence of conditions satisfies the required condition, iff, it is $1$ for a positive set of indices.) However, I do not see an immediate way to generalize this to such partial orders which satisfy $\vert \mathbb{P} \vert \ge \kappa$. $\endgroup$ – Not Mike May 8 at 9:23
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    $\begingroup$ @NotMike also true for posets like $Add(\omega, \kappa)$ or $Random(\omega, \kappa)$ $\endgroup$ – Otto May 12 at 12:30
  • $\begingroup$ @Otto indeed, it's true for any c.c.c. poset of size $\kappa$ which adds sufficiently many splitting reals. $\endgroup$ – Not Mike May 12 at 17:27
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    $\begingroup$ It also seems to be true for any $\kappa$-length finite support iteration of c.c.c forcings of size $<\kappa$. $\endgroup$ – Otto May 15 at 16:10
  • $\begingroup$ @Otto I agree and the finite support iteration could be of any length. Also, if we assume, for brevity, that $P$ is a complete ccc boolean algebra, then the answer is positive when we replace "infinite" by "uncountable". The proof is similar to the one for finite support iteration. $\endgroup$ – Ashutosh May 16 at 14:23

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