Ccc forcings and measurable cardinals

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 $$D$$.

• 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 $\vert \mathbb{P} \vert \ge \kappa$. – Not Mike May 8 at 9:23
• @NotMike also true for posets like $Add(\omega, \kappa)$ or $Random(\omega, \kappa)$ – Otto May 12 at 12:30
• @Otto indeed, it's true for any c.c.c. poset of size $\kappa$ which adds sufficiently many splitting reals. – Not Mike May 12 at 17:27
• It also seems to be true for any $\kappa$-length finite support iteration of c.c.c forcings of size $<\kappa$. – Otto May 15 at 16:10
• @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. – Ashutosh May 16 at 14:23