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Given $\mathbb{B} = \langle B, \wedge, \vee, \neg, 0, 1 \rangle$ an atomless complete Boolean algebra that has a $< \mkern-4mu \kappa$-closed dense subset and is $\kappa^+$-c.c., we define a forcing notion $\mathbb{P} = \langle P, \leq \rangle$, as follows:

  • a condition $p \in P$ is any partial function from $B^-$ into $B^- \times B^-$ (where $B^- = B - \{1\}$) such that

    1. $|dom(p)| < \kappa$,
    2. whenever we have $a, b \in dom(p)$, $a \vee b \neq 1$, $p(a) = (a_0, a_1)$ and $p(b) = (b_0, b_1)$, we must also have $a_0 \wedge b_1 = 0$ and $a_1 \wedge b_0 = 0$,
    3. (optional) for all $a \in dom(p)$, if $p(a) = (a_0, a_1)$ then $a_0 \vee a_1 = a$.

  • $p \leq q$ iff $p$ extends $q$ as a function.

What can we say about the c.c.-ness of $\mathbb{P}$? Also, what additional assumptions on $\mathbb{B}$ and on $\kappa$ are sufficient to make $\mathbb{P}$ $\kappa^+$-c.c.?

EDIT: I try to describe better what the conditions of the forcing entail, and made requirement 3. optional if it helps with answering the questions without that requirement. This is motivated by comments from Andreas Blass.

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  • $\begingroup$ If your Boolean algebra is complete, then it is $\lambda$ closed for any $\lambda$. $\endgroup$
    – Jonathan Schilhan
    Commented Oct 14, 2018 at 18:05
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    $\begingroup$ @Jonathan, incorrect. In fact quite the opposite is true. A complete boolean algebra is never countably closed. This is because we can form a countable maximal antichain by taking joins of parts of any maximal antichain. If $\{ b_n : n < \omega \}$ is such a partition, then let $a_n = \bigvee_{m\geq n} b_m$. Then the sequence of $a_n$'s gives a descending chain with no lower bound. $\endgroup$
    – Monroe Eskew
    Commented Oct 14, 2018 at 19:07
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    $\begingroup$ Can you say a bit about what you want your forcing to do? $\endgroup$
    – Monroe Eskew
    Commented Oct 14, 2018 at 19:21
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    $\begingroup$ Trying to be more specific about my previous comment: Suppose that $a,b,c$ are pairwise disjoint elements of $B$ whose join is $1$, and let $x$ be any element of $B$. Then there's a condition $p$ with domain $\{a,b,c\}$ that sends $a$ to $(a\land x,a\land\neg x)$ and similarly for $b$ and $c$. It seems to me that this $p$ will be an atom in the separative quotient of your forcing; all its extensions in your forcing just add more pairs $(m\land x,m\land\neg x)$ for various $m$ and the same $x$, so they're all compatible. What am I missing? $\endgroup$
    – Andreas Blass
    Commented Oct 14, 2018 at 23:33
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    $\begingroup$ In the earlier question that you linked to in your comment here, I don't see any requirement that the pieces $b_0,b_1$ into which $f$ splits an element $b$ must have join $b$. But in the present question, requirement 2 includes $a_0\lor a_1=a$. Is that difference intentional? $\endgroup$
    – Andreas Blass
    Commented Oct 15, 2018 at 0:15

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