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Let $B$ be a complete Boolean algebra with bottom and top elements $0$ and $1$. A set $D \subseteq B$ is open if for all $a \in D$ and $b \leq a$, $b \in D$, it is dense if for all $a \in B$ there is $b \leq a$ with $b \in D$, and it is a partition if for all $a,b \in D$, $a\land b=0$, but $\bigvee D = 1$.

It is a standard ZFC result that for all open* dense $D \subseteq B\setminus \{0\}$ there is $A \subseteq D$ that is a partition of $B$. I am interested in the possibility of proving a slightly weaker version in ZF.

Question. If $D \subseteq B$ is open dense, then is there a partition $A$ such that, for all $a \in A$, $\{b \in B \mid b < a\} \subseteq D$? ($A \subseteq D$ would be sufficient, but I suspect this may be impossible without AC)

Ideally, this would be uniformly constructible over the open dense sets. I have tried some routes, e.g. taking $A=\{ \bigvee\{a \in D \mid a \land b \neq 0\}\mid b \in D\}$, or $A = \{ \bigwedge \{a \notin D \mid a > b\} \mid b \in D\}$, but I have not bee able to prove that they work (and I suspect that they may not).

*Open except that $0 \notin D$.

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  • $\begingroup$ I suspect a counter example can be cooked from the ccc forcing that collapses $\omega_1$, by looking at the obvious dense set there in the boolean completion. It will define an uncountable antichain, but it should lie "above" the dense set. $\endgroup$ Commented Jun 7 at 11:35
  • $\begingroup$ Let $B = \{0,1\}$ with the usual order. Consider $D = \{0\}$, then $D$ is open, it is dense, but there is no $A\subseteq D$ such that $A$ is a partition of $B$ - which would contractict your standard ZFC result... what did I get wrong? $\endgroup$ Commented Jun 10 at 12:59
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    $\begingroup$ @DominicvanderZypen I forgot to specify that $D \neq \{0\}$! I'm thinking of the algebras as notions of forcing, so I was implicitly omitting $0$ in my head. $\endgroup$ Commented Jun 10 at 13:21
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    $\begingroup$ Calliope, you don't just want to say $D\neq\{0\}$, but rather require $a,b\neq 0$ in your definition of density. $\endgroup$ Commented Jun 10 at 13:31
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    $\begingroup$ Or, state the definition for arbitrary posets, and then apply it to $B\smallsetminus\{0\}$ rather than $B$. I think this is closest to the intended meaning. $\endgroup$ Commented Jun 10 at 14:24

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Over ZF - Foundation, this principle (and your slight weakening) is equivalent to the Axiom of Multiple Choice, which asserts that for every family $\mathcal{F}$ of nonempty sets, there is a function $f$ which maps each $X \in \mathcal{F}$ to a finite nonempty subset of $X.$ Notably, this implies AC holds in the well-founded universe.

$\Leftarrow$ This is a minor modification of the standard transfinite construction. Let $f$ be a multiple choice function on $\mathcal{P}(B) \setminus \{\emptyset\}.$ We construct a continuous $\subseteq$-chain $\langle A_{\alpha} \rangle$ of disjoint subsets of $D$ with strictly increasing suprema.

Let $A_0=\emptyset.$ Given $A_{\alpha},$ let $x_{\alpha}= 1 - \bigvee A_{\alpha}.$ If $x_{\alpha} \neq 0,$ then construct $A_{\alpha+1}$ by adjoining to $A_{\alpha}$ all minimal Boolean combinations of $f({x_{\alpha}\downarrow} \cap D)$ which are in $D.$

$\Rightarrow$ Fix a family $\mathcal{F}$ of infinite sets. For each $X \in \mathcal{F},$ let $(M_X, \nu_X)$ be the standard measure algebra for $2^X.$ Consider the simple product of Boolean algebras (in the terminology of Fremlin) $B=\prod_{X \in \mathcal{F}} M_X.$ Note that $B$ is equivalent as a forcing to the lottery sum of the random forcings associated to each $X \in \mathcal{F}.$

Let $D=\bigcup_{X \in \mathcal{F}} \{a < 1_X: \nu_X(a) \le \frac{1}{2}\}.$ Let $A$ be a partition such that for all $a \in A,$ $a \downarrow \in D.$ We immediately have $A \subseteq D.$

Fix $X \in \mathcal{F}.$ Let $A_X = A \cap 1_X \downarrow.$ Clearly, $\bigsqcup A_X = 1_X.$ Let $A'_X \subseteq A_X$ be the finite set of points on which $\nu_X$ is maximized. For each $a \in A'_X,$ there is a finite nonempty set $S_a$ of points $x \in X$ for which $\nu_X(a \triangle a|_x)$ is maximized. The calculation $a|_x$ takes place inside $M_X$ and denotes the event that the symmetric difference of $r \subset X$ with $\{x\}$ satisfies the event $a.$

The function $f(X)=\bigcup_{a \in A'_X} S_a$ is as desired.

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