# Example of Boolean Algebra that satisfies distributive law but violates complete distributive law

More precisely, I'm interested to know the example of Boolean Algebra $B$, such that for any $a, b, c \in B$, $a \cap (b \cup c) = (a \cap b) \cup (a \cap c)$, but there exists $\{ P_{ij}:i\in I, j \in J\} \subseteq B$, $\bigwedge_{i \in I}\vee_{j \in J}P_{ij} \neq \bigvee_{a \in J^I}\wedge_{i \in I}P_{ia(i)}$.

Added: As reminded by Hagen von Eitzen's comment, I'm also interested in whether Axiom of Choice plays a role in complete distributive law. Is it really necessary to well-order all the functions of $J^I$in the RHS of complete distributive law to make it meaningful? In the finite case, we usually specify an order of subformulaes in order to calculte it. Is it true for infinite operations?

In our case, the problem boils down to whether $I$ and $J$ are well-ordered, since we can define a lexicographic order in $J^I$ induced by the order of $I$ and $J$..

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The main problem seems to be that the existence of $a\in J^I$ may require the Axiom of Choice. –  Hagen von Eitzen Jan 20 '13 at 13:23
Note that $\kappa$-distributivity is equivalent to not adding functions from $\kappa$ to $V$ in the forcing extension. So taking the completion of any notion of forcing that adds e.g. subsets to cardinals is an example to what you are looking for. –  Apostolos Jan 20 '13 at 14:50
I'm not sure what you're after. Are you looking for a model in which every distributive Boolean algebra is completely distributive? –  Asaf Karagila Jan 20 '13 at 15:07

Consider the partial order $P$ consisting of all finite partial functions $p : \omega \to \omega$, and let $B$ denote the regular-open algebra of $P$ (giving $P$ the partial-order topology). This is a complete Boolean algebra.
Fixing $W \subseteq \omega$ of size $\omega$, for each $i \in W$ define $$\begin{gather} b_{i,0} = \{ p \in P : i \in \mathrm{dom} (p) , p(i) = 0 \} \\ b_{i,1} = - b_{i,0} = \{ p \in P : i \in \mathrm{dom} (p), p(i) \neq 0 \} \end{gather}$$ Clearly $b_{i,0} + b_{i,1} = 1$ for all $i$, and therefore $$\prod_{i \in W} \sum_{n < 2} b_{i,n} = 1.$$
Suppose, however, that $f : W \to 2$ is given. If $\prod_{i \in W} b_{i , f(i)} \neq 0$ then there is a $p \in P$ such that $p \in b_{i , f(i)}$ for all $i \in W$, which is impossible as $p$ must have finite domain. Therefore $\prod_{i \in W} b_{i , f(i)} = 0$ for all $f : W \to 2$, and thus $\sum_f \prod_{i \in W} b_{i , f(i)} = 0$.