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I'm aware of the following elementary result on Boolean algebras. Let $\{p_1, ..., p_n\} \subseteq B$ be a finite subset of $B$. Let $2^n$ be the set of all functions from $\{1,..., n\}$ to $\{0, 1\}$. For any $\sigma \in 2^n$, let $p^\sigma = \sigma(1)p_1 \land \sigma(2)p_2 \land ... \land \sigma(n)p_n$, where $0p_i = -p_i$ (completement of $p_i$) and $1p_i = p_i$. Then, $\{p^\sigma \, | \, \sigma \in 2^n\}$ is a partition of unity.

My question is: does the infinite version of this result hold on complte Boolean algebras? In the following sense: Let $B$ be a complete Boolean algebra. Let $\{p_i \, | \, i \in I\} \subseteq B$ be an infinite subset of $B$ (indexed by $I$). Let $2^I$ be the set of all functions from $I$ to $\{0, 1\}$. From any $\sigma \in 2^I$, let $p^\sigma = \bigwedge\limits_{i \in I}\sigma(i)p_i$, where $0p_i = -p_i$ and $1p_i = p_i$. Question: is $\{p^\sigma \, | \, \sigma \in 2^I\}$ a partition of unity?

Severine Climacus
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    No, there are counterexamples in non-atomic complete Boolean algebras. For example, in the measure algebra of $[0,1]$ (i.e., Borel sets modulo sets of Lebesgue measure $0$), let $p_n$ be (the equivalence class of) the set of numbers whose $n$-th binary digit is $1$. Then all the intersections $p^sigma$ are the zero element of the algebra. – Andreas Blass Jun 01 '22 at 01:17

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