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Let $\mathfrak A$ be a subset of $\mathrm{Pow}(\mathbb N)$, the powerset of $\mathbb N$. Assume that $\mathfrak A$ is a complete Boolean algebra in the induced order, i.e., the inclusion order. Does it follow that $\mathfrak A$ is atomic?

A complete Boolean algebra $\mathfrak A$ is said to be atomic in case every nonzero element $A \in \mathfrak A$ is above a minimal nonzero element. We do not assume that the suprema and infima in $\mathfrak A$ are also those in $\mathrm{Pow}(\mathbb N)$. However, in the case of interest, $A_1 \wedge A_2 = 0$ does imply that $A_1 \cap A_2 = \emptyset$, for all $A_1, A_2 \in \mathfrak A$.

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The answer is negative. Let $A$ be the completion of the denumerable atomless BA $B$. Then $A$ is complete and atomless. $A$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$. In fact, $B$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$, and by Sikorski's extension theorem, this embedding can be extended to an embedding of $A$ into $\mathrm{Pow}(\omega)$. $B$ can be embedded in $\mathrm{Pow}(\omega)$ because $\mathrm{Pow}(\omega)$ has an independent subset of size $\omega$. Even of size $2^\omega$.

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    $\begingroup$ Welcome to MathOverflow! $\endgroup$
    – Asaf Karagila
    Nov 8, 2020 at 10:07
  • $\begingroup$ Thanks a lot for your answer. I think that it would be very valuable for me if you would fill in some of the details of the argument. For example, how do you use Sikorski's extension theorem to obtain an embedding of $A$ into $\mathrm{Pow}(\mathbb N)$? $\endgroup$ Nov 8, 2020 at 21:57

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