Complete Boolean algebras of subsets of $\mathbb N$

Question

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 $A_1, A_2 \in \mathfrak A$.

Answer 1

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 $2^\omega$.