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A Boolean algebra is superatomic if its every subalgebra has an atom. I'm trying to determine whether $P(\omega)$, i.e. the power set algebra of the set of all natural numbers (finite ordinals) $\omega$, can be superatomic.

In the algebra $P(\omega)$, the singletons are atoms. But it is not immediately clear whether there are atoms in every subalgebra of $P(\omega)$.

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closed as off-topic by Namaste, JonMark Perry, José Carlos Santos, Tom-Tom, Mostafa Ayaz Mar 6 '18 at 22:13

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$\mathcal P(\omega)$ is not superatomic. There is a bijection (there are many) between $\omega$ and the set $\mathbb Q$ of all rational numbers. Thus it suffices to show $\mathcal P(\mathbb Q)$ is not superatomic.

Consider the set of all bounded intervals in $\mathbb Q$ without a rational least upper bound or greatest lower bound (for example, $\{ q\in\mathbb Q : \sqrt 2 < q < \pi \}$).

Let $\mathcal A$ be the set of all unions of finitely many such intervals. Then $\mathcal A$ is closed under intersections and unions, so when $\mathcal A$ is partially ordered by inclusion, then meet and join are just intersection and union.

There are no atoms in $\mathcal A$.

PS: Perhaps I should add that unbounded intervals need to be included since this set needs to be closed under complementation.

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