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 $P(\omega)$.
$\mathcal P(\omega)$ 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.
