Atomless Boolean algebra

Remarks

The one-element algebra, with $0 = 1$, is atomless in a trivial way. Every other atomless Boolean algebra is infinite, since in a finite algebra with $0 \ne 1$ a minimal element of $B \setminus \set{0}$ is an atom.

Example: let $B$ be the algebra of finite unions of half-open intervals $[s, t)$ with $0 \le s < t \le 1$, together with $\emptyset$, as subsets of $[0, 1)$. A nonzero $a \in B$ contains some $[s, t)$ with $s < t$; then $b = [s, (s + t)/2)$ lies in $B$ and satisfies $\emptyset \ne b \subsetneq a$.

Non-example: the power set of a nonempty set $X$ is not atomless, since each singleton $\set{x}$ is an atom.