Let $\Omega$ be any set and let $\mathcal A$ be an algebra of sets in $\Omega$. An element $E\in \mathcal A$ is said to be an atom if there is no non-empty element $A\in \mathcal A$ such that $A\subsetneq E$. (By convention we allow $\emptyset$ also to be an atom.)
I am wondering if the following is true:
Let $\mathcal A$ be an algebea of sets in a set $\Omega$. Let $\mathcal S\subseteq \mathcal A$ be such that the members of $\mathcal S$ are pairwise disjoint, and the algebra generated by $\mathcal S$ is same as $\mathcal A$. Then $\mathcal S$ contains all the non-empty atoms of $\mathcal A$.
EDIT: I think that under the above hypothesis we can further claim that $\mathcal S$ has nothing else but atoms of $\mathcal A$.
Can somebody confirm if the above is true?
Thanks.
It’s false if $\Omega$ is finite. Suppose that $\Omega=\{x_0,\ldots,x_n\}$, where $n\ge 1$, $\mathscr{A}=\wp(\Omega)$, and
$$\mathscr{S}=\big\{\{x_k\}:1\le k\le n\big\}\;.$$
For $k=1,\ldots,n$ let $C_k=\Omega\setminus\{x_k\}$; clearly each $C_k$ is in the algebra generated by $\mathscr{S}$, and $\bigcap_{k=1}^nC_k=\{x_0\}$, a non-empty atom of $\mathscr{A}$.
In general the algebra generated by the pairwise disjoint family $\mathscr{S}$ is isomorphic to the algebra of finite and cofinite subsets of $\mathscr{S}$. If $\mathscr{A}$ is infinite, $\mathscr{S}$ must also be infinite, and in that case the atoms of the algebra generated by $\mathscr{S}$ are precisely the elements of $\mathscr{S}$: they correspond to the singleton subsets of $\mathscr{S}$, which are the atoms of the algebra of finite and cofinite subsets of $\mathscr{S}$. Thus, $\mathscr{S}$ must consist precisely of the atoms of $\mathscr{A}$, which must be the algebra generated by its atoms.