Each Element of an Algebra can be Partitioned into "Atoms" 
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.)

(My definition of an atom is different from the standard definition of an atom as used in measure theory.)
Question: Can each element of $\mathcal A$ be partitioned into atoms?
If $\mathcal A$ was generated by finitely many members of $\Omega$, the answer is clearly in the affirmative.
I am unable to prove (or disprove) this in the case where $\mathcal A$ is not generated by finitely many members of $\Omega$.
Can somebidy help?
 A: No, not necessarily.
If this is true, we say that the algebra is atomic. But not every Boolean algebra is atomic, or even has atoms.
Consider for example $\mathcal P(\Bbb N)$ as a Boolean algebra, and $\rm fin$ as the ideal of finite sets there. If you haven't seen that the quotient of a Boolean algebra by an ideal is a Boolean algebra, this is a good exercise. Now $\cal B=P(\Bbb N)/\rm fin$ is a Boolean algebra, and $[A]\leq_\mathcal B[A']$ if and only if $A\setminus A'$ is a finite set.
So given any $[A]$ in $B$, if it is non-zero then $A$ is infinite, and we can split $A$ into two infinite subsets, $A_0$ and $A_1$, in which case $[A_0]<_\mathcal B[A]$.
But wait, you say, this is a general Boolean algebra! And the question is about a field of sets. But rest assured, the Stone Representation theorem tells us, amongst other things, that every Boolean algebra is isomorphic to a field of sets. And if $B$ a field of sets isomorphic to $\cal B$ above, then $B$ itself cannot have any atoms either.
(One nice thing about atomless Boolean algebras is that their first order theory is complete. To see why, you can prove via a back-and-forth argument that every two countable atomless Boolean algebras are isomorphic. This means that the theory is $\aleph_0$ categorical which in turns implies completeness.)
