As I mentioned elsewhere, I'm working on Schimmerling's A Course on Set Theory. One of the nice features of the book (for me, anyway) is the addition of some interesting exercises on Boolean algebras. Now there's one such exercise which has been bugging me for a while. In chapter 2, the last exercise is about the quotient algebra $\mathcal{P}(\omega)/Finite$, which is defined as follows. First, define an equivalence relation $E$ on $\mathcal{P}(\omega)$ such that $x E y$ iff their symmetric difference is finite. Then, taking $[x]_E$ to be $x$'s equivalence class, we define the Boolean operations as:
$[x]_E \vee [y]_E = [x \cup y]_E$
$[x]_E \wedge [y]_E = [x \cap y]_E$
$\neg [x]_E = [\omega - x]_E$
$\bot = [\varnothing]_E$
$\top = [\omega]_E$
(Obviously, one must show that these operations are well-defined, but I'll leave this out, as it's the first part of the exercise and relatively easy)
Now, the thing is, part 2 of the exercise ask us to prove that this algebra is atomless. I have sketched a proof, but it makes use of a crucial assumption that is not mentioned anywhere in the book (well, up to that exercise, anyway), so I'd like to know if there's an alternative proof in which that assumption isn't used. I'll highlight the assumption in the proof below.
Suppose there is a $[y]_E \in \mathcal{P}(\omega)/E$ which is different from $[\varnothing]_E$. We know from the above that $y$ is infinite. Since every infinite set is the union of two disjoint infinite sets, choose one of those as $x$. Then $x \subseteq y$, that is, $[x]_E \preceq [y]_E$ and $x$ is infinite, that is, $x \not \in [\varnothing]_E$. Since $y - x$ is infinite (for the other disjoint set is also infinite), we have that $[x]_E \not = [y]_E$. Therefore, for every $[y]_E \in \mathcal{P}(\omega)/E$ which is different from $[\varnothing]_E$, we can construct a smaller element $[x]_E$ such that $[x]_E \not = \bot$ and $[x]_E \not = [y]_E$, i.e. no element $[y]_E$ is an atom. Therefore, the quotient algebra $\mathcal{P}(\omega)/E$ is atomless.
Any thoughts?
Well, since all those sets are subsets of $\omega$, you can just prove that this is possible directly. Let $A\subseteq\omega$ be infinite, then there is a unique order preserving bijection between $A$ and $\omega$ (incidentally, it's also the Mostowski collapse of $A$). Now pick $A_0$ and $A_1$ to be those with are sent to the even and odd integers respectively. Since $A$ is infinite, both these sets must be infinite, as wanted.
However, in the absence of choice, it is consistent that there are infinite sets which cannot be partitioned into two infinite subsets. If $A$ is such set, then $\mathcal P(A)/E=\{[A]_E,[\varnothing]_E\}$ which has an atom. So you can't really dispense this assumption in the general case, but the above shows that it is easily proven.
(And any other proof would essentially amount to proving the above statement. Since it's really equivalent to saying that $\mathcal P(\omega)/E$ is atomless.)