Let $F$ be a sigma algebra such that every element of $F$ is the union of two disjoint nonempty sets also in $F$. Prove that $F$ is uncountable. I can create a sequence of distinct sets and show that $F$ is countably infinite. I'm looking to create a power set of a countably infinite set, I suppose, but I'm not used to wading so deep into set theory. I am studying Bass's book on graduate analysis to prepare for a class this Fall. This is exercise 2.6.
 A: By induction, we can construct a sequence $(A_n)_{n\geqslant 1}$ of non-empty disjoint elements of $\mathcal F$. Indeed, since $\Omega$ is not empty, then 
it can be written as $\Omega=A\cup B$ with $A,B\in\mathcal F$ two non-empty disjoint elements. Then define $A_1:=A$ and then work with $B$. This set can be written as $B'\cup B'$; choose $A_2:=B'$ and work with $B''$, etc...
Then define the injective map 
$$\iota\colon 2^{\mathbb N}\to\mathcal F,I\mapsto \bigcup_{i\in I}A_i.$$
A: I suppose the wording should run "the union of at least two disjoint nonempty sets," or the $\sigma$-algebra $\{\{\},X\}$ would qualify. Given that assumption, you can include an infinite binary tree, i.e. one with nodes $a_{i,j}$ where $j\leq 2^i$ and $i$ running over $\mathbb{N}$, into your algebra. This is uncountable because it's in bijection with infinite sequences of $1$ and $0$.
Then map $a_{0,0}$ to the maximal element $X$ of your $\sigma$-algebra. $X$ is a disjoint union $Y\sqcup Z\sqcup...$-let those be your $a_{1,0}$ and $a_{1,1}$. Continue, using the inductive hypothesis that each $a_{i,j}$ is disjoint from sets at its own level $i$, disjoint from or properly contained in those at earlier levels, and vice versa. Then in decomposing you'll never pick a previously chosen set again, or you'd have chosen some $a_{i,j}\subset a_{i,j'}$, contradicting the hypothesis.
A: EDIT 1: In fact, as pointed out in the comments, such $\sigma$-algebras do exist (I had originally claimed a proof that they didn't) and the method below simply gives a different way of solving the problem.
EDIT 2: In fact, I now realise that Zorn's lemma is not required at all.
Suppose $\mathcal{F}$ is as described, and countable. Let $x \in X$ and define $S = \{ F \in \mathcal{F} : x\in F\}$. This is non-empty (as $X \in S$) and is countable since $\mathcal{F}$ is countable. Then $S^*$ defined to be the intersection over all sets in $S$ lies in $\mathcal{F}$ (since $\mathcal{F}$ is a $\sigma$-algebra) and it contains $x$ since each element of $S$ does. But by hypothesis we can write $S^*$ as a union of two disjoint non-empty sets $A$ and $B$. Without loss of generality, $A$ contains $x$ and is strictly contained in $S^*$. But $A$ lies in $S$ and so contains $S^*$, giving a contradiction.
A: So I found a solution in the same vein of @Kevin Carlson that managed to deal with @Andreas Blass 's concerns about undefined intersections.
Let $A \in \mathcal{A}$ 
Then we can decompose $A$ into disjoint subsets (And decompose those subsets) to form a tree structure where each level consists of disjoint nodes, whose pairwise unions form the nodes in the layer above. 
Now the nodes can be assigned an address as a binary real $k \in [0,1] $ where $1$ means descend left and $0$ means descend right.
We can now uniquely identify every real number with an element of $A$ of the form by defining the following operations:
$$ Q_0 (A, B) = A \cap B^c $$  (intuitively meaning remove B from A)
$$ Q_1 (A, B) = A \cup B $$  (intuitively meaning add B to A)
Now every real number on the unit interval corresponds to a countable infinite descending path on this tree, where we alternate between removing and adding sets as we travel down the tree. 
I'm going to refrain from defining that formally here (although if someone posts in the comments i'll be happy to add it in) 
