About writing a countable family of sets in terms of pairwise disjoint sets I was wondering if the following statement is true:
Let $\mathfrak{a}$ be a countable collection of sets. Does this imply that there is a countable collection $\mathfrak{b}$ of pairwise disjoint sets such that every set in $\mathfrak{a}$ is a finite union of sets in $\mathfrak{b}$?
Is there a book/reference where I can find the answer? Or is it easy to prove?
 A: I do not know a reference, but believe that your claim is false:
Consider $$\mathfrak{a}= \{A_k|k\in\mathbf{N}\},\mbox{ where } A_k = \mathbf{N} \setminus \{k\}.$$
Now assume that there is a countable family of sets and denote them as $$\mathfrak{b}= \{B_k|k\in\mathbf{N}\}$$ such that all the required properties all hold.
It is suffices to show that each $B_k$ must be a singleton (why?).
To this end, suppose there are two distinct numbers $n$ and $m$ such that
$n, m \in B_k$. Because the elements of $\mathfrak{b}$ are pairwise disjoint sets and $n \in A_m$, it follows that $B_k \subset A_m$. However, this means $m \in\mathbf{N}\setminus \{m\}$, which is a contradiction.
A: You cannot ensure it in general. Let $\mathscr{A}=\{A_n:n\in\Bbb N\}$. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a pairwise disjoint family such that each $A_n$ is the union of finitely many members of $\mathscr{B}$, and let $X=\bigcup\mathscr{A}=\bigcup\mathscr{B}$. For distinct $x,y\in X$ say that $\mathscr{A}$ separates $x$ and $y$ if there is an $n\in\Bbb N$ such that $A_n$ contains exactly one of $x$ and $y$. It’s clear that if $\mathscr{A}$ separates $x$ and $y$, then $x$ and $y$ must belong to distinct members of $\mathscr{B}$. 
Suppose that $Y$ is an infinite subset of $X$ such that $\mathscr{A}$ separates $y_0$ and $y_1$ whenever $y_0$ and $y_1$ are distinct elements of $Y$; then $Y$ cannot be the union of finitely many members of $\mathscr{B}$. Thus, to construct a counterexample we need only build a family $\mathscr{A}$ that separates the members of an infinite subset $Y$ of $\bigcup\mathscr{A}$, and take the family $\{Y\}\cup\mathscr{A}$.
This is easily done. Taking $\Bbb N$ as the underlying set $X$, for instance, we can set $A_n=\Bbb N\setminus\{n\}$: if $m,n\in\Bbb N$ with $m\ne n$, then $A_m\cap\{m,n\}=\{n\}$, so $\mathscr{A}=\{A_k:k\in\Bbb N\}$ separates $m$ and $n$. Or we can let $Y=\{2^k:k\in\Bbb N\}$, $A=\Bbb N\setminus Y$, $A_n=A\cup\{2^n\}$ for each $n\in\Bbb N$, and $$\mathscr{A}=\{Y\}\cup\{A_n:n\in\Bbb N\}\;;$$
if $m,n\in\Bbb N$ with $m\ne n$, then $A_m\cap\{2^m,2^n\}=\{2^m\}$.
