# convert union of sets into union of disjoint sets

A union of finite sets can always be converted to a union of finite disjoint sets.

I was wondering if a union of countable sets can always be converted to a union of countable disjoint sets? If yes how to do that?

Thanks!

Update with Paul's question:

"I wonder about the natural extension of this question to uncountable unions. Can we always refine an arbitrary set {Aα} to {Bα} (i.e. with Bα⊂Aα for each α) such that ⋃αBα=⋃αAα? The answer below uses countability pretty blatantly and I can't think of a way to do it without. "

Thanks!

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I wonder about the natural extension of this question to uncountable unions. Can we always refine an arbitrary set $\{A_\alpha\}$ to $\{B_\alpha\}$ (i.e. with $B_\alpha\subset A_\alpha$ for each $\alpha$) such that $\bigcup_\alpha B_\alpha=\bigcup_\alpha A_\alpha$? The answer below uses countability pretty blatantly and I can't think of a way to do it without. –  Paul VanKoughnett Oct 28 '10 at 1:55
I also would like to know about that! I will add it to my post. –  Tim Oct 28 '10 at 2:03
@Paul: You could use a well-ordering of the index set to do the same thing in the uncountable case. But I don't know what can be said without the axiom of choice. –  Jonas Meyer Oct 28 '10 at 2:05

Given $A_1,A_2,\ldots$, let $B_1=A_1$, and for $n\gt1$, $B_n=A_n\setminus\cup_{k=1}^{n-1}A_k$. Then for $m\lt n$, $B_m$ is contained in $A_m$ while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint. You can check that $\cup_{k=1}^n B_k=\cup_{k=1}^n A_k$ for all $n$, and $\cup_{k=1}^\infty B_k=\cup_{k=1}^\infty A_k$.
If you have a collection of sets indexed by an arbitrary set, then assuming the axiom of choice you can put a well-order on the index set. Given $\{A_i\}_{i\in I}$, where $(I,\lt)$ is a well-ordered set, we can define $\{B_i\}_{i\in I}$ as follows. Let $B_{i_0}=A_{i_0}$, where $i_0$ is the least element of $I$, and for $i\gt i_0$, $B_i=A_i\setminus\cup_{j\lt i}A_j$.