Let $A_1,A_2,\dotsc$ be a collection of sets, which may or may not already be disjoint, and define $B_1=A_1,B_{n+1}=A_{n+1}\setminus\bigcup_{i\le n}A_i$ for $n\ge1$. Then $B_1,B_2,\dotsc$ is disjoint and $\bigcup^\infty A_i=\bigcup^\infty B_i$. Now I'm curious if this works given an arbitrary collection of sets $A_i$, $i\in I$, where $I$ is ANY index set. I'm thinking, why not take steal the structure of the original index set for the new one?
Define $J=I$ and $B_j=A_i$ for some $j\in J$ and any $i\in I$, then define $B_k=A_l\setminus\bigcup_j B_j$ where for every new $B_k$ defined, $l\in I$ s.t. $l$ not already chosen and $j$ ranges over the $B_j$ already defined. Then $B_j$, $j\in J$, is disjoint and $\bigcup_{i\in I}A_i=\bigcup_{j\in J}B_j$.
But all this seems very sketchy. For $I=\mathbb{N}$, the method of "disjointizing" $A_i$ seems to work in the that defining the sets $B_i$ is "complete" (so the $B_i$'s exist), i.e. in the sense that we somehow exhaust $i\in\mathbb{N}$, and that the $B_i$'s are unique, since the next $B_{n+1}$ is uniquely determined by the $\le$ relation on $\mathbb{N}$. (I believe we can un-unique them by matching $J=I$ if the second method if it works). But an index set $I$ could have zero structure and any cardinality, and then we need a method to keep track of already chosen elements.
My question is: Does the second method work, i.e. do the $B_i$'s exist? If they do, I don't think they're unique, but it doesn't matter since all we want is disjointness.
Best Answer
The property of $\mathbb{N}$ which allows us to do what you want is the fact that it is well-ordered. For those not familiar:
Definition: An order $\leqslant$ on a set $A$ is a well-ordering if it satisfies the following conditions:
There is a theorem, the Well-Ordering Theorem, which is equivalent to the axiom of choice, which states that every set can be well-ordered. Using this theorem we may "disjointize" any collection of sets.
Let $\mathcal A= \{A_i\}_{i\in I}$ be a collection of sets indexed by the the set $I$. We will construct the sets $B_i$, which have the desired property that $$\bigcup_{i\in I} B_i=\bigcup_{i\in I} A_i,$$ using a process called transfinite induction. Let $\leqslant$ be a well-ordering of $I$. Let $a$ be the least element of $I$ and write $B_a=A_a.$ Now let $i\in I$ be such that for each $j< i$ we have constructed pairwise disjoint sets $B_j$ from the collection $\mathcal{A}.$ Then let $B_i=A_i\setminus \bigcup_{j<i} A_j$. It is clear that $B_i\cap B_j=\varnothing$ for each $j<i$. Since $B_i\subseteq A_i$ for each $i\in I$, we have $$\bigcup_{i\in I} B_i\subseteq\bigcup_{i\in I} A_i.$$ To see the reverse inclusion, observe that for each $x\in\bigcup_{i\in I} A_i$, the set $C_x=\{i\in I\;|\;x\in A_i\}\subseteq I$ has a least element $j$ and that $x\in B_j,$ so $$\bigcup_{i\in I} B_i\supseteq\bigcup_{i\in I} A_i,$$ giving equality of the two sets.
At first it may seem that transfinite induction works even with sets which are only totally ordered, such as the closed unit interval $[0,1].$ However, this is not the case. To see this, I recommend reading the proof that transfinite induction works. Intuitively, without a well-ordering, one has trouble "moving on to the next element" while performing the induction. I used the first chapter of Munkres' Topology, but there may be better sources. I enjoyed Munkres because he goes through a good amount of material, and the supplementary exercises at the end of nearly every chapter are challenging and illuminating.