Rearrangement of Countable Unions?

Question

Given a family of countable sets $\{ A_i \}$, define $X = \cup_{i \geq 1} A_i$.

Under what conditions (with proof) on $A_i$, is it possible to rearrange the sets in this countable union?

An analogy with series leads us to the rearrangement theorem which says that a series can be rearranged iff it is absolutely convergent. Is there some other rearrangement theorem for unions?

Source of this problem is $\sigma$-algebras: Often when checking for the countable union is closed condition, we rearrange the union (without performing any check) - I was confused as to if this is always allowed?

Answer 1

$\bigcup_{i\ge 1}A_i$ is by definition simply

$$\left\{x:\exists i\in\Bbb Z^+\,(x\in A_i)\right\}\,,$$

which clearly does not depend on the order in which the sets are enumerated. You can simply let $\mathscr{A}=\{A_i:i\in\Bbb Z^+\}$; then

$$\bigcup_{i\ge 1}A_i=\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}\,,$$

with no reference to the enumeration at all.

Unlike the sum of an infinite series, the union of an infinite collection of sets is not a limit: it is simply a certain set of objects, namely, those that belong to at least one member of the collection.

Answer 2

Something loosely analogous to series re-arrangement: Let $S=(A_i)_{i\in \Bbb N}$ be a sequence of sets. It is not required that $A_i\ne A_j$ when $i\ne j.$

We define $\lim\sup S=\cap_{i\in \Bbb N} \cup_{j\ge i} A_j.$

We define $\lim\inf S=\cup_{i\in \Bbb N}\cap_{j\ge i}A_j.$

Then $\lim\sup S\supseteq\lim\inf S.$

Now if $f:\Bbb N\to\Bbb N$ is a bijection, let $B_i=A_{f(i)}$ and let $T=(B_i)_{i\in \Bbb N}.$ Then $$\lim\sup S=\lim\sup T \;\text { and }\; \lim\inf S=\lim\inf T.$$

This is because $x\in \lim\sup S$ iff $\{i\in \Bbb N: x\in A_i\}$ is infinite, and because $x\in \lim\inf S$ iff $\{i\in \Bbb N: x\not\in A_i\}$ is finite.