Does countable $\cup\cap$ of a sequence of sets equal countable $\cap\cup$ of some subsequence? Let $A = \cup_{i=1}^{\infty}\cap_{j=1}^{\infty} A_{j}^{i}$.  Can $A$ be written $\cap_{i=1}^{\infty}\cup_{j=1}^{\infty} A_{k(i,j)}^{l(i,j)}$  Where $l,k$ are some maps $\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$?  Trying to prove it got overwhelming, so was just wondering if anybody already knows the answer.
 A: The set $\mathbb Q$ of rational numbers is the union of countably many intersections of countably many open intervals in the real line $\mathbb R$.  Specifically, it is the union of countably many one-element sets, each of which is the intersection of a countable decreasing sequence of open intervals.  I claim that $\mathbb Q$ is not the intersection of countably many unions of countably many open intervals. The reason is that each of those unions would be a dense open set in $\mathbb R$, dense because it includes $\mathbb Q$ and open because it's a union of open intervals. By the Baire category theorem, every intersection of countably many dense open sets of real numbers is of the second Baire category, whereas $\mathbb Q$, being the union of countably many one-element sets, is of first category.
A: Here is a more set-theoretic answer. Consider the sets: $$B_j^i := \{f:\mathbb{N} \rightarrow \mathbb{N}:f(i) \neq j\}.$$
Let $A_j^i = \{i\} \cup B_j^i$, where $i \in \mathbb{N}$. Then $A = \mathbb{N}$, since any function must be eliminated from the LHS. Now, look at the RHS. If it were $A$, then each of the intersections in the rearrangement must contain $i \in \mathbb{N}$, and so in particular we have to have some $j$ with $i \in A_{k(i,j)}^{l(i,j)}.$ Let this $j$ be denoted $p(i)$. So $i = l(i,p(i))$. (Since only sets with top index $i$ contain $i$.) Now let $f$ be any function satisfying $f(i) \neq k(i,p(i))$ for all $i$.Then $f \in A_{k(i,p(i))}^{i} = A_{k(i,p(i))}^{l(i,p(i))}$. So this $f$ lies in the RHS.
