When do union and intersection commute? Suppose we're given a family of sets $A_{mn}$. When does
$$
\bigcap_{m\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}A_{mn}=\bigcup_{n\in\mathbb{N}}\bigcap_{m\in\mathbb{N}}A_{mn}?
$$
For instance, if $A_{mn}=(\frac{m}{n},\infty)$ then the left side is $(0,\infty)$ and the right side is $\varnothing$.
 A: They distribute but they don't commute.
To make things simple, suppose we have just four sets $A_{ij}, i=1\ldots 2, j=1\ldots 2$.  On the left side you have 
$$ L = \bigcap_{m=1}^2 \bigcup_{n=1}^2 A_{mn} = (A_{11} \cup A_{12}) \cap (A_{21} \cup A_{22})$$
On the right you have 
$$ R = \bigcup_{n=1}^2 \bigcap_{m=1}^2 A_{mn} = (A_{11} \cap A_{21}) \cup (A_{12} \cap A_{22})$$
Using the distributive law in $L$ gives you
$$ L = (A_{11} \cap A_{21}) \cup (A_{11} \cap A_{22}) \cup (A_{12} \cap A_{21}) \cup (A_{12} \cap A_{22})$$
So $R$ is missing the terms $A_{11} \cap A_{22}$ and $A_{12} \cap A_{21}$.
In order for $L=R$, those must be contained in some of the terms that are present.
Similarly in the infinite case, the condition for equality is that
 $\bigcap_{m=1}^\infty A_{m,n_m} \subseteq \bigcup_{n} \bigcap_{m} A_{mn}$ for every sequence $n_1, n_2, \ldots$ of positive integers.
A: I know it is an old topic, but I was computing $\left(\bigcup_{i=1}^m A_i\right)\setminus\left(\bigcup_{j=1}^n B_j\right)$ and found something similar to this question. One can calculate $\left(\bigcup_{i=1}^m A_i\right)\setminus\left(\bigcup_{j=1}^n B_j\right)=\bigcap_{j=1}^n\left(\left(\bigcup_{i=1}^m A_i\right)\setminus B_j\right)=\bigcap_{j=1}^n \bigcup_{i=1}^m (A_i\setminus B_j)$. But one can also calculate $\left(\bigcup_{i=1}^m A_i\right)\setminus\left(\bigcup_{j=1}^n B_j\right)=\bigcup_{i=1}^m\left(A_i\setminus \bigcup_{j=1}^n B_j\right)=\bigcup_{i=1}^m\bigcap_{j=1}^n(A_i\setminus B_j)$. Then we conclude that $\bigcap_{j=1}^n \bigcup_{i=1}^m (A_i\setminus B_j)=\bigcup_{i=1}^m\bigcap_{j=1}^n(A_i\setminus B_j)$.
