# Interchanging finite union of finite intersection of sets

I would like to exchange the set operation :

$$\cup_{i=1}^m\cap_{j=1}^{n_i} A_{i,j}$$ to be $$\cap \cup A_{i,j}$$ but it is a bit confusing to keep track of the index, and deduce the formula. Compare to summation case $$\sum_i \sum_j = \sum_j \sum_i$$ in finite sum case. I think interchanging $\cup, \cap$ might not be the same as interchanging $\sum \sum$.

Can anyone provide the way to see the formula to interchanging $$\cup \cap \ \mbox{to} \ \cap \cup$$ of the set $A_{i,j}$ above.

Thank you.

Here is the actual problem I would like to solve : Define $$G = \{\sqcup_{i=1}^m\ (\cap_{j=1}^{n_i} A_{i,j}) |\ \mbox{either} \ A_{i,j} \ \mbox{or} \ A_{i,j}^c \in \scr{C}\}$$ where $\scr{C}$ is a given non-empty collection of sets. ($\sqcup$ indicate disjoint union operation)

I would like to show that $G$ is an algebra. I struck with the step that if $A \in G$, then $A^c \in G$ which involving interchanging union and intersection. (Actually, I have to show that $G = A(\scr{C})$ (the algebra generated by $\scr{C}).$

• Which def'n of disjoint union are you using? For the disjoint union of $A$ and $B,$ I use $(A\times \{0\}) \cup (B\times \{1\})$. – DanielWainfleet Jan 23 '16 at 22:18
• @user254665 I use $$A \sqcup B = A \cup B \ \mbox{with} \ A \cap B = \phi$$ – Both Htob Jan 23 '16 at 23:16