2
$\begingroup$

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 $\scr{C}).$

$\endgroup$
  • $\begingroup$ Which def'n of disjoint union are you using? For the disjoint union of $A$ and $B,$ I use $(A\times \{0\}) \cup (B\times \{1\})$. $\endgroup$ – DanielWainfleet Jan 23 '16 at 22:18
  • $\begingroup$ @user254665 I use $$A \sqcup B = A \cup B \ \mbox{with} \ A \cap B = \phi$$ $\endgroup$ – Both Htob Jan 23 '16 at 23:16
2
$\begingroup$

In general, you cannot exchange unions and intersections. To compare with your other example, it looks more like exchanging a sum and a product than two sums...

$\endgroup$
  • $\begingroup$ I add the detail of the actual problem I have to solve under the question. It might turn out that my approach to this question is not appropriate, and I do not need to interchange union and intersection. $\endgroup$ – Both Htob Jan 23 '16 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.