Suppose there is a nonempty set $A_n^i$ that is indexed over $\omega$, the natural numbers. Can I say the following is true?

$$\bigcup_{i \in \omega} \left\{\bigcap_{n\in \omega} A_n^i\right\} = \bigcap_{n\in \omega}\left\{ \bigcup_{i \in \omega} A_n^i\right\}$$

Can anyone give me some idea as to whether nor not I would be able to interchange the union and intersection?


2 Answers 2


Note that


iff $\exists i\in\omega\,\forall n\in\omega\,(x\in A_n^i)$, while


iff $\forall n\in\omega\,\exists i\in\omega\,(x\in A_n^i)$; the latter condition is on the face of it easier to satisfy, so you should look for an example in which


Specifically, we might try to construct the sets $A_n^i$ so that there is some element $a\in A_n^n$ for all $n\in\omega$, which will ensure that


but so that there is no $i\in\omega$ such that $a\in A_n^i$ for all $n\in\omega$. This is easy: for each $i\in\omega$ make sure that $a\in A_n^i$ iff $n=i$. Thus, we can let

$$A_n^i=\begin{cases} \{a\},&\text{if }n=i\\ \varnothing,&\text{otherwise}\;. \end{cases}$$





  • $\begingroup$ That makes perfect sense. Thank you! $\endgroup$
    – Maria
    Oct 31, 2016 at 16:47
  • $\begingroup$ @Maria: You’re welcome! $\endgroup$ Oct 31, 2016 at 16:47

This is not possible in general. Even for finite unions and intersections you can have $$ (A_1\cap A_2)\cup(B_1\cap B_2)\subsetneqq(A_1\cup B_1)\cap(A_2\cup B_2). $$ Take for example $A_2=A_1^c$ (the complement in $X\neq\emptyset$) and $B_1=A_2,\;B_2=A_1$; then $$ A_1\cap A_2=B_1\cap B_2=\emptyset,\qquad A_1\cup B_1=A_2\cup B_2=X. $$


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