Prove $\bigcap (A_n \cup B_n) = \left(\bigcap A_n\right)\cup\left(\bigcap B_n\right)$ in the last post I got a perfect answer to an inclusion:
$$\bigcap (A_n \cup B_n) \supset \left(\bigcap A_n\right)\cup\left(\bigcap B_n\right)$$
But now I have to do two more things:


*

*Example when this inclusion becomes an equality? I guess it happens in a very specific case, but cannot find any. I thought of empty sets for $A_n$ or $B_n$, but doesn't seem to have worked.

*Prove, that this inclusion becomes an equality for descending sets; that is, $\forall_{n\in \Bbb N} A_{n+1} \subset A_n$ and same for $B_n$? I get the idea behind it, of course $\bigcap A_n =\lim\limits_{n\to \infty} A_n$ but what else?
 A: If $A_{n+1}\subset A_n$ and $B_{n+1}\subset B_n\,\forall n\in\Bbb N$, the RHS becomes $\left(\lim\limits_{n\to\infty}A_n\right)\cup\left(\lim\limits_{n\to\infty}B_n\right)$. 
Using the distributive properties 


*

*$X\cap(Y\cup Z)=(X\cap Y)\cup(X\cap Z)$ and

*$X\cup(Y\cap Z)=(X\cup Y)\cap(X\cup Z)$, 
we have \begin{align}(A_i\cup B_i)\cap(A_{i+1}\cup B_{i+1})&=((A_i\cup B_i)\cap A_{i+1})\cup((A_i\cup B_i)\cap B_{i+1})\\&=((A_{i+1}\cap A_i)\cup(A_{i+1}\cap B_i))\cup((B_{i+1}\cap A_i)\cup(B_{i+1}\cap B_i))\\&=((A_{i+1}\cap B_i)\cup A_{i+1})\cup((A_i\cap B_{i+1})\cup B_{i+1})\\&=A_{i+1}\cup B_{i+1}\end{align} since $X\cap Y\subseteq Y\implies(X\cap Y)\cup Y=Y$. Hence the LHS becomes \begin{align}\bigcap(A_n\cup B_n)&=(A_1\cup B_1)\cap(A_2\cup B_2)\cap(A_3\cup B_3)\cap\cdots\\&=(A_2\cup B_2)\cap(A_3\cup B_3)\cap\cdots\\&=(A_3\cup B_3)\cap\cdots\\&=\cdots\\&=\left(\lim\limits_{n\to\infty}A_n\right)\cup\left(\lim\limits_{n\to\infty}B_n\right)\\&=\left(\bigcap A_n\right)\cup\left(\bigcap B_n\right).\end{align}
A: For the first part: If you take all sets to be empty, you trivially get equality. The assertion more generally becomes true if either all $(A_n)_n$ or all $(B_n)_n$ are empty. Then both sides either read $\bigcap A_n$ or $\bigcap B_n$ respectively. But all this cases already covered by the second assertion.

For the second part: Here’s the mental image you should have: Make two lines of dots, one for each sequence of sets $(A_n)_n$ and $(B_n)_n$. Whenever an element is an $A_n$ or $B_n$ make an × at the according place and – otherwise. For a given element $x$, a line may look like this:

$A_•$ ×––××××–×…
$B_•$ –×––××–×–…

indicating that $x$ is

*

*in $A_1$, not in $A_2$, not in $A_3$, but in $A_4$ …

*not in $B_1$, but in $B_2$, but not in $B_3$, …

Translate for an element being in $\bigcap A_n ∪ \bigcap B_n$ or $\bigcap (A_n ∪ B_n)$ into this mental image. What does it mean when $(A_n)_n$ and $(B_n)_n$ are descending sequences of sets?
To formally prove this, let $x ∈ \bigcap (A_n ∪ B_n)$ and then assume it’s not in $\bigcap A_n$. What can you conclude?

(1) Being in $\bigcap (A_n ∪ B_n)$ in general looks something like this:

$A_•$ ×––××–×…
$B_•$ –×××–×–…

(2) Being in $\bigcap A_n ∪ \bigcap B_n$ looks like one of these three patterns:

$A_•$ ×××××××… | –––––––… | ×××××××…
$B_•$ –––––––… | ×××××××… | ×××××××…

(3) If $(A_n)_n$ and $(B_n)_n$ are both descending, the only possible membership patterns for these sequences of sets look something like this:

$A_•$ ××××––––…
$B_•$ ×–––––––…

This is because if an element is in $A_4$, say, it automatically is in $A_3$, $A_2$ and $A_1$ as well.
Now, is it possible for an element to look like (1) and (3), but not like (2)?
A: The equality you want to show is false in general, the inclusion you have is the best you can do.
E.g. Let $A_n = \{1\}$ for $n$ even, $A_n= \{0\}$ for $n$ odd.
and $B_n = \{0\}$ for $n$ even, $B_n=\{1\}$ for $n$ odd.
Then for any $n$, $A_n \cup B_n = \{0,1\}$ so $\bigcap_n (A_n \cup B_n)=\{0,1\}$ as well, while $\bigcap_n A_n = \bigcap B_n = \emptyset$, hence $(\bigcap_n A_n) \cup (\bigcap_n B_n ) = \emptyset$.
If the $A_n$ and the $B_n$ are decreasing sequences in inclusion, things are better: suppose $x \in A_n \cup B_n$ for all $n$, and suppose $x \notin \bigcap_n A_n$ and also $x \notin \bigcap_n B_n$. This gives us some $n_1$ such that $x \notin A_{n_1}$ and some $n_2$ such that $x \notin B_{n_2}$. Then define $m=\max(n_1,n_2)+1$ and by assumption $x \in A_m \cup B_m$, and if $x \in A_m$ by decreasingness we have $x \in A_{n_1}$, contradiction. And if $x \in B_m$, $x \in B_{n_2}$, also a contradiction. So for decreasing sequences, the equality holds.
