Proving $\bigcap_{i\in\Bbb{N}}(A_i\cup B_i) = (\bigcap_{i\in\Bbb{N}}A_i) \cup (\bigcap_{i\in\Bbb{N}}B_i)$ for nonincreasing set series I am having some problems with the following:

Prove that if  $A_0 \supseteq A_1 \supseteq A_2 \supseteq ...$ and $B_0 \supseteq B_1 \supseteq B_2 \supseteq ...$ then $\bigcap_{i\in\Bbb{N}}(A_i\cup B_i) = (\bigcap_{i\in\Bbb{N}}A_i) \cup (\bigcap_{i\in\Bbb{N}}B_i)$

Although it is quite easy to show that $\bigcap_{i\in\Bbb{N}}(A_i\cup B_i) \supseteq (\bigcap_{i\in\Bbb{N}}A_i) \cup (\bigcap_{i\in\Bbb{N}}B_i)$ by unwinding the definitions, I can't find any way to prove the opposite. I tried experimenting with the fact that $\limsup_{i\to\infty} A_i=\liminf_{i\to\infty}A_i=\lim_{i\to\infty}A_i$ (following from the fact the series is nonincreasing, same for $B_i$), but it doesn't seem to help any bit except for adding another layer of abstraction. I thought that maybe if $\lim_{i\to\infty}(\oldstyle{A_i\cup B_i})=\lim_{i\to\infty}(\oldstyle{A_i})\cup\lim_{i\to\infty}(\oldstyle{B_i})$ for any converging $\oldstyle{(A_n)_{n\in\Bbb{N}} ,(B_n)_{n\in\Bbb{N}}}$ were true, this problem would be solvable without much hassle. However I haven't found anything about any equality like this (and I am not sure if it is true anyways).
 A: This will be a more or less a refinement of what Robert said in the other answer, since i feel there was a big skip in logic.
We assume that $A_{0} \supseteq A_{1} \supseteq A_{2} \supseteq ...$ and $B_{0} \supseteq B_{1} \supseteq B_{2} \supseteq ...$
I assume that you have proved $\bigcap_{i \in \mathbb{N}}(A_i \cup B_i) \supseteq (\bigcap_{i \in \mathbb{N}} A_i) \cup (\bigcap_{i \in \mathbb{N}} B_i) $, so i will prove that $\bigcap_{i \in \mathbb{N}}(A_i \cup B_i) \subseteq (\bigcap_{i \in \mathbb{N}} A_i) \cup (\bigcap_{i \in \mathbb{N}} B_i)$. 
Suppose that $x \in \bigcap_{i \in \mathbb{N}}(A_i \cup B_i) $. This means that ($\forall i \in \mathbb{N})(x \in A_i \lor x \in B_i)$. Like Robert said, this means that $x \in A_i$ for an infinite number of $i$, or $x \in B_i$ for an infinite number of $i$ (or both). 
Now suppose $x \in A_i$ for an infinite number of $i$. We want to show that $x \in \bigcap_{i \in \mathbb{N}} A_i$. Assume for contradiction that there exists a $k \in \mathbb{N}$ such that $x \notin A_k$. Since $x \in A_i$ for an infinite number of $i$, we can find a $k' > k$ such that $x \in A_{k'}$. However, since $k' > k$, we have that $A_{k'} \subseteq A_k$, and therefore must $x \in A_k$, which is our contradiction. Since there can't exist  a $k$ such that $x \notin A_k$, $x$ must be an element of $A_i$ for all $i$, which in other words means that $x \in \bigcap_{i \in \mathbb{N}}A_i$.
This can be done equivalently if $x \in B_i$ for an infinite number of $i$. Since one of those statements must be true, we can conclude that $x \in  (\bigcap_{i \in \mathbb{N}} A_i) \cup (\bigcap_{i \in \mathbb{N}} B_i)$, which also means that $\bigcap_{i \in \mathbb{N}}(A_i \cup B_i) \subseteq (\bigcap_{i \in \mathbb{N}} A_i) \cup (\bigcap_{i \in \mathbb{N}} B_i)$.
A: If $x \in \bigcap (A_i \cup B_i)$, then $\forall i~(x \in A_i \lor x \in B_i)$.  Thus, either $x \in A_i$ for infinitely many $i$ or $x \in B_i$ for infinitely many $i$.  But if, say, $x \in A_i$ for infinitely many $i$, then $A_i$ nested $\Rightarrow x \in \cap A_i$.  If it's not true that $x \in A_i$ for infinitely many $i$, then $x \in B_i$ for infinitely many $i$ so that, by the same reasoning, $x \in \cap B_i$.
Thus, $x \in \cap A_i$ or $x \in \cap B_i$, so $x \in \bigcup ((\cap A_i) \cup (\cap B_i))$ and we are done.
