Does the limit of the union (intersection) of 2 sets equal the union (intersection) of the limits? Suppose we have two sequences of sets $A_n$ and $B_n$ such that $A_n \to A$ and $B_n \to B$
i.) Does $A_n \bigcup B_n \to A \bigcup B$
and
ii.) Does $A_n \bigcap B_n \to A \bigcap B$
If not, is there a counterexample? 
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I don't think the limit should hold. 
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For example, 
$\liminf_{n\to\infty} A_n \bigcup B_n \neq \liminf_{n\to\infty} A_n \bigcup \liminf_{n\to\infty} B_n$.
This is apparent if we take $A_n = \{{(-1)^n}\}$ and $B_n = \{{(-1)^{n+1}}\}$.
The same $A_n$ and $B_n$ show that $\limsup_{n\to\infty} A_n \bigcap B_n \neq \limsup_{n\to\infty} A_n \bigcap \limsup_{n\to\infty} B_n$.
On the other hand, per Proof that lim sup of union equals union of lim sup, the relation holds for $\limsup$ in the case of union and $\liminf$ in the case of intersection. I've also worked through both of those cases but am stuck here, and I'm not sure what I'm missing.
 A: Took a step back, thought about the problem more, and it became clear the limits distribute.
Useful Facts 


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*$\liminf_{n \to \infty} A_n \subset \limsup_{n \to \infty} A_n$
since $\omega \in A_n$ $\forall n\geq N$ $\implies \omega \in A_n$ i.o.


*$\limsup_{n \to \infty} A_n \cap B_n \subset \limsup_{n \to \infty} A_n \cap \limsup_{n \to \infty} B_n$
since $\omega \in A_n \cap B_n$ i.o. $\implies \omega \in A_n$ i.o. and $\omega \in B_n$ i.o.


*$\liminf_{n \to \infty} A_n \cup \liminf_{n \to \infty} B_n \subset \liminf_{n \to \infty} A_n \cup B_n$
since $\omega \in A_n$ or $\omega \in B_n$ $\forall n \geq N$ $\implies$ $\omega \in A_n \cup B_n$ $\forall n \geq N$


*$\limsup_{n \to \infty} A_n \cup B_n = \limsup_{n \to \infty} A_n \cup \limsup_{n \to \infty} B_n$
(see link above)


*$\liminf_{n \to \infty} A_n \cap B_n = \liminf_{n \to \infty} A_n \cap \liminf_{n \to \infty} B_n$
since $\omega \in A_n \cap B_n$ $\forall n \geq N$, then $\omega \in A_n$ and $\omega \in B_n$ $\forall n \geq N$
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Proofs
I. If $A_n \to A$ and $B_n \to B$, then $\liminf_{n \to \infty} A_n \cup B_n = A \cup B = \limsup_{n \to \infty} A_n \cup B_n$
$$A \cup B = \liminf_{n \to \infty} A_n \cup \liminf_{n \to \infty} B_n \subset \liminf_{n \to \infty} A_n \cup B_n \subset \limsup_{n \to \infty} A_n \cup B_n = \limsup_{n \to \infty} A_n \cup \limsup_{n \to \infty} B_n = A \cup B$$
$$\implies \liminf_{n \to \infty} A_n \cup B_n = A \cup B = \limsup_{n \to \infty} A_n \cup B_n$$
By (3), (1), and (4)
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II. If $A_n \to A$ and $B_n \to B$, then $\liminf_{n \to \infty} A_n \cap B_n = A \cap B = \limsup_{n \to \infty} A_n \cap B_n$
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$$A \cap B = \liminf_{n \to \infty} A_n \cap \liminf_{n \to \infty} B_n = \liminf_{n \to \infty} A_n \cap B_n \subset \limsup_{n \to \infty} A_n \cap B_n \subset \limsup_{n \to \infty} A_n \cap \limsup_{n \to \infty} B_n = A \cap B$$
$$\implies \liminf_{n \to \infty} A_n \cap B_n = A \cap B = \limsup_{n \to \infty} A_n \cap B_n$$
By (5), (1), and (2)
