Let $A_n$ and $B_n$ be a sequence of sets such that $A_n$ converges to $A$ and $B_n$ converges to $B$. Is it true that:
(i) $A_n \cup B_n$ converges to $A \cup B$?
(ii) $A_n \cap B_n$ converges to $A \cap B$?
I want to say yes for (i) and no for (ii) but I am having trouble dealing with examples of $A_n$ and $B_n$ since they depend on $n$.
For general sequences of sets $A_n$ and $B_n$ (not necessarily convergent), you can prove that $$ (\liminf A_n) \cup (\liminf B_n) \subseteq \liminf (A_n\cup B_n). $$ (One-line proof: if $x$ is in all but finitely many of the $A_n$, then it's in all but finitely many of the $A_n\cup B_n$.) Similarly, one can prove that $$ \limsup (A_n\cup B_n) \subseteq (\limsup A_n) \cup (\limsup B_n). $$ (One-line proof: if $x$ is in infinitely many of the $A_n\cup B_n$, then it's either in infinitely many of the $A_n$ or in infinitely many of the $B_n$.) In particular, if $A_n$ converges to $A$ and $B_n$ converges to $B$, then these combine to give $$ A \cup B \subseteq \liminf (A_n\cup B_n) \subseteq \limsup (A_n\cup B_n) \subseteq A\cup B, $$ which shows that $A_n\cup B_n$ converges to $A\cup B$.
Statement (ii) is also true: it can be proved in a similar way, or it can actually be derived from (i) by taking complements.