I know the property that for sets $A_n,B_n$
$$
\limsup_{n\to\infty}(A_n\cup B_n)=\limsup_{n\to\infty}(A_n)\cup \limsup_{n\to\infty}(B_n)
$$
holds.
I'm curious if it also holds that
$$
\liminf_{n\to\infty}(A_n\cup B_n)=\liminf_{n\to\infty}(A_n)\cup \liminf_{n\to\infty}(B_n)?
$$
Best Answer
$\limsup_{n \to \infty} (A_n \cup B_n)$ contains elements in $A_n \cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $\liminf$ suggests you need to consider intersections instead of unions.
$\liminf_{n \to \infty} (A_n \cap B_n)$ contains elements in $A_n \cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$\liminf_{n \to \infty} (A_n \cap B_n) = \liminf_{n \to \infty} (A_n) \cap \liminf_{n \to \infty} (B_n).$$