Proof: Limit superior intersection How to prove $\limsup(\{A_n \cup B_n\}) = \limsup(\{A_n\}) \cup \limsup(\{B_n\})$? Thanks!
 A: Use the definition, and double inclusion; that is, show that every element of $\limsup(A\cup B)$ must be either an element of $\limsup(A)$ or of $\limsup(B)$; then show that every element of $\limsup(A)$ must be in $\limsup(A\cup B)$ and that every element of $\limsup(B)$ must be in $\limsup(A\cup B)$.
Of course, one must assume that you mean your "$A$" to be a sequence of sets and your "$B$" to likewise be a sequence of sets... Otherwise, what you write does not really make much sense  (limit superior and limit inferior of a single set is not usually defined).
A: Another nice way is to use characteristic functions:
The map $\chi : \mathcal{P}(\Omega) \to \{0,1\}^\Omega$ assigns to every subset of $\Omega$ its characteristic function.


*

*$\chi$ is bijective.

*$\chi$ is continuous, i.e. $\chi_{\lim\sup_{n\to\infty} A_n} = \lim\sup_{n\to\infty}\, \chi_{A_n}$ (pointwise limit)

*$\chi$ is a homomorphism, i.e. $\chi_{A \cup B} = \chi_A + \chi_B - \chi_A \chi_B$


Now your question reduces to the computation of an ordinary limit.
