How to prove $\limsup\limits{} \left(A_n \cap B_n\right)\subseteq\limsup\limits{} \left(A_n \right) \cap \limsup\limits{}\left(B_n \right) $?

Question

Prove that for any collections $\{A_n\}$ and $\{B_n\}$, $$\limsup\limits{} \left(A_n \cap B_n\right)\subseteq\limsup\limits{} \left(A_n \right) \cap \limsup\limits{}\left(B_n \right) $$

Any hints please? Right now, i have following relation as the beginning:

$$\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} \left(A_k \cap B_k\right) \subseteq \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} \left(A_k \right) \cap \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} \left(B_k \right)$$

I can see it intuitively. Should i start it by using substitution?

Answer 1

Remember that $\lim \sup A_n$ is the set of elements that belong to an infinite number of $A_n$,s. So if

$$\omega \in \lim \sup \ (A_k \cap B_k),$$

we know that $\omega$ belongs to $(A_k \cap B_k)$ for an infinite number of $k$'s. We conclude that $\omega$ belongs to an infinite number of $A_k$'s AND belongs to an infinite number of $B_k's$. Therefore, $\omega \in \lim \sup (A_k) \ \cap \ \lim \sup \ (B_k)$. $$$$