# Intersection of languages using closure properties

1. $L_1 \in \mathrm{CFL} \cap L_2\in \mathrm{CSL}$ is $\mathrm{CSL}$ because every $\mathrm{CFL}$ is $\mathrm{CSL}$ and by applying closure property of $\mathrm{CSL}$ under intersection, it's $\mathrm{CSL}$.

2. $L_1 \in \mathrm{ RL} \cap L_2 \in \mathrm{CFL}$ is same as $\mathrm{CFL} \cap \mathrm{CFL}$ . As $\mathrm{CFL}$ is not closed under intersection, we promote $\mathrm{CFL}$ to $\mathrm{CSL}$. So $\mathrm{CSL} \cap\mathrm{CSL}$ is $\mathrm{CSL}$.

But, we know that 2nd is guaranteed to be $\mathrm{CFL}$ (also $\mathrm{CSL}$).

Why I am not able to arrive at the answer ($\mathrm{CFL}$) using closure properties for 2nd case ? Can't we apply closure properties when considering regular languages?

In general, if we want to know the family in which the intersection (or whatever operation $\oplus$) of two languages belongs, then we look for the smallest families $\mathrm{X}$ and $\mathrm{Y}$ that contain the two languages and for which we know an applicable closure property.