# Regular language under intersection and complement confusion

I know that regular languages are closed under closure properties. But, for example, we know if $$L$$ is regular, then its complement $$L^\complement$$ is also regular. If we have $$L_1$$ and $$L_2$$ as regular then $$L_1 \cap L_2 = L_3$$ makes $$L_3$$ regular. Now for the case of when $$L$$ is not regular, then its complement $$L^\complement$$ is also not regular. But when $$L_1$$ and $$L_2$$ are not regular then $$L_1 \cap L_2 = L_3$$ doesn't make $$L_3$$ not regular, why? does being closed mean for non regularity too?

• Just a simple observation. Take any two language, be it non-regular, with no common strings. Their intersection must be $\emptyset$ which is regular. In general, I think the intersection can be quite arbitrary depending on the languages you are intersecting, so it can either result to a regular language or not. I am not 100% sure here so I am just making this a comment. Nov 11, 2022 at 3:47
• cs.stackexchange.com/q/14462/755
– D.W.
Nov 11, 2022 at 6:37

Let $$L_1 = \{aa^nb^n \mid n \ge 0\}$$ and $$L_2 = \{ba^nb^n \mid n \ge 0\}$$. Then $$L_1 \cap L_2 = \emptyset$$ is regular, but neither $$L_1$$ nor $$L_2$$ are regular.