I'm trying to prove that$\exists L_1, L_2 : L_1$ and $L_2$ are context-free languages $\land\;L_1 \cap L_2 = L_3$ is an undecidable language.I know that context-free languages are not closed under intersection.This means that I can produce an $L_3$, which is undecidable.An example would be $L_1 = \{a^n | n \in \mathbb{N}\} \cap L_2 = \{0\} = \emptyset$.Is this a correct proof?If not, how can I prove this theorem?Is the empty language decidable?
Can an intersection of two context-free languages be an undecidable language?
computability;context free;undecidability
Context-free languages are decidable, and decidable languages areclosed under intersection.So, though the intersection of two CF languages may not be CF, it is decidable.Remarks on your example:$\emptyset=\{\}\neq$ $\{0\}$$L_1\cap\emptyset=\emptyset$  which is context-free.You cannot prove your claim, because it is wrongthe empty language is decidable: the answer is always no, this string is not in the empty set.