Can an intersection of two context-free languages be an undecidable language?

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?

Context-free languages are decidable, and decidable languages are
closed 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 wrong

-
the empty language is decidable: the answer is always "no, this string is not in the empty set".

StackExchange Computer Science Q#28531, answer score: 9