Languages

Problem:

Mixing it up: Let $F$ denote some finite language, $R$ denote some regular language, $C$ denote some context-free language, $D$ denote some decidable language, $E$ denote some
recognizable language, and $N$ denote some non-recognizable language. For each one of the
following statements, prove whether it always holds, sometimes holds, or never holds:


$N - D$ is recognizable

Short answer: never

Support I could enumerate $N - D$, I can also enumerate $D$ because $D$ is decidable. Now I can dovetail to enumerate $N = (N - D) \cup D$, therefore I should not be able to enumerate $N - D$.

$RE$ is context free 

Short answer: sometimes

A finite language is recursive enumerable and also context free.
A recursively enumerable language include $a^n b^n c^n$, which is not context free.

$N \cap F$ is decidable

Short answer: always

$N \cap F$ is finite, and a finite language is always decidable.

$E \cup F$ is recognizable

Short answer: sometimes

E is a recognizable language, it could be finite. F is finite, so E union F could be finite, and a finite language is regular.

E could be $a^n b^n$, and $F$ could be the empty language, so $E \cup F$ is $a^n b^n$ and is not regular.

$N^*$ is non recognizable

Short answer: always

If $N*$ is recognizable, we will have a Turing machine that halt on $N*$. We can modify that machine such that whenever it reach a final state, any input will go to a state that just consume the input and do nothing and never halt, that would become a recognizer for $N$.
Therefore $N^*$ is not recognizable.

$N \cup C$ is finite

Short answer: never

If $N \cup C$ is finite, then $N$ is finite, but finite language is always recognizable.