show that this decidable set $C$ exists

Question

I came across this problem which says that given disjoint sets $A$ and $B$ s.t $\bar{A}$ and $\bar{B}$ are both computably enumerable (c.e.), there exists a decidable set $C$ s.t. $A \subseteq C$ and $C \cap B = \emptyset$.

I think one way to construct $C$ is to show that $\bar{A}-\bar{B}$ is c.e., but is the set difference $\bar{A}-\bar{B}$ for this particular case c.e.?

Answer 1

No, it is not necessarily recursively enumerable. There are languages that are recursively enumerable but not recursive. Thus, their complement is not recursively enumerable. From that, you should be able to prove that the answer to the question in the final sentence of your post is no (I'll let you fill in the details from there).