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 $ A \cap B = \emptyset$ .

I tried out an attempt, but there’s a part of the attempt of which am not so sure… So here’s my attempt:

Since both $ \bar{A}$ and $ \bar{B}$ are c.e. and the fact that $ A$ and $ B$ are disjoint, we have $ \bar{A}-\bar{B}=B$ . We claim that $ B$ is also c.e. (and this is the part which am not sure of). To show that $ \bar{A}-\bar{B}=B$ is c.e., we can have this enumerator $ E$ for $ \bar{A}-\bar{B}$ :

$ E$ : ignore any input

enumerate words $ \{s_1,s_2,s_3,…,s_k\} \in \Sigma^*$ lexicographically from $ k=1$ to $ \infty$

run recognizer $ M_{\bar{A}}$ on input words $ \{s_1,s_2,s_3,…,s_k\}$ for $ k$ steps

run recognizer $ M_{\bar{B}}$ on input words $ \{s_1,s_2,s_3,…,s_k\}$ for $ k$ steps

if both $ M_{\bar{A}}$ and $ M_{\bar{B}}$ accept an input word $ s \in \{s_1,s_2,s_3,…,s_k\}$ , print $ s$

Assuming that $ E$ correctly enumerates $ \bar{A}-\bar{B}$ , since $ B=\bar{A}-\bar{B}$ , we have that $ B$ is c.e.. Since both $ B$ and $ \bar{B}$ are c.e., it follows that $ B$ is decidable.

So to construct $ C$ , we can just set $ \bar{B}=C$ . It follows that $ A \subseteq C$ and $ C$ is decidable.

Is this attempt correct or am I missing something?