What is co-Turing recognizable, how do I prove complement of two languages is decidable using Co-Turing concepts?

Question

Let L1 and L2 be two languages such that there exist no string w that belongs to both L1 and L2. I am struggling on How to Prove that, if L1 and L2 are both co-Turing-recognizable, there exists a decidable language A such that L1 ⊆ A and L2 ⊆ A`. A` - complement of A

Answer 1

We may assume that neither of L1 nor L2 is decidable, since if either is, the solution is trivial (let A = L1 or A' = L2 if L1 or L2 is decidable, respectively). In particular, neither L1 nor L2 is Turing-recognizable.

Given that, A must be equal to the set L1 with some more elements added to it (it must have at least the elements in A1 if it is to be a superset). Because L2 is a subset of A', none of the elements added to L1 to form A can be in L2. Furthermore, we must add infinitely many items since adding finitely many items cannot render A decidable where L1 is not.

Split up the stuff not in L1 or L2 into two languages R1 and R2 such that those languages have nothing in common and every string is in exactly one of L1, L2, R1 and R2. Furthermore, choose R1 and R2 so that R1 is co-Turing-recognizable, R2 is Turing-recognizable and both sets are infinite. Let A = L1 U R1. Now, A' = L2 U R2.

1. A is co-Turing-recognizable. If w is not in L1, we can eventually recognize that fact. If w is not in R1, we can decide that fact. Therefore, we can eventually recognize that w is in neither.

2. L2 is c-Turing-recognizable. If w is not in L2, we can eventually recognize that fact. If it's not in L2, then it's either in A or R2. But we can decide whether w is in R2 since R2 is decidable. Therefore, if we recognize that w is not in L2 and decide it's not in R2, we have recognized that w is in A. Therefore, A is Turing-recognizable.

3. We saw in 1 that A is co-Turing-recognizable and in 2 that A is Turing-recognizable. Therefore, A is decidable. Consequently, A' is decidable.

Note that we sort of waved our hands there when we "split up" stuff not in L1 or L2 into two infinite languages, one co-Turing-recognizable and the other co-Turing-recognizable. It seems like it's safe to assume that in any infinite language, there must be exist a proper subset of that language which is recognizable but not decidable. You might want to look that up and/or prove separately to verify. Proof idea: the elements of any infinite set could be put into lexicographic order, in which case there is a bijection with the language of all strings over the alphabet; because there are such recognizable but undecidable languages over the set of all strings, so too must there be recognizable but undecidable languages over this set of strings. It might be important to note that (L1 U L2)' is recognizable as that might be required to make any argument rigorous.