My question is this:
Let Y and N be two languages over a fixed alphabet Σ such that both Y and the complement of N are Turing-recognisable.
Prove that if N ⊆ Y , then there exists a decidable language S that splits Y and N, that is: N ⊆ S ⊆ Y.
I already have a theorem which says if a language and its complement are both Turing-recognisable, then the language is decidable. I don't quite know how to apply that to S in this proof though.
Your help is appreciated, thanks!