Given a non recursive set $A$, is it always true for a set of odd or for a set of even numbers of the set $A$ to be non recursive?

Question

Let's say we have a set $A$, which is non recursive, a set $B = \{x\in A|x\text{ is odd}\}$ and a set $C = \{x \in A|x\text{ is even}\}$.

Is it always true that at least one of $B$ and $C$ is non recursive as well?

If both of them were recursive, wouldn't that mean that $A$ is recursive as well, since $A = B \cup C$ and recursion is conserved through the union operation? So as a result, not only is that possible, that is actually necessary.

I have tried looking up something on the topic of the union of a non recursive and a recursive sets on the internet, but haven't found anything.

Is my judgement correct or am I missing something?

Edit: Is it always true, that at least one of $B$ and $C$ is non recursive as well, rather than "Is it possible"

Answer 1

If both $B$ and $C$ were to be recursive, then their characteristic functions $\chi_B$, and $\chi_C$ would be recursive, and thus since $\chi_A = \chi_B + \chi_C$ plus the fact that recursive functions are closed under addition, $\chi_A$ would be recursive and therefore $A$ too.

When saying $\chi_A = \chi_B + \chi_C$ I used the fact that $B \cup C = A$ and the fact that they are disjoint, but, if the later does not hold you still have $\chi_A(x) = \min(\chi_B(x) + \chi_C(x),0)$ and $\min$ is also recursive.

Edit: fixed phrasing on second paragraph.

Answer 2

Your judgement is correct, at least one of B or C must be non-recursive, since recursive sets are closed under (finite) union. This is fairly simple to prove, as given a procedure for deciding B and a procedure for deciding C, one can create a procedure for deciding B union C = A that is simply to run the procedure for B and the procedure for C, and return true if either of them do and false otherwise.