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I know that if decidability of $A \cap B$ and $A \cup B$ doesn’t guarantee the decidability of any of $A$ or $B$. We can prove that:

ATM is not decidable. Since decidable languages are closed under complementation, ATM' is also not decidable. But $\textrm{ATM} \cup \textrm{ATM}' = \Sigma^*$ and $\textrm{ATM} \cap \textrm{ATM}' = \varnothing$ both are decidable.

But what about recognizability?

Rémi
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Luis Ramirez
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Then answer is no.

The language $A$ of Turing machines that halt on every input is not recognizable, but its complement $A^c$ is also not recognizable. Then both $A \cup A^c = \Sigma^*$ and $A \cap A^c = \varnothing$ are recognizable, but neither are $A$ nor $A^c$.

Rémi
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