# Are the undecidable languages closed under complement?

Are the undecidable languages closed under complement? How can the answer be proved?

• Prove the contrapositive: the complement of a decidable language is decidable. – Yuval Filmus Sep 4 '17 at 10:10
• Try to prove a more general fact: For any self-inverse function $f : A \to A$, set $B \subseteq A$ is closed against $P$ if and only if $\overline{B} = A \setminus B$ is closed against $P$. – Raphael Sep 4 '17 at 10:34