User blog comment:Vel!/Formal logic challenge/@comment-1605058-20140710070627/@comment-1605058-20140710072845

Note on why it cannot be bijection: no such bijection is computable (but it's nontrivial). We can also show it directly - if \(\phi\) is bijection, there is unique \(k\) with \(\phi(k)(n)=n\) (identity function), and we can decide \(A=\{k\}\) easily.