the complement of A_TM
The complement of A_TM, written co-A_TM, is the set of all (M, w) pairs where M does NOT accept w — either M rejects, or M loops forever and never accepts. It is the exact opposite roster of A_TM, formed by flipping every yes to a no. Studying it turns the abstract decidable / recognizable / co-recognizable picture into something concrete: co-A_TM is the textbook example of a language that is not even recognizable.
The argument is short and pretty. There is a theorem that a language L is decidable if and only if BOTH L and its complement are Turing-recognizable. We already know A_TM is recognizable (the universal machine accepts whenever M accepts) but undecidable (the diagonal proof). Now suppose, for contradiction, that co-A_TM were also recognizable. Then A_TM and co-A_TM would both be recognizable, and the theorem would make A_TM decidable — contradicting what we just proved. Therefore co-A_TM is unrecognizable.
There is a satisfying intuition for why confirming co-A_TM membership is hopeless. To accept a pair (M, w) in co-A_TM you would need to confirm that M never accepts w, but 'never' is the trouble: at any finite moment M might still be running and could yet accept on the very next step. You can confirm acceptance by waiting long enough to see it, but you can never confirm permanent non-acceptance by waiting, because no finite wait rules out a future acceptance. That asymmetry is exactly why A_TM is recognizable and co-A_TM is not.
Suppose M loops forever on w without accepting. Then (M, w) is in co-A_TM, but a recognizer watching M run sees only an endless computation: it can never be sure M will not accept on some future step, so it can never safely say yes.
co-A_TM is unrecognizable: if it were recognizable, A_TM would be decidable, a contradiction.
co-A_TM is co-recognizable (its complement A_TM is recognizable) but not recognizable. Being co-recognizable is not the same as being recognizable.