Turing machine accepts two different strings

I am having hard time to proving this problem

$C=\{\langle M \rangle \mid M \text{ is a Turing Machine , } L(M) \text { only contains two different strings}\}$

some ideas that i have tried are :

1. i have to prove if $C$ is not recursively enumerable. I know if $A$ is recursively enumerable, and $co\text{-}A$ is recursively enumerable, then $A$ is recursive. so the first idea is proving if $co\text{-}C$ is recursively enumerable, but i stuck in how to find $u$ and $v$ so that $co\text{-}C$ will not accept.

2. as co-halting problem is not recursively enumerable, i think $C$ can be reducible from co-halting problem. then if TM $M$ that recognizes $C$ can be use to solve co-halting problem that it proves the problem, but i don't really understand how exactly to prove it.

3. the idea is the same with #2, but using $A_{TM}$.

Can anyone give me some explanation this problem, or what the best way to prove the problem.

thank you a lot. EDIT : the problem is how to prove $C$ is not recursively enumerable

migrated from stackoverflow.comNov 11 '17 at 17:58

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• Your "problem" is a set of machines; what exactly are you trying to prove? – Scott Hunter Nov 5 '17 at 0:18
• oh, i'm sorry, the problem is how to prove that C is not r.e as i edited the question. thanks – paiman Nov 5 '17 at 1:30
• This seems relevant: math.stackexchange.com/a/38018 – Scott Hunter Nov 7 '17 at 18:32
• Hint: Prove that co-C is r.e. ( run M on input 1,2,3,... and as soon has it accepts more than ...); then apply Rice ... – Vor Nov 11 '17 at 19:41