Is $L = \{ \langle \langle \ M\ \rangle \rangle \ | \ M \ \text{does not accept}\ 010 \} $ Turing recognizeable?

Question

I'm working on the following problem:

Is the following language Turing recognizable (recursively enumerable) ?

$$L = \{ \langle \langle \ M\ \rangle \rangle \ | \ M \ \text{does not > accept}\ 010 \} $$

The way I see it: Suppose that a machine $M$ loops forever on $010$. If a $TM$ recognizes $L$, it should accept $M$ in that case. But that means that it should know if $M$ loops forever or not, which is not possible. So, $L$ is not Turing recognizable.

Is my proof correct, and can it be more formal?

Answer 1

You are presenting an argument which falls short of being a proof. In particular, it is not clear why a Turing machine recognizing $L$ should know whether $M$ loops forever or not; indeed, it is not so clear what do you mean by know in this context.

Here is one way a proof could go. Suppose that $L$ were r.e. The language of Turing machines which do accept $010$ is also r.e. By running both machines in parallel, we can decide whether a given Turing machine accepts $010$, i.e., we could solve the halting problem, which we know is undecidable. Therefore $L$ cannot be r.e.