Why is the language containing the Turing machines which only accept their own encoding not applicable to the diagonalization proof?

Question

I saw this question and asked myself why the original problem is not solvable through diagonalization. Let $$L = \bigl\{\langle M \rangle \mid L(M) = \{\langle M\rangle\}\bigr\}$$ Take the complement $\overline{L}$. Suppose there is a Turing Machine $M$ which decides this language. We get the following two cases:

• $\langle M \rangle \in \overline L \implies M \text{ accepts } \langle M \rangle \implies \langle M \rangle \notin \overline{L}$

• $\langle M \rangle \notin \overline L \implies M \text{ does not accept } \langle M \rangle \implies \langle M \rangle \in \overline{L}$

Is this a valid proof?

Answer 1

If $\langle M \rangle \in \overline{L}$ then by definition of $M$, $M$ accepts $\langle M \rangle$, and so $\langle M \rangle \in L(M)$. This is not the same as $L(M) = \{\langle M \rangle\}$, and so we cannot conclude that $\langle M \rangle \in L$. It could be, for example, that $M$ accepts other inputs.

If $\langle M \rangle \notin \overline{L}$ then by definition of $M$, $M$ does not accept $\langle M \rangle$, and so $\langle M \rangle \notin L(M)$. In particular, $L(M) \neq \{ \langle M \rangle \}$, and so by definition of $L$, $\langle M \rangle \in \overline{L}$. So in this case we do get a contradiction.

The conclusion is that $\langle M \rangle \in \overline{L}$, and this does not directly lead to any contradiction.