Decidability of language of TMs which accept only their Gödel number [duplicate]

Question

I am trying to prove that $L = \{\langle M \rangle \mid L(M) = \{\langle M \rangle \}\}$ is undecidable, where $\langle M \rangle$ is the code of the TM $M$, and $L(M)$ the language recognized by $M$.

I think I can't use Rice's theorem, so I tried to find a reduction. The halting problem for example does not help me in this case. Do you have any idea how to prove it?

Answer 1

The recursion theorem states that if $A(x,y)$ is a two-input Turing machine then there is a Turing machine $M$ such that $M(x) = A(x,\langle M \rangle)$. Moreover, $M$ can be constructed from $A$ effectively (i.e., using an algorithm). Using this, we can reduce the halting problem to your problem. Given a Turing machine $B$ and an input $z$, construct a new Turing machine $A(x,y)$ which acts as follows:

• If $x = y$ then simulate $B$ on $z$.
• Otherwise, enter an infinite loop.

Use the recursion theorem to construct a new machine $M$ that on input $x$ acts as follows:

• If $x = \langle M \rangle$ then simulate $B$ on $z$.
• Otherwise, enter an infinite loop.

Then $M \in L$ iff $B$ accepts $z$.