Is Universality/Completeness Problem of Turing Machines RE or non-RE?

Question

Consider $Universality_{TM} = \{<M> | \space M \space is \space TM \space and \space L(M)=\sum^{*} \}$
According to Rice Theorem, I know that $Universality_{TM}$ is undecidable.
But is $Universality_{TM}$ Recursively Enumerable or non-RE?

I think it is non-RE, because we can't simulate M and check for every string $w$ whether w is in L(M) since there are infinite strings and also M may loops on some w. But this is informal understanding.
I am in search of formal proof (probably using Reduction).

Answer 1

Let $L_{\in} = \{\langle M, x\rangle \mid M \text{ is a TM and }x\in L(M)\}$. It is well known that $L_{\in}$ is undecidable. However, it is RE: simulate the computation of $x$ on $M$ and wait for it to halt and accept.

Since $L_{\in}$ is undecidable and RE, that means that its complement $\overline{L_{\in}}$ is not RE. Now, let us show that $\overline{L_{\in}}$ is many-one reducible to $L_u = \{\langle M\rangle\mid M \text{ is a TM and }L(M) = \Sigma^*\}$:

Let $\langle M, x\rangle$ be an input. We create a Turing machine $M_x$ such that on input $y$:

• $M_x$ simulates the $|y|$ first steps of computation of $M$ on input $x$;
• if this simulation halts and $M$ accepts $x$, then $M_x$ loops infinitely;
• otherwise, $M_x$ halts and accepts.

Now, let us distinguish:

• if $\langle M_x\rangle\in L_u$, that means that $M_x$ always halts on any input $y$, which means that the computation of $M(x)$ never halts, so $\langle M, x\rangle \in \overline{L_{\in}}$;
• if $\langle M_x \rangle \notin L_u$, that means that there exists $y$ such that $M_u(y)$ never halts, so the computation of $M(x)$ halts and accepts after $|y|$ steps, so $\langle M, x\rangle \notin \overline{L_{\in}}$.

Since $\overline{L_{\in}}$ is not RE and $\overline{L_{\in}}\leqslant_m L_u$, that means that $L_u$ is not RE.

Exercise: prove that $\overline{L_u}$ is not RE.