Is L={⟨M⟩∣L(M)∈P}in RE, when P is a non-trivial property and Z∈P is infinite and in RE or R?

Question

Let $Z \subseteq \Sigma^*$ be an infinite language such that $Z \in \text{RE}$ or $Z \in \text{R}$.
Let $P \subseteq RE$ be a non-trivial language property

Now define the language:

$$ L = \{\langle M \rangle \mid L(M) \in P \} $$

Question:
If $Z \in P$ is infinite (and either in RE or in R), is $L \in \text{RE}$?
In other words, does the fact that $Z$ is infinite and recognizable (or decidable) help in making $L$ semi-decidable?

I know that by Rice's theorem $L\notin R$, but I am really sure It's also not in RE

Any clarifications, counterexamples, or connections to Rice’s theorem would be appreciated.

Answer 1

The answer is negative, and the following is a counter-example. The property $P_1 = \{ L: L\in \text{RE and $L$} \text{ is infinite}\}$ is such that $L_{P_1} = \{ \langle M\rangle: L(M)\in P_1\}$ is not in $\text{RE}$. Yet, the property $P_2 = \{ L: L\in \text{RE and $L$} \text{ is non-empty}\}$ is such that $L_{p_2}\in \text{RE}$. Both properties contain $Z$, yet one is in $\text{RE}$ and the other is not. In fact, one can show that if a nontrivial semantic property $P$ does not contain the empty language, then $A_{TM} \leq_m L_P$. The latter gives us more information than Rice’s theorem, yet does not give us “all the information”. Specifically, in the above example, both $P_1$ and $P_2$ do not contain the empty language, and the fact that $A_{TM} \leq L_{P_i}$ does not give us any information on whether $L_{P_i}$ is in $\text{RE}$, it only tells us that $L_{P_i}$ is not in $\text{coRE}$.

This may interest you: when the universal language does not belong to the property, then we can conclude that language is not in $\text{RE}$, but your question is far from this case as well. See here.