I am trying to think of a question that discusses the relationship between RE, coRE and R. Namely-
is it true that for all For every language 1∉ RE there exists a language 2∉ coRE such that 1∪2∉R ?
I am not sure whether this claim is true or not. I was trying to think of examples using disjoint alphabets but so far with little success.
I'd love to get some guidance/an explanation why the claim is true/not true.
Thank you!
Since $L_1$ is not computably enumerable, it has to be infinite. It follows that $L_1$ has continuumsly many sublanguages, of which only countably many are co-computably-enumerable. So we can pick $L_2 \subseteq L_1$ as non-co.c.e., and then find that $L_1 \cup L_2 = L_1$ is not decidable.
In fact, the assumption on $L_1$ can be dropped. Specifically, it holds that for every language $L_1\subseteq \Sigma_1^*$ there exists a language $L_2\notin \text{coRE}$ such that $L_1 \cup L_2 \notin \text{R}$. The proof is quite simple and is based on the intuition, you mentioned, of manipulating the alphabets:
Let $K \subseteq \Sigma_2^*$ be some language not in $\text{coRE}$. Then, the language $L_2 = 0\cdot K$, where $0\notin \Sigma_1 \cup \Sigma_2$ is a fresh letter, is such that $L_1 \cup L_2 \notin \text{R}$. Indeed, this follows by a simple reduction: if by contradiction there is machine $T$ that decides $L_1 \cup L_2 = L_1 \cup 0 \cdot K$, then we can define machine $M$ that decides $K$ as follows. The machine $M$ on input $x$, simulates the run of $T$ on $0\cdot x$ and answers the same. Indeed, $M$ accepts $x$ iff $T$ accepts $0\cdot x$ iff $x \in K$. Note that the last equivalence follows from the fact that $0\notin \Sigma_1$ and thus $0\cdot x$ cannot be in $L_1$.
