I'm working on the following problem: let $W\_e$ be the computably enumerable set which is the domain of the $e$-th Turing program, and $K$ be the Halting problem, at which level of the arithmetic hierarchy does each of the following sets appear?

1. $A\_1=\{e: W\_e$ contains no even numbers $\}$
2. $A\_2=\{e: W\_e$ contains some element of $K\;\}$
3. $A\_3=\{e: W\_e$ contains every element of $K\;\}$
4. $A\_4=\{e: W\_e$ is infinite $\}$

I think $A\_1 \in \Pi\_1$ because it can written as $A\_1=\{\exists x \in W\_e (\forall n \neg (x=2n) )\}$, i.e. one universal and one bounded quantifier. I'm a little bit worried that the bounded set is c.e., does it change things?

Since $K \in \Sigma\_1$ and $A\_2=\{e: \exists x \in W\_e (x \in K)\}$, $A\_2$ is in $\Sigma\_1$, but it's not $\Pi\_1$ so it's not $\Delta\_1$.

$A\_3=\{e: \forall x \in K (x \in W\_e)\}$ so it's $\Pi\_1$

I'm not sure what to do with $A\_4$

I would also like to view these sets in the context of Turing machine, but can't wrap my head around it.