When learning basic computability theory we are usually given as examples of arithmetic sets sets which are complete for their level of the arithmetic hierarchy (like the halting set, the set of indices of total functions etc.). Our teachers are of course quick to point out that not all sets are complete in this way. However, one could still be led to believe that arithmetically complex sets have quite a lot of computational power. Surely this cannot be the case.
Can you give me an example of an arithmetic set $A$ of some (precise) complexity $\Sigma^0_n$ (or $\Pi^0_n$ or whatever) such that $A$ does not compute all the sets of lower complexity? Can you give me one where $A$ does not compute the halting problem?
I should point out that a friend (who likes models of PA) has already given me an example by constructing a particular model of PA whose standard system contains a completion of PA but omits the halting problem. While this example is perfectly fine, it feels like the problem should have a much more elementary solution.
If you take the usual construction of a Turing incomplete real between $0$ and $0'$, via the Kleene-Post method, the real $\xi$ that is constructed is automatically $\Delta^0_2$, and in particular $\Sigma^0_2$, although $\xi$ does not compute $0'$ by construction. In the construction, we can also ensure $\xi$ is not $\Sigma^0_1$, so that the complexity is strictly $\Delta^0_2$.
In practice, every type of real made by a priority construction can additionally be made $\Sigma^0_n$ for sufficiently high $n$, where $n$ is large enough to cover all the oracles needed to make the construction effective. So the literature can be mined for other examples of oddly behaved arithmetical sets.