Let $e$ be an index of an oracle Turing machine program and $k$ be some natural number. Let us say that a subset of $\mathbb N$ is *arithmetic* if it is definable in the model $\langle \mathbb N,+,\cdot,<,0,1\rangle$. Now suppose that there is a non-arithmetic oracle $A$ such that $\Psi_e^A(k)\uparrow$. Is it possible that $\Psi_e^B(k)\downarrow$ with every *infinite* arithmetic oracle $B$?

Note that it is not possible if we remove the key assumption that $B$ must be infinite. Suppose there is some non-arithmetic oracle $A$ so that $\Psi_e^A(k)\uparrow$. We can then consider the infinite binary tree $T$ consisting of all finite sequences $s$ such that $\Psi_e^B(k)$ does not halt with any oracle $B$ extending $s$ in less than length of $s$ many steps. The tree $T$ is arithmetic and therefore must have an arithmetic path $P$ (this follows from the proof of König's Lemma). Clearly $\Psi_e^P(k)\uparrow$. But of course $P$ can be finite! In this argument, there is no obvious way to guarantee that $P$ is infinite because it is possible to have an infinite arithmetic binary tree without an arithmetic path with infinitely many 1s.