 Solution

Rice theorem says that every nontrivial extensional property of programs computing partial functions is undecidable. Equivalently, a nontrivial index set cannot be a computable set.

Let $A$ be such an index set. Replacing $A$ by its complement if necessary, assume that the nowhere-defined function has no index in $A$. Since the property is nontrivial, choose an index $a\in A$. If membership in $A$ were decidable, we could decide the diagonal halting problem as follows. For each $x$, defineBy the S-m-n theorem, a total computable function $q$ produces an index $q(x)$ for ${\psi }_x$. If $f_{x,1}(x)$ diverges, then ${\psi }_x$ is nowhere defined and $q(x)\notin A$. If it halts, then ${\psi }_x\simeq f_{a,1}$ and extensionality gives $q(x)\in A$. A decider for $A$ would therefore decide whether $f_{x,1}(x)$ halts, a contradiction. Hence $A$ is undecidable.