My problem is as follows: Show that a set $A$ of expressions is effectively enumerable iff there is a decidable set $B$ of pairs $\langle\alpha, n\rangle$ (consisting of an expression $\alpha$ and an integer $n$) such that $A=\operatorname{dom}B$. (The $\operatorname{dom}B$ is the set of all objects $x$ such that $\langle x, y \rangle \in B$ for some $y$.)
My problem is that the textbook says that this problem is very difficult problem... But I cannot understand why it is so difficult. Since effective enumerable set of expressions is just a list, why not just take a list with the integer subscribed? which makes the problem easy... Please help me understand the problem.
