Iranian Mathematical Olympiad

Define $f(A) = \max(A)$ when $A$ is finite. Evidently, $f$ is $A$-predictor when $A$ is finite. We extend $f$ to all subsets of $\mathbb{N}$. We say two subsets $A,B$ are equivalent if $B$ is derived from $A$ by adding and deleting a finite number of elements; i.e. $A\Delta B$ is finite. This is an equivalence relation. By the Axiom of Choice, we can select an element from each class of equivalency. For an arbitrary proper subset $A$, let $S_A$ be the selected element from the class of $A$. So, $A\Delta S_A$ is finite. Define $f(A) = \max(A\Delta S_A)$ when $A \neq S_A$ and define $f(S_A)$ arbitrarily. We claim this function is $A$-predictor for all subsets $A \subseteq \mathbb{N}$.

Let $x$ be a natural number such that $x \notin A$. $A$ and $A \cup \{x\}$ are equivalent, so $S_A = S_{A \cup \{x\}}$. So
$$f(A \cup \{x\}) = f(A\Delta \{x\}) = \max((A\Delta \{x\})\Delta S_A) = \max((A\Delta S_A)\Delta \{x\}).$$
For $x > \max(A\Delta S_A)$ we have $(A\Delta S_A)\Delta \{x\} = (A\Delta S_A) \cup \{x\}$. therefore $f(A \cup \{x\}) = x$ and the claim is proved. □