Let $X$ be a smooth projective curve over an algebraically closed field $k$. Let $S$ and $T$ be two disjoint infinite subsets of $X(k)$. Does there exist a morphism $f:X \to \mathbb{P}^1$ and two infinite subsets $A$ and $B$ of $\mathbb{P}^1(k)$ such that the preimages of $A$ and $B$ by $f$ lie inside $S$ and $T$ respectively?


Edit: As pointed out by Dragon, this may fail when $k$ is countable, but I would like to know the answer in the general case.



Partitions of algebraic curves and maps to $\mathbb{P}^1$
No in case $k$ is countable of characteristic zero by a simple diagonal argument. Assume $X$ does not have genus $0$ for simplicity. Number the maps $f\_1, f\_2, \ldots$. For each $n$ by induction choose $S\_n$ and $T\_n$ finite disjoint subsets of $X(k)$ such that $f\_i$, $i \leq n$ maps an element of $S\_n$ and $T\_n$ to the same point of $\mathbf{P}^1$. Given $S\_n$ and $T\_n$ you can choose $S\_{n + 1}$ and $T\_{n + 1}$ by adding a single point $P$ to $S\_n$ and $Q$ to $T\_n$ with $P \not = Q$ and $P, Q \not \in S\_n \cup T\_n$ with $f\_{n + 1}(P) = f\_{n + 1}(Q)$. This is possible as almost all fibres of $f\_{n + 1}$ have cardinality $> 1$. Then take $S = \bigcup S\_n$ and $T = \bigcup T\_n$.


