A question about "small" sets of algebraic numbers

Question

For the sake of brevity let me use the following terms. A subset $X$ of $\bar{\mathbb{Q}}$ will be called "small" if for any number field $K$, the intersection $X\cap K$ is finite. Similarly, a set $Y\subset \bar{\mathbb{Q}}$ will be called "large" if for any number field $K$, $Y$ contains all but finitely many elements of $K.$ My question is whether the following claim is true.

Claim: Let $f\in \bar{\mathbb{Q}}[x, y]$ be a polynomial which is non constant in $y,$ and let $X\subset \bar{\mathbb{Q}}$ be a "small" set. Then the set $Y=\lbrace y\in \bar{\mathbb{Q}}$: there exists $x\in X$ such that $f(x,y)=0\rbrace$ is not "large".

Answer 1

Not if $f(x,y)$ cuts out a curve of genus at least $2$.

Consider such a polynomial $f(x,y)$. Choose an ordering of number fields, such as by discriminant. Consider the set $X = \{ x \in \overline{\mathbb Q} :$ there exists $y \in \overline{\mathbb Q}$ such that $f(x,y)=0$ and $\mathbb Q(y)$ comes before $\mathbb Q(x)$.

Then $Y$ contains the set of all $y$ such that there exists $x \in \overline{\mathbb Q}$ such that $f(x,y)=0$ and $\mathbb Q(x)$ comes after $\mathbb Q(y)$.

It is sufficient to prove that the first set is small. Then by symmetry, and the fact that for all but finitely many $y$ there is at least one such $x$, the second set is large.

But this follows from Mordell's conjecture. Given a number field $K$, there are only finitely many number fields that come before $K$ or a subfield of $K$. For each such field $L$, there are finitely many pairs $(x,y) \in K \times L$ with $f(x,y)=0$, because all such pairs are $KL$-points of the curve $f(x,y)=0$. Adding those all together, only finitely many values of $x$ occur.