# The field closure of a countable union of countable fields is countable?

If $K_1 \subset K_2\cdots \subset K_n \subset \cdots$ is a tower of countable fields then their union $\bigcup_n K_n$ is a countable field.

If $\{K_a\}$ is a countable family, but not a tower, of countable subfields of a field $F$, the field closure $\langle \bigcup_a K_a\rangle$ in $F$ is countable? Note that here the field closure is not an algebraic extension of fields.

This question came from The exponential extension of $\mathbb{Q}$ is a proper subset of $\mathbb{C}$?.

Take the union of the fields $K$, it is a countable set. Now define by induction a sequence of sets, whose union is a field containing $K$. At each step you gather all the products and sums and inverses of elements from the previous step.
Now since $K$ is countable the first step of the induction is countable, and by induction all the sets generated are countable. Finally, the union of these is countable and it is not hard to see it is a subfield of $F$.
Hint: look for a tower of subfields of $F$ whose union is the field you're looking for.