= Solution
For a <countable set> $A$, the <polynomials> of <polynomial degree> at most $d$ with coefficients in $A$ are described by coefficient tuples in $A^{d+1}$. The <finite Cartesian power of a countable set> result makes this a <countable set>. Taking the union over $d\geq0$ shows there are only countably many such <polynomials> altogether; restricting to the nonzero <polynomials> preserves countability.
A nonzero <polynomial> of <polynomial degree> $d$ has at most $d$ distinct <roots of a polynomial>. To see the bound, a root $r$ gives the factorization $P(x)=(x-r)Q(x)$ by the <factor theorem>, and each other root is a root of $Q$, whose degree is $d-1$; induction starts with a nonzero constant, which has no roots. Thus $\phi(A)$ is a <countable union of countable sets>, each of them finite, and is a <countable set>.
Starting with the given <countable set> $A_0$, <mathematical induction> now shows that every $A_n=\phi(A_{n-1})$ is countable. A final application of the <countable union of countable sets> theorem gives
$$
\boxed{\bigcup_{n=1}^{\infty}A_n\text{ is countable}.}
$$
This countability argument does not require that the sequence of <sets> be increasing.
Back to article page
