Cardinality of a set of polynomial Given a set $A = \{x \in \mathbb{R}: \exists m \in \mathbb{N}, \exists b_0, \ldots, b_m \in \mathbb{Z} \text{ with } b_m \neq 0 \text{ and } b_mx^m + \cdots + b_0 = 0\}$. Im trying to find cardinality of A. One of the hints given were to write the set as a set of unions to get rid of quantifiers. Although I have no idea what sets would you even take a union of. Another thing that I thought about is that if $m$ is fixed, the by rational root theorem there are $m$ roots of the equation so would that imply that its cardinality is at most $m$? What would I have to include in the proof to make it formal since my arguments are just informal observations? Also just to confirm, in the set A, the $m$ is fixed right?
EDIT: would the unions would be the union of sets $A_i$ for $i \in \{0,1,\ldots, m\}$ where $A_i = \{x \in \mathbb{R}: i \in \mathbb{N}, b_0, \ldots, b_i \in \mathbb{Z} \text{ with } b_i \neq 0 \text{ and } b_ix^i + \cdots + b_0 = 0\}$? The idea being that the unions of $A_i$ will have overlaps so it is at most $m$.
 A: Let's define a set $P_m=\{b_0+b_1x+\cdots+b_mx^m: b_m\ne0 \text{ and $b_i\in \mathbb Z$ for $i=1,2,\cdots,m$}\}$ 

Define $f_m:P_m\to \mathbb Z\times\mathbb Z\times\cdots \times\mathbb Z$ ($m+1$ times) by $f_m(b_0+b_1x+\cdots+b_mx^m)=(b_0,b_1,\cdots,b_m)$. Now prove that $f_m$ is bijection and this proves that $P_m$ is countable.
Therefore, $P=\cup_{m=1}^\infty P_m$ is also countable. So let $P=\{\beta_1(x), \beta_2(x),\cdots\}$, where $\beta_i(x)$'s are polynomials. 
With this understanding, your $A=\cup_{i=1}^\infty\{x\in \mathbb R: \beta_i(x)=0\}$. 
Now Use this hint to prove $A$ is countable: Every polynomial of degree $m\in \mathbb N$ has $m$ roots, which may be complex or real.
A: There is a countable set o polynomials with integer coefficients, so the union of their sets of roots is at most countable. On the other hand, each rational number is a root of such a polynomial (of degree 1, but also of whatever degree you want), so the number of such roots is at least countable. So, countable it is.
