I am currently self-studying Baby Rudin and I have written what I think is a solution to Exercise 2 of Chapter 2. It reads:
$\textbf{Exercise 2}:$ A complex number $z$ is said to be algebraic if there are integers $a_0,a_1,\dots,a_n$, not all zero, such that $$a_nz^n+a_{n-1}z^{n-1}+\dots+a_1z+a_0 = 0.$$Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$, there are finitely many equations satisfying $$n+|a_0|+|a_1|+\dots+|a_n| = N.$$
My proof does not use this hint, hence I am wondering if it is valid. It uses Theorem 2.12 from Rudin which states that a countable union of countable sets is countable. Here's the proof:
Proof.Consider the set (where $n \in \mathbb{N}$) $$E_n = \{p \in \mathbb{Z}[x] : \deg(p) = n\}.$$ Since $E_n$ is in bijection with $\mathbb{Z}^n$ (the map $(p = a_nx^n+\dots+a_0)\mapsto(a_n,\dots,a_0)$ is the obvious one), $E_n$ is countable.Now, consider $$S_n = \bigcup_{p \in E_n} \{z \in \mathbb{C} : p(z) = 0\}.$$ Each polynomial $p(x)$ in $E_n$ has at most $n$ roots in $\mathbb{C}$, so $\{z \in \mathbb{C} : p(z) = 0\}$ is finite. Since $E_n$ is countable, by Theorem 2.12, $S_n$ is countable. (It can be noted that, in abstract algebra language, $S_n$ contains each algebraic complex number whose minimal polynomial degree is $n$).Now, consider $$A = \bigcup_{n=1}^{\infty} S_n$$
By Theorem 2.12, $A$ is countable, and it is clear that every algebraic number is contained in this set. Hence, the algebraic numbers are countable. $\blacksquare$
Any help is appreciated. Thank you!