The question asks me to prove that the set of all algebraic integers are countable.
Now the hint given in Rudin is :
$N=|a_0|+|a_1|+...+|a_n|+n$. The equation has finitely many solutions.
Now, the sum of modulo of the coefficients of the equation, and the degree of the equation is always a positive quantity.
So, fixing n, there are finitely many ways in which I can change the coefficients to get $N$. Since $ N \in Z$. The infinite subset of a countable set is countable and the countable union of countable sets is also countable.So from here I can conclude that there are finitely many polynomials with integer coefficients.
By the fundamental theorem of algebra there are only roots to a polynomial equation. So can I conclude from here that the total number of algebraic numbers is countable. I think I have confused myself in using the hint. Some suggestions would be useful. I did go through the previous post similar to this, but the approach was different I saw.
Let's try adding some notation, it can make many topology problems easier. Let $$A_n = \{x: a_nx^n + a_{n-1}x^{n-1}+\ldots + a_0\hspace{4mm}\text{and}\hspace{4mm} a_0,\ldots, a_n\in\mathbb{Z}\}$$ So $A_n$ is the set of algebraic numbers who are the solution to a polynomial with coefficients from $\mathbb{Z}$ and is of order at most $n$. It follow that $A = \bigcup_{n\in\mathbb{N}}A_n$ will be the set of all algebraic numbers. If you show that $A_n$ is finite then since the countable union of finite sets is countable, then $A$ will be countable.
You have the idea that $A_n$ is countable but the wording could be improved. Using the hint notice that there is a finite number of polynomials of degree $n$ whose roots could belong in $A_n$. Next, each of these polynomials have a finite number of roots. It follows that $A_n$ must be finite. Hence $A$ is countable.