# Proof that there are countable many algebraic numbers

I understand that this question has been asked before (or variants of it at least), but most of the proofs I have seen seem to be quite complicated and I wanted to clarify if this proof is valid.

Firstly, the book I am reading defines an algebraic number as:

A number is algebraic if it is the root of a polynomial with integer coefficients

Now consider the set $S_n$ which denotes the roots of all the polynomials with integer coefficients with degree $n$, if $x_0$ is a root of this polynomial and also a root of a another polynomial of degree $n' > n$ then $x_0$ is always contained in the lower set and not repeated. Clearly, there is a injection between the polynomials $S_n$ and $\mathbb{Z}^{n+1}$ (for example, where the elements of $\mathbb{Z}^{n+1}$ denote the coefficients of the polynomial), so $S_n$ is countable. There are additionally a countably infinite number of $S_i$'s (i.e. $S_1, S_2, \ldots , S_n$).

Additionally by definition, $$\bigcup_{i=1}^n S_i = A, \text{ where } A \text{ denotes the set of all algebraic numbers}$$

Since the union of a countably infinite number of countably infinite sets is also countably infinite, the set of algebraic numbers is countable.

• Close : you are counting the polynomials, now you have to count their roots :-) – Nicolas FRANCOIS Nov 16 '16 at 6:08
• @NicolasFRANCOIS, can you please enlighten me on this? I can't quite think of a way to rigorously extend this argument to that – q.Then Nov 16 '16 at 6:11
• Well, a polynomial of degree n has at most n roots.... – fleablood Nov 16 '16 at 6:27

This is fine. Most of the proofs of this statement just make precise the injections and/or show that the union of countably-infinite countable sets are countable. Mild note is that there is an injection between $S_n$ and $\mathbb{Z}^{n+1}$ and not $\mathbb{Z}^n$.