# Showing that set the of algebraic numbers are countable [Proof verification]

Consider the polynomial with (variable) integer coefficients : $$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$$, where $$x$$ varies over the field of complex numbers. The set of all zeros of the polynomial gives the set of all algebraic numbers.

Now, the number of roots of a polynomial of $$n$$ th degree has at most $$n$$ roots. Again, [from the hint of Rudin], we can write only in a finite number of ways $$n +|a_0| +|a_1|+...|a_n|= N \in \mathbb{N}$$, i.e, the combination of $$a_i$$ 's summing up to $$N$$ is a finite set; the permutation of those $$a_i$$ 's also form a countable set. Now, construct the set

$$S_n^N =\{ x,\ a_i,\ n, \ N : a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n = 0$$ and $$n +|a_0| +|a_1|+...+|a_n|= N\}$$, which is evidently finite from the aforementioned arguments.

Now, $$\bigcup \limits_{n=1}^{\infty} S_n^N$$ is at most countable, being a collection of countable sets. As we further do the union $$\bigcup \limits_{N=1}^{\infty}[ \bigcup \limits_{n=1}^{\infty} S_n^N ]$$ we get again a countable set, which equals the set of all algebraic numbers.

Is this legit? [ although not fully original ]

Thus the number of roots of $$\cup_n$$P(n) = k × k = k.