Show that the set of algebraic numbers is countable

Question

Is this proof correct to show that the set of algebraic numbers is countable?

1. Show as a lemma that the infinite disjoint union of countable sets is countable
2. show that $\mathbb{Q}[t]_{deg \leq n}$ is countable $\forall$ n

As $\mathbb{Q}[t] = \bigcup_{i=0}^{\infty} \mathbb{Q}[t]_{deg \leq i}$ , $\mathbb{Q}[t]$ is countable by the lemma.

3. Show that $$E = \bigsqcup_{p_i \in \mathbb{Q}[t] \backslash \{0\}}^{\infty} A_i $$ . With $A_i , i \in \{0,...,n\}$ the set of roots of the polynomial $p$ is countable

(E is a disjoint union)

4. Conclude by saying that is the $A_i$ are disjoint, we are done and if they aren't, you can rewrite them as $A'_i$ that form a disjoint union equal to the union of $A_i$

Thank you!

Answer 1

Show as a lemma that the infinite disjoint union of countable sets is countable

Strictly speaking, you have to replace "infinite" by countable. Once you've done this, I think the proof is quite trivial - for example you can arrange these elements in a square array.

Also, though the above result is true, it doesn't apply straight away to your case because, as you may have probably noted yourself, the root sets of two different polynomials aren't necessarily disjoint. However, you simply have to realize that "disjoint" is actually stronger than you need - if the countability holds for disjoint union, it must also hold for non-disjoint union. (Hint: enlarge the aforementioned square array where needed and define a surjection from $\Bbb Z^2$ to the elements.)

Answer 2

The algebraic numbers are roots of polynomials and there are $d$ of them for degree $d$. The polynomials are enumerable (as integer tuples of enumerable length), hence their roots are enumerable as well.

Answer 3

The countable union of countable sets is countable. There are only countably many polynomials of degree $n$ with rational coefficients. Each has a finite number of roots (FTA).