Is $\cup_{n=0}^{\infty} P_{n}=\{(a_0,\cdots, a_n) : a_k \in \mathbb{N}\}$ countable? The orginal question is: 

A complex number z is said to be algeraic if there are integers $a_0, \cdots,a_n$, not all zero, such that $a_0z^n + a_1z^{n-1}+ \cdots+a_n=0.$ Prove that the set of all algebraic numbers is countable.


proof(sketch) $\:$ Define $P_{n}=\{(a_0,\cdots, a_n) : a_0,\cdots,a_n \in \mathbb{Z^{+}}\}$ . Since $\mathbb{Z^{n+1}}$ is countable, we get $P_{n}$ is countable for all $n \in \mathbb{Z^{+}}$. The set of all polynominals with integer cofficients is $\cup_{n=0}^{\infty}P_{n}$. Since $P_{n}$ is countable for all $n \in \mathbb{N}$,  we have $\cup_{n=0}^{\infty}P_{n}$ is countable.


But there's an contradiction : 

If $\cup_{n=0}^{\infty}P_{n}$ is countbale, we have $\mathbb{Z^{+}} \times \mathbb{Z^{+}} \times \mathbb{Z^{+}} \times \cdots$ is countable.Which contradicts the fact that the infinite-tuples is uncountable.


My question is :

Is $\cup_{n=0}^{\infty}P_{n}$ countable?

If it is countable,  what's wrong with the contradiction?

If it is uncountable, what's wring with my provement?
 A: 
Is $\cup_{n=0}^\infty P_n countable? 

Yes, and your proof is correct (if the unproved assertions are assumed to be known) -- although note that you probably want $a_0$ specifically to be non-zero, to avoid double counting.

If it is countable, what's wrong with the contradiction? 

The set $P = \cup_{n=0}^\infty P_n$ contains all finite sequences of integers. The set $\mathbb Z^\infty = \mathbb Z \times \mathbb Z \times \cdots$ contains infinite sequences of integers. You can identify $P$ with a subset of $\mathbb Z^\infty$ by identifying a sequence $(a_0, \ldots, a_n)$ with the sequence $(a_n, a_{n-1}, \ldots, a_0, 0, 0, 0, \ldots)$, but no element of $P$ corresponds to non-terminating sequences like $(1, 1, 1, \ldots)$ or $(1, 2, 3, \ldots)$.
A: $\Bbb Z^+ \times \Bbb Z^+\times \cdots$ is strictly larger than $\bigcup P_n$. For instance, the former contains the infinite sequence $(1, 1, 1, \ldots)$. Look closer into this, and you will see that there is no contradiction.
Yes, $\bigcup P_n$ is countable.
A: The set of all $n$-tuples of natural numbers with varying $n\geq 0$ is countable.
For this, consider the bijection $\bigcup_n {\Bbb N}^n\rightarrow{\Bbb N}_0$ where $(a_1,\ldots,a_n)$ is mapped to $\prod_i p_i^{a_i}$ and $(p_i)_i$ is the infinite sequence of prime numbers; here the empty word is mapped to 0. 
Since there is a bijection between the set of integers and the set of natural numbers, this result holds also in your setting. 
