I was reading through a proof (section 16)
Let $\mathbb{Z}_0 = \mathbb{Z}$ \ $\{0\}$ and let $\mathbb{Z}_n = \mathbb{Z}^{n-1} \times \mathbb{Z}_0$ for $n \in \mathbb{N}_+ $ The set $\mathbb{Z}_n$ is countably infinite for each $n$. Let $C=\bigcup_{1}^{\infty}\mathbb{Z}^{n}$ Think of $C$ as the set of coefficients and note that $C$ is countable infinite. Let $P$ denote the set of polynomials of degree $1$ or more, with integer coefficients. The function $(a_0,a_1,…,a_n) \rightarrow a_0x+a_1x+…+a_nx^n$ maps $C$ onto $P$, hence $P$ is countable. For $p \in P$, let $A_P$ denote the set of roots of $P$. A polynomials of degree $n$ has at most $n$ roots, by the fundamental theorem of algebra, so in particular $A_p$ is finite for each $p \in P$. Finally, note that $\mathbb{A}=\bigcup_{p \in P}A_{p}$ and so $\mathbb{A}$ is countable. Of course $\mathbb{N} \subset \mathbb{A}$ so $\mathbb{A}$ is infinite
The step defining $C$ is what confuses me. I understand that for any finite $k$, $C=\bigcup_{1}^{k}\mathbb{Z}_{n}$ is countable as the finite product of countably infinite sets is countable. But as I understand, the infinite product of sets that include at least 2 elements is uncountable.
What is the distinction here that makes $C$ countable in this case?
Note: I've seen many proofs for the Algebraic numbers being countable, I'm specifically interested in understanding this "limit" for products of sets in relation to cardinality. I'm having a difficult time understanding the distinction of $\mathbb{Z}^{n-1} \times \mathbb{Z}_0$ where $n \rightarrow \infty$ as opposed to $\mathbb{Z}^{\mathbb{N}}$ which is obviously uncountable
Best Answer
$\mathbb{Z}_n$ consists precisely of the finite sequences of integers with $n$ entries, and last entry not equal to $0$.
If you define $C_k = \cup_{1}^k\mathbb{Z}_k$, then $C_k$ consists precisely of the finite sequences of integers, each of length at most $k$, and last entry not equal to $0$.
Now, what is $C=\cup_{1}^{\infty}\mathbb{Z}_k$?
First, note that it is not a "limit". You can't write $C=\lim_{n\to\infty}C_n$, because "taking the limit" is not an operation you do with sets, at least not without a very precise definition which you have not provided.
Rather, $C$ is a union. A union of infinitely many sets, but still just a union. What does a union of sets contain? It contains precisely the elements that are in at least one of the sets in the union. That is: $$x\in C\iff \text{there exists a }k\text{ such that }x\in \mathbb{Z}_k.$$
So $C=\cup_{i=1}^{\infty}\mathbb{Z}_k$ consists precisely of the elements that are in at least one of the $\mathbb{Z}_k$. That is: the elements of $C$ are precisely the finite sequences of integers with last entry nonzero. Why? Because it contains everything in $C_1$ (sequences of length 1), in $C_2$ (sequences of length 2), in $C_3$ (sequences of length 3), etc. And nothing else.
No element of $C$ has infinitely many terms: they all have finitely many terms. You never "get" to $\mathbb{Z}^{\mathbb{N}}$: no element of $\mathbb{Z}^{\mathbb{N}}$ (that is, no infinite sequence of integers) lies in $C$, because no $\mathbb{Z}_k$ contains any infinite sequences.
The set $C$ is countable because there are only countably many sequences of length $1$, and countably many sequences of length $2$, and countably many sequences of length $3$, and so on, and this countable union of countable sets exhausts all of $C$. You are correct that the set of infinite sequences is uncountable, but that collection is not at play here.