Let $M_i$ be the free $\mathbb Z$ module $\mathbb Z,$ and let $M$ be the direct product

Question

Here is the question I am trying to solve:

Let $M_i$ be the free $\mathbb Z$ module $\mathbb Z,$ and let $M$ be the direct product $\Pi_{i\in \mathbb Z^+}M_i.$ Each element of $M$ can be written uniquely in the form $(a_1, a_2,a_3, \dots )$ with $a_i \in \mathbb Z$ for all $i.$ Let $N$ be the submodule of $M$ consisting of all such tuples with only finitely many nonzero $a_i.$ Assume $M$ is a free $\mathbb Z$ module with basis $\mathcal B.$

$(a)$ Show that $N$ is countable.

My idea is:

I should show that there is a bijection from the set of natural numbers to $N.$ I am guessing that the following map in the following picture might work (I found it online) :

But I am not sure what is $N^+$ in the solution, where should be the beside $\phi_n$? Why the domain is $\mathbb Z^n$?

Could anyone help me answer these questions please?

Answer 1

I think $N^+$ is simply the positive natural numbers and $\phi_n$ is an inclusion of $\mathbb{Z}^n$ into $N$, so we end up writing $N$ as a countable union of (countable) images of $\mathbb{Z}^n$. If you want a bijection of $N$ with a countable set, click below

Alternatively, consider sending $(a_1, a_2, \dots)$ to the continued fraction $$a_1+\frac{1}{a_2+\frac{1}{\dots}}$$ when $a_i\geq 0$ (terminate the fraction appropriately when you encounter a tail of zeroes). This gives a nonnegative rational number when only finitely many $a_i$s are nonzero and is bijective (note that $a_1$ is the floor of the continued fraction...) Using a bijection $\mathbb{Z}\to\mathbb{N}$, one can get an bijection $N\to\mathbb{Q}_{\geq 0}$ showing that $N$ is countable.