Here is the question that I am trying to understand its solution:
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.
Here is a solution I found online:
I think the map has typos in it, like, what is the $i$ beside $\phi_n$? why we are increasing the index of $i$ by $1$? why $i$ is starting from $0$ even though the first index of $a$ is $1$ in the definition of $M$?
Can someone help me correct all the typos in this map please?
Technically this is not an answer to the question, since it doesn‘t resolve the many difficulties the proposed solution has. But somehow I felt like proving the statement anyway.
Note that $N$ is in bijection to the set of functions $$\operatorname{Map}_f(\Bbb N, \Bbb Z) = \{f:\Bbb N\rightarrow \Bbb Z\mid f(n)=0 \text{ for all but finitely many }n\}$$ of functions with finite support.
So we might just show the following more general statement: Given two countable sets $X,Y$ and a distinguished element $0\in Y$ the set $\operatorname{Map}_f(X,Y)$ of finitely supported functions is countable.
First note that for any finite set $F$ the set $\operatorname{Map}(F,Y)$ is countable.
Then note that for countable $X$ the set $\mathcal{P}_f(X)$ of finite subsets is countable. This is the special case $Y=\{0,1\}$ of what we want to show, but we better not argue like that. Luckily we can prove it by hand: Writing $\binom{X}{k}$ for the set of subsets of cardinality exactly $k$ we find that $$\mathcal{P}_f(X) = \bigcup\limits_{k\in\Bbb N_0} \binom{X}{k}$$ is countable as countable disjoint union of countable sets.
Finally the bijection $$\operatorname{Map}_f(X,Y) \cong \bigcup\limits_{I\in\mathcal{P}_f(X)} \operatorname{Map}(I,Y\setminus\{0\})$$ implies our claim.