Is this set of functions countable? I want to know if the set of functions $F=\{f:\mathbb{Z}\to\mathbb{Z}:f(n)\neq 0 \text{ for finitely many n}\}$.
I haven't done a lot of progress really, but considering define the sets $A_n=\{f: \mathbb{Z}\to \mathbb{Z}\ : f(n)=0\}$. If I could prove that each $A_n$ is countable then the union of them (which is the set $F$) would be countable, but how can I prove the countability of $A_n$?

Hint: First see how many functions do you have satisfying:
$f:\mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(n) \neq 0$ for exactly one $n$. This can easily be seen to be countable. Then $f(n) \neq 0$ for finitely many $n$ is just $f(n) \neq 0$  for exactly 1 $n$ union $f(n) \neq 0$ for exactly 2 $n$ , union.....
Alternately, think of these functions as the " bi- infinite sequences" $(...,a_1,a_2,...)$ where all but finitely many $a_i's$ are zero.
