Cardinality of all sequences of non-negative integers with finite number of non-zero terms. (NBHM 2012)

Question

Consider the set $S$ of all sequences of non-negative integers with finite number of non-zero terms.

1. Is the set $S$ countable or not?
2. What is the cardinality of the set $S$ if it is not countable?

My intuition is the set is countable. The sequence has only finitely many non-zero terms. For any fixed $N$ consider the set $A_N$ which contains all sequences $\{a_n\}$ s.t. $a_k = 0$ $\forall$ $k >N$. The set $A_N$ is countable as the first $N$ terms of a sequences can be filled up by non-negative integers in a countable number of ways. So $A_N$ is countable and $S$ is a countable union of countable sets. So $S$ is countable.

I do not know if it is true or false. If it is false please identify the mistake. Thank you for your help.

Please suggest me a book where I shall get sufficient number of such type of problem to clear basic ideas on cardinal number.

If it has been already discussed please post the like and reply the reference request.

Answer 1

You can think a sequence as a finite subset of $\mathbb{N}^2$: for all $a_n\neq 0$, take the point $(n,a_n)$. This way, you can inject $S$ in the set $\mathcal{P'}(\mathbb{N}^2)$ (with $\mathcal{P'}$ I mean the set of finite subsets) and this is in bijection with $\mathcal{P'}(\mathbb{N})$. But $\mathcal{P'}(\mathbb{N})$ is countable, because it is a countable union of countable sets (subsets with $0$ elements, subsets with $1$ element, subsets with $2$ elements...).

Answer 2

Let $\mathbb{N}$ be the set of non-negative integers. If $s=s_0,s_1,s_2,\dots,s_n,\dots$ is a sequence in $S$, define $\psi(s)$ by $$\psi(s)=\left(\prod_{i=0}^\infty p_i^{s_i}\right)-1,$$ where $p_i$ is the $i$-h prime. By the Unique Factorization Theorem, $\psi$ is a bijection from $S$ to $\mathbb{N}$.

Answer 3

Put a decimal point in front of any of these sequences and you have a rational number between 0 and 1. The set is countable. There is a bit more to it than just that. You will need to consider repetitions. You end up with a countable number of equivalence classes that are each at most countable. 1,1,0,0,0,0... becomes .110000... and 11,0,0,0,0... also becomes .110000...