Countable set can be listed in a sequence Let $S$ be a countable set, i.e. exists bijection between $\mathbb{N}$ and $S$. Why elements of $S$ can be listed in a sequence? 
EDIT: I guess that bijection is not crucial. It's sufficient that mapping (namely $f$) is surjective, i.e. for any $T\in S$ exists $n\in \mathbb{N}$ such that $T=f(n)$. Hence $S$ can be arrabged in sequence $\{f(1), f(2), \dots\}$. Am I right?
 A: A sequence $s = (x_{n} \mid n \in \mathbb N)$ is 'the same' as a
function $f \colon \mathbb N \to \operatorname{ran}(s)$. Depending on
your exact definitions of sequence and function they may not
be the same set (using my definition, they are actually the same set),
but they are certainly mutually definable. I.e. given a sequence
$s = (x_{n} \mid n \in \mathbb N)$ we may define a function $f \colon
\mathbb N \to \operatorname{ran}(s)$ by letting $f(n) := s_{n}$ and
conversely, given a function $f \colon \mathbb N \to X$, we may define
a sequence $s =(x_{n} \mid n \in \mathbb N)$ by letting $s_{n} :=
f_{n}$. This way, converting a function to a sequence and then converting the resulting sequence to a function gives back the original function and likewise if we start with a sequence, convert it into a function and back to a sequence.
In particular, given a bijection $f \colon \mathbb N \to X$, we may
list all elements of $X$ via the sequence $(f(n) \mid n \in \mathbb N)$.
A: Because the way we formalize the concept of "infinite sequence" in set theory is as a function with domain $\mathbb N$.
Therefore "$S$ can be listed in a sequence" means, by definition, that there is a function $f:\mathbb N\to S$ that hits each element of $S$ -- and hits it exactly once, because we suppose every element of $S$ to appear only once in the sequence.
But such an $f$ is exactly a bijection between $\mathbb N$ and $S$.
