Can the inverse of an element of a countable set approach infinity? Say we have a countable set $A$. 
$f:\mathbb{N}\to A$. 
Can we say that there exists at least one element $a\in A$ such that $f^{-1}(a)$ is greater than any number $n\in \mathbb{N}$ we choose. Remember that $a$ is an element we explicitly know. 
 A: Assuming $f$ is injective, $f^{-1}(a)$ is just a natural number, say $m$, because the domain of $f$ is $\mathbb{N}$. That is, $m \in \mathbb{N}$ and $f(m) = a$. But then $m+1 > m$, and in fact there are only finitely many numbers below $m$, so it's very much not the case that $f^{-1}(a)$ is greater than any natural number we choose.
Slogan: "if a number is greater than every natural number, then it isn't a natural number".
However, if you'd asked if $f^{-1}(a)$ can be made arbitrarily large by varying $a$, then this is true: for each $n \in \mathbb{N}$, $f(n) \in A$ and then $f^{-1}(f(n)) = n$. We can make $n$ as large as we like.
A: It exists. Let $f: \Bbb N \to A$ such that $f(\Bbb N)=a$. Then for any $n \in \Bbb N$, there exists an element $m$ of $\Bbb N=f^{-1}(a)$ such that $m > n$.
In general, I'm sorry to say No. 

For example, $f$ is an injection. Then $f^{-1}(a) \in \Bbb N$. It cannot be greater any element of $\Bbb N$.

A: Approaching infinity does not mean being transfinite. Furthermore, the natural numbers are ordered without a maximum, so being "larger than any other number" means that you're not a natural number.
If we do not require $f$ to be injective, however, it is possible that $f^{-1}(a)$ is an unbounded set of integers. Namely, the following statement is true:$$\newcommand{\textsf}[1]{\mathsf{\text{#1}}}\textsf{For all }{\it n}\textsf{ there exists }{\it m}\textsf{ such that}\colon {\it n<m}\textsf{ and }f(m)=a.$$
In this sense, $f^{-1}(a)$ does "approach" infinity. It is possible, that $f\colon\Bbb N\to A$ is such that every $a\in A$ has an infinite preimage. At least, of course, when $A$ is countable.
