Let $A$ be infinite, meaning that there is no injection from $A$ to $\{1,...,n\}$ for all $n\in \mathbb{N}$ and assume there exists $f:A\rightarrow \mathbb{N}$ which is injective. I am trying to show there must be an bijection from $A$ to $\mathbb{N}$.
My train of thought so far has been:
Assume there is no bijection from $A$ to $\mathbb{N}$. Then $f$ is not surjective. Let $i \in \mathbb{N}$ be the smallest element of $\mathbb{N}$ which is not mapped to $\mathbb{N}$ by $f$. This gives rise to an injective map $f_i:A \rightarrow \mathbb{N}\setminus \{i\}$. If $f_i$ were surjective this means for all $j \in \mathbb{N}\setminus \{i\}$ there exists some $a_j$ in $A$ such that $f_i(a_j)=j$. Define $g:\mathbb{N}\setminus \{i\} \rightarrow \mathbb{N}$ by $g(n)=n$ for $i
$f_P:A \rightarrow \mathbb{N}\setminus \{P\},$
where $P$ is a finite set of elements of $\mathbb{N}$.
I really don't know if this is going anywhere though. Is this train of thought correct, and (if so) is there a way to extend it to a proof? If not, how would one show this?
Note: this is not homework,at least not my homework. I found it in some lecture notes I was reading.