Prove that if an injection exists such that $f: A\to \mathbb{N}$, the set $A$ is countable

Question

I know that since an injection, $f: A \rightarrow \mathbb{N}$ exists, that $|A| \leq |\mathbb{N}|$. That's as far as I've gotten.

The definition for "countable" from my book states that "A set is considered countable if the cardinality of the set is a subset of the natural numbers, $|S| \subseteq \mathbb{N}$".

I'm not sure how to transition from the step I'm on (or if it's even right) to the part given in the definition.

Answer 1

Since there is an injection into the naturals, then clearly there's a bijection into a subset of the naturals, call it $B$, then since $f : A \to B$ is a bijection we have $|A| = |B|$, and since $B \subset \Bbb{N}$ we have $|A| = |B| \leq |\Bbb{N}|$, any set $B$ with the last relation is countable, thus $A$ is countable.

Answer 2

one way to prove this could be to work in reverse, i.e. if $\exists g:\Bbb N\to A$ where $g$ is a surjection, then $A$ is countable, so simply let $g=f^{-1}$ for all elements mapped to by $f$, and let $g(x)=a\in A$ (for one fixed element $a\in A$) for all elements that are not mapped.