A nonempty set is countable iff it is the image of a function whose domain is a nonempty countable set

Question

A nonempty set is countable iff it is the image of a function whose domain is a nonempty countable set

Attempt:

Let $A$ be non empty and countable. If $A$ is countably infinite, we can find a bijection $f: \mathbb{N} \to A$. If I understand the problem correctly, I ${\bf need}$ to build a function $g: B \to A $ where $B$ is countable and $g(B) = A$. But, I have no idea how to proceed ? Can someone lead into the right direction?

The converse I believe is more manageable. Suppose there is some map whose domain is countable set then the image is also countable either finite or infinite... qed

I am still stuck on $\implies$ direction.

Answer 1

The definition of a countable set is all sets $S$ such that there exists an function $f: S \to \mathbb N$ which is injective.

To construct such a function $g: \mathbb N \to S$ ($\mathbb N$ is clearly countable), simply take the inverse of $f$ where it is defined, and map everything else to a random element of $S$. Since $f$ maps all elements of $S$ to a natural, you'll end up getting $S$ as your range.

Answer 2

I might be doing something circular, but:

I think you can just pick $B = \mathbb N$ and $g=f$ for $A$ countably infinite.

For $A$ finite of cardinality $n \ge 1$, choose $B=\{1, 2, ..., n\}$ and $g$ is any bijection or inj/surj between $A$ and $B$.