Set of Finite Subsets of Countable Set is Countable/Proof 1

Theorem
Let $A$ be a countable set.

Then the set of all finite subsets of $A$ is countable.

Proof
By the definition of a countable set, there exists an injection $g: A \to \N$.

Let $\mathcal F$ denote the set of all finite subsets of $A$.

Let $f: \mathcal F \to \N$ be the mapping defined by:
 * $\displaystyle f \left({F}\right) = \prod_{k \mathop \in g \left({F}\right)} p_{k+1}$

where $p_n$ denotes the $n$th prime number.

We define $f \left({\varnothing}\right) = 1$, the vacuous product.

Let $F, G \in \mathcal F$, and suppose that $f \left({F}\right) = f \left({G}\right)$.

By Prime Decomposition of Integers is Unique, it follows that $g \left({F}\right) = g \left({G}\right)$.

By Preimage of Image of Injection, it follows that $F = g^{-1} \left({g \left({F}\right)}\right) = g^{-1} \left({g \left({G}\right)}\right) = G$.

That is, $f$ is an injection.