Set of Finite Subsets of Countable Set is Countable/Proof 2

Theorem
Let $A$ be a countable set.

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

Proof
By definition of a countable set, there is an injection $f: A \to \N$.

Let $\le_f$ be the ordering induced by $f$ on $A$; by Injection Induces Well-Ordering, it is a well-ordering.

Let $A^{\text{fin}}$ be the set of finite subsets of $A$.

For $n \in \N$, denote with $A^{(n)}$ the set of subsets of $A$ that have precisely $n$ elements.

For $A' \in A^{(n)}$, we write $A' = \left\{{a'_1, a'_2, \ldots, a'_n}\right\}$ uniquely so that $i < j$ implies $a'_i <_f a'_j$.

Then we define $f^{(n)}: A^{(n)} \to \N$ by:


 * $f^{(n)} \left({A'}\right) := \displaystyle \prod_{i \mathop = 1}^n p_i^{f \left({a'_i}\right)}$

where $p_i$ denotes the $i$th prime.

By the Fundamental Theorem of Arithmetic, if $f^{(n)} \left({A'}\right) = f^{(n)} \left({A''}\right)$, then:


 * $\left({f \left({a'_1}\right), \ldots, f \left({a'_n}\right)}\right) = \left({f \left({a_1}\right), \ldots, f \left({a_n}\right)}\right)$

and as $f$ is injective, also:


 * $\left({a'_1, \ldots, a'_n}\right) = \left({a_1, \ldots, a_n}\right)$

and in particular $A' = A''$.

Thus $f^{(n)}: A^{(n)} \to \N$ is an injection, and so $A^{(n)}$ is countable.

Now define $f^{\text{fin}}: A^{\text{fin}} \to \N \times \N$ by:


 * $f^{\text{fin}} \left({A'}\right) := \left({n, f^{(n)} \left({A'}\right)}\right)$

where $n = \left\vert{A'}\right\vert$ is the cardinality of $A'$.

Suppose that $f^{\text{fin}} \left({A'}\right) = f^{\text{fin}} \left({A''}\right)$.

Then by Equality of Ordered Pairs, $n = \left\vert{A'}\right\vert = \left\vert{A''}\right\vert$, and:


 * $f^{(n)} \left({A'}\right) = f^{(n)} \left({A''}\right)$

Since $f^{(n)}$ was already shown to be an injection, it follows that $A' = A''$.

Thus $f^{\text{fin}}$ is also an injection.

By Cartesian Product of Natural Numbers, $\N \times \N$ is countable.

The result follows from Injection to Countable Set.