Set of Finite Subsets of Countable Set is Countable

Theorem
Let $A$ be a countable set.

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

Proof
Let $A^{(n)}$ be the set of subsets of $A$ with no more than $n$ elements.

Thus:
 * $A^{(0)} = \left\{{\varnothing}\right\}$
 * $A^{(1)} = A^{(0)} \cup \left\{{\left\{{a}\right\}: a \in A}\right\}$

and $\forall n \ge 0$:
 * $A^{(n+1)} = \left\{{a^{(n)} \cup a^{(1)}: a^{(n)} \in A^{(n)} \land a^{(1)} \in A^{(1)}}\right\}$

Let $A$ be countable.


 * $A^{(1)}$ is countable, as its cardinality is $1 + \left|{A}\right|$.

Suppose $A^{(n)}$ is countable.

Then by Union of Countable Sets of Sets, so $A^{(n+1)}$ also countable.

The result follows by induction.