Set of Finite Subsets of Countable Set is Countable/Proof 3

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 us verify by induction that each $A^{(n)}$ is countable.

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.

By induction, each $A^{(n)}$ is countable.

Denote with $A^f$ the set of finite subsets of $A$.

It is apparent that every finite subset is in some $A^{(n)}$, and so:


 * $A^f = \displaystyle \bigcup_{n \mathop \in \N} A^{(n)}$

The result follows from Countable Union of Countable Sets is Countable.