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

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

Thus:


 * $A^{\paren 0} = \set \O$
 * $A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$

and $\forall n \ge 0$:
 * $A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in A^{\paren n} \land a^{\paren 1} \in A^{\paren 1} }$

Let us verify by induction that each $A^{\paren n}$ is countable.

Let $A$ be countable.


 * $A^{\paren 1}$ is countable, as its cardinality is $1 + \card A$.

Suppose $A^{\paren n}$ is countable.

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

By induction, each $A^{\paren n}$ is countable.

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

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


 * $A^f = \ds \bigcup_{n \mathop \in \N} A^{\paren n}$

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