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

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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, $\le_f$ is a well-ordering.

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

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

For $A' \in A^{\left({n}\right)}$, 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 $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ by:

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

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

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

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

As $f$ is injective, also:

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

and in particular $A' = A$.

Thus $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ is an injection.

So $A^{\left({n}\right)}$ is countable.

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

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

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:

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

Since $f^{\left({n}\right)}$ 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 with Itself is Countable, $\N \times \N$ is countable.

The result follows from Domain of Injection to Countable Set is Countable.

$\blacksquare$