Set of Finite Subsets of Countable Set is Countable

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Theorem

Let $A$ be a countable set.


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


Proof 1

By the definition of a countable set, there exists an injection $g: A \to \N$.


Let $\mathcal F$ denote the set of all finite subsets of $A$.

Let $f: \mathcal F \to \N$ be the mapping defined by:

$\displaystyle \map f F = \prod_{k \mathop \in \map g F} p_{k + 1}$

where $p_n$ denotes the $n$th prime number.

We define $\map f \O = 1$, the vacuous product.


Let $F, G \in \mathcal F$, and suppose that $\map f F = \map f G$.

By Expression for Integer as Product of Primes is Unique:

$\map g F = \map g G$

By Subset equals Preimage of Image iff Mapping is Injection:

$F = \map {g^{-1} } {\map g F} = \map {g^{-1} } {\map g G} = G$

That is, $f$ is an injection.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Proof 3

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.

$\blacksquare$