I am looking for a beautiful way of showing the following basic result in elementary set theory:
If $A$ is a countable set then the set of finite subsets of $A$ is
countable.
I proved it as follows but my proof is somewhat fugly so I was wondering if there is a neat way of showing it:
Let $|A| \le \aleph_0$. If $A$ is finite then $P(A)$ is finite and hence countable. If $|A| = \aleph_0$ then there is a bijection $A \to \omega$ so that we may assume that we are talking about finite subsets of $\omega$ from now on. Define a map $\varphi: [A]^{<\aleph_0} \to (0,1) \cap \mathbb Q$ as $B \mapsto \sum_{n \in \omega} \frac{\chi_B (n)}{2^n}$. Then $\varphi$ is injective hence the claim follows. (The proof of which is what is the core-fugly part of the proof and I omit it.).
Best Answer
If $A$ is an infinite countable set, $S$ is the set of finite subsets of $A$ and $S_k$ is the set of $k$-element subsets of $A$, then
$$ |S| = \sum_{k \in \mathbb{N}} |S_k| \leq \sum_{k \in \mathbb{N}} |A|^k \leq \sum_{k \in \mathbb{N}} \aleph_0 = |\mathbb{N}| \cdot \aleph_0 = \aleph_0 $$