The finite powerset is countable

I've tried to prove the following theorem: the set of all finite subsets of a countable set is countable. My proof is different from that suggested by my professor and I'm therefore trying to understand if it is correct.

To prove such a claim I have to show the existence of a bijection between $A$ and $\mathcal{P}_f(A)$(the set of all finite subsets). Given that $A$ is countable ex hypothesi I can write it as $\{a_1, a_2,\cdots, a_n,\cdots\}$.

At this point I want to write $\mathcal{P}_f(A)$ in an appropriate way so that I can show the pattern of the bijection. This way would be the following:

$$\mathcal{P}_f(A)=\{\emptyset, \{a_1\}, \{a_2\}, \{a_2, a_1\}, \{a_3\}, \{a_3, a_2\}, \{a_3,a_1\}, \{a_3, a_2, a_1\},\cdots, \{a_n\}, \{a_n, a_{n-1}\}, \{a_n, a_{n-2}\},\cdots, \{a_n, a_1\}, \{a_n, a_{n-1}, a_{n-2}\},\cdots\{a_n,a_{n-1},\cdots, a_1\}\}$$

Shown that I can write $\mathcal{P}_f(A)$ with such a patter is my proof complete? I think so since, to prove that $2\mathbb{Z}$ is countable is enough to show that I can write it as $\{0, 1, -1, 2, -2, \cdots\}$ and therefore assigning a natural number in progressive order.

Is my solution correct? If not, why? Thanks in advance for any answer.

Solution 1:

From comment:

Essentially that works, though perhaps you could describe the pattern more precisely.

One possibility might be to say that

• if you consider $b_1,b_2,\ldots$
• with $b_{2^m+k}=\{a_{m+1}\} \cup b_{k}$ where $1≤k≤2^m$ and
• with $b_1=\emptyset$

then every element of $\mathcal{P}_f(A)$ appears in the sequence and nothing else does; since the sequence is countable (has an obvious bijection with $\mathbb N$) so too is $\mathcal{P}_f(A)$.