For this problem, my proof was:
If we want to express out the set of all the finite subsets, $F$.
$F = \{\{n_{1}\},\{n_{1},n_{2}\},\{n_{1},n_{2},n_{3}\},\cdots\}$ with $n_{1} \in \mathbb{N}$, $n_{2} \in \mathbb{N} – \{n_{1}\}$, $n_{3} \in \mathbb{N} – \{n_{1},n_{2}\}$.
Notice that the first element of the $F$ above comes with the only one element $\in \mathbb{N}$, the second element of $F$ comes with only two elements $\in \mathbb{N} \cdots$.
There are countably infinite number of the first element of $F$, and countably infinite number of the second element of $F$(since countably infinite $\times$ countably infinite gives countably infinite). Then the cardinality of $F$ is the sum of finite number of countably infinite, which is countably infinite.
Therefore, we can conclude that set of all finite subsets of $\mathbb{N}$ is a countably finite set.
However, I realized that the proof above although got the rough idea but is very informal and actually made use of P&C.
Now I think it is possible to do a proof with mapping: For example:
$\{n_{1},n_{2}\} \hookrightarrow (n_{1},n_{2}) \in \mathbb{N} \times \mathbb{N}$.
However, I found it difficult to phrase it in this approach.
Hope some one can help me to improve my proof. Thank you!
Best Answer
Hint: Show that the set of finite subsets is countable, to see this, consider $S(n)$ the set of subsets of length $n$, it is countable, thus $\cup_{n\in N}S(n)$ is countable. this implies its complementary as the same cardinality than $P(N)$ the set of subsets of $N$ which is uncountable.