How to prove that the set of all not-empty and finite subsets of $\mathbb{N}$ is countable? The set of all subsets of $\mathbb N$ is $ \mathcal{P}{(\mathbb N)} $ right?
So how can I prove that $ \mathcal{P}{(\mathbb N)} $ is countable?
 A: Let $A$ be a finite subset of $\mathbb N$. Write
$$ A = \{ a_1, a_2, \dots , a_n\}. $$
Let $B$ denote the set of all such finite subsets and define $f : B \rightarrow \mathbb N$ according to :
$$ \text{for } A = \{a_1, a_2, \dots , a_n\} : f(A) = 2^{a_1} + 2^{a_2} + \dots + 2^{a_n}$$
Then it is easy to see that $f$ is an injection of $B$ into $\mathbb N$, so $B$ must be countable.
A: Hint.  There are only finitely many subsets of $\Bbb N$ whose elements add up to a given sum $n$.
A: $$\aleph_0\le|\{A\subseteq\mathbb N: |A|<\aleph_0\}|=\Big|\bigcup_{n\in\mathbb N}\{A\subseteq\mathbb N: |A|=n\}\Big|\le\aleph_0\cdot\sup_{n\in\mathbb N}\aleph_0^n=\aleph_0\cdot\sup_{n\in\mathbb N}\aleph_0=\aleph_0$$
A: \begin{align}
\text{All nonempty subsets of } \{1\}: & \{1\} \\
\text{All nonempty subsets of } \{1,2\}: & \{1\}, \{2\}, \{1,2\} \\
\text{All nonempty subsets of } \{1,2,3\}: & \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \\
\vdots \qquad \vdots \qquad \vdots \qquad{} & {} \qquad \vdots \qquad \vdots \qquad \vdots
\end{align}
Now count them, starting on the first row above: one, two, three, four, etc., skipping duplicates.
