Show that:
The set of all finite subsets of $\mathbb{N}$ is a countable set
My attempt:
Lets define the set $M:=\lbrace K : K \subset \mathbb{N} \wedge |K|<\infty \rbrace$
We now show that $|M|=|\aleph_0|$
For that task we define $A_n:=\lbrace K:K\in M\wedge |K|=n\rbrace$
We show $A_n$ is countable for all $n \in \mathbb{N}$
[IB:]
For $n=1$
$A_1$ is obviously countable
[IH:]
We have shown the statement is true for a certain $n$. We no show its true for $n+1$.
[IS:]
$A_{n+1}=A_n\cup\lbrace a\,\cup k:a \in A_n,k\in \mathbb{N} \rbrace$
Since (IH:) $A_n$ is countable we can define a sequence $(a_n)$ which has $A_n$ as underlying set.
Now we define:
$$\tau_k:\mathbb{N}\longrightarrow D_k:n \mapsto a_n \cup k$$
with $k \in \mathbb{N} \vee k=\varnothing$ and $D_k:=\lbrace a_n \cup k:n\in \mathbb{N} \rbrace $
$\tau_k$ is definitly surjective $\forall k \in \mathbb{N}$
which means $D_k$ is a countable set $\forall k\in \mathbb{N}$
$\Longrightarrow \bigcup\limits_{k\in \mathbb{N}} D_k=A_{n+1}$ is countable
which finishes the induction and tells us $A_n$ is countable $\forall n \in \mathbb{N}$
This implies the union $\bigcup\limits_{n\in \mathbb{N}} A_n=M$ is also countable
$\Box$
Would be great of someone could look over it and tell me if im correct, and if not, what to improve :) thank you very much
