Prove that $|\{\,f\in \Bbb N \rightarrow \Bbb N\,|\,\forall x,y\in\Bbb N.(x \,-y\in\Bbb N_3) \, \rightarrow \,(f(x) = f(y))\}| = \aleph_0\,$, where $\Bbb N_3 = \{3k|k\in\Bbb N\}$
I understand that the $f$ function must be such that $\{\lambda n\in\Bbb N.n\bmod 3+i\,|\,i \in \Bbb R\}$ under these conditions.
So let's say that to prove the cardinality of $\aleph_0$ i define a function $f\in \Bbb N \rightarrow (\Bbb N \rightarrow \Bbb N)$, $f=\lambda n\in \Bbb N.\{\lambda n\in\Bbb N.n\bmod 3+i\,|\,i \in \Bbb R\}$, will the inverse bijective function $f^{-1}$ be $f^{-1} = \lambda g\in \Bbb N \rightarrow \Bbb N.n$, how can i define $f^{-1}$ to return the first variable in the ordered pair ?
As pointed out in a comment the bijection to $\mathbb N^3$ should be obvious.
You have probably seen arguments that $|\mathbb N|=|\mathbb Z|$ and $|\mathbb N|=|\mathbb Q|$, the latter is typically done by identifying $\mathbb Q$ with a subset of $\mathbb Z^2$ or often just that $\{q\in\mathbb Q\mid q>0\}$ can be identified with a subset of $\mathbb N^2$. You need a similar construction to prove that $|\mathbb N^3|=|\mathbb N|$, but actually writing that bijection down is a bit of work.