Let $g:\Bbb N\times\Bbb N \to\Bbb N$ biyective s.t. $g(n, m) = 2^{n-1}(2m-1)$. Show that there exists a famaily set $\{C_n : n \in\Bbb N\}$ of subsets of $\Bbb N$, infinites, disjoints pairwise and the union of all of them is $\Bbb N$.
Since $g$ is bijective, then $g^{-1}$ exists. I tried constructing $C_k = \{g^{-1}(k)\}$ for all $k\in\Bbb N$. But the problem with that is that $C_k$ has two elements, say $a$ and $b$ such that $g(a,b) = k$. I considered using one of both elements, but still you could get that by using another natural number, so they are no longer disjoints. What am I missing?
Define $C_n=g(\{n\}\times\Bbb N)$. Since $g$ is injective and$$n\ne m\implies(\{n\}\times\Bbb N)\cap(\{m\}\times\Bbb N)=\emptyset,$$the $C_n$'s are disjoint. Since $g$ is surjective, $\bigcup_{n\in\Bbb N}C_n=\Bbb N$. And, since $\{n\}\times\Bbb N$ is infinite, each $C_n$ is infinite.