Let $g:\mathbb N\to\mathbb N$ be a non-increasing function (which means $x\leq y\implies g(x)\geq g(y)\ \forall x,y\in \mathbb N$).
I'm interested in the cardinality of the set which contains all such functions like $g$. My educated guess would be that this set is countable, but how can I prove it?
Given such $g$, let $m=\min\{\,g(n)\mid n\in\Bbb N,\}$ and $n_0=\min\{\,n\in\Bbb N\mid g(n)=m\,\}$. Then $g(n)=m$ for all $n\ge n_0$ and we see that $g$ is essentially given by the finite sequence $g(0),\ldots, g(n_0)$. The set of finite sequences of natural numbers is well-known to be countable.