Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 8E / b

Suppose all functions $\mathbb{N}\to \mathbb{N}$ could be listed as $f_1,f_2,\dots$. DefineThen $g$ differs from $f_n$ at $n$ for every $n$, so it is absent from the list. This Cantor's diagonal argument proves that ${\mathbb{N}}^{\mathbb{N}}$ is an uncountable set.

$A$ is uncountable. For each binary sequence $({\epsilon }_j)$, defineThis is a nondecreasing function, and different binary sequences give different functions, so the uncountable set of binary sequences injects into $A$.

$B$ is countable. Every nonincreasing sequence of positive integers can decrease only finitely many times and is therefore eventually constant. Such a sequence is specified by a finite sequence of positive integers, and the set of all finite integer sequences is a countable union of countable sets. Hence $B$ is countable.