TITLE: Is the sets of all maps from $\mathbb{N}$ to $\mathbb{N}$ countable?
QUESTION [6 upvotes]: Is the sets of all maps from $\mathbb{N}$ to $\mathbb{N}$ countable? 
Attempts: I think it is uncountable. Consider $P(\mathbb{N})$ where $P(A)$ denote as the power sets of A. We know that $P(\mathbb{N})$ is uncountable. Moreover, by the axiom of choice, for any non empty set, we know that there exist a function $f: P(\mathbb{N})\to \mathbb{N}$ . Note that $P(\mathbb{N})$ have uncountable many elements and since each $f$ belongs to the set of all mapping from $\mathbb{N}$ to $\mathbb{N}$ so there exist infinite uncountably elements and hence this set are uncountable. Is my attempt work??

REPLY [7 votes]: Inside the set of functions $\def\N{\Bbb N}\N\to\N$ is the subset of functions $\N\to\{0,1\}$ which are precisely the characteristic functions of subsets of $\N$ (as indicated in this answer), and which subsets you already know to be uncountable un number. So certinaly $\N^\N$ is uncountable.