I need to find the cardinality of the following set:
$$ A = \{f \in \mathbb N \to \mathbb N|\forall n, m \in \mathbb N.(n \geq m) \to (f(n) \leq f (m))\} $$
I'm not sure if $A$ is countable or not. I tried to find an injective function from $P(N)$ to $A$ but couldn't find one.
Any ideas?
Yes, A is countable. A function $f \in A$ is unique identified by $f(1)$, and its values on a (finite) set of arguments where it changes (the size of that set can be at most $f(1)$, because is it non-increasing). Call this set $V$.
Let $p_i$ denote the i-th prime. Each $f \in A$ maps to a unique natural number $x$ which can be constructed as follows: start with $x=2^{f(1)}$, and for each $v \in V$, multiply $x$ by $p_v^{f(v)}$.
This provides an injective function from $A$ to the natural numbers; the other direction is trivial. So $A$ is a countable set.