Is $\{f \colon \mathbb{N}\mapsto\mathbb{N}: f \text{ is a bijection}\}$ numerable?

Question

Let $A :=\{f \colon \mathbb{N}\mapsto\mathbb{N}: f \text{ is a bijection}\}$. Is $A$ numerable?

I don't know how to attack this problem, but I think that this set is not numerable. However, I can't see how to start the demonstration.

Answer 1

Given $X\subset \Bbb N$ such that both $X$ and $\Bbb N\setminus X$ are infinite, we can define an $f\in A$ by letting $f$ map the $k$th smallest element of $X$ to the $k$th smallest element of $\Bbb N\setminus X$ and vice versa. To see that $|A|\ge|\mathcal P(\Bbb N)|$, note that for every subset $Y$ of $\Bbb N$ we can pick $X=3Y\cup (3\Bbb N+1)$ to obtain a different $f$ for every choice of $Y$.

Answer 2

Hint: Find an injection of the powerset $\mathcal P(\mathbb N)$ into $A$. Now apply Cantor's theorem.