Proof that $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ is not countable by diagonalization. So, as the title says, I want to prove that $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ isn't countable - by diagonalization.
How can I create such $f$ that I cannot count?
edit:
Let's assume by contradiction that  $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ is indeed countable.
Lets mark:
$f_0 = (a_{00},a_{01},...,a_{0n....}$
$f_1 = (a_{10},a_{11},...,a_{1n....}$
$f_2 = (a_{20},a_{21},...,a_{2n....}$
and so on...
Lets create $F' = (a_0,a_1,a_2,...,a_n...$ such that $a_{i}=\begin{cases}
a_{ii}+1 &amp; if\;a_{ii}+1\notin \left \{ a_1,...,a_{i-1} \right \}\\
max\left \{ a_1,...,a_{i-1} \right \}+1 &amp; otherwise
\end{cases}$
Is this build sufficient?
Note that $|S|\ge2^{\Bbb \aleph_0}\gt\aleph_0$  The second inequality by Cantor's diagonalization argument.

The first is not that hard to establish:  given a subset of $\Bbb N$, define an injective map by setting it equal to the identity on the subset, and extending to all of $\Bbb N$ by an element of the symmetric group on the complement with no fixed points.
Note that $|S|\ge2^{\Bbb \aleph_0}\gt\aleph_0$  The second inequality by Cantor's diagonalization argument.

The first is not that hard to establish:  given a subset of $\Bbb N$, define an injective map by setting it equal to the identity on the subset, and extending to all of $\Bbb N$ by an element of the symmetric group on the complement with no fixed points.