The set of natural number functions is uncountable I thought the set of natural number functions would be of the same cardinality as the countably infinite product of $\mathbb{N}$, which is countable.
Each natural number function can be identified with an infinite-tuple of $\mathbb{N}$ by letting the $i$th entry be the image of the number $i$ under the function.
 A: The cardinality of the natural number functions is $\mathbb{N}^{\mathbb{N}} \geq 2^{\mathbb{N}}$ and $2^{\mathbb{N}}$ is uncountable
A: Every real number in $(0,1)$ has either one or two decimal representations. (Some have two, like $0.1000\ldots$ which is equal to $0.0999\ldots$.) But any way, each real number in $(0,1)$ gives at least one decimal representation. And that decimal representation can be viewed as a function from $\mathbb{N}$ to $\mathbb{N}$ by taking the $n$th digit after the decimal. (Actually, it's more restrictive: from $\mathbb{N}$ to $\{0,1,2,3,4,5,6,7,8,9\}$.)
For example, $0.72465\ldots$ can be used to define a function $$\begin{align}1&\mapsto7\\ 2&\mapsto2\\ 3&\mapsto4\\ 4&\mapsto6\\ 5&\mapsto5\\ \vdots&\phantom{\mapsto{}}\vdots\end{align}$$
So the set of functions from $\mathbb{N}$ to $\mathbb{N}$ contains a subset that is as large as the real interval $(0,1)$. (And seemingly even stronger, so does the set of functions from $\mathbb{N}$ to $\{0,1,2,3,4,5,6,7,8,9\}$. And seemingly even stronger still, if we used binary instead of decimal, so does the set of functions from $\mathbb{N}$ to $\{0,1\}$.)
A: Cantors diagonalisation method: If it is countable then it is $F=\{f_1,f_2,\ldots\}$. But the function $g(i):=f_i(i)+1$ is not in $F$. A contradiction to the initial assumption. 
Proof:
For every  $i $ we have
$$g(i)=f_i (i)+1 \implies g (i)\ne f_i (i) \implies g \ne f_i $$
and so $g \notin F$.
A: The countably infinite product of $\mathbb{N}$ is not countable, I believe, by Cantor's diagonal argument.
A: Let's consider the following set: $F = \{ f:D \in 2^\mathbb{N} \rightarrow \mathbb{N} \ | \ f(x) = x \}$;
Notice how there is a function $f$ for each possible set of natural numbers residing inside $2^\mathbb{N}$, which will allow us to draw the following conclusion: $|F| = |2^\mathbb{N}|$.
Now, observe how $F$ is a much smaller set than $\mathbb{N}^\mathbb{N}$(because it contains only functions of the form $f(x)=x$), but even so, it's uncountable. Therefore $\mathbb{N}^\mathbb{N}$ is evidently uncountable since it's a superset of $F$.
$F \subseteq \mathbb{N}^\mathbb{N}, |F| = |2^\mathbb{N}| \implies |\mathbb{N}^\mathbb{N}|=|2^\mathbb{N}|  \ \ \ \square $
