Unbounded family of functions from $\mathbb{N}$ to $\mathbb{N}$ Definitions:


*

*For functions $f,g:\mathbb{N}\to \mathbb{N}$, we say $f\leq ^{*} g$ if there exist $n\in\mathbb{N}$ such that for $n\leq m$ we have $f(m)\leq g(m)$

*Family $F$ of functions from $\mathbb{N}$ to $\mathbb{N}$ is unbounded if for every function $g:\mathbb{N}\to \mathbb{N}$, exist $f\in F$ such that $f\leq ^{*} g$ isn't holds.
Question:

Show that an unbounded family $F$ of functions from $\mathbb{N}$ to
  $\mathbb{N}$ is  uncountable.

Thank you for any help! 
 A: Instead of showing that every unbounded family is uncountable, let's try the (equivalent) contraposition: Show that every countable family fails to be unbounded.
Let $F$ be a countable family of functions $\mathbb N\to \mathbb N$.
That is, there exists a bijection $\mathbb N\to F$, or simply put, we can enumerate the elements of $F$ like this:
$$\tag1F=\{f_n\mid n\in\mathbb N\}.$$
This $F$ would be unbounded if for every $g\colon\mathbb N\to \mathbb N$ there exists an $f\in F$ such that $f\le^* g$ is false.
Thus we have to show that there exists $g\colon\mathbb N\to \mathbb N$ such that for all $f\in F$ the statement $f\le^* g$ is true.
Simply define
$$\tag2g(n)=\max\{f_k(n)\mid k\le n\}.$$
Now let $f$ be an arbitrary element of $F$.
We have to show $f\le^* g$.
By $(1)$, we have $f=f_n$ for some $n$.
Then using $(2)$ 
$$\tag3g(m)=\max\{f_k(m)\mid k\le m\}\ge f(m)\quad \text{for }m\ge n$$
because $f$ occurs among the $f_k$ with $k\le m$.
But $(3)$ is just the definition of $f\le^* g$. $_\square$
