proving relations between set of functions i'm having problem with this question. i'll explain what i've done after i ask it.
let F be the set of all functions $N\to N$. K is a relation over F that is defined like this: for every $f,g \in F$ $(f,g) \in K$ iff for every $n\in N$  $f(n) \leq g(n)$
a)are there maximal elements for the relation K? is there a greatest element? prove.
b)are there minimal elements for the relation K? is there a smallest element?
c)(where i'm having most of the trouble) prove that for every $f \in F$ there's $g \in F$ that covers f, and prove that for every $f \in F$ there exist more than one such g.
what i've done:
a)let $h,j \in F$ and for every $x \in N$ $h(x) \leq j(x)$, so $(f,g) \in K$, thus there exists some h(x) that would greater than or equal to j(x), but that is not possible since j(x) must be greater or equal, so such number could not exist in N and thus there is no greatest value since for every $h(x) \leq j(x)$. from that we can also infer that the maximum value might be given j(x)
b)same thing, but the opposite to show minimal and smallest element while concidering 0 is the smallest and the minimal is $h(x)$ iff for every $x \in N$ $h(x) \leq j(x)$, thus h(x) must be less or equal to j(x) with the limit of 0(the smallest element).
c)let $x \in N$ and let $g_x \in F$ be defined like this: $g_x(k)=\begin{cases}
f(k),&\text{if }k\ne m\\
f(k)+1,&\text{if }k=m\;.
\end{cases}$ 
so $f \leq g_x(k)$ and $f \neq g_x$. now if we say that there exists a $h \in F$ so that $h \neq f$ , then if $k=m$ -> then $g(k)=f(k)+1$ because F is defined to be from N->N and there can't be $f(k)≤h(k)≤g_x(k)$ and if it's not equal, then by the information given in the question it's bigger and most cover. (i don't know how to prove that more of one  like that exists).
i've done my best efforts, please help me if you can.
 A: a) There is no maximal element. To prove this, we suppose that $f:\mathbb{N}\to\mathbb{N}$ is maximal and define $g:\mathbb{N}\to\mathbb{N}$ by
$$
g(n) = \begin{cases}
f(1)+1 & \text{if}\ n=1 \\
f(n) & \text{otherwise}.
\end{cases}
$$
Then $f\ne g$ and $(f,g)\in K$. This contradiction shows that there are no maximal elements. Hence there is no greatest element, as a greatest element must be maximal.
b) There is a smallest element. Consider $f:\mathbb{N}\to\mathbb{N}$ given by
$f(n)=1$ for every $n\in\mathbb{N}$. Then $(f,g)\in K$ for every $g:\mathbb{N}\to\mathbb{N}$. Consequently this element is also minimal.
c) Suppose $f:\mathbb{N}\to\mathbb{N}$. For each $k\in\mathbb{N}$, define $g_k:\mathbb{N}\to\mathbb{N}$ by
$$
g_k(n) = \begin{cases}
f(k) + 1 & \text{if}\ n=k \\
f(n) & \text{otherwise}.
\end{cases}
$$
We claim that $\{g_k \mid k\in\mathbb{N}\}$ is a countable set of covers of $f$.
If $k\ne j$, then $g_k(k)=f(k)+1>f(k)=g_j(k)$. Thus these functions are distinct and hence $\{g_k \mid k\in\mathbb{N}\}$ is countable.
It remains to show that each $g_k$ is a cover of $f$. Fix $k\in\mathbb{N}$. Certainly $(f,g_k)\in K$ and $f\ne g_k$. Suppose $h:\mathbb{N}\to\mathbb{N}$ such that $(f,h),(h,g_k)\in K$. Then for each $n\ne k$, we have
$$
f(n) \leq h(n) \leq g_k(n)=f(n).
$$
Hence $h(n)=f(n)=g_k(n)$ for every $n\ne k$. Now
$$
f(k) \leq h(k) \leq g_k(k) = f(k)+1
$$
implies that either $h(k)=f(k)$ or $h(k)=g_k(k)$. In the first case we have $h=f$, and in the second we have $h=g_k$. Therefore $g_k$ covers $f$.
