F is a group that includes all functions from N to N
K is relation over F.
For f,g ∈ F: (f,g) ∈ K iff ∀ n∈N, f(n)≤g(n).
Obviously K is Partially ordered set and not Total Order.
My problem is with proving that: for ∀ f∈F ∃ g ∈ F that covers f. and that for ∀ f∈F there is more than one g that covers f.
Can anyone help me to prove this? Thanks :)
HINT: Given $f\in F$, consider the following functions: for each $m\in\Bbb N$ let $g_m\in F$ be defined by
$$g_m(k)=\begin{cases} f(k),&\text{if }k\ne m\\ f(k)+1,&\text{if }k=m\;. \end{cases}$$