Consider the set $A=\{n\in\mathbb{N} \mid n\geq 2\}.$ Let $A\times A$ be an ordered set defined by: $(a,b)\preceq (c,d)$ iff $a\mid c$ and $b\leq d$. I have proved that there is no maximal, greatest, first element.
Now I'm trying to find how much minimal elements there are. I think that there are infinite number of minimal elements but don't know how to prove it.
Note that all minimal elements must have $2$ in the second place, as $(a,2) \prec (a,b)$ for $b \gt 2$. What $c$ do not have numbers that divide them?
Added: the minimal elements will be all pairs of the form $(p,2)$ for $p$ a prime. These will be minimal because any $(a,b) \preceq (p,2)$ must have $a|p$ so $a=p$ and $b \le 2$ so $b=2$. Any other $(a,b)$ will not be minimal because either $a$ will have a proper divisor or there will be a number strictly less than $b$. In either case there will be a pair that is strictly less.