Let $(\mathbb Z \times \mathbb Z^*, \pi)$ be a poset defined as follows:
$\begin{aligned} (a,b) \pi(c,d)\Leftrightarrow (a,b) = (c,d) \text{ or } r(a,b)
whereas $r(x,y)$ is the reminder of the division of $x$ by $y$.
Let $X=\{(3,2),(12,5),(6,2),(11,9)\} \subset \mathbb Z \times \mathbb Z^*$ and find maximum, minimum, maximal elements, minimal elements, upper bounds, lower bounds, supremum and infimum.
I noticed that in gereral $\nexists(a,b)\in \mathbb Z \times \mathbb Z^* : r(a,b) <0$ and the only elements that have $0$ as reminder are in $S=\{(na,a)\in \mathbb Z \times \mathbb Z^* : n\in \mathbb Z\}$.
Generally speaking, let $P$ be a (partially) ordered set, and let $A$ be a subset of $P$ then we say $y\in P$ is a lower bound for $A$ if and only if $y\leq a$ for all $a\in A$.
As I come to find all the lower bounds for my specific poset, do I need to look for $(x,y)\in \mathbb Z \times \mathbb Z^*$ so that $(x,y)\leq (a,b)$ for all $(a,b)\in X$, meaning looking for $(x,y)\in\mathbb Z \times \mathbb Z^* : r(x,y) \leq r(a,b)$ or shall I substitute $\pi$ for $\leq$ used in the general definition and therefore look for $(x,y)\in \mathbb Z \times \mathbb Z^*$ so that $(x,y) \pi (a,b)$?
If the former is true then all lower bounds are in $S$, otherwise if the latter is true then there are no lower bounds because by very definition of $\pi$, if $(a,b)\neq(x,y)$ and $r(a,b)=r(x,y)$ then $(a,b)$ and $(x,y)$ are not comparable. Is that correct?