I would like to know if my understanding of the concept of a poset is correct.
From what I've learnt from the class:
A poset must be transitive, reflexive, and antisymmetric. Am I right?
Therefore, I have just tried out the following example.
$$ A=\{1,2,5,10\}, B=\{1,2,3,6,9,18\}$$
When I calculate $A\times B$, I get:
$$A\times B=\{(1,1),(1,2),(1,3),(1,6),\\ (1,9),(1,18),(2,1),(2,2),(2,3),(2,6),(2,9),(2,18),\\ (5,1),(5,2),(5,3),(5,6),(5,9),(5,18),(10,1),(10,2),\\(10,3),(10,6),(10,9),(10,18)\}$$
Please correct me if I'm wrong.
Thx
A poset is not to be confused with the Cartesian product of two sets $A, B$. It is set under an ordering relation, and it is this relation that is a subset of $A\times B$ (if and only if $A = B$) which must satisfy the three properties (reflexivity, antisymmetry, and transitivity) to be deemed an ordering relation.
Note, e.g., that $(3, 3) \notin A\times B$, so $A\times B$ is not reflexive, and hence cannot be an ordering relation.
Can you find a subset of $A\times B$ which satisfies reflexivity, antisymmetry, and transitivity? You can only find such a subset of a Cartesian product is of the form $A\times A$, of some set $A$. Since clearly, as given, $A\neq B$, there can be no such ordering relation which is a subset of $A\times B$, if for no other reason than reflexivity will invariably fail.
Exercise: Try using your posted sets $A, B$ to determine $A\times A$ and $B\times B$. Then using the ordering relation of divisibility, determine which ordered pairs, in each Cartesian product separately, belong to $O_1\subseteq A\times A$, and $O_2 \subseteq B\times B$, respectively. You'll find that divisibility is indeed and ordering relation, and that $A, B$ indeed are posets.