# Show that a particular set is a poset

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

• I'm not sure what your cartesian product is supposed to have to do with the idea of a poset. Is there an order on $A$ and $B$ (perhaps something to do with divisibility)? And are you trying to use that to make $A\times B$ into a poset? Or is your question something else? Apr 15, 2014 at 15:09
• @MarkBennet, My example question is:Let (D10,|) and (D18,|) be two partially ordered sets Apr 15, 2014 at 15:19
• @titanfall: You may need to backtrack first to the concepts of binary relation and partial order, as a poset (partially ordered set) is a set with a partial order, the latter of which is a reflexive, antisymmetric and transitive binary relation.
– J W
Apr 15, 2014 at 17:08

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.

• I still dun really get it ;-( Apr 15, 2014 at 15:23
• I added a suggestion for an exercise in my post. Give it a try, and check in with me a little later. Apr 15, 2014 at 15:27
• A={1,2,5,10} B={1,2,3,6,9,18} P1={(1,1),(1,2),(1,5),(1,10),(2,1),(2,2),(2,5),(2,10),(5,1),(5,2),(5,5),(5,10),(10,1),(10,2),(10,5),(10,10)} P2={(1,1),(1,2),(1,3),(1,6),(1,9),(1,18),(2,1),(2,2),(2,3),(2,6),(2,9),(2,18),(3,1),(3,2),(3,3),(3,6),(3,9),(3,18),(6,1),(6,2),(6,3),(6,6),(6,9),(6,18),(9,1),(9,2),(9,3),(9,6),(9,9),(9,18),(18,1),(18,2),(18,3),(18,6),(18,9),(18,18)} Apr 15, 2014 at 16:01
• P1={(1,1),(1,2),(1,5),(1,10),(2,2),(2,10),(5,5),(5,10),(10,10)} P2={(1,1),(1,2),(1,3),(1,6),(1,9),(1,18),(2,2),(2,6),(2,18),(3,3),(3,6),(3,9),(3,18),(6,6),(6,18),(9,9),(9,18),(18,18)} Like dis? Apr 15, 2014 at 16:25
• Plz correct me if I'm wrong. Btw,i suddenly came across a question on my mind.. if P1 x P2...could it be Matrix of P1 x Matrix of P2? Apr 15, 2014 at 16:31