Proving a simple partially ordered set

I am losing my mind over this:

(a) The relation $A=\{(1,1),(2,2),(3,3),(4,4),(3,2),(2,1),(3,1),(4,1)\}$ on the set $S=\{1,2,3,4\}.$

I'm having trouble figuring out if it's reflexive, symmetric, antisymmetric and transitive, because I don't know which ordered pairs to use (I know it has to be antisymmetric, not symmetric, but I want to try to understand all of them).

Like, I want to say it's reflexive, because for every element $S,$ we have an ordered pair that is $a\leq b$ and $b \leq a$, which are $(1,1), (2,2), (3,3), (4,4)$

no idea how to find out if it's transitive.

You are correct it is reflexive.Not symmetric as$(2,1)\in A$ but $(1,2) \notin A$. transitive is satisfied as there is only one pair in $A$ such that $(a,b)\in A$ and $(b,c) \in A$i.e.$(3,2)$ and $(2,1)$.for antisymmetric it is vacuosly true as there are no elements in $A$ such that both $(a,b)$ and $(b,a) \in A$
• Check the conditions of partially ordered set once more.It says $(a,a)$ should be in $A$ forall $a$.For symmetry if $(a,b)\in A$ then we must have $(b,a)\in A$ – Learnmore Nov 10 '14 at 5:35