Consider the strict order binary relation $xRy$ over the set $A$ with the following properties:
irreflexive $\forall x \in A \neg (xRx$)
asymmetric $\forall xy \in A (xRy \Rightarrow \neg yRx)$
transitive $\forall xyz \in A (xRy \land yRz \Rightarrow xRz)$
Non Empty Set $A \neq \emptyset$
Non Empty Relation $R \neq \emptyset$
What is minimum size of the set $A$ over which the relation $R$ is defined?
I need to know the minimum size/power to use as a base case in an inductive proof of a strict order.
Here is my attempt to find the smallest set and relation with these properties. Examining $\left\vert{A}\right\vert$ for values 1,2, and 3.
$\left\vert{A}\right\vert=1, A=\{a\}$ and there is only possible relation is $R=\{(a,a)\}$ which is not irreflexive
$\left\vert{A}\right\vert=2, A=\{a,b\}$ and there are only two mutually exclusive possible relations $R=\{(a,b)\}$ or $R=\{(b,a)\}$ each of which gives a relation that is asymmetric, vacuously irreflexive and vacuously transitivity. It has power $R^1$
$\left\vert{A}\right\vert=3, A=\{a,b,c\}$ then $R=\{(a,b),(b,c),(a,c)\}$ which is asymmetric, irreflexive and transitivity
It seems to me $\left\vert{A}\right\vert=3$ is the only case where true (as opposed vacuous) transitivity holds. It has power $R^2$ or ($R \circ R$). $xRy$ is intended to formally represent a relation IsOnTopOf where one object is physically placed on top of another. Which base case is best suited to this situation? I am not sure whether or not I can use the vacuously irreflexive and vacuously transitivity as a base case for an inductive proof