I have a relation on the set $A = \{a,b,c,d\}$:
$$R_1=\{(d,c),(c,a),(b,d),(d,a),(a,a),(b,c),(b,a)\}$$
I need to determine whether this relation has the trichotomy property or not?
If by any chance you do not know about the trichotomy property then here is the definition:
Let $R$ be a relation on a set $A$. Then $R$ has the trichotomy property over $A$ if and only if, for each $x,y∈A$ exactly one of the following holds:
$$xRy \quad x=y \quad yRx$$
And here is the definition of transitivity: if $xRy$ and $yRz$, then $xRz$.
So is the relation $R_1$ is trichotomic and transitive or not?
Thanks And Please Help.
I find much easier to reason this kind of problems with a drawing of the relations in hand.
Here the arrows $x\to y$ means $(x,y)\in R$
From the digraph, it is clear that the trichotomy property holds, since every pair of nodes has one and only one edge between them.
Transitivity also holds, since every time we have $(x,y)$ and $(y,z)$ we also have $(x,z)$ (See how there are "triangles" between the connected nodes?).
It might be enlightening to see some graphs in which the properties do not hold.
In the previous digraph we for example do not have trichotomy, since we both have that $(b,c)$ and $(c,b)$. (As an addendum, this violates antisymmetry also, which is a necessary condition for trichotomy).