Suppose I have the relation R = {(0, 0), (1, 1), (2, 2), (3, 3)} on the set S = {(0, 1, 2, 3)}.
I know the digraph for this would just have a loop around the 0, 1, 2, and 3 vertices with no arcs connecting any of them. I know it is reflexive because all vertices have loops. It is symmetric because there are no single arcs, only double arcs or no arcs at all (in this case, no arcs at all is true). It is antisymmetric because there are no double arcs, only single arcs or no arcs at all (in this case, no arcs at all is true). But why is it transitive?
I think the reason is similar to why no arcs makes the relation symmetric and antisymmetric, but I am not entirely confident on why exactly no arcs also makes it transitive. Is there a proof that can show this or...?
Transitivity says that if $x\mathrel{R}y$ and $y\mathrel{R}z$, then $x\mathrel{R}z$. For your relation $R$ the hypothesis that $x\mathrel{R}y$ and $y\mathrel{R}z$ is true only when $x=y=z$, and in that case the conclusion $x\mathrel{R}z$ is also true. You can make the same argument in general when there are no arcs between distinct vertices: $x\mathrel{R}y$ and $y\mathrel{R}z$ is true only when $x=y=z$ and $x\mathrel{R}x$, and when that’s the case, $x\mathrel{R}z$. When $x=y=z$ but $x\not\mathrel{R}x$, the hypothesis is false, so the fact that the conclusion $x\mathrel{R}z$ is false is irrelevant.