# Example of a relation which is reflexive, transitive, but not symmetric and not antisymmetric

I'm trying to think of a simple example of a two coordinate $$(a,b)\in R$$ relation which is reflexive, transitive, but not symmetric and not antisymmetric over $$\mathbb{N}$$ (meaning $$R\subseteq\mathbb{N}\times\mathbb{N}$$).

I can't seem to think of one. I would be glad to see some suggestions without actually proving them. I just struggling to think of an example.

• See my comments on what symmetry / antisymmetry mean from a graphical point of view here. Here is another post of mine from a few years ago that gives similar graph theoretical interpretations of reflexivity and transitivity. Armed with those interpretations and the idea of drawing graphs, try coming up with an example. – JMoravitz Dec 14 '19 at 14:52
• Take $R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(1,3)\}$ – almagest Dec 14 '19 at 14:55

Consider $$\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4)\}$$ over $$\{1,2,3,4\}$$. It is not symmetric because $$3\sim4$$ but not $$4\sim3$$ and it is not antisymmetric because $$1\sim2$$ and $$2\sim1$$ but $$1\neq2$$.
If you want to extend that to all of $$\mathbb N$$, you can just do $$\{(i,i)\mid i\in\mathbb N\}\cup\{(1,2),(2,1),(3,4)\}$$ for the same reason.
Actually, almagest did inspire me to think of a less contrived example over $$\mathbb N$$: $$R=\left\{(a,b)\in\mathbb N^2\mid \left\lfloor\frac a2\right\rfloor \le \left\lfloor\frac b2\right\rfloor\right\}$$
The question asks to find a preorder on $$\mathbb{N}$$ that is neither an equivalence relation nor a partial order. One such relation is the relation $$R$$ where $$(m,n) \in R$$ iff $$m$$ and $$n$$ are both even, or $$m$$ and $$n$$ are both odd, or $$m$$ is even and $$n$$ is odd. This is not an equivalence relation because, assuming that the natural numbers include zero, $$(0,1) \in R$$, but $$(1,0) \not\in R$$. It is also not a partial order, because $$(2,4)$$ and $$(4,2)$$ are both in $$R$$, for example.