Reflexive but not Transitive relation

Question

What is an example of a relation $\mathscr{R}$ on a set $S$ such that $\mathscr{R}$ is reflexive but not transitive?

Here is what I have come up with.

Let $S = \mathbb{Z}$. Then let $\mathscr{R} = \{(x,y) \in S\times S|(x > 0 \land y>0) \lor (x < 0 \land y < 0)\}$

Because $(2,2) \in \mathscr{R}$ (i.e reflexive). But if $(-2,0) \in \mathscr{R}$ and $(0,2) \in \mathscr{R}$ but $(-2,2) \notin \mathscr{R}$ (i.e. NOT transitive)

Can anyone verify whether I have answered the question correctly or provide a much simpler example for a relation that is reflexive but not transitive on $\mathbb{Z}$?

Answer 1

The relation $R=\{(1,1),(1,2),(2,2),(2,3),(3,3)\}$ on the set $\{1,2,3\}$ is reflexive and not transitive.

If you want the relation to be on the set of integers, cheat as follows: consider the relation $R=\{(1,2),(2,3)\}\cup\{(n,n):n\in\mathbb Z\}$.

Answer 2

You’ve almost answered it correctly. The problem is that $0$ is neither positive nor negative, so $\langle -2,0\rangle$ and $\langle 0,2\rangle$ are not actually in the relation $R$ as you defined it. However, you can modify the definition just a little to get a relation that works fine.

You want $\langle m,n\rangle\in R$ if either $m\ge 0$ and $n\ge 0$, or $m\le 0$ and $n\le 0$. A slick way to say this is that $\langle m,n\rangle\in R$ if and only if $mn\ge 0$. Clearly $n^2\ge 0$ for all $n\in\Bbb Z$, so $R$ is reflexive. And since we’ve now included $0$ with both the positive and the negative integers, we now do have $\langle -2,0\rangle\in R$ and $\langle 0,2\rangle\in R$, and of course $\langle -2,2\rangle\notin R$.