# Around binary relations on sets

We are considering here binary relations on a set $$A$$. Let’s recall that a binary relation $$R$$ on $$A$$ is a subset of the cartesian product $$R \subseteq A \times A$$. The statement $$(x,y) \in R$$ is read as $$x$$ is $$R$$-related to $$y$$ and also denoted by $$x R y$$.

Some importants properties of a binary relation $$R$$ are:

reflexive
For all $$x \in A$$ it holds $$x R y$$
irreflexive
For all $$x \in A$$ it holds not $$x R y$$
symmetric
For all $$x,y \in A$$ it holds that if $$x R y$$ then $$y R x$$
antisymmetric
For all $$x,y \in A$$ if $$x R y$$ and $$y R x$$ then $$x=y$$
transitive
For all $$x,y,z \in A$$ it holds that if $$x R y$$ and $$y R z$$ then $$x R z$$

A relation that is reflexive, symmetric and transitive is called an equivalence relation. Let’s see that being reflexive, symmetric and transitive are independent properties.

### Symmetric and transitive but not reflexive

We provide two examples of such relations. For the first one, we take for $$A$$ the set of the real numbers $$\mathbb R$$ and the relation $R = \{(x,y) \in \mathbb R^2 \, | \, xy >0\}.$ $$R$$ is symmetric as the multiplication is also symmetric. $$R$$ is also transitive as if $$xy > 0$$ and $$yz > 0$$ you get $$xy^2 z >0$$. And as $$y^2 > 0$$, we have $$xz > 0$$ which means that $$x R z$$. However, $$R$$ is not reflexive as $$0 R 0$$ doesn’t hold.

For our second example, we take $$A= \mathbb N$$ and $$R=\{(1,1)\}$$. It is easy to verify that $$R$$ is symmetric and transitive. However $$R$$ is not reflexive as $$n R n$$ doesn’t hold for $$n \neq 1$$. Continue reading Around binary relations on sets