Is any relation that is reflexive also symmetric and also transitive?

Question

For an example relation $R$ on $\{1,2\}$ of $\{(1,1), (2,2)\}$.

This relation is reflexive, but is it also symmetric and transitive?

It appears to be symmetric because $(1,1)$ is present as is $(1,1)$, and $(2,2)$ is present as is $(2,2)$, in other words, if $x = 1$ and $y = 1$, then $(x,y)$ is present and $(y,x)$ is present.

Following from that, it also appears to be transitive because for $x = 1, y = 1$ and $z = 1$, $(x,y)$ is present, $(y,z)$ is present, and so is $(x,z)$.

Is there some restriction on symmetric and transitive relations in that the elements of the ordered pair cannot be equal?

Answer 1

No, you're only considering the diagonal of the set, which is always an equivalence relation. But what if you took $R=\{(1,1),(2,2),(2,1)\}$? It's still a valid relation, it's reflexive on $\{1,2\}$ but it's not symmetric since $(1,2)\not\in R$. The point is you can have more than just pairs of form $(x,x)$ in your relation.

Answer 2

The diagonal of a Cartesian product is always an equivalence relation (trivially), and also an order(!) relation (also trivially). The mucky aspect that intervenes in the definition of equivalence and order relations is the conditional connective: as opposed to reflexivity, symmetry, anti-symmetry, and transitivity are stated in conditional terms. As such, unlike reflexivity, neither of the latter three forces an ontological commitment, hence any ties between the two sets (the source and the target of the Cartesian product) besides the diagonal are optional.