An example of symmetric transitive relation that is not reflexive on a set of natural numbers $\mathbb{N}$

Question

An example of symmetric transitive relation that is not reflexive on a set of natural numbers $\mathbb{N}$. My guess is that such relation does not exist, but I don't know how to prove it.

Answer 1

Define $$R = \{(m, n)\mid m, n\in \mathbb N \;\text{ and }\;m, n \text{ are both even}\}$$

Then, for every odd integer $t\in \mathbb N$, $(t, t)\notin R$, hence reflexivity fails. Recall that for reflexivity to hold, it must be the case that for every $n\in \mathbb N$, $(n, n) \in \mathbb N$. No exceptions.

However, it is easy to verify that $R$ is both symmetric and transitive.

Answer 2

We consider the set $\mathbb N$ of all natural numbers.Then the relation:

$a$ ~ $b$ $\iff a=b\neq1$ is transitive, symmetric, but not reflexive

Answer 3

$\varnothing$.

It's transitive and symmetric by vacuous arguments. But it is not reflexive since $(n,n)\notin\varnothing$ for all $n\in\Bbb N$!