3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It comes down to having an isolated element, one that is not related to anything, including itself. Thus reflexivity fails. $\endgroup$ – hardmath Jul 20 '14 at 15:47
6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

$\varnothing$.

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

$\endgroup$
  • $\begingroup$ Definitely my favorite answer. $\endgroup$ – Martin Brandenburg Jul 20 '14 at 15:10
  • $\begingroup$ Trivial examples strike back! (At the end of the example we learn that Darth Trivius is in fact the father of Luck Setwalker.) $\endgroup$ – Asaf Karagila Jul 20 '14 at 15:11
0
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.