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.
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It comes down to having an isolated element, one that is not related to anything, including itself. Thus reflexivity fails. – hardmath Jul 20 '14 at 15:47
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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.
amWhy
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$\varnothing$.
It's transitive and symmetric by vacuous arguments. But it is not reflexive since $(n,n)\notin\varnothing$ for all $n\in\Bbb N$!
Asaf Karagila
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Trivial examples strike back! (At the end of the example we learn that Darth Trivius is in fact the father of Luck Setwalker.) – Asaf Karagila Jul 20 '14 at 15:11
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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
Marm
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