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Theorem: If a relation is symmetric and transitive then it is reflexive.

Proof. Let R be a symmetric and transitive relation. Take elements x,y satisfying x R y. Then y R x (since R is symmetric), and so by the transitive law x R x. So R is reflexive.

I'm trying to show this is an incorrect proof to an incorrect theorem but I'm confused for something. Can a and c ever equal each other? So, aRb and bRa -> aRa

kiwizor
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  • @JMoravitz oh I always thought they had to be unique. That changes my view on relations thanks. – kiwizor Jan 13 '23 at 14:03

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$a,b,c$ do not need to be unique. The relation $\{(1,2),(2,1)\}$ is not transitive because with $a=c=1$ and $b=2$ you do have $(a,b)$ and $(b,c)$ in the relation but $(a,c)$ is not.

Further, you do not need to have any examples of $a,b,c$ such that $(a,b)$ and $(b,c)$ are both in the relation in the first place. Scenarios where you do not have any such choices of $a,b,c$ are also considered transitive.

As to the question of whether symmetry and transitivity together imply reflexivity, they do not. Consider the relation $\emptyset$ over some nonempty domain.

JMoravitz
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  • @Jean-ArmandMoroni well... sure... but you could say that about anything. That isn't all that interesting. Here, an example of such a third property is that every element appears in at least one of the pairs in the relation. The punchline is that this doesn't need to be the case. – JMoravitz Jan 13 '23 at 14:07
  • We could say that symmetry plus transitivity plus a third property imply reflexivity. This third property is: $\forall x, \exists y, x R y$. – Jean-Armand Moroni Jan 13 '23 at 14:08
  • The comment to which you replied was incomplete, sorry. (And you correctly guessed what I was about to say). When I write comments on a smartphone, I always end up hitting the "Add comment" button by mistake... – Jean-Armand Moroni Jan 13 '23 at 14:11