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While going through previous discussions, I saw this answer: https://math.stackexchange.com/a/442/969638. After reading through the answer, I was wondering if seriality and transitivity implied reflexivity?

I feel like the answer is yes, since every element is related to some other element, there's going to be some cycle, and using transitivity we can then get a relation between an element and itself...? But I'm not sure though.

Dwight Shrute
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No, of course not: take the usual strict order on the natural numbers.

Naïm Camille Favier
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  • Ah, yeah, I forgot to mention that it was restricted to finite sets, sorry! As a follow-up question: in the case of finite sets, how would we go about proving why the implication would be true? – Dwight Shrute Jul 28 '24 at 14:50
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    Actually this is not even true for finite sets: consider the relation $R$ on ${0,1}$ with $0 R 1$ and $1 R 1$. – Naïm Camille Favier Jul 28 '24 at 15:28
  • Okay, I see. And if we had the condition that we must have seriality, transitivity, and irreflexivity on a finite set, would this be impossible? And if so, how would I go about showing whether it's possible or not? – Dwight Shrute Jul 28 '24 at 16:40
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    Indeed, a serial and transitive relation on a finite set must have some $x$ with $xRx$. You can prove this by applying seriality $n$ times to $x_0 \in [n]$ and applying the pigeonhole principle. – Naïm Camille Favier Jul 28 '24 at 16:49