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Given a finite set $S$ and a partial order $P \subseteq S \times S$, can we always enlarge $P$ to a total order $T$ on $S$?

Tian Vlasic
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Fantas Anadolou
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    What do you mean by “of smallest size”? – Michael Weiss Jun 15 '23 at 12:41
  • It's a finite set so I guess this is clear. But just in case, for $T$ the "smallest total closure", and $T'$ any other total closure, we necessarily have $T \subseteq T'$. – Fantas Anadolou Jun 15 '23 at 12:54
  • If the underlying set is the same, all orders have the same size. If you mean the cardinality of the set ${(a,b):a\leq b}$, then that is constant, for a fixed underlying set. – amrsa Jun 15 '23 at 12:56
  • @amrsa: just to stress, this is a question about finite sets so there are no subtle issues relating to cardinality. – Fantas Anadolou Jun 15 '23 at 13:00
  • @FantasAnadolou Yes, I understood that. And I keep my statement. Also, it's very relevant the question from Michael Weiss: at what do you refer when you talk about size of a total order? Again, see my previous comment. – amrsa Jun 15 '23 at 13:03
  • @amrsa: then I think I am confused: A total order $T$ is a relation, hence a subset of $S \times S$. We say that $T$ is smaller than another relation $T'$ if $T \subseteq T'$. What is the problem with this statement? I am sorry if I am missing something obvious. – Fantas Anadolou Jun 15 '23 at 13:08
  • The problem is that if $T$ and $T'$ are total orders on the same set $S$, then $T \subseteq T'$ implies $T = T'$, that is, one can't be larger than the other. Think about it. It's not difficult to justify. – amrsa Jun 15 '23 at 13:12
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    Of course!!!! Thanks for pointing this out. I will edit the question accordingly. – Fantas Anadolou Jun 15 '23 at 13:18
  • Even in the infinite case: Szpilrajn extension theorem. – amrsa Jun 15 '23 at 13:25
  • See also https://en.wikipedia.org/wiki/Topological_sorting – lhf Jun 15 '23 at 13:44
  • I reject the suggested edit, because yes, "extending" is the right terminology, but the reason OP probably has not found existing questions is because they don't know the right terminology making it more difficult to google. Leaving this with the "wrong" terminology will act as a jump off point for people with the same problem in the future – Felix Benning Jun 15 '23 at 15:43
  • https://math.stackexchange.com/q/271003/445105 does this answer your question? – Felix Benning Jun 15 '23 at 15:44

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