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I know, that

The smallest possible reflexive relation on a non-empty set is the diagonal ordered pairs of Cartesian product.

The largest possible reflexive relation on a non-empty set is the entire Cartesian product.

Likewise, The smallest possible anti-symmetric relation on a non-empty set is a null set.

But, I am not able to discern...

The largest possible anti-symmetric relation on a non-empty set?

P.S. I am not asking about- Total number of possible anti-symmetric relation.

Ubi.B
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    Maximal partial orders on a set $S$ are not unique unless $S$ has $\leq 1$ element (take the opposite ordering, for example), so no "the largest". – user10354138 Jul 30 '20 at 06:59
  • @user10354138 your comment suffice my need. Yep, there doesn't exist unique largest subset of Cartesian product of give set that represent it. – Ubi.B Jul 30 '20 at 08:24

1 Answers1

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Let $A=\{1,2,3\}$. Then both $R=\{(1,1)\}$ and $S=\{(1,2), (1,3)\}$ are anti-symmetric but $R \not\subset S$ and $S \not\subset R$. So no largest anti-symmetric relation can exist.

Anurag A
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  • The reason $R \not \subset S$ and $S \not \subset R$ is because there's a bigger antisymmetric relation than that, I think, unless I misunderstood something (per my own answer). Granted there is no singular "unique" antisymmetric relation, but we can at least quantify the number of such relations and even offer nice-looking samples (again, if my own understanding is correct). – PrincessEev Jul 30 '20 at 07:06
  • @EeveeTrainer OP doesn't want the number of such relations as stated in the question. – Anurag A Jul 30 '20 at 07:07
  • Relation S can also contain all reflexive pairs and in that case R will be proper subset of S. – Ubi.B Jul 30 '20 at 08:22
  • @Ubihatt The point is that the "largest" set having a property is the one that has all the sets with that same property as it's subsets which cannot happen here. Perhaps this example can help you. Consider $R={(1,2)}$ and $S={(2,1)}$, then both are anti-symmetric but cannot be subsets of each other. – Anurag A Jul 30 '20 at 12:28