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So if there exists an asymmetric relation within a set, does it also entail that there will be an antisymmetric relation in that same set?

If so, then it is possible to find out whether a set antisymmetric solely from knowing whether a set has a symmetric relation or not (because the absence of a symmetric relation entails a asymmetric relation).

Is this correct?

Are there any other ways (such as this) to determine whether a set is antisymmetric or not?

Thanks for reading.

Asaf Karagila
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Wade Griffin
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Consider these definitions:

  • Asymmetric relation: Given a set $S$ and a dyadic relation $R \in S\times S$, $R$ is asymmetric if it is never the case that for any ordered pair $(x, y)$ in $R$, the pair $(y, x)$ is in $R$.

As a consequence of this definition, any pair $(x,y)$ where $x=y$ is forbidden in an asymmetric relation. That is, it is always irreflexive.

  • Antisymmetric relation: Given a set $S$ and a dyadic relation $R \in S\times S$, $R$ is antisymmetric if, whenever $(x, y)$ and $(y, x)$ are in $R$, then it is necessarily the the case that $x=y$.

A relation is asymmetric if and only if it is both irreflexive and (vacuously) antisymmetric.

  • Diagonal or identity relation: $\{(x, y) | x,y \in S \land x = y\}$. It is always reflexive $-$ indeed, it is the reflexive closure of the empty relation.

You can always generate a "maximal antisymmetric closure" of an asymmetric relation in $S\times S$ by making the union of that asymmetric relation with the diagonal relation in $S\times S$. The result is a reflexive antisymmetric relation $-$ which is asymmetric if and only if $S$ is empty (since that relation would have to be simultaneously irreflexive and reflexive).

Antonielly
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There always exists at least one antisymmetric relation on a set, $\emptyset$.

It is meaningless to speak of a set being symmetric or antisymmetric. You can only call a relation on a set symmetric, not the set itself. And, of course, just because a relation on a set is not symmetric, that does not mean it is antisymmetric.

5xum
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  • But doesn't the falsehood of a symmetric relation entail the truth of a asymmetric relation, and if there exists an asymmetric relation on a set then it must also have an antisymmetric relation? I forgot to mention in the question, but i'm talking about binary sets for these relations, if that matters. – Wade Griffin Sep 17 '14 at 07:56
  • @WadeGriffin As I already said: Every set has a antisymmetric relation. I don't understand what you are asking. – 5xum Sep 17 '14 at 07:58