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I've been asked to explain why "a binary relation, $R$, that is irreflexive and transitive will be anti-symmetric." I started to think about the question by coming up with some relations that I understood to be both irreflexive and transitive. One of those was $R = \{(x,y), (y,z), (z,x), (x,z), (z,y), (y,x)\}$.

This relation confuses me. If reflexivity is defined as $\forall x \in A(xRx)$ and there are no elements of the form $(x,x)$ contained within $R$, then it seems that $R$ is irreflexive. Transitivity is defined as $\forall x,y,z \in A (xRy \land yRz \to xRz)$. $R$ also appears to be transitive. How then can I show that all binary relations which are irreflexive and transitive are anti-symmetric, when the definition of symmetry ($\forall x,y \in A (xRy \to yRx)$) seems to hold for $R$?

Is it correct for me to believe that $R$ is irreflexive, transitive and also symmetric?

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  • $\begingroup$ Your $R $ isn't transitive since you have $(x,y) $ and $(y,x) $ but not $(x,x) $. $\endgroup$ – Mark S. Nov 20 '16 at 2:51
  • $\begingroup$ Oh! ... That makes sense. $\endgroup$ – paleto-fuera-de-madrid Nov 20 '16 at 2:52
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$R$ is not transitive, although it may appear so at first glance if drawn as a triangle with edges between vertices. As Mark S. pointed out in the comments, $(x,y) \in R$ and $(y,x) \in R$, but $(x,x) \notin R$. The relation would have to be reflexive to allow for this.

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