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Was asked to Work out transitive closure of: R = {(1, 1),(1, 3),(2, 2),(2, 1),(3, 3),(4, 4),(4, 3),(4, 2)}

I did using Warshall's, getting: R*={(1,1)(1,3)(2,1)(2,2)(2,3)(3,3)(4,1)(4,2)(4,3)(4,4)}

Is R* antisymmetric?

I understand antisymmetric means if (a,b) exists and (b,a) exists then a=b.

But I am confused here, since there are symmetric elements here too: (1,2)(2,1)

Does the exclusion of (1,2)(3,1)(3,2)(1,4) etc make R* antisymmetric or symmetric or neither?

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  • $\begingroup$ Critics concerning title: being antisymmetric applies on relations (wich are specific sets). Not on sets in general. $\endgroup$ – drhab Oct 26 '15 at 12:41
  • $\begingroup$ Quite right, apologies. Will amend $\endgroup$ – ak1652 Oct 26 '15 at 12:44
  • $\begingroup$ @drhab relations are subsets of the cartesian product of a set with itself. So the poster is right telling "antisymmetric set". $\endgroup$ – user279325 Oct 26 '15 at 13:50
  • $\begingroup$ @user279325 I know that relations are specific sets (as mentioned in my former comment) but that is not a justification to speak of "antisymmetric sets". Likewise we do not speak about positive complex numbers. Using such terminology gives rise to senseless questions as: is set $\{\varnothing\}$ antisymmetric? $\endgroup$ – drhab Oct 26 '15 at 14:13
  • $\begingroup$ @drhab you're right. Mistake from me. Although the empty relation (corresponding to $\emptyset$ is antisymmetric. $\endgroup$ – user279325 Oct 26 '15 at 14:29
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Checking element by element you see it's antisymmetric. If (2,1) and (1,2) exist, there would be a problem, but it's not the case.

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  • $\begingroup$ Many thanks. If for example (1,2) existed, would it be neither symmetric nor anti symmetric? $\endgroup$ – ak1652 Oct 26 '15 at 12:21
  • $\begingroup$ If (1,2) exists, it's no antisymmetric because $1\neq2$, and it's no symmetric since $(1,3)$ exists, but $(3,1)$ doesn't exist. $\endgroup$ – user279325 Oct 26 '15 at 13:47

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