Given the relation $\{(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)\}\,$ determine whether it is reflexive, transitive, symmetric, or anti-symmetric.

I found this set to be reflexive and symmetric. But not transitive and anti-symmetric.

Would it be correct to say that this set would be anti-symmetric if we remove either the element $(1,2)$ or $(2,1)$?

Also, the solution claims this set to be transitive. But I found it not to be so, due to the reasoning that $(2,3)$ and $(3,4)$ is not in the set.

Is my understanding of these ideas correct? Thank you.

  • $\begingroup$ No, the given relation is transitive, since for every $aRb$ and $bRc$, we have $aRc$ in the relation, where $a,b,c\in\{1,2,3,4\}$. $\endgroup$ – Prasun Biswas Apr 1 '15 at 9:41
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    $\begingroup$ Transitivity would require $(2,3)$ to be in the set only if there were some $n\in\{1,2,3,4\}$ such that $(2,n)$ and $(n,3)$ were in the set. There is no such $n$, however, so transitivity says nothing about the pair $(2,3)$. $\endgroup$ – Brian M. Scott Apr 1 '15 at 9:42
  • $\begingroup$ Just to clarify, on which set is the relation made? Is it $R:A\mapsto A$ where $A=\{1,2,3,4\}$ $\endgroup$ – Prasun Biswas Apr 1 '15 at 9:43
  • $\begingroup$ @PrasunBiswas Yes, sorry I forgot to include it. $\endgroup$ – user265675 Apr 1 '15 at 9:44
  • $\begingroup$ @Meryll Note that that detail is of paramount importance. $\endgroup$ – Git Gud Apr 1 '15 at 9:47

It is reflexive if this is a relation over the set $\{1,2,3,4\}$, and yes, the relation is symmetric.

Yes, If we remove $(1,2)$ or $(2,1)$ then it is anti-symmetric.

The relation is transitive, we do not need $(2,3)$ and $(3,4)$ to be in the set. Especially there is no pairs in the relation $(2,x)$ and $(x,3)$, which is what we would need in order to force $(2,3)$ to be in the relation due to transitivity.

  • $\begingroup$ So following that, even if we remove $(2,1)$ from the set, it can still remain transitive? Is this correct? $\endgroup$ – user265675 Apr 1 '15 at 9:53
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    $\begingroup$ Yes. Since for the remaining "weird" relation $(1,2)$ we do not have anything on the form $(2,x)$ except for $(2,2)$ which is left in the relation. Note though that if we remove (1,1) from our original relation, you do not only lose reflexivity, but also transitivity since $(1,2) $ and $(2,1)$ implies $(1,1)$. $\endgroup$ – Ove Ahlman Apr 1 '15 at 10:58

is reflexive as (a,a);a=1(1)4 exist,is symmetric as (1,2) nd (2,1) exist and transitive as corrosponding term of (2,1),(1,2)- (1,1) exist in this relation.

  • $\begingroup$ It's rather difficult to understand this answer, as written. Please do what you can to make it clearer. $\endgroup$ – Cameron Buie Aug 4 '15 at 2:36

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