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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.

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  • $\begingroup$ 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
    Commented Apr 1, 2015 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
    Commented Apr 1, 2015 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
    Commented Apr 1, 2015 at 9:43
  • $\begingroup$ @PrasunBiswas Yes, sorry I forgot to include it. $\endgroup$
    – user265675
    Commented Apr 1, 2015 at 9:44
  • $\begingroup$ @Meryll Note that that detail is of paramount importance. $\endgroup$
    – Git Gud
    Commented Apr 1, 2015 at 9:47

3 Answers 3

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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.

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  • $\begingroup$ So following that, even if we remove $(2,1)$ from the set, it can still remain transitive? Is this correct? $\endgroup$
    – user265675
    Commented Apr 1, 2015 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
    Commented Apr 1, 2015 at 10:58
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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.

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    $\begingroup$ It's rather difficult to understand this answer, as written. Please do what you can to make it clearer. $\endgroup$
    – Cameron Buie
    Commented Aug 4, 2015 at 2:36
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It is reflexive as {(1,1),(2,2),(3,3),(4,4)} are present in the set assuming it is a relation of {1,2,3,4} (original set).

Taking (1,1) as an example.

All (x,x) are symmetric too as: if x = 1, y = 1. (x,y) ∈ R and (y,x) ∈ R.

All (x,x) are symmetric too as: if x = 1, y = 1, z = 1 (x,y) ∈ R, (y,z) ∈ R and (x,z) ∈ R.

Here, (1,2) and (2,1) ∈ R, so the only other value that is not reflexive is taken care of by being symmetric.

Taking, (1,2)∈ R, (2,1) ∈ R, (1,1) ∈ R.
taking (2,1) ∈ R, (1,2) ∈, (2,2) ∈ R.

Checking for you, you will see all values have transitive relations existential.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jan 25, 2022 at 16:46

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