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The question is to find an example of a set $S$ and three relations $R_1$, $R_2$, and $R_3$ on it, such that $R_1$ is reflexive but not transitive, $R_2$ is transitive but not symmetric and $R_3$ is symmetric but not reflexive.

This is what I have done: Let set $S=\{a,b,c,d\}$ and the three realations are as follows:

  • $R_1 = \{(a,a),(b,b),(c,c),(d,d)\}$
  • $R_2 = \{(a,b),(b,c),(c,d)\}$
  • $R_3 = \{(a,b),(b,a),(c,d),(d,c)\}$

If there is a set which is wrong and not satisfying what is needed can anyone help?

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  • $\begingroup$ I think you asked this earlier before deleting soon after. Please don't do that. A user could be almost finished typing out an answer to your question; in this case, that user was me. If you wish to change some aspect of the question, please edit it. If at any point you are unsure of how to do something on this site, it would be better for you to ask in a comment. $\endgroup$ – Michael Albanese Sep 3 '13 at 1:47
  • $\begingroup$ The first is not right, your relation is transitive, in a trivial way. Whenever $(x,y)$ is in $R_1$, and $(y,z)$ is in $R_1$, we have $(x,z)$ is in $R_1$. You could spoil transitivity by adding $(a,b)$ and $(b,c)$, and that's all. $\endgroup$ – André Nicolas Sep 3 '13 at 1:50
  • $\begingroup$ Is R1 reflexive ? $\endgroup$ – user91745 Sep 3 '13 at 1:58
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$R_1$ is reflexive, but it is also transitive. To see that it is transitive, let $(x, y), (y, z) \in R_1$. By your definition of $R_1$, $x = y$ and $y = z$. Therefore $(x, z) = (x, y) = (y, z) = (y, y) \in R_1$. To fix this, add a pair of elements of the form $(x, y)$ and $(y, z)$ with $x \neq y$, $y \neq z$, and $x \neq z$ but do not add $(x, z)$; this breaks transitivity.

$R_2$ is not symmetric, but it is not transitive (on any pair of elements). For example, $(a, b), (b, c) \in R_2$ but $(a, c) \not\in R_2$. Try adding in the necessary elements, for example $(a, c)$. Note, after including the first lot of elements you may need to then reexamine your relation and see if there is anything else that needs to be added so that your relation is transitive.

$R_3$ is symmetric but not transitive as desired. Interestingly, a symmetric, non-reflexive relation will also be non-transitive. Can you see why?

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  • $\begingroup$ Ultrapedantry: a symmetric, non-reflexive relation will either be non-transitive or empty :) (Extrasuperpedantry: unless your carrier set is empty, in which case the empty (indeed, only) relation on it is reflexive) $\endgroup$ – Ben Millwood Sep 5 '13 at 22:24
  • $\begingroup$ @BenMillwood: Of course, you are correct. I should have considered the empty relation. $\endgroup$ – Michael Albanese Sep 5 '13 at 22:58
  • $\begingroup$ "Should have" is a bit strong. Your answer is probably better off without mentioning it, but I couldn't resist. $\endgroup$ – Ben Millwood Sep 6 '13 at 0:14
  • $\begingroup$ What if I let R3 to be an empty set... will still satisfy the condition for R3???? $\endgroup$ – user91745 Sep 15 '13 at 9:40

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