Find the smallest relation containing the relation

$$R=\{ (1,2),(2,1),(2,3),(3,4),(4,1) \}$$ that is

  • Reflexive and transitive

  • Reflexive, transitive and symmetric

Well this seems easy to do. However, I'm not sure whether the question is meant to find the (for the first part) the reflexive and transitive closures, or is it something else?

If it's a closure case, then the first part would be: $$R=\{ (1,2),(2,1),(2,3),(3,4),(4,1),(1,1),(2,2),(3,3),(4,4),(1,3),(2,4),(3,1),(4,2)\}$$

But this doesn't seem right for some reason and just wanted to clarify what the question is asking.

  • 1
    $\begingroup$ Your interpretation of the question is correct. The $R$ you propose isn't correct: $(1,3)\in R\land (3,4)\in R$, but $(1,4)\not \in R$. You shouldn't use the same letter for the closure. You're using $R$ with two different meanings. $\endgroup$ – Git Gud Nov 30 '13 at 23:52
  • $\begingroup$ Great! I guess the "smallest" doesn't seem so small after all. $\endgroup$ – Dimitri Nov 30 '13 at 23:54
  • $\begingroup$ That interpretation looks right to me. Also, in the future, you can edit your previous, identical question: math.stackexchange.com/questions/577802/… $\endgroup$ – Henry Swanson Nov 30 '13 at 23:55
  • $\begingroup$ If the relation is 'connected' in that you can reach every 'node' from every other 'node', then if the relation is transitive, it must contain every pair. $\endgroup$ – copper.hat Nov 30 '13 at 23:57
  • $\begingroup$ @HenrySwanson Yes, I knew there was a previous question related to this but was more concerned about the definitions of the properties. I just wasn't sure how to "follow up" on the same question though.. $\endgroup$ – Dimitri Dec 1 '13 at 0:00

You’re missing $\langle 1,4\rangle,\langle 3,2\rangle$, and $\langle 4,3\rangle$; the first is required by $\langle 1,2\rangle$ and $\langle 2,4\rangle$, the second by $\langle 3,4\rangle$ and $\langle 4,2\rangle$, and the last by $\langle 4,1\rangle$ and $\langle 1,3\rangle$, for instance. The reflexive, transitive closure of $R$ is in fact $\{1,2,3,4\}\times\{1,2,3,4\}$. (And you should not call it $R$, as that name is already in use for the original relation.)

  • $\begingroup$ Yes, I keep forgetting to check that for every new ordered pair I add to my relation, I must also check for transitivity and symmetry in there as well. $\endgroup$ – Dimitri Dec 1 '13 at 0:05
  • $\begingroup$ @Dimitri: Note that in this particular case symmetry comes for free with transitivity. $\endgroup$ – Brian M. Scott Dec 1 '13 at 0:07
  • $\begingroup$ By the looks of it you're right. So does the order matter in cases like these? It seems as though transitivity should be done the same time as the reflexive. Or am I just over-thinking this? $\endgroup$ – Dimitri Dec 1 '13 at 0:11
  • $\begingroup$ @Dimitri: When you compute the reflexive, transitive closure, it makes no difference whether you add the pairs $\langle x,x\rangle$ first or last. $\endgroup$ – Brian M. Scott Dec 1 '13 at 0:13
  • $\begingroup$ Yes, very true. Thank you once again. $\endgroup$ – Dimitri Dec 1 '13 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.