# Smallest relation for reflexive, symmetry and transitivity

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.

• 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. Commented Nov 30, 2013 at 23:52
• Great! I guess the "smallest" doesn't seem so small after all. Commented Nov 30, 2013 at 23:54
• That interpretation looks right to me. Also, in the future, you can edit your previous, identical question: math.stackexchange.com/questions/577802/… Commented Nov 30, 2013 at 23:55
• 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. Commented Nov 30, 2013 at 23:57
• @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.. Commented Dec 1, 2013 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.)
• @Dimitri: When you compute the reflexive, transitive closure, it makes no difference whether you add the pairs $\langle x,x\rangle$ first or last. Commented Dec 1, 2013 at 0:13