# Symmetric closure of the reflexive closure of the transitive closure of a relation

Give an example to show that when the symmetric closure of the reflexive closure of the transitive closure of a relation is formed, the result is not necessarily an equivalence relation.

My attempt at a solution:

$$R = \{(2,1),(2,3)\}$$.

Transitive closure: $$\{(2,1),(2,3)\}$$.

Reflexive closure: $$\{(1,1),(2,1),(2,2),(2,3),(3,3)\}$$.

Symmetric closure: $$\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)\}$$.

Since the set is missing $$(1,3)$$ and $$(3,1)$$ to be transitive, it is not an equivalence relation.

I am not sure if this is correct.

• It is indeed correct. – Brian M. Scott Nov 30 '15 at 8:11
• Could you please clarify why R is already transitive at the beginning? – Philippe Nov 30 '15 at 23:40
• Transitivity says that if there are elements $a,b,c$ (any of which can be equal) such that $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$, then $\langle a,c\rangle\in R$. This condition is vacuous – says nothing – if there are no elements $a,b$, and $c$ such that $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$. That’s the case here: the hypothesis of the if-then is never satisfied, so the then part is never invoked. To put it differently, if $R$ were not transitive, there would have to be $a,b$, and $c$ such that $\langle a,b\rangle\in R$ and ... – Brian M. Scott Dec 1 '15 at 4:21
• ... $\langle b,c\rangle\in R$, but $\langle a,c\rangle\notin R$. And there certainly aren’t, since there aren’t even $a,b$, and $c$ such that $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$. – Brian M. Scott Dec 1 '15 at 4:22
• Very clear explanation. Thank you very much! – Philippe Dec 2 '15 at 4:32