# Suppose $R$ is a relation on $A$. Prove that if $R$ is reflexive then $R \subseteq R \circ R$. Counterexample?

Problem: Suppose $$R$$ is a relation on $$A$$. Prove that if $$R$$ is reflexive then $$R \subseteq R \circ R$$. Counterexample: Let $$A = \{1,2\}$$ and $$R = \{(1,1),(2,2),(1,2)\}$$. Then $$R \circ R = \{(1,2)\}$$. Obviously $$R\nsubseteq R \circ R$$. This is an exercise I saw in a book, so I assume this is not a valid counterexample. Either I don't understand the problem or the author made some kind of a mistake. Which one is it?

• Probably you have to check if $R◦R = \{ (1,2) \}$ is right ... See Composition of relations. Mar 23, 2015 at 13:41
• If the containment had gone the other way, i.e., if it had said $R \circ R \subseteq R$, then this would be equivalent to $R$ being transitive. But the way it is worded, saying $R \subseteq R \circ R$, it actually is a correct statement, given that $R$ is reflexive. In my answer, initially I was too hasty and wrote something wrong; I have edited it now, to avoid further confusion. Mar 23, 2015 at 14:20

It looks like your calculation of $R \circ R$ in this case is incorrect and that your example is not actually a counterexample to the statement that if $R$ is reflexive then $R \subseteq R \circ R$.