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

Question

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?

Answer 1

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 $R \subseteq R \circ R$.

Answer 2

There are more pairs in the R ◦ R of your example. Think of what you can compose from (1,1) and then (2,2). This will show you why your example is not a counterexample.

In addition, there is a pretty straightforward proof for the initial claim : just take an element of R , make use of the reflexive property and you will show that the element belongs to R ◦ R as well.