Exercise 3.3

Suppose that $C$ is a relation on the set $A$. This argument is perfectly valid for any $a,b\in A$ such that $\mathit{aCb}$, which is to say that we can conclude that $\mathit{aCa}$ in this case (and by the same argument $\mathit{bCb}$). However, reflexivity requires $\mathit{aCa}$ to hold for every $a\in A$. So if there is no $b\in A$ such that $\mathit{aCb}$ then the above argument cannot be applied and we cannot conclude that $\mathit{aCa}$. In this case, the element $a$ is effectively not involved in the relation at all.

This is perhaps best illustrated with an example: suppose that $A=\left\{1,2,3,4\right\}$ and

It is easy to verify that $C$ is both symmetric and transitive on $A$ but it is clearly not reflexive since $(4,4)\notin C$. One can also observe how $4$ is not involved in the relation at all and, if it were, it would have to be that $(4,4)\in C$ if $C$ were to remain symmetric and transitive.