"Give a relation that satifies the condition:"
Symmetric and transitive but not reflexive.
This is what I gave:
R = {(x,y), (y,z), (z,x), (y,x), (z,y), (x,z)}
I was told this was not transitive. However, my book's definition of transitive is the following:
R is transitive if for every x, y, and z, xRy and yRz implies xRz
What did I do incorrectly?
At the risk of confusing you even more, here is a well known fake theorem.
"Theorem". If a relation is symmetric and transitive, it must be reflexive.
"Proof". Suppose $R$ is symmetric and transitive. Take any element $x$. Take an element $y$ such that $x\,R\,y$. Since $R$ is symmetric we have $y\,R\,x$. But then since $R$ is transitive and $x\,R\,y$ and $y\,R\,x$, we have $x\,R\,x$. As this is true for all $x$, the relation $R$ is reflexive.
If you can figure out what is wrong with this argument, it may guide you in the direction of the example you need. Good luck!
Let us use $\{a,b,c\}$ as the set your relation is over to avoid confusion with the $x,y,z$ in the definitions of reflexive, symmetric, transitive. To be reflexive you would have to have $\{(a,a),(b,b),(c,c)\} \subset R$ Your relation corresponds to $\{(a,b),(b,a),(a,c)(c,a),(b,c),(c,b)\}$ so your relation is certainly not reflexive. In the definition of transitive, it is not required that $x,y,z$ be distinct elements, so I can say that if the relation is transitive $(aRb\ \& \ bRa) \implies aRa$. By inspection this is false, so the relation is not transitive.
To rescue this, assume $aRb$. Then by symmetry you have $bRa$. By transitivity, then we need $aRa$ Does this suggest a way out?
