Is $R=\left \{ (a,a),(a,b),(b,a),(b,b),(c,c),(c,d),(d,c),(d,d) \right \}$ an equivalence relation on $X$? Let $X= \left \{ a,b,c,d \right \}$ and $R=\left \{ (a,a),(a,b),(b,a),((b,b),(c,c),(c,d),(d,c),(d,d) \right \}$. I want to show that $R$ is an equivalence relation on $X$.
My work:
$R$ is reflexive: If $a\in X$, $aRa$ since $(a,a) \in R$.
$R$ is symmetric: If $a,b \in X$, then $(a,b) \in R \Leftrightarrow (b,a) \in R$.
$R$ is not transitive: If $a,b,c \in X$, then $aRb$ and $bRc$ implies $aRc$. But $(b,c)$ is not an element of $R$ and $(a,c)$ is not an element of $R$. 
So $R$ is not an equivalence relation. But I am supposed to show that $R$ is an equivalence relation on $X$. For transitivity should $a\neq b\neq c$?
 A: Please note that $R$ is transitive if $(x,y) \in R$ and $(y,z) \in R \Rightarrow (x,z) \in R$
Now as $(a,b) \in R$ and $(b,c) \notin R$ you need not check for $(a,c)$ for this pair.
So whenever you find any pair of elements $(x,y)$ and $(y,z)$ both belonging to $R$ then only you check whether $(x,z)$ is in $R$ or not.
A: It doesn't matter whether (b,c) $\in$ R or not. Had it been the case that (a,b) $\in$ R and (b,c) $\in$ R, but (a,c) $\notin$ R, then I would conclude that R is not transitive. In this case, (b,c) $\notin$ R, so I dont worry. The relation is, by default transitive. You have done for reflexive and symmetric. So it is an equivalence relation.
A: for a relation to be an equivalence relation it must be reflexive, symmetric and transitive.
$R$ is reflexive if $ \forall x (x,x) \in R $ and $(a,a),(b,b),(c,c),(d,d) \in R$ so it is refexive
$R$ is symmetric if $ \forall x y \left( (x,y) \in R \to(y,x) \in R \right) $test this yourself.
$R$ is transitive if $ \forall x yz \left( \left((x,y) \in R \land (y,z) \in R \to (x,z) \in R \right) \right) $ test this also yourself
If your set pases all these tests it is an equivalence relation
Good luck
