The best and the most reliable order to satisfy properties of equivalence relation is in the given order
=> Reflexive Closure-->Symmetric Closure-->Transitivity closure
The reason for this assertion is that like for instance if you are following the order
=> Transitivity closure-->Reflexive Closure-->Symmetric Closure
(just like has been asked) ,you may just end up with elements that you added for symmetric closure not being accounted for transitivity as has been shown in the example given in question which has been cited here for reference
R={(2,1),(2,3)} .
Transitive closure: {(2,1),(2,3)}.
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}.
Since the set is missing (1,3) and (3,1) to be transitive, it is not an equivalence relation.
But if you follow the order of satisfying Reflexive Closure first,then Symmetric Closure and at last Transitivity closure,then the equivalence property is satisfied as shown.
R={(2,1),(2,3)} .
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}
Transitive closure:{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,2),(3,1),(3,3)}
Best Answer
A relation is transitive if whenever $x\sim y$ and $y\sim z$ then $x\sim z$.
Notice however, that if the relation is transitive and $x\sim y$, $y\sim z$ and $z\sim w$ as well, this further implies that $x\sim z$ and $z\sim w$ which implies that $x\sim w$.
When finding the transitive closure of a relation, you need to add all ordered pairs $(x,y)$ where you can write a sequence of pairs that already exist in the relation $(x,a_1),(a_1,a_2),\dots,(a_{n-1},a_n),(a_n,y)$ regardless of length.
In the relation above, you can write $2\mapsto 0\mapsto 3$, so you will need to add the pair $(2,3)$. You can also travel $2\mapsto 0\mapsto 3\mapsto 2$, so the pair $(2,2)$ will also need to be added.
Similarly for the others that you already found and $(3,3)$. However, the pair $(1,1)$ is not going to be added as there exist no pairs where $1$ is the second entry (nothing leads to it). As such, it doesn't cause any problem for the transitivity of the relation.