Give an example to show that when the symmetric closure of the reflexive closure of the transitive closure of a relation is formed, the result is not necessarily an equivalence relation.
My attempt at a solution:
$R = \{(2,1),(2,3)\}$.
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.
I am not sure if this is correct.
Best Answer
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)}