Question

Do we necessarily get an equivalence relation when we form the transitive closure of the symmetric closure of the reflexive closure of a relation?

Short answer

It is true that the transitive closure of the symmetric closure of the reflexive closure of a relation is an equivalence relation.

Step-by-step solution

Given data

Let \(R\) be any relation whatsoever. Let \(S\) be the symmetric closure of the reflexive closure of \(R\).\(y\)

Concept  used of Equivalence relation

An equivalence relation is a binary relation that is reflexive, symmetric and transitive

Show  the equivalence relation

By definition, \(S\) is both reflexive and symmetric.

The transitive closure \(T\) of \(S\) is obtained as follows: \(x\) is related to \(y\) in \(T\) if there exists a sequence of elements \(p,q, \ldots ,w\) such that \(x\) is related to \(p\), \(p\) is related to \(q, \ldots ,w\) is related to \(y\).

Let \(x\) be related to \(y\) and be related to \(z\)in \(T\).

Then combining the sequence of elements connecting \(x\) and \(y\)and\(y\)and\(z\) as above we see that \(x\) is related to \(z\).

In other words, \(T\) is transitive.

\(T\)is already reflexive and symmetric.

Thus, we have shown that the transitive closure of the symmetric closure of the reflexive closure of (any) relation \(R\) is an equivalence relation.