Let be a relation on a non-empty set . Define, explicitly in terms of , the relation which is the symmetric closure of . You should prove that your relation is indeed the smallest symmetric relation on containing as a subset.
What I've tried
Let $R$ be a relation on a non-empty set $A$
Thus $x,y \in A | xRy$
Let $S$ be a symmetric closure of $R$.
$ S = R \cup R^{-1}$
$S$ = {$(x,y) \in A$x$A| x R y$} $\cup$ {$(y,x)\in A$x$A|xRy$}
Thus $S$ is symmetric by inspection.
1- Suppose $S'$ is a smaller symmetrical closure
2- $\exists a,b \in S|$it is not part of S' and are symmetric.
3- However from above we have shown that S contains all symmetrical pairs of elements (as it is a symmetric closure).
by 2 and 3, we have reached a contradiction.
Thus S is the smallest symmetric relation on A containing R as a subset.
Is the symmetric closure of R defined as the smallest symmetric relation containing R?
You neglected to give an explicit formula for symmetric closure.
All you did was to show that the smallest is unique.
Exercise. Prove S = R $\cup$ R$^{-1}$ is symmetric.
In addition, prove that if K is a symmetric relation for the same set of elements, that K is a subset of S.