Let $A$ be a commutative ring. A multiplicatevely closed subset $S$ in $A$ is called saturated if for all $a,b \in A$, $ab \in S$ implies $a\in S$ and $b\in S$.
Prove that
(i) For a multiplicatevely closed subset $S\subseteq A$, prove that $\bar{S} =\{a\in A$ | there exists $b\in A $ with $ab\in S\}$ is a multiplicatevely closed set,
(ii) $S\subseteq \bar{S}$,
(iii) $ \bar{\bar{S}}=S$, i.e., $\bar{S} $ is saturated.
Work: I have done (i). For (ii) I took one element in $S$ but I have no reason to believe why it must always lie in $\bar{S}$, can you explain why it should always hold?
For (iii) I tried to prove that $\bar{S}$ is saturated but failed: if I have $ab\in \bar{S}$ and I have to prove that $a$, $b$ lie in $\bar{S}$. $ab\in \bar{S}$ implies that there exists $c\in A$ such that $abc\in \bar{S}$. $abc\in \bar{S}$ implies that there exists $d$ in $A$ such that $abcd \in S$. But how would I prove that $a, b\in \bar{S}$?
So, please suggest some hints for (ii) and (iii).