Let $\mathcal R$ be a $\sigma$-ring, then: $\{E\subset X; E\cap F \in \mathcal R\text{ for every } F \in \mathcal R\}$ is a $\sigma$-algebra. I'm trying to Solve the following question:   

Let $X$ be a non empty set and  $\mathcal R$ be a  $\sigma$-ring from subset of $X$. Prove that: $$S=\{E\subset X; E\cap F \in \mathcal R\text{ for every } F \in \mathcal R\}$$ is a $\sigma$-algebra.  

I can show that $S$ is closed under the countable union of it's subsets. But how I can show that if $E\in S$, then $E^c \in S$?  

Let: $E\in S$. For every $F\in \mathcal R$, we know that $E\cap F \in \mathcal R$. So, $F-(E\cap F) \in \mathcal R$. And it's easy to see that, $E^c \cap F=F-(E\cap F)$. Thus, $E^c \cap F \in \mathcal R$. So, $E^c\in S$.
