Stable set by intersection and by finite union i'm reading a classical book in measure theory, and there is something i don't get. I would say it is a missprint but since the book is famous, probably there is something i don't get, and i need some other opinion.
They claim the following in one of their proofs : Let $\mathcal{E}$ be a set of subsets of a set $E$ so that $\emptyset \in \mathcal{E}$, and denote $\mathcal{E}'$ the smallest set of subsets of $E$ among those which contain $\mathcal{E}$ and are stable by finite union and by any intersection.
Let $\mathcal{F}$ be the smallest set of subsets of $E$ among those which contain $\mathcal{E}$ and are stable by finite union. Then $\mathcal{E}'$ is the smallest set of subsets of $E$ among those which contain $\mathcal{F}$ and are stable by any intersection.
Do you agree that this claim is clearly wrong or do i miss something ?
 A: Call $$\Sigma = \{ X \subseteq E: \mathcal{E} \subseteq X, \mbox{ stable under finite unions and all intersections}\}$$
$$\Omega = \{ X \subseteq E: \mathcal{E} \subseteq X, \mbox{ stable under finite unions}\}$$
then clearly $\Sigma \subseteq \Omega$, so
$$\mathcal{F} = \bigcap \Omega \subseteq \bigcap \Sigma = \mathcal{E}'$$
Moreover, $\mathcal{E}'$ is stable under finite unions and all intersections, and contains $\mathcal{F}$.
Now, let $Y$ be stable under finite unions and all intersections, and containing $\mathcal{F}$. We need to show that $Y$ contains $\mathcal{E}'$.
But now, $\mathcal{E} \subseteq \mathcal{F} \subseteq Y$, so $Y \in \Sigma$. This implies that $\mathcal{E}' \subseteq Y$.
A: The statement is true, but I needed some time to see it aswell.
Let $S$ be the smallest set containing $\mathcal{F}$ which is stable by any intersection. We claim that $S$ is then stable by finite union which proves the statement.
Of course it suffices to show, that for $A,B\in S$ we have $A\cup B\in S$.
Let $A,B\in S$. Then there exist $A_i,B_j \in \mathcal{F}$ with $i\in I, j\in J$, s.t. $A = \bigcap_{i\in I} A_i$ and $\bigcap_{j\in J} B_j = B$. This is by minimality of $S$.
Now we have $\displaystyle A\cup B = \left(\bigcap_{i\in I}A_i\right)\cup \left(\bigcap_{j\in J}B_j\right) = \bigcap_{i\in I,j\in J} (A_i\cup B_j)$.
Since $A_i,B_j\in\mathcal{F}$ and $\mathcal{F}$ is stable by finite unions we have $A_i\cup B_j\in\mathcal{F}$ for all $i,j$, hence have $A\cup B\in S$ as intersection of sets in $\mathcal{F}$.
