Let $A$ be a $\sigma$-algebra generated by $B$.
Is it true that: $\bar x=\underset{C\in A|x\in C}{\cap} C=\underset{C \in B|x\in C}{\cap} C$ ? An exercises seems to be using it as a fact but I can't see why it works.
I can see that $\underset{C\in A|x\in C}{\cap} C \subset \underset{C \in B|x\in C}{\cap} C$
since $B \subset A$, but I don't know how to show the other inclusion.
I thought about a set in $E\in A \setminus C$ such that $x \in E$ but $z \notin E$ with $z \in \underset{C \in B|x\in C}{\cap} C$ in order to find a contradiction, but I don't see any, mostly because I don't know anything about how $A$ is constructed other than the fact it is the smallest $\sigma$-algebra which contains $B$.
Edit:
Ok, I guess this can't be true since the $\sigma$-algebra generated by $A$ over $E$ is $\{\emptyset, A,A^c,E\}$
If $x$ is in $E$ but not in $A$, then $E\space \cap\space A^c =A^c=E$ which is not logical if $A$ isn't the empty set... I must have misunderstood what I read then
Answering my own question so that it does not remain unanswered:
As the edit points out in the OP, the proposition is false.
Since there is a slight clash in terms of notations between the beginning and the end of the OP, I will henceforth consider $(A)$ to be the $\sigma-$algebra generated by the set $A \subset E$.
Let $x \in E$ such that $x \notin A$ (where $E$ denotes the set on which the sigma-algebra generated by $A$ is defined, and assuming $A \neq E$).
The $\sigma-$algebra generated by $A$ is $(E, A, A^c, \emptyset)$ (assuming $A \neq \emptyset)$.
The proposition in the OP implies $\bar x=\underset{C\in (A)|x\in C}{\cap} C=E\cap A^c=A^c=\underset{C \in \left\lbrace A \right\rbrace|x\in C}{\cap} C=E$ which is nonsensical whenever $A \neq \emptyset$.