Measure theory and ring of sets

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I encounter some trouble by proving something. We are considering $X$ a set and $S \subset \mathcal{P}(X)$ closed by finite intersection and $\mathcal{R}$ the ring generated by $S$. We denote by $\pi$ the smallest system $S \subset \pi \subset \mathcal{P}(X)$ which verify :

  1. $\pi$ is closed by finite disjoint unions
  2. $\pi$ is closed by difference $A \setminus B$, $B \subset A$

We want to show that $\pi = \mathcal{R}$.

The inclusion $\pi \subset \mathcal {R}$ is ok. But now, it's written that we consider :

$$C_1 = \{ A \in \mathcal {R} \; | \; A \cap B \in \pi \quad \forall B \in S\}$$ $$C_2 = \{ B \in \mathcal {R} \; | \; A \cap B \in \pi \quad \forall A \in \mathcal {R}\}$$

And we want to show that $C_1 = \mathcal{R}$ to then show that $C_2 = \mathcal {R}$ which will permit to conclude.

But I don't succeed to prove it. Actually, we want to show that $C_1$ is a ring which contain $S$, if we do so, then it will be ok. So, I tried to do it :

  • $C_1$ is not empty cause it contains $S$ (as $S$ is closed by finite intersection, and $S \subset \pi$).
  • Now, let's show that for all $A, B \in C_1$, $A \setminus B = A \cap \overline{B} \in C_1$. Let $E \in S$, we want to show $A \cap \overline{B} \cap E \in \pi$. Actually, we have $A \cap E \in \pi$ and same for $B$, so we have $A \cap \overline{B} \cap E = F \cap \overline{B} = F \setminus B$ with $B, F \in \pi$. But we don't have necessarily $B \subset F$, so why this intersection (or difference) will be in $\pi$ ?

And same to prove that $C_1$ is closed by finite union, cause if I consider $A, B \in C_1$ and $E \in S$, I want to show that $(A \cup B) \cap E = (A \cap E) \cup (B \cap E) = (A \cap E) \cup (B \setminus A \cap E) \in \pi$. That's a finite disjoint union, but how we could know that $B \setminus A$ is on $\pi$ ? It would be if $A \subset B$, but it's not necessarily the case...

Actually, if we had replaced the condition 2. for $\pi$ by the set difference (i.e $A \setminus B = A \cap \overline{B} \in \pi \; \forall A, B \in \pi$), it would have worked, but I have a doubt here...

Someone could help me, please ?