Let $\mathcal{C}$ be any collection of subsets of a set $X$ and let $E \subseteq X$. Let $\mathcal{C} \cap E := \{C \cap E : C \in \mathcal{C}\}$. Let $\mathcal{C}$ be any collection of subsets of a set $X$ and $E \subseteq X$. Define
$$
\mathcal{C} \cap E := \{C \cap E :  C \in \mathcal{C}\}.
$$
and let $\mathcal{A}(\mathcal{C})$ denote the algebra generated by the elements in $\mathcal{C}$.

*

*Show that $\mathcal{C} \cap E \subseteq \mathcal{A}(\mathcal{C}) \cap E$ and deduce that $\mathcal{A}(\mathcal{C}\cap E) \subseteq \mathcal{A}(\mathcal{C}) \cap E$.


*Define $\mathcal{F} := \{A \subseteq X : A \cap E \in \mathcal{A}(\mathcal{C}\cap E)\}$.
Then, $\mathcal{F}$ is an algebra of subsets of $X$, $\mathcal{C} \subseteq \mathcal{F}$ and $\mathcal{F}\cap E = \mathcal{A}(\mathcal{C} \cap E)$.


*Using 1 and 2, deduce that $\mathcal{A}(\mathcal{C})\cap E = \mathcal{A}(\mathcal{C} \cap E)$.

I have already obtained the result of the first one. And also that $\mathcal{F}$ is an algebra. But it is not very clear to me, how to obtain the rest of part 2, i.e. that:
$$
\mathcal{C} \subseteq \mathcal{F} \  \text{ and } \ \mathcal{F} \cap E = \mathcal{A}(\mathcal{C} \cap E).
$$

 A: To check that $\mathcal{C} \subseteq \mathcal{F}$ you need to take any $A \in \mathcal{C}$ and prove that $A \in \mathcal{F}$. Well, this is basically pushing the definitions:
$$
A \in \mathcal{C} \Rightarrow A \cap E \in \mathcal{C}\cap E \Rightarrow A \cap E \in \mathcal{A}(\mathcal{C}\cap E ) \Rightarrow A \in \mathcal{F}
$$
Do you see why each implication is true?
Next, to prove that $\mathcal{F} \cap E = \mathcal{A}(\mathcal{C}\cap E)$, notice that by definition of $\mathcal{F}$ one automatically has
$$
\mathcal{F} \cap E \subseteq \mathcal{A}(\mathcal{C}\cap E)
$$
It suffices to prove the other inclusion. You already checked in part 1 that $\mathcal{A}(\mathcal{C}\cap E) \subseteq \mathcal{A}(\mathcal{C})\cap E$. Since $\mathcal{C} \subseteq \mathcal{F}$ and you also checked that $\mathcal{F}$ is an algebra we have $\mathcal{A}(\mathcal{C}) \subseteq \mathcal{F}$. Putting all together gives
$$
\mathcal{A}(\mathcal{C}\cap E) \subseteq \mathcal{A}(\mathcal{C})\cap E \subseteq \mathcal{F} \cap E
$$
yielding the desired inclusion.
