Exercise 1.4.2 (Restriction of a Boolean algebra)

We verify the criteria from the Definition 1.4.1 (Boolean algebra).

(i)

(Empty set)
Since $B$ is a Boolean algebra, we have $\varnothing \in B$. Therefore $\varnothing =\varnothing \cap Y\in B{\text{↾}}_Y$.

(ii)

(Complement)
Let $E{\text{↾}}_Y\in B{\text{↾}}_Y$. Then, for some $E\in B$ we have $E{\text{↾}}_Y=E\cap Y$. Since $B$ is a Boolean algebra, $X∕E\in B$. Therefore, $(X∕E)\cap Y\in B{\text{↾}}_Y$. But $$B{\text{↾}}_Y\ni (X∕E)\cap Y=(X\cap Y)∕(E\cap Y)=Y∕(E\cap Y)=Y∕E{\text{↾}}_Y$$

which is the complement of $E{\text{↾}}_Y$ in the space $Y$. Thus, the complement of $E{\text{↾}}_Y$ with respect to $Y$ is also contained in $B{\text{↾}}_Y$.

(iii)

(Finite unions)
Let $E{\text{↾}}_Y,F{\text{↾}}_Y\in B{\text{↾}}_Y$. Then for some $E,F\in B$ we have $E{\text{↾}}_Y=E\cap Y$ and $F{\text{↾}}_Y=F\cap Y$. Then $E\cup F\in B$ by definition, and $B{\text{↾}}_Y\ni (E\cup F)\cap Y=(E\cap Y)\cup (F\cap Y)=E{\text{↾}}_Y\cup f{\text{↾}}_Y$.