Exercise 1.4.6 (Intersection of Boolean algebras)

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

(i)

(Empty set)
The empty set is contained in ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$ since for all $\alpha \in I$ we have $\varnothing \in B_{\alpha }$ by definition.

(ii)

(Complement)
Pick an arbitrary $E\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$; in other words, for all $\alpha \in I:E\in B_{\alpha }$. By definition of the Boolean algebra, we also must have for all $\alpha \in I:E^c\in B_{\alpha }$. Thus, $E^c\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$.

(iii)

(Finite unions)
Pick an arbitrary $E,F\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$. Again, since by definition of the Boolean algebra for all $\alpha \in I:E\cup F\in B_{\alpha }$, we also have $E\cup F\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$.

Now let $C$ be another Boolean algebra that is coarser than all $B_{\alpha }$, i.e., $\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in I:C\subseteq B_{\alpha }$. Then $C$ is also coarser than their intersection $C\subseteq {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in I}{B}_{\alpha }$ by the set-theoretic laws.