Proposition 13.1.4. Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then:
For any $A, B \in \alg$, $A \cap B \in \alg$.
For any $A, B \in \alg$, $A \setminus B \in \alg$.
If $\alg$ is a $\sigma$-algebra, then:
For any $\seq{A_n}\in \alg$, $\bigcap_{n \in \natp}A_{n} \in \alg$.
Proof. (1, 1’): $\bigcap_{i \in I}A_{i} = \braks{\bigcup_{i \in I}A_i^c}^{c}$.
(2): $A \setminus B = A \cap B^{c}$.$\square$