Proof, [Lemma 8.2.4, Coh13]. Let $\cf \subset 2^{X}$ be the smallest subset of $2^{X}$ satisfying the lemma. Since $\cb_{X}$ satisfies the lemma, $\cf \subset \cb_{X}$. On the other hand, let

then by definition, $\cf_{0}$ is closed under complements. Let $\seq{A_n}\subset \cf_{0}$, then

Since $\cf_{0} \subset \cf$ is closed under complements, $\seq{A_n^c}\subset \cf$ as well. By (2) and (3), $\bigcup_{n \in \natp}A_{n} \in \cf$ and $\bigcap_{n \in \natp}A_{n}^{c} \in \cf$. Thus $\bigcup_{n \in \natp}A_{n} \in \cf$ as well. By (1), $\cf_{0}$ is a $\sigma$-algebra that contains all open subsets of $X$, so $\cf \supset \cf_{0} \supset \cb_{X}$.$\square$