Is it true that $\sigma(\cap_{n \in \mathbb{N}} \mathcal{C}_{n}) = \cap_{n \in \mathbb{N}} \sigma(\mathcal{C}_{n})$?

Question

Is it true that if $\Omega$ is a set and $(\mathcal{C}_{n})_{n \in \mathbb{N}}$ is a decreasing sequence of classes of subsets of $\Omega$, then the sigma-algebra generated by the intersection of the classes $\mathcal{C}_{n}$ is equal to the intersection of the sigma-algebras generated by each individual class $\mathcal{C}_{n}$?

Answer 1

No. Take $\mathcal{C}_n$ to be balls in $\mathbb{R}$ of radius at most $1/n$. Then $\bigcap_{n \in \mathbb{N}}\mathcal{C}_n = \emptyset$, while the $\sigma$-algebra generated by each $\mathcal{C}_n$ is the standard $\sigma$-algebra on $\mathbb{R}$.