# If $E \in \sigma(\mathcal{C})$ then there exists a countable subset $\mathcal{C}_0 \subseteq \mathcal{C}$ with $E \in \sigma(\mathcal{C}_0)$

Given a collection of sets $\mathcal{C}$ and $E$ an element in the $\sigma$-algebra generated by $\mathcal{C}$, how do I show that $\exists$ a countable subcollection $\mathcal{C_0} \subset \mathcal{C}$ such that $E$ is an element of the $\sigma$-algebra, $\mathcal{A}$ generated by $\mathcal{C_0}$?

The hint says to let $H$ be the union of all $\sigma$-algebras generated by countable subsets of $\mathcal{C}$….although I don’t know why.

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A good strategy would be to prove that $H$ is equal to $\sigma(\mathcal C)$, the $\sigma$-algebra generated by $\mathcal C$. It should be clear that $H \subset \sigma(\mathcal C)$. For the reverse inclusion, it is enough to show that $H$ is, in fact, a $\sigma$-algebra. Check each of the axioms! At some point, you will have to use the fact that a countable union of countable sets is countable.