# Nice arrangement of open sets in $\sigma$-algebras

Let $X$ be a topological space and $\mathcal{E}$ be a topological base for $X$. Let us denote Bor$(\mathcal{E})$, by the smallest $\sigma$-algebra containing $\mathcal{E}$.

Q. Let $O$ be an open set contained in Bor$(\mathcal{E})$. Does there exists a sequence of open sets $\{O_n\}\subseteq \mathcal{E}$ with $O=\cup O_n$?

Not necessarily. Let $X$ be an uncountable set with the discrete topology, and let $\mathcal{E}$ be the collection of singletons, which is a base for the topology, since every set is a union of singletons.
The $\sigma$-algebra generated by the singletons, what you call $\text{Bor}(\mathcal{E})$, is the algebra consisting exactly of the countable and co-countable sets. But notice that if $Y$ is a co-countable subset of $X$, then it is open and in $\text{Bor}(\mathcal{E})$, but it is not the union of countably many singletons. So this is a negative instance of your desired property.