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


Q. Assume that Bor$(\mathcal{E})=$Bor$(\tau)$. Let $O$ be an open in $X$. Does there exists a sequence of 
open sets $\{O\_n\}\subseteq \mathcal{E}$ with $O=\cup O\_n$? 



Nice representation of open sets in $\sigma$-algebras in certain circumstances
At least under MA+$\neg$CH the answer is negative. It is known that under MA+$\neg$CH the real line contains an uncountable set A such that every subset of A is Borel in A (moreover, it is of type $F\_\sigma$ and $G\_\delta$). 


Now consider the product $X=A\times\mathbb Q$ and let $\mu$ be the (metrizable separable) topology of the space X. Next, consider the topology $\tau$ on $X$, generated by the base $$\mathcal E=\mu\cup\{\{(x,0)\}:x\in A\}.$$


It is easy to see that the $\sigma$-algebra, generated by the base $\mathcal E$ coincides with the Borel $\sigma$-algebra, which coincides with the algebra of all subset of $X$. On the other hand, the open set $A\times\{0\}\in \tau$ cannot be represented as the countable union of basic set from the family $\mathcal E$.


