What is the minimum $ \sigma$-algebra that contains open intervals with rational endpoints

Let $\mathcal C:=\{(a,b),a,b\in\Bbb Q\}$. Denote $\mathcal R$ the $\sigma$-ring generated by $\mathcal C$ and $\mathcal A$ the $\sigma$-algebra generated by $\mathcal C$.

Let $a<b$ two real number, $\{a_n\}$ and $\{b_n\}$ two sequence of rational numbers, $a_n\downarrow a$ and $b_n\uparrow b$. Then $$(a,b)=\bigcup_{n\geq 1}(a_n,b_n)$$ hence $(a,b)\in\mathcal R$ and $(a,b)\in\mathcal A$. Since we can write each open set as a countable union of intervals, $\mathcal A$ contains the Borel $\sigma$-algebra, and since it's the smallest $\sigma$-algebra containing $\mathcal C$, it's actually the Borel $\sigma$-algebra.

We have $\mathbb R=\bigcup_{n\geq 1}\underbrace{(-n,n)}_{\in\mathcal C}$ hence $\Bbb R\in\mathcal R$. Hence $\mathcal R$ is stable by countable union, complementation and contains the whole real line: it's a $\sigma$-algebra. By the same argument, $\mathcal R=\mathcal B(\Bbb R)$.