Prove that $\mathbb{Q}$ belongs to the Borel $\sigma$-algebra of $\mathbb{R}$ Prove that the set of rational numbers $\mathbb{Q}$ belongs to the Borel $\sigma$-algebra of $\mathbb{R}$.
I'm unsure about how to prove this. I'm also unsure about the difference between a σ-algebra and a Borel σ-algebra.
 A: As mentioned in the coomments, a $\sigma$ algebra is a set of subsets containing the empty set and closed under countable unions and complements. The Borel $\sigma$ algebra is a specific one attached to $\mathbb R$. It is described abstractly as the smallest $\sigma$ algebra containing all the open subsets of $\mathbb R$. You can check that this makes sense by proving that the arbitrary intersection of $\sigma$- algebras must be a $\sigma$-algebra. Then define the Borel $\sigma$-algebra as the intersection of all $\sigma$-algebras which contain the open subsets of $\mathbb R$.
This is an elegant but somewhat intractable definition. Can we actually name any Borel sets (Borel sets are elements of the Borel $\sigma$ algebra)? Well all we know to start with is that any open set must be Borel. Thus, the complement of any open set must be Borel, i.e. closed sets are Borel. But as closed sets are Borel, any countable union of closed sets must be Borel. In particular, any singleton $\{a\}$ is closed, therefore Borel. Any countable set is a countable union of points, which are all Borel, so any countable set must therefore also be Borel. As $\mathbb Q$ is countable, it is therefore Borel.
