Can I always represent a Borel set as a countable sum of open and closed sets?

Question

I would like to use this fact in the proof and it seems to me to follow directly from the definition of the σ-algebra and the Borel set, but with the definition by intersection of all σ-algebras containing all open sets I have doubts. If it matters I only need it on $\mathbb{R}^n$ with Euclidean topology.

Answer 1

For a simple counterexample, the set of irrational numbers is one. Because it has no interior, if it can be written as a countable union of open sets and closed sets, then it is $F_\sigma$ (countable union of closed sets), so its complement, the set of rational numbers, is $G_\delta$ (countable intersection of open sets). Note that as rationals are dense, it will then be a countable intersection of open dense sets. But then you can intersect these with $\mathbb{R} \setminus \{q\}$ for all $q \in \mathbb{Q}$, which form a countable collection of open dense sets, to obtain the empty set. So a countable intersection of open dense sets is empty, which contradicts the Baire category theorem.