Boolean Closure and Borel sets

Question

Denote the boolean closure of a family of sets $\mathcal S$ by $\mathcal B(\mathcal F)$, then in a metric space it is well known that $\mathcal B(\mathcal F) = \mathcal B(\mathcal G) = \mathcal B(\mathcal F \cup \mathcal G) \subseteq F_{\sigma} \cap G_{\delta}$. Now my question, is

1. $\mathcal F \cup \mathcal G$, i.e. the family of all open or closed sets, closed under boolean operations, i.e. $\mathcal B(\mathcal F \cup \mathcal G) \subseteq \mathcal F \cup \mathcal B$?
2. Is the family $\mathcal F_{\sigma} \cup \mathcal G_{\delta}$ closed under boolean operations, i.e. $\mathcal B(\mathcal F_{\sigma} \cup \mathcal G_{\delta}) \subseteq F_{\sigma} \cup G_{\delta}$?

Answer 1

1. $\mathcal{F} \cup \mathcal{G}$ is not closed under Boolean operations. Consider $( -1 , 0 ) \cup [ 0 , 1 ]$.

   The fact that $\mathcal{B} ( \mathcal{F} ) \subseteq F_\sigma \cap G_\delta$ rests heavily on the fact that closed subsets of metric spaces are G_δ (and so open sets are F_σ).

2. Similarly, the $F_\sigma \cup G_\delta$ is not closed under Boolean operations: consider the union of two sets "living" on different intervals in $\mathbb{R}$, one which is G_δ but not F_σ, and the other which is F_σ but not G_δ (for example, $( - \infty , 0 ) \setminus \mathbb{Q}$ and $( 0 , + \infty ) \cap \mathbb{Q}$).