Using the following definition:
A set $A \subseteq \mathbb{R}$ is called an $F_{\sigma}$ set if it can be written as the countable union of closed sets. A set $B \subseteq \mathbb{R}$ is called a $G_{\delta}$ set if it can be written as the countable intersection of open sets.
I can show that the countable union of $F_{\sigma}$ sets is an $F_{\sigma}$ set and the countable intersection of $G_{\delta}$ sets is a $G_{\delta}$ set. However, how can I show that the finite intersection of $F_{\sigma}$ sets is an $F_{\sigma}$ set and likewise, the finite union of $G_{\delta}$ sets is a $G_{\delta}$ set?
Best Answer
Let $F_1, \dots, F_n$ be $F_{\sigma}$ sets, then there are closed sets $\{F_{1j}\}_{j=1}^{\infty}, \dots, \{F_{nj}\}_{j=1}^{\infty}$ such that $F_i = \bigcup_{j=1}^{\infty}F_{ij}$ for $i = 1, \dots, n$. Then
$$\bigcup_{i=1}^n F_i = \bigcup_{i=1}^n\bigcup_{j=1}^{\infty}F_{ij} = \bigcup_{j=1}^{\infty}\bigcup_{i=1}^nF_{ij} = \bigcup_{j=1}^{\infty}F'_j$$
where $F'_j = \bigcup_{i=1}^nF_{ij}$. As $F'_j$ is a finite union of closed sets it is closed. Therefore $\bigcup_{i=1}^n F_i$ is a countable union of closed sets, so it is an $F_{\sigma}$ set.
Likewise for $G_{\delta}$ (or take complements).