2

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?

SwiftMo
  • 1,047
  • 1
  • 13
  • 23
  • 1
    This question (http://math.stackexchange.com/questions/290884/finite-unions-and-intersections-f-sigma-and-g-delta-sets?rq=1 ) seems to provide the answer you're after. – postmortes Jan 06 '14 at 13:22

2 Answers2

2

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).

Michael Albanese
  • 99,526
1

Hint Finite intersections of opens sets are open, and finite unions of closed sets are closed.

Pedro
  • 122,002