G𝛿 and Fσ Sets

We now give definitions for two important types of subsets of $\mathbb{R}$ - $G_{\delta}$ sets and $F_{\sigma}$ sets.

Definition: A subset $G$ of $\mathbb{R}$ is a $G_{\delta}$-set if $G$ is a countable intersection of open sets.

Note that each open interval $(a, b) \subset \mathbb{R}$ is a $G_{\delta}$-set.

For example, the set of irrational numbers $I$ is a $G_{\delta}$ set. Note that the set of rational numbers $\mathbb{Q}$ is countable, that is, $\mathbb{Q} = \{ q_1, q_2, ... \}$. For each $n \in \mathbb{N}$ let $Q_n = (-\infty, q_n) \cup (q_n, \infty)$. Then each $Q_n$ is an open set. The set of irrational numbers can be given by:

(1)

\begin{align} \quad I = \bigcap_{n=1}^{\infty} (-\infty, q_n, q_n, \infty) = \bigcap_{n=1}^{\infty} Q_n \end{align}

So $I$ is a countable intersection of open sets which shows that $I$ is a $G_{\delta}$ set.

Definition: A subset $F$ of $\mathbb{R}$ is an $F_{\sigma}$-set if $F$ is a countable union of closed sets.

Note that each closed interval $[a, b] \subset \mathbb{R}$ is an $F_{\sigma}$-set.

Theorem 1: $G$ is a $G_{\delta}$ set if and only if $G^c$ is an $F_{\sigma}$ set.

• Proof: Let $G$ be a $G_{\delta}$ set. Then $\displaystyle{G = \bigcap_{n=1}^{\infty} A_n}$ where each $A_n$ is an open set in $\mathbb{R}$. Thus the complement of $G$ is given by:

(2)

\begin{align} \quad G^c = \left ( \bigcap_{n=1}^{\infty} A_n \right )^c = \bigcup_{n=1}^{\infty} A_n^c \end{align}

• Since $A_n$ is open, $A_n^c$ is closed. Thus $G^c$ is a countable union of closed sets. From above, we see that $G$ is a $G_{\delta}$ set if and only if $G^c$ is an $F_{\sigma}$ set. $\blacksquare$

Theorem 2: If $F \subset \mathbb{R}$ is countable then $F$ is an $F_{\sigma}$ set.

• Proof: Since $F$ is countable, $F = \{ f_1, f_2, ... \}$. For each $n \in \mathbb{N}$ let $F_n = \{ f_n \}$. Then each $F_n$ is a singleton set which is closed and:

(3)

\begin{align} \quad F = \bigcup_{n=1}^{\infty} F_n \end{align}

• The above representation shows that $F$ is a countable union of closed sets, so $F$ is an $F_{\sigma}$ set. $\blacksquare$

Corollary 3: The set of rational numbers $\mathbb{Q}$ is an $F_{\sigma}$ set.

• Proof: $\mathbb{Q}$ is countable so by the previous theorem, $\mathbb{Q}$ is an $F_{\sigma}$ set. $\blacksquare$