Example of a set that is $F_\sigma$ but not $G_\delta$

Question

Can anyone give me an example? My textbook offers a hint to let be a denumerable set that is dense, although it doesn't say dense in what.

Answer 1

The set of rational numbers in the reals.

Answer 2

Let $\Bbb Q=\{q_n:n\in \Bbb N\}.$ Let $F=\{O_n\}$ be a family of open subsets of $\Bbb R$ such that $\cap F\supset \Bbb Q.$ Let $G=\{O_n\backslash \{q_n\}:n\in \Bbb N\}.$ By the Baire Category Theorem, $\cap G$ is dense in $\Bbb R.$

So $(\cap F)\backslash \Bbb Q=\cap G\ne \emptyset.$ So $\cap F\ne \Bbb Q.$

So $\Bbb Q$ is not a $G_{\delta}$ subset of $\Bbb R.$ But $\Bbb Q$ is an $F_{\sigma}$ subset of $\Bbb R$ because $\Bbb Q=\cup_{n\in \Bbb N}\{q_n\}.$

If $S$ is dense in $\Bbb R$ and $S$ is a $G_{\delta}$ subset of $\Bbb R$ then by the above technique, $S$ is uncountable. So any countable dense subset of $\Bbb R$ is $F_{\sigma}$ but not $G_{\delta}.$

Answer 3

Your text sounds like Stein's Real Analysis.

They want you to use the Baire Category Theory (https://en.wikipedia.org/wiki/Baire_category_theorem) with $\mathbb{Q}$. Indeed, as $\mathbb{Q}$ is countable, it is a $F_{\sigma}$. Now we argue by contradiction to show that $\mathbb{Q}$ is not a $G_{\delta}$. Indeed, suppose for contradiction that $\mathbb{Q}$ is a $G_{\delta}$. Then there exists open sets $\alpha_i$ such that $\mathbb{Q} = \bigcap_{k=1}^{\infty} \alpha_i$.

Now let $\{Q_i: Q_i \in \mathbb{Q}, i \in \mathbb{N}\}$ be an explicit enumeration of $\mathbb{Q}$ and consider $ A_{k} : =(-\infty, Q_k) \cup (Q_k, \infty)$, then $\overline{A_k} = \mathbb{R}$, so we see $\overline{\bigcap_{k=1}^{\infty} A_k} =\mathbb{R}$. So by Baire Category Theory, one sees that $\overline{\bigcap_{k=1}^{\infty} A_k \cap \bigcap_{k=1}^{\infty} \alpha_k} = \mathbb{R}$, but $\bigcap_{k=1}^{\infty} A_k \cap \bigcap_{k=1}^{\infty} \alpha_k = \mathbb{Q} \cap \mathbb{Q}^c = \emptyset$, so we have $\overline{\emptyset} = \mathbb{R}$, which is a contradiction. Therefore, $\mathbb{Q}$ is not a $G_{\delta}$.