I read a section of a book and it made mention of the set of rationals not being a $G_\delta$. However, it gave no proof. I read on wikipedia about using contradiction, but it made use of the Baire category theorem, which is unfamiliar to me.
I was wondering if anyone could offer me a different proof; perhaps using the fact that the complement of $G_\delta$ is $F_\sigma$.

Thanks.

I suspect that just about any proof that doesn’t directly use the Baire category theorem either uses a consequence of it or proves a special case of it. I’ve chosen the second course.

Suppose that $\mathbb{Q} = \bigcap\limits_{k\in\omega}V_k$, where each $V_k$ is open in the usual topology on $\mathbb{R}$. Clearly each $V_k$ is dense in $\mathbb{R}$. Let $\mathbb{Q}=\{q_k:k\in\omega\}$ be an enumeration of the rationals, and for each $k\in\omega$ let $W_k=V_k\setminus \{q_k\}$; clearly each $W_k$ is dense and open in $\mathbb{R}$, and $\bigcap\limits_{k\in\omega}W_k = \varnothing$.

Let $(a_0,b_0)$ be any non-empty open interval such that $[a_0,b_0]\subseteq W_0$. Given a non-empty open interval $(a_k,b_k)$, let $r_k=\frac14(b_k-a_k)$; clearly $a_k<a_k+r_k<b_k-r_k<b_k$. Since $W_{k+1}$ is dense and open, there is a non-empty open interval $(a_{k+1},b_{k+1})$ such that $$(a_{k+1},b_{k+1}) \subseteq [a_{k+1},b_{k+1}] \subseteq W_{k+1}\cap (a_k+r_k,b_k-r_k),$$ and the construction can continue.

For $k\in\omega$ let $J_k = [a_k,b_k] \subseteq W_k$. For each $k \in \omega$ we have $J_k \supseteq J_{k+1}$, so $\{J_k:k\in\omega\}$ is a decreasing nest of non-empty closed intervals. Let $J = \bigcap\limits_{k\in\omega}J_k$; $J\subseteq J_k \subseteq W_k$ for each $k\in\omega$, so $J \subseteq \bigcap\limits_{k\in\omega}W_k = \varnothing$. But the nested intervals theorem guarantees that $J \ne \varnothing$, so we have a contradiction. Thus, $\mathbb{Q}$ cannot be a $G_\delta$-set in $\mathbb{R}$.
