Set of Rational Numbers is not G-Delta Set in Reals

Theorem
Let $\Q$ be the set of rational numbers.

Let $\struct {\R, \tau}$ be the real number line under the usual (Euclidean) topology.

Then $\Q$ is not a $G_\delta$ set in $\R$.

Proof
$\Q$ is a $G_\delta$ set in $\R$.

Let $\Q = \displaystyle \bigcap_{i \mathop \in \N} V_i$.

Since Rational Numbers are Countably Infinite, there exists an enumeration of $\Q$.

Write $\Q = \set {q_i}_{i \mathop \in \N}$.

Define $U_i = V_i \setminus \set {q_i}$.

We show that $U_i$ is dense in $\R$.


 * Let $A \subseteq \R$ be an open set of $\struct {\R, \tau}$.


 * From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau}$.


 * So there exists $\openint a b \subseteq A$ for some $a < b$.


 * One of $q_i \le a$, $q_i > a$ must hold.


 * Suppose $q_i \le a$.


 * By Between two Real Numbers exists Rational Number we have:


 * $\exists r \in Q: a < r < b$.


 * Since $r > a \ge q_i$:


 * $r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.


 * Suppose $q_i > a$.


 * By Between two Real Numbers exists Rational Number we have:


 * $\exists r \in Q: a < r < \min \set {q_i, b}$.


 * Since $r < q_i$:


 * $r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.


 * In both cases we see that $r \in \openint a b \cap U_i \subseteq A \cap U_i$.


 * Therefore $A \cap U_i \ne \O$.


 * Thus $U_i$ is dense in $\R$.

Also we have:

By Real Number Line is Complete Metric Space and Baire Category Theorem, $\struct {\R, \tau}$ is a Baire space.

By definition of Baire Space, $\displaystyle \bigcap_{i \mathop \in \N} U_i$ is dense in $\R$.

But Empty Set is Nowhere Dense.

This is a contradiction.

Hence $\Q$ is not a $G_\delta$ set in $\R$.