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

## Theorem

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

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

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

## Proof

Aiming for a contradiction, suppose $\Q$ 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:

 $\displaystyle \bigcap_{i \mathop \in \N} U_i$ $=$ $\displaystyle \bigcap_{i \mathop \in \N} \paren {V_i \setminus \set {q_i} }$ $\displaystyle$ $=$ $\displaystyle \paren {\bigcap_{i \mathop \in \N} V_i} \setminus \paren {\bigcup_{i \mathop \in \N} \set {q_i} }$ Difference with Union $\displaystyle$ $=$ $\displaystyle \Q \setminus \Q$ $\displaystyle$ $=$ $\displaystyle \O$ Set Difference with Self is Empty Set

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 $\R$.

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