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

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

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:

\(\ds \bigcap_{i \mathop \in \N} U_i\) \(=\) \(\ds \bigcap_{i \mathop \in \N} \paren {V_i \setminus \set {q_i} }\)
\(\ds \) \(=\) \(\ds \paren {\bigcap_{i \mathop \in \N} V_i} \setminus \paren {\bigcup_{i \mathop \in \N} \set {q_i} }\) Difference with Union
\(\ds \) \(=\) \(\ds \Q \setminus \Q\)
\(\ds \) \(=\) \(\ds \O\) 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, $\ds \bigcap_{i \mathop \in \N} U_i$ is dense in $\R$.

But Empty Set is Nowhere Dense.

This is a contradiction.


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

$\blacksquare$


Sources

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