$\mathbf{Q}$ as the countable intersection of open sets? $\def\Q{\mathbf{Q}}$
$\def\N{\mathbf{N}}$
$\def\Z{\mathbf{Z}}$
$\def\I{\mathbf{I}}$
$\def\Fs{F_\sigma}$
$\def\Gd{G_\delta}$
Baire's Theorem tells us that $\Q$ is not a $\Gd$ set. However the set of integers $\Z$ is $\Gd$ by the following collection of sets:
$$\Z = \bigcap_{n \in \N}\bigcup_{m \in \Z}{{\left(m-\frac{1}{n},m+\frac{1}{n} \right)}}$$
I want to understand why a similar cover for $\Q$ fails to work. Baire's Theorem, although a valid explanation, is somewhat philosophically unsatisfying.
$$\Q \overset{?}{=} \bigcap_{n \in \N}\bigcup_{q \in \Q}{{\left(q-\frac{1}{n},q+\frac{1}{n} \right)}}$$
To look at what it would mean if there was an irrational number $x \notin \Q$ in this set: If it is in this set, then it was initially in some interval centred at some rational number $q$. But as $n$ gets arbitrarily large, eventually the neighbourhood around $q$ will become small enough that it will no longer contain $x$. This almost feels like it could qualify as a contradiction but I guess it's still possible for $x$ to be in infinitely many intervals. It just looks very counter-intuitive. If there was a direct way to construct an irrational $x \notin \Q$ that is in this set, then that would be philosophically satisfying.
Final Remark. To anyone that wants to appeal to the length of the set or Lebesgue measure $\neq 0$ or something related to that,

*

*I am not familiar with any of these notions formally, so I'll likely get more confused by it.

*I could modify the construction like this:

$$\Q \overset{?}{=} \bigcap_{n \in \N}\bigcup_{r=1}^{\infty}{{\left(q_r-\frac{1}{n \cdot 2^r},q_r+\frac{1}{n \cdot 2^r} \right)}}$$
where $\left\{q_1,q_2,q_3 \dots \right\}$ is an enumeration of $\Q$. This set looks to have zero "length" (by my informal understanding).
 A: Try and follow a proof of the Baire Category theorem. Let $\langle O_n:n\in\mathbb{N}\rangle$ be a sequence of open sets that contain $\mathbb{Q}$, and let $\langle q_n:n\in\mathbb{N}\rangle$ enumerate $\mathbb{Q}$.
Take a closed interval $[a_1,b_1]$ (with ($a_1<b_1$) inside $O_1$ and such that $q_1\notin[a_1,b_1]$.
Keep going recursively: given $[a_n,b_n]$ (with $a_n<b_n$ of course), note that $O_{n+1}\cap(a_n,b_n)\neq\emptyset$. Take $[a_{n+1},b_{n+1}]$ inside that intersection with $a_{n+1}<b_{n+1}$ and $q_{n+1}$ not in the interval.
In the end the intersection $\bigcap_n[a_n,b_n]$ is nonempty, contained in $\bigcap_nO_n$ and disjoint from $\mathbb{Q}$ (because the intersection is an interval disjoint from $\mathbb{Q}$ it contains just one irrational number).
You can make this quite constructive: the $a_{n+1}$ and $b_{n+1}$ can be rational numbers and they can form the first pair in some enumeration.
This idea is quite flexible: you can have $2^{n-1}$ disjoint intervals at stage $n$ and take two disjoint intervals in each of their intersections with $O_{n+1}$. In this way you'll find a copy of the Cantor set in $\bigcap_nO_n$ that is disjoint from $\mathbb{Q}$.
