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$\def\Q{\mathbf{Q}}$ $\def\N{\mathbf{N}}$ $\def\Z{\mathbf{Z}}$ $\def\I{\mathbf{I}}$ $\def\Fs{F_\sigma}$ $\def\Gd{G_\delta}$

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,

  1. I am not familiar with any of these notions formally, so I'll likely get more confused by it.
  2. 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).

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  • $\begingroup$ $\mathbb{Z}$ is closed. Every closed subset of a complete metric space is a $G_\delta$ set (this is a result by Alexandroff). $\endgroup$
    – Mittens
    Commented Feb 1, 2022 at 5:35
  • $\begingroup$ This question and solutions may answer your last comment. It links measure and category. $\endgroup$
    – Mittens
    Commented Feb 1, 2022 at 5:39
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    $\begingroup$ For each $n$ we have $\bigcup_{q\in\mathbb{Q}}(q-{1\over n}, q+{1\over n})=\mathbb{R}$, so in fact $$\bigcap_{n\in\mathbb{N}}\bigcup_{q\in\mathbb{Q}}(q-{1\over n}, q+{1\over n})=\bigcap_{n\in\mathbb{N}}\mathbb{R}=\mathbb{R}.$$ $\endgroup$ Commented Feb 1, 2022 at 5:42
  • $\begingroup$ One remark that might help your intuition is that a set of real numbers can have zero "length" (measure) yet still be uncountable and have infinitely many irrational numbers—the Cantor set is one example. $\endgroup$ Commented Feb 1, 2022 at 5:56
  • $\begingroup$ @GregMartin The Cantor set was proved to have irrational numbers since it was proved to be perfect and that non-empty perfect sets are uncountable. Is there some nice rigorous proof to show that my last constructed set isn't countable? $\endgroup$ Commented Feb 1, 2022 at 9:13

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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 $\mathbb{Q}$.

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  • $\begingroup$ I read and followed the logic of the proof, but it now looks like if you replace $\mathbf{Q}$ with $\mathbf{Z}$, the arguement doesn't change! Why can't this arguement be used to prove that $\mathbf{Z}$ can't be $G_\delta$? I have a feeling it's got to do with the fact that $\mathbf{Q}$ is dense in $\mathbf{R}$ but I can't put my finger on the logic there. $\endgroup$ Commented Feb 7, 2022 at 6:31
  • $\begingroup$ Could you also expand a bit on the last idea you posted: "In this way you'll find a copy of the Cantor set." You totally lost me there. $\endgroup$ Commented Feb 7, 2022 at 6:33
  • $\begingroup$ @JackBrady with $\mathbb{Z}$ the argument doesn't work because now the open sets need not be dense, so you don't have as much liberty to choose the intervals. Concretely, the intersection of $[a_n,b_n]$ and $O_{n+1}$ may be disjoint $\endgroup$
    – Saúl RM
    Commented Apr 9, 2022 at 17:49

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