# $\mathbf{Q}$ 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,

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).

• $\mathbb{Z}$ is closed. Every closed subset of a complete metric space is a $G_\delta$ set (this is a result by Alexandroff). Commented Feb 1, 2022 at 5:35
• This question and solutions may answer your last comment. It links measure and category. Commented Feb 1, 2022 at 5:39
• For each $n$ we have $\bigcup_{q\in\mathbb{Q}}(q-{1\over n}, q+{1\over n})=\mathbb{R}$, so in fact $$\bigcap_{n\in\mathbb{N}}\bigcup_{q\in\mathbb{Q}}(q-{1\over n}, q+{1\over n})=\bigcap_{n\in\mathbb{N}}\mathbb{R}=\mathbb{R}.$$ Commented Feb 1, 2022 at 5:42
• 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. Commented Feb 1, 2022 at 5:56
• @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? Commented Feb 1, 2022 at 9:13

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 ($$a_1) inside $$O_1$$ and such that $$q_1\notin[a_1,b_1]$$.

Keep going recursively: given $$[a_n,b_n]$$ (with $$a_n of course), note that $$O_{n+1}\cap(a_n,b_n)\neq\emptyset$$. Take $$[a_{n+1},b_{n+1}]$$ inside that intersection with $$a_{n+1} 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 $$\mathbb{Q}$$.

• 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. Commented Feb 7, 2022 at 6:31
• 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. Commented Feb 7, 2022 at 6:33
• @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 Commented Apr 9, 2022 at 17:49