How would one go about proving that the rationals are not the countable intersection of open sets?

Question

I'm trying to prove that the rationals are not the countable intersection of open sets, but I still can't understand why

$$\bigcap_{n \in \mathbf{N}} \left\{\left(q - \frac 1n, q + \frac 1n\right) : q \in \mathbf{Q}\right\}$$

isn't a counter-example. Any ideas? Thanks!

Do you mean to write $ \displaystyle\bigcap_{n \in \mathbf{N}} \displaystyle\bigcup_{q\in \mathbb{Q}} (q - 1/n, q + 1/n)$ ? — jspecter, Sep 01 '11 at 01:31
If not you are taking the intersection of a bunch of pairwise disjoint sets of open subsets of $\mathbb{R}.$ — jspecter, Sep 01 '11 at 01:35
possible duplicate of Proof That $\mathbb{R} \setminus \mathbb{Q}$ Is Not an $F_{\sigma}$ Set — t.b., Sep 01 '11 at 07:41

Willie Wong · Accepted Answer · 2016-11-28T02:24:48.333

33

If by $\{ (q-1/n, q+1/n): q\in \mathbf{Q}\}$ you mean $$ \bigcup_{q\in\mathbf{Q}} (q-1/n, q+1/n) $$ you will find that for each fixed $n$, that set is equal to $\mathbf{R}$, independently of $n$. So the intersection you wrote down is equal to $\mathbf{R}$

The usual proof that $\mathbf{Q}$ is not a countable intersection of open sets uses the Baire Category Theorem. The Irrational numbers can be written as a countable intersection of open, dense subsets:

$$ \mathbf{R}\setminus \mathbf{Q} = \bigcap_{q\in \mathbf{Q}} \mathbf{R}\setminus\{q\} $$

Since $\mathbf{Q}$ is dense, any open set containing it would necessarily be an open dense subset of $\mathbf{R}$. So if $\mathbf{Q}$ were to be able to be written as a countable intersection $\cap_{n\in \mathbf{N}}A_n$ with each $A_n$ open, $\mathbf{Q}$ would be a countable intersection of open, dense subsets of $\mathbf{R}$.

But then we would have $$ \emptyset = (\mathbf{R}\setminus\mathbf{Q}) \cap \mathbf{Q} $$ is a countable intersection of open dense subsets, which contradicts the Baire Category Theorem.

edited Nov 28 '16 at 02:24

answered Sep 01 '11 at 01:41

Willie Wong

73,139

What about $\bigcap_{n=1}^{\infty} \bigcup_{k=1}^{\infty} \left( q_k - \frac1{2^{k+n}},q_k + \frac1{2^{k+n}} \right)$? For each fixed $n$, the union is not $\mathbb{R}$ since the length of each union is $\frac1{2^{n-1}}$. – Adhvaitha Sep 01 '11 at 01:57
1

Very concise. Thanks! – Elchanan Solomon Sep 01 '11 at 01:58
@Adhvaitha: see the construction given by Robert Israel. – Willie Wong Sep 01 '11 at 13:29
1

This generalizes: if $(X,d)$ is a complete metric space with no isolated points and if $D$ is a dense countable subset, then $D$ is not a $G_\delta$. – Pedro May 31 '14 at 23:23
You sure you don't mean: $ \bigcup_{q\in\mathbf{Q}} (q-1/n, q+1/n) $, in your second line, @WillieWong? – Nov 27 '16 at 04:12
@chell: thanks. fixed. – Willie Wong Nov 28 '16 at 02:24
Simplest answer on this topic I've encountered. – Aelx Oct 26 '21 at 09:42

score 18 · Answer 2 · answered Sep 01 '11 at 03:37

The right way to do this is via the Baire Category Theorem, but it may be helpful to also give a more-or-less explicit construction. Let $U_n, \ n=1,2,\ldots$ be a sequence of open sets, each containing the rationals $\mathbb Q$. I will find a member of $\bigcap_{n=1}^\infty U_n$ that is not in $\mathbb Q$. Let $r_n, \ n=1,2,\ldots$ be an enumeration of the rationals. I will construct two sequences $a_n$ and $b_n, \ n= 1,2,\ldots$ such that $a_n < a_{n+1} < b_{n+1} < b_n$, $(a_n, b_n) \subset U_n$, and $r_n \notin (a_n, b_n)$. Namely, given $a_n$ and $b_n$, we can find a rational $c \ne r_{n+1}$ in $(a_n, b_n)$ and take $a_{n+1} = c - \delta$ and $b_{n+1} = c + \delta$ where $\delta>0$ is small enough that $(c-\delta, c+\delta) \subset U_{n+1}$, $c - \delta > a_n$, $c + \delta < b_n$ and $\delta < |r_{n+1} - c|$. Now the sequence $a_n$ is bounded above and increasing, so it has a limit $L$. By construction, $L \in (a_n, b_n) \subset U_n$ for each $n$, i.e. $L \in \bigcap_{n=1}^\infty U_n$, Moreover, $L \ne r_n$ for all $n$, so $L \notin \mathbb Q$, as required.

Well done! I think for beginners,an explicit construction-assuming it's not too complicated and ugly-is much more informative then the big machinery of modern analysis. — Mathemagician1234, Sep 01 '11 at 04:30
@Robert Israel Sir Proof is elegant .But I had doubt .You had constructed nested sequence of open interval $(a_n,b_n)$ But What is argument to show that there infinite intersection must be nonempty? Because I know only about Cantor Intersection Theorem Which tell about Nonempty closed bounded interval.Where is I am Missing? — Curious student, May 04 '18 at 04:46
The limit $L$ of the left endpoints of the intervals is in the intersection. — Robert Israel, May 04 '18 at 07:02

How would one go about proving that the rationals are not the countable intersection of open sets?

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