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

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!

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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 1 '11 at 1:31
If so then the intersection is equal to $\mathbb{R}.$ –  jspecter Sep 1 '11 at 1:34
If not you are taking the intersection of a bunch of pairwise disjoint sets of open subsets of $\mathbb{R}.$ –  jspecter Sep 1 '11 at 1:35

If by $\{ (q-1/n, q+1/n): q\in \mathbf{Q}\}$ you mean $$\bigcup_{q\in\mathbf{Q}} (q-1/n, 1+1/n)$$ 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.

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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 1 '11 at 1:57
Very concise. Thanks! –  Isaac Solomon Sep 1 '11 at 1:58
@Adhvaitha: see the construction given by Robert Israel. –  Willie Wong Sep 1 '11 at 13:29
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 Tamaroff May 31 at 23:23

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.

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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 1 '11 at 4:30