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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
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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
    
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2 Answers

up vote 11 down vote accepted

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
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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
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