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From Baire category theorem, we see that $\mathbb{Q}$ can not be a $G_{\delta}$. But consider the following construction:

Let us consider $\mathbb{Q}\cap [0,1]$, putting all the elements in the set in a sequence, denoted $\{a_n\}$. We define $V_i=\bigcup_{j}[a_j-1/2^{i+j},a_j+1/2^{i+j}]\cap [0,1].$ Notice that $\mathbb{Q}\subset V_i$.

So we define $V=\bigcap_{i} V_i.$ We have $\mathbb{Q}\subset V$, and $V$ is a zero-measure set.

Although it is clear that $V\neq \mathbb{Q}$, I cannot find any irrational number in $V$. Is there someway to find an irrational number that is in V, which proves $V\neq \mathbb{Q}$?

Thank you.

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    Isn't a proof of Baire's theorem for complete metric spaces usually constructive? I'd just run the argument for your example. – 2011-10-07

2 Answers 2

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For an "explicit" construction, consider $z = \sum_{i=1}^\infty 2^{-k_i}$ , where given $k_1 < \ldots < k_n$, if $\sum_{i=1}^n 2^{-k_i} = a_{m(n)}$, $k_{n+1} = k_n + 1 + 2 m(n)$. Note that $a_{m(n)} < z < a_{m(n)} + 2^{1-k_{n+1}} < a_{m(n)} + 2^{-2m(n)}$ so $z \in V_{m(n)}$. Since $m(n) \to \infty$ as $n \to \infty$, $z$ is in all the $V_i$. Its base-2 expansion contains infinitely many 1's but also has arbitrarily large gaps between the 1's, so can't be eventually periodic, and thus $z$ must be irrational.

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    $m(n)$ is the index $j$ such that $a_j = \sum_{i=1}^n 2^{-k_i}$. – 2014-04-18
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Since $V$ depends heavily on the specific enumeration of $\mathbb{Q}\cap[0,1]$, any explicit identification of a particular irrational number in $V$ must also depend on the enumeration. I’m not at all sure that such an explicit identification is possible even starting from some very nice, fully specified enumeration of $\mathbb{Q}\cap[0,1]$.

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    Well, if you wanted some specific irrational$x$to appear in the intersection, then you could at least reverse engineer an enumeration of the rationals which does the job. Just find an enumeration $q_1,q_2,q_3,\ldots$ such that |q_n - x| < 1/4^n for, say, every odd $n$. – 2011-10-07