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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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  • $\begingroup$ Isn't a proof of Baire's theorem for complete metric spaces usually constructive? I'd just run the argument for your example. $\endgroup$ – Mike F Oct 7 '11 at 6:36
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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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  • $\begingroup$ Yes, you are right! So brilliant solution. Thank you! $\endgroup$ – Jiajun Wu Oct 7 '11 at 18:56
  • $\begingroup$ What is $m(n)$ here? $\endgroup$ – MJD Apr 17 '14 at 16:19
  • $\begingroup$ $m(n)$ is the index $j$ such that $a_j = \sum_{i=1}^n 2^{-k_i}$. $\endgroup$ – Robert Israel Apr 18 '14 at 1:44
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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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    $\begingroup$ 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 $n$. $\endgroup$ – Mike F Oct 7 '11 at 7:13

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