Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Isn't a proof of Baire's theorem for complete metric spaces usually constructive? I'd just run the argument for your example. –  Mike F Oct 7 '11 at 6:36

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
    
Yes, you are right! So brilliant solution. Thank you! –  Jiajun Wu Oct 7 '11 at 18:56
    
What is $m(n)$ here? –  MJD Apr 17 at 16:19
    
$m(n)$ is the index $j$ such that $a_j = \sum_{i=1}^n 2^{-k_i}$. –  Robert Israel Apr 18 at 1:44

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 $\mathbb{Q}\cap[0,1]$.

share|improve this answer
3  
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 $n$. –  Mike F Oct 7 '11 at 7:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.