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I have couple of questions related to the properties of real numbers.

  1. My first question is as follows. Let $S_{\epsilon} = \displaystyle \bigcup_{k=1}^{\infty} \left( q_k-\frac{\epsilon}{2^{k+1}},q_k+\frac{\epsilon}{2^{k+1}} \right)$, where all the rationals are listed as $\{q_1,q_2,\ldots,q_n,\ldots\}$. The length of this set is bounded by $\epsilon$. This means there are a lot of irrationals not in the set. How do I go about explicitly constructing an irrational number not in the set? The irrational number will depend on the way I list the rationals but once the list is given I should be able to construct an irrational number not in the set.
  2. My second question is motivated from this question. I came to know that the set of rationals is not a $G_{\delta}$ set. However let us consider this. Let $$S_n = \bigcup_{k=1}^{\infty} \left(q_k - \frac{\epsilon}{2^{k+n+1}},q_k + \frac{\epsilon}{2^{k+n+1}} \right).$$ Clearly, $S_n$ is an open set and the length of $S_n$ is bounded by $\displaystyle \frac{\epsilon}{2^{n}}$. Let $$S = \bigcap_{n=1}^{\infty} S_n.$$ $S$ is a $G_{\delta}$ set and the length of $S$ is zero. Further, $\mathbb{Q} \subseteq S$. What other numbers are in $S$? How do I explicitly construct a number in $S \backslash \mathbb{Q}$? If there are no other numbers i.e. if $\mathbb{Q} = S$, then doesn't it imply that $\mathbb{Q}$ is a $G_{\delta}$ set?

Thanks, Adhvaitha

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    $\epsilon$ is just some arbitrarily small number. I can take it to be any number. I don't think it matters.2011-09-01
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    The length of the first interval - the one corresponding to $k=1$ - is already $\epsilon$, so I think your set $S_{\epsilon}$ is of length between $\epsilon$ and $2\epsilon$.2011-09-01
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    Yes. I have changed it.2011-09-01
  • 0
    Still, the length is not correct. @Gerry's point was that, apart from the obvious estimates that the length is between the length of the first interval and twice that, there is no universal formula for it (in particular not epsilon which, at the moment, is an upper bound).2011-09-01
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    Ok. Yes. Edited. But it is not the main thing in the problem.2011-09-01
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    General note: just because you can show that a set is "large" does not mean that it is feasible to construct elements of it. For instance, there are sets $S$ of real numbers such that $\mathbb{R} \smallsetminus S$ is known to have measure zero, but it is an open problem to construct--with proof--a real number in $S$.2011-09-03

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