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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    $\begingroup$ $\epsilon$ is just some arbitrarily small number. I can take it to be any number. I don't think it matters. $\endgroup$
    – Adhvaitha
    Sep 1, 2011 at 3:42
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    $\begingroup$ 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 $2\epsilon$. $\endgroup$ Sep 1, 2011 at 4:44
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    $\begingroup$ Yes. I have changed it. $\endgroup$
    – Adhvaitha
    Sep 1, 2011 at 4:55
  • $\begingroup$ 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). $\endgroup$
    – Did
    Sep 1, 2011 at 7:21
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    $\begingroup$ Ok. Yes. Edited. But it is not the main thing in the problem. $\endgroup$
    – Adhvaitha
    Sep 1, 2011 at 7:27

2 Answers 2


In response to your first question, any algebraic irrational is not in $S_\varepsilon$ for some $\varepsilon > 0$ if we label the rationals in increasing order of the sum of their numerators and denominators. If $x\in S_{1/i}$ for all $i\ge1$, then there exists an infinite sequence of rationals $q_{a_1}, q_{a_2}, ...$ such that $x \in \left(q_{a_i}-\frac{1/i}{2^{a_i}},q_{a_i}+\frac{1/i}{2^{a_i}}\right)$. Our choice of ordering guarantees that for all $q_k$ within, say, a distance of $1$ from $x$, that $k$ is polynomially related to the denominator of $q_k$. Therefore, the error of $q_{a_i}$ in approximating $x$ is at most $2^{-poly(denominator(q_{a_i}))}$. By Liouville's theorem, this sequence of rationals approximates $x$ too well for $x$ to be algebraic.

This also answers your second question. Any transcendental $x$ with good enough rational approximants will be in $S$.

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    $\begingroup$ The first sentence can't possibly be right, since even the first interval in the union will contain an algebraic number, indeed, an algebraic irrational (which may be what you meant), indeed, infinitely many algebraic irrationals. $\endgroup$ Sep 1, 2011 at 4:36
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    $\begingroup$ Thank you for catching the typo. I have edited my post to clarify the claim. By considering algebraic irrationals in the interval $(n,n+1)$ for large enough $n$, one can construct a valid algebraic irrational not in $S_\varepsilon$ for a fixed $\varepsilon$. I may add this to the post later. $\endgroup$
    – Perce
    Sep 1, 2011 at 4:58

I'll answer the first question by considering a slightly different problem where it is easier to explicitly construct an irrational number not in $S_\epsilon$.

Namely, consider the interval $(0,1)$. Arrange the rational numbers $q_k$ inside it and "widen" each to a length $1/10^k$. In other words, define intervals

$$ I_k := \left(q_k-\frac12\frac1{10^k}, q_k + \frac12\frac1{10^k}\right) .$$

Obviously, the set

$$ S := \bigcup_{k=1}^\infty I_k .$$

contains all rational numbers, but it has length at most $1/9$, so there has to exist a number $x\in (0,1)-S$.

To construct this number, observe the following: if we divide the interval (0,1) into ten equal parts $(0,1/10)$, $(1/10,2/10)$, etc. up to $(9/10,1)$, then the first interval $I_1$ may only meet at most two of these parts. In other words, the number $x$ is contained in one of the parts, say $(3/10,4/10)$. Now, we can again subdivide this part into ten equal parts and likewise find that the interval $I_2$ may only intersect two of them.

Repeating this procedure, we obtain $x$ as a limit point of a sequence of nested intervals. Moreover, this construction makes it very easy to write down the decimal expansion of $x$. For instance, if we arrange the rational numbers as

$$ q_k = 1/2, 1/3, 2/3, 1/4, \dots $$

then we can choose $x = 0.3170\dots$ for example.

  • $\begingroup$ Are you using a particular scheme to generate 0.3170... or just picking any interval not intersected last section for each digit? $\endgroup$
    – Casebash
    Oct 18, 2018 at 12:54

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