TITLE: Constructing a number not in $\bigcup\limits_{k=1}^{\infty} (q_k-\frac{\epsilon}{2^k},q_k+\frac{\epsilon}{2^k})$
QUESTION [17 upvotes]: I have couple of questions related to the properties of real numbers.

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.
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

REPLY [8 votes]: 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.