Constructing a number not in $\bigcup\limits_{k=1}^{\infty} (q_k-\frac{\epsilon}{2^k},q_k+\frac{\epsilon}{2^k})$

Question

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

Answer 1

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 $S$.

Answer 2

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.