Does there exist a real number a given distance from each rational number?

Question

Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $x\in \mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?

This problem was inspired by an easier version of the problem where we assume the stronger condition that $\displaystyle\sum_{n=1}^\infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.

Let $\Omega$ denote the set of $x\in \mathbb{R}$ that do not satisfy the given property; we claim that $\Omega \neq \mathbb{R}$ and so an $x\in \mathbb{R}$ with given property exists. Indeed, $\Omega=\{x\in\mathbb{R} \ | \ \exists \ n\in \mathbb{N} \ \mathrm{such \ that} \ |x-r_n| \leq a_n\}=\displaystyle\bigcup_{n=1}^\infty \ [r_n-a_n,r_n+a_n]$
whose Lebesgue measure $\lambda(\Omega)\leq \displaystyle\sum_{n=1}^\infty \lambda([r_n-a_n,r_n+a_n])=\displaystyle\sum_{n=1}^\infty 2a_n < \infty$, so $\Omega \neq \mathbb{R}$.

Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.

Answer 1

No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).

(At the same time, given a sequence $(a_i)_{i\in\mathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $\pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)

For simplicity, let's look at $[0,\infty)$ rather than $\mathbb{R}$ (this doesn't make a substantive difference).

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The key point is the following picture: we chop $[0,\infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $\infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[\sum_{0<j\le i}{1\over j}, (\sum_{0<j\le i}{1\over j})+{1\over i+1})$. Note that each $B_k$ has "diameter" ${1\over k+1}$, and hence if $q\in B_k$ then the ball around $q$ with radius ${2\over k+1}$ covers $B_k$.

Now fix any enumeration of rationals $E=(r_i)_{i\in\mathbb{N}}$, and pick a sequence $n_i$ ($i\in\mathbb{N}$) of naturals such that:

• $n_i<n_{i+1}$, and

• $r_{n_i}\in B_i$.

Such a sequence must exist since $B_i\cap\mathbb{Q}$ is always infinite.

Finally, let $A=(a_i)_{i\in\mathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2\over i+1}$ for all $i$. The set $$\{(r_{n_i}-a_i, r_{n_i}+a_i): i\in\mathbb{N}\}$$ covers $[0,\infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.

Answer 2

The answer is no. Consider the following open cover of $\Bbb R$:

$$\mathcal U=\{ (H_{2n} , H_{2n+4}) \ | n \in \Bbb N_{\ge 1}\} \cup \{ (-H_{2n+4} , -H_{2n}) \ | n \in \Bbb N_{\ge 1}\} \cup \{ (-3, 3) \}$$ where $$H_n = \sum_{j=1}^n \frac 1j$$ is the $n$-th harmonic number. Note that for all $\varepsilon >0$ only finitely many elements $U \in \mathcal U$ satisfy $\lambda (U) > \varepsilon$.

Let $\{ U_{n} \}_{n \ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= \lambda (U_n)$ converges to $0$.

Let $\{ r_n \}_{n \ge 1}$ be an enumeration of rational numbers such that for all $n$ $$r_{n} \in U_n$$ holds. Recall that $a_n$ is rational, and it converges to $0$.

Now, for all $x \in \Bbb R$ there exists $n \ge 1$ such that $x \in U_n$: thus $|x-r_n| < a_n$.