What is the largest sequence of open balls around rational numbers that does not cover $\mathbb R$?

Let $$\mathbb Q = \{q_1,q_2,\dots\}$$ be the rational numbers. Let $$B_n = \{x \in \mathbb R: |x-q_n| < 2^{-n}\}$$ be the open ball of radius $$2^{-n}$$ around $$q_n$$.

Then the Lebesgue measure of $$U=\cup_{n=1}^\infty B_n$$ is at most $$\lambda(\cup_{n=1}^\infty B_n) \le \sum_{n=1}^\infty \lambda(B_n) = 1$$

In particular, $$U \subsetneq \mathbb R$$. But this is surprising since $$\mathbb Q$$ is dense in $$\mathbb R$$!

Now let $$C_n = \{x \in \mathbb R: |x-q_n| < c_n\}$$ where the $$c_n$$ are positive numbers in $$\mathbb R$$. Obviously, if $$\sum_{n=1}^\infty c_n < \infty$$ the same argument as above shows that $$\cup_{n=1}^\infty C_n \subsetneq \mathbb R$$. But what about nonsummable $$(c_n)$$? For example, if $$c_n = \frac{1}{n}$$, do we still get $$\cup_{n=1}^\infty C_n \subsetneq \mathbb R$$ or will $$\cup_{n=1}^\infty C_n$$ now cover $$\mathbb R$$?

More generally, given a fixed enumeration of the rationals, can we characterize a "slowest-declining" monotonous sequence $$(c_n)$$ such that $$\cup_{n=1}^\infty C_n \subsetneq \mathbb R$$?

• This certainly depends on the enumeration. For any sequence $c_n$ with infinite sum there is an enumeration for which $C_n$ cover $\mathbb R$, while for any $c_n$ tending to zero there is an enumeration for which they do not. Feb 7, 2020 at 12:32
• Feb 7, 2020 at 12:34

In case $$\sum c_n = +\infty$$, the answer will depend on the enumeration $$q_n$$.
As long as $$c_n \to 0$$, there is an enumeration $$(q_n)$$ of the rationals so that $$\sqrt{2} \notin \bigcup_{n=1}^\infty (q_n - c_n,q_n+c_n)$$.
And as long as $$\sum c_n = +\infty$$, there is an enumeration $$(q_n)$$ of the rationals so that $$\mathbb R = \bigcup_{n=1}^\infty (q_n - c_n,q_n+c_n)$$.