6
$\begingroup$

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 $\cup_{n=1}^\infty C_n \subsetneq \mathbb R$?

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Wojowu
    Feb 7, 2020 at 12:32
  • $\begingroup$ See Relative null-ness and this 22 November 2017 comment. $\endgroup$ Feb 7, 2020 at 12:34

1 Answer 1

3
$\begingroup$

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 $\mathbb R = \bigcup_{n=1}^\infty (q_n - c_n,q_n+c_n)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .