Let $(r_n)_{n \ge 1}$ be an enumeration of the rationals. Consider the union $A := \cup_n (r_n-\frac{1}{n^2},r_n+\frac{1}{n^2})$. It is unclear a-priori whether $A$ covers the real line, since although the rationals are dense in the reals, the $\frac{1}{n^2}$'s might shrink too fast. However, using measure theory, it is very easy to see $A$ does not cover much: indeed, $m(A) \le \sum_n m((r_n-\frac{1}{n^2},r_n+\frac{1}{n^2})) = \sum_n \frac{2}{n^2} = \frac{\pi^2}{3}$.

Since this argument relies much on the convergence of $\sum_n \frac{1}{n^2}$, I am wondering whether $B := \cup_n (r_n-\frac{1}{n},r_n+\frac{1}{n})$ covers the whole real line, or what portion of it? Does the amount covered depend on the enumeration we choose?

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    $\begingroup$ I am quite sure you can find some enumeration of rationals such that $B$ does not cover the whole real line. The interesting question is whether $B$ may cover the real line for some enumeration. $\endgroup$ – Crostul Dec 6 '16 at 8:34
  • $\begingroup$ Yea, you could probably take an enumeration where for every $n$ not a perfect square, $r_n \in [0,1]$ and for every $n = m^2$, $r_n \not \in [0,1]$, so that we could apply the first argument to outside $[0,1]$. $\endgroup$ – mathworker21 Dec 6 '16 at 8:47
  • $\begingroup$ You can enumerate $\Bbb Q$ in such a way that $ B=\Bbb R.$ You can also enumerate $\Bbb Q$ in such a way that $ B$ has finite measure $\endgroup$ – DanielWainfleet Nov 4 '18 at 8:32
  • $\begingroup$ @DanielWainfleet How do you know you can enumerate $\mathbb{Q}$ in such a way that $B = \mathbb{R}$? $\endgroup$ – mathworker21 Nov 4 '18 at 8:33
  • $\begingroup$ @mathworker21. I have posted an answer to that part of the Q. $\endgroup$ – DanielWainfleet Nov 4 '18 at 10:39

Let $q_3\in \Bbb Q \cap (5/6,1).$

Recursively, for $j\in \Bbb Z^+$ let $q_{3(j+1)} \in \Bbb Q \cap (q_{3j}+\frac {1}{3j}-\frac {1}{6(j+1)}, q_{3j}+\frac {1}{3j}).$

Then $0<q_{3i}< q_{3j}$ when $i<j,$ and we have $\cup_{j=1}^n(-\frac {1}{3j}+q_{3j},\frac {1}{3j}+q_{3j}\supset [1,1+ \sum_{j=1}^n\frac {1}{6j}).$

So $\cup_{j\in \Bbb Z^+}(-\frac {1}{3j}+q_{3j},\frac {1}{3j}+q_{3j})\supset [1,\infty).$

Similarly we can find a discrete $\{q_{3j-1}:j\in \Bbb Z^+\}\subset \Bbb Q$ such that $q_{3j+2}<q_{3j-1}<0$ and $\cup_{j\in \Bbb Z^+}(-\frac {1}{3j-1}+q_{3j-1},\frac {1}{3j-1}+q_{3j-1})\supset (-\infty,-1].$

Let $q_1=0.$

Since the set $S=\{q_1\}\cup \{q_{3j}:j\in \Bbb Z^+\}\cup \{q_{3j-1}:j\in \Bbb Z^+\}$ is discrete, the set $\Bbb Q$ \ $S$ is infinite so we can enumerate $\Bbb Q$ \ $S=\{q_{3j+1}:j\in \Bbb Z^+\}.$

And we have $\cup_{j\in \Bbb Z^+}(-1/j+q_j,1/j+q_j)=\Bbb R.$

We can also enumerate $\Bbb Q$ in a different way, to make $\cup_{n\in \Bbb N}(-1/n+q_n,1/n+q_n)$ a set of finite measure.

  • $\begingroup$ Sorry for late response. I don't think you necessarily enumerated all of the rationals. For example, we can have $C = \{1\}, A = \{2,4,6,8,\dots\}, B = \{3,5,7,9,\dots\}$ in which case you can't get the rationals not covered by $A$ or $B$ by rationals coming from $C$. How would you fix this? I don't think you can just throw in the rest of the rationals wherever you please $\endgroup$ – mathworker21 Mar 3 at 17:51
  • $\begingroup$ I will look into this shortly. $\endgroup$ – DanielWainfleet Mar 4 at 10:38
  • $\begingroup$ time is not this short $\endgroup$ – mathworker21 Mar 25 at 7:03

Okay, so let's first cover the case, that $r_n$ does not have to enumerate all the rational numbers. Then you can make an enumeration of the rationals between (0,1) and B will obviously not cover the reals.

Now you can replace the elements $r_{n^2} $ with an enumeration of all the rationals which don't cover the reals because of your first argument. You will have a few duplicates in your sequence though. But you can fix that by skipping all the rationals in (0,1) which are not enumerated by a squared natural number.

This argument also works for balls with radius $\frac{1}{log(n)}$ or worse. You just replace the $exp(n^2)$ elements instead

For a sequence that covers the reals, take $r_n=\frac{1}{2n}$ then you will cover all the positive reals. Halving/quartering the distance frees up subsequences which you can use to cover the negative reals or all the rationals if you want.

(Okay I will never again type an answer on a phone...)


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