# Union of Balls around Rationals

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?

• 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. – Crostul Dec 6 '16 at 8:34
• 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 $[0,1]$. – mathworker21 Dec 6 '16 at 8:47
• 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 – DanielWainfleet Nov 4 '18 at 8:32
• @DanielWainfleet How do you know you can enumerate $\mathbb{Q}$ in such a way that $B = \mathbb{R}$? – mathworker21 Nov 4 '18 at 8:33
• @mathworker21. I have posted an answer to that part of the Q. – DanielWainfleet Nov 4 '18 at 10:39

## 2 Answers

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 $$0 when $$i 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 $$q_{3j+2} and $$\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.

• 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 – mathworker21 Mar 3 at 17:51
• I will look into this shortly. – DanielWainfleet Mar 4 at 10:38
• time is not this short – 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...)