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Today one of my students asked an interesting question, which I was unable to answer. Concretely:

Let $(z_n)_{n\in \mathbb N^+}$ be an enumeration of $\mathbb Q+\mathrm i\mathbb Q$. The question is, whether $(B_{1/n}(z_n))_{n\in \mathbb N^+}$ is a cover of $\mathbb C$, i.e. if $$ \mathbb C\subseteq \bigcup_{n=1}^\infty B_{1/n}(z_n). $$ It is intuitively clear (?) that this should hold. For every $z\in \mathbb C$ there exists a subsequence $(z_{n_k})_{k\in \mathbb N}$ converging to $z$, since $\mathbb Q+\mathrm i\mathbb Q$ is dense in $\mathbb C$. I am however unable to show that we can find $k\in \mathbb N$, such that $|z-z_{n_k}|<\frac{1}{n_k}$. It's somewhat awkward, since we have the same index $n_k$ on both sides and I am not sure how to deal with this.

Thanks for any insights!

Nico Zimmer
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2 Answers2

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Note that the sum of the areas covered by the balls (disregarding overlaps) is $$\sum_{n=1}^{\infty}\frac {\pi}{n^2}=\frac {\pi^3}6,$$ which is considerably less than the area of the complex plane.

Xander Henderson
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lulu
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  • Thank you very much! My intuition was apparently fooling me. – Nico Zimmer Jul 12 '23 at 16:56
  • This was exactly the answer I was writing. – Xander Henderson Jul 12 '23 at 16:56
  • @XanderHenderson I'm just happy to see a (relatively) natural computation resulting in the appearance of $\pi^3$. – lulu Jul 12 '23 at 16:56
  • @NicoZimmer Yeah, intuition isn't your friend in this sort of thing. – lulu Jul 12 '23 at 16:57
  • I like the phrasing here "which is considerably less than the area of the complex plane". Well, about infinitely many times less, give or take :). –  Jul 12 '23 at 16:58
  • @NicoZimmer Not sure what level your students are at, but if they haven't seen the computation of the sum of the square reciprocals, it's not needed here. The sum clearly converges and that's enough. – lulu Jul 12 '23 at 17:00
  • @lulu: We're doing multivariable real analysis, so students should be well aware of this series. I feel somewhat ashamed now though, seeing how wrong I really was. – Nico Zimmer Jul 12 '23 at 17:02
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    Here is an interesting extension to the problem: suppose that the ball centered at $z_n$ has volume $1/n$. The series no longer converges. What now? :P – Xander Henderson Jul 12 '23 at 17:04
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    @NicoZimmer These things are famously unintuitive. My advice is to set aside intuition entirely when approaching them. – lulu Jul 12 '23 at 17:04
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    @XanderHenderson Right, just take radius $\frac 1{\sqrt n}$ and the simple argument evaporates. Of course, if you can select the enumeration, the same argument holds. Just put all the square free indexed rationals inside the unit circle and the rest can't possibly cover the complement of circle. – lulu Jul 12 '23 at 17:07
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    @lulu My solution was to put the powers of 2 outside the unit circle, but yes, same idea. – Xander Henderson Jul 12 '23 at 17:08
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In comments, it is further asked that we cover $\mathbb{Q}+i\mathbb{Q}$ with $B_{\frac1{\sqrt{n}}}(z_n)$.

Enumerate $(z_n\in\mathbb{Q}+i\mathbb{Q}:n\ge1)=\mathbb{Q}+i\mathbb{Q}$ so that $|z_n-e|\gt\frac1{\sqrt{n}}$. Then $$ e\not\in\bigcup_{n=1}^\infty B_{\frac1{\sqrt{n}}}(c_n)\tag1 $$

Details of the Enumeration

Define a sequence of enumerations of annuli for $k\ge1$: $$ W_k=\left(w_{k,j}\in\mathbb{Q}+i\mathbb{Q}:j\ge1\text{ and }\sqrt{\frac2{k^2+k}}\lt|w_{k,j}-e|\le\sqrt{\frac2{k^2-k}}\right)\tag2 $$ where $$ \bigcup_{k=1}^\infty W_k=\mathbb{Q}+i\mathbb{Q}\tag3 $$ Put these enumerations into a single enumeration by diagonalization: $$ z_n=w_{k,j}\tag4 $$ where $$ \begin{align} m&=\left\lfloor\frac{1+\sqrt{8n-7}}2\right\rfloor\tag{5a}\\ k&=n-\frac{m^2-m}2\tag{5b}\\[9pt] j&=m-k+1\tag{5c} \end{align} $$ Here are tables of some results from $(5)$: $$ \begin{array}{} \begin{array}[t]{c|c|c|c} n & m & k & j \\ \hline 1 & 1 & 1 & 1 \\ 2 & 2 & 1 & 2 \\ 3 & 2 & 2 & 1 \\ 4 & 3 & 1 & 3 \\ 5 & 3 & 2 & 2 \\ 6 & 3 & 3 & 1 \\ 7 & 4 & 1 & 4 \\ 8 & 4 & 2 & 3 \\ 9 & 4 & 3 & 2 \\ 10 & 4 & 4 & 1 \end{array}& \begin{array}[t]{c|c|c|c} n & m & k & j \\ \hline 11 & 5 & 1 & 5 \\ 12 & 5 & 2 & 4 \\ 13 & 5 & 3 & 3 \\ 14 & 5 & 4 & 2 \\ 15 & 5 & 5 & 1 \\ 16 & 6 & 1 & 6 \\ 17 & 6 & 2 & 5 \\ 18 & 6 & 3 & 4 \\ 19 & 6 & 4 & 3 \\ 20 & 6 & 5 & 2 \\ 21 & 6 & 6 & 1 \end{array} \end{array}\tag6 $$

Note that $n=\frac{k^2+k}2+j-1$. This means that the first time that an element of $W_k$ is seen is when $n=\frac{k^2+k}2$. Thus, $$ n\lt\frac{k^2+k}2\implies z_n\in\bigcup\limits_{j=1}^{k-1}W_j\tag7 $$ For any $n\ge1$, there is a $k\ge2$ so that $\frac{k^2-k}2\le n\lt\frac{k^2+k}2$. Therefore, $$ \begin{align} |z_n-e| &\gt\sqrt{\frac2{k^2-k}}\tag{8a}\\ &\ge\frac1{\sqrt{n}}\tag{8b} \end{align} $$ Explanation:
$\text{(8a):}$ apply $(2)$ and $(7)$
$\text{(8b):}$ $n\ge\frac{k^2-k}2$

robjohn
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  • So basically, whenever $r_n\to 0 (n\to\infty)$ you can choose the enumeration $z_n$ of $\mathbb Q+i\mathbb Q$ such that $e$ is not covered? I guess this is pretty much what you are claiming here. Would be really nice to expand that to a full proof. –  Jul 12 '23 at 18:22
  • I think that any monotonically increasing function from $[0,\infty)$ onto itself can replace the square root in $(2)$ and $(8)$. – robjohn Jul 14 '23 at 17:26