# Covering each point of the plane with circles three times

The following problem arose from an Italian discussion group: I am not so sure about the optimal tags for the question, so feel free to improve them.

Definition: we say that $$E\subseteq\mathbb{R}^2$$ admits a cover with degree $$d$$ if there is a family $$\mathscr{F}$$ of distinct circles (with positive radii) such that every point of $$E$$ belongs to exactly $$d$$ circles in $$\mathscr{F}$$.

Preliminary results:

1. There is a cover of $$\mathbb{R}^2\setminus\{O\}$$ with degree $$1$$ given by concentric circles;
2. There is no cover of $$\mathbb{R}^2$$ with degree $$1$$, due to the accumulation point of any chain of nested circles;
3. There is a cover of $$\mathbb{R}^2$$ with degree $$2$$, for instance the one given by the unit circles centered at $$(2n,y)$$ for any $$n\in\mathbb{Z},y\in\mathbb{R}$$;
4. There is a cover of $$\mathbb{R}^2\setminus\{O\}$$ with degree $$2$$, for instance the one given by the circles tangent to the sides of a quadrant.

Now the actual question, which I did not manage to tackle (yet):

Is there a cover of $$\mathbb{R}^2$$ with degree $$3$$?

I believe the answer is negative: I (unsuccessfully) tried to prove that any degree-3 cover admits a degree-2 subcover, in order to get a contradictory degree-1 cover by switching to the complement. Any insight is welcome.

Small update: actually it is not possible to extract a degree-2 subcover from a (hypothetical) degree-3 cover. Once a circle $$\Gamma_1$$ is removed from a degree-3 cover, we are forced to remove a circle contained in the interior of $$\Gamma_1$$ and this leads to a chain of circles converging to a point.

If a cover with degree 3 exists, every chain of contained circles has a finite number of elements. The elements of $$\mathscr{F}$$ form a POset: we may say that $$\Gamma_1 < \Gamma_2$$ if $$\Gamma_1$$ is contained in the interior of $$\Gamma_2$$. This also allows to assign a parity to each element of $$\mathscr{F}$$ and each point of the plane, according to the length of a maximal chain contained in the circle / surrounding such point.

Further thoughts: let us assume that a degree-3 cover of $$\mathbb{R}^2$$ exists. By the chain argument there is a disk $$D$$ such that $$\partial D\in\mathscr{F}$$ and every circle covering $$\mathring{D}$$ meets $$\partial D$$ at two non-antipodal points. Let us name $$\mathscr{f}$$ the family of circles covering $$\mathring{D}$$. We have that $$\partial D$$ is covered three times, so the circles traversing each point of $$\partial D$$ are $$\partial D$$ itself and two elements of $$\mathscr{f}$$. These couples of circles "exiting" from any point of $$S^1=\partial D$$ have to cover three times any point of $$\mathring{D}$$, which is pretty strange. Many sub-questions came to my mind:

1. Is it possible to partition $$\mathscr{f}$$ into two/three subfamilies of disjoint circles?
2. Is it possible to use the descending chain condition as a substitute for continuity, then apply some topological trick?
3. Are Tucker's lemma, Euler characteristic or the theory of planar graphs useful in some way?

Starting from any $$P_0\in\partial D$$ we may take $$P_{-1}$$ and $$P_1$$ as the points of $$\partial{D}$$ connected to $$P_0$$ via the elements of $$\mathscr{f}$$ through $$P_0$$, then define $$P_{-2},P_{2},P_{-3},P_{3},\ldots$$ in the same way. This gives a partition of $$\partial D$$ into finite cycles (a cycle of length $$2$$ might occur if $$P_1=P_{-1}$$) and "infinite cycles" with a numerability of elements. Infinite cycles have limit points, which are troublesome. On the other hand also a partition into finite cycles only does not seem to stand any chance of covering any point of $$\mathring{D}$$ exactly three times.

Yet another measure-theoretic thought. Let $$S^1=\partial D$$ (which we may assume to have radius $$1$$), let $$P\in S^1$$. Two elements of $$\mathscr{f}$$ go through $$P$$: let $$L(P)$$ be the total length of the arcs given by the intersections with $$\mathring{D}$$. Let us assume that $$L:S^1\to (0,4\pi)$$ is an integrable function. By integrating $$L$$ over $$S^1$$ we have that each arc in $$\mathscr{f}\cap\mathring{D}$$ is counted twice, hence $$\int_{S_1} L = 2\int_{\text{arcs in }\mathscr{f}} 1=2\cdot 3\text{Area}(D) = 6\pi$$ and the average length of an arc in $$\mathscr{f}$$ is $$\frac{3}{2}$$. It follows that most of the arcs of an "integrable 3-cover" of $$\mathring{D}$$ are pretty short, forcing a concentration of the arcs near the boundary of $$D$$. This violation of uniformity probably leads to the fact that if a cover of $$\mathbb{R}^2$$ with degree $$3$$ exists, it is not integrable.

• It doesn't answer your question, but isn't the union of covers 1 and 3 a degree 3 cover of $\mathbb{R}^2 \backslash \{O\}$? Jul 15, 2021 at 18:10
• @JohnWaylandBales: yes it is, by suitably overlapping 1. and 3. you get a degree-3 cover of the punctured plane. You also have covers of $\mathbb{R}^2$ with any even degree, by overlapping instances of 2. with different radii. Jul 15, 2021 at 18:13
• Perhaps investigate whether a 3-cover of the plane by circles implies a 3-cover by straight lines. Just spitballing. Jul 15, 2021 at 18:21
• @JohnWaylandBales: a 3-cover by lines is trivial: it is enough to consider all the horizontal, vertical or diagonal (meaning parallel to $y=x$) lines. Jul 15, 2021 at 18:23
• Can you get a degree-n covering of the punctured plane by covering $\Bbb R^2$ with $n$ families of lines and then doing a bilinear transformation of the complex plane that exchanges 0 and $\infty$?
– MJD
Oct 28, 2021 at 1:13

Below I construct a cover of $$\mathbb{R}^2$$ with degree $$3$$. The idea is to build an almost-"degree 2 cover" of $$\mathbb{R}^2$$ for which a single point is covered three times (using variations on the trick from the degree 2 cover given in the question), then to balance things out by adding the degree $$1$$ cover of $$\mathbb{R}^2 - \{O\}$$.
Lemma: Let $$U$$ be an open disk and let $$P$$ be a point on its boundary. Then there is a family $$\mathscr{F}$$ of circles contained in $$U \cup \{P\}$$ such that each point of $$U$$ belongs to exactly two circles of $$\mathscr{F}$$, while $$P$$ belongs to exactly one circle of $$\mathscr{F}$$.
Proof: For $$a > 0$$, let $$L(a)$$ be the line given by the equation $$x = a$$, and let $$C(a)$$ be the circle of radius $$a$$ centered at $$(a, 0)$$, so $$C(a)$$ passes through the origin $$O$$. Assume without loss of generality that $$P = O$$ and that $$U$$ is the open disk bounded by $$C(1/2)$$. Now, for $$0 < a < b$$, let $$\mathscr{G}(a, b)$$ be the family of circles of radius $$\frac{b-a}{2}$$ with centers on the line $$L(\frac{b+a}{2})$$, so the circles of $$\mathscr{G}(a, b)$$ are contained within the closed strip bounded by $$L(a)$$ and $$L(b)$$, and each point of $$L(a)$$ and $$L(b)$$ belongs to exactly one circle of $$\mathscr{G}(a, b)$$, while each point in the interior of the strip belongs to exactly two circles of $$\mathscr{G}(a, b)$$. Define $$\mathscr{G} = \bigcup_{n=1}^\infty \mathscr{G}\left(1 + \frac{1}{n+1}, 1 + \frac{1}{n} \right) \cup \bigcup_{n=1}^\infty \mathscr{G}\left(2 + \frac{1}{n+1}, 2 + \frac{1}{n} \right) \cup \bigcup_{n=3}^\infty \mathscr{G}(n, n+1)$$ so that, if we define $$V = \bigcup_{a > 1} L(a) = \{(x, y) : x > 1\}$$, all circles in $$\mathscr{G}$$ are contained in $$V$$, and each point in $$V$$ belongs to exactly two circles in $$\mathscr{G}$$, except for those points in $$L(2)$$, each of which belongs to exactly one circle in $$\mathscr{G}$$. Now, under inversion with respect to the unit circle centered at the origin, $$L(a)$$ maps to $$C(\frac{1}{2a}) - \{O\}$$, $$V$$ maps to $$U$$, and each circle in $$\mathscr{G}$$ maps to a circle in a new family $$\mathscr{G}'$$ of circles contained in $$U$$. By the above, each point in $$U$$ belongs to exactly two circles in $$\mathscr{G}'$$, except for those points on $$C(1/4) - \{O\}$$, each of which belongs to exactly one circle in $$\mathscr{G}'$$. To conclude, we define $$\mathscr{F} = \mathscr{G}' \cup \{C(1/4)\}$$, which has the desired properties.
Proposition: There is a cover of $$\mathbb{R}^2$$ with degree $$3$$.
Proof: Define $$C'(r)$$ to be the circle of radius $$r$$ centered at the origin. Let $$P = (2, 0)$$ and let $$U$$ be the open disk bounded by $$C'(2)$$. For each $$n \geq 1$$, let $$\mathscr{H}_n$$ be the family of unit circles with centers on $$C'(2n+1)$$, so if we let $$E_n$$ be the interior of the region bounded by $$C'(2n)$$ and $$C'(2n+2)$$, then each circle in $$\mathscr{H}_n$$ is contained in $$C'(2n) \cup E_n \cup C'(2n+2)$$, and each point in $$E_n$$ belongs to exactly two circles in $$\mathscr{H}_n$$, while each point on $$C'(2n)$$ or $$C'(2n+2)$$ belongs to exactly one circle in $$\mathscr{H}_n$$. Letting $$\mathscr{H} = \bigcup_{n=1}^\infty \mathscr{H}_n$$, we see that each point in $$\mathbb{R}^2 - (C'(2) \cup U)$$ belongs to exactly two circles in $$\mathscr{H}$$, each point in $$C'(2)$$ belongs to exactly one circle in $$\mathscr{H}$$, and each point in $$U$$ belongs to zero circles in $$\mathscr{H}$$. To cover $$U$$, let $$\mathscr{F}$$ be the family of circles given by the Lemma applied to $$U$$ and $$P$$, and define $$\mathscr{F}' = \mathscr{F} \cup \mathscr{H} \cup \{C'(2)\}$$, so each point of $$\mathbb{R}^2$$ belongs to exactly two circles in $$\mathscr{F}'$$, except $$P$$ which belongs to exactly three (since it lies on $$C'(2)$$). But none of the circles in $$\mathscr{F}'$$ are centered at $$P$$, so we can extend the family $$\mathscr{F}'$$ to a covering of $$\mathbb{R}^2$$ with degree $$3$$ by adding all circles centered at $$P$$.