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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.

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  • $\begingroup$ It doesn't answer your question, but isn't the union of covers 1 and 3 a degree 3 cover of $\mathbb{R}^2 \backslash \{O\}$? $\endgroup$ Jul 15, 2021 at 18:10
  • $\begingroup$ @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. $\endgroup$ Jul 15, 2021 at 18:13
  • $\begingroup$ Perhaps investigate whether a 3-cover of the plane by circles implies a 3-cover by straight lines. Just spitballing. $\endgroup$ Jul 15, 2021 at 18:21
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    $\begingroup$ @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. $\endgroup$ Jul 15, 2021 at 18:23
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    $\begingroup$ 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 $\infty$? $\endgroup$
    – MJD
    Oct 28, 2021 at 1:13

1 Answer 1

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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 $P$.

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  • $\begingroup$ Great answer! As far as I can tell, everything works. $\endgroup$
    – Milten
    Oct 31, 2021 at 20:30

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