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1) Euclidian plane $\mathbb{R}^2$ is not a union of disjoint circumferences (assuming point is not a circumference of radius $0$).

2) If we exclude 1 point from plane, concentric circumferences with center at this point form a desired partition.

3) Choose $n \geq 2$ distinct points $p_1, \dots, p_n$. Is $\mathbb{R}^2 \setminus \{p_1, \dots, p_n\}$ a union of disjoint circumferences?

I believe 3) is not, but I don't know how to prove it.

Here's my proof of (1).

Suppose such partition exist. Let's construct a sequence of compact circles $A_1 \supset A_2 \supset \dots$, such that $r_n \to 0$, where $r_n$ denotes the radius of $A_n$. Let $A_1$ be any circumference with its interior. Having chosen $A_n$, choose circumference passing through its center. Let $A_{n+1}$ be a union of this circumference with its interior. Obviously, $r_{n+1} < r_n/2$. Thus, $r_n \to 0$.

By Cantor's intersection theorem $$\bigcap_{n=1}^{\infty} A_n \neq \varnothing.$$ Choose $p \in \bigcap_{n=1}^{\infty} A_n$. If $p$ is a boundary point of some $A_n$ then $p \notin A_{n+1}$. Hence, $p$ is an interior point of all $A_n$ and boundary point of some other circle. This circle should be contained in all $A_n$ (otherwise circumferences would intersect), which is impossible because $r_n \to 0$.

Any ideas on (3)? Thanks in advance.

(1) and (2) are included for completeness.

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    $\begingroup$ Your question is not quite clear. Are you saying that proving the statements 1),2) and answering 3) with a proof are the problems given to you, and you want us to a) confirm or disconfirm your proof of 1), and b) give ideas on how to answer and prove statement 3)? $\endgroup$ – Rory Daulton Jan 13 '16 at 20:56
  • $\begingroup$ For (3), consider this image: upload.wikimedia.org/wikipedia/commons/thumb/5/56/… $\endgroup$ – Neal Jan 13 '16 at 21:01
  • $\begingroup$ @RoryDaulton, I posted 1) and 2) because I wanted 3) to seem more natural. $\endgroup$ – edubrovskiy Jan 13 '16 at 22:03
  • $\begingroup$ @Neal, I believe this is a partition of plane minus a line. $\endgroup$ – edubrovskiy Jan 13 '16 at 22:04
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Ah, this was a really fun problem! Here goes, hope I didn't mess anything up. :)

I'll do the case just $n=2$, and leave the generalization up to you. Color a circle magenta if it contains $p_1$ but not $p_2$ and color a circle cyan if it contains $p_2$ but not $p_1$. Color $p_1$ itself magenta and $p_2$ itself cyan as well. Finally, color a circle neon yellow if it contains both $p_1$ and $p_2$. By repeating the argument in (1) there are no circles enclosing neither $p_1$ nor $p_2$. Hence every point is either magenta, cyan, or neon yellow.

Now note that given any magenta circle, its interior is completely magenta. Actually, the magenta circles can be totally ordered by inclusion (since they can't intersect). So we consider two cases:

  • If there is a maximal magenta circle (i.e. a magenta circle not contained in any other magenta circle) then the set of all magenta points is just a closed disk.
  • If there is no maximal magenta circle, then the set of magenta points can also be expressed as the union over all magenta circles of their interiors. This is a union of open sets, so it is itself open.

We conclude the set of magenta points is either a closed disk or an open set. Similarly for the set of cyan points. Moreover, each set of points is convex.

To finish the problem:

  • Suppose there are no neon yellow points. If the magenta points form a closed disk, then the cyan points are $\mathbb R^2$ minus a disk which is not convex. Contradiction. So the magenta points must be open. Similarly the cyan points must be open. But $\mathbb R^2$ is connected, so it can't be written as the union of two open sets.
  • Now suppose there are neon yellow points. We claim there is a neon yellow circle minimal by inclusion. If not, then repeat the argument of (1) to get a contradiction, since any neon yellow circle must have diameter the distance from $p_1$ to $p_2$. So we can find a neon yellow circle $\mathscr C$ whose interior is all magenta and cyan. Now repeat the argument of the previous part, replacing $\mathbb R^2$ by the interior of $\mathscr C$.

Remark: There is a ``near miss'' (for $n=2$) if you look at one of the families of Apollonian circles: it's possible to cover $\mathbb R^2$ minus two points and a line.

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  • $\begingroup$ Actually, the magenta circles can be totally ordered by inclusion (since they can't intersect). Why is it impossible to have 2 disjoint magenta circles inside another magenta circle? $\endgroup$ – edubrovskiy Jan 13 '16 at 22:20
  • $\begingroup$ Now suppose there are neon yellow points. We claim there is a neon yellow circle minimal by inclusion. Why set of neon yellow points should contain any circles? $\endgroup$ – edubrovskiy Jan 13 '16 at 22:47
  • $\begingroup$ No two magenta circles can be disjoint, because the interiors of their disks are supposed to be have a common point $p_1$. $\endgroup$ – Evan Chen Jan 13 '16 at 23:09
  • $\begingroup$ The neon yellow points are also formed by unions of circles, namely those circles which enclose both $p_1$ and $p_2$. They are also totally ordered by inclusion. $\endgroup$ – Evan Chen Jan 13 '16 at 23:10
  • $\begingroup$ I don't see why point is neon yellow iff it's magenta and cyan. $\endgroup$ – edubrovskiy Jan 14 '16 at 12:49

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