Does there exist the set of balls(may be not disjoint) $X=\{B_i\subset\mathbb{R^2};i\in I\}$, satisfing following properties?(Note that the ball has a positive real radius)

  1. Let the set of all lines in plane to be $L$. For each $l \in L,\ l\cap B_i \ne \varnothing $ for some $i\in I$.
  2. $\{N_l=\text{the number of balls which the line} \ l \ \ \text{meets}\mid l\in L\}$ is bounded.

(This does not mean the number of ball which line meets is finite. This statement is stronger.)

Is there such a set??

Edit: I exchage "transversal" with "meet". Here is a original problem posted in Math Stack Exchange. https://math.stackexchange.com/questions/1387435/there-is-a-no-set-which-every-line-meets-the-ball

  • 2
    $\begingroup$ This was previously asked on Math Stack Exchange: math.stackexchange.com/questions/1387435/… . $\endgroup$ – Kevin P. Costello Aug 15 '15 at 4:22
  • $\begingroup$ But there was a wrong answer. Therefore I post the question here. $\endgroup$ – user148928 Aug 15 '15 at 4:39
  • $\begingroup$ What does it mean "transwersals"? Intersects? $\endgroup$ – Alexandre Eremenko Aug 15 '15 at 5:23
  • $\begingroup$ We can interpret this word as "meet". $\endgroup$ – user148928 Aug 15 '15 at 7:57
  • 5
    $\begingroup$ Even if you don't like the answer that was given before, you should still link to the other question. This avoids duplication of effort. It lets people read the discussion of the earlier version, where you clarified, for example, that you mean closed balls. $\endgroup$ – Douglas Zare Aug 15 '15 at 8:04

No, such $X$ does not exist.

Assume the contrary. Take a point $O$ which is outside all disks and a circle $\omega$ centered at $O$. We implement polarity with respect to this circle (thus, in what follows we consider only lines not passing through $O$).

Take any ball $B_i\in X$. The poles of all lines tangent to $B_i$ form a hyperbola $h_i$ with a focus at $O$; then the poles of all lines intersecting $B_i$ form the set $H_i$ consisting of $h_i$ and of the interior points of its two branches. Let $A_i$ be the cone bounded by the asymptotes of $h_i$ and containing $H_i$ (we assume that $A_i$ contains no points of the asymptotes except for $O\in A_i$). Notice here that in any closed subcone of $A_i$, all points which are sufficiently far from $O$ lie in $H_i$.

The fact that $X$ satisfies the required properties yields that the sets $H_i$ cover the plane (except for $O$) but in a bounded number of layers. We will show that this is impossible.

Take an arbitrary $A_{i_1}$ and some its closed subcone $S_1$. Some interior point $x_1$ of $S_1$ is not covered by $H_{i_1}$, so it is covered by some $H_{i_2}$. Take a closed subcone $S_2$ of $S_{i_1}\cap A_{i_2}$ and proceed similarly. Thus, on the $(k+1)$th step we choose an interior point $x_k\in S_k$ not covered by $H_{i_1},\dots,H_{i_k}$ (all points close enough to $O$ are such), take $H_{i_{k+1}}$ covering $x_k$, and take a closed subcone $S_{k+1}$ of $S_k\cap A_{i_{k+1}}$.

Finally, for every $k$, all points in $S_{k}$ which are sufficiently far from $O$ are covered by each of $H_{i_1},\dots,H_{i_k}$, so some point is covered by arbitrarily many of the $H_i$.

  • $\begingroup$ Does there exist a point which is outside all disks? $\endgroup$ – user148928 Sep 15 '15 at 8:25
  • $\begingroup$ Surely yes. Otherwise an arbitrary line is covered by balls; but every ball covers only a finite interval, so there are infinitely many balls meeting the line. $\endgroup$ – Ilya Bogdanov Sep 15 '15 at 9:43
  • $\begingroup$ For me is not clear why $S_{i_m} \cap A_{i_{m+1}}$ is nonempty. $\endgroup$ – Arseniy Akopyan Apr 8 '16 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.