3
$\begingroup$

Define an open $k$-sector of a disk as the portion between two radii separated by an angle of $2\pi/k$, but open along the two radii (and closed along the circle boundary). Call a set a sub-$k$-sector if it is a closed set and a subset of an open $k$-sector. So an open $2$-sector is a half-disk minus the diameter, and a sub-$2$-sector is a closed set that fits inside this set.


  SubSectorCovers
          Left: Coverage by three sub-$2$-sectors. Right: Four sub-$4$-sectors. Diameters are dotted lines.
It is clear that one can position three sub-$2$-sectors so that they cover a disk (above, left).

Q1. Is it correct that three sub-semiballs suffice to cover a ball in $\mathbb{R}^3$? What is the generalization to $\mathbb{R}^d$?

By a sub-semiball I mean the natural analog: a closed subset of a half-ball minus its bounding plane.

Q2. Does it require six sub-$4$-sectors to cover a disk?

I would like to believe that $k{+}1$ sub-$k$-sectors suffice—-so little of the quadrant is missing!—but the drawing above suggests otherwise.


Update. Here is my attempt to illustrate Douglas Zare's answer to Q2.
          Sub4Sectors


$\endgroup$
4
  • $\begingroup$ How about the center? Perhaps the definitions need minor adjustments? $\endgroup$
    – John B
    Dec 24, 2015 at 14:56
  • $\begingroup$ @JohnB: The center is not part of an open sector, because the two bounding radii are subtracted from the sector. For the half-disk, the entire diameter is subtracted, including the center. $\endgroup$ Dec 24, 2015 at 15:00
  • $\begingroup$ Understood, but you say that "It is clear that three sub-2-sectors suffice to cover a disk". However, no sub-2-sector contains the center. $\endgroup$
    – John B
    Dec 24, 2015 at 15:11
  • $\begingroup$ @JohnB: Ah. I meant that one can position three sub-$2$ sectors so that they cover the disk, as in the drawing. I will try to clarify. $\endgroup$ Dec 24, 2015 at 15:15

2 Answers 2

3
$\begingroup$

On Q1. The intersection of sub-semiball with the boundary sphere of the ball cannot contain two opposite points. Thus, bu the Borsuk--Ulam(--Lusternik--Shnirelman) theorem it is impossible to cover this sphere by three such closed sets.

Let me now finish the consideration of Q2 started by Douglas. For $k\geq 7$, again $k+1$ sub-$k$-sectors suffice!

Take $k-1$ (not sub!) $k$-sectors centered at the center of the disk with very small overlaps. Taking their appropriate sub-$k$-sectors we cover everything except a very small "star" around the center and, roughly speaking, a $(k-\varepsilon)$-sector of a bit larger radius. Now put the $k$th sub-$k$-sector so as to cover the star and most of the $(k-\varepsilon)$-sector, without some part around its boundary arc. This arc, together with its boundary, can now be covered by the last sub-$k$-sector.

You may see an example for $k=7$ below. I tried to do my best, but it seems that one needs to be very precise in order to cover the central star; still, it's obviously possible.

Eight sub-7-sectors

$\endgroup$
7
  • $\begingroup$ Edited, so as to address the rest of Q2. $\endgroup$ Dec 24, 2015 at 17:26
  • $\begingroup$ Good point about sub-semiball; I took the liberty of adopting your terminology. $\endgroup$ Dec 24, 2015 at 18:04
  • $\begingroup$ OK, I've deleted this part of text. $\endgroup$ Dec 24, 2015 at 18:12
  • $\begingroup$ After placing the $k-1$ sub-sectors, what remains has angular extent $2\pi/k + \epsilon$. The $k$-th subsector can cover the center star, but will leave a neighborhood of a radius, and a neighborhood of $2\pi/k + \epsilon$ circular arc uncovered. I don't see how this can be covered by one more sub-sector. It seems two are needed...? $\endgroup$ Dec 25, 2015 at 14:55
  • $\begingroup$ No, if we shift it towards the center, it will leave just a neighborhood of the arc. OK, I'll try to make a corresponding picture by myself (sorry, perhaps with a worse quality...). $\endgroup$ Dec 25, 2015 at 15:25
3
$\begingroup$

I'll address question $2$: Since the distance between $(1,0)$ and $(0,1)$ is $\sqrt{2} \gt 1$, you can cover a neighborhood of a radius with a sub-$4$-sector. The remainder of the disk can be broken into four sub-$4$-sectors by dividing it radially. The same idea lets you cover a disk with four sub-$3$-sectors or six sub-$5$-sectors.

When $k \ge 6$, a sub-$k$-sector has diameter less than the radius of the circle. Any sub-$6$-sector containing the center can't contain a point on the boundary. Any sub-$6$-sector contains less than $2\pi/6$ radians of the boundary, so it takes at least seven to cover the boundary. So, at least eight sub-$6$-sectors are needed to cover the disk.

This does not prove that it requires $k+2$ sub-$k$-sectors to cover a disk for $k \gt 6$, but that many suffices by covering a neighborhood of a radius using two sub-$k$-sectors, then dividing the remainder radially.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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