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2 added lower bound

Edit 2012.05.31 I decided not to wait any longer for Noam Elkies. Here is my idea of a lower bound argument. It can probably be extended to open balls; I prefer to use compactness and closed balls for simplicity.

Let there be a covering of the closed unit ball by finitely many closed balls of radius 1/2. Any covering ball which contains the center, call it c, of the unit ball contains at most one point, call it p, on the boundary B of the unit ball. Since B minus p is open with respect to B, p is contained in one of the other covering balls which does not contain c. So we can assume the boundary B is covered by balls none of which contain c. The covering now has a finite number of balls which cover B plus at least one more ball covering c, and perhaps others.

Now replace the covering above with a new (perhaps identical) covering: shift each ball toward or away from c so as to maximize its intersection with B. This places each covering ball center at distance sqrt(3)/2 from c. B is still covered, and this new covering along with a ball of radius 1/2 placed with its center also at c is another (perhaps the same) covering with the same or fewer number of balls. Thus the posted problem is (essentially) the same as optimally covering B with caps of spherical radius of 30 degrees.

Elsewhere I noted Neil Sloane had a cover of a 3-d sphere with 20 caps each of radius slightly less than 30 degrees. I now claim anupper bound of 21 for the posted problem. Assuming Sloane's expertise with sphere packing, I expect 21 to be an exact bound.You can ask him for the covering number for dimensions greater than 3.END Edit 2012.05.31

1

Here is an idea which should generalize to dimensions 2 and greater. I will start with dimension 2.

Let us place a circle of radius 1/2 in the center of the radius 1 ball. We will place most, if not all, of the rest of the balls at a distance such that the center of the small ball is sqrt(3)/2 from the center of the large ball. This placement is chosen so that the angle of arc cut out of the two concentric circles is the same, which turns out to be 60 degrees. Now a convexity argument should show that every thing between the 60 degree arc on the small circle and the corresponding arc on the large circle will be covered by the same ball. The general covering problem is now reduced to a covering of the surface of the smaller (or the larger) sphere by circular caps which extend 60 degrees of arc

For n=2, this is a matter of taking the ratio 360/60. For n=3, I propose 6 caps around the equator, and for each hemisphere 6 more caps appropriately spaced with centers at latitude 30 degrees, and 6 more at latitude 60 degrees, sharing central longitude lines with the equatorial circles. Even if I messed up and two polar circles are needed, that gives a total of 33 spheres, but I think 31 balls suffice.

I am not familiar with higher dimensional sphere coverings, so I'll let someone else take over. I imagine that someone else can come up with a lower bound based on this style of arrangement. (Hey Noam Elkies, care to try out more dimensions?)

If Joseph understands this, maybe we will be graced with a few illustrations of it.

Gerhard "Ask Me! About System Design" Paseman, 2012.05.27