I am going to assume that the question is if there are values of n for which K is empty for all possible selections of the a's and P's. I assume this because for n>=3 if you make the radius of the second "sphere" larger than twice the radius of the first then they have no intersection and thus K is empty.
if n=1 then we have just a single "sphere" namely two points on the number line for K.
If n=2, then the first and second circles intersect in one point if a2=2*a1, 2 points if a2<2*a1 and zero points if a2>2*a1. Thus K is not always empty.
Now assume n>=3. Imagine we start building the set K but starting with K initially being the entire first sphere. We then pick a point in K and a radius and then redefine K as the intersection of the current K with the new "sphere". Now each time we do this K remains a "sphere" just simply being of lower dimension (with the empty set having zero dimension), thus by the time we add the last circle we need K to be at least 1 dimensional. Now at each step, we can minimize the amount of dimension reduction by keeping the radius constant. Thus if we say that a1=a2=...=an=r then each redefinition of K creates a "sphere" of 1 less dimension and reduces the radius by a factor of sqrt(3)/2. However, if at any point the reduced radius of K is less than r/2 then the next iteration of K will be empty.
Thus we need r*(sqrt(3)/2)^(n-2)>=r/2
which translates to n<=6
thus for n>=7 K is empty for all possible selections
Posted by Daniel
on 2015-11-11 14:20:46