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Intersecting Hyperspheres?
We say that a set S in R^n is D-bounded, D > 0, if, for all x,y ∈ S, |x-y| <= D. We say that S is maximally D-bounded if, for any z ∉ S, S ∪ {z} is not D-bounded.
It is trivial to prove that any maximally D-bounded set is necessarily closed and bounded. My challenge is this: Prove that, for any maximally D-bounded set S, there exists a subset B of S, of Lebesgue Measure zero, such that S is equal to the intersection of all circles C of radius D with center b ∈ B.
circles = hyperspheres. New I picked that title for a reason :)
*knew
2 Answers
- ?Lv 71 decade agoFavorite Answer
I think we can take B to be the boundary of S
The boundary is included in S, since S is closed
B is n-1 dimensional, so it has Lebesgue measure 0
Then S=intersection disks with radius D with centers in B
Clearly S is included in intersection since S is D-bounded
To prove the intersection is included in S, we use the maximality.
Because if there is z not in S but in intersection,
the supremum of distance between z and elements in S,
is attained by a distance from z and an element in the boundary.
So S U {z} is D-bounded, contradiction with the maximality.
btw those sets are more like hyperdisks, not hyperspheres
- gianlinoLv 71 decade ago
Well, boundaries of closed sets can be really awful of fractal dimension, anything and what not.
So I guess the fact that S is maximally bounded should be used to show that the boundary behaves nicely.
Edit ok here S is convex so no problem. You might even be able to find B such that m(B) = 0 where m is the n-1 dimensional Lebesgue measure.
