If a set of points in the plane contains one point in each convex region of area 1, then can it have finite density?

what is the density of the points? In my understanding, it means the average number of points in unit square. If each convex region is unit square, this problem is trivial and the answer is yes. In the general case, I guess the statement is true.

  • $\begingroup$ Defining density as the "average number of points in a unit square" seems a bit vague, and something more precise might be needed. But the main problem is this: not all convex regions of area 1 are unit squares, and you have to have a point in every convex region of area 1! $\endgroup$ – James Cranch Apr 14 '14 at 11:22
  • $\begingroup$ How about defining the "average density" as $\lim_{r \to \infty} F_r/B_r$ where $F_r$ are the number of points within distance $r$ from the origin, and $B_r$ the area of a circle with radius $r$? $\endgroup$ – Per Alexandersson Apr 14 '14 at 11:34
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    $\begingroup$ The question is a duplicate of mathoverflow.net/q/3307/806 because of John's theorem (which implies that every convex set can be sandwiched between two triangles of approximately same area). To best of my knowledge, the problem of constructing a linear-sized net for the family of all convex sets (equiv. triangles) with respect to the Lebesgue measure on [0,1]^2 is still open. $\endgroup$ – Boris Bukh Apr 14 '14 at 12:19
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    $\begingroup$ Why is this called "dead flies"? $\endgroup$ – Nate Eldredge Apr 14 '14 at 19:01
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    $\begingroup$ corollary isn't the same as duplicate $\endgroup$ – guest Apr 14 '14 at 20:45