Every point in 3D-space is colored either red, green or blue. Let R (resp. G and B) be the set of distances between red (resp. green and blue) points. Prove that at least one of R, G, or B, consists of all the non-negative real numbers.

NOTATION:

  R* = { P in E^3 | P is colored red }
  G* = { P in E^3 | P is colored green }
  B* = { P in E^3 | P is colored blue }

  R = { d(P,Q) | P,Q in R* } 
  G = { d(P,Q) | P,Q in G* } 
  B = { d(P,Q) | P,Q in B* }

  X << Y denotes X is a subset of Y

  X ^ Y denotes the intersection of sets X and Y

  X v Y denotes the union of sets Y and Y

  X - Y  denotes the set difference of sets X and Y

  sphere(X,y) denotes the sphere with center X
              and radius y 

PROOF (Others can flesh out the details):

  Assume r in [0,inf) - R,
         g in [0,inf) - G, and
         b in [0,inf) - B

  WOLOG let 0 < b <= g <= r.

  Pick a point L in R*. Then sphere(L,r) << G* v B*.

  Case A: sphere(L,r) << G*

          Pick a point M in sphere(L,r). Then

          sphere(M,g) ^ sphere(L,r) << G*

          This is a contradiction since it is a 
          space circle containing colorless points.

  Case B: sphere(L,r) << B*

          Pick a point M in sphere(L,r). Then

          sphere(M,b) ^ sphere(L,r) << B*

          This is a contradiction since it is a 
          space circle containing colorless points.

  Case C: (sphere(L,r) ^ B*) and (sphere(L,r) ^ G*)
          are both non-empty
           
          Pick a point M in sphere(L,r) ^ G*. Then

          sphere(M,g) ^ sphere(L,r) << B*. Pick a

          point N in sphere(M,g) ^ sphere(L,r). Then

          sphere(N,b) ^ sphere(M,g) ^ sphere(L,r) << B*

          This is a contradiction since it is a set 
          containing two colorless points.

  Therefore, at least one of the sets R, G, or B must
  equal [0,inf).

 Oops - Back slashes don't work to well.

To clean up the details:

Case 1: 0 < b <= g <= r

 This is the case discribed above.

Case 2: 0 <= g <= r

 This case is described in Case A above.

Case 3: 0 <= 0 <= r

 No points colored green or blue. Clearly,

 R = [0,inf)

Case 4: 0 <= 0 <= 0

 All points are colorless. Contradicts problem statement.

Edited on August 25, 2006, 9:41 pm

Edited on August 25, 2006, 9:49 pm

Edited on August 27, 2006, 11:54 am

Posted by Bractals on 2006-08-25 17:23:07