We can find M>0, such that each of the three sets have elements in (0,M], to simplify our notation, let X,Y,Z be the intersection of A,B,C with (0,M] respectively, and let x,y,z be the corresponding supremums. Clearly, (0,M] = union(X,Y,Z)
Without loss of generality, we may assump x<=y<=z=M, and let Y1= intsec(Y,(x,M]) and Z1=(Z,(x,M])
case 1. x < y. Then both Y1 and Z1 are non empty, and (x,M] = union(Y1,Z1)
so, for any x1 in X, we can find y1 in Y1 and z1 in Z1, such that |y1-z1| < x1, and by construction, y1 >= x1, z1 >= x1, so x1,y1,z1 can form a triangle.
case 2. x=y, then inf(Z1) = x, so we can find x1,y1,z1 from X,Y,Z respecively, so that they are close to x, with maximum difference less than x/9. Then it can be easily shown that x1,y1,z1 can be the lengths of the edges of a triangle.
Without loss of generality, we may assump x<=y<=z=M, and let Y1= intsec(Y,(x,M]) and Z1=(Z,(x,M])
case 1. x < y. Then both Y1 and Z1 are non empty, and (x,M] = union(Y1,Z1)
so, for any x1 in X, we can find y1 in Y1 and z1 in Z1, such that |y1-z1| < x1, and by construction, y1 >= x1, z1 >= x1, so x1,y1,z1 can form a triangle.
case 2. x=y, then inf(Z1) = x, so we can find x1,y1,z1 from X,Y,Z respecively, so that they are close to x, with maximum difference less than x/9. Then it can be easily shown that x1,y1,z1 can be the lengths of the edges of a triangle.
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