Suppose that I drew an infinite number of disjoint closed curves in the plane (such as circles, squares, etc.). Suppose that I then tell you that there is one curve for each positive real number.

You would not have too much trouble believing my assertions at this point. For example, I could have drawn all circles with center at the origin. They are all disjoint, and for each positive real number x, there is a corresponding circle - namely, the circle of radius x.

But suppose that I also tell you that all the curves I drew were figure eights. Can you believe my assertions now?

(A figure eight is a curve in the plane obtained from the basic "8" shape by any combination of translation, rotation, expansion, or shrinking.)

if we assume that the area of any 8-curve is non-0 (it has a set with a positive area embedded inside it), and since there is no way to fill a whole component of the 8 curve with other curves (some space will always remain unfilled) then for any curve there will always be a point with rational coordinates which is inside the curve but not inside any other curve that's embedded in one of its compartments and therefore that means that we can define a mapping from the rational coordinates onto the set of curves hence the set is countable, and I think that proves that it cannot be done.

One can be mistaken to think that it can be accomplished by drawing a "fractal" in which every component holds another curve in it recursivly and thus there is 2^N curves, but it is not actually true because a sequence of embedded curve does not lead to a curve - each curve is embedded under a "finite number" of iterations, in other words, the number of "lines" in a recursive fractal is countable.

Posted by ronen on 2005-02-22 21:40:31