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.)

(In reply to re: solution - the number of curves must be countable(?) by David Shin)

First of all, david was right because the compartment of the curve can be filled without any area left - consider the following fractal - when we assume that the shape of eight is a "square" shape hence all of the lines in the curve are perpendicular to each other - like a digital clock eight - define a fractal as such: for each eight, devide each compartment into a 3x3 grid with 9 equal-aread regtangles. draw an 8-curve in the middle cell of the grid, and treat all of the other 9 minus 1 cells as we have done with the whole compartment - hence devide them to 3x3 grid etc... fill every compartment of every 8-curve recursively. one can prove that this method leaves no point "exclusive" to only one shape.

BUT - the number of eights is countable no matter what algorythm is given to fill the curve's compartments. consider this following proof by negation:

suppose there is an algorythm that describes how the 8's should be drawn. now in each compartment of a curve we can devide the area of the compartment to two sections. in one section we can draw the embedded curves just as the algorythm describes - only we rescale everything by 0.5 - in other words - no matter how the algorhytm proposes to fill the compartment with other curves, fill only the bottom half of the compartment in exactly the same way, but rescale everything to fit in half-compartment rather that a full one. it is obvious that by applying this "patch" to the algorythm, one can draw the exact same number of eights, but leaving a non-zero area "exclusive" to each eight, hence a mapping from rationals to curves can be defined...

this problem has a lot to do with a famous fractal called the Cantor-set, in which one recursivly removes thirds from the unit interval, only there, the set is closed, hence contains the limits of sequences of its points, and by drawing eights, one cannot accomplish a fractal that will also be a closed set.

Posted by ronen on 2005-02-24 15:04:04