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(3): I think it's... by David Shin)

You're really forcing me to be very explicit in the details of my little proof--not that that's a bad thing.

Yes, I do realize that the two loops of each figure 8 do not have to be equal in size.  However, it seems that each loop must enclose a positive area, which is what makes this different from circles.

Perhaps I can slightly modify my proof to avoid circular assumptions.

Let's assume that we can enclose uncountably many figure 8s in a single largest one.  If this is tre, then we should be able to fit just as many even when we limit the area used to only the left compartments.  The right compartments each enclose a region with positive area.  There is one empty right compartment for every figure 8.  As a result, we will have uncountably many disjoint regions in a plane.  But there cannot be uncountably many disjoint regions in a plane (this is the same reason that we cannot reach uncountable infinity when no figure 8 encloses another).

Thanks to Avin, I know that the above two cases are not the only two cases that need proofs.  Lastly, I must prove the case where each figure 8 is enclosed by uncountably many bigger ones, and there is a single smallest one.  It seems that the above proof should work for this case as well, but I am not yet sure.

Posted by Tristan on 2005-02-17 23:42:23