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 I think it's... by Tristan)

I agree with your proof that if no two 8s overlap, then this is impossible. However, I don't see the two possibilities as the only two possibilities. You say the only other possibility is if all figure 8s are enclosed within a single figure 8. Yet this is not the case even with the circles - all of the circles are not enclosed within a single circle; rather for any circle, there exist infinitely many circles enclosed by it and infinitely many circles enclosing it. I suspect if this is possible with the figure 8s, the same situation must occur.

The difficulty with the figure 8s is that the central point of each 8 must be at a different location, since the central point of each 8 is part of the 8 and thus must be disjoint from all other 8s (so np_rt, I don't think your solution works).

I've been stuck thinking of how to prove/disprove the possibility of this though. One approach at disproving it could be:

Given any infinite configuration of disjoint 8s on the plane. Consider the central point of each 8 in polar notation (r,theta). Create a linear ordering of these points by sorting ascending theta first, then r (e.g. (1,90)<(1,180)<(2,90)).

If we can somehow always map these points to either the natural numbers or the rational numbers, we have a proof. My thought in using polar notation is that if we could prove that the centers of the 8s cannot be arbitrarily close to each other, we know there is a minimum distance between any two 8s, and therefore we could show that the linear ordering defined above is a well ordering, thus proving the cardinality of the 8s equivalent to the natural numbers.

Edited on February 16, 2005, 1:31 pm

Posted by Avin on 2005-02-16 13:30:14