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: I think it's... by Avin)

You're right, I forgot a third case, where all figure 8s are enclosed within infinitely others.  This is a separate case from the second case, since only one needs to be able to reach uncountable infinity.  If we remove all the circles with radius greater than 1, or less than 1, there would still be uncountably many.  I think this case is very similar to the second one, and I'm currently wondering whether my proof can be used for either.

Response to David Shin:
I think my proof does not at all work with circles.  First, let's establish that a countable number of countable sets cannot combine to form an uncountable set, am I right?  Therefore, we only need to consider the left compartment of each figure 8, and there is no need to put two or more side by side in a compartment.

At each step, we add a figure 8 into the current compartment, then the left compartment of the figure 8 just added becomes the current one.  At each step, the current compartment shrinks to a fraction of its previous area, a fraction that is below 1.  The area of the current compartment need not converge at 0, but we should be able to count the number of times the compartment shrinks.  It is possible to find two figure 8s with no figure 8 separating them.  This does not apply to the case of the circles, because in each step of adding a new circle, the remaining area does not shrink at all.

Should I be wrong, I think my error would be that I limit my thinking to a step by step process.

Posted by Tristan on 2005-02-17 03:15:00