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.)

...no.

When I first saw this in the queue, I thought it looked easy, but I had apparently violated the word "disjoint"--which makes the puzzle very trivial--and so it's back to square one.  I guess I have to read more carefully.

Now that that's been cleared up, I think that the assertions are impossible.  I can't come up with a complete proof, but I can come up with some convincing arguments.

First off, there are two options.  We can either have all the figure 8s enclosed within a single figure 8, or they cover the infinite plane without enclosing one another.  For the assertions to be possible, at least one of these options must be able to reach uncountable infinity.  If neither reaches it, there is no way to combine them to reach uncountable infinity--countable and countable make uncountable?--doesn't make any sense.  Now I can argue that in each case, uncountable infinity is unreachable.

Each figure 8 must enclose at least a tiny amount of area.  If no figure 8 encloses any other figure 8s, we would be dividing the plane into many regions.  Is there any way for the number of regions in an infinite plane be equivalent to the number of reals?  Maybe if the regions were points, but they're not.  You can count the regions, one by one, making them countable.

Now, let's say that all the figure 8s are enclosed in a single figure 8, which covers a finite area.  To be more space efficient, imagine each figure 8 to be shaped more like a theta or even a pair of rectangles.  Now, the figure 8 has two compartments in which it can enclose other 8s.  For every figure 8 we put into a compartment, we divide the compartment into at least two new compartments.  Each compartment must have contain at least a small area.  The compartments do not necessarily approach area zero as infinite figure 8s are nested--we can probably make the lower compartment take up a bigger fraction of space as we nest more and more 8s, making the area converge above 0.  However, it seems we can just count the number of times the compartments are divided--it's always countable.  This may be the hole in my argument, but I just don't see how it could be otherwise.

Posted by Tristan on 2005-02-15 23:56:25