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Figure Eights (Posted on 2005-02-15) Difficulty: 5 of 5
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.)

See The Solution Submitted by David Shin    
Rating: 4.2000 (5 votes)

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Some Thoughts A thought | Comment 29 of 34 |

It seems to me that one thing we know for sure is that given any two eights, at least one of them is going to have a "lobe" that is disjoint from both lobes of the other.

So, for each eight, pick a point from each of its lobes. No other eight can share this pair, right? Doesn't this allow us to do what Tristan has been grappling at? If we know that no lobe is "flat", then can't we always pick points with rational coordinates? And thus we will have a one-to-one correspondence between the set of eights  and a countable set of pairs of points (being a subset of QXQ).

Now I will wait for the example that shows I am wrong :-)

Edited on February 21, 2005, 12:48 pm
  Posted by owl on 2005-02-21 04:50:08

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