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| Figure Eights (Posted on 2005-02-15) |
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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.)
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Submitted by David Shin
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Rating: 4.2000 (5 votes)
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Solution:
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The assertions cannot all be true.
Suppose C is a set of disjoint figure 8's in the plane. Call a point (x,y) rational if both x and y are rational. Because the rationals are countable, so is the set of rational points in the plane and, moreover, so is the set of pairs (P,Q) where P and Q are rational points in the plane (this is because the Cartesian product of two countable sets is itself countable, which can be proved by a standard diagonalization argument). Recall that the rationals are dense in the reals, which implies that the rational points are dense in the plane; that is, any disk in the plane contains a rational point.
Now, to each figure 8 in C associate a pair (P,Q) of rational points by choosing P inside one loop of the 8 and Q inside the other. The interior of a loop contains a disk, and so such rational points can be found. Because the figure 8's cannot cross each other, two figure 8's cannot be assigned the same pair (P,Q). In other words, this assignment of pairs is a one-to-one correspondence of C with a subset of a countable set. So C is itself countable. |
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