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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 re(11): I think it's... - Elaboration | Comment 28 of 34 |
(In reply to re(10): I think it's... - Elaboration by David Shin)

I was thinking it was impossible for any figure 8 to be enclosed in infinitely many right compartments, just as it is for the left compartments.

But as I recall, that wasn't one of the things I proved, I merely proved that there would be countable number of 8s if right compartments are unused.

In fact, I just thought up a counterexample: there can be a figure 8 enclosed in (countably) infinitely many right compartments, even if there is a single largest figure 8.  The counterexample goes as follows:

The largest figure 8 has a small figure 8 in its right compartment.  We can draw a middle-sized figure 8 with its right compartment around the smaller 8, and its left compartment half the size of that of the small 8.  We can draw another figure 8 with its right compartment around the middle-sized 8, and its left compartment half the size of that of the middle-sized 8.  We can do this infinitely (not infinite induction since though there will be infinite 8s, the infinitieth step is not taken).  The end result will be that the smallest has infinite right compartments enclosing it.  The empty left compartments, while there are aleph null, take a finite amount of space.

Perhaps I can build on this counterexample to a solution that shows that the assertions are possible, quickly invalidating my developing proof.

  Posted by Tristan on 2005-02-19 17:28:58

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