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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): I think it's... | Comment 20 of 34 |
(In reply to re(2): I think it's... by Tristan)

Tristan,

First of all, a couple points:

1.  You are correct that a countable union of countable sets is itself countable.

2.  Not sure if you or others realize this, but the two loops of the figure eight need not be equal in size.

Now, in response to your last post:

You seem to be assuming that of all figure eights contained within the loop of some bigger one, there is a largest one.  You then argue that this largest figure eight takes up a fraction of the loop that is strictly below 1.

Again, you can use the same argument in the circles case.  Assume that of all the circles drawn inside one larger circle, there is a largest one.  This circle takes up a fraction of the larger circle that is strictly below 1.

Of course, you see, it is the assumption that is faulty in both cases.  It might be the case that within one figure eight, there is an infinite sequence of figure eights where each encloses the previous one.

  Posted by David Shin on 2005-02-17 19:31:39

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