All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
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(4): I think it's... | Comment 21 of 34 |
(In reply to re(3): I think it's... by David Shin)

You're really forcing me to be very explicit in the details of my little proof--not that that's a bad thing.

Yes, I do realize that the two loops of each figure 8 do not have to be equal in size.  However, it seems that each loop must enclose a positive area, which is what makes this different from circles.

Perhaps I can slightly modify my proof to avoid circular assumptions.

Let's assume that we can enclose uncountably many figure 8s in a single largest one.  If this is tre, then we should be able to fit just as many even when we limit the area used to only the left compartments.  The right compartments each enclose a region with positive area.  There is one empty right compartment for every figure 8.  As a result, we will have uncountably many disjoint regions in a plane.  But there cannot be uncountably many disjoint regions in a plane (this is the same reason that we cannot reach uncountable infinity when no figure 8 encloses another).

Thanks to Avin, I know that the above two cases are not the only two cases that need proofs.  Lastly, I must prove the case where each figure 8 is enclosed by uncountably many bigger ones, and there is a single smallest one.  It seems that the above proof should work for this case as well, but I am not yet sure.

  Posted by Tristan on 2005-02-17 23:42:23

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information