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Zero Content Proofs

  1. Dec 1, 2007 #1
    [​IMG]

    Well...I have no idea about this question. I don't even know where to start, and I am having terrible panic on these types of proof.

    Can someone please explain and guide me through? Believe it or not, my textbook (which is horrible) has absolutely no examples on it, and it expects me to be able to do in the exercises after the section, this is making me feeling really deseparate...

    I really need some help on this topic! So if someone understands it well, please help me out!

    Thanks a million!
     
    Last edited: Dec 1, 2007
  2. jcsd
  3. Dec 1, 2007 #2

    EnumaElish

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    What is v(Ui)?
     
  4. Dec 1, 2007 #3

    Dick

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    Just draw a picture of the diamond. You can cover it with four squares with side length 1 for a total area of 4. If you use squares with sides of length 1/2, you can cover it with eight of them, for a total area of 2. If you use sixteen with side 1/4, total area 1. Extrapolate.
     
  5. Dec 1, 2007 #4

    quasar987

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    To Enuma:

    v(U_i) is the "volume" of U_i. So if U = [a_1,b_1] x ... x [a_n,b_n], then vol(U)= (b_1-a_1)...(b_n-a_n).
     
  6. Dec 1, 2007 #5
    But the definition says FINITE cover of rectangles, then how can I extrapolate to infinity?

    How can do a prove on this rigorously?

    I am really lost on this topic, and I hope somebody can provide more help...

    Thanks!
     
  7. Dec 1, 2007 #6

    Dick

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    The more rectangles you use the smaller the area. You can stop when the area is less than epsilon. Describing a cover whose area is less than epsilon is a rigorous proof.
     
  8. Dec 2, 2007 #7
    I sort of get the idea now.

    But how can I describe the reatangles mathematically and choose an appropriate partition?


    Also, do I need 4 separate cases (one for each side of the diamond) for this proof?
     
    Last edited: Dec 2, 2007
  9. Dec 2, 2007 #8

    EnumaElish

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    I'll illustrate with squares with each side = s.

    To cover each side of the diamond with k squares, you need to solve ("length of one side of the diamond"/k)^2 = 2s^2 for s.
     
  10. Dec 2, 2007 #9
    Hi,

    Why do I have to solve for s? How can this help?
     
  11. Dec 2, 2007 #10
    Can I just restrict my attention to one side, say x>0, y>0, prove that it has zero cotent, and argue that the other 3 sides also has zero content due to SYMMETRY?
     
  12. Dec 2, 2007 #11

    EnumaElish

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    In response to your "But how can I describe the reatangles mathematically and choose an appropriate partition?"
     
  13. Dec 2, 2007 #12

    EnumaElish

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    I think so.
     
  14. Dec 2, 2007 #13
    Um...why do you need to solve the left side? And why do you need a "2" on the right side?

    Sorry...I don't see where this comes from...
     
  15. Dec 2, 2007 #14

    EnumaElish

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    Each side of your diamond is a diagonal, and can be thought of as a diagonal of a square. Say it has length = L.

    To cover L with k squares, the diagonal of each of the k squares has to be L/k. Are you with me so far?
     
  16. Dec 2, 2007 #15
    Yes, I am with you so far!
     
  17. Dec 2, 2007 #16

    Dick

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    Ok, so what's the area of one of those squares? What's the total area of all of them?
     
  18. Dec 2, 2007 #17
    Is it L^2 / (2k) ?

    How does this help?
     
  19. Dec 2, 2007 #18

    Dick

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    If that's the total area of all of them, then yes. What happens as k->infinity?
     
  20. Dec 2, 2007 #19
    But wait a second...the definition says FINITE cover of rectangles, then how be k be infinity?
     
  21. Dec 2, 2007 #20

    EnumaElish

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    Dick is making the (intuitive) point that you can get the area of the cover as close to zero as you like; because in the limit (k ---> infinity), it has zero area. In other words, for a given epsilon, you can make the cover smaller than epsilon by pushing k as close to infinity as you like.
     
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