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Bolzano-Weierstrass Theorem

  1. Oct 7, 2005 #1
    Let S be an infinite and bounded subset of R. Thus S has a point of accumulation.

    Proof: Let T be the set of reals such that for every t E T there are infinitely many elements of S larger than t. Let M be such that -M<S0<M for all S0 E S.

    The set T is nonempty and bounded and hence it has a supremum say A.

    Proof of claim: A is an accumulation point of S.

    Can someone please help me with the proof of the claim or give it to me rather? Never did I expect such a course that demanded so much memorization.

    Thanks
     
  2. jcsd
  3. Oct 8, 2005 #2

    HallsofIvy

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    "Can someone please help me with the proof of the claim or give it to me rather? "

    Why in the world should we want to prevent you from learning it yourself? We have no reason to dislike you that much! If I were to give you a hint, it would be to look up the definition of "accumulation point" in your textbook. 90% of "proving" things is using the definitions.

    I am also, by the way, moving this to "College Homework". If it isn't homework, or reviewing for a test, I can't imagine why you would be doing it!
     
  4. Oct 8, 2005 #3

    Astronuc

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  5. Oct 8, 2005 #4
    "I am also, by the way, moving this to "College Homework". If it isn't homework, or reviewing for a test, I can't imagine why you would be doing it!"

    I have a slight problem with this. I am currently doing an independent study in topology with Munkres and also Mendelson, and my main strategy when I go through the book is trying to prove the major theorems that are outlined in the book by myself first, and then looking up how they do them if I can't do them and I'm really stuck. This person may just be a little more stubborn and would want a little hint that perhaps he doesn't want to go to the book for.
     
  6. Oct 8, 2005 #5

    saltydog

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    Hey Nusc, here try this: An Analysis text with ALL the answers (about $200.00 bucks, I don't set the price).

    "Undergraduate Analysis", by Serge Lang

    "Problems and Solutions for UnderGraduate Analysis", by R. Sharkarchi

    Hey, if anybody talks to Gale, tell her to do the same thing.:smile:
     
  7. Oct 8, 2005 #6
    I don't want anything too rigorous.

    Suppose that E>0 is given. Then if n is large enough, we have A-E<an<=bn<A+E. Between an and bn there are infinitely many distinct elements of S. So the interval (A+E,A-E) contains points of S that are not equal to A. Hence A is an accumulation point.
     
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