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BolzanoWeierstrass Theorem 
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#1
Oct705, 07:46 PM

P: 776

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
Oct805, 07:52 AM

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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! 


#3
Oct805, 08:36 AM

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References
http://mathworld.wolfram.com/Bolzano...ssTheorem.html http://en.wikipedia.org/wiki/Bolzano...strass_theorem http://planetmath.org/?op=getobj&from=objects&id=2129 


#4
Oct805, 10:36 AM

P: 174

BolzanoWeierstrass Theorem
"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. 


#5
Oct805, 12:53 PM

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P: 1,593

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. 


#6
Oct805, 02:26 PM

P: 776

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


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