Can anyone provide a proof for the Bolzano-Weierstrass Theorem claim?

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In summary, the proof shows that the supremum A of the set T is also an accumulation point of the infinite and bounded subset S of R. This is proven by showing that the interval (A+E,A-E), where E is a given positive number, contains infinitely many distinct elements of S. Therefore, A is an accumulation point of S.
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Nusc
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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
 
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  • #2
"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
"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
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:
 
  • #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.
 

1. What is the Bolzano-Weierstrass Theorem?

The Bolzano-Weierstrass Theorem is a fundamental theorem in real analysis that states that any bounded sequence of real numbers has a convergent subsequence. It is named after the mathematicians Bernhard Bolzano and Karl Weierstrass.

2. What does "bounded sequence" mean in the context of the Bolzano-Weierstrass Theorem?

A bounded sequence is a sequence of real numbers that is limited or confined within a certain range. In other words, the terms in the sequence do not exceed a specific upper or lower bound.

3. How is the Bolzano-Weierstrass Theorem used in mathematics?

The Bolzano-Weierstrass Theorem is used to prove the existence of the limit of a bounded sequence. It is also a key tool in proving other important theorems in real analysis, such as the Heine-Borel Theorem and the Cauchy Convergence Criterion.

4. Can the Bolzano-Weierstrass Theorem be applied to sequences of complex numbers?

Yes, the Bolzano-Weierstrass Theorem can also be applied to sequences of complex numbers. However, in this case, the sequence must be bounded in both the real and imaginary parts.

5. Is the Bolzano-Weierstrass Theorem a sufficient condition for convergence?

No, the Bolzano-Weierstrass Theorem is not a sufficient condition for convergence. It only guarantees the existence of a convergent subsequence within a bounded sequence, but it does not guarantee that the entire sequence will converge.

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