Bolzano-Weierstrass Theorem

[Nusc]

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

[HallsofIvy]

"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!

[Astronuc]

References

http://mathworld.wolfram.com/Bolzano-WeierstrassTheorem.html

http://en.wikipedia.org/wiki/Bolzano-Weierstrass_theorem

http://planetmath.org/?op=getobj&from=objects&id=2129

[philosophking]

"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.

[saltydog]

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:

[Nusc]

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.