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...ssTheorem.html

http://en.wikipedia.org/wiki/Bolzano...strass_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.

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.