Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck in my one of the homework problems, the question is like the following:

Let $(x_n)$ be a bounded sequence, and let $c$ be the greatest cluster point of $(x_n)$:

(a) Prove that for every $\epsilon > 0 $ there is $N$ such that for $n > N$ we have $x_n < c + \epsilon.\;$ (Hint: use the Bolzano-Weierstrass theorem.)

(b) Let $b_m = \text{sup}\{x_n : n >=m\};\; b = \text{lim}\; b_m$. Prove that $b \le c.\;$ (Hint: use (a).)

Can anyone give me a hand please? Thanks

share|improve this question
1  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  amWhy Nov 11 '12 at 23:53
1  
This is the verbatim reproduction of the text of a homework you were given to do. What did you try? Where are you stuck? –  Did Nov 11 '12 at 23:54
    
This is my homework and I am stuck. I am not sure how to use Bolzano Weierstrass thm here. Can you please help me do it? –  user49065 Nov 11 '12 at 23:58
add comment

1 Answer 1

Hints:

(a) Try the contrapositive: suppose there is an $\epsilon>0$ such that infinitely many $x_n$'s satisfy $x_n\ge c+\epsilon$, then this determines a subsequence that has a cluster point $\ge c+\epsilon$.

(b) For all $\epsilon>0$ we have $b_m < c+\epsilon$ for almost all $m$ (i.e., for $m>N$ for a fixed $N\in\Bbb N$). Show that the sequence $(b_m)$ is bounded and monotonic. Thus, it has a limit, and so its limit satisfies $b\le c+\epsilon$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.