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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

i am stuck in one of my 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).)

For part a), I tried to show the contrapositive, i supposed suppose there is an ϵ>0 such that infinitely many xn's satisfy xn≥c+ϵ, then this determines a subsequence that has a cluster point ≥c+ϵ. But I cannot completely explain this

For part b, I see that For all ϵ>0 we have bm < c+ϵ for almost all m (i.e., for m>N for a fixed N∈N). So i have to show that the sequence (bm) is bounded and monotonic. Thus, it has a limit, and so its limit satisfies b≤c+ϵ. But again, i could not fully explain it.

Can somebody give me a hand? Thanks

share|cite|improve this question

You're on the right track for part (a). If you have a cluster point for a subsequence of $(x_n)$, do you have a cluster point for the main sequence? If so, and that cluster point is $\geq c+\epsilon$ and $\epsilon>0$, than what does this say about $c$? You're nearly there.

For part (b), you're also very close. If $x\leq y+\epsilon$ for all $\epsilon>0$, then what can you conclude about $x$ in relation to $y$?

share|cite|improve this answer

Your Answer


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.