# bounded sequence, cluster point, Bolzano-Weierstrass…

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

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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 $y$?