# Mathematical Analysis-Cluster points-Bolzano Weistrass THM

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

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(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 $b\le c+\epsilon$.