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I have an exam tomorrow and could not solve the question below. Could you please help me?

Regards

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 $b_m = \sup\{x_n : n \ge m\}$, $b = \lim b_m$. Prove that $b \le c^∗$. (Hint: use (a).)

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1 Answer 1

up vote 2 down vote accepted

(a) Fix $\varepsilon>0$. Then there are finitely many $n$ such that $x_n\geq c^*+\varepsilon$; indeed, if that were not the case, the subsequence $\{x_n:\ x_n\geq c^*+\varepsilon\}$ has a cluster point by Bolzano-Weierstrass, and this cluster point would be $\geq c^*+\varepsilon>c^*$, contradicting the fact that $c^*$ is the greatest cluster point.

So let $N$ be the biggest $n$ such that $x_n\geq c^*+\varepsilon$; thus, for every $n>N$, $x_n<c^*+\varepsilon$.

(by the way, the point $c^*$, is the "limit superior" of the sequence, denoted $\limsup_n x_n$)

(b) First, $b$ exists because the sequence $b_m$ is monotone and bounded. Given $\varepsilon>0$, by (a) we have an $N$ such that $n>N$ implies $x_n<c^*+\varepsilon$; so $b_m<c^*+\varepsilon$ if $m>N$. Thus $b\leq c^*+\varepsilon$. As $\varepsilon$ was arbitrary, $b\leq c^*$.

Comment: actually, one always has $b=c^*$

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Should not it be $\geq c^*+\varepsilon>c^*$ rather than $\geq c^*\varepsilon>c^*$ ? –  Amadeus Bachmann Nov 5 '12 at 8:26
    
Yes, thanks for noticing the typo :) –  Martin Argerami Nov 5 '12 at 13:50

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