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If $A \subseteq l_\infty$, and $A=\{l\in l_\infty: |l_n| \le b_n \}$, where $b_n$ is a sequence of real, non-negative numbers, then if $A$ is compact subset of $X$ it must mean that $\lim (b_n) = 0$.

I tried doing this by contradicition, if $A$ is compact, it means that it is closed subset in $X$, which implies it is complete, but if we assume $\lim(b_n) \neq0$ I should maybe be able to show $\exists$ a Cauchy sequence for which this sequence converges outside of $A$. However, I can't think of any counterexample. Am I doing this wrong?

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    While closedness is not neccessary anyway it is easy to show. Consider continuous functional $f_n:\ell_\infty\to\mathbb{C}:x\mapsto x_n$ and note that the set $\{z\in\mathbb{C}:|z|\leq b_n\}$ is closed hence the set $f_n^{-1}(\{z\in\mathbb{C}:|z|\leq b_n\})$ is closed in $\ell_\infty$. Note $A=\bigcap\limits_{n=1}^\infty f_n^{-1}(\{z\in\mathbb{C}:|z|\leq b_n\})$, so it is closed as intersection of closed sets.2012-10-18

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If $b_n$ does not converge to $0$ then there exists $\varepsilon>0$ and a subsequence $b_{n_k}$ such that $b_{n_k}>\varepsilon$. Therefore the sequence $ x_k=\underbrace{(0..\varepsilon..0..)}_{\text{ position }n_k} $ is contained in $A$ and has no convergent subsequence in $\ell^\infty$ (the distance between any two elements $x_i,x_j$ is $\varepsilon>0$).

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    It is easy to prove that any convergent sequence is Cauchy by using the definition of the limit. Intuitively, any two elements of the sequence sufficiently close to the limit are close to each other.2012-10-19