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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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The set is closed regardless of whether or not $b_n\to 0$, so that is the wrong approach. I recommend contraposition, assuming that $(b_n)$ does not converge to zero and using this to show that $A$ contains a sequence of points whose pairwise distances are bounded below, hence $A$ is not sequentially compact. –  Jonas Meyer Oct 18 '12 at 6:00
    
How is it obvious that the set is closed? –  Simon Sehayek Oct 18 '12 at 6:01
    
I didn't say it is obvious, but have you tried to show that it is closed? If a sequence in $A$ converges to $x$, can you show that $|x_n|\leq b_n$ for all $n$? (You don't need to show that it is closed to solve this problem.) –  Jonas Meyer Oct 18 '12 at 6:04
    
So should I use the fact that if $\lim (b_n) \neq 0$ then $\exists$ subsequence of $b_n$, $b_{n_k}$ s.t. $b_{n_k} \geq \epsilon_o \forall n_k$ –  Simon Sehayek Oct 18 '12 at 6:06
    
Atreyu: Yes, at least I would. –  Jonas Meyer Oct 18 '12 at 6:07

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up vote 3 down vote accepted

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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Sorry, this may be a stupid question but what is the relation between the definition of convergence and any pairwise points in the sequence having distance bounded below? Is it because $x_k$ has no limit points? –  Simon Sehayek Oct 18 '12 at 15:13
    
@Atreyu: $(x_k)$ has no Cauchy subsequence, hence no convergent subsequence. –  Jonas Meyer Oct 18 '12 at 18:35
    
Wait a complete metric space satisfies that any sequence converges iff it is Cauchy? I thought it was just every Cauchy sequence must converge... –  Simon Sehayek Oct 19 '12 at 2:15
    
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. –  Beni Bogosel Oct 19 '12 at 10:30

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