# Compact subsets in $l_\infty$

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

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 $\varepsilon>0$).
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