Compactness in $\ell_{\infty}$ Somebody know a compactness result for subsets of $\ell_{\infty}$? I am looking for non-trivial or not well known results about this topic.
Thanks in advance for your comments.
 A: $\newcommand{\N}{\mathbb N}$
Since notation for a sequence of sequences is awkward I'm going to consider $\ell_\infty$ to be the space of all bounded functions $f:\N\to\mathbb C$.
It seems to me that $S\subset\ell_\infty$ is compact if and only if (i) $S$ is closed and bounded and (ii) the elements of $S$ are almost determined by their values at finitely many points, in the following sense:


(ii) For every $\epsilon>0$ there exist $\delta>0$ and $N$ such that if $f,g\in S$ and $|f(j)-g(j)|<\delta$ for $j=1,2\dots, N$ then $||f-g||_\infty<\epsilon$.


Suppose that (i) and (ii) hold and $(f_n)\subset S$. Since $S$ is bounded there is a pointwise convergent subsequence: $f_{n_j}(k)\to  f(k)$ for all $k$, with $f\in\ell_\infty$. Now (ii) implies that $(f_{n_j})$ is uniformly Cauchy, hence uniformly convergent.
Conversely, suppose $S$ is compact. Then $S$ is certainly closed and bounded. If (ii) fails then there exist $\epsilon>0$ and $(f_n),(g_n)\subset S$ with $$|f_n(j)-g_n(j)|<1/n,\quad j=1,2\dots, n,\quad(*)$$
but $||f_n-g_n||_\infty\ge\epsilon$. Now there exists a pair of subsequences $(f_{n_j}), (g_{n_j})$ and $f,g\in\ell_\infty$ with $||f_{n_j}-f||_\infty\to0$ and $||g_{n_j}-g||_\infty\to0$. And now (*) implies that $f=g$, contradicting $||f_n-g_n||_\infty\ge\epsilon$.
