Characterization of Compactness in $\ell^\infty$ The metric space $\ell^\infty$ consists of all bounded real valued sequences with the metric is $ d(x,y) = \sup_{i \in \mathbb{N}} | \xi_i - \eta_i | $, where $x = (\xi_i), y = (\eta_i)$. I have been wondering about compactness in this space. I know that this space is isometric with $C(\beta \mathbb{N})$ and so compactness in $\ell^\infty$ comes down to the Arzela-Ascoli theorem in $C(\beta \mathbb{N})$. However, I barely understand what a continuous function is on $\beta \mathbb{N}$, and so I'm nowhere near understanding what equicontinuity would be. Besides, I am more interested in what equicontinuity in $C(\beta \mathbb{N})$ would mean for the corresponding sequences in $\ell^\infty$. 
So my question is this:
What is a characterization of compactness in $\ell^\infty$ which only concerns itself with conditions on the sequences? In other words, what are the necessary and sufficient conditions to place on a family of sequences in $\ell^\infty$ to ensure the family is compact?
 A: I dont know if this one helps you but since $\ell^{\infty}$ is complete if you take a compact $K\subset \ell^{\infty}$ then $K$ is closed and totally bounded. Conversly , if you take $K\subset \ell^{\infty}$ closed and totally bounded , since $\ell^{\infty}$ is complete metric space it follows that $K$ is also complete, hence $K$ is complete and totally bounded hence compact.
So the question comes down to whether a closed subset $K$ of $\ell^{\infty}$ is totally bounded.
The same applies in $C_{0}$ then there is a characterisation of $K\subset C_0 $ closed and totally bounded $\Longleftrightarrow$ for every $\epsilon>0$ there exist $n_{0} \in \Bbb N$ such that for every $x=(\xi_{k})\in K$ and for every $k\geq n_{0}$ then $|\xi_{k}|\leq \epsilon.$
Its intresting if there is any similar for the $\ell^{\infty}$.
I hope these thoughts will help you to tackle the problem , if you find anything let me know ! 
A: I'm going to write the elements of $\ell_\infty$ as functions instead of sequences, to nicify the notation.
One characterization of compactness is this:


$S\subset\ell_\infty$ is compact if and only if it is closed and bounded and has the following property: If $(f_n)$ is a sequence in $S$ and $f_n(j)\to f(j)$ for every $j$ then $||f_n-f||\to0$.


Suppose $S$ satisfies those three conditions and $(f_n)$ is a sequence in $S$. Then there is a subsequence, which I'm going to write as $(g_n)$, such that $g_n(j)\to g(j)$ for every $j$. Since $S$ is bounded, $g\in\ell_\infty$, and the hypothesis shows that $||g_n-g||\to0$. So $(f_n)$ has a convergent subsequence.
Conversely, suppose $S$ is compact, $(f_n)$ is a sequence in $S$ and $f_n(j)\to f(j)$ for every $j$. In general in a compact metric space if $x\not\to  x$ then $(x_n)$ has a subsequence converging to $y\ne x$. But clearly no subsequence of $(f_n)$ can converge to anything but $f$; hence $f_n\to f$.
You could make that proof look fancier by involing Banach-Alaoglu. Note that the unit ball of $\ell_\infty$ is weak* metrizable, since $\ell_1$ is separable. And note that if $||f_n||_\infty\le 1$ then $f_n\to f$ weak* if and only if $f_n(j)\to f(j)$ for every $j$. So the result says that $S$ is compact if and only if $S$ is closed and bounded and the identity map on $S$ is weak*-to-norm continuous.
Edit: The rest of this is wrong. The error is that a norm-closed set need not be weak* closed. Leaving it here instead of deleting it  in case anyone wants to try to fix it. (I believe it's correct for convex sets...)
Hmm, just noticed that that works in any measure space, or at least any measure space such that $(L^1)^*=L^\infty$:


In any such measure space, $S\subset L^\infty$ is compact if and only if $S$ is closed and bounded and the identity map on $S$ is weak*-to-norm continuous.


Proof: One direction is obvious, since if $S$ is closed and bounded it is weak* compact.
Suppose $S$ is compact. Then $S$ is certainly closed and bounded. The identity map on $L^\infty$ is norm-to-weak* continuous. Hence standard compact-Hausdorff-automatic-continuity-of-the-inverse magic shows that the identity map from $S$ with the norm topology to $S$ with the weak* topology is a homeomorphism.
Not that it's clear whether that weak*-to-norm continuity holds in a given example - I'm not claiming any of this is actually useful...
