# weak vs. norm compactness in $\ell_1$

So I'm trying to show that weakly compact sets in $\ell_1$ are norm-compact. I've already proven that weak sequential convergence implies norm convergence. I think the idea I want to go with is to take some collection $\mathcal{F}$ (satisfying the finite intersection property) in my weakly compact set $K$ and note that by weak compactness I have $\bigcap_{F \in \mathcal{F}} \overline{F}^w \neq \varnothing$. I want to show in fact that $\bigcap_{F \in \mathcal{F}} \overline{F} \neq \varnothing$. To do this I suppose I want to show $\overline{F}^w = \overline{F}$ (i.e. the weak closure and norm closure always agree). My first idea was to prove that $\ell_1$'s weak topology was first countable, but this turned out to be false. So my current idea is to show that the ball of $\ell_1$ under the weak topology is first-countable (I know it's not metrizable). I feel like if I can show this, then I can "inflate" the ball to include $K$ and then use its first countability to prove that $\overline{F}^w$ is determined by limits of sequences in $F$ and that we need not worry about nets in $F$, from which it will follow (because weak seq. conv. is the same as norm conv.) that $\overline{F}^w = \overline{F}$. Any hints would be much appreciated.

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Let $F \subset \ell_1$ be weakly compact. By Eberlein–Šmulian theorem, $F$ is weakly sequentially compact. But weakly convergence and strongly convergence are equivalent in $\ell_1$, so $F$ is sequentially compact, and so compact, for the norm topology of $\ell_1$.
The main step is to prove that the weak topology is metrizable on a weakly compact subset $F$. For $n \geq 1$, let $x_n' : x \mapsto \sum\limits_{k \geq 1} x(k)y_n(k) \in (\ell_1)^*$ where $y_n=( \underset{n}{\underbrace{0,...,0}},1,1,...)$. Because $\bigcap\limits_{n\geq 1} \text{ker}(x_n') = \{0\}$, $d : (x,y) \mapsto \sum\limits_{k \geq 1} \frac{1}{2^k}|x_k'(x-y)|$ defines a metric on $F$. You can show that the identity from $F$ with the weak topology to $F$ with the metric topology is continuous. And because $F$ is weakly compact, you deduce that the identity is an homeomorphism. So the weak topology is indeed metrizable.
Can you give a proof for this particular case without invoking Eberlein–Šmulian? $\pmb{c}_0$ is separable. – dx7hymcxnpq Nov 2 '12 at 18:53
You can prove that in $\ell_1$, the weak topology is metrizable on a weakly compact subset; I added some elements of proof. – Seirios Nov 2 '12 at 20:51