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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 $\ell_1$.

EDIT: The proof that weakly compactness implies weakly sequentially compactness is not very difficult. I use this article.

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.

Then, you can conclude as compactness and sequentially compactness are equivalent in metrizable spaces.

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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

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